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Question 1 of 120

2026 Practice Test B Math, question 24

Question

A chemical decays in such a way that the amount left at the end of each week is 20% less than the amount at the beginning of that same week. What percentage of the original amount is left after two weeks?

  • A40%
  • B60%
  • C64%
  • D80%
Show the answer

Correct answer: C

Let x be the original amount of the chemical. It loses 20% after each week, which means 80% of the chemical remains at the end of each week. End of first week: 0.80x0.80x

At the end of the second week, 80% of the amount left at the end of the first week remains.

End of second week: 0.80(0.80x)=0.64x0.80(0.80x) = 0.64x or 64%

Question analysis

A chemical loses 20% of the amount present at the start of each week, and students find what percentage of the original amount is left after two weeks. It tests applying a percent decrease more than once. The question turns on which amount each week's loss is measured against.

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 A typical grade 7 percent task applies one increase or decrease to one amount. This one applies a weekly decrease twice in a row, an early form of exponential decay that goes beyond typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
2 steps
Reading level
Question and choices: about grade 8, 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

Two equal decreases in a row feel like one decrease of twice the size, and that is the trap: 20% and 20% suggest 40% gone. The stem states the rule, each loss measured from that week’s start, in one quick phrase. Working week by week from 100 shows the second loss is 16, not 20, which separates the answer from the choice four points away, while stopping after one week or reporting the loss costs the point more simply.

  1. A very close wrong answer major Choosing the answer

    The choices 60% and 64% sit four percentage points apart, so a rough estimate cannot choose between them; only seeing that the second 20% is taken from 80, not 100, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Turning words into math moderate Understanding the question

    The phrase “the amount at the beginning of that same week” sets each week’s base, so the second week’s 20% is measured from what was left after the first week. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Variables instead of numbers minor Understanding the question

    No starting amount is given, so students must pick one, such as 100 grams, or work with a percent of a percent, 0.8 × 0.8. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Use a starting amount of 100 so that percents become units, find what remains after the first week, then apply the same 20% loss to that remainder.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of an amount
part=p100×amount\text{part} = \frac{p}{100} \times \text{amount}
p is the percent, and amount is the whole you take the percent of. Percent means out of 100, so 20% is 20/100. Multiply to find the part. Be sure the amount is the right whole. Here, each week uses the amount at the start of that week.
Percent left after a loss
percent left=100%−percent lost\text{percent left} = 100\% - \text{percent lost}
The whole amount is 100%. If 20% goes away, the rest stays. So 80% of each week’s starting amount is left at the end of that week. Keeping 80% is the same as multiplying by 0.8.

  1. Sort out the question

    Each week the chemical loses 20% of what it has. The key words are “the amount at the beginning of that same week.” So each week’s 20% comes from that week’s starting amount. No amount is given. You need the percent of the original amount that is left after two weeks.

  2. Pick an easy starting amount

    Pretend the chemical starts at 100 grams. With 100, the number of grams left is the same as the percent left. Any starting amount gives the same percent, so choose the easiest one.

  3. Work out the first week

    Use the percent formula. The first week starts with 100 grams:

    20100×100=20\frac{20}{100} \times 100 = 20

    The chemical loses 20 grams, so 100−20=80100 - 20 = 80 grams are left. That is 80% after one week.

    Watch out: 80% is only the first week. Stopping here gives choice D.

  4. Work out the second week

    The second week starts with 80 grams, not 100. So take 20% of 80:

    20100×80=16\frac{20}{100} \times 80 = 16

    This time the chemical loses only 16 grams, so 80−16=6480 - 16 = 64 grams are left.

    Watch out: taking away 20 grams again uses the original amount. That gives 60%, choice B. Choice A, 40%, adds the two losses. But the question asks what is left, and only 36% was lost.

  5. Read the answer as a percent

    WeekStart (grams)Loss (grams)Left (grams)
    11002080
    2801664

    Another way: each week keeps 80%, or 0.8, of its start. Two weeks gives 0.8×0.8=0.640.8 \times 0.8 = 0.64, which is 64%.

    64 grams out of the original 100 grams is 64%. So the answer is C, 64%.

Check your answer

Think about the size of the second loss. The second week starts with less, so it loses less than 20 grams. That means more than 60% is left. It does lose something, so less than 80% is left. Of the four choices, only 64% is between 60% and 80%.

More questions like this

The original DOE question
Question 24 as shown on the DOE’s digital test, page 96 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B Math, page 96, question 24.

Question 2 of 120

2026 Practice Test B Math, question 28

Question

A circle has a radius of 5 inches. What is the area of the circle in terms of π\pi?

Use π\pi to express your answer.

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: 25π

Solution:

25π25\pi

The area of a circle is its radius squared times π\pi.

A=πr2A = \pi r^2

r=5r = 5

A=π(5)2A = \pi(5)^2

A=25πA = 25\pi

Question analysis

This circle question gives a radius of 5 inches and asks for the area in terms of π, typed into a box rather than chosen. Students recall and apply the formula for the area of a circle. The formula is not printed on the test, and the area and circumference formulas are easy to mix up.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the area of a circle from its radius is the grade 7 circle standard applied directly, in one routine calculation. Typing the answer in terms of π, with no choices to compare, adds a little risk.
Estimated difficulty
Straightforward: two steps or one careful reading.
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

This is one formula and one square, so the question comes down to recall and to the form of the answer. The area and circumference formulas both pair π with the radius, and 2πr gives a tidy value with π in it that looks just as right. The other losses are about form: replacing π with 3.14, which the instructions rule out twice, or dropping π after squaring 5.

  1. Something you must already know moderate Working it out

    The formula A = πr² is not printed with the question, and the circumference formula 2πr uses the same r and π, so recalling the wrong one gives a value that looks just as finished. (Anderson & Krathwohl, 2001)

  2. An answer in an unexpected form minor Choosing the answer

    The answer is built on a keypad with a π key and must keep π as a symbol; a decimal worked out with 3.14 is not accepted, and no choices show the expected form. (Enright, Morley & Sheehan, 2002)

Strategy: Recall the area formula rather than the circumference formula, square the radius before multiplying, and leave π as a symbol instead of replacing it with 3.14.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle
A=πr2A = \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 (pi) is a fixed number, a little more than 3. r² means r times r. Square the radius first, then multiply by π. The area is in square units.
Circumference of a circle
C=2πrC = 2 \pi r
C is the circumference, the distance around the circle. r is the radius, and π is the same number as above. This formula measures a length, not the space inside. It is here so you can tell the two formulas apart.

  1. Sort out the question

    The circle has a radius of 5 inches. The radius is the distance from the center to the edge. You need the area, the space inside the circle. In terms of π means π stays in the answer as a symbol. Do not change it to 3.14.

  2. Choose the area formula

    The area formula is A=πr2A = \pi r^2. It uses the radius squared.

    Watch out: the circumference formula, C=2πrC = 2\pi r, also uses π and the radius. It gives 2×π×5=10π2 \times \pi \times 5 = 10\pi, the distance around the circle in inches. That is not the area.

  3. Square the radius

    Squaring a number means multiplying it by itself:

    r2=52=5×5=25r^2 = 5^2 = 5 \times 5 = 25

    Watch out: squaring is not doubling. 5×2=105 \times 2 = 10 would lead to 10π again.

  4. Multiply by π

    Put 25 into the formula in place of r²:

    A=π×25=25πA = \pi \times 25 = 25\pi

    The area is 25π square inches. Write the number first and then π, the way you write 3x for 3 times x.

    Watch out: leave π as π. Using 3.14 gives 78.5, which is not accepted. Dropping π leaves 25, which is not the area.

  5. Enter it with the π key

    On the test, the keypad has a π key in its bottom left corner. Press 2, then 5, then the π key. You do not need a times sign. Enter only the answer, with no A = and no units. On this site, type 25π or 25pi.

    So you would enter 25π.

Check your answer

Estimate. π is a little more than 3, so 25π is a little more than 75 square inches. A square drawn around the circle is 10 inches on each side, so its area is 100 square inches. The circle fits inside that square, so its area must be less than 100. It is.

More questions like this

The original DOE question
Question 28 as shown on the DOE’s digital test, page 100 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B Math, page 100, question 28.

Question 3 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: 3×21=633 \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.
What makes it hard
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The counting principle
total=m×n\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
pairs=n×(n−1)2\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.

  1. 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.

  2. 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.

    ToppingPairs withPairs
    12, 3, 4, 5, 6, 76
    23, 4, 5, 6, 75
    34, 5, 6, 74
    45, 6, 73
    56, 72
    671

    6+5+4+3+2+1=216 + 5 + 4 + 3 + 2 + 1 = 21

    There are 21 different pairs of toppings.

  3. Check the pairs with the formula

    Use the pairs formula with 7 toppings:

    7×62=422=21\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 42×3=12642 \times 3 = 126, choice D.

  4. Bring in the sizes

    Each of the 21 topping pairs can come in any of the 3 sizes. Use the counting principle:

    21×3=6321 \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.

  5. 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.

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=6321 + 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 4 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+50y = 20x + 50

An equation of the form px+q=rpx + 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+50y = 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.
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.

  1. 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)

  2. 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

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=(rate)x+(fee)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.

  1. 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.

  2. Work out a few totals

    Find the total for a few months by hand. Each month adds one more $20:

    MonthsMonthly charges ($)Total ($)
    11×20=201 \times 20 = 2050+20=7050 + 20 = 70
    22×20=402 \times 20 = 4050+40=9050 + 40 = 90
    33×20=603 \times 20 = 6050+60=11050 + 60 = 110

    The $50 shows up once in every total. The $20 shows up once for each month.

  3. Match each number to its job

    Use the fee-and-rate model, y=(rate)x+(fee)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+20y = 50x + 20 charges $50 a month. y=70xy = 70x charges the fee every month. y=20xy = 20x forgets the fee.

  4. Write the equation

    Put the rate and the fee into the model:

    y=20x+50y = 20x + 50

    Here 20x means 20 times x. An equal form is fine too, such as y=50+20xy = 50 + 20x, because you can add in either order.

  5. 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.

Check your answer

Test the equation with the totals you found. For 1 month, y=20×1+50=70y = 20 \times 1 + 50 = 70, and Amiya pays $50 + $20 = $70. For 2 months, y=20×2+50=90y = 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 5 of 120

2026 Practice Test C Math, question 1

Question

What is the prime factorization of 756?

  • A22⋅32⋅212^{2} \cdot 3^{2} \cdot 21
  • B22⋅3⋅7⋅92^{2} \cdot 3 \cdot 7 \cdot 9
  • C22⋅7⋅272^{2} \cdot 7 \cdot 27
  • D22⋅33⋅72^{2} \cdot 3^{3} \cdot 7
Show the answer

Correct answer: D

Determine the prime factors of 756 using a factor tree:

A factor tree for 756: 756 splits into 2 and 378; 378 into 2 and 189; 189 into 3 and 63; 63 into 3 and 21; and 21 into 3 and 7.

The prime factors are 2, 2, 3, 3, 3, and 7, or 2⋅2⋅3⋅3⋅3⋅72 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 7, which can be expressed using exponents as 22⋅33⋅72^{2} \cdot 3^{3} \cdot 7.

Question analysis

This number question asks for the prime factorization of 756, written with exponents. Students break the number into prime factors, with a factor tree or repeated division, and match the result to one of four products. Every choice multiplies back to 756, so the real test is whether each factor shown is actually prime.

NYS standards
  • 6.NS.4 common factors and multiples
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Factor trees and exponent notation are grade 6 skills, and 756 splits quickly by small divisors. What adds a little is that every choice multiplies out to 756, so checking the product alone settles nothing.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 11, 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 factoring itself is quick: halving twice and dividing by 3 three times leaves 7. The question is built so that multiplying back proves nothing, since every choice equals 756, and three of them simply stop the factor tree at a composite number. Students who check only the product, or who take an odd factor such as 21 or 27 for a prime, pick a factorization that is unfinished.

  1. A very close wrong answer moderate Choosing the answer

    2² · 3² · 21 shares the right answer's 2² and differs only in 3² · 21 against 3³ · 7, and both equal 756, so neither an estimate nor a product check separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Students must know that a prime factorization may hold only primes, and must spot that 21, 9 and 27 are composite even though none of them is even. (Anderson & Krathwohl, 2001)

Strategy: Divide 756 by 2 while the result is even, then by 3 while it divides evenly, and keep going until every factor left is prime; then compare factor by factor, not by product.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Writing a repeated factor with an exponent
an=a×a×⋯×aa^{n} = a \times a \times \cdots \times a
a is the number being multiplied, called the base. n is the exponent, the small raised number. It tells how many times a appears in the product. So 2² means 2 × 2, and 3³ means 3 × 3 × 3. Exponents are a short way to write a factor that repeats.

  1. Sort out the question

    You need the prime factorization of 756. That means writing 756 as a product of prime numbers only. A product is the answer to a multiplication. A prime number has exactly two factors, 1 and itself. Examples are 2, 3, 5 and 7.

    All four choices multiply to 756. So multiplying a choice back does not pick the answer. You must check that every factor is prime.

  2. Divide by 2 while you can

    A factor tree splits a number into two factors, again and again. Start with the smallest prime, 2. 756 is even, so 2 goes into it:

    756=2×378756 = 2 \times 378

    378 is even too:

    378=2×189378 = 2 \times 189

    189 is odd, so 2 does not go into it. You have pulled out two 2s.

  3. Divide by 3 while you can

    Add the digits of 189: 1+8+9=181 + 8 + 9 = 18. A number can be divided by 3 when its digits add up to a number that 3 goes into. 3 goes into 18, so it goes into 189:

    189=3×63189 = 3 \times 63

    63=3×2163 = 3 \times 21

    21=3×721 = 3 \times 7

    7 is prime, so the tree stops. You have pulled out three 3s and one 7.

  4. Write the factors with exponents

    Collect every prime you pulled out:

    756=2×2×3×3×3×7756 = 2 \times 2 \times 3 \times 3 \times 3 \times 7

    Use the exponent rule. Two 2s make 222^{2}, and three 3s make 333^{3}. The choices use a raised dot, ·, for times. So 756 is 22⋅33⋅72^{2} \cdot 3^{3} \cdot 7.

  5. Choose the answer

    Watch out: each wrong choice is a factor tree that stopped too soon. A ends in 21, which is 3 × 7. B has 9, which is 3 × 3. C has 27, which is 3 × 3 × 3. Being odd does not make a number prime.

    Only D uses primes alone. So the answer is D, 22⋅33⋅72^{2} \cdot 3^{3} \cdot 7.

Check your answer

Multiply your answer back: 22×33=4×27=1082^{2} \times 3^{3} = 4 \times 27 = 108, and 108×7=756108 \times 7 = 756. Then look at each factor. 2, 3 and 7 are all prime. D passes both tests.

More questions like this

The original DOE question
Question 1 as shown on the DOE’s digital test, page 74 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 74, question 1.

Question 6 of 120

2026 Practice Test C Math, question 4

Question

A person buys a used car with a down payment of $1,200 and makes monthly payments of $275 for 3 years. What is the total amount the person pays for the car?

  • A$2,025
  • B$9,900
  • C$11,100
  • D$13,500
Show the answer

Correct answer: C

Determine the total cost of the car by adding the down payment to the total monthly payments over 3 years.

There are 12 months in a year, so multiply (12 months/year × 3 years) to get a total of 36 months. Determine the total monthly payments by multiplying the monthly payment ($275) × the total number of months (36):

275×36=9,900275 \times 36 = 9{,}900

Determine the total cost of the car by adding the down payment ($1,200) to the total monthly payments ($9,900):

9,900+1,200=11,1009{,}900 + 1{,}200 = 11{,}100

The total cost of the car is $11,100.

Question analysis

A used car is bought with a down payment of $1,200 followed by monthly payments of $275 for 3 years, and students find the total paid. The work is to count the payments, multiply, and add the down payment. Mixing up years and months, or forgetting the money paid up front, leads to the wrong totals among the choices.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 A total made of one payment up front plus equal monthly payments is a short whole-number calculation, simpler than grade 7 word problems with percents or negatives. The only conversion is from years to months.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 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 situation is familiar and the method direct, so this is one of the easier questions on the test. Points are lost in two places: turning 3 years into 36 payments, and remembering the money paid up front after the long multiplication. Each slip, like paying the down payment once a year, leads to a listed total, so checking that the answer holds exactly one $1,200 and 36 payments of $275 is worth the seconds.

  1. Much to keep track of minor Working it out

    The $1,200 down payment is paid once while the $275 repeats every month, so it has to be set aside while 36 payments are totaled and then added at the end. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Error-prone arithmetic minor Working it out

    The product 275 × 36 = 9,900 is a two-digit multiplication done by hand; splitting it as 275 × 4 = 1,100 and then 1,100 × 9 makes it quick and safe. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Turn 3 years into a number of months, multiply the monthly payment by it, and add the down payment once at the end.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing years into months
months=12×years\text{months} = 12 \times \text{years}
Each year has 12 months. Multiply the number of years by 12 to get the number of months. You need months here because a payment is made every month.
Total cost with a down payment
total=down payment+monthly payment×months\text{total} = \text{down payment} + \text{monthly payment} \times \text{months}
The down payment is money paid once, at the start. The monthly payment is paid every month. Multiply it by the number of months to get all the monthly payments. Then add the down payment only once, because it is paid only once.

  1. Sort out the question

    The person pays $1,200 once, at the start. This is the down payment. Then the person pays $275 every month for 3 years. You need the total amount paid for the car.

    The payments come once a month, but the time is given in years. So change the years into months first.

  2. Change 3 years into months

    Use the months formula:

    months=12×3=36\text{months} = 12 \times 3 = 36

    So the person makes 36 monthly payments.

    Watch out: multiplying $275 by 3 counts only 3 payments. That leads to choice A, $2,025.

  3. Add up the monthly payments

    Multiply $275 by 36. A quick way is to split 36 into 4 × 9:

    275×4=1,100275 \times 4 = 1{,}100

    1,100×9=9,9001{,}100 \times 9 = 9{,}900

    So the 36 payments add up to $9,900.

    Watch out: stopping here gives choice B, $9,900. The down payment is still missing.

  4. Add the down payment once

    Use the total-cost formula:

    total=1,200+9,900=11,100\text{total} = 1{,}200 + 9{,}900 = 11{,}100

    The person pays $11,100 in all. Adding $1,200 for every year counts it three times. That gives choice D, $13,500.

    So the answer is C, $11,100.

Check your answer

Estimate. $275 is a little less than $300, and 36×300=10,80036 \times 300 = 10{,}800. Add the $1,200: 10,800+1,200=12,00010{,}800 + 1{,}200 = 12{,}000. The real total should be a little less than $12,000. $11,100 fits.

More questions like this

The original DOE question
Question 4 as shown on the DOE’s digital test, page 77 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 77, question 4.

Question 7 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?

  • A36
  • B18
  • C6
  • D5
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∘3x^{\circ} and 2x∘2x^{\circ} is equal to 180°:

3x+2x=1803x + 2x = 180

Solve for x:

5x=1805x = 180

5x5=1805\frac{5x}{5} = \frac{180}{5}

x=36x = 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.
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.

  1. 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)

  2. 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

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∘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°.

  1. 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.

  2. 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=1803x + 2x = 180

    Watch out: the total is 180°, not 90°. Using 90 gives 5x = 90 and x = 18, choice B.

  3. Solve for x

    Three x’s plus two x’s make five x’s:

    5x=1805x = 180

    Divide both sides by 5 to get x alone:

    x=1805=36x = \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.

  4. 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 3×36=1083 \times 36 = 108 degrees. The bottom one is 2×36=722 \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 8 of 120

2026 Practice Test C Math, question 10

Question

Two parallel lines, n and p, rise from lower left to upper right, and both cross a horizontal line m. Where line n crosses line m, the angle above line m, between the left part of m and the upper part of n, is marked r°. Farther right, a vertical segment runs from line p down to line m, meeting m at a right angle (marked with a small square); the angle at the top, between line p and the vertical segment, is marked 73°.

In the figure above, n∥pn \parallel p. Straight line m intersects both line n and line p. What is the value of r?

  • A17
  • B117
  • C163
  • D173
Show the answer

Correct answer: C

The triangle formed by line p, line m, and the line segment connecting the two contains a 73° angle, a right angle (90°), and a missing angle measuring x°. The sum of the interior angles of a triangle is 180°, so begin by finding the value of x:

180∘=73∘+90∘+x180^{\circ} = 73^{\circ} + 90^{\circ} + x

180∘=163∘+x180^{\circ} = 163^{\circ} + x

17∘=x17^{\circ} = x

Line m is a straight line, so it has a measure of 180°. The 17° angle from the triangle and the adjacent angle outside the triangle are supplementary, so they have a sum of 180°. Let s represent the measure of the adjacent angle outside the triangle. The value of s can be determined by subtracting:

s=180∘−17∘s = 180^{\circ} - 17^{\circ}

s=163∘s = 163^{\circ}

Given that n and p are parallel lines and that they have been cut by a transversal, line m, then corresponding angles are congruent. Angles r and s are corresponding angles, so the value of r is congruent to the value of s, which is 163°.

Question analysis

Two parallel lines, n and p, cross a straight line m, and a segment drawn at a right angle to m forms a small triangle with lines m and p, where a 73° angle is marked. Students find the angle r at line n. It combines a triangle's angle sum, a straight angle and parallel lines, and the angle wanted sits away from the marked ones.

NYS standards
  • 8.G.5 angle sums, transversals, angle-angle similarity
  • 7.G.5 angle relationships and equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 angle problems use supplementary, vertical and adjacent angles. This one also needs the angle sum of a triangle and the angles formed by parallel lines and a transversal, grade 8 facts, chained across a busy figure.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 3, 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 fact needs only a subtraction, but the student has to plan a route from the 73° angle at line p over to r at line n. The first number the route gives, the triangle's 17° angle, is listed, and it is the acute partner of r rather than r itself. The figure helps here: r is drawn plainly obtuse, and 163 is also 90 + 73, the triangle's exterior angle, which confirms it.

  1. Reading a figure, table or graph moderate Understanding the question

    The 73° sits between line p and the vertical segment, not between p and m, and r is the obtuse angle above m on the left of line n, far from every marked angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Three facts chain together: the triangle's angles sum to 180°, the angles along straight line m sum to 180°, and corresponding angles at the parallel lines n and p are equal. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Something you must already know minor Working it out

    The angle pairs formed by a line crossing two parallel lines are a grade 8 topic, and the question uses them without naming them. (Anderson & Krathwohl, 2001)

Strategy: Find the third angle of the right triangle, use the straight line m to get the angle beside it, then carry that angle over to line n using the parallel lines.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles of a triangle
a+b+c=180∘a + b + c = 180^{\circ}
a, b and c are the three angles inside any triangle. They always add up to 180°. If you know two of them, subtract their sum from 180° to find the third.
Angles on a straight line
a+b=180∘a + b = 180^{\circ}
a and b are two angles side by side on a straight line. Together they make a half turn, which is 180°. Use it to find the angle next to one you already know.
Corresponding angles
n∥p⇒corresponding angles are equaln \parallel p \Rightarrow \text{corresponding angles are equal}
The symbol ∥ means is parallel to. When line m crosses two parallel lines, it makes the same angles at each crossing. Angles in the same position at the two crossings are called corresponding angles. They are equal.

  1. Sort out the question

    Lines n and p are parallel, and line m crosses both. Near line p, a straight-up segment meets line m. The small square shows a right angle, 90°. Line p, line m and that segment make a small triangle. Its top angle is 73°.

    You need r, the wide angle where line n crosses line m.

  2. Find the triangle’s third angle

    The third angle is where line p crosses line m. Call it x. Use the triangle rule:

    73+90+x=18073 + 90 + x = 180

    x=180−163=17x = 180 - 163 = 17

    So that angle is 17°.

    Watch out: this is not r. Stopping here gives choice A, 17.

  3. Use the straight line m

    The 17° angle sits above line m, on the right of line p. Beside it, on the left of line p, is a wide angle. Call it s. The two angles fill the straight line m, so they add to 180°:

    s=180−17=163s = 180 - 17 = 163

    So s is 163°.

    Watch out: borrow carefully. 180 − 17 is 163, not 173, choice D.

  4. Carry the angle to line n

    Angle s is above line m and to the left of line p. Angle r is above line m and to the left of line n. They are in the same position, so they are corresponding angles. Since n and p are parallel:

    r=s=163r = s = 163

    So the answer is C, 163.

Check your answer

Look at the figure. Angle r is wider than a right angle, so it must be more than 90°. That rules out 17. Also, the narrow angle beside r, on the right of line n, matches the 17° angle at line p. And 163+17=180163 + 17 = 180, as it should on a straight line.

More questions like this

The original DOE question
Question 10 as shown on the DOE’s digital test, page 83 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 83, question 10.

Question 9 of 120

2026 Practice Test C Math, question 12

Question

A spinner has 5 congruent sections labeled as shown. One side of a fair coin is heads, and the other side is tails.

A spinner shaped like a regular pentagon, divided into 5 congruent triangular sections that meet at the center, where an arrow points straight up. Going clockwise from the upper left, the sections are labeled 1, 2, 1, 4 and 3. Beside the spinner is a coin (a dime) showing its heads side, captioned Heads.

The spinner is spun one time, and the coin is flipped. What is the probability that the spinner lands on 1 and the coin lands on tails?

Select the place on the number line that represents the probability.

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

00.10.20.30.40.50.60.70.80.91

Show the answer

Correct answer: the point 0.2

The spinner has five equal-sized sections. Two of the sections are labeled with a 1. So, the probability of landing on a 1 is two out of five, or 2/5.

A coin has two sides, one heads and one tails. So, the probability of landing on tails is one out of two, or 1/2.

The compound probability of the spinner landing on a 1 and the coin landing on tails can be calculated by multiplying 2/5 and 1/2 as shown:

25⋅12=210=0.2\frac{2}{5} \cdot \frac{1}{2} = \frac{2}{10} = 0.2

Question analysis

A spinner with five equal sections is spun once and a fair coin is flipped. Students find the probability that the spinner lands on 1 and the coin shows tails, then select that point on a number line from 0 to 1. Counting the right sections and combining the two events correctly are the tricky parts.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 6.SP.6 probability from 0 to 1
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the chance of two independent events together is grade 7 compound probability, here with one spinner and one coin. Counting sections from the picture and placing the result on a number line add two small steps.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 59 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 probability itself takes one multiplication of two simple fractions; the losses come before and after it. Before, the spinner has to be searched for every section labeled 1, and the coin's chance brought in as well. After, 1/5 has to be rewritten as 2/10 to find its point, since only 0, 1/2 and 1 are labeled. Checking that the result is smaller than both separate chances catches an addition.

  1. Reading a figure, table or graph moderate Understanding the question

    The two sections labeled 1, upper left and right, are not side by side, and the number line labels only 0, 1/2 and 1, so its nine points must be read as tenths. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Nothing says that the chance of two independent events both happening is the product of their chances, 2/5 × 1/2, rather than their sum. (Anderson & Krathwohl, 2001)

Strategy: Count the sections labeled 1 to get the spinner's chance, take the coin's chance of tails, multiply the two because both must happen, and write the result in tenths.

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).
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event
P=number of results you wanttotal number of resultsP = \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. The bottom counts all possible results. It works when every result is equally likely, like equal sections of a spinner.
Probability that two things both happen
P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)
A and B are two events that do not affect each other, like a spin and a coin flip. P(A) and P(B) are their chances. Multiply them to get the chance that both happen. Needing both makes the chance smaller, never bigger.

  1. Sort out the question

    The spinner has 5 congruent sections. Congruent means the same size and shape, so each is equally likely. The coin has a heads side and a tails side. You need the chance that the spinner lands on 1 and the coin lands on tails.

  2. Find the spinner’s chance

    Two sections show 1: the upper left one and the one on the right. So 2 of the 5 equal sections give a 1:

    P(1)=25P(1) = \frac{2}{5}

    Watch out: the two 1s are not next to each other. Counting only one gives 1/5 here, and 0.1 at the end.

  3. Find the coin’s chance

    A fair coin is equally likely to land on either side. Tails is 1 of its 2 sides:

    P(tails)=12P(\text{tails}) = \frac{1}{2}

  4. Multiply the two chances

    The spin and the flip do not affect each other, and both must happen. So multiply:

    25×12=210=0.2\frac{2}{5} \times \frac{1}{2} = \frac{2}{10} = 0.2

    Watch out: stopping at 2/5 gives 0.4. Adding gives 25+12=910\frac{2}{5} + \frac{1}{2} = \frac{9}{10}, or 0.9. Both are wrong.

  5. Select the point

    On the test, the number line labels only 0, 1/2 and 1. Between 0 and 1 are 9 targets, one at each tenth. 0.2 is the second target to the right of 0. On this site, choose 0.2 from the Point menu.

    So the answer is the point at 0.2.

Check your answer

List every result. The spinner has 5 sections and the coin has 2 sides, so there are 5×2=105 \times 2 = 10 results, all equally likely. Just 2 of them are a 1 with tails, one for each section labeled 1. That is 2 out of 10, or 0.2.

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 C Math, page 85, question 12.

Question 10 of 120

2026 Practice Test C Math, question 15

Question

A grocery store sells 1-gallon containers of milk for $3.99. The store also sells orange juice in a 6-pack of 5.5-fluid-ounce bottles for $1.79. Suppose the store wants to sell its orange juice in gallon containers instead. To the nearest dollar, how much more would a gallon of orange juice cost than a gallon of milk?

Enter your answer in the space.

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

Show the answer

Correct answer: 3

To determine the cost of a gallon of orange juice, begin by finding the total number of ounces in the 6-pack of bottles:

5.5 oz×6=33 oz5.5\text{ oz} \times 6 = 33\text{ oz}

So, 33 oz of orange juice costs $1.79.

There are 128 oz in 1 gallon. Set up a proportion using the $1.79 cost per 33 oz of orange juice and the unknown cost (x) per 1 gallon of orange juice:

33 oz$1.79=128 ozx\frac{33\text{ oz}}{\$1.79} = \frac{128\text{ oz}}{x}

Cross multiply to solve for x in the proportion:

(33)×(x)=($1.79)×(128)(33) \times (x) = (\$1.79) \times (128)

33x=$229.1233x = \$229.12

x≈$6.94x \approx \$6.94

So, the orange juice costs $6.94 per gallon, and the milk costs $3.99 per gallon.

Subtract the cost per gallon of milk from the cost per gallon of orange juice.

$6.94−$3.99=$2.95\$6.94 - \$3.99 = \$2.95

Round the difference to the nearest dollar.

$2.95 rounds to $3.

Question analysis

A store sells milk by the gallon for $3.99 and orange juice in six-packs of 5.5-fluid-ounce bottles for $1.79. Students work out what a gallon of the juice would cost at the same price per ounce and how much more that is than the milk, to the nearest dollar. The ounces in a gallon must be known, and several steps involve decimals.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3d converting measurement units
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Unit prices are grade 6 and 7 work, but this chains five steps with decimals and needs a fact the question does not give, the number of fluid ounces in a gallon. Typing a rounded difference, with no choices, adds risk.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
5 steps
Reading level
Question and choices: about grade 5, 60 words.
Corrected from the DOE handbook
  • A rounded value in the explanation is now marked as approximate (≈ instead of =).
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 calculation is long, and its key fact is missing from the page. Students who know the 128 fluid ounces still face a division by 33 that does not come out evenly, which tempts them to round the price per ounce to 5 cents and lose about 50 cents a gallon. Keeping the exact quotient to the end, or scaling 33 ounces up by about 3.9, keeps the difference near $2.95, which rounds to 3.

  1. Something you must already know moderate Working it out

    The question never says that a gallon holds 128 fluid ounces, yet the whole comparison depends on it; the 6-pack holds only 6 × 5.5 = 33 ounces. (Anderson & Krathwohl, 2001)

  2. Many steps moderate Working it out

    Five moves follow one another: the ounces in a pack, the scale-up to 128 ounces, the division by 33, the subtraction of $3.99 and the rounding to a whole dollar. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    1.79 × 128 = 229.12 and 229.12 ÷ 33 are long decimal calculations without a calculator, and a slip in either moves the price of a gallon. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  4. An answer in an unexpected form minor Choosing the answer

    The rounded difference is typed with no choices, so an unrounded $2.95, or a figure rounded too soon, looks just as finished as the whole-dollar answer. (Enright, Morley & Sheehan, 2002)

Strategy: Find the ounces in one six-pack, scale its price up to the 128 fluid ounces in a gallon, subtract the milk's price, and round only at the end.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation goes from 33x = $229.12 to “x = $6.94”. But 229.12 ÷ 33 = 6.9430…, so $6.94 is the cost per gallon rounded to the cent. The site writes x ≈ $6.94. Nothing else changes: $6.94 − $3.99 = $2.95, which rounds to $3, and the unrounded cost gives the same answer.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fluid ounces in a gallon
fluid ounces=128×gallons\text{fluid ounces} = 128 \times \text{gallons}
A gallon holds 128 fluid ounces. This is a fact you must know, because the question does not give it. A fluid ounce measures an amount of liquid. Multiply gallons by 128 to change them into fluid ounces.
Cross-multiplying a proportion
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
A proportion says two ratios are equal. Here each ratio is ounces over dollars. a and c are the tops, and b and d are the bottoms. Multiply each top by the other bottom, and the two products are equal. That turns the proportion into an equation.

  1. Sort out the question

    Milk costs $3.99 for 1 gallon. Orange juice comes in a 6-pack of 5.5-fluid-ounce bottles for $1.79. You need how much more a gallon of juice would cost than a gallon of milk. Round to the nearest dollar.

    The question doesn’t say how big a gallon is. You must know this fact: 1 gallon is 128 fluid ounces.

  2. Find the ounces in a 6-pack

    Each of the 6 bottles holds 5.5 fluid ounces:

    6×5.5=336 \times 5.5 = 33

    So $1.79 buys 33 fluid ounces of juice.

  3. Set up a proportion

    Let x be the cost in dollars of a gallon, 128 fluid ounces, of juice. The price per ounce stays the same. So ounces over dollars is the same for the pack and the gallon:

    331.79=128x\frac{33}{1.79} = \frac{128}{x}

    Cross-multiply:

    33×x=1.79×128=229.1233 \times x = 1.79 \times 128 = 229.12

  4. Find the cost of a gallon

    Divide both sides by 33:

    x=229.12÷33≈6.94x = 229.12 \div 33 \approx 6.94

    The sign ≈ means about equal. The exact value is 6.943…, so a gallon of juice costs about $6.94.

    Watch out: don’t round too early. One ounce costs about $0.054. Rounding that to $0.05 makes a gallon $6.40, and your answer becomes 2.

  5. Subtract the milk’s price

    Find how much more the juice costs:

    6.94−3.99=2.956.94 - 3.99 = 2.95

    A gallon of juice costs $2.95 more than a gallon of milk.

    Watch out: entering 7, the juice’s price, forgets to subtract the milk.

  6. Round and enter it

    The question asks for the nearest dollar. $2.95 is between $2 and $3. Its cents, 95, are 50 or more, so it rounds up to $3. Do not enter 2.95.

    Type only the number in the box. So you would enter 3.

Check your answer

Estimate. 128 ounces is almost 4 packs, since 4×33=1324 \times 33 = 132. Four packs cost 4×1.79=7.164 \times 1.79 = 7.16 dollars, so a gallon costs a bit less than that. Milk costs about $4. About $7 minus $4 is $3, which matches.

More questions like this

The original DOE question
Question 15 as shown on the DOE’s digital test, page 88 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 88, question 15. This is the DOE’s version, before the corrections listed in the analysis.

Question 11 of 120

2026 Practice Test C Math, question 17

Question

A number line from 0 to 8 with a tick and a label at every whole number. Four points are marked on it: M at 1, R at 3, S at 6 and N at 7.

What is the distance between the midpoints of MN\overline{MN} and RS\overline{RS}?

  • A12\frac{1}{2} unit
  • B1 unit
  • C1121\frac{1}{2} units
  • D2 units
Show the answer

Correct answer: A

MN\overline{MN} begins at 1 and ends at 7. Determine the length of MN\overline{MN}:

7−1=67 - 1 = 6

The midpoint of a line segment is located exactly halfway between the two endpoints. Since MN\overline{MN} is 6 units in length, the two endpoints are 6 units apart. Determine the location of the midpoint, in relation to each endpoint, by dividing 6 by 2:

6÷2=36 \div 2 = 3

The location of the midpoint of MN\overline{MN} is 3 units from either endpoint. Find the midpoint by either adding 3 to the location of M (1) or subtracting 3 from the location of N (7):

1+3=41 + 3 = 4

7−3=47 - 3 = 4

The midpoint of MN\overline{MN} is located at 4.

RS\overline{RS} begins at 3 and ends at 6. Determine the length of RS\overline{RS}:

6−3=36 - 3 = 3

Since RS\overline{RS} is 3 units in length, the two endpoints are 3 units apart. Determine the location of the midpoint, in relation to each endpoint, by dividing 3 by 2:

3÷2=1.53 \div 2 = 1.5

The location of the midpoint of RS\overline{RS} is 1.5 units from either endpoint. Find the midpoint by either adding 1.5 to the location of R (3) or subtracting 1.5 from the location of S (6):

3+1.5=4.53 + 1.5 = 4.5

6−1.5=4.56 - 1.5 = 4.5

The midpoint of RS\overline{RS} is located at 4.5.

Determine the distance between the two midpoints by subtracting:

4.5−4=0.54.5 - 4 = 0.5

0.5 is equivalent to 12\frac{1}{2}, making the distance between the two midpoints 12\frac{1}{2} unit.

Question analysis

A number line from 0 to 8 shows points M, R, S and N, and students find the distance between the midpoint of segment MN and the midpoint of segment RS. It tests locating points and finding distances on a number line. One segment sits inside the other, and several lengths in the figure can be mistaken for the distance wanted.

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 with whole-number positions. Two midpoints must be located before their distance can be found, so the task takes three short steps rather than one.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 17 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 positions are whole numbers, so the midpoints come from simple averages, (1 + 7) ÷ 2 and (3 + 6) ÷ 2. The difficulty is keeping several lengths apart: the half-lengths of the two segments, the gaps at their ends, and the distance actually asked for, half a unit. Marking both midpoints on the number line before subtracting makes the small gap visible.

  1. Reading a figure, table or graph minor Understanding the question

    Segment RS lies inside MN, from 3 to 6, and its midpoint, 4 1/2, falls between two ticks of the number line rather than on a labeled point. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Two midpoints have to be found, at 4 and 4 1/2, before their distance is taken, and the half-length of RS met on the way, 1 1/2, is among the choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find each midpoint as the average of its segment's two endpoints, then subtract the smaller midpoint from the larger one.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint on a number line
midpoint=a+b2\text{midpoint} = \frac{a + b}{2}
a and b are the numbers at the two ends of a segment. The midpoint is the point exactly halfway between them. Adding the ends and dividing by 2 finds their average, which is the middle.
Distance on a number line
distance=larger−smaller\text{distance} = \text{larger} - \text{smaller}
The distance between two points on a number line is how many units apart they are. Subtract the smaller number from the larger one, so the distance is never negative.

  1. Sort out the question

    The number line shows M at 1, R at 3, S at 6 and N at 7. Segment MN goes from 1 to 7. Segment RS goes from 3 to 6. A midpoint is the point exactly halfway along a segment. You need the distance between the two midpoints.

  2. Find the midpoint of MN

    Use the midpoint formula with the ends 1 and 7:

    1+72=82=4\frac{1 + 7}{2} = \frac{8}{2} = 4

    The midpoint of MN is at 4, which is 3 units from each end.

  3. Find the midpoint of RS

    Use the same formula with the ends 3 and 6:

    3+62=92=4.5\frac{3 + 6}{2} = \frac{9}{2} = 4.5

    The midpoint of RS is at 4.5. That is halfway between the marks for 4 and 5, where no point is labeled.

  4. Subtract the two midpoints

    Use the distance formula:

    4.5−4=0.54.5 - 4 = 0.5

    0.5 is one half, so the midpoints are 1/2 unit apart.

    Watch out: half of MN is 3, and half of RS is 1 1/2. Subtracting those gives 1 1/2, choice C. But they are half-lengths, not positions. And 2, choice D, is only the distance from M to R.

  5. Choose the answer

    The midpoints, 4 and 4.5, are close together because RS sits near the middle of MN. So the answer is A, 1/2 unit.

Check your answer

Find each midpoint another way: walk half the length from the left end. MN is 7−1=67 - 1 = 6 units long, and 1+3=41 + 3 = 4. RS is 6−3=36 - 3 = 3 units long, and 3+1.5=4.53 + 1.5 = 4.5. Same midpoints, 1/2 unit apart.

More questions like this

The original DOE question
Question 17 as shown on the DOE’s digital test, page 90 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 90, question 17.

Question 12 of 120

2026 Practice Test C Math, question 21

Question

In a fruit basket, the ratio of apples to oranges is the same as the ratio of cherries to walnuts. If there are 6 oranges, 16 cherries, and 48 walnuts, how many apples are there?

  • A2
  • B3
  • C18
  • D38
Show the answer

Correct answer: A

The ratios of the items are equal, so they can be expressed using a proportion.

Let a represent the number of apples, o the number of oranges, c the number of cherries, and w the number of walnuts. Write a proportion showing apples:oranges is equal to cherries:walnuts:

ao=cw\frac{a}{o} = \frac{c}{w}

Substitute the given values for oranges (6), cherries (16), and walnuts (48):

a6=1648\frac{a}{6} = \frac{16}{48}

Cross multiply to form an equation to solve for the number of apples (a):

(a×48)=(16×6)(a \times 48) = (16 \times 6)

Solve for the value of a:

48a=9648a = 96

a=2a = 2

Question analysis

In a fruit basket, the ratio of apples to oranges equals the ratio of cherries to walnuts, and students find the number of apples from the counts of oranges, cherries and walnuts. It tests setting up and solving a proportion. Matching each quantity to its place in the ratio matters, since the wrong order gives a very different count.

NYS standards
  • 6.RP.3 ratio and rate reasoning
  • 6.RP.1 ratio concepts and language
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Equal ratios with whole numbers are grade 6 work, and the numbers divide evenly. Grade 7 proportion problems are usually longer; here the only care is keeping apples with oranges in the same order as cherries with walnuts.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 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 divide evenly, since 16/48 is 1/3, so the work is in the setup. Writing the two ratios in the order the sentence gives them makes the apples a third of the oranges; writing either one upside down makes them three times the oranges. A quick sense check helps: cherries are fewer than walnuts, so apples must be fewer than oranges.

  1. Turning words into math minor Understanding the question

    The sentence pairs apples with oranges and cherries with walnuts in the same order, so a/6 = 16/48; reading either pair the other way turns one ratio upside down. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Four kinds of fruit are named and three counts given, and the unknown apples must be matched with the smaller count, the cherries, not the larger, the walnuts. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Write apples over oranges equal to cherries over walnuts, simplify the known ratio, and scale it to match the 6 oranges.

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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios
applesoranges=cherrieswalnuts\frac{\text{apples}}{\text{oranges}} = \frac{\text{cherries}}{\text{walnuts}}
A ratio compares two amounts. Writing it as a fraction puts the first thing named on top. The question says the two ratios are the same, so the two fractions are equal. Keep the same order in both.

  1. Sort out the question

    A ratio compares two amounts. The ratio of apples to oranges equals the ratio of cherries to walnuts. You know there are 6 oranges, 16 cherries and 48 walnuts. You need the number of apples.

  2. Write the proportion

    Put the first thing of each pair on top. Apples go with oranges, and cherries go with walnuts. Let a be the number of apples:

    a6=1648\frac{a}{6} = \frac{16}{48}

    Watch out: keep the order the same. Writing 48 over 16 flips the second ratio. That gives 18, choice C.

  3. Simplify the known ratio

    16 goes into both 16 and 48. Divide the top and the bottom by 16:

    16÷1648÷16=13\frac{16 \div 16}{48 \div 16} = \frac{1}{3}

    So there is 1 cherry for every 3 walnuts. There must also be 1 apple for every 3 oranges.

  4. Find the apples

    Now the proportion is:

    a6=13\frac{a}{6} = \frac{1}{3}

    The bottom 6 is 3 × 2, so the top is 1 × 2. That makes a=2a = 2.

    Watch out: 48÷16=348 \div 16 = 3 is choice B, but that is not the number of apples. And 38 comes from using differences instead of ratios.

    So the answer is A, 2.

Check your answer

Compare the ratios. Apples to oranges is 2 to 6. Cherries to walnuts is 16 to 48. In both, the second number is 3 times the first: 2×3=62 \times 3 = 6 and 16×3=4816 \times 3 = 48. So the ratios are equal.

More questions like this

The original DOE question
Question 21 as shown on the DOE’s digital test, page 94 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 94, question 21.

Question 13 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 3:5:43 : 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 5×9=455 \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=6327 + 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.
What makes it hard
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Using the parts of a ratio
actual number=parts×value of one part\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.

  1. 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.

  2. Find the value of one part

    Drawings are 5 parts, and there were 45 drawings:

    45÷5=945 \div 5 = 9

    So each part of the ratio stands for 9 works.

  3. Find each group

    Multiply each group’s parts by 9:

    TypePartsActual number
    Paintings33×9=273 \times 9 = 27
    Drawings55×9=455 \times 9 = 45
    Photographs44×9=364 \times 9 = 36

    The drawings row gives 45, which matches the question.

  4. Add only the two groups asked for

    Add the paintings and the photographs:

    27+36=6327 + 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.

Check your answer

Paintings and photographs are 3+4=73 + 4 = 7 parts. Each part is 9 works, so 7×9=637 \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 14 of 120

2026 Practice Test C Math, question 32

Question

A box plot titled MONTHLY HIGH TEMPERATURES IN NEW YORK above a number line labeled Temperature (°F), from 36 to 88 with a labeled tick every 4. The left whisker starts at about 39, just left of the tick for 40. The box runs from 48 to about 78, between the ticks for 76 and 80, with the median line at 64. The right whisker ends at 84.

The box plot shows the average monthly high temperatures in New York City for 12 months. What is the difference between the range and the interquartile range of the temperature data?

Enter your answer in the space.

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

Show the answer

Correct answer: 15

Determine the range by subtracting the minimum value on the box plot from the maximum value on the box plot:

Maximum value =84∘ F= 84^{\circ}\text{ F}, Minimum value =39∘ F= 39^{\circ}\text{ F}

84∘ F−39∘ F=45∘ F84^{\circ}\text{ F} - 39^{\circ}\text{ F} = 45^{\circ}\text{ F}

Determine the interquartile range by subtracting the lower quartile value on the box plot from the upper quartile value on the box plot:

Upper Quartile Value =78∘ F= 78^{\circ}\text{ F}, Lower Quartile Value =48∘ F= 48^{\circ}\text{ F}

78∘ F−48∘ F=30∘ F78^{\circ}\text{ F} - 48^{\circ}\text{ F} = 30^{\circ}\text{ F}

Determine the difference between the range (45° F) and the interquartile range (30° F):

45∘ F−30∘ F=15∘ F45^{\circ}\text{ F} - 30^{\circ}\text{ F} = 15^{\circ}\text{ F}

Question analysis

A box plot of New York City's average monthly high temperatures for 12 months sits above a number line labeled every 4 degrees. Students type the difference between the range and the interquartile range of the data. Two of the plot's values fall between labeled ticks, so careful reading of the scale decides the answer.

NYS standards
  • 7.SP.1 box plots and interquartile range
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Reading a box plot and finding its interquartile range are grade 7 skills. Here two of the five values fall between ticks spaced 4 apart, and the range, the IQR and their difference must all be found and typed with no choices.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 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

The method is standard, but the key rests on reading the scale between its labels. The axis is ticked only every 4 degrees, and the minimum and the upper quartile land off the ticks, so rounding either one to the nearest label moves the result by a degree or two with nothing to flag it. Students who find the range correctly can also stop there, entering 45, the larger of the two spreads, instead of the difference the question asks for.

  1. Reading a figure, table or graph major Understanding the question

    Two of the five values fall between ticks spaced 4 degrees apart: the left whisker ends just short of the 40 tick, at 39, and the box’s right edge sits midway between 76 and 80, at 78. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Four values are read and combined in three subtractions, the range, the interquartile range and then their difference, so one misread value changes the final entry. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Something you must already know minor Working it out

    Neither measure is defined on the page: the range spans the whiskers, and the interquartile range is the width of the box, the upper quartile minus the lower. (Anderson & Krathwohl, 2001)

  4. An answer in an unexpected form minor Choosing the answer

    The result is typed with no choices to compare, so an entry one or two degrees off, from rounding a value to the nearest tick, looks just as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Read the minimum, both quartiles and the maximum, judging where each mark falls between the ticks, then subtract for the range and for the IQR, and subtract those two results.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range
range=maximum−minimum\text{range} = \text{maximum} - \text{minimum}
The maximum is the greatest value, at the right end of the right whisker. The minimum is the least value, at the left end of the left whisker. The range tells how far the whole set of data spreads.
Interquartile range (IQR)
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
Q3 is the upper quartile, the right edge of the box. Q1 is the lower quartile, the left edge of the box. The IQR is the width of the box. It tells how far the middle half of the data spreads.

  1. Sort out the question

    A box plot shows data with five numbers. The two whiskers are the lines on each side of the box. Their ends are the minimum, the least value, and the maximum, the greatest. The box runs from the lower quartile to the upper quartile. These are the middles of the lower half and the upper half of the data. The line inside the box is the median, the middle value.

    You need the range minus the interquartile range (IQR), in degrees.

  2. Read the values between the ticks

    The ticks on the number line are 4 degrees apart. So each degree is one quarter of the space between two ticks. Three values sit right on ticks: the box’s left edge at 48, the median at 64, and the right whisker’s end at 84.

