Practice Test

Numbers taken from the question

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Question 1 of 120

2026 Practice Test B Math, question 13

Question

A basket contains 3 apples, 4 oranges, 5 bananas, and 8 pears.

If one piece of fruit is randomly pulled from the basket, what is the probability that it will be a pear?

Select the place on the number line to plot the point.

On the test you click a point on the number line; here, choose it from the menu.

Point

Show the answer

Correct answer: the point 0.4

The total number of fruits is:

$3(\text{apples}) + 4(\text{oranges}) + 5(\text{bananas}) + 8(\text{pears}) = 20\text{ fruits}$

The probability of picking a pear is:

$\frac{8\ (\text{pears})}{20\ (\text{total})} = \frac{8}{20} \text{ or } 0.4$

Question analysis

A basket holds apples, oranges, bananas and pears, and students find the probability that a piece of fruit pulled at random is a pear, then plot it as a point on a number line from 0 to 1. It tests simple probability as a part of a whole. With no choices to pick from, the fraction has to be turned into tenths to find its place.

NYS standards: 6.SP.8a uniform probability models
6.SP.6 probability from 0 to 1
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of a simple event from counts is grade 6 work in New York. This one adds four whole numbers, forms one fraction and plots it among marked tenths, a shorter and more routine task than grade 7 probability.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Adding the four counts to 20 and writing 8 over 20 is routine, so most students succeed. The losses come from the number line: its decimal labels make it easy to read the 8 pears as 0.8, and a part-to-part fraction such as 8/12 falls between marks and has to be rounded, which should prompt a second look. Simplifying 8/20 to 2/5 and then to 4/10 makes the mark plain.

Something you must already know minor Working it out
The question does not say that a probability is the favorable count over the whole count, here the 8 pears over all 20 pieces of fruit rather than over the other 12. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The number line is labeled only in tenths from 0 to 1, so 8/20 has to be rewritten as a decimal before the right mark can be chosen. (Enright, Morley & Sheehan, 2002)

Strategy: Add up all the fruit for the total, write the number of pears over that total, rewrite the fraction in tenths, and select that mark on the line.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

0.8 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Takes the 8 pears straight to the mark labeled 0.8, never dividing by the 20 pieces of fruit in the basket.
0.7 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Compares the 8 pears with the 12 other pieces of fruit, 8/12 or about 0.67, a part-to-part ratio, and plots the nearest tenth.
0.6 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Puts the 12 apples, oranges and bananas over the 20 pieces of fruit, plotting the chance of not getting a pear instead of getting one.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event: $P = \frac{\text{number of results you want}}{\text{total number of results}}$; P is the probability, the chance that something happens. It goes from 0 (never) to 1 (always). The top counts the results you want, here pears. The bottom counts every possible result, here every piece of fruit. It works when each result is equally likely, like grabbing fruit without looking.
Equal fractions: $\frac{a}{b} = \frac{a \div n}{b \div n}$; a is the top of a fraction, and b is the bottom. n is any number except 0. Dividing the top and the bottom by the same number gives an equal fraction. Use it to get 10 on the bottom, which is easy to write as a decimal.

Sort out the question
The basket has 3 apples, 4 oranges, 5 bananas and 8 pears. You pull out one piece without looking. You need the probability that it is a pear. Then you plot that number as a point. The number line runs from 0 to 1, with a mark at every tenth.
Count all the fruit
The bottom of the fraction is every piece of fruit you could pull out. Add all four kinds:
$3 + 4 + 5 + 8 = 20$
There are 20 pieces of fruit in all.
Write the probability as a fraction
Use the probability formula. There are 8 pears out of 20 pieces of fruit:
$P = \frac{8}{20}$
Watch out: the bottom is all the fruit, not just the other fruit. $\frac{8}{12}$ compares the pears with the 12 fruits that are not pears. It is about 0.67, and it leads you to the mark at 0.7. And 12 over 20 is the chance of not getting a pear.
Change it to tenths
The number line is marked in tenths. So rewrite the fraction with 10 on the bottom. Divide the top and the bottom by 2:
$\frac{8}{20} = \frac{8 \div 2}{20 \div 2} = \frac{4}{10} = 0.4$
Four tenths is written 0.4.
Watch out: the 8 pears do not make 0.8. That skips dividing by the 20 pieces of fruit.
Plot the point
Start at 0 and count four marks to the right: 0.1, 0.2, 0.3, 0.4. On the test, you click the mark labeled 0.4. On this site, you choose 0.4 from the menu.
So the answer is the point at 0.4.

Check your answer

The chance of not getting a pear is $\frac{12}{20} = 0.6$ . A pear and not a pear cover every result, so the two must add to 1: $0.4 + 0.6 = 1$ . Also, pears are less than half of the fruit, so the point should be left of 0.5. It is.

More questions like this

The original DOE question

Question 13 as shown on the DOE’s digital test, page 85 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 85, question 13.

Question 2 of 120

2026 Practice Test B Math, question 31

Question

A pizza shop offers a choice of 3 sizes (small, medium, and large) and 7 different toppings. Different pizzas can be created by changing the size and the choice of toppings. If Cody wants to order a pizza with exactly 2 different toppings, how many different pizzas can he create?

A6
B21
C63
D126

Show the answer

Correct answer: C

Figure out how many different topping pairs are possible. Use 1, 2, 3, 4, 5, 6, 7 to represent the toppings, and create a list of possible pairs:

1,2; 1,3; 1,4; 1,5; 1,6; 1,7
2,3; 2,4; 2,5; 2,6; 2,7
3,4; 3,5; 3,6; 3,7
4,5; 4,6; 4,7
5,6; 5,7
6,7

So there are 21 different topping combinations for one pizza.

Since there are 3 pizza sizes, multiply the total number of combinations by 3 to get the total number of different pizzas Cody can create: $3 \times 21 = 63$ .

Question analysis

Cody orders a pizza with exactly 2 of 7 toppings in one of 3 sizes, and students count how many different pizzas he could create. It tests counting combinations, where the order of the toppings does not matter, combined with a second choice. The challenge is to count every possible pizza exactly once.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 In grade 7, compound events are usually ordered pairs, such as a coin with a number cube. Here an unordered pair of toppings is chosen from seven and then matched with three sizes, a combination count beyond typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question packs two counts into one sentence, each with its own trap. Order does not matter among the toppings, while the size multiplies every pair, and students tend to remember one of these and forget the other. Listing pairs systematically, each topping matched only with later ones, keeps the first count honest, and the stem’s “changing the size and the choice of toppings” is the reminder that size counts too.

Combines ideas moderate Working it out
Two counts are joined: the unordered pairs of toppings from 7, then the 3 sizes each pair can come in, and a mistake in either count changes the total. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
Students must know that a pizza with mushroom and onion is the same as one with onion and mushroom, so the ordered count 7 × 6 has to be halved. (Anderson & Krathwohl, 2001)

Strategy: Number the toppings 1 to 7, list each pair once by matching every topping only with higher-numbered ones, count the list, then bring in the choice of size.

Traps

Stops one step early: A choice is a number from partway through the solution.
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 3 sizes by the 2 toppings Cody wants, using the number of toppings on the pizza instead of the number of possible pairs.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Lists or counts the 21 pairs of toppings correctly, then stops without giving each pair its 3 possible sizes.
D Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: moderate
Counts ordered pairs, 7 × 6 = 42, so each pair of toppings appears twice, and then multiplies by the 3 sizes to reach 126.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The counting principle: $\text{total} = m \times n$; m is the number of ways to make one choice, and n is the number of ways to make a second choice. Every first choice can go with every second choice, so you multiply. Here, every pair of toppings can come in every size.
Pairs when order does not matter: $\text{pairs} = \frac{n \times (n - 1)}{2}$; n is the number of items to choose from. There are n ways to pick a first item and n − 1 ways to pick a different second one. That counts each pair twice, once in each order. Dividing by 2 counts each pair once.

Sort out the question
Each of Cody’s pizzas has one size and exactly 2 different toppings. There are 3 sizes and 7 toppings. Pizzas differ by “changing the size and the choice of toppings.” So a small and a large with the same toppings are two different pizzas. You need to count every pizza once.
List the topping pairs
Order does not matter: onion with peppers is the same pizza as peppers with onion. So number the toppings 1 to 7, and pair each one only with bigger numbers. Then no pair shows up twice.
Topping Pairs with Pairs
1 2, 3, 4, 5, 6, 7 6
2 3, 4, 5, 6, 7 5
3 4, 5, 6, 7 4
4 5, 6, 7 3
5 6, 7 2
6 7 1
$6 + 5 + 4 + 3 + 2 + 1 = 21$
There are 21 different pairs of toppings.
Check the pairs with the formula
Use the pairs formula with 7 toppings:
$\frac{7 \times 6}{2} = \frac{42}{2} = 21$
Watch out: 7 × 6 = 42 counts every pair twice, once as 1 then 2 and once as 2 then 1. Keeping 42 leads to $42 \times 3 = 126$ , choice D.
Bring in the sizes
Each of the 21 topping pairs can come in any of the 3 sizes. Use the counting principle:
$21 \times 3 = 63$
Watch out: stopping at 21 leaves out the sizes, which gives choice B. Choice A, 6, multiplies the 3 sizes by the 2 toppings on one pizza. That does not count pizzas.
Give the answer
There are 21 topping pairs, and each one comes in 3 sizes. Cody can create 63 different pizzas. So the answer is C, 63.

Topping	Pairs with	Pairs
1	2, 3, 4, 5, 6, 7	6
2	3, 4, 5, 6, 7	5
3	4, 5, 6, 7	4
4	5, 6, 7	3
5	6, 7	2
6	7	1

Check your answer

Count one size at a time. A small pizza can have any of the 21 pairs, so there are 21 small pizzas. There are also 21 medium and 21 large ones: $21 + 21 + 21 = 63$ . That is more than the 21 pairs but fewer than 126, so it is reasonable.

More questions like this

The original DOE question

Question 31 as shown on the DOE’s digital test, page 103 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 103, question 31.

Question 3 of 120

2026 Practice Test B Math, question 46

Question

Amiya subscribes to an online streaming service. She pays a onetime setup fee of $50, plus $20 per month for service.

Write an equation that can be used to determine the total amount Amiya will pay, y, for x months of service.

Enter your equation in the space provided. Enter only the equation.

On the test you build the answer on an on-screen keypad; here, type it. Use ^ for a power, sqrt( ) or √ for a root, pi for π, and <= or >= for ≤ and ≥.

Show the answer

Correct answer: y = 20x + 50

Solution: $y = 20x + 50$

An equation of the form $px + q = r$ can be used to represent the total amount Amiya will pay.

Using the information from the scenario, the onetime setup fee of $50 is the constant term (q), and the $20 per month of service is the coefficient (p) of the number of months (x).

So, the equation that can be used to determine the total amount Amiya will pay, y, for x months of service is $y = 20x + 50$ .

Question analysis

Amiya pays a one-time setup fee plus a monthly charge for a streaming service, and students type an equation for her total cost y after x months. It tests writing a linear equation from a word problem with a starting amount and a rate. The two dollar amounts play different roles, so which one goes with x has to be thought through.

NYS standards: 7.EE.4 equations and inequalities from problems
6.EE.9 dependent and independent variables
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an equation for a fixed fee plus a steady monthly charge is a familiar grade 7 task, and the numbers are whole. Typing the full equation, with both variables and no choices to compare, makes it less forgiving.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 51 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are small and whole, so the work is in the roles. The $50 is paid once and the $20 every month, and a student who places them in the order they appear writes the fee as the monthly rate; others leave the fee out entirely. Checking the equation at x = 1 and x = 2, where Amiya has paid $70 and then $90, shows at once whether each amount is in the right place.

Turning words into math moderate Understanding the question
Only the words “onetime setup fee” and “per month” tell which amount is multiplied by x and which is added once, and the $50 comes first in the sentence. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The whole equation, with y, an equals sign and x, is built on a keypad with no choices, so a reversed or incomplete equation looks as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Work out Amiya's total for 1 month and for 2 months by hand, see which amount is added again each month, and build an equation that gives both totals.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

y = 20x Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Multiplies the $20 monthly charge by x months but never adds the one-time $50 setup fee, so the equation gives only the cost of service.
y = 50x + 20 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Attaches x to the $50 fee and adds the $20 once, putting the numbers into y = mx + b in the order they appear rather than by which one repeats.
y = 70x Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the $50 fee and the $20 charge into $70 and multiplies by the months, which would charge the setup fee again every month.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a fee and a rate: $y = (\text{rate})x + (\text{fee})$; y is the total cost in dollars, and x is the number of months. The rate is the amount paid every month. You multiply it by x because it is paid once for each month. The fee is paid only once, so you add it once, no matter how many months.

Sort out the question
Amiya pays two kinds of charges:
- a setup fee of $50, paid one time only;
- a charge of $20 for every month of service.
y is her total cost in dollars, and x is the number of months. You need an equation that links them. The answer is an equation, not a number.
Work out a few totals
Find the total for a few months by hand. Each month adds one more $20:
Months Monthly charges ($) Total ($)
1 $1 \times 20 = 20$ $50 + 20 = 70$
2 $2 \times 20 = 40$ $50 + 40 = 90$
3 $3 \times 20 = 60$ $50 + 60 = 110$
The $50 shows up once in every total. The $20 shows up once for each month.
Match each number to its job
Use the fee-and-rate model, $y = (\text{rate})x + (\text{fee})$ . The rate is the amount that repeats each month, $20. It multiplies the months, so it becomes 20x. The fee, $50, is paid once, so it is added once.
Watch out: the $50 comes first in the question, but it is not the rate. $y = 50x + 20$ charges $50 a month. $y = 70x$ charges the fee every month. $y = 20x$ forgets the fee.
Write the equation
Put the rate and the fee into the model:
$y = 20x + 50$
Here 20x means 20 times x. An equal form is fine too, such as $y = 50 + 20x$ , because you can add in either order.
Enter it on the keypad
Type the whole equation, including y =. Without it, you have an expression, not an equation. On the test, the keypad has x, y and = keys. Press y, =, 2, 0, x, +, 5, 0. You do not need a times sign or dollar signs. On this site, type it the same way.
So you would enter y = 20x + 50.

Months	Monthly charges ($)	Total ($)
1	$1 \times 20 = 20$	$50 + 20 = 70$
2	$2 \times 20 = 40$	$50 + 40 = 90$
3	$3 \times 20 = 60$	$50 + 60 = 110$

Check your answer

Test the equation with the totals you found. For 1 month, $y = 20 \times 1 + 50 = 70$ , and Amiya pays $50 + $20 = $70. For 2 months, $y = 20 \times 2 + 50 = 90$ . Both match. For 0 months, y = 50, just the setup fee.

More questions like this

The original DOE question

Question 46 as shown on the DOE’s digital test, page 118 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 118, question 46.

Question 4 of 120

2026 Practice Test C Math, question 6

Question

$A parallelogram that leans to the right. The angle at its upper left vertex is labeled 3x°, and the angle at its lower left vertex is labeled 2x°.$

In the parallelogram above, what is the value of x?

Show the answer

Correct answer: A

In a parallelogram, any two adjacent angles are supplementary, so the sum of the two given angles is equal to 180°.

Write an equation showing that the sum of $3x^{\circ}$ and $2x^{\circ}$ is equal to 180°:

$3x + 2x = 180$

Solve for x:

$5x = 180$

$\frac{5x}{5} = \frac{180}{5}$

$x = 36$

Question analysis

A parallelogram is drawn with the two angles at the ends of its left side labeled 3x° and 2x°, and students find the value of x. It tests the angle facts of a parallelogram and solving a simple equation. Knowing what the two angles must add up to is the key fact, since the figure gives no other number.

NYS standards: 7.G.5 angle relationships and equations
8.G.5 angle sums, transversals, angle-angle similarity
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 The angle fact needed, about consecutive angles of a parallelogram, rests on grade 8 parallel-line reasoning. Once it is known, the work is one short equation, like a typical grade 7 problem about angle sums.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the angle fact is in place, 5x = 180 is a one-line equation, so the question tests recall more than computation. Students who use 90° for the pair land on half the right value, and the two smallest choices come only from combining the labels' 3 and 2. A sense check helps: x = 36 makes the angles 108° and 72°, one obtuse and one acute, just as the drawing shows.

Something you must already know moderate Working it out
The figure gives no number of degrees, only 3x° and 2x°, so everything rests on recalling that the two angles at the ends of one side of a parallelogram add to 180°. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The marked angles sit at the upper and lower ends of the left side, consecutive rather than opposite, which is what makes them supplementary instead of equal. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide what two angles that share a side of a parallelogram add up to, set 3x + 2x equal to that total, solve for x, and check that the two angles fit the figure.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses 90° as the total of the two angles, the complementary total, so 5x = 90 gives 18, though consecutive angles of a parallelogram add to 180°.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the two coefficients from the labels, 3 × 2 = 6, without using any angle fact or writing an equation.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Combines 3x + 2x into 5x and gives the coefficient 5, stopping before dividing the 180° total by it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Neighboring angles of a parallelogram: $a + b = 180^{\circ}$; a and b are two angles of a parallelogram that share a side, like the top left and bottom left angles. They always add to 180°. Here is why. The four angles add to 360°, and opposite angles are equal. So each pair of neighbors is half of 360°.

Sort out the question
The figure is a parallelogram. That is a four-sided shape whose opposite sides are parallel, so they never meet. Two angles are marked: 3x° at the upper left and 2x° at the lower left. You need the value of x.
Write an equation for the angles
The two marked angles are at the two ends of the left side. So they are neighbors, not opposite angles. Use the rule that neighbors add to 180°:
$3x + 2x = 180$
Watch out: the total is 180°, not 90°. Using 90 gives 5x = 90 and x = 18, choice B.
Solve for x
Three x’s plus two x’s make five x’s:
$5x = 180$
Divide both sides by 5 to get x alone:
$x = \frac{180}{5} = 36$
Watch out: stopping at 5, the number in front of x, gives choice D. Multiplying 3 × 2 = 6 gives choice C, but that uses no angle fact.
Choose the answer
The value of x is 36. It is a number, not the size of an angle. The angles themselves are 3 × 36 and 2 × 36 degrees.
So the answer is A, 36.

Check your answer

Find the two angles. The top one is $3 \times 36 = 108$ degrees. The bottom one is $2 \times 36 = 72$ degrees. They add to 180°. In the figure, the top angle is wide, more than 90°. The bottom one is narrow, less than 90°. That matches.

More questions like this

The original DOE question

Question 6 as shown on the DOE’s digital test, page 79 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 79, question 6.

Question 5 of 120

2026 Practice Test C Math, question 9

Question

Maria rides her bicycle to school at a constant speed of 15 miles per hour. If the distance to school is 6 miles, how many minutes will it take Maria to get to school?

Show the answer

Correct answer: C

Maria’s rate of speed is given as 15 miles per hour. Since 1 hour is equivalent to 60 minutes, her rate of speed, in miles per minute, can be determined.

$\frac{15\text{ miles}}{1\text{ hour}} = \frac{15\text{ miles}}{60\text{ minutes}} = \frac{1\text{ mile}}{4\text{ minutes}}$ , or $\frac{1}{4}$ mile per minute.

Therefore, it takes Maria 4 minutes to ride 1 mile.

The distance to school is 6 miles, so it will take $6 \times 4 = 24$ minutes to ride to school.

Question analysis

Maria bikes 6 miles to school at a constant 15 miles per hour, and students find how many minutes the ride takes. It tests using a rate to find a time and converting hours to minutes. With the speed given per hour and the question asking, in bold, for minutes, keeping track of units is the main hurdle.

NYS standards: 6.RP.3b unit rate problems
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Distance, speed and time with whole numbers is grade 6 rate work. The only extra demand is an answer in minutes when the speed is per hour, a conversion grade 7 students make routinely.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are small, but the speed is per hour and the answer is wanted in minutes, so students have to choose a route through the units. Finding the minutes per mile first, 4 at 15 miles an hour, turns the rest into a single multiplication. The wrong choices come from using a number in the stem as the answer, from mixing the 60 minutes with the 6 miles, or from rounding 0.4 hour to a half hour.

Something you must already know minor Working it out
Nothing on the page says that time is distance divided by speed or that an hour has 60 minutes, and the bold word minutes requires both. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
6 ÷ 15 is less than 1, 0.4 hour, and turning it into minutes means 0.4 × 60; the cleaner route, 60 ÷ 15 = 4 minutes a mile, avoids decimals. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find how many minutes Maria needs to ride one mile at 15 miles per hour, then multiply by the number of miles to school.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the 60 minutes in an hour by the 6 miles, 60 ÷ 6 = 10, as if the ride took a full hour, and ignores the 15 mph speed.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the 15 in 15 miles per hour as the number of minutes, a number copied from the question without any calculation.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Sees that 6 miles is a little under half of 15, rounds the time of 0.4 hour up to half an hour, and gives 30 minutes.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Minutes for each mile: $\text{minutes per mile} = 60 \div \text{miles per hour}$; Miles per hour tells how many miles you ride in one hour, which is 60 minutes. Sharing the 60 minutes evenly among those miles gives the minutes for each mile. This works when the speed stays the same the whole time.
Distance, speed and time: $\text{time} = \text{distance} \div \text{speed}$; Distance is how far you go, here in miles. Speed is how far you go in one hour. Dividing the distance by the speed gives the time in hours.

Sort out the question
Maria rides at a constant speed of 15 miles per hour. Constant means it never changes. School is 6 miles away. You need the time in minutes, not hours. You must know that one hour has 60 minutes.
Find the minutes for one mile
15 miles per hour means Maria rides 15 miles in 60 minutes. Use the minutes-per-mile formula:
$60 \div 15 = 4$
So each mile takes Maria 4 minutes.
Find the minutes for 6 miles
Each of the 6 miles takes 4 minutes:
$6 \times 4 = 24$
So the ride takes 24 minutes.
Watch out: 15 is the speed, not a time, so choice B is a trap. And $60 \div 6 = 10$ , choice A, ignores the speed.
Choose the answer
Another way is the time formula. $6 \div 15 = 0.4$ , so the ride takes 0.4 of an hour. In minutes, that is $0.4 \times 60 = 24$ .
Watch out: rounding 0.4 hour up to half an hour gives choice D, 30.
So the answer is C, 24.

Check your answer

Check with the speed. 24 minutes is $\frac{24}{60} = \frac{2}{5}$ of an hour. In a full hour Maria rides 15 miles. So in 2/5 of an hour she rides $\frac{2}{5} \times 15 = 6$ miles. That is exactly the distance to school.

More questions like this

The original DOE question

Question 9 as shown on the DOE’s digital test, page 82 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 82, question 9.

Question 6 of 120

2026 Practice Test C Math, question 11

Question

Allison has 5 stamp albums with 576 stamps in each album. If she transfers her stamp collection to 6 albums, each holding 378 stamps, how many stamps will be left over?

A198
B612
C620
D632

Show the answer

Correct answer: B

Determine the total number of stamps in 5 albums holding 576 stamps:

$5 \times 576 = 2{,}880$

Determine the total number of stamps in 6 albums holding 378 stamps:

$6 \times 378 = 2{,}268$

Subtract 2,268 from 2,880 to find the number of stamps that will be left over:

$2{,}880 - 2{,}268 = 612$

Question analysis

Allison moves a stamp collection from 5 albums of 576 stamps into 6 albums that hold 378 each, and students find how many stamps are left over. It tests multiplying and subtracting large whole numbers in a word problem. The question gives no total, so the collection has to be counted before the new albums' capacity is taken away.

NYS standards: 7.EE.3 multi-step rational number problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Whole-number multiplication and subtraction in a three-step word problem is easier than grade 7 work with fractions, decimals or negatives. Because the products are large, a slip or a rounded estimate easily gives another choice.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The plan is clear from the stem: count the stamps, then take away what the six new albums hold, so the question tests careful arithmetic. Its choices punish the estimating that saves time elsewhere, since rounding either factor shifts the result by only a few tens, onto another listed number. Students who skip the album counts altogether and compare one album with one album find 198 waiting.

A very close wrong answer moderate Choosing the answer
612, 620 and 632 lie within 20 of one another, so rounding 576 and 378 to friendly numbers before multiplying lands on a wrong choice; only exact products separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Two products of three-digit numbers, 5 × 576 = 2,880 and 6 × 378 = 2,268, must be done by hand and then subtracted with regrouping. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply to find the total number of stamps, multiply again to find how many the six new albums hold, subtract, and check the last digit of each product.

Traps

Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Subtracts 378 from 576, comparing one old album with one new album, and ignores that there are 5 old albums and 6 new ones.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Rounds 576 to 580 and 378 to 380 before multiplying, so 2,900 − 2,280 = 620, an estimate rather than the exact number left over.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: weak
Rounds only 576 up to 580, getting 2,900 stamps, then subtracts the exact 2,268, so the count comes out 20 too high.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total in equal groups: $\text{total} = \text{number of groups} \times \text{amount in each group}$; Here the groups are albums. The amount in each group is how many stamps one album holds. When every group is the same size, multiplying is a fast way to add them all up.

Sort out the question
Allison has 5 albums with 576 stamps in each. She moves all her stamps into 6 new albums. Each new album holds 378 stamps. Some stamps will not fit. You need the number left over.
Count all her stamps
Use the equal-groups formula. Split 576 into 500 + 76 to make it easy:
$5 \times 576 = 5 \times 500 + 5 \times 76$
$2{,}500 + 380 = 2{,}880$
Allison has 2,880 stamps.
Count what the new albums hold
Do the same for the 6 new albums. Split 378 into 300 + 78:
$6 \times 378 = 6 \times 300 + 6 \times 78$
$1{,}800 + 468 = 2{,}268$
The new albums hold 2,268 stamps in all.
Subtract to find the leftovers
The stamps left over are the ones she has minus the ones that fit:
$2{,}880 - 2{,}268 = 612$
Watch out: $576 - 378 = 198$ , choice A, compares just one old album with one new one. Rounding to 580 and 380 first gives 620, choice C. Rounding only 576 gives 632, choice D. Use the exact numbers.
So the answer is B, 612.

Check your answer

Add the leftovers to the stamps that fit: $2{,}268 + 612 = 2{,}880$ . That is exactly the number Allison started with. So no stamps were lost or counted twice.

More questions like this

The original DOE question

Question 11 as shown on the DOE’s digital test, page 84 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 84, question 11.

Question 7 of 120

2026 Practice Test C Math, question 24

Question

In a showing of an artist’s works, the ratio of the number of paintings, drawings, and photographs shown is 3:5:4, respectively. If the number of drawings shown was 45, what is the total number of photographs and paintings that were shown?

A36
B63
C84
D108

Show the answer

Correct answer: B

The ratio of paintings to drawings to photographs is $3 : 5 : 4$ .

First, determine the number of actual paintings and photographs.

Since the number of actual drawings is 45, each ratio is multiplied by 9, since $5 \times 9 = 45$ .

Type	Ratio	Actual Number
Paintings	3	27
Drawings	5	45
Photographs	4	36

Then, add the actual number of paintings and photographs.

$27 + 36 = 63$

Question analysis

An art showing has paintings, drawings and photographs in the ratio 3:5:4, and 45 drawings were shown. Students find the total number of photographs and paintings. It tests using a three-part ratio to find actual amounts, and the question asks for two of the three groups together, not any single group or the whole show.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scaling a ratio from one known part is grade 6 and 7 work. A three-part ratio, and a question about two of the three groups combined, make it a three-step task typical of grade 7 ratio problems.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 10, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The scale factor comes from one division, 45 ÷ 5 = 9, so the risk lies in what to do with it. Two tempting results are listed: the photographs alone, and the whole show with the drawings added in. Multiplying the 3 + 4 = 7 parts for paintings and photographs by 9 gives the total in one step and avoids both.

Much to keep track of moderate Working it out
The ratio has three parts, the known count belongs to the middle part, drawings, and the question wants the first and third parts combined, so three roles must be tracked. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math minor Understanding the question
The word respectively assigns 3, 5 and 4 to paintings, drawings and photographs in that order, and the question then names two of them in the other order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find how many works each ratio part stands for from the drawings, multiply it by the parts for paintings and for photographs, and add only those two.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the photographs, 4 × 9 = 36, and stops before adding the 27 paintings that the question also asks about.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 7 parts for paintings and photographs by 12, the total of the ratio's parts, instead of by 9.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds every group correctly but adds the 45 drawings too, giving the whole show, 108, rather than the paintings and photographs.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Using the parts of a ratio: $\text{actual number} = \text{parts} \times \text{value of one part}$; The ratio 3:5:4 splits the show into equal parts. Parts is how many parts a group has. Each part stands for the same number of works. Find that number from the group you know, then multiply each group’s parts by it.

Sort out the question
The ratio of paintings to drawings to photographs is 3:5:4. Respectively means in that order: 3 for paintings, 5 for drawings and 4 for photographs. There were 45 drawings. You need the number of photographs and paintings together.
Find the value of one part
Drawings are 5 parts, and there were 45 drawings:
$45 \div 5 = 9$
So each part of the ratio stands for 9 works.
Find each group
Multiply each group’s parts by 9:
Type Parts Actual number
Paintings 3 $3 \times 9 = 27$
Drawings 5 $5 \times 9 = 45$
Photographs 4 $4 \times 9 = 36$
The drawings row gives 45, which matches the question.
Add only the two groups asked for
Add the paintings and the photographs:
$27 + 36 = 63$
Watch out: 36 alone, choice A, leaves out the paintings. Adding the 45 drawings too gives 108, choice D, the whole show.
So the answer is B, 63.

Type	Parts	Actual number
Paintings	3	$3 \times 9 = 27$
Drawings	5	$5 \times 9 = 45$
Photographs	4	$4 \times 9 = 36$

Check your answer

Paintings and photographs are $3 + 4 = 7$ parts. Each part is 9 works, so $7 \times 9 = 63$ . That matches. Also, 63 to 45 is the same as 7 to 5, just as the ratio says.

More questions like this

The original DOE question

Question 24 as shown on the DOE’s digital test, page 97 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 97, question 24.

Question 8 of 120

2026 Practice Test C Math, question 29

Question

A smoothie shop uses a consistent ratio of fruit to yogurt in its recipes. The table shows the proportional relationship between the number of cups of fruit (x) and the number of cups of yogurt (y) in different smoothie batches.

SMOOTHIE RECIPE

Cups of Fruit x	Cups of Yogurt y
6	1.5
8	2.0
10	2.5
12	3.0

What is the constant of proportionality of the number of cups of yogurt to the number of cups of fruit?

Enter your answer in the space.

Fill-in question: work out the answer and type it as a number.

Show the answer

Correct answer: 0.25

To find p, the constant of proportionality of the number of cups of yogurt, y, to the number of cups of fruit, x, use the equation $y = px$ .

The table shows that there are 1.5 cups of yogurt for 6 cups of fruit.

$1.5 = 6p$

$\frac{1.5}{6} = \frac{6p}{6}$

$\frac{1}{4} = 0.25 = p$

The constant of proportionality is 0.25.

Question analysis

A smoothie shop's recipe table lists cups of fruit and cups of yogurt for four batches in a proportional relationship. Students type the constant of proportionality of yogurt to fruit. It tests finding a unit rate from a table, and since fruit is listed first while the question names yogurt first, the ratio is easy to take upside down.

NYS standards: 7.RP.2b constant of proportionality
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the constant of proportionality in a table is a core grade 7 skill, and these values are friendly decimals. Typing the answer with no choices, and taking the ratio in the order the question gives, are the added demands.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
An answer in an unexpected form minor
Solution: 1 step
Reading level: Question and choices: about grade 6, 75 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Any row gives the answer with one division, and the values divide evenly, so the points are lost on direction. Fruit fills the first column and comes first in the stem’s own ratio, which invites dividing it by the yogurt. The table also climbs by a steady half cup of yogurt, a number that looks like a constant but ignores the 2 cups of fruit added each time. Checking that 0.25 times each fruit value gives its yogurt value settles it.

Turning words into math moderate Understanding the question
The opening sentence speaks of a ratio of fruit to yogurt, but the question asks for the constant of yogurt to fruit, so the division is y ÷ x, against the table’s column order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The constant is typed into a box with no choices, so 4, the ratio taken the other way round, is a believable entry that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: Check which quantity the question names first, divide that column by the other in any row, and confirm the quotient is the same in a second row.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

4 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides cups of fruit by cups of yogurt, 6 ÷ 1.5 = 4, following the stem’s “ratio of fruit to yogurt” and turning the requested constant upside down.
0.5 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Takes the steady rise of 0.5 cup of yogurt from row to row as the constant, ignoring that the fruit rises by 2 cups each time.
1.5 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives 1.5, the first yogurt value in the table, as if the constant were an amount in one batch rather than a ratio of two amounts.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $k = \frac{y}{x}$; k is the constant of proportionality. y and x are two matching amounts from the same row of the table. In a proportional relationship, y = kx, so y is always k times x. That is why dividing y by x gives the same k in every row.

Sort out the question
The table lists cups of fruit (x) and cups of yogurt (y) for four batches. You need the constant of proportionality of yogurt to fruit.
That is the number of cups of yogurt for each 1 cup of fruit. It is also the number you multiply x by to get y.
Divide yogurt by fruit
Use the first row: 6 cups of fruit and 1.5 cups of yogurt. Yogurt goes on top, because the question says yogurt to fruit:
$k = \frac{y}{x} = \frac{1.5}{6} = 0.25$
So there is 0.25 cup, or one quarter of a cup, of yogurt for each cup of fruit.
Watch out: the first sentence says fruit to yogurt, but the question asks the other way. Dividing $6 \div 1.5 = 4$ turns the answer upside down.
Check two more rows
The constant must be the same in every row. Test two more:
$\frac{2.0}{8} = 0.25 \qquad \frac{3.0}{12} = 0.25$
They match. Watch out: the yogurt goes up by 0.5 cup from row to row. That 0.5 is not the constant. The fruit goes up by 2 cups at the same time, and $0.5 \div 2 = 0.25$ .
Enter the number
Type only the number in the box, with no units. You can type 0.25 or .25. On this site, 1/4 counts too.
So you would enter 0.25.

Check your answer

Multiply each amount of fruit by 0.25. $6 \times 0.25 = 1.5$ , $8 \times 0.25 = 2$ , $10 \times 0.25 = 2.5$ and $12 \times 0.25 = 3$ . Every result matches the yogurt column. Also, each batch has less yogurt than fruit, so the constant should be less than 1.

More questions like this

The original DOE question

Question 29 as shown on the DOE’s digital test, page 102 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 102, question 29.

Question 9 of 120

2026 Practice Test C Math, question 33

Question

A bag contains 40 tiles that are either red, green, or blue. A tile is selected at random, the color is recorded, and the tile is put back in the bag. This process is carried out 120 times, and 18 of those times a red tile is selected. Based on the information, what is the most likely number of red tiles in the bag?

Show the answer

Correct answer: A

First, determine what fraction of the sample were red tiles, given 18 of the 120 tiles drawn were red.

$\frac{18 \div 6}{120 \div 6} = \frac{3}{20}$

To predict how many of the actual 40 tiles in the bag are red, multiply 40 by the fraction of red tiles in the sample.

$\frac{3}{20}(40) = \frac{120}{20} = 6$

Question analysis

A bag holds 40 red, green and blue tiles, and a tile is drawn and put back 120 times, with red coming up 18 times. Students predict the most likely number of red tiles in the bag. It tests using experimental results to estimate what is in the bag, and the draws outnumber the tiles, which can confuse the scaling.

NYS standards: 6.SP.7 long-run relative frequency
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Predicting from repeated trials is grade 6 probability, and grade 7 students make similar estimates from samples. The twist is scaling down, from 120 draws to a bag of 40 tiles, rather than up to a larger group.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 61 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

There are more draws than tiles, which is what makes this unusual: the 120 draws must be scaled down to a bag of 40, not up to a bigger group. Red came up 18 times because each tile went back in the bag, so 18 is not a count of red tiles. Since there are three draws for every tile, dividing the 18 red draws by 3 gives the same estimate as taking 3/20 of 40.

Turning words into math minor Understanding the question
The 18 red draws come from 120 draws with the tile put back each time, so they give a share, 3/20, to take of the 40 tiles, not a count of red tiles. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Students must know that the share of red in repeated draws estimates the share of red tiles in the bag; the stem never says to multiply 40 by that share. (Anderson & Krathwohl, 2001)

Strategy: Write the red draws as a fraction of all 120 draws, simplify it, and take that same fraction of the 40 tiles in the bag.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that red came up in 18 of 120 draws, or 15%, and gives 15 as the count, stopping before taking that share of the 40 tiles.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives 18, the number of times red was drawn, as the number of red tiles, though each tile went back and could be drawn many times.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working from 18 red draws in 120 gives 30 red tiles, three quarters of the bag, when red came up in fewer than one draw in six.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from repeated draws: $\text{red tiles} \approx \frac{\text{red draws}}{\text{all draws}} \times \text{tiles in the bag}$; The fraction is the share of draws that came up red. That share is your best guess for the share of tiles that are red. Multiply it by the number of tiles in the bag to predict the red tiles. The sign ≈ means about equal.

Sort out the question
The bag has 40 tiles, each red, green or blue. A tile is drawn, its color is written down, and it goes back in the bag. This happens 120 times, and red comes up 18 times.
You need the most likely number of red tiles in the bag.
See why 18 is not the answer
Each tile goes back in the bag after it is drawn. So the same red tile can be drawn again and again. That means 18 counts draws, not tiles.
Watch out: choice C, 18, mixes up the number of red draws with the number of red tiles.
Find the share of red draws
Red came up 18 times out of 120. Simplify by dividing the top and bottom by 6:
$\frac{18}{120} = \frac{3}{20}$
As a decimal, that is 0.15, or 15%.
Watch out: 15 is a percent here, not a number of tiles. Stopping now gives choice B.
Take that share of the bag
Use the same share of the 40 tiles in the bag:
$\frac{3}{20} \times 40 = \frac{120}{20} = 6$
So about 6 of the 40 tiles are red.
Give the answer
The best prediction is 6 red tiles. The other 34 tiles are probably green or blue.
So the answer is A, 6.

Check your answer

There were 120 draws for 40 tiles. That is 3 draws for every tile. So divide the red draws by 3: $18 \div 3 = 6$ . Same answer. Choice D, 30 red tiles, would be three quarters of the bag. Then red would come up far more than 18 times.

More questions like this

The original DOE question

Question 33 as shown on the DOE’s digital test, page 106 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 106, question 33.

Question 10 of 120

2026 Practice Test C Math, question 36

Question

A deck of 52 playing cards contains 13 hearts. Sara has 12 cards from this deck in her hand, 3 of which are hearts. The other cards remain in the deck. What is the probability that a card drawn at random from the remainder of the deck will be a heart?

A $\frac{1}{10}$
B $\frac{1}{4}$
C $\frac{3}{13}$
D $\frac{1}{3}$

Show the answer

Correct answer: B

First, determine the number of cards in the remaining deck:

$52 - 12 = 40$

Next, determine the number of hearts left in the remaining deck:

$13 - 3 = 10$

Last, determine the probability of drawing a heart from the remaining deck:

$\frac{10\text{ hearts remaining}}{40\text{ total cards remaining}} = \frac{10}{40} = \frac{1}{4}$

Question analysis

A standard deck of 52 playing cards has 13 hearts, and Sara holds 12 cards, 3 of them hearts. Students find the probability that a card drawn at random from the rest of the deck is a heart. Her hand changes both the number of hearts and the number of cards left, so both counts must be updated.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Finding a simple probability is grade 6 content. This one adds a step many grade 7 problems skip: the cards in Sara's hand must be taken out of both the hearts and the whole deck first.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 51 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The careful route takes two subtractions and a fraction: 10 hearts among 40 cards. The numbers are kind, because Sara’s hand is a quarter hearts, just like the full deck, so a student who ignores her hand and divides 13 by 52 also reaches 1/4. The wrong choices come from describing her hand instead of the deck, or from comparing hearts with the cards that are not hearts.

Much to keep track of minor Working it out
Four counts are in play, 52 cards, 13 hearts, Sara’s 12 cards and her 3 hearts, and both the hearts and the total must be reduced before the fraction is formed. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math minor Understanding the question
“The other cards remain in the deck” and “the remainder of the deck” point to the 40 cards Sara is not holding, not to the full 52 or to her hand. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work out how many cards are left in the deck and how many of them are hearts, then write the hearts left over the cards left and simplify.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the 10 hearts left in the deck to the 10 in 1/10, never dividing them by the 40 cards that remain.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Puts Sara’s 3 hearts over the deck’s 13, or her 12 cards over 52, both 3/13, which describes her hand rather than the cards left.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Compares the 10 hearts left with the 30 other cards left, a part-to-part ratio of 1 to 3, instead of with all 40 cards.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance something happens. The top counts the cards that are hearts. The bottom counts every card you could draw. Each card is equally likely to be drawn, so the fraction gives the chance.

Sort out the question
A full deck has 52 cards, and 13 of them are hearts. Sara holds 12 cards, and 3 of those are hearts. The rest of the cards stay in the deck.
You need the chance that a card drawn from the rest of the deck is a heart. Her hand changes both counts.
Count the cards left
Take Sara’s 12 cards out of the deck:
$52 - 12 = 40$
So 40 cards are left in the deck.
Count the hearts left
Sara holds 3 of the 13 hearts. Take those out:
$13 - 3 = 10$
So 10 hearts are left in the deck.
Find the probability
Use the probability formula. Put the hearts left over all the cards left, then simplify:
$P = \frac{10}{40} = \frac{1}{4}$
Watch out: comparing the 10 hearts with the 30 cards that are not hearts gives $\frac{1}{3}$ , choice D. The bottom must be all 40 cards. Choice C, $\frac{3}{13}$ , describes Sara’s hand, not the deck. Choice A, $\frac{1}{10}$ , uses the 10 hearts but forgets the 40 cards.
Give the answer
The chance of drawing a heart from the rest of the deck is 1 in 4.
So the answer is B, $\frac{1}{4}$ .

Check your answer

10 is one quarter of 40, because $4 \times 10 = 40$ . Here is another check. Sara’s hand is 3 hearts out of 12, which is also one quarter. Taking a hand that is one quarter hearts out of a deck that is one quarter hearts leaves one quarter hearts.

More questions like this

The original DOE question

Question 36 as shown on the DOE’s digital test, page 109 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 109, question 36.

Question 11 of 120

2026 Practice Test C Math, question 38

Question

This formula can be used to determine p, the number of pages that will be in a book that contains w words, with an average of r words per page, and f extra pages in the front for the title and publishing information.

$p = \frac{1}{r}w + f$

If the formula is solved for w, it can be used to find the number of words in a book with p pages. What will the formula be if it is solved for w?

A $w = r(p - f)$
B $w = \frac{1}{r}p + f$
C $w = f + rp$
D $w = r + p + f$

Show the answer

Correct answer: A

Apply the properties of equality to solve for w.

$p = \frac{1}{r}w + f$

Subtract f from both sides of the equation.

$p - f = \frac{1}{r}w$

Multiply both sides by r.

$r(p - f) = r\left(\frac{1}{r}w\right)$

$r(p - f) = w$

$w = r(p - f)$

Question analysis

A formula gives the number of pages p in a book from its number of words w, the average words per page r, and the extra front pages f. Students choose the formula solved for w. It tests rearranging a formula with inverse operations, and with only letters there are no numbers to check, so the order of the steps matters.

NYS standards: 8.EE.7 linear equations in one variable
7.EE.4a two-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve two-step equations such as 5 = x/4 + 2 with numbers. Rearranging a formula with four letters to solve for a different one is grade 8 work, with no numbers to check against.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Many steps minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 82 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The formula is short, but solving it for w means undoing two operations in reverse order, with letters only. The tempting shortcut multiplies p by r and leaves the f untouched, which yields a formula that looks as finished as the right one. Swapping p and w without undoing anything is the other quick trap. Substituting a small book, 5,000 words at 250 a page with 4 front pages, shows which formula gives back 5,000.

Variables instead of numbers moderate Understanding the question
Four letters and no numbers mean each choice has to be judged by its structure, unless students pick values such as r = 250, w = 5,000 and f = 4 to test it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Two inverse operations must come in order, subtracting f before multiplying by r; multiplying first means the f would have to be multiplied by r as well. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Treat it like a two-step equation with numbers: undo what was done to w last, then the step before, applying each operation to the whole of the other side.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats solving for w as trading the places of p and w, keeping the 1/r and the + f exactly where they were.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Multiplies p by r to undo the 1/r but leaves the + f alone, neither subtracting it first nor multiplying it by r too.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the other three letters, r + p + f, keeping the formula’s letters but none of its multiplication or division.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying by the reciprocal: $r \times \frac{1}{r} = 1$; r is any number except 0. The reciprocal of r is 1/r, the number flipped upside down. A number times its reciprocal is always 1. So multiplying by r undoes the 1/r in front of w and leaves w by itself.

Sort out the question
The formula is $p = \frac{1}{r}w + f$ . Here p is the number of pages, w the number of words, r the words on each page, and f the extra front pages.
You need the formula solved for w. That means w alone on one side of the equal sign.
See what happens to w
Read the formula like a recipe. Start with w. First it is multiplied by $\frac{1}{r}$ , which is the same as dividing by r. Then f is added.
To get w back, undo the steps in reverse order. Take away f first. Then undo the $\frac{1}{r}$ .
Subtract f from both sides
Taking f from both sides keeps them equal:
$p - f = \frac{1}{r}w$
Now the only thing left to undo is the $\frac{1}{r}$ .
Multiply both sides by r
Use the reciprocal rule. Multiply both whole sides by r:
$r(p - f) = r \times \frac{1}{r}w = w$
So $w = r(p - f)$ .
Watch out: $w = f + rp$ , choice C, multiplies only p by r. The f must be taken away before you multiply. Or, if you multiply first, every term gets multiplied by r.
Give the answer
Choice B only trades the places of p and w, and it undoes nothing. Choice D just adds the letters.
So the answer is A, $w = r(p - f)$ .

Check your answer

Try a book with 5,000 words, 250 words a page and 4 front pages. The first formula gives $p = 5{,}000 \div 250 + 4 = 24$ pages. Choice A gives back the words: $250 \times (24 - 4) = 5{,}000$ . Choice C gives $4 + 250 \times 24 = 6{,}004$ , which is wrong.

More questions like this

The original DOE question

Question 38 as shown on the DOE’s digital test, page 111 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 111, question 38.

Question 12 of 120

2026 Practice Test C Math, question 40

Question

In basketball, a player can earn 1 point for making a free throw and 2 points for making another basket. Marcus makes a mean of 8 free throws per game, and he scores a total of at least 25 points per game. The inequality below shows this relationship, where b represents the number of 2-point baskets Marcus makes.

$2b + 8 \geq 25$

What is the number of 2-point baskets Marcus needs to make if he wants to score at least his minimum total points for a game and make his mean number of free throws?

A $b \leq 8$
B $b \geq 9$
C $b \geq 16$
D $b \leq 25$

Show the answer

Correct answer: B

Solve the inequality $2b + 8 \geq 25$ .

$2b + 8 - 8 \geq 25 - 8$

$2b \geq 17$

$\frac{2b}{2} \geq \frac{17}{2}$

$b \geq 8.5$

Since the number of baskets made must be a whole number, the fewest 2-point baskets, b, he must make is 9.

Question analysis

In basketball, Marcus makes a mean of 8 one-point free throws a game and wants at least 25 points, shown by the inequality 2b + 8 ≥ 25. Students find how many 2-point baskets he needs. It tests solving a two-step inequality and interpreting the solution in context, since a player cannot make part of a basket.

NYS standards: 7.EE.4b two-step inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving a two-step inequality from a word problem is a grade 7 standard. Here the inequality is already written, so the extra demand is interpreting its solution as a count of baskets, typical of grade 7 tasks.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: An answer in an unexpected form minor
Complex sentences minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 92 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The inequality is handed over, so the solving is two routine moves; the judgment comes after. Dividing 17 by 2 leaves a half, and since a basket cannot be split, the boundary has to move up to 9: eight baskets give 2(8) + 8 = 24 points, one short of 25. Students who flip the symbol while dividing, or who move the 8 without changing its sign, still find a choice to match.

An answer in an unexpected form minor Choosing the answer
Solving gives b ≥ 8.5, which no choice shows; it has to be recognized as b ≥ 9, because Marcus cannot make half a basket. (Enright, Morley & Sheehan, 2002)
Complex sentences minor Understanding the question
The 33-word question restates the given inequality as “at least his minimum total points” and “make his mean number of free throws”, wording harder than the math it describes. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Undo the added 8 and then the 2 to solve for b, then ask which whole numbers of baskets actually reach 25 points, testing the boundary value in the original inequality.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reverses the inequality when dividing by 2, as if dividing always flipped it, so b ≤ 8.5, and whole baskets give b ≤ 8.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves the 8 to the right without changing its sign, so 2b ≥ 33 and b ≥ 16.5, then drops the half to match b ≥ 16.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 25-point total from the inequality as a limit on baskets and writes at most, a number copied from the question rather than solved for.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Solving an inequality: $\text{if } A \geq B, \text{ then } A - C \geq B - C \text{ and } \frac{A}{2} \geq \frac{B}{2}$; A and B are the two sides. The sign ≥ means greater than or equal to. Taking the same amount C from both sides keeps the sign the same. Dividing both sides by a positive number, like 2, also keeps it the same. Only a negative number flips it.

Sort out the question
A free throw is worth 1 point, and each other basket is worth 2 points. Marcus makes 8 free throws, which give 8 points. The letter b is the number of 2-point baskets, so they give 2b points.
He wants at least 25 points in all. That is the inequality $2b + 8 \geq 25$ . An inequality compares two amounts. The sign ≥ means greater than or equal to. You need the values of b that work.
Subtract 8 from both sides
Undo the added 8 first:
$2b + 8 - 8 \geq 25 - 8$
$2b \geq 17$
Watch out: adding 8 instead gives $2b \geq 33$ . That leads to choice C, which is wrong.
Divide both sides by 2
2b means 2 times b. Divide both sides by 2. Since 2 is positive, the sign stays the same:
$\frac{2b}{2} \geq \frac{17}{2}$
$b \geq 8.5$
Watch out: do not flip the sign here. Flipping it gives $b \leq 8.5$ , which leads to choice A.
Think about whole baskets
Marcus cannot make half a basket. So test the whole numbers near 8.5:
$2 \times 8 + 8 = 24$
$2 \times 9 + 8 = 26$
Eight baskets give 24 points, one short of 25. Nine baskets give 26 points, which is enough. So he needs 9 or more baskets.
Give the answer
The number of baskets is 9 or more, written $b \geq 9$ . Choice D, $b \leq 25$ , just copies the 25 from the points.
So the answer is B, $b \geq 9$ .

Check your answer

Try a bigger number, such as b = 12. Then $2 \times 12 + 8 = 32$ points, which is at least 25. Any number of baskets above 9 also works, and that is what ≥ says. Below 9, the total stays at 24 points or fewer.

More questions like this

The original DOE question

Question 40 as shown on the DOE’s digital test, page 113 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 113, question 40.

Question 13 of 120

2026 Practice Test C Math, question 41

Question

Elijah is at the school bookstore. He wants to buy a water bottle for $12.50 and some notebooks that cost $4.75 each. He has at most $35.00 to spend. Write an inequality that represents the number of notebooks, n, Elijah can purchase.

Enter your answer in the space provided. Enter only your answer.

On the test you build the answer on an on-screen keypad; here, type it. Use ^ for a power, sqrt( ) or √ for a root, pi for π, and <= or >= for ≤ and ≥.

Show the answer

Correct answer: 12.50 + 4.75n ≤ 35.00

The question uses the phrase “at most,” which means “less than or equal to” (≤).

Since notebooks cost $4.75 each and Elijah will buy n notebooks, the total cost just for the notebooks is found by multiplying $4.75 by n, or $4.75n$ .

Combining the cost of a water bottle ($12.50) with the cost of n notebooks, the inequality is $12.50 + 4.75n \leq 35.00$ .

This may also be written as $4.75n + 12.50 \leq 35.00$ .

Question analysis

Elijah has at most $35.00 to buy one water bottle and some notebooks at the school bookstore, with a price given for each. Students type an inequality for the number of notebooks, n, he can buy. It tests writing an inequality from a word problem. The two prices play different roles, and the phrase "at most" decides the inequality symbol.

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an inequality for a budget with a one-time cost and a cost per item is typical grade 7 work. The answer is typed, so both the symbol for "at most" and the role of each price must be right.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 50 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The prices are given in the order they are used, so the expression for the total is the easier half. The points are lost on the symbol: “at most” has to be heard as no more than, with equality allowed, and the keypad offers all four inequality signs side by side. Pricing one and two notebooks with the bottle, $17.25 and $22.00, shows which amount repeats and confirms the total.

Turning words into math moderate Understanding the question
The $12.50 bottle is bought once and the $4.75 is paid for each of n notebooks, and “at most $35.00” has to become ≤, which allows a total of exactly $35.00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The whole inequality, with n, its symbol and both prices, is built on a keypad offering ≤, <, ≥ and >, so a wrong symbol looks as complete as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Price a few cases by hand, such as 1 and 2 notebooks with the water bottle, to see which amount repeats, then decide whether a total of exactly $35.00 is allowed.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

12.50 + 4.75n ≥ 35.00 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Builds the total correctly but reads “at most” as at least, writing ≥, which would mean Elijah must spend $35.00 or more.
12.50 + 4.75n < 35.00 Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: moderate
Builds the total correctly but uses a strict less-than sign, leaving out a total of exactly $35.00, which “at most” allows.
17.25n ≤ 35.00 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the bottle’s price to each notebook’s, $12.50 + $4.75 = $17.25, and multiplies by n, as if a bottle came with every notebook.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a one-time cost: $\text{total} = \text{one-time cost} + (\text{price each}) \times n$; The one-time cost is paid once, no matter what. The price each is paid once for every item. n is the number of items. You multiply the price each by n because you pay it n times.
At most: $\text{total} \leq \text{limit}$; The sign ≤ means less than or equal to. The total can be under the limit or exactly equal to it, but never over it. Use it for words like at most and no more than.

Sort out the question
Elijah buys one water bottle for $12.50. He also buys n notebooks at $4.75 each. He can spend at most $35.00.
You need an inequality for n. An inequality compares two amounts with a sign such as ≤ instead of =.
Price a few cases
Work out the cost for a few numbers of notebooks:
Notebooks Total ($)
1 $12.50 + 4.75 = 17.25$
2 $12.50 + 9.50 = 22.00$
3 $12.50 + 14.25 = 26.75$
The $12.50 shows up once. The $4.75 shows up once for each notebook.
Write the total cost
Use the total cost formula. The bottle is the one-time cost, and each notebook costs $4.75:
$12.50 + 4.75n$
Here 4.75n means 4.75 times n.
Watch out: do not add the two prices first. $17.25n$ charges for a new bottle with every notebook.
Choose the sign
At most $35.00 means the total can be less than $35.00 or exactly $35.00. It cannot be more. So use ≤:
$12.50 + 4.75n \leq 35.00$
Watch out: ≥ means at least, the opposite. And < leaves out a total of exactly $35.00, which Elijah is allowed to spend.
Enter it on the keypad
Type the whole inequality. Nothing is printed before the box. On the test, press 1, 2, ., 5, 0, +, 4, ., 7, 5, the n key, the ≤ key, then 3, 5, ., 0, 0. The ≤ key is in the fourth row. Skip the dollar signs. On this site, type <= for ≤.
So you would enter 12.50 + 4.75n ≤ 35.00. An equal form, such as 4.75n + 12.50 ≤ 35.00, counts too.

Notebooks	Total ($)
1	$12.50 + 4.75 = 17.25$
2	$12.50 + 9.50 = 22.00$
3	$12.50 + 14.25 = 26.75$

Check your answer

Test some numbers. For 4 notebooks, $12.50 + 4.75 \times 4 = 31.50$ , which is not over 35.00. For 5 notebooks, $12.50 + 4.75 \times 5 = 36.25$ , which is too much. So Elijah can buy up to 4 notebooks. That makes sense for a budget of $35.00.

More questions like this

The original DOE question

Question 41 as shown on the DOE’s digital test, page 114 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 114, question 41.

Question 14 of 120

2026 Practice Test C Math, question 46

Question

A rain gauge shows that 10 centimeters of rain fell yesterday. If 1 inch is equal to approximately 2.5 centimeters, about how many inches of rain fell yesterday?

A1
B2.5
C4
D10

Show the answer

Correct answer: C

Set up a proportion to solve:

$\frac{1\text{ in.}}{2.5\text{ cm}} = \frac{x\text{ in.}}{10\text{ cm}}$

$2.5x = 10$

$x = 10 \div 2.5$

$x = 4$

10 centimeters is approximately 4 inches.

Question analysis

A rain gauge shows that 10 centimeters of rain fell, and 1 inch is about 2.5 centimeters. Students find about how many inches of rain fell. It tests converting between metric and customary units with a given conversion fact, and the main risk is converting in the wrong direction or reusing a number from the question.

NYS standards: 6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting units with a given conversion fact is grade 6 content, and this needs a single division with friendly numbers, lighter than grade 7 work with rates and percents.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Solution: 1 step
Reading level: Question and choices: about grade 10, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

A single division answers the question, and its numbers are friendly: four lengths of 2.5 centimeters make 10. Every wrong choice is simply a number from the stem, so the only real risk is reaching for a given value instead of converting. Inches are longer than centimeters, so the rainfall in inches has to be less than 10 but more than the 1 inch that 2.5 centimeters make.

Turning words into math minor Understanding the question
The fact “1 inch is equal to approximately 2.5 centimeters” has to be turned into a division, how many 2.5-centimeter lengths fit into 10, rather than a multiplication. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Ask how many groups of 2.5 centimeters fit into the 10 centimeters of rain, since each group of 2.5 centimeters makes one inch.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks the 1 from the conversion fact, 1 inch for 2.5 centimeters, reading the given fact as the answer instead of applying it to 10 centimeters.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 2.5, the number of centimeters in one inch, straight from the conversion fact rather than finding how many inches 10 centimeters make.
D Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: moderate
Keeps the 10 from the rain gauge, as if centimeters and inches were the same size, so no conversion is done at all.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Converting centimeters to inches: $\text{inches} = \text{centimeters} \div 2.5$; About 2.5 centimeters make 1 inch. So each group of 2.5 centimeters is 1 inch. Dividing the centimeters by 2.5 counts how many of those groups fit. That count is the number of inches.

Sort out the question
The rain gauge shows 10 centimeters (cm) of rain. You know that 1 inch is about 2.5 cm. You need the rain in inches.
Decide whether to multiply or divide
An inch is longer than a centimeter. So it takes fewer inches than centimeters to measure the same rain. The answer must be less than 10. That tells you to divide, not multiply.
Watch out: choice D, 10, keeps the same number, as if inches and centimeters were the same size.
Divide by 2.5
Use the conversion formula:
$10 \div 2.5 = 4$
You can see it by adding: $2.5 + 2.5 + 2.5 + 2.5 = 10$ . Four groups of 2.5 cm fit into 10 cm. So 10 cm is about 4 inches.
Give the answer
Watch out: choices A and B, 1 and 2.5, are just the numbers in the conversion fact. They do not use the 10 cm at all.
About 4 inches of rain fell. So the answer is C, 4.

Check your answer

Change back. Each inch is 2.5 cm, so 4 inches is $4 \times 2.5 = 10$ cm. That is what the gauge showed. Also, 4 inches is less than 10, as it should be, because inches are longer.

More questions like this

The original DOE question

Question 46 as shown on the DOE’s digital test, page 119 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 119, question 46.

Question 15 of 120

2026 Practice Test C Math, question 47

Question

The height of a plant, in millimeters, is proportional to the number of days the plant has grown. The plant has grown 20 millimeters in 8 days. What is the constant of proportionality, in millimeters per day?

A0.4
B2.5
C12
D20

Show the answer

Correct answer: B

The constant of proportionality is determined by dividing the number of millimeters by the number of days:

$\frac{20\text{ mm}}{8\text{ days}} = \frac{20}{8} = 2.5\text{ mm/day}$

Question analysis

A plant's height in millimeters is proportional to the number of days it has grown, and it grew 20 millimeters in 8 days. Students find the constant of proportionality in millimeters per day. It tests finding a unit rate, and the order of the division decides whether the rate comes out in the units asked for.