    • The left whisker starts just left of the 40 tick. It sits about one quarter of the way back toward 36. So the minimum is 40−1=3940 - 1 = 39 degrees.
    • The box’s right edge is halfway between 76 and 80. Halfway is 78 degrees.

  3. Find the range

    Use the range formula:

    range=84−39=45\text{range} = 84 - 39 = 45

    The data spread over 45 degrees.

    Watch out: rounding the minimum to the 40 tick gives a range of 44. Then the final answer comes out 14, which is wrong.

  4. Find the IQR

    Use the IQR formula with the two edges of the box:

    IQR=78−48=30\text{IQR} = 78 - 48 = 30

    The middle half of the data spreads over 30 degrees.

    Watch out: reading the box’s right edge as 76 gives an IQR of 28. Then the final answer comes out 17.

  5. Subtract the IQR from the range

    The question asks for the difference between the two:

    45−30=1545 - 30 = 15

    Watch out: 45 is only the range. Do not stop there. Type only the number, with no degree sign.

    So you would enter 15.

Check your answer

The range covers the whole plot, and the IQR covers only the box. So their difference is the two whiskers added together. The left whisker is 48−39=948 - 39 = 9 degrees long. The right whisker is 84−78=684 - 78 = 6 degrees long. 9+6=159 + 6 = 15, which matches.

More questions like this

The original DOE question
Question 32 as shown on the DOE’s digital test, page 105 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 105, question 32.

Question 15 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?

  • A6
  • B15
  • C18
  • D30
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.

18÷6120÷6=320\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.

320(40)=12020=6\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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from repeated draws
red tiles≈red drawsall draws×tiles in the bag\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.

  1. 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.

  2. 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.

  3. Find the share of red draws

    Red came up 18 times out of 120. Simplify by dividing the top and bottom by 6:

    18120=320\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.

  4. Take that share of the bag

    Use the same share of the 40 tiles in the bag:

    320×40=12020=6\frac{3}{20} \times 40 = \frac{120}{20} = 6

    So about 6 of the 40 tiles are red.

  5. 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÷3=618 \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 16 of 120

2026 Practice Test C Math, question 35

Question

If 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a, what is the value of a in terms of b?

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: a = 10b

To find the value of a in terms of b, solve 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a for a. Begin by distributing 4 on the left side of the equation:

(4×a)+(4×2b)=28b+2a(4 \times a) + (4 \times 2b) = 28b + 2a

4a+8b=28b+2a4a + 8b = 28b + 2a

Then, get the a terms on one side and the b terms on the other:

4a−2a=28b−8b4a - 2a = 28b - 8b

Solve for the value of a in terms of b by combining like terms and then dividing each side by 2:

2a=20b2a = 20b

a=10ba = 10b

Question analysis

An algebra question gives an equation with a and b on both sides, one side written as 4 times a sum in parentheses, and students type a in terms of b after a printed "a =". It tests the distributive property and solving linear equations. The answer is typed, so a slip in distributing or moving terms is not caught by any choices.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve equations like 4(x + 2) = 28 with numbers. This one has letters on both sides, needs the distributive property and like terms, and asks for a in terms of b, which is grade 8 algebra.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 4, 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

Every step is routine, yet there are four of them, each with a classic slip, and the typed answer catches none. Multiplying the 4 into only the a, or carrying 8b across without changing its sign, gives another whole multiple of b that looks finished. Testing b = 1, where 4(a + 2) and 28 + 2a must come out equal, confirms an entry quickly.

  1. Variables instead of numbers moderate Understanding the question

    Both letters appear on both sides, and the answer must stay a multiple of b, so there is no number to check an entry against unless students invent one. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Many steps moderate Working it out

    Four moves run in a chain, distributing the 4, gathering the a terms, gathering the b terms and dividing by 2, and an error at any one carries into the typed result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. An answer in an unexpected form minor Choosing the answer

    The answer is typed after a printed “a =”, with no choices, so 20b, the line before dividing by 2, can be entered without anything on screen questioning it. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute the 4 to both terms in the parentheses, gather the a terms on one side and the b terms on the other, then divide by the coefficient of a.

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.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
k(x+y)=kx+kyk(x + y) = kx + ky
k is the number outside the parentheses. x and y are the terms inside. The number outside multiplies every term inside, not just the first one. It works because 4 groups of (x + y) hold 4 x’s and 4 y’s.
Doing the same thing to both sides
if A=B, then A−C=B−C and A2=B2\text{if } A = B, \text{ then } A - C = B - C \text{ and } \frac{A}{2} = \frac{B}{2}
A and B are the two equal sides of an equation. C is any amount. Taking the same amount from both sides, or dividing both by the same number, keeps them equal, like a balanced scale.

  1. Sort out the question

    The equation is 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a. You need a in terms of b. That means a alone on one side, with only b and numbers on the other.

    The answer box already shows a =, so you type only what comes after it.

  2. Multiply out the parentheses

    Use the distributive property. The 4 multiplies both terms inside:

    4×a+4×2b=28b+2a4 \times a + 4 \times 2b = 28b + 2a

    4a+8b=28b+2a4a + 8b = 28b + 2a

    Watch out: multiplying only the a by 4 gives 4a + 2b. That leads to the wrong answer 13b.

  3. Gather the a terms

    Like terms are terms with the same letter, such as 4a and 2a. Put the a terms on the left. Subtract 2a from both sides:

    4a−2a+8b=28b4a - 2a + 8b = 28b

    2a+8b=28b2a + 8b = 28b

  4. Gather the b terms

    Now subtract 8b from both sides:

    2a=28b−8b2a = 28b - 8b

    2a=20b2a = 20b

    Watch out: moving 8b without changing its sign gives 28b+8b=36b28b + 8b = 36b. That leads to 18b, which is wrong.

  5. Divide both sides by 2

    2a means 2 times a. Divide both sides by 2 to get a alone:

    a=20b2=10ba = \frac{20b}{2} = 10b

    Watch out: 20b is equal to 2a, not a. Do not stop there.

  6. Enter it after the printed a =

    On the test, the box already shows a =. Press 1, then 0, then the b key. You do not need a times sign. Do not type a = again. On this site, type 10b in the box.

    So you would enter 10b, which makes a = 10b.

Check your answer

Pick b = 1, so a = 10. The left side is 4(10+2)=4×12=484(10 + 2) = 4 \times 12 = 48. The right side is 28+2×10=4828 + 2 \times 10 = 48. The sides match, so 10b works.

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 C Math, page 108, question 35.

Question 17 of 120

2026 Practice Test D Math, question 1

Question

What is the prime factorization of 3,575?

  • A5⋅11⋅135 \cdot 11 \cdot 13
  • B52⋅11⋅135^{2} \cdot 11 \cdot 13
  • C5⋅7155 \cdot 715
  • D52⋅1435^{2} \cdot 143
Show the answer

Correct answer: B

By observation, 3,575 ends in 75; therefore, 3,575 is divisible by 25.

3,575÷25=1433{,}575 \div 25 = 143

Test 143 for factors and eliminate 3, 5, and 7, but 11 is a factor.

143÷11=13143 \div 11 = 13; both 11 and 13 are prime numbers.

The first factor, 25, is not prime but is the product of primes, 5×5=525 \times 5 = 5^{2}.

The prime factorization of 3,575 is 52×11×135^{2} \times 11 \times 13.

Question analysis

This number question asks for the prime factorization of 3,575, chosen from four products of factors. Students divide by primes, using divisibility tests, until only primes remain. The four products look much alike, so the work must be carried all the way through, keeping count of every prime divided out along the way.

NYS standards
  • 6.NS.4 common factors and multiples
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Breaking a number into primes is grade 6 number work that relies on divisibility tests. The number here is larger than in most class examples, but the work is still a few divisions, lighter than typical grade 7 problems.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 11, 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

Three of the four choices multiply back to 3,575, so checking a product rules out only the one that equals 715. The question turns on whether 143 is prime: it survives the quick tests for 2, 3 and 5, and the factor 11 is easy to miss. Students who keep dividing until every factor is prime, and who write a prime as often as it divides, finish in under a minute.

  1. A very close wrong answer moderate Choosing the answer

    5² · 143 matches the right factorization in its first factor and multiplies back to 3,575 just as well, so a product check cannot separate them; only splitting 143 into 11 × 13 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Nothing on the page defines a prime factorization or gives a test for 11, and 143 passes the familiar quick tests for 2, 3 and 5, so it looks prime. (Anderson & Krathwohl, 2001)

Strategy: Divide by the smallest prime that goes in evenly, again and again, writing down every prime you use until the quotient reaches 1, then compare that list with each choice.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisibility by 25
last two digits 00,25,50 or 75→divisible by 25\text{last two digits } 00, 25, 50 \text{ or } 75 \rightarrow \text{divisible by } 25
A number is divisible by 25 when 25 goes into it with no remainder. You only need to look at the last two digits. If they are 00, 25, 50 or 75, the number is divisible by 25. This works because 100 is 4 × 25, so the hundreds never leave a remainder.
Writing a repeated factor with an exponent
a2=a×aa^{2} = a \times a
a is any number. The small raised 2 is an exponent. It tells you how many times a is used as a factor, so 5² means 5 × 5 = 25. Use an exponent in a prime factorization when the same prime shows up more than once.

  1. Sort out the question

    A prime number is a whole number greater than 1 whose only factors are 1 and itself. Examples are 2, 3, 5, 7, 11 and 13. The prime factorization of 3,575 writes it as a product (the answer to a multiplication) of prime numbers only. So you must break 3,575 down until every factor is prime.

  2. Divide by 25 first

    3,575 ends in 75. So it is divisible by 25:

    3,575÷25=1433{,}575 \div 25 = 143

    That means 3,575 = 25 × 143. But 25 is not prime. It is 5 × 5, which you can write as 5².

  3. Test 143 for prime factors

    Now see if 143 splits. Try the primes in order. 143 is odd, so 2 does not work. Its digits add to 1 + 4 + 3 = 8, which is not a multiple of 3, so 3 does not work. It does not end in 0 or 5, so 5 does not work. Since 7 × 20 = 140, 143 ÷ 7 leaves a remainder of 3, so 7 does not work. Now try 11:

    143÷11=13143 \div 11 = 13

    So 143 = 11 × 13, and both 11 and 13 are prime.

    Watch out: 143 looks prime, but it is not. Stopping at 5² · 143 leads to choice D.

  4. Put all the primes together

    Collect every prime you found. The 5 appears twice, then 11 and 13 once each. The dot · means times, just like ×.

    3,575=5×5×11×13=52⋅11⋅133{,}575 = 5 \times 5 \times 11 \times 13 = 5^{2} \cdot 11 \cdot 13

    Watch out: write the 5 twice, since it divides 3,575 twice. Writing it once gives 5 · 11 · 13, choice A, which equals only 715. Choice C, 5 · 715, stops too early, because 715 ends in 5 and splits further.

    So the answer is B, 52⋅11⋅135^{2} \cdot 11 \cdot 13.

Check your answer

Multiply the primes back together: 25 × 11 = 275, and 275 × 13 = 3,575. That is the number you started with. Also, 5, 11 and 13 are all prime, so nothing can be split further.

More questions like this

The original DOE question
Question 1 as shown on the DOE’s digital test, page 74 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 74, question 1.

Question 18 of 120

2026 Practice Test D Math, question 5

Question

In a school raffle, 1 out of every 25 students wins a prize. Each prize includes a randomly selected trading card from a set of 100 different cards, labeled C1 through C100, with each card equally likely to be chosen. If one student is selected at random, what is the probability that the student both wins a prize and receives Card C17?

Write an expression that represents the probability that a student wins the prize and receives Card C17.

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: 1/25 · 1/100

To calculate the joint probability, first determine the probability of each independent event.

The probability that a randomly selected student wins the prize is 125\frac{1}{25}.

The probability that a randomly selected student will receive Card C17 is 1100\frac{1}{100}, because Card C17 is 1 card out of 100 labeled cards.

The probability that a student wins the prize and receives Card C17 is the product of those two probabilities, 125⋅1100\frac{1}{25} \cdot \frac{1}{100}.

Question analysis

In a school raffle, 1 out of every 25 students wins a prize, and each prize includes one of 100 equally likely trading cards. Students type an expression for the probability that a randomly chosen student both wins and gets one particular card. It tests the probability of two independent events, and combining the two chances correctly is the main hurdle.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 6.SP.8a uniform probability models
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students find compound probabilities, usually by listing outcomes for coins or number cubes. These two stages have 25 and 100 outcomes, far too many pairs to list, so a rule must be used, and the answer is typed as an expression.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 8, 87 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 chance is stated almost directly, 1 out of 25 and 1 out of 100, so the question tests one rule: both events must happen, and the chances multiply. The word “and” pulls students toward adding, which gives a result larger than either chance, a sign that something is wrong. Others write only the card's chance and forget that the student must first win a prize.

  1. Something you must already know moderate Working it out

    The question never says that the chance of two independent events both happening is the product of their chances, so students must bring that rule to the 1/25 and the 1/100 themselves. (Anderson & Krathwohl, 2001)

  2. An answer in an unexpected form minor Choosing the answer

    The answer is typed as an expression with no choices to compare, so a sum of the two chances, which looks just as finished, gets no warning on screen. (Enright, Morley & Sheehan, 2002)

Strategy: Write each event's chance as a fraction, then apply the rule for two independent events that must both happen, and check that the result is smaller than either chance alone.

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.
  • Stops one step early: A choice is a number from partway through the solution.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome
P=outcomes you wantall possible outcomesP = \frac{\text{outcomes you want}}{\text{all possible outcomes}}
P is the probability, the chance that something happens, from 0 to 1. The top counts the outcomes you want. The bottom counts all the outcomes that could happen. Use it when every outcome is equally likely, like one card picked from 100.
Probability that two independent events both happen
P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)
A and B are two events. Independent means one does not change the chance of the other. P(A) is the chance of A, and P(B) is the chance of B. Multiply them to get the chance that both happen. The result is smaller than either chance, since both things must happen.

  1. Sort out the question

    Two things must both happen. First, the student wins a prize. Second, the card in that prize is Card C17. You need an expression for that chance. An expression is a math phrase made of numbers and signs, with no equal sign. You do not have to work it out to one number.

  2. Find the chance of winning

    The question says “1 out of every 25 students wins a prize.” One student is the part that wins, and 25 is the whole group. So:

    P(win)=125P(\text{win}) = \frac{1}{25}

  3. Find the chance of Card C17

    Each prize has one card, picked at random from 100 cards. Every card is equally likely. Card C17 is 1 of those 100 cards:

    P(C17)=1100P(\text{C17}) = \frac{1}{100}

  4. Multiply the two chances

    Winning a prize does not change which card is inside it. So the two events are independent. To find the chance that both happen, multiply:

    P(win and C17)=125⋅1100P(\text{win and C17}) = \frac{1}{25} \cdot \frac{1}{100}

    The dot · means times. Worked out, this is 1/2500.

    Watch out: the word and does not mean add here. 1/25 + 1/100 is bigger than either chance, but needing both things must be less likely. Writing only 1/100 forgets that the student must win a prize first.

  5. Enter the expression

    On the test, press the fraction key, a small box over a small box. Type 1 on top and 25 on the bottom. Move the cursor out of the fraction with the → key. Press the dot key for times. Then use the fraction key again for 1 over 100. On this site, type 1/25 * 1/100 or 1/25 · 1/100. The value 1/2500 also counts.

    So you would enter 1/25 · 1/100.

Check your answer

Picture 2,500 students. 1 in 25 wins, so about 100 students win a prize. Of those 100, about 1 gets Card C17. That is 1 student out of 2,500, which matches 1/25 · 1/100 = 1/2500. It is smaller than 1/25 and 1/100, as it should be.

More questions like this

The original DOE question
Question 5 as shown on the DOE’s digital test, page 78 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 78, question 5.

Question 19 of 120

2026 Practice Test D Math, question 7

Question

A middle school library has a shelf with 40 books, each labeled with a different number from 201 to 240. One book is selected at random. What is the probability, as a percentage, of selecting book number 223?

Enter your answer in the space provided.

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

Show the answer

Correct answer: 2.5%

The probability that book number 223 will be selected at random from 40 books is 1 out of 40. To find the probability as a percentage, convert the fraction to a decimal.

The chance of selecting one specific book is 140=0.025\frac{1}{40} = 0.025.

Convert the decimal to a percentage: 0.025×100=2.5%0.025 \times 100 = 2.5\%

Question analysis

A library shelf holds 40 books numbered 201 to 240, and one is picked at random. Students find the probability of picking one particular book and type it as a percentage. It tests simple probability and converting a fraction to a percent; the book numbers play no part, and the decimal point needs care.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the probability of one equally likely outcome is grade 6 content. The only further step is rewriting the fraction as a percent, which needs care with place value, so it is lighter than grade 7 compound probability.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 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 book numbers, 201 to 240, and the chosen number 223 play no part, since every book is equally likely and the shelf holds 40. The chance is 1 in 40, and the real demand is the form: changing it to a percent means a decimal with a zero in the tenths place, then a shift of two places. Students who know that 1/4 is 25% can check that 1/40 must be a tenth of that.

  1. An answer in an unexpected form moderate Choosing the answer

    Only the bold word percentage sets the form, since the empty box shows no % sign, so the probability 1/40 must be turned into 2.5, not entered as the fraction or as 0.025. (Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Dividing 1 by 40 gives 0.025, with a zero right after the decimal point, and losing that zero turns the percent into 25. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the books, write the chance of picking one particular book as a fraction, then turn the fraction into a decimal and multiply by 100 to get the percent.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome
P=outcomes you wantall possible outcomesP = \frac{\text{outcomes you want}}{\text{all possible outcomes}}
P is the probability, the chance of an event. The top counts the outcomes you want. The bottom counts all the equally likely outcomes. Use it when each outcome has the same chance, like picking one book at random.
Changing a fraction to a percent
percent=ab×100\text{percent} = \frac{a}{b} \times 100
a/b is the fraction. Percent means out of 100. Multiplying by 100 tells you how many out of 100 the fraction is. For example, 1/2 × 100 = 50, so 1/2 is 50%.

  1. Sort out the question

    The shelf has 40 books, numbered 201 to 240. One book is picked at random, so every book has the same chance. You need the chance of picking book 223, written as a percentage. Book 223 is on the shelf, since 223 is between 201 and 240.

  2. Write the probability as a fraction

    Only 1 book is number 223, and there are 40 books in all:

    P=140P = \frac{1}{40}

    The numbers on the labels do not matter beyond this. They just give each book its own name.

  3. Change the fraction to a decimal

    Divide 1 by 40. It helps to start from 1/4 = 0.25. Since 40 is ten times 4, 1/40 is one tenth of 1/4:

    1÷40=0.0251 \div 40 = 0.025

    Watch out: there is a 0 right after the decimal point. Losing it gives 0.25 and then 25, which is wrong.

  4. Change the decimal to a percent

    Multiply by 100 to get a percent. That moves the decimal point two places to the right:

    0.025×100=2.50.025 \times 100 = 2.5

    So the chance is 2.5%.

    Watch out: 0.025 is the probability as a decimal, and 1/40 is the same chance as a fraction. The question asks for a percentage, so neither one is the answer here.

  5. Enter the number

    The box shows no % sign. Type 2.5. Adding the sign, as 2.5%, also counts.

    So you would enter 2.5.

Check your answer

2.5% means 2.5 out of every 100. Out of 40 books, that is 40 × 0.025 = 1 book, the one numbered 223. Also, one book out of 40 should be a small chance, and 2.5% is small.

More questions like this

The original DOE question
Question 7 as shown on the DOE’s digital test, page 80 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 80, question 7.

Question 20 of 120

2026 Practice Test D Math, question 15

Question

111+122+133\frac{1}{11} + \frac{1}{22} + \frac{1}{33}

What is the value of the expression shown above?

  • A122\frac{1}{22}
  • B111\frac{1}{11}
  • C16\frac{1}{6}
  • D322\frac{3}{22}
Show the answer

Correct answer: C

To add fractions, find a common denominator and rename the fractions using the common denominator. The least common denominator for 11, 22, and 33 is 66. After adding the fractions, the final fraction can be reduced to lowest terms.

111=1×611×6=666\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}

122=1×322×3=366\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}

133=1×233×2=266\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}

666+366+266=1166=16\frac{6}{66} + \frac{3}{66} + \frac{2}{66} = \frac{11}{66} = \frac{1}{6}

Question analysis

This fractions question asks for the value of 1/11 + 1/22 + 1/33. Students find a common denominator for three fractions, rename each one, add and simplify. The denominators are all multiples of 11, and adding the tops and bottoms straight across, or forgetting to rename a numerator, leads to tempting wrong choices.

NYS standards
  • 7.NS.1d properties for adding and subtracting
  • 6.NS.4 common factors and multiples
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding fractions with unlike denominators is taught before grade 7. Finding a common denominator for three related numbers and simplifying is short, routine work, lighter than typical grade 7 problems with signed rational numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 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 checks whether students rename fractions before adding, and each wrong choice comes from a shortcut that skips that step or stops early. Seeing that 22 and 33 are multiples of 11 points to 66 as the common denominator. A size check helps too: the sum must be more than 1/11, its largest term, which rules out two of the wrong choices at once.

  1. Error-prone arithmetic moderate Working it out

    Three unlike denominators must be renamed over 66, not over the product of all three, and the numerators 6, 3 and 2 have to be found one by one before adding. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. An answer in an unexpected form minor Choosing the answer

    The sum comes out as 11/66, which must be reduced to match a choice; its simplest form has a denominator, 6, unlike any in the expression. (Enright, Morley & Sheehan, 2002)

Strategy: Notice that 22 and 33 are multiples of 11, rename all three fractions over their least common multiple, add the numerators, and simplify 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.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a/b is a fraction, and n is any number except 0. Multiplying the top and the bottom by the same n gives an equal fraction. It works because n/n is 1, and multiplying by 1 changes nothing. Use it to give fractions the same bottom.
Adding fractions with the same denominator
ad+bd+cd=a+b+cd\frac{a}{d} + \frac{b}{d} + \frac{c}{d} = \frac{a + b + c}{d}
d is the denominator, the bottom number, which tells the size of each piece. a, b and c are the numerators, the top numbers, which count the pieces. When the pieces are the same size, add the tops and keep the bottom.

  1. Sort out the question

    You need to add three fractions: 1/11, 1/22 and 1/33. Their denominators (bottom numbers) are all different. You can only add fractions when the pieces are the same size. So first give all three the same denominator.

  2. Find a common denominator

    A common denominator is a number that 11, 22 and 33 all divide into. Notice that 22 = 2 × 11 and 33 = 3 × 11. So the smallest number that all three divide into is:

    2×3×11=662 \times 3 \times 11 = 66

  3. Rename each fraction

    Multiply the top and bottom of each fraction by the number that makes the bottom 66:

    111=1×611×6=666\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}

    122=1×322×3=366\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}

    133=1×233×2=266\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}

  4. Add the tops

    Now every piece is a sixty-sixth. Add the tops and keep the bottom:

    666+366+266=1166\frac{6}{66} + \frac{3}{66} + \frac{2}{66} = \frac{11}{66}

    Watch out: never add the bottoms. Adding straight across gives 3/66 = 1/22, choice A. Putting the three 1s over 33 gives 3/33 = 1/11, choice B.

  5. Simplify the fraction

    To simplify, divide the top and bottom by a number that goes into both. Here 11 goes into 11 and into 66:

    1166=11÷1166÷11=16\frac{11}{66} = \frac{11 \div 11}{66 \div 11} = \frac{1}{6}

    Watch out: add all three fractions. Stopping after the first two gives 3/22, choice D.

    So the answer is C, 16\frac{1}{6}.

Check your answer

Estimate with decimals. 1/11 is about 0.091, 1/22 is about 0.045 and 1/33 is about 0.030. They add to about 0.166. And 1/6 is 0.1666…, a match. The other choices are all 0.136 or less.

More questions like this

The original DOE question
Question 15 as shown on the DOE’s digital test, page 88 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 88, question 15.

Question 21 of 120

2026 Practice Test D Math, question 19

Question

A store manager purchases a piece of furniture for $456.00. To determine the selling price, the manager increases the purchase cost by 125%. A customer buys the furniture and pays an additional 8% sales tax. How much does the customer pay for the furniture?

  • A$615.60
  • B$627.48
  • C$1,108.08
  • D$1,846.80
Show the answer

Correct answer: C

Given a purchase price of $456.00, the 125% increase in price is calculated as:

456×1.25=570456 \times 1.25 = 570

The selling price equals the purchase price plus the increase, or:

456+570=1,026456 + 570 = 1{,}026

The sales tax on the selling price is:

1,026(0.08)=82.081{,}026(0.08) = 82.08

The customer pays the selling price plus the sales tax:

1,026+82.08=1,108.081{,}026 + 82.08 = 1{,}108.08

Question analysis

A store manager buys a piece of furniture for $456.00 and raises the price by 125% to set the selling price, and the customer also pays 8% sales tax. Students find the total the customer pays. It tests percent increase and sales tax in sequence, and reading an increase of more than 100% correctly is the main challenge.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students find markups and sales tax, usually with increases under 100%. An increase of more than 100% is unusual and easy to misread, and two percent changes must be applied in order, which goes beyond routine grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 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

A percent above 100 is unusual in grade 7 work, and it is the center of the question: the markup more than doubles the price. Students who treat “by 125%” as “125% of” still find a listed answer after adding tax, and so do students who add $125. A rough check helps: the selling price must be more than twice $456, about $1,000, and the tax adds less than a tenth of that.

  1. Turning words into math moderate Understanding the question

    “Increases the purchase cost by 125%” adds 125% of $456 to the $456 itself, a multiplier of 2.25; reading it as 125% of the cost leaves the original price out. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps moderate Working it out

    Four steps in a row, the increase, the selling price, the tax and the total, each depend on the one before, and two of them are easy to merge or skip. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    Taking 8% of $1,026 means multiplying by 0.08, and a decimal point one place off, 0.8, makes the tax ten times too large. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Turn each percent change into a multiplier, remembering that an increase by a percent is added to the original amount, apply the markup and then the tax, and estimate to check the size.

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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
p% of n=p100×np\% \text{ of } n = \frac{p}{100} \times n
n is the starting amount, and p is the percent. Percent means out of 100, so 125% is 125/100 = 1.25 and 8% is 8/100 = 0.08. Multiply n by that decimal to find the part.
Price after an increase
new price=old price+increase\text{new price} = \text{old price} + \text{increase}
The old price is what you start with. The increase is the amount added on. To increase by a percent, find that percent of the old price, then add it to the old price. Sales tax is added the same way.

  1. Sort out the question

    The manager pays $456.00 for the furniture. The selling price is that cost increased by 125%. The customer then pays 8% sales tax on top. Sales tax is an extra charge, a percent of the price. You need the total the customer pays.

  2. Find the increase

    Find 125% of $456. As a decimal, 125% is 1.25, which is 1 and a quarter. A quarter of 456 is 114:

    456×1.25=456+114=570456 \times 1.25 = 456 + 114 = 570

    The increase is $570.

  3. Find the selling price

    Increases by means the increase is added to the cost:

    456+570=1,026456 + 570 = 1{,}026

    The selling price is $1,026.

    Watch out: $570 is only the increase. Using it as the price gives $615.60, choice A. And 125% is a percent, not $125. Adding $125 leads to $627.48, choice B.

  4. Find the sales tax

    As a decimal, 8% is 0.08. Find 8% of the selling price:

    1,026×0.08=82.081{,}026 \times 0.08 = 82.08

    The tax is $82.08.

    Watch out: 8% is 0.08, not 0.8. Using 0.8 makes the tax ten times too big and gives $1,846.80, choice D.

  5. Add the tax to the price

    The customer pays the selling price plus the tax:

    1,026+82.08=1,108.081{,}026 + 82.08 = 1{,}108.08

    So the answer is C, $1,108.08.

Check your answer

Estimate. The price more than doubles: $456 plus $570 is about $1,000. Tax of 8% adds about $80. So the total should be a bit over $1,080, and only $1,108.08 is close. Or multiply in one go: 456 × 2.25 × 1.08 = 1,108.08.

More questions like this

The original DOE question
Question 19 as shown on the DOE’s digital test, page 92 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 92, question 19.

Question 22 of 120

2026 Practice Test D Math, question 23

Question

Which situation involves quantities that combine to equal zero?

  • Areceiving $5 as a gift and then giving $5 to a friend
  • Bbuying a book for $10 and then buying lunch for $10
  • Cgetting on an elevator at the ground floor, rising 3 floors, and then rising 3 more floors
  • Dstarting in a cave 20 feet below the ground, climbing up 20 feet, and then climbing up 20 feet more
Show the answer

Correct answer: A

When looking for a situation where quantities combine to make zero, find two values with the same absolute value where one value is positive and one value is negative.

In the situation in option A, receiving a gift of $5 is represented by +5+5, and giving a friend $5 is represented by −5-5. These values combine to equal zero because +5−5=0+5 - 5 = 0.

Question analysis

This question gives four everyday situations involving money, an elevator and a cave, and asks which one involves quantities that combine to make zero. It tests representing gains and losses, or ups and downs, with positive and negative numbers. Equal numbers alone do not make zero, so the direction of each amount must be read carefully.

NYS standards
  • 7.NS.1a opposite quantities making 0
  • 6.NS.5 positive and negative numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Describing situations where opposite quantities make zero is named in the grade 7 standards, and this is its most direct form: one reading of four everyday situations, easier than typical grade 7 work with signed numbers.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
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

The idea that a quantity and its opposite add to zero is familiar, and the right situation states it plainly. The wrong choices are built from equal amounts: two of them repeat a number but move in one direction, and the cave climb reaches zero partway before going 20 feet past it. Writing each situation as a sum of signed numbers takes a few seconds and settles it.

  1. Turning words into math minor Understanding the question

    Each situation must be turned into signed numbers, receiving or rising as plus and giving or buying as minus, before its total can be judged. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Every choice must be checked minor Choosing the answer

    Every choice repeats one amount, $5, $10, 3 floors or 20 feet, so equal numbers alone do not identify the answer; the direction of each move has to be checked. (Embretson & Wetzel, 1987)

Strategy: Write each situation as signed numbers, using + for gains or upward moves and − for losses or downward moves, add them, and see which total is zero.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Opposites add to zero
a+(−a)=0a + (-a) = 0
a is any number, and −a is its opposite: the same size, but the other sign. A number and its opposite always add to 0. On a number line, they sit the same distance from 0 on opposite sides. Equal moves in opposite directions cancel out.

  1. Sort out the question

    You need the situation whose amounts add up to zero. Give each amount a sign. Use + for getting money or going up. Use − for spending money or going down. Then add the amounts in each choice. The direction matters, not just the numbers.

  2. Write each choice with signs

    Turn each situation into signed numbers and add them up:

    ChoiceAmountsTotal
    A+5, then −50
    B−10, then −10−20
    C+3, then +3+6
    Dstart at −20, then +20, then +20+20

  3. Look at the tricky choices

    In B, buying a book and buying lunch both take money away. Equal amounts cancel only when they go in opposite directions. In C, rising and rising again both go up, so you end 6 floors up.

    Watch out: in D, starting at −20 and climbing 20 feet does reach 0, ground level. But then you climb 20 feet more. You end at +20, 20 feet above the ground, so D is wrong.

  4. Choose the zero total

    Only choice A has two opposite amounts of the same size. Getting $5 is +5, and giving $5 away is −5:

    5+(−5)=05 + (-5) = 0

    So the answer is A: receiving $5 as a gift and then giving $5 to a friend.

Check your answer

Think about where you end up. In A, you end with exactly the money you started with, so the total change is zero. In every other choice, you end with less money, higher up, or above the ground.

More questions like this

The original DOE question
Question 23 as shown on the DOE’s digital test, page 96 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 96, question 23.

Question 23 of 120

2026 Practice Test D Math, question 27

Question

A birdseed mixture is 25\frac{2}{5} sunflower seeds. The cost of the sunflower seeds is $0.10 per ounce. At this rate, what is the cost of the sunflower seeds in 25 pounds of this birdseed? (Note: 1 lb = 16 oz.)

  • A$10.00
  • B$16.00
  • C$40.00
  • D$160.00
Show the answer

Correct answer: B

First, calculate the amount of sunflower seeds in 25 pounds of this birdseed by using ratio reasoning to find that there are 10 pounds of sunflower seeds in the birdseed.

25=1025\frac{2}{5} = \frac{10}{25}

Then, use ratio reasoning to find the cost of the sunflower seeds per pound.

($0.101 ounce)(16 ounces1 pound)=$1.60\left(\frac{\$0.10}{1 \text{ ounce}}\right)\left(\frac{16 \text{ ounces}}{1 \text{ pound}}\right) = \$1.60

The cost of the sunflower seeds is $1.60 per pound.

Finally, find the total cost using ratios by multiplying the unit cost per pound by 10 pounds.

$1.601 pound×10 pounds1=$16.00\frac{\$1.60}{1 \text{ pound}} \times \frac{10 \text{ pounds}}{1} = \$16.00

The first two steps above could be done in the other order; that is, find $1.60 per pound first and 10 pounds of sunflower seeds second.

Question analysis

A birdseed mixture is two-fifths sunflower seeds, which cost 10 cents an ounce, and students find the cost of the sunflower seeds in 25 pounds of the mix. It combines a fraction of a whole, a pounds-to-ounces conversion and a unit price. Each step's result looks like a possible answer, so keeping track of which quantity is in hand matters.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3d converting measurement units
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Multistep ratio problems with a unit conversion are typical grade 7 work. This one chains three short steps, a fraction of the mixture, pounds to ounces and a price per ounce, and the numbers are chosen to come out evenly.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
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

Each step is routine, but the steps can be done in more than one order, and every partial result looks like a price once a dollar sign is put in front of it. Each wrong choice leaves something out: the 2/5, the price per ounce, or the conversion and the price together. Labelling each number with its unit, pounds, ounces or dollars, as it is found keeps the chain straight and shows when it is finished.

  1. Many steps moderate Working it out

    Three operations are chained: 2/5 of 25 pounds gives 10 pounds of sunflower seeds, the note turns them into 160 ounces, and $0.10 an ounce prices them; stopping at 10 pounds or 160 ounces, or skipping the 2/5, gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Combines ideas minor Working it out

    A fraction of a mixture, a pounds-to-ounces conversion and a price per ounce meet in one question, and the price is per ounce while the amount is given in pounds. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Find the weight of the sunflower seeds first, change it to ounces with the note, then multiply by the price per ounce; or turn the price into dollars per pound first.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A fraction of an amount
part=fraction×whole\text{part} = \text{fraction} \times \text{whole}
The whole is the full amount, here 25 pounds of birdseed. The fraction tells how much of it you want. To find 2/5 of something, split it into 5 equal parts and take 2 of them. Multiplying does both at once.
Changing pounds to ounces
ounces=pounds×16\text{ounces} = \text{pounds} \times 16
The note says 1 pound is 16 ounces. So each pound turns into 16 ounces. An ounce is smaller than a pound, so you get a bigger number.
Total cost
cost=number of units×price of one unit\text{cost} = \text{number of units} \times \text{price of one unit}
The units here are ounces. The price of one unit is $0.10 for one ounce. Every ounce costs the same, so you multiply the number of ounces by the price of one.

  1. Sort out the question

    You know:

    • The birdseed is 25\frac{2}{5} sunflower seeds.
    • There are 25 pounds of birdseed.
    • Sunflower seeds cost $0.10 per ounce.
    • 1 pound = 16 ounces.

    You need the cost, in dollars, of just the sunflower seeds. The price is per ounce, but the weight is in pounds. So you will need to change pounds to ounces.

  2. Find the weight of sunflower seeds

    Only 2/5 of the birdseed is sunflower seeds. Find 2/5 of 25 pounds:

    25×25=505=10\frac{2}{5} \times 25 = \frac{50}{5} = 10

    Another way: 25 ÷ 5 = 5, and 2 fifths is 2 × 5 = 10. So there are 10 pounds of sunflower seeds.

    Watch out: 10 is a number of pounds, not dollars. Stopping here and calling it $10.00 leads to choice A.

  3. Change pounds to ounces

    The price is per ounce, so turn the 10 pounds into ounces. Each pound is 16 ounces:

    10×16=16010 \times 16 = 160

    There are 160 ounces of sunflower seeds.

    Watch out: 160 is a number of ounces, not dollars. Calling it $160.00 leads to choice D.

  4. Find the cost

    Each ounce costs $0.10. Multiply:

    160×0.10=16.00160 \times 0.10 = 16.00

    The sunflower seeds cost $16.00.

    Watch out: skipping the 2/5 prices all 25 pounds. That is 400 ounces, and 400×0.10=40.00400 \times 0.10 = 40.00, choice C.

    So the answer is B, $16.00.

Check your answer

Do it in another order. One pound of sunflower seeds costs 16×0.10=1.6016 \times 0.10 = 1.60 dollars. Ten pounds cost 10×1.60=16.0010 \times 1.60 = 16.00 dollars. You get $16.00 again.

More questions like this

The original DOE question
Question 27 as shown on the DOE’s digital test, page 100 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 100, question 27.

Question 24 of 120

2026 Practice Test D Math, question 29

Question

Karl has one red spinner and one blue spinner. Each spinner is divided into 4 equal sections, numbered 1 through 4. He spins each spinner once and writes down the number that each lands on. What is the probability that the two numbers, when multiplied together, will have 4 as a product?

  • A116\frac{1}{16}
  • B316\frac{3}{16}
  • C47\frac{4}{7}
  • D716\frac{7}{16}
Show the answer

Correct answer: B

There are four possible outcomes for each spin of the spinner:

1, 2, 3, 4

The possible outcomes of two spins are:

11121314212223243132333441424344\begin{array}{cccc} 11 & 12 & 13 & \boxed{14} \\ 21 & \boxed{22} & 23 & 24 \\ 31 & 32 & 33 & 34 \\ \boxed{41} & 42 & 43 & 44 \end{array}

Of the 16 possible outcomes, only 3 pairings have a product of 4: 14, 22, and 41.

Therefore, the probability that the two numbers resulting from two consecutive spins have a product of 4 is 316\frac{3}{16}.

Question analysis

Karl spins a red spinner and a blue spinner, each with four equal sections numbered 1 to 4, and multiplies the two results. This compound probability question asks for the chance that the product is 4. Students build the sample space of outcomes, and listing every pair rather than the ones that come to mind is the safe route.

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
At grade 7 Finding a compound probability from a list or table of equally likely outcomes is standard grade 7 work. Two four-section spinners give a small sample space that is quick to write out, and the event needs one careful scan of the products.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 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

The question is decided by the count of favorable outcomes, not by any calculation. Students who think of 4 only as 2 × 2 find one outcome, and those who think in factor pairs find two, which is not a listed choice and should send them back to the table. A 4-by-4 grid of products, red down the side and blue across, shows all three cells at once.

  1. Something you must already know moderate Working it out

    Nothing in the stem says that red 1 with blue 4 and red 4 with blue 1 are separate outcomes, or that two four-part spinners give 16 equally likely pairs. (Anderson & Krathwohl, 2001)

  2. Much to keep track of minor Working it out

    All 16 pairs must be scanned for a product of 4, and the two pairs that use 1 and 4 are easy to miss beside the obvious 2 times 2. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Make a 4-by-4 table with the red spinner's numbers down the side and the blue spinner's across the top, write each product in its cell, and count the cells that match.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event
P=number of outcomes you wanttotal number of outcomesP = \frac{\text{number of outcomes you want}}{\text{total number of outcomes}}
P is the probability, the chance that something happens. An outcome is one possible result. This works when every outcome is equally likely, like landing on equal sections of a spinner.
Counting outcomes of two actions
total outcomes=a×b\text{total outcomes} = a \times b
a is the number of results for the first action, and b is the number for the second. Each result of the first can go with each result of the second, so you multiply.

  1. Sort out the question

    Karl spins two spinners, one red and one blue. Each has 4 equal sections, numbered 1 to 4. He multiplies the two numbers. You need the probability that the product, the answer to a multiplication, is 4.

  2. Count all the outcomes

    The red spinner has 4 results, and so does the blue one. Each red result can go with each blue result:

    4×4=164 \times 4 = 16

    So there are 16 outcomes, all equally likely. Red 1 with blue 4 is not the same as red 4 with blue 1. They come from different spinners.

  3. Make a table of products

    Write each product in a table. Red goes down the side and blue goes across the top.

    Red × Blue1234
    11234
    22468
    336912
    4481216

    Three cells show 4: red 1 with blue 4, red 2 with blue 2, and red 4 with blue 1.

    Watch out: thinking of 4 only as 2 × 2 finds one outcome. That gives 1/16, choice A.

  4. Find the probability

    Use the probability rule. Three outcomes give 4, out of 16 in all:

    P=316P = \frac{3}{16}

    Watch out: 7 outcomes have a 4 on at least one spinner. But that counts pairs that show a 4, not pairs whose product is 4. That mistake gives 7/16, choice D.

    So the answer is B, 316\frac{3}{16}.

Check your answer

List the ways to multiply two numbers from 1 to 4 and get 4: 1×41 \times 4, 2×22 \times 2 and 4×14 \times 1. That is 3 ways. The table has 16 cells, and 13 of them are not 4. So a small chance like 3/16 makes sense.

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 D Math, page 102, question 29.

Question 25 of 120

2026 Practice Test D Math, question 33

Question

A car costing $24,000 is subject to a sales tax of 8%. If Bindu made a $10,000 down payment on this car, what is the total amount she has left to pay?

  • A$14,000
  • B$15,620
  • C$15,920
  • D$25,920
Show the answer

Correct answer: C

First, find the sales tax. Then, add the sales tax to the cost of the car. Finally, subtract the down payment to find out how much she has left to pay.

(0.08)(24,000)=1,920(0.08)(24{,}000) = 1{,}920

$24,000+$1,920=$25,920\$24{,}000 + \$1{,}920 = \$25{,}920

$25,920−$10,000=$15,920\$25{,}920 - \$10{,}000 = \$15{,}920

Question analysis

Bindu buys a $24,000 car with an 8% sales tax and makes a $10,000 down payment, and students find how much she still has to pay. This percent word problem combines finding a tax, adding it to the price and subtracting a payment. Every step must be done in order, since leaving one out gives an amount that looks just as reasonable.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Adding sales tax to a price is a standard grade 7 percent task. This one also subtracts a down payment at the end, three routine steps with round numbers, much like a typical classroom problem.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 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 method is the standard one for tax on a purchase, so the question checks order and completeness. Forgetting the tax or forgetting the down payment each produces an amount as reasonable-looking as the answer, and a careless 8 × 24 produces one just $300 below it. Multiplying the price by 1.08 in one step, which gives $25,920, is a second route that catches a slip in the tax.

  1. Many steps minor Working it out

    Three operations come in a fixed order, 8% of $24,000, then adding the tax, then subtracting the $10,000, and leaving out the first or the last gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Finding 8% of $24,000 comes down to 8 × 24 = 192 with two zeros added, and a lost carry in that product turns $1,920 into $1,620. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find 8% of the price and add it, or multiply the price by 1.08, to get the full cost with tax, then subtract what Bindu has already paid.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}
Percent means out of 100. So 8% is 8/100, or 0.08. To find a percent of a number, change the percent to a decimal and multiply it by the whole.
Amount left to pay
amount left=price+tax−down payment\text{amount left} = \text{price} + \text{tax} - \text{down payment}
The price plus the tax is the full cost. A down payment is money paid at the start. Subtract it, because that part is already paid.

  1. Sort out the question

    You know:

    • The car costs $24,000.
    • The sales tax is 8%. Sales tax is extra money added to a price.
    • Bindu paid $10,000 at the start. That is the down payment.

    You need the amount Bindu still has to pay. Work in order: tax, then full cost, then subtract.

  2. Find the sales tax

    Change 8% to the decimal 0.08. Multiply it by the price:

    0.08×24,000=1,9200.08 \times 24{,}000 = 1{,}920

    A quick way: 1% of 24,000 is 240, so 8% is 8×240=1,9208 \times 240 = 1{,}920. The tax is $1,920.

    Watch out: in 8 × 24 you carry a 3 from 8 × 4 = 32. Dropping it gives 162, so the tax looks like $1,620. That leads to $15,620, choice B.

  3. Find the full cost

    Add the tax to the price:

    24,000+1,920=25,92024{,}000 + 1{,}920 = 25{,}920

    The car costs $25,920 with tax.

    Watch out: this is not the answer yet. Stopping here gives $25,920, choice D. Bindu already paid part of it.

  4. Subtract the down payment

    Bindu already paid $10,000. Take it away:

    25,920−10,000=15,92025{,}920 - 10{,}000 = 15{,}920

    She has $15,920 left to pay.

    Watch out: leaving out the tax gives 24,000−10,000=14,00024{,}000 - 10{,}000 = 14{,}000, choice A.

    So the answer is C, $15,920.

Check your answer

Use a shortcut for the full cost. The price plus 8% more is 108% of the price, or 1.08 times it: 24,000×1.08=25,92024{,}000 \times 1.08 = 25{,}920. Then 25,920−10,000=15,92025{,}920 - 10{,}000 = 15{,}920. Same answer.

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 D Math, page 106, question 33.

Question 26 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?

  • A4
  • B6
  • C12
  • D24
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.
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.

  1. 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)

  2. 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)

  3. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting with two choices
number of groups=ways to pick the man×ways to pick the women\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.

  1. 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.

  2. 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.

  3. 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 4×3=124 \times 3 = 12 pairs, each one twice. Then doubling for the men gives 24, choice D.

  4. Add a man to each pair

    Each pair of women can go with man 1 or man 2. Use the counting rule:

    6×2=126 \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=126 + 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 27 of 120

2026 Practice Test D Math, question 36

Question

Leah has set a goal to read a total of 500 pages of a textbook. The table shows the book page numbers for the reading she did during three days.

DAILY READING LOG

Day Pages Read
1 pages 45–90, inclusive
2 pages 100–135, inclusive
3 pages 145–190, inclusive

What percentage of her 500-page goal did Leah read on these three days? Round your answer to the nearest tenth.

Enter your answer in the space.

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

Show the answer

Correct answer: 25.6%

First, determine the number of pages read each day. 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 day.

Day 1: (90−45)+1=46(90 - 45) + 1 = 46 pages

Day 2: (135−100)+1=36(135 - 100) + 1 = 36 pages

Day 3: (190−145)+1=46(190 - 145) + 1 = 46 pages

Leah read a total of 128 pages, since 46+36+46=12846 + 36 + 46 = 128.

To determine the percent, express 128 pages out of 500 pages as a fraction.

128500=128÷4500÷4=32125\frac{128}{500} = \frac{128 \div 4}{500 \div 4} = \frac{32}{125}

32÷125=0.256=25.6%32 \div 125 = 0.256 = 25.6\%

Question analysis

Leah's daily reading log lists three ranges of textbook pages, each marked inclusive, and her goal is 500 pages. This percent question asks what percentage of the goal she read over the three days, rounded to the nearest tenth and typed in. Counting the pages in each inclusive range correctly is the tricky step before the percent.

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 Finding what percent one number is of another is grade 6 and 7 work. This adds three inclusive page ranges to count, which takes more care than subtracting, then a rounding step, all typed with no choices to check against.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 3, 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

Counting pages in a range is the whole difficulty; the percent at the end is routine once the total is right. Subtracting first from last feels natural, and here it even rewards the student with a round 25%, so nothing looks wrong. Checking the rule on a tiny range, pages 1 to 3, which is 3 pages and not 2, fixes the count, and the instruction to round to the nearest tenth hints that the answer is not a whole number.

  1. Something you must already know moderate Working it out

    Every row says inclusive, but nothing says that pages 45 to 90 hold 90 − 45 + 1 = 46 pages; subtracting alone loses one page on each of the three days. (Anderson & Krathwohl, 2001)

  2. Many steps moderate Working it out

    Three page counts, a sum, a division by 500 and a change to a percent come in sequence, and a one-page error in each count carries through to the end. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. An answer in an unexpected form minor Choosing the answer

    The answer is typed with no choices, and the miscounted total, 125 pages, gives a tidy 25% that looks more finished than the right value with its tenth. (Enright, Morley & Sheehan, 2002)

Strategy: Count each day's pages carefully, testing your method on a short range such as pages 1–3, add the three days, divide by 500, and write the result as a percent to the nearest tenth.

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).
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting pages in an inclusive range
pages=last page−first page+1\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
percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100
The part is the amount you have, here the pages read. The whole is the amount you compare with, here the 500-page goal. Dividing gives a decimal. Multiplying by 100 turns it into a percent, a number out of 100.

  1. Sort out the question

    Leah’s goal is 500 pages. The log shows three days:

    • Day 1: pages 45 to 90
    • Day 2: pages 100 to 135
    • Day 3: pages 145 to 190

    Each range is inclusive. You need the percent of 500 pages she read, rounded to the nearest tenth.

  2. Count the pages each day

    Use the rule for an inclusive range: last page minus first page, plus 1.

    90−45+1=4690 - 45 + 1 = 46

    135−100+1=36135 - 100 + 1 = 36

    190−145+1=46190 - 145 + 1 = 46

    Watch out: subtracting alone gives 45, 35 and 45 pages. That is 125 pages, which is exactly 25%. A round answer like that can hide the mistake.

  3. Add the three days

    Add the pages from all three days to get her total:

    46+36+46=12846 + 36 + 46 = 128

    Leah read 128 pages in all. This is the part you compare with the goal.

  4. Change to a percent

    Divide the pages read by the goal, then multiply by 100:

    128500×100=0.256×100=25.6\frac{128}{500} \times 100 = 0.256 \times 100 = 25.6

    Leah read 25.6% of her goal. That already ends in tenths, so no rounding is needed.

    Watch out: 0.256 is a decimal, not a percent. And 26 rounds to a whole number, not to the nearest tenth.