NYS standards: 7.RP.2b constant of proportionality
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a constant of proportionality from one pair of values is a basic grade 7 skill. With one division and the units named in the question, it asks less than most grade 7 problems.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 7, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One division gives the constant, and the question itself names the order by asking for millimeters per day. The tempting slip is dividing the smaller number by the larger, 8 by 20, out of habit, which gives a rate in days per millimeter. A plant that grows 20 millimeters in 8 days grows more than 2 millimeters a day, a quick sense check on any result.

Something you must already know minor Working it out
Students must know that the constant is the height divided by the days; the units in the question, millimeters per day, name the order of the division. (Anderson & Krathwohl, 2001)

Strategy: Use the units in the question, millimeters per day, as the instruction: divide the millimeters grown by the number of days.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides the days by the millimeters, 8 ÷ 20 = 0.4, which gives days per millimeter, the rate turned upside down.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts the days from the millimeters, 20 − 8 = 12, combining the two numbers without thinking about a rate.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the 20 millimeters the plant grew over all 8 days, the total growth, rather than the growth for each single day.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $k = \frac{y}{x}$; Here y is the height in millimeters, and x is the number of days. k is the growth in one day, in millimeters per day. In a proportional relationship, y = kx, so dividing y by x gives k.

Sort out the question
The plant’s height is proportional to the number of days it grows. It grew 20 millimeters (mm) in 8 days.
You need the constant of proportionality in millimeters per day. Per day means for each day. So you need how much the plant grows in one day.
Divide millimeters by days
The units tell you the order: millimeters per day means millimeters divided by days. Use the formula:
$k = \frac{20}{8} = \frac{5}{2} = 2.5$
So the plant grows 2.5 mm each day.
Avoid the tempting slips
Watch out: $8 \div 20 = 0.4$ , choice A, is days per millimeter. That is the rate upside down.
Choice C, 12, comes from $20 - 8$ . Subtracting does not make a rate. Choice D, 20, is the growth over all 8 days, not one day.
Give the answer
The constant of proportionality is 2.5 mm per day.
So the answer is B, 2.5.

Check your answer

Multiply back. At 2.5 mm a day for 8 days, the plant grows $2.5 \times 8 = 20$ mm. That matches. Also, 2 mm a day would give only 16 mm, and 3 mm a day would give 24 mm. So the rate must be between 2 and 3.

More questions like this

The original DOE question

Question 47 as shown on the DOE’s digital test, page 120 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 120, question 47.

Question 16 of 120

2026 Practice Test D Math, question 2

Question

What is the value of z in $\frac{x}{10} + \frac{y - x}{5} = \frac{z}{10}$ ?

A1
B5
C $y$
D $2y - x$

Show the answer

Correct answer: D

First, multiply both sides of the equation by 10 to eliminate the fractions. In other words, multiply each term by 10.

$\frac{x}{10} + \frac{y - x}{5} = \frac{z}{10}$

$10\left(\frac{x}{10} + \frac{y - x}{5}\right) = 10\left(\frac{z}{10}\right)$

$\frac{10x}{10} + \frac{10(y - x)}{5} = \frac{10z}{10}$

$x + 2(y - x) = z$

Then, distribute and combine like terms to find the value for z.

$x + 2(y - x) = z$

$x + 2y - 2x = z$

$2y - x = z$

Question analysis

This algebra question gives an equation with three letters and fractions, x over 10 plus (y − x) over 5 equals z over 10, and asks for the value of z. Students clear the denominators, distribute and combine like terms. The two fractions on the left have different denominators, which is where the usual slip happens.

NYS standards: 7.EE.1 linear expressions with rational coefficients
7.EE.2 rewriting expressions to show relationships
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students add and expand linear expressions with fraction coefficients, usually in one letter. Here three letters appear, one numerator is a difference, and the result is an expression rather than a number, which adds abstraction but no new skill.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 6, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The equation has three fractions and three letters, and the work is three routine moves: multiply through by 10, distribute, collect the x terms. Most errors come from the middle term, whose denominator 5 differs from the other two; treating it as tenths makes z simply y. Students who test values should avoid x = y, since that pair makes y and 2y − x agree, and x = y = 1 even makes the number 1 fit.

Variables instead of numbers moderate Understanding the question
The answer is z in terms of x and y, not a number, yet two choices are plain numbers, 1 and 5, which tempt students who expect the equation to give z a single value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic moderate Working it out
Clearing the denominators turns (y − x)/5 into 2(y − x), and the −2x from distributing must then combine with the lone x to leave −x, two places where a factor or a sign slips. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply every term on both sides by 10 so no fractions remain, distribute across the parentheses, and combine the x terms; substituting small numbers for x and y checks the result.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: weak
Tests x = 1 and y = 1, gets 1/10 + 0 = z/10, so z = 1, and takes the choice that reads 1; any pair with x ≠ y rules it out.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Expects z to be a number and takes the 5 from the middle denominator, a number copied from the equation rather than the result of any step.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the numerators x and y − x over 10 as if (y − x)/5 were already in tenths, so the left side becomes y/10 and z = y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides: $\text{if } a = b \text{, then } 10 \times a = 10 \times b$; a and b are the two sides of an equation. Multiplying both sides by the same number keeps them equal, like adding the same weight to both pans of a balance. Here 10 is a good choice, because it clears away every fraction.
The distributive property: $a(b - c) = a \times b - a \times c$; a is the number outside the parentheses, and b and c are the terms inside. Multiply a by each term inside, and keep the minus sign between them. It works because a groups of (b − c) is a groups of b minus a groups of c.

Sort out the question
The equation has three letters, x, y and z. You need z written using x and y. The answer will not be one number, because x and y can be anything. The fractions make the work messy, so clear them first.
Multiply every term by 10
The bottoms are 10, 5 and 10, and each one divides into 10. So multiply every term by 10:
$10 \times \frac{x}{10} + 10 \times \frac{y - x}{5} = 10 \times \frac{z}{10}$
10 times x/10 is x. Since 10 ÷ 5 = 2, 10 times (y − x)/5 is 2(y − x). And 10 times z/10 is z.
$x + 2(y - x) = z$
Distribute the 2
Use the distributive property. Multiply the 2 by both y and x:
$2(y - x) = 2y - 2x$
So the equation becomes:
$x + 2y - 2x = z$
Combine like terms
Like terms have the same letter. Here x and −2x are like terms. Since 1 − 2 = −1, they combine to −x:
$x - 2x = -x$
That leaves:
$2y - x = z$
Watch out: do not add the tops of x/10 and (y − x)/5 straight away. Their bottoms are different. Doing it anyway gives z = y, choice C. Choices A and B are plain numbers, but z changes when x and y do.
So the answer is D, $2y - x$ .

Check your answer

Pick numbers. Let x = 2 and y = 7. The left side is 2/10 + 5/5 = 0.2 + 1 = 1.2. Choice D gives z = 2 × 7 − 2 = 12, and 12/10 = 1.2. Both sides match.

More questions like this

The original DOE question

Question 2 as shown on the DOE’s digital test, page 75 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 75, question 2.

Question 17 of 120

2026 Practice Test D Math, question 6

Question

The cost of shipping a package is $12, plus $2 for each additional pound over 3 pounds. Which expression represents the total cost, in dollars, of shipping a package that weighs p pounds? Assume that $p \geq 3$ .

A $12 + 2(p + 3)$
B $12 + 2(p - 3)$
C $14p + 3$
D $14p - 3$

Show the answer

Correct answer: B

Since the additional cost per pound is not added until the package weighs more than 3 pounds, the total cost for 1, 2, or 3 pounds is $12.

1 pound is $12
2 pounds is $12
3 pounds is $12

Since p is greater than or equal to 3 pounds, the total cost for:

4 pounds is

$12 + (4 - 3)(2) = 14$

5 pounds is

$12 + (5 - 3)(2) = 16$

p pounds is

$12 + (p - 3)(2) = 12 + 2(p - 3)$

You may try to write the expression without generating the pattern by recognizing that subtracting 3 from the number of pounds for the package before multiplying by $2 and then adding $12 results in the total cost.

Question analysis

Shipping a package costs $12, plus $2 for each additional pound over 3 pounds, and students choose the expression for the total cost of a package that weighs p pounds. It tests writing an algebraic expression from a word problem. The per-pound charge starts only after the first 3 pounds, and turning that phrase into algebra is the tricky part.

NYS standards: 6.EE.6 variables representing unknown numbers
6.EE.2a writing algebraic expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Writing an expression from a cost description is grade 6 algebra. Because the per-pound charge starts only after the first 3 pounds, it is less routine than a fee plus a simple rate, in line with typical grade 7 demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards modeling the fee in two parts: a flat $12 that covers the first 3 pounds, and $2 for each pound beyond them. The two choices that start with 12 look equally reasonable until a weight is tried, and the two that begin with 14p come from merging the fee with the per-pound charge. Testing p = 3, which must cost exactly $12, rules out three choices in one step.

Turning words into math moderate Understanding the question
“$2 for each additional pound over 3 pounds” must become 2(p − 3), a charge on the pounds beyond the first 3, and the word “over” invites writing p + 3 instead. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
12 + 2(p + 3) and 12 + 2(p − 3) differ only in the sign inside the parentheses, so their shape cannot separate them; only a test weight or a careful reading can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
The weight is the letter p, so there is no cost to work out and compare; choosing a weight such as 5 pounds, which costs $16, turns every choice into a number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Test a weight such as 5 pounds: work out its cost from the description, then substitute 5 for p in each choice and keep the one that gives the same cost.

Traps

Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Reads “over 3 pounds” as adding 3, so the $2 charge applies to p + 3 pounds instead of only the pounds above the first 3.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Merges the $12 and the $2 into $14 for every pound and tacks on the 3 from the stem, building an expression from its numbers without a model.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Charges $14 for every pound by merging the fee with the $2 rate, then subtracts 3 for the first 3 pounds, mixing dollars with pounds.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost with a base fee and a rate: $\text{total} = \text{base fee} + \text{rate} \times \text{extra pounds}$; The base fee is the set price, $12, which covers the first 3 pounds. The rate is the price for each pound after that, $2. The extra pounds are the pounds over 3, which is p − 3. Use it when a price has a fixed part plus a part per unit.

Sort out the question
Shipping costs $12 to start. Each pound over 3 pounds adds $2. The package weighs p pounds, and p is at least 3. You need an expression for the total cost in dollars.
Count the extra pounds
Only the pounds above 3 cost extra. To find them, take 3 away from the weight:
$\text{extra pounds} = p - 3$
For example, a 5-pound package has 5 − 3 = 2 extra pounds.
Watch out: over 3 pounds means more than 3. It does not tell you to add 3. Writing p + 3 leads to choice A.
Build the expression
Each extra pound costs $2, so the extra pounds cost 2(p − 3) dollars. Add the $12 base fee:
$\text{total} = 12 + 2(p - 3)$
Watch out: the $12 is paid once, not for every pound. Choices C and D charge $14 for every pound, which mixes up the fee and the rate.
Test each choice with 5 pounds
A 5-pound package costs $12 plus two $2 charges, which is $16. Put p = 5 into each choice:
Choice Value when p = 5
A 12 + 2(8) = 28
B 12 + 2(2) = 16
C 14(5) + 3 = 73
D 14(5) − 3 = 67
Only B gives $16. So the answer is B, $12 + 2(p - 3)$ .

Choice	Value when p = 5
A	12 + 2(8) = 28
B	12 + 2(2) = 16
C	14(5) + 3 = 73
D	14(5) − 3 = 67

Check your answer

Try p = 3, the smallest weight allowed. There are no extra pounds, so it should cost just $12. Choice B gives 12 + 2(0) = 12. Try p = 4: 12 + 2(1) = 14, which is $12 plus one $2 charge.

More questions like this

The original DOE question

Question 18 of 120

2026 Practice Test D Math, question 12

Question

$A circle with center P. Points R and T are on the circle at the ends of a horizontal diameter through P, and point S is on the circle above it. Segment SP is labeled 12. A vertical segment from S meets RT at a right angle, dividing RT into a part labeled x, from R to that point, and a part labeled 5, from that point to T.$

P is the center of the circle above, and R, S, and T are points on the circle. What is the value of x?

Show the answer

Correct answer: D

$\overline{\mathrm{SP}}$ represents the radius of the circle and has length 12.

Both $\overline{\mathrm{RP}}$ and $\overline{\mathrm{PT}}$ are radii and also have length 12.

Therefore, the length of $\overline{\mathrm{RT}}$ , which is a diameter, is 24 and is also represented by $x + 5$ .

$x + 5 = 24$

$x = 19$

Question analysis

A circle with center P has points R, S and T on it, a segment SP labeled 12, and a vertical segment from S that splits RT into a piece labeled x and a piece labeled 5. Students find the value of x. It tests basic circle facts, and the challenge is seeing which lengths in the figure must be equal.

NYS standards: 7.G.4 circle area and circumference
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 circle work centers on area and circumference formulas. This needs only the fact that every radius has the same length, but spotting which segments are radii in a figure with extra markings is less routine than using a formula.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
No obvious method minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 24 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The figure gives one measurement on a slanted radius and two labels on the diameter, and the right angle at the foot of the vertical segment is a decoy. Once RP and PT are seen as radii, RT is 24, and x is that diameter less the 5. Students who never connect the slanted 12 to the horizontal line fall back on the labeled numbers, adding 12 and 5, or take x to end at the center.

Reading a figure, table or graph moderate Understanding the question
The 12 labels the slanted segment SP, not the horizontal one, and the arrow for x runs from R past the center P to the foot of the vertical segment, so x is neither a radius nor a whole diameter. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
No obvious method minor Working it out
The right-angle mark and the triangle under S suggest the Pythagorean theorem, which is no help; the way in is seeing that RP and PT are radii as well. (Schoenfeld, 1985)

Strategy: Mark every segment from P to the circle as 12, notice that R and T are the ends of a diameter, then write an equation for that diameter using x and the 5.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Information you don't need: The question gives numbers or facts that play no part in the answer.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Reads x as the radius RP, as if its arrow ended at the center P, though it runs past P to the foot of the vertical segment.
B Trusting the picture (Lesh, Post & Behr, 1987)
Reveals a surface strategy Expected pull: weak
Judges x from the drawing, where it looks about three times as long as the piece labeled 5, instead of working from the radius of 12.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Adds the two labeled numbers, 12 + 5 = 17, without placing the 12 on the diameter, as if x were a radius plus the 5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radius and diameter: $d = 2r$; r is the radius, the distance from the center to any point on the circle. d is the diameter, the distance straight across the circle through the center. A diameter is two radii laid end to end, one on each side of the center, so it is twice as long.

Sort out the question
P is the center of the circle, and R, S and T are on the circle. Segment SP is 12. Segment RT goes straight across through P. A short line from S meets RT at a right angle. It splits RT into x, from R to that point, and 5, from that point to T. You need x.
See that SP is a radius
SP runs from the center P to the point S on the circle. So SP is a radius, and the radius is 12. Every radius of a circle has the same length. RP and PT also run from the center to the circle, so RP = 12 and PT = 12.
Find the diameter RT
RT passes through the center, so it is a diameter. Use the formula with r = 12:
$\mathrm{RT} = 2r = 2 \times 12 = 24$
RT is 24 long. It is the radius RP plus the radius PT.
Solve for x
RT is also split into x and 5, so:
$x + 5 = 24$
Subtract 5 from both sides:
$x = 24 - 5 = 19$
Watch out: x runs from R past the center P, so it is longer than a radius. Reading it as 12, choice A, stops at P. Adding 12 + 5 = 17, choice C, puts the 12 in the wrong place.
Work from the numbers, not the picture
The figure is not drawn to scale. In it, x looks about three times as long as 5, which suggests 15, choice B. Trust the labels instead. The right triangle under S is not needed at all.
So the answer is D, 19.

Check your answer

The point where the short line meets RT is 19 from R. P is 12 from R, so that point is 19 − 12 = 7 past the center. Then 7 + 5 = 12, which is PT, a radius. Everything fits.

More questions like this

The original DOE question

Question 12 as shown on the DOE’s digital test, page 85 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 85, question 12.

Question 19 of 120

2026 Practice Test D Math, question 16

Question

The table shows the proportional relationship between x cups of oatmeal and y ounces of raisins in a bread recipe.

BREAD RECIPE

Cups of Oatmeal (x)	Ounces of Raisins (y)
2	1.0
3	1.5
6	3.0

What is the constant of proportionality of the number of ounces of raisins to the number of cups of oatmeal?

Enter your answer in the space.

Fill-in question: work out the answer and type it as a number.

Show the answer

Correct answer: 0.5

To find p, the constant of proportionality of ounces of raisins, y, to cups of oatmeal, x, use the equation $y = px$ .

The table shows that there is 1 ounce of raisins for 2 cups of oatmeal.

$1 = 2p$

$\frac{1}{2} = \frac{2p}{2}$

$0.5 = p$

The constant of proportionality is 0.5.

Question analysis

A table for a bread recipe lists cups of oatmeal and the ounces of raisins that go with them, in a proportional relationship. Students type the constant of proportionality of raisins to oatmeal. It tests reading a unit rate from a table, and the order in which the question names the two quantities matters.

NYS standards: 7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a constant of proportionality from a table is a grade 7 standard, and here a single division in one row gives it. The demand is lighter than typical grade 7 work; the order of the ratio in the wording needs the most care.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math moderate
An answer in an unexpected form minor
Solution: 1 step
Reading level: Question and choices: about grade 6, 56 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every row gives the same ratio, so the division itself is easy; the question tests which quantity goes on top. Because oatmeal comes first in the table and its numbers are the larger ones, dividing cups by ounces feels natural and produces a clean whole number. Writing the relationship as y = kx, raisins equal k times oatmeal, and checking it on a second row such as 3 cups and 1.5 ounces fixes the order.

Turning words into math moderate Understanding the question
The phrase “of the number of ounces of raisins to the number of cups of oatmeal” fixes the order y ÷ x, while the table lists oatmeal first, inviting 2 ÷ 1.0. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The constant is typed in with no choices, so an upside-down ratio of 2, a whole number, looks as finished as the decimal the table actually gives. (Enright, Morley & Sheehan, 2002)

Strategy: Divide an ounces-of-raisins value by the cups of oatmeal in the same row, in the order the question names them, and check that every row gives the same value.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

2 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Divides cups of oatmeal by ounces of raisins, 2 ÷ 1.0, turning the ratio upside down: cups per ounce instead of ounces per cup.
1 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Reads the first raisin amount, 1.0 ounce, as the constant without dividing it by the 2 cups of oatmeal in the same row.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $y = kx \quad \text{so} \quad k = \frac{y}{x}$; x and y are two amounts that grow together at the same rate. k is the constant of proportionality, the number you always multiply x by to get y. Here y is ounces of raisins and x is cups of oatmeal, so k is ounces of raisins per cup of oatmeal.

Sort out the question
The table shows cups of oatmeal, x, and ounces of raisins, y. Proportional means the raisins are always the same multiple of the oatmeal. That multiple is the constant of proportionality. The question asks for raisins to oatmeal. So you want ounces of raisins for each cup of oatmeal: y ÷ x.
Divide y by x in one row
Use the first row of the table: 2 cups of oatmeal go with 1.0 ounce of raisins.
$k = \frac{y}{x} = \frac{1.0}{2} = 0.5$
So there is 0.5 ounce of raisins for each cup of oatmeal.
Check the other rows
A constant must be the same in every row:
$\frac{1.5}{3} = 0.5 \qquad \frac{3.0}{6} = 0.5$
Every row gives 0.5, so the rule is $y = 0.5x$ .
Watch out: the table lists oatmeal first, which tempts you to divide 2 by 1.0. That gives 2, which is cups per ounce, the ratio upside down. And 1.0 alone is just one raisin amount, not the constant.
Enter the answer
Type the constant in the box as a number, with no units. On this site, 0.5, .5 and 1/2 all count.
So you would enter 0.5.

Check your answer

Multiply each oatmeal amount by 0.5: 2 × 0.5 = 1.0, 3 × 0.5 = 1.5 and 6 × 0.5 = 3.0. Each result matches the raisins in its row. Also, every raisin number is smaller than its oatmeal number, so the constant must be less than 1.

More questions like this

The original DOE question

Question 16 as shown on the DOE’s digital test, page 89 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 89, question 16.

Question 20 of 120

2026 Practice Test D Math, question 22

Question

Seth grows strawberries in his garden.

He picked $2\frac{1}{2}$ cups of strawberries on Monday.
He picked 4 cups of strawberries on Tuesday.
He ate $\frac{3}{4}$ cup of strawberries for breakfast each day on Wednesday, Thursday, and Friday.

How many cups of strawberries did Seth have left?

A $3\frac{1}{2}$
B $4\frac{1}{4}$
C6
D9

Show the answer

Correct answer: B

When Seth picks strawberries, the value represents a positive number, and when he eats the strawberries, the value represents a negative number. Write an expression and then evaluate the expression, where the negative value is multiplied by 3 because he ate the same amount on 3 different days.

$2\frac{1}{2} + 4 + (3)\left(-\frac{3}{4}\right)$

$2\frac{1}{2} + 4 - 2\frac{1}{4}$

$6\frac{1}{2} - 2\frac{1}{4}$

$6\frac{2}{4} - 2\frac{1}{4}$

$4\frac{1}{4}$

If the multiplication is confusing, evaluate the expression below which represents the sum for the 5 days with each day listed separately.

$2\frac{1}{2} + 4 + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right)$

$4\frac{1}{4}$

Question analysis

Seth picked strawberries on Monday and Tuesday, in amounts given as a mixed number and a whole number, then ate the same fraction of a cup at breakfast on three days. Students find how many cups he has left. It tests adding, multiplying and subtracting fractions in a short word problem, and the eaten amount repeats each day.

NYS standards: 7.NS.3 four operations with rational numbers
7.EE.3 multi-step rational number problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding mixed numbers and multiplying a fraction by a whole number are skills from before grade 7. The three steps are short and the numbers friendly, so the demand is below typical grade 7 work with signed rational numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The bulleted list lays out the story day by day, so setting up the problem is easy; the points are lost in the fractions. The number of breakfast days, 3, sits right beside the 3/4 cup, and subtracting 3 cups instead of three quarters of a cup each day is the tempting shortcut. Writing every amount in fourths, 26/4 picked and 9/4 eaten, makes the last step a whole-number subtraction.

Error-prone arithmetic moderate Working it out
Three breakfasts of 3/4 cup make 9/4 = 2 1/4 cups, and taking that from 6 1/2 means renaming the half as 2/4, two fraction steps where a slip is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Picking adds and eating subtracts, and “each day on Wednesday, Thursday, and Friday” means the 3/4 cup counts three times, not once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add the two amounts picked, multiply the breakfast amount by the number of days, and subtract that from the total; writing every amount in fourths keeps the arithmetic simple.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Subtracts 3 cups, the number of breakfast days, instead of 3 × 3/4 = 2 1/4 cups, so 6 1/2 − 3 = 3 1/2.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives exactly 6; it is near the 6 1/2 cups picked and mostly catches students who drop the fractions or guess.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the 2 1/4 cups eaten to the 6 1/2 cups picked instead of subtracting, 8 3/4 cups, and takes the nearest choice, 9.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total of a repeated amount: $\text{total} = \text{number of times} \times \text{amount each time}$; Use it when the same amount happens again and again. Here Seth eats 3/4 cup on each of 3 days, so the number of times is 3 and the amount each time is 3/4 cup.
A whole number times a fraction: $n \times \frac{a}{b} = \frac{n \times a}{b}$; n is a whole number, and a/b is a fraction. Multiply the whole number by the top only, and keep the bottom. It works because n groups of a pieces make n × a pieces, and each piece is still size 1/b.

Sort out the question
Seth picked 2 1/2 cups on Monday and 4 cups on Tuesday. Then he ate 3/4 cup each day on Wednesday, Thursday and Friday. That is 3 days of eating. Picking adds strawberries, and eating takes them away. You need the number of cups he has left.
Add what he picked
Add the two days of picking:
$2\frac{1}{2} + 4 = 6\frac{1}{2}$
He picked 6 1/2 cups in all, before he ate any.
Find how much he ate
He ate 3/4 cup on each of 3 days. Multiply:
$3 \times \frac{3}{4} = \frac{9}{4} = 2\frac{1}{4}$
To change 9/4 to a mixed number, note that 4 fits into 9 twice, with 1 left over. He ate 2 1/4 cups.
Watch out: he ate 3/4 cup three times, not 3 cups. Subtracting 3 gives 3 1/2, choice A.
Take away what he ate
Write the half as fourths, so both fractions have the same bottom: 1/2 = 2/4. Then subtract the wholes and the fourths:
$6\frac{1}{2} - 2\frac{1}{4} = 6\frac{2}{4} - 2\frac{1}{4} = 4\frac{1}{4}$
Watch out: eating takes away. Adding the 2 1/4 cups gives 8 3/4, and the nearest choice is 9, choice D.
So the answer is B, $4\frac{1}{4}$ cups.

Check your answer

Use decimals. 2.5 + 4 = 6.5 cups picked. 3 × 0.75 = 2.25 cups eaten. Then 6.5 − 2.25 = 4.25 cups, which is 4 1/4.

More questions like this

The original DOE question

Question 22 as shown on the DOE’s digital test, page 95 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 95, question 22.

Question 21 of 120

2026 Practice Test D Math, question 25

Question

A bus trip takes 9 hours if the mean speed is 50 miles per hour. How many hours would this trip take if the mean speed was 45 miles per hour?

Show the answer

Correct answer: B

First, find the length of the bus trip in miles by multiplying the time, 9 hours, by the average (mean) speed, 50 miles per hour, to get 450 miles.

Then, divide the total distance, 450 miles, by the average (mean) speed, 45 miles per hour, to get the total time:

$450 \div 45 = 10\text{ hours}$

Question analysis

A bus trip takes 9 hours at a mean speed of 50 miles per hour, and students find how long the same trip takes at 45 miles per hour. It tests speed, distance and time. Working out the fixed distance first is the key, since guessing how the time should change leads many students astray.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Speed, distance and time problems are grade 6 rate work, and this one uses whole numbers in two steps. Its only twist is how the time responds when the speed changes, which makes it easier than typical grade 7 rate problems.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two short calculations solve it, but only after noticing that speed and time move in opposite directions: a slower bus takes longer. Students who scale the 9 hours by 45/50, as in a direct proportion, get a little over 8 hours, a result that should feel wrong for a slower trip. Finding the distance first turns the question into one division with no direction to decide.

No obvious method moderate Working it out
The question never mentions the trip's length, yet the 450-mile distance is what stays fixed; setting up a proportion between hours and speeds, the usual reflex, points the wrong way. (Schoenfeld, 1985)
Something you must already know minor Working it out
Students must bring distance = speed × time themselves, to turn 9 hours at 50 miles per hour into miles before dividing by the new speed. (Anderson & Krathwohl, 2001)

Strategy: Find the trip's length in miles from the first speed and time, since that length stays the same, then divide it by the new mean speed.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Treats time as proportional to speed, scaling 9 hours by 45/50 to 8.1, about 8, though a slower bus needs more time, not less.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 12 hours; it needs a slip such as 540 miles for 9 × 50, so it mostly draws guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the 5-mile-per-hour drop in speed to the 9 hours, 9 + 5 = 14, combining numbers from the question without a model of the trip.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, speed and time: $d = r \times t$; d is the distance traveled. r is the rate, or speed, such as miles per hour. t is the time. It works because at 50 miles each hour, 9 hours covers 9 groups of 50 miles.
Finding the time: $t = d \div r$; This is the same rule turned around. d is the distance and r is the speed. Divide the distance by the speed to find t, the number of hours. Use it when you know how far and how fast.

Sort out the question
A bus trip takes 9 hours at a mean speed of 50 miles per hour. The mean speed is the average speed over the whole trip. You need the time for the same trip at 45 miles per hour. The length of the trip does not change, so find it first.
Find the length of the trip
Use distance equals speed times time, with a speed of 50 miles per hour and a time of 9 hours:
$d = 50 \times 9 = 450$
The trip is 450 miles long.
Find the new time
Now the bus goes 45 miles per hour over the same 450 miles. Divide the distance by the speed:
$t = 450 \div 45 = 10$
The trip takes 10 hours.
Check that it makes sense
A slower bus needs more time for the same trip. 10 hours is more than 9 hours, so it fits.
Watch out: time does not shrink with speed. Scaling 9 hours by 45/50 gives about 8 hours, choice A, which would make the slower bus arrive sooner. Adding the drop of 5 miles per hour to 9 hours gives 14, choice D, which has no reason behind it.
So the answer is B, 10 hours.

Check your answer

Check the distance both ways. At 50 miles per hour for 9 hours, the bus goes 50 × 9 = 450 miles. At 45 miles per hour for 10 hours, it goes 45 × 10 = 450 miles. Same trip, same distance.

More questions like this

The original DOE question

Question 25 as shown on the DOE’s digital test, page 98 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 98, question 25.

Question 22 of 120

2026 Practice Test D Math, question 35

Question

XYZ Corporation hired 6 people as summer workers. Of these 6 people, 3 will be hired as permanent employees in the fall. If 2 of the 6 people are male, how many of the possible groups of 3 people hired in the fall will include only 1 male?

Show the answer

Correct answer: C

An organized list can be used to determine the number of groups of 3 people that include only 1 male.

Let the letters A, B, C, and D represent the females, and let the numbers 1 and 2 represent the males.

Then, create a list of all possible combinations of 3 people with only 1 male.

AB1, AB2, AC1, AC2, AD1, AD2

BC1, BC2, BD1, BD2

CD1, CD2

There are 12 possible groups with only 1 male.

Question analysis

A company hired 6 summer workers, 2 of them male, and will keep 3 as permanent employees in the fall. This counting question asks how many possible groups of 3 include only 1 male. Students count combinations where order does not matter, and the bold word only, with two kinds of workers to combine, makes careful listing essential.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students list outcomes with organized lists and tree diagrams, usually ordered ones. Counting unordered groups of three under an "only one" condition, built from two kinds of people, goes beyond that typical work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Something you must already know minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question hides a two-stage count in a hiring story. Students tend to get one stage right and forget the other: they count the pairs of women and stop, or they count ordered picks and so find every group twice. Naming the women A to D and the men 1 and 2, then listing the pairs of women alphabetically with each man attached, makes every group visible and checkable.

Combines ideas moderate Working it out
Two counts are joined: the 6 ways to pick 2 of the 4 women, then the 2 ways to pick the man, and the total is their product. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
Students must already know that a group is the same whichever order its members are named, so picking one woman and then another repeats the reverse pick. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The bold word only means exactly one man, so the other two places in each group of 3 must go to two of the 4 women, a step the stem leaves unsaid. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Name the women A to D and the men 1 and 2, list every pair of women without repeats, then attach each man to each pair and count the groups.

Traps

Stops one step early: A choice is a number from partway through the solution.
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 2 men by the 2 places left for women, 2 × 2 = 4, using numbers from the question without listing a single group.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Counts the 6 pairs of women correctly but stops before pairing each one with either of the 2 men, which doubles the count.
D Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: moderate
Counts ordered picks of the two women, 4 × 3 = 12 instead of 6, so each group appears twice, then doubles for the men to reach 24.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting with two choices: $\text{number of groups} = \text{ways to pick the man} \times \text{ways to pick the women}$; Each way of picking the man can go with each way of picking the two women. So you multiply the two counts. It is like outfits: 2 shirts with 6 pairs of pants make 12 outfits.

Sort out the question
Six people work for the summer. Two are male, so the other 4 are female. Three will be hired in the fall.
You need the number of groups of 3 with only 1 male. That means exactly 1 man and 2 women. In a group, order does not matter. Ann, Bea and Carl is the same group as Carl, Ann and Bea.
Give each person a name
Call the women A, B, C and D. Call the men 1 and 2. Every group you want has 2 letters and 1 number, like AB1.
List the pairs of women
List each pair once. Start with A, then pair B only with letters after it, and so on:
- AB, AC, AD
- BC, BD
- CD
That is 6 pairs. BA is not new, since it is the same pair as AB.
Watch out: counting the order gives $4 \times 3 = 12$ pairs, each one twice. Then doubling for the men gives 24, choice D.
Add a man to each pair
Each pair of women can go with man 1 or man 2. Use the counting rule:
$6 \times 2 = 12$
Watch out: stopping at the 6 pairs of women leaves out the men. That gives 6, choice B.
So the answer is C, 12.

Check your answer

Write out every group. With man 1: AB1, AC1, AD1, BC1, BD1, CD1. With man 2: AB2, AC2, AD2, BC2, BD2, CD2. Each list has 6 groups, and $6 + 6 = 12$ . No group repeats, and each has just one man.

More questions like this

The original DOE question

Question 35 as shown on the DOE’s digital test, page 108 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 108, question 35.

Question 23 of 120

2026 Practice Test D Math, question 41

Question

The distance an eagle flies is proportional to the time the eagle spends flying. An eagle flies 90 miles in 3 hours. What is the constant of proportionality for this relationship?

Show the answer

Correct answer: B

The constant of proportionality is the ratio of y (90) to x (3). $90 \div 3 = 30$ . So the constant of proportionality is 30.

Question analysis

An eagle flies 90 miles in 3 hours, and the distance it flies is proportional to its flying time. This proportional relationship question asks for the constant of proportionality. Students need to know that the constant is the unit rate, here miles per hour, and find it from the two values given.

NYS standards: 7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 students find constants of proportionality in tables, graphs and equations. This one is a single division of miles by hours given in a sentence, the same unit rate work students do in grade 6.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 7, 29 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One division answers the question, so it is among the easiest on the test. The wrong choices are the two numbers in the stem and one unrelated value, so a student who knows what the constant means is not tempted. Checking that the constant times 3 hours gives back the 90 miles confirms it.

Something you must already know minor Working it out
Students must know that the constant of proportionality is the unit rate, distance over time, here the miles flown in a single hour; the stem gives only the term. (Anderson & Krathwohl, 2001)

Strategy: Divide the distance by the time, because the constant of proportionality is the distance flown in one hour, then check by multiplying it by the 3 hours.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the 3 hours printed in the stem as the constant, instead of the miles flown in each hour.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working from 90 miles and 3 hours leads to 60, which may echo the 60 minutes in an hour, so it mostly attracts guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives the 90 miles, the distance for the whole 3-hour flight, as if the eagle covered it in a single hour.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = kx \text{, so } k = \frac{y}{x}$; x is one amount, here the time in hours, and y is the other, here the distance in miles. k is the constant of proportionality. It is the same for every pair, and it equals y divided by x. Here it is the miles flown in 1 hour.

Sort out the question
The eagle flies 90 miles in 3 hours. Proportional means the distance is always the same number times the time. That number is the constant of proportionality. You need to find it.
Divide the distance by the time
Use the formula. The distance is y and the time is x:
$k = \frac{90}{3} = 30$
The constant is 30. It means the eagle flies 30 miles in each hour.
Avoid the numbers in the question
Watch out: 3 is the number of hours, choice A. And 90 is the distance for all 3 hours, choice D. Neither is the distance for 1 hour.
So the answer is B, 30.

Check your answer

Multiply back. At 30 miles each hour, the eagle flies $30 \times 3 = 90$ miles in 3 hours. That matches the question. Hour by hour, it goes 30, 60, then 90 miles.

More questions like this

The original DOE question

Question 24 of 120

2026 Practice Test D Math, question 42

Question

A student reads $\frac{1}{12}$ of a book in $\frac{1}{3}$ hour. What fraction of the book can the student read in 1 hour?

A $\frac{1}{36}$
B $\frac{1}{4}$
C $\frac{5}{12}$
D $\frac{12}{3}$

Show the answer

Correct answer: B

To find the fraction of the book the student can read in 1 hour, find the unit rate per hour. First, set up the ratio.

$\frac{\frac{1}{12} \text{ book}}{\frac{1}{3} \text{ hour}}$

Then use division to find the unit rate.

$\frac{\frac{1}{12}}{\frac{1}{3}} = \frac{1}{12} \div \frac{1}{3} = \frac{1}{12} \times \frac{3}{1} = \frac{3}{12} = \frac{1}{4}$

Another way to solve the problem is to set up the ratio and then multiply both the numerator and denominator by 3, since $\frac{1}{3}$ hour times 3 equals 1 hour.

$\frac{3 \times \frac{1}{12} \text{ book}}{3 \times \frac{1}{3} \text{ hour}} = \frac{\frac{1}{4} \text{ book}}{1 \text{ hour}}$

Question analysis

A student reads 1/12 of a book in 1/3 of an hour, and this rate question asks what fraction of the book the student can read in a whole hour. It tests finding a unit rate when both amounts are fractions. Deciding whether to multiply or divide, and which fraction goes on top, decides the result.

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of fractions is the grade 7 standard itself. This is a short example with unit fractions, where scaling the time up to a whole hour is as quick as dividing.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Thinking in thirds of an hour makes this quick: an hour holds three of them, so the student reads three twelfths of the book. Students who reach for a fraction rule instead must choose between multiplying and dividing, and between two orders of division. A sense check rules out two wrong choices at once: a whole hour must cover more than 1/12 of the book, and no more than the whole book.

Error-prone arithmetic minor Working it out
Both amounts are unit fractions, and 1/12 ÷ 1/3 means multiplying by 3, which gives 3/12 that still has to be simplified to match a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase “fraction of the book … in 1 hour” asks for book per hour, so the fraction of the book is divided by the time, not the other way round. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Ask how many thirds of an hour make 1 hour, then scale the fraction of the book by the same factor; or divide the fraction of the book by the time.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 1/12 by 1/3 instead of dividing, getting 1/36, which would mean reading less of the book in an hour than in a third of one.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two fractions, 1/12 + 4/12 = 5/12, combining an amount of the book with an amount of time as if they were alike.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Divides the time by the amount read, 1/3 ÷ 1/12 = 12/3, which is hours per book, the rate turned upside down.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{unit rate} = \frac{\text{part of the book}}{\text{hours}}$; A unit rate tells how much happens in 1 unit, here 1 hour. Put the part of the book on top and the time on the bottom. Dividing finds the part of the book for one whole hour.
Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Flip only the fraction you divide by.

Sort out the question
The student reads $\frac{1}{12}$ of a book in $\frac{1}{3}$ of an hour. You need the part of the book read in 1 whole hour. That is a unit rate.
Scale up to one hour
One hour holds three thirds of an hour. So in 1 hour the student reads 3 times as much:
$3 \times \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$
To simplify 3/12, divide the top and bottom by 3. The student reads 1/4 of the book in 1 hour.
Check with the unit rate
Divide the part of the book by the time. Multiply by the reciprocal of 1/3, which is 3/1:
$\frac{1}{12} \div \frac{1}{3} = \frac{1}{12} \times \frac{3}{1} = \frac{3}{12} = \frac{1}{4}$
Both ways give 1/4.
Avoid the wrong operations
Watch out: multiplying gives $\frac{1}{12} \times \frac{1}{3} = \frac{1}{36}$ , choice A. That is less than 1/12. But in a full hour, the student must read more than in a third of an hour. Dividing the time by the book gives 12/3, choice D. That is 4 whole books.
So the answer is B, $\frac{1}{4}$ .

Check your answer

If the student reads 1/4 of the book each hour, the whole book takes 4 hours. In 1/3 of an hour, the student reads $\frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$ of the book. That matches the question.

More questions like this

The original DOE question

Question 42 as shown on the DOE’s digital test, page 115 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 115, question 42.

Question 25 of 120

2026 Practice Test D Math, question 44

Question

A bakery charges $1.25 for each cupcake and $2.50 for a serving tray. The equation describes p, the price of c cupcakes and a serving tray, not including tax.

$p = 2.50 + 1.25c$

If Samantha bought some cupcakes and a serving tray for a total of $22.50, not including tax, how many cupcakes did she buy?

Show the answer

Correct answer: C

Using the specified equation $p = 2.50 + 1.25c$ , determine the number of cupcakes Samantha bought by substituting the total paid ($22.50) for p, and solve the equation for the value of c, the number of cupcakes bought.

$22.50 = 2.50 + 1.25c$

$22.50 - 2.50 = 1.25c$

$20.00 \div 1.25 = c$

$16 = c$

Question analysis

A bakery's price for c cupcakes and a serving tray is given by the equation p = 2.50 + 1.25c, and Samantha paid $22.50 before tax. This equation question asks how many cupcakes she bought. Students substitute the total and solve a two-step equation with decimal money amounts, undoing the steps in the right order.

NYS standards: 7.EE.4a two-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving a two-step equation of the form px + q = r in a money context is core grade 7 algebra. The equation is given, so the work is substituting the total and dividing by a decimal, a routine example.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 50 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The equation is supplied, so the question is a two-step solve with money. The tray’s $2.50 has to come off the total before dividing, and students who divide the whole $22.50 by $1.25 get a whole number that matches a choice, so nothing warns them. Substituting the chosen number back into p = 2.50 + 1.25c is a quick, decisive check.

Error-prone arithmetic minor Working it out
After subtracting the tray, $20 must be divided by $1.25, a decimal divisor that is easiest handled as 2,000 ÷ 125 or by noting that 4 cupcakes cost $5. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute the total for p, subtract the tray's cost from both sides, then divide by the price of one cupcake; check by putting your number back into the equation.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Notes that 4 cupcakes cost $5.00 and that the $20 left after the tray holds four such amounts, then gives the 4 groups instead of 16 cupcakes.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the $22.50 total by the $2.50 tray price, combining two numbers from the question without using the equation.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the whole $22.50 by $1.25 a cupcake, forgetting that $2.50 of the total paid for the serving tray.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides: $\text{if } a = b \text{, then } a - n = b - n \text{ and } \frac{a}{n} = \frac{b}{n}$; a and b are the two sides of an equation, and n is any number (not zero when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it level. That is how you get c, the number of cupcakes, by itself.

Sort out the question
The equation $p = 2.50 + 1.25c$ gives the price. p is the price in dollars, and c is the number of cupcakes. The 2.50 is the tray, and 1.25 is the price of one cupcake. Samantha paid $22.50. You need c.
Put in the total
Samantha paid $22.50 in all, so put 22.50 in place of p. Now the equation has only one unknown, c:
$22.50 = 2.50 + 1.25c$
Subtract the tray
Take 2.50 away from both sides. That removes the cost of the tray:
$22.50 - 2.50 = 1.25c$
$20.00 = 1.25c$
So $20.00 went to cupcakes alone.
Watch out: dividing all of $22.50 by $1.25 gives 18, choice D. That forgets the tray.
Divide by the price of one cupcake
Divide both sides by 1.25:
$c = 20.00 \div 1.25 = 16$
An easy way: 4 cupcakes cost $4 \times 1.25 = 5.00$ dollars. $20 is 4 groups of $5. So Samantha bought $4 \times 4 = 16$ cupcakes.
Watch out: 4 is the number of $5 groups, not cupcakes. That is choice A.
So the answer is C, 16.

Check your answer

Put 16 back into the equation. $1.25 \times 16 = 20.00$ , and $2.50 + 20.00 = 22.50$ . That is what Samantha paid, so 16 is right.

More questions like this

The original DOE question

Question 44 as shown on the DOE’s digital test, page 117 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 117, question 44.

Question 26 of 120

2026 Practice Test D Math, question 47

Question

$A tall right rectangular prism standing on end. The bottom front edge is labeled 2 in., the bottom edge running back on the right is labeled 2 in., and the height is labeled 6 in. Right-angle marks are drawn at the lower left corner of the front face and at the front left corner of the top face.$

What shape will result from a horizontal slice of the figure above?

Aa square that has 2-inch sides
Ba triangle that has 2-inch sides
Ca rectangle that has a 2-inch side and a 6-inch side
Da triangle that has one 2-inch side and two 6-inch sides

Show the answer

Correct answer: A

Based on the diagram, the figure shown is a right rectangular prism with a square base. A horizontal slice of the figure will be a plane parallel to the square base, as in this example.

$The same prism, 2 in. by 2 in. by 6 in., drawn with its hidden edges dashed and a horizontal slice partway up: a square parallel to the base and the same size as the base.$

Since the horizontal slice is parallel to the base, it will be congruent to the base, which is a square with 2-inch sides.

Question analysis

A figure shows a right rectangular prism measuring 2 inches by 2 inches by 6 inches, standing on end. This geometry question asks what two-dimensional shape a horizontal slice of the prism makes. It tests visualizing cross-sections of solids, using the labeled lengths to name both the shape and its size.

NYS standards: 7.G.3 slicing three-dimensional solids
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Describing the shape made by slicing a prism parallel to its base is exactly the grade 7 standard. This is its simplest case, a single slice of a labeled right prism, typical of a classroom example.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 2, 46 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

A horizontal cut runs parallel to the ground, and this prism stands on a square base, so the slice copies the base. The main danger is picturing the cut along the tall front face, which gives a 2-by-6 rectangle, the shape a vertical slice makes. The two triangles need a slanted cut through a corner and catch only students guessing.

Reading a figure, table or graph minor Understanding the question
The prism is drawn standing on end, 6 inches tall on a 2-by-2 base, so the cut has to be pictured in a flat drawing where the tall front face is the most visible shape. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Students must know that a slice parallel to the base of a prism has the same shape and size as the base, which here is the square at the bottom. (Anderson & Krathwohl, 2001)

Strategy: Picture a flat cut across the prism parallel to the ground, ask which face of the prism that cut runs parallel to, and read that face's edge lengths from the labels.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Expects a slice to cut off a corner and make a triangle, as a cut through a pyramid might, not a cut parallel to this square base.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Describes a vertical slice, which runs down the 6-inch height and makes a 2-by-6 rectangle, mixing up horizontal and vertical.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Builds a triangle from the labels in the figure, one 2-inch and two 6-inch sides, a shape no horizontal cut of this prism can make.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A slice parallel to the base: $\text{slice parallel to the base} \cong \text{base}$; The sign ≅ means congruent: the same shape and the same size. A prism has the same shape all the way up, like a stack of matching cards. So a cut parallel to its base, at any height, is a copy of the base.

Sort out the question
The figure is a right rectangular prism, a box shape whose faces are all rectangles. It stands on end. The two edges along the bottom are labeled 2 in. The tall edge is labeled 6 in.
You need the shape you get from a horizontal slice. Horizontal means flat, like the floor or a tabletop.
Find the base
The base is the face the prism stands on. The two bottom edges you can see are both labeled 2 in. The right-angle mark on the top face shows its corners are square. The top and the bottom match. So the base is a square with 2-inch sides.
Picture the slice
A horizontal slice cuts straight across the prism, parallel to the base. The walls of the prism go straight up, so the slice is a copy of the base. It is a square with 2-inch sides, at whatever height you cut.
The 6-inch height does not matter. It only tells how tall the stack of squares is.
Rule out the other shapes
Watch out: a 2-inch by 6-inch rectangle, choice C, comes from a vertical slice. That cut goes straight down, through the 6-inch height. A triangle, choice B or D, needs a slanted cut through a corner.
So the answer is A, a square that has 2-inch sides.

Check your answer

Picture the slice sliding up and down. At the bottom, it matches the base, 2 in. by 2 in. At the top, it matches the top face, the same square. In between, the walls are straight, so its size never changes.

More questions like this

The original DOE question

Question 27 of 120

2026 Practice Test D Math, question 50

Question

This table shows the number of birds that came to a bird feeder on Monday and Tuesday.

BIRDS AT BIRD FEEDER

Day	Number of Birds
Monday	30
Tuesday	24

What was the percent decrease in the number of birds from Monday to Tuesday?

A6%
B20%
C24%
D25%

Show the answer

Correct answer: B

To determine the percent decrease from Monday to Tuesday, first calculate the decrease in the number of birds at the feeder.

$30 - 24 = 6$

The difference in the number of birds at the feeder divided by the number of birds on Monday will yield the percent decrease from Monday to Tuesday.

$\frac{6}{30} = 0.20$

0.20 represents 20%.

There was a 20% decrease from Monday to Tuesday.

Question analysis

A table shows that 30 birds came to a bird feeder on Monday and 24 on Tuesday. This percent change question asks for the percent decrease from Monday to Tuesday. Students find the change and compare it with one day's count, and choosing the right day as the base decides the result.

NYS standards: 7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Percent decrease is core grade 7 work, and this is a plain example: one subtraction and one division with friendly numbers. The only real demand is measuring the change against the right day's count.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The subtraction is trivial, so the question is about the base. Both counts sit in the table, and dividing by the later one gives an equally neat percent that is easy to accept. Saying “a decrease from Monday” and putting Monday’s 30 under the change avoids it, and checking that 20% of 30 is 6 confirms the result.

A very close wrong answer minor Choosing the answer
Dividing the drop of 6 by Monday’s 30 gives 20% and by Tuesday’s 24 gives 25%, two round, listed percents only 5 points apart, so only the choice of base separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that a percent decrease is the change divided by the starting amount, here Monday’s count, a rule the question does not state. (Anderson & Krathwohl, 2001)

Strategy: Subtract to find how many fewer birds came on Tuesday, divide by Monday's count, the starting amount, and change the fraction to a percent.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Finds the decrease of 6 birds and reads it as 6 percent, as if the count had dropped from 100 rather than from 30.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Reads Tuesday’s 24 birds as a percent, taking a number straight from the table without working out the change at all.
D Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Divides the decrease of 6 by Tuesday’s 24 birds instead of Monday’s 30, measuring the change from the later day: 6/24 = 25%.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent decrease: $\text{percent decrease} = \frac{\text{old} - \text{new}}{\text{old}} \times 100$; Old is the starting amount, here Monday’s count. New is the later amount, Tuesday’s count. The top is how much it dropped. You divide by the starting amount because a change is measured from where you started.

Sort out the question
The table shows 30 birds on Monday and 24 birds on Tuesday. You need the percent decrease from Monday to Tuesday. From Monday means Monday is the starting amount.
Find the drop
Subtract Tuesday’s count from Monday’s:
$30 - 24 = 6$
There were 6 fewer birds.
Watch out: 6 is a number of birds, not a percent. Calling it 6% leads to choice A.
Compare the drop with Monday
Use the formula. Divide the drop by Monday’s count, then multiply by 100:
$\frac{6}{30} \times 100 = 0.20 \times 100 = 20$
The number of birds went down by 20%.
Watch out: dividing by Tuesday’s 24 gives $\frac{6}{24} = 0.25$ , or 25%, choice D. Measure the change from the starting day. Choice C, 24%, just copies a number from the table.
So the answer is B, 20%.

Check your answer

Take 20% of Monday’s count: $0.20 \times 30 = 6$ . Then $30 - 6 = 24$ , which is Tuesday’s count. The answer checks out.

More questions like this

The original DOE question

Question 50 as shown on the DOE’s digital test, page 123 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 123, question 50.

Question 28 of 120

2024 Test A Math, question 2

Question

Angle M and angle R are supplementary. The measure of angle R is 5 times the measure of angle M. What is the measure of angle R, in degrees?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 150

Angle M and angle R are supplementary, so the sum of the angles is 180°.

Let m equal the measure of angle M, and let r equal the measure of angle R.

$r + m = 180$

The measure of angle R is 5 times the measure of angle M.

$r = 5m$

$5m + m = 180$

$6m = 180$

$m = \frac{180}{6} = 30$

$r = 5m = 5(30) = 150$

Therefore, the measure of angle R is 150°.

Question analysis

A geometry grid-in about two supplementary angles, where one angle is five times the size of the other. Students must recall what supplementary means, write an equation linking the two angles, and solve it. The question asks for the larger angle, so stopping at the smaller one is the usual slip.

NYS standards: 7.G.5 angle relationships and equations
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students regularly write and solve simple equations for unknown angles using supplementary and complementary facts. This is a two-step version of exactly that task, with whole numbers and no figure to read.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once supplementary is read as a 180° total, the problem is a one-variable equation with whole numbers. Students lose the point in two ways: by stopping at the smaller angle, which the equation produces first, or by using 90° for the total. Checking that the gridded angle is the larger of the pair, and that the two angles add to 180°, catches both.

Something you must already know moderate Working it out
The stem never says what supplementary means; the 180° total has to be recalled, and a student who confuses it with complementary's 90° halves every angle. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The sentence "The measure of angle R is 5 times the measure of angle M" has to become R = 5M, so the 180° splits into six equal parts, not five. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
As a grid-in there is nothing to compare against, so gridding angle M's 30°, the first value the equation gives, looks just as finished as angle R. (Enright, Morley & Sheehan, 2002)

Strategy: Call angle M one part and angle R five parts; together the six parts make the supplementary total, so find one part and then multiply for angle R.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

30 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Solves 6m = 180 correctly for angle M, 30°, and grids it instead of multiplying by 5 to get angle R, the angle the question asks for.
75 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses 90° for supplementary, the complementary total, so 6m = 90 gives m = 15 and angle R = 75, half its real size.
36 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides 180 by the 5 in the stem, combining the two numbers without an equation that gives angle M its own share of the 180°.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Supplementary angles: $M + R = 180$; M and R are the sizes of two angles, in degrees. Supplementary means the two angles add up to 180 degrees. Put side by side, they make a straight line. Do not mix this up with complementary, which means a total of 90 degrees.

Sort out the question
Two angles, M and R, are supplementary. Angle R is 5 times as big as angle M. You need the measure of angle R, in degrees. Measure means its size. Notice that the question asks for R, not M.
Write the facts as equations
Supplementary angles add up to 180°.
$M + R = 180$
The sentence “The measure of angle R is 5 times the measure of angle M” becomes this:
$R = 5 \times M$
Watch out: supplementary means 180°, not 90°. Using 90° gives R = 75, which is wrong.
Find angle M
Put $5 \times M$ in place of R. Then the equation has only one unknown, M.
$M + 5 \times M = 180$
$6 \times M = 180$
$M = 180 \div 6 = 30$
Think of 6 equal parts. M is 1 part and R is 5 parts. So angle M is 30°.
Find angle R
Angle R is 5 times angle M.
$R = 5 \times 30 = 150$
Angle R is 150°. Watch out: do not stop at M. Gridding 30 gives the wrong angle. So you would grid 150.

Check your answer

Add the two angles: $30 + 150 = 180$ . So they are supplementary. Also, $150 \div 30 = 5$ , so R really is 5 times M. Both facts in the question check out.

More questions like this

The original DOE question

Question 59 as printed on page 75 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 75, question 59. Open page 75 of the handbook (PDF)

Question 29 of 120

2024 Test A Math, question 11

Question

Jar 1 and Jar 2 each contain $\frac{1}{2}$ cup of water. If $\frac{1}{4}$ of the water in Jar 1 is poured into Jar 2, how much water is now in Jar 2?

A $\frac{1}{8}$ cup
B $\frac{1}{4}$ cup
C $\frac{5}{8}$ cup
D $\frac{3}{4}$ cup

Show the answer

Correct answer: C

Jar 1 starts with $\frac{1}{2}$ cup of water. If $\frac{1}{4}$ of its water is poured into Jar 2, then the amount of water added to Jar 2 is $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ cup.

Jar 2 starts with $\frac{1}{2}$ cup of water. If $\frac{1}{8}$ cup of water is added to it, then the amount of water now in Jar 2 is $\frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8}$ cup.

Question analysis

A fraction word problem about two jars that each hold half a cup of water; a quarter of the water in the first jar is poured into the second. Students find how much water the second jar now holds. Reading exactly what the quarter is a quarter of is the detail that matters.

NYS standards: 7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Multiplying and adding positive fractions is review by grade 7, and the problem takes two short steps. It is easier than typical grade 7 work with negative numbers or longer rational-number problems, apart from careful reading.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are friendly halves, quarters and eighths, and the work is two steps. The question turns on one phrase: the quarter is a fraction of Jar 1's water, not a measured quarter cup, and reading it the second way gives a choice that looks right. Students who also remember that Jar 2 already held water finish correctly.

Turning words into math moderate Understanding the question
The phrase "1/4 of the water in Jar 1" means a quarter of Jar 1's half cup, an eighth of a cup, not a quarter cup; the word of calls for multiplying. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Adding the 1/8 cup to Jar 2's 1/2 cup needs the common denominator 8, so 1/2 becomes 4/8 before the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the amount poured by taking one quarter of Jar 1's half cup, then add that amount to the half cup already in Jar 2.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the amount poured, 1/4 of 1/2 cup = 1/8 cup, and stops there instead of adding it to the half cup already in Jar 2.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks the fraction 1/4 straight from the question, as if the share poured were the amount of water now in Jar 2.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Reads 1/4 of the water as 1/4 cup and adds it to Jar 2's half cup, 1/2 + 1/4 = 3/4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of an amount: $\frac{a}{b} \text{ of } N = \frac{a}{b} \times N$; N is the whole amount, and a over b is the part of it you want. In math, the word of means multiply. For example, half of 10 is one half times 10, which is 5.
Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the tops, and b and d are the bottoms. Multiply the tops together and the bottoms together. Taking a part of a part makes the pieces smaller, so the bottom gets bigger.
Adding fractions with the same bottom: $\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}$; a and b are the tops, and d is the shared bottom, called the denominator. If the bottoms match, add the tops and keep the bottom. If they do not match, rewrite one fraction first.

Sort out the question
Each jar starts with $\frac{1}{2}$ cup of water. You pour “ $\frac{1}{4}$ of the water in Jar 1” into Jar 2. You need how much water is in Jar 2 after that, in cups.
Find how much water is poured
Jar 1 has $\frac{1}{2}$ cup. You pour $\frac{1}{4}$ of it. Of means multiply.
$\frac{a}{b} \text{ of } N = \frac{a}{b} \times N$
$\frac{1}{4} \times \frac{1}{2} = \frac{1 \times 1}{4 \times 2} = \frac{1}{8}$
You pour $\frac{1}{8}$ cup. Watch out: a quarter of the water is not a quarter cup. It is a quarter of a half cup. Adding $\frac{1}{4}$ cup gives $\frac{3}{4}$ cup, choice D.
Make the bottoms match
Now add the poured water to the $\frac{1}{2}$ cup already in Jar 2. The bottoms are 2 and 8, so rewrite the half with 8 on the bottom. Multiply its top and bottom by 4.
$\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}$
Add the water in Jar 2
$\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}$
$\frac{4}{8} + \frac{1}{8} = \frac{5}{8}$
Jar 2 now holds $\frac{5}{8}$ cup. Watch out: $\frac{1}{8}$ cup is only the water poured in. Stopping there gives choice A. So the answer is C, $\frac{5}{8}$ cup.

Check your answer

Jar 2 started with a half cup and got a little more. So it should hold just over a half cup, and $\frac{5}{8}$ is. Also, the water in both jars still adds up to 1 cup: Jar 1 has $\frac{3}{8}$ cup left, and $\frac{3}{8} + \frac{5}{8} = 1$ .

More questions like this

The original DOE question

Question 68 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 68. Open page 78 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 30 of 120

2024 Test A Math, question 12

Question

If n is a whole number, and 0.01 is between $\frac{1}{n}$ and $\frac{1}{n + 2}$ , what is the value of n?

Show the answer

Correct answer: D

$0.01 = \frac{1}{100}$ , which is between $\frac{1}{101}$ and $\frac{1}{99}$ .

Therefore, $n + 2 = 101$ and $n = 99$ .

Question analysis

A number sense question: 0.01 lies between 1 over n and 1 over n + 2, where n is a whole number, and students find n. It tests comparing a decimal with unit fractions and knowing how a fraction changes as its denominator grows. The unusual form, with n in two denominators, is what slows students down.

NYS standards: 6.NS.7a inequality as relative position
6.EE.5 solutions and substitution
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students compare fractions and decimals routinely, but not with an unknown in the denominators. The comparison is simple once 0.01 is written as a fraction; the unfamiliar setup is the main demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The setup looks unfamiliar, with n in two denominators, but one rewrite makes it routine: 0.01 is 1/100, so the two denominators must sit on either side of 100. The three small choices fail a quick check, since every fraction from 1/1 to 1/4 is far above 0.01, and 0 cannot be a denominator at all. Students who test the choices instead of rewriting the decimal spend a little longer but reach the same place.