  5. Enter the answer

    The box has no % sign, but the question asks for a percentage, so the number in the box is the percent. Type it with its decimal point. Typing 25.6% also counts on this site. So you would enter 25.6.

Check your answer

Estimate. 10% of 500 pages is 50 pages. 128 pages is a bit more than 2 × 50 = 100, so the answer is between 20% and 30%. Then check exactly: 0.256×500=1280.256 \times 500 = 128 pages. It matches.

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 D Math, page 109, question 36.

Question 28 of 120

2026 Practice Test D Math, question 37

Question

What 3-digit number between 630 and 650 is divisible by both 5 and 8?

Enter your answer in the space.

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

Show the answer

Correct answer: 640

The final digit of the three-digit number is 0. Each of the digits 0 through 9, inclusive, must be considered.

To narrow the choices, first consider that the number between 630 and 650 must be divisible by both 5 and 8.

A number divisible by 5 must end in 0 or 5. Therefore, the only possibilities are 0 and 5.

  • If the number ended in 0, the whole number 640 must also be divisible by 8:

    640÷8=80 and 640÷5=128640 \div 8 = 80 \text{ and } 640 \div 5 = 128

    Both result in a remainder of 0.

  • If the number ended in 5, the whole numbers 635 and 645 must also be divisible by 8:

    635÷8=79R3 and 635÷5=127635 \div 8 = 79R3 \text{ and } 635 \div 5 = 127

    645÷8=80R5 and 645÷5=129645 \div 8 = 80R5 \text{ and } 645 \div 5 = 129

    Dividing 635 by 5 results in a remainder of 0, but dividing 635 by 8 results in a remainder of 3.

    Dividing 645 by 5 results in a remainder of 0, but dividing 645 by 8 results in a remainder of 5.

Therefore, 640 is the only whole number between 630 and 650 that is divisible by both 5 and 8.

Question analysis

This number question asks for the 3-digit number between 630 and 650 that is divisible by both 5 and 8, typed into a box. Students use divisibility rules or common multiples to find it. The work is short, but the number has to pass both tests, so checking only one divisor is the easy slip.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Common multiples are grade 6 content. Divisibility by 5 limits the last digit, and only a few numbers lie between 630 and 650, so this is a quick check, shorter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
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

Few numbers need checking, so most students succeed; the risk is testing only one divisor. Students who start from multiples of 8 meet 632 first, and those who start from multiples of 5 can slip on 645 ÷ 8. Using the least common multiple, 40, shows at once that the number must be a multiple of 40, which leaves one candidate in the range.

  1. Much to keep track of minor Working it out

    Each candidate must pass two tests, divisible by 5 and by 8, and both ends of the range, 630 and 650, pass the first test but fail the second. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Something you must already know minor Working it out

    The shortcut that a multiple of 5 ends in 0 or 5 is assumed, and it cuts the search to 635, 640 and 645 before any division by 8. (Anderson & Krathwohl, 2001)

Strategy: Use divisibility by 5 to limit the last digit, then check only those few numbers in the range for divisibility by 8.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisible by 5
last digit=0 or 5\text{last digit} = 0 \text{ or } 5
Divisible means it divides evenly, with no remainder. A whole number is divisible by 5 only when its last digit is 0 or 5. That is because counting by 5s always lands on a number ending in 0 or 5.
Least common multiple
LCM(5,8)=5×8=40\text{LCM}(5, 8) = 5 \times 8 = 40
The least common multiple, or LCM, is the smallest number that both 5 and 8 divide evenly. 5 and 8 share no factor except 1, so their LCM is their product. Any number divisible by both is a multiple of 40.

  1. Sort out the question

    You need a 3-digit number between 630 and 650. It must be divisible by 5 and also by 8. Between means 630 and 650 themselves do not count.

  2. Use the rule for 5

    A number divisible by 5 ends in 0 or 5. Between 630 and 650, only three numbers do:

    635,640,645635, \; 640, \; 645

    Only these need checking for 8.

  3. Test each number for 8

    Divide each one by 8 and look at the remainder, the amount left over:

    • 635 ÷ 8 = 79, with 3 left over.
    • 640 ÷ 8 = 80, with nothing left over.
    • 645 ÷ 8 = 80, with 5 left over.

    Only 640 divides evenly by 8.

    Watch out: 645 passes the test for 5 but fails for 8. And 632, which is 8×798 \times 79, passes for 8 but ends in 2, so it fails for 5.

  4. Enter the number

    640 passes both tests. Type the whole number in the box.

    Watch out: 40 is the smallest number divisible by 5 and 8, but it is not between 630 and 650.

    So you would enter 640.

Check your answer

Use the LCM. A number divisible by both 5 and 8 must be a multiple of 40. 40×16=64040 \times 16 = 640. The multiples next to it are 600 and 680, which are outside the range. So 640 is the only one.

More questions like this

The original DOE question
Question 37 as shown on the DOE’s digital test, page 110 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 110, question 37.

Question 29 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+1.25cp = 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?

  • A4
  • B9
  • C16
  • D18
Show the answer

Correct answer: C

Using the specified equation p=2.50+1.25cp = 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+1.25c22.50 = 2.50 + 1.25c

22.50−2.50=1.25c22.50 - 2.50 = 1.25c

20.00÷1.25=c20.00 \div 1.25 = c

16=c16 = 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.
What makes it hard
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.

  1. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides
if a=b, then a−n=b−n and an=bn\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.

  1. Sort out the question

    The equation p=2.50+1.25cp = 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.

  2. 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+1.25c22.50 = 2.50 + 1.25c

  3. Subtract the tray

    Take 2.50 away from both sides. That removes the cost of the tray:

    22.50−2.50=1.25c22.50 - 2.50 = 1.25c

    20.00=1.25c20.00 = 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.

  4. Divide by the price of one cupcake

    Divide both sides by 1.25:

    c=20.00÷1.25=16c = 20.00 \div 1.25 = 16

    An easy way: 4 cupcakes cost 4×1.25=5.004 \times 1.25 = 5.00 dollars. $20 is 4 groups of $5. So Samantha bought 4×4=164 \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. 1.25×16=20.001.25 \times 16 = 20.00, and 2.50+20.00=22.502.50 + 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 30 of 120

2026 Practice Test D Math, question 45

Question

Marcus and Isla are saving money to buy concert tickets.

  • They decide to save the same amount each week.
  • They already have $8 saved.
  • After 3 weeks, the two have $26 saved in total.

How much money, in dollars, is being saved by Marcus and Isla each week?

Enter your answer in the space.

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

Show the answer

Correct answer: 6

Define the variable:

Let w represent the amount of money Marcus and Isla save per week.

Construct an equation:

Start with $8.

Marcus and Isla save w each week for 3 weeks, which can be expressed as 3⋅w3 \cdot w or 3w3w.

The total amount saved can be represented by the expression 8+3w8 + 3w.

After 3 weeks, Marcus and Isla have $26. This can be represented by the equation 8+3w=268 + 3w = 26.

Solve the equation:

8+3w=268 + 3w = 26

3w=183w = 18 (subtracting 8 from both sides)

w=6w = 6 (dividing both sides by 3)

Marcus and Isla are saving $6 per week.

Question analysis

Marcus and Isla already have $8 toward concert tickets, save the same amount each week, and have $26 in all after 3 weeks. Students type how many dollars the two of them save each week. It tests writing and solving a two-step equation, keeping the money saved beforehand apart from the weekly savings.

NYS standards
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 the grade 7 standard. This one uses small whole numbers and two steps, a typical classroom example, with no choices to compare.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 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

Two inverse steps solve it, subtracting the starting $8 and dividing by the 3 weeks, and most mistakes drop one of them. The $18 saved over three weeks is the natural stopping place, and dividing $26 by 3 without removing the $8 gives an awkward value that should prompt a second look. Adding three weeks of the answer to $8 and getting $26 confirms it.

  1. Turning words into math minor Understanding the question

    The bullet points hold a start amount, “already have $8 saved”, and a total after three weeks, “$26 saved in total”, which must become 8 + 3w = 26. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. An answer in an unexpected form minor Choosing the answer

    The amount is typed with no choices, so stopping at the 3-week total of $18 leaves a whole number that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: Take away the amount saved before the weeks began, then share what is left equally among the weeks; check by adding three weeks of savings to the starting amount.

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.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total savings
total=start+weeks×amount each week\text{total} = \text{start} + \text{weeks} \times \text{amount each week}
The start is the money saved before the weeks begin. The amount each week is added once for every week, so you multiply it by the number of weeks. The total is all the money together.
Doing the same to both sides
if a=b, then a−c=b−c and ac=bc\text{if } a = b \text{, then } a - c = b - c \text{ and } \frac{a}{c} = \frac{b}{c}
a and b are the two sides of an equation, and c is any number (not zero when you divide). Doing the same thing to both sides keeps them equal. That is how you get the unknown by itself.

  1. Sort out the question

    You know:

    • Marcus and Isla already have $8.
    • They save the same amount each week.
    • After 3 weeks they have $26 in all.

    You need the amount the two of them save together each week.

  2. Write an equation

    Let w be the dollars saved each week. Three weeks add 3w dollars to the $8 they started with:

    8+3w=268 + 3w = 26

  3. Subtract the starting money

    Take 8 away from both sides:

    3w=26−8=183w = 26 - 8 = 18

    They saved $18 over the 3 weeks.

    Watch out: 18 is the amount for all 3 weeks, not for each week. Do not stop here.

  4. Divide by the number of weeks

    Divide both sides by 3:

    w=18÷3=6w = 18 \div 3 = 6

    They save $6 each week.

    Watch out: dividing all $26 by 3 gives about 8.67. That forgets the $8 saved before. Halving 6 to get 3 is wrong too: the question asks what the two save together.

  5. Enter the answer

    Type only the number in the box, with no dollar sign and no units. So you would enter 6.

Check your answer

Count up week by week. Start at $8. After week 1: 8+6=148 + 6 = 14. After week 2: 14+6=2014 + 6 = 20. After week 3: 20+6=2620 + 6 = 26. That is the $26 in the question.

More questions like this

The original DOE question
Question 45 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 D Math, page 118, question 45.

Question 31 of 120

2026 Practice Test D Math, question 46

Question

A company charges $3.00 to ship a 4-pound package. The cost includes a base fee of $1.50 plus $0.50 for each pound over the first pound. The table below illustrates how the cost depends on the total weight of the package. Complete the table by matching each total weight to its correct total shipping cost.

Move the correct answer to each box. Not all answers will be used.

SHIPPING COSTS

Total Weight (pounds) Total Cost to Ship (dollars)
2
2.50
6
5.50

On the test you drag tiles into the boxes; here, choose each box’s tile from its menu. Each tile can be used once. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Show the answer

Correct answer: blank 1: 2; blank 2: 3; blank 3: 4; blank 4: 9

Total Weight (pounds) Total Cost to Ship (dollars)
2 2
3 2.50
6 4
9 5.50

Let x represent the weight of the package in pounds. The first pound is included in the base fee of $1.50, so (x−1)(x - 1) represents the number of pounds that cost $0.50 each.

The cost equation can be written as Cost=$1.50+$0.50(x−1)\text{Cost} = \$1.50 + \$0.50\,(x - 1)

  • The cost for a 2-pound package is $1.50+$0.50(2−1)=$1.50+$0.50(1)=$2.00\$1.50 + \$0.50(2 - 1) = \$1.50 + \$0.50(1) = \$2.00
  • The cost for a 6-pound package is $1.50+$0.50(6−1)=$1.50+$0.50(5)=$1.50+$2.50=$4.00\$1.50 + \$0.50(6 - 1) = \$1.50 + \$0.50(5) = \$1.50 + \$2.50 = \$4.00

  • The weight of a package that costs $2.50 to ship can be found by solving

    $2.50=$1.50+$0.50(x−1)\$2.50 = \$1.50 + \$0.50(x - 1)

    $1.00=$0.50(x−1)\$1.00 = \$0.50(x - 1)

    2=x−12 = x - 1

    x=3 poundsx = 3 \text{ pounds}

  • The weight of a package that costs $5.50 to ship can be found by solving

    $5.50=$1.50+$0.50(x−1)\$5.50 = \$1.50 + \$0.50(x - 1)

    $4.00=$0.50(x−1)\$4.00 = \$0.50(x - 1)

    8=x−18 = x - 1

    x=9 poundsx = 9 \text{ pounds}

To check:

Given that the total cost is $3.00 for the initial package

$3.00=$1.50+$0.50(x−1)\$3.00 = \$1.50 + \$0.50(x - 1)

$1.50=$0.50(x−1)\$1.50 = \$0.50(x - 1)

3=x−13 = x - 1

x=4x = 4 pounds fits the linear pattern of the table

Question analysis

A shipping company charges a base fee plus 50 cents for each pound over the first, and a 4-pound package costs $3.00. Students drag number tiles into a table to complete two costs and two weights, with some tiles left unused. It tests building and using a linear rule, and the wording about the first pound must be applied in every row.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.9 dependent and independent variables
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve word problems with equations like p(x + q) = r. This table needs that rule four times, twice run backward from a cost to a weight, and the phrase about the first pound must be applied every time.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 4, 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

The rule is stated in words and confirmed by the 4-pound example, so the work is applying it consistently in both directions. Charging every pound breaks the $3.00 example at once, since 4 pounds would cost $3.50, and models that ignore the wording leave costs or weights with no matching tile. A shortcut helps: each extra pound adds $0.50 to the $3.00, so the table can be filled by counting up or down from 4 pounds.

  1. Turning words into math moderate Understanding the question

    The phrase “$0.50 for each pound over the first pound” means the first pound is covered by the $1.50 fee, so a package of x pounds pays for only x − 1 of them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Four boxes must all be right, two found forward from a weight and two backward from a cost, and the backward rows need the first pound added back. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Write the rule as the base fee plus $0.50 times the pounds after the first, check it against the 4-pound example, then use it forward for the costs and backward for the weights.

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.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Shipping cost
C=1.50+0.50×(w−1)C = 1.50 + 0.50 \times (w - 1)
C is the total cost in dollars, and w is the weight in pounds. The $1.50 base fee is paid once, and it covers the first pound. Each pound after the first costs $0.50. There are w − 1 of those pounds.

  1. Sort out the question

    The base fee of $1.50 covers the first pound. Each pound over the first adds $0.50. A 4-pound package costs $3.00.

    You fill four boxes: the costs for 2 and 6 pounds, and the weights that cost $2.50 and $5.50. Each tile is used once at most, and four tiles are not used.

  2. Test the rule on 4 pounds

    Check the rule with the example. A 4-pound package has 3 pounds over the first:

    C=1.50+0.50×(4−1)=1.50+1.50=3.00C = 1.50 + 0.50 \times (4 - 1) = 1.50 + 1.50 = 3.00

    That is $3.00, so the rule fits.

    Watch out: charging $0.50 for every pound, the first one too, gives $3.50 for 4 pounds. That does not match.

  3. Find the two costs

    For 2 pounds, 1 pound is over the first:

    1.50+0.50×1=2.001.50 + 0.50 \times 1 = 2.00

    For 6 pounds, 5 pounds are over the first:

    1.50+0.50×5=1.50+2.50=4.001.50 + 0.50 \times 5 = 1.50 + 2.50 = 4.00

    The costs are the tiles 2 and 4.

    Watch out: 6 pounds is 3 times 2 pounds, but the cost is not 3 times $2. The base fee is paid once. Tripling gives the tile 6. Leaving out the base fee gives the tile 2.5.

  4. Work back to the two weights

    For $2.50, take off the base fee: 2.50−1.50=1.002.50 - 1.50 = 1.00. At $0.50 a pound, $1.00 pays for 2 extra pounds. Add the first pound: 2+1=32 + 1 = 3 pounds.

    For $5.50: 5.50−1.50=4.005.50 - 1.50 = 4.00. $4.00 pays for 8 extra pounds. Add the first pound: 8+1=98 + 1 = 9 pounds.

    Watch out: forgetting to add back the first pound gives 2 and 8.

  5. Fill in the four boxes

    From top to bottom, the boxes get 2, 3, 4 and 9. The tiles 2.5, 3.5, 6 and 7 are not used. On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Each extra pound adds $0.50, so count from 4 pounds at $3.00. Going down: 3 pounds cost $2.50 and 2 pounds cost $2.00. Going up: 5 pounds $3.50, 6 pounds $4.00, 7 pounds $4.50, 8 pounds $5.00 and 9 pounds $5.50. All four boxes match.

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 D Math, page 119, question 46.

Question 32 of 120

2026 Practice Test D Math, question 49

Question

Anise has a piece of cloth measuring 45 inches by 75 inches. She wants to cut it into squares of equal size, without any cloth left over. What is the greatest possible side length, in inches, of one of those squares?

  • A3
  • B5
  • C15
  • D45
Show the answer

Correct answer: C

Given that each piece will be a square, the length and width will be the same. The greatest possible length can be determined by finding the greatest common factor of the length and width of the original piece of cloth.

Begin by finding the prime factorization of the length (75 in.) and width (45 in.).

45=3×3×545 = 3 \times 3 \times 5

75=3×5×575 = 3 \times 5 \times 5

The common factors can be seen as 3×53 \times 5.

The greatest possible side length, in inches, will be 3×5=153 \times 5 = 15.

Question analysis

Anise wants to cut a 45-inch by 75-inch piece of cloth into equal squares with nothing left over. This factors question asks for the greatest possible side length of each square, in inches. Students must see that the side has to divide both lengths evenly, then find the largest number that does.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Greatest common factors of two numbers up to 100 are grade 6 content. The cloth setting asks students to recognize that this is what is needed, but the work is one short factor search, below typical grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 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

Once the side is seen as a common factor, the work is short, since 45 and 75 share both 3 and 5. Students who spot one shared factor and stop reach a listed choice, and those who think only of the largest square that fits use the full 45-inch width. Checking that 45 and 75 both divide by the chosen side, and that no larger number does, closes it.

  1. Turning words into math minor Understanding the question

    The words “squares of equal size, without any cloth left over” mean the side must divide both 45 and 75 exactly; nothing names the greatest common factor. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Finding the greatest common factor of 45 and 75 calls for prime factorization or a list of factors, a grade 6 skill the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Write each length as a product of primes, multiply the primes they share, and check that both 45 and 75 divide evenly by the result.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor
GCF=product of the prime factors both numbers share\text{GCF} = \text{product of the prime factors both numbers share}
A factor divides a number evenly. The greatest common factor, or GCF, is the biggest number that divides both. Write each number as a product of primes, numbers whose only factors are 1 and themselves. Then multiply the primes they share.

  1. Sort out the question

    The cloth is 45 inches by 75 inches. It is cut into equal squares, with no cloth left over. You need the greatest side the squares can have.

    No cloth left over means the side must fit evenly into 45 and into 75. So the side is a common factor of both. The greatest one is the GCF.

  2. Break each number into primes

    Write each length as a product of prime numbers:

    45=3×3×545 = 3 \times 3 \times 5

    75=3×5×575 = 3 \times 5 \times 5

    The two lists share one 3 and one 5.

  3. Multiply the shared primes

    Multiply the primes the two numbers share:

    GCF=3×5=15\text{GCF} = 3 \times 5 = 15

    The squares can be 15 inches on a side.

    Watch out: 3 and 5 are each common factors, choices A and B. But they are not the greatest. Use every prime the numbers share.

  4. Rule out the full width

    Watch out: a 45-inch square, choice D, fits the width. But 75 ÷ 45 = 1 with 30 left over. That leaves a strip of cloth 45 inches by 30 inches.

    So the answer is C, 15.

Check your answer

Divide each length by 15: 45÷15=345 \div 15 = 3 and 75÷15=575 \div 15 = 5. So the cloth cuts into 3 rows of 5 squares, with nothing left. Both divisions come out even.

More questions like this

The original DOE question
Question 49 as shown on the DOE’s digital test, page 122 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 122, question 49.

Question 33 of 120

2024 Test A Math, question 4

Question

Mr. Chan’s lawn grows 2182\frac{1}{8} inches every 2 weeks. He mows his lawn every 2 weeks and cuts off the top 1341\frac{3}{4} inches of lawn. If Mr. Chan’s lawn was 4 inches tall at the beginning of the season, how many inches tall, in decimal form, is Mr. Chan’s lawn after 8 weeks?

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

Show the answer

Correct answer: 5.5

The lawn grows 2182\frac{1}{8} inches every 2 weeks. Find the total number of inches it grows in 8 weeks.

8÷2=48 \div 2 = 4

4×2184 \times 2\frac{1}{8}

4×2.125=8.54 \times 2.125 = 8.5

The lawn was 4 inches tall at the beginning of the season.

8.5+4=12.58.5 + 4 = 12.5

Mr. Chan mows his lawn every 2 weeks and cuts off the top 1341\frac{3}{4} inches. Find the total number of inches that are cut off in 8 weeks.

8÷2=48 \div 2 = 4

4×1344 \times 1\frac{3}{4}

4×1.75=74 \times 1.75 = 7

The height after 8 weeks is the total number of inches grown minus the total number of inches cut off.

12.5−7=5.512.5 - 7 = 5.5

Question analysis

This multistep word problem follows Mr. Chan's lawn over 8 weeks: it grows a mixed number of inches every 2 weeks, and he mows a mixed number of inches off on the same schedule. Students combine the growth, the cuts and the starting height and give the final height as a decimal. Leaving out one of those three pieces is the usual error.

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
At grade 7 Grade 7 students solve multi-step problems with mixed numbers and convert between fractions and decimals. This one strings four such steps together, including counting the 2-week periods, which is typical grade 7 work done carefully.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 3, 49 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

No single step is hard, but three parts must all reach the final height: the starting 4 inches, four rounds of growth and four rounds of mowing. Leaving out any one of them gives a clean decimal that can be gridded without a second thought. Working one 2-week period at a time, in which the lawn gains 3/8 inch net, keeps the bookkeeping short.

  1. Many steps moderate Working it out

    Four pieces have to be combined: count the four 2-week periods in 8 weeks, total the growth, total the mowing, and start from the lawn's first 4 inches. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic moderate Working it out

    The amounts are mixed numbers with unlike denominators, 2 1/8 and 1 3/4, which must be multiplied by 4 or subtracted and then gridded as a decimal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Turning words into math minor Understanding the question

    The phrase "every 2 weeks" governs both the growing and the mowing, so 8 weeks means four rounds of each, not eight. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  4. An answer in an unexpected form minor Choosing the answer

    As a grid-in it gives no choices, so a height that leaves out the starting 4 inches, or leaves out the mowing, looks as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Find the net change for one 2-week period by subtracting the cut from the growth, multiply by the number of such periods in 8 weeks, and add the starting height.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook’s explanation says “Mr. Chen mows his lawn every 2 weeks.” The question calls him Mr. Chan throughout, so the site now reads “Mr. Chan.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mixed number to decimal
abc=a+b÷ca\frac{b}{c} = a + b \div c
A mixed number has a whole part, a, and a fraction part, b over c. To write it as a decimal, divide b by c to get the fraction part as a decimal. Then add the whole part. For example, one and a half is 1 + 0.5 = 1.5.
Repeated amount
total=n×a\text{total} = n \times a
n is how many times something happens. a is the amount each time. Adding the same amount n times is the same as multiplying it by n.

  1. Sort out the question

    The lawn starts at 4 inches tall. Every 2 weeks it grows 2182\frac{1}{8} inches. Every 2 weeks Mr. Chan also cuts off 1341\frac{3}{4} inches. You need the height after 8 weeks, as a decimal. Three things matter: the start, the growing and the cutting.

  2. Count the 2-week periods

    Both the growing and the cutting happen every 2 weeks. So count how many 2-week periods fit in 8 weeks.

    8÷2=48 \div 2 = 4

    There are 4 periods. The lawn grows 4 times and is cut 4 times. Watch out: do not count 8 rounds, one for each week. That leads to the wrong answer 7.

  3. Change the mixed numbers to decimals

    The answer must be a decimal, so change both amounts now.

    218=2+1÷8=2.1252\frac{1}{8} = 2 + 1 \div 8 = 2.125

    134=1+3÷4=1.751\frac{3}{4} = 1 + 3 \div 4 = 1.75

    Each period, the lawn grows 2.125 inches and loses 1.75 inches.

  4. Add up the growing

    The lawn grows 2.125 inches, 4 times.

    total=n×a\text{total} = n \times a

    total=4×2.125=8.5\text{total} = 4 \times 2.125 = 8.5

    It grows 8.5 inches in all. Add the starting height of 4 inches.

    4+8.5=12.54 + 8.5 = 12.5

    Without mowing, the lawn would be 12.5 inches tall. Watch out: do not forget the starting 4 inches. Leaving them out leads to 1.5.

  5. Add up the cutting

    Mr. Chan cuts off 1.75 inches, 4 times.

    total=4×1.75=7\text{total} = 4 \times 1.75 = 7

    He cuts off 7 inches in all.

  6. Subtract the cutting

    12.5−7=5.512.5 - 7 = 5.5

    The lawn is 5.5 inches tall after 8 weeks. Watch out: 12.5 is not the answer. It leaves out the mowing. So you would grid 5.5.

Check your answer

Work one period at a time. Each period the lawn gains 2.125−1.75=0.3752.125 - 1.75 = 0.375 inch. In 4 periods that is 4×0.375=1.54 \times 0.375 = 1.5 inches. Add the starting 4 inches: 4+1.5=5.54 + 1.5 = 5.5. Same answer.

More questions like this

The original DOE question
Question 61 as printed on page 76 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 76, question 61. Open page 76 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 34 of 120

2024 Test A Math, question 6

Question

Mr. Jones has 550 goats, which is 10% more than Mr. King has. How many more goats does Mr. Jones have than Mr. King?

  • A50
  • B55
  • C495
  • D500
Show the answer

Correct answer: A

First, set up a proportion to determine x, the number of goats Mr. King has.

KingJones=x550\frac{\text{King}}{\text{Jones}} = \frac{x}{550}

x550=100%100%+10%\frac{x}{550} = \frac{100\%}{100\% + 10\%}

Then solve the proportion for x.

x550=100%100%+10%\frac{x}{550} = \frac{100\%}{100\% + 10\%}

550(100%)=x(100%+10%)550(100\%) = x(100\% + 10\%)

550(100%)=x(110%)550(100\%) = x(110\%)

550(1)=x(1.10)550(1) = x(1.10)

550=1.10x550 = 1.10x

550÷1.10=(1.10÷1.10)x550 \div 1.10 = (1.10 \div 1.10)x

500=x500 = x

Finally, determine the difference between the number of goats Mr. Jones has, 550, and the number of goats Mr. King has, 500.

550−500=50550 - 500 = 50

Mr. Jones has 50 more goats than Mr. King.

Question analysis

A percent word problem: Mr. Jones has 550 goats, which is 10% more than Mr. King has, and students find how many more goats Mr. Jones has. It asks students to work with a percent increase when only the larger amount is known. The tricky part is deciding which amount the 10% is taken of.

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 Grade 7 students usually find a percent increase from a known original amount. This runs the task backward: the increased amount is given, and the original must be recovered before the difference can be found, a step many do not expect.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 1, 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 arithmetic is one division and one subtraction, so the question is really about the base of a percent. The sentence gives the bigger number and calls it 10% more, and the reflex is to take 10% of the number on the page. That shortcut lands on a listed choice five goats from the answer. Students who find King's herd first, 550 ÷ 1.1, still have to remember that the question wants the difference, not the herd.

  1. Turning words into math major Understanding the question

    The clause "which is 10% more than" makes King's unknown herd the base of the 10%, so 550 is 110% of his number rather than the amount the 10% is taken of. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    The choices 50 and 55 are only 5 goats apart, and both are 10% of a herd; only taking the percent of King's 500 rather than of 550 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Mr. Jones's 550 is 110% of Mr. King's number, so divide 550 by 1.10 to find Mr. King's goats, then subtract that from 550.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=original×(1+p100)\text{new} = \text{original} \times \left(1 + \frac{p}{100}\right)
The original is the amount you start from, the one the percent is taken of. p is the percent of increase. The new amount is all of the original, which is 100%, plus p% more. For 10% more, you multiply by 1.10.

  1. Sort out the question

    Mr. Jones has 550 goats. That is “10% more than Mr. King has”. You need how many more goats Mr. Jones has than Mr. King. That is the difference between the two herds, not the size of a herd.

  2. Decide what the 10% is of

    The words “10% more than Mr. King” mean the 10% is a part of Mr. King’s herd. So Mr. King’s herd is the original. Mr. Jones has all of Mr. King’s goats, 100%, plus 10% more. That is 110% of Mr. King’s herd. Watch out: do not take 10% of 550. That gives 55, choice B. It uses the wrong herd. Taking those 55 away from 550 gives 495, choice C, which is also wrong.

  3. Find Mr. King’s herd

    Call Mr. King’s number of goats K. Use the percent increase formula with p = 10. The new amount is 550.

    new=original×(1+p100)\text{new} = \text{original} \times \left(1 + \frac{p}{100}\right)

    550=K×1.10550 = K \times 1.10

    K=550÷1.10=500K = 550 \div 1.10 = 500

    Mr. King has 500 goats. Watch out: 500 is Mr. King’s herd, choice D. It is not the difference.

  4. Find the difference

    550−500=50550 - 500 = 50

    Mr. Jones has 50 more goats than Mr. King. So the answer is A, 50.

Check your answer

Check the 10%. 10% of 500 is 0.10×500=500.10 \times 500 = 50, and 500+50=550500 + 50 = 550, Mr. Jones’s herd. So 550 really is 10% more than 500. If King had 495, then 10% more would be about 545, not 550.

More questions like this

The original DOE question
Question 63 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 2025 Admissions, page 77, question 63. Open page 77 of the handbook (PDF)

Question 35 of 120

2024 Test A Math, question 7

Question

If 2yx−y2x=□2x\frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x} and x≠0x \neq 0, what expression is represented by □\square?

  • Ayy
  • B2y2y
  • C3y3y
  • D4y4y
Show the answer

Correct answer: C

Use properties of equations to determine the missing expression:

2yx−y2x=□2x\frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}

22∙2yx−y2x=□2x\frac{2}{2} \bullet \frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}

4y2x−y2x=□2x\frac{4y}{2x} - \frac{y}{2x} = \frac{\square}{2x}

4y−y2x=□2x\frac{4y - y}{2x} = \frac{\square}{2x}

3y2x=□2x\frac{3y}{2x} = \frac{\square}{2x}

The missing expression is 3y3y.

Question analysis

An algebra question with fractions: two fractions with x and 2x in their denominators are subtracted, and students find the numerator that makes the difference equal a fraction over 2x. It tests rewriting a fraction over a common denominator when the denominators contain a variable. Adjusting only part of a fraction is the common mistake.

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 subtract fractions with unlike denominators and combine like terms, but rarely see a variable in a denominator. The steps are the same two as with numbers, so the demand is typical once the form is familiar.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 13, 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

This is subtraction of fractions with unlike denominators, written with letters. The box over 2x shows the target denominator, which helps, but each wrong choice matches a familiar fraction error: subtracting the tops before matching the bottoms, or rewriting the first fraction and stopping. Students who rewrite 2y/x as 4y/2x and then subtract y from the new numerator finish in seconds.

  1. Variables instead of numbers moderate Understanding the question

    The denominators are x and 2x, so the common denominator is built with a letter, and there is no number to check unless students substitute values such as x = 1 and y = 2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Only the first fraction changes: 2y/x becomes 4y/2x with top and bottom both doubled, while y/2x already has the denominator the box sits over. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Rewrite the first fraction over 2x by multiplying its top and bottom by 2, then subtract the numerators, which now share a denominator.

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.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top of the fraction, b is the bottom, and k is any number except zero. Multiplying the top and bottom by the same number does not change the value. It is like multiplying by k over k, which is 1.
Subtracting fractions with the same bottom
ad−bd=a−bd\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. When two fractions have the same bottom, their pieces are the same size. So subtract the tops and keep the bottom.

  1. Sort out the question

    The box, □\square, stands for a missing expression. An expression is a math phrase made of numbers and letters, like 5y. You need what goes in the box to make the two sides equal. x≠0x \neq 0 means x is not zero, so no fraction divides by zero.

  2. Pick a shared bottom

    You can only subtract fractions when their bottoms match. The bottoms here are x and 2x. The box sits over 2x, so use 2x for both. The second fraction, y2x\frac{y}{2x}, already has it.

  3. Rewrite the first fraction over 2x

    To turn x into 2x, multiply by 2. Multiply the top by 2 as well, so the value stays the same.

    ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}

    2yx=2y×2x×2=4y2x\frac{2y}{x} = \frac{2y \times 2}{x \times 2} = \frac{4y}{2x}

    Watch out: this is only the first fraction. Stopping here and choosing 4y gives choice D.

  4. Subtract the tops

    ad−bd=a−bd\frac{a}{d} - \frac{b}{d} = \frac{a - b}{d}

    4y2x−y2x=4y−y2x=3y2x\frac{4y}{2x} - \frac{y}{2x} = \frac{4y - y}{2x} = \frac{3y}{2x}

    4y take away y is like 4 apples take away 1 apple. That leaves 3y. Watch out: do not subtract the tops, 2y − y, before the bottoms match. That gives y, choice A.

  5. Match the box

    The left side equals 3y2x\frac{3y}{2x}. The right side is □2x\frac{\square}{2x}. The bottoms are the same, so the tops must be equal. The box is 3y. So the answer is C, 3y3y.

Check your answer

Try numbers. Let x = 1 and y = 2. The left side is 41−22=4−1=3\frac{4}{1} - \frac{2}{2} = 4 - 1 = 3. With 3y in the box, the right side is 3×22×1=62=3\frac{3 \times 2}{2 \times 1} = \frac{6}{2} = 3. Both sides are 3, so 3y works.

More questions like this

The original DOE question
Question 64 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 2025 Admissions, page 77, question 64. 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 36 of 120

2024 Test A Math, question 8

Question

Square PQRS, with Q at the top left, R at the top right, P at the bottom left and S at the bottom right, and a right-angle mark at S. Point S is marked as the center of a circle that passes through P and R. The quarter of the circle inside the square, bounded by sides SP and SR and the arc from P to R, is shaded.

In the figure above, PQRS is a square. Point S is the center of the circle, and points P and R are on the circle. If the area of the square is 4 square centimeters, what is the area, in square centimeters, of the shaded quarter of the circle?

  • Aπ4\frac{\pi}{4}
  • Bπ\pi
  • C2π2\pi
  • D4π4\pi
Show the answer

Correct answer: B

The area of square PQRS is 4 square centimeters. So

s2=4s^{2} = 4

s2=4\sqrt{s^{2}} = \sqrt{4}

s=2 cms = 2\text{ cm}

Sides PS and RS are each a radius of the circle. So the radius of the circle is 2 cm.

The shaded region is 14\frac{1}{4} of the circle. The area of a circle is A=πr2A = \pi r^{2}. So the area of the shaded region, in square centimeters, is

14A=14πr2=\frac{1}{4}A = \frac{1}{4}\pi r^{2} =

14π(22)=\frac{1}{4}\pi(2^{2}) =

14π(4)=π\frac{1}{4}\pi(4) = \pi

Question analysis

A geometry question with a figure: square PQRS has one corner at the center of a circle, and the quarter of the circle inside the square is shaded. Given the area of the square, students find the area of the shaded quarter circle. The key is connecting the square's side to the circle's radius.

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
At grade 7 Circle area is grade 7 content, and this question adds only one linking step from the square's area to the radius. Typical grade 7 circle work gives the radius directly; here it takes one short extra inference.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 4, 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

The question chains three short facts: from the square's area to its side, from the side to the radius, and from the whole circle to a quarter of it. None is hard, but each wrong choice marks a broken link, most often the whole circle's area, which looks like a finished answer. Pointing to S as the center before computing anything settles the radius.

  1. Reading a figure, table or graph moderate Understanding the question

    The radius is never labeled: students must see that S is both a corner of the square and the circle's center, so sides PS and RS are radii. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Neither the circle area formula, π times the radius squared, nor the step from a square's area to its side length is given. (Anderson & Krathwohl, 2001)

  3. Combines ideas minor Working it out

    Square and circle are linked in one chain: the area 4 gives a side of 2, the side is the radius, and the shaded part is a quarter of π × 2². (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Take the square root of the square's area to get its side, which is also the radius, then take one quarter of pi times the radius squared.

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.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square
A=s2A = s^2
A is the area, the space inside. s is the length of one side, and s² means s × s. All four sides of a square are equal, so its area is side times side.
Area of a circle
A=πr2A = \pi r^2
A is the area of the whole circle. r is the radius: the distance from the center to the edge. π, called pi, is a number a little more than 3, about 3.14. Multiply π by r × r. This works for every circle.

  1. Sort out the figure

    PQRS is a square with an area of 4 sq cm. Point S is the center of the circle. Points P and R are on the circle. The shaded part is a quarter of the circle, inside the square. You need the shaded area, in square centimeters. The answer will have π in it.

  2. Find the side of the square

    Use the square formula. The area is 4, so find the number that times itself gives 4.

    A=s2A = s^2

    4=s24 = s^2

    s=2s = 2

    Each side is 2 cm long, because 2×2=42 \times 2 = 4.

  3. See that the side is the radius

    Side SP goes from the center, S, to point P on the circle. So SP is a radius. So is side SR. The radius is 2 cm. Watch out: the side is a radius, not a diameter, the full width of the circle. Using a radius of 1 leads to π4\frac{\pi}{4}, choice A.

  4. Find the area of the whole circle

    A=πr2A = \pi r^2

    A=π×22=π×4=4πA = \pi \times 2^2 = \pi \times 4 = 4\pi

    The whole circle has an area of 4π sq cm. Watch out: this is the whole circle. Stopping here gives choice D.

  5. Take one quarter

    The shaded part is one quarter of the circle, so take one quarter of 4π.

    14×4π=π\frac{1}{4} \times 4\pi = \pi

    The shaded area is π sq cm. So the answer is B, π\pi.

Check your answer

π is about 3.14, so the shaded part is about 3.14 sq cm. The whole square is 4 sq cm. In the figure, the shaded part fills most of the square, but not all. So a bit less than 4 makes sense. The choice 4π is about 12.6, far too big.

More questions like this

The original DOE question
Question 65 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 2025 Admissions, page 77, question 65. Open page 77 of the handbook (PDF)

Question 37 of 120

2024 Test A Math, question 11

Question

Jar 1 and Jar 2 each contain 12\frac{1}{2} cup of water. If 14\frac{1}{4} of the water in Jar 1 is poured into Jar 2, how much water is now in Jar 2?

  • A18\frac{1}{8} cup
  • B14\frac{1}{4} cup
  • C58\frac{5}{8} cup
  • D34\frac{3}{4} cup
Show the answer

Correct answer: C

Jar 1 starts with 12\frac{1}{2} cup of water. If 14\frac{1}{4} of its water is poured into Jar 2, then the amount of water added to Jar 2 is 12×14=18\frac{1}{2} \times \frac{1}{4} = \frac{1}{8} cup.

Jar 2 starts with 12\frac{1}{2} cup of water. If 18\frac{1}{8} cup of water is added to it, then the amount of water now in Jar 2 is 12+18=48+18=58\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.
What makes it hard
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of an amount
ab of N=ab×N\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
ab×cd=a×cb×d\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
ad+bd=a+bd\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.

  1. Sort out the question

    Each jar starts with 12\frac{1}{2} cup of water. You pour “14\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.

  2. Find how much water is poured

    Jar 1 has 12\frac{1}{2} cup. You pour 14\frac{1}{4} of it. Of means multiply.

    ab of N=ab×N\frac{a}{b} \text{ of } N = \frac{a}{b} \times N

    14×12=1×14×2=18\frac{1}{4} \times \frac{1}{2} = \frac{1 \times 1}{4 \times 2} = \frac{1}{8}

    You pour 18\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 14\frac{1}{4} cup gives 34\frac{3}{4} cup, choice D.

  3. Make the bottoms match

    Now add the poured water to the 12\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.

    12=1×42×4=48\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}

  4. Add the water in Jar 2

    ad+bd=a+bd\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}

    48+18=58\frac{4}{8} + \frac{1}{8} = \frac{5}{8}

    Jar 2 now holds 58\frac{5}{8} cup. Watch out: 18\frac{1}{8} cup is only the water poured in. Stopping there gives choice A. So the answer is C, 58\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 58\frac{5}{8} is. Also, the water in both jars still adds up to 1 cup: Jar 1 has 38\frac{3}{8} cup left, and 38+58=1\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 38 of 120

2024 Test A Math, question 18

Question

Each number in a sequence is formed by doubling the previous number and then adding 1. If the 9th number in the sequence is 63, what is the 10th number minus the 7th number?

  • A96
  • B111
  • C112
  • D127
Show the answer

Correct answer: C

First, write an equation based on the rule described in the item. Let x be the previous number in the sequence and y be the number in the sequence. The ninth number in the sequence is 63.

2x+1=y2x + 1 = y

Determine the 10th number in the sequence, y, by making x=63x = 63 and solving:

2(63)+1=y2(63) + 1 = y

126+1=y126 + 1 = y

127=y127 = y

Determine the 8th number in the sequence, x, by making y=63y = 63 and solving:

2x+1=632x + 1 = 63

Subtract one from both sides of the equation.

2x+(1−1)=63−12x + (1 - 1) = 63 - 1

2x=622x = 62

2x2=622\frac{2x}{2} = \frac{62}{2}

x=31x = 31

Determine the 7th number in the sequence, x, by making y=31y = 31 and solving:

2x+1=312x + 1 = 31

2x+(1−1)=31−12x + (1 - 1) = 31 - 1

2x=302x = 30

2x2=302\frac{2x}{2} = \frac{30}{2}

x=15x = 15

Subtract the 7th number in the sequence from the 10th number:

127−15=112127 - 15 = 112

The difference is 112.

Question analysis

A sequence question: each number is found by doubling the one before it and adding 1, and the 9th number is 63. Students find the 10th number minus the 7th. That means running the rule forward once and backward twice, while keeping track of which position each number holds.

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
Above grade 7 Grade 7 students solve two-step equations but rarely have to run a rule backward through a sequence. This mixes forward and backward steps with position counting, so it asks for more planning than a typical grade 7 task.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 7, 32 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

Applying the rule forward is routine; running it backward is not, and the question needs both. Students must also count positions carefully, since the 9th number is given and the 7th is two steps back, not one. Each listed wrong answer stops or slips at a different stage, so writing the four numbers in a row, 15, 31, 63 and 127, before subtracting is the reliable way.

  1. No obvious method moderate Working it out

    The rule is stated only forward, double and add 1, but the 7th and 8th numbers come before the given 9th, so the rule must be undone: subtract 1, then halve. (Schoenfeld, 1985)

  2. Many steps moderate Working it out

    One forward step reaches the 10th number, 127, two backward steps reach the 8th and 7th, 31 and 15, and a final subtraction finishes. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Much to keep track of minor Working it out

    Four positions and their values, 7th to 10th, have to be kept straight, and the 8th number is easy to mistake for the 7th. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Apply the rule to 63 to get the 10th number; to go back a position, subtract 1 and halve, doing it twice to reach the 7th; then subtract.

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).
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: Finding the 7th number, the handbook writes “2x + (1 − 1 = 31 − 1)”, with the closing parenthesis after the whole equation, so the line reads as an equation inside an expression (and 1 − 1 = 31 − 1 is false). The site now reads 2x + (1 − 1) = 31 − 1, the same step the handbook writes correctly for the 8th number (2x + (1 − 1) = 63 − 1). It leads to 2x = 30 and x = 15 as before.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule
y=2x+1y = 2x + 1
A sequence is a list of numbers that follows a rule. x is a number in the list, and y is the next number. To get the next number, double x and then add 1.
Going back one number
x=(y−1)÷2x = (y - 1) \div 2
y is a number in the list, and x is the one just before it. To go back, undo the rule in reverse order. The rule doubles and then adds 1. So first subtract 1, then divide by 2.

  1. Sort out the question

    Each number is “formed by doubling the previous number and then adding 1”. Previous means the one just before. The 9th number is 63. You need the 10th number minus the 7th number. So you go forward one number and back two numbers.

  2. Find the 10th number

    Use the rule with x = 63.

    y=2x+1y = 2x + 1

    y=2×63+1=126+1=127y = 2 \times 63 + 1 = 126 + 1 = 127

    The 10th number is 127. Watch out: remember to add 1. Doubling only gives 126 and leads to 111, choice B.

  3. Plan how to go back

    The 7th number comes before the 9th, so you must undo the rule. The rule doubled, then added 1. Undo the last step first: subtract 1. Then undo the doubling: divide by 2.

    x=(y−1)÷2x = (y - 1) \div 2

  4. Find the 8th number

    x=(63−1)÷2=62÷2=31x = (63 - 1) \div 2 = 62 \div 2 = 31

    The 8th number is 31. Watch out: 31 is the 8th number, not the 7th. Using it gives 127−31=96127 - 31 = 96, choice A.

  5. Find the 7th number

    Step back once more, from the 8th number. Use the same rule with y = 31.

    x=(31−1)÷2=30÷2=15x = (31 - 1) \div 2 = 30 \div 2 = 15

    The 7th number is 15.

  6. Subtract the 7th from the 10th

    127−15=112127 - 15 = 112

    Watch out: 127 alone is choice D. You still have to subtract the 7th number. So the answer is C, 112.

Check your answer

Go forward from 15 using the rule. 2×15+1=312 \times 15 + 1 = 31, then 2×31+1=632 \times 31 + 1 = 63, then 2×63+1=1272 \times 63 + 1 = 127. You land on 63 as the 9th number, so the list is right. And 127−15=112127 - 15 = 112.

More questions like this

The original DOE question
Question 75 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 75. Open page 80 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 39 of 120

2024 Test A Math, question 21

Question

A coordinate grid with origin O, the x-axis and y-axis each marked 5, 10, 15 and 20, and grid lines every 5 units. One line segment joins the points (0, 20) and (20, 0), and another joins (0, 15) and (10, 0); all four points are marked with dots. The region between the two segments, with corners at (0, 15), (0, 20), (20, 0) and (10, 0), is shaded.

What is the area, in square units, of the shaded region shown in the figure above?

  • A75
  • B125
  • C150
  • D200
Show the answer

Correct answer: B

The area of the shaded region, in square units, is the area of the triangle created by the shaded and unshaded regions minus the area of the triangle created by the unshaded region.

The area of the triangle created by the shaded and unshaded regions is 200 square units.

12bh=12(20)(20)=12(400)=200\frac{1}{2}bh = \frac{1}{2}(20)(20) = \frac{1}{2}(400) = 200

The area of the triangle created by the unshaded region is 75 square units.

12bh=12(10)(15)=12(150)=75\frac{1}{2}bh = \frac{1}{2}(10)(15) = \frac{1}{2}(150) = 75

Subtract to find the area of the shaded region.

200−75=125200 - 75 = 125

Question analysis

A coordinate geometry question: two line segments cross a grid, one from (0, 20) to (20, 0) and one from (0, 15) to (10, 0), and the region between them is shaded. Students find the shaded area. The shape is not a standard figure, so they need to see it as one triangle with another removed.

NYS standards
  • 6.G.1 area of polygons
  • 6.G.3 polygons in the coordinate plane
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Area of triangles is grade 6 content; finding a composite area by subtraction is the kind of task grade 7 students meet regularly. Points on the axes make every length easy to read.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 6, 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 shaded shape has no name, so the question is about seeing it as a difference of two right triangles whose legs lie on the axes. Once that is seen, both areas are one-line calculations. The wrong choices come from stopping after either triangle or from pairing a base from one segment with a height from the other.

  1. Reading a figure, table or graph moderate Understanding the question

    All four lengths come from where the two segments meet the axes, at 20 and 15 on the y-axis and at 10 and 20 on the x-axis, read against grid lines 5 units apart. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. No obvious method minor Working it out

    The shaded region is a four-sided shape with no formula of its own; it has to be seen as the large triangle with the small one removed. (Schoenfeld, 1985)

  3. Something you must already know minor Working it out

    The area of a triangle, one half of base times height, is assumed and not given. (Anderson & Krathwohl, 2001)

Strategy: Find the area of the large right triangle under the outer segment and subtract the area of the smaller triangle under the inner segment, reading the legs from the axes.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \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.

  1. Sort out the figure

    The grid lines are 5 units apart. Two slanted segments run from the y-axis down to the x-axis. The outer one goes from 20 on the y-axis to 20 on the x-axis. The inner one goes from 15 on the y-axis to 10 on the x-axis. The shaded region is the strip between them. You need its area, in square units.

  2. See the shape as two triangles

    The shaded region has four sides, and it has no simple formula. But look at the big triangle under the outer segment, with its corner at O. It is the shaded strip plus the small white triangle. So the shaded area is the big triangle minus the small one.

  3. Find the area of the big triangle

    The big triangle has a right angle at O. Its base runs along the x-axis from 0 to 20, so it is 20 units. Its height runs up the y-axis from 0 to 20, so it is 20 units.

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×20×20=200A = \frac{1}{2} \times 20 \times 20 = 200

    Watch out: this is the whole big triangle. Stopping here gives 200, choice D.

  4. Find the area of the small triangle

    The small triangle also has its right angle at O. Its base is 10 units, from 0 to 10 on the x-axis. Its height is 15 units, from 0 to 15 on the y-axis.

    A=12×10×15=75A = \frac{1}{2} \times 10 \times 15 = 75

    Watch out: 75 is the white triangle, choice A, not the shaded part. Also, pair the base and height of the same triangle. Mixing 20 and 15 gives 150, choice C.

  5. Subtract to find the shaded area

    Take the small white triangle away from the big triangle.

    200−75=125200 - 75 = 125

    The shaded area is 125 square units. So the answer is B, 125.

Check your answer

In the figure, the shaded strip looks bigger than the white triangle. So the shaded area should be more than half of the big triangle, which is 200÷2=100200 \div 2 = 100. 125 is more than 100. And 125+75=200125 + 75 = 200, the whole big triangle.