Variables instead of numbers moderate Understanding the question
The unknown sits in two denominators, n and n + 2, so students must reason about which unit fractions lie on either side of 0.01 rather than solve an equation. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Two facts are assumed: 0.01 equals 1/100, and a unit fraction shrinks as its denominator grows, so 1/99 is just above 0.01 and 1/101 just below. (Anderson & Krathwohl, 2001)

Strategy: Rewrite 0.01 as a unit fraction, recall that a larger denominator gives a smaller fraction, and test each choice in both 1/n and 1/(n + 2).

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats 1/0 as 0, so 0.01 seems to lie between 0 and 1/2; but division by zero is undefined, so n cannot be 0.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 1 from 0.01 or from 1/n as the value of n, without checking that 0.01 is far below both 1 and 1/3.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 2 from n + 2, or the two decimal places of 0.01, without testing 1/2 and 1/4, which are both far above 0.01.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions: $\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}$; A unit fraction has 1 on top, like one fifth. a and b are positive whole numbers on the bottom. A bigger bottom cuts the whole into more pieces, so each piece is smaller. So the fraction with the bigger bottom is the smaller number.

Sort out the question
n is a whole number: 0, 1, 2, 3 and so on. The number 0.01 is between $\frac{1}{n}$ and $\frac{1}{n + 2}$ . Between means one fraction is bigger than 0.01 and the other is smaller. You need the value of n.
Write 0.01 as a fraction
0.01 means one hundredth.
$0.01 = \frac{1}{100}$
Now all three numbers are fractions with 1 on top. That makes them easy to compare.
Find fractions on each side
With 1 on top, a bigger bottom makes a smaller fraction.
$\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}$
So $\frac{1}{99}$ is a little more than $\frac{1}{100}$ , and $\frac{1}{101}$ is a little less.
$\frac{1}{101} < \frac{1}{100} < \frac{1}{99}$
The bottoms 99 and 101 are 2 apart, just like n and n + 2.
Match the fractions to n
The smaller bottom is n, and the larger bottom is n + 2.
$n = 99 \text{ and } n + 2 = 101$
Both fit. Watch out: small choices do not work. For n = 1, the fractions are 1 and $\frac{1}{3}$ . Both are far bigger than 0.01. For n = 0, $\frac{1}{0}$ is not a number, because you can’t divide by zero. So the answer is D, 99.

Check your answer

Write the fractions as decimals. $\frac{1}{99} \approx 0.0101$ , a bit more than 0.01. $\frac{1}{101} \approx 0.0099$ , a bit less. The sign ≈ means about. So 0.01 sits right between them.

More questions like this

The original DOE question

Question 69 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 69. Open page 78 of the handbook (PDF)

Question 31 of 120

2024 Test A Math, question 25

Question

$Points M, N, P and Q, each marked with a dot, lie in that order from left to right on a line. A dimension arrow from M to N is labeled x cm, and a longer dimension arrow from M to P is labeled y cm. No lengths are marked for NP, PQ or MQ.$

In the figure above, N is the midpoint of $\overline{MQ}$ . Which segment has length $(2x - y)$ centimeters?

A $\overline{PQ}$
B $\overline{NP}$
C $\overline{MQ}$
D $\overline{MP}$

Show the answer

Correct answer: A

These facts are given:

N is the midpoint of $\overline{MQ}$ .
$\overline{MN}$ has a length of x centimeters.
$\overline{MP}$ has a length of y centimeters.

Use the first two facts to determine that $\overline{MQ}$ has a length of $2x$ centimeters. So $(2x - y)$ can be represented by $MQ - MP$ , which is $\overline{PQ}$ .

Question analysis

A geometry and algebra question with a labeled figure: points M, N, P and Q lie on a line, MN is x cm, MP is y cm, and N is the midpoint of MQ. Students find which segment has length 2x − y. They must turn the midpoint fact into an expression and read a difference of lengths on the figure.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.EE.6 variables representing unknown numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students evaluate and rewrite expressions but rarely read one as a length on a figure. Combining the midpoint idea from geometry with a difference of expressions, and no numbers, makes this more abstract than typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Reading a figure, table or graph moderate
Solution: 2 steps
Reading level: Question and choices: about grade 9, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question reverses the usual task: instead of finding a length, students match an expression to a segment. The midpoint fact turns into MQ = 2x, and taking away the y of MP leaves the piece from P to Q. The wrong choices reward stopping at 2x, reading only the labeled arrows, or matching the letter y; picking x = 5 and y = 7 lets students check with numbers.

Variables instead of numbers moderate Understanding the question
Every length is a letter: MN is x, MP is y, and the target is the expression 2x − y, so the matching segment cannot be found by measuring. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph moderate Understanding the question
The figure marks only MN and MP with arrows; the length of MQ comes from the stem's midpoint fact, not from the drawing, and has to be written as 2x. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the midpoint to write the length of the whole segment in terms of x, or pick sample values for x and y, then see which segment's length matches 2x − y.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Leaves out the midpoint fact, so the only difference it can read from the arrows is NP = y − x, which it takes for 2x − y.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Uses the midpoint to find that MQ is 2x centimeters and picks that segment, stopping before subtracting the y of MP.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Sees y in the expression and picks MP, the segment labeled y, without working out the rest of 2x − y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint: $MQ = 2 \times MN$; N is the midpoint of segment MQ, the point exactly in the middle. MN is the half from M to N, and MQ is the whole segment. The whole is two equal halves, so it is 2 times MN.
Adding segments: $MP + PQ = MQ$; When points lie in order on a line, the short segments add up to the long one. MP runs from M to P, and PQ runs from P to Q. Put end to end, they make MQ.

Sort out the figure
Points M, N, P and Q lie in that order on a line. Segment MN is x cm long, and segment MP is y cm long. N is the midpoint of $\overline{MQ}$ , the point exactly halfway from M to Q. You need the segment that is $2x - y$ cm long.
Find the length of MQ
MN is half of MQ, so MQ is twice as long.
$MQ = 2 \times MN = 2x$
MQ is 2x cm long. Watch out: 2x is only part of the expression. Stopping here gives MQ, choice C.
Take MP away from MQ
Subtracting y from 2x means taking MP away from MQ. MP covers the part from M to P. What is left of MQ is the piece from P to Q.
$MP + PQ = MQ$
$PQ = MQ - MP = 2x - y$
So PQ is $2x - y$ cm long. Watch out: do not pick MP just because y appears. MP is y, choice D. NP is $y - x$ , choice B. So the answer is A, $\overline{PQ}$ .

Check your answer

Try numbers. Say x = 5, so MN = 5 and MQ = 10. Say y = 7, so MP = 7. Then PQ is $10 - 7 = 3$ . And $2x - y = 10 - 7 = 3$ . They match. NP would be $7 - 5 = 2$ , which does not.

More questions like this

The original DOE question

Question 82 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 82, question 82. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 32 of 120

2024 Test A Math, question 27

Question

On Wednesday, a baker produced 100 more loaves of bread than were produced on Tuesday. On Thursday, the baker produced 50 fewer loaves than were produced on Tuesday. If the total number of loaves produced on all three days was 230, how many loaves were produced on Wednesday?

A60
B80
C120
D160

Show the answer

Correct answer: D

Let t represent the number of loaves of bread produced on Tuesday.

First, determine the number of loaves of bread produced on Tuesday, t, using the number of loaves produced on Wednesday, $t + 100$ , the number produced on Thursday, $t - 50$ , and the total number produced on these three days, 230:

$t + t + 100 + t - 50 = 230$

$t + t + t + 100 - 50 = 230$

$3t + 50 = 230$

$3t + (50 - 50) = 230 - 50$

$3t = 180$

$(3 \div 3)t = 180 \div 3$

$t = 60$

Next, determine the number of loaves of bread produced on Wednesday, $t + 100$ , when $t = 60$ :

$t + 100 = 60 + 100 = 160$

The number of loaves of bread produced on Wednesday is 160.

Question analysis

A word problem about a baker's bread over three days: Wednesday's output is 100 loaves more than Tuesday's, Thursday's is 50 fewer than Tuesday's, and the total is 230. Students find the number of loaves made on Wednesday. Writing every day in terms of Tuesday, and answering for the right day, are the keys.

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation of the form px + q = r from a word problem is a core grade 7 standard. This is a typical three-step version, with one negative adjustment and an extra step after solving.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 45 words.
Corrected from the DOE handbook: A misplaced parenthesis in the explanation’s working was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Writing all three days in terms of Tuesday turns the story into 3t + 50 = 230, a routine equation. The trap comes after it is solved: the value found is Tuesday's, and it is a listed choice. Students who label t as Tuesday from the start, and reread the question before choosing, add the 100 loaves and are done.

Turning words into math moderate Understanding the question
Both comparisons are made with Tuesday, 100 more loaves on Wednesday and 50 fewer on Thursday, so the three days become t, t + 100 and t − 50. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
The equation's unknown is Tuesday's count, while the question asks for Wednesday's, so one more step remains after the equation is solved. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let t be Tuesday's loaves, write Wednesday and Thursday in terms of t, set the sum equal to 230, solve for t, and add 100.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Solves 3t + 50 = 230 for Tuesday's loaves, t = 60, and gives that instead of Wednesday's t + 100.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Subtracts both 100 and 50 from the total, 230 − 100 − 50 = 80, combining the numbers without writing each day in terms of Tuesday.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds Tuesday's 60, then gets Wednesday by subtraction, 230 − 60 − 50, taking Thursday as 50 loaves instead of 60 − 50 = 10.

What was corrected from the DOE handbook, and why

Explanation: The handbook writes “3t + (50 − 50 = 230 − 50)”, with the closing parenthesis after the whole equation, so the line reads as an equation inside an expression (and 50 − 50 = 230 − 50 is false). The site now reads 3t + (50 − 50) = 230 − 50: subtract 50 from both sides, giving 3t = 180 as before.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c$; a and b are the two sides of an equation, and c is any number. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. The same is true if you divide both sides by the same number that is not zero.

Sort out the question
The baker made bread on Tuesday, Wednesday and Thursday. Wednesday had 100 more loaves than Tuesday. Thursday had 50 fewer loaves than Tuesday. All three days add up to 230 loaves. You need the number of loaves made on Wednesday.
Write each day using Tuesday
Both other days are compared with Tuesday. So call Tuesday’s loaves t.
Day Loaves
Tuesday $t$
Wednesday $t + 100$
Thursday $t - 50$
100 more means add 100. 50 fewer means subtract 50.
Write the equation and combine
$t + (t + 100) + (t - 50) = 230$
Group the t’s together and the plain numbers together. Three t’s make 3t, and $100 - 50 = 50$ .
$3t + 50 = 230$
Watch out: do not just subtract 100 and 50 from 230. That gives 80, choice B, and skips writing each day with t.
Solve for Tuesday
Subtract 50 from both sides. Then divide both sides by 3.
$\text{if } a = b \text{, then } a - c = b - c$
$3t + 50 - 50 = 230 - 50$
$3t = 180$
$t = 180 \div 3 = 60$
Tuesday had 60 loaves. Watch out: 60 is Tuesday, choice A. The question asks about Wednesday.
Find Wednesday’s loaves
Wednesday had 100 more loaves than Tuesday, so add 100 to t.
$t + 100 = 60 + 100 = 160$
Wednesday had 160 loaves. So the answer is D, 160.

Day	Loaves
Tuesday	$t$
Wednesday	$t + 100$
Thursday	$t - 50$

Check your answer

Find all three days. Tuesday is 60, Wednesday is 160, and Thursday is $60 - 50 = 10$ . Add them: $60 + 160 + 10 = 230$ . That is the total in the question, so 160 is right.

More questions like this

The original DOE question

Question 84 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 84. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 33 of 120

2024 Test A Math, question 29

Question

Which graph represents the solution to the inequality $x + 4 \geq 3$ ?

A $Number line from −4 to 7 with a tick at every integer. A closed dot at −1, with the line shaded from −1 to the right, ending in an arrow.$
B $Number line from −4 to 7 with a tick at every integer. A closed dot at 1, with the line shaded from 1 to the right, ending in an arrow.$
C $Number line from −4 to 7 with a tick at every integer. A closed dot at 3, with the line shaded from 3 to the right, ending in an arrow.$
D $Number line from −4 to 7 with a tick at every integer. A closed dot at 4, with the line shaded from 4 to the right, ending in an arrow.$

Show the answer

Correct answer: A

Use the properties of equations to solve the inequality for x.

$x + 4 \geq 3$

Subtract 4 from both sides of the inequality.

$x + 4 - 4 \geq 3 - 4$

$x \geq -1$

The graph of $x \geq -1$ will have a point at −1 and will include all values that are greater than, or to the right of, −1. The greater than or equal sign is used, which indicates that the solution includes −1. So the point at −1 will be solid. Option A is the correct answer.

Question analysis

This inequality question asks which number line graph shows the solution set of x + 4 ≥ 3. Students solve a one-step inequality by undoing the addition, then match the result to a graph. All four graphs use a solid dot and an arrow pointing right, so the whole question rests on where the shaded ray begins.

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 inequality work is usually two-step, such as 3x − 5 ≥ 10, and asks students to choose an open or closed dot and a direction. This one needs a single subtraction, and all four graphs share a closed dot and a rightward arrow.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 13, 9 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Solving takes a single subtraction, and the graphs differ only in where the solid dot sits, so the question comes down to 3 − 4. Students who lose the negative sign land on the graph that starts at 1, while those who never isolate x pick a graph that starts at a number printed in the inequality. Testing 0, which satisfies 0 + 4 ≥ 3, shows that the shading must include 0.

Reading a figure, table or graph minor Understanding the question
All four graphs share a closed dot and a rightward arrow, so the only thing to read is where the dot sits, against tick labels printed with raised minus signs. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
Undoing the + 4 means computing 3 − 4, the one step that crosses zero; a dropped minus sign moves the dot from −1 to 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract 4 from both sides to get x alone, then find the graph whose solid dot sits at that number; test one shaded value in the original inequality to confirm.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Computes 4 − 3 instead of 3 − 4, or drops the minus sign of −1, and starts the shaded ray at positive 1.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Starts the ray at 3, the number on the right side of the inequality, as if x ≥ 3, without subtracting the 4.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Starts the ray at 4, the number added to x, reading it off the inequality instead of solving for x.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an inequality balanced: $\text{if } a \geq b \text{, then } a - c \geq b - c$; a and b are the two sides of an inequality, and c is any number. The sign ≥ stands for greater than or equal to. Taking the same number away from both sides keeps the comparison true, like taking the same weight off both pans of a scale. Use it to get x by itself.

Sort out the question
An inequality is a math sentence that compares two amounts. The sign $\geq$ means is greater than or equal to. So $x + 4 \geq 3$ says that x plus 4 is 3 or more.
You need the graph that shows every number x that makes this true. All four graphs have a solid dot and an arrow to the right. Only the place where the dot sits is different.
Get x by itself
Right now 4 is added to x. To undo adding 4, subtract 4 from both sides.
$\text{if } a \geq b \text{, then } a - c \geq b - c$
$x + 4 - 4 \geq 3 - 4$
$x \geq -1$
So x is −1 or any number greater than −1. Watch out: 3 − 4 is −1, not 1. Dropping the minus sign puts the dot at 1, choice B. Putting the dot at 3 or 4 means you read a number off the inequality without solving. That leads to C or D.
Match the solution to a graph
On a number line, greater numbers are to the right. So the shading starts at −1 and goes right. The arrow shows that it keeps going.
The dot is solid, or filled in, because of the or equal to part. A solid dot means −1 itself is a solution, a number that makes the inequality true.
Graph A has a solid dot at −1 and is shaded to the right. So the answer is A, the graph with a solid dot at −1.

Check your answer

Test a number from the shaded part, such as 0. $0 + 4 = 4$ , and 4 is more than 3. Now test the dot: $-1 + 4 = 3$ , and 3 is equal to 3, so −1 works. A number left of the dot, −2, gives $-2 + 4 = 2$ . That is less than 3, so −2 should not be shaded, and it isn’t.

More questions like this

The original DOE question

Question 86 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 86. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 34 of 120

2024 Test A Math, question 36

Question

CELL PHONE SALES BY COLOR

Color	Percentage of Cell Phones Sold
White	$8k$
Black	30
Blue	$30 - 2k$
Red	$k + 5$
Total	100

The table above shows cell phone sales by color. What percentage of the cell phones sold were blue?

A18%
B20%
C22%
D28%

Show the answer

Correct answer: B

First, use the information given in the table and properties of equations to determine the value of k:

$8k + 30 + 30 - 2k + k + 5 = 100$

$8k - 2k + k + 30 + 30 + 5 = 100$

$7k + 65 = 100$

$7k + (65 - 65) = 100 - 65$

$7k = 35$

$k = 5$

Then, determine the percentage of cell phones sold that were blue by finding the value of the expression $30 - 2k$ (given in the table), when $k = 5$ :

$30 - 2k = 30 - 2(5) =$

$30 - 10 = 20$

The percentage of cell phones sold that were blue is 20%.

Question analysis

A table of cell phone sales by color gives each color's percentage as an expression in k, such as 8k for white and 30 − 2k for blue, with a total of 100. Students write and solve an equation for k, then find the percentage for blue. Combining terms with a negative coefficient, and remembering that k itself is not the answer, both need care.

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve two-step equations that are written out for them. Here the equation must be built from a table whose percentages total 100, with like terms to combine, and the value found must then be substituted back.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question hides an equation in a table, and the key insight is that the four percentages must add to 100. After that the algebra is routine, and no listed choice rewards stopping early, since neither k nor any other color's percentage appears. The wrong choices sit near the answer or use the blue expression with k ignored, so students who cannot set up the equation are left guessing.

Many steps moderate Working it out
The answer needs a chain: add the four expressions, set the sum to 100, collect 8k − 2k + k into 7k, solve for k, then substitute into 30 − 2k. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph minor Understanding the question
No equation is printed; the table's Total row of 100 is what turns the four percentages written in k into one equation to solve. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
Collecting like terms includes the negative coefficient in the blue row, −2k, and the constants 30, 30 and 5 have to be gathered too. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add the four color expressions and set the sum equal to 100, combine like terms to solve for k, then substitute k into the blue row's expression.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common slip gives k = 6 from 7k + 65 = 100, the value 18% would need; it sits 2 points below the answer and mostly draws guesses.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No usual error in collecting terms or solving yields k = 4, which 22% requires, so this near neighbor of the answer mostly attracts guessers.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Never solves for k and reads the blue entry 30 − 2k as 30 − 2, as if k were 1, giving 28.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Combining like terms: $ak + bk = (a + b)k$; k is the unknown number. a and b are the numbers in front of k, called coefficients. Terms with the same letter are called like terms. You can add their coefficients, because a groups of k plus b groups of k make a + b groups of k.
Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two sides of an equation, and c is any number (not 0 when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it balanced. Use it to undo each step and get k alone.

Sort out the table
The table gives the percent of phones sold in each color. Some entries use k, a letter that stands for an unknown number. 8k means 8 times k. You need the percent that were blue, $30 - 2k$ . To find it, you first need k.
Write an equation for the total
Every phone sold is one of the four colors. So the four percents add up to the total, 100, as the last row of the table shows.
$8k + 30 + (30 - 2k) + (k + 5) = 100$
Combine like terms
Terms with k in them are called like terms, so they can be combined. Group the k terms together and the plain numbers together. A k on its own means 1k.
$ak + bk = (a + b)k$
$8k - 2k + k = (8 - 2 + 1)k = 7k$
$30 + 30 + 5 = 65$
So the equation becomes $7k + 65 = 100$ . Watch out: the blue row takes away 2k. Keep its minus sign, or the k terms will be wrong.
Solve for k
Undo the adding first, then the multiplying.
$\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c$
$7k + 65 - 65 = 100 - 65$
$7k = 35$
$k = 35 \div 7 = 5$
So k is 5. But k is not the answer. You still need the blue row.
Find the blue percent
Put 5 in for k in the blue row.
$30 - 2k = 30 - 2 \times 5 = 30 - 10 = 20$
So 20% of the phones sold were blue. Watch out: 2k means 2 times k, not 2. Reading $30 - 2k$ as $30 - 2$ gives 28%, choice D.
So the answer is B, 20%.

Check your answer

Put k = 5 into every row. White: $8 \times 5 = 40$ . Black: 30. Blue: 20. Red: $5 + 5 = 10$ . Now add them: $40 + 30 + 20 + 10 = 100$ . The total matches the table, so k = 5 is right.

More questions like this

The original DOE question

Question 93 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 85, question 93. Open page 85 of the handbook (PDF)

Question 35 of 120

2024 Test A Math, question 46

Question

A graph shows the proportional relationship between the number of test questions a student gets correct, x, and the student’s test score, y. The ordered pair $\left(1, \frac{5}{4}\right)$ is on the graph. What does the y-coordinate of the ordered pair represent in this relationship?

AThe test will last $1\frac{1}{4}$ hours.
BEach test question is worth $1\frac{1}{4}$ points.
CAn average student can answer 5 questions in 4 minutes.
DA student who answers 5 questions correctly will earn 4 points.

Show the answer

Correct answer: B

The graph represents a proportional relationship, which means the y-intercept of the graph is (0, 0).

This means that the point $\left(1, \frac{5}{4}\right)$ represents the unit rate for this situation.

In this situation, x represents the number of questions answered correctly, and y represents the number of points received.

Therefore, for each question answered correctly, the student receives $\frac{5}{4}$ , or $1\frac{1}{4}$ , points.

Question analysis

A proportional relationship links the number of test questions a student gets right, x, to the test score, y, and the point (1, 5/4) lies on its graph. Students decide what the y-coordinate of that point means in the situation. The choices reuse the numbers 5, 4 and 1 1/4 in different ways, so the meaning of each coordinate must be kept straight.

NYS standards: 7.RP.2d points on proportional graphs
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Explaining what a point on the graph of a proportional relationship means is named in the grade 7 standards. This item is a direct case, with the graph described in words rather than drawn and only one point to interpret.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Variables instead of numbers minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 73 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question checks one idea: when x is 1, the y-coordinate is the score for one correct answer, the unit rate. The pitfall is reading 5/4 as two separate numbers, 5 and 4, and attaching them to questions and points the wrong way round. Ruling out the choices about hours and minutes is quick, since the relationship never mentions time.

Variables instead of numbers minor Understanding the question
No graph is drawn: the relationship and the point (1, 5/4) exist only in words, so students must say for themselves that x counts correct answers and y counts points. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The point gives 5/4, but two choices write it as 1 1/4 and two split it into 5 and 4, so the right reading must be recognized in another form. (Enright, Morley & Sheehan, 2002)

Strategy: Say in words what x and y stand for, then read the ordered pair as a sentence: when x has the first value, y has the second.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Attaches 1 1/4 to hours of testing, matching the number while ignoring that y measures the test score, not time.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Reads 5/4 as 5 questions in 4 minutes, using the numbers 5 and 4 with units that the relationship never measures.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Reads 5/4 as 5 correct answers for 4 points, turning the rate upside down; the point means 1 correct answer earns 5/4 points.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = k \times x$; x and y are two amounts that change together, here right answers and the test score. k is the constant of proportionality: how much y goes with 1 of x. In a proportional relationship, y is always the same number times x. So when x is 1, y equals k.

Sort out the question
x is the number of questions a student gets right. y is the student’s test score, in points. An ordered pair such as $\left(1, \frac{5}{4}\right)$ lists x first and y second. The y-coordinate is the second number, $\frac{5}{4}$ . You need what that number means.
Read the point as a sentence
In the point, x is 1 and y is $\frac{5}{4}$ . Put those into words: a student who gets 1 question right scores $\frac{5}{4}$ points.
Use the proportional rule
The relationship is proportional. So the score is always the same number times the questions right.
$y = k \times x$
$\frac{5}{4} = k \times 1$
So k is $\frac{5}{4}$ . Every right answer adds the same $\frac{5}{4}$ points. That makes $\frac{5}{4}$ the points for each question.
Write it as a mixed number
A mixed number is a whole number plus a fraction. Four quarters make 1 whole, so five quarters are 1 whole and 1 quarter.
$\frac{5}{4} = 1\frac{1}{4}$
Each question right is worth $1\frac{1}{4}$ points.
Match the choices
Choice B says each question is worth $1\frac{1}{4}$ points. That matches. Watch out: choice D splits $\frac{5}{4}$ into 5 and 4. But 5 right answers earn $5 \times \frac{5}{4} = 6\frac{1}{4}$ points, not 4. Choices A and C talk about time, which this relationship does not measure.
So the answer is B, each test question is worth $1\frac{1}{4}$ points.

Check your answer

Try another point. A student with 4 right answers scores $4 \times \frac{5}{4} = 5$ points. That fits B, since four questions at $1\frac{1}{4}$ points each make 5 points. It does not fit D, which says 5 questions earn 4 points.

More questions like this

The original DOE question

Question 103 as printed on page 88 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 88, question 103. Open page 88 of the handbook (PDF)

Question 36 of 120

2024 Test A Math, question 47

Question

In a survey of 200 adults in the town of Waskegon, 45 reported reading the online version of the Waskegon Bulletin the previous day. If 25,000 adults live in Waskegon, which number is the best estimate of the number of adults who read the online version of the Waskegon Bulletin the previous day?

A5,600
B9,000
C11,300
D24,800

Show the answer

Correct answer: A

In the survey, 45 out of 200 adults reported reading the online version of the Waskegon Bulletin, and there are 25,000 adults living in Waskegon. Set up a proportion to find the number of adults who read the online version.

$\frac{45}{200} = \frac{x}{25{,}000}$

$x = \frac{(45)(25{,}000)}{200} =$

$(45)(125) = 5{,}625$

Rounded to the nearest 100, this number is 5,600.

Question analysis

In a survey of 200 adults in Waskegon, 45 said they read the online Waskegon Bulletin the day before, and the town has 25,000 adults. This sampling question asks for the best estimate of how many adults in town read it. Students scale the sample's rate up to the population with a proportion and round.

NYS standards: 6.SP.1b samples and populations
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Using a sample to estimate a count for a whole population is typical grade 7 statistics and ratio work. This version takes one proportion with large numbers and a rounding step, about the usual grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 13, 50 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The idea is a single proportion, and the choices are far apart, so a reasonable estimate is enough. The main trap is treating 45 out of 200 as 45 percent, which doubles the estimate; the other wrong choices combine the survey numbers without a proportion. Noting that 45 is a bit under a quarter of 200, so the town figure must be a bit under a quarter of 25,000, confirms the choice.

Turning words into math minor Understanding the question
The survey's 45 out of 200 is a rate to scale up to the town's 25,000 adults, a fraction of the sample rather than a percent. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
The scale factor 25,000 ÷ 200 = 125 and the product 45 × 125 = 5,625 involve large numbers, and the result must then be rounded to the nearest hundred. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Set the survey's fraction, 45 out of 200, equal to the unknown number out of 25,000, then scale up: 25,000 is 125 times 200.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Multiplies the two survey numbers, 45 × 200 = 9,000, combining numbers from the question without setting up a proportion.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Treats 45 out of 200 as 45 percent, as if 100 adults were surveyed: 0.45 × 25,000 = 11,250, about 11,300.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts the 200 surveyed adults from the 25,000 in town, giving the number of adults who were not surveyed.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportion: $\frac{a}{b} = \frac{c}{d}$; A proportion says two ratios are equal. A ratio compares two numbers by dividing. Here a out of b in the survey matches c out of d in the whole town. Use it when a small group stands for a bigger group.

Sort out the question
A survey asked 200 adults, and 45 of them read the online paper. The town has 25,000 adults. You need a good estimate of how many adults in the whole town read it. Assume the town is like the survey. Then the same fraction of adults read it.
Set up a proportion
A proportion says two fractions are equal. Let x be the number of readers in the town. The fraction of readers is the same in both groups.
$\frac{a}{b} = \frac{c}{d}$
$\frac{45}{200} = \frac{x}{25{,}000}$
Watch out: multiplying 45 by 200 gives 9,000, choice B. That mixes the numbers without a plan.
Find the scale factor
The scale factor is how many times bigger the town is than the survey group.
$25{,}000 \div 200 = 125$
The town is 125 times as big. So multiply the top, 45, by 125 too.
Scale up the readers
$x = 45 \times 125$
Break 125 into 100 and 25.
$45 \times 100 = 4{,}500$
$45 \times 25 = 1{,}125$
$4{,}500 + 1{,}125 = 5{,}625$
About 5,625 adults read it. Watch out: 45 out of 200 is not 45%. A percent is out of 100. Using 45% gives about 11,300, choice C.
Round to the nearest hundred
The choices are rounded, since this is an estimate. 5,625 is between 5,600 and 5,700. Its last two digits, 25, are less than 50, so round down to 5,600.
So the answer is A, 5,600.

Check your answer

45 out of 200 is a bit less than 50 out of 200, which is one fourth. One fourth of 25,000 is 6,250. So the answer should be a bit less than 6,250. 5,600 fits, and the other choices are all much bigger.

More questions like this

The original DOE question

Question 104 as printed on page 88 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 88, question 104. Open page 88 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 37 of 120

2024 Test A Math, question 48

Question

A hiker plans on hiking 17 miles in 3 days. Which equation describes the relationship between the number of days hiked, x, and the number of miles traveled, y?

A $y = \frac{3}{17}x$
B $y = 3x$
C $y = \frac{17}{3}x$
D $y = 17x$

Show the answer

Correct answer: C

There is a proportional relationship between the number of miles traveled, y, and the number of days hiked, x.

$\frac{y}{x} = \frac{17}{3}$

$y = \frac{17}{3}x$

So the equation $y = \frac{17}{3}x$ describes the relationship.

Question analysis

A hiker plans to cover 17 miles in 3 days, and this proportional relationship question asks for the equation linking days hiked, x, to miles traveled, y. Students find the constant rate and write it as an equation of the form y = kx. Putting the right quantity on top of the rate is the main decision.

NYS standards: 7.RP.2c proportional relationship equations
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an equation y = kx for a proportional relationship is a core grade 7 skill, and this is the textbook form. The only extra demand is keeping the rate's direction straight: miles per day or days per mile.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Writing y = kx is routine, so the question rests on putting the right number on top of the rate. Two choices use one number from the stem with no division at all, and one inverts the rate. Substituting x = 3 into each equation is the fastest check, since only one of them gives y = 17.

A very close wrong answer moderate Choosing the answer
The coefficients 17/3 and 3/17 use the same two numbers, one the other inverted, so only checking that 3 days gives 17 miles separates the two equations. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math minor Understanding the question
With x as days and y as miles, "17 miles in 3 days" becomes miles per day, 17 ÷ 3, the number that multiplies x. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the constant of proportionality as miles per day, since y is miles and x is days, then write y equals that constant times x; check with x = 3.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides days by miles, 3/17, the number of days per mile, turning the rate upside down.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Uses the 3 days as the rate, writing y = 3x, a number from the question with no division.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Uses the 17 total miles as if they were covered in one day, writing y = 17x without dividing by the 3 days.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $k = \frac{y}{x}$; x and y are two amounts that change together. k is the constant of proportionality: how much y goes with 1 of x. In a proportional relationship, y divided by x gives the same k every time.
Proportional relationship: $y = kx$; kx means k times x. In a proportional relationship, y is always k times x. Here x is the number of days and y is the miles. k is the miles for each day. Use it to write the equation once you know k.

Sort out the question
The hiker plans 17 miles in 3 days. x is the number of days, and y is the number of miles. You need an equation that links them. Assume the hiker goes the same distance each day. Then y is always the same number times x.
Find the miles for each day
That number is the constant of proportionality, k. It is miles divided by days.
$k = \frac{y}{x}$
$k = \frac{17}{3}$
That is $5\frac{2}{3}$ miles each day. Watch out: miles go on top, because y is miles. $\frac{3}{17}$ is days per mile, upside down. That is choice A.
Write the equation
$y = kx$
$y = \frac{17}{3}x$
Watch out: 17 is the miles for all 3 days, not for 1 day. So $y = 17x$ , choice D, is too fast. And $y = 3x$ , choice B, uses the days as the rate.
So the answer is C, $y = \frac{17}{3}x$ .

Check your answer

Put in x = 3, for 3 days: $y = \frac{17}{3} \times 3 = 17$ . That is 17 miles, which matches the plan. Try choice A the same way: $\frac{3}{17} \times 3 = \frac{9}{17}$ . That is less than 1 mile, far too short.

More questions like this

The original DOE question

Question 105 as printed on page 88 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 88, question 105. Open page 88 of the handbook (PDF)

Question 38 of 120

2024 Test A Math, question 55

Question

The table below shows the number of cups of red paint and blue paint used to make a purple paint mixture.

PURPLE PAINT

Cups of Red Paint	Cups of Blue Paint
1	1.5
4	6
11	y

Based on the relationship between the number of cups of red paint and the number of cups of blue paint, what is the value of y?

A10.5
B13
C16.5
D24

Show the answer

Correct answer: C

Divide the number of cups of blue paint by the number of cups of red paint to find the constant of proportionality.

$\frac{1.5}{1} = \frac{6}{4} = 1.5$

Multiply to find the value of y.

$11 \times 1.5 = 16.5$

Question analysis

A table gives cups of red and blue paint for a purple mixture: 1 and 1.5, 4 and 6, and 11 and an unknown y. This ratio table question asks for y based on the relationship between the two colors. The tricky part is noticing whether the relationship works by multiplying or by adding.

NYS standards: 7.RP.2b constant of proportionality
6.RP.3a tables of equivalent ratios
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a constant of proportionality from a table is grade 7 work, but here one row already pairs 1 cup of red with its cups of blue, so a single multiplication finishes it, lighter than typical grade 7 tasks.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 57 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The relationship is multiplicative, 1.5 cups of blue for every cup of red, and one multiplication finishes the question. The trap is additive thinking: red grows by 7 cups from the second row to the third, and adding the same 7 to the blue column gives a listed answer. Testing a rule on every row exposes the additive one, since blue is 0.5 more than red in the first row but 2 more in the second.

Reading a figure, table or graph minor Understanding the question
The table's first row, 1 cup of red to 1.5 cups of blue, hands over the constant of proportionality; the other two rows only confirm it. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the row with 1 cup of red paint to read the cups of blue per cup of red, multiply that rate by 11, and check the rate against the row with 4 cups.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Scales only the jump from 4 to 11 cups of red, 7 × 1.5 = 10.5, and forgets to add the 6 cups of blue already there.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Thinks additively: red goes up 7 cups from 4 to 11, so adds the same 7 to the 6 cups of blue, getting 13.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the two numbers in the second row, 4 × 6 = 24, without finding how the two columns are related.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $k = \frac{y}{x}$; x and y are two amounts that change together. k is the constant of proportionality: how much y goes with 1 of x. In a proportional relationship, y divided by x gives the same k every time.
Proportional relationship: $y = k \times x$; x is the cups of red paint, and y is the cups of blue paint. k is the cups of blue for each cup of red. In a proportional relationship, y is always k times x. Use it to find y for any amount of red.

Sort out the table
The table shows cups of red and blue paint for the same purple. 1 cup of red goes with 1.5 cups of blue. 4 cups of red go with 6 cups of blue. You need y, the cups of blue for 11 cups of red. The same color means the paints mix at the same rate in every row.
Find the blue for each cup of red
$k = \frac{y}{x}$
$k = \frac{1.5}{1} = 1.5 \quad \text{and} \quad k = \frac{6}{4} = 1.5$
Both rows give 1.5. So there are always 1.5 cups of blue for each cup of red. This number is the constant of proportionality.
Use the rate for 11 cups
$y = k \times x$
$y = 1.5 \times 11 = 16.5$
Think of it as 11 cups plus half of 11 cups: $11 + 5.5 = 16.5$ . So y is 16.5 cups of blue.
Avoid the traps and answer
Watch out: red goes up 7 cups from 4 to 11. Adding the same 7 cups to the 6 cups of blue gives 13, choice B. Paint mixes by multiplying, not by adding. Watch out: $7 \times 1.5 = 10.5$ , choice A, is only the extra blue for the 7 extra cups of red.
So the answer is C, 16.5.

Check your answer

Divide to check: $16.5 \div 11 = 1.5$ . That matches the other rows. Also, every row has more blue than red, so y must be more than 11. That rules out 10.5.

More questions like this

The original DOE question

Question 112 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 90, question 112. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 39 of 120

2024 Test A Math, question 57

Question

An ice cream shop sells 16 cups of ice cream for $48 and 4 cups for $12. There is a proportional relationship between the number of cups of ice cream and the cost. What is the constant of proportionality for this relationship?

A3 cups per dollar
B3 dollars per cup
C4 cups per dollar
D4 dollars per cup

Show the answer

Correct answer: B

There is a proportional relationship between the cost and the number of cups of ice cream.

Divide the cost by the number of cups of ice cream to find the constant of proportionality.

$48\text{ dollars} \div 16\text{ cups}$

$= 3\text{ dollars per cup}$

$12\text{ dollars} \div 4\text{ cups}$

$= 3\text{ dollars per cup}$

The constant is the cost per cup, 3 dollars.

Question analysis

An ice cream shop sells 16 cups for $48 and 4 cups for $12, and the cost is proportional to the number of cups. This proportional relationship question asks for the constant of proportionality, with its units. The choices pair two numbers with two kinds of units, so the units of the rate matter as much as the number.

NYS standards: 7.RP.2b constant of proportionality
6.RP.2 unit rate concept
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 students find constants of proportionality in tables, graphs and equations. Here one division gives the number, as in grade 6 unit pricing, and the real demand is attaching the right units to it.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 50 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One division gives the number, so the question is really about units. The cost depends on the number of cups, and dividing $48 by 16 cups gives the dollars for each cup; reversing the label describes a shop where one dollar buys three cups. The second offer, 4 cups for $12, is a built-in check that the rate stays the same.

A very close wrong answer moderate Choosing the answer
Two choices carry the number 3, "3 cups per dollar" and "3 dollars per cup", so finding 48 ÷ 16 is not enough; the units decide. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that dividing dollars by cups gives dollars per cup: a quotient's units read as the top quantity per the bottom one. (Anderson & Krathwohl, 2001)

Strategy: Divide one quantity by the other for one offer, check that the other offer gives the same quotient, and label it with the numerator's units per the denominator's units.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Gets 3 from 48 ÷ 16 but labels it cups per dollar, turning the rate upside down; at these prices a dollar buys only 1/3 of a cup.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the two cup amounts, 16 ÷ 4 = 4, and attaches cups per dollar, a ratio of like quantities given made-up units.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Divides the two costs, $48 ÷ $12 = 4, or the two cup counts, and calls the result dollars per cup.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $k = \frac{y}{x}$; In a proportional relationship, y is always the same number, k, times x. Here x is the number of cups and y is the cost in dollars. So k is y divided by x. Its units are the top’s units per the bottom’s units: dollars per cup.

Sort out the question
16 cups cost $48, and 4 cups cost $12. The cost is proportional to the cups. That means the cost is always the same number times the number of cups. That number is the constant of proportionality. You need it with the right units.
Divide cost by cups
For each offer, divide the cost in dollars by the number of cups.
$k = \frac{y}{x}$
$k = \frac{48}{16} = 3 \quad \text{and} \quad k = \frac{12}{4} = 3$
Both offers give 3. So the constant is 3.
Name the units
Dollars are on top and cups are on the bottom. So the units are dollars per cup. 3 dollars per cup means each cup costs $3. Watch out: two choices use 3. Choice A, 3 cups per dollar, flips the rate. Then 16 cups would cost only about $5, not $48.
Avoid the traps and answer
Watch out: dividing two costs, $48 \div 12$ , or two cup counts, $16 \div 4$ , gives 4. That compares like amounts, so it is not a cost for each cup. It leads to choice C or D.
So the answer is B, 3 dollars per cup.

Check your answer

Multiply back. At 3 dollars per cup, 16 cups cost $3 \times 16 = 48$ , which is $48. And 4 cups cost $3 \times 4 = 12$ , which is $12. Both match the shop’s prices.

More questions like this

The original DOE question

Question 114 as printed on page 91 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 91, question 114. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 40 of 120

2024 Test B Math, question 8

Question

If $\frac{9}{2x}=\frac{3y}{8}$ , where $x\neq 0$ and $y\neq 0$ , what is the product of x and y?

Show the answer

Correct answer: C

$\frac{9}{2x}=\frac{3y}{8}$

$9\bullet 8=72$

$2x\bullet 3y=6xy$

$72=6xy$

$12=xy$

Question analysis

An algebra question gives a proportion with x in one denominator and y in one numerator, 9 over 2x equal to 3y over 8, and asks for the product of x and y. Neither variable can be found on its own, so students must clear the fractions and treat the product as a single quantity.

NYS standards: 8.EE.7 linear equations in one variable
6.EE.2b parts of an expression
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students cross-multiply numeric proportions to find one unknown. Here two variables sit in the proportion and the question wants their product, a literal-equation idea from grade 8 that many grade 7 students have not met.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question looks harder than it is, because it asks for a product rather than for x or y. Cross-multiplying turns the proportion into 72 = 6xy, and dividing by 6 finishes it, so the only real hurdle is trusting that xy can be solved for as one quantity. The wrong choices come from grabbing numbers along the way, not from a nearly complete solution.

No obvious method moderate Working it out
Neither x nor y can be found alone, so students must see that cross-multiplying gives 72 = 6xy, where the product xy appears as one block to solve for. (Schoenfeld, 1985)
Variables instead of numbers minor Understanding the question
Both unknowns stay letters and only their product is asked for, so there is no value of x or y to check a result against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Cross-multiply to clear both fractions, collect the numbers on one side and x and y together on the other, then divide so that only the product is left.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Matches the denominators, 2x = 8, to get x = 4 and gives the value of x instead of the product of x and y.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the 6 in 72 = 6xy, the product of the coefficients 2 and 3, as the answer without dividing 72 by it.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 2 and the 8 from the two denominators, combining numbers in the equation instead of cross-multiplying.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a, b, c and d are numbers or letters, and b and d are not zero. When two fractions are equal, multiply each top by the other bottom. The two products are equal. It works because it is the same as multiplying both sides by b and by d.

Sort out the question
The equation says that two fractions are equal:
$\frac{9}{2x} = \frac{3y}{8}$
Here 2x means 2 times x, and 3y means 3 times y. You need the product of x and y. A product is the answer to a multiplication. So you need $x \times y$ , written xy. You do not need x or y alone.
Cross-multiply the fractions
An equation without fractions is easier to work with. So cross-multiply.
$\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$
$9 \times 8 = 2x \times 3y$
Watch out: do not just set the bottoms equal. Writing 2x = 8 gives x = 4, choice A. That is x, not the product.
Multiply out both sides
On the left, $9 \times 8 = 72$ . On the right, multiply the numbers and keep the letters together.
$2x \times 3y = 2 \times 3 \times x \times y = 6xy$
So the equation is now:
$72 = 6xy$
The xy is the product you want, all in one piece.
Divide both sides by 6
xy is being multiplied by 6. To undo that, divide both sides by 6.
$xy = 72 \div 6 = 12$
Watch out: 6 is just the number in front of xy. Choosing it gives B.
So the answer is C, 12.

Check your answer

Pick x = 4. Then y must be 3, so that xy = 12. Left side: $\frac{9}{2 \times 4} = \frac{9}{8}$ . Right side: $\frac{3 \times 3}{8} = \frac{9}{8}$ . The two sides match, so a product of 12 works.

More questions like this

The original DOE question

Question 65 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 200, question 65. Open page 200 of the handbook (PDF)

Question 41 of 120

2024 Test B Math, question 14

Question

Karen reads at the rate of 60 pages per day. Martina reads 25% more pages per day than Karen. If the two students begin 1,500-page reading assignments on the same day, how many days sooner will Martina complete the assignment than Karen?

Show the answer

Correct answer: B

First, determine the rate at which Martina reads, in pages per day, using the rate at which Karen reads, 60 pages per day.

$60+60\bullet\frac{25}{100}=$

$60+60\bullet 0.25=$

$60+15=75$

Next, determine the number of days it will take each person to complete the 1,500-page reading assignment.

Karen: $\frac{1{,}500\text{ pages}}{k\text{ days}}=\frac{60\text{ pages}}{1\text{ day}}$

$60k=1{,}500$

$(60\div 60)k=1{,}500\div 60$

$k=25\text{ days}$

Martina: $\frac{1{,}500\text{ pages}}{m\text{ days}}=\frac{75\text{ pages}}{1\text{ day}}$

$75m=1{,}500$

$(75\div 75)m=1{,}500\div 75$

$m=20\text{ days}$

Finally, determine the difference between the two numbers of days.

$25-20=5$

So Martina will complete the assignment 5 days sooner than Karen.

Question analysis

A reading-rate word problem: Karen reads 60 pages a day, and Martina reads 25% more pages a day. Both start a 1,500-page assignment on the same day, and students find how many days sooner Martina finishes. It combines a percent increase with rates and times, and asks for a difference in days.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually stop after one increase or decrease. This one turns the percent into a new rate, finds two finishing times and compares them, four steps that go beyond a typical grade 7 task.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Combines ideas moderate
Turning words into math minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is easy on its own; the difficulty is in carrying the chain to the end. Three of the choices are numbers a student meets or makes along the way: the extra pages per day, a quotient of two rates, or a rate built on 25 pages instead of 25%. Writing both finishing times down before subtracting keeps the target, a number of days, in view.

Many steps moderate Working it out
Four operations chain together: 25% of 60, Martina's 75 pages, the two finishing times 1,500 ÷ 60 and 1,500 ÷ 75, and finally their difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Combines ideas moderate Working it out
A percent increase in a daily rate has to be turned into a change in days, and 25% more pages a day does not mean 25% fewer days. (Embretson, 1998; Daniel & Embretson, 2010)
Turning words into math minor Understanding the question
“25% more pages per day than Karen” makes Martina's rate 1.25 × 60, and “how many days sooner” asks for a difference of two times, not of two rates. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Increase 60 by 25% to get Martina's daily pages, divide 1,500 by each girl's daily rate to get her number of days, then subtract the two numbers of days.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides Karen's 60 pages a day by Martina's 15 extra pages a day, combining two rates without finding how long either girl takes.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Adds 25 pages instead of 25%, giving Martina 85 pages a day, so she needs 1,500 ÷ 85 ≈ 17.6, or 18 days, and 25 − 18 = 7.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that Martina reads 25% of 60 = 15 more pages each day and stops there, giving a number of pages instead of days.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{new} = \text{old} + \frac{p}{100} \times \text{old}$; Old is the starting amount, and p is the percent it grows by. First find p percent of the old amount. That is the extra. Then add the extra to the old amount to get the new amount.
Time from a rate: $\text{days} = \text{total pages} \div \text{pages per day}$; Total pages is the whole assignment. Pages per day is how fast someone reads. Dividing tells how many days of reading fit into the whole job.

Sort out what you know
Karen reads 60 pages per day. Martina reads 25% more pages per day than Karen. Each girl must read 1,500 pages. You need how many days sooner Martina finishes. That is the difference between their numbers of days.
Find Martina’s pages per day
25% more means Karen’s 60 pages plus 25% of 60.
$\text{new} = \text{old} + \frac{p}{100} \times \text{old}$
$60 + \frac{25}{100} \times 60 = 60 + 15 = 75$
Martina reads 75 pages per day. Watch out: 15 is extra pages, not days. Stopping here gives 15, choice D. Adding 25 pages instead of 25% gives 85 pages and leads to 7, choice C.
Find each girl’s number of days
Divide the whole assignment by each girl’s pages per day.
$\text{days} = \text{total pages} \div \text{pages per day}$
$\text{Karen: } 1{,}500 \div 60 = 25$
$\text{Martina: } 1{,}500 \div 75 = 20$
Karen needs 25 days, and Martina needs 20 days.
Subtract the days
“How many days sooner” asks for the difference.
$25 - 20 = 5$
Martina finishes 5 days sooner. Watch out: dividing 60 by 15 mixes two rates and gives 4, choice A.
So the answer is B, 5.

Check your answer

Multiply back. Karen: $25 \times 60 = 1{,}500$ pages. Martina: $20 \times 75 = 1{,}500$ pages. Both read all 1,500 pages, and 25 days is 5 days more than 20 days.

More questions like this

The original DOE question

Question 71 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 202, question 71. Open page 202 of the handbook (PDF)

Question 42 of 120

2024 Test B Math, question 17

Question

A bowl contains 12 green candies, 4 yellow candies, and some red candies. The probability of choosing a green candy at random is twice as great as the probability of choosing a red candy at random. What is the probability of choosing a yellow candy at random?

A $\frac{2}{11}$
B $\frac{2}{9}$
C $\frac{1}{4}$
D $\frac{3}{11}$

Show the answer

Correct answer: A

The probability of choosing a green candy is twice as great as the probability of choosing a red candy. There are 12 green candies. Let r = the number of red candies.

$12=2r$

$r=6$

So there are 6 red candies. 12 green candies + 4 yellow candies + 6 red candies = 22 total candies.

There are 4 yellow candies out of a total of 22 candies. So the probability of choosing a yellow candy is

$\frac{4}{22}=\frac{2}{11}$ .

Question analysis

A probability question about a bowl of 12 green candies, 4 yellow candies and an unknown number of red ones. A green candy is twice as likely to be picked as a red one, and students find the chance of picking a yellow candy. The missing red count has to be worked out before any probability can be found.

NYS standards: 6.SP.8a uniform probability models
6.EE.7 one-step equations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Simple probability is grade 6 content in New York, and grade 7 problems usually give every count. Working backward from a comparison of two probabilities to a missing count, then forward to a third probability, is more than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Combines ideas moderate
Solution: 3 steps
Reading level: Question and choices: about grade 7, 49 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question withholds the red count, and everything depends on recovering it from the comparison of probabilities. Students who cannot make that step tend to leave red out altogether, while those who find 6 red candies can still slip at the end by using the wrong total or the wrong color. Writing all three counts and the total of 22 before forming any fraction prevents both endings.

Turning words into math moderate Understanding the question
“Twice as great as the probability of choosing a red candy” must become 12 = 2 × the red count, which works only because both probabilities share one total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Combines ideas moderate Working it out
An unknown count has to be found from a ratio before any probability can be written, joining proportional reasoning with probability as a part over the total. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Turn twice as likely into a statement about numbers of candies to find the red count, add all three colors for the total, then put the yellow count over that total and simplify.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Finds the 6 red candies but compares the 4 yellow with the 18 green and red ones, 4/18, a part-to-part ratio instead of part to whole.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Leaves out the red candies, whose number is not given, and divides the 4 yellow by the 16 green and yellow candies.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds the 6 red candies and the total of 22 correctly but gives the chance of red, 6/22 = 3/11, instead of yellow.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of the kind you want}}{\text{total number}}$; P is the probability, the chance that something happens. The top is how many candies are the kind you want. The bottom is how many candies there are in all. Each candy is just as likely to be picked, so this fraction gives the chance.

Sort out what you know
The bowl has 12 green candies, 4 yellow candies and some red candies. You are not told how many are red. Let r be the number of red candies.
The chance of picking green is twice the chance of picking red. You need the probability (the chance) of picking yellow.
Compare green and red
Write each chance with the probability formula.
$P = \frac{\text{number of the kind you want}}{\text{total number}}$
Green and red have the same bottom number, the total number of candies. So if the chance of green is twice the chance of red, there are twice as many green candies as red ones.
$12 = 2 \times r$
Find the red candies and the total
$r = 12 \div 2 = 6$
There are 6 red candies. Now add up all the candies.
$12 + 4 + 6 = 22$
There are 22 candies in all. Watch out: do not leave out the red candies just because their number was not given. Using 16 as the total gives 1/4, choice C.
Find the chance of yellow
There are 4 yellow candies out of 22.
$P = \frac{4}{22} = \frac{2}{11}$
Divide the top and the bottom by 2 to simplify. Watch out: the question asks for yellow, not red. 6/22 = 3/11 is the chance of red, choice D. The bottom must be all 22 candies. Using 18 gives 2/9, choice B.
So the answer is A, $\frac{2}{11}$ .

Check your answer

Check the clue. The chance of green is $\frac{12}{22}$ , and the chance of red is $\frac{6}{22}$ . Twelve is twice 6, so green is twice as likely. Also, the three chances add up to 1: $\frac{12}{22} + \frac{4}{22} + \frac{6}{22} = \frac{22}{22} = 1$ .

More questions like this

The original DOE question

Question 74 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 74. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 43 of 120

2024 Test B Math, question 19

Question

Serena was studying the number of water stations for different lengths of races. The table below shows the data she collected.

WATER STATIONS IN RACES

Length of Race (mi)	Number of Water Stations
3	2
12	8
18	12

Which equation represents the relationship between the length of the race, x miles, and the number of water stations, y?

A $y=\frac{2}{3}x$
B $y=\frac{3}{2}x$
C $y=x+6$
D $y=x-1$

Show the answer

Correct answer: A

There is a proportional relationship between the length of the race, x miles, and the number of water stations, y. The constant of proportionality is

$\frac{y}{x}=\frac{2}{3}$

$y=\frac{2}{3}x$

Therefore, the equation $y=\frac{2}{3}x$ represents this relationship.

Question analysis

A table from a study of races shows the number of water stations for races of 3, 12 and 18 miles. Students choose the equation that links the race length x to the number of stations y. It tests reading a relationship from a table and expressing it as an equation in x and y.

NYS standards: 7.RP.2c proportional relationship equations
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing an equation for a relationship given in a table is a central grade 7 skill. With three clean rows and whole numbers, this is a typical grade 7 exercise.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 57 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The table is small and its numbers are friendly, so the question comes down to which way the ratio runs and whether every row is checked. One wrong equation fits only the first row and another fits none, so testing a single row is not enough. Putting the 12-mile race into each equation, which should give 8 stations, leaves only one rule standing.

A very close wrong answer moderate Choosing the answer
The coefficients 2/3 and 3/2 use the same two digits, and only dividing stations by miles, 2 ÷ 3, rather than miles by stations, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
The table lists miles first and stations second, while each equation puts stations, y, alone on the left, so every row must be read as an (x, y) pair. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Test each equation on all three rows of the table, not just the first; only a rule that fits every row represents the relationship.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Checks only one number: Tries one value in each choice and takes the first that works, when that value happens to work for a wrong choice too; a second value would rule it out.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides race length by water stations, 3 ÷ 2, turning the ratio upside down, so the rule gives more stations than miles.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Builds an addition rule from the gap of 6 between 18 and 12 in the last row, with the sign reversed, and it fits no row of the table.
D Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: moderate
Checks only the first row, where 3 − 1 = 2 works, without testing the 12- and 18-mile races, where it gives 11 and 17.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship: $y = kx$; x is the race length in miles, and y is the number of water stations. k is the constant of proportionality, the number of stations for each mile. Find it with y ÷ x. If every row gives the same k, the table is proportional.

Sort out the table
Each row of the table is a pair of numbers. The first is the race length, x, in miles. The second is the number of water stations, y. You need the equation that works for every row.
Divide y by x in each row
See if every row has the same ratio of stations to miles. Divide y by x.
Miles (x) Stations (y) y ÷ x
3 2 $\frac{2}{3}$
12 8 $\frac{8}{12} = \frac{2}{3}$
18 12 $\frac{12}{18} = \frac{2}{3}$
Every row gives 2/3. So the relationship is proportional: y is always the same number times x.
Write the equation
$y = kx$
Here k is the ratio you found, $\frac{2}{3}$ .
$y = \frac{2}{3}x$
Watch out: divide stations by miles, not miles by stations. 3 ÷ 2 gives 3/2, choice B. That rule would give more stations than miles.
Test the other choices and answer
Watch out: test every row, not just the first one. Choice D, y = x − 1, works for 3 miles. But it gives 11 stations for 12 miles, not 8. Choice C, y = x + 6, gives 9 stations for 3 miles, not 2.
So the answer is A, $y = \frac{2}{3}x$ .

Miles (x)	Stations (y)	y ÷ x
3	2	$\frac{2}{3}$
12	8	$\frac{8}{12} = \frac{2}{3}$
18	12	$\frac{12}{18} = \frac{2}{3}$

Check your answer

Put each race length into $y = \frac{2}{3}x$ . $\frac{2}{3} \times 3 = 2$ , $\frac{2}{3} \times 12 = 8$ and $\frac{2}{3} \times 18 = 12$ . All three match the table.

More questions like this

The original DOE question

Question 76 as printed on page 204 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 204, question 76. Open page 204 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 44 of 120

2024 Test B Math, question 23

Question

Each loaf of bread a baker makes requires $\frac{1}{5}$ ounce of salt. How many loaves can be made if 40 ounces of salt are used?

A8
B20
C45
D200

Show the answer

Correct answer: D

Each loaf of bread requires $\frac{1}{5}$ ounce of salt. Therefore, the baker can make 5 loaves of bread with 1 ounce of salt. Multiply to find the number of loaves that can be made with 40 ounces of salt:

$5\times 40=200$

Question analysis

A baking word problem: each loaf of bread needs 1/5 ounce of salt, and the baker uses 40 ounces. Students find how many loaves that much salt makes, working with a whole number and a unit fraction. Deciding whether to multiply or divide is the whole question.

NYS standards: 6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Dividing a whole number by a fraction in a word problem is a grade 6 standard. This takes one step with a unit fraction, simpler than typical grade 7 rational number problems.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Solution: 1 step
Reading level: Question and choices: about grade 3, 24 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The only decision is whether to multiply or divide, and thinking about one ounce first settles it: one ounce holds five fifths, so it makes five loaves. Students who multiply 40 by 1/5 get a small number, but a tiny amount of salt per loaf should give many loaves, not few. That reasonableness check leaves a single choice.

Turning words into math minor Understanding the question
“How many loaves” from 40 ounces at 1/5 ounce each asks how many fifths fit in 40, a division that many students turn into multiplying by 1/5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work out how many loaves one ounce of salt makes, then scale up to 40 ounces; dividing 40 by 1/5 gives the same count.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 40 by 1/5 instead of dividing, finding a fifth of the salt rather than how many fifths of an ounce it holds.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Knows to multiply 40 by 5 but drops a zero, writing the product as 20 instead of 200.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the 5 from 1/5 to the 40 ounces, combining the numbers in the problem without a model of the situation.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of equal groups: $\text{groups} = \text{total} \div \text{size of one group}$; Total is the whole amount you have. Size of one group is how much each one uses. Dividing tells how many groups fit into the total. Here one group is the salt for one loaf.
Dividing by a unit fraction: $a \div \frac{1}{n} = a \times n$; a is any number. 1 over n is a unit fraction, a fraction with 1 on top. Dividing by it asks how many nths fit into a. Each whole holds n of them, so you multiply by n.

Sort out what you know
Each loaf uses $\frac{1}{5}$ ounce of salt. The baker uses 40 ounces of salt. You need how many loaves that makes. That means how many fifths of an ounce fit into 40 ounces.
Decide to divide
You are splitting the salt into equal groups of 1/5 ounce. Each group makes one loaf. So divide the total salt by the salt for one loaf.
$\text{groups} = \text{total} \div \text{size of one group}$
$\text{loaves} = 40 \div \frac{1}{5}$
Watch out: do not multiply by 1/5. $40 \times \frac{1}{5} = 8$ , choice A, is a fifth of the salt, not a number of loaves.
Divide by the fraction
Each ounce holds 5 fifths, so 1 ounce makes 5 loaves. Dividing by 1/5 is the same as multiplying by 5.
$a \div \frac{1}{n} = a \times n$
$40 \div \frac{1}{5} = 40 \times 5 = 200$
The salt makes 200 loaves. Watch out: keep both zeros. 20, choice B, drops one.
So the answer is D, 200.

Check your answer

Multiply back. 200 loaves at $\frac{1}{5}$ ounce each use $200 \times \frac{1}{5} = 40$ ounces. That is all the salt. Also, each loaf uses less than 1 ounce, so there should be more loaves than ounces. 200 is more than 40.

More questions like this

The original DOE question

Question 80 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 205, question 80. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 45 of 120

2024 Test B Math, question 26

Question

If 1 sind = 0.75 plunk, how many sinds are equivalent to 8 plunks, rounded to the nearest hundredth?

A1.33
B6.00
C7.25
D10.67

Show the answer

Correct answer: D

Set up a proportion:

$\frac{1\text{ sind}}{0.75\text{ plunk}}=\frac{x\text{ sinds}}{8\text{ plunks}}$

$1(8)=0.75x$

$8=\frac{75}{100}x$

$8=\frac{3}{4}x$

$8\left(\frac{4}{3}\right)=x$

$\frac{32}{3}=x$

$10\frac{2}{3}=x$

$10.67\approx x$

Question analysis

A unit conversion question with made-up units: 1 sind equals 0.75 plunk, and students find how many sinds equal 8 plunks, rounded to the nearest hundredth. With unfamiliar unit names there is no everyday sense to lean on, so deciding whether to multiply or divide by 0.75 is the hard part.

NYS standards: 6.RP.3d converting measurement units
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Converting units with ratio reasoning is grade 6 content, and made-up unit names are a familiar SHSAT twist. Working with a decimal rate and rounding to hundredths put it at about the level of typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Made-up units remove the everyday sense that usually tells students which way a conversion goes, so the relationship must be reasoned out. The most tempting error multiplies by 0.75 and goes the wrong way; another finds the rate for one plunk and stops. A size check settles it: each sind is less than a plunk, so the number of sinds must exceed 8.

Turning words into math moderate Understanding the question
“1 sind = 0.75 plunk” says a sind is smaller than a plunk, so 8 plunks hold more than 8 sinds; seeing that decides between dividing and multiplying by 0.75. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
8 ÷ 0.75 is easiest as 8 × 4/3 = 32/3, and the repeating decimal 10.666… must then be rounded to the nearest hundredth. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Set up a proportion with sinds over plunks on both sides, put the 8 plunks in the matching place, solve for the sinds, and round to the nearest hundredth.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds how many sinds make one plunk, 1 ÷ 0.75 ≈ 1.33, and stops before scaling up to 8 plunks.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 8 by 0.75 instead of dividing, converting in the wrong direction and finding the plunks in 8 sinds.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts 0.75 from 8, combining the two numbers in the question without setting up a conversion.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a, b, c and d are numbers or letters, and b and d are not zero. When two fractions are equal, multiply each top by the other bottom. The two products are equal. It works because it is the same as multiplying both sides by b and by d.
Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is any number, and b over c is the fraction you divide by. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down. It asks how many of that fraction fit into a.

Sort out what you know
Sinds and plunks are made-up units, like inches and feet. One sind equals 0.75 plunk, so a sind is smaller than a plunk. You need how many sinds equal 8 plunks. Round to the nearest hundredth, two places after the decimal point.
Decide if the answer is more than 8
A sind is only 3/4 of a plunk. So it takes more than 8 sinds to make 8 plunks. The answer must be more than 8. Watch out: $8 \times 0.75 = 6$ , choice B, goes the wrong way. That is the number of plunks in 8 sinds.
Set up a proportion
A proportion says that two ratios are equal. Put sinds on top and plunks on the bottom on both sides. Let x be the number of sinds.
$\frac{1}{0.75} = \frac{x}{8}$
$\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$
$1 \times 8 = 0.75 \times x$
So $0.75x = 8$ .
Divide by 0.75
0.75 is the same as $\frac{3}{4}$ . Divide both sides by it. To divide by a fraction, multiply by its reciprocal.
$a \div \frac{b}{c} = a \times \frac{c}{b}$
$x = 8 \div \frac{3}{4} = 8 \times \frac{4}{3} = \frac{32}{3} = 10.666\ldots$
Watch out: 1.33, choice A, is the number of sinds in just 1 plunk. Keep going to 8 plunks.
Round to the nearest hundredth
Look at the third digit after the point, the thousandths digit. It is 6, which is 5 or more. So round the hundredths digit up, from 6 to 7.
$10.666\ldots \approx 10.67$
So the answer is D, 10.67.

Check your answer

Multiply back. Each sind is 0.75 plunk, so 10.67 sinds are $10.67 \times 0.75 \approx 8.00$ plunks. That matches the 8 plunks. The answer is also more than 8, as it must be.

More questions like this

The original DOE question

Question 83 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 206, question 83. Open page 206 of the handbook (PDF)

Question 46 of 120

2024 Test B Math, question 28

Question

A community center has two 30-member exercise groups. The director recorded the ages of each group’s members in the box plots.

$Two box plots titled AGES OF GROUPS over one number line from 15 to 33, with a tick at every whole number and labels at 15, 18, 21, 24, 27, 30 and 33. Group P: whiskers from 17 to 33, box from 19 to 27, median line at 24. Group Q: whiskers from 15 to 29, box from 17 to 26, median line at 21.$

Which statement about the two groups can be verified based on the box plots?

ABoth groups have at least one member who is exactly 29 years old.
BGroup P has more members who are at least 17 years old than Group Q has.
CBoth groups have an equal number of members who are between 21 and 24 years old.
DGroup Q has fewer members who are between 19 and 27 years old than Group P has.

Show the answer

Correct answer: B

Groups P and Q have an equal number of members. All the members of Group P are 17 or older, but only about $\frac{3}{4}$ of the members of Group Q are 17 or older.

Question analysis

Box plots show the ages of the members of two 30-member exercise groups at a community center, Group P and Group Q. Students decide which statement comparing the groups can be verified from the plots alone. The challenge is knowing what a box plot guarantees about individual ages and counts, and what it leaves unknown.

NYS standards: 7.SP.1 box plots and interquartile range
7.SP.4 comparative inferences about populations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students read box plots and compare medians and spreads. Deciding which claims two box plots can prove, from what the five key values guarantee about counts, is a harder reasoning task than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 3, strategic thinking · mainly evaluate
What makes it hard: Every choice must be checked major
Something you must already know moderate
Reading a figure, table or graph moderate
Solution: 3 steps
Reading level: Question and choices: about grade 5, 91 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a question about evidence: the right statement is the only one the box plots guarantee, while the others may or may not be true. Students who treat the plots as pictures of every member, assuming ages fill each whisker or spread evenly through each box, find support for claims that cannot be checked. Asking what must hold for any 30 ages with these summaries separates the verifiable claim from the merely plausible ones.

Every choice must be checked major Choosing the answer
Three of the statements are not false but unverifiable, so each must be tested against what the five-number summaries guarantee, and none can be dismissed at a glance. (Embretson & Wetzel, 1987)
Something you must already know moderate Working it out
Students must know that the minimum and maximum are actual ages, that each quarter of a plot holds about a quarter of the 30 members, and that a whisker shows nothing in between. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph moderate Understanding the question
The scale is labeled only every 3 years, so values such as 17, 19, 26 and 29 have to be read from unlabeled ticks on two separate plots. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: List each group's minimum, first quartile, median, third quartile and maximum, then test every statement against only what those five values guarantee about the 30 ages.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Sees that Group Q's maximum is exactly 29 and that 29 also lies in Group P's upper whisker, assuming someone in Group P is that age.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the two medians, 21 and 24, and assumes the stretch between them holds the same number of members in both groups.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Knows Group P's box from 19 to 27 holds half its members, then assumes ages spread evenly so Group Q has fewer there, which its plot cannot show.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Members in each part of a box plot: $\text{members in each part} \approx \frac{1}{4} \times n$; n is the number of members in a group, here 30. A box plot splits the data into four parts at the minimum, Q1, the median, Q3 and the maximum. Each part holds about one fourth of the data. It does not show how the ages spread out inside a part.

Read the two box plots
A box plot shows five numbers. The minimum is the youngest age, and the maximum is the oldest. Q1 is the left edge of the box, and Q3 is the right edge. The median is the line inside the box.
Group Min Q1 Median Q3 Max
P 17 19 24 27 33
Q 15 17 21 26 29
Each group has 30 members.
Know what can be verified
To verify a statement means to know for sure that it is true. A box plot shows only five numbers, not every age. The minimum and maximum are real ages of real members. The other ages are hidden inside the box and whiskers.
Check choice B
In Group P, the youngest member is 17. So all 30 members are at least 17. In Group Q, the youngest member is 15, and 15 is under 17. So at most 29 members of Group Q are at least 17.
Group P has 30 such members, and Group Q has 29 or fewer. So choice B is true for sure.
Rule out choices A and C
A: Group Q’s oldest member is 29, so Group Q has a 29-year-old. But in Group P, 29 is inside the right whisker. A whisker does not show single ages. Watch out: 29 is in Group P’s range, but maybe no one in Group P is exactly 29.
C: The plots do not show how many members are between 21 and 24 in either group.
Rule out choice D and answer
Group P’s box, from 19 to 27, is two of its four parts.
$\text{members in each part} \approx \frac{1}{4} \times 30 = 7.5$
So about 15 members of Group P are between 19 and 27. But 19 and 27 fall in the middle of Group Q’s parts, and ages inside a part can bunch up in any way. Group Q could have more or fewer. Watch out: D assumes the ages are spread out evenly, and a box plot never tells you that.
So the answer is B.

Group	Min	Q1	Median	Q3	Max
P	17	19	24	27	33
Q	15	17	21	26	29

Check your answer

Choice B uses only the two minimums, which are real ages. Group P’s youngest member is 17, and Group Q’s is 15. That alone proves it. Each other choice needs ages inside a box or whisker, which a box plot does not show.

More questions like this

The original DOE question

Question 85 as printed on page 207 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 207, question 85. Open page 207 of the handbook (PDF)

Question 47 of 120

2024 Test B Math, question 29

Question

Jar Q contains 12 balls, and Jar R contains 8 balls. The probability of drawing a yellow ball at random from Jar Q is $\frac{1}{3}$ . The probability of drawing a yellow ball at random from Jar R is $\frac{3}{4}$ . All the balls are poured into an empty jar, Jar S. What is the probability of drawing a yellow ball at random from Jar S?

A $\frac{1}{4}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{4}{7}$

Show the answer

Correct answer: C

Jar Q contains 12 balls, and the probability of drawing a yellow ball at random is $\frac{1}{3}$ .

$\frac{1}{3}\times\frac{4}{4}=\frac{4}{12}$

There are 4 yellow balls in Jar Q.

Jar R contains 8 balls, and the probability of drawing a yellow ball at random is $\frac{3}{4}$ .

$\frac{3}{4}\times\frac{2}{2}=\frac{6}{8}$

There are 6 yellow balls in Jar R.

Jar S contains all the balls from Jar Q and Jar R.

$12+8=20$

There are 20 balls in Jar S.

$4+6=10$

There are 10 yellow balls in Jar S.

$\frac{10}{20}\div\frac{10}{10}=\frac{1}{2}$

The probability of drawing a yellow ball at random from Jar S is $\frac{1}{2}$ .

Question analysis

Two jars hold different numbers of balls and give different chances of drawing a yellow ball, and then everything is poured into a third jar. Students find the probability of drawing yellow from the combined jar. The tricky part is that the jars are different sizes, so their probabilities cannot be merged without first counting the yellow balls.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 probability tasks usually give the counts and ask for a fraction. Here each jar's probability must be turned back into a count of yellow balls before the jars are pooled, a reverse step that typical classwork rarely asks for.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 66 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light; the hurdle is seeing that chances from jars of different sizes cannot be combined as fractions. Every wrong choice works on 1/3 and 3/4 directly, by multiplying them, by keeping one, or by adding tops and bottoms, while the right path runs through counts of balls. Asking how many yellow balls each jar holds before anything is poured avoids all three.

Turning words into math moderate Understanding the question
Each jar's probability, 1/3 of 12 balls and 3/4 of 8 balls, has to be turned back into a count of yellow balls, 4 and 6, before the two jars can be combined. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The straight-across 4/7, about 0.57, sits just above the right 1/2, and both lie between the jars' chances of 1/3 and 3/4, so a rough estimate does not rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Turn each jar's probability into a number of yellow balls by multiplying it by that jar's total, then divide all the yellow balls by all the balls.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Multiplies the jars' chances, 1/3 × 3/4 = 1/4, the rule for two draws that must both be yellow, though only one ball is drawn from the mixture.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Repeats Jar Q's 1/3 from the stem, as if the bigger jar's chance carried over to the mixture unchanged.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the fractions straight across, (1 + 3)/(3 + 4) = 4/7, treating 1/3 and 3/4 as if they were counts of balls rather than chances.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P(\text{yellow}) = \frac{\text{yellow balls}}{\text{all balls}}$; P(yellow) is the probability, or chance, of drawing a yellow ball. It is a number from 0 to 1. The top counts the yellow balls. The bottom counts all the balls. It works because every ball is equally likely to be drawn.

Sort out what you know
Jar Q holds 12 balls. The probability, or chance, of drawing yellow from it is $\frac{1}{3}$ .
Jar R holds 8 balls. The chance of drawing yellow from it is $\frac{3}{4}$ .
All the balls are poured into Jar S. You need the chance of drawing yellow from Jar S.
Count the yellow balls in Jar Q
The jars are different sizes, so you can’t mix their chances directly. Count the yellow balls instead. The chance tells what fraction of the balls are yellow. So multiply it by the number of balls.
$\text{yellow balls} = \frac{1}{3} \times 12 = 4$
Jar Q has 4 yellow balls.
Count the yellow balls in Jar R
Do the same for Jar R. Three fourths of its 8 balls are yellow.
$\text{yellow balls} = \frac{3}{4} \times 8 = 6$
Jar R has 6 yellow balls.
Count the balls in Jar S
Jar S gets every ball from both jars. Add all the balls. Then add the yellow balls.
$12 + 8 = 20$
$4 + 6 = 10$
Jar S has 20 balls, and 10 of them are yellow.
Find the chance for Jar S
Use the probability formula with the Jar S counts.
$P(\text{yellow}) = \frac{10}{20} = \frac{1}{2}$
Watch out: Adding the tops and the bottoms, 1 + 3 over 3 + 4, gives $\frac{4}{7}$ , choice D. Those numbers are parts of fractions, not balls. Multiplying $\frac{1}{3} \times \frac{3}{4}$ gives $\frac{1}{4}$ , choice A. That is the chance of two yellow draws, one from each jar.
So the answer is C, $\frac{1}{2}$ .

Check your answer

Half of the 20 balls is 10, and 4 + 6 = 10 yellow balls. That matches. Also, the mixed jar’s chance should land between the two jars’ chances, $\frac{1}{3}$ and $\frac{3}{4}$ . One half does.

More questions like this

The original DOE question

Question 86 as printed on page 207 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 207, question 86. Open page 207 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 48 of 120

2024 Test B Math, question 40

Question

The value of a particular car decreases at a constant rate. If the car is worth $15,000 three years after its original purchase date and $10,000 five years after its original purchase date, what was the value of the car on its original purchase date?

A$18,000
B$20,000
C$22,500
D$25,000

Show the answer

Correct answer: C

The value of the car decreased $5,000 in two years: $15,000 to $10,000.

$5{,}000\div 2=2{,}500$

Therefore, the value of the car decreases $2,500 each year. After the first 3 years, the car’s value had decreased $7,500.

So $7{,}500+15{,}000=22{,}500$ .

Question analysis

A car loses value at a constant rate, and its worth three years and five years after purchase is given. Students find what the car was worth on the day it was bought. It tests rate of change and initial value in a linear relationship, and the tricky part is counting the right number of years when working back.

NYS standards: 8.F.4 modeling with linear functions
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Finding a starting value from two later values of a steady change is grade 8 linear-function work. A grade 7 student can reason with a yearly rate, but extending it back to the purchase date goes beyond typical grade 7 problems.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 10, 43 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is a straight-line pattern read backward, and its numbers are friendly; the difficulty is counting years correctly. The $5,000 drop happens over 2 years, not 5, and the purchase lies 3 years before the $15,000 value, not 2. Each miscount gives a listed answer, so sketching a timeline from year 0 to year 5 before computing pays off.

Turning words into math moderate Understanding the question
The phrases "three years after its original purchase date" and "five years after" must become points at year 3 and year 5 on a timeline whose start, year 0, holds the unknown value. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Three moves are chained: the $5,000 drop over 2 years, the $2,500 yearly rate, and three years of that rate added back to $15,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the yearly drop from the change between the two given years, then add that drop back once for each year between the purchase and year 3.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides the $5,000 drop by 5, the year of the second value, instead of the 2 years between year 3 and year 5, then adds back 3 × $1,000.
B Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Adds back the $5,000 two-year drop once, which gives the value at year 1, one year short of the 3 years back to the purchase.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two given values, $15,000 + $10,000, combining numbers from the question without a model of the steady decrease.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rate of change: $\text{rate} = \frac{\text{change in value}}{\text{change in time}}$; change in value is how many dollars the value went up or down. change in time is how many years passed. rate is the change for one year. A constant rate means the change is the same every year.

Sort out the question
The car’s value “decreases at a constant rate.” That means the car loses the same number of dollars each year. Year 0 is the day it was bought.
Year Value
0 ?
3 $15,000
5 $10,000
You need the value at year 0.
Find the drop for one year
From year 3 to year 5 is 2 years. In that time, the value fell by $15{,}000 - 10{,}000 = 5{,}000$ dollars.
$\text{rate} = \frac{5{,}000}{2} = 2{,}500$
The car loses $2,500 each year. Watch out: Divide by the 2 years between the values, not by 5. Dividing by 5 leads to $18,000, choice A.
Go back to year 0
From year 0 to year 3 is 3 years. The car lost $2,500 in each of them. So add that amount back 3 times.
$15{,}000 + 3 \times 2{,}500 = 15{,}000 + 7{,}500 = 22{,}500$
Watch out: Adding back only one $5,000 gives $20,000, choice B. That is the value at year 1, not year 0.
So the answer is C, $22,500.

Year	Value
0	?
3	$15,000
5	$10,000

Check your answer

Count forward from $22,500, taking off $2,500 each year. Year 1 is $20,000. Year 2 is $17,500. Year 3 is $15,000. Year 4 is $12,500. Year 5 is $10,000. Both given values match.

More questions like this

The original DOE question

Question 97 as printed on page 210 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 97. Open page 210 of the handbook (PDF)

Question 49 of 120

2024 Test B Math, question 47

Question

Three different colors of paint are available to paint the window frame, the door, and the wall of a house. If no two parts of the house may be painted the same color, how many different ways are there to paint the three parts of the house?

Show the answer

Correct answer: B

One method of solving the problem is to use a table to represent the sample space. Let the three colors be red, blue, and green.

PAINT COLORS

Window Frame	Door	Wall
Red	Blue	Green
Red	Green	Blue
Blue	Red	Green
Blue	Green	Red
Green	Red	Blue
Green	Blue	Red

There are 6 possible combinations of colors for the window frame, the door, and the wall where no two parts are painted the same color.

Question analysis

Three paint colors are available for a house's window frame, door and wall, and no two of the three parts may be painted the same color. Students count the different ways to paint them. It tests counting arrangements with a sample space, and the rule against repeating a color is what shapes the count.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Counting arrangements with an organized list is grade 7 work, and this one has only three parts and three colors. It is typical practice, with the no-repeat rule as the one condition to keep in mind.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

With three colors and three parts, the six arrangements can be listed in well under a minute, so the question rewards either a quick organized list or the counting principle. The main loss comes from ignoring the rule that no two parts match, which turns 3 × 2 × 1 into 3 × 3 × 3. The other wrong choices simply combine the two 3s in the stem.

Something you must already know minor Working it out
The counting principle is assumed: 3 color choices for the window frame, 2 left for the door, then 1 for the wall, multiplied together. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The rule that "no two parts of the house may be painted the same color" means each color is used exactly once, which removes one choice at every step. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Count the color choices for the window frame, then for the door with one color already used, then for the wall, and multiply; or list the arrangements.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the number of colors, 3, as the number of ways, without arranging the colors across the three parts.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Multiplies the 3 colors by the 3 parts, 3 × 3 = 9, a product of the stem's numbers rather than a count of arrangements.
D Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Allows any color on each part, 3 × 3 × 3 = 27, ignoring the rule that a color used once cannot be used again.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle: $\text{total ways} = a \times b \times c$; a is the number of choices for the first part, b for the second and c for the third. Multiply them to get every possible way. It works because each choice for one part can go with each choice for the next part.

Sort out the question
There are 3 colors and 3 parts to paint: the window frame, the door and the wall. No two parts may be the same color. So each color is used exactly once.
You need the number of different ways to paint the three parts.
Count the choices for each part
Paint the parts in order. Each color you use is gone for the next part.
- Window frame: any of the 3 colors, so 3 choices.
- Door: one color is used, so 2 choices are left.
- Wall: only 1 color is left, so 1 choice.
Multiply the choices
Use the counting principle. Each choice for the window frame goes with each choice for the door, and so on. So you multiply.
$\text{total ways} = 3 \times 2 \times 1 = 6$
Watch out: $3 \times 3 \times 3 = 27$ , choice D, lets a color be used twice. $3 \times 3 = 9$ , choice C, just multiplies the numbers in the question.
So the answer is B, 6.

Check your answer

List them. Use R for red, B for blue and G for green, in the order frame, door, wall: RBG, RGB, BRG, BGR, GRB and GBR. That is 6 ways, and each one uses all three colors once.

More questions like this

The original DOE question

Question 104 as printed on page 212 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 104. Open page 212 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 50 of 120

2024 Test B Math, question 51

Question

Rita used 125 small solid cubes to build a large solid cube. The measure of each edge of the small cubes equals 1 inch. What is the measure of each edge of the large cube, in inches?

A5
B25
C41.33
D125

Show the answer

Correct answer: A

The volume of a cube with a side length of s is

$V=s^{3}$

Since 125 small cubes were used to build the large cube, $125=s^{3}$ .

Each side of the large cube is made up of 5 small cubes.

$125=5^{3}$

Since 5 small cubes make up each edge of the large cube, and the edge of each small cube measures 1 inch, the measure of each edge of the large cube is $1\times 5=5\text{ inches}$ .

Question analysis

Rita stacks 125 one-inch cubes into a single large solid cube. Students find the length of each edge of the large cube in inches. It tests volume and perfect cubes with a building-block model, and the care needed is in telling apart the cubes along an edge, across a face and in the whole solid.

NYS standards: 6.G.5 perfect squares and perfect cubes
6.G.2 volume with fractional edge lengths
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Building a large cube from unit cubes is the model grade 6 uses to introduce perfect cubes. It is a single step with small numbers, easier than typical grade 7 geometry, which works with composite solids and formulas.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 3, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is one of the easiest questions on the test, since 125 is a familiar cube and the small cubes make the large one easy to picture as 5 rows of 5 in 5 layers. The wrong choices name a layer instead of an edge, a division by 3 for three dimensions, or the total count itself, and multiplying 5 × 5 × 5 as a check rules all of them out.

Something you must already know minor Working it out
Students must see that 125 one-inch cubes make a volume of 125 cubic inches, and know or find that 5 × 5 × 5 = 125. (Anderson & Krathwohl, 2001)

Strategy: Think of the large cube as equal rows, columns and layers of small cubes, and find the number that, used three times as a factor, gives 125.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Gives the number of small cubes in one layer or on one face, 5 × 5 = 25, instead of the number along one edge.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides 125 by 3 for the cube's three dimensions instead of taking a cube root; that gives 41.67, and 41.33 is the only choice near it.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives back the 125 from the stem, the count of small cubes and the volume in cubic inches, rather than an edge length.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a cube: $V = s^3$; V is the volume, the space inside, in cubic units. s is the length of one edge. s³ means s × s × s. A cube’s length, width and height are all s, so length × width × height becomes s × s × s.

Sort out the question
Rita stacks 125 small cubes into one big solid cube. Each small cube has edges of 1 inch. You need the length of one edge of the big cube, in inches.
Find the big cube’s volume
Each small cube is $1 \times 1 \times 1 = 1$ cubic inch. The big cube is solid, with no gaps. So its volume is 125 cubic inches.
Find the edge length
$V = s^3$
$125 = s \times s \times s$
Find the number that makes 125 when you use it three times. Try 5.
$5 \times 5 \times 5 = 25 \times 5 = 125$
So 5 small cubes sit along each edge. Each one is 1 inch long, so the edge is 5 inches.
Rule out the other choices
Watch out: 25, choice B, is the number of small cubes in one layer, 5 rows of 5. 125, choice D, is the number of cubes, or the volume. Dividing 125 by 3 does not give an edge either. That is about 41.67, near choice C.
So the answer is A, 5.

Check your answer

Picture building it. A row of 5 cubes is 5 inches long. Five rows make a layer of 25 cubes. Five layers make $25 \times 5 = 125$ cubes. That matches.

More questions like this

The original DOE question

Question 108 as printed on page 213 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 108. Open page 213 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 51 of 120

2024 Test B Math, question 52

Question

$Two open boxes, labeled A and B. Box A holds two balls, numbered 2 and 6. Box B holds three balls, numbered 3, 4 and 5.$

Five numbered balls are placed in two boxes as shown in the figure above. Davina picks one ball at random from Box B and places it in Box A. Then Yusuf picks one ball from Box A without looking. What is the probability that Yusuf will pick a ball with an odd number?

A $\frac{2}{9}$
B $\frac{1}{3}$
C $\frac{2}{5}$
D $\frac{2}{3}$

Show the answer

Correct answer: A

The probability of Davina picking an odd-numbered ball from Box B is $\frac{2}{3}$ . If Davina places an odd-numbered ball in Box A, then the probability of Yusuf picking an odd-numbered ball from Box A is $\frac{1}{3}$ . The probability of both of these events happening is the product of the two probabilities.

$\frac{2}{3}\times\frac{1}{3}=\frac{2}{9}$

Question analysis

Two boxes hold five numbered balls, and Davina moves one ball at random from Box B into Box A before Yusuf draws one from Box A. Students find the probability that Yusuf's ball shows an odd number. It tests two-stage probability, and the tricky part is that the second draw depends on which ball was moved.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 compound probability lists outcomes for independent stages such as coins or spinners. Here the first random move changes what is in the second box, so students must follow two linked stages, a dependence typical grade 7 work does not include.
Estimated difficulty: Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Thinking it demands: Depth of knowledge 3, strategic thinking · mainly analyze
What makes it hard: No obvious method major
Something you must already know minor
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 57 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two chance events are hidden in what reads like a question about a single draw. Every listed wrong answer is a reasonable single-stage answer, either one of the two factors on its own or a count of odd balls across both boxes. Listing the nine equally likely pairs, Davina's ball and then Yusuf's, shows that only two of them end with an odd ball and makes the structure visible without any formula.

No obvious method major Working it out
The stem asks one question about Yusuf's draw, but the answer depends on Davina's earlier random pick; the key is that Box A starts with only even balls, so Yusuf can draw odd only by drawing the ball she moved. (Schoenfeld, 1985)
Something you must already know minor Working it out
Multiplying the chances of the two stages, 2/3 for an odd ball moving and 1/3 for Yusuf then drawing it, is a rule students must bring themselves. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The numbers appear only in the figure, Box A holding 2 and 6 and Box B holding 3, 4 and 5, so the odd balls have to be spotted there. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: List every equally likely pair of a ball moved by Davina and a ball picked by Yusuf, and count the pairs that end with an odd ball.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds Yusuf's chance of drawing the odd ball once one is in Box A, 1 out of 3, leaving out the chance that Davina moves an odd ball at all.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Counts the odd balls in the figure, 3 and 5, over all five balls, 2/5, ignoring that Yusuf draws only from Box A.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Gives the chance that Davina moves an odd ball, 2 out of 3, the first stage only, without multiplying by Yusuf's 1/3 chance of drawing it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of wanted outcomes}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row: $P(A \text{ then } B) = P(A) \times P(B \text{ after } A)$; A and B are two events that happen one after the other. P(A) is the chance of the first. P(B after A) is the chance of the second, counting only what is left after the first. You multiply because the second must also happen inside the share of times the first did.

Read the figure
Box A holds balls numbered 2 and 6. Box B holds balls numbered 3, 4 and 5.
An odd number does not divide evenly by 2. So the only odd balls are 3 and 5, and both start in Box B. Box A starts with no odd balls at all.
Spot the two chance events
Two things happen by chance, one after the other:
1. Davina picks one of the 3 balls in Box B and moves it to Box A.
2. Then Yusuf picks one of the 3 balls now in Box A.
Yusuf gets an odd ball only if both go right. Davina must move an odd ball. Then Yusuf must pick that very ball.
The question asks only about Yusuf. But his chance depends on Davina’s pick, which comes first.
Find the chance for Davina’s pick
Box B has 3 balls, and 2 of them, the 3 and the 5, are odd.
$P(\text{odd ball moved}) = \frac{2}{3}$
If she moves the 4 instead, Box A holds only even balls. Then Yusuf can’t get an odd ball at all.
Find the chance for Yusuf’s pick
Say Davina moved an odd ball. Box A now holds 3 balls: the 2, the 6 and that one odd ball.
$P(\text{Yusuf picks it}) = \frac{1}{3}$
Watch out: $\frac{1}{3}$ alone is choice B. It forgets that Davina might move the 4.
Multiply the two chances
Both events must happen, so use the rule for events in a row.
$P(\text{odd}) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$
Watch out: $\frac{2}{3}$ alone, choice D, stops after Davina’s pick. $\frac{2}{5}$ , choice C, counts 2 odd balls out of all 5. But Yusuf picks only from Box A.
So the answer is A, $\frac{2}{9}$ .

Check your answer

List all 9 equally likely outcomes. Davina can move 3, 4 or 5. Then Yusuf picks one of 3 balls.

Ball moved	Box A then	Odd picks
3	2, 6, 3	1
4	2, 6, 4	0
5	2, 6, 5	1

That is 2 odd picks out of 9, or $\frac{2}{9}$ .

More questions like this

The original DOE question

Question 109 as printed on page 214 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 109. Open page 214 of the handbook (PDF)

Question 52 of 120

2023 Test A Math, question 1

Question

Ms. Li opened a retirement account with a deposit of $2,500. This account earns 4% simple interest annually. How many years will it take her to earn $500 on her $2,500 deposit?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 5

Simple interest (y) is calculated by multiplying the initial deposit (p), the interest rate (r), and time (t):

$prt = y$

$(2{,}500)(0.04)t = 500$

$100t = 500$

$t = 5$

Question analysis

Ms. Li deposits $2,500 in a retirement account that pays 4% simple interest each year, and this grid-in asks how many years it takes to earn $500 in interest. Students find one year's interest and work out how many years are needed. It turns the usual simple interest question around, asking for the time rather than the interest.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Simple interest is a grade 7 percent application, usually asked as how much interest builds up over a given time. This one gives the interest and asks for the time, a short reverse step with friendly numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Turning words into math minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once one year's interest is known, $100, the rest is counting how many years make $500, so the arithmetic is light. The question turns the usual simple interest problem around, giving the interest and asking for the time, and that reversal is where students stall or grid a percent instead of a number of years. Writing 4% as 0.04, not 0.4, keeps the yearly interest from coming out ten times too big.

Something you must already know minor Working it out
The stem says 4% simple interest annually but never says that each year adds the same 4% of the $2,500 deposit, $100, so the yearly amount has to come from memory. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The question gives the interest, $500, and asks for the time, so the familiar formula runs backward: 2,500 × 0.04 × t = 500 has to be solved for t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
As a grid-in it shows no choices, so 20, the interest as a percent of the deposit, can be gridded with nothing on the page to say it is not a number of years. (Enright, Morley & Sheehan, 2002)

Strategy: Work out one year's interest as 4% of $2,500, then find how many of those yearly amounts add up to the $500 of interest.

Traps

Stops one step early: A choice is a number from partway through the solution.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

20 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that the $500 of interest is 20% of the $2,500 deposit and grids 20, stopping before dividing by the 4% earned each year.
0.5 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Writes 4% as 0.4 instead of 0.04, so one year's interest looks like $1,000 and the $500 seems to take only half a year.
125 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the $500 of interest by the 4 in 4%, combining two numbers from the stem without first finding 4% of the deposit.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest: $I = P \times r \times t$; I is the interest, the extra money the account pays you. P is the deposit, the money you put in. r is the yearly rate as a decimal. t is the time in years. Simple means each year’s interest is figured on the deposit only, so every year adds the same amount.
Percent to decimal: $p\% = \frac{p}{100}$; p is the percent. Percent means out of 100. So divide by 100 to write it as a decimal. For example, 4% is 4 ÷ 100, which is 0.04.

Sort out the question
Ms. Li puts $2,500 into an account. This first amount is called the deposit. Each year the account pays 4% simple interest. Interest is extra money the bank pays you for keeping your money there. You need the number of years it takes to earn $500 of interest. The answer is a number of years, not dollars and not a percent.
Find one year’s interest
First write 4% as a decimal. 4% means 4 out of 100.
$4\% = \frac{4}{100} = 0.04$
Now use the formula for 1 year, so t = 1.
$I = P \times r \times t$
$I = 2{,}500 \times 0.04 \times 1 = 100$
The account earns $100 each year. Watch out: 4% is 0.04, not 0.4. Using 0.4 makes one year’s interest $1,000, and you would grid 0.5 years.
Find how many years make $500
Simple interest adds the same $100 every year. So the time t must make the interest $500.
$2{,}500 \times 0.04 \times t = 500$
$100 \times t = 500$
$t = 500 \div 100 = 5$
It takes 5 years. Watch out: $500 is 20% of $2,500. But 20 is a percent, not a number of years. Don’t grid 20. So you would grid 5.

Check your answer

Count up one year at a time: $100, $200, $300, $400, $500. That is 5 years. Or put t = 5 back into the formula: $2{,}500 \times 0.04 \times 5 = 500$ . That matches the $500 in the question.

More questions like this

The original DOE question

Question 58 as printed on page 75 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 75, question 58. Open page 75 of the handbook (PDF)

Question 53 of 120

2023 Test A Math, question 9

Question

A revolving sign makes 1 complete revolution every 90 minutes. If the sign starts moving at 2:30 p.m., at what time will the sign complete 8 revolutions?

A1:00 a.m.
B2:30 a.m.
C4:00 p.m.
D12:00 midnight

Show the answer

Correct answer: B

First, use a proportion to find out how many minutes it will take the sign to complete 8 revolutions. Let x equal the total number of minutes:

$\frac{x}{8} = \frac{90}{1}$

$x = 720$

Next, convert 720 minutes into hours and minutes:

$\frac{720}{60} = 12\text{ hours}$

Finally, 12 hours after 2:30 p.m. is 2:30 a.m.

Question analysis

A revolving sign turns once every 90 minutes and starts at 2:30 in the afternoon. Students find the clock time when it finishes 8 full revolutions. The question combines a rate with converting minutes to hours, and then adding the elapsed time to the starting time on a 12-hour clock.

NYS standards: 6.RP.3b unit rate problems
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 A constant rate, a multiplication and a conversion from minutes to hours are grade 6 skills. The only step beyond routine work is adding the elapsed time to a starting clock time.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is routine, so most points are lost in keeping elapsed time and clock time apart. The 12 hours is a duration, and the clock reading that ends it is 12 hours after 2:30 in the afternoon, not 12:00. A sense check helps: 12 hours after any time shows the same digits on the clock with a.m. and p.m. swapped.

Many steps minor Working it out
Three moves: 8 × 90 = 720 minutes, 720 ÷ 60 = 12 hours, then 12 hours counted on from the 2:30 start, which crosses midnight into the morning. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Counting hours past midnight means switching from p.m. to a.m. on a 12-hour clock, which the stem leaves to the student; the choices mix both halves of the day. (Anderson & Krathwohl, 2001)

Strategy: Multiply 90 minutes by 8 to get the total time, change it to hours, and then count that many hours forward from 2:30 p.m.

Traps

Stops one step early: A choice is a number from partway through the solution.
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Counts only 7 revolutions, 630 minutes or 10½ hours, as if the first turn were already done at the start, and lands at 1:00 a.m.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds a single revolution's 90 minutes to the 2:30 start, giving 4:00 in the afternoon, without multiplying by the 8 revolutions.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds that 8 revolutions take 12 hours and picks 12:00 midnight, reading the elapsed time as a clock time instead of counting on from 2:30.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total time: $\text{total time} = \text{time for one} \times \text{how many}$; Time for one is how long one revolution takes. How many is the number of revolutions. Each one takes the same time, so multiply to get the total time.
Minutes to hours: $\text{hours} = \text{minutes} \div 60$; There are 60 minutes in 1 hour. So to change minutes to hours, divide by 60. Each group of 60 minutes makes 1 hour.

Sort out the question
The sign turns all the way around once every 90 minutes. One full turn is called a revolution. The sign starts at 2:30 p.m. You need the clock time when it finishes 8 revolutions.
Find the total time in minutes
Each revolution takes 90 minutes, and there are 8 of them.
$\text{total time} = 90 \times 8 = 720$
The 8 revolutions take 720 minutes. Watch out: use all 8 revolutions. Counting only 7 gives 630 minutes and 1:00 a.m., choice A. Adding just one 90 minutes gives 4:00 p.m., choice C.
Change the minutes to hours
Clock times are easier to count in hours. There are 60 minutes in an hour.
$\text{hours} = 720 \div 60 = 12$
The sign turns for 12 hours.
Count forward from 2:30 p.m.
Start at 2:30 p.m. and add 12 hours. When you pass midnight, p.m. changes to a.m. So 12 hours after 2:30 p.m. is 2:30 a.m. the next morning. Watch out: 12 hours is an amount of time, not a clock time. Reading it as 12:00 midnight gives choice D. So the answer is B, 2:30 a.m.

Check your answer

From 2:30 p.m. to 2:30 a.m. is one full trip around the clock, 12 hours. 12 hours is $12 \times 60 = 720$ minutes. And $720 \div 90 = 8$ , the number of revolutions in the question.

More questions like this

The original DOE question

Question 66 as printed on page 77 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 66. Open page 77 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 54 of 120

2023 Test A Math, question 10

Question

$A number line from 0 to 10 with a tick and a label at every whole number and arrows at both ends. Point W is marked with a dot at 0, and point X is marked with a dot at 2.$

Points Y and Z are not shown on the number line above. If X is the midpoint of $\overline{WY}$ , and Y is the midpoint of $\overline{WZ}$ , where on the number line would point Z be located?

Show the answer

Correct answer: D

First, find where Y is located on the number line. Point X is the midpoint of $\overline{WY}$ . Since there are 2 units between W and X, there must also be 2 units between X and Y. Thus, Y is located at 4 on the number line.

Point Y is the midpoint of $\overline{WZ}$ . Use the same reasoning as above. Since there are 4 units between W and Y, there must be 4 units between Y and Z. So, Z is located at 8 on the number line.

Question analysis

A number line shows point W at 0 and point X at 2, and two more points, Y and Z, are described but not drawn. Students use two midpoint facts, one after the other, to find where Z lies. Each midpoint fact is about a different segment, so keeping track of which point is the middle one matters.

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Midpoints on a number line rest on distance, which grade 7 students find as a difference. Here every point lies at a whole number of zero or more, so the work is two short steps with no negative numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each midpoint statement doubles a distance from W, so the problem is two short doublings, 2 to 4 and then 4 to 8. The risk is stopping halfway, since Y's position is found first and appears among the choices, or reusing X as an endpoint in the second statement. Rereading which point the question asks about, and checking that Y is halfway between W and Z, catches both.

Many steps minor Working it out
The two midpoint statements must be used in order: the first places Y at 4, and only then can the second, about segment WZ, place Z. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph minor Understanding the question
Only W at 0 and X at 2 are marked on the 0-to-10 line, and the overbars name WY and WZ as segments, so Y and Z have to be placed from the statements alone. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the first midpoint to place Y as far beyond X as X is from W, then use the second midpoint the same way to place Z beyond Y.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives 2, the position of point X, the only labeled point besides W, without using either midpoint statement.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Uses the first statement to place Y at 4 and answers with Y's position instead of going on to locate Z, the point the question asks about.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Treats Y as the midpoint of XZ instead of WZ, so Z lands 2 units past Y, at 6, continuing the even steps from W to X to Y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint of a segment: $AM = MB$; A and B are the ends of a segment, a straight piece of a line. M is its midpoint, the point exactly halfway. AM is the distance from A to M, and MB is the distance from M to B. A midpoint splits a segment into two equal parts, so these are equal.

Sort out the question
The number line shows W at 0 and X at 2. Points Y and Z are not shown. X is the midpoint of segment WY. Y is the midpoint of segment WZ. A midpoint is the point exactly halfway along a segment. You need the number where Z sits.
Place point Y
X is halfway from W to Y. So X to Y is as long as W to X.
$WX = 2 - 0 = 2$
$XY = WX = 2$
So Y is 2 units past X.
$2 + 2 = 4$
Point Y is at 4. Watch out: this is Y, not Z. Stopping here gives 4, choice B.
Place point Z
Now Y is halfway from W to Z. W to Y is 4 units, so Y to Z is also 4 units.
$YZ = WY = 4 - 0 = 4$
$4 + 4 = 8$
Watch out: the second midpoint is for segment WZ, not XZ. Taking just 2 more units from Y puts Z at 6, choice C. So the answer is D, 8.

Check your answer

Check both midpoints. Halfway from 0 to 4 is $(0 + 4) \div 2 = 2$ , which is X. Halfway from 0 to 8 is $(0 + 8) \div 2 = 4$ , which is Y. Both facts in the question are true.

More questions like this

The original DOE question

Question 67 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 67. Open page 78 of the handbook (PDF)

Question 55 of 120

2023 Test A Math, question 27

Question

A basket contains red balls, green balls, and white balls. There are 12 red balls in the basket. The probability of randomly choosing a red ball is 1 in 3. If the probability of randomly choosing a green ball is 1 in 4, how many green balls are in the basket?

Show the answer

Correct answer: C

First, use the information about the red balls to calculate the total number of balls in the basket:

$\frac{1}{3} = \frac{12}{x}$	Cross multiply.
$x = 36$

There are a total of 36 balls in the basket. The probability of selecting a green ball is 1 in 4. Use that information to calculate how many green balls are in the basket:

$\frac{g}{36} = \frac{1}{4}$	Cross multiply.
$4g = 36$	Apply the multiplicative inverse property; divide both sides of the equation by 4.
$g = 9$

There are 9 green balls in the basket.

Question analysis

A basket holds red, green and white balls, including 12 red ones, and the chances of drawing red and green are 1 in 3 and 1 in 4. Students find how many green balls are in the basket. Getting the total from the red count is the step students skip when they apply a probability to the wrong number.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Simple probability is grade 6 content, but here it runs backward: a count and a probability give the total, which then gives another count. Two linked steps like this are typical of grade 7 work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 46 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The basket's total is never stated, and that missing number is the whole problem: it has to be recovered from the red balls before the green probability means anything. Applying the green chance straight to the 12 red balls is the quick wrong move. Checking that 12 red and 9 green fit in 36 balls with 15 white left over confirms the setup.

Turning words into math minor Understanding the question
The sentence "The probability of randomly choosing a red ball is 1 in 3" has to become 12 = one third of the total, which makes the basket hold 36 balls. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Two linked steps: the red balls give the total, 12 × 3 = 36, and only then does the green probability give a count, 36 ÷ 4 = 9. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use the red balls to find the total, since 12 is one third of it, then take one fourth of that total for the green balls.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Takes one fourth of the 12 red balls, 12 ÷ 4 = 3, applying the green probability to the only count the stem gives instead of to the total.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No use of the two probabilities gives 8, which is one third of the 24 balls that are not red, so it mostly catches guesses.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Scales the 12 red balls by 4/3 instead of 3/4, taking the 1-in-4 chance to mean more balls than the 1-in-3 chance: 12 × 4/3 = 16.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{balls of one color}}{\text{all the balls}}$; P is the probability, the chance of picking that color. Each ball is equally likely to be picked. So the chance is the fraction of all the balls that have that color. A chance of 1 in 3 means one third of the balls.

Sort out the question
The basket has red, green and white balls. There are 12 red balls. The chance of picking red is 1 in 3. The chance of picking green is 1 in 4. You need the number of green balls. You don’t know the total number of balls yet, so find it first.
Find the total number of balls
1 in 3 means one third of all the balls are red. So 12 is one third of the total. Call the total number of balls b.
$\frac{12}{b} = \frac{1}{3}$
$b = 12 \times 3 = 36$
There are 36 balls in all.
Find the number of green balls
1 in 4 means one fourth of all 36 balls are green. Call the number of green balls g.
$\frac{g}{36} = \frac{1}{4}$
$g = 36 \div 4 = 9$
Watch out: take one fourth of all 36 balls, not of the 12 red ones. $12 \div 4 = 3$ gives choice A. Multiplying 12 by $\frac{4}{3}$ gives 16, choice D. So the answer is C, 9.

Check your answer

A 1 in 4 chance is smaller than a 1 in 3 chance. So there should be fewer green balls than red ones, and 9 is less than 12. Check: $\frac{9}{36} = \frac{1}{4}$ . The rest, $36 - 12 - 9 = 15$ , are white.

More questions like this

The original DOE question

Question 84 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 84. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 56 of 120

2023 Test A Math, question 29

Question

Integer x is evenly divisible by 3. Which expression below is also evenly divisible by 3?

A $2x + 1$
B $3x - 5$
C $4x - 1$
D $4x + 6$

Show the answer

Correct answer: D

Since integer x is evenly divisible by 3, substitute 3 for x in the answer options to determine which expression is also divisible by 3:

Let $x = 3$ .

Option A:

$2x + 1 = 2(3) + 1 = 7$

Option B:

$3x - 5 = 3(3) - 5 = 4$

Option C:

$4x - 1 = 4(3) - 1 = 11$

Option D:

$4x + 6 = 4(3) + 6 = 18$

Option D is the only value that is divisible by 3.

Question analysis

This number properties question says an integer x is divisible by 3 and asks which of four linear expressions in x is also divisible by 3, whatever multiple of 3 the integer is. Students can test a multiple of 3 in each expression or think about remainders. Testing a value that is not a multiple of 3 can make a wrong expression look right.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.4 common factors and multiples
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students rewrite and evaluate linear expressions, but they seldom ask whether one is divisible by 3 for every allowed value. One well-chosen substitution settles it, so the demand is about typical for grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked minor
Solution: 2 steps
Reading level: Question and choices: about grade 13, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The algebra is light; the question turns on choosing a test value that obeys the condition. The habit of trying x = 1 backfires here, because 1 is not a multiple of 3 and it makes two wrong expressions come out divisible. Students who substitute 3, or who see that 4x and 6 are both multiples of 3 whenever x is, settle it quickly, and a second multiple such as 6 confirms it.

Variables instead of numbers moderate Understanding the question
Nothing pins x down except that it is a multiple of 3, so each expression must be judged for every such x, and any test value has to be a multiple of 3 itself, such as 3 or 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked minor Choosing the answer
The four expressions differ only in their coefficients and constants, from 2x + 1 to 4x + 6, so none can be dismissed at a glance; each needs a test or a look at its terms. (Embretson & Wetzel, 1987)

Strategy: Substitute a multiple of 3, such as 3 or 6, for x in every expression and keep the one whose value divides evenly by 3; a second multiple confirms it.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Tests x = 1, which is not a multiple of 3, finds that 2(1) + 1 = 3 divides evenly, and stops at the first choice; with x = 3 it gives 7.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Sees the 3 in 3x and takes the whole expression to be a multiple of 3, overlooking that subtracting 5 spoils it: with x = 3 it gives 4.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Tries a value that is not a multiple of 3, such as x = 4, where 4x − 1 = 15 divides evenly, though x = 3 gives 11.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding multiples of 3: $3a + 3b = 3(a + b)$; a and b are whole numbers, so 3a and 3b are multiples of 3. A multiple of 3 is a number that 3 divides evenly, with nothing left over. Adding two multiples of 3 always gives another multiple of 3, because you can pull the 3 out of both parts.

Sort out the question
“Evenly divisible by 3” means 3 goes into the number with nothing left over. Such a number is called a multiple of 3, like 3, 6, 9 or 12.
You know that x is a multiple of 3. You need the expression that is a multiple of 3 every time, whatever multiple of 3 x is.
Test x = 3 in every choice
Pick a number for x that follows the rule. The easiest one is 3. Put it into each expression.
Choice Work Divisible by 3?
A $2(3) + 1 = 7$ No
B $3(3) - 5 = 4$ No
C $4(3) - 1 = 11$ No
D $4(3) + 6 = 18$ Yes
Watch out: your test number must be a multiple of 3. With x = 1, choice A gives 3 and looks right. That leads to the wrong choice, A.
See why D always works
One test could be lucky, so see why D works for every x. Since x is a multiple of 3, 4x is one too. For example, 4 × 3 = 12 and 4 × 6 = 24. The number 6 is also a multiple of 3, since 6 = 3 × 2.
$3a + 3b = 3(a + b)$
So 4x + 6 is a multiple of 3 plus a multiple of 3. That is always a multiple of 3.
Rule out the other choices
Watch out: choice B starts with 3x, but then it subtracts 5. The 5 is not a multiple of 3, so it spoils the result, and 3(3) − 5 = 4. Choices A and C also add or take away a number that is not a multiple of 3.
So the answer is D, $4x + 6$ .

Choice	Work	Divisible by 3?
A	$2(3) + 1 = 7$	No
B	$3(3) - 5 = 4$	No
C	$4(3) - 1 = 11$	No
D	$4(3) + 6 = 18$	Yes

Check your answer

Try two more multiples of 3. With x = 6, choice D gives $4(6) + 6 = 30$ , and 30 ÷ 3 = 10. With x = 9, it gives $4(9) + 6 = 42$ , and 42 ÷ 3 = 14. Both divide evenly, so D keeps working.

More questions like this

The original DOE question

Question 57 of 120

2023 Test A Math, question 32

Question

Mei-Ling is one of 6 members of a committee. If 2 members of that committee are selected to go to a conference, how many of the possible pairs of members would include Mei-Ling?

Show the answer

Correct answer: A

Call the 6 members A, B, C, D, E, and M (for Mei-Ling). The possible combinations are AB, AC, AD, AE, AM, BC, BD, BE, BM, CD, CE, CM, DE, DM, and EM. Of those pairs, 5 include Mei-Ling.

Question analysis

This counting question describes a six-member committee sending two members to a conference and asks how many of the possible pairs include Mei-Ling. Students count combinations in which order does not matter. The traps are counting a pair twice, pairing Mei-Ling with herself, or multiplying the numbers in the question instead of listing the pairs.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students count outcomes with organized lists and tree diagrams. Here the list of pairs is short and a single condition, that one named person belongs to the pair, filters it, which matches routine grade 7 counting work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The count is small enough to list, so the question rewards a clear picture of what a pair is: two different people, in no particular order. The tempting errors count Mei-Ling as her own partner, count each pair twice, or multiply the two numbers in the stem. Writing the five pairs out, M with each of A through E, takes seconds and rules out all three.

Turning words into math minor Understanding the question
Pairs that include Mei-Ling are simply Mei-Ling with each of the other 5 members, a count of her possible partners, though the phrase “possible pairs” invites counting all 15 pairs. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Listing pairs by letter, as the explanation does with A to E and M, means tracking which pairs are already written, so that AM and MA are not both counted. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Label the members with letters and list every pair that contains Mei-Ling, remembering that a pair is two different people and that order does not matter.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts all 6 committee members as possible partners, forgetting that Mei-Ling cannot be paired with herself, one more than the real count.
C Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: moderate
Counts ordered pairs, Mei-Ling with a partner and that partner with Mei-Ling, as different, so each of the 5 pairs is counted twice.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 6 members by the 2 people chosen, 6 × 2 = 12, a product of the stem's numbers that counts no actual pairs.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of pairs in a group: $\text{pairs} = \frac{n \times (n - 1)}{2}$; n is the number of people in the group. Each person can pair up with the n − 1 others. Multiplying counts every pair twice, once from each partner’s side. So you divide by 2.

Sort out the question
The committee has 6 members, and Mei-Ling is one of them. Two members go to the conference together. That makes a pair. You don’t need every possible pair. You only need the pairs that include Mei-Ling.
Find Mei-Ling’s possible partners
A pair with Mei-Ling is Mei-Ling plus one other member. Call Mei-Ling M, and call the other members A, B, C, D and E. Her pairs are M with A, M with B, M with C, M with D and M with E. That is 5 pairs.
Watch out: she can’t be paired with herself. Counting all 6 members as partners gives 6, choice B.
Don’t count a pair twice
Mei-Ling with A is the same pair as A with Mei-Ling. The same two people go either way. Watch out: counting both orders doubles the count to 10, choice C. Multiplying 6 × 2 = 12 gives choice D, but that does not count any pairs.
Check against all the pairs
Count every pair in the committee.
$\text{pairs} = \frac{n \times (n - 1)}{2}$
$\text{pairs} = \frac{6 \times 5}{2} = 15$
Each pair has 2 people, so the 15 pairs have 30 name spots. All 6 members get the same share: $30 \div 6 = 5$ . So the answer is A, 5.

Check your answer

List all 15 pairs: AB, AC, AD, AE, AM, BC, BD, BE, BM, CD, CE, CM, DE, DM, EM. Now circle the ones with M in them: AM, BM, CM, DM, EM. There are exactly 5.

More questions like this

The original DOE question

Question 89 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 89. Open page 83 of the handbook (PDF)

Question 58 of 120

2023 Test A Math, question 35

Question

$A number line with tick marks every 2 units, labeled −6, −4, −2, 0, 2, 4 and 6. Point J is the dot at −4 and point K is the dot at 6.$

On the number line above, point L (not shown) is located on line segment JK so that $\mathrm{JL} = \frac{2}{3}\mathrm{LK}$ . What is the position of point L?

A−2
B0
C2
D4

Show the answer

Correct answer: B

$\mathrm{JK} = 6 - (-4) = 10$ units. Let $x = \mathrm{JL}$ . Then $\mathrm{LK} = 10 - x$ . Substitute these expressions in the given equation:

$x = \frac{2}{3}(10 - x)$

Solve for x:

$x = \frac{2}{3}(10 - x)$	Apply the multiplicative inverse property; multiply both sides of the equation by 3.
$3x = 2(10 - x)$	Apply the distributive property; multiply the 2 by 10 and x.
$3x = 20 - 2x$	Apply the additive inverse property; add $2x$ to both sides of the equation.
$5x = 20$	Apply the multiplicative inverse property; divide both sides of the equation by 5.
$x = 4$

So, point L is 4 units to the right of point J: $-4 + 4 = 0$ . Point L is located at 0 on the number line.

Question analysis

This number line question puts J at −4 and K at 6 and asks for the position of a point L on segment JK where JL is two thirds of LK. Students find the segment's length, divide it in a 2-to-3 ratio, and turn a distance into a coordinate. Reading the ratio backwards, or giving a distance instead of a position, leads to wrong answers.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Distances on a number line that crosses zero are grade 7 rational-number work, and splitting a length in a given ratio is grade 7 proportional reasoning. Turning a fraction of one part into a split, then into a position, makes a typical multistep task.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Reading a figure, table or graph minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The ratio is stated between two unknown pieces, which is less familiar than a fraction of a whole, and the final step turns a length into a position. Students who stop at JL's length, read 2/3 the wrong way round, or use its 2 as a distance all reach listed points. Sketching five equal shares of 2 units from −4 to 6, two for JL and three for LK, makes the position plain.

Turning words into math moderate Understanding the question
JL = 2/3 LK compares the two pieces with each other, not a piece with the whole JK, so the 10 units must be split in the ratio 2 : 3, giving JL 4 and LK 6. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
J's position, −4, and K's, 6, are read from the number line, whose labeled ticks run every 2 units, and L itself is not drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Three steps follow in order: the length JK = 6 − (−4), the length JL from the ratio, and finally L's position counted from J rather than from 0. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length of JK, split it into five equal shares with two for JL and three for LK, then count JL's length to the right from J.

Traps

Stops one step early: A choice is a number from partway through the solution.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the 2 in 2/3 as the length of JL and moves 2 units right from J at −4, without relating JL to LK or to JK.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Turns the ratio upside down, making LK two thirds of JL, so JL is 6 and L lands at 2, nearer K than J.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds JL = 4 correctly but gives that length as L's position, stopping before counting the 4 units from J at −4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line: $\text{distance} = \text{right point} - \text{left point}$; Subtract the position on the left from the position on the right. The answer is how many units apart the two points are. A distance is never negative.
Splitting a length in a ratio: $\text{one part} = \frac{\text{whole length}}{a + b}$; A ratio a : b compares two pieces. If a length is split in the ratio a : b, think of it as a + b equal parts. The first piece gets a of them and the second gets b. Divide the whole length by a + b to find one part.

Sort out the question
Read the number line. Point J is at −4 and point K is at 6. Point L sits somewhere between them. JL is the length from J to L, and LK is the length from L to K. The rule says JL is $\frac{2}{3}$ of LK. You need the position of L.
Find the length of JK
$\text{distance} = \text{right point} - \text{left point}$
K is on the right, so subtract J’s position from K’s.
$\mathrm{JK} = 6 - (-4) = 6 + 4 = 10$
Taking away −4 is the same as adding 4. J and K are 10 units apart.
Turn the rule into equal parts
JL is $\frac{2}{3}$ of LK. So for every 2 units in JL, LK has 3 units. That means JK is made of 2 + 3 = 5 equal parts. JL gets 2 parts and LK gets 3.
$\text{one part} = \frac{\text{whole length}}{a + b}$
$\text{one part} = \frac{10}{2 + 3} = 2$
Each part is 2 units long.
Find the two lengths
$\mathrm{JL} = 2 \times 2 = 4$
$\mathrm{LK} = 3 \times 2 = 6$
Watch out: JL is $\frac{2}{3}$ of LK, not the other way round. Flipping the ratio makes JL 6 and puts L at 2, choice C.
Find the position of L
L is 4 units to the right of J. Start at J’s position, −4, and move 4 to the right.
$-4 + 4 = 0$
Watch out: 4 is the length of JL, not where L sits. Answering 4 is choice D. So the answer is B, 0.

Check your answer

Put L at 0 and test the rule. JL = 0 − (−4) = 4, and LK = 6 − 0 = 6. Then $\frac{2}{3} \times 6 = 4$ , which is JL. Also 4 + 6 = 10, the whole length JK.

More questions like this

The original DOE question

Question 92 as printed on page 84 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 92. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 59 of 120

2023 Test A Math, question 36

Question

Josef and Mai divided some stamps between themselves. Josef got 60% of the stamps. If Josef received 500 more stamps than Mai, how many stamps did Josef receive?

A300
B800
C1,500
D2,500

Show the answer

Correct answer: C

Let j be the number of stamps Josef has. Then Mai has $j - 500$ stamps. The total number of stamps is $j + (j - 500) = 2j - 500$ . Since Josef has 60% of the stamps, set up an equation to solve for j:

$\frac{j}{2j - 500} = \frac{60}{100}$	Cross multiply.
$100j = 60(2j - 500)$	Apply the distributive property; multiply the 60 by $2j$ and 500.
$100j = 120j - 30{,}000$	Apply the additive inverse property; subtract $120j$ from both sides of the equation.
$-20j = -30{,}000$	Apply the multiplicative inverse property; divide both sides of the equation by −20.
$j = 1{,}500$

Question analysis

This percent word problem says Josef and Mai split some stamps, Josef got 60% of them, and he received 500 more than Mai; it asks how many stamps Josef received. Students must connect a difference in counts to a difference in percents. Stopping at the total, or taking a percent of 500, gives tempting wrong answers.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here neither share nor the total is known; students must link the 500-stamp gap to the gap between two percents before finding the total and then Josef's part.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
No obvious method moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 26 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is easy once the 500 is seen as the gap between two percents of one total; seeing that is the hard part. The tempting move is to apply the 60% to the only number given, the 500, which treats a difference as a whole. Students who do find the total must still take Josef's share of it, since the total waits among the choices. Checking that 1,500 and 1,000 differ by 500 confirms the answer.

Turning words into math moderate Understanding the question
The stem describes one split in two ways, Josef's 60% of the total and his 500-stamp lead over Mai, and the link, that the lead is 60% − 40% = 20% of the total, is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
No obvious method moderate Working it out
No equation is suggested: the quick route treats the 500 as 20 percentage points and scales up to 100%, while the handbook's route, j over 2j − 500, needs cross-multiplying and a negative coefficient. (Schoenfeld, 1985)
Many steps minor Working it out
Even the short route takes three moves, the 20% gap, the 2,500 total and Josef's 60% of it, and the total is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: If Josef has 60%, Mai has the other 40%, so the 500-stamp difference is a known percent of the total; use it to find the total, then take Josef's share.