More questions like this

The original DOE question
Question 78 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 81, question 78. Open page 81 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 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 MQ\overline{MQ}. Which segment has length (2x−y)(2x - y) centimeters?

  • APQ\overline{PQ}
  • BNP\overline{NP}
  • CMQ\overline{MQ}
  • DMP\overline{MP}
Show the answer

Correct answer: A

These facts are given:

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

Use the first two facts to determine that MQ\overline{MQ} has a length of 2x2x centimeters. So (2x−y)(2x - y) can be represented by MQ−MPMQ - MP, which is PQ\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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint
MQ=2×MNMQ = 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=MQMP + 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.

  1. 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 MQ\overline{MQ}, the point exactly halfway from M to Q. You need the segment that is 2x−y2x - y cm long.

  2. Find the length of MQ

    MN is half of MQ, so MQ is twice as long.

    MQ=2×MN=2xMQ = 2 \times MN = 2x

    MQ is 2x cm long. Watch out: 2x is only part of the expression. Stopping here gives MQ, choice C.

  3. 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=MQMP + PQ = MQ

    PQ=MQ−MP=2x−yPQ = MQ - MP = 2x - y

    So PQ is 2x−y2x - y cm long. Watch out: do not pick MP just because y appears. MP is y, choice D. NP is y−xy - x, choice B. So the answer is A, PQ\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=310 - 7 = 3. And 2x−y=10−7=32x - y = 10 - 7 = 3. They match. NP would be 7−5=27 - 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 41 of 120

2024 Test A Math, question 34

Question

In a certain state, the sales tax rate increased from 7.0% to 7.5%. What was the increase in the sales tax on a $200 item?

  • A$1
  • B$10
  • C$14
  • D$15
Show the answer

Correct answer: A

The increase in sales tax percentage is

7.5%−7.0%=0.5%7.5\% - 7.0\% = 0.5\%

0.5% of $200 is

(0.5%)(200)=(0.005)(200)=1(0.5\%)(200) = (0.005)(200) = 1

Question analysis

This percent word problem says a state's sales tax rate rose from 7.0% to 7.5% and asks how much more tax is charged on a $200 item. Students find the change in the rate and apply it to the price, or compute both taxes and compare. Converting a fraction of a percent to a decimal is where errors creep in.

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 Sales tax is a standard grade 7 percent topic, usually finding the tax or the total on one price. This one compares two tax rates whose difference is a fraction of a percent, where placing the decimal point is the main hazard.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 22 words.
Corrected from the DOE handbook
  • A dash printed in place of a minus sign 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 arithmetic is tiny once 0.5% is written correctly, so the question turns on the decimal point. A percent below 1 is unfamiliar, and moving the point one place instead of two gives a choice ten times too large. Students can sidestep it by finding both taxes, $14 and $15, and subtracting, though either tax on its own is also a listed choice.

  1. Error-prone arithmetic moderate Working it out

    The change in rate is half of one percent, and writing 0.5% as a decimal moves the point two places, to 0.005; one place gives 0.05 and a tenfold error. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The question asks for "the increase in the sales tax", a difference in dollars, while the stem's first sentence is about the two rates themselves. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Subtract the old rate from the new rate, then take that percent of $200, writing it as a decimal carefully; or find both taxes and subtract.

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.
  • 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

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes the subtraction with an em dash, “7.5% — 7.0% = 0.5%”. The site uses a minus sign: 7.5% − 7.0% = 0.5%. Nothing else changes; 0.5% of $200 is $1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. Percent means out of 100, so p percent is p over 100. The whole is the amount you take the percent of, here the $200 price. The part is the result. To write a percent as a decimal, move the decimal point two places left.

  1. Sort out the question

    Sales tax is extra money added to a price. The tax rate went from 7.0% to 7.5%. The item costs $200. You need “the increase in the sales tax” in dollars. That is how much more tax you pay now.

    Watch out: the tax at the old rate is $14, choice C. The tax at the new rate is $15, choice D. Neither one is the increase.

  2. Find how much the rate went up

    Subtract the old rate from the new rate.

    7.5−7.0=0.57.5 - 7.0 = 0.5

    The rate went up 0.5%. That is half of one percent.

  3. Find 0.5% of $200

    The extra tax is 0.5% of the price. Use the percent formula with p = 0.5 and the whole = 200.

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    part=0.5100×200=0.005×200=1\text{part} = \frac{0.5}{100} \times 200 = 0.005 \times 200 = 1

    The tax goes up by $1. Watch out: 0.5% is 0.005, not 0.05. Moving the point only one place gives $10, choice B.

    So the answer is A, $1.

Check your answer

Find both taxes and subtract. Old tax: 0.07×200=140.07 \times 200 = 14, so $14. New tax: 0.075×200=150.075 \times 200 = 15, so $15. The difference is 15−14=115 - 14 = 1, or $1. Both ways give the same answer.

More questions like this

The original DOE question
Question 91 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 91. Open page 85 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 42 of 120

2024 Test A Math, question 37

Question

Number line with points X, Y and Z marked from left to right; Y is closer to X than to Z. Above the line, a dimension arrow from X to Y is labeled 1 and one from Y to Z is labeled 2.

On the number line above, the distance between X and Y is 1 unit, and the distance between Y and Z is 2 units. What is the distance, in units, between Y and the midpoint of X and Z?

  • A12\frac{1}{2}
  • B1
  • C1121\frac{1}{2}
  • D3
Show the answer

Correct answer: A

First, find the distance between X and Z by adding the distance between X and Y to the distance between Y and Z.

1+2=31 + 2 = 3

Since the distance between X and Z is 3, the midpoint between X and Z is 1121\frac{1}{2} units from X.

The midpoint between X and Z is 1121\frac{1}{2} units from X, and the distance between X and Y is 1. The distance between Y and the midpoint of X and Z is 112−1=121\frac{1}{2} - 1 = \frac{1}{2}.

Question analysis

This number line question shows points X, Y and Z, with 1 unit from X to Y and 2 units from Y to Z, and asks how far Y is from the midpoint of X and Z. Students find the length of XZ, locate its midpoint, and compare it with Y. Each step produces a tempting number, so stopping early is the usual mistake.

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 Grade 7 number line problems usually involve negative numbers and fractions as coordinates. This one has no coordinates, only two whole-number distances, and needs three short steps: a total length, a midpoint and a difference.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 39 words.
Corrected from the DOE handbook
  • An imprecise statement about the midpoint 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 arithmetic is light, but every intermediate result is waiting among the choices, so the risk is stopping before the last subtraction. Students also have to keep two midpoints apart: the one asked about lies between X and Z, not halfway along the 2-unit stretch from Y. Marking X at 0 and the midpoint at 1.5 on the drawing makes the half-unit gap visible.

  1. Reading a figure, table or graph minor Understanding the question

    The number line has no coordinates, only dimension arrows labeled 1 and 2, so students must assign positions themselves, such as X at 0, Y at 1 and Z at 3. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Three short results come in sequence, the length XZ, the midpoint's distance from X, then its distance from Y, and each is also a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Place X at 0 so that Y and Z get positions from the given distances, find the midpoint of X and Z as half their distance, then compare it with Y's position.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says “the midpoint between X and Z is 1½” (twice). A midpoint is a point, and this number line has no numbers on it, so the 1½ is the midpoint’s distance from X (half of XZ = 3). Both sentences now say the midpoint is “1½ units from X.” The rest is unchanged: Y is 1 unit from X, so the distance between Y and the midpoint is 1½ − 1 = ½.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding lengths on a line
XZ=XY+YZXZ = XY + YZ
X, Y and Z are points on one line, with Y between X and Z. XY and YZ are the lengths of the two short pieces. XZ is the length of the whole. The two pieces join end to end to make the whole.
Midpoint of a segment
distance from an end to the midpoint=length of the segment2\text{distance from an end to the midpoint} = \frac{\text{length of the segment}}{2}
A segment is the straight piece of a line between two points, its ends. The midpoint is the point exactly halfway between the ends. It splits the segment into two equal halves. So it is half the length away from each end.

  1. Sort out the picture

    The points go X, then Y, then Z, from left to right. X to Y is 1 unit, and Y to Z is 2 units. You need the midpoint of X and Z, the point halfway between them. Then you need how far Y is from that midpoint. The line has no numbers, so measure everything from X.

  2. Find the length from X to Z

    XZ=XY+YZXZ = XY + YZ

    XZ=1+2=3XZ = 1 + 2 = 3

    X to Z is 3 units. Watch out: 3 is the whole length, choice D. It is not a distance to the midpoint.

  3. Place the midpoint

    The midpoint of X and Z is half of 3 units from X.

    distance from an end to the midpoint=length of the segment2\text{distance from an end to the midpoint} = \frac{\text{length of the segment}}{2}

    32=112\frac{3}{2} = 1\frac{1}{2}

    So the midpoint is 1121\frac{1}{2} units to the right of X. Watch out: this is its distance from X, not from Y. Stopping here gives choice C.

  4. Measure from Y to the midpoint

    Y is 1 unit to the right of X. The midpoint is 1121\frac{1}{2} units to the right of X. Subtract to find the gap between them.

    112−1=121\frac{1}{2} - 1 = \frac{1}{2}

    Watch out: use the midpoint of X and Z, not of Y and Z. Halving the 2 units from Y to Z gives 1, choice B.

    So the answer is A, 12\frac{1}{2} unit.

Check your answer

Give each point a number, with X at 0. Then Y is at 1 and Z is at 3. Halfway between 0 and 3 is 1121\frac{1}{2}. From Y at 1 to 1121\frac{1}{2} is 12\frac{1}{2}. This fits the picture too: Y sits a little left of the middle of XZ.

More questions like this

The original DOE question
Question 94 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 2025 Admissions, page 86, question 94. 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). This is the DOE’s version, before the corrections listed in the analysis.

Question 43 of 120

2024 Test A Math, question 44

Question

A tank with a 500-gallon capacity currently contains 75 gallons of water. Additional water is poured into this tank at a rate of 5 gallons per minute. After 45 minutes of adding water, what percentage of the tank’s total capacity will be filled? (Assume that there is no loss of water from the tank.)

  • A45%
  • B55%
  • C60%
  • D70%
Show the answer

Correct answer: C

If 5 gallons of water is added to the tank per minute for 45 minutes, then 5×45=2255 \times 45 = 225 gallons is added to the tank.

The tank started with 75 gallons of water in it. After the 45 minutes, it contains 75+225=30075 + 225 = 300 gallons, assuming no loss of water.

The tank has a capacity of 500 gallons, and 300500=0.6\frac{300}{500} = 0.6.

Rewrite decimal 0.6 as a percentage:

0.6=610=60100=60%.0.6 = \frac{6}{10} = \frac{60}{100} = 60\%.

Question analysis

A 500-gallon tank holds 75 gallons, and water flows in at 5 gallons per minute for 45 minutes. This rate and percent word problem asks what percentage of the tank's capacity will then be filled. Students find the water added, the new amount in the tank and its percent of 500, keeping track of every quantity of water.

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 Grade 7 students solve multistep rate and percent problems. This one combines a pouring rate, a starting amount and a percent of capacity in three short steps with friendly numbers, which matches typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 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 friendly, so the question tests whether students track every quantity of water. Forgetting the 75 gallons that were there from the start is the costly slip, because the share of the tank that was poured in is itself a listed choice. A quick check, that 300 gallons is more than half of 500 but well under three quarters, confirms the answer.

  1. Much to keep track of minor Working it out

    The 75 gallons already in the tank, given in the first sentence, must be held aside through the rate calculation and added back before dividing. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Many steps minor Working it out

    The chain runs from 5 gallons a minute for 45 minutes to 225 gallons, then to 300 gallons in the tank, then to 300 out of 500 as a percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply the rate by the time to get the water added, account for the water already in the tank, and divide the total by the 500-gallon capacity.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Amount from a rate
amount=rate×time\text{amount} = \text{rate} \times \text{time}
The rate is how much water goes in each minute, here 5 gallons. The time is how many minutes it runs. Multiply them to get the amount added. It works because the same amount goes in every minute.
Percent of a whole
percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100
The part is the amount you have, and the whole is the full amount. Dividing gives what fraction of the whole you have. Multiplying by 100 turns it into a percent, which means out of 100.

  1. Sort out the question

    The tank holds at most 500 gallons. That is its capacity. It already has 75 gallons in it. Water is added at 5 gallons per minute for 45 minutes. You need the percent of the tank that will be full.

  2. Find the water added

    The same amount of water goes in every minute, so multiply the rate by the time.

    amount=rate×time\text{amount} = \text{rate} \times \text{time}

    amount=5×45=225\text{amount} = 5 \times 45 = 225

    So 225 gallons are poured in during the 45 minutes.

  3. Find the total water in the tank

    Add the water that was already there.

    75+225=30075 + 225 = 300

    The tank holds 300 gallons after 45 minutes. Watch out: don’t forget the first 75 gallons. The 225 gallons alone fill 45% of the tank, choice A.

  4. Find the percent full

    percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100

    percent=300500×100=35×100=60\text{percent} = \frac{300}{500} \times 100 = \frac{3}{5} \times 100 = 60

    The tank is 60% full. Watch out: find the part that is full, not the part that is empty. Choice B, 55%, is the empty part if you also forget the first 75 gallons.

    So the answer is C, 60%.

Check your answer

10% of 500 gallons is 50 gallons. The tank has 300 gallons, which is 6 groups of 50. So it is 60% full. Also, 300 gallons is less than 500, so the tank never overflows.

More questions like this

The original DOE question
Question 101 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 2025 Admissions, page 87, question 101. Open page 87 of the handbook (PDF)

Question 44 of 120

2024 Test A Math, question 45

Question

Misha wants to use ribbon to make 2 straps for a backpack. The ribbon costs $5.00 a yard. If each strap requires 1141\frac{1}{4} yards of ribbon, how much will Misha pay for the ribbon (not including tax)?

  • A$4.00
  • B$6.25
  • C$11.25
  • D$12.50
Show the answer

Correct answer: D

Each strap requires 1141\frac{1}{4} yards of ribbon, and ribbon costs $5.00 a yard.

114×5=54×51=254=6.251\frac{1}{4} \times 5 = \frac{5}{4} \times \frac{5}{1} = \frac{25}{4} = 6.25, so the ribbon for each strap costs $6.25.

Misha wants to make 2 straps.

6.25×2=12.506.25 \times 2 = 12.50, so the cost of the ribbon for 2 straps is $12.50.

Question analysis

Misha needs ribbon for 2 backpack straps, each using 1 1/4 yards, and ribbon costs $5.00 a yard. This cost word problem asks how much she will pay for the ribbon before tax. Students multiply a mixed number by a price per yard and by the number of straps; it is short, but both the mixed number and the two straps need care.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Multiplying a mixed number by a whole number and finding a total cost are grade 5 and 6 skills. With two short steps and friendly numbers, this is easier than typical grade 7 problems with percents or negatives.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 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 is short and the numbers are friendly; points are lost by stopping at one strap or by mishandling the mixed number. Pricing a single strap gives a listed choice, as does doubling only the whole-number part of 1 1/4. Finding the total ribbon first, 2 1/2 yards, before multiplying by the price keeps both errors away.

  1. Error-prone arithmetic minor Working it out

    The length per strap is a mixed number, 1 1/4 yards, so the arithmetic is easiest as 2 × 1.25 = 2.5 yards, then 2.5 × $5.00. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Much to keep track of minor Working it out

    The number of straps comes in the first sentence and the length of each in the last, so the 2 has to be carried through to the cost. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find the total ribbon first by doubling the length for one strap, then multiply the yards by $5.00; writing 1 1/4 as 1.25 keeps the arithmetic simple.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost
cost=price per unit×number of units\text{cost} = \text{price per unit} \times \text{number of units}
Price per unit is the cost of one unit, here $5.00 for one yard. Number of units is how many you buy, here yards of ribbon. Multiply them to get the total before tax.

  1. Sort out the question

    Misha makes 2 straps. Each strap needs 1141\frac{1}{4} yards of ribbon. A yard is a length of 3 feet. Ribbon costs $5.00 a yard. You need the total cost for both straps, with no tax.

  2. Find the total ribbon

    First change the mixed number to a decimal. One quarter is 0.25, so 1141\frac{1}{4} is 1.25. Then double it for 2 straps.

    2×1.25=2.52 \times 1.25 = 2.5

    Misha needs 2.5 yards of ribbon. Watch out: double the whole mixed number, the 1 and the quarter. Doubling only the 1 gives 2142\frac{1}{4} yards and $11.25, choice C.

  3. Find the cost

    cost=price per unit×number of units\text{cost} = \text{price per unit} \times \text{number of units}

    cost=5×2.5=12.5\text{cost} = 5 \times 2.5 = 12.5

    The ribbon costs $12.50. Watch out: this is for both straps. One strap’s ribbon costs $6.25, choice B. Don’t stop there.

    So the answer is D, $12.50.

Check your answer

Estimate. Each strap needs a bit more than 1 yard, so both need a bit more than 2 yards. Two yards cost $10, so the answer should be a bit more than $10. $12.50 fits. Also, 2 yards cost $10 and the extra half yard costs $2.50, which makes $12.50.

More questions like this

The original DOE question
Question 102 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 2025 Admissions, page 87, question 102. 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 45 of 120

2024 Test A Math, question 53

Question

Carlos has $350 in a savings account that earns 5% simple interest each year. How much will he have in the account after 1 year, if there is no money withdrawn?

  • A$17.50
  • B$175.00
  • C$367.50
  • D$525.00
Show the answer

Correct answer: C

Find 5% of 350.

(5%)(350)=(0.05)(350)=17.5(5\%)(350) = (0.05)(350) = 17.5

350+17.5=367.50350 + 17.5 = 367.50

Question analysis

Carlos has $350 in a savings account earning 5% simple interest a year, and this interest question asks how much will be in the account after 1 year with no withdrawals. Students find 5 percent of the balance and use it. Reading whether the question wants the interest or the account's new balance matters most.

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
Below grade 7 Simple interest is a grade 7 percent topic, but one year of interest is just 5 percent of the balance added on. Typical grade 7 problems run for several years or ask for the rate, so this is lighter.
Estimated difficulty
Routine: one familiar step, little to misread.
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

This is a one-year interest question with friendly numbers, so most students who know what interest is will finish it. The two ways to lose the point are stopping at the interest, which is listed, and writing 5% as 0.5, which gives a listed interest and a listed balance of its own. A sensible balance is only a little more than $350, which rules out every wrong choice.

  1. Something you must already know minor Working it out

    Simple interest for one year is the rate times the balance, 5% of $350, added to what is already there; the stem does not spell this out. (Anderson & Krathwohl, 2001)

  2. Error-prone arithmetic minor Working it out

    Writing 5% as 0.05, not 0.5, is the one place-value decision, and 0.05 × 350 = 17.5 must then be read as $17.50. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find 5 percent of $350 by multiplying by 0.05, then decide whether the question wants the interest alone or the amount in the account.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest
I=P×r×tI = P \times r \times t
I is the interest, the extra money the bank pays you. P is the principal, the money you put in. r is the yearly rate as a decimal. t is the time in years. Simple means you earn interest only on the money you started with.

  1. Sort out the question

    Carlos starts with $350. The account earns 5% simple interest each year. Interest is extra money the bank adds to your account. You need the total in the account after 1 year. That is the $350 plus the interest.

  2. Change the percent to a decimal

    Percent means out of 100, so 5% is 5 out of 100.

    5100=0.05\frac{5}{100} = 0.05

    Watch out: 5% is 0.05, not 0.5. Using 0.5 makes the interest $175, which leads to B or D.

  3. Find the interest

    Put in P = 350, r = 0.05 and t = 1 year.

    I=P×r×tI = P \times r \times t

    I=350×0.05×1=17.5I = 350 \times 0.05 \times 1 = 17.5

    The interest for one year is $17.50. That is the money added to the account.

  4. Add the interest to the savings

    350+17.50=367.50350 + 17.50 = 367.50

    After 1 year, Carlos has $367.50. Watch out: $17.50 is only the interest, choice A. The question asks how much is in the account.

    So the answer is C, $367.50.

Check your answer

10% of $350 is $35. 5% is half of that, $17.50, so the interest is right. The total should be a little more than $350, since 5% is small. $367.50 fits. $525.00 would mean the account grew by half.

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 2025 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 46 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.

1.51=64=1.5\frac{1.5}{1} = \frac{6}{4} = 1.5

Multiply to find the value of y.

11×1.5=16.511 \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.
What makes it hard
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.

  1. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality
k=yxk = \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×xy = 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.

  1. 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.

  2. Find the blue for each cup of red

    k=yxk = \frac{y}{x}

    k=1.51=1.5andk=64=1.5k = \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.

  3. Use the rate for 11 cups

    y=k×xy = k \times x

    y=1.5×11=16.5y = 1.5 \times 11 = 16.5

    Think of it as 11 cups plus half of 11 cups: 11+5.5=16.511 + 5.5 = 16.5. So y is 16.5 cups of blue.

  4. 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: 7×1.5=10.57 \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÷11=1.516.5 \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 47 of 120

2024 Test A Math, question 56

Question

Number line from −3 to 8 with a tick at every integer. Point P is at −3, point Q is at 1 and point R is at 7, each marked with a dot.

On the number line above, what is the distance, in units, between the midpoint of PQ\overline{PQ} and the midpoint of QR\overline{QR}?

  • A3
  • B4
  • C5
  • D6
Show the answer

Correct answer: C

To find the midpoint of a segment, subtract the value on the left from the value on the right and then divide by 2. Then add the result to the value on the left of the segment.

Find the midpoint of PQ\overline{PQ}:

1−(−3)=41 - (-3) = 4

4÷2=24 \div 2 = 2

2+(−3)=−12 + (-3) = -1

Find the midpoint of QR\overline{QR}:

7−1=67 - 1 = 6

6÷2=36 \div 2 = 3

3+1=43 + 1 = 4

The distance between −1 and 4 on the number line is 5 units.

Question analysis

A number line marks points P, Q and R at labeled integer positions, with P to the left of zero. This midpoint and distance question asks how far apart the midpoints of segments PQ and QR are. Students locate both midpoints and then measure the distance between them, which takes care with negative positions.

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 Grade 7 students measure distances on a number line as the difference of two positions, including across zero. Here two midpoints must be located first, but all three points sit on labeled integers, which keeps it at grade 7 level.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 11, 21 words.
Corrected from the DOE handbook
  • A reversed instruction 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 points sit on labeled integers, so the midpoints can be found by counting, and the question is mostly about finishing carefully. Dropping the negative in 4 − (−1), stopping at the second midpoint, and counting tick marks instead of spaces each give a listed choice. Marking −1 and 4 on the figure and counting the unit spaces between them confirms the distance.

  1. Error-prone arithmetic minor Working it out

    P sits at −3, so the midpoint of PQ is (−3 + 1) ÷ 2 = −1, and the final distance, 4 − (−1), subtracts a negative number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Three results come in sequence, the midpoint of PQ, the midpoint of QR, then the gap between them, and the second of them, 4, is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Read the positions of P, Q and R, average each pair of endpoints to place the two midpoints, then subtract the left midpoint from the right one.

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.)
  • 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).

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook’s rule says to “subtract the value on the right from the value on the left and then divide by 2,” but its own working does the opposite: 1 − (−3) = 4 for PQ and 7 − 1 = 6 for QR, the right value minus the left. Following the rule as printed gives (−3 − 1) ÷ 2 = −2 and (1 − 7) ÷ 2 = −3, which would put the midpoints at −5 and −2, outside the segments, and lead to a distance of 3, a wrong choice. The site now says “subtract the value on the left from the value on the right.” The midpoints, −1 and 4, and the distance between them, 5 units, are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint on a number line
midpoint=left end+right end−left end2\text{midpoint} = \text{left end} + \frac{\text{right end} - \text{left end}}{2}
The left end and right end are the numbers at the two ends of a segment. Right end minus left end is the segment’s length. The midpoint is halfway, so it is half the length to the right of the left end.
Distance on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
On a number line, numbers grow as you move right. The distance between two points is the larger number minus the smaller one. That is the right point minus the left point, so a distance is never negative.

  1. Read the number line

    P is at −3, Q is at 1 and R is at 7. A bar over two letters, such as PQ\overline{PQ}, means the segment between them. A segment is the straight piece of line between two points. You need the midpoint of each segment, the point exactly halfway between its ends. Then you need the distance between those two midpoints.

  2. Find the midpoint of PQ

    The left end is −3 and the right end is 1.

    midpoint=left end+right end−left end2\text{midpoint} = \text{left end} + \frac{\text{right end} - \text{left end}}{2}

    midpoint=−3+1−(−3)2=−3+42=−3+2=−1\text{midpoint} = -3 + \frac{1 - (-3)}{2} = -3 + \frac{4}{2} = -3 + 2 = -1

    So the midpoint of PQ is at −1. Note that 1−(−3)1 - (-3) is 1+31 + 3, which is 4.

  3. Find the midpoint of QR

    The left end is 1 and the right end is 7.

    midpoint=1+7−12=1+62=1+3=4\text{midpoint} = 1 + \frac{7 - 1}{2} = 1 + \frac{6}{2} = 1 + 3 = 4

    So the midpoint of QR is at 4. Watch out: 4 is where a midpoint is, not the answer. Stopping here gives choice B.

  4. Measure between the midpoints

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    distance=4−(−1)=4+1=5\text{distance} = 4 - (-1) = 4 + 1 = 5

    Watch out: keep the minus sign on −1. Using 4−1=34 - 1 = 3 gives choice A. Also count the spaces, not the tick marks. From −1 to 4 there are 5 spaces but 6 marks, and 6 is choice D.

    So the answer is C, 5 units.

Check your answer

Each midpoint should sit in the middle of its segment. −1 is 2 units from both −3 and 1. 4 is 3 units from both 1 and 7. There is also a shortcut. The gap between the midpoints is half of PR: 7−(−3)2=102=5\frac{7 - (-3)}{2} = \frac{10}{2} = 5.

More questions like this

The original DOE question
Question 113 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 113. Open page 91 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 48 of 120

2024 Test B Math, question 1

Question

35(2x+5)−2x\frac{3}{5}(2x+5)-2x

After the expression above is simplified, what is the coefficient of x, expressed as a decimal?

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

Show the answer

Correct answer: -0.8

First, distribute 35\frac{3}{5} over (2x+5)(2x+5) to get the expression 65x+3−2x\frac{6}{5}x+3-2x.

Express −2x-2x as −105x-\frac{10}{5}x and then add like terms:

65x+3+(−105x)=−45x+3\frac{6}{5}x+3+\left(-\frac{10}{5}x\right)=-\frac{4}{5}x+3.

The coefficient of x is −45-\frac{4}{5}.

When expressed as a decimal,

−45=−0.8-\frac{4}{5}=-0.8.

Question analysis

An algebra grid-in on simplifying an expression: three fifths times the quantity 2x + 5, minus 2x. Students distribute a fraction, combine the x terms and report the coefficient of x as a decimal. Multiplying by a fraction, the subtraction that follows, and gridding the decimal correctly all leave room for slips.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.2b parts of an expression
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Expanding and simplifying linear expressions with fraction coefficients is a core grade 7 skill, and this is a typical short example. Writing the coefficient as a decimal in the grid adds care but no new idea.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 17 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 distribution itself is routine; points are lost after it. Students who stop at 6/5x forget the −2x outside the parentheses, and those who combine the x terms correctly can still drop the sign of a negative result. Rewriting 2x as 10/5x keeps the subtraction in fifths, and substituting x = 1 into both forms confirms the coefficient before gridding.

  1. Error-prone arithmetic moderate Working it out

    Multiplying 2x by 3/5 gives 6/5x, and taking away 2x means rewriting it as 10/5x; the result is a negative fraction that must then become a decimal with its sign. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. An answer in an unexpected form minor Choosing the answer

    There are no choices to check against, and the coefficient must be gridded as a decimal with its minus sign, so −4/5 has to become −0.8. (Enright, Morley & Sheehan, 2002)

Strategy: Multiply both terms inside the parentheses by 3/5, rewrite 2x in fifths so the x terms share a denominator, combine them, and convert that fraction to a decimal.

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.)
  • Stops one step early: A choice is a number from partway through the solution.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. To multiply out the parentheses, multiply each term inside by a. It works because a groups of b + c is a groups of b plus a groups of c.
Multiplying a fraction by a whole number
ab×c=a×cb\frac{a}{b} \times c = \frac{a \times c}{b}
a over b is the fraction, and c is the whole number. Multiply only the top number, the numerator, by c. The bottom number, the denominator, stays the same. The size of each piece does not change. You just have more pieces.

  1. Sort out the question

    You start with this expression. An expression is a math phrase with numbers, letters and signs, but no equals sign.

    35(2x+5)−2x\frac{3}{5}(2x + 5) - 2x

    To simplify it means to write it in its shortest form. The coefficient of x is the number multiplied by x. In 7x, the coefficient is 7. You need the coefficient of x, written as a decimal.

  2. Multiply out the parentheses

    The 3/5 multiplies everything inside the parentheses. Use the distributive property.

    a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c

    35(2x+5)=35×2x+35×5\frac{3}{5}(2x + 5) = \frac{3}{5} \times 2x + \frac{3}{5} \times 5

    Multiply only the top of 3/5 by 2: 35×2=65\frac{3}{5} \times 2 = \frac{6}{5}. Also, 35×5=3\frac{3}{5} \times 5 = 3. Now the whole expression is:

    65x+3−2x\frac{6}{5}x + 3 - 2x

    Watch out: do not double the bottom number too. Writing 6/10 leads to −1.4.

  3. Write 2x in fifths

    There are now two x terms: 65x\frac{6}{5}x and −2x-2x. To combine them, they need the same denominator (bottom number). One whole is 5/5, so 2 wholes are 10/5.

    2x=105x2x = \frac{10}{5}x

    Watch out: the −2x is outside the parentheses, but it still counts. Stopping at 6/5 and gridding 1.2 is a common mistake.

  4. Combine the x terms

    Six fifths take away ten fifths leaves negative four fifths.

    65x−105x=−45x\frac{6}{5}x - \frac{10}{5}x = -\frac{4}{5}x

    The 3 has no x, so it stays as it is. The simplified expression is −45x+3-\frac{4}{5}x + 3. So the coefficient of x is −45-\frac{4}{5}. Watch out: 6 − 10 is negative. Dropping the minus sign gives 0.8.

  5. Change it to a decimal

    A fraction means the top divided by the bottom: 4÷5=0.84 \div 5 = 0.8. Keep the minus sign.

    −45=−0.8-\frac{4}{5} = -0.8

    So you would grid -0.8, with the minus sign in its own column.

Check your answer

Pick an easy value, x = 1, and put it into both forms. The first form: 35×7−2=4.2−2=2.2\frac{3}{5} \times 7 - 2 = 4.2 - 2 = 2.2. The simplified form: −0.8+3=2.2-0.8 + 3 = 2.2. Both give 2.2, so the simplifying worked.

More questions like this

The original DOE question
Question 58 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 2025 Admissions, page 198, question 58. Open page 198 of the handbook (PDF)

Question 49 of 120

2024 Test B Math, question 3

Question

Scale drawing of a garden shaped like a rectangle with its lower right corner cut out; all six sides are horizontal or vertical. The top side is labeled 13 cm and the left side 6 cm. From the top right corner, the right side, labeled 3 cm, goes down to an unlabeled horizontal side running left, then an unlabeled vertical side runs down to the bottom side, which is labeled 8 cm.

The figure above shows a scale drawing of a garden, where 1 centimeter represents 2.5 meters. What is the perimeter of the actual garden in meters?

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

Show the answer

Correct answer: 95

First, find the perimeter of the figure.

Two sides do not have labels. The top of the figure measures 13 cm and the bottom shows a length measuring 8 cm, so the horizontal side without a label measures 5 cm (13−8)(13-8).

The left side of the figure measures 6 cm and the right side shows a length measuring 3 cm, so the vertical side without a label measures 3 cm (6−3)(6-3).

So the total perimeter of the scale drawing is

13+3+5+3+8+6=38 cm13+3+5+3+8+6=38\text{ cm}

Use the scale to find the perimeter of the actual garden:

1 cm2.5 m=38 cmp m\frac{1\text{ cm}}{2.5\text{ m}}=\frac{38\text{ cm}}{p\text{ m}}

p=2.5×38=95p=2.5\times 38=95

Question analysis

A grid-in on a scale drawing of an L-shaped garden, where each centimeter stands for 2.5 meters and two sides carry no labels. Students find the perimeter of the drawing, then convert it to the real garden's perimeter in meters. The unlabeled sides are easy to leave out.

NYS standards
  • 7.G.1 scale drawings
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Scale drawings are grade 7 content, and a typical task converts one or two given lengths. Here two missing sides must be worked out and the perimeter assembled before the scale applies, more steps but nothing beyond grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 9, 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 hides the work: it labels four of the six sides, and the missing two are short pieces of the notch that are easy to overlook. Once all six are found, the scale step is a single multiplication, but it must go the right way, since the real garden is larger than the drawing. Tracing the outline and counting six sides guards against the most common loss.

  1. Reading a figure, table or graph moderate Understanding the question

    Two sides of the notch carry no labels, so students must find them from the opposite sides: 13 − 8 = 5 cm across and 6 − 3 = 3 cm down. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas minor Working it out

    The perimeter of an L-shaped drawing has to be joined to a scale of 2.5 meters per centimeter, two separate skills in one answer. (Embretson, 1998; Daniel & Embretson, 2010)

  3. An answer in an unexpected form minor Choosing the answer

    With no choices, nothing flags a perimeter that is missing two sides or still in centimeters; the student must check the six-sided total alone. (Enright, Morley & Sheehan, 2002)

Strategy: Use the opposite sides to work out the two unlabeled lengths, add all six sides of the drawing, then multiply the total in centimeters by 2.5 meters per centimeter.

Traps

  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter
P=sum of all the side lengthsP = \text{sum of all the side lengths}
P is the perimeter, the distance all the way around a shape. Add the length of every side, once each. It works for any shape with straight sides, even one with a corner cut out.
Using a scale
actual length=drawing length×s\text{actual length} = \text{drawing length} \times s
s is the scale, the number of meters that 1 centimeter on the drawing stands for. Here s is 2.5. Each centimeter on the drawing is 2.5 meters in real life, so you multiply.

  1. Sort out what you know

    The figure is a scale drawing, a small copy of the real garden. Each 1 cm on the drawing stands for 2.5 m in the real garden. You need the perimeter of the real garden in meters. The perimeter is the distance all the way around.

    The shape has 6 sides. Four are labeled: 13 cm, 3 cm, 8 cm and 6 cm. The two sides around the cut-out corner have no labels.

  2. Find the missing side across

    Look at the top side, 13 cm. Below it, the bottom side (8 cm) and the short unlabeled side across the notch together reach just as far. So the missing side is the difference.

    13−8=513 - 8 = 5

    The missing side across is 5 cm.

  3. Find the missing side going down

    Now look at the left side, 6 cm. On the right, the 3 cm side and the unlabeled side going down together make the same height.

    6−3=36 - 3 = 3

    The missing side going down is 3 cm.

  4. Add all six sides

    P=sum of all the side lengthsP = \text{sum of all the side lengths}

    P=13+3+5+3+8+6=38P = 13 + 3 + 5 + 3 + 8 + 6 = 38

    The perimeter of the drawing is 38 cm. Watch out: adding only the four labeled sides gives 30 cm. That leads to the wrong answer 75.

  5. Change to real meters

    Each centimeter stands for 2.5 meters, so multiply by 2.5.

    actual length=drawing length×s\text{actual length} = \text{drawing length} \times s

    actual length=38×2.5=76+19=95\text{actual length} = 38 \times 2.5 = 76 + 19 = 95

    Here 38×2=7638 \times 2 = 76 and 38×0.5=1938 \times 0.5 = 19. The real perimeter is 95 meters. Watch out: 38 is still in centimeters, so do not grid it. Dividing by 2.5 gives 15.2, which makes the garden smaller than its drawing.

    So you would grid 95.

Check your answer

Slide the two notch sides out to the corner. The distance around stays the same, so it equals a 13 cm by 6 cm rectangle: 2×(13+6)=382 \times (13 + 6) = 38. Also, 95 meters is more than 38, which fits, because each centimeter stands for more than 1 meter.

More questions like this

The original DOE question
Question 60 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 2025 Admissions, page 198, question 60. Open page 198 of the handbook (PDF)

Question 50 of 120

2024 Test B Math, question 4

Question

At one location, the temperature increased from −7° F to 4° F between 5:00 a.m. and 9:00 a.m. By 11:00 a.m., the temperature was 3.5 times the 9:00 a.m. temperature. What was the total increase in temperature between 5:00 a.m. and 11:00 a.m., in degrees Fahrenheit?

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

Show the answer

Correct answer: 21

The 11:00 a.m. temperature is 3.5 times the 9:00 a.m. temperature. The 9:00 a.m. temperature was 4° F.

4×3.5=144\times 3.5=14

The temperature increased from −7° F at 5:00 a.m. to 14° F at 11:00 a.m., which is an increase of 21° F.

Question analysis

A temperature word problem with negative numbers: the temperature rises from −7° F to 4° F by 9:00 a.m., and by 11:00 a.m. it is 3.5 times the 9:00 a.m. reading. Students find the total rise from 5:00 to 11:00 a.m. Subtracting a negative starting temperature is where most slips happen.

NYS standards
  • 7.NS.3 four operations with rational numbers
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Using signed numbers for temperature change is standard grade 7 work. This adds a decimal multiplier and asks for a change rather than a temperature, two short steps, which keeps it at a typical grade 7 level.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 5, 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

Finding the 11:00 a.m. temperature is a single multiplication; the points are lost in the subtraction that follows. A rise that starts below zero has two parts, 7 degrees up to zero and 14 more above it, and students who treat −7 as 7 lose the first part. The times in the stem also invite measuring from 9:00 a.m. instead of 5:00 a.m., which covers only the last two hours.

  1. Error-prone arithmetic moderate Working it out

    The last step is 14 − (−7): subtracting the negative starting temperature means adding 7, and the part of the rise below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    “3.5 times the 9:00 a.m. temperature” multiplies the reading of 4°, not the 11-degree rise, and “total increase” runs from the 5:00 a.m. reading, not from 9:00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Much to keep track of minor Working it out

    Three times of day carry three temperatures, −7°, 4° and 14°, and the answer compares the first with the last, not either of them with the middle one. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find the 11:00 a.m. temperature from the 9:00 a.m. one, then subtract the 5:00 a.m. temperature from it, remembering that subtracting a negative number adds its size.

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.)
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Change in temperature
change=end−start\text{change} = \text{end} - \text{start}
End is the last temperature, and start is the first one. Subtract start from end to find how much it went up. It works with negative numbers too. A positive change means the temperature rose.
Subtracting a negative number
a−(−b)=a+ba - (-b) = a + b
a is any number, and −b is a negative number. Taking away a negative is the same as adding its size. On a thermometer, going from −7 up to 0 is 7 degrees, so those 7 degrees get added.

  1. Sort out what you know

    There are three times and three temperatures, in degrees Fahrenheit (°F).

    • 5:00 a.m.: −7°F, which is 7 degrees below zero
    • 9:00 a.m.: 4°F
    • 11:00 a.m.: 3.5 times the 9:00 a.m. temperature

    You need the total increase from 5:00 a.m. to 11:00 a.m. That is how much warmer it got from the first time to the last.

  2. Find the 11:00 temperature

    “3.5 times the 9:00 a.m. temperature” means 3.5 times 4°F.

    3.5×4=143.5 \times 4 = 14

    At 11:00 a.m. it was 14°F. Watch out: 14 is a temperature, not the increase. Gridding 14 stops too early.

  3. Subtract the start from the end

    The increase runs from 5:00 a.m. to 11:00 a.m. So it starts at −7°F and ends at 14°F.

    change=end−start\text{change} = \text{end} - \text{start}

    change=14−(−7)\text{change} = 14 - (-7)

    Watch out: start at 5:00 a.m., not at 9:00 a.m. Starting from 4°F gives only 10.

  4. Subtract the negative number

    Taking away a negative number is the same as adding its size.

    a−(−b)=a+ba - (-b) = a + b

    14−(−7)=14+7=2114 - (-7) = 14 + 7 = 21

    The temperature rose 21°F. Watch out: writing 14 − 7 = 7 loses the minus sign. It skips the 7 degrees the air warmed just to reach zero.

    So you would grid 21.

Check your answer

Count up on a thermometer. From −7°F up to 0°F is 7 degrees. From 0°F up to 14°F is 14 more. 7+14=217 + 14 = 21. Or add the two parts of the morning: it rose 11 degrees by 9:00 and 10 more by 11:00, and 11+10=2111 + 10 = 21.

More questions like this

The original DOE question
Question 61 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 2025 Admissions, page 199, question 61. Open page 199 of the handbook (PDF)

Question 51 of 120

2024 Test B Math, question 5

Question

The figure below is a triangular prism. The lateral sides are rectangles.

A triangular prism with a triangle on the top and bottom and three rectangular sides. On the top triangle, a dashed segment labeled 8 in. runs from the far vertex to the opposite side and meets it at a right angle. The bottom edge below that opposite side is labeled 4 in., and the vertical edge at its left end is labeled 2 in.

What is the volume of the figure, in cubic inches?

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

Show the answer

Correct answer: 32

The volume of a triangular prism is found by multiplying the area of the triangular base by the height of the prism.

The area of the triangular base of the prism is 12bh=12(4)(8)=16\frac{1}{2}bh=\frac{1}{2}(4)(8)=16 square inches.

The height of the triangular prism is 2 inches.

So the volume of the triangular prism, in cubic inches, is 16∙2=3216\bullet 2=32.

Question analysis

A geometry grid-in: find the volume, in cubic inches, of a triangular prism drawn with its triangular faces on top and bottom. The drawing gives a 4-inch side of the triangle, a dashed 8-inch height drawn to that side, and a 2-inch edge. Students must recall the volume formula and match each length to its role.

NYS standards
  • 7.G.6 area, surface area, and volume
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the volume of a right triangular prism is named in the grade 7 standards, and this is a direct two-step case. The only extra demand is reading a slanted drawing to see which length plays which role.
Estimated difficulty
Straightforward: two steps or one careful reading.
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 rewards knowing that a prism's volume is its base area times its height, then matching each length to its job. The drawing lies on its side, with a short 2-inch height and the triangle flat on top, which makes the base harder to spot. Most lost points come from the one half: leaving it out doubles the volume, and stopping at the triangle's area gives half of it.

  1. Something you must already know moderate Working it out

    Neither formula is given: the volume is the area of the triangular base times the prism's height, and that area needs the one half in 1/2 × base × height. (Anderson & Krathwohl, 2001)

  2. Reading a figure, table or graph minor Understanding the question

    The prism lies with its triangles on top and bottom, so the 2-inch vertical edge is its height, while the dashed 8-inch line is the triangle's height to the 4-inch side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. An answer in an unexpected form minor Choosing the answer

    There are no choices, so a product that forgets the one half, or one that stops at the area of the triangle, gives no warning. (Enright, Morley & Sheehan, 2002)

Strategy: Find the area of the triangular face as one half times base times height, using the 4-inch side and the dashed 8-inch height, then multiply by the 2-inch edge of the prism.

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.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, one side of the triangle. h is the height, the straight distance from that side to the opposite corner, at a right angle to the side. You take half because a triangle is half of a rectangle with the same base and height.
Volume of a prism
V=B×HV = B \times H
V is the volume, the space inside. B is the area of the base, the shape at each end. H is the height of the prism, the distance between the two ends. It works because the prism is like a stack of layers, each the shape of the base.

  1. Sort out the figure

    A prism is a solid with two matching ends joined by flat sides. Here the ends are triangles, on the top and bottom. The other sides are rectangles. You need the volume, the space inside, in cubic inches.

    The dashed line on the top triangle is 8 in. It meets one side of the triangle at a right angle (the small square). The bottom edge under that side is 4 in. The edge going straight up is 2 in.

  2. Find the triangle’s base and height

    The side that the dashed line meets is the top edge of a rectangle face. Opposite sides of a rectangle are equal. So that side is 4 in., the same as the bottom edge under it.

    So the triangle has a base of 4 in. Its height is the dashed line, 8 in., because it meets the base at a right angle.

  3. Find the area of the triangle

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×4×8=16A = \frac{1}{2} \times 4 \times 8 = 16

    The triangle’s area is 16 square inches. Watch out: 16 is only the area of one end. Stopping here gives the wrong answer 16.

  4. Multiply by the prism’s height

    The height of the prism is the distance between the two triangles. That is the 2 in. edge.

    V=B×HV = B \times H

    V=16×2=32V = 16 \times 2 = 32

    The volume is 32 cubic inches. Watch out: do not forget the one half. 4×8×2=644 \times 8 \times 2 = 64 is the volume of a box, not of this prism.

    So you would grid 32.

Check your answer

Picture a box that is 4 in. by 8 in. by 2 in. Its volume is 4×8×2=644 \times 8 \times 2 = 64 cubic inches. The triangle has half the area of a 4 by 8 rectangle, so the prism holds half as much as the box. Half of 64 is 32.

More questions like this

The original DOE question
Question 62 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 2025 Admissions, page 199, question 62. Open page 199 of the handbook (PDF)

Question 52 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?

  • A4
  • B5
  • C7
  • D15
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∙25100=60+60\bullet\frac{25}{100}=

60+60∙0.25=60+60\bullet 0.25=

60+15=7560+15=75

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

Karen: 1,500 pagesk days=60 pages1 day\frac{1{,}500\text{ pages}}{k\text{ days}}=\frac{60\text{ pages}}{1\text{ day}}

60k=1,50060k=1{,}500

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

k=25 daysk=25\text{ days}

Martina: 1,500 pagesm days=75 pages1 day\frac{1{,}500\text{ pages}}{m\text{ days}}=\frac{75\text{ pages}}{1\text{ day}}

75m=1,50075m=1{,}500

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

m=20 daysm=20\text{ days}

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

25−20=525-20=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.
What makes it hard
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=old+p100×old\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
days=total pages÷pages per day\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.

  1. 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.

  2. Find Martina’s pages per day

    25% more means Karen’s 60 pages plus 25% of 60.

    new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}

    60+25100×60=60+15=7560 + \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.

  3. Find each girl’s number of days

    Divide the whole assignment by each girl’s pages per day.

    days=total pages÷pages per day\text{days} = \text{total pages} \div \text{pages per day}

    Karen: 1,500÷60=25\text{Karen: } 1{,}500 \div 60 = 25

    Martina: 1,500÷75=20\text{Martina: } 1{,}500 \div 75 = 20

    Karen needs 25 days, and Martina needs 20 days.

  4. Subtract the days

    “How many days sooner” asks for the difference.

    25−20=525 - 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: 25×60=1,50025 \times 60 = 1{,}500 pages. Martina: 20×75=1,50020 \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 53 of 120

2024 Test B Math, question 24

Question

What is the value of (157−167)(347−367)\frac{\left(1\frac{5}{7}-1\frac{6}{7}\right)}{\left(3\frac{4}{7}-3\frac{6}{7}\right)}?

  • A−12-\frac{1}{2}
  • B−17-\frac{1}{7}
  • C17\frac{1}{7}
  • D12\frac{1}{2}
Show the answer

Correct answer: D

First, subtract to simplify.

157−167347−367=−17−27=−17÷−27\frac{1\frac{5}{7}-1\frac{6}{7}}{3\frac{4}{7}-3\frac{6}{7}}=\frac{-\frac{1}{7}}{-\frac{2}{7}}=-\frac{1}{7}\div -\frac{2}{7}

Follow the process for dividing fractions:

(−17)×(−72)=714=12\left(-\frac{1}{7}\right)\times\left(-\frac{7}{2}\right)=\frac{7}{14}=\frac{1}{2}

Question analysis

A fractions question: find the value of a quotient whose top is 1 5/7 minus 1 6/7 and whose bottom is 3 4/7 minus 3 6/7. Students subtract mixed numbers that share a denominator and then divide one difference by the other. Keeping track of positive and negative signs through both steps is the main hurdle.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students subtract and divide rational numbers, including negatives, and use the sign rules for quotients. This item chains three such operations with easy sevenths, a typical grade 7 exercise.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 10, 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 mixed numbers look heavy, but the whole parts cancel and only sevenths remain. What decides the question is sign handling: each difference is negative, and one negative divided by another is positive. Students who stop after the top, subtract instead of divide, or keep a minus sign each find their result among the choices.

  1. Error-prone arithmetic moderate Working it out

    Both differences are negative, 1 5/7 − 1 6/7 = −1/7 and 3 4/7 − 3 6/7 = −2/7, and dividing them needs both the sign rule and a reciprocal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Two subtractions must be finished before the division, and a sign lost in either one carries into the final quotient. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out the top and the bottom separately, writing each difference as a signed fraction in sevenths, then divide by multiplying the top by the reciprocal of the bottom and apply the sign rule.

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.)
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Only the second fraction flips.
Dividing two negative numbers
(−a)÷(−b)=a÷b(-a) \div (-b) = a \div b
a and b are positive numbers, and b is not zero. A negative number divided by a negative number gives a positive answer. The two minus signs cancel. Multiplying two negatives works the same way.

  1. Sort out the problem

    The long fraction bar means divide. So you need the top difference divided by the bottom difference. A difference is the answer to a subtraction. Work out the top and the bottom first, then divide.

  2. Work out the top

    Both numbers on top have the same whole number, 1. So the wholes cancel, and you only need 57−67\frac{5}{7} - \frac{6}{7}. You take away more than you have, so the answer is negative.

    157−167=57−67=−171\frac{5}{7} - 1\frac{6}{7} = \frac{5}{7} - \frac{6}{7} = -\frac{1}{7}

  3. Work out the bottom

    Both numbers on the bottom have the whole number 3, so those cancel too.