Traps

Stops one step early: A choice is a number from partway through the solution.
Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: moderate
Takes 60% of 500, applying the percent to the difference between the two shares as if the 500 stamps were the whole collection.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes 60% of 500 to get 300 and then adds the 500 extra stamps, combining the stem's numbers without a model of the two shares.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Sees that the 500-stamp gap is 20% of the total, finds the total of 2,500, and stops before taking Josef's 60% of it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{p}{100} \times \text{whole}$; p is the percent. The whole is the total amount, and the part is the piece you want. Percent means “out of 100,” so 60% is 60 out of every 100. Use it to find a part, or work backward to find the whole.

Sort out the question
Josef and Mai split some stamps. Josef got 60% of them. He also got 500 more stamps than Mai. You need how many stamps Josef got. You don’t know the total yet, so find it first.
Find Mai’s percent
All the stamps make 100%. Josef has 60%, so Mai has the rest.
$100\% - 60\% = 40\%$
Mai has 40% of the stamps.
Find what the 500 stamps are
Josef’s extra 500 stamps are the gap between his share and Mai’s share.
$60\% - 40\% = 20\%$
So 500 stamps are 20% of all the stamps.
Find the total number of stamps
$\text{part} = \frac{p}{100} \times \text{whole}$
$500 = \frac{20}{100} \times \text{whole}$
$\text{whole} = 500 \div \frac{20}{100} = 500 \times 5 = 2{,}500$
There are 2,500 stamps in all. Watch out: that is everyone’s stamps, choice D. Don’t stop here.
Find Josef’s share
$\text{part} = \frac{60}{100} \times 2{,}500 = 1{,}500$
Watch out: taking 60% of 500 gives 300, choice A. The 500 is the gap between them, not the total. So the answer is C, 1,500 stamps.

Check your answer

Mai gets 40% of 2,500, which is 1,000 stamps. Josef’s 1,500 minus Mai’s 1,000 is 500, just as the question says. And 1,500 + 1,000 = 2,500, the whole amount.

More questions like this

The original DOE question

Question 93 as printed on page 84 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 93. Open page 84 of the handbook (PDF)

Question 60 of 120

2023 Test A Math, question 37

Question

$A circle with a shaded sector between two radii, one pointing straight up and one pointing up and to the right. The central angle of the shaded sector is marked 45°.$

The shaded sector of the circle shown above has an area of $18\pi$ square feet. What is the circumference of the circle?

A $144\pi$ ft
B $24\pi$ ft
C $18\pi$ ft
D $9\pi$ ft

Show the answer

Correct answer: B

The area of the entire circle is $\pi r^{2}$ .

The fraction of the circle that is shaded is $\frac{45}{360} = \frac{1}{8}$ . So, the area of the sector is $\frac{1}{8}\pi r^{2}$ . Use that information to find r:

$\frac{1}{8}\pi r^{2} = 18\pi$

$r^{2} = 144$

$r = 12$

Use r to calculate the circumference of the circle:

$c = 2\pi r = 2\pi(12) = 24\pi$

The circumference is $24\pi$ feet.

Question analysis

This circle question shows a shaded 45° sector with an area of 18π square feet and asks for the circumference of the whole circle. Students work backward: from the sector to the full area, from the area to the radius, and from the radius to the circumference. Mixing up area and circumference, or stopping partway, gives wrong choices.

NYS standards: 7.G.4 circle area and circumference
6.G.5 perfect squares and perfect cubes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 circle problems usually give the radius and ask for area or circumference. This one runs backward from the area of a 45° sector, needing the sector's share of the circle, a square root and a switch to circumference: four linked steps.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Something you must already know moderate
Reading a figure, table or graph minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 30 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question moves from a part of the circle's area to the whole circle's length, so area and circumference both appear and are easy to swap. Many students scale the sector up to 144π and stop, giving an area when a length was asked. Solving r² = 144 also needs a square root, after which 2πr is routine. Asking at the end whether the result is a length, measured in feet rather than square feet, catches the main trap.

Combines ideas moderate Working it out
Three ideas chain together: the 45° sector is 1/8 of the circle, so the whole area is 8 × 18π = 144π, then the radius comes from the area, and the circumference from the radius. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know moderate Working it out
Neither πr² nor 2πr is given, nor that a full turn is 360°, so every link of the chain rests on formulas students must already know. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The stem says only “shaded sector”; its 45° angle is marked at the center in the figure, the one place the fraction of the circle comes from. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Work out what fraction of the circle a 45° sector is, scale its area up to the whole circle, solve for the radius, and only then use the circumference formula.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Scales the sector up to the whole circle, 8 × 18π = 144π, and gives that area, though the question asks for the circumference.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Repeats the sector's area from the stem, 18π, as if the shaded part's area were the circle's circumference.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No step with the 45° angle or the circle formulas gives 9π, which is half the sector's area, so it mostly attracts guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a circle in a sector: $\text{fraction of the circle} = \frac{\text{angle}}{360}$; A sector is a slice of a circle, like a slice of pizza. Its angle is the angle at the center, in degrees. A full turn around the center is 360 degrees. So the slice is that fraction of the whole circle.
Area of a circle: $A = \pi r^{2}$; A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π, called pi, is a number a little more than 3. r² means r times r.
Circumference of a circle: $C = 2\pi r$; C is the circumference, the distance around the circle. r is the radius and π is pi. Two radii make the diameter, the distance across, so this is the same as π times the diameter.

Sort out the question
The shaded slice has a 45° angle at the center. Its area is $18\pi$ square feet. You need the circumference, the distance around the whole circle. For that you need the radius, so work back to it.
Find what fraction is shaded
$\text{fraction of the circle} = \frac{\text{angle}}{360}$
$\frac{45}{360} = \frac{1}{8}$
The shaded slice is one eighth of the circle. Eight slices like it would fill the circle.
Find the area of the whole circle
$8 \times 18\pi = 144\pi$
The whole circle has an area of $144\pi$ square feet. Watch out: this is an area, not the circumference. Stopping here gives choice A.
Find the radius
$A = \pi r^{2}$
$\pi r^{2} = 144\pi$
$r^{2} = 144$
$r = 12$
Divide both sides by π. Then ask what number times itself makes 144. It is 12, so the radius is 12 feet.
Find the circumference
$C = 2\pi r$
$C = 2 \times \pi \times 12 = 24\pi$
Leave π as it is, because the choices are written with π. The circumference is $24\pi$ feet. So the answer is B, $24\pi$ ft.

Check your answer

Work forward. A radius of 12 feet gives an area of $\pi \times 12 \times 12 = 144\pi$ square feet. One eighth of that is $144\pi \div 8 = 18\pi$ square feet, the shaded area in the question.

More questions like this

The original DOE question

Question 94 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 94. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 61 of 120

2023 Test A Math, question 47

Question

Which percentage is closest in value to 0.0099?

A0%
B0.1%
C1%
D100%

Show the answer

Correct answer: C

Since 0.0099 is equal to 0.99%, the answer option closest to that value is 1%.

Question analysis

This percent question asks which of four percentages is closest in value to the decimal 0.0099. Students convert a small decimal to a percent by multiplying by 100, then compare it with choices that differ by factors of ten. Moving the decimal point the wrong number of places, or reading the digits 99 as a percent, leads to wrong answers.

NYS standards: 6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting a decimal to a percent is grade 6 work. This decimal has two zeros before its first digit, so the point must move carefully, but it is a single conversion followed by picking the nearest choice, below typical grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 7 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One conversion answers the question, and the choices are spaced by powers of ten, so the work is all in the decimal places. A decimal this small looks like nothing, and students who forget to multiply by 100 see it as nearly 0%. Others move the point the wrong number of places or read the digits 99 as a percent. Seeing that 0.01 is exactly 1% puts 0.0099 just below it.

Error-prone arithmetic minor Working it out
The decimal 0.0099 has two zeros before its digits, so moving the point two places gives 0.99, and miscounting the places lands on 0.1% or 99%. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
Students must know that a percent is a number of hundredths, so 0.0099 is multiplied by 100 to become 0.99%, a rule the question takes for granted. (Anderson & Krathwohl, 2001)

Strategy: Multiply the decimal by 100 to write it as a percent, moving the point two places to the right, then choose the percent nearest to that value.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reads 0.0099 as if it were already a percent, 0.0099%, which is nearest 0%, never multiplying by 100 to change the decimal into a percent.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Moves the decimal point one place instead of two, turning 0.0099 into about 0.1, and writes that as 0.1%, ten times too small.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Sees the digits 99, thinks of 99 percent, and picks the nearest choice, ignoring the zeros that make 0.0099 less than one hundredth.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimal to percent: $\text{percent} = \text{decimal} \times 100$; A percent is a number of hundredths. To change a decimal into a percent, multiply by 100. That moves the decimal point two places to the right. For example, 0.25 × 100 = 25, so 0.25 is 25%.

Sort out the question
The number 0.0099 is a decimal, but the choices are percents. To compare them, change 0.0099 into a percent first. Then find the choice nearest to it.
Change the decimal to a percent
$\text{percent} = \text{decimal} \times 100$
$0.0099 \times 100 = 0.99$
So 0.0099 is 0.99%. The decimal point moved two places to the right. Watch out: 0.0099 is not already a percent. Reading it as 0.0099% points to 0%, choice A.
Find the nearest choice
Find how far 0.99% is from each choice.
Choice Distance from 0.99%
A, 0% 0.99
B, 0.1% 0.89
C, 1% 0.01
D, 100% 99.01
Watch out: moving the point only one place gives about 0.1, choice B. So the answer is C, 1%.

Choice	Distance from 0.99%
A, 0%	0.99
B, 0.1%	0.89
C, 1%	0.01
D, 100%	99.01

Check your answer

Work the other way. 1% is 1 out of 100, or 0.01. And 0.0099 is just a tiny bit less than 0.0100. So 0.0099 is very close to 1%, just a little less.

More questions like this

The original DOE question

Question 104 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 87, question 104. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 62 of 120

2023 Test A Math, question 50

Question

A rectangular concrete driveway is 30 feet long, 8 feet wide, and 6 inches thick. What is the volume of the concrete?

A44 cu ft
B48 cu ft
C120 cu ft
D240 cu ft

Show the answer

Correct answer: C

First convert 6 inches to 0.5 foot. Then, calculate the volume, in cubic feet:

$30 \times 8 \times 0.5 = 120$

Question analysis

This volume word problem describes a concrete driveway 30 feet long, 8 feet wide and 6 inches thick, and asks for the volume of the concrete in cubic feet. Students convert inches to feet and multiply three dimensions. Mixing units, or stopping at the area of the driveway's surface, is the main way to go wrong.

NYS standards: 6.G.2 volume with fractional edge lengths
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Volume of a rectangular prism with a fractional edge is grade 6 content. The only extra step is changing inches to feet before multiplying, so it is lighter than grade 7 volume work with composite or triangular prisms.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 2, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Volume of a box is familiar, so the question is decided by the mix of units: two dimensions in feet and one in inches. Multiplying the three numbers as printed gives 1,440, which is not a choice, and that should send students back to convert. The listed wrong answers instead come from leaving out the thickness or combining the numbers without the volume rule.

Something you must already know minor Working it out
The stem gives no formula and no conversion, so students must bring both the volume rule, length times width times height, and the 12 inches in a foot. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
The thickness becomes the fraction 1/2 foot, so the product is 30 × 8 × 1/2, or, if the inches are kept, 1,440 divided by 12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Change 6 inches to a fraction of a foot first so that all three measurements share a unit, then multiply length, width and thickness.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the three numbers in the stem, 30 + 8 + 6 = 44, instead of multiplying the dimensions.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies only the width and the thickness, 8 × 6 = 48, combining two of the stem's numbers and leaving out the 30-foot length.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the area of the driveway's surface, 30 × 8 = 240, and stops before multiplying by the half-foot thickness.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Inches to feet: $\text{feet} = \text{inches} \div 12$; There are 12 inches in 1 foot. So to change inches into feet, divide the number of inches by 12.
Volume of a rectangular box: $V = l \times w \times h$; V is the volume, the space inside a box shape. l is the length, w is the width and h is the height, here the thickness. All three must use the same unit. The answer is in cubic units, such as cubic feet.

Sort out the question
The driveway is a flat box of concrete. It is 30 feet long, 8 feet wide and 6 inches thick. You need its volume, the space the concrete fills, in cubic feet. The thickness is in inches, so change it to feet first.
Change the thickness to feet
$\text{feet} = \text{inches} \div 12$
$6 \div 12 = \frac{1}{2}$
So 6 inches is half a foot. Now all three measures are in feet.
Multiply the three measures
$V = l \times w \times h$
$V = 30 \times 8 \times \frac{1}{2} = 240 \times \frac{1}{2} = 120$
The volume is 120 cubic feet. Watch out: 30 × 8 = 240 is only the area of the top surface, in square feet. Stopping there gives choice D. Adding 30 + 8 + 6 = 44 gives choice A.
So the answer is C, 120 cu ft.

Check your answer

Think of it another way. A slab 30 feet by 8 feet and a full foot thick would hold 240 cubic feet. This one is only half a foot thick, so it holds half as much, 120 cubic feet. Using 6 instead of one half would give 1,440, far too big.

More questions like this

The original DOE question

Question 107 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 89, question 107. Open page 89 of the handbook (PDF)

Question 63 of 120

2023 Test A Math, question 53

Question

At West College, students are randomly assigned to one of 20 dormitories and one of 6 dining rooms. Kharleen likes 6 of the dormitories and 2 of the dining rooms. What is the probability that she is assigned to both a dormitory and a dining room that she likes?

A10%
B12%
C19%
D38%

Show the answer

Correct answer: A

Kharleen likes $\frac{6}{20} = \frac{3}{10}$ of the dormitories. She likes $\frac{2}{6} = \frac{1}{3}$ of the dining rooms. Multiply those fractions together to calculate the probability of Kharleen being assigned to both a dormitory and a dining room that she likes: $\frac{3}{10} \times \frac{1}{3} = \frac{1}{10} = 10\%$ .

Question analysis

This probability word problem says students at West College are randomly assigned one of 20 dormitories and one of 6 dining rooms, and Kharleen likes 6 dormitories and 2 dining rooms. It asks for the chance that she gets both a dormitory and a dining room she likes. Students combine two independent events and write the result as a percent.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the chance that two independent events both happen is grade 7 compound probability. Each event's chance comes straight from the counts given, and joining them and writing the result as a percent is a typical grade 7 task.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 45 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns on one rule, multiplying the chances of two independent events. Students who know it reduce the fractions first and finish in seconds; students who don't tend to combine the counts, and 6 × 2 = 12 reads naturally as 12%. The other two choices match no common working, so the real contest is between the product of the counts and the product of the chances.

Something you must already know moderate Working it out
The rule that the chance of two independent events both happening is the product of their chances, 3/10 × 1/3, is assumed; nothing in the stem says to multiply. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The phrase “both a dormitory and a dining room that she likes” describes two separate random assignments that must both go her way, so the two chances are combined, not compared. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The choice 12% sits only 2 points above 10%, so a rough sense that the answer is small cannot separate them; only computing 3/10 × 1/3 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the chance of a liked dormitory and the chance of a liked dining room as fractions, multiply them because both must happen, and convert the product to a percent.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Multiplies the numbers of liked options, 6 × 2 = 12, and reads that product as 12%, without dividing by the 20 × 6 possible assignments.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No combination of 6/20 and 2/6 gives 19%, a value between the right answer and either single chance, so it mostly draws estimates and guesses.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 38%, which is larger than either single chance, 30% or about 33%, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all outcomes}}$; P is the probability, the chance that something happens. The top counts the outcomes you want. The bottom counts every outcome that could happen. Each outcome must be equally likely.
Probability of two separate events: $P(\text{A and B}) = P(\text{A}) \times P(\text{B})$; A and B are two events that don’t affect each other, like two separate random picks. P(A) is the chance of A, and P(B) is the chance of B. The chance that both happen is the two chances multiplied.

Sort out the question
Kharleen gets one of 20 dorms, or dormitories, where students live. She also gets one of 6 dining rooms. Each is picked at random. She likes 6 dorms and 2 dining rooms. You need the chance that she gets a dorm she likes and a dining room she likes.
Find the chance of a liked dorm
$P = \frac{\text{outcomes you want}}{\text{all outcomes}}$
$P(\text{dorm}) = \frac{6}{20} = \frac{3}{10}$
She likes 6 of the 20 dorms. Divide the top and bottom by 2 to simplify.
Find the chance of a liked dining room
Use the same formula for the dining rooms.
$P(\text{dining room}) = \frac{2}{6} = \frac{1}{3}$
She likes 2 of the 6 dining rooms. Divide the top and bottom by 2.
Multiply the two chances
The two picks are separate. One doesn’t change the other.
$P(\text{A and B}) = P(\text{A}) \times P(\text{B})$
$\frac{3}{10} \times \frac{1}{3} = \frac{3}{30} = \frac{1}{10}$
Both must happen, so the chance is smaller than either one alone.
Change to a percent
$\frac{1}{10} = \frac{10}{100} = 10\%$
Watch out: multiplying 6 × 2 = 12 and calling it 12% skips dividing by all the possible assignments. That is choice B. So the answer is A, 10%.

Check your answer

Count every possible assignment. There are $20 \times 6 = 120$ dorm and dining room pairs, all equally likely. She likes $6 \times 2 = 12$ of them. Then $\frac{12}{120} = \frac{1}{10}$ , which is 10%, the same answer.

More questions like this

The original DOE question

Question 110 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 110. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 64 of 120

2023 Test B Math, question 11

Question

$n = 0.13 + \frac{207}{100}$

What is the value of n expressed as a fraction?

A $\frac{1}{3}$
B $\frac{11}{5}$
C $\frac{9}{4}$
D $\frac{23}{10}$

Show the answer

Correct answer: B

$n = 0.13 + \frac{207}{100} =$

$\frac{13}{100} + \frac{207}{100} =$

$\frac{220}{100} = \frac{11}{5}$

Question analysis

The number n equals 0.13 plus 207/100, and students must write n as a single fraction. This question mixes a decimal with a fraction in hundredths. Converting both to the same form, adding, and reducing to lowest terms are the steps, and the four fractions offered look quite different at first glance.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting decimals to fractions and simplifying are skills taught before grade 7. Both numbers here are in hundredths, so the addition is immediate and only the final reduction takes thought.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: An answer in an unexpected form minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Converting 0.13 to hundredths makes this a single addition, 13 + 207 = 220 hundredths, so the question tests little beyond reducing a fraction. Two of the wrong choices sit close to 2.2 but come from no common working, and the third only echoes the digits of 0.13. A student who checks each choice as a decimal finds only one equal to 2.2.

An answer in an unexpected form minor Choosing the answer
The sum comes out as 220/100 or 2.2, but the choices are in lowest terms, so it must be reduced, or each choice converted, before anything matches. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
One term is a decimal and the other a fraction; writing 0.13 as 13/100 puts both in hundredths, and a misplaced digit there changes the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write 0.13 as 13/100 so both terms are in hundredths, add the numerators, then divide the top and bottom by their greatest common factor.

Traps

Answer in another form: The right answer is written differently from the way you worked it out.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Echoes the digits of 0.13, reading it as the fraction 1/3, and ignores the 207/100 altogether.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 9/4, which is 2.25; adding 13 and 207 hundredths gives 2.2, so this mostly draws guesses near the sum.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Nothing leads to 23/10 except 13 + 207 coming out as 230, a slip with no reason behind it, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths: $\text{a decimal with two places} = \frac{\text{its digits}}{100}$; The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths. Write its digits over 100. For example, 0.47 is 47/100.
Adding fractions with the same bottom: $\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$; a and b are the tops. c is the shared bottom, the denominator. When the bottoms match, the pieces are the same size. So add the tops and keep the bottom.

Sort out the question
n equals 0.13 plus $\frac{207}{100}$ . One part is a decimal and one is a fraction. You need n as a fraction. The choices are in lowest terms, which means they can’t be simplified any more.
Write the decimal as a fraction
The 3 in 0.13 is in the hundredths place. So 0.13 is 13 hundredths.
$0.13 = \frac{13}{100}$
Now both numbers are in hundredths. Watch out: 0.13 is not $\frac{1}{3}$ . That only copies the digits, choice A.
Add the hundredths
Use the rule for adding fractions. The bottoms match, so add the tops and keep the 100 on the bottom.
$\frac{13}{100} + \frac{207}{100} = \frac{13 + 207}{100} = \frac{220}{100}$
Simplify the fraction
$\frac{220}{100}$ is not a choice, so simplify it. Find the greatest number that divides both 220 and 100. That is 20.
$\frac{220 \div 20}{100 \div 20} = \frac{11}{5}$
So the answer is B, $\frac{11}{5}$ .

Check your answer

Change both to decimals. $\frac{11}{5} = 11 \div 5 = 2.2$ . And $0.13 + 2.07 = 2.20$ . They match. Choices C and D are close, but $\frac{9}{4} = 2.25$ and $\frac{23}{10} = 2.3$ .

More questions like this

The original DOE question

Question 68 as printed on page 194 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 68. Open page 194 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 65 of 120

2023 Test B Math, question 16

Question

$A wheel: a circle with a small solid hub at its center and 15 straight spokes running from the hub to the rim, equally spaced around it.$

This wheel has 15 spokes, equally spaced. What is the measure of the angle between adjacent spokes?

A30°
B24°
C15°
D12°

Show the answer

Correct answer: B

The total number of degrees in a circle is 360. The measure of the angle between adjacent spokes is $\frac{360}{15} = 24^{\circ}$ .

Question analysis

A wheel has 15 spokes, equally spaced around its hub, as the figure shows. Students find the measure of the angle between two neighboring spokes. The question tests knowing how many degrees make a full turn around a point and sharing them equally among the spokes.

NYS standards: 7.G.5 angle relationships and equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding an angle by dividing a full turn into equal parts is early angle measurement. It takes one division, and the only fact needed is the number of degrees in a circle, so it sits below grade 7.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 5, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One fact and one division answer this question, so it is among the easiest on the test. The wrong choices catch students who use a straight angle’s 180° for the full turn, who take the number of spokes as the angle, or who guess a familiar angle. Multiplying the answer by 15 to get back to 360° is a quick check.

Something you must already know minor Working it out
The question never says that a full turn is 360°; knowing it, and that the 15 spokes split it into 15 equal angles, is the whole method. (Anderson & Krathwohl, 2001)

Strategy: The spokes split the full turn around the hub into 15 equal angles, so divide the number of degrees in a circle by 15.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working with 15 spokes gives 30°, which is 360 ÷ 12, the angle between hours on a clock, so it mostly draws guesses at a familiar angle.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the number of spokes, 15, as the angle in degrees, matching a number in the question without any calculation.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Divides 180 by 15, using a straight angle’s 180° as the full turn around the hub instead of 360°.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal angles around a point: $\text{each angle} = 360 \div \text{number of equal angles}$; One full turn around a point is 360 degrees. If lines from the point split that turn into equal angles, each angle gets the same share. So divide 360 by how many angles there are.

Sort out the question
The wheel has 15 spokes. They all start at the hub, the center of the wheel, and they are equally spaced. Adjacent spokes are spokes right next to each other. You need the angle between two adjacent spokes.
Count the angles around the hub
Go once around the hub. Between each spoke and the next is one angle. So 15 spokes make 15 angles, all the same size. Together they make one full turn, which is 360 degrees.
Watch out: a full turn is 360 degrees, not 180. Using 180 gives 12 degrees, choice D.
Share the full turn
Use the formula for equal angles.
$\text{each angle} = 360 \div \text{number of equal angles}$
$\text{each angle} = 360 \div 15 = 24$
Each angle is 24 degrees. Watch out: 15 is the number of spokes, not the size of the angle, so choice C is wrong. So the answer is B, 24°.

Check your answer

Multiply back: 15 angles of 24 degrees make $15 \times 24 = 360$ degrees, a full turn. Also, a clock’s 12 hour marks are 30 degrees apart. This wheel has more spokes than that, so each angle must be smaller than 30 degrees.

More questions like this

The original DOE question

Question 73 as printed on page 195 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 73. Open page 195 of the handbook (PDF)

Question 66 of 120

2023 Test B Math, question 17

Question

$m \bullet t = 72$

For the equation above, what is the sum of the values of m when $t = 1$ , when $t = 2$ , and when $t = 3$ ?

A12
B72
C132
D135

Show the answer

Correct answer: C

First, find the value of m for each value of t:

For $t = 1$ :

$1m = 72$

$m = 72$

For $t = 2$ :

$2m = 72$

$m = 36$

For $t = 3$ :

$3m = 72$

$m = 24$

Then, add those three values together:

$72 + 36 + 24 = 132$

Question analysis

The equation m • t = 72 links two variables, and the raised dot means multiplication. Students find m when t is 1, 2 and 3, and then add those three values of m. Each value is a one-step division, and the trap is combining the t values before dividing.

NYS standards: 6.EE.7 one-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Solving equations like 3m = 72 is grade 6 work. This item repeats that step three times and adds the results, more bookkeeping than difficulty, so it is below typical grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each value of m comes from one division, so the work is bookkeeping rather than algebra. The trap is structural: adding 1, 2 and 3 first and dividing once feels efficient but answers a different question, since m does not change in step with t. Writing the three values of m in a row before adding them keeps the steps separate.

Turning words into math minor Understanding the question
“The sum of the values of m when t = 1, when t = 2, and when t = 3” asks for three separate solutions added together, not one solution at a combined t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Three divisions, 72 ÷ 1, 72 ÷ 2 and 72 ÷ 3, and an addition make four small steps, each easy but each a chance to drop a value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Divide 72 by each value of t in turn to get three separate values of m, then add the three quotients.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Adds the three t values to get 6 and divides 72 by 6, as if the sum of the m values were m at the sum of the t values.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives back 72, the constant in the equation and the value of m only when t = 1, without the values for t = 2 and t = 3.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 135, which is 3 more than the right sum; it would need a slip such as 72 ÷ 3 = 27, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division undoes multiplication: $\text{if } a \times b = c \text{, then } a = c \div b$; a, b and c are numbers, and b is not 0. If you know a product and one of the numbers that made it, divide to find the other one. For example, 4 × 5 = 20, so 4 = 20 ÷ 5.

Sort out the question
The equation is $m \bullet t = 72$ . The dot means times. You need m three times: when t is 1, when t is 2 and when t is 3. Then add the three values of m.
Get m by itself
Since m times t is 72, m is 72 divided by t.
$m = 72 \div t$
Now put in each value of t, one at a time.
Find m for each value of t
Each time, divide 72 by the value of t. A table keeps the three answers in order.
Value of t Value of m
1 $m = 72 \div 1 = 72$
2 $m = 72 \div 2 = 36$
3 $m = 72 \div 3 = 24$
Add the three values
$72 + 36 + 24 = 132$
Watch out: don’t add the t values first. $72 \div (1 + 2 + 3) = 12$ is choice A, and it is not the same thing. Also, 72 alone is only the first value, choice B. So the answer is C, 132.

Value of t	Value of m
1	$m = 72 \div 1 = 72$
2	$m = 72 \div 2 = 36$
3	$m = 72 \div 3 = 24$

Check your answer

Check each pair by multiplying: $72 \times 1 = 72$ , $36 \times 2 = 72$ and $24 \times 3 = 72$ . All three pairs make 72. Then add again: 72 + 36 = 108, and 108 + 24 = 132.

More questions like this

The original DOE question

Question 74 as printed on page 196 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 74. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 67 of 120

2023 Test B Math, question 19

Question

Orange juice costs $2 per gallon, and grapefruit juice costs $3 per gallon. A recent shipment of equal amounts of the two types of juice costs a total of $250. How many gallons of orange juice are in the shipment?

A50
B62.5
C100
D125

Show the answer

Correct answer: A

Let x represent the number of gallons of orange juice. Since the shipment contains equal amounts of the two types of juice, x also represents the number of gallons of grapefruit juice. Write an equation to solve for x:

$2x + 3x = 250$

$5x = 250$

$x = 50$

Question analysis

Orange juice costs $2 a gallon and grapefruit juice costs $3 a gallon, and a shipment with equal amounts of each costs $250. Students find how many gallons of orange juice are in the shipment. The word problem leads to a simple equation; the key phrase is equal amounts of juice, not equal amounts of money.

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation from a short word problem is standard grade 7 work. This one has a single unknown, since both juices come in the same number of gallons, and only two short steps.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Solution: 3 steps
Reading level: Question and choices: about grade 6, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The equation is short once the key phrase is read as equal gallons, since one gallon of each juice costs $5 together and $250 buys 50 such pairs. Read as equal spending, the same phrase leads to a listed answer that is not a whole number of gallons. The other trap is giving the gallons of both juices together when only orange juice is asked for.

Turning words into math moderate Understanding the question
“Equal amounts of the two types of juice” means the same number of gallons of each, x, so the total cost is 2x + 3x, not two equal halves of $250. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Price one gallon of each juice together, then see how many such pairs $250 buys; that is the number of gallons of each kind.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Reads “equal amounts” as equal spending, $125 on each juice, and divides $125 by $2 a gallon to get 62.5.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the right count for each juice but gives the whole shipment, 50 + 50 = 100 gallons, instead of the orange juice alone.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides $250 by 2, the orange juice price or the number of juices, combining numbers from the question without equal gallons of each.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost: $\text{cost} = \text{price per gallon} \times \text{number of gallons}$; The price per gallon is what one gallon costs. Multiply it by the number of gallons to get what you pay in all. For example, 3 gallons at $2 a gallon cost $6.

Sort out the question
Orange juice costs $2 a gallon. Grapefruit juice costs $3 a gallon. The shipment has equal amounts of each, so it has the same number of gallons of each juice. It costs $250 in all. You need the number of gallons of orange juice.
Write the cost of each juice
Let x be the number of gallons of each juice. Use the cost formula for each one.
$\text{cost} = \text{price per gallon} \times \text{number of gallons}$
$\text{orange juice cost} = 2 \times x = 2x$
$\text{grapefruit juice cost} = 3 \times x = 3x$
Add the costs and solve
The two costs add up to $250. Two x’s plus three x’s make five x’s.
$2x + 3x = 250$
$5x = 250$
$x = 250 \div 5 = 50$
There are 50 gallons of each juice.
Give the answer
Watch out: equal amounts means equal gallons, not equal dollars. Spending $125 on each juice gives 62.5 gallons, choice B. Also, 100 gallons is both juices together, choice C. So the answer is A, 50.

Check your answer

Find each cost. 50 gallons of orange juice cost $50 \times 2 = 100$ dollars. 50 gallons of grapefruit juice cost $50 \times 3 = 150$ dollars. Together, $100 + 150 = 250$ dollars, the total in the question.

More questions like this

The original DOE question

Question 76 as printed on page 196 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 76. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 68 of 120

2023 Test B Math, question 25

Question

The town of Waterville covers an area of 50 square miles. In 2005, its population was 84,000. In 2006, its population had decreased so that the average population per square mile was equal to 1,500. What was the total decrease in population from 2005 to 2006?

A7,500
B8,400
C9,000
D9,200

Show the answer

Correct answer: C

Let p represent the population of Waterville in 2006. The population per square mile is:

$\frac{p}{50} = 1{,}500$

Solve for p:

$p = 1{,}500(50) = 75{,}000$

The population decrease from 2005 to 2006 is $84{,}000 - 75{,}000 = 9{,}000$ .

Question analysis

The town of Waterville covers 50 square miles and had 84,000 people in 2005. By 2006 the population had dropped so that there were 1,500 people per square mile on average. Students find how much the population decreased, turning the population density into a total population before subtracting.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Using a rate such as people per square mile is grade 6 content. Here it is one step inside a two-step problem with large round numbers, which matches typical grade 7 multi-step rate work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question gives a total for 2005 but only a rate for 2006, so the two years cannot be compared until the rate becomes a population. After that, one subtraction finishes it. Students who stop at the 2006 population, or who reach for a number shaped like 84,000, land on listed choices, while the near miss above the answer comes from no sensible working.

Turning words into math moderate Understanding the question
“The average population per square mile was equal to 1,500” is a density, which has to become a total, 1,500 × 50 = 75,000, before it can be compared with 84,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
The 2006 population is only a midpoint; subtracting it from 84,000 is a second step, and the question asks for that decrease, not either population. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply 1,500 people per square mile by 50 square miles to get the 2006 population, then subtract it from the 2005 population.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the 2006 population, 1,500 × 50 = 75,000, and stops; 7,500, one zero short, is the choice whose digits match it.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes a tenth of the 2005 population, 84,000 ÷ 10, a number built from the question rather than from the density.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 9,200; it lies just above the right decrease and needs an arithmetic slip with no reason behind it, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Population per square mile: $\text{people per square mile} = \frac{\text{population}}{\text{area}}$; The population is the number of people who live in a place. The area is its size, here in square miles. Dividing gives the average number of people for each square mile. To get back the population, multiply people per square mile by the area.

Sort out the question
Waterville covers 50 square miles. In 2005, 84,000 people lived there. In 2006, there were 1,500 people per square mile, on average. You need how many fewer people lived there in 2006 than in 2005.
Find the 2006 population
Let p be the 2006 population. Use the formula.
$\text{people per square mile} = \frac{\text{population}}{\text{area}}$
$1{,}500 = \frac{p}{50}$
To undo dividing by 50, multiply both sides by 50.
$p = 1{,}500 \times 50 = 75{,}000$
So 75,000 people lived there in 2006.
Find the decrease
A decrease is how much a number went down. Subtract the 2006 population from the 2005 population.
$84{,}000 - 75{,}000 = 9{,}000$
The population went down by 9,000 people.
Give the answer
Watch out: 75,000 is the 2006 population, not the decrease. Choice A, 7,500, looks like it but has one zero fewer. The question asks how much the population dropped. So the answer is C, 9,000.

Check your answer

Work with people per square mile instead. In 2005, there were $84{,}000 \div 50 = 1{,}680$ people per square mile. In 2006, there were 1,500. That is 180 fewer per square mile, and $180 \times 50 = 9{,}000$ fewer people.

More questions like this

The original DOE question

Question 82 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 82. Open page 199 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 69 of 120

2023 Test B Math, question 27

Question

$\mathrm{N} = \{6, 7, \ldots, 21, 22\}$

Set L (not shown) is a list of 25 consecutive integers. The median of the integers in set N above is equal to the least integer in set L. What is the median of the integers in set L?

Show the answer

Correct answer: B

First, find the median of set N. There are 17 integers in set N, so the 9th integer is the median. The 9th integer is 14.

The least integer in set L is equal to the median of set N, so the least integer in set L is 14.

There are 25 integers in set L, so the median is the 13th integer. That integer is 26.

Question analysis

Set N lists the consecutive integers from 6 to 22, and set L is a list of 25 consecutive integers whose least member equals the median of N. Students find the median of set L. It chains two median questions together, and counting positions in a run of consecutive integers invites off-by-one slips.

NYS standards: 6.SP.5c range, center, and overall pattern
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Medians of short lists are grade 6 content. This item hides both lists, one behind set notation with an ellipsis and one not shown at all, and links them, so it needs more planning and careful counting than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
No obvious method moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains two medians, and both depend on counting members correctly in runs of consecutive integers. The easy slip is between position and distance: the 13th member of L is 12 steps above its first, not 13. With four consecutive choices, every counting error has a home, so writing out the first and last members of L, 14 and 38, and averaging them is the safest check.

A very close wrong answer moderate Choosing the answer
The choices 25, 26, 27 and 28 are consecutive, so a count that is off by one position lands on a neighboring choice with nothing to flag it. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
No obvious method moderate Working it out
Set L is never listed; its median has to be found from its least member, 14, and its size, 25, by working out that the middle is the 13th member. (Schoenfeld, 1985)
Reading a figure, table or graph minor Understanding the question
Set N is written with an ellipsis, {6, 7, …, 21, 22}, so its 17 members must be counted, and 22 − 6 = 16 undercounts by one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find N's median by averaging its least and greatest members. L's median is its middle member, so work out that member's position and count up from L's least member.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives 25, the number of integers in set L stated in the question, instead of finding which integer sits in the middle.
C Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts 13 places up from 14 instead of 12, mixing up the median’s position, 13th, with its distance from the first integer.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 28, which is twice N’s median of 14 and matches no count along set L, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting the integers in a list: $\text{count} = \text{last} - \text{first} + 1$; This counts the whole numbers in a list that goes up by 1, like 6, 7, 8. Subtracting gives the number of steps from the first number to the last. You add 1 because the first number counts too.
Position of the median: $\text{position} = \frac{\text{count} + 1}{2}$; The count is how many numbers are in a list that is in order. When the count is odd, the median is at this position. It has the same number of values before it and after it.

Sort out the question
Set N is 6, 7, and so on up to 22. The dots stand for every integer in between. Integers are whole numbers. Set L has 25 consecutive integers, one right after another. Its least number equals the median of N. You need the median of L.
Count the numbers in set N
$\text{count} = \text{last} - \text{first} + 1$
$22 - 6 + 1 = 17$
Set N has 17 integers. Watch out: $22 - 6 = 16$ misses one, because the first number counts too.
Find the median of set N
$\text{position} = \frac{\text{count} + 1}{2}$
$\text{position} = \frac{17 + 1}{2} = 9$
The median is the 9th number. 6 is the 1st, so the 9th is 8 steps up: 6 + 8 = 14. The median of N is 14.
Find the middle place in set L
So set L starts at 14 and has 25 integers. Find the position of its median.
$\text{position} = \frac{25 + 1}{2} = 13$
The median of L is its 13th number.
Count up to the 13th number
14 is the 1st number. The 13th is 12 steps after it.
$14 + 12 = 26$
Watch out: the 13th number is 12 steps up, not 13. Adding 13 gives 27, choice C. And 25 is how many numbers L has, not its median. So the answer is B, 26.

Check your answer

Set L runs from 14 to $14 + 24 = 38$ . For consecutive integers, the median is halfway between the ends: $(14 + 38) \div 2 = 26$ . That matches. The same trick gives N’s median: $(6 + 22) \div 2 = 14$ .

More questions like this

The original DOE question

Question 84 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 84. Open page 199 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 70 of 120

2023 Test B Math, question 28

Question

The sum of the numbers x, y, and z is 40. The ratio of x to y is 1:2, and the ratio of y to z is 2:5. What is the value of y?

Show the answer

Correct answer: D

The ratio of x to y is 1:2, so $\frac{x}{y} = \frac{1}{2}$ or $x = \frac{1}{2}y$ . Similarly, the ratio of y to z is 2:5, so $\frac{y}{z} = \frac{2}{5}$ . Solve for z in terms of y:

$5y = 2z$

$\frac{5}{2}y = z$

Use the given sum of x, y, and z to find the value of y:

$x + y + z = 40$

$\frac{1}{2}y + y + \frac{5}{2}y = 40$

$4y = 40$

$y = 10$

Question analysis

This ratio question gives two ratios that share y: x to y is 1:2 and y to z is 2:5, and x, y and z total 40. Students find the value of y. The numbers in a ratio count parts, not amounts, so the work is to combine the ratios and find what one part is worth.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Most grade 7 ratio tasks split a total using one ratio. Here two ratios linked through y must first become a single three-part ratio; since y's term already matches, that step is short and the problem stays at grade 7.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Variables instead of numbers minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards seeing that the two ratios share y’s term, which makes one three-part ratio. From there, one part is worth 5 and y is two parts. The wrong choices are numbers that live inside the ratios, and adding all four ratio numbers, which counts y twice, also leads to the part count, so that choice draws students along two paths.

No obvious method moderate Working it out
Nothing says to merge the ratios, but y is 2 in both, so x : y : z becomes 1 : 2 : 5 at once; the explanation’s route through fractions of y is slower. (Schoenfeld, 1985)
Variables instead of numbers minor Understanding the question
x, y and z appear only in two ratios and a sum of 40, so the ratio numbers 1, 2 and 5 are easy to mistake for the values themselves. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Adding the 8 parts, dividing 40 by 8 to get 5 per part, and taking y’s 2 parts are three steps, and the part count is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Write x : y : z as one ratio using y's matching term, add the parts, divide 40 by that sum to get one part, and multiply by y's parts.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes y’s term in the ratios, 2, as y’s actual value, never using the sum of 40.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two 2s that stand for y in the two ratios, 2 + 2 = 4, treating ratio terms as amounts.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Combines the ratios into 1 : 2 : 5 and gives the total number of parts, 1 + 2 + 5 = 8, instead of y’s share of the 40.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Splitting a total by a ratio: $\text{one part} = \text{total} \div \text{number of parts}$; Each number in a ratio counts parts that are all the same size. Add the numbers in the ratio to find how many parts make up the total. Dividing the total by that number gives the size of one part. Each amount is its number of parts times that size.

Sort out the question
x, y and z add up to 40. A ratio compares two amounts. The ratio of x to y is 1:2, so y is twice x. The ratio of y to z is 2:5. You need the actual value of y, not its number in a ratio.
Combine the two ratios
y is 2 in both ratios. So both ratios count the same size of part, and they join into one.
$x : y : z = 1 : 2 : 5$
So x is 1 part, y is 2 parts and z is 5 parts.
Find the total number of parts
$1 + 2 + 5 = 8$
The 40 is split into 8 equal parts. Watch out: 8 counts parts. It is not y, so choice C is wrong.
Find the size of one part
$\text{one part} = \text{total} \div \text{number of parts}$
$\text{one part} = 40 \div 8 = 5$
The 40 is shared equally among the 8 parts. So each part is worth 5.
Find y
y is 2 parts, so multiply the size of one part by 2.
$y = 2 \times 5 = 10$
Watch out: the 2 in the ratio counts parts, so y is not 2, choice A. Adding the two 2s to get 4 is wrong too, choice B. So the answer is D, 10.

Check your answer

Find all three: x = 5, y = 10 and z = 25. They add to $5 + 10 + 25 = 40$ . x to y is 5 to 10, which is 1:2. y to z is 10 to 25, which is 2:5.

More questions like this

The original DOE question

Question 85 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 85. Open page 199 of the handbook (PDF)

Question 71 of 120

2023 Test B Math, question 31

Question

If $xy=30$ , what is the value of $10\div x$ in terms of y?

A $\frac{1}{10}y$
B $\frac{1}{3}y$
C $3y$
D $10y$

Show the answer

Correct answer: B

Solve $xy=30$ , the given equation, for x:

$xy=30$

$x=\frac{30}{y}$

Then:

$10\div x=10\div\frac{30}{y}=10\times\frac{y}{30}=\frac{1}{3}y$

Question analysis

An algebra question: given that x times y equals 30, students express 10 divided by x using only y. It tests solving for one variable and dividing by an algebraic fraction. Working with letters instead of numbers, and dividing in the right order, are what trip students up.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.1 dividing fractions by fractions
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students rewrite expressions that have numbers and one variable. This one links two variables through their product and asks for a quotient, with no numbers given to check an answer against.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The algebra is three short moves, but each is a place to turn a fraction upside down: solving xy = 30 for x, dividing 10 by that fraction, and reducing 10/30. Students who reach for the bigger-over-smaller division they know from whole numbers land on 3y. Choosing two numbers with a product of 30, such as 3 and 10, and testing every choice settles the question in seconds.

Variables instead of numbers moderate Understanding the question
x and y are tied only by their product, 30, and the answer must be an expression in y, so there is no number to check unless students pick a pair such as x = 3 and y = 10. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Substituting x = 30/y turns 10 ÷ x into 10 divided by a fraction, which must be flipped to 10 × y/30 and then reduced. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The answer, (1/3)y, and the choice 3y use the same 3, one dividing y and one multiplying it, so a student unsure whether 10 ÷ 30 or 30 ÷ 10 applies sees both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Solve xy = 30 for x, substitute that fraction into 10 ÷ x, and divide by multiplying by its reciprocal; or choose values of x and y with a product of 30 and test each choice.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Writes the 10 from 10 ÷ x under y as one tenth of y, never using the product xy = 30 to link the two letters.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Replaces x with 30/y correctly but divides 30 by 10 instead of 10 by 30, taking the larger number over the smaller, and gets 3y.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats dividing by x as multiplying by y, as if x and y were reciprocals, though their product is 30, not 1, which gives 10y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is the number you divide. b over c is the fraction you divide by: b is its top and c is its bottom. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. For example, 6 ÷ 1/2 = 6 × 2 = 12, because 12 halves fit in 6.

Sort out the question
$xy$ means x times y. So x times y is 30. You need $10 \div x$ “in terms of y.” That means your answer can use y and numbers, but no x.
Solve for x
To swap x out, first get x by itself. Divide both sides of $xy = 30$ by y.
$x = \frac{30}{y}$
This says x is 30 divided by y.
Put the fraction in place of x
Replace x in $10 \div x$ with this fraction.
$10 \div x = 10 \div \frac{30}{y}$
Now use the rule for dividing by a fraction. Flip the fraction and multiply.
$10 \div \frac{30}{y} = 10 \times \frac{y}{30} = \frac{10y}{30}$
Simplify the fraction
Both 10 and 30 can be divided by 10.
$\frac{10y}{30} = \frac{y}{3} = \frac{1}{3}y$
Watch out: 10 ÷ 30 is not 3. Dividing the bigger number by the smaller one gives 3y, choice C. Also, dividing by x is not the same as multiplying by y. That only works when xy is 1, not 30. It leads to choice D.
So the answer is B, $\frac{1}{3}y$ .

Check your answer

Pick numbers that multiply to 30, like x = 3 and y = 10. Then $10 \div x = 10 \div 3 = \frac{10}{3}$ . Choice B gives $\frac{1}{3} \times 10 = \frac{10}{3}$ . They match. Choice C would give 30, and choice D would give 100.

More questions like this

The original DOE question

Question 88 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 88. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 72 of 120

2023 Test B Math, question 32

Question

$A number line with arrows at both ends and three points marked on it. From left to right, point P is at 1 1/3, point Q is at 3, and point R is at 5 1/6.$

On the number line above, how many units longer is $\overline{QR}$ than $\overline{PQ}$ ?

A $\frac{1}{6}$
B $\frac{1}{2}$
C $\frac{2}{3}$
D3

Show the answer

Correct answer: B

First, find the lengths of the given segments:

$\mathrm{QR}=5\frac{1}{6}-3=2\frac{1}{6}$

$\mathrm{PQ}=3-1\frac{1}{3}=1\frac{2}{3}$

Then, find the difference:

$2\frac{1}{6}-1\frac{2}{3}=$

$\frac{13}{6}-\frac{5}{3}=$

$\frac{13}{6}-\frac{10}{6}=$

$\frac{3}{6}=\frac{1}{2}$

Question analysis

A number line shows three points labeled with mixed numbers and a whole number, and students find how many units longer one segment is than another. It tests subtracting mixed numbers with unlike denominators to find distances. Regrouping when the fraction being subtracted is the larger one is where most errors happen.

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding distances between points on a number line is grade 7 work. Here the points are mixed numbers with different denominators, so finding two distances and their difference makes it a typical but careful grade 7 task.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is three subtractions, so most errors are in the fractions rather than the plan. Taking 1 1/3 from a whole number such as 3 tempts students to subtract only the whole parts and keep the 1/3, which reverses which segment looks longer. Rewriting every coordinate in sixths, 8/6, 18/6 and 31/6, turns the work into whole-number subtraction of numerators.

Error-prone arithmetic moderate Working it out
Both lengths subtract mixed numbers with different denominators, 5 1/6 − 3 and 3 − 1 1/3, and the second needs regrouping before the two results are compared in sixths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph minor Understanding the question
The three coordinates come from the labels under P, Q and R, and the segments QR and PQ must each be measured from Q at 3, the point they share. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find each length by subtracting the left endpoint from the right one, writing the mixed numbers in sixths, then subtract the shorter length from the longer.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts only the whole numbers in 3 − 1 1/3 and keeps the 1/3, giving PQ as 2 1/3, then compares it with 2 1/6 and gets 1/6.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2/3; it is only the fraction part of PQ’s length, 1 2/3, so it mostly draws guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 3, the coordinate of point Q, a number read off the line rather than a difference between two lengths.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Length on a number line: $\text{length} = \text{right point} - \text{left point}$; The length of a segment is the distance between its two end points. Subtract the number on the left from the number on the right. Subtracting tells how many units you move to get from one point to the other.
Mixed number to improper fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; a is the whole number part. b over c is the fraction part, with b on top and c on the bottom. Each whole holds c parts, so a wholes hold a × c parts. Then add the b extra parts. An improper fraction has a top bigger than its bottom.

Read the number line
Point P is at $1\frac{1}{3}$ . Point Q is at 3. Point R is at $5\frac{1}{6}$ .
The bar over QR means segment QR, the piece of the line from Q to R. You need how much longer QR is than PQ. So find both lengths, then subtract.
Find the length of QR
R is on the right, and Q is on the left. Use the length rule.
$\text{QR} = 5\frac{1}{6} - 3 = 2\frac{1}{6}$
QR is $2\frac{1}{6}$ units long.
Find the length of PQ
Q is on the right, and P is on the left.
$\text{PQ} = 3 - 1\frac{1}{3}$
Take it in two parts. First, 3 − 1 = 2. Then take away the extra $\frac{1}{3}$ . That gives $2 - \frac{1}{3} = 1\frac{2}{3}$ . So PQ is $1\frac{2}{3}$ units long.
Watch out: Don’t just do 3 − 1 and keep the $\frac{1}{3}$ . That gives $2\frac{1}{3}$ , which is too long. It leads to choice A.
Write both lengths in sixths
The fractions have different bottoms, 6 and 3. Change both lengths to sixths so you can subtract. Use the mixed number rule.
$2\frac{1}{6} = \frac{2 \times 6 + 1}{6} = \frac{13}{6}$
$1\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3} = \frac{10}{6}$
Subtract the lengths
Take the shorter length away from the longer one.
$\frac{13}{6} - \frac{10}{6} = \frac{3}{6} = \frac{1}{2}$
QR is $\frac{1}{2}$ unit longer than PQ.
So the answer is B, $\frac{1}{2}$ .

Check your answer

Write every point in sixths. P is $\frac{8}{6}$ , Q is $\frac{18}{6}$ and R is $\frac{31}{6}$ . QR is 31 − 18 = 13 sixths. PQ is 18 − 8 = 10 sixths. The difference is 3 sixths, which is one half.

More questions like this

The original DOE question

Question 89 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 89. Open page 200 of the handbook (PDF)

Question 73 of 120

2023 Test B Math, question 35

Question

In Center City, the ratio of people to dogs is $4:1$ , and the ratio of dogs to cats is $3:2$ . What is the ratio of people to cats?

A $2:1$
B $7:3$
C $6:1$
D $12:1$

Show the answer

Correct answer: C

Take each ratio and solve for dogs:

$\frac{p}{d}=\frac{4}{1}$

$4d=p$

$d=\frac{1}{4}p$

And

$\frac{d}{c}=\frac{3}{2}$

$d=\frac{3}{2}c$

Set the two expressions for d equal to each other and solve for $\frac{p}{c}$ :

$\frac{1}{4}p=\frac{3}{2}c$

$p=6c$

$\frac{p}{c}=\frac{6}{1}$

So, the ratio of people to cats is $6:1$ .

Question analysis

A ratio question about Center City, where the ratio of people to dogs and the ratio of dogs to cats are both given. Students link the two ratios through the dogs to find the ratio of people to cats. It tests combining ratios, a step that goes wrong when the given numbers are paired up as written.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 ratio problems usually work with a single ratio. Here two ratios share the dogs but give them different numbers of parts, so one ratio must be rescaled before people and cats can be compared.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards seeing that the dogs link the two ratios but are counted differently in each. Reading the people’s 4 and the cats’ 2 straight off the page gives a tidy 2 : 1, and it is the trap most students fall into. Rewriting people to dogs as 12 : 3 makes one chain, 12 : 3 : 2, from which people to cats is 12 : 2.

No obvious method moderate Working it out
Dogs are 1 part in the first ratio and 3 parts in the second, so the ratios cannot be joined until 4 : 1 is rescaled to 12 : 3; nothing in the stem says to do this. (Schoenfeld, 1985)
Much to keep track of minor Working it out
Three groups, people, dogs and cats, run through two ratios, and students must keep track of which number belongs to which group while scaling. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Rewrite both ratios with the same number for dogs, so that people to dogs to cats becomes one three-part ratio, then read off people to cats and simplify.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Pairs the people’s 4 from the first ratio with the cats’ 2 from the second, 4 : 2 = 2 : 1, without first matching the dogs’ parts.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two ratios term by term, (4 + 3) : (1 + 2) = 7 : 3, treating ratio terms as amounts that can be combined.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Multiplies people per dog by the dogs’ 3, 4 × 3 = 12, and stops before dividing by the cats’ 2, giving 12 : 1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent ratios: $a : b = (a \times k) : (b \times k)$; a and b are the two amounts in a ratio, in order. k is any number you choose, except 0. Multiplying both amounts by the same number keeps the comparison the same. For example, 4 : 1 and 12 : 3 both mean 4 people for every dog.

Sort out the question
A ratio compares two amounts. People to dogs is 4 : 1. That means 4 people for every 1 dog. Dogs to cats is 3 : 2. That means 3 dogs for every 2 cats.
You need people to cats. Dogs show up in both ratios, so they are the link.
Make the dog numbers match
The dogs are 1 in the first ratio and 3 in the second. To join the ratios, the dog numbers must be the same. So multiply both parts of the first ratio by 3.
$4 : 1 = (4 \times 3) : (1 \times 3) = 12 : 3$
Now there are 12 people for every 3 dogs.
Join the two ratios
Both ratios now have 3 dogs. So you can write all three groups in one ratio.
$\text{people} : \text{dogs} : \text{cats} = 12 : 3 : 2$
For every 12 people, there are 3 dogs and 2 cats.
Read off and simplify
People to cats is 12 : 2. Divide both parts by 2.
$12 : 2 = (12 \div 2) : (2 \div 2) = 6 : 1$
Watch out: Taking the 4 people and the 2 cats straight from the problem gives 4 : 2 = 2 : 1, choice A. That skips matching the dogs. Stopping at 12 : 1, without using the 2 cats, gives choice D.
So the answer is C, 6 : 1.

Check your answer

Build a small town. Start with 2 cats. The second ratio gives 3 dogs. The first ratio gives 4 people for each dog, so 4 × 3 = 12 people. That is 12 people for 2 cats, or 6 people for each cat.

More questions like this

The original DOE question

Question 92 as printed on page 201 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 92. Open page 201 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 74 of 120

2023 Test B Math, question 36

Question

The side length of square S is shorter than the side length of square T by 1 centimeter. The area of square S is 9 square centimeters. What is the difference, in square centimeters, between the area of square T and the area of square S?

Show the answer

Correct answer: B

Let x represent the side length of square T, in centimeters. Then the side length of square S is $(x-1)$ .

The area of square S is:

$(x-1)(x-1)=9$

$(x-1)^{2}=9$

$x-1=3$

$x=4$

So the side length of square T is 4 cm, and the area of square T is 16 sq cm.

The difference between the areas, in square centimeters, is $16-9=7$ .

Question analysis

Two squares differ in side length by 1 centimeter, and the smaller one has an area of 9 square centimeters. Students find how much greater the larger square's area is. It tests the link between a square's side and its area, and the extra area is easy to miscount when picturing the border that the longer sides add.

NYS standards: 6.G.5 perfect squares and perfect cubes
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 The area of a square and recognizing 9 as a perfect square are grade 6 ideas. This takes three short steps with small whole numbers, lighter than grade 7 work on composite figures, though the first step runs backward from area to side.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every number is small, so the work is keeping track of which quantity the question wants: a difference of areas, not a side or a single area. The added area forms an L-shaped strip around square S, and students who picture it as two strips 4 centimeters long count the corner square twice. Finding both areas, 9 and 16, and subtracting avoids reasoning about the strip at all.

Many steps minor Working it out
The work runs from area to side and back: the square root of 9 gives S’s side, adding 1 gives T’s, squaring gives 16, and subtracting gives the difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The phrase “shorter than the side length of square T by 1 centimeter” describes S from T’s point of view, so T’s side is S’s side plus 1, not minus 1. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the side of square S from its area, add 1 centimeter to get the side of square T, find that square's area, and subtract the smaller area from the larger.

Traps

Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds the side of square S, 3 centimeters, from its area of 9 and gives that length instead of a difference between two areas.
C Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: moderate
Pictures the added area as two 4-by-1 strips along two sides of square S, 8 in all, counting the 1-by-1 corner where they overlap twice.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the stem’s two numbers, 9 + 1 = 10, as if a side 1 centimeter longer made the area 1 square centimeter larger, and gives that area.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s \times s$; A is the area, the space inside the square, in square units. s is the length of one side. All four sides of a square are equal, so the area is the side times itself. To go backward from an area, find the number that times itself gives the area.

Sort out the two squares
Square S has an area of 9 square centimeters. Its side is 1 centimeter shorter than the side of square T. You need the area of square T minus the area of square S, in square centimeters.
Find the side of square S
Use the area formula backward. Which number times itself makes 9?
$3 \times 3 = 9$
So each side of square S is 3 centimeters. This number is called the square root of 9.
Find the side of square T
Square T’s side is 1 centimeter longer than square S’s side.
$3 + 1 = 4$
Each side of square T is 4 centimeters.
Find the area of square T
Use the area formula with a side of 4 centimeters.
$A = 4 \times 4 = 16$
The area of square T is 16 square centimeters.
Subtract the areas
Take the smaller area away from the larger one.
$16 - 9 = 7$
The difference is 7 square centimeters.
Watch out: Picturing the new area as two strips, each 4 by 1, gives 8, choice C. But those strips share a 1-by-1 corner, so it gets counted twice. Also, 3 is a side length, not a difference of areas. That is choice A.
So the answer is B, 7 square centimeters.

Check your answer

Draw square T and mark square S in one corner. The extra space is an L shape. It is a 3-by-1 strip, another 3-by-1 strip and a 1-by-1 corner. That is 3 + 3 + 1 = 7 square centimeters. It matches.

More questions like this

The original DOE question

Question 93 as printed on page 201 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 93. Open page 201 of the handbook (PDF)

Question 75 of 120

2023 Test B Math, question 43

Question

A certain square puzzle has an area of 64 square inches, and a cubic block has a volume of 64 cubic inches. What is the ratio of the length of a side of the puzzle to the length of a side of the block?

A $1:3$
B $1:2$
C $2:3$
D $2:1$

Show the answer

Correct answer: D

The length of one side of the puzzle is $\sqrt{64}=8$ inches. The length of one side of the block is $\sqrt[3]{64}=4$ inches.

The ratio of the side of the puzzle to the side of the block is $8:4$ , which reduces to $2:1$ .

Question analysis

A square puzzle has an area of 64 square inches, and a cube-shaped block has a volume of 64 cubic inches. Students find the ratio of a side of the puzzle to a side of the block. Both figures are described by the number 64, but turning an area and a volume back into lengths takes different steps, and the order of the ratio matters.

NYS standards: 8.EE.2 square and cube roots
6.G.5 perfect squares and perfect cubes
6.RP.1 ratio concepts and language
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 area and volume problems go from side lengths to an area or a volume. Going backward needs a square root and a cube root, and cube roots are grade 8 content, although 64 is both a perfect square and a perfect cube.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Combines ideas minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 46 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Because 64 is both a perfect square and a perfect cube, the computation is easy once students know to undo squaring and cubing. The risk is at the end: with sides of 8 and 4 inches in hand, writing the block first turns the ratio upside down. Students who never find the side lengths are left with ideas such as two dimensions against three.

Something you must already know moderate Working it out
Neither formula is given: the puzzle’s side is the number whose square is 64 and the block’s edge the number whose cube is 64, and cube roots are grade 8 work. (Anderson & Krathwohl, 2001)
Combines ideas minor Working it out
A square and a cube share the number 64 in different units, so two different inverse operations feed one ratio, which must list the puzzle’s side first. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Find the side of the square from its area and the edge of the cube from its volume by testing small whole numbers, then write the puzzle's side first and simplify.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 1 : 3; it makes the puzzle’s side the shorter one, the reverse of 8 and 4 inches, so it mostly draws guesses.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: moderate
Finds both lengths correctly, 8 and 4 inches, but writes the block’s side first, turning the ratio upside down to 1 : 2.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the square to 2 and the cube to 3, the powers behind square inches and cubic inches, and writes 2 : 3 without finding any length.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s \times s$; A is the area, in square units. s is the length of one side. All sides of a square are equal, so the area is the side times itself. To find the side from the area, ask which number times itself gives the area.
Volume of a cube: $V = s \times s \times s$; V is the volume, the space inside, in cubic units. s is the length of one edge. A cube’s length, width and height are all equal, so the volume is the edge used three times. To go backward, ask which number used three times gives the volume.