    347−367=47−67=−273\frac{4}{7} - 3\frac{6}{7} = \frac{4}{7} - \frac{6}{7} = -\frac{2}{7}

    Watch out: do not stop after the top. −17-\frac{1}{7}, choice B, is only part of the problem.

  4. Divide the top by the bottom

    Both results are negative. The two minus signs cancel.

    (−a)÷(−b)=a÷b(-a) \div (-b) = a \div b

    −17÷(−27)=17÷27-\frac{1}{7} \div \left(-\frac{2}{7}\right) = \frac{1}{7} \div \frac{2}{7}

    Now flip the second fraction and multiply.

    ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

    17×72=714=12\frac{1}{7} \times \frac{7}{2} = \frac{7}{14} = \frac{1}{2}

  5. Avoid the traps and answer

    Watch out: keeping a minus sign gives −12-\frac{1}{2}, choice A. Subtracting the bottom from the top, instead of dividing, gives 17\frac{1}{7}, choice C.

    So the answer is D, 12\frac{1}{2}.

Check your answer

Think in sevenths. The top is −1 seventh and the bottom is −2 sevenths. That is like (−1)÷(−2)=12(-1) \div (-2) = \frac{1}{2}. Or multiply back: 12×(−27)=−17\frac{1}{2} \times \left(-\frac{2}{7}\right) = -\frac{1}{7}, which is the top.

More questions like this

The original DOE question
Question 81 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 81. Open page 205 of the handbook (PDF)

Question 54 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:

1 sind0.75 plunk=x sinds8 plunks\frac{1\text{ sind}}{0.75\text{ plunk}}=\frac{x\text{ sinds}}{8\text{ plunks}}

1(8)=0.75x1(8)=0.75x

8=75100x8=\frac{75}{100}x

8=34x8=\frac{3}{4}x

8(43)=x8\left(\frac{4}{3}\right)=x

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

1023=x10\frac{2}{3}=x

10.67≈x10.67\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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying
ab=cd means a×d=b×c\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÷bc=a×cba \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.

  1. 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.

  2. 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: 8×0.75=68 \times 0.75 = 6, choice B, goes the wrong way. That is the number of plunks in 8 sinds.

  3. 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.

    10.75=x8\frac{1}{0.75} = \frac{x}{8}

    ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c

    1×8=0.75×x1 \times 8 = 0.75 \times x

    So 0.75x=80.75x = 8.

  4. Divide by 0.75

    0.75 is the same as 34\frac{3}{4}. Divide both sides by it. To divide by a fraction, multiply by its reciprocal.

    a÷bc=a×cba \div \frac{b}{c} = a \times \frac{c}{b}

    x=8÷34=8×43=323=10.666…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.

  5. 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…≈10.6710.666\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 10.67×0.75≈8.0010.67 \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 55 of 120

2024 Test B Math, question 31

Question

If y=4xy=4x, what is the value of 3y+2(3y+5)−x3y+2(3y+5)-x in terms of x?

  • A8x+108x+10
  • B32x+1032x+10
  • C35x+1035x+10
  • D36x+1036x+10
Show the answer

Correct answer: C

First, substitute 4x4x for each y in the expression:

3y+2(3y+5)−x=3y+2(3y+5)-x=

3(4x)+2[3(4x)+5]−x3(4x)+2[3(4x)+5]-x

Then simplify the expression:

3(4x)+2[3(4x)+5]−x3(4x)+2[3(4x)+5]-x

12x+2(12x+5)−x12x+2(12x+5)-x

12x+24x+10−x12x+24x+10-x

12x+24x−x+1012x+24x-x+10

35x+1035x+10

The simplified expression is 35x+1035x+10.

Question analysis

Given that y equals 4x, students rewrite the expression 3y + 2(3y + 5) − x using only x. It tests substitution, the distributive property and combining like terms. Errors creep in when a term is dropped or when x terms and y terms are merged before y has been replaced.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Substituting an expression for a variable and simplifying with the distributive property is standard grade 7 work. Putting the substitution inside parentheses calls for care, but the steps are routine for students who combine like terms fluently.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
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

Substitution and distribution are routine here, so points are lost to bookkeeping rather than method. The expression has three x pieces, 12x, 24x and a lone − x, and the last is easy to overlook after the larger terms. The other two wrong choices come from collecting 3y, 6y and x as if they were like terms, a reason to replace y with 4x before simplifying anything.

  1. A very close wrong answer moderate Choosing the answer

    Every choice ends in + 10, and 36x + 10 differs from 35x + 10 by a single x, so only remembering the last − x in the expression separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    The substitution sits inside nested grouping, 2[3(4x) + 5], so the 2 must reach both the 12x and the 5 before the x terms are collected. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Replace every y with 4x in parentheses before doing anything else, then distribute the 2 and collect the x terms and the constant separately.

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.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. Terms are the parts joined by plus or minus signs. Multiply a by each term inside. It works because 2 groups of (12x + 5) hold 2 of the 12x and 2 of the 5.

  1. Sort out the question

    y=4xy = 4x means y is always 4 times x. You need to rewrite the expression with no y in it. That is what “in terms of x” means. The answer will use only x and numbers.

  2. Replace each y with 4x

    Put 4x in parentheses everywhere you see y. The parentheses keep the multiplying clear.

    3y+2(3y+5)−x=3(4x)+2(3(4x)+5)−x3y + 2(3y + 5) - x = 3(4x) + 2(3(4x) + 5) - x

    3(4x) means 3 times 4x, which is 12x. It shows up twice.

    =12x+2(12x+5)−x= 12x + 2(12x + 5) - x

  3. Multiply out the parentheses

    Use the distributive property. The 2 must multiply both the 12x and the 5.

    2(12x+5)=2×12x+2×5=24x+102(12x + 5) = 2 \times 12x + 2 \times 5 = 24x + 10

    The whole expression is now:

    12x+24x+10−x12x + 24x + 10 - x

  4. Combine the like terms

    Like terms have the same letter part. Here the x terms are 12x, 24x and −x. The −x means take away one x.

    12x+24x−x=35x12x + 24x - x = 35x

    The 10 has no x, so it stays as it is. The result is 35x+1035x + 10.

    Watch out: Don’t forget the −x at the end. Stopping at 12x + 24x gives 36x + 10, choice D. Also, swap in 4x before you combine. Adding 3y, 6y and −x as if they were all y’s mixes unlike terms. It leads to choice B.

    So the answer is C, 35x+1035x + 10.

Check your answer

Test a number. Let x = 1, so y = 4. The original expression gives 3(4)+2(3×4+5)−1=12+34−1=453(4) + 2(3 \times 4 + 5) - 1 = 12 + 34 - 1 = 45. Your answer gives 35(1)+10=4535(1) + 10 = 45. They match. Choice D would give 46.

More questions like this

The original DOE question
Question 88 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 2025 Admissions, page 208, question 88. 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 56 of 120

2024 Test B Math, question 33

Question

If x=−4x=-4, what is the sum of x and 10% of |x||x|?

  • A−4.4
  • B−3.6
  • C0.4
  • D4.4
Show the answer

Correct answer: B

First, determine 10% of |x||x|.

0.10∙|x|=0.10\bullet|x|=

0.10∙|−4|=0.10\bullet|-4|=

0.10∙4=0.40.10\bullet 4=0.4

Then determine the sum of x and 10% of |x||x|.

−4+0.4=−3.6-4+0.4=-3.6

The sum is −3.6.

Question analysis

With x equal to a negative number, students find the sum of x and 10 percent of the absolute value of x. It tests absolute value, a simple percent and adding numbers with different signs. The tricky part is keeping track of which term is negative and which is not.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students add positive and negative decimals and find simple percents. This joins those skills with an absolute value in two short steps, so it matches typical practice, with the sign of each term as the main hazard.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 11 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 short, but it asks students to treat the same number two ways: x keeps its negative sign while |x| drops it. Each wrong choice gets the sign wrong in one place, by taking 10% of −4, by making both terms positive, or by stopping at the positive 0.4. Writing the two terms separately, −4 and +0.4, before adding prevents each slip.

  1. Error-prone arithmetic moderate Working it out

    The sum mixes a negative integer with a positive decimal, −4 + 0.4, and the result has to keep the negative sign while moving 0.4 toward zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    Students must know that |−4| = 4, so the 10% is taken of a positive number even though x itself is negative. (Anderson & Krathwohl, 2001)

Strategy: Evaluate the absolute value first, take 10 percent of that positive number, and then add the result to x itself, keeping x's negative sign.

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.)
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value
|a|=distance from a to 0|a| = \text{distance from } a \text{ to } 0
The two bars mean absolute value. a is any number. Its absolute value is how far it is from 0 on the number line. A distance is never negative, so |−4| = 4.
Percent of a number
part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}
The whole is the amount the percent is taken of. The percent tells how many out of every 100. The part is the piece you get. Use it to find a percent of any number.

  1. Sort out the question

    You know that x = −4. You need the sum of two things: x itself, and 10% of |x|. A sum means you add.

    The bars around x mean absolute value. So the 10% is taken of |x|, not of x.

  2. Find the absolute value

    −4 is 4 steps from 0 on the number line. So its absolute value is 4. An absolute value is never negative.

    |x|=|−4|=4|x| = |-4| = 4

  3. Take 10 percent of it

    part=10100×4=0.4\text{part} = \frac{10}{100} \times 4 = 0.4

    So 10% of |x| is 0.4.

    Watch out: If you take 10% of x instead, you get −0.4. Then −4 + (−0.4) = −4.4, choice A. The bars make the number positive first.

  4. Add x and the result

    Now add x, which is still −4. Its own minus sign stays.

    −4+0.4=−3.6-4 + 0.4 = -3.6

    Start at −4 on the number line. Move 0.4 to the right, toward 0. You land on −3.6.

    Watch out: Stopping at 0.4 gives choice C. Using 4 in place of x gives 4.4, choice D.

    So the answer is B, −3.6.

Check your answer

Adding a small positive number should move −4 a little toward 0. The answer −3.6 is just to the right of −4, so that fits. Also, −3.6−(−4)=0.4-3.6 - (-4) = 0.4, the amount you added.

More questions like this

The original DOE question
Question 90 as printed on page 209 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 90. Open page 209 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 57 of 120

2024 Test B Math, question 36

Question

The integers (n−1)(n-1), n, and (n+1)(n+1) are factors of 168. What is the greatest possible value of n?

  • A3
  • B7
  • C14
  • D84
Show the answer

Correct answer: B

First, determine all possible factors of 168. One method of doing this is to use a table to list pairs of factors.

Factors of 168
1 168
2 84
3 56
4 42
6 28
7 24
8 21
12 14

Then determine the group of three sequential factors (n−1)(n-1), n, and (n+1)(n+1) where n is as large as possible. This group is 6, 7, and 8, where n=7n=7.

Question analysis

Three consecutive integers, written as n − 1, n and n + 1, must all be factors of 168, and students find the greatest possible value of n. It tests factors and consecutive integers. The catch is that a number can be a factor while its neighbors are not, so every member of the run needs checking.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Listing factors is grade 6 work, but here students must read n − 1, n and n + 1 as three consecutive integers and search the factor list for the largest such run. Pairing algebraic notation with factors matches grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 7, 17 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 factors of 168 are listed in order, 1, 2, 3, 4, 6, 7, 8, 12 and up, the runs of three in a row stand out, and the last of them is 6, 7, 8. Students lose points by stopping at the first run they see or by checking only n itself. Testing each choice with its two neighbors, starting from the largest, settles it in under a minute.

  1. Variables instead of numbers minor Understanding the question

    The stem names the numbers only as (n − 1), n and (n + 1), which has to be read as three whole numbers in a row that all divide 168. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    Each candidate must pass three divisibility tests at once, for itself and for both neighbors, while the search continues for the greatest such n. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: List the factor pairs of 168 in order, then look along the list for three whole numbers in a row; or test each choice together with its two neighbors.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pair
a×b=168a \times b = 168
a and b are whole numbers. If a times b equals 168, then a and b are both factors of 168. A factor divides a number with no remainder. Listing every pair, from small a upward, finds every factor.

  1. Sort out the question

    A factor of 168 divides 168 with no remainder. The numbers n − 1, n and n + 1 are three whole numbers in a row, such as 6, 7 and 8. The middle one is n.

    All three must be factors of 168. You need the greatest n that works.

  2. List the factor pairs of 168

    Try each whole number from 1 up. When it divides 168, write it with its partner. Stop when the pairs meet, after 12 and 14.

    Small factorPartner
    1168
    284
    356
    442
    628
    724
    821
    1214

  3. Look for three in a row

    Here are all the factors in order:

    1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168

    The runs of three in a row are 1, 2, 3, then 2, 3, 4, then 6, 7, 8. No other three factors sit side by side.

  4. Pick the greatest middle number

    The middle numbers of the runs are 2, 3 and 7. The greatest is 7, so n = 7.

    Watch out: Stopping at the run 2, 3, 4 gives 3, choice A. Keep looking. Also, 14 and 84 are factors, but their neighbors are not. Neither 13 nor 15 divides 168, so 14, choice C, fails. For 84, choice D, neither 83 nor 85 is a factor.

    So the answer is B, 7.

Check your answer

Divide 168 by each number in the run: 168÷6=28168 \div 6 = 28, 168÷7=24168 \div 7 = 24 and 168÷8=21168 \div 8 = 21. All three are whole numbers, so 6, 7 and 8 are all factors.

More questions like this

The original DOE question
Question 93 as printed on page 209 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 93. Open page 209 of the handbook (PDF)

Question 58 of 120

2024 Test B Math, question 39

Question

A closed rectangular box drawn in three dimensions. Its height is labeled 20 cm, the bottom edge running back on the left is labeled 40 cm, and the bottom front edge is labeled 50 cm.

What is the total surface area, in square centimeters, of the closed box shown above?

  • A3,800
  • B5,600
  • C7,600
  • D40,000
Show the answer

Correct answer: C

To find surface area, use the formula A=2(hw+hl+lw)A=2(hw+hl+lw):

A=2[(20)(40)+(20)(50)+(50)(40)]A=2[(20)(40)+(20)(50)+(50)(40)]

A=2(800+1,000+2,000)A=2(800+1{,}000+2{,}000)

A=2(3,800)=7,600A=2(3{,}800)=7{,}600

Question analysis

A drawing of a closed box gives its length, width and height in centimeters, and students find its total surface area in square centimeters. It tests surface area of a rectangular prism. The work is short, but every face of the box has to be accounted for, and surface area must be kept apart from volume.

NYS standards
  • 7.G.6 area, surface area, and volume
  • 6.G.4 nets and surface area
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Surface area of a rectangular prism is a grade 7 standard, and this box has whole-number dimensions shown in a drawing. Three face areas, a sum and the opposite faces make it typical practice rather than a stretch.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 8, 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

Finding the three face areas, 800, 1,000 and 2,000, is quick; the question tests whether students account for all six faces. The wrong choices are what the box would give with only three faces, with the top left off, or with the dimensions multiplied into a volume. Listing the faces in pairs before adding is the safest route.

  1. Something you must already know minor Working it out

    No surface-area formula is given; students must know that a box's six faces come in three matching pairs, 20 × 40, 20 × 50 and 40 × 50. (Anderson & Krathwohl, 2001)

  2. Reading a figure, table or graph minor Understanding the question

    The drawing shows only three faces and labels 20, 40 and 50 cm on separate edges, so the hidden faces and their dimensions have to be pictured. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the areas of the three different faces from the labeled length, width and height, add them, and double the sum to cover the opposite faces.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Surface area of a box
SA=2(lw+lh+wh)SA = 2(lw + lh + wh)
SA is the surface area, the total area of all six faces. l is the length, w is the width and h is the height. lw, lh and wh are the areas of the three different faces. You double because each face has a matching face on the opposite side.

  1. Sort out the question

    The box is a rectangular prism, a box shape with six rectangle faces. Its length is 50 cm, its width is 40 cm and its height is 20 cm.

    The box is closed, so it has a top. Surface area is the total area of all the outside faces, in square centimeters.

  2. Find the three different faces

    The faces come in matching pairs. Find the area of one face from each pair. Multiply its two edges.

    FacesEdgesArea of one (sq cm)
    Front and back50 × 201,000
    Left and right40 × 20800
    Top and bottom50 × 402,000

  3. Add the faces and double

    Use the formula. Add the three areas. Then double the sum to cover the matching faces you can’t see.

    SA=2(50×40+50×20+40×20)SA = 2(50 \times 40 + 50 \times 20 + 40 \times 20)

    SA=2(2,000+1,000+800)=2×3,800=7,600SA = 2(2{,}000 + 1{,}000 + 800) = 2 \times 3{,}800 = 7{,}600

    The surface area is 7,600 square centimeters.

  4. Avoid the traps

    Watch out: Stopping at 3,800 counts only three faces, choice A. Leaving off the top, as for an open box, gives 5,600, choice B. Multiplying all three edges gives 40,000, choice D. That is the volume, in cubic centimeters.

    So the answer is C, 7,600.

Check your answer

Add all six faces one by one: 1,000+1,000+800+800+2,000+2,000=7,6001{,}000 + 1{,}000 + 800 + 800 + 2{,}000 + 2{,}000 = 7{,}600. Also check the units. Area multiplies two lengths, so it is in square centimeters. Multiplying three lengths gives cubic units, which is volume.

More questions like this

The original DOE question
Question 96 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 96. Open page 210 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

2024 Test B Math, question 43

Question

A community garden had 50 members in 2001. In each of the next three years, the number of members increased by 40% over the number of members from the previous year. How many members were there in 2004? (Round to the nearest whole number.)

  • A70
  • B98
  • C110
  • D137
Show the answer

Correct answer: D

To find out how many members there were in 2002, calculate 40% of 50 and then add it to 50.

50×0.40=2050\times 0.40=20

20+50=70 members20+50=70\text{ members}

To find out how many members there were in 2003, calculate 40% of 70 and then add it to 70.

70×0.40=2870\times 0.40=28

28+70=98 members28+70=98\text{ members}

To find out how many members there were in 2004, calculate 40% of 98 and then add it to 98.

98×0.40=39.298\times 0.40=39.2

The result is 39.2, which is rounded to 39 to keep members to the nearest whole number.

39+98=137 members39+98=137\text{ members}

Question analysis

A community garden has 50 members in 2001, and membership grows by 40 percent each year over the year before. Students find the number of members in 2004, rounded to a whole number. It tests repeated percent increase, where each of the three rounds of growth must start from the previous year's total.

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 increase is grade 7 content, and each step here is a routine one. Repeating it three times, each time on the new total, adds length and bookkeeping but no new idea, so it matches grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 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

Compound growth is the whole question: each year's 40% is taken of a larger number, so the increases grow from 20 to 28 to about 39. Students who add 20 three times treat the growth as simple, and those who lose count of the years stop at an earlier total that is also listed. Multiplying by 1.4 three times, or checking that the third increase is the largest, separates the right count from the rest.

  1. Many steps moderate Working it out

    Three yearly increases have to be applied in turn, 50 to 70 to 98 to 137.2, each taking 40% of the new total rather than of the original 50. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The phrase "increased by 40% over the number of members from the previous year" sets a new base each year, the wording that separates compound from simple growth. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    The third year's 40% of 98 is 39.2, the first decimal in the chain, and the rounding instruction applies only to the final count. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply the membership by 1.4 once for each year from 2002 through 2004, carrying the unrounded value forward, and round only at the end.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}
old is last year’s number of members. p is the percent increase, here 40. new is this year’s number. You add p percent of the old amount to the old amount. Each year, the new number becomes next year’s old number.

  1. Sort out the question

    In 2001 the garden had 50 members. In each of the next three years, 2002, 2003 and 2004, the members grew by 40%.

    The 40% is of last year’s members, so the base changes every year. You need the number for 2004, rounded to the nearest whole number.

  2. Find the members in 2002

    40% of 50 is 20. Add it to 50.

    new=50+40100×50=50+20=70\text{new} = 50 + \frac{40}{100} \times 50 = 50 + 20 = 70

    There were 70 members in 2002.

  3. Find the members in 2003

    Now 70 is the old number. Take 40% of 70, not of 50.

    new=70+40100×70=70+28=98\text{new} = 70 + \frac{40}{100} \times 70 = 70 + 28 = 98

    There were 98 members in 2003.

  4. Find the members in 2004

    Now 98 is the old number.

    new=98+40100×98=98+39.2=137.2\text{new} = 98 + \frac{40}{100} \times 98 = 98 + 39.2 = 137.2

    Round to the nearest whole number. 137.2 is closer to 137 than to 138, so there were 137 members.

    Watch out: Adding 20 members every year uses 40% of 50 each time. That gives 110, choice C. Stopping after one or two years gives 70 or 98, choices A and B.

    So the answer is D, 137.

Check your answer

A 40% rise is the same as multiplying by 1.4. So 50×1.4×1.4×1.4=50×2.744=137.250 \times 1.4 \times 1.4 \times 1.4 = 50 \times 2.744 = 137.2, which rounds to 137. It matches.

More questions like this

The original DOE question
Question 100 as printed on page 211 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 100. Open page 211 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 60 of 120

2024 Test B Math, question 46

Question

The least positive integer evenly divisible by the first 6 positive integers is 60. What is the least positive integer evenly divisible by the first 8 positive integers?

  • A1,680
  • B840
  • C480
  • D420
Show the answer

Correct answer: B

Each option is divisible by 1, 2, 3, 4, 5, and 6.

Option A:

1,680 is evenly divisible by the first 8 positive integers.

Option B:

840 is evenly divisible by the first 8 positive integers and is less than 1,680.

Option C:

480 is not evenly divisible by 7.

480÷7≈68.57480\div 7\approx 68.57

Option D:

420 is not evenly divisible by 8.

420÷8=52.5420\div 8=52.5

Option B (840) is the correct answer.

Question analysis

Knowing that 60 is the smallest number divisible by each of 1 through 6, students find the smallest positive integer divisible by each of 1 through 8. It tests least common multiples and prime factors. The challenge is adding only the factors that 7 and 8 really need, so the result works for all eight numbers and is still the least.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Least common multiples are grade 6 content, usually for two small numbers. Extending the given result from six numbers to eight means deciding what 7 and 8 add, more reasoning than typical classwork but no new content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 13, 25 words.
Corrected from the DOE handbook
  • A rounded decimal in the explanation is now marked as approximate (≈ instead of =).
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 given fact, 60 for the first six integers, is a head start, but only for students who see what 7 and 8 add. Multiplying 60 by 7 is needed; multiplying by all of 8 is not, since 60 already supplies two of its three factors of 2. Students who multiply by the new numbers whole get answers that are too large or that miss one divisor, so checking the smaller choices first is the efficient route.

  1. Every choice must be checked moderate Choosing the answer

    Both 1,680 and 840 are divisible by 7 and by 8, so testing the choices is safe only from the smallest up; 420 fails for 8 and 480 for 7. (Embretson & Wetzel, 1987)

  2. Something you must already know minor Working it out

    Building on 60 needs prime factors: 7 adds a new prime, while 8 = 2 × 2 × 2 needs only one more 2, since 60 already holds two. (Anderson & Krathwohl, 2001)

Strategy: Start from 60 and ask which extra prime factors 7 and 8 require, or test the choices from the smallest up for divisibility by both 7 and 8.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: To show that 480 is not evenly divisible by 7, the handbook writes 480 ÷ 7 = 68.57. The quotient is really 68.571428…, a decimal that never ends, so 68.57 is only a rounded value. The site writes 480 ÷ 7 ≈ 68.57, as the handbook itself does for other rounded results in this test. The point of the step is unchanged: the quotient is not a whole number, so 480 is not evenly divisible by 7 and is not divisible by all of the first 8 positive integers.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Evenly divisible
a÷b=a whole numbera \div b = \text{a whole number}
a is the number you test, and b is the number you divide by. If the answer is a whole number, with no remainder, a is evenly divisible by b. Then b is a factor of a.

  1. Sort out the question

    The first 8 positive integers are 1, 2, 3, 4, 5, 6, 7 and 8. “Evenly divisible” means dividing leaves no remainder.

    You need the least, or smallest, number that all 8 of them divide. You are told that 60 is the smallest for 1 to 6. So build on 60.

  2. Bring in the 7

    60 is not divisible by 7. And 7 is prime, which means its only factors are 1 and 7. So the new number needs a factor of 7. Multiply 60 by 7.

    60×7=42060 \times 7 = 420

  3. Bring in the 8

    8 is 2×2×22 \times 2 \times 2, so the number needs three 2s as factors. But 420 has only two: 420=2×2×105420 = 2 \times 2 \times 105, and 105 is odd. Multiply by one more 2.

    420×2=840420 \times 2 = 840

    Now 840÷8=105840 \div 8 = 105 and 840÷7=120840 \div 7 = 120. Both are whole numbers.

  4. Test the choices from smallest up

    Choice D: 420÷8=52.5420 \div 8 = 52.5. That is not a whole number, so 420 fails.

    Choice C: 480÷7≈68.57480 \div 7 \approx 68.57. The ≈ sign means about, since the decimal never ends. It is not a whole number, so 480 fails.

    Choice B: 840 works for 7 and 8. It also works for 1 to 6, since 840=60×14840 = 60 \times 14.

    Watch out: 1,680, choice A, works too. But it is twice 840, so it is not the least.

    So the answer is B, 840.

Check your answer

Divide 840 by each number from 1 to 8. You get 840, 420, 280, 210, 168, 140, 120 and 105. Every answer is a whole number, so 840 works for all eight.

More questions like this

The original DOE question
Question 103 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 103. Open page 212 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 61 of 120

2024 Test B Math, question 48

Question

Two triangles. Triangle MNP has N at the top and base MP labeled 10 cm; sides MN and NP each have a tick mark, showing they are equal. The smaller triangle XYZ has Y at the top, base XZ labeled 6 cm, and sides XY and YZ each labeled 9 cm.

Triangle MNP is similar to triangle XYZ, and MN=NP\mathrm{MN}=\mathrm{NP}. What is the perimeter of triangle MNP?

  • A25 cm
  • B30 cm
  • C36 cm
  • D40 cm
Show the answer

Correct answer: D

Similar geometric figures have sides that are proportional.

106=x9\frac{10}{6}=\frac{x}{9}

53=x9\frac{5}{3}=\frac{x}{9}

x=(5)(9)3=15x=\frac{(5)(9)}{3}=15

Because MN=NP\mathrm{MN}=\mathrm{NP}, both unknown sides are equal to 15. Therefore, the three sides of MNP are 15, 15, and 10. The perimeter of a triangle is the sum of its three side lengths.

15+15+10=4015+15+10=40

Question analysis

Two isosceles triangles are drawn, one with all its side lengths labeled and the other with only its base labeled, and the triangles are similar. Students find the perimeter of triangle MNP in centimeters. It tests similar figures and proportional sides, and the catch is that two sides must be found from the scale and the equal-sides marks.

NYS standards
  • 8.G.4 similarity and dilations
  • 7.G.1 scale drawings
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Similar triangles are grade 8 geometry, though grade 7 scale drawings use the same proportional reasoning. Students also need the equal-sides marks to find a missing side, a combined step beyond routine grade 7 scale work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 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 chains a scale factor to a perimeter, and both links have a trap. Some students scale by adding, since the base grew by 4, and others find the 15 cm leg correctly but then add only some of the sides. Writing all three sides of the large triangle, 15, 15 and 10, before adding is what the question rewards.

  1. Reading a figure, table or graph moderate Understanding the question

    The figure carries every number: tick marks show MN = NP, the base MP is 10 cm, and triangle XYZ has 9 cm legs on a 6 cm base. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Students must know that corresponding sides of similar triangles share one scale factor, here 10/6 = 5/3, though the stem never says so. (Anderson & Krathwohl, 2001)

  3. Error-prone arithmetic minor Working it out

    The scale factor 5/3 is a fraction, and 9 × 5/3 = 15 is easier to get right by dividing 9 by 3 first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Compare the two bases to get the scale factor, multiply a leg of the small triangle by it, then use the equal sides to add up the perimeter.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Similar triangles
MNXY=MPXZ\frac{\mathrm{MN}}{\mathrm{XY}} = \frac{\mathrm{MP}}{\mathrm{XZ}}
Similar triangles have the same shape but can differ in size. MN matches XY, and MP matches XZ. Each side of the big triangle is the same number of times as long as its matching side. So these two fractions are equal.
Perimeter of a triangle
P=a+b+cP = a + b + c
P is the perimeter, the distance all the way around. a, b and c are the lengths of the three sides. Add them to go once around the shape.

  1. Sort out the question

    In triangle MNP, the base MP is 10 cm. The tick marks on MN and NP show those two sides are equal. In triangle XYZ, the base XZ is 6 cm, and XY and YZ are each 9 cm.

    Similar means the same shape, with every side scaled by the same amount. You need the perimeter of MNP, the sum of its three sides.

  2. Match the sides

    The letters line up in order: M with X, N with Y, and P with Z. So the base MP matches the base XZ. The side MN matches XY, and NP matches YZ.

  3. Find the scale factor

    The scale factor is how many times bigger MNP is. Compare the two bases.

    MPXZ=106=53\frac{\mathrm{MP}}{\mathrm{XZ}} = \frac{10}{6} = \frac{5}{3}

    Every side of MNP is 53\frac{5}{3} times as long as its matching side in XYZ.

  4. Find the equal sides

    Use the similar triangles formula with XY = 9.

    MN9=53\frac{\mathrm{MN}}{9} = \frac{5}{3}

    MN=9×53=3×5=15\mathrm{MN} = 9 \times \frac{5}{3} = 3 \times 5 = 15

    Divide 9 by 3 first to keep the numbers small. MN is 15 cm. Since MN = NP, NP is 15 cm too.

    Watch out: The base grew by 4 cm, but adding 4 cm to each side is wrong. That gives 13 cm sides and 36, choice C. Similar shapes grow by multiplying.

  5. Add the three sides

    P=15+15+10=40P = 15 + 15 + 10 = 40

    The perimeter is 40 cm. Watch out: Leaving out one 15 cm side gives 25, choice A. Leaving out the base gives 30, choice B.

    So the answer is D, 40 cm.

Check your answer

The small triangle’s perimeter is 9 + 9 + 6 = 24 cm. Perimeters grow by the same scale factor as the sides. So 24×53=4024 \times \frac{5}{3} = 40. It matches.

More questions like this

The original DOE question
Question 105 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 105. Open page 212 of the handbook (PDF)

Question 62 of 120

2024 Test B Math, question 50

Question

Chris has 1 red shirt, 2 white shirts, and 3 black shirts. Chris picks a shirt at random. Without putting the first shirt back, he picks a second shirt at random. What is the probability that the first shirt is black and the second shirt is red?

  • A112\frac{1}{12}
  • B110\frac{1}{10}
  • C15\frac{1}{5}
  • D710\frac{7}{10}
Show the answer

Correct answer: B

Use the joint probability formula: P (black shirt first) • P (red shirt second, given black shirt first).

Initially there are 3 black shirts out of 6 total shirts, so the probability of picking a black shirt first is 36\frac{3}{6}.

Once a black shirt is picked first, there is 1 red shirt out of 5 total shirts remaining.

So the probability of picking a red shirt second, given that a black shirt was picked first, is 15\frac{1}{5}.

Then P (black shirt first) • P (red shirt second, given black shirt first)
=36∙15=330=110=\frac{3}{6}\bullet\frac{1}{5}=\frac{3}{30}=\frac{1}{10}.

Question analysis

Chris has red, white and black shirts and picks two at random, one after the other, without putting the first back. Students find the probability that the first shirt is black and the second is red. It tests the probability of two dependent events, where the second pick has one fewer shirt to choose from.

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 is usually found with a list or tree of equally likely outcomes. Only two picks are involved, but adjusting the second pick for the shirt already taken goes beyond typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 48 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 dependent picks with small numbers make this a textbook question, but the choices punish each half-step. Students who leave the first shirt in the drawer get a fraction just below the right one, students who find only the second pick's chance stop at a listed value, and students who add the two chances get a probability far too large. Writing both fractions, 3/6 and 1/5, and multiplying is the whole method.

  1. Something you must already know moderate Working it out

    The rule for two dependent picks, the first pick's chance times the second's given the first, is assumed, along with the drop to 5 shirts for the second pick. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    1/10 and 1/12 differ only in the second pick's denominator, 5 shirts or 6, so estimating cannot separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Turning words into math minor Understanding the question

    "Without putting the first shirt back" sets the second pick's total at 5 shirts, and the order, black first then red, fixes which fraction comes first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the chance of black on the first pick, then the chance of red on the second pick with one shirt fewer, and multiply the two.

Traps

  • Without replacement: In a second pick there is one fewer item, which changes the probability.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of wanted outcomestotal number of outcomesP = \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 then B)=P(A)×P(B after A)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.

  1. Sort out the question

    Chris has 1 red, 2 white and 3 black shirts. That is 6 shirts. He picks one and keeps it out. Then he picks a second shirt.

    You need the chance that the first shirt is black and the second is red. Both must happen, in that order.

  2. Find the chance of black first

    3 of the 6 shirts are black. Each shirt is equally likely to be picked.

    P(black first)=36P(\text{black first}) = \frac{3}{6}

    That is one half.

  3. Find the chance of red second

    A black shirt is now out and not put back. That leaves 5 shirts: 1 red, 2 white and 2 black.

    P(red second)=15P(\text{red second}) = \frac{1}{5}

    Watch out: The bottom is 5, not 6. Using 6 gives 112\frac{1}{12}, choice A, as if the shirt went back.

  4. Multiply the two chances

    Use the rule for events in a row. The second pick must go right too, inside the half of the time the first one did.

    P(black, then red)=36×15=330=110P(\text{black, then red}) = \frac{3}{6} \times \frac{1}{5} = \frac{3}{30} = \frac{1}{10}

    Watch out: Adding the chances gives 710\frac{7}{10}, choice D. Stopping at 15\frac{1}{5} gives choice C.

    So the answer is B, 110\frac{1}{10}.

Check your answer

Count every ordered pair of picks. There are 6 choices for the first and 5 for the second, so 6×5=306 \times 5 = 30 pairs. Pairs with black then red: 3 black shirts times 1 red shirt is 3. So the chance is 330=110\frac{3}{30} = \frac{1}{10}.

More questions like this

The original DOE question
Question 107 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 107. Open page 213 of the handbook (PDF)

Question 63 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?

  • A29\frac{2}{9}
  • B13\frac{1}{3}
  • C25\frac{2}{5}
  • D23\frac{2}{3}
Show the answer

Correct answer: A

The probability of Davina picking an odd-numbered ball from Box B is 23\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 13\frac{1}{3}. The probability of both of these events happening is the product of the two probabilities.

23×13=29\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.
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.

  1. 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)

  2. 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)

  3. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of wanted outcomestotal number of outcomesP = \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 then B)=P(A)×P(B after A)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.

  1. 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.

  2. 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.

  3. Find the chance for Davina’s pick

    Box B has 3 balls, and 2 of them, the 3 and the 5, are odd.

    P(odd ball moved)=23P(\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.

  4. 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(Yusuf picks it)=13P(\text{Yusuf picks it}) = \frac{1}{3}

    Watch out: 13\frac{1}{3} alone is choice B. It forgets that Davina might move the 4.

  5. Multiply the two chances

    Both events must happen, so use the rule for events in a row.

    P(odd)=23×13=29P(\text{odd}) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}

    Watch out: 23\frac{2}{3} alone, choice D, stops after Davina’s pick. 25\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, 29\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 movedBox A thenOdd picks
32, 6, 31
42, 6, 40
52, 6, 51

That is 2 odd picks out of 9, or 29\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 64 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=yprt = y

(2,500)(0.04)t=500(2{,}500)(0.04)t = 500

100t=500100t = 500

t=5t = 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.
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.

  1. 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)

  2. 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)

  3. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest
I=P×r×tI = 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%=p100p\% = \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.

  1. 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.

  2. Find one year’s interest

    First write 4% as a decimal. 4% means 4 out of 100.

    4%=4100=0.044\% = \frac{4}{100} = 0.04

    Now use the formula for 1 year, so t = 1.

    I=P×r×tI = P \times r \times t

    I=2,500×0.04×1=100I = 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.

  3. Find how many years make $500

    Simple interest adds the same $100 every year. So the time t must make the interest $500.

    2,500×0.04×t=5002{,}500 \times 0.04 \times t = 500

    100×t=500100 \times t = 500

    t=500÷100=5t = 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: 2,500×0.04×5=5002{,}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 65 of 120

2023 Test A Math, question 3

Question

Solve for x:

7x+3−2(2x+1)=137x + 3 - 2(2x + 1) = 13

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

Show the answer

Correct answer: 4

7x+3−2(2x+1)=137x + 3 - 2(2x + 1) = 13 Apply the distributive property; multiply the −2 by 2x2x and 1.
7x+3−4x−2=137x + 3 - 4x - 2 = 13 Combine like terms.
3x+1=133x + 1 = 13 Apply the additive inverse property; subtract 1 from both sides of the equation.
3x=123x = 12 Apply the multiplicative inverse property; divide both sides of the equation by 3.
x=4x = 4

Question analysis

This grid-in gives the equation 7x + 3 − 2(2x + 1) = 13 and asks students to solve for x. It tests the distributive property with a negative factor, combining like terms and solving a two-step equation. The minus sign in front of the parentheses has to reach both terms inside, a step many students rush.

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 expand expressions with negative factors and solve equations such as p(x + q) = r. This equation needs both skills plus combining like terms, which is routine work with one negative sign that is easy to lose.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
4 steps
Reading level
Question and choices: about grade 3, 4 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 standard multi-step linear equation, and the work is short once the parentheses are cleared. Everything rests on distributing −2 correctly; the later steps, combining 7x with −4x and 3 with −2, then undoing the + 1 and the × 3, are routine. Substituting 4 back into the left side, 28 + 3 − 18 = 13, is a fast check that catches a sign slip.

  1. Error-prone arithmetic moderate Working it out

    The −2 in front of (2x + 1) must reach both terms, giving −4x − 2; the second product, −2 × 1, is the one most often written as +2 or left as +1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. An answer in an unexpected form minor Choosing the answer

    As a grid-in with no choices, a value such as 3 or 8/3 from a distributing slip looks just as finished as the right one; only substituting back exposes it. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute −2 to both terms inside the parentheses, combine the x terms and the constants, then undo the addition and the multiplication in turn.

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.
  • Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. Multiply each term inside by a. If a is negative, every product gets its sign too.
Keeping an equation balanced
if a=b, then a−c=b−c and a÷c=b÷c\text{if } a = b \text{, then } 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.

  1. Sort out the equation

    You need the value of x that makes 7x+3−2(2x+1)=137x + 3 - 2(2x + 1) = 13 true. x is the unknown number. The tricky part is the −2 in front of the parentheses. It multiplies everything inside them.

  2. Multiply out the parentheses

    Use the distributive property with a = −2, b = 2x and c = 1.

    −2(2x+1)=(−2)×2x+(−2)×1=−4x−2-2(2x + 1) = (-2) \times 2x + (-2) \times 1 = -4x - 2

    So the equation becomes this:

    7x+3−4x−2=137x + 3 - 4x - 2 = 13

    Watch out: the −2 must multiply the 1 too. Leaving + 1 gives x = 3. Writing + 2 instead of − 2 gives x=83x = \frac{8}{3}.

  3. Combine like terms

    Like terms are terms of the same kind. Put the x terms together. Put the plain numbers together.

    7x−4x=3x7x - 4x = 3x

    3−2=13 - 2 = 1

    3x+1=133x + 1 = 13

  4. Get x by itself

    Undo the + 1 first. Subtract 1 from both sides, so the equation stays balanced.

    3x+1−1=13−13x + 1 - 1 = 13 - 1

    3x=123x = 12

    3x means 3 times x. So divide both sides by 3.

    x=12÷3=4x = 12 \div 3 = 4

    Watch out: don’t stop at 3x = 12. 12 is three times the answer. So you would grid 4.

Check your answer

Put x = 4 back into the first equation: 7×4+3−2(2×4+1)=28+3−2×9=31−18=137 \times 4 + 3 - 2(2 \times 4 + 1) = 28 + 3 - 2 \times 9 = 31 - 18 = 13. That matches the 13 on the right side, so x = 4 is right.

More questions like this

The original DOE question
Question 60 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 60. Open page 75 of the handbook (PDF)

Question 66 of 120

2023 Test A Math, question 4

Question

Line l is vertical and line n is horizontal, and a small square marks the right angle where they meet. Line m rises from the lower left to the upper right; it crosses line l below line n and crosses line n to the right of line l. Where m crosses l, the angle between the lower part of l and the lower-left part of m is marked 35°. Where m crosses n, the angle between the right part of n and the upper-right part of m is marked x°.

In the figure above, line l is perpendicular to line n. What is the value of x?

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

Show the answer

Correct answer: 55

Since vertical angles are congruent, the right triangle in the figure has acute angles with measures of 35° and x°. The interior angles of a triangle sum to 180°, so set up an equation to solve for x:

90+35+x=18090 + 35 + x = 180

125+x=180125 + x = 180

x=55x = 55

Question analysis

Three lines cross in a figure: line l is perpendicular to line n, and a slanted line m meets them at a marked 35° angle and an unknown angle x. This geometry grid-in asks for the value of x. Students must see the right triangle the lines form and carry the marked angles into it using vertical angles.

NYS standards
  • 8.G.5 angle sums, transversals, angle-angle similarity
  • 7.G.5 angle relationships and equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 angle problems use vertical, supplementary and complementary angles. This one also needs the angle sum of a triangle, which New York places in grade 8, and the triangle is not drawn as a separate shape, so students must spot it.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 4, 17 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 hides a small right triangle between the three lines, and the question is easy once students see it. Its angles are the right angle, the vertical partner of the 35° angle and the vertical partner of x, so x is what the right angle and 35° leave of 180°. Students who never spot the triangle tend to guess a direct relationship between the two marked angles, which no angle fact supports.

  1. Reading a figure, table or graph moderate Understanding the question

    Neither marked angle is inside the right triangle that lines l, m and n form: the 35° opens below the point where m crosses l, and x opens above line n, so each must be carried in as a vertical angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know moderate Working it out

    The stem states only that l is perpendicular to n; students must supply that vertical angles are equal and that the three angles of a triangle add to 180°. (Anderson & Krathwohl, 2001)

  3. An answer in an unexpected form minor Choosing the answer

    As a grid-in it offers no choices, so 125, the right angle plus the 35° angle, can be gridded with nothing to signal the missing subtraction. (Enright, Morley & Sheehan, 2002)

Strategy: Find the triangle formed by the three lines, use vertical angles to move the 35° and x° angles inside it, and apply the triangle's angle sum with the right angle.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Vertical angles
angle a=angle b\text{angle } a = \text{angle } b
When two lines cross, they make an X. The angles straight across from each other in the X are called vertical angles. a and b are one such pair. They are always equal, because each one plus the angle between them makes a straight line.
Angles of a triangle
A+B+C=180A + B + C = 180
A, B and C are the three angles inside any triangle, in degrees. They always add up to 180°. Use it when you know two of the angles and need the third.

  1. Sort out the figure

    Line l is perpendicular to line n. Perpendicular means they meet at a right angle, 90°. The small square in the figure marks it. Line m crosses both lines. The three lines make a triangle with a right angle. You need x, the angle marked x° where m crosses n.

  2. Move the 35° angle into the triangle

    The 35° angle is below the point where m crosses l. So it is outside the triangle. Lines l and m make an X there. The triangle’s angle is straight across from the 35° angle. They are vertical angles, so they are equal. The triangle’s angle at that corner is 35°.

  3. Move the x° angle into the triangle

    The x° angle is above line n, so it is outside the triangle too. Lines m and n make another X. The triangle’s angle at that crossing is straight across from x°. So the third angle of the triangle is also x°. Watch out: line m does not cross l and n at the same angle. Lines l and n are not parallel, so x is not 35.

  4. Use the triangle’s angle sum

    The triangle’s three angles are 90°, 35° and x°.

    A+B+C=180A + B + C = 180

    90+35+x=18090 + 35 + x = 180

    125+x=180125 + x = 180

    x=180−125=55x = 180 - 125 = 55

    Watch out: 125 is only the two known angles added together. You still subtract it from 180. Gridding 125 is wrong. So you would grid 55.

Check your answer

In a right triangle, the two smaller angles add up to 90°. That is because the right angle uses 90° of the 180°. Check: 35+55=9035 + 55 = 90. Also, in the figure the x° angle looks a bit bigger than the 35° angle, and 55 is.

More questions like this

The original DOE question
Question 61 as printed on page 76 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 76, question 61. Open page 76 of the handbook (PDF)

Question 67 of 120

2023 Test A Math, question 5

Question

The mean value of 8 numbers is 17. Three of these numbers (9, 11, and 20) are discarded. What is the mean of the 5 remaining numbers?

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

Show the answer

Correct answer: 19.2

First, find the sum of the original 8 numbers. Let x equal the sum of those numbers and use the mean formula to solve for x:

x8=17\frac{x}{8} = 17

x=136x = 136

Next, subtract the 3 discarded numbers from the sum:

136−9−11−20=96136 - 9 - 11 - 20 = 96

Finally, calculate the mean of the remaining 5 numbers by dividing the new sum, 96, by 5:

965=19.2\frac{96}{5} = 19.2

Question analysis

Eight numbers have a mean of 17, and three of them, 9, 11 and 20, are removed. This statistics grid-in asks for the mean of the five numbers that are left. Students have to rebuild the total from the mean before removing anything, and then divide by the new count rather than the old one.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Finding a mean is grade 6 content, but working back from a mean to a total, removing values and averaging again is a typical grade 7 multistep task. As a grid-in, it offers no choices to check a wrong total against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 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

A mean cannot be adjusted directly when numbers are removed, so the question rewards switching from the mean to the total and back. The removed numbers, 9, 11 and 20, average less than 17, so the new mean must rise above 17; that quick check rules out an answer such as 12. Points are lost by dividing by the wrong count or by stopping at the new total.

  1. Many steps moderate Working it out

    Three operations come in a fixed order: multiply the mean by 8 to get 136, subtract 9, 11 and 20 to leave 96, then divide by the numbers that remain. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    The count changes partway through, from 8 numbers to 5, and the divisor at the end must be the new count, not the 8 from the first sentence. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. An answer in an unexpected form minor Choosing the answer

    The answer, 19.2, is a decimal that must be gridded with its point, and with no choices a total of 96 or a mean over 8 numbers looks complete. (Enright, Morley & Sheehan, 2002)

Strategy: Turn the mean into a total by multiplying by 8, subtract the three discarded numbers, and divide what is left by the count of numbers that remain.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean, or average
mean=sum of the numbershow many numbers\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}
The mean is the average. The sum is what you get when you add all the numbers. Divide it by how many numbers there are. The mean is the value each number would have if the total were shared out equally.
Sum from a mean
sum=mean×how many numbers\text{sum} = \text{mean} \times \text{how many numbers}
This is the mean formula turned around. If you know the mean and how many numbers there are, multiply them to get the sum. It works even when you don’t know the numbers themselves.

  1. Sort out the question

    There are 8 numbers. Their mean, or average, is 17. You don’t know what the 8 numbers are. Three of them, 9, 11 and 20, are discarded, or thrown away. That leaves 5 numbers. You need the mean of those 5 numbers.

  2. Find the sum of all 8 numbers

    You can’t add the 8 numbers, because you don’t know them. But the mean tells you their sum.

    sum=mean×how many numbers\text{sum} = \text{mean} \times \text{how many numbers}

    sum=17×8=136\text{sum} = 17 \times 8 = 136

    The 8 numbers add up to 136.

  3. Take away the three discarded numbers

    Subtract the three numbers that were thrown away.

    136−9−11−20=136−40=96136 - 9 - 11 - 20 = 136 - 40 = 96

    The 5 numbers left add up to 96. Watch out: 96 is a sum, not a mean, so gridding 96 is wrong. Also, dividing 136 by 5 without subtracting first gives 27.2, which is wrong.

  4. Divide by the 5 numbers left

    Now use the mean formula for the numbers that are left.

    mean=sum of the numbershow many numbers\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}

    mean=965=19.2\text{mean} = \frac{96}{5} = 19.2

    Watch out: divide by 5, not by 8. Only 5 numbers are left. Dividing by 8 gives 12. So you would grid 19.2, with the decimal point in its own column.

Check your answer

The three numbers thrown away add to 40, so their mean is about 13. That is less than 17. Removing small numbers should make the mean go up, and 19.2 is more than 17. Also, 19.2×5=9619.2 \times 5 = 96, the sum of the 5 numbers left.

More questions like this

The original DOE question
Question 62 as printed on page 76 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 76, question 62. Open page 76 of the handbook (PDF)

Question 68 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:

x8=901\frac{x}{8} = \frac{90}{1}

x=720x = 720

Next, convert 720 minutes into hours and minutes:

72060=12 hours\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
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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total time
total time=time for one×how many\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
hours=minutes÷60\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.

  1. 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.

  2. Find the total time in minutes

    Each revolution takes 90 minutes, and there are 8 of them.

    total time=90×8=720\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.

  3. Change the minutes to hours

    Clock times are easier to count in hours. There are 60 minutes in an hour.

    hours=720÷60=12\text{hours} = 720 \div 60 = 12

    The sign turns for 12 hours.

  4. 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 12×60=72012 \times 60 = 720 minutes. And 720÷90=8720 \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 69 of 120

2023 Test A Math, question 25

Question

Line h is horizontal, with arrows at both ends. Line j rises from the lower left to the upper right, and line k rises steeply from the bottom, crossing line h to the right of line j; lines j and k meet at point Q above line h. Where j crosses h, the angle above h and to the left of j is marked 135°. Where k crosses h, the angle below h and to the right of k is marked 100°. At Q, the angle between the parts of lines k and j above Q is marked x°.