Sort out the shapes
The puzzle is a square with an area of 64 square inches. The block is a cube with a volume of 64 cubic inches. A cube is a box with every edge the same length. You need the ratio of the puzzle’s side to the block’s side, with the puzzle first.
Find the puzzle’s side
Use the area formula backward. Which number times itself makes 64?
$8 \times 8 = 64$
The puzzle’s side is 8 inches. This number is the square root of 64.
Find the block’s side
Use the volume formula backward. Which number used three times makes 64? Try small numbers. 3 × 3 × 3 = 27 is too small.
$4 \times 4 \times 4 = 64$
The block’s side is 4 inches. This number is the cube root of 64.
Write and simplify the ratio
Put the puzzle first, as the question asks. Then divide both parts by 4.
$8 : 4 = (8 \div 4) : (4 \div 4) = 2 : 1$
Watch out: Writing the block’s side first gives 1 : 2, choice B. The order in a ratio matters.
So the answer is D, 2 : 1.

Check your answer

Check both sides. 8 × 8 = 64 square inches, and 4 × 4 × 4 = 64 cubic inches. The puzzle’s side is longer, so its number in the ratio must be bigger. Only choice D puts the bigger number first.

More questions like this

The original DOE question

Question 100 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 100. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 76 of 120

2023 Test B Math, question 44

Question

In a certain class, Rosa will take a total of 4 tests. She has already taken 2 of the tests and earned scores of 81 and 83. What is the least possible score Rosa can earn on the third test and still be able to finish the class with an average score of 85 on all 4 tests? (Assume that test scores can range from 0 to 100.)

Show the answer

Correct answer: A

Let x represent Rosa’s score on the third test, and let y represent her score on the fourth test.

$\frac{81+83+x+y}{4}=85$

$164+x+y=340$

$x+y=176$

If Rosa scores 100 on the fourth test, then the least score she can earn on the third test is 76.

Question analysis

Rosa has scores of 81 and 83 on the first two of four tests and wants a mean of 85 on all four. Students find the lowest score she can earn on the third test and still reach that average, with scores capped at 100. Two tests are still unknown, so the problem needs a plan, not a single equation.

NYS standards: 6.SP.5c range, center, and overall pattern
6.EE.8 writing simple inequalities
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 mean problems usually find one missing score. Here two tests remain, and the least possible score depends on how high the other unknown score is allowed to go, a planning step beyond typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 60 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Most students can find that Rosa needs 176 more points, but the question is about the least third score, which only comes from pushing the fourth score to its limit. Splitting the 176 evenly gives a natural-looking 88 that ignores the word least. The parenthesis about the 0 to 100 range is the key information, so it pays to ask what each stated limit is for.

No obvious method moderate Working it out
The least third score depends on how high the fourth can go; only the parenthesis “test scores can range from 0 to 100” tells students to set the fourth test at 100. (Schoenfeld, 1985)
Many steps minor Working it out
The 340 points needed, the 164 already earned, the 176 still needed and the 100 allowed on the last test make four steps, each a chance to lose track. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply the target mean by 4 for the total points needed, subtract the two scores so far, then decide how the remaining points can be split so the third score is as low as possible.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the mean of the two scores Rosa already has, (81 + 83) ÷ 2 = 82, a number that answers nothing the question asks.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds the 176 points still needed and splits them evenly, 88 on each test, missing that the word least lets the fourth test carry more.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 90; leaving out the fourth test gives (81 + 83 + x) ÷ 3 = 85 and x = 91, the nearest route, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average: $\text{average} = \frac{\text{sum of the scores}}{\text{number of scores}}$; The sum is all the scores added together. The number of scores is how many there are. The average, or mean, shares the total equally among them. Turned around, the sum equals the average times the number of scores.

Sort out the question
Rosa takes 4 tests. She scored 81 and 83 on the first two. She wants an average of 85 on all 4. You need the lowest score she can get on test 3 and still reach that average. Each score is from 0 to 100.
Find the total points needed
Turn the average formula around. The sum is the average times the number of tests.
$\text{sum of the scores} = 85 \times 4 = 340$
Rosa needs 340 points in all.
Find the points still needed
Subtract the points she already has.
$81 + 83 = 164$
$340 - 164 = 176$
Tests 3 and 4 must add up to 176 points.
Make test 4 as high as possible
To make test 3 as low as it can be, test 4 must carry as many points as it can. The highest score is 100.
$176 - 100 = 76$
So the lowest score on test 3 is 76.
Watch out: Splitting 176 evenly gives 88 on each test, choice C. That works, but it is not the least score. And 82 is just the average of her first two tests, choice B.
So the answer is A, 76.

Check your answer

Add all four scores: 81 + 83 + 76 + 100 = 340. Then 340 ÷ 4 = 85, the average she wants. A score of 75 on test 3 would need 101 on test 4, which is not possible.

More questions like this

The original DOE question

Question 101 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 101. Open page 203 of the handbook (PDF)

Question 77 of 120

2023 Test B Math, question 45

Question

The Barnes family and the Ramirez family each have a pizza of equal size. The first pizza is cut into 18 equal slices, and the second pizza is cut into 15 equal slices. If the Barnes family eats 11 slices from the first pizza, what is the greatest number of whole slices from the second pizza that the Ramirez family can eat without eating a greater percentage of a pizza than the Barnes family ate?

Show the answer

Correct answer: D

The fraction of the pizza that the Barnes family eats is $\frac{11}{18}$ . Let x represent the number of slices of pizza that the Ramirez family eats. Then the fraction of the pizza that the Ramirez family eats is $\frac{x}{15}$ . Write an inequality and solve for x:

$\frac{x}{15}\leq\frac{11}{18}$

$x\leq\frac{11\times 15}{18}$

$x\leq\frac{165}{18}$

$x\leq 9\frac{1}{6}$

So, the greatest number of whole slices of pizza the Ramirez family can eat is 9.

Question analysis

Two pizzas of the same size are cut into 18 and 15 equal slices, and one family eats 11 slices of the first. Students find the most whole slices of the second pizza the other family can eat without eating a larger share. It tests comparing fractions with different denominators and rounding to a whole number in the right direction.

NYS standards: 7.EE.4b two-step inequalities
6.RP.3 ratio and rate reasoning
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 inequality problems usually ask for the greatest whole number that fits a budget. This one wraps that idea in a long stem about shares of two pizzas sliced differently, with several conditions to keep straight at once.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Complex sentences moderate
Turning words into math moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 11, 72 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The math is one inequality between fractions, but the long final sentence hides it, so students who skim fall back on comparing slice counts. Reasoning that the second pizza has 3 fewer slices, so the Ramirez family should eat 3 fewer, is the strongest pull. Comparing 11/18 with x/15, or turning 11/18 into about 61% and taking 61% of 15, keeps the comparison about shares of a pizza.

Complex sentences moderate Understanding the question
The question is one 42-word sentence ending in a nested condition, “without eating a greater percentage of a pizza than the Barnes family ate”, a reading hurdle rather than more math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)
Turning words into math moderate Understanding the question
The limit of no “greater percentage of a pizza” must become the inequality x/15 ≤ 11/18, comparing fractions of two pizzas cut into different numbers of slices, not slice counts. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Solving gives x ≤ 11 × 15 ÷ 18 = 165/18, a division by 18 that leaves 9 1/6, which must then be cut back to whole slices. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the Barnes family's share as a fraction of their pizza, then find the largest whole number of fifteenths that does not exceed it, by cross-multiplying or by converting to decimals.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds that 9 slices is the limit but gives the slices left on the second pizza, 15 − 9 = 6, answering a different question.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts the stem’s numbers, 18 − 11 = 7, the slices left on the Barnes family’s pizza, without comparing shares of the two pizzas.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reasons that the second pizza has 3 fewer slices, so the Ramirez family should eat 3 fewer, 11 − 3 = 8, comparing by subtraction instead of by fractions.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a pizza: $\text{fraction eaten} = \frac{\text{slices eaten}}{\text{slices in the pizza}}$; The top counts the slices eaten. The bottom counts all the equal slices the pizza was cut into. The fraction tells how much of the whole pizza was eaten. Use fractions to compare pizzas cut into different numbers of slices.

Sort out the question
The two pizzas are the same size. The Barnes pizza has 18 slices, and they eat 11. The Ramirez pizza has 15 slices, so its slices are bigger. You need the most whole slices the Ramirez family can eat without eating a bigger share of their pizza.
Find the Barnes family’s share
Their share is the slices they ate out of all the slices in their pizza. Use the fraction formula.
$\text{fraction eaten} = \frac{11}{18}$
The Barnes family ate $\frac{11}{18}$ of their pizza.
Write an inequality
Let x be the number of Ramirez slices. Their share is $\frac{x}{15}$ . It must be no more than the Barnes share. The sign ≤ means less than or equal to.
$\frac{x}{15} \leq \frac{11}{18}$
Solve for x
Multiply both sides by 15 to get x by itself.
$x \leq \frac{11 \times 15}{18} = \frac{165}{18}$
Now divide. 18 goes into 165 nine times, since 18 × 9 = 162. There are 3 left over.
$x \leq 9\frac{3}{18} = 9\frac{1}{6}$
Pick the whole number
x must be a whole number no bigger than $9\frac{1}{6}$ . So the most is 9 slices. 10 slices would be too many.
Watch out: The Ramirez pizza has 3 fewer slices, but that does not mean they eat 3 fewer. Their slices are bigger. Eating 11 − 3 = 8 slices is choice C, which is not the most they can eat.
So the answer is D, 9.

Check your answer

Use decimals. The Barnes share is $11 \div 18 \approx 0.611$ . Nine Ramirez slices are $9 \div 15 = 0.6$ , just under it. Ten slices are $10 \div 15 \approx 0.667$ , too much. So 9 is the most.

More questions like this

The original DOE question

Question 102 as printed on page 204 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 102. Open page 204 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 78 of 120

2023 Test B Math, question 47

Question

Four slips of paper numbered 1, 2, 3, and 4 were placed in a bag:

One slip was selected from the bag, and the number recorded.
Without replacement, a second slip was chosen.
This process continued until no more slips remained in the bag.
One part of a tree diagram representing the sample space is shown.

$A partial tree diagram. The top number, 1, branches down to 2, 3, and 4. Below 2 are 3 and 4, below 3 are 2 and 4, and below 4 are 2 and 3. The 4 under the 2 has one branch leading down to an empty box.$

Which number should be placed in the box to complete this part of the tree diagram?

Show the answer

Correct answer: C

The “without replacement” statement means that the bag contained four slips for the first draw, three slips for the second draw, two slips for the third draw, and one slip for the last draw. Following the branch from the top and navigating down to the box, the 1 was chosen first, the 2 was chosen second, and the 4 was chosen third. The only remaining number is 3. The element would be 1243.

Question analysis

Four slips numbered 1 to 4 are drawn from a bag one at a time without replacement, and part of a tree diagram of the outcomes is shown. Students decide which number belongs in the empty box at the end of one branch. It tests reading a tree diagram and understanding what drawing without replacement means.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Tree diagrams for compound events are a grade 7 topic, but here the tree is already drawn and only one box is empty. Following a single branch to that box is shorter than typical grade 7 work with sample spaces.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 68 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Only one branch matters, and once it is read as the slips drawn so far, 1, 2 and 4, the last slip left in the bag is the only one missing. Errors come from not counting the top 1 as a draw, from forgetting that a drawn slip stays out, or from copying a nearby number in the diagram instead of following the path.

Reading a figure, table or graph minor Understanding the question
The box hangs below the branch 1, then 2, then 4, and students must read that path from the top as the order of the first three draws. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Follow the branch from the top down to the box, list the slips already drawn on that path, and name the one still left in the bag.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: moderate
Treats the 1 at the top as the start of the diagram rather than the first slip drawn, so 1 seems still to be in the bag.
B Without replacement (Fischbein & Schnarch, 1997)
Reveals a misconception Expected pull: weak
Lets slip 2, already drawn second on this branch, come up again, missing that a slip drawn without replacement stays out of the bag.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Copies 4, the number directly above the box, reading the diagram as a pattern of numbers rather than as the order of the draws.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Sort out the question
The bag holds four slips: 1, 2, 3 and 4. Slips are drawn one at a time. “Without replacement” means a drawn slip does not go back in the bag. So each number can be drawn only once.
Read the tree diagram
A tree diagram shows the orders the draws can happen in. Each row going down is one draw. The 1 at the top is the first slip drawn. The numbers below it are the choices for the second draw, and so on.
Follow the path to the box
Start at the top and go down the lines to the box.
- First draw: 1
- Second draw: 2
- Third draw: 4
- Fourth draw: the box
Slips 1, 2 and 4 are already out of the bag.
Find the slip that is left
The bag started with 1, 2, 3 and 4. Only slip 3 has not been drawn. So it must be the fourth draw.
Watch out: The 1 at the top counts as a draw, so 1 is not left. That trap is choice A. Slip 2 was already drawn, so choice B can’t be right. And 4 is just the number above the box, choice D.
So the answer is C, 3.

Check your answer

Read the whole path: 1, 2, 4, 3. Each number from 1 to 4 shows up exactly once. That is just what should happen when no slip goes back in the bag.

More questions like this

The original DOE question

Question 104 as printed on page 204 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 104. Open page 204 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 79 of 120

2023 Test B Math, question 49

Question

A rectangular swimming pool of uniform depth is 20 yards long and 14 yards wide. If the water is 6 feet deep, what is the volume of water in the pool, in cubic yards?

A40
B286
C560
D1,680

Show the answer

Correct answer: C

First, change 6 feet to yards:

$6\text{ feet}\times\frac{1\text{ yard}}{3\text{ feet}}=2\text{ yards}$

Then, multiply the dimensions of the pool to find the volume, in cubic yards:

$20\times 14\times 2=560$

Question analysis

A rectangular swimming pool is 20 yards long and 14 yards wide, and the water is 6 feet deep. Students find the volume of water in cubic yards. It tests the volume of a rectangular prism, and the depth is in a different unit from the other measurements, so it has to be converted first.

NYS standards: 6.G.2 volume with fractional edge lengths
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 The volume of a rectangular prism is grade 5 and 6 content, and the conversion here gives a whole number. Grade 7 volume work is usually longer, with composite or triangular prisms; here the only extra step is one unit conversion.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The only real trap is the mixed units: 20 and 14 are yards, 6 is feet, and the answer is wanted in cubic yards. Multiplying the three numbers as printed gives a listed choice three times too large. Underlining the unit of each measurement before multiplying, and converting the odd one out, takes seconds.

Combines ideas minor Working it out
The length and width are in yards but the depth is in feet, so a conversion, 6 feet = 2 yards, must come before the volume formula is used. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
Neither 3 feet = 1 yard nor volume = length × width × depth is given; both are assumed in a stem that asks for cubic yards. (Anderson & Krathwohl, 2001)

Strategy: Change the depth from feet to yards first, so that all three measurements are in yards, then multiply the length, width and depth.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the three measurements, 20 + 14 + 6 = 40, with no conversion and no volume formula, combining the stem’s numbers.
B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Finds the area of the water’s surface, 20 × 14 = 280, then adds the depth of 6 instead of multiplying by it.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Multiplies 20 × 14 × 6 without changing 6 feet to 2 yards, mixing yards with feet and getting a volume three times too large.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a rectangular box: $V = l \times w \times h$; V is the volume, the space inside, in cubic units. l is the length, w is the width, and h is the height or depth. All three must be in the same unit. Multiplying them counts the unit cubes that fill the box.
Feet to yards: $\text{yards} = \text{feet} \div 3$; There are 3 feet in 1 yard. So to change feet to yards, divide the number of feet by 3. A yard is longer than a foot, so you get a smaller number.

Sort out the measurements
The pool is shaped like a box. Uniform depth means the water is equally deep everywhere. The pool is 20 yards long and 14 yards wide. The water is 6 feet deep. You need the volume, the space the water fills, in cubic yards.
Change the depth to yards
Two sides are in yards, but the depth is in feet. All three must use the same unit before you multiply. The answer is in cubic yards, so change feet to yards.
$6 \div 3 = 2$
The water is 2 yards deep.
Find the volume
Use the volume formula with all three measurements in yards.
$V = 20 \times 14 \times 2 = 560$
The pool holds 560 cubic yards of water.
Give the answer
Watch out: Multiplying 20 × 14 × 6 mixes yards with feet. It gives 1,680, choice D, which is three times too big.
So the answer is C, 560 cubic yards.

Check your answer

Check the units. Yards times yards times yards gives cubic yards, just what the question asks for. Also, 1,680 ÷ 3 = 560. So using 6 feet unchanged makes the answer exactly three times too big.

More questions like this

The original DOE question

Question 106 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 106. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 80 of 120

2023 Test B Math, question 51

Question

A number, n, is divided by 25. If the quotient is 10 with a remainder of 5, what is the quotient expressed as a decimal?

A10.20
B10.25
C10.50
D10.75

Show the answer

Correct answer: A

A quotient of 10 with a remainder of 5 means the quotient is

$10+\frac{5}{25}=10\frac{1}{5}=10.20$

Question analysis

A number is divided by 25, giving a quotient of 10 with a remainder of 5, and students write the quotient as a decimal. It tests what a remainder means and how it becomes part of a decimal quotient. Connecting the remainder to the decimal part of the answer is the key idea.

NYS standards: 6.NS.2 multi-digit division
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Division with remainders and decimal quotients are grade 6 skills. This takes two short steps, with no word problem to translate, though students must know what a remainder means once a quotient is written as a decimal.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know moderate
Solution: 2 steps
Reading level: Question and choices: about grade 5, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question checks one idea: a remainder is a fraction of the divisor, not a string of digits to place after the point. Students who write the remainder 5 as tenths, or set the divisor 25 after the point, land on listed choices. Rebuilding n = 25 × 10 + 5 = 255 and dividing by 25 confirms the decimal.

Something you must already know moderate Working it out
A remainder of 5 means 5 out of the 25 needed for another whole, so it becomes 5/25 = 0.2, a link between remainders and decimals the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Rebuild n from the quotient and the remainder, then divide it by 25 to a decimal; or write the remainder as a fraction of the divisor.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Writes the divisor, 25, after the decimal point, setting the stem’s numbers 10 and 25 side by side instead of turning the remainder into a fraction.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Writes the remainder 5 as the tenths digit, 10.5, believing a remainder becomes the digits after the decimal point.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 10.75; it treats the remainder as three quarters of the divisor, which nothing in the stem suggests, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Quotient with a remainder: $\text{quotient} = q + \frac{r}{d}$; d is the divisor, the number you divide by. q is the whole-number part of the quotient, the answer to the division. r is the remainder, what is left over. The leftover still has to be divided by d, so it becomes the fraction r over d.

Sort out the words
n is divided by 25. So 25 is the divisor, the number you divide by. The quotient is the answer to a division. Here it is 10 with a remainder of 5. The remainder is the amount left over. You need the exact quotient as a decimal.
Turn the remainder into a fraction
The 5 left over still has to be divided by 25. So it becomes the fraction $\frac{5}{25}$ . Use the formula.
$\text{quotient} = 10 + \frac{5}{25}$
Change the fraction to a decimal
Simplify first. Divide the top and bottom by 5. Then write the fraction in hundredths.
$\frac{5}{25} = \frac{1}{5} = \frac{20}{100} = 0.20$
So the quotient is 10 + 0.20 = 10.20.
Give the answer
Watch out: A remainder is not the digits after the decimal point. Writing the 5 after the point gives 10.50, choice C. Writing the 25 after it gives 10.25, choice B.
So the answer is A, 10.20.

Check your answer

Rebuild n. Ten 25s plus the 5 left over is $25 \times 10 + 5 = 255$ . Now divide: $255 \div 25 = 10.2$ . That matches 10.20.

More questions like this

The original DOE question

Question 108 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 108. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 81 of 120

2023 Test B Math, question 57

Question

A box contains 14 red marbles, 10 black marbles, and 16 white marbles. How many more white marbles need to be added to the box so that the probability of randomly drawing a white marble is $\frac{3}{5}$ ?

Show the answer

Correct answer: D

Let w represent the number of additional white marbles.

Set the probability of drawing a white marble equal to $\frac{3}{5}$ and solve the equation for w:

$\frac{16+w}{14+10+16+w}=\frac{3}{5}$

$\frac{16+w}{40+w}=\frac{3}{5}$

$5(16+w)=3(40+w)$

$80+5w=120+3w$

$2w=40$

$w=20$

Question analysis

A box holds 14 red, 10 black and 16 white marbles, and students find how many white marbles must be added so that the chance of drawing a white one is 3/5. Working backward from a target probability to a number of marbles is the unusual step.

NYS standards: 6.SP.8a uniform probability models
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 probability questions usually give the counts and ask for a probability. This runs the other way: the target probability is given, and the number to add changes both the white count and the total.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The tempting shortcut is to take 3/5 of the 40 marbles already in the box, which forgets that the new white marbles join the total. Students who set up the probability with the unknown in both the top and the bottom solve it in a few lines, and testing choices works too: adding 20 gives 36 white of 60, exactly 3/5. The other two choices are counts copied from the stem.

Turning words into math moderate Understanding the question
“How many more white marbles need to be added” puts the unknown in two places, (16 + w)/(40 + w), because each white marble added also raises the total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
The unknown number of added marbles has to be carried as a letter through a proportion, 5(16 + w) = 3(40 + w), before it can be found. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Totaling the marbles, setting up the proportion, cross-multiplying and solving 80 + 5w = 120 + 3w make four steps, unless the choices are tested one by one. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Let w be the number of white marbles added, write the new probability with w in both the top and the bottom, set it equal to 3/5 and solve; or test each choice.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals partial knowledge Expected pull: strong
Takes 3/5 of the 40 marbles now in the box, 24, and subtracts the 16 white ones, forgetting that added marbles raise the total.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 10, the number of black marbles in the stem, without setting up the new probability or testing it.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 14, the number of red marbles in the stem, without checking that adding 14 white marbles gives 30 of 54, not 3/5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P(\text{white}) = \frac{\text{white marbles}}{\text{all marbles}}$; P(white) is the probability, or chance, of drawing a white marble. The top counts the white marbles. The bottom counts every marble in the box. It works because each marble is equally likely to be drawn.
Cross multiplication: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a and c are the tops. b and d are the bottoms. When two fractions are equal, each top times the other bottom gives the same product. It works because you multiply both sides by both bottoms. Use it to clear fractions from an equation.

Sort out the box
The box has 14 red, 10 black and 16 white marbles.
$14 + 10 + 16 = 40$
That is 40 marbles. You will add some white marbles. Call that number w. You need w so that the chance of drawing white is $\frac{3}{5}$ .
Write the new probability
Adding w white marbles changes both counts. There will be 16 + w white marbles and 40 + w marbles in all. Use the probability formula.
$\frac{16 + w}{40 + w} = \frac{3}{5}$
Watch out: Taking $\frac{3}{5}$ of 40 gives 24, and 24 − 16 = 8, choice A. That forgets that the new marbles also make the box bigger.
Cross multiply
Fractions are hard to solve with. Clear them with cross multiplication.
$5 \times (16 + w) = 3 \times (40 + w)$
Then multiply each outside number by both terms inside the parentheses.
$80 + 5w = 120 + 3w$
Solve for w
Subtract 3w from both sides. Then subtract 80 from both sides.
$2w = 40$
Divide both sides by 2.
$w = 20$
You need to add 20 white marbles.
So the answer is D, 20.

Check your answer

Add 20 white marbles. Then there are 16 + 20 = 36 white marbles and 40 + 20 = 60 marbles in all. Divide the top and bottom by 12: $\frac{36}{60} = \frac{3}{5}$ . It works.

More questions like this

The original DOE question

Question 114 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 114. Open page 208 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 82 of 120

2022 Test A Math, question 28

Question

The table shows the income and expenses in a bicycle shop on one day.

BICYCLE SHOP TRANSACTIONS

Description	Income	Expense
Paid employees		$\$120 \times 4$
Sold 2 bicycles	$\$400 \times 2$
Sold clothing	$95
Bought new door		$500
Sold bike parts	p

The sum of all transactions was $70. What is the value of p?

A−$275
B−$85
C$70
D$155

Show the answer

Correct answer: D

The sum of all transactions ($70) includes the income data (positive values) and the expense data (negative values). Write an equation to determine the value of p by adding all the income data and subtracting all the expense data:

$400 \bullet 2 + 95 + p - 120 \bullet 4 - 500 = 70$

Simplify the expression to determine the value of p:

$800 + 95 + p - 480 - 500 = 70$

$895 + p - 980 = 70$

$p - 85 = 70$

$p = 155$

Question analysis

A table lists one day of income and expenses at a bicycle shop, including two products to work out and an unknown amount, p, from selling bike parts. Given that all the transactions add up to $70, students find p. They must treat income as positive and expenses as negative, then solve an equation.

NYS standards: 7.EE.3 multi-step rational number problems
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve multi-step money problems with positive and negative amounts. This table adds an unknown income entry, so students treat income as positive and expenses as negative before solving a one-step equation.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The table turns a one-variable equation into bookkeeping: each entry must be multiplied out and given a sign by its column before anything is added. The known entries come to a loss of $85, so the bike parts must cover that loss and still leave $70, which is why p is larger than the day's total. Students who stop at the running total, or who copy the $70, never use the equation at all.

Turning words into math moderate Understanding the question
“The sum of all transactions was $70” has to be read with income positive and expenses negative, giving 800 + 95 + p − 480 − 500 = 70. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
Two cells hold products, $120 × 4 and $400 × 2, that must be worked out, and p sits in the Income column beside the sales, so it is added. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
The known entries total 895 − 980 = −85, a negative running sum, so p must exceed 70 by 85 rather than fall short of it. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work out the two products in the table, add the income entries and subtract the expenses, set that total plus p equal to 70, and solve for p.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single mistake gives −275; it takes counting the $95 clothing sale as an expense and then stopping at the known total, 800 − 95 − 480 − 500.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Totals the known entries correctly, 895 − 980 = −85, and gives that, stopping before solving for the p that brings the sum to $70.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Picks $70, the stated sum of all the transactions, as the value of the bike parts, without setting up an equation.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of all transactions: $\text{sum} = \text{income} - \text{expenses}$; A transaction is money coming in or going out. Income is money that comes in, so it counts as plus. Expenses are money that goes out, so they count as minus. The sum is what is left, like money added to and taken from a wallet.
Keeping an equation balanced: $\text{if } a = b \text{, then } a + c = b + c \text{, } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

Sort out the question
The table lists money that came into the shop, called income. It also lists money that went out, called expenses. The bike parts brought in p dollars, an amount you don’t know. All the income minus all the expenses is $70. You need p.
Work out the products in the table
Two cells in the table show multiplication.
$120 \times 4 = 480$
$400 \times 2 = 800$
So paying the employees cost $480. The 2 bicycles brought in $800.
Write the equation
The income is $800, $95 and p. The expenses are $480 and $500. p sits in the Income column, so it is added.
$\text{income} - \text{expenses} = 70$
$800 + 95 + p - 480 - 500 = 70$
Combine the known numbers
$800 + 95 = 895 \qquad 480 + 500 = 980$
$895 - 980 = -85$
The known entries come to −85 dollars. That is a loss of $85. Now the equation is shorter:
$p - 85 = 70$
Watch out: −85 is choice B, but it is not p. You still have to solve for p.
Solve for p
Add 85 to both sides. This keeps the equation balanced.
$p - 85 + 85 = 70 + 85$
$p = 155$
The bike parts brought in $155. Watch out: $70 is the total for the day, not the bike parts. Don’t just copy it, choice C. So the answer is D, $155.

Check your answer

Put 155 back in. Income: $800 + 95 + 155 = 1{,}050$ dollars. Expenses: $480 + 500 = 980$ dollars. $1{,}050 - 980 = 70$ . That matches. And p should be more than 70, because the bike parts had to cover the $85 loss first.

More questions like this

The original DOE question

Question 85 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 86, question 85. Open page 86 of the handbook (PDF)

Question 83 of 120

2022 Test A Math, question 35

Question

$A graph titled INFORMATION ENTERED, with Time (hr) on the x-axis from 0 to 2, labeled at 0.5, 1, 1.5 and 2, with a grid line every 0.25 hour, and Keystrokes on the y-axis from 0 to 12,000, labeled and with a grid line every 1,000. The origin is labeled O. A straight line starts at the origin and rises to the right, passing through (0.5, 3,500), (1, 7,000) and (1.5, 10,500), and ends in an arrow just past 12,000.$

A clerk enters information on a computer. The graph shows the relationship between time, in hours, and the number of keystrokes the clerk makes. What is the meaning of the point (1.5, 10500) on the graph?

AThe clerk makes 10,500 keystrokes in each 1 hr of entering information.
BThe clerk makes 10,500 keystrokes in each 1.5 hr of entering information.
CThe clerk makes 15,750 keystrokes in each 1 hr of entering information.
DThe clerk makes 15,750 keystrokes in each 1.5 hr of entering information.

Show the answer

Correct answer: B

The x-axis represents the time, in hours, the clerk enters information into the computer, and the y-axis represents the number of keystrokes the clerk makes.

The x-value in (1.5, 10500) gives 1.5 hours as the amount of time the clerk enters information.

The y-value gives 10,500 as the number of keystrokes made by the clerk.

Since the graph is a proportional line, the clerk’s rate is 10,500 keystrokes in each 1.5 hours of entering information.

Question analysis

A straight-line graph through the origin shows how many keystrokes a clerk makes over time, in hours. Students choose the statement that explains what the point (1.5, 10500) means in this situation. Each choice pairs a number of keystrokes with a span of time, so reading both coordinates and their units correctly decides it.

NYS standards: 7.RP.2d points on proportional graphs
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Explaining what a point on the graph of a proportional relationship means in its situation is a named grade 7 skill. This is a short, direct example with a clear graph, though the four statements look alike and must be read closely.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Reading a figure, table or graph moderate
A very close wrong answer minor
Solution: 1 step
Reading level: Question and choices: about grade 6, 74 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Nothing needs computing; the question checks that a point on a proportional graph is read as a pair, a time and a count together. Students trained to look for a unit rate expect every statement about a graph to be per hour, and the two statements with 15,750 come from forcing that idea with a product. The graph itself shows 7,000 keystrokes at 1 hour, a figure no statement offers, so every per-hour statement fails.

Reading a figure, table or graph moderate Understanding the question
The stem gives the point (1.5, 10500), and the axis labels must say what each number is: 1.5 hours of time along x and 10,500 keystrokes up y. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer minor Choosing the answer
The two statements with 10,500 keystrokes differ only in the time, each 1 hr or each 1.5 hr, so the x-coordinate alone separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Use the axis labels to say in words what each coordinate of the point stands for, the x-value in hours and the y-value in keystrokes, then find the statement that says the same.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reads 10,500 from the point but assumes any point on a rate graph gives the amount for 1 hour, ignoring the x-coordinate, 1.5.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies 10,500 by 1.5 to get 15,750 and calls it an hourly rate, though the hourly rate is 10,500 ÷ 1.5 = 7,000.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the two coordinates, 10,500 × 1.5 = 15,750, and pairs that product with the 1.5 hours from the point, combining numbers without a model.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A point on a graph: $(x, y)$; A point is written as a pair in parentheses. x, the first number, tells how far across, along the bottom axis. y, the second number, tells how far up, along the side axis. The axis labels tell what each number measures.
Unit rate from a point: $\text{unit rate} = \frac{y}{x}$; Here y is the number of keystrokes and x is the time in hours. Dividing gives the keystrokes for 1 hour. On a straight line that starts at (0, 0), every point gives the same unit rate.

Sort out the question
The graph shows how many keystrokes a clerk makes over time. A keystroke is one press of a key. You need to say what the point (1.5, 10500) means in words. Each choice pairs a number of keystrokes with an amount of time. The axis labels tell you what each number in the point stands for.
Read the x-value
The first number in a point is the x-value. The bottom axis is labeled Time (hr). So x = 1.5 means 1.5 hours of work. On the graph, 1.5 is halfway between 1 and 2.
Read the y-value
The second number is the y-value. The side axis is labeled Keystrokes. So y = 10,500 means 10,500 keystrokes. Put the two together. In 1.5 hours, the clerk makes 10,500 keystrokes.
Use the straight line
The line is straight and starts at O, the point (0, 0). That means the clerk types at a steady speed. So every 1.5 hours of work brings 10,500 keystrokes. Watch out: the point does not give a count for 1 hour. Its x-value is 1.5, not 1. Choice A treats 10,500 as the count for 1 hour.
Rule out the 15,750 choices
15,750 is $10{,}500 \times 1.5$ . Multiplying the two numbers in a point means nothing here. The count for 1 hour comes from dividing:
$\text{unit rate} = \frac{10{,}500}{1.5} = 7{,}000$
That is not 15,750, so C is wrong. D is wrong too, since the graph shows 10,500 at 1.5 hours. So the answer is B: 10,500 keystrokes in each 1.5 hr.

Check your answer

Look at the graph at 1 hour. The line crosses 7,000 keystrokes there. At 1.5 hours it is at 10,500. Since $7{,}000 \times 1.5 = 10{,}500$ , the line fits choice B. No choice gives 7,000 for 1 hour, so every per-hour choice is wrong.

More questions like this

The original DOE question

Question 92 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 89, question 92. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 84 of 120

2022 Test A Math, question 36

Question

When the three-digit whole number 27 ? is divided by 3 or by 4, the remainder is zero. What is the missing digit?

Show the answer

Correct answer: D

Each of the options 0, 2, 3, and 6 must be considered:

Option A: In the case of 0, the three-digit whole number is 270. Divide 270 by both 3 and 4:

$270 \div 3 = 90 \text{ (or } 90R0\text{)}$

$270 \div 4 = 67\frac{1}{2} \text{ (or } 67R2\text{)}$

Dividing 270 by 3 results in a remainder of 0, but dividing 270 by 4 results in a remainder of 2, so 0 cannot be the missing digit.

Option B: In the case of 2, the three-digit whole number is 272. Divide 272 by both 3 and 4:

$272 \div 3 = 90\frac{2}{3} \text{ (or } 90R2\text{)}$

$272 \div 4 = 68 \text{ (or } 68R0\text{)}$

Dividing 272 by 4 results in a remainder of 0, but dividing 272 by 3 results in a remainder of 2, so 2 cannot be the missing digit.

Option C: In the case of 3, the three-digit whole number is 273. Divide 273 by both 3 and 4:

$273 \div 3 = 91 \text{ (or } 91R0\text{)}$

$273 \div 4 = 68\frac{1}{4} \text{ (or } 68R1\text{)}$

Dividing 273 by 3 results in a remainder of 0, but dividing 273 by 4 results in a remainder of 1, so 3 cannot be the missing digit.

Option D: In the case of 6, the three-digit whole number is 276. Divide 276 by both 3 and 4:

$276 \div 3 = 92 \text{ (or } 92R0\text{)}$

$276 \div 4 = 69 \text{ (or } 69R0\text{)}$

Dividing 276 by 3 results in a remainder of 0, and dividing 276 by 4 results in a remainder of 0, so 6 is the missing digit.

Question analysis

A three-digit number begins with 27 and is missing its last digit, and it divides evenly by both 3 and 4. Students find the missing digit. It tests factors, multiples and divisibility rules; a digit that works for one divisor may fail the other, so both tests must be applied.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Divisibility by 3 and by 4 is grade 6 number sense. Testing four possible digits against two conditions is quick, much like a grade 6 exercise, as long as both divisors are actually checked.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Every choice must be checked minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every candidate passes at least one of the two tests, which makes the question a matter of finishing the checks rather than finding a method. Students who test only by 3, or only by 4, find a fitting number and stop. Seeing that a number divisible by both 3 and 4 is divisible by 12, and that 276 = 12 × 23, settles it in one step.

Turning words into math minor Understanding the question
The words “divided by 3 or by 4, the remainder is zero” mean that both divisions come out even, though or can be read as either one being enough. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Every choice must be checked minor Choosing the answer
Each candidate, 270, 272, 273 and 276, has to be tested against two divisors, and three of them pass one test, so stopping after one check is costly. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
The divisibility tests are not given: a number is divisible by 3 when its digit sum is, and by 4 when the number formed by its last two digits is. (Anderson & Krathwohl, 2001)

Strategy: Use the divisibility tests: a number is divisible by 4 when its last two digits are, and by 3 when its digit sum is. Keep only a digit that passes both.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Checks only division by 3, which 270 passes with a digit sum of 9, and never tests 4, though 270 ÷ 4 leaves a remainder of 2.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Checks only division by 4, which 272 passes because 72 is a multiple of 4, but its digit sum is 11, so 272 ÷ 3 leaves 2.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the missing digit to the divisor 3 in the question; 273 is a multiple of 3 but leaves a remainder of 1 when divided by 4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Test for dividing by 3: $\text{digit sum divides by 3} \rightarrow \text{number divides by 3}$; The digit sum is what you get when you add up a number’s digits. Divides by 3 means it divides evenly, with a remainder of 0. If the digit sum divides evenly by 3, so does the whole number. Use it to test a big number quickly.
Test for dividing by 4: $\text{last two digits divide by 4} \rightarrow \text{number divides by 4}$; Look only at the number made by the last two digits. If it divides evenly by 4, so does the whole number. This works because every 100 divides evenly by 4, so only the last two digits matter.

Sort out the question
The number is 27 followed by a missing digit. Remainder zero means the division comes out even, with nothing left over. The remainder is zero when the number is “divided by 3 or by 4.” That means both divisions come out even. So the number must pass both tests. The choices give four digits to try: 0, 2, 3 and 6.
Test each number for 3
Add the digits of each number.
Number Digit sum Divides by 3?
270 $2 + 7 + 0 = 9$ yes
272 $2 + 7 + 2 = 11$ no
273 $2 + 7 + 3 = 12$ yes
276 $2 + 7 + 6 = 15$ yes
So 270, 273 and 276 pass. 272 fails.
Test each number for 4
Look at the last two digits: 70, 72, 73 and 76.
$70 \div 4 = 17 \text{ remainder } 2 \qquad 72 \div 4 = 18$
$73 \div 4 = 18 \text{ remainder } 1 \qquad 76 \div 4 = 19$
So 272 and 276 pass. 270 and 273 fail.
Keep the digit that passes both
Only 276 passes both tests. Check by dividing:
$276 \div 3 = 92 \qquad 276 \div 4 = 69$
Watch out: 0 and 3 pass only the test for 3, so A and C fail. 2 passes only the test for 4, so B fails. And 3 is not right just because the question says 3. So the answer is D, 6.

Number	Digit sum	Divides by 3?
270	$2 + 7 + 0 = 9$	yes
272	$2 + 7 + 2 = 11$	no
273	$2 + 7 + 3 = 12$	yes
276	$2 + 7 + 6 = 15$	yes

Check your answer

A number that divides evenly by both 3 and 4 also divides evenly by 12, since $3 \times 4 = 12$ . And $12 \times 23 = 276$ . So 276 is a multiple of 12. None of 270, 272 or 273 is.

More questions like this

The original DOE question

Question 93 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 89, question 93. Open page 89 of the handbook (PDF)

Question 85 of 120

2022 Test A Math, question 44

Question

Report	Subject	Topic Choices
#1	Geography	Kenya, Thailand, Chile, Greece, New Zealand
#2	English	R. Frost, L. Hughes, N. Giovanni
#3	Science	1997 Mars mission
#4	Math	π, regular polyhedrons

Anita is assigned to report on 1 topic in each of the 4 subjects shown in the table above. How many different sets of 4 reports are possible, according to the choices available to her?

Show the answer

Correct answer: C

One report is assigned for each of 4 subjects, and each subject has a specified number of topic choices:

5 choices for Geography
3 choices for English
1 choice for Science
2 choices for Math

The basic counting principle (product rule) applies since the number of choices for each subject is independent of the choices for the other subjects. The number of different sets of 4 reports possible is the product.

$5 \times 3 \times 1 \times 2 = 30$

Question analysis

A table lists the topics Anita may choose from for reports in geography, English, science and math. She reports on one topic in each subject, and students count how many different sets of four reports are possible. It tests the counting principle; reading how many topics each row offers, including a row with only one, comes first.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Counting the outcomes of several independent choices is grade 7 work, usually done with a tree diagram or a list. Four subjects make a full tree impractical, so a counting shortcut is needed, but each count is small and read from the table.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 52 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the four counts are read from the table, the question is a single product, so the real test is recognizing that independent choices multiply. Students who add the counts get 11, which is not offered, a useful signal that the method is wrong. Science's single topic multiplies by 1 and changes nothing, but it still has to be read as one choice, not as none.

Something you must already know moderate Working it out
The counting principle is not stated: choosing one topic from each subject means multiplying the numbers of choices, 5 × 3 × 1 × 2, not adding them. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The topic lists must be counted by their commas, and two rows mislead: Science offers a single topic, and Math's “π, regular polyhedrons” is two topics, not one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Count the topic choices in each row of the table, reading the commas carefully, then apply the counting principle, since Anita picks one topic from every subject.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Adds the topic counts instead of multiplying and leaves out Science's single topic as no real choice, 5 + 3 + 2 = 10.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Counts the orders in which four reports could be written, 4 × 3 × 2 × 1 = 24, instead of the ways to choose one topic per subject.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 4 subjects by the 11 topics listed in all, 4 × 11 = 44, combining numbers without a model of the choices.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle: $\text{number of ways} = a \times b \times c \times d$; a, b, c and d are the numbers of choices for each separate pick. When you make one pick from each group, and the groups don’t affect each other, multiply the numbers of choices. Each choice in one group can go with every choice in the others.

Sort out the question
Anita writes 4 reports, one for each subject. For each subject she picks 1 topic from that row of the table. A set of 4 reports is one full pick: one topic for each subject. You need how many different sets are possible.
Count the topics in each row
Use the commas to count the topics.
- Geography: Kenya, Thailand, Chile, Greece, New Zealand. That is 5.
- English: Frost, Hughes, Giovanni. That is 3.
- Science: only the 1997 Mars mission. That is 1.
- Math: π and regular polyhedrons. That is 2.
New Zealand is two words but one country, so it is one topic.
Multiply the choices
Use the counting principle. Each of the 5 geography topics can go with each of the 3 English topics, and so on.
$5 \times 3 \times 1 \times 2 = 30$
Science has just one topic, so every set has it. Multiplying by 1 changes nothing. Watch out: don’t add the counts. Adding them without Science, $5 + 3 + 2 = 10$ , gives choice A.
Rule out the other choices
$4 \times 3 \times 2 \times 1 = 24$ counts the orders for writing 4 reports. The question does not ask about order, so 24, choice B, is wrong. $4 \times 11 = 44$ multiplies the subjects by all the topics. That is choice D, and it has no meaning here. So the answer is C, 30.

Check your answer

Build it up. Geography and English alone give $5 \times 3 = 15$ pairs, such as Kenya with Frost. Science adds only one topic, so there are still 15. Math doubles it, π or polyhedrons, so $15 \times 2 = 30$ sets.

More questions like this

The original DOE question

Question 101 as printed on page 92 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 92, question 101. Open page 92 of the handbook (PDF)

Question 86 of 120

2022 Test A Math, question 50

Question

Nadia paid $1.00 to mail a letter. The cost to mail a letter is $0.55 for the first ounce of weight plus $0.15 for each additional ounce. What was the weight, in ounces, of the letter Nadia mailed?

A1.4
B1.5
C3
D4

Show the answer

Correct answer: D

Let x represent the number of ounces Nadia’s letter weighed. The first ounce costs $0.55, so $(x - 1)$ represents the number of ounces that cost $0.15 each.

Given that the total cost is $1.00, the equation that describes the situation can be written and solved as shown.

$0.55 + 0.15(x - 1) = 1.00$

$0.55 + 0.15x - 0.15 = 1.00$

$0.15x = 1.00 - 0.55 + 0.15$

$0.15x = 0.60$

$\frac{0.15x}{0.15} = \frac{0.60}{0.15}$

$x = 4$

So the letter Nadia mailed weighed 4 ounces.

Question analysis

Nadia pays $1.00 to mail a letter, and postage costs $0.55 for the first ounce plus $0.15 for each additional ounce. Students find the weight of the letter in ounces. It tests writing and solving a two-step equation from a pricing rule in which the two prices apply to different ounces.

NYS standards: 7.EE.4a two-step equations
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve word problems that lead to equations such as px + q = r. This one takes that form once each price is matched to the right ounces, a step that makes it a little more demanding than routine practice.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Pricing the letter ounce by ounce, $0.55, $0.70, $0.85, $1.00, reaches the total at the fourth ounce without any algebra. The equation route works too, but the first ounce is easy to lose: dividing the $0.45 left by $0.15 counts only the ounces after the first. The two decimal choices come from merging the two prices or from treating dollars as ounces.

Turning words into math moderate Understanding the question
“$0.55 for the first ounce of weight plus $0.15 for each additional ounce” makes the cost 0.55 + 0.15(x − 1), so the $0.15 covers every ounce except the first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
The money is in decimals, $1.00 − $0.55 = $0.45 and then $0.45 ÷ $0.15 = 3, and dividing by a decimal invites a misplaced point. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the cost of a letter weighing x ounces, $0.55 plus $0.15 for each ounce after the first, set it equal to $1.00 and solve, then check by pricing the letter ounce by ounce.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Merges the two prices into $0.70 for every ounce and divides $1.00 by $0.70, about 1.43, which rounds to 1.4.
B Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Counts the first ounce, then treats the $0.45 left over as 0.45 ounce, getting 1.45 ounces, which rounds to 1.5.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the 3 additional ounces that the $0.45 left over pays for, 0.45 ÷ 0.15 = 3, and stops before adding the first ounce.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost to mail a letter: $C = 0.55 + 0.15 \times (w - 1)$; C is the cost in dollars. w is the weight in ounces. The first ounce costs $0.55. Each ounce after the first costs $0.15, and there are w − 1 of them. Subtracting 1 leaves out the first ounce, which is already paid for.
Keeping an equation balanced: $\text{if } a = b \text{, then } a + c = b + c \text{, } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

Sort out the question
Nadia paid $1.00 to mail a letter. The first ounce costs $0.55. Each additional ounce, meaning each ounce after the first, costs $0.15. You need the weight of the letter in ounces.
Write the equation
Let w be the weight in ounces. The first ounce is 1 of the w ounces. So the extra ounces number w − 1. The cost is $1.00.
$0.55 + 0.15 \times (w - 1) = 1.00$
Take off the first ounce’s cost
Subtract 0.55 from both sides. This keeps the equation balanced.
$0.15 \times (w - 1) = 1.00 - 0.55 = 0.45$
So $0.45 pays for the extra ounces.
Find the extra ounces
Divide both sides by 0.15. Multiply both numbers by 100 to make them whole.
$w - 1 = 0.45 \div 0.15 = 45 \div 15 = 3$
There are 3 extra ounces after the first one.
Add back the first ounce
$w = 3 + 1 = 4$
The letter weighed 4 ounces. Watch out: 3 counts only the ounces after the first. It is choice C. Also, don’t join $0.55 and $0.15 into one price. $1.00 \div 0.70$ is about 1.4, choice A. So the answer is D, 4 ounces.

Check your answer

Price the letter one ounce at a time. 1 ounce: $0.55. 2 ounces: $0.70. 3 ounces: $0.85. 4 ounces: $1.00. The cost reaches $1.00 at 4 ounces. That matches.

More questions like this

The original DOE question

Question 107 as printed on page 95 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 95, question 107. Open page 95 of the handbook (PDF)

Question 87 of 120

2022 Test A Math, question 51

Question

COST OF ORANGES

Number of Oranges	Cost
4	$1.80
6	$2.70
9	$4.05

The table shows the cost of oranges. What is the unit rate for the cost per orange?

A$0.45
B$0.90
C$1.35
D$1.80

Show the answer

Correct answer: A

The unit rate (cost per orange) can be determined by dividing the cost of x oranges by x.

$\frac{1.80}{4} = \frac{2.70}{6} = \frac{4.05}{9} = 0.45$

The unit rate is $0.45.

Question analysis

A table shows what 4, 6 and 9 oranges cost, in dollars and cents. Students find the unit rate, the cost of a single orange. It tests identifying a unit rate in a proportional table, and the tricky part is keeping straight which column holds the cost and which holds the number of oranges.

NYS standards: 7.RP.2b constant of proportionality
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a unit rate in a table belongs to grade 7 proportional relationships, but here it takes one division with money in a clearly proportional table, so the task is easier than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 3, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Any row gives the rate in one division, and the three rows agree, so the question is easy once unit rate is read as the cost of one orange. The traps are differences between rows, which look like a change per orange but cover two or three oranges, and the first cost in the table, which is for four. Dividing a second row confirms the result.

Reading a figure, table or graph minor Understanding the question
The table lists 4, 6 and 9 oranges but never a single orange, so the cost of one has to be computed from a row, such as $1.80 ÷ 4. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The term unit rate is not defined; students must know it means the cost of one orange, a row's cost divided by its number of oranges. (Anderson & Krathwohl, 2001)

Strategy: Pick any row and divide its cost by its number of oranges; doing the same with a second row confirms that the rate stays the same.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Subtracts the first two costs, $2.70 − $1.80 = $0.90, the price of 2 more oranges, without dividing by the 2 oranges it covers.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Subtracts the last two costs, $4.05 − $2.70 = $1.35, the price of 3 more oranges, rather than the price of one.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes $1.80, the first cost in the table, which is what 4 oranges cost together, not one.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{unit rate} = \frac{\text{cost}}{\text{number of oranges}}$; The unit rate is the cost of one item, here one orange. The cost is what a group of oranges costs together. Divide by the number of oranges in that group. Any row of the table gives the same rate, because each orange costs the same.

Sort out the question
The table gives the cost of 4, 6 and 9 oranges. It never shows the cost of one orange. A unit rate is the amount for one unit. Here that means the cost of one orange. You need to find it.
Divide one row
Use the first row: 4 oranges cost $1.80.
$\text{unit rate} = \frac{\text{cost}}{\text{number of oranges}}$
$\text{unit rate} = \frac{1.80}{4} = 0.45$
One orange costs $0.45. Watch out: $1.80 is the cost of 4 oranges, not one. Don’t pick choice D.
Check with the other rows
$\frac{2.70}{6} = 0.45 \qquad \frac{4.05}{9} = 0.45$
Every row gives $0.45, so the rate stays the same. Watch out: $2.70 - 1.80 = 0.90$ is the cost of 2 more oranges, choice B. $4.05 - 2.70 = 1.35$ is the cost of 3 more, choice C. Neither is for one orange. So the answer is A, $0.45.

Check your answer

Multiply back: $0.45 \times 4 = 1.80$ and $0.45 \times 9 = 4.05$ . Both match the table. Even a difference works once you divide: $0.90 \div 2 = 0.45$ .

More questions like this

The original DOE question

Question 108 as printed on page 95 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 95, question 108. Open page 95 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 88 of 120

2022 Test B Math, question 4

Question

The total cost of a book is given by the expression $1.08x$ , where x is the price of the book before tax. The tax rate is m%. What is the value of m?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 8

The price of a book before tax is x and after tax is $1.08x$ . Subtract to determine the tax rate.

$1.08x - x = 0.08x$ , so the tax rate is 0.08.

The tax rate m% can be written as $\frac{m}{100}$ .

Solve for m:

$\frac{m}{100} = 0.08$

$100\left(\frac{m}{100}\right) = 100(0.08)$

$m = 8$

Question analysis

The total cost of a book, tax included, is given by the expression 1.08x, where x is the price before tax, and the tax rate is m%. Students find the value of m and type it in. It tests reading a percent increase from an algebraic expression, and the rate has to be entered as a number of percent, not as a decimal.

NYS standards: 7.EE.2 rewriting expressions to show relationships
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Reading a percent increase from an expression such as 1.08x is a named grade 7 idea. It takes one short step, but students must turn a decimal into a percent and type the value with no choices to check.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: An answer in an unexpected form moderate
Something you must already know minor
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The expression already holds the answer, since 1.08x is the price, 1x, plus a tax of 0.08x. The work lies in pulling the tax apart from the price and then writing it in the form the stem asks for, the number in front of a percent sign. Students who stop at the decimal, or who read the whole 1.08 as 108 percent, grid a value that looks reasonable. Trying a price of $100, which costs $108 with tax, makes the 8 visible.

An answer in an unexpected form moderate Choosing the answer
The rate is asked for as the m in m%, so the 0.08 found from the expression must become 8, and the grid takes 0.08 or 108 just as readily. (Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Nothing in the stem says that multiplying a price by 1.08 adds 8% of it; students must split 1.08x into 1x for the price and 0.08x for the tax. (Anderson & Krathwohl, 2001)
Variables instead of numbers minor Understanding the question
The price is the letter x, never a dollar amount, so the tax has to be read from the coefficient 1.08 rather than worked out from a number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Split 1.08x into the price before tax plus the tax added to it, then write the tax part's coefficient as a percent.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

0.08 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the tax as 0.08x and enters the decimal rate 0.08, one step short of multiplying by 100 to give m as a number of percent.
108 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Reads 1.08 as 108% and enters it, giving the whole cost as a percent of the price instead of only the tax added to it.
1.08 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Copies the coefficient 1.08 from the expression as the rate, without separating the original price, 1x, from the tax added on top.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction: $m\% = \frac{m}{100}$; Percent means out of 100. So m% is m hundredths, the fraction m over 100. To change a decimal to a percent, multiply it by 100. To change a percent to a decimal, divide it by 100.

Sort out the question
x is the price of the book before tax. The total cost with tax is 1.08x, which means 1.08 times x. The tax rate is m%. You need the number m.
The tax rate is the percent of the price that is added on as tax.
Split the total into price and tax
The total is the price plus the tax. Split 1.08 into 1 and 0.08.
$1.08x = 1x + 0.08x = x + 0.08x$
The x part is the price you started with. So the extra 0.08x is the tax. The tax is 0.08 times the price.
Change the decimal to a percent
The rate is 0.08 as a decimal. Use the formula and set it equal to 0.08.
$\frac{m}{100} = 0.08$
To get m alone, multiply both sides by 100.
$m = 0.08 \times 100 = 8$
So the tax rate is 8%.
Give the answer
The question asks for m, the number in front of the % sign. That is 8.
Watch out: 0.08 is the rate as a decimal, one step short. And 1.08, or 108%, counts the price as well as the tax. So you would grid 8.

Check your answer

Try a price of $100. The total is $1.08 \times 100 = 108$ dollars. The tax is 108 − 100 = 8 dollars. And $8 out of $100 is 8%. So m = 8.

More questions like this

The original DOE question

Question 61 as printed on page 198 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 198, question 61. Open page 198 of the handbook (PDF)

Question 89 of 120

2022 Test B Math, question 30

Question

A robot moves forward at the rate of 9 steps every 6 seconds. If each step is x feet long, what must x be in order for the robot to travel exactly 270 feet per minute?

Show the answer

Correct answer: A

First, convert steps per second to steps per minute.

$\frac{9\text{ steps}}{6\text{ seconds}}=\frac{90\text{ steps}}{60\text{ seconds}}=\frac{90\text{ steps}}{1\text{ minute}}$

Then, use ratio reasoning to determine that each step must be 3 feet since 90 times 3 is 270 (or 270 divided by 90 is 3).

$\frac{90\text{ steps}}{1\text{ minute}}\times\frac{3\text{ feet}}{1\text{ step}}=\frac{270\text{ feet}}{1\text{ minute}}$

$\frac{270\text{ feet}}{1\text{ minute}}\div\frac{90\text{ steps}}{1\text{ minute}}=$

$\frac{270\text{ feet}}{1\text{ minute}}\times\frac{1\text{ minute}}{90\text{ steps}}=$

$\frac{3\text{ feet}}{1\text{ step}}$

Question analysis

A robot moves forward 9 steps every 6 seconds, and each step is x feet long. Students find the step length that makes the robot travel exactly 270 feet per minute. It tests rates and unit conversion: the rate is given per 6 seconds but the target is per minute, so the units must be matched before solving.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3d converting measurement units
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 rate problems often give one rate and ask for a total. This one links steps per time, feet per step and feet per minute, with seconds to convert, a typical but full multistep ratio problem.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are friendly, but the rate is given per 6 seconds and the target per minute, so the units have to meet first. A minute holds ten 6-second stretches, so the robot takes 90 steps in it, and 270 feet over 90 steps settles the length. Students who divide 270 by the 9 steps forget that those steps take only 6 seconds, and those who scale 270 down to 6 seconds stop at 27 feet, a distance rather than a step length.

Much to keep track of moderate Working it out
Three rates are in play, 9 steps per 6 seconds, x feet per step and 270 feet per minute, and their units must be lined up before any two are combined. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Many steps minor Working it out
The rate is first scaled from 6 seconds to 60, giving 90 steps a minute, and only then compared with the 270 feet covered in that minute. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Change the rate to steps per minute first, since 6 seconds goes into a minute evenly, then ask what step length makes that many steps cover 270 feet.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 9 steps by the 6 seconds, 54, and divides 270 by it, combining the stem’s numbers without a model of the rate.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Scales 270 feet per minute down to 27 feet per 6 seconds but stops before dividing by the 9 steps taken in that time.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides 270 feet by 9 steps, as if the robot took 9 steps a minute, never converting the 6-second rate to a minute.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rate: $\text{rate} = \frac{\text{amount}}{\text{time}}$; A rate compares an amount with the time it takes. Here the robot’s rate is 9 steps for every 6 seconds. To compare two rates, first write them for the same length of time, such as one minute, which is 60 seconds.
Distance from steps: $\text{distance} = \text{number of steps} \times \text{length of each step}$; Every step covers the same length. So the distance is the number of steps times the length of one step. Here the distance and the step length are both in feet.

Sort out the question
The robot takes 9 steps every 6 seconds. Each step is x feet long. You need the x that makes the robot go exactly 270 feet per minute.
The two rates use different times: 6 seconds and 1 minute. Match the times before you compare them.
Find the steps in one minute
One minute is 60 seconds. Find how many 6-second stretches fit in a minute.
$60 \div 6 = 10$
In each stretch, the robot takes 9 steps. So in one minute it takes 10 times as many.
$9 \times 10 = 90$
The robot takes 90 steps per minute.
Find the length of each step
In one minute, those 90 steps must cover 270 feet. Use the distance formula.
$\text{distance} = \text{number of steps} \times \text{length of each step}$
$270 = 90 \times x$
To get x alone, divide both sides by 90.
$x = 270 \div 90 = 3$
Each step must be 3 feet long.
Give the answer
Watch out: dividing 270 by 9 gives 30, choice D. That treats 9 steps as a whole minute’s steps, but they take only 6 seconds. Cutting 270 feet per minute down to 27 feet per 6 seconds and stopping gives 27, choice C. That is a distance, not a step length. So the answer is A, 3.

Check your answer

Check the units. 90 steps per minute times 3 feet per step gives $90 \times 3 = 270$ feet per minute. That is the goal. Or think in 6 seconds: 9 steps of 3 feet is 27 feet, and 10 of those stretches make 270 feet.

More questions like this

The original DOE question

Question 87 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 205, question 87. Open page 205 of the handbook (PDF)

Question 90 of 120

2022 Test B Math, question 35

Question

READING ASSIGNMENTS

Assignment	Pages Assigned
I	3–17, inclusive
II	25–38, inclusive
III	45–60, inclusive

Over a two-week period, students are assigned a total of 250 pages to read from a science textbook. The table above shows the first three reading assignments within the larger assignment. What percentage of the 250 pages is the total number of pages in these first three assignments?

A16.8%
B18.0%
C22.8%
D24.0%

Show the answer

Correct answer: B

First determine the pages read for each assignment. Since the pages per assignment are inclusive, subtract the first page from the last page assigned and then add 1 to determine the pages read for each assignment.