Lines j and k intersect at point Q, and h is a straight line. What is the value of x?

  • A55
  • B45
  • C35
  • D30
Show the answer

Correct answer: C

First, find the measures of the three angles inside the triangle. The lower left angle is supplementary to the one marked 135°, so the measure of that angle is 180−135=45180 - 135 = 45. The angle on top is a vertical angle with the one marked x°, so the measure of that angle is also x°. The angle on the lower right is a vertical angle with the one marked 100°, so the measure of that angle is also 100°. Since the measures of the three interior angles of a triangle sum to 180°, set up an equation and solve for x:

45+x+100=18045 + x + 100 = 180

145+x=180145 + x = 180

x=35x = 35

Question analysis

Lines j and k cross at point Q above a straight line h, forming a triangle, with a 135° angle marked where j meets h and a 100° angle where k meets h. Students find the angle x at Q. Each marked angle lies outside the triangle, so it has to be turned into an inside angle before the triangle's angle sum can be used.

NYS standards
  • 8.G.5 angle sums, transversals, angle-angle similarity
  • 7.G.5 angle relationships and equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students use supplementary and vertical angles to find unknown angles. This figure needs both, applied at three different vertices, plus the angle sum of a triangle, which New York teaches in grade 8.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 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

Every angle the solution needs has to be carried into the triangle from outside, and the two marked angles need different treatment. The 135° is supplementary to the triangle's angle, but the 100° is vertical to it, so subtracting both from 180° out of habit builds the wrong triangle. Checking which side of each line a marked angle sits on, not just how near the triangle it is, separates the two cases.

  1. Reading a figure, table or graph major Understanding the question

    None of the three marked angles is inside the triangle formed by h, j and k: the 135° is outside on the left, the 100° is below line h, and x is above Q. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Two different angle facts are needed: the 135° gives a supplementary interior angle of 45°, while the 100° and x carry into the triangle unchanged as vertical angles. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Something you must already know minor Working it out

    Supplementary angles, vertical angles and the triangle's 180° sum are all assumed, and the question names none of them. (Anderson & Krathwohl, 2001)

Strategy: Mark the three angles of the triangle formed by lines h, j and k: one is supplementary to 135°, one is vertical to 100°, and the one at Q is vertical to x°.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line
a+b=180a + b = 180
a and b are two angles side by side on a straight line. Together they make a half turn, a straight angle. So they add up to 180°. Angles like this are called supplementary.
Vertical angles
angle a=angle b\text{angle } a = \text{angle } b
When two lines cross, they make an X. The angles straight across from each other in the X are called vertical angles. a and b are one such pair. They are always equal, because each one plus the angle between them makes a straight line.
Angles of a triangle
A+B+C=180A + B + C = 180
A, B and C are the three angles inside any triangle, in degrees. They always add up to 180°. Use it when you know two of the angles and need the third.

  1. Sort out the figure

    Lines h, j and k make a triangle. Its corners are where j crosses h, where k crosses h, and point Q. The figure marks two angles outside the triangle, 135° and 100°. You need x, the angle marked above Q.

  2. Find the angle where j crosses h

    The 135° angle is above h and to the left of j. The triangle’s angle is right next to it, on the other side of j. Together they make the straight line h. Call the triangle’s angle a.

    a+b=180a + b = 180

    135+a=180135 + a = 180

    a=180−135=45a = 180 - 135 = 45

    So that corner of the triangle is 45°. Watch out: this is only one corner, not x. Stopping here gives 45, choice B.

  3. Find the angle where k crosses h

    The 100° angle is below h and to the right of k. The triangle’s angle is above h and to the left of k. Lines h and k make an X there, and these two angles are straight across from each other. They are vertical angles, so the triangle’s angle is 100°. Watch out: it is not 180 − 100 = 80. That angle is beside the 100°, not across from it. Using 80 gives 55, choice A.

  4. Find the angle at Q

    The x° angle is above Q, between the top parts of lines j and k. The triangle’s angle at Q is between the parts of j and k below Q. These two angles are straight across from each other. So they are vertical angles, and the triangle’s angle at Q is also x°.

  5. Use the triangle’s angle sum

    A+B+C=180A + B + C = 180

    45+100+x=18045 + 100 + x = 180

    145+x=180145 + x = 180

    x=180−145=35x = 180 - 145 = 35

    The three angles of the triangle must add up to 180°. Only x is unknown. So the answer is C, 35.

Check your answer

Add the triangle’s three angles: 45+100+35=18045 + 100 + 35 = 180. Also, the angle at Q looks narrow in the figure, much less than a right angle of 90°. 35° fits that.

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 2024 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 70 of 120

2023 Test A Math, question 28

Question

It took Lars 2 hours to ride his bicycle 48 kilometers. What was his average speed in miles per hour? (Use the approximation 1 mile = 1.6 kilometers.)

  • A1.5
  • B15.0
  • C30.0
  • D38.4
Show the answer

Correct answer: B

First, calculate Lars’s average speed in kilometers per hour:

482=24\frac{48}{2} = 24

Then, convert kilometers to miles:

m24=11.6\frac{m}{24} = \frac{1}{1.6} Cross multiply.
m(1.6)=24(1)m(1.6) = 24(1) Apply the multiplicative inverse property; divide both sides of the equation by 1.6.
m=241.6=15m = \frac{24}{1.6} = 15

Question analysis

Lars rides 48 kilometers in 2 hours, and this rate question asks for his average speed in miles per hour, using 1 mile = 1.6 kilometers. Students find a unit rate and convert it to other units. Deciding whether to multiply or divide by 1.6, and remembering the 2 hours, are where students go wrong.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a speed and converting units with a given rate are grade 6 ratio skills. The question takes two short steps with friendly numbers, lighter than typical grade 7 rate problems with fractional quantities.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 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

The question tests the direction of a conversion more than the arithmetic. A mile is longer than a kilometer, so a distance has fewer miles than kilometers, and multiplying by 1.6 runs the conversion backward. The other trap is converting the 48 kilometers to 30 miles and forgetting that the ride took 2 hours, which gives a distance instead of a speed.

  1. Combines ideas minor Working it out

    A rate and a unit conversion are combined: 48 ÷ 2 = 24 kilometers per hour, then 24 ÷ 1.6 = 15 miles per hour, and the two steps can come in either order. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Error-prone arithmetic minor Working it out

    Dividing by 1.6 means dividing 240 by 16; a decimal point in the wrong place makes the speed ten times too small, and that value is offered. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the speed in kilometers per hour first, then divide by 1.6, since a mile is longer than a kilometer and the number of miles must be smaller.

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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average speed
speed=distancetime\text{speed} = \frac{\text{distance}}{\text{time}}
Distance is how far you go. Time is how long it takes. Speed is how far you go in each unit of time, such as each hour. Average speed shares the whole trip out evenly over the time.
Kilometers to miles
miles=kilometers÷1.6\text{miles} = \text{kilometers} \div 1.6
A kilometer is a unit of length used in most countries. 1 mile is about 1.6 kilometers, so a mile is longer. To find how many miles, find how many groups of 1.6 kilometers fit. So divide by 1.6.

  1. Sort out the question

    Lars rode 48 kilometers in 2 hours. You need his average speed in miles per hour. That means miles, not kilometers. Use the fact that 1 mile is about 1.6 kilometers.

  2. Find the speed in kilometers per hour

    Divide the distance by the time.

    speed=482=24\text{speed} = \frac{48}{2} = 24

    Lars went 24 kilometers each hour. This speed is still in kilometers, so you are not done yet.

  3. Change kilometers to miles

    Divide by 1.6. To make it easier, multiply both numbers by 10 first. Then you divide by the whole number 16.

    24÷1.6=240÷16=1524 \div 1.6 = 240 \div 16 = 15

    Watch out: divide by 1.6, don’t multiply. A mile is longer, so there are fewer miles than kilometers. 24×1.6=38.424 \times 1.6 = 38.4 is choice D. Dividing 24 by 16 gives 1.5, choice A, with the decimal point in the wrong place.

  4. Name the answer

    Lars rode 15 miles each hour. Watch out: changing 48 kilometers to 30 miles is only part of the work. 30 is a distance, choice C. You still have to divide by the 2 hours. So the answer is B, 15.0 miles per hour.

Check your answer

Work backward. At 15 miles per hour for 2 hours, Lars rides 30 miles. And 30×1.6=4830 \times 1.6 = 48 kilometers, the distance in the question. Also, 15 is smaller than 24, as it should be, since miles are longer than kilometers.

More questions like this

The original DOE question
Question 85 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 85. Open page 82 of the handbook (PDF)

Question 71 of 120

2023 Test A Math, question 34

Question

Parallelogram PQRS, with P at the bottom left, Q at the top left, R at the top right and S at the bottom right. The angle at P is marked 56°, the angle at Q is marked x°, and the angle at S is marked y°. The angle at R is not marked.

In the parallelogram above, what is the value of x+yx + y?

  • A112
  • B124
  • C148
  • D248
Show the answer

Correct answer: D

In a parallelogram, opposite angles are congruent, so the measure of the unmarked angle is 56°. The sum of the interior angles of a parallelogram is equal to 360°. Use that information to solve the problem:

x+y+56+56=360x + y + 56 + 56 = 360

x+y+112=360x + y + 112 = 360

x+y=248x + y = 248

Question analysis

This geometry question shows parallelogram PQRS with a 56° angle at P and unknown angles x and y at Q and S, and asks for the sum x + y. Students need angle facts about parallelograms: opposite angles match, and the four angles total 360°. Stopping after finding only one angle, or adding the wrong pair, are the common mistakes.

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 parallelogram facts needed, equal opposite angles and a 360° angle sum, rest on grade 8 reasoning about parallel lines and triangles. Once they are known, the arithmetic is short, about the size of a typical grade 7 angle problem.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 10, 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

Two short angle facts finish the question, but the figure marks three of the four corners, and the unmarked one at R is easy to forget. Students who find one obtuse angle stop with half the total, and those who add the two 56° angles give the part already known. Checking that the four angles, 56, 124, 56 and 124, add to 360 confirms the sum.

  1. Something you must already know moderate Working it out

    The figure gives one angle and no rules: students must bring both that opposite angles of a parallelogram are equal and that a quadrilateral's angles total 360°, or that neighboring angles add to 180°. (Anderson & Krathwohl, 2001)

  2. Reading a figure, table or graph minor Understanding the question

    The labels x and y sit at Q and S, opposite corners, while the corner R across from the 56° angle at P carries no label at all. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Mark the angle at R equal to the angle at P, since opposite angles of a parallelogram are equal, then take both away from the total of a quadrilateral's four angles.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Opposite angles of a parallelogram
∠P=∠R\angle P = \angle R
A parallelogram is a four-sided shape whose opposite sides are parallel. P and R are opposite corners: they don’t share a side. Opposite angles of a parallelogram are always equal.
Angles of a four-sided shape
∠P+∠Q+∠R+∠S=360\angle P + \angle Q + \angle R + \angle S = 360
P, Q, R and S are the four corners. Their angles always add up to 360 degrees. A line from one corner to the opposite corner cuts the shape into two triangles, and each triangle’s angles add up to 180 degrees.
Neighboring angles of a parallelogram
∠P+∠Q=180\angle P + \angle Q = 180
P and Q are corners next to each other, joined by one side. In a parallelogram, their angles add up to 180 degrees. The same is true of any two neighboring corners.

  1. Sort out the figure

    PQRS is a parallelogram. The angle at corner P is 56°. The angle at Q is x°, and the angle at S is y°. The angle at R has no label. You need x + y, the angles at Q and S added together.

  2. Find the angle at R

    R is across from P, so they are opposite corners.

    ∠P=∠R\angle P = \angle R

    ∠R=56\angle R = 56

    So the angle at R is 56°, the same as at P.

  3. Use the 360° total

    ∠P+∠Q+∠R+∠S=360\angle P + \angle Q + \angle R + \angle S = 360

    56+x+56+y=36056 + x + 56 + y = 360

    x+y+112=360x + y + 112 = 360

    x+y=360−112=248x + y = 360 - 112 = 248

    Watch out: 112 is the two angles you already know, not x + y. Answering 112 is choice A.

  4. Find x and y one at a time

    Q is next to P, so use the neighbor rule.

    ∠P+∠Q=180\angle P + \angle Q = 180

    56+x=180,x=12456 + x = 180, \quad x = 124

    S is also next to P, so y = 124 too. Then 124+124=248124 + 124 = 248, the same total. Watch out: stopping at one angle, 124, gives choice B. The question asks for both added together.

    So the answer is D, 248.

Check your answer

Add all four angles: 56+124+56+124=36056 + 124 + 56 + 124 = 360. That is the right total for a four-sided shape. Also, the angles at Q and S look wider than a right angle in the figure, and 124° is wider than 90°.

More questions like this

The original DOE question
Question 91 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 91. Open page 84 of the handbook (PDF)

Question 72 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 JL=23LK\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

JK=6−(−4)=10\mathrm{JK} = 6 - (-4) = 10 units. Let x=JLx = \mathrm{JL}. Then LK=10−x\mathrm{LK} = 10 - x. Substitute these expressions in the given equation:

x=23(10−x)x = \frac{2}{3}(10 - x)

Solve for x:

x=23(10−x)x = \frac{2}{3}(10 - x) Apply the multiplicative inverse property; multiply both sides of the equation by 3.
3x=2(10−x)3x = 2(10 - x) Apply the distributive property; multiply the 2 by 10 and x.
3x=20−2x3x = 20 - 2x Apply the additive inverse property; add 2x2x to both sides of the equation.
5x=205x = 20 Apply the multiplicative inverse property; divide both sides of the equation by 5.
x=4x = 4

So, point L is 4 units to the right of point J: −4+4=0-4 + 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.
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
distance=right point−left point\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
one part=whole lengtha+b\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.

  1. 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 23\frac{2}{3} of LK. You need the position of L.

  2. Find the length of JK

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    K is on the right, so subtract J’s position from K’s.

    JK=6−(−4)=6+4=10\mathrm{JK} = 6 - (-4) = 6 + 4 = 10

    Taking away −4 is the same as adding 4. J and K are 10 units apart.

  3. Turn the rule into equal parts

    JL is 23\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.

    one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}

    one part=102+3=2\text{one part} = \frac{10}{2 + 3} = 2

    Each part is 2 units long.

  4. Find the two lengths

    JL=2×2=4\mathrm{JL} = 2 \times 2 = 4

    LK=3×2=6\mathrm{LK} = 3 \times 2 = 6

    Watch out: JL is 23\frac{2}{3} of LK, not the other way round. Flipping the ratio makes JL 6 and puts L at 2, choice C.

  5. 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=0-4 + 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 23×6=4\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 73 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−500j - 500 stamps. The total number of stamps is j+(j−500)=2j−500j + (j - 500) = 2j - 500. Since Josef has 60% of the stamps, set up an equation to solve for j:

j2j−500=60100\frac{j}{2j - 500} = \frac{60}{100} Cross multiply.
100j=60(2j−500)100j = 60(2j - 500) Apply the distributive property; multiply the 60 by 2j2j and 500.
100j=120j−30,000100j = 120j - 30{,}000 Apply the additive inverse property; subtract 120j120j from both sides of the equation.
−20j=−30,000-20j = -30{,}000 Apply the multiplicative inverse property; divide both sides of the equation by −20.
j=1,500j = 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.
What makes it hard
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.

  1. 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)

  2. 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)

  3. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\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.

  1. 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.

  2. Find Mai’s percent

    All the stamps make 100%. Josef has 60%, so Mai has the rest.

    100%−60%=40%100\% - 60\% = 40\%

    Mai has 40% of the stamps.

  3. Find what the 500 stamps are

    Josef’s extra 500 stamps are the gap between his share and Mai’s share.

    60%−40%=20%60\% - 40\% = 20\%

    So 500 stamps are 20% of all the stamps.

  4. Find the total number of stamps

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    500=20100×whole500 = \frac{20}{100} \times \text{whole}

    whole=500÷20100=500×5=2,500\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.

  5. Find Josef’s share

    part=60100×2,500=1,500\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 74 of 120

2023 Test A Math, question 43

Question

Kim jogs 8 kilometers in 1 hour 40 minutes. At that rate, how many meters does she jog per minute?

  • A0.08
  • B80
  • C800
  • D8,000
Show the answer

Correct answer: B

First, calculate Kim’s jogging time in minutes:

1 hour 40 minutes = 100 minutes

Then, convert kilometers to meters:

8 kilometers = 8,000 meters

Now simplify the ratio to find the answer:

8,000100=80\frac{8{,}000}{100} = 80

Question analysis

This rate question says Kim jogs 8 kilometers in 1 hour 40 minutes and asks how many meters she jogs per minute. Students convert a mixed time to minutes and kilometers to meters before finding a unit rate. Skipping a conversion, or stopping before dividing, gives answers that are off by factors of ten or more.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Unit rates and metric conversions are grade 6 topics. This needs two conversions, hours and minutes to minutes and kilometers to meters, then one division with whole numbers, simpler than grade 7 rate problems built on fractions.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 17 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 rate itself is one division; what makes the question work is that the distance and the time both arrive in the wrong units. Converting only the time gives a speed in kilometers per minute, and converting only the distance and stopping gives the whole trip in meters. Writing the units beside every number, meters over minutes, shows when the answer is ready.

  1. Something you must already know minor Working it out

    The conversions are not given: 1 kilometer is 1,000 meters and 1 hour is 60 minutes, so 8 kilometers becomes 8,000 meters and 1 hour 40 minutes becomes 100 minutes. (Anderson & Krathwohl, 2001)

  2. Many steps minor Working it out

    Two conversions, one for distance and one for time, come before the single division, and skipping the distance conversion leaves an answer a thousand times too small. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the time entirely in minutes and the distance entirely in meters, then divide distance by time; check that the answer's units are meters per minute.

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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Kilometers to meters
meters=kilometers×1,000\text{meters} = \text{kilometers} \times 1{,}000
Each kilometer is 1,000 meters, so multiply the number of kilometers by 1,000 to get meters. The word part kilo means a thousand.
Rate
rate=distancetime\text{rate} = \frac{\text{distance}}{\text{time}}
The rate, or speed, is how far you go in one unit of time. Distance is how far, and time is how long. Dividing shares the distance equally among the minutes, so you get the distance for one minute.

  1. Sort out the question

    Kim jogs 8 kilometers in 1 hour 40 minutes. You need her speed in meters per minute: how many meters she jogs in one minute. The question gives kilometers and hours, so change both units first.

  2. Change the time to minutes

    One hour is 60 minutes. Add the extra 40 minutes.

    60+40=10060 + 40 = 100

    Kim jogs for 100 minutes.

  3. Change the distance to meters

    meters=kilometers×1,000\text{meters} = \text{kilometers} \times 1{,}000

    8×1,000=8,0008 \times 1{,}000 = 8{,}000

    Kim jogs 8,000 meters. Watch out: 8,000 is the whole distance, not meters per minute. Stopping here gives choice D.

  4. Divide the distance by the time

    rate=distancetime\text{rate} = \frac{\text{distance}}{\text{time}}

    rate=8,000100=80\text{rate} = \frac{8{,}000}{100} = 80

    She jogs 80 meters each minute. Watch out: if you skip the change to meters, you get 8 ÷ 100 = 0.08 kilometer per minute, choice A. So the answer is B, 80 meters per minute.

Check your answer

Work backward. At 80 meters a minute, 100 minutes gives 80×100=8,00080 \times 100 = 8{,}000 meters. That is 8 kilometers, just as the question says.

More questions like this

The original DOE question
Question 100 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 2024 Admissions, page 86, question 100. 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 75 of 120

2023 Test A Math, question 45

Question

A rectangle. Point R is on the top side, near its right end. Line segment RS runs down and to the right from R, crosses the right side of the rectangle, and ends at point S outside it. At R, the angle between the top side, to the left of R, and segment RS is marked 130°. Where RS crosses the right side, the angle outside the rectangle between the right side, above the crossing, and the part of RS going on to S is marked y.

The diagram above shows line segment RS intersecting a rectangle. What is the measure of angle y?

  • A140°
  • B130°
  • C50°
  • D40°
Show the answer

Correct answer: A

First, find the measures of the interior angles of the triangle formed by the line segment and the rectangle. The top angle on the right is a vertex of the rectangle, so the measure of that angle is 90°. The top angle on the left is supplementary to the given angle of 130°. So, the measure of that angle is 180−130=50180 - 130 = 50. The measure of the lower angle is 40°, since 180−90−50=40180 - 90 - 50 = 40. Angle y is supplementary to the lower angle of 40°, so the measure of angle y is 180−40=140180 - 40 = 140.

Question analysis

This geometry question shows line segment RS cutting across a corner of a rectangle, with a 130° angle marked where RS meets the top side, and asks for angle y where it crosses the right side. Students combine supplementary angles, the right angle of a rectangle and the angle sum of a triangle. Stopping at an angle inside the triangle is the main trap.

NYS standards
  • 8.G.5 angle sums, transversals, angle-angle similarity
  • 7.G.5 angle relationships and equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 angle problems use supplementary, vertical and adjacent angles. This one adds the angle sum of a triangle, which New York places in grade 8, and chains three facts around a small triangle that students must first spot in the figure.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 7, 17 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 angle facts are basic, but the route runs through a triangle the figure does not point out, and the answer lies outside it. Students who stop inside the triangle give its 40° angle or its 50° angle, and those who assume y matches the marked 130° skip the work entirely. Checking that y and the triangle's angle beside it form a straight line, 140 + 40 = 180, confirms the result.

  1. Reading a figure, table or graph moderate Understanding the question

    The small triangle cut off at the rectangle's top right corner is never outlined or labeled, its 90° corner is unmarked, and y sits outside it, between RS and the rectangle's right side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Three steps follow in order, a straight angle at R, the triangle's angle sum, and a straight angle at the right side, and the first two results are both listed choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Something you must already know minor Working it out

    The rectangle's corner must be taken as 90° although no right-angle mark is drawn, and the 180° totals for a straight line and a triangle are assumed. (Anderson & Krathwohl, 2001)

Strategy: Look at the small triangle cut off at the rectangle's corner: get its angle at R from the straight top side, use the 90° corner and the triangle's angle sum, then use the straight right side.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line
a+b=180a + b = 180
a and b are two angles side by side that together make a straight line. A straight line makes an angle of 180 degrees. So if you know one of them, subtract it from 180 to get the other.
Angles in a triangle
a+b+c=180a + b + c = 180
a, b and c are the three angles inside any triangle. They always add up to 180 degrees. If you know two of them, subtract both from 180 to find the third.
Outside angle of a triangle
outside angle=a+b\text{outside angle} = a + b
An outside angle sits next to a triangle’s angle, along a straight line. a and b are the two inside angles at the other corners. The outside angle equals their sum.

  1. Sort out the figure

    Segment RS starts at R on the top side of the rectangle. It goes down to the right, crosses the right side, and ends at S. Call the rectangle’s top right corner P. Call the point where RS crosses the right side Q. Then R, P and Q make a small triangle. Angle y is at Q, outside the rectangle.

  2. Find the triangle’s angle at R

    The 130° angle and the triangle’s angle at R sit side by side along the top side, a straight line.

    a+b=180a + b = 180

    130+∠R=180130 + \angle \mathrm{R} = 180

    ∠R=180−130=50\angle \mathrm{R} = 180 - 130 = 50

    Watch out: 50° is only the first step. Stopping here gives choice C.

  3. Find the triangle’s angle at Q

    P is a corner of the rectangle, so its angle is 90°.

    a+b+c=180a + b + c = 180

    50+90+∠Q=18050 + 90 + \angle \mathrm{Q} = 180

    ∠Q=180−140=40\angle \mathrm{Q} = 180 - 140 = 40

    Watch out: y is outside the triangle, so this 40° angle, choice D, is not y.

  4. Find angle y

    Angle y and the triangle’s 40° angle at Q sit side by side along RS, a straight line.

    a+b=180a + b = 180

    y+40=180y + 40 = 180

    y=140y = 140

    So the answer is A, 140°.

Check your answer

Use the outside angle rule. Angle y is outside the triangle at Q. The two inside angles at the other corners are 50° and 90°, so y=50+90=140y = 50 + 90 = 140. Also, y looks wider than a right angle in the figure, so it must be more than 90°.

More questions like this

The original DOE question
Question 102 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 102. 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 76 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:

30×8×0.5=12030 \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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Inches to feet
feet=inches÷12\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×w×hV = 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.

  1. 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.

  2. Change the thickness to feet

    feet=inches÷12\text{feet} = \text{inches} \div 12

    6÷12=126 \div 12 = \frac{1}{2}

    So 6 inches is half a foot. Now all three measures are in feet.

  3. Multiply the three measures

    V=l×w×hV = l \times w \times h

    V=30×8×12=240×12=120V = 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 77 of 120

2023 Test A Math, question 57

Question

Chanelle selects a marble from a container and then returns it to the container. When she does this 3 times, the probability of choosing a red marble all 3 times is 1216\frac{1}{216}. Based on this information, what is the probability of Chanelle choosing a red marble the first time she selects a marble?

  • A172\frac{1}{72}
  • B136\frac{1}{36}
  • C18\frac{1}{8}
  • D16\frac{1}{6}
Show the answer

Correct answer: D

Let 1x\frac{1}{x} be the probability of Chanelle selecting a red marble the first time. The probability of selecting a red marble three times is 1x×1x×1x\frac{1}{x} \times \frac{1}{x} \times \frac{1}{x}, which is given as 1216\frac{1}{216}. Since 216=6×6×6216 = 6 \times 6 \times 6, the value of x is 6, and the probability of Chanelle selecting a red marble the first time is 16\frac{1}{6}.

Question analysis

This probability question says Chanelle draws a marble and puts it back, three times, and the chance of red all three times is 1/216. It asks for the chance of red on her first draw. Students reverse the multiplication rule for repeated independent draws, which means finding a cube root, and dividing by 3 is the tempting shortcut.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 8.EE.2 square and cube roots
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Multiplying probabilities for repeated draws with replacement is grade 7 compound probability. Running it backward, from three draws to one, needs a cube root, a grade 8 idea, though the numbers are friendly and the work is short once set up.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 55 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 three draws are seen as one chance cubed, the question is a cube root, and 216 is a perfect cube many students know. The risk lies in undoing the cube the wrong way: dividing by 3 because there are three draws, or stopping at 36 after one division by 6. A quick check, 1/6 × 1/6 × 1/6 = 1/216, confirms the answer.

  1. Something you must already know moderate Working it out

    Students must know that three draws with the marble returned each time multiply, p × p × p, and must recognize 216 as 6 × 6 × 6; the stem states neither. (Anderson & Krathwohl, 2001)

  2. No obvious method minor Working it out

    The usual direction runs from one draw to three; here the three-draw chance is given and the bold first asks for one draw, so students work backward and undo a cube. (Schoenfeld, 1985)

Strategy: Call the chance of red on one draw p, so three draws with replacement give p × p × p; then find the number whose cube equals the given probability.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of repeated picks
P(red 3 times)=p×p×pP(\text{red 3 times}) = p \times p \times p
p is the chance of red on one pick. Chanelle puts the marble back each time, so every pick has the same chance p, and one pick doesn’t change the next. The chance of red on all three picks is the three chances multiplied.

  1. Sort out the question

    Chanelle picks a marble and puts it back, 3 times. The chance of red all 3 times is 1216\frac{1}{216}. You need the chance of red on just the first pick. Call that chance p.

  2. Write the three picks as a product

    She puts each marble back, so the container is the same every time. Each pick has the same chance, p.

    P(red 3 times)=p×p×pP(\text{red 3 times}) = p \times p \times p

    p×p×p=1216p \times p \times p = \frac{1}{216}

  3. Find the number that fits

    You need a fraction that, multiplied by itself three times, makes 1216\frac{1}{216}. Its top is 1, since 1 × 1 × 1 = 1. For the bottom, find a number that times itself three times makes 216.

    6×6×6=36×6=2166 \times 6 \times 6 = 36 \times 6 = 216

    So the bottom is 6, and p is 16\frac{1}{6}.

  4. Avoid the tempting shortcuts

    Watch out: dividing 216 by 3 gives 72, choice A. Three picks means multiplying three times, not dividing by 3. Also, 136\frac{1}{36} is the chance of red twice in a row, choice B.

    So the answer is D, 16\frac{1}{6}.

Check your answer

Multiply it out: 16×16×16=1216\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216}. That matches the question. Choice A fails fast: 172×172\frac{1}{72} \times \frac{1}{72} is already far smaller than 1216\frac{1}{216}.

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 2024 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 78 of 120

2023 Test B Math, question 1

Question

LAST YEAR’S TOTAL SALES

Vehicles Number
Four-door cars 5,850
Two-door cars 2,250
Trucks 900
Total 9,000

The table above shows the number of each type of vehicle sold at a dealership last year. If 1,000 vehicles will be sold next month, what is the best estimate (based on last year’s sales) of the number of two-door cars that will be sold?

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

Show the answer

Correct answer: 250

Let x represent the number of two-door cars that will be sold next month. Use the information in the table to set up a proportion:

x1,000=2,2509,000\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000}

x=(2,250)(1,000)9,000=2,2509=250x = \frac{(2,250)(1,000)}{9{,}000} = \frac{2{,}250}{9} = 250

Question analysis

A table shows last year's sales at a car dealership: four-door cars, two-door cars and trucks, 9,000 vehicles in all. This grid-in asks for the best estimate of how many two-door cars will be sold next month if 1,000 vehicles are sold. Students set up a proportion from the table; most of its rows are not needed.

NYS standards
  • 6.RP.3 ratio and rate reasoning
  • 6.SP.7 long-run relative frequency
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 proportion problems often take several steps or awkward numbers. This one scales a single share from a table to a new total, with numbers that divide cleanly, which is closer to grade 6 ratio work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 55 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 share is seen as 2,250 out of 9,000, a quarter, the arithmetic is light, which makes this one of the easier grid-ins. Points are lost by entering the share itself as a percent, by reading the wrong row of the table, or by building a monthly average when the question supplies next month’s total. With no choices to compare, checking that the entry is a quarter of 1,000 is the only safeguard.

  1. Turning words into math minor Understanding the question

    The phrase “best estimate (based on last year’s sales)” has to become a proportion: two-door cars keep the same share of next month’s 1,000 vehicles as of last year’s 9,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Reading a figure, table or graph minor Understanding the question

    The table lists three vehicle types and a total; only the two-door row, 2,250, and the 9,000 total are needed, so the four-door and truck rows are distractions. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find what fraction of last year's 9,000 vehicles were two-door cars, simplify it, then take that fraction of the 1,000 vehicles expected next month.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportion
partwhole=partwhole\frac{\text{part}}{\text{whole}} = \frac{\text{part}}{\text{whole}}
A proportion says two fractions are equal. Each fraction here is a part over its whole: two-door cars over all vehicles sold. Use it when a share stays the same while the total changes. Then you can find the missing part.

  1. Sort out the question

    The table shows last year’s sales. In all, 9,000 vehicles were sold, and 2,250 of them were two-door cars. Next month, 1,000 vehicles will be sold. You need about how many of them will be two-door cars.

    “Based on last year’s sales” means two-door cars keep the same share of all sales. A share is a part out of a whole. The four-door and truck rows don’t matter here.

  2. Find last year’s share of two-door cars

    Last year, 2,250 out of 9,000 vehicles were two-door cars. Write that as a fraction and simplify it.

    2,2509,000=14\frac{2{,}250}{9{,}000} = \frac{1}{4}

    This works because 2,250×4=9,0002{,}250 \times 4 = 9{,}000. So 1 out of every 4 vehicles sold was a two-door car.

  3. Set up the proportion and solve

    Let x be the number of two-door cars next month. Next month’s share must match last year’s share.

    x1,000=2,2509,000=14\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000} = \frac{1}{4}

    To get x alone, multiply both sides by 1,000.

    x=14×1,000=250x = \frac{1}{4} \times 1{,}000 = 250

    So about 250 two-door cars will be sold.

  4. Give the answer

    Watch out: one fourth is the same as 25%. Gridding 25 gives the percent, not the number of cars. Also be sure to use the two-door row. The four-door row would give 650. So you would grid 250.

Check your answer

Try it another way. Next month’s 1,000 vehicles are 19\frac{1}{9} of last year’s 9,000. So take 19\frac{1}{9} of the 2,250 two-door cars: 2,250÷9=2502{,}250 \div 9 = 250. Both ways give 250 cars.

More questions like this

The original DOE question
Question 58 as printed on page 191 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 58. Open page 191 of the handbook (PDF)

Question 79 of 120

2023 Test B Math, question 3

Question

A kindergarten teacher has 72 crayons, 60 pencils, and 84 sheets of paper to distribute to the children in her class. If each child receives an equal number of each item and there are no items remaining, what is the greatest possible number of children in the class?

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

Show the answer

Correct answer: 12

The greatest possible number of children in the class is the greatest common factor of 72, 60, and 84. Write the prime factorization of each number:

72=23×3272 = 2^{3} \times 3^{2}

60=22×3×560 = 2^{2} \times 3 \times 5

84=22×3×784 = 2^{2} \times 3 \times 7

Then, find the factors that all three have in common, which is 22×3=122^{2} \times 3 = 12.

Question analysis

A kindergarten teacher shares 72 crayons, 60 pencils and 84 sheets of paper so that every child gets the same number of each item, with nothing left over. This grid-in asks for the greatest possible number of children in the class. Students must recognize a greatest common factor problem and not confuse it with a least common multiple.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Greatest common factors are grade 6 content, normally of two numbers. This one uses three numbers and a word setting that must be translated, but the factoring is short, so the demand stays below typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 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 story has to be recognized as a sharing problem before any arithmetic starts, and the word greatest points to the largest divisor shared by all three counts. Students who spot a small shared factor can stop too soon, while those who think of a common multiple head the wrong way entirely. Checking that 72, 60 and 84 all divide by the entry, and that no larger number does, settles it.

  1. Turning words into math moderate Understanding the question

    Nothing names a greatest common factor; “each child receives an equal number of each item and there are no items remaining” means the class size must divide 72, 60 and 84 exactly. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Finding the greatest common factor of three numbers, by prime factorization or by listing factors, is a grade 6 method the question assumes rather than explains. (Anderson & Krathwohl, 2001)

Strategy: Write 72, 60 and 84 as products of primes, multiply the primes all three share, and check that each amount divides evenly by the result.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor
GCF=product of the shared prime factors\text{GCF} = \text{product of the shared prime factors}
A factor of a number divides it with nothing left over. The GCF is the greatest number that is a factor of every number in the group. A prime number has only 1 and itself as factors, like 2, 3, 5 and 7. Multiply the primes that all the numbers share, as many times as each number has them.

  1. Sort out the question

    The teacher has 72 crayons, 60 pencils and 84 sheets of paper. Every child gets the same number of each item, with none left over.

    So the number of children must divide 72, 60 and 84 with no remainder. A number like that is a common factor. You need the greatest one.

  2. Break each number into primes

    Write each number as a product of primes. A product is the answer to a multiplication.

    72=2×2×2×3×372 = 2 \times 2 \times 2 \times 3 \times 3

    60=2×2×3×560 = 2 \times 2 \times 3 \times 5

    84=2×2×3×784 = 2 \times 2 \times 3 \times 7

    You can check each one by multiplying it back out.

  3. Pick out the shared primes

    Look for the primes that appear in all three lists. Each list has at least two 2s and one 3. Only 72 has a third 2 and a second 3. The 5 and the 7 appear in one list each.

    GCF=2×2×3=12\text{GCF} = 2 \times 2 \times 3 = 12

    Watch out: keep every shared prime. Using only one 2 gives 6. Leaving out the 3 gives 4.

  4. Give the answer

    Watch out: 2,520 is the smallest number that all three numbers divide into. That is the least common multiple, a different idea. The class can’t have 2,520 children when there are only 60 pencils. So you would grid 12.

Check your answer

Share the items among 12 children. Each child gets 72÷12=672 \div 12 = 6 crayons, 60÷12=560 \div 12 = 5 pencils and 84÷12=784 \div 12 = 7 sheets. Nothing is left over. The numbers 6, 5 and 7 share no factor but 1, so no bigger class works.

More questions like this

The original DOE question
Question 60 as printed on page 191 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 60. Open page 191 of the handbook (PDF)

Question 80 of 120

2023 Test B Math, question 4

Question

Mr. Smith opened a retirement account with a deposit of $900. This account earns 5% simple interest annually. How many years will it take for his $900 deposit to earn $360 interest?

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

Show the answer

Correct answer: 8

Calculate simple interest (y) by multiplying the initial deposit (p) by the interest rate (r) by the number of years (t):

prt=yprt = y

900(0.05)t=360900(0.05)t = 360

45t=36045t = 360

t=8t = 8

Question analysis

Mr. Smith deposits $900 in a retirement account that earns 5% simple interest each year. This grid-in asks how many years it takes for the deposit to earn $360 in interest. Students find one year's interest and then how many years add up to the total, and the percent must become the right decimal.

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 grade 7 percent work, and a typical problem asks for the interest earned over a given time. This one runs the formula backward to find the time, one extra step, with friendly numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 7, 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

The numbers are friendly once one year’s interest is found, so most of the difficulty is setting up the right relationship. Students who divide $360 by 5 treat the rate as a dollar amount, and students who compare $360 with $900 stop at a percent. Multiplying back, eight years of $45, confirms the $360.

  1. Something you must already know minor Working it out

    Simple interest is the deposit times the rate times the years; the stem does not give this rule, and it must be known to see that each year adds the same $45. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The question gives the total interest, $360, and asks for the time, so the usual interest calculation has to be run backward, as 45t = 360. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Writing 5% as 0.05, not 0.5 or 0.005, is the one place-value decision, and a slip there changes the yearly interest by a power of ten. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work out the interest for one year, 5 percent of $900, then divide the $360 target by that yearly amount.

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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest
I=P×r×tI = P \times r \times t
I is the interest, the money the account earns. P is the deposit you start with. r is the yearly rate, written as a decimal. t is the number of years. With simple interest, every year earns the same amount, because it is always based on the first deposit.

  1. Sort out the question

    Mr. Smith puts $900 into an account. It earns 5% simple interest each year. Interest is extra money the bank pays you for keeping your money there. You need the number of years until the interest adds up to $360.

  2. Write 5% as a decimal

    Percent means “out of 100.” So 5% is 5 out of 100.

    5100=0.05\frac{5}{100} = 0.05

    Watch out: 5% is not 0.5 or 0.005. Using 0.005 makes a year’s interest $4.50, and that leads to 80 years.

  3. Find the interest for one year

    Each year the account earns 5% of $900.

    900×0.05=45900 \times 0.05 = 45

    So it earns $45 every year. Watch out: 5% is not $5 a year. Dividing 360 by 5 gives 72, which is wrong.

  4. Use the formula to find the years

    Put the numbers into the formula. I is 360, P is 900 and r is 0.05.

    I=P×r×tI = P \times r \times t

    360=900×0.05×t360 = 900 \times 0.05 \times t

    360=45×t360 = 45 \times t

    To undo “times 45,” divide both sides by 45.

    t=360÷45=8t = 360 \div 45 = 8

  5. Give the answer

    It takes 8 years. Watch out: $360 is 40% of $900. Gridding 40 stops too soon, because each year adds only 5%. So you would grid 8.

Check your answer

Add up the interest year by year. $45 a year for 8 years is 45×8=36045 \times 8 = 360 dollars. Also, 8 years at 5% a year is 40%, and 40% of $900 is $360.

More questions like this

The original DOE question
Question 61 as printed on page 192 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 61. Open page 192 of the handbook (PDF)

Question 81 of 120

2023 Test B Math, question 5

Question

Solve the equation for x:

(21)(14)x(49)(48)=0.875\frac{(21)(14)x}{(49)(48)} = 0.875

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

Show the answer

Correct answer: 7

(21)(14)x(49)(48)=0.875\frac{(21)(14)x}{(49)(48)} = 0.875

(2149)(1448)x=0.875\left(\frac{21}{49}\right)\left(\frac{14}{48}\right)x = 0.875

(37)(724)x=0.875\left(\frac{3}{7}\right)\left(\frac{7}{24}\right)x = 0.875

(18)x=0.875\left(\frac{1}{8}\right)x = 0.875

x=(0.875)(8)=7x = (0.875)(8) = 7

Question analysis

This equation puts x in a fraction with (21)(14) on top and (49)(48) below, set equal to the decimal 0.875. The grid-in asks for the value of x. Students do best to cancel common factors before multiplying anything out, and it helps to know the fraction that 0.875 equals.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.2c properties for multiplying and dividing
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Seventh graders solve equations like ax = b, usually with a simple coefficient. Here the coefficient is a product of four numbers that is best cancelled first, and a decimal has to be matched with a fraction.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 6 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-unknown equation dressed in large numbers, and the work is in reducing them without slipping. Students who cancel early find that the whole coefficient is one eighth and that 0.875 is seven eighths, which makes the last step mental. Those who multiply out face a four-digit product and a decimal division, and with nothing to check against, an early slip goes unnoticed.

  1. Error-prone arithmetic moderate Working it out

    Multiplied out, the coefficient is 294 over 2,352 and the right side is the decimal 0.875; both invite slips unless 21/49 and 14/48 are cancelled first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. No obvious method minor Working it out

    The quick route pairs 21 with 49 and 14 with 48 before multiplying, so that 3/7 × 7/24 collapses to 1/8; the layout of the fraction does not suggest that pairing. (Schoenfeld, 1985)

  3. An answer in an unexpected form minor Choosing the answer

    The result is entered with no choices, so a coefficient’s denominator or a quotient taken the wrong way looks as complete as the right value. (Enright, Morley & Sheehan, 2002)

Strategy: Cancel before multiplying: reduce 21 over 49 and 14 over 48, multiply the simplified fractions, then undo that coefficient, writing 0.875 as a fraction if it helps.

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).
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying fractions
ab×cd=a×cb×d\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}
a and c are the tops, or numerators. b and d are the bottoms, or denominators. Multiply the tops together and the bottoms together. It also works backward: you can split one fraction into two smaller ones. That lets you simplify before the numbers get big.

  1. Sort out the question

    You need the value of x. On top, 21, 14 and x are multiplied. The bottom is 49 times 48. The right side is 0.875.

    Multiplying 49 by 48 makes a big number. It is easier to simplify first.

  2. Split the fraction into two

    Use the rule for multiplying fractions backward. Pair 21 with 49, and 14 with 48.

    21×1449×48=2149×1448\frac{21 \times 14}{49 \times 48} = \frac{21}{49} \times \frac{14}{48}

    Now the equation says these two fractions times x equal 0.875.

  3. Simplify each fraction

    To simplify, divide the top and bottom by the same number. Both 21 and 49 divide by 7. Both 14 and 48 divide by 2.

    2149=371448=724\frac{21}{49} = \frac{3}{7} \qquad \frac{14}{48} = \frac{7}{24}

    Watch out: 48 ÷ 2 is 24, not 12. Writing 712\frac{7}{12} leads to 3.5.

  4. Multiply the simple fractions

    37×724=3×77×24=324=18\frac{3}{7} \times \frac{7}{24} = \frac{3 \times 7}{7 \times 24} = \frac{3}{24} = \frac{1}{8}

    The 7 on top and the 7 on the bottom cancel, since 7 ÷ 7 = 1. Then divide 3 and 24 by 3 to get 18\frac{1}{8}. So the equation becomes:

    18×x=0.875\frac{1}{8} \times x = 0.875

  5. Solve for x

    One eighth of x is 0.875. To undo dividing by 8, multiply both sides by 8.

    x=0.875×8=7x = 0.875 \times 8 = 7

    Watch out: 8 is not the answer, and don’t divide 0.875 by 8. So you would grid 7.

Check your answer

0.875 is the same as 78\frac{7}{8}. One eighth of 7 is 78\frac{7}{8}, so x = 7 works. You can also multiply it all out: 21×14×7=2,05821 \times 14 \times 7 = 2{,}058 and 49×48=2,35249 \times 48 = 2{,}352. Then 2,058÷2,352=0.8752{,}058 \div 2{,}352 = 0.875.

More questions like this

The original DOE question
Question 62 as printed on page 192 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 62. Open page 192 of the handbook (PDF)

Question 82 of 120

2023 Test B Math, question 12

Question

The perimeter of a 12-sided polygon is 100 centimeters. The length of one side is 25 centimeters. The length of another side is 35 centimeters. The remaining sides are equal in length to one another. What is the length of each of the remaining sides, in centimeters?

  • A4
  • B5
  • C20
  • D40
Show the answer

Correct answer: A

Let x represent the length of one of the 10 remaining sides:

25+35+10x=10025 + 35 + 10x = 100

60+10x=10060 + 10x = 100

10x=4010x = 40

x=4x = 4

Question analysis

A 12-sided polygon has a perimeter of 100 centimeters: one side is 25 cm long, another is 35 cm, and the rest are all equal to one another. Students find the length of each remaining side. This perimeter problem turns into a two-step equation, and the tricky part is counting the sides that are left.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving a two-step equation from a short description is standard grade 7 work. The perimeter setting is simple; the only extra care is counting the sides that remain after the two given ones.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 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 perimeter and the two named sides make the subtraction obvious, so the real decision comes after it: what to divide the leftover 40 centimeters by. Students who stop at the leftover length, or who lose track of how many sides remain, reach a listed choice. Writing 25 + 35 + 10x = 100 keeps the count of 10 in view.

  1. Turning words into math minor Understanding the question

    “The remaining sides are equal in length to one another” hides the count: 12 sides minus the two given ones leaves 10 equal sides to share the rest of the perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Subtracting 25 and 35 from 100, counting the sides left, and dividing are three steps, and the first result, 40, is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Subtract the two known sides from the perimeter, count how many sides are left, and share the remaining length equally among them.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a polygon
P=the sum of all the side lengthsP = \text{the sum of all the side lengths}
P is the perimeter, the total distance around a shape. A polygon is a flat shape with straight sides. A sum is the answer to an addition. So add the lengths of every side to get P.

  1. Sort out the question

    The shape is a polygon with 12 sides. Its perimeter is 100 centimeters. One side is 25 centimeters, and another is 35 centimeters. The other sides all have the same length. You need that length.

  2. Count the equal sides

    Two of the 12 sides are already known. The rest are the equal sides.

    12−2=1012 - 2 = 10

    So there are 10 equal sides. Call the length of each one x.

  3. Write an equation for the perimeter

    Use the perimeter formula. All 12 sides add up to 100. Ten sides of length x add up to 10x, which means 10 times x.

    P=the sum of all the side lengthsP = \text{the sum of all the side lengths}

    25+35+10x=10025 + 35 + 10x = 100

  4. Solve for x

    60+10x=10060 + 10x = 100

    10x=100−60=4010x = 100 - 60 = 40

    The 10 equal sides add up to 40 centimeters. Watch out: 40 is all ten sides together, not one side. Stopping here gives choice D. Now share the 40 among the 10 sides.

    x=40÷10=4x = 40 \div 10 = 4

  5. Give the answer

    Each remaining side is 4 centimeters long. Watch out: sharing the 40 centimeters between only 2 sides gives 20, choice C. That treats the shape as if it had 4 sides, but it has 12. So the answer is A, 4.

Check your answer

Add all 12 sides: 25+35+10×4=25+35+40=10025 + 35 + 10 \times 4 = 25 + 35 + 40 = 100. That is the perimeter the question gives, so 4 centimeters is right.

More questions like this

The original DOE question
Question 69 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 69. Open page 194 of the handbook (PDF)

Question 83 of 120

2023 Test B Math, question 14

Question

Each number in a sequence is 3 more than twice the number that comes just before it. If 93 is a number in the sequence, what number comes just before it?

  • A92
  • B90
  • C48
  • D45
Show the answer

Correct answer: D

Let n represent a number in the sequence, and let x represent the number that comes just before n:

n=3+2xn = 3 + 2x

93=3+2x93 = 3 + 2x

90=2x90 = 2x

45=x45 = x

Question analysis

In a sequence, each number is 3 more than twice the number just before it, and 93 is one of the numbers. Students find the number that comes just before 93. That means running the rule backward, undoing the adding and the doubling in reverse 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 Grade 7 students routinely solve two-step equations. The demand here is translating the wordy sequence rule into one, or running the rule backward in the right order, rather than any harder content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 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 arithmetic is two short steps, so the question rests on reading a wordy rule correctly and running it in reverse. Each wrong choice marks a different place to go wrong: ignoring the rule, stopping after the subtraction, or undoing the adding with more adding. Checking the answer forward, twice the number plus 3, confirms it reaches 93.

  1. Turning words into math moderate Understanding the question

    “Each number … is 3 more than twice the number that comes just before it” must become 2x + 3 = 93, with x the earlier number, not the later one. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method minor Working it out

    The rule is stated forward, from a number to the next, but the question runs backward, so each operation has to be undone in reverse order: subtract 3, then halve. (Schoenfeld, 1985)

Strategy: Call the earlier number x, write the rule as an expression in x set equal to 93, and solve it by undoing each operation in reverse order.

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.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule
n=2x+3n = 2x + 3
x is any number in the sequence, a list that follows a rule. n is the number right after it. Twice means times 2, so twice x is 2x. Then 3 more means add 3. The same rule links every number to the one after it.

  1. Sort out the question

    A sequence is a list of numbers that follows a rule. Here, each number is “3 more than twice the number that comes just before it.” 93 is in the list. You need the number right before 93. Call it x.

  2. Write the rule as an equation

    Twice x is 2x. Three more than that is 2x + 3. The rule says this equals the next number, which is 93.

    n=2x+3n = 2x + 3

    93=2x+393 = 2x + 3

  3. Undo the adding of 3

    The rule doubles x first, then adds 3. To go backward, undo the steps in reverse order. First, subtract 3 from both sides.