Assignment I: $(17-3)+1=15$

Assignment II: $(38-25)+1=14$

Assignment III: $(60-45)+1=16$

The student was assigned a total of 45 pages, since $15+14+16=45$ .

To determine the percent, express 45 pages out of 250 pages as a fraction.

$\frac{45}{250}=\frac{45\div 5}{250\div 5}=\frac{9}{50}$

$\frac{9}{50}=\frac{18}{100}=18\%$

Question analysis

A table lists the first three reading assignments from a science textbook as page ranges marked inclusive, out of 250 pages assigned over two weeks. Students find what percentage of the 250 pages the three assignments make up in total. It tests percents, but the harder part comes first: counting the pages in a range when both end pages are included.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Expressing one number as a percent of another is familiar from grades 6 and 7. Here that number must first be counted from three inclusive page ranges, a step most grade 7 percent problems leave out.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Many steps moderate
Reading a figure, table or graph minor
Solution: 5 steps
Reading level: Question and choices: about grade 6, 56 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Counting pages in a range is the real test; the percent at the end is routine once the three counts are right. Subtracting first page from last feels natural, and the 42 pages it gives lead straight to a listed choice. The table also invites shortcuts, the span from page 3 to page 60 or the last page alone, and each lands on a choice too. Checking the rule on a tiny range, pages 1 to 3, which hold 3 pages and not 2, fixes the count.

Something you must already know moderate Working it out
Each row says inclusive, but nothing says that pages 3 to 17 hold 17 − 3 + 1 = 15 pages; subtracting alone loses one page from each of the three assignments. (Anderson & Krathwohl, 2001)
Many steps moderate Working it out
Three page counts, their sum, a fraction of the 250 pages and a change to a percent come in sequence, and a miscount in the first step carries through to the choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph minor Understanding the question
The table’s ranges leave gaps, pages 18 to 24 and 39 to 44, so the first and last pages listed, 3 and 60, do not bound the pages actually assigned. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Work out how many pages each range holds when both end pages count, total the three, then write that total as a fraction of 250 and change it to a percent.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Subtracts the first page from the last in each row, 14 + 13 + 15 = 42 pages, leaving out one end of each inclusive range, and 42/250 = 16.8%.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Treats pages 3 to 60 as one stretch, 60 − 3 = 57, counting the unassigned pages between assignments, and 57/250 = 22.8%.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the last assigned page, 60, as the number of pages read, as if the assignments covered pages 1 to 60, and 60/250 = 24%.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting pages in an inclusive range: $\text{pages} = \text{last page} - \text{first page} + 1$; Inclusive means the first and last pages are both read. Subtracting counts the steps from the first page to the last. That misses one page, so add 1. Test it: pages 1 to 3 are 3 pages, and 3 − 1 + 1 = 3.
Percent of a total: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the amount you have, here the pages in the three assignments. The whole is the amount you compare with, here 250 pages. Dividing gives a fraction or decimal. Multiplying by 100 turns it into a percent, a number out of 100.

Sort out the question
The whole assignment is 250 pages. The table gives three page ranges:
- I: pages 3 to 17
- II: pages 25 to 38
- III: pages 45 to 60
Each range is inclusive. That means its first and last pages are read too. You need the percent of 250 pages these three make up.
Count the pages in each range
Use the rule: last page minus first page, plus 1.
$17 - 3 + 1 = 15$
$38 - 25 + 1 = 14$
$60 - 45 + 1 = 16$
Watch out: subtracting alone gives 14 + 13 + 15 = 42 pages. That leads to 16.8%, choice A.
Add the three ranges
$15 + 14 + 16 = 45$
The three assignments hold 45 pages in all.
Watch out: don’t count pages 3 to 60 as one stretch. Pages 18 to 24 and 39 to 44 are not assigned. That mistake gives 57 pages and 22.8%, choice C. Using 60 pages gives 24.0%, choice D.
Change to a percent
$\text{percent} = \frac{45}{250} \times 100$
Simplify first. Divide the top and bottom by 5, then multiply them both by 2.
$\frac{45}{250} = \frac{9}{50} = \frac{18}{100}$
That is 18 out of 100, or 18%. So the answer is B, 18.0%.

Check your answer

Work backward. 10% of 250 pages is 25 pages, so 1% is 2.5 pages. Then 18% is $18 \times 2.5 = 45$ pages. That matches the count of 45.

More questions like this

The original DOE question

Question 92 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 206, question 92. Open page 206 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 91 of 120

2020 Test A Math, question 36

Question

CELL PHONE SALES BY COLOR

Color	Percent of Cell Phones Sold
White	$8k$
Black	30
Blue	$30 - 2k$
Red	$k + 5$
Total	100

The table above shows cell phone sales by color. What percent of the cell phones sold were blue?

A18%
B20%
C22%
D28%

Show the answer

Correct answer: B

First, use the information given in the table and properties of equations to determine the value of k.

$8k + 30 + 30 - 2k + k + 5 = 100$

$8k - 2k + k + 30 + 30 + 5 = 100$

$7k + 65 = 100$

$7k + (65 - 65) = 100 - 65$

$7k = 35$

$k = 5$

Then determine the percentage of cell phones sold that were blue by finding the value of the expression $30 - 2k$ (given in the table) when $k = 5$ .

$30 - 2k = 30 - 2(5) = 30 - 10 = 20$

The percentage of cell phones sold that were blue is 20%.

Question analysis

This algebra question gives cell phone sales by color in a table, with three of the percents written in terms of k: 8k for white, 30 − 2k for blue and k + 5 for red, while black is 30 and the total is 100. Students build an equation, find k, and use it to work out the blue percent, taking care with the negative term.

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 classwork usually supplies the equation. Here students must notice that the four color entries make up the whole, write and simplify a sum with a negative term, solve it, and evaluate one expression: familiar steps, a little longer than usual.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 3, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the Total row becomes an equation the algebra is short, so the question mostly tests whether students see that step. Finding k is a halfway point but not a listed choice, which removes the usual early-stop trap. Students who cannot build the equation are left with the blue entry itself, and reading 2k as 2 gives the only wrong answer with a real route; the other two sit two points either side of the key and match no common slip.

Many steps moderate Working it out
Four moves in a row: add the four color entries, set the sum equal to the Total row's 100, solve 7k + 65 = 100 for k, then put k into the blue entry. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph minor Understanding the question
No equation is printed; it comes from seeing that the table's four percent entries, three of them written in k, must add up to the Total of 100. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
Collecting like terms means keeping the blue row's negative term, 8k − 2k + k = 7k, and gathering the constants 30, 30 and 5 into 65. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Since the color percents make up the whole, set the sum of the four entries equal to 100, gather the k terms and the constants, solve for k, and evaluate the blue entry.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Would need k = 6, which makes the four entries total 107 rather than 100; no usual slip in setting up or solving gives that, so it draws guesses.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Would need k = 4, which makes the entries sum to only 93; no common error in combining terms or solving produces it, so it mostly catches guessers.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Skips solving and reads the blue entry 30 − 2k as 30 − 2, treating k as if it were 1, which gives 28.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Combining like terms: $ak + bk = (a + b)k$; k is the unknown number. a and b are the numbers multiplied by k, called coefficients. Terms with the same letter are like terms. You may add their coefficients, since a groups of k plus b groups of k make a + b groups of k.
Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c$; a and b stand for the two sides of an equation. c is any number, but not 0 when you divide. An equation works like a scale in balance. If you do the same thing to both sides, it stays balanced. Use this to get k by itself.

Sort out the table
The table gives the percent of phones sold in each color. Three of the entries use k, a letter that stands for an unknown number. $8k$ means 8 times k.
You need the percent that were blue, $30 - 2k$ . To find it, you first need k.
Write an equation for the total
Every phone sold was one of the four colors. So the four percents must add up to the whole, 100. The Total row says so.
$8k + 30 + (30 - 2k) + (k + 5) = 100$
Combine like terms
Group the k terms together and the plain numbers together. A k on its own means 1k.
$ak + bk = (a + b)k$
$8k - 2k + k = (8 - 2 + 1)k = 7k$
$30 + 30 + 5 = 65$
Now the equation is $7k + 65 = 100$ . Watch out: the blue row takes away 2k. Keep its minus sign, or the k terms come out wrong.
Solve for k
Subtract 65 from both sides. Then divide both sides by 7.
$\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c$
$7k + 65 - 65 = 100 - 65$
$7k = 35$
$k = 35 \div 7 = 5$
So k is 5. That is not the answer yet. You still need the blue row.
Find the blue percent
Put 5 in place of k in the blue row.
$30 - 2k = 30 - 2 \times 5 = 30 - 10 = 20$
So 20% of the phones sold were blue. Watch out: 2k means 2 times k, not just 2. Reading $30 - 2k$ as $30 - 2$ gives 28%, choice D.
So the answer is B, 20%.

Check your answer

Put k = 5 into every row. White: $8 \times 5 = 40$ . Black: 30. Blue: 20. Red: $5 + 5 = 10$ . Now add them: $40 + 30 + 20 + 10 = 100$ . The total matches the table, so k = 5 and 20% are right.

More questions like this

The original DOE question

Question 93 as printed on page 92 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2021 Admissions, page 92, question 93. Open page 92 of the handbook (PDF)

Question 92 of 120

2019 Test A Math, question 18

Question

Joseph is 5 feet 9 inches tall, and Roberto is 6 feet 4.5 inches tall. What is the difference in height, in inches, between Roberto and Joseph?

A4.5
B5.5
C6.5
D7.5

Show the answer

Correct answer: D

First, convert their heights from feet and inches to inches, by multiplying the number of feet by 12 and adding the inches:

Roberto: $6\text{ feet} + 4.5\text{ inches}$

$6(12\text{ inches}) + 4.5\text{ inches} =$

$72\text{ inches} + 4.5\text{ inches} = 76.5\text{ inches}$

Joseph: $5\text{ feet} + 9\text{ inches}$

$5(12\text{ inches}) + 9\text{ inches} =$

$60\text{ inches} + 9\text{ inches} = 69\text{ inches}$

Then, subtract Joseph’s height from Roberto’s height: $76.5 - 69 = 7.5$

Question analysis

Joseph is 5 feet 9 inches tall and Roberto is 6 feet 4.5 inches tall, and students find the difference in their heights in inches. It tests converting mixed feet-and-inches measurements and subtracting with a decimal. Subtracting the feet and the inches separately is risky, because Joseph's 9 inches exceed Roberto's 4.5 and a foot holds 12 inches.

NYS standards: 6.RP.3d converting measurement units
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting feet to inches is grade 6 ratio work, and the subtraction involves one decimal. Grade 7 measurement problems usually chain more steps or use a scale; this one needs two conversions and one subtraction.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 23 words.
Corrected from the DOE handbook: A misspelled name in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The heights mix two units, and the inch parts run the wrong way, 4.5 against 9, so subtracting part by part goes wrong unless a foot is regrouped as 12 inches. Converting both heights to inches first avoids the borrow entirely. The choices sit a single inch apart, so a rough estimate cannot stand in for the exact working, and a student who treats feet as tens of inches lands squarely on a wrong choice.

Error-prone arithmetic moderate Working it out
Subtracting 5 feet 9 inches from 6 feet 4.5 inches needs a borrow, since 4.5 inches is less than 9, and the borrowed foot holds 12 inches, not the 10 that decimal habits suggest. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The stem never states that a foot is 12 inches, yet both heights must be turned into inches, 76.5 and 69, before the difference can be taken. (Anderson & Krathwohl, 2001)

Strategy: Change both heights to inches by multiplying the feet by 12 and adding the extra inches, then subtract the shorter height from the taller one.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Subtracts the two inch figures alone, 9 − 4.5 = 4.5, taking the smaller from the larger and leaving the feet out of the comparison entirely.
B Units or scale (King et al., 2004)
Reveals a misconception Expected pull: strong
Treats feet as tens of inches, reading the heights as 64.5 and 59, or borrowing a foot as 10 inches, so 64.5 − 59 = 5.5.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No conversion or borrow gives 6.5, one inch short of the true gap; only a miscount such as taking Joseph as 70 inches reaches it, so it mostly catches guesses.

What was corrected from the DOE handbook, and why

Explanation: The handbook’s explanation ends “Then, subtract Joseph’s height from Robert’s height: 76.5 − 69 = 7.5” (p. 133, question 75). The taller boy is Roberto, as the question and the explanation’s own “Roberto:” line call him, so the site now reads “Roberto’s height.” Nothing else changes: Roberto is 6(12) + 4.5 = 76.5 inches, Joseph is 5(12) + 9 = 69 inches, and the difference is 7.5 inches.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing feet and inches to inches: $\text{total inches} = 12 \times \text{feet} + \text{extra inches}$; There are 12 inches in 1 foot. Multiply the feet by 12 to turn them into inches. Then add the extra inches. Now the whole height is in one unit, inches, so two heights can be compared by subtracting.

Sort out the question
Joseph is 5 feet 9 inches tall. Roberto is 6 feet 4.5 inches tall. You need the difference in their heights, in inches. The difference is how much taller Roberto is, so you subtract. Each height mixes two units, feet and inches. Change both to inches first. The question doesn’t say it, but 1 foot is 12 inches.
Change Roberto’s height to inches
Six feet is 72 inches. Then add the extra 4.5 inches.
$\text{total inches} = 12 \times \text{feet} + \text{extra inches}$
$\text{total inches} = 12 \times 6 + 4.5 = 72 + 4.5 = 76.5$
Roberto is 76.5 inches tall.
Change Joseph’s height to inches
Do the same for Joseph. Five feet is 60 inches. Then add the extra 9 inches.
$\text{total inches} = 12 \times 5 + 9 = 60 + 9 = 69$
Joseph is 69 inches tall.
Subtract the two heights
Subtract Joseph’s height from Roberto’s height, the shorter from the taller.
$76.5 - 69 = 7.5$
Roberto is 7.5 inches taller.
Watch out: a foot is 12 inches, not 10. Treating it as 10 turns the heights into 64.5 and 59, and $64.5 - 59 = 5.5$ , choice B. Subtracting only the inches, $9 - 4.5 = 4.5$ , leaves out the feet and gives choice A.
So the answer is D, 7.5 inches.

Check your answer

Count up from Joseph’s height. From 5 feet 9 inches, 3 more inches reach 6 feet. From 6 feet, 4.5 more inches reach Roberto’s height. In all, $3 + 4.5 = 7.5$ inches. That matches.

More questions like this

The original DOE question

Question 75 as printed on page 100 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 100, question 75. Open page 100 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 93 of 120

2019 Test A Math, question 21

Question

Anya contributed $1,200 toward the purchase of a $2,000 computer. Her brother contributed $240 toward the same computer. Her parents provided the rest of the money for the computer. What percentage of the total cost of the computer did Anya’s parents pay?

A24%
B28%
C32%
D36%

Show the answer

Correct answer: B

First, calculate the amount of money Anya’s parents contributed:

$2{,}000 - 1{,}200 - 240 = 560$ .

Then, find what percentage of $2,000 is $560:

$\frac{560}{2{,}000} = 0.28 = 28\%$

Question analysis

Anya puts $1,200 and her brother $240 toward a $2,000 computer, and their parents pay the rest. Students find what percent of the cost the parents paid. It is a two-step percent word problem in which the percent describes an amount the stem never states.

NYS standards: 6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Writing a part as a percent of a whole is grade 6 work. Finding the parents' share first is quick subtraction, and a total of $2,000 makes the division easy, so this is lighter than grade 7's multistep percent problems.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two easy steps, a subtraction and a division by $2,000, make up the whole solution, so the risk lies in skipping the first. The parents' share is never stated, and a student hunting the stem for a percent finds the brother's $240, whose digits echo one of the choices. The other two wrong percents match no working with the stated amounts and catch only guesses.

Turning words into math minor Understanding the question
“Her parents provided the rest of the money” hides the amount the question is about: $2,000 − $1,200 − $240 = $560 must be found before any percent. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Subtract both contributions from $2,000 to find the parents' share, then divide that share by $2,000 and change the decimal to a percent.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Turns the brother's $240 into 24%, matching the digits of a number in the stem instead of finding the parents' $560 share of the $2,000.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No subtraction or division of $1,200, $240 and $2,000 gives 32%, which would mean the parents paid $640, so it mostly catches guesses.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Matches no working with the stated amounts: 36% of $2,000 is $720, a share that no combination of the two contributions leaves for the parents.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a total: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the amount you are asking about, here what the parents paid. The whole is the full amount, the cost of the computer. Dividing gives the fraction of the cost. Multiplying by 100 turns that fraction into a percent, a number out of 100.

Sort out the question
The computer costs $2,000. Anya paid $1,200. Her brother paid $240. Her parents paid the rest. “The rest” means whatever was still left to pay. You need the parents’ part as a percent of the whole $2,000. The question never gives the parents’ amount, so find it first.
Find what the parents paid
Take Anya’s money and her brother’s money away from the full cost.
$2{,}000 - 1{,}200 - 240 = 560$
The parents paid $560.
Watch out: the brother’s $240 has the same digits as 24%, choice A. But the question asks about the parents, not the brother.
Write the parents’ share as a percent
Use the percent formula. The part is $560, and the whole is $2,000.
$\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$
$\text{percent} = \frac{560}{2{,}000} \times 100 = \frac{28}{100} \times 100 = 28$
To simplify, divide the top and the bottom of the fraction by 20. That gives 28 out of 100, which is 28%. So the answer is B, 28%.

Check your answer

Find everyone’s share. Anya paid 1,200 out of 2,000, which is 60%. Her brother paid 240 out of 2,000, which is 12%. The parents paid 28%. Together, $60 + 12 + 28 = 100$ . The three shares make up the whole cost, so 28% fits.

More questions like this

The original DOE question

Question 78 as printed on page 100 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 100, question 78. Open page 100 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 94 of 120

2019 Test A Math, question 48

Question

Anita played 3 games and had a mean score of 140. Tariq played 2 games and had a mean score of 90. What was the mean of the scores for all 5 of these games?

A100
B115
C120
D230

Show the answer

Correct answer: C

The sum of the scores of Anita’s 3 games is $3(140) = 420$ . The sum of the scores of Tariq’s 2 games is $2(90) = 180$ . Add those two sums and divide by 5 to find the mean:

$\frac{420 + 180}{5} = \frac{600}{5} = 120$

Question analysis

Anita averaged 140 points over 3 games and Tariq averaged 90 over 2 games. This mean, or average, question asks for the mean score of all 5 games combined. The two means stand for different numbers of games, so students have to think about the totals behind them.

NYS standards: 6.SP.5c range, center, and overall pattern
6.SP.3 measures of center and variation
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Means are grade 6 content. Combining two groups of different sizes requires turning each mean back into a total, a step that grade 7 test items often demand and many students skip.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 1, 30 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The quick move, averaging 140 and 90, gives a listed choice, so the question rewards students who stop to ask what each mean stands for. Anita's three games carry more weight than Tariq's two, which pushes the combined mean above the 115 midpoint toward 140. Turning both means into totals and dividing by all 5 games is only one step longer than the shortcut.

Turning words into math moderate Understanding the question
Each mean stands for a total that must be rebuilt first: 140 over 3 games is 420 points and 90 over 2 games is 180, before the 600 is shared among 5 games. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
The rule that a mean times its count gives the total is never stated, and it is what shows that two means over different numbers of games cannot simply be averaged. (Anderson & Krathwohl, 2001)
A very close wrong answer minor Choosing the answer
115 and 120 both lie between the two means, so a reasonableness check passes both; only the weighting, 3 games at 140 against 2 at 90, lifts the true mean above the midpoint. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Multiply each player's mean by the games that player played to get two point totals, add them, and divide by all 5 games.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No weighting of 140 and 90 gives 100, which sits nearer Tariq's lower mean though Anita played more games, so it mostly catches guesses.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Averages the two means, (140 + 90) ÷ 2 = 115, giving Tariq's 2 games the same weight as Anita's 3.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two means, 140 + 90 = 230, and never divides, giving a value above both players' means, which no mean of their games can be.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean (average): $\text{mean} = \frac{\text{sum of the scores}}{\text{number of scores}}$; The mean, or average, shares a total evenly. The sum of the scores is all the scores added up. The number of scores is how many games there are. Dividing tells what each game would score if all the games scored the same.
Total from a mean: $\text{sum of the scores} = \text{mean} \times \text{number of scores}$; This is the mean formula turned around. If you know the mean and how many scores there are, multiply them to get the total. It works because the mean is the same share given to every game.

Sort out the question
Anita played 3 games, and her mean score was 140. Tariq played 2 games, and his mean score was 90. The mean is the average: add the scores and divide by how many there are. You need the mean of all 5 games together. You don’t know any single score, but you can find each player’s total.
Find Anita’s total
Use the mean formula turned around. Anita’s mean is 140, over 3 games.
$\text{sum of the scores} = \text{mean} \times \text{number of scores}$
$\text{sum of the scores} = 140 \times 3 = 420$
Anita scored 420 points in her 3 games.
Find Tariq’s total
Do the same for Tariq. His mean is 90, over 2 games.
$\text{sum of the scores} = 90 \times 2 = 180$
Tariq scored 180 points in his 2 games.
Find the mean of all 5 games
Add the two totals. Then divide by all 5 games.
$\text{mean} = \frac{\text{sum of the scores}}{\text{number of scores}}$
$\text{mean} = \frac{420 + 180}{5} = \frac{600}{5} = 120$
The mean of all 5 games is 120 points.
Watch out: don’t just average the two means. $(140 + 90) \div 2 = 115$ , choice B, counts Tariq’s 2 games as much as Anita’s 3. Adding the means, $140 + 90 = 230$ , forgets to divide. That is choice D.
So the answer is C, 120.

Check your answer

The mean must be between 90 and 140. It should be closer to 140, because Anita played more games. 120 is. Also, Anita’s 3 games average 20 points above 120, which is 60 points in all. Tariq’s 2 games average 30 points below 120, also 60 in all. The two balance, so 120 is the middle.

More questions like this

The original DOE question

Question 105 as printed on page 105 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 105, question 105. Open page 105 of the handbook (PDF)

Question 95 of 120

2019 Test B Math, question 10

Question

A recipe uses $\frac{1}{4}$ teaspoon of salt and $2\frac{1}{2}$ teaspoons of sugar. What is the ratio of salt to sugar in this recipe?

A $1:20$
B $1:10$
C $1:5$
D $1:4$

Show the answer

Correct answer: B

$\frac{1}{4} : 2\frac{1}{2}$

$\frac{1}{4} : \frac{5}{2}$

$4\left(\frac{1}{4}\right) : 4\left(\frac{5}{2}\right)$

$1 : 10$

Question analysis

A recipe calls for 1/4 teaspoon of salt and 2 1/2 teaspoons of sugar, and students write the ratio of salt to sugar in whole numbers. It tests ratios whose terms are a fraction and a mixed number. Both terms have to be rescaled together, and the mixed number is where slips creep in.

NYS standards: 7.RP.1 unit rates with fractions
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Ratios of fractions are grade 7 content, and this one is routine: convert a single mixed number, then scale both terms up to whole numbers. It asks for no further use of the ratio.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

A ratio of a fraction to a mixed number is unfamiliar enough that students reach for shortcuts, and each shortcut has a choice waiting. Pairing numerators after converting, or reading the salt's 1/4 as a ratio by itself, never compares the two amounts properly. Scaling both terms by the same factor, 4, or asking how many quarter teaspoons fit into 2 1/2 teaspoons, keeps the comparison intact.

Error-prone arithmetic moderate Working it out
The sugar is a mixed number, so 2 1/2 must become 5/2, and both terms must then be multiplied by 4, or 5/2 divided by 1/4, without dropping a denominator. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The recipe gives a fraction and a mixed number, but every choice is a whole-number ratio, so 1/4 : 2 1/2 has to be recognized in its scaled form. (Enright, Morley & Sheehan, 2002)

Strategy: Change 2 1/2 to 5/2, then divide the salt amount by the sugar amount; the quotient, written as a fraction with numerator 1, gives the ratio directly.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Multiplies both terms by 4 but takes 4 × 5/2 as 20, multiplying the numerator and forgetting to divide by the 2.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Rewrites 2 1/2 as 5/2 and pairs the numerators, 1 : 5, as if 1/4 and 5/2 shared a denominator.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Reads the 1/4 teaspoon of salt as the ratio 1 : 4 itself, never bringing the 2 1/2 teaspoons of sugar into the comparison.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing a mixed number to a fraction: $2\frac{1}{2} = \frac{2 \times 2 + 1}{2}$; A mixed number is a whole number plus a fraction. Multiply the whole number by the bottom number to count the halves in the wholes. Then add the top number, and keep the same bottom. Two wholes are 4 halves, so 2 and a half is 5 halves.
Equivalent ratios: $a : b = (a \times k) : (b \times k)$; a and b are the two amounts in a ratio, in order. k is any number except 0. Multiplying both amounts by the same number keeps the comparison the same. Use it to turn the fractions in a ratio into whole numbers.

Sort out the question
The recipe uses $\frac{1}{4}$ teaspoon of salt and $2\frac{1}{2}$ teaspoons of sugar. A ratio compares two amounts. “Salt to sugar” means salt first, then sugar. Every choice is a ratio of whole numbers. So you need to turn the two amounts into whole numbers that compare the same way.
Change the mixed number to a fraction
A mixed number is hard to work with. Write the sugar as a fraction.
$2\frac{1}{2} = \frac{2 \times 2 + 1}{2}$
$2\frac{1}{2} = \frac{4 + 1}{2} = \frac{5}{2}$
So the ratio of salt to sugar is $\frac{1}{4} : \frac{5}{2}$ .
Multiply both parts by 4
Multiply both parts by the same number to clear the fractions. Use 4, because both bottom numbers, 4 and 2, go into 4.
$a : b = (a \times k) : (b \times k)$
$\frac{1}{4} : \frac{5}{2} = \left(\frac{1}{4} \times 4\right) : \left(\frac{5}{2} \times 4\right) = 1 : 10$
For the sugar, $5 \times 4 = 20$ , and then $20 \div 2 = 10$ .
Watch out: forgetting to divide by 2 gives 1 : 20, choice A. Pairing the top numbers, 1 : 5, choice C, compares quarters with halves. And 1 : 4, choice D, is only the salt amount. It leaves out the sugar.
So the answer is B, 1 : 10.

Check your answer

Think in quarter teaspoons. The salt is 1 quarter teaspoon. One teaspoon is 4 quarters, so the sugar is $2.5 \times 4 = 10$ quarter teaspoons. That is 1 quarter of salt for every 10 quarters of sugar, or 1 : 10.

More questions like this

The original DOE question

Question 67 as printed on page 178 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 178, question 67. Open page 178 of the handbook (PDF)

Question 96 of 120

2019 Test B Math, question 26

Question

$A number line with arrows at both ends and four tick marks. From left to right, the ticks are labeled 0 (below the line), x and y (above the line), and x + y (above the line) with 24 below it. A bracket above the line marks the distance from x to y as 4.$

According to the number line above, what is the value of y?

Show the answer

Correct answer: C

According to the number line, $y - x = 4$ and $x + y = 24$ . Add the two equations together and solve for y:

$\begin{array}{l} y - x = 4 \\ + x + y = 24 \\ \hline 2y = 28 \end{array}$

$y = 14$

Question analysis

A number line marks 0, x, y and x + y, with x + y at 24 and a bracket showing that x and y are 4 units apart. Students find the value of y. The question combines reading a number line with solving for two unknowns, and the bracket and the sum have to be used together.

NYS standards: 7.EE.4a two-step equations
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Two unknowns usually call for grade 8 systems, but here one unknown is 4 more than the other, so a single grade 7 equation of the form px + q = r is enough. Reading those facts from a number line adds a step.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Variables instead of numbers minor
No obvious method minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The number line holds all the information but none of it as an equation, and the drawing is not to scale, so the positions must come from the labels. Students who split 24 evenly stop at the point midway between x and y, and those who take the bracket's 4 for a position subtract it from 24. Writing the two facts down and checking the result, 10 + 14 = 24 with a gap of 4, closes off every listed slip.

Reading a figure, table or graph moderate Understanding the question
The two facts needed, y − x = 4 and x + y = 24, are never written; they come from the bracket over x and y and from the 24 under the tick marked x + y. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Variables instead of numbers minor Understanding the question
Neither x nor y has a value, and the point x + y is itself built from both, so nothing can be placed until the two facts are combined. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
No obvious method minor Working it out
Two unknowns usually call for a system of equations, which grade 7 students have not met; they must instead write y as x + 4, or see that x and y sit 2 units either side of 12. (Schoenfeld, 1985)

Strategy: Write the two facts the labels give, y − x = 4 and x + y = 24, then add the equations or replace y with x + 4; the drawing is not to scale, so don't measure.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the labeled distance between x and y as the value of y, a number read straight off the figure rather than worked out.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Halves the sum 24 to get 12, the point midway between x and y, and stops before adding half of the 4-unit gap.
D Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: moderate
Takes the 4 above x as the position of x rather than the gap from x to y, so y = 24 − 4 = 20.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line: $\text{distance} = \text{right point} - \text{left point}$; On a number line, numbers get bigger to the right. The distance between two points is how many units apart they are. Subtract the left number from the right one, so the distance is never negative.
Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c$; An equation says two sides are equal, like a balanced scale. a and b are the two sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal. Use it to get the letter by itself.

Sort out the question
The number line shows 0, x, y and x + y. You need the value of y. The drawing is not to scale, though the question doesn’t say so. That means the spacing on the page doesn’t match the numbers. If you measure, y looks like it is near 15. So don’t measure. Work only from the labels.
Write the labels as equations
An equation is a math sentence that says two amounts are equal. The bracket marked 4 runs from x to y, and y is to the right.
$\text{distance} = \text{right point} - \text{left point}$
$y - x = 4$
So y is 4 more than x, or y = x + 4. The tick marked x + y has 24 under it, so x + y = 24.
Watch out: the 4 is the distance from x to y. It is not where x is, and it is not y. Taking x as 4 gives $24 - 4 = 20$ , choice D. Taking y as 4 is choice A.
Put x + 4 in place of y
Two unknown letters are hard to solve at once. y is the same as x + 4, so swap it into x + y = 24. Now there is only one letter.
$x + (x + 4) = 24$
$2x + 4 = 24$
Here 2x means 2 times x. Two x’s plus 4 make 24.
Solve for x
Subtract 4 from both sides. Then divide both sides by 2. Doing the same to both sides keeps the equation balanced.
$\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c$
$2x + 4 - 4 = 24 - 4$
$2x = 20$
$x = 20 \div 2 = 10$
So x is 10.
Find y
y is 4 more than x.
$y = x + 4 = 10 + 4 = 14$
So y is 14.
Watch out: half of 24 is 12, choice B. That is the point halfway between x and y. y is 2 more than that.
So the answer is C, 14.

Check your answer

Put both numbers back. $10 + 14 = 24$ , so x + y lands at 24. $14 - 10 = 4$ , so x and y are 4 apart. Both labels work. Notice that y is 14, not the 15 or so that the drawing suggests.

More questions like this

The original DOE question

Question 83 as printed on page 181 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 181, question 83. Open page 181 of the handbook (PDF)

Question 97 of 120

2018 Test A Math, question 4

Question

A survey asked students what pets they have. Based on the results, the following statements are all true.

20 students have cats.
23 students have dogs.
3 students have both dogs and cats.
5 students have no dogs or cats.

How many students were surveyed?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 45

Let x = number of students with only cats as pets.

Let y = number of students with only dogs as pets.

Calculate x and y using the given information:

There are 20 students who have cats, and of those 20 students, 3 have both cats and dogs.

Thus,

$x=20-3=17$

There are 23 students who have dogs, and of those 23 students, 3 have both cats and dogs.

Thus,

$y=23-3=20$

To find the total number of students surveyed, add the number of students who only have cats (x), the number of students who only have dogs (y), the number of students who have both (3), and the number of students who have neither (5):

$3+5+x+y=8+17+20=45$

Question analysis

A grid-in counting question about a survey of students' pets: some have cats, some dogs, some both, and some neither. Students find how many were surveyed by sorting everyone into groups without counting anyone twice. The students who own both pets are the trap in this classic Venn diagram problem.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 classes organize outcomes in lists and tables, but overlapping groups like these are rarely taught directly. The arithmetic is simple whole-number addition and subtraction; the demand lies in spotting the double count.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Much to keep track of minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is small; the trap is in the wording. Because each bullet reads like a separate group, students tend to add the numbers as listed and count the students with both pets twice. A Venn diagram with the overlap filled in first solves the problem, and the remaining risk is forgetting that students with no pets were surveyed too.

Turning words into math moderate Understanding the question
The statement "20 students have cats" already includes the 3 who have both pets, and so does "23 students have dogs", which the separate bulleted list hides. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Four facts, two overlapping groups, their overlap and an outside group of 5 with neither pet, all have to be placed before anything is added. (Sweller, 1988; Carpenter, Just & Shell, 1990)
An answer in an unexpected form minor Choosing the answer
As a grid-in it gives no choices, so a total that counts the 3 twice or leaves out the 5 looks just as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Draw two overlapping circles: put the both-pets group in the middle, subtract it from each pet total to fill the outer parts, then add every region and the students with neither.

Traps

Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

48 Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: strong
Adds 20 + 23 + 5 = 48, counting the 3 students with both pets once among the cat owners and again among the dog owners.
40 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Removes the overlap correctly, 20 + 23 − 3 = 40, but counts only the pet owners, leaving out the 5 surveyed students with no dogs or cats.
51 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds all four numbers in the list, 20 + 23 + 3 + 5 = 51, treating the both-pets group as an extra group of students.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting two groups that overlap: $T = C + D - B + N$; T is the total number of students. C is the number with cats, and D is the number with dogs. B is the number with both pets, and N is the number with neither. Students with both are counted in C and again in D, so you subtract B once. Then add N.

Sort out the four facts
You need the number of students who were surveyed. Here is what you know:
- 20 students have cats.
- 23 students have dogs.
- 3 students have both dogs and cats.
- 5 students have no dogs or cats.
The tricky part: the 3 students with both pets are already inside the 20 and inside the 23.
Split the pet owners into groups
Picture two circles that overlap, one for cats and one for dogs. The overlap holds the 3 students with both pets. Take them out of each pet total to find who has only one kind of pet.
$20 - 3 = 17$
So 17 students have only cats.
$23 - 3 = 20$
So 20 students have only dogs.
Add every group once
Now every student is in exactly one group.
Group Students
Only cats 17
Only dogs 20
Both pets 3
Neither pet 5
$17 + 20 + 3 + 5 = 45$
The formula gives the same total.
$T = 20 + 23 - 3 + 5 = 45$
Avoid the traps and answer
Watch out: adding $20 + 23 + 5$ counts the 3 students with both pets twice. That gives 48. Watch out: the 5 students with no pets were surveyed too. Leaving them out gives 40. Adding all four numbers in the list gives 51, which is also wrong. So you would grid 45.

Group	Students
Only cats	17
Only dogs	20
Both pets	3
Neither pet	5

Check your answer

Check each fact against your groups. Cat owners: $17 + 3 = 20$ . Dog owners: $20 + 3 = 23$ . Both: 3. Neither: 5. Every fact matches, so 45 students were surveyed.

More questions like this

The original DOE question

Question 61 as printed on page 81 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 61. Open page 81 of the handbook (PDF)

Question 98 of 120

2018 Test A Math, question 8

Question

In a certain school, course grades range from 0 to 100. Adrianna took 4 courses and her mean course grade was 90. Roberto took 5 courses. If both students have the same sum of course grades, what was Roberto’s mean?

Show the answer

Correct answer: A

The sum of Adrianna’s course grades equals 4 times the mean (average) of her grades:

90 × 4 = 360

Roberto has the same sum (360) as Adrianna. Find the mean of his course grades:

360 ÷ 5 = 72

Question analysis

This averages word problem compares two students' course grades. One student's number of courses and mean grade are given, and a second student has the same total over a different number of courses. Students work back from a mean to a sum, then find a new mean, while the stated grade range plays no part.

NYS standards: 6.SP.5c range, center, and overall pattern
6.SP.3 measures of center and variation
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a mean is grade 6 content, and this needs only the step back from a mean to a total and one division. Typical grade 7 statistics work compares two data sets, which asks more.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 2, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns on one idea, that a mean times the number of items gives the total, and the numbers are easy. Students who skim "the same sum" as "the same mean" carry Adrianna's 90 straight across, and the 0-to-100 range in the first sentence offers a second number to misuse. Converting to the shared total of 360 first makes the rest a single division.

Turning words into math moderate Understanding the question
"Both students have the same sum of course grades" is the bridge: Adrianna's mean of 90 must be turned into a total, 4 × 90, before it says anything about Roberto. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
The link between a mean and a total, sum = mean × number of courses, is not given and has to be used in both directions. (Anderson & Krathwohl, 2001)

Strategy: Multiply the first student's mean by her number of courses to get the shared total, then divide that total by Roberto's number of courses.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Uses 100, the top of the grade range, in place of Adrianna's mean of 90: 4 × 100 = 400, and 400 ÷ 5 = 80.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Assumes an equal sum means an equal mean and gives Adrianna's 90, ignoring that Roberto's 360 points are spread over 5 courses, not 4.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds a fifth grade of 100, the top of the stated range, to Adrianna's total: (360 + 100) ÷ 5 = 92.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean (average): $M = \frac{S}{n}$; M is the mean, also called the average. S is the sum, all the values added together. n is how many values there are. Dividing shares the total out equally, so the mean is what each value would be if they were all the same.
Sum from the mean: $S = M \times n$; This is the mean formula turned around. S is the sum, M is the mean, and n is how many values there are. If n values were each equal to the mean, they would add up to M times n. That is the same as the real sum.

Sort out the question
Adrianna took 4 courses, and her mean grade was 90. The mean is the average. Roberto took 5 courses. The key fact is that both have “the same sum of course grades”. The sum is all the grades added together. You need Roberto’s mean. The 0 to 100 range is not needed.
Find Adrianna’s total
You know her mean and her number of courses. Use the sum formula.
$S = M \times n$
$S = 90 \times 4 = 360$
Her grades add up to 360 points.
Find Roberto’s mean
Roberto’s grades also add up to 360 points. But he spreads them over 5 courses. Use the mean formula.
$M = \frac{S}{n}$
$M = \frac{360}{5} = 72$
His mean is 72.
Avoid the traps and answer
Watch out: the same sum does not mean the same mean. Roberto has more courses sharing the points, so his mean is lower. Choosing 90 is choice C. Watch out: don’t use 100, the top grade, in place of 90. That leads to 80, choice B. So the answer is A, 72.

Check your answer

More courses share the same 360 points, so Roberto’s mean must be lower than 90. 72 is lower. Multiply back: $5 \times 72 = 360$ , which matches Adrianna’s total.

More questions like this

The original DOE question

Question 65 as printed on page 82 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 82, question 65. Open page 82 of the handbook (PDF)

Question 99 of 120

2018 Test A Math, question 12

Question

$Square pyramid with apex R over the square base MNPQ; corner N is at the back, and its edges are dashed as hidden. Segment RS runs down face QRP from R to point S on base edge QP, with a right-angle mark at S.$

In the pyramid above, each triangular face has the same area, and the base MNPQ is a square that measures 8 centimeters on each side. If the length of $\overline{\mathrm{RS}}$ = 6 centimeters, what is the surface area of the pyramid excluding the base?

A48 sq cm
B96 sq cm
C128 sq cm
D160 sq cm

Show the answer

Correct answer: B

Line segment $\overline{\mathrm{RS}}$ is the altitude, or height, of triangle QRP. The length of $\overline{\mathrm{QP}}$ is 8 cm. Use that information to find the area of triangle QRP:

$A=\frac{1}{2}bh=\frac{1}{2}(8)(6)=24$ sq cm.

There are 4 congruent triangles in the pyramid, so the surface area is

4 × 24 = 96 sq cm

Question analysis

A geometry question about the surface area of a square pyramid, leaving out its base. The base edges are 8 cm and segment RS, the height of one triangular face drawn from the top to a base edge, is 6 cm. Students find one face's area and multiply by the number of faces, reading excluding the base carefully.

NYS standards: 7.G.6 area, surface area, and volume
6.G.4 nets and surface area
6.G.1 area of polygons
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Surface area of a right pyramid from its faces is named in the grade 7 standards, and this is a direct two-step case. It matches typical grade 7 work, though the drawing's slant height must be read correctly.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 49 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once RS is read as the height of a face, the computation is short: one triangle, then four of them. Points go to the figure and to one word. Students who count only the faces drawn in front, or who include the square base, reach listed choices, and the bold "excluding" is all that separates the lateral area from the total. The statement that every face has the same area is what makes multiplying by four valid.

Reading a figure, table or graph moderate Understanding the question
The drawing shows two lateral faces in front and two behind dashed edges, and RS, with its right-angle mark at S, is the height of face QRP, not of the pyramid. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The triangle area formula, one half times base times height, and the fact that a square pyramid has four lateral faces are both assumed, not given. (Anderson & Krathwohl, 2001)

Strategy: Find the area of one triangular face with one half times base times height, using RS as the height, then multiply by the four congruent faces.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: moderate
Finds one face as ½ × 8 × 6 = 24 but counts only the two faces drawn in front, 2 × 24 = 48, missing the two hidden ones.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Uses only the 8 cm edge, as both base and height of each face, 4 × ½ × 8 × 8 = 128, and never uses RS.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Finds the four faces, 96 sq cm, then adds the 8 × 8 = 64 sq cm base that the question says to exclude.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.

Sort out the pyramid
The pyramid sits on a square base, MNPQ. Each side of the square is 8 cm. Four triangle faces rise from the base and meet at the top point, R. Every face has the same area. Segment RS is 6 cm. You need the area of the four faces, “excluding the base”. Excluding means leaving it out.
See what RS measures
RS runs from R down face QRP to point S on edge QP. The small square at S marks a right angle. So RS is the height of triangle QRP. The base of that triangle is QP, one side of the square, so it is 8 cm.
Find the area of one face
Use the triangle formula with base 8 cm and height 6 cm.
$A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times 8 \times 6 = 24$
One face has an area of 24 sq cm.
Add the four faces
All four faces are the same, so multiply by 4.
$4 \times 24 = 96$
Watch out: only two faces are drawn in front. The other two hide behind the dashed lines. Counting just two faces gives 48, choice A. Watch out: don’t add the base, $8 \times 8 = 64$ . That gives 160, choice D. So the answer is B, 96 sq cm.

Check your answer

Each face fits inside an 8 by 6 rectangle, with an area of 48 sq cm. The triangle is half of that, 24 sq cm, and four faces make 96 sq cm. That is less than 160, which would include the base.

More questions like this

The original DOE question

Question 69 as printed on page 83 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 83, question 69. Open page 83 of the handbook (PDF)

Question 100 of 120

2018 Test A Math, question 15

Question

1 dollar = 7 lorgs

1 dollar = 0.5 dalt

Kevin has 140 lorgs and 16 dalts. If he exchanges the lorgs and dalts for dollars according to the rates above, how many dollars will he receive?

A$28
B$52
C$182
D$282

Show the answer

Correct answer: B

Use proportions to make the conversions:

Lorgs to dollars:

$\frac{140}{x}=\frac{7}{1}$

$7x=140$

$x=\$20$

Dalts to dollars:

$\frac{16}{x}=\frac{0.5}{1}$

$0.5x=16$

$x=\$32$

Total dollars = 20 + 32 = $52

Question analysis

A unit conversion word problem with made-up currencies, lorgs and dalts, each with an exchange rate to dollars. Students convert two amounts into dollars and add them. One rate is a decimal less than 1, which makes it easy to multiply where dividing is needed.

NYS standards: 6.RP.3d converting measurement units
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Converting units with ratio reasoning is grade 6 content. Two conversions and a sum make it typical grade 7 work, with the decimal rate as the step most likely to trip students up.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 29 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The two conversions look alike but run in different directions in students' minds: 140 lorgs at 7 per dollar is plainly a division, while 0.5 dalt per dollar looks like a price. Most wrong answers come from the dalts. The safest approach is to ask how many dollars one coin is worth, a seventh of a dollar for a lorg and 2 dollars for a dalt, before multiplying.

Turning words into math moderate Understanding the question
"1 dollar = 0.5 dalt" means a dalt is worth 2 dollars, but read word by word it suggests a dalt is half a dollar, inviting multiplication by 0.5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Dividing 16 by 0.5 means dividing by a decimal less than 1, which makes the result larger, a step many students do the other way by habit. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Read each rate as coins per dollar, divide each coin amount by its own rate, and only then add the two dollar amounts.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Converts the lorgs correctly, 140 ÷ 7 = 20, but multiplies the dalts by the rate, 16 × 0.5 = 8, and adds to get $28.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies each coin amount by the other coin's rate, 140 × 0.5 + 16 × 7 = 70 + 112 = 182, combining the numbers without a model.
D Units or scale (King et al., 2004)
Reveals a slip Expected pull: weak
Divides each amount by the other coin's rate, 140 ÷ 0.5 = 280 and 16 ÷ 7 ≈ 2.3, and rounds the total to $282.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing coins into dollars: $d = \frac{c}{r}$; d is the number of dollars. c is how many coins you have. r is the rate: how many of those coins make 1 dollar. Each dollar uses up r coins, so dividing tells how many dollars the coins fill.

Sort out the rates
The rates tell how many coins make 1 dollar. 1 dollar = 7 lorgs. 1 dollar = 0.5 dalt, which is half a dalt. Kevin has 140 lorgs and 16 dalts. You need how many dollars he gets for all of them.
Change the lorgs to dollars
It takes 7 lorgs to make 1 dollar. So divide the lorgs by 7.
$d = \frac{c}{r}$
$d = \frac{140}{7} = 20$
The lorgs are worth 20 dollars.
Think about what a dalt is worth
Half a dalt makes 1 dollar. So a whole dalt makes 2 dollars. A dalt is worth more than a dollar, not less. Watch out: don’t read the rate as a dalt being worth half a dollar. That leads to $16 \times 0.5 = 8$ and a total of $28, choice A.
Change the dalts to dollars
Use the same formula, with a rate of 0.5 dalt for each dollar.
$d = \frac{16}{0.5} = 32$
Dividing by one half is the same as multiplying by 2, so this is $16 \times 2 = 32$ . The dalts are worth 32 dollars.
Add the two amounts
Add the dollars from the lorgs and the dollars from the dalts.
$20 + 32 = 52$
Kevin gets 52 dollars. So the answer is B, $52.

Check your answer

Work backward. 20 dollars trade for $20 \times 7 = 140$ lorgs. 32 dollars trade for $32 \times 0.5 = 16$ dalts. Both match what Kevin has, so $52 is right.

More questions like this

The original DOE question

Question 72 as printed on page 84 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 72. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 101 of 120

2018 Test A Math, question 16

Question

A box of colored pencils contains exactly 6 red pencils. The probability of choosing a red pencil from the box is $\frac{2}{7}$ . How many of the pencils in the box are not red?

Show the answer

Correct answer: B

Let $x$ be the total number of colored pencils in the box. Set up a proportion to find $x$ :

$\frac{2}{7}=\frac{6}{x}$

$2x=42$

$x=21$

If there are 6 red pencils, then the number of pencils that are not red is

21 − 6 = 15.

Question analysis

A probability word problem: a box holds exactly 6 red pencils, and the chance of drawing a red one is 2/7. Students work backward from the probability to the number of pencils in the box, then answer the question actually asked, which is about the pencils that are not red.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Simple probability is grade 6 content in New York. Reversing it to find a total is a step beyond plugging counts into a fraction, which puts this at about the level of typical grade 7 work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 32 words.
Corrected from the DOE handbook: A spelling mistake in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Students have to treat a probability as a ratio and scale it up: 6 red pencils is three times the 2 in 2/7. The common loss is choosing the total number of pencils, which appears naturally halfway through the work and is a listed choice. Reading the 7 as a pencil count, or scaling by 6 instead of 3, leads to the other wrong answers.

Turning words into math moderate Understanding the question
"The probability of choosing a red pencil from the box is 2/7" has to become a part-to-whole ratio: 2 of every 7 pencils are red, so the 6 red pencils are 2 parts. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
The question assumes students know that a probability is favorable outcomes over all outcomes, here used in reverse to find the total from the red count. (Anderson & Krathwohl, 2001)

Strategy: Since 2 parts out of 7 are red, find how many pencils one part stands for, then count the parts that are not red.

Traps

Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Reads 2/7 as 2 red pencils among exactly 7 and takes 7 − 2 = 5, ignoring the stated 6 red pencils.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Solves 2/7 = 6/x to get 21 pencils in all and stops before subtracting the 6 red ones that the bold not excludes.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals partial knowledge Expected pull: weak
Treats the 6 red pencils as one share of the 7 instead of two, so each share is 6 and the 5 non-red shares make 30.

What was corrected from the DOE handbook, and why

Explanation: The handbook says “the number of pencils that are not read is 21 − 6 = 15”. “read” now reads “red”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{favorable}}{\text{total}}$; P is the probability, the chance that something happens. Favorable is the number of ways to get what you want, here a red pencil. Total is the number of all possible picks, here every pencil in the box. It works when every pick is equally likely.

Sort out the question
A probability is a number that tells how likely something is. The box has exactly 6 red pencils. The chance of picking a red one is $\frac{2}{7}$ . You need the number of pencils that are not red.
Read the probability as parts
$\frac{2}{7}$ means 2 out of every 7 pencils are red. Picture the box split into 7 equal parts. Red pencils fill 2 of the parts. The other 5 parts are not red. Watch out: the box does not hold just 7 pencils. It has 6 red ones alone. Taking $7 - 2 = 5$ is choice A.
Find the size of one part
The 6 red pencils fill 2 parts. Divide to find one part.
$6 \div 2 = 3$
Each part is 3 pencils. So the whole box has $7 \times 3 = 21$ pencils.
Check with the probability formula
Put the counts into the formula to make sure.
$P = \frac{\text{favorable}}{\text{total}} = \frac{6}{21} = \frac{2}{7}$
Divide the top and bottom of 6/21 by 3 to get 2/7. It matches.
Count the pencils that are not red
$21 - 6 = 15$
That is the same as 5 parts of 3 pencils each. Watch out: 21 is all the pencils, choice C. The question asks only for the ones that are not red. So the answer is B, 15.

Check your answer

Red and not red should make the whole box: $6 + 15 = 21$ . Red is less likely than not red, since 2/7 is less than half. 6 red against 15 not red fits that.

More questions like this

The original DOE question

Question 73 as printed on page 84 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 73. Open page 84 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 102 of 120

2018 Test A Math, question 17

Question

The sum of the numbers x, y, and z is 50. The ratio of x to y is 1:4, and the ratio of y to z is 4:5. What is the value of y ?

Show the answer

Correct answer: D

Since both ratios have $y$ in common, solve for $x$ and $z$ in terms of $y$ in both equations. Using $x:y=1:4$ , solve for $x$ in terms of $y$ .

$\frac{x}{y}=\frac{1}{4}$

$x=\frac{y}{4}$

Using the ratio $y:z=4:5$ , solve for $z$ in terms of $y$ :

$\frac{y}{z}=\frac{4}{5}$

$z=\frac{5}{4}y$

The question states $x+y+z=50$ . Substitute from the two equations above and solve for $y$ .

$\frac{1}{4}y+y+\frac{5}{4}y=50$

$\frac{10}{4}y=50$

$10y=200$

$y=20$

Question analysis

A ratio problem with three unknowns: x, y and z add to 50, x to y is 1 to 4, and y to z is 4 to 5. Students merge the two ratios into one three-part ratio and use the sum to find y. The step that is easy to skip is turning ratio parts into actual values.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 ratio problems usually involve one ratio. Here two ratios share a term and must be merged before the total is used, though the shared term already matches, which keeps it near typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Variables instead of numbers minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards noticing that y has the same term in both ratios, which turns two ratios into one three-part ratio in a single move. Students who miss this often write x and z in terms of y, as the explanation does, which works but is slower. The wrong choices are numbers taken from the ratios themselves, so a student who never scales up to the total of 50 finds an answer waiting.

No obvious method moderate Working it out
The two ratios have to be merged into one, x : y : z = 1 : 4 : 5, which works only because y is 4 in both; nothing in the wording says to combine them. (Schoenfeld, 1985)
Variables instead of numbers minor Understanding the question
x, y and z are given only through two ratios and their sum of 50, so there is no actual count to anchor the work until the parts are scaled. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Adding the parts, dividing 50 by their total and scaling y's 4 parts are three steps, and the count of parts after the first step is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Since y is 4 in both ratios, write x : y : z as one combined ratio, find the value of one part from the total of 50, and scale up.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes y's term in the ratios, 4, as y's actual value, never using the sum of 50.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two 4s that stand for y in the two ratios, 4 + 4 = 8, treating ratio terms as amounts.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Combines the ratios into 1 : 4 : 5 and gives the total number of parts, 1 + 4 + 5 = 10, instead of y's share of the 50.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sharing a total in a ratio: $\text{one part} = \text{total} \div \text{number of parts}$; In a ratio like 3:2, the amounts are made of equal-sized parts. Add the numbers in the ratio to count the parts. Divide the total by that count to get the size of one part. Then multiply by how many parts each amount has.

Sort out the question
The three numbers x, y and z add up to 50. The ratio of x to y is 1:4. The ratio of y to z is 4:5. A ratio compares sizes. You need the actual value of y.
Join the two ratios
y is 4 in both ratios. So the two ratios use the same size parts, and you can join them into one.
$x : y : z = 1 : 4 : 5$
This means x is 1 part, y is 4 parts, and z is 5 parts.
Count the parts
$1 + 4 + 5 = 10$
There are 10 parts in all. Watch out: 10 is the number of parts, not the value of y. Picking it is choice C.
Find the size of one part
Use the sharing formula. The 10 parts share the total of 50.
$\text{one part} = 50 \div 10 = 5$
Each part is worth 5.
Find y
y is 4 parts.
$y = 4 \times 5 = 20$
Watch out: the 4 in the ratio counts parts. It is not the value of y. Picking 4 is choice A. So the answer is D, 20.

Check your answer

Find all three: x = 5, y = 20, z = 25. They add up: $5 + 20 + 25 = 50$ . x to y is 5 to 20, which is 1:4. y to z is 20 to 25, which is 4:5. Everything fits.

More questions like this

The original DOE question

Question 74 as printed on page 84 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 74. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 103 of 120

2018 Test A Math, question 19

Question

In Centerville, 45% of the population is female, and 60% of the population commutes to work daily. Of the total Centerville population, 21% are females who commute to work daily. What percentage of the total Centerville population are males who do not commute to work daily?

A15%
B16%
C24%
D39%

Show the answer

Correct answer: B

Create a table with the information provided in the problem and use subtraction to fill in the rest of the table:

	Female	Male	TOTAL
Commutes to work	21%	39% (60 − 21)	60%
Does not commute to work	24% (45 − 21)	16% (40 − 24)	40% (100 − 60)
TOTAL	45%		100%

16% of the population is male and does not commute to work.

Question analysis

A percent problem about overlapping groups in a town's population: the percents who are female, who commute daily, and who are female commuters are given. Students find the percent who are male and do not commute. A two-way table helps, since several of its cells are easy to confuse with the one asked for.

NYS standards: 7.SP.8b sample spaces for compound events
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually take a percent of one quantity. This needs two overlapping categories organized in a two-way table, which is not in New York's grade 7 standards, with three subtractions and two tempting cells.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: No obvious method moderate
Much to keep track of moderate
Turning words into math moderate
Solution: 3 steps
Reading level: Question and choices: about grade 12, 43 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question gives three overlapping percents and asks for the one group none of them describes directly. Without a table, students tend to subtract the given numbers in some order, and several orders yield a listed choice. The last trap is the word "not", since the males who do commute are one cell away. Filling every cell of a two-by-two table before answering takes under a minute and leaves nothing to guess.

No obvious method moderate Working it out
Nothing in the wording points to a two-way table, yet the four groups, female or male by commuting or not, fall into place only once one is drawn. (Schoenfeld, 1985)
Much to keep track of moderate Working it out
Three given percents, 45, 60 and 21, plus their complements, 55 and 40, have to be kept straight while the cells are filled in by subtraction. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math moderate Understanding the question
"Of the total Centerville population, 21% are females who commute" is a share of everyone, not of the females, and belongs in a single cell of the table. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Make a two-by-two table of female and male against commutes and does not commute, fill in the totals, and complete each cell by subtraction.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Subtracts one given percent from another, 60 − 45 = 15, without placing the 21% of female commuters anywhere.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Fills in the first empty cell, females who do not commute, 45 − 21 = 24, and stops before using it to reach the males' cell.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Finds the males who commute, 60 − 21 = 39, missing the word not in the question.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts add up to the whole: $a + b = W$; In a two-way table, every row and every column splits a group into two parts. a and b are the two parts, and W is the whole group. If you know the whole and one part, subtract to find the other part.

Sort out the question
Every number is a percent of the whole town. 45% of people are female. 60% commute, which means they travel to work every day. 21% are females who commute. You need the percent who are “males who do not commute to work daily”.
Make a two-way table
A two-way table sorts people two ways at once: female or male, and commutes or not. Put in what you know. The whole town is 100%.
Female Male Total
Commute 21% ? 60%
Do not commute ? ? ?
Total 45% ? 100%
Fill in the two totals
Everyone is female or male, so the males are what is left.
$100\% - 45\% = 55\%$
Everyone either commutes or does not.
$100\% - 60\% = 40\%$
So 55% are male, and 40% do not commute.
Find the males who commute
Use the commute row. All commuters are 60%, and the female commuters are 21%.
$60\% - 21\% = 39\%$
So 39% are males who commute. Watch out: the question says do not commute. Picking 39% is choice D.
Find the males who do not commute
Use the male column. All males are 55%, and the males who commute are 39%.
$55\% - 39\% = 16\%$
Watch out: 24% is the females who do not commute, $45\% - 21\%$ . That is choice C. So the answer is B, 16%.

	Female	Male	Total
Commute	21%	?	60%
Do not commute	?	?	?
Total	45%	?	100%

Check your answer

Find the same cell another way. Females who do not commute: $45\% - 21\% = 24\%$ . All who do not commute: 40%. So males who do not commute: $40\% - 24\% = 16\%$ . The answers match.

More questions like this

The original DOE question

Question 76 as printed on page 84 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 76. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 104 of 120

2018 Test A Math, question 23

Question

SCORES ON MATH QUIZ

Score	Number of Students
85	4
75	4
65	2

What is the mean score of the 10 students in the table above?

A22.5
B75
C77
D85

Show the answer

Correct answer: C

For each row, multiply the number of students by the score. Then add those together and divide by the total number of students to find the mean (average) of the 10 students.

$\frac{85(4)+75(4)+65(2)}{10}=\frac{340+300+130}{10}=\frac{770}{10}=77$

Question analysis

This statistics question gives a frequency table of quiz scores: three scores and how many of the 10 students earned each. Students find the mean score of all 10 students. Each score has to count as many times as students earned it, which is easy to overlook when only three scores are listed.

NYS standards: 6.SP.5c range, center, and overall pattern
6.SP.5a number of observations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Computing a mean is grade 6 work, and this weighted version from a table takes two short steps. It is a little easier than typical grade 7 statistics, which compares whole data sets.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 2, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The procedure is routine once students see that the table lists counts, not separate students. The tempting shortcut is to average the three scores in the first column, which gives a listed choice close to the right one. Multiplying each score by its count, then dividing by 10, takes well under a minute.

A very close wrong answer moderate Choosing the answer
77 and 75 are only 2 points apart, so a rough sense of the middle score cannot choose between them; only weighting each score by its 4, 4 or 2 students does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
The table pairs each score with a number of students, so the 10 students are 4, 4 and 2, and each score must be weighted by its count. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Multiply each score by its number of students, add the three products, and divide by the 10 students.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the three scores, 85 + 75 + 65 = 225, and divides by the 10 students, an average lower than every score in the table.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Averages the three listed scores, (85 + 75 + 65) ÷ 3 = 75, ignoring that 4, 4 and 2 students earned them.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks the top score, which 4 students earned, instead of computing a mean of all 10 scores.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean (average): $M = \frac{S}{n}$; M is the mean, also called the average. S is the sum, all the values added together. n is how many values there are. Dividing shares the total out equally, so the mean is what each value would be if they were all the same.