    93−3=2x93 - 3 = 2x

    90=2x90 = 2x

    Watch out: 90 is twice the number, not the number itself. Stopping here gives choice B. And don’t add 3, which leads to 48, choice C.

  4. Undo the doubling

    Now divide both sides by 2.

    x=90÷2=45x = 90 \div 2 = 45

    Watch out: “just before” means just before in this list. It does not mean one less than 93, so 92 is wrong. So the answer is D, 45.

Check your answer

Run the rule forward from 45. Twice 45 is 90, and 3 more is 93: 2×45+3=932 \times 45 + 3 = 93. So 45 really does come just before 93.

More questions like this

The original DOE question
Question 71 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 71. Open page 195 of the handbook (PDF)

Question 84 of 120

2023 Test B Math, question 18

Question

A rectangle whose bottom side is labeled 16 ft and whose left side is labeled 8 ft. It is shaded gray except for a white triangle inside it: the triangle’s vertex is at the middle of the left side, and its base lies along the right side and is labeled 4 ft.

The diagram above shows a rectangle with an inscribed triangle. The triangle’s vertex and base touch the sides of the rectangle as shown. What is the total area of the shaded portion?

  • A32 sq ft
  • B64 sq ft
  • C96 sq ft
  • D128 sq ft
Show the answer

Correct answer: C

First, find the area of the rectangle, in square feet:

16×8=12816 \times 8 = 128

Then, find the area of the triangle:

12×4×16=32\frac{1}{2} \times 4 \times 16 = 32

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle:

128−32=96 sq ft128 - 32 = 96\text{ sq ft}

Question analysis

A 16-foot by 8-foot rectangle has a white triangle inside it, with one vertex at the middle of the left side and a 4-foot base along the right side. Students find the total area of the shaded part. They subtract the triangle's area, whose height is not labeled and must be read from the figure.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Composite areas built from rectangles and triangles appear in grades 6 and 7. This one is typical, but the triangle lies on its side, so students must see which length is its height rather than read it from a label.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 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 shaded region is a rectangle with a triangle cut out, so the plan is standard, but the triangle is turned so that its height runs across the figure. Students who see that the height is the 16-foot side have two short areas to find and one subtraction. The listed wrong answers stop at either area or leave out the one half.

  1. Reading a figure, table or graph moderate Understanding the question

    The triangle lies on its side: its 4-foot base runs along the right edge, so its height is the rectangle’s full 16-foot length, which the figure labels only as the bottom side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    The rectangle’s area, the triangle’s area and their difference are three separate results, and the first two, 128 and 32, are themselves listed choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the rectangle's area, find the triangle's area from its base on the right side and its height across the rectangle, then subtract.

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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle
A=l×wA = l \times w
A is the area, the space inside a flat shape, in square units. l is the length and w is the width. It works because the rectangle is filled by w rows of unit squares, with l squares in each row.
Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side you measure from. h is the height, the distance from the base to the opposite 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.

  1. Sort out the figure

    The figure is a rectangle 16 feet long and 8 feet tall. Inside it is a white triangle. Its tip touches the middle of the left side. Its base lies along the right side and is 4 feet long. The shaded part is the gray area around the triangle. You need its area.

    Plan: find the rectangle’s area, then take away the triangle’s area.

  2. Find the area of the rectangle

    A=l×wA = l \times w

    A=16×8=128A = 16 \times 8 = 128

    The rectangle’s area is 128 square feet. Watch out: this includes the white triangle. Stopping here gives choice D.

  3. Find the triangle’s height

    The triangle lies on its side. Its base is the 4-foot piece on the right side, which runs up and down. The height must meet the base at a right angle, a square corner. So the height runs straight across, from the base to the tip on the left side. That is the full length of the rectangle, 16 feet.

  4. Find the area of the triangle

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×4×16=32A = \frac{1}{2} \times 4 \times 16 = 32

    The triangle’s area is 32 square feet. Watch out: don’t forget the 12\frac{1}{2}. Using 4×16=644 \times 16 = 64 leaves 64 for the shaded part, choice B.

  5. Subtract to find the shaded area

    128−32=96128 - 32 = 96

    The shaded part is 96 square feet. Watch out: 32 is the white triangle, not the shaded part, so choice A is wrong. So the answer is C, 96 sq ft.

Check your answer

The triangle’s base, 4 feet, is half the rectangle’s height of 8 feet. Its height is the full 16 feet. So the triangle is half of half the rectangle, or one fourth of it. One fourth of 128 is 32, and 128−32=96128 - 32 = 96.

More questions like this

The original DOE question
Question 75 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 75. Open page 196 of the handbook (PDF)

Question 85 of 120

2023 Test B Math, question 23

Question

FRUIT SOLD
AT STORE XYZ

Fruit Number Sold
Apples 8
Bananas 6
Cherries 12
Oranges 6
Peaches ?
Total 50

Peaches made up what percentage of the total number of pieces of fruit sold at Store XYZ, according to the table above?

  • A18%
  • B28%
  • C36%
  • D64%
Show the answer

Correct answer: C

First, find the number of peaches sold:

50−(8+6+12+6)=50 - (8 + 6 + 12 + 6) =

50−32=1850 - 32 = 18

Then, find the percentage:

1850=0.36=36%\frac{18}{50} = 0.36 = 36\%

Question analysis

A table lists the fruit sold at Store XYZ, apples, bananas, cherries and oranges, with the number of peaches shown as a question mark and a total of 50. Students find what percentage of the fruit sold were peaches. They first work out the missing count, then turn it into a percent of the total.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a missing value and writing a part as a percent of a whole are grade 6 skills. With a total of 50, the percent step is quick, so the task is lighter than typical grade 7 percent problems.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 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 total of 50 makes the percent step easy, since each piece of fruit is 2 percent, but the question first hides the count it needs. Students who find the 18 peaches and stop, or who give the share of the four listed fruits, reach listed choices. Checking that the peaches’ percent and the other fruits’ percent add to 100 confirms the answer.

  1. Reading a figure, table or graph minor Understanding the question

    The table gives four counts, a question mark for peaches and a total of 50, so the missing count must be found by subtraction before any percent. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Adding 8, 6, 12 and 6, subtracting from 50, and turning 18 out of 50 into a percent are three steps, and the count of 18 is itself a listed percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the four known counts, subtract from 50 to get the peaches, then double that count, since each piece out of 50 is 2 percent.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a total
percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100
The part is the amount you are asking about, here the peaches. The whole is the total, here all 50 pieces of fruit. Dividing gives the fraction of the whole. Multiplying by 100 turns it into a percent, which means out of 100.

  1. Sort out the question

    The table shows how many of each fruit Store XYZ sold. The number of peaches is missing, shown by a question mark. The total is 50 pieces of fruit. You need the percent of the total that were peaches.

  2. Add the fruits you know

    8+6+12+6=328 + 6 + 12 + 6 = 32

    Apples, bananas, cherries and oranges add up to 32 pieces of fruit. That is every fruit except the peaches.

  3. Find the number of peaches

    The peaches are whatever is left of the 50.

    50−32=1850 - 32 = 18

    So 18 peaches were sold. Watch out: 18 is a count, not a percent. Choosing 18% is choice A.

  4. Turn the count into a percent

    percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100

    1850×100=0.36×100=36\frac{18}{50} \times 100 = 0.36 \times 100 = 36

    Peaches were 36% of the fruit sold. Watch out: 32 out of 50 is 64%, choice D. That is the other four fruits, not the peaches. So the answer is C, 36%.

Check your answer

Out of 50 pieces, each piece is 2% of the total, because 100÷50=2100 \div 50 = 2. So 18 peaches are 18×2=3618 \times 2 = 36 percent. Also, 36% for peaches plus 64% for the rest makes 100%.

More questions like this

The original DOE question
Question 80 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 2024 Admissions, page 198, question 80. Open page 198 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 86 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:

p50=1,500\frac{p}{50} = 1{,}500

Solve for p:

p=1,500(50)=75,000p = 1{,}500(50) = 75{,}000

The population decrease from 2005 to 2006 is 84,000−75,000=9,00084{,}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
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.
What makes it hard
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Population per square mile
people per square mile=populationarea\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.

  1. 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.

  2. Find the 2006 population

    Let p be the 2006 population. Use the formula.

    people per square mile=populationarea\text{people per square mile} = \frac{\text{population}}{\text{area}}

    1,500=p501{,}500 = \frac{p}{50}

    To undo dividing by 50, multiply both sides by 50.

    p=1,500×50=75,000p = 1{,}500 \times 50 = 75{,}000

    So 75,000 people lived there in 2006.

  3. Find the decrease

    A decrease is how much a number went down. Subtract the 2006 population from the 2005 population.

    84,000−75,000=9,00084{,}000 - 75{,}000 = 9{,}000

    The population went down by 9,000 people.

  4. 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÷50=1,68084{,}000 \div 50 = 1{,}680 people per square mile. In 2006, there were 1,500. That is 180 fewer per square mile, and 180×50=9,000180 \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 87 of 120

2023 Test B Math, question 33

Question

A line segment with point W at its left end and point Y at its right end. Point X lies on the segment between W and Y, closer to Y.

The distance from W to X is 70% of the distance from W to Y. If the distance from X to Y is 15 miles, what is the distance, in miles, from W to X?

  • A10.5
  • B35
  • C45
  • D50
Show the answer

Correct answer: B

Since the distance from W to X is 70% of the distance from W to Y, the distance from X to Y is 30% of the distance from W to Y. Let d represent the distance from W to Y. Then 15 miles is 30% of d.

0.30d=150.30d=15

d=150.30=50d=\frac{15}{0.30}=50

So, the distance from W to X, in miles, is 50−15=3550-15=35.

Question analysis

A percent question with a line segment: point X lies between W and Y, the distance from W to X is 70% of the whole, and the distance from X to Y is 15 miles. Students find the distance from W to X. Matching the 15 miles to the right percent before finding the whole is the step students most often skip.

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 Grade 7 percent problems often give a part and its percent and ask for the whole. This one gives the distance for the part the percent does not describe and asks for the other part, so the whole has to be found on the way.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 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 percent and the given distance describe different pieces of the road, which is the whole difficulty. Students who apply the 70% straight to the 15 miles get a small number fast, and students who solve correctly for the whole of WY must still remember to take XY away. Checking that 35 is 70% of 50 confirms the answer and catches both slips.

  1. Turning words into math moderate Understanding the question

    The 70% describes WX, but the 15 miles given is XY, so the sentence has to be recast as “XY is 30% of WY” before the 15 miles can be used. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Finding the complement, 30%, then the whole, 15 ÷ 0.30 = 50, then the part, 50 − 15, gives three results, and the middle one is itself a choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out what percent of WY the 15-mile stretch from X to Y must be, use it to find the whole of WY, then find the part from W to X.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}
The whole is the full amount. The part is a piece of it. The percent tells how many hundredths of the whole the part is. For example, 30% means 30 out of every 100, or 0.30. Use it whenever a percent describes a piece of a whole.

  1. Sort out the distances

    Point X sits between W and Y. So the trip from W to Y has two pieces: W to X, then X to Y.

    W to X is 70% of the whole distance from W to Y. X to Y is 15 miles. You need the distance from W to X, in miles.

  2. Find the percent for X to Y

    The whole distance from W to Y is 100%. W to X uses 70% of it. The rest is X to Y.

    100−70=30100 - 70 = 30

    So the 15 miles from X to Y is 30% of the whole distance.

  3. Find the whole distance

    Let d be the distance from W to Y. Use the percent formula. The part is 15, and the percent is 30.

    15=30100×d=0.30×d15 = \frac{30}{100} \times d = 0.30 \times d

    Divide both sides by 0.30 to get d by itself.

    d=15÷0.30=50d = 15 \div 0.30 = 50

    The distance from W to Y is 50 miles.

  4. Find the distance from W to X

    Take the 15 miles from X to Y away from the whole.

    50−15=3550 - 15 = 35

    W to X is 35 miles.

    Watch out: 50 miles is the whole distance from W to Y, choice D. You still have to subtract. Also, the 70% is of the whole distance, not of the 15 miles. Taking 70% of 15 gives 10.5, choice A.

    So the answer is B, 35 miles.

Check your answer

Is 35 really 70% of 50? 0.70×50=350.70 \times 50 = 35. Yes. And 35 + 15 = 50, the whole distance. The figure also shows X closer to Y. So W to X should be longer than 15 miles, and it is.

More questions like this

The original DOE question
Question 90 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 90. 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 88 of 120

2023 Test B Math, question 34

Question

There are 3 different highways from City W to City X, 4 different highways from City X to City Y, and 3 different highways from City Y to City Z. How many different routes are there for a trip from City W to City X to City Y to City Z?

  • A10
  • B12
  • C24
  • D36
Show the answer

Correct answer: D

Multiply the numbers of highways between each pair of cities:

3×4×3=363\times 4\times 3=36

Question analysis

A counting question about highways: there are 3 highways from City W to City X, 4 from X to Y and 3 from Y to Z. Students count the different routes from W to Z through X and Y. It tests the counting principle, and deciding whether to add or multiply the numbers of highways is the key.

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 Grade 7 students build tree diagrams and lists for compound events. This item needs only the counting principle across three stages, with no restriction on the choices, which makes it shorter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 11, 48 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 a route is seen as one highway on each of three stretches, the count is a single product, 3 × 4 × 3. The wrong choices come from adding the counts, from stopping after two stretches, and from adding two-stretch products. A small tree for the first two stretches shows why the numbers multiply: each of the 3 highways to City X branches into 4.

  1. Something you must already know minor Working it out

    The counting principle, multiplying the 3, 4 and 3 highway choices because each route uses one highway on every stretch, is assumed; the stem never says to multiply. (Anderson & Krathwohl, 2001)

Strategy: A route uses one highway on each of the three stretches, so count the choices for each stretch and multiply; a small tree for two stretches shows why.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle
total ways=a×b×c\text{total ways} = a \times b \times c
a, b and c are the numbers of choices for the first, second and third parts of a task. Every choice for one part can go with every choice for the others. So you multiply. Use it when you make one choice at each stage and want to count all the ways.

  1. Sort out the trip

    A route goes from City W to X, then to Y, then to Z. It uses exactly one highway on each of three stretches:

    • W to X: 3 highways
    • X to Y: 4 highways
    • Y to Z: 3 highways

    You need to count all the different routes.

  2. Count routes from W to Y

    Start small. Pick any of the 3 highways to City X. From there, you can take any of the 4 highways to City Y. So each first highway leads to 4 routes.

    3×4=123 \times 4 = 12

    There are 12 routes from City W to City Y.

  3. Add the last stretch

    Each of those 12 routes can end with any of the 3 highways to City Z. So multiply again. This is the counting principle with all three stretches.

    3×4×3=12×3=363 \times 4 \times 3 = 12 \times 3 = 36

    There are 36 routes in all.

  4. Avoid the common traps

    Watch out: Adding 3 + 4 + 3 = 10 counts highways, not routes. That gives choice A. Stopping at 12 forgets the last stretch, choice B. Adding 12 + 12 = 24 mixes two partial counts, choice C. A route is built by choosing one highway after another, so you multiply.

    So the answer is D, 36.

Check your answer

Picture one highway from W to X. After it, there are 4 × 3 = 12 ways to finish the trip. Each of the 3 highways from W to X gives 12 ways. That makes 12 + 12 + 12 = 36 routes.

More questions like this

The original DOE question
Question 91 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 91. Open page 201 of the handbook (PDF)

Question 89 of 120

2023 Test B Math, question 35

Question

In Center City, the ratio of people to dogs is 4:14:1, and the ratio of dogs to cats is 3:23:2. What is the ratio of people to cats?

  • A2:12:1
  • B7:37:3
  • C6:16:1
  • D12:112:1
Show the answer

Correct answer: C

Take each ratio and solve for dogs:

pd=41\frac{p}{d}=\frac{4}{1}

4d=p4d=p

d=14pd=\frac{1}{4}p

And

dc=32\frac{d}{c}=\frac{3}{2}

d=32cd=\frac{3}{2}c

Set the two expressions for d equal to each other and solve for pc\frac{p}{c}:

14p=32c\frac{1}{4}p=\frac{3}{2}c

p=6cp=6c

pc=61\frac{p}{c}=\frac{6}{1}

So, the ratio of people to cats is 6:16:1.

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.
What makes it hard
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent ratios
a:b=(a×k):(b×k)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.

  1. 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.

  2. 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=(4×3):(1×3)=12:34 : 1 = (4 \times 3) : (1 \times 3) = 12 : 3

    Now there are 12 people for every 3 dogs.

  3. Join the two ratios

    Both ratios now have 3 dogs. So you can write all three groups in one ratio.

    people:dogs:cats=12:3:2\text{people} : \text{dogs} : \text{cats} = 12 : 3 : 2

    For every 12 people, there are 3 dogs and 2 cats.

  4. Read off and simplify

    People to cats is 12 : 2. Divide both parts by 2.

    12:2=(12÷2):(2÷2)=6:112 : 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 90 of 120

2023 Test B Math, question 42

Question

1 gallon = 16 cups

1 cup = 8 fluid ounces

If 1 gallon = 2x2^{x} fluid ounces, what is the value of x?

  • A4
  • B7
  • C8
  • D64
Show the answer

Correct answer: B

If 1 gallon is equal to 16 cups, and 1 cup is equal to 8 fluid ounces, then

1 gallon=1\text{ gallon}=

(16)(8 fluid ounces)=(16)(8\text{ fluid ounces})=

128 fluid ounces128\text{ fluid ounces}

128=27128=2^{7}

1 gallon=27 fluid ounces1\text{ gallon}=2^{7}\text{ fluid ounces}

So, x=7x=7.

Question analysis

This question gives two facts, 16 cups in a gallon and 8 fluid ounces in a cup, and says a gallon equals 2 to the power x fluid ounces. Students find the exponent x. It combines a unit conversion with writing a number as a power of 2, and the exponent is easy to confuse with a multiplier.

NYS standards
  • 6.RP.3d converting measurement units
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Converting units with given facts is grade 6 work, and this conversion is a single multiplication. The twist is writing the result as a power of 2, which calls for fluency with powers that grade 7 work rarely asks for.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 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 conversion is one multiplication, so the question really tests whether students can name the exponent for 128. Stopping at the cups, where 16 = 2⁴, or treating 2 to the power x as 2 times x gives a listed choice. Writing 16 as 2⁴ and 8 as 2³ and adding the exponents is the quickest route, and doubling from 1 seven times is a sure check.

  1. Combines ideas minor Working it out

    A unit conversion, 16 cups × 8 fluid ounces = 128 fluid ounces, has to be joined to exponents by recognizing 128 as a power of 2. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Something you must already know minor Working it out

    Students must know or build the powers of 2 up to 128, and see that the exponent counts factors of 2, not the number of terms in a list starting at 1. (Anderson & Krathwohl, 2001)

Strategy: Convert 1 gallon to fluid ounces by multiplying the two facts, then keep halving the result, or keep doubling from 2, counting how many factors of 2 it takes.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • 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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Meaning of an exponent
2x=2×2×⋯×22^x = 2 \times 2 \times \cdots \times 2
The 2 is the base, the number being multiplied. x is the exponent, the small raised number. It tells how many 2s are multiplied together. For example, 2 to the third power means 2 × 2 × 2, which is 8.
Multiplying powers of the same base
2a×2b=2a+b2^a \times 2^b = 2^{a + b}
a and b are exponents. The first power has a twos multiplied together. The second has b more twos. Multiplying puts all the twos together, so you add the exponents.

  1. Sort out the question

    A gallon is 16 cups. A cup is 8 fluid ounces. A gallon is also 2x2^x fluid ounces. You need x, the exponent. It tells how many 2s are multiplied together.

  2. Change a gallon to fluid ounces

    Each of the 16 cups holds 8 fluid ounces. So multiply.

    16×8=12816 \times 8 = 128

    One gallon is 128 fluid ounces.

  3. Write 128 as a power of 2

    Now you need 2x=1282^x = 128. Keep multiplying by 2, and count the 2s.

    Number of 2s1234567
    Product248163264128

    It takes seven 2s to make 128.

    27=1282^7 = 128

    So x = 7.

  4. Give the answer

    Watch out: A gallon is 16 cups, and 16=2416 = 2^4. Stopping there gives 4, choice A. And 2x2^x does not mean 2 times x. Solving 2 × x = 128 gives 64, choice D.

    So the answer is B, 7.

Check your answer

Use the exponent rule. 16 is 242^4 and 8 is 232^3. So 16×8=24×23=24+3=2716 \times 8 = 2^4 \times 2^3 = 2^{4 + 3} = 2^7. The exponent is 7 again.

More questions like this

The original DOE question
Question 99 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 99. Open page 203 of the handbook (PDF)

Question 91 of 120

2018 Test A Math, question 1

Question

Parallelogram PQRS with side PS on a horizontal line that extends past P and S, and side QR at the top. At S, the angle between side SR and the line extended to the right of S is marked 72°. At Q, a vertical line rises above the figure at a right angle to side QR (right-angle mark), and angle x° is marked on the left of Q, between that vertical line and side QP.

In the figure above, PQRS is a parallelogram. What is the value of xx?

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

Show the answer

Correct answer: 162

First, find the measure of angle PQR. The measure of angle PQR is equal to the measure of angle PSR.

m∠PSR = 180 − 72 = 108.

So, the measure of angle PQR is also 108.

108 + 90 + x = 360

198 + x = 360

x = 162

Question analysis

A grid-in geometry question about angles around a parallelogram PQRS. A 72° angle sits outside corner S on the extended base, and at corner Q a right angle and the unknown angle x complete the turn around the point. Students chain three angle facts, and the angle asked for lies outside the shape, not inside it.

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
Above grade 7 Grade 7 angle work usually applies one or two facts, such as supplementary or vertical angles, to find an angle. This chains three, and needs the equal opposite angles of a parallelogram, which rest on grade 8 parallel-line facts.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 7, 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 arithmetic is two subtractions; the difficulty is seeing which angles the marks refer to. Students lose points by treating the 72° as the parallelogram's own angle at S, or by stopping at angle PQR, the most natural number to grid. The question rewards tracing a route from S to Q before computing anything, and checking at the end that x is the angle outside the shape that the arc marks.

  1. Reading a figure, table or graph major Understanding the question

    The 72° is marked outside the parallelogram, between side SR and the base extended past S, and x sits outside corner Q, between QP and a vertical ray that meets QR at a right-angle mark. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Three angle facts have to be chained: a straight line gives the angle inside S, opposite angles of a parallelogram carry it to Q, and the angles around Q make a full 360° turn. (Embretson, 1998; Daniel & Embretson, 2010)

  3. An answer in an unexpected form minor Choosing the answer

    As a grid-in there are no choices to check against, so a student who stops at angle PQR, inside the parallelogram, gets no signal that the job is unfinished. (Enright, Morley & Sheehan, 2002)

Strategy: Use the straight line at S to find the parallelogram's angle there, carry it to the opposite corner Q, then take that angle and the right angle away from 360.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line
a+b=180a + b = 180
a and b are two angles side by side on a straight line. Together they make a half turn, so they always add up to 180 degrees. Use it when an angle you need shares a straight line with an angle you know.
Opposite angles of a parallelogram
∠PQR=∠PSR\angle PQR = \angle PSR
The symbol ∠ means angle. ∠PQR is the angle at corner Q, between sides QP and QR. ∠PSR is the angle at the opposite corner, S. In a parallelogram, opposite corners always have equal angles, because the shape leans the same way at both.
Angles around a point
a+b+c=360a + b + c = 360
a, b and c are angles that fit around one point with no gaps and no overlaps. Going all the way around is a full turn, which is 360 degrees. So the angles add up to 360.

  1. Sort out what the figure shows

    PQRS is a parallelogram. That is a four-sided shape whose opposite sides are parallel, which means they never meet. You need the value of x, in degrees.

    Look closely at the marks. The 72° angle is outside the shape, at S. It sits between side SR and the base line, which keeps going past S. At Q, a vertical line meets side QR. The small square means a right angle, 90°. The angle x is outside the shape too. It runs from that vertical line around to side QP.

  2. Find the angle inside corner S

    Angle PSR is the shape’s own angle at S. It sits right next to the 72° angle, on the same straight base line. So the two angles add up to 180°.

    a+b=180a + b = 180

    ∠PSR+72=180\angle PSR + 72 = 180

    ∠PSR=180−72=108\angle PSR = 180 - 72 = 108

    So angle PSR is 108°. Watch out: 72° is not the inside angle at S. Using 72 as the inside angle leads to the wrong answer 198.

  3. Carry the angle over to Q

    Corners Q and S are opposite each other. In a parallelogram, opposite angles are equal. So angle PQR matches angle PSR.

    ∠PQR=∠PSR=108\angle PQR = \angle PSR = 108

    So angle PQR is 108°. Watch out: this is not x. Angle PQR is inside the shape, and x is outside it. Stopping here and gridding 108 is the most common mistake.

  4. Add the angles around Q

    Three angles fill the space all the way around point Q. They are the right angle (90°), angle PQR (108°) and x. Together they make a full turn.

    a+b+c=360a + b + c = 360

    90+108+x=36090 + 108 + x = 360

    198+x=360198 + x = 360

    Watch out: don’t skip the right angle at Q. Leaving out the 90 gives 252.

  5. Solve for x

    Subtract 198 from 360 to find what is left for x.

    x=360−198=162x = 360 - 198 = 162

    So x is 162°. So you would grid 162.

Check your answer

Add the three angles at Q: 90+108+162=36090 + 108 + 162 = 360. That is a full turn. Also, the arc for x is wider than a right angle but not as wide as a straight line. So x should be between 90° and 180°, and 162° is.

More questions like this

The original DOE question
Question 58 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 58. Open page 81 of the handbook (PDF)

Question 92 of 120

2018 Test A Math, question 2

Question

The owner of a tree farm plants pine trees and oak trees in a ratio of 8:38:3. How many oak trees are planted if 264 pine trees are planted?

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

Show the answer

Correct answer: 99

Let xx be the number of oak trees when 264 pine trees are planted. Set up a proportion and solve for xx:

x264=38\frac{x}{264}=\frac{3}{8}

8x=7928x=792

x=99x=99

Question analysis

A ratio word problem about a tree farm that plants pine and oak trees in a fixed ratio of 8 to 3. Given how many pine trees are planted, students find the matching number of oak trees and grid it in. The main risk is pairing each number with the wrong kind of tree.

NYS standards
  • 6.RP.3a tables of equivalent ratios
  • 6.RP.1 ratio concepts and language
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 This is a one-ratio proportion with whole numbers, the kind of problem grade 6 students solve with equivalent ratios. A typical grade 7 task adds rates with fractions or a multistep percent; this has neither.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 4, 28 words.
Corrected from the DOE handbook
  • An arithmetic slip 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

This is a one-ratio scaling problem with friendly numbers: 264 splits into 8 equal parts of 33. The only real risk is matching 264 to the wrong term of 8:3, and a quick sense check catches it, since oak trees are the smaller group and must number fewer than 264. Students who grid the one-part value or the total of both kinds of tree have answered a question that was not asked.

  1. Turning words into math minor Understanding the question

    The ratio 8:3 lists pine first, so the 264 pine trees match the 8 and the oak count matches the 3; reading the terms in the other order flips the setup. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. An answer in an unexpected form minor Choosing the answer

    As a grid-in it offers no choices, so an upside-down proportion that gives 704 oak trees, far more than the 264 pines, has to be caught by the student's own check. (Enright, Morley & Sheehan, 2002)

Strategy: Divide the number of pine trees by the pine part of the ratio to get the size of one part, then multiply by the oak part.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook goes from x/264 = 3/8 to “8x = 762”. Cross-multiplying gives 8x = 3 × 264 = 792, so that line now reads 8x = 792. The handbook’s final answer, x = 99, was already right (792 ÷ 8 = 99).

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a and b are the two numbers in a ratio, written as a fraction. n is any number you choose, except zero. Multiplying both numbers by the same n gives an equal ratio. This is how you scale a small ratio up to real counts.

  1. Sort out the ratio

    A ratio compares two amounts. Here the ratio of pine trees to oak trees is 8:3. Pine comes first, so pine goes with the 8. Oak goes with the 3. So for every 8 pine trees, there are 3 oak trees.

    The farm planted 264 pine trees. You need the number of oak trees.

  2. Find the size of one part

    Think of the trees in equal groups, called parts. Pine trees fill 8 parts, and oak trees fill 3 parts. The 264 pine trees are the 8 parts, so divide by 8.

    264÷8=33264 \div 8 = 33

    Each part is 33 trees. Watch out: 33 is one part, not the number of oak trees. Don’t stop here.

  3. Find the number of oak trees

    Oak trees fill 3 parts. Multiply the part size by 3.

    3×33=993 \times 33 = 99

    You can see the same thing with equal ratios. Multiply both numbers of 8:3 by 33.

    83=8×333×33=26499\frac{8}{3} = \frac{8 \times 33}{3 \times 33} = \frac{264}{99}

    So 264 pine trees go with 99 oak trees.

  4. Avoid the traps and answer

    Watch out: if you match 264 to the 3 instead of the 8, you flip the ratio. That gives 704 oak trees. Watch out: the question asks only for oak trees. Adding the 264 pines gives 363. So you would grid 99.

Check your answer

The ratio 8:3 says there are fewer oak trees than pine trees. 99 is less than 264, so that fits. Also, divide both counts by 33: 264÷33=8264 \div 33 = 8 and 99÷33=399 \div 33 = 3. That gives back 8:3.

More questions like this

The original DOE question
Question 59 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 59. Open page 81 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 93 of 120

2018 Test A Math, question 3

Question

For what value of ww is 4w=2w−84w=2w-8?

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

Show the answer

Correct answer: -4

4w=2w−84w=2w-8

2w=−82w=-8

w=−4w=-4

Question analysis

A one-line algebra grid-in: find the value of w that makes 4w equal to 2w minus 8. Students solve a linear equation with the variable on both sides by collecting the w terms and dividing. Errors come from handling the subtraction sign or stopping before w stands alone.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.NS.2b rational number division
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 Equations with the variable on both sides are formally a grade 8 topic, but this one takes two short steps with whole numbers. It asks about as much as a typical grade 7 two-step equation.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 7, 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 of the easiest questions on the test: collect the w terms on one side and divide. Errors come from carelessness rather than method, such as dropping the negative sign or gridding −8 after the first step because it already looks like a finished answer. Substituting the result back into both sides of the equation takes seconds and catches both.

  1. Error-prone arithmetic minor Working it out

    The only constant is −8, so the answer is negative; dividing −8 by the coefficient 2 is the step where the sign is most often lost. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. An answer in an unexpected form minor Choosing the answer

    The grid-in needs the negative sign filled in its own column, and there are no choices to show that the answer lies below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Subtract 2w from both sides so the variable appears only once, divide by the remaining coefficient, and check by substituting into both sides.

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.)
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced
if a=b, then a−c=b−c\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 or term. 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 add the same thing, or divide both sides by the same number that is not zero.

  1. Sort out the equation

    An equation says two things are equal. You need the value of w that makes 4w4w equal to 2w−82w - 8. Here 4w means 4 times w, and 2w means 2 times w. The answer may be a negative number, so keep track of the signs.

  2. Get w on one side

    w is on both sides. Subtract 2w from both sides. Then w is only on the left, and the equation stays balanced.

    4w−2w=2w−8−2w4w - 2w = 2w - 8 - 2w

    2w=−82w = -8

    Now 2 times w is −8.

  3. Divide both sides by 2

    To undo times 2, divide both sides by 2. A negative number divided by a positive number is negative.

    w=−8÷2=−4w = -8 \div 2 = -4

    Watch out: keep the minus sign. Dropping it gives 4. Also, don’t stop at 2w=−82w = -8 and grid −8. So you would grid -4, with the minus sign in its own column.

Check your answer

Put −4 back in for w. Left side: 4×(−4)=−164 \times (-4) = -16. Right side: 2×(−4)−8=−8−8=−162 \times (-4) - 8 = -8 - 8 = -16. Both sides are −16, so −4 works.

More questions like this

The original DOE question
Question 60 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 60. Open page 81 of the handbook (PDF)

Question 94 of 120

2018 Test A Math, question 6

Question

The set of possible values of mm is {5,7,9}\left \{ 5,7,9 \right \}. What is the set of possible values of kk if 2k=m+32k=m+3?

  • A{3,4,5}\left \{ 3,4,5 \right \}
  • B{4,5,6}\left \{ 4,5,6 \right \}
  • C{8,10,12}\left \{ 8,10,12 \right \}
  • D{10,14,18}\left \{ 10,14,18 \right \}
Show the answer

Correct answer: B

2k=m+32k=m+3

so

k=m+32k=\cfrac{m+3}{2}

Substitute each value of m to find the values of k:

k=5+32=82=4k=\cfrac{5+3}{2}=\cfrac{8}{2}=4

k=7+32=102=5k=\cfrac{7+3}{2}=\cfrac{10}{2}=5

k=9+32=122=6k=\cfrac{9+3}{2}=\cfrac{12}{2}=6

The set k is { 4, 5, 6}.

Question analysis

An algebra question about substitution: m can be 5, 7 or 9, and an equation links k to m. Students find every possible value of k by substituting each value of m and solving. The curly-brace set notation can look unfamiliar, and each value has to be carried all the way through the solve.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Each substitution leads to a one-step grade 6 equation, easier than a typical grade 7 two-step problem. What adds demand is doing it three times and reading the answer as a set of values, which grade 7 rarely uses.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 9, 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 equation is simple, and the main hazard is stopping one step short: m + 3 gives 8, 10 and 12, a listed choice even though those are values of 2k. Checking a single value of m is the other risk, since k = 4 and k = 5 both appear in the first two sets. Students who substitute all three values and halve each result finish quickly.

  1. Variables instead of numbers minor Understanding the question

    There is no single unknown: m stands for the set {5, 7, 9}, and the answer is the matching set for k, so the equation has to be applied to three values. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the equation so k equals m plus 3, all divided by 2, then evaluate it for each of the three values of m.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Getting k by itself
k=m+32k = \frac{m + 3}{2}
k and m are the letters from the equation 2k = m + 3. The fraction bar means divide. If 2 times k equals m + 3, then k is half of m + 3. You get this by dividing both sides of the equation by 2.

  1. Sort out the question

    The letter m can be 5, 7 or 9. The curly braces { } show a set, which is a list of numbers. The equation 2k=m+32k = m + 3 links k to m. You need every value k can have, one for each value of m.

  2. Get k by itself

    2k means 2 times k. To undo times 2, divide both sides by 2. Doing the same thing to both sides keeps the equation true.

    k=m+32k = \frac{m + 3}{2}

  3. Try each value of m

    Put each value of m into the formula. Add 3 first, then divide by 2.

    k=5+32=82=4k = \frac{5 + 3}{2} = \frac{8}{2} = 4

    k=7+32=102=5k = \frac{7 + 3}{2} = \frac{10}{2} = 5

    k=9+32=122=6k = \frac{9 + 3}{2} = \frac{12}{2} = 6

    Watch out: 8, 10 and 12 are the values of m + 3, which equal 2k, not k. Stopping before you divide by 2 gives choice C.

  4. Pick the matching set

    The values of k are 4, 5 and 6. Watch out: try all three values of m. If you only try m = 5, you get k = 4, and choice A has a 4 too. So the answer is B, {4, 5, 6}.

Check your answer

Put each k back into the equation. 2×4=8=5+32 \times 4 = 8 = 5 + 3. 2×5=10=7+32 \times 5 = 10 = 7 + 3. 2×6=12=9+32 \times 6 = 12 = 9 + 3. All three work, so the set is right.

More questions like this

The original DOE question
Question 63 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 63. Open page 82 of the handbook (PDF)

Question 95 of 120

2018 Test A Math, question 14

Question

Which number line below shows the solution to the inequality −4<x2<2-4< \frac{x}{2}< 2 ?

  • ANumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −2 and 1, with the segment between them shaded.
  • BNumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −4 and 2, with the segment between them shaded.
  • CNumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −8 and 4, with the line shaded to the left of −8 and to the right of 4.
  • DNumber line from −10 to 10 with a tick at every integer, labeled at −10, −5, 0, 5 and 10. Open circles at −8 and 4, with the segment between them shaded.
Show the answer

Correct answer: D

Multiply each term by 2 to eliminate the fraction, and isolate xx:

−4(2)<(x2)(2)<2(2)-4(2)< \left(\frac{x}{2}\right)(2)< 2(2)

−8<x<4-8< x< 4

Therefore, xx must be between −8 and 4.

Question analysis

An inequality question: choose the number line that shows the solution of a compound inequality in which x divided by 2 lies between −4 and 2. Students undo the division in every part and match the graph's endpoints and shading. The four number lines differ in their endpoints and in which part of the line is shaded.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve two-step inequalities and graph them on a number line. This compound inequality needs only one operation applied to all three parts at once, so it matches typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 11 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 one step, multiplying through by 2, so most of the work is matching the result to a picture. The choices pair each common slip with a graph: the bounds left as written, the bounds halved, and the right bounds shaded on the wrong side. Reading the open circles against the labeled ticks at −5 and 5 settles it.

  1. Reading a figure, table or graph moderate Understanding the question

    The four number lines are labeled only at −10, −5, 0, 5 and 10, so open circles at −8 and 4, or at −4 and 2, have to be located by counting ticks. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Error-prone arithmetic minor Working it out

    All three parts of −4 < x/2 < 2 must be multiplied by 2, and the negative bound −4 becomes −8; halving instead gives −2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply all three parts of the inequality by 2 to isolate x, then pick the graph whose open circles and shading match the new bounds.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • 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.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying an inequality by a positive number
if a<b, then a×2<b×2\text{if } a < b \text{, then } a \times 2 < b \times 2
a and b are numbers, and the less-than sign means the left side is smaller. If you multiply both sides by the same positive number, like 2, the smaller side stays smaller. So the sign keeps pointing the same way. A negative number would flip it, but 2 is positive.

  1. Sort out the inequality

    An inequality compares two amounts that may not be equal. The sign << means less than. Here x2\frac{x}{2} means x divided by 2. The inequality says x divided by 2 is between −4 and 2. You need the values of x itself. Then you pick the number line that shows them.

  2. Multiply all three parts by 2

    To undo dividing by 2, multiply by 2. Multiply all three parts, so the whole statement stays true.

    −4×2<x2×2<2×2-4 \times 2 < \frac{x}{2} \times 2 < 2 \times 2

    −8<x<4-8 < x < 4

    Watch out: multiply, don’t divide. Dividing gives −2 and 1, choice A. Leaving the ends at −4 and 2 is choice B.

  3. Read what the answer means

    x is greater than −8 and less than 4. So x is any number between −8 and 4. x cannot equal −8 or 4 exactly. On a number line, that means open circles at −8 and 4, with the part between them shaded.

  4. Find the matching number line

    The number lines are labeled only at −10, −5, 0, 5 and 10. Count the tick marks. −8 is 3 ticks left of −5, and 4 is 1 tick left of 5. Watch out: choice C has circles at −8 and 4 but shades the parts outside them. So the answer is D.

Check your answer

Test a number from the shaded part, like 0. 0÷2=00 \div 2 = 0, and 0 is between −4 and 2. Now test 6, which is outside. 6÷2=36 \div 2 = 3, and 3 is not less than 2. So only the part between −8 and 4 works.

More questions like this

The original DOE question
Question 71 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 71. Open page 84 of the handbook (PDF)

Question 96 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 27\frac{2}{7} . How many of the pencils in the box are not red?

  • A5
  • B15
  • C21
  • D30
Show the answer

Correct answer: B

Let xx be the total number of colored pencils in the box. Set up a proportion to find xx:

27=6x\frac{2}{7}=\frac{6}{x}

2x=422x=42

x=21x=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.
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.

  1. 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)

  2. 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

The wrong answers, classified

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=favorabletotalP = \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.

  1. 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 27\frac{2}{7}. You need the number of pencils that are not red.

  2. Read the probability as parts

    27\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=57 - 2 = 5 is choice A.

  3. Find the size of one part

    The 6 red pencils fill 2 parts. Divide to find one part.

    6÷2=36 \div 2 = 3

    Each part is 3 pencils. So the whole box has 7×3=217 \times 3 = 21 pencils.

  4. Check with the probability formula

    Put the counts into the formula to make sure.

    P=favorabletotal=621=27P = \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.

  5. Count the pencils that are not red

    21−6=1521 - 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=216 + 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 97 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.
What makes it hard
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.

  1. 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)

  2. 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)

  3. 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

The wrong answers, classified

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=Wa + 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.

  1. 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”.

  2. 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%.

    FemaleMaleTotal
    Commute21%?60%
    Do not commute???
    Total45%?100%

  3. Fill in the two totals

    Everyone is female or male, so the males are what is left.

    100%−45%=55%100\% - 45\% = 55\%

    Everyone either commutes or does not.

    100%−60%=40%100\% - 60\% = 40\%

    So 55% are male, and 40% do not commute.

  4. Find the males who commute

    Use the commute row. All commuters are 60%, and the female commuters are 21%.

    60%−21%=39%60\% - 21\% = 39\%

    So 39% are males who commute. Watch out: the question says do not commute. Picking 39% is choice D.

  5. 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%55\% - 39\% = 16\%

    Watch out: 24% is the females who do not commute, 45%−21%45\% - 21\%. That is choice C. So the answer is B, 16%.

Check your answer

Find the same cell another way. Females who do not commute: 45%−21%=24%45\% - 21\% = 24\%. All who do not commute: 40%. So males who do not commute: 40%−24%=16%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 98 of 120

2018 Test A Math, question 20

Question

Mrs. Cranston bought five bottles of water for $0.90 each and 8 pounds of meat. She paid a total of $26.90 for these items, not including tax. What was the price per pound of the meat?

  • A$2.80
  • B$3.25
  • C$14.40
  • D$22.40
Show the answer

Correct answer: A

Let xx be the price per pound for the meat. Set up an equation to show what Mrs. Cranston spent:

5(0.90)+8x=26.905(0.90)+8x=26.90

4.50+8x=26.904.50+8x=26.90

8x=22.408x=22.40

x=2.80x=2.80

The price per pound for the meat is $2.80.

Question analysis

This money word problem has a shopper buying five bottles of water at a set price plus 8 pounds of meat, for a known total. Students write and solve a two-step equation to find the price per pound of the meat. The decimals call for care, and the question wants a price per pound rather than a total.

NYS standards
  • 7.EE.4a two-step equations
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Word problems that lead to an equation like px + q = r are the core of grade 7 algebra. This is a standard example with money decimals and no extra twist.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 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

This is a two-step money problem: take away the water, then divide by the pounds. The listed wrong answers come from doing one of those steps wrongly or not at all, especially stopping at the total cost of the meat, which looks like a finished dollar amount. Checking that 8 pounds at the chosen price plus $4.50 makes $26.90 confirms the answer.

  1. Turning words into math minor Understanding the question

    "five bottles of water for $0.90 each" has to become 5 × 0.90 before the total is split, and the unknown is a price per pound, not the meat's total cost. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Every amount is a decimal of dollars, so 26.90 − 4.50 and 22.40 ÷ 8 must be lined up and worked carefully without a calculator. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the cost of all five bottles from the total, then divide what is left by the 8 pounds of meat.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost
C=p×qC = p \times q
C is the total cost. p is the price of one item, or of one pound. q is the quantity, how many items or pounds you buy. Each one costs p, so q of them cost p times q.

  1. Sort out the question

    Mrs. Cranston bought five bottles of water “for $0.90 each”. She also bought 8 pounds of meat. Everything cost $26.90 in all. You need the price per pound of the meat, the cost of just 1 pound.

  2. Find the cost of the water

    Each bottle costs $0.90, and there are 5 bottles. Use the cost formula.

    C=p×qC = p \times q

    C=0.90×5=4.50C = 0.90 \times 5 = 4.50

    The water cost $4.50. Watch out: each means every bottle costs $0.90. Taking away just one $0.90 leads to $3.25, choice B.

  3. Find the cost of the meat

    Take the water away from the total. What is left paid for the meat.

    26.90−4.50=22.4026.90 - 4.50 = 22.40

    All 8 pounds of meat cost $22.40. Watch out: this is the cost of all the meat, not of one pound. Stopping here gives choice D.

  4. Find the price per pound

    Use the cost formula again. Now C is 22.40 and q is 8 pounds. Divide to find p.

    22.40=p×822.40 = p \times 8

    p=22.40÷8=2.80p = 22.40 \div 8 = 2.80

    Each pound costs $2.80. So the answer is A, $2.80.

Check your answer

Put it all back together. Meat: 8×2.80=22.408 \times 2.80 = 22.40. Water: 4.50. Total: 22.40+4.50=26.9022.40 + 4.50 = 26.90. That is exactly what she paid.

More questions like this

The original DOE question
Question 77 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 77. Open page 85 of the handbook (PDF)

Question 99 of 120

2018 Test A Math, question 22

Question

1 sind = 4 lorgs

2 plunks = 5 dalts

5 sinds = 2 harps

1 plunk = 3 harps

A nation has five types of coins: sinds, dalts, lorgs, harps, and plunks. The relationship between the coins is shown above. Which coin is most valuable?

  • Asind
  • Bdalt
  • Charp
  • Dplunk
Show the answer

Correct answer: D

1 sind = 4 lorgs,

so 1 sind > 1 lorg.

2 harps = 5 sinds,

so 1 harp > 1 sind.

1 plunk = 3 harps,

so 1 plunk > 1 harp,

meaning that 1 plunk > 1 sind and 1 lorg.

2 plunks = 5 dalts,

so 1 plunk > 1 dalt.

Therefore, the plunk is the most valuable.

Question analysis

This reasoning question invents a money system: four exchange rates link five made-up coins. Students compare the coins step by step to find which one is worth the most. The rates run in different directions, so it is easy to misjudge which side of an equation holds the more valuable coin.

NYS standards
  • 6.RP.1 ratio concepts and language
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 6 and 7 ratio work converts between two units at a time. Chaining four invented rates and comparing values with no common unit given is less familiar, though the arithmetic stays simple.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 1, 38 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 is little arithmetic, but the relationships have to be read the right way round and chained. Students who think a larger count means a more valuable coin rank the coins backward, and students who stop after one or two links pick a coin that beats only its neighbor. Writing each coin's worth in lorgs, 4 for a sind, 10 for a harp, 12 for a dalt and 30 for a plunk, makes the ranking plain.

  1. Turning words into math moderate Understanding the question

    In "2 plunks = 5 dalts" the coin with the larger count is worth less, the reverse of what the bigger number suggests, and the same holds in all four equations. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of moderate Working it out

    Five coins and four exchange rates have to be linked into one ranking; dalts connect only through plunks, and lorgs only through sinds. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. No obvious method minor Working it out

    Nothing says how to compare coins that never appear in the same equation; converting every coin to lorgs, or chaining inequalities, has to be chosen. (Schoenfeld, 1985)

Strategy: Rewrite every coin in terms of lorgs, which only ever appear as the smaller unit, following the rates one link at a time, then compare.

Traps

  • Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Value of a group of coins
V=n×vV = n \times v
V is what a group of coins is worth. n is the number of coins, and v is what one coin is worth. Here every value is counted in lorgs, the smallest coin. Using one unit for all the coins lets you compare them.

  1. Sort out the question

    There are five coins and four trading rules. You need the coin that is worth the most. Read each rule carefully. In 2 plunks = 5 dalts, it takes more dalts to match. So each dalt is worth less than a plunk. More coins on one side means each of those coins is worth less.

  2. Pick one coin to measure with

    Measure every coin in lorgs, the way you might measure every length in inches. The first rule says 1 sind = 4 lorgs. So one sind is worth 4 lorgs.

  3. Find a harp’s value in lorgs

    The rule 5 sinds = 2 harps links harps to sinds. First find what 5 sinds are worth.

    V=5×4=20V = 5 \times 4 = 20

    So 2 harps are worth 20 lorgs, and 1 harp is worth half of that.

    20÷2=1020 \div 2 = 10

    One harp is worth 10 lorgs.

  4. Find a plunk’s value in lorgs

    The rule 1 plunk = 3 harps says a plunk is worth 3 harps.

    V=3×10=30V = 3 \times 10 = 30

    One plunk is worth 30 lorgs.

  5. Find a dalt’s value in lorgs

    The rule 2 plunks = 5 dalts links dalts to plunks. Two plunks are worth 2×30=602 \times 30 = 60 lorgs. So 5 dalts are worth 60 lorgs too.

    60÷5=1260 \div 5 = 12

    One dalt is worth 12 lorgs.

  6. Compare all five coins

    CoinValue in lorgs
    lorg1
    sind4
    harp10
    dalt12
    plunk30

    Watch out: 5 dalts trade for only 2 plunks, so a dalt is worth less. Reading that rule backward leads to choice B. Watch out: the harp beats the sind, but the plunk beats the harp. Stopping early gives choice C. So the answer is D, plunk.

Check your answer

Test each rule with the lorg values. 2 plunks = 60 lorgs, and 5 dalts = 60 lorgs. 5 sinds = 20 lorgs, and 2 harps = 20 lorgs. 1 plunk = 30 lorgs, and 3 harps = 30 lorgs. Every rule balances.

More questions like this

The original DOE question
Question 79 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 79. Open page 85 of the handbook (PDF)

Question 100 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 xx be the number of second-, third-, and fourth-year students.

Then the total number of students in the college is

663+x663+x

Set up a proportion and solve for xx:

151=663+x179\frac{15}{1}=\frac{663+x}{179}

663+x=179(15)663+x=179(15)

663+x=2685663+x=2685

x=2022x=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.
What makes it hard
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.

  1. 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)

  2. 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

The wrong answers, classified

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×fs = 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.

  1. 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.

  2. Find every student in the college

    Multiply the teachers by 15.

    s=15×179s = 15 \times 179

    Break it into easy pieces: 10×179=1,79010 \times 179 = 1{,}790 and 5×179=8955 \times 179 = 895.