Sort out the table
The table lists 3 scores and how many students got each one. 4 students scored 85. 4 students scored 75. 2 students scored 65. That is $4 + 4 + 2 = 10$ students. You need the mean, or average, of all 10 scores.
Find the sum of all 10 scores
Each score counts once for every student who got it. So multiply each score by its number of students.
$85 \times 4 = 340$
$75 \times 4 = 300$
$65 \times 2 = 130$
Add the three results.
$340 + 300 + 130 = 770$
All 10 scores add up to 770 points.
Divide by the number of students
$M = \frac{S}{n}$
$M = \frac{770}{10} = 77$
Watch out: averaging just the three scores gives $(85 + 75 + 65) \div 3 = 75$ , choice B. That acts as if only one student got each score. So the answer is C, 77.

Check your answer

More students scored 85 than 65. So the mean should be a little above the middle score, 75. 77 fits. A mean must also be between the lowest and highest scores, 65 and 85. That rules out 22.5.

More questions like this

The original DOE question

Question 80 as printed on page 85 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 80. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 105 of 120

2018 Test A Math, question 27

Question

The faculty of a certain four-year college consists of 179 teachers. There are 663 first-year students. The student-to-faculty ratio for the entire college is 15 to 1. What is the total number of second-, third-, and fourth-year students?

A1,989
B2,022
C2,652
D2,685

Show the answer

Correct answer: B

Let $x$ be the number of second-, third-, and fourth-year students.

Then the total number of students in the college is

$663+x$

Set up a proportion and solve for $x$ :

$\frac{15}{1}=\frac{663+x}{179}$

$663+x=179(15)$

$663+x=2685$

$x=2022$

Question analysis

A ratio word problem about a four-year college with 179 teachers, 663 first-year students and a student-to-faculty ratio of 15 to 1. Students use the ratio to find the total number of students, then remove the first-year class. The phrase four-year invites the wrong assumption that every year is the same size.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Using a ratio to find a total and then subtracting a part is a standard two-step grade 7 ratio problem. The three-digit multiplication and the four-year detail add little beyond typical classwork.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question hinges on what the ratio describes: the whole college, not one class of students. Students who see that multiply, subtract and are done; those who stop at the total, or assume each year matches the 663 first-years, find their number listed. Because the choices come in close pairs, a careless multiplication can also land on a wrong one.

Turning words into math moderate Understanding the question
"The student-to-faculty ratio for the entire college is 15 to 1" covers students in all four years, so 15 × 179 is every student, and the 663 first-years must come out. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
179 × 15 and the subtraction of 663 are multi-digit computations by hand, and the choices come in pairs only 33 apart. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply the 179 faculty by 15 to get the number of all students, then subtract the 663 first-year students to leave the other three years.

Traps

Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Ignores the ratio and assumes each of the other three years also has 663 students, 663 × 3 = 1,989.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Assumes all four years have 663 students and multiplies 663 × 4 = 2,652, which also counts the first-year class the question leaves out.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Multiplies 179 × 15 = 2,685 to get every student in the college and stops before subtracting the 663 first-year students.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Using a ratio to find an amount: $s = 15 \times f$; s is the number of students, and f is the number of faculty, the teachers. A ratio of 15 to 1 means 15 students for every 1 teacher. So each teacher goes with 15 students, and you multiply the teachers by 15.

Sort out the question
The faculty means the teachers. The college has 179 of them. The ratio of students to faculty is 15 to 1 “for the entire college”. So there are 15 students for every teacher, counting all four years. 663 students are first-years. You need the second-, third- and fourth-year students together.
Find every student in the college
Multiply the teachers by 15.
$s = 15 \times 179$
Break it into easy pieces: $10 \times 179 = 1{,}790$ and $5 \times 179 = 895$ .
$1{,}790 + 895 = 2{,}685$
There are 2,685 students in all four years.
Take away the first-years
$2{,}685 - 663 = 2{,}022$
That leaves 2,022 second-, third- and fourth-year students. Watch out: 2,685 still includes the first-years. Stopping there gives choice D. Watch out: don’t assume each year has 663 students. That leads to 1,989, choice A. So the answer is B, 2,022.

Check your answer

Add back: $2{,}022 + 663 = 2{,}685$ students. Then $2{,}685 \div 179 = 15$ . That is 15 students for each teacher, the ratio in the question.

More questions like this

The original DOE question

Question 84 as printed on page 86 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 84. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 106 of 120

2018 Test A Math, question 30

Question

Today, Tien’s age is $\frac{1}{4}$ of Jordan’s age. In 2 years, Tien’s age will be $\frac{1}{3}$ of Jordan’s age. How old is Jordan today?

A4 years old
B6 years old
C12 years old
D16 years old

Show the answer

Correct answer: D

First, set up an equation to express Tien’s age ( $T$ ) and Jordan’s age ( $J$ ) today:

$T=\frac{1}{4}J$

Two years from now, Tien’s age will be $T+2$ , and Jordan’s age will be $J+2$ .

Set up an equation about the relationship between Tien’s age and Jordan’s age in two years:

$T+2=\frac{1}{3}(J+2)$

Solve the above equation for $T$ :

$T=\frac{1}{3}(J+2)-2$

Now set the two equations equal to each other and solve for $J$ :

$\frac{1}{4}J=\frac{1}{3}(J+2)-2$

$\frac{1}{4}J=\frac{1}{3}J-\frac{4}{3}$

$-\frac{1}{12}J=-\frac{4}{3}$

$J=-\frac{4}{3}\left(-\frac{12}{1}\right)$

$J=16$

Question analysis

This age word problem says Tien is one fourth of Jordan's age today and will be one third of Jordan's age in two years, then asks for Jordan's current age. Students must turn two fractional relationships into equations and solve them together. Both people age by the same two years, and it is easy to solve for the wrong person.

NYS standards: 8.EE.8c systems in real-world problems
8.EE.7b linear equations with rational coefficients
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 equations have one unknown on one side, like 3x + 5 = 20. This one links two unknown ages with fractional coefficients and leads to a variable on both sides, which is grade 8 algebra.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math major
Error-prone arithmetic moderate
Solution: 4 steps
Reading level: Question and choices: about grade 1, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The difficulty is in the setup: two fractional relationships at two different times, with both people aging. Once the equations are right, the algebra is short, but it ends with two ages and only one is asked for. Testing the choices against both conditions is a reliable alternative that avoids the algebra entirely.

Turning words into math major Understanding the question
The second sentence must become T + 2 = (J + 2)/3; writing J/3, or forgetting that Jordan also ages two years, breaks the equation. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic moderate Working it out
Substituting J/4 for T leaves J/4 + 2 = (J + 2)/3, an equation with fractions on both sides that must be cleared, usually by multiplying by 12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Let Jordan's age be J, write Tien's age as J/4, and set J/4 + 2 equal to one third of (J + 2); or test each choice against both conditions.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Solves the equations correctly but reports Tien's age today, 4, which is one fourth of Jordan's 16, instead of Jordan's age.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Gives Tien's age two years from now, 4 + 2 = 6, answering for the wrong person and the wrong time.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Picks 12 because it is a multiple of both 4 and 3, but then Tien would be 3 and later 5, not a third of 14.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property: $a \times (b + c) = a \times b + a \times c$; a, b and c are any numbers. To multiply a sum by a number, multiply each part of the sum, then add. It works because a groups of (b + c) is the same as a groups of b plus a groups of c.

Sort out what you know
Today, Tien’s age is $\frac{1}{4}$ of Jordan’s age. That means Jordan is 4 times as old as Tien.
In 2 years, both of them will be 2 years older. Then Tien’s age will be $\frac{1}{3}$ of Jordan’s. So Jordan will be 3 times as old as Tien. You need Jordan’s age today.
Name the ages with a letter
Let t stand for Tien’s age today. Jordan is 4 times as old, so his age is 4t, which means 4 times t. Add 2 years to each age:
Person Today In 2 years
Tien $t$ $t + 2$
Jordan $4t$ $4t + 2$
Watch out: Jordan gets 2 years older too. Do not forget to add 2 to his age.
Write an equation for later
In 2 years, Jordan’s age will be 3 times Tien’s age. Write that as an equation:
$4t + 2 = 3 \times (t + 2)$
Use the distributive property on the right side. Multiply both t and 2 by 3:
$4t + 2 = 3t + 6$
Solve for Tien’s age
Take 3t away from both sides. Doing the same thing to both sides keeps them equal:
$t + 2 = 6$
Now take 2 away from both sides:
$t = 4$
So Tien is 4 years old today. Watch out: 4 is Tien’s age, not Jordan’s. Choice A, 4 years old, is a trap.
Find Jordan’s age
Jordan is 4 times as old as Tien:
$4 \times 4 = 16$
So the answer is D, 16 years old.

Person	Today	In 2 years
Tien	$t$	$t + 2$
Jordan	$4t$	$4t + 2$

Check your answer

Put 16 back into the story. Today, Tien is 16 ÷ 4 = 4. In 2 years, Tien is 6 and Jordan is 18. Is 6 one third of 18? Yes, because 18 ÷ 3 = 6. Choice C fails: Tien would be 3, then 5, and 5 is not a third of 14.

More questions like this

The original DOE question

Question 87 as printed on page 87 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 87. Open page 87 of the handbook (PDF)

Question 107 of 120

2018 Test A Math, question 34

Question

There are 6 different cookies on a plate. Aiden will choose 2 of these cookies to pack in his lunch. How many different pairs of 2 cookies can he choose from the 6?

Show the answer

Correct answer: B

Create a list of the possible pairs. Let the cookies be named A, B, C, D, E, and F.

AB, AC, AD, AE, AF

BC, BD, BE, BF

CD, CE, CF

DE, DF

There are a total of 15 possible pairs of cookies that Aiden can choose.

Question analysis

This counting question asks how many different pairs of cookies Aiden can choose from six different cookies on a plate. Students must count combinations, where taking the same two cookies in a different order is not a new pair. The challenge is that the order of picking does not matter, so each pair must be counted once.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students list the outcomes of compound events with organized lists and tree diagrams. Counting pairs is close to that work, but students must notice that the order of choosing does not matter.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of moderate
Solution: 2 steps
Reading level: Question and choices: about grade 4, 29 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Students who know the multiplication principle are the most at risk, because it counts each pair twice. The question rewards noticing that a pair has no order, either by listing systematically or by halving 6 × 5. The list of 15 pairs takes under a minute, so there is no time pressure to justify a shortcut.

Much to keep track of moderate Working it out
Listing all pairs by hand means tracking which have been written so none is repeated or skipped, which an organized list such as AB to AF handles. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Name the cookies with letters and list the pairs systematically: the first cookie with every later one, then the second with every later one, and so on.

Traps

Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
Without replacement: In a second pick there is one fewer item, which changes the probability.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Multiplies the 6 cookies by the 2 being chosen, a product of the stem's numbers that does not count any pairs.
C Overlapping groups (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Counts ordered picks, 6 choices then 5, so every pair such as AB and BA is counted twice, giving 30 instead of half that.
D Without replacement (Fischbein & Schnarch, 1997)
Reveals a misconception Expected pull: weak
Uses 6 × 6, as if the first cookie went back on the plate, which counts order and also allows the same cookie twice.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of pairs: $\frac{n \times (n - 1)}{2}$; n is how many things you choose from. Each of the n things can pair with the n − 1 others. That counts every pair twice, once from each side, so you divide by 2.

Sort out the question
There are 6 different cookies. Aiden picks 2 of them. A pair is just two cookies together.
Order does not matter. Picking cookie 1 then cookie 2 gives the same pair as cookie 2 then cookie 1. You need to count the different pairs.
List the pairs in order
Number the cookies 1 to 6. Pair cookie 1 with every cookie after it. Then pair cookie 2 with every cookie after it, and so on. Only pairing with later numbers stops you from listing a pair twice.
Start with Pairs How many
1 1-2, 1-3, 1-4, 1-5, 1-6 5
2 2-3, 2-4, 2-5, 2-6 4
3 3-4, 3-5, 3-6 3
4 4-5, 4-6 2
5 5-6 1
Add up the pairs
Each row has one fewer pair than the row above it. That is because one fewer cookie comes after it. Add the last column of the table:
$5 + 4 + 3 + 2 + 1 = 15$
There are 15 different pairs.
Check with the pairs formula
Put n = 6 into the formula:
$\frac{6 \times 5}{2} = \frac{30}{2} = 15$
Watch out: 6 × 5 = 30 counts each pair twice, once for each order. Stopping there gives choice C. And 6 × 6 = 36, choice D, lets Aiden pick the same cookie twice. So the answer is B, 15.

Start with	Pairs	How many
1	1-2, 1-3, 1-4, 1-5, 1-6	5
2	2-3, 2-4, 2-5, 2-6	4
3	3-4, 3-5, 3-6	3
4	4-5, 4-6	2
5	5-6	1

Check your answer

Think of handshakes. Each of 6 people shakes hands with the 5 others. That looks like 30 handshakes, but each handshake has two people in it, so it was counted twice. 30 ÷ 2 = 15, which matches the list.

More questions like this

The original DOE question

Question 91 as printed on page 87 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 91. Open page 87 of the handbook (PDF)

Question 108 of 120

2018 Test A Math, question 39

Question

Suppose $M=\frac{w}{x}$ , $N=\frac{y}{z}$ , and $w,x,y$ , and $z$ do not equal 0. What is $\frac{M}{N}$ in terms of $w,x,y$ , and $z$ ?

A $\frac{wx}{yz}$
B $\frac{wy}{xz}$
C $\frac{wz}{xy}$
D $\frac{xy}{wz}$

Show the answer

Correct answer: C

It is easier to rewrite

$\frac{M}{N}$

$M\div N$

since they are both fractions.

$M\div N=\frac{w}{x}\div \frac{y}{z}=\frac{w}{x}\times \frac{z}{y}=\frac{wz}{xy}$

Question analysis

This algebra question defines M and N as fractions built from the letters w, x, y and z, and asks for M divided by N in terms of those letters. Students must divide one fraction by another using only variables. Without numbers to check against, it is easy to multiply instead of divide or flip the wrong fraction.

NYS standards: 6.NS.1 dividing fractions by fractions
7.NS.2c properties for multiplying and dividing
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students divide numeric fractions such as 2/3 ÷ 4/5. This item applies the same rule with four letters instead of numbers, which adds abstraction but no new steps.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 10, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is fraction division in disguise, and students who recognize M/N as M ÷ N need one rule. The difficulty is that every choice is built from the same four letters, so a wrong rule still produces something that looks like a reasonable answer. Substituting small numbers, such as w = 1, x = 2, y = 3 and z = 4, checks the result in seconds.

Variables instead of numbers moderate Understanding the question
All four letters stand for unknown nonzero numbers, so dividing w/x by y/z must be done symbolically, with no value to check the result against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
All four choices arrange w, x, y and z as one fraction, and xy/wz is wz/xy turned over, so only carrying out the division in the right order tells them apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Rewrite M/N as M ÷ N, then multiply by the reciprocal of the second fraction; substituting small numbers for the letters checks the result.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Stacks M's letters over N's letters, writing wx over yz, without carrying out any fraction division.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies M by N straight across, top times top and bottom times bottom, instead of dividing.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Flips the first fraction instead of the second, computing x/w × y/z, which gives N divided by M.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a, b, c and d are numbers that are not 0. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). It works because dividing by a number is the same as multiplying by 1 over that number.
Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the tops and b and d are the bottoms. Multiply the tops together and the bottoms together. Use it after you flip the second fraction.

Sort out what you know
M is the fraction w over x. N is the fraction y over z. The letters stand for numbers that are not 0.
The fraction bar in $\frac{M}{N}$ means M divided by N. So you must divide one fraction by another.
Write it as a division
Rewrite the big fraction as a division. Then put in the fractions for M and N. This lets you use the rule for dividing fractions:
$\frac{M}{N} = M \div N = \frac{w}{x} \div \frac{y}{z}$
Flip the second fraction
Use the rule for dividing fractions. Keep the first fraction. Flip only the second one, and change divide to multiply:
$\frac{w}{x} \times \frac{z}{y}$
Watch out: flipping the first fraction instead gives choice D, which is N ÷ M. Multiplying without flipping at all gives choice B.
Multiply across
Multiply the tops and the bottoms:
$\frac{w \times z}{x \times y} = \frac{wz}{xy}$
Here wz means w times z. So the answer is C, $\frac{wz}{xy}$ .

Check your answer

Try small numbers. Let w = 1, x = 2, y = 3 and z = 4. Then M = 1/2 and N = 3/4. So M ÷ N = 1/2 × 4/3 = 4/6. Choice C gives (1 × 4) ÷ (2 × 3) = 4/6 too. They match.

More questions like this

The original DOE question

Question 96 as printed on page 88 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 88, question 96. Open page 88 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 109 of 120

2018 Test A Math, question 41

Question

$Bar graph titled "Number of Schools in Five Cities". The vertical axis, Number of Schools, runs from 0 to 10 in steps of 1; the horizontal axis is City. The bars show M 8, N 5, P 9, Q 6 and R 4 schools.$

The graph above shows the number of schools per city for five small cities. Cities M and N each have 500 students per school. City P has 400 students per school. Cities Q and R each have 700 students per school. Which of the five cities has the greatest number of students?

ACity M
BCity P
CCity Q
DCity R

Show the answer

Correct answer: C

Take each city’s number of schools and multiply by the number of students. It is not necessary to calculate all 5 of these.

Cities M and N have the same number of students per school, so just calculate the number of students in City M because it has more schools than City N.

The same goes for Q and R — only Q needs to be calculated because it has more schools than R.

M= 8 × 500 = 4,000

P = 9 × 400 = 3,600

Q = 6 × 700 = 4,200

City Q has the greatest number of students.

Question analysis

This graph question shows the number of schools in each of five small cities and gives the number of students per school for each city. Students multiply schools by students per school to compare the total number of students. Comparing only the bars, or only the rates, can mislead.

NYS standards: 6.RP.3b unit rate problems
6.SP.5 summarizing data sets
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Reading a bar graph and multiplying is familiar from grade 6. Grade 7 data tasks often add percents or compare distributions; this one needs only a few whole-number products, so it is easier than typical grade 7 work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Much to keep track of minor
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 57 words.
Corrected from the DOE handbook: A misstatement in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Neither the tallest bar nor the highest rate wins, which is the point of the question: total students depend on both. The explanation's shortcut, comparing only M, P and Q because N and R each share a rate with a taller bar, cuts the work to three products. The top two totals are close, so the products must actually be computed rather than estimated.

A very close wrong answer moderate Choosing the answer
City M's 4,000 students and City Q's 4,200 differ by only 200, so judging from bar heights and rates cannot separate them; only computing 8 × 500 and 6 × 700 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Much to keep track of minor Working it out
The rates come in three groups in the stem, 500 for M and N, 400 for P, 700 for Q and R, and each must be matched to its bar. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Reading a figure, table or graph minor Understanding the question
The number of schools comes only from the bar heights, read against gridlines from 0 to 10; all five bars end exactly on a gridline. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: For each city, multiply the number of schools read from the graph by its students per school and compare the products; cities that share a rate need only their school counts compared.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Computes M's 8 × 500 = 4,000 and P's 9 × 400 = 3,600 and stops, never working out Q's 6 × 700 = 4,200.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Picks the tallest bar, P's nine schools, which answers which city has the most schools rather than the most students.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks a city with the stem's highest rate, 700 students per school, without noticing that R has only 4 schools to Q's 6.

What was corrected from the DOE handbook, and why

Explanation: The handbook says “Cities M and N have the same number of students”. They have the same number of students per school (500), not the same total: M has 8 × 500 = 4,000 and N has 5 × 500 = 2,500, which is why only City M, with more schools, needs to be calculated. The sentence now says “the same number of students per school”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total students in a city: $S = n \times s$; S is the total number of students in a city. n is the number of schools, read from the bar. s is the number of students in each school. You multiply because every school in that city has the same number of students.

Read the graph
The bar graph shows how many schools each city has. Read the top of each bar against the numbers on the left. M has 8, N has 5, P has 9, Q has 6 and R has 4.
The question gives the students per school. You need the city with the most students in all.
Multiply for each city
Use the formula for each city:
City Schools Students per school Total students
M 8 500 4,000
N 5 500 2,500
P 9 400 3,600
Q 6 700 4,200
R 4 700 2,800
Compare the totals
The biggest total is 4,200, for City Q.
Watch out: City P has the tallest bar, but that means the most schools, not the most students. That leads to choice B. City M’s 4,000 is close, so do not stop before you work out Q. That is choice A.
So the answer is C, City Q.

City	Schools	Students per school	Total students
M	8	500	4,000
N	5	500	2,500
P	9	400	3,600
Q	6	700	4,200
R	4	700	2,800

Check your answer

A shortcut: M and N both have 500 per school, and M has more schools, so M beats N. Q and R both have 700, and Q beats R. So only M, P and Q matter: 4,000, 3,600 and 4,200. Q still wins.

More questions like this

The original DOE question

Question 98 as printed on page 89 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 89, question 98. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 110 of 120

2018 Test A Math, question 43

Question

$\frac{w}{x}=\frac{y}{z}$

In the equation above, $w,x,y$ , and $z$ are positive numbers. Which of these is equal to $z$ ?

A $x$
B $xy$
C $\frac{w}{xy}$
D $\frac{xy}{w}$

Show the answer

Correct answer: D

Solve the equation for $z$ :

$\frac{w}{x}=\frac{y}{z}$

$wz=xy$

$z=\frac{xy}{w}$

Question analysis

This algebra question gives the proportion w/x = y/z, with all four letters positive, and asks which expression equals z. Students must solve the equation for one letter in terms of the others. With only letters on the page, it is easy to multiply when you should divide or to stop one step early.

NYS standards: 8.EE.7 linear equations in one variable
6.EE.7 one-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students cross-multiply numeric proportions such as 3/4 = 9/x. Solving a proportion made only of letters for one of them is a literal equation, a grade 8 algebra skill, though it takes just two steps.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 10, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method, cross-multiply and divide, is the same one students use on numeric proportions, but letters hide whether each step went the right way. The two most attractive wrong choices are both one step from the answer: one never divides, and the other divides upside down. A quick check with 2/3 = 4/6, where z = 6, settles it.

Variables instead of numbers moderate Understanding the question
The proportion has only letters, so cross-multiplying to wz = xy and dividing by w must be done without numbers to show whether w goes on top or bottom. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
The choices w/(xy) and xy/w use the same three letters with the fraction turned over, so only dividing in the right direction after wz = xy separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Cross-multiply to clear both fractions, then divide both sides by whatever multiplies z; check with a true proportion such as 2/3 = 4/6.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches z with x because both sit in a denominator, picking a letter from the equation without solving for z.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Cross-multiplies to wz = xy but stops there, giving xy without dividing both sides by w.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Cross-multiplies correctly but divides the wrong way round, putting w over xy instead of xy over w.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a, b, c and d are numbers that are not 0. When two fractions are equal, each top times the other bottom gives the same answer. It works because you multiply both sides by both bottoms, which clears the fractions.

Sort out what you need
The equation says w over x equals y over z. All four letters are positive numbers. You need to get z alone on one side. Then the other side shows what z equals.
Cross-multiply the fractions
Multiply each top by the other fraction’s bottom:
$w \times z = x \times y$
So wz = xy, and z is no longer on the bottom. Watch out: stopping here and picking xy gives choice B. z is not alone yet.
Get z alone
z is being multiplied by w. To undo that, divide both sides by w. Doing the same to both sides keeps them equal:
$\frac{w \times z}{w} = \frac{x \times y}{w}$
$z = \frac{xy}{w}$
Watch out: dividing the wrong way round gives w over xy, choice C. So the answer is D, $\frac{xy}{w}$ .

Check your answer

Use a true pair of equal fractions: 2/3 = 4/6. So w = 2, x = 3, y = 4 and z = 6. Choice D gives 3 × 4 ÷ 2 = 6, which is z. Choice C gives 2 ÷ 12, which is not 6.

More questions like this

The original DOE question

Question 100 as printed on page 89 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 89, question 100. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 111 of 120

2018 Test A Math, question 45

Question

A metal square used in an electronic device must have a thickness of 0.02 inch, with an allowable error of 1 percent. What is the greatest allowable thickness of the metal square?

A0.0002 in.
B0.02 in.
C0.0202 in.
D0.03 in.

Show the answer

Correct answer: C

Find 1% of 0.02:

$(0.02)(\frac{1}{100})=0.0002$

The greatest allowable thickness would be

$0.02+0.0002=0.0202$ inch.

Question analysis

This percent question says a metal square must be 0.02 inch thick with an allowable error of 1 percent, and asks for the greatest allowable thickness. Students must find 1 percent of a small decimal and add it to the target thickness. Keeping track of decimal places is what makes it tricky.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Percent error and tolerance problems appear in grade 7, so the idea is on level. The demand comes from taking a percent of a tiny decimal and adding the result, where a slipped decimal place is easy to miss.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math moderate
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The idea of tolerance is simple, but the numbers are tiny, and the question turns on what the 1 percent is a percent of. Students who read it as 0.01 inch land on a choice that looks sensible, and those who find the error correctly may stop before adding it. Writing 0.02 × 1.01 in one step avoids both.

Error-prone arithmetic moderate Working it out
One percent of 0.02 is 0.0002, two more decimal places, and adding it to 0.02 needs the places lined up to get 0.0202. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math moderate Understanding the question
The phrase "allowable error of 1 percent" must be read as 1 percent of 0.02 inch, added for the greatest thickness, not 1 percent as a fixed 0.01 inch. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The choices 0.0202, 0.02 and 0.0002 are built from the same digits, and the first two differ by only 1 percent, so only careful counting of decimal places tells them apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Find 1 percent of 0.02 by dividing by 100, counting decimal places carefully, then add that error to 0.02 to get the upper limit.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the allowable error, 1 percent of 0.02 = 0.0002 inch, and stops without adding it to the target thickness.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the target thickness from the stem, 0.02 inch, ignoring that the allowed error permits a slightly thicker square.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Treats 1 percent as 0.01 inch, a percent of 1 inch rather than of 0.02 inch, and adds it to get 0.03.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $x = \frac{p}{100} \times n$; x is the part you want. p is the percent. n is the whole amount. Percent means out of 100, so p percent of a number is p hundredths of it.

Sort out what you know
The square should be 0.02 inch thick. An allowable error of 1 percent means it may be off by up to 1 percent of 0.02 inch. It may be that much thinner or thicker.
You need the greatest allowed thickness, so look at the thicker side.
Find 1 percent of 0.02
$x = \frac{1}{100} \times 0.02 = 0.0002$
Dividing by 100 moves the decimal point two places to the left. So the error is 0.0002 inch.
Watch out: 1 percent here is not 0.01 inch. It is 1 percent of 0.02 inch. Adding 0.01 gives 0.03, which is choice D.
Add the error to the thickness
The thickest square is the right size plus the error. Line up the decimal points:
$0.0200 + 0.0002 = 0.0202$
Watch out: 0.0002 is only the error, choice A. You must add it to 0.02. So the answer is C, 0.0202 in.

Check your answer

0.0202 is just a tiny bit more than 0.02, which makes sense for a small error. Check it: 0.0202 − 0.02 = 0.0002. And 0.0002 ÷ 0.02 = 0.01, which is 1 percent.

More questions like this

The original DOE question

Question 102 as printed on page 90 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 102. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 112 of 120

2018 Test A Math, question 47

Question

If $3n$ is a positive even number, how many odd numbers are in the range from $3n$ up to and including $3n+5$ ?

Show the answer

Correct answer: B

Since $3n$ is even, then $3n+1$ must be odd.

Thus $3n+3$ and $3n+5$ are also odd.

So there are a total of 3 numbers in this range that are odd.

Question analysis

This number properties question says 3n is a positive even number and asks how many odd numbers lie from 3n up to and including 3n + 5. Students must decide which integers in that span are odd. The coefficient in 3n is a distraction; only the fact that 3n is even matters.

NYS standards: 6.EE.2b parts of an expression
6.EE.6 variables representing unknown numbers
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students work with expressions like 3n + 5 but rarely reason about whether they are even or odd. The idea is simple once the integers are written out, so the demand is about typical for grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Something you must already know minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 12, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Writing the six integers from 3n to 3n + 5 and marking them even, odd, even, odd, even, odd answers the question in seconds. The traps come from not writing them: dropping the last one, which "up to and including" keeps in, or reading the 5 in 3n + 5 as a count. Picking any even value for 3n, such as 6, makes the check concrete.

Variables instead of numbers moderate Understanding the question
No value of n is given, so students must reason about parity: since 3n is even, 3n + 1, 3n + 3 and 3n + 5 are the odd ones. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
The parity rules, even plus odd is odd and even plus even is even, are assumed; the coefficient 3 is a distraction once 3n is said to be even. (Anderson & Krathwohl, 2001)
A very close wrong answer minor Choosing the answer
The choices 2, 3, 4 and 5 are consecutive counts, so 2 and 3 cannot be told apart by sizing up; only checking whether 3n + 5 is inside the range separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write out every integer from 3n to 3n + 5, mark 3n as even, alternate even and odd from there, and count the odd ones.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Leaves out 3n + 5, stopping before the last number although "up to and including" keeps it, and counts only 3n + 1 and 3n + 3.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Counts 3n itself as odd because its coefficient 3 is odd, ignoring the stem's statement that 3n is even, and adds it to the three.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the 5 in 3n + 5 as the count, subtracting the ends instead of listing the six integers and picking out the odd ones.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Even plus odd: $\text{even} + \text{odd} = \text{odd}$; An even number splits into pairs. An odd number has one left over. Adding them keeps that one left over, so the sum is odd. So adding 1, 3 or 5 to an even number gives an odd number.
Even plus even: $\text{even} + \text{even} = \text{even}$; Two numbers that each split into pairs still split into pairs when you add them. So adding 2 or 4 to an even number gives an even number.

Sort out what you know
You are told that 3n is even. You do not need to know n itself. You need to count the odd numbers from 3n to 3n + 5.
“Up to and including” means 3n + 5 counts too.
List all the numbers
Go up by 1 each time, from 3n to 3n + 5. That gives six numbers:
$3n,\ 3n+1,\ 3n+2,\ 3n+3,\ 3n+4,\ 3n+5$
Mark each as even or odd
3n is even. Adding 1, 3 or 5 makes it odd. Adding 2 or 4 keeps it even.
Number Even or odd
$3n$ even
$3n+1$ odd
$3n+2$ even
$3n+3$ odd
$3n+4$ even
$3n+5$ odd
Watch out: 3n is even because the question says so. The 3 being odd does not matter. Counting 3n as odd gives choice C, 4.
Count the odd numbers
The odd ones are 3n + 1, 3n + 3 and 3n + 5. That is 3 numbers.
Watch out: leaving out 3n + 5 gives 2, choice A. It is the last number, and it is included. So the answer is B, 3.

Number	Even or odd
$3n$	even
$3n+1$	odd
$3n+2$	even
$3n+3$	odd
$3n+4$	even
$3n+5$	odd

Check your answer

Try a real number. If n = 2, then 3n = 6, which is even. The numbers from 6 to 11 are 6, 7, 8, 9, 10 and 11. The odd ones are 7, 9 and 11. That is 3.

More questions like this

The original DOE question

Question 104 as printed on page 90 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 104. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 113 of 120

2018 Test A Math, question 51

Question

A sports store has a container of handballs: 4 blue, 5 red, 8 yellow, 9 white, and 11 green. If one ball is picked from the container at random, what is the probability that it will be yellow?

A $\frac{1}{37}$
B $\frac{1}{8}$
C $\frac{8}{37}$
D $\frac{8}{29}$

Show the answer

Correct answer: C

The total number of handballs in the container is

$4+5+8+9+11=37$

Since there are 8 yellow handballs, the probability of selecting a yellow handball is

$\frac{8}{37}$

Question analysis

This probability question describes a container of handballs in five colors and asks the chance that one ball picked at random is yellow. Students must add all the balls to get the total and compare the yellow count to it. Using the wrong total, such as only the balls that are not yellow, is the easy mistake.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of a single event as favorable outcomes over total outcomes is grade 6 content. This item takes one addition and one fraction, so it is easier than typical grade 7 compound probability work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-event probability: add the five colors to get 37 balls and put the 8 yellow ones over that total. The only real decision is the denominator, and the wrong choices offer the non-yellow count, the yellow count, and a numerator of 1. Students who say "favorable over total" to themselves avoid all three.

A very close wrong answer minor Choosing the answer
The choices 8/37 and 8/29 share the numerator 8 and differ only in the denominator, all 37 balls against the 29 that are not yellow. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Add up every ball in the container to get the total number of outcomes, then put the number of yellow balls over that total.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds the 37 balls correctly but puts 1 on top, giving the chance of picking one particular ball rather than any yellow one.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Puts 1 over the 8 yellow balls, a fraction built from the yellow count alone that ignores the other 29 balls.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Compares the 8 yellow balls with the 29 that are not yellow, a part-to-part ratio, instead of with all 37 balls.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{f}{t}$; P is the probability, the chance that something happens. f is the number of favorable outcomes, the ones you want. t is the total number of outcomes. It works when every outcome is equally likely, like picking one item at random.

Sort out what you know
The container has 4 blue, 5 red, 8 yellow, 9 white and 11 green balls. One ball is picked at random, so every ball has the same chance. You need the chance it is yellow.
Count all the balls
Add up every color, because any ball could be picked:
$4 + 5 + 8 + 9 + 11 = 37$
There are 37 balls in all. That is the total number of outcomes.
Put yellow over the total
8 of the balls are yellow. Those are the favorable outcomes:
$P = \frac{8}{37}$
Watch out: $\frac{8}{29}$ compares yellow with the 29 balls that are not yellow. Probability compares the part with all the balls, so choice D is wrong. So the answer is C, $\frac{8}{37}$ .

Check your answer

8 out of 37 is about 0.22, a little more than 1 in 5. That makes sense, since yellow is a fair share of the balls. Choice A, 1/37, is the chance of picking one certain ball, not any yellow one.

More questions like this

The original DOE question

Question 108 as printed on page 91 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 91, question 108. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 114 of 120

2018 Test A Math, question 52

Question

Each week, Leon has fixed expenses of $1,250 at his furniture shop. It costs him $150 to make a chair in his shop, and he sells each chair for $275. What is Leon’s profit if he makes and sells 25 chairs in 1 week?

A$1,875
B$2,500
C$3,125
D$4,375

Show the answer

Correct answer: A

Each chair costs Leon $150 to make, and he sells the chair for $275.

His profit is found by subtracting the cost from the price:

$275 − $150 = $125 per chair

If Leon makes and sells 25 chairs in a week, his initial profit is

25 × $125 = $3,125

However, Leon has additional fixed expenses of $1,250 per week, so this cost must also be subtracted to arrive at the profit. His final profit is

$3,125 − $1,250 = $1,875

Question analysis

This profit word problem gives Leon's weekly fixed expenses at his furniture shop, the cost to make each chair and its selling price, then asks his profit for 25 chairs in one week. Students must find the profit on each chair, multiply, and account for the fixed expenses. Forgetting the fixed expenses, or adding them, are the main slips.

NYS standards: 7.EE.3 multi-step rational number problems
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Multi-step money problems are standard grade 7 work. This one has three whole-number steps and a familiar profit idea, so it matches typical grade 7 demand, with the fixed cost as the detail to remember.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is whole-number and friendly, so the question is about bookkeeping: which amounts are income, which are costs, and which happen once a week. The fixed expenses are named first in the stem and are easy to forget by the time the per-chair profit is multiplied. A final check that revenue minus all costs, 6,875 − 3,750 − 1,250, gives the same result confirms it.

Much to keep track of moderate Working it out
Three dollar amounts play different roles, a weekly fixed expense, a cost per chair and a price per chair, and each must enter the profit with the right sign. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
The bold "profit" is not defined; students must know it means money taken in minus every cost, including costs that do not depend on the number of chairs. (Anderson & Krathwohl, 2001)

Strategy: Subtract the cost from the price to get the profit per chair, multiply by the number of chairs, then subtract the week's fixed expenses.

Traps

Stops one step early: A choice is a number from partway through the solution.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the $150 cost by 25 chairs and subtracts the $1,250, treating the cost of making a chair as money earned.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the profit on 25 chairs at $125 each, $3,125, and stops before subtracting the $1,250 fixed expenses.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds the $1,250 fixed expenses to the $3,125 chair profit instead of subtracting them, treating an expense as income.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Profit: $P = I - C$; P is the profit, the money Leon keeps. I is the income, the money he takes in from sales. C is the total cost: every dollar he spends, including fixed expenses. Profit is what is left after all costs are paid.

Sort out what you know
Leon pays $1,250 each week no matter what. These are fixed expenses, like rent. Each chair costs $150 to make, and he sells each one for $275.
He makes and sells 25 chairs. Profit is the money left after paying all the costs.
Find the profit on one chair
Each chair sells for $275 but costs $150 to make. Subtract to see what he keeps:
$275 - 150 = 125$
He keeps $125 on each chair he sells.
Find the profit on 25 chairs
$25 \times 125 = 3{,}125$
The chairs bring in $3,125 more than they cost to make. Watch out: $3,125 is choice C, but it is not the final profit. The fixed expenses still have to be paid.
Subtract the fixed expenses
$P = 3{,}125 - 1{,}250 = 1{,}875$
Watch out: the $1,250 is money spent, so subtract it. Adding it gives $4,375, choice D. So the answer is A, $1,875.

Check your answer

Work it the long way. Sales: 25 × $275 = $6,875. Costs: 25 × $150 = $3,750, plus $1,250, which makes $5,000. Profit: $6,875 − $5,000 = $1,875. Same answer.

More questions like this

The original DOE question

Question 109 as printed on page 91 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 91, question 109. Open page 91 of the handbook (PDF)

Question 115 of 120

2018 Test A Math, question 56

Question

SERVINGS OF FRUITS AND VEGETABLES

Number of Servings of Fruits and Vegetables	Number of Students
0	5
1	7
2	3
3	4
4	0
5	1

There are 20 students in a class. The frequency table above shows the number of students in this class who ate 0, 1, 2, 3, 4, or 5 servings of fruits and vegetables yesterday. What is the mean number of servings of fruits and vegetables eaten yesterday per student in this class?

A $1\frac{1}{2}$
B $3$
C $3\frac{1}{3}$
D $4$

Show the answer

Correct answer: A

First, determine the total number of servings of fruits and vegetables that the students ate by multiplying the number of servings by the number of students in each row of the table.

Then add that column to get the total number of servings:

Number of Servings of Fruits and Vegetables	Number of Students	Number of Servings × Number of Students
0	5	0
1	7	7
2	3	6
3	4	12
4	0	0
5	1	5

Total: 30

Calculate the mean by dividing the total number of servings of fruits and vegetables by the total number of students:

$\frac{30}{20}=1\frac{1}{2}$

Question analysis

This statistics question uses a frequency table showing how many of 20 students ate 0, 1, 2, 3, 4 or 5 servings of fruits and vegetables, and asks for the mean servings per student. Students must weight each serving amount by its number of students, add, and divide. Averaging the wrong column of the table is the common mistake.

NYS standards: 6.SP.5c range, center, and overall pattern
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Finding a mean is grade 6 content; doing it from a frequency table, where each value must be weighted by its count, is the multi-step version grade 7 students meet regularly.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 64 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The answer is smaller than every wrong choice, which surprises students who average the table's columns directly. The question rewards seeing that each row stands for several students, so the total servings come from weighting each value by its count. The stem's reminder that there are 20 students points to the right divisor.

Reading a figure, table or graph moderate Understanding the question
The frequency table has a value column and a count column; the mean needs each value multiplied by its count, so the two columns play different roles. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Six products, 0, 7, 6, 12, 0 and 5, must be found and added to 30 before dividing by the 20 students. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply each serving amount by its number of students, add those products to get the total servings, then divide by the 20 students.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Averages the serving amounts 1 through 5 listed in the table, (1 + 2 + 3 + 4 + 5) ÷ 5 = 3, without weighting by the students.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Averages the Number of Students column, 20 students over 6 rows, finding the mean count per row instead of the servings per student.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the 20 students by 5, the largest serving amount, combining two numbers from the page without a model of the mean.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean: $m = \frac{S}{n}$; m is the mean, or average. S is the sum of all the values, here the total servings eaten. n is the number of values, here the number of students. The mean shares the total equally, as if every student ate the same amount.

Sort out the table
This is a frequency table. Each row says how many students ate a certain number of servings. For example, the second row says 7 students ate 1 serving each.
There are 20 students. You need the mean, or average, number of servings per student.
Find the servings in each row
In each row, multiply the servings by the number of students:
Servings Students Servings × Students
0 5 0
1 7 7
2 3 6
3 4 12
4 0 0
5 1 5
Add up all the servings
$0 + 7 + 6 + 12 + 0 + 5 = 30$
The class ate 30 servings in all. The students add up to 20, as the question says: 5 + 7 + 3 + 4 + 0 + 1 = 20.
Divide by the number of students
$m = \frac{30}{20} = 1\frac{10}{20} = 1\frac{1}{2}$
Watch out: do not just average the serving numbers 1 to 5. That gives 3, choice B, and ignores how many students ate each amount. Dividing 20 students by 6 rows gives choice C.
So the answer is A, $1\frac{1}{2}$ .

Servings	Students	Servings × Students
0	5	0
1	7	7
2	3	6
3	4	12
4	0	0
5	1	5

Check your answer

12 of the 20 students ate 0 or 1 serving. Only 5 ate 3 or more. So the mean should be small, between 1 and 2. $1\frac{1}{2}$ fits. Choices B, C and D are all 3 or more, which is too big.

More questions like this

The original DOE question

Question 113 as printed on page 92 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 92, question 113. Open page 92 of the handbook (PDF)

Question 116 of 120

2018 Test A Math, question 57

Question

A paste used to cover a billboard is made by mixing the following ingredients by weight: 4 parts powder, 3 parts water, 2 parts resin, and 1 part hardener. To cover one billboard requires 30 pounds of this paste. How many total pounds of resin are required to cover 4 billboards?

Show the answer

Correct answer: C

The ratio is $4:3:2:1$ , so the total parts is 10.

Since there are two parts resin, the fraction of resin is

$\frac{2}{10}=\frac{1}{5}$

So the amount of resin in 30 lb of paste (for 1 billboard) is

$\frac{1}{5}\times 30=6$ lb.

For 4 billboards, that would be 6 × 4 = 24 lb.

Question analysis

This ratio word problem gives a billboard paste recipe of 4 parts powder, 3 parts water, 2 parts resin and 1 part hardener by weight, with 30 pounds of paste per billboard. It asks how many pounds of resin four billboards need. Students must find resin's share of the total parts and scale up to four billboards.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Part-to-whole ratio problems with several ingredients are typical grade 7 work. This one adds a final scaling step for four billboards, which keeps it a routine multistep ratio problem at grade 7 level.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 45 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The recipe lists four ingredients but only resin's share matters, so the work is part over whole and then scaling. The question asks for four billboards only in the last sentence, after the per-billboard amount has been set up, which is why stopping early is the main risk. Finding the total paste first, 120 pounds, and taking a fifth of it reaches the answer in one step.

Many steps minor Working it out
Add the parts to 10, take resin's 2/10 of 30 pounds, then multiply by 4 billboards: three short steps. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the ratio parts to get the total, find resin's fraction of one batch of paste, take that fraction of 30 pounds, and multiply by the number of billboards.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the resin for one billboard, 2/10 of 30 = 6 pounds, and stops without multiplying by four billboards.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Treats resin's 2 parts as 2 pounds and multiplies by 4 billboards, ignoring the 30 pounds of paste each one needs.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Uses powder's 4 parts instead of resin's 2, finding 4/10 of 120 pounds, which is the powder needed for four billboards.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction from a ratio: $f = \frac{p}{T}$; f is the fraction of the mix that is one ingredient. p is that ingredient’s number of parts. T is the total number of parts, found by adding all the ratio numbers. Use it to turn a ratio into a fraction of the whole.

Sort out what you know
The paste is 4 parts powder, 3 parts water, 2 parts resin and 1 part hardener, by weight. A part is just an equal share of the mix.
One billboard needs 30 pounds of paste. You need the pounds of resin for 4 billboards.
Find resin’s fraction of the paste
$4 + 3 + 2 + 1 = 10$
The mix has 10 parts in all, and resin is 2 of them:
$f = \frac{2}{10} = \frac{1}{5}$
So resin is one fifth of the paste. Watch out: 2 parts is not 2 pounds. Treating it that way gives 2 × 4 = 8, choice B.
Find the resin for one billboard
$\frac{1}{5} \times 30 = 6$
One billboard needs 6 pounds of resin. Watch out: 6 is choice A, but that covers only one billboard.
Multiply by four billboards
Each billboard needs the same 6 pounds of resin, so multiply by 4:
$6 \times 4 = 24$
Four billboards need 24 pounds of resin. So the answer is C, 24.

Check your answer

Work it another way. Four billboards need 4 × 30 = 120 pounds of paste. One fifth of 120 is 24. Choice D, 48, is the powder, which has 4 parts out of 10.

More questions like this

The original DOE question

Question 114 as printed on page 92 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 92, question 114. Open page 92 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 117 of 120

2018 Test B Math, question 1

Question

How many 5-digit numbers can be created using the digits 2, 3, 5, 7, and 8 without repeating any digits within that 5-digit number?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 120

There are 5 choices for the first digit, 4 choices for the second digit, 3 choices for the third digit, 2 choices for the fourth digit, and 1 choice for the final digit.

The total number of possibilities is

5 × 4 × 3 × 2 × 1 = 120

Question analysis

This grid-in counting question asks how many different 5-digit numbers can be made from five given digits when no digit may repeat. It tests the counting principle for arrangements rather than arithmetic. The catch is the no-repeat rule: each place has one fewer digit to choose from, and ignoring that gives a far larger count.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 probability work builds sample spaces with lists and tree diagrams for two or three stages. This needs the multiplication principle with a shrinking number of choices across five places, which listing cannot reach.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 11, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is tiny; the difficulty is knowing that arrangements are counted by multiplying, and that the no-repeat rule removes one digit for each later place. Students who have met the counting principle finish in seconds, while those who start listing numbers run out of time or guess a product they half remember. With no choices to check against, the 3,125 that comes from ignoring the rule looks as convincing as the right count.

Something you must already know moderate Working it out
Nothing in the stem names the counting principle: students must already know to multiply the digits left for each place, 5 × 4 × 3 × 2 × 1, because listing 5-digit numbers by hand cannot finish. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in there are no choices to compare with, so a count of 3,125 or 25 cannot be caught by noticing that it is missing from a list. (Enright, Morley & Sheehan, 2002)

Strategy: Fill the five places from left to right, write how many digits are still available for each place, and multiply those five counts together.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.

The wrong answers, classified

3125 Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Uses the counting principle but lets digits repeat, multiplying 5 × 5 × 5 × 5 × 5 instead of using one fewer digit for each later place.
25 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Multiplies the five digits by the five places, 5 × 5, as if each digit simply took each position once.
15 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Sees that the choices shrink from 5 to 1 but adds them, 5 + 4 + 3 + 2 + 1, instead of multiplying.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle: $\text{total ways} = a \times b \times c \times \ldots$; a is the number of choices for the first spot, b for the second, c for the third, and so on. Multiply them to get every possible way. It works because each choice for one spot can go with each choice for the next spot.

Sort out the question
You have five digits: 2, 3, 5, 7 and 8. A digit is one of the symbols 0 to 9. You build 5-digit numbers from them.
“Without repeating” means each digit is used only once in a number. So 23578 is allowed, but 22578 is not.
You need to find how many different numbers you can make.
Count the choices for each spot
Think of the number as five empty spots. Fill them from left to right.
- First spot: any of the 5 digits, so 5 choices.
- Second spot: one digit is used up, so 4 are left.
- Third spot: 3 are left.
- Fourth spot: 2 are left.
- Last spot: only 1 is left.
Multiply the choices together
Use the counting principle. Each choice for one spot works with every choice for the next spot. So you multiply. You do not add.
$\text{total ways} = 5 \times 4 \times 3 \times 2 \times 1 = 120$
There are 120 different numbers.
Watch out: Adding 5 + 4 + 3 + 2 + 1 gives 15, which is far too small. Using 5 choices for every spot gives 3,125. That count lets digits repeat, which the question does not allow.
So you would grid 120.

Check your answer

Try a smaller case you can list. With the digits 2, 3 and 5, the rule gives $3 \times 2 \times 1 = 6$ numbers. The list is 235, 253, 325, 352, 523 and 532. That is 6, so the rule works.

More questions like this

The original DOE question

Question 58 as printed on page 153 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 58. Open page 153 of the handbook (PDF)

Question 118 of 120

2018 Test B Math, question 11

Question

At 1:00 p.m. one day, the temperature was 8 degrees above zero. During the rest of the day, the temperature fell 3 degrees per hour. What was the temperature at 7:00 p.m. that day?

A−13°
B−10°
C−7°
D5°

Show the answer

Correct answer: B

7:00 p.m. is 6 hours after 1:00 p.m. Calculate the number of degrees the temperature dropped in 6 hours:

3 × 6 = 18 degrees

Subtract that from the starting point (8 degrees) to find the solution:

8 − 18 = −10 degrees

Question analysis

This integer word problem starts with a temperature 8 degrees above zero at 1:00 p.m. and has it fall 3 degrees every hour. Students find the temperature at 7:00 p.m. by counting hours and subtracting into negative numbers. Getting the number of hours between the two times right is the main hurdle.

NYS standards: 7.NS.3 four operations with rational numbers
7.NS.1b rational number addition
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students use positive and negative numbers for real changes such as temperature. This is a standard three-step example of that skill, with a counting slip as its only real trap.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light; the question turns on counting the hours from 1:00 to 7:00 correctly. Counting the clock times instead of the gaps between them gives seven hours, starting the drop an hour late gives five, and both errors are built into the choices. Subtracting the times, 7 − 1, is safer than counting on fingers.

Error-prone arithmetic minor Working it out
The total drop of 18 degrees is larger than the starting 8 degrees, so the subtraction crosses zero and the result must be negative. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the hours that pass from 1:00 to 7:00 by subtracting the times, multiply by the 3-degree drop, and subtract that total from 8.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts the clock times 1, 2, 3, 4, 5, 6, 7 as seven hours, so subtracts 21 degrees from 8.
C Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Starts the drop at 2:00 instead of 1:00, counting five hours and subtracting 15 degrees from 8.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts a single 3-degree drop from 8, combining the two numbers in the question without counting the hours.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total change: $\text{change} = \text{rate} \times \text{time}$; The rate is how much the temperature changes each hour, here 3 degrees down. The time is the number of hours. Multiply them to get the total change, because the same drop happens every hour.

Write down what you know
At 1:00 p.m. it was 8 degrees above zero. Each hour after that, it dropped 3 degrees.
You need the temperature at 7:00 p.m.
Count the hours
Subtract the clock times: 7 − 1 = 6. So 6 hours pass.
Watch out: Do not count the clock numbers 1, 2, 3, 4, 5, 6, 7. That is 7 numbers but only 6 hours. Using 7 hours gives −13°, choice A. Starting the count at 2:00 gives 5 hours and −7°, choice C.
Find the total drop
The temperature drops the same amount each hour, so multiply.
$\text{change} = \text{rate} \times \text{time} = 3 \times 6 = 18$
The temperature fell 18 degrees in all.
Subtract from the start
Start at 8 degrees and take away the 18-degree drop.
$8 - 18 = -10$
The drop is bigger than 8, so the temperature goes below zero. A temperature below zero is a negative number. It is 10 degrees below zero.
So the answer is B, −10°.

Check your answer

Count down hour by hour: 8 at 1:00, 5 at 2:00, 2 at 3:00, −1 at 4:00, −4 at 5:00, −7 at 6:00 and −10 at 7:00. You land on −10.

More questions like this

The original DOE question

Question 68 as printed on page 155 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 155, question 68. Open page 155 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 119 of 120

2018 Test B Math, question 12

Question

A bag contains 75 marbles that are red, blue, or green. The ratio of red to blue marbles is $15:7$ , and the ratio of blue to green marbles is $7:3$ . If 2 blue marbles are removed and replaced with 2 green marbles, what will be the new ratio of red to green marbles?

A $3:1$
B $5:1$
C $15:3$
D $45:11$

Show the answer

Correct answer: D

The ratio of red to blue to green is 15 : 7 : 3.

Find the proportion of blue marbles. Add the numbers of the ratio and use the total sum as the denominator:

$\frac{7}{15+7+3}=\frac{7}{25}$

Find the proportion of green marbles:

$\frac{3}{25}$

Since there are a total of 75 marbles, the number of blue marbles is

$\frac{7}{25}\times 75=21$

The number of green marbles is

$\frac{3}{25}\times 75=9$

The number of red marbles is

$75-21-9=45$

If 2 blue marbles are removed and replaced with 2 green marbles, the number of blue marbles is now 19 and the number of green marbles is now 11. The ratio of red marbles to green marbles is 45 : 11.

Question analysis

This ratio problem describes a bag of 75 marbles with a red-to-blue ratio and a blue-to-green ratio that share the blue term, then swaps 2 blue marbles for 2 green ones. Students combine the ratios, find the actual number of each color, and form a new ratio. The swap at the end is what turns a routine ratio exercise into a long one.

NYS standards: 7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 ratio problems usually involve one ratio and one scaling step. This links two ratios through a shared term, converts to counts and then changes the counts, a longer chain than typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Much to keep track of minor
Turning words into math minor
Solution: 5 steps
Reading level: Question and choices: about grade 5, 54 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is easy because both ratios share the 7 for blue, so the work is about keeping the chain going without stopping early. The main danger is doing the swap on the ratio numbers or forgetting it altogether. Two of the wrong choices, 5:1 and 15:3, are the same ratio, so a student who notices that can rule out both before doing any arithmetic.

Many steps moderate Working it out
The chain is long: merge 15:7 and 7:3 into 15:7:3, find that 25 parts make 75 marbles, scale to 45, 21 and 9, then add 2 to the green count. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Three colors' counts must be tracked through the swap, which lowers blue and raises green while red stays where it was. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math minor Understanding the question
The phrase “2 blue marbles are removed and replaced with 2 green marbles” must be read as changing the actual counts, not the ratio numbers 7 and 3. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Merge the two ratios into one red-to-blue-to-green ratio, scale it so the parts total 75, make the swap on those counts, then write red to green.

Traps

Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the ratio's 3 as the actual number of green marbles and adds the 2 swapped marbles to it, getting 15:5, or 3:1.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the counts 45 red and 9 green correctly but forgets the swap, giving the original ratio 45:9 in lowest terms.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Reads red and green straight off the combined ratio 15:7:3, with no counts and no swap; it equals 5:1 anyway.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Value of one part: $\text{one part} = \frac{\text{total}}{\text{sum of the ratio numbers}}$; A ratio like 15:7:3 splits a total into equal parts. The sum of the ratio numbers is how many parts there are. Dividing the total by that sum tells how many marbles are in one part.

Write down what you know
There are 75 marbles: red, blue and green. A ratio compares two amounts. Red to blue is 15:7. Blue to green is 7:3.
Then 2 blue marbles are taken out, and 2 green ones are put in. You need the new ratio of red to green.
Join the two ratios
Blue is 7 in both ratios. So you can join them into one ratio.
$\text{red} : \text{blue} : \text{green} = 15 : 7 : 3$
This means for every 15 red marbles, there are 7 blue and 3 green.
Find the size of one part
The ratio has $15 + 7 + 3 = 25$ parts in all. The 75 marbles fill those 25 parts.
$\text{one part} = \frac{\text{total}}{\text{sum of the ratio numbers}} = \frac{75}{25} = 3$
Each part is 3 marbles.
Count each color
Multiply each ratio number by 3 to get the real counts.
$15 \times 3 = 45 \qquad 7 \times 3 = 21 \qquad 3 \times 3 = 9$
So there are 45 red, 21 blue and 9 green. They add to 75, so the counts are right.
Make the swap
Take out 2 blue: $21 - 2 = 19$ blue. Put in 2 green: $9 + 2 = 11$ green. Red does not change. It is still 45.
The new ratio of red to green is 45:11. It cannot be simplified, because 45 and 11 share no factor except 1.
Watch out: 45:9 is the ratio before the swap. It simplifies to 5:1, choice B. Choice C, 15:3, is that same old ratio. Make the swap with the real counts.
So the answer is D, 45:11.

Check your answer

Count the marbles after the swap: $45 + 19 + 11 = 75$ . The total is still 75, as it should be. A swap moves marbles in and out but does not change the total.

More questions like this

The original DOE question

Question 69 as printed on page 155 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 155, question 69. Open page 155 of the handbook (PDF)

Question 120 of 120

2018 Test B Math, question 13

Question

DESSERT CHOICES

Dessert	Number of Times Ordered
Cookies	42
Pie	23
Cake	47
Ice Cream	48

The table above shows the number of times that different desserts were ordered at a restaurant. Based on this information, what is the probability of a customer ordering ice cream as a dessert?

A25%
B30%
C40%
D48%

Show the answer

Correct answer: B

The total number of desserts ordered is

42 + 23 + 47 + 48 = 160

The probability that ice cream was chosen is

$\frac{48}{160}=\frac{3}{10}=30\%$ .

Question analysis

This probability question gives a table of how many times four desserts were ordered at a restaurant and asks how likely a customer is to order ice cream, as a percent. Students total the orders and divide. The trap is using a raw count, or the wrong total, in place of the full number of desserts ordered.

NYS standards: 6.SP.8b probability models from observed frequencies
6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a probability from observed frequencies is grade 6 content in New York. Here it takes one sum, one division and a conversion to a percent with friendly numbers, lighter than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 45 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a direct relative-frequency question with friendly numbers, and most students finish it quickly. The risk is using a number from the table as if it were already a percent, or treating the four desserts as equally likely. Remembering that a probability compares one count with the total of all counts prevents both mistakes.

Reading a figure, table or graph minor Understanding the question
The four counts sit in a table, and the probability needs all of them: the ice cream row gives the numerator and the sum of the column gives the denominator. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
Adding 42 + 23 + 47 + 48 by hand and then simplifying 48/160 to 3/10 are small steps where a careless total spoils the percent. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add the four counts in the table to get all the orders, divide the ice cream count by that total, and convert the fraction to a percent.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the four desserts as equally likely, one chance in four, ignoring how often each was actually ordered.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Divides 48 by the 112 orders of the other desserts instead of all 160, getting about 43%, and takes the nearest choice.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Reads the 48 ice cream orders as 48 percent without dividing by the total number of desserts ordered.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all outcomes}}$; P is the probability, the chance that something happens. The top counts the outcomes you want, here ice cream orders. The bottom counts every outcome, here all dessert orders. It works because each order counts the same.
Fraction to percent: $\text{percent} = \text{fraction} \times 100$; Percent means out of 100. Multiply a fraction or a decimal by 100 to write it as a percent. For example, 3/10 times 100 is 30, so 3/10 is 30%.

Sort out the table
The table shows how many times each dessert was ordered. Ice cream was ordered 48 times.
You need the chance that a customer orders ice cream. The orders in the table show how often each dessert is picked.
Find the total orders
Add all four rows of the table.
$42 + 23 + 47 + 48 = 160$
There were 160 dessert orders in all. This is the bottom of the fraction, because it counts every order.
Write the probability
Put the counts into the probability formula. The top is the 48 ice cream orders. The bottom is all 160 orders.
$P = \frac{\text{outcomes you want}}{\text{all outcomes}} = \frac{48}{160}$
Simplify the fraction. Both numbers can be divided by 16.
$\frac{48 \div 16}{160 \div 16} = \frac{3}{10}$
Change to a percent
$\frac{3}{10} \times 100 = 30$
The probability is 30%.
Watch out: 48 orders is not 48%, choice D. You must divide by the total. Also, the four desserts are not equally popular. So the chance is not 1 out of 4, or 25%, choice A.
So the answer is B, 30%.

Check your answer

Work backward: $0.30 \times 160 = 48$ . That matches the ice cream row. Also, 48 is a bit less than a third of 160, and 30% is a bit less than a third.

More questions like this

The original DOE question

Question 70 as printed on page 155 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 155, question 70. Open page 155 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).