    1,790+895=2,6851{,}790 + 895 = 2{,}685

    There are 2,685 students in all four years.

  3. Take away the first-years

    2,685−663=2,0222{,}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=2,6852{,}022 + 663 = 2{,}685 students. Then 2,685÷179=152{,}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 101 of 120

2018 Test A Math, question 29

Question

A car is traveling 55 miles per hour, and 1 mile = 5,280 feet. Which of the following calculations would give the car’s speed in feet per second?

  • A55⋅52801\frac{55\cdot 5280}{1}
  • B55⋅52803600\frac{55\cdot 5280}{3600}
  • C55⋅36005280\frac{55\cdot 3600}{5280}
  • D55⋅528060\frac{55\cdot 5280}{60}
Show the answer

Correct answer: B

Divide the rate by the number of seconds in an hour. (Since there are 60 minutes in an hour and 60 seconds in a minute, multiply 60 × 60 = 3,600 seconds in an hour):

553600\frac{55}{3600} miles per second

Multiply by the number of feet in a mile (5,280):

55⋅52803600\frac{55\cdot 5280}{3600} feet per second

Question analysis

This unit conversion question gives a car's speed of 55 miles per hour and asks which calculation expresses it in feet per second. Students have to change miles to feet and hours to seconds in one expression. The tricky part is converting both units at once and keeping each conversion factor on the correct side of the fraction bar.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students usually convert one unit at a time, such as feet to inches. This item converts both parts of a rate at once, miles per hour into feet per second, and asks students to recognize the right setup.
Estimated difficulty
Straightforward: two steps or one careful reading.
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

Nothing has to be calculated; the work is judging which setup turns miles per hour into feet per second. Students who convert only part of the way find a choice waiting, whether they stop at feet per hour or go from hours to minutes instead of seconds. Tracking which unit each factor cancels, rather than computing, is what the question rewards.

  1. A very close wrong answer moderate Choosing the answer

    The answer, 55 · 5,280/3,600, and the choice 55 · 5,280/60 are written identically except for the denominator, so only knowing that an hour has 3,600 seconds, not 60, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    The stem gives 5,280 feet per mile but not the 3,600 seconds in an hour, which students must build from 60 × 60 themselves. (Anderson & Krathwohl, 2001)

Strategy: Write the speed as 55 miles over 1 hour and multiply by unit fractions such as 5,280 feet over 1 mile, checking that miles and hours cancel and only feet per second remain.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Speed
r=dtr = \frac{d}{t}
r is the speed, or rate. d is the distance traveled. t is the time it took. Speed tells how far something goes in one unit of time, so you divide the distance by the time.

  1. Sort out what you know

    The car goes 55 miles in 1 hour. One mile is 5,280 feet. You need the speed in feet per second. That means how many feet the car goes in 1 second.

    The choices are setups, not numbers. So you only need to pick the right setup. You do not have to work out the final number.

  2. Change the miles to feet

    Each mile is 5,280 feet. So 55 miles is 55 groups of 5,280 feet. Multiply to get the distance in feet:

    55×5,28055 \times 5{,}280

    That is how many feet the car goes in 1 hour. Watch out: if you stop here, you have feet per hour, not feet per second. That is choice A.

  3. Change the hour to seconds

    The question gives you miles to feet. It does not give you hours to seconds, so you build that yourself. An hour has 60 minutes, and each minute has 60 seconds:

    60×60=3,60060 \times 60 = 3{,}600

    So 1 hour is 3,600 seconds. Watch out: 60 is the number of minutes in an hour, not seconds. Dividing by 60 gives feet per minute. That is choice D.

  4. Divide the distance by the time

    Use the speed formula. The distance is 55 × 5,280 feet. The time is 3,600 seconds:

    r=55×5,2803,600r = \frac{55 \times 5{,}280}{3{,}600}

    Feet on top and seconds on the bottom give feet per second. Watch out: choice C turns both changes upside down, so the miles and hours never cancel. So the answer is B, 55⋅5,2803,600\frac{55 \cdot 5{,}280}{3{,}600}.

Check your answer

Estimate. 55 × 5,280 is about 290,000 feet in an hour. Divide by 3,600 seconds and you get about 80 feet each second. A car on a highway covering 80 feet in one second sounds right.

More questions like this

The original DOE question
Question 86 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 86. 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 102 of 120

2018 Test A Math, question 33

Question

Johan leased a car for three years. He paid a one-time fee of $1,000, and an additional $300 per month for the full three years. At the end of the three years, what is the total amount Johan paid for leasing this car?

  • A$1,900
  • B$4,600
  • C$10,800
  • D$11,800
Show the answer

Correct answer: D

Three years is 36 months (12 × 3).

Set up an expression to find the total amount Johan paid:

1,000+300(36)=$11,8001{,}000+300(36)=\$11{,}800

Question analysis

This word problem about leasing a car combines a one-time fee with a monthly payment over three years and asks for the total amount paid. Students must convert years to months, multiply by the monthly payment and add the fee. Leaving out one piece of the cost, or mixing years with months, is the usual slip.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 This is a short total-cost calculation with whole numbers, easier than typical grade 7 word problems, which bring in percents, negative numbers or fractions. The only care needed is turning years into months.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
3 steps
Reading level
Question and choices: about grade 4, 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

This is the easiest kind of multi-step word problem: whole numbers, a familiar money situation and a direct method. The only places to slip are changing three years into 36 months and remembering the one-time fee after the long multiplication. Each slip has its own choice, so a quick check that both costs are in the total is worth the seconds.

  1. Much to keep track of minor Working it out

    The $1,000 fee is paid once while the $300 repeats monthly, so it must be held aside while 36 × 300 is computed. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. A very close wrong answer minor Choosing the answer

    The choices $11,800 and $10,800 differ by only $1,000 out of about $11,000, so a rough total cannot separate them; only including the one-time fee does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Count the months in the lease first, multiply the monthly payment by that number, and then add the one-time fee.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a fee
T=f+m×nT = f + m \times n
T is the total paid. f is the one-time fee, paid only once. m is the monthly payment. n is the number of months. You multiply first because the monthly payment repeats every month. Then you add the fee once.

  1. Sort out what you know

    Johan paid $1,000 one time. He also paid $300 every month for 3 years. You need the total he paid.

    The payments are counted by the month, but the time is given in years. So first change the years to months.

  2. Count the months

    Each year has 12 months:

    3×12=363 \times 12 = 36

    So Johan made 36 monthly payments. Watch out: multiplying $300 by 3 years instead of 36 months gives $900. Adding the fee then gives $1,900, which is choice A.

  3. Add up the monthly payments

    Multiply the monthly payment by the number of months:

    300×36=10,800300 \times 36 = 10{,}800

    The monthly payments add up to $10,800.

  4. Add the one-time fee

    Use the formula. The fee goes in once:

    T=1,000+300×36=1,000+10,800=11,800T = 1{,}000 + 300 \times 36 = 1{,}000 + 10{,}800 = 11{,}800

    Watch out: $10,800 is choice C. It leaves out the $1,000 fee. So the answer is D, $11,800.

Check your answer

Round to check. $300 a month for 36 months is a bit more than $10,000. Add the $1,000 fee and you get a bit more than $11,000. $11,800 fits. Choices A and B are far too small for three years of payments.

More questions like this

The original DOE question
Question 90 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 90. 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 103 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
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.
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.

  1. 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)

  2. 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)

  3. 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

The wrong answers, classified

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×sS = 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.

  1. 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.

  2. Multiply for each city

    Use the formula for each city:

    CitySchoolsStudents per schoolTotal students
    M85004,000
    N55002,500
    P94003,600
    Q67004,200
    R47002,800

  3. 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.

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 104 of 120

2018 Test A Math, question 43

Question

wx=yz\frac{w}{x}=\frac{y}{z}

In the equation above, w,x,yw,x,y , and zz are positive numbers. Which of these is equal to zz?

  • Axx
  • Bxyxy
  • Cwxy\frac{w}{xy}
  • Dxyw\frac{xy}{w}
Show the answer

Correct answer: D

Solve the equation for zz:

wx=yz\frac{w}{x}=\frac{y}{z}

wz=xywz=xy

z=xywz=\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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying
ab=cd means a×d=b×c\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.

  1. 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.

  2. Cross-multiply the fractions

    Multiply each top by the other fraction’s bottom:

    w×z=x×yw \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.

  3. 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:

    w×zw=x×yw\frac{w \times z}{w} = \frac{x \times y}{w}

    z=xywz = \frac{xy}{w}

    Watch out: dividing the wrong way round gives w over xy, choice C. So the answer is D, xyw\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 105 of 120

2018 Test A Math, question 44

Question

Number line with points W, X, Y and Z marked from left to right. W is at −10 and Z is at 8; a tick mark at 0 lies between X and Y. X and Y have no values given.

On the number line above, points WW, XX, YY, and ZZ are integers, and WX:XY:YZ=4:2:3WX:XY:YZ = 4:2:3. What is the value of WY\overline{WY}?

  • A8
  • B11
  • C12
  • D18
Show the answer

Correct answer: C

Convert the ratios into fractions of WZWZ. Use the sum of the ratios for the denominator.

WX:XY:YZ=4:2:3WX:XY:YZ=4:2:3

WX=44+2+3WZ=49WZWX=\frac{4}{4+2+3}WZ=\frac{4}{9}WZ

XY=24+2+3WZ=29WZXY=\frac{2}{4+2+3}WZ=\frac{2}{9}WZ

The part of WZWZ that is WYWY is the sum of those fractions:

WY=49WZ+29WZ=69WZ=23WZWY=\frac{4}{9}WZ+\frac{2}{9}WZ=\frac{6}{9}WZ=\frac{2}{3}WZ

Find the length of WZWZ:

WZ=8−(−10)=18WZ=8-(-10)=18

The value of WYWY is 23(18)=12\frac{2}{3}(18)=12.

Question analysis

This number line question places integer points W, X, Y and Z on a line and gives the ratio of the segments WX, XY and YZ as 4 to 2 to 3. Students find the whole length from W to Z, split it by the ratio, and add the parts that make up WY. Mixing up a part with the whole is the main risk.

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
Above grade 7 Grade 7 students find distances on a number line and use part-to-whole ratios, usually separately. This item joins them: a negative endpoint, a three-term ratio and a choice of which segments to combine.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 10, 21 words.
Corrected from the DOE handbook
  • Equations in the explanation that mixed up a fraction of a length with the length itself were 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

Each step is routine, but the question keeps asking for a different piece of the line: the whole WZ, the part WX, and finally WY, which is two parts added together. Two of the wrong choices are exactly those other pieces. Students who mark the actual positions, X at −2 and Y at 2, can check the ratio before answering.

  1. Combines ideas moderate Working it out

    A three-part ratio, 4:2:3, must be applied to a length found from a negative endpoint, 8 − (−10) = 18, joining ratio and number-line distance. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Reading a figure, table or graph moderate Understanding the question

    Only W at −10 and Z at 8 are labeled, and the drawing is not to scale: Y is drawn nearer 5 than 2, so reading WY from the picture fails. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    Find WZ, split it into 9 equal parts of 2, then add the 4 parts of WX and the 2 parts of XY. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length WZ from the endpoints on the number line, divide it by the total number of ratio parts, and multiply by the parts from W to Y.

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.
  • 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

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes WX = 4/9, XY = 2/9 and WY = 2/3, and then says “The value of WY is 2/3(18) = 12”. Those fractions are parts of WZ, not the lengths themselves (WX is 8 and WY is 12), so WY cannot be both 2/3 and 12. The lines now read WX = 4/9 WZ, XY = 2/9 WZ and WY = 2/3 WZ. The method and the answer are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
d=b−ad = b - a
d is the distance between two points. b is the point on the right and a is the point on the left. Subtracting left from right always gives a positive distance. It works even when a is negative.
Size of one part of a ratio
u=Lpu = \frac{L}{p}
u is the size of one part. L is the whole length being shared. p is the total number of parts, found by adding the numbers in the ratio. Every part is the same size, so you share the length equally.

  1. Sort out what you know

    W is at −10 and Z is at 8. X and Y are between them. The ratio 4:2:3 means WX is 4 parts, XY is 2 parts and YZ is 3 parts. All the parts are the same size.

    You need the length WY, from W to Y. The drawing is not exact, so use the numbers, not the picture.

  2. Find the whole length WZ

    d=8−(−10)=8+10=18d = 8 - (-10) = 8 + 10 = 18

    WZ is 18 units long. Subtracting a negative number is the same as adding.

  3. Find the size of one part

    Add the ratio numbers to get the number of parts:

    4+2+3=94 + 2 + 3 = 9

    Share the 18 units equally among 9 parts:

    u=189=2u = \frac{18}{9} = 2

    Each part is 2 units long.

  4. Find WY

    WY is WX and XY together, so it has 4 + 2 = 6 parts:

    6×2=126 \times 2 = 12

    Watch out: WX alone is 4 × 2 = 8, which is choice A. And 18 is the whole line, choice D. So the answer is C, 12.

Check your answer

Place the points. W is at −10. X is 8 units to the right, at −2. Y is 4 more, at 2. Z is 6 more, at 8, which matches its label. And the distance from −10 to 2 is 2 − (−10) = 12.

More questions like this

The original DOE question
Question 101 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 101. Open page 89 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 106 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)(1100)=0.0002(0.02)(\frac{1}{100})=0.0002

The greatest allowable thickness would be

0.02+0.0002=0.02020.02+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.
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
x=p100×nx = \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.

  1. 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.

  2. Find 1 percent of 0.02

    x=1100×0.02=0.0002x = \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.

  3. 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=0.02020.0200 + 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 107 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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Profit
P=I−CP = 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.

  1. 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.

  2. Find the profit on one chair

    Each chair sells for $275 but costs $150 to make. Subtract to see what he keeps:

    275−150=125275 - 150 = 125

    He keeps $125 on each chair he sells.

  3. Find the profit on 25 chairs

    25×125=3,12525 \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.

  4. Subtract the fixed expenses

    P=3,125−1,250=1,875P = 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 108 of 120

2018 Test A Math, question 54

Question

Number line with points J, K, L and M marked from left to right. Only K has a value: 3/8.

On the number line above, JK=312JK=3\frac{1}{2} , JM=934JM=9\frac{3}{4} , and LM=118LM=1\frac{1}{8} . What is the position of point LL?

  • A5185\frac{1}{8}
  • B5145\frac{1}{4}
  • C5125\frac{1}{2}
  • D6146\frac{1}{4}
Show the answer

Correct answer: C

First, use JK=312JK=3\frac{1}{2} to find the location of JJ:

38−J=312\frac{3}{8}-J=3\frac{1}{2}

J=38−312=−318J=\frac{3}{8}-3\frac{1}{2}=-3\frac{1}{8}

Now, use JM=934JM=9\frac{3}{4} to find the location of MM:

M−(−318)=934M-\left(-3\frac{1}{8}\right)=9\frac{3}{4}

M+318=934M+3\frac{1}{8}=9\frac{3}{4}

M=934−318=658M=9\frac{3}{4}-3\frac{1}{8}=6\frac{5}{8}

Finally, use LM=118LM=1\frac{1}{8} to find the location of LL:

658−L=1186\frac{5}{8}-L=1\frac{1}{8}

L=658−118=548=512L=6\frac{5}{8}-1\frac{1}{8}=5\frac{4}{8}=5\frac{1}{2}

Question analysis

This number line question gives the lengths JK, JM and LM as mixed numbers and asks for the position of point L. Students must chain three distance steps, moving left from K to J, right to M and back to L, with fractions in eighths and a negative position. Losing track of where K sits is the main danger.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students add and subtract rational numbers and find distances on a number line one at a time. This item chains three such steps with mixed numbers and a negative position, more than typical grade 7 work demands.
Estimated difficulty
Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Solution
4 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

Every number is a mixed fraction, one position is negative, and the figure's even spacing hides how long KL really is. The distance route, KL = 9 3/4 − 3 1/2 − 1 1/8, avoids the negative entirely but ends one step short of a position unless K's 3/8 is added back. That last step is the one most students skip, which is why the question sits at the top of the scale.

  1. A very close wrong answer major Choosing the answer

    The choices 5 1/8, 5 1/4 and 5 1/2 lie within 3/8 of one another, so no estimate separates them; only exact eighths and adding K's 3/8 pick out the right one. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic moderate Working it out

    Every step mixes halves, quarters and eighths, and placing J means 3/8 − 3 1/2 = −3 1/8, a subtraction that crosses zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Reading a figure, table or graph moderate Understanding the question

    Only K is labeled, and the points are drawn evenly spaced although KL is over four times LM, so the picture gives no help with sizes. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  4. Many steps minor Working it out

    Three linked moves, K to J, J to M and M back to L, each depend on the last, so an early slip carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Start from K's labeled position, subtract JK to place J, add JM to place M, then subtract LM, keeping every fraction in eighths.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
d=b−ad = b - a
d is the distance between two points. b is the point on the right and a is the point on the left. Subtracting left from right always gives a positive distance. It works even when a is negative.
Equal fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top number (numerator) and b is the bottom number (denominator). Multiplying both by the same number k gives an equal fraction. It is like cutting each piece into k smaller pieces. Use it to give fractions the same bottom before you add or subtract.

  1. Sort out what you know

    Only K has a value: 38\frac{3}{8}. JK means the distance from J to K. You know JK = 3123\frac{1}{2}, JM = 9349\frac{3}{4} and LM = 1181\frac{1}{8}.

    You need the position of L, the number where L sits. The points are drawn evenly spaced, but the distances are not equal. So use the numbers, not the picture.

  2. Write the fractions in eighths

    The fractions use halves, fourths and eighths. Give them all a bottom of 8:

    312=348934=9683\frac{1}{2} = 3\frac{4}{8} \qquad 9\frac{3}{4} = 9\frac{6}{8}

    Watch out: 34\frac{3}{4} is 68\frac{6}{8}, not 48\frac{4}{8}. That slip leads to choice B.

  3. Find the distance from K to M

    JM is JK and KM put together. So take JK away from JM:

    968−348=6289\frac{6}{8} - 3\frac{4}{8} = 6\frac{2}{8}

    KM is 6286\frac{2}{8}, or 6146\frac{1}{4}. Watch out: this is a distance, not where L sits. Choice D stops here.

  4. Find the distance from K to L

    KM is KL and LM put together. So take LM away:

    628−118=5186\frac{2}{8} - 1\frac{1}{8} = 5\frac{1}{8}

    KL is 5185\frac{1}{8}. Watch out: this is still a distance. Choice A stops here.

  5. Find the position of L

    L is 5185\frac{1}{8} to the right of K, and K is at 38\frac{3}{8}. Add:

    38+518=548=512\frac{3}{8} + 5\frac{1}{8} = 5\frac{4}{8} = 5\frac{1}{2}

    So the answer is C, 5125\frac{1}{2}.

Check your answer

Place every point. J is 3483\frac{4}{8} left of K, at −318-3\frac{1}{8}. M is 9689\frac{6}{8} right of J, at 6586\frac{5}{8}. L is 1181\frac{1}{8} left of M, at 5485\frac{4}{8}, which is 5125\frac{1}{2}. Same answer.

More questions like this

The original DOE question
Question 111 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 111. Open page 91 of the handbook (PDF)

Question 109 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?

  • A6
  • B8
  • C24
  • D48
Show the answer

Correct answer: C

The ratio is 4:3:2:14:3:2:1, so the total parts is 10.

Since there are two parts resin, the fraction of resin is

210=15\frac{2}{10}=\frac{1}{5}

So the amount of resin in 30 lb of paste (for 1 billboard) is

15×30=6\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.
What makes it hard
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.

  1. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction from a ratio
f=pTf = \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.

  1. 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.

  2. Find resin’s fraction of the paste

    4+3+2+1=104 + 3 + 2 + 1 = 10

    The mix has 10 parts in all, and resin is 2 of them:

    f=210=15f = \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.

  3. Find the resin for one billboard

    15×30=6\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.

  4. Multiply by four billboards

    Each billboard needs the same 6 pounds of resin, so multiply by 4:

    6×4=246 \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 110 of 120

2018 Test B Math, question 5

Question

Four straight lines cross at point P: one horizontal, one vertical and two slanted. Above the horizontal line, from right to left: a 60° angle between the horizontal line and the slanted line rising to the right, a 30° angle between that line and the vertical line, an unlabeled angle between the vertical line and the slanted line rising to the left, and the angle y° between that line and the horizontal line. Below the horizontal line, a 25° angle lies between the vertical line and the slanted line falling to the right.

Four straight lines intersect at point P as shown above. What is the value of yy ?

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

Show the answer

Correct answer: 65

Call the missing angle in the top half of the figure xx.

The sum of the four angles on the top of the figure is equal to 180°.

x+y+30+60=180x+y+30+60=180

Since xx is a vertical angle with the 25° angle, then xx is also 25°. Use that to solve for yy.

25+y+30+60=18025+y+30+60=180

y+115=180y+115=180

y=65y=65

Question analysis

This grid-in angle question shows four straight lines crossing at a single point P and asks for the size of an unknown angle y. Students combine vertical angles with the fact that angles along a straight line add to 180 degrees. One angle they need is unlabeled and must be found from its vertical partner.

NYS standards
  • 7.G.5 angle relationships and equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Using vertical angles and straight angles to write an equation for an unknown angle is a core grade 7 geometry task. Four lines create more angles than usual, but the reasoning stays at grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 4, 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 arithmetic is one subtraction; the work is in seeing the figure correctly. Students have to notice that a fourth angle sits between the 30° angle and y, recognize it as the vertical partner of the 25° angle, and then use the straight line. The drawing is not to scale: y is drawn close to the 60° angle across the vertical line, which rewards anyone who computes rather than eyeballs.

  1. Reading a figure, table or graph major Understanding the question

    Four lines make eight angles around P but only three are labeled; above the horizontal line sit 60°, 30°, an unlabeled angle and y°, and that unlabeled angle is easy to overlook. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    The solution joins two facts: the unlabeled angle is vertical to the 25° angle below P, and the four angles above the horizontal line make a straight angle of 180°. (Embretson, 1998; Daniel & Embretson, 2010)

  3. An answer in an unexpected form minor Choosing the answer

    As a grid-in, nothing confirms the result, and the drawing shows y at about 60°, so a reading of the picture produces a number that looks right. (Enright, Morley & Sheehan, 2002)

Strategy: Pick a straight line through P and list the angles on one side of it; replace the unlabeled one with the labeled angle vertical to it, then subtract from 180.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line
a+b+c+d=180a + b + c + d = 180
a, b, c and d are angles that sit side by side along one side of a straight line, all meeting at one point. Together they make a half turn, which is 180 degrees. This works for any number of angles, not just four.
Vertical angles
∠1=∠2\angle 1 = \angle 2
When two lines cross, they make an X. Angle 1 and angle 2 are opposite each other across the crossing point. Angles like these are called vertical angles. They are always equal.

  1. Look closely at the figure

    Four lines cross at point P. Look above the flat, or horizontal, line. From right to left there are four angles: 60°, 30°, an angle with no label, and y°.

    Below the horizontal line there is a 25° angle. You need to find y.

  2. Find the unlabeled angle

    The unlabeled angle sits between the up-and-down, or vertical, line and the slanted line rising to the left.

    That slanted line keeps going through P and down to the right. The 25° angle sits between that same line and the vertical line, below P.

    So the two angles are opposite each other across P. They are vertical angles, so they are equal. The unlabeled angle is 25°.

  3. Add the angles above the line

    The four angles above the horizontal line fill one side of a straight line. So they add up to 180°.

    a+b+c+d=180a + b + c + d = 180

    y+25+30+60=180y + 25 + 30 + 60 = 180

    y+115=180y + 115 = 180

  4. Solve for y

    Take 115 away from 180 to get y alone.

    y=180−115=65y = 180 - 115 = 65

    Watch out: Do not skip the unlabeled angle. Using only 30 and 60 gives 90. Do not stop at the sum, 115, either. And do not guess from the picture. Angle y looks like 60°, but the drawing is not exact.

    So you would grid 65.

Check your answer

Add all four angles above the horizontal line: 65+25+30+60=18065 + 25 + 30 + 60 = 180. They fill a straight line exactly, so 65 fits.

More questions like this

The original DOE question
Question 62 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 62. Open page 153 of the handbook (PDF)

Question 111 of 120

2018 Test B Math, question 8

Question

M=3N=P4=Q+5=R7>0M=3N=\frac{P}{4}=Q+5=\frac{R}{7}> 0

Based on the statement above, which variable has the greatest value?

  • AMM
  • BNN
  • CPP
  • DRR
Show the answer

Correct answer: D

Break the equations apart to each equal MM:

M=3NM=3N

M=P4M=\frac{P}{4}

M=Q+5M=Q+5

M=R7M=\frac{R}{7}

Pick a number to substitute into the equations, and solve the equations to find the values of MM, NN, PP, QQ, and RR.

Let M=2M=2. Since all the equations are equal to 2, substitute 2 to find each variable.

M=3NM=3N

2=3N2=3N

23=N\frac{2}{3}=N

M=P4M=\frac{P}{4}

2=P42=\frac{P}{4}

8=P8=P

M=Q+5M=Q+5

2=Q+52=Q+5

−3=Q-3=Q

M=R7M=\frac{R}{7}

2=R72=\frac{R}{7}

14=R14=R

Variable RR has the greatest value.

Question analysis

This algebra question sets five expressions equal to one another and to a positive number, then asks which variable has the greatest value. It tests whether students can rewrite each variable in terms of one shared value and compare them. What makes it tricky is that there are no numbers to compute, so each variable's size has to be reasoned out.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students rewrite expressions to see how quantities relate, usually one equation at a time. This chain of five linked expressions, with no numbers to compute and a positivity condition to use, is a reasoning task beyond routine grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 10, 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 question rewards a strategy, not a calculation: once the shared value is named, each variable follows in one line and R is clearly largest. Without that step, students judge by how the expressions look, and the look is backwards, since the 3 in 3N makes N small and the 7 under R makes R large. The positivity condition at the end is easy to skip but is what fixes the order.

  1. Variables instead of numbers major Understanding the question

    There are no numbers to compute: five variables are tied together in the one chain M = 3N = P/4 = Q + 5 = R/7, and each must be compared through the shared value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. No obvious method moderate Working it out

    Nothing says how to compare; the efficient move is to name the common value, or pick one such as 12, and rewrite each variable, which students must think of themselves. (Schoenfeld, 1985)

  3. Much to keep track of minor Working it out

    Five relations and the condition > 0 must be held at once; the condition matters, since with a negative common value the order of M, N, P and R would reverse. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Call the shared value k, a positive number, and rewrite each variable as an expression in k; or substitute a convenient positive value such as 12 and compare the results.

Traps

  • Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing division
ab=c means a=b×c\frac{a}{b} = c \text{ means } a = b \times c
a is the top of the fraction, b is the bottom, and c is the result. Multiplying undoes dividing. So if a divided by b is c, then a is b times c.
Undoing multiplication
b×a=c means a=c÷bb \times a = c \text{ means } a = c \div b
b is the number multiplied by a, and c is the result. Dividing undoes multiplying. So to get a alone, divide c by b.

  1. Understand the chain

    The statement is a chain of equal signs. It says five expressions all have the same value. The last part says that value is greater than 0, so it is positive.

    You need to find which letter has the largest value.

  2. Pick a value to test

    All five expressions are equal, so give them one shared value. Pick a positive number that 3 divides evenly. That way N will be a whole number.

    Let the shared value be 12. Then M=12M = 12, and each of the other four expressions also equals 12.

  3. Solve for each letter

    Undo each operation to find each letter. Divide to undo times. Multiply to undo divide. Subtract to undo plus.

    3N=12, so N=12÷3=43N = 12 \text{, so } N = 12 \div 3 = 4

    P4=12, so P=12×4=48\frac{P}{4} = 12 \text{, so } P = 12 \times 4 = 48

    Q+5=12, so Q=12−5=7Q + 5 = 12 \text{, so } Q = 12 - 5 = 7

    R7=12, so R=12×7=84\frac{R}{7} = 12 \text{, so } R = 12 \times 7 = 84

  4. Compare the values

    The values are M = 12, N = 4, P = 48, Q = 7 and R = 84. R is the largest.

    Watch out: 3N does not make N big. It means 3 times N equals M, so N is only one third of M. That trap points to choice B. Also, P is big, but R is bigger. Check every letter before you pick choice C.

    So the answer is D, R.

Check your answer

Think without numbers. R divided by 7 is M, so R is 7 times M. P is only 4 times M. N is a third of M, and Q is 5 less than M. Since M is positive, 7 times M is the biggest.

More questions like this

The original DOE question
Question 65 as printed on page 154 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 65. Open page 154 of the handbook (PDF)

Question 112 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 15:715:7, and the ratio of blue to green marbles is 7:37:3. If 2 blue marbles are removed and replaced with 2 green marbles, what will be the new ratio of red to green marbles?

  • A3:13:1
  • B5:15:1
  • C15:315:3
  • D45:1145:11
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:

715+7+3=725\frac{7}{15+7+3}=\frac{7}{25}

Find the proportion of green marbles:

325\frac{3}{25}

Since there are a total of 75 marbles, the number of blue marbles is

725×75=21\frac{7}{25}\times 75=21

The number of green marbles is

325×75=9\frac{3}{25}\times 75=9

 The number of red marbles is

75−21−9=4575-21-9=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.
What makes it hard
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.

  1. 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)

  2. 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)

  3. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Value of one part
one part=totalsum of the ratio numbers\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.

  1. 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.

  2. Join the two ratios

    Blue is 7 in both ratios. So you can join them into one ratio.

    red:blue:green=15:7:3\text{red} : \text{blue} : \text{green} = 15 : 7 : 3

    This means for every 15 red marbles, there are 7 blue and 3 green.

  3. Find the size of one part

    The ratio has 15+7+3=2515 + 7 + 3 = 25 parts in all. The 75 marbles fill those 25 parts.

    one part=totalsum of the ratio numbers=7525=3\text{one part} = \frac{\text{total}}{\text{sum of the ratio numbers}} = \frac{75}{25} = 3

    Each part is 3 marbles.

  4. Count each color

    Multiply each ratio number by 3 to get the real counts.

    15×3=457×3=213×3=915 \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.

  5. Make the swap

    Take out 2 blue: 21−2=1921 - 2 = 19 blue. Put in 2 green: 9+2=119 + 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=7545 + 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 113 of 120

2018 Test B Math, question 14

Question

What is the least common multiple of 24, 6, and 18?

  • A36
  • B48
  • C72
  • D144
Show the answer

Correct answer: C

Since 18 and 24 are both multiples of 6, find the least common multiple of only 18 and 24.

Multiples of 18:

18, 36, 54, 72…

Multiples of 24:

24, 48, 72…

The least common multiple of 6, 18, and 24 is 72.

Question analysis

This number properties question asks for the least common multiple of three numbers: 24, 6 and 18. Students can list multiples or use prime factors. Each candidate has to be checked against all three numbers, and it must be the smallest number that works, not just any shared multiple.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Least common multiples are grade 6 content, where New York's standard uses two numbers up to 12. This uses three larger numbers, but it is still a short factor task below typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 8 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

Finding a least common multiple is routine, and listing multiples of 24 reaches the answer by the third one. The wrong choices are built to catch incomplete checking: each passes the test for some of the numbers but fails for one, or passes for all and is simply not the smallest. Testing every candidate against all three numbers is the whole skill.

  1. Every choice must be checked minor Choosing the answer

    Each wrong choice passes part of the test: 36 is a multiple of 6 and 18, 48 of 6 and 24, and 144 of all three, so every candidate needs checking against all three numbers. (Embretson & Wetzel, 1987)

  2. Something you must already know minor Working it out

    Students must know what “least common multiple” means and see that 6 can be dropped, since it already divides both 24 and 18. (Anderson & Krathwohl, 2001)

Strategy: Note that 6 divides both 24 and 18, so only those two matter; list multiples of 24 and stop at the first one that 18 also divides.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiples of a number
n×1, n×2, n×3, …n \times 1,\ n \times 2,\ n \times 3,\ \ldots
n is any whole number. Its multiples are what you get when you multiply n by 1, 2, 3 and so on. The least common multiple of some numbers is the smallest number that is a multiple of every one of them.

  1. Understand the question

    A multiple of a number is that number times 1, 2, 3, and so on. A common multiple is a multiple of every number in the list.

    You need the least, or smallest, common multiple of 24, 6 and 18.

  2. Drop the 6

    6 goes evenly into 24, since 6×4=246 \times 4 = 24. It also goes into 18, since 6×3=186 \times 3 = 18.

    So any multiple of 24 is already a multiple of 6. You only need to work with 24 and 18.

  3. List the multiples of 24

    Start with the bigger number, since its list grows faster. Check each multiple to see if 18 divides it evenly.

    • 24×1=2424 \times 1 = 24, and 24 ÷ 18 is not a whole number.
    • 24×2=4824 \times 2 = 48, and 48 ÷ 18 is not a whole number.
    • 24×3=7224 \times 3 = 72, and 72÷18=472 \div 18 = 4. It works.

  4. Give the answer

    72 is the first multiple of 24 that 18 also divides. It is a multiple of 6 too, since 72÷6=1272 \div 6 = 12.

    Watch out: 36 works for 6 and 18, but not for 24, so choice A is wrong. 144 works for all three, but it is not the smallest.

    So the answer is C, 72.

Check your answer

Test 72 with all three numbers: 72÷24=372 \div 24 = 3, 72÷6=1272 \div 6 = 12 and 72÷18=472 \div 18 = 4. All are whole numbers. The smaller choices, 36 and 48, each fail for one of the numbers.

More questions like this

The original DOE question
Question 71 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 71. Open page 155 of the handbook (PDF)

Question 114 of 120

2018 Test B Math, question 18

Question

PRICES FOR NEWSPAPER ADVERTISING

Page Space Price
14\frac{1}{4} page $200
12\frac{1}{2} page $350
full page $600

The table above shows prices for newspaper advertising. A store purchased 14\frac{1}{4} pages, 12\frac{1}{2} pages, and full pages of page space in equal numbers for a total of $11,500. What is the total amount of page space the store purchased?

  • A1341\frac{3}{4} pages
  • B10 pages
  • C161216\frac{1}{2} pages
  • D171217\frac{1}{2} pages
Show the answer

Correct answer: D

The question says that an equal number (xx) of each type of page space was purchased. To find the number of each type of page space that was purchased, multiply the price per type by xx and set it equal to the total amount spent, then solve for xx:

200x+350x+600x=11,500200x+350x+600x=11,500

1,150x=11,5001,150x=11,500

x=10x=10

Thus, the store purchased 10 units of each type of page space. To find the total amount of page space purchased, multiply each type of page space by 10, and add:

(10×14 page)+(10×12 page)+(10×1 page)\left(10\times\frac{1}{4}\text{ page}\right)+\left(10\times\frac{1}{2}\text{ page}\right)+(10\times 1\text{ page})

=1712 pages=17\frac{1}{2}\text{ pages}

Question analysis

This word problem gives a table of newspaper advertising prices for quarter-page, half-page and full-page ads, and says a store bought equal numbers of each for a set total. Students write and solve an equation for how many ads of each size were bought. The question wants total page space, which takes a second calculation with fractions.

NYS standards
  • 7.EE.4a two-step equations
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students write and solve equations like px = r from word problems. This adds a table, combines three prices into one coefficient and needs a second step with fractions after solving, a longer chain than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 5, 59 words.
Corrected from the DOE handbook
  • An arithmetic slip 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

Solving for the number of ads is the obvious goal, and finding it feels like the finish, but the question asks for page space, which needs one more pass through the table. The awkward wording about buying pages in equal numbers slows students before any arithmetic starts. The wrong choices mark each place to stop: one set of ads, the number of each ad, or a slip in adding the fractional pages.

  1. Many steps major Working it out

    The work has two phases: turn the money into a count, 1,150x = 11,500 so x = 10, then turn the count into space by multiplying 10 by 1/4, 1/2 and 1 and adding. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math moderate Understanding the question

    The phrase “purchased 1/4 pages, 1/2 pages, and full pages of page space in equal numbers” has to be read as the same number of ads of each size, so the three prices add to one coefficient. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Reading a figure, table or graph minor Understanding the question

    Both columns of the table are needed: the price column builds the equation, and the page-space column turns the count of ads into the final total. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Let x be the number of each ad size, add the three prices to write one equation for the total, solve it, then multiply x by the page space in one set of ads.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook adds 200x + 350x + 600x and writes 1,500x = 11,500. The sum is 1,150x, and the handbook’s next line, x = 10, only follows from 1,150x = 11,500. The site shows 1,150x = 11,500.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of equal sets
number of sets=total spentcost of one set\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}}
A set here means one ad of each size. The cost of one set is what one of each costs together. Dividing the total money by it tells how many sets were bought, because every set costs the same.

  1. Sort out the question

    The store bought the same number of each ad size: quarter page, half page and full page. It spent $11,500 in all.

    You need the total page space. That is not the number of ads, and not the money.

  2. Find the cost of one set

    Think of one quarter page, one half page and one full page as a set. The store bought equal numbers of each, so it bought whole sets.

    200+350+600=1,150200 + 350 + 600 = 1{,}150

    One set costs $1,150.

  3. Find how many sets

    number of sets=total spentcost of one set=11,5001,150=10\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}} = \frac{11{,}500}{1{,}150} = 10

    The store bought 10 sets. That means 10 ads of each size.

    Watch out: 10 is the number of each ad, not the page space. Stopping here gives choice B.

  4. Find the page space in one set

    Add the sizes in one set. Write them in quarters so they have the same bottom number.

    14+12+1=14+24+44=74\frac{1}{4} + \frac{1}{2} + 1 = \frac{1}{4} + \frac{2}{4} + \frac{4}{4} = \frac{7}{4}

    One set is 74\frac{7}{4}, or 1341\frac{3}{4}, pages.

    Watch out: 1341\frac{3}{4} pages is only one set. That is choice A.

  5. Multiply by the number of sets

    There are 10 sets, so multiply the space in one set by 10.

    10×74=704=171210 \times \frac{7}{4} = \frac{70}{4} = 17\frac{1}{2}

    So the answer is D, 171217\frac{1}{2} pages.

Check your answer

Add the sizes another way. 10 quarter pages make 2.5 pages, not 1.5. 10 half pages make 5 pages. 10 full pages make 10 pages. Then 2.5+5+10=17.52.5 + 5 + 10 = 17.5, the same answer.

More questions like this

The original DOE question
Question 75 as printed on page 156 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 75. Open page 156 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 115 of 120

2018 Test B Math, question 19

Question

How many positive odd numbers satisfy the inequality 3x+8≤923x+8\leq 92 ?

  • A13
  • B14
  • C17
  • D28
Show the answer

Correct answer: B

Solve the inequality for x.

3x + 8 ≤ 92

3x ≤ 84

x ≤ 28

The positive odd numbers less than or equal to 28 are 1, 3, 5, …, 25, and 27. There are 14 of them.

Question analysis

This question asks how many positive odd numbers satisfy the two-step inequality 3x + 8 ≤ 92. Students solve the inequality and then count the odd whole numbers in the solution. The counting after the algebra is where errors creep in, especially at the end of the list.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving a two-step inequality is grade 7 work. Asking for a count of the odd solutions adds a number-property step and an endpoint check, but the total demand stays close to a typical grade 7 problem.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 16, 9 words.
Corrected from the DOE handbook
  • The explanation’s description of the solutions 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

Solving the inequality is routine, and that is the trap: x ≤ 28 looks like an answer and 28 is among the choices. The real task is counting odd numbers from 1 to 27, where the usual shortcut, (27 − 1) ÷ 2, misses the first term. Pairing each odd number with the even number after it, up to 28, gives the count directly and avoids the fencepost slip.

  1. A very close wrong answer moderate Choosing the answer

    13 and 14 differ by one, and a rough count of the odd numbers up to 27 cannot separate them; only counting both ends, 1 and 27, gives the exact 14. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Combines ideas minor Working it out

    Solving 3x + 8 ≤ 92 is only half the work; counting the positive odd whole numbers up to 28 is a separate number-sense task. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Solve the inequality as you would an equation, keeping the direction of the sign, then count odd numbers from 1 up to the boundary by pairing or halving.

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).
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook solves the inequality to x ≤ 28 and then lists “the positive odd numbers less than 28.” The solutions include 28 itself, so the site says “less than or equal to 28.” The count does not change, because 28 is even: the odd numbers are still 1, 3, 5, …, 27, which is 14 numbers.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Solving an inequality
if a≤b, then a−c≤b−c\text{if } a \leq b \text{, then } a - c \leq b - c
a, b and c are numbers. The sign ≤ means less than or equal to. You may subtract the same number from both sides. You may also divide both sides by the same positive number. The sign stays the same either way.
Counting odd numbers
odd numbers from 1 to n=n2\text{odd numbers from 1 to } n = \frac{n}{2}
n is an even number. Split the numbers from 1 to n into pairs: 1 and 2, 3 and 4, and so on. There are n ÷ 2 pairs, and each pair holds exactly one odd number.

  1. Sort out the question

    You need to count the positive odd numbers that make 3x+8≤923x + 8 \leq 92 true.

    The sign ≤ means is less than or equal to. Positive means greater than 0. Odd numbers are 1, 3, 5, 7, and so on.

  2. Solve the inequality

    Solve it like an equation. Subtract 8 from both sides. Then divide both sides by 3. Dividing by a positive number keeps the sign the same.

    3x+8≤923x + 8 \leq 92

    3x≤843x \leq 84

    x≤28x \leq 28

    Watch out: 28 is the largest x allowed, not the count. That trap is choice D.

  3. List the odd numbers

    The positive odd numbers up to 28 are 1, 3, 5, and so on, up to 25 and 27. The number 28 is even, so it does not count.

  4. Count them by pairing

    Split 1 to 28 into pairs: 1 and 2, 3 and 4, and so on, up to 27 and 28. Each pair has one odd number.

    odd numbers from 1 to 28=282=14\text{odd numbers from 1 to 28} = \frac{28}{2} = 14

    Watch out: Working out (27−1)÷2=13(27 - 1) \div 2 = 13 counts the jumps of 2 from 1 to 27, not the numbers. You must add 1 for the first number. That mistake gives choice A.

    So the answer is B, 14.

Check your answer

Test the edges. For x = 27: 3×27+8=893 \times 27 + 8 = 89, which is at most 92, so it works. For x = 29: 3×29+8=953 \times 29 + 8 = 95, which is too big. So 27 is the last odd number that works.

More questions like this

The original DOE question
Question 76 as printed on page 156 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 76. Open page 156 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 116 of 120

2018 Test B Math, question 20

Question

If 36y=4x\frac{36}{y}=4x, what is the value of xx when y=3y=3?

  • A3
  • B4
  • C9
  • D12
Show the answer

Correct answer: A

Substitute 3 for yy and solve for xx:

36y=4x\frac{36}{y}=4x

363=4x\frac{36}{3}=4x

12=4x12=4x

3=x3=x

Question analysis

This equation question gives 36 divided by y equal to 4x and asks for the value of x at one given value of y. Students substitute, simplify the fraction and solve a one-step equation. It is short, so the risk lies in stopping one step early or dividing by the wrong number.

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 Substituting a value and solving a one-step equation are grade 6 skills. With two quick steps and whole numbers, this is lighter than the multistep equations typical of grade 7.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 8, 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

This is substitution followed by a one-step equation, with small whole numbers throughout, so most students answer correctly. The only real risk is stopping at 12, which is the value of 4x rather than of x. Writing 12 = 4x as its own line before reading the choices makes the last division hard to skip.

  1. Variables instead of numbers minor Understanding the question

    The variable y sits in a denominator, 36/y, which looks less familiar than an equation in one unknown, though substituting y = 3 removes it at once. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Replace y with its value, simplify the left side to a single number, and then divide by the coefficient of x.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing multiplication
a×x=b means x=b÷aa \times x = b \text{ means } x = b \div a
a is the number multiplied by x, and b is the result. Division undoes multiplication. So to get x alone, divide both sides by a.

  1. Write down what you know

    The equation is 36y=4x\frac{36}{y} = 4x. You are told y = 3. You need the value of x.

    The letter y is on the bottom of a fraction. Putting in its value turns the fraction into a plain number.

  2. Put in the value of y

    Replace y with 3.

    363=4x\frac{36}{3} = 4x

    The fraction means 36 divided by 3, which is 12.

    12=4x12 = 4x

    Watch out: 12 is the value of 4x, not x. Stopping here gives choice D.

  3. Divide both sides by 4

    4x means 4 times x. To undo multiplying by 4, divide both sides by 4.

    x=12÷4=3x = 12 \div 4 = 3

    So the answer is A, 3.

Check your answer

Put x = 3 and y = 3 back in. The left side is 36÷3=1236 \div 3 = 12. The right side is 4×3=124 \times 3 = 12. Both sides match, so x = 3 is right.

More questions like this

The original DOE question
Question 77 as printed on page 157 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 77. Open page 157 of the handbook (PDF)

Question 117 of 120

2018 Test B Math, question 21

Question

 Points XX , YY, and ZZ are on a straight line, and YY is between XX and ZZ . Length YZ=35XY\overline{YZ}=\frac{3}{5}\overline{XY}, and length XY=20\overline{XY}=20 centimeters. What is the length of XZ\overline{XZ}?

  • A12 cm
  • B24 cm
  • C30 cm
  • D32 cm
Show the answer

Correct answer: D

Since XY=20\overline{XY}=20 cm, use that to find YZ\overline{YZ} :

YZ=35XY=35(20)=12 cm\overline{YZ}=\frac{3}{5}\overline{XY}=\frac{3}{5}(20)=12\text{ cm}

XZ=XY+YZ=20+12=32 cm\overline{XZ}=\overline{XY}+\overline{YZ}=20+12=32\text{ cm}

Question analysis

This segment question places points X, Y and Z on a line with Y between the others, gives XY as 20 centimeters, and says YZ is three fifths of XY. Students find the length of the whole segment XZ. Keeping track of which segment the question asks for matters more than the arithmetic.

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 Finding a fraction of a length and adding segments takes only two short steps with friendly numbers. Grade 7 rational-number problems are usually longer or involve negatives, so this sits below grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 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

There is no figure, so students who sketch the three points see at once that XZ is made of the two pieces XY and YZ. The arithmetic, three fifths of 20, is friendly. Most errors come from answering with the piece that was just calculated instead of the whole segment the question asks for.

  1. Turning words into math minor Understanding the question

    The condition “Y is between X and Z” has to become XZ = XY + YZ, and the overbar notation for segment lengths may be unfamiliar to some students. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find three fifths of 20 to get YZ, then add the two parts, since Y lies between X and Z.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a number
ab of n=n÷b×a\frac{a}{b} \text{ of } n = n \div b \times a
n is the starting amount. b is the bottom of the fraction: split n into b equal pieces. a is the top: take a of those pieces. Here n is 20, b is 5 and a is 3.
Adding segment lengths
XZ=XY+YZ\overline{XZ} = \overline{XY} + \overline{YZ}
X, Y and Z are points on one line, with Y between X and Z. XY and YZ are the lengths of the two short pieces. XZ is the whole length. The two pieces join end to end to make the whole.

  1. Picture the line

    Points X, Y and Z are on one straight line, with Y between X and Z. So the order is X, then Y, then Z.

    A segment is the straight piece between two points. The bar over two letters means the length of that segment.

    XY is 20 cm. You need XZ, the whole length.

  2. Find the length of YZ

    YZ is three fifths of XY. To find three fifths of 20, first find one fifth. Then take 3 of those.

    35 of 20=20÷5×3=4×3=12\frac{3}{5} \text{ of } 20 = 20 \div 5 \times 3 = 4 \times 3 = 12

    So YZ is 12 cm.

    Watch out: 12 cm is only one piece of the line. Stopping here gives choice A.

  3. Add the two pieces

    Y is between X and Z, so the two pieces join to make the whole.

    XZ=XY+YZ=20+12=32\overline{XZ} = \overline{XY} + \overline{YZ} = 20 + 12 = 32

    XZ is 32 cm.

    So the answer is D, 32 cm.

Check your answer

Check the fraction: 1220=35\frac{12}{20} = \frac{3}{5}, so YZ really is three fifths of XY. Also, XZ must be longer than XY, since it holds XY and more. 32 is longer than 20.

More questions like this

The original DOE question
Question 78 as printed on page 157 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 78. Open page 157 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 118 of 120

2018 Test B Math, question 27

Question

Nicole’s age now is three times Carmen’s age. If Carmen will be 17 in two years, how old was Nicole 5 years ago?

  • A38 yr
  • B40 yr
  • C45 yr
  • D50 yr
Show the answer

Correct answer: B

First, find out how old Nicole and Carmen are now.

Let NN = Nicole’s age now. Let CC = Carmen’s age now.

C+2=17C+2=17

C=15C=15 (Carmen’s age now)

N=3CN=3C

N=3(15)=45N=3(15)=45 (Nicole’s age now)

N−5=45−5=40N-5=45-5=40 (Nicole’s age 5 years ago)

Question analysis

This age word problem says Nicole is three times as old as Carmen now, and that Carmen will be 17 in two years. Students work out both present ages and then go back five years for Nicole. Keeping the times straight is the tricky part: two years ahead for Carmen, five years back for Nicole.

NYS standards
  • 7.EE.4 equations and inequalities from 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 write simple equations from word problems. Age problems like this one take three short steps, and the shifts in time are the only real hurdle, so the demand is typical of grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 1, 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 arithmetic is short, but the question moves through three points in time, and only one of them carries the three-times relationship. Students who find Nicole's present age often stop there, since it is a choice. Labeling each age with its time, now or then, keeps the two-year and five-year shifts with the right person and in the right direction.

  1. Turning words into math moderate Understanding the question

    Three points in time appear, now, in two years and 5 years ago, and “three times Carmen's age” holds only for now, so each shift must be applied to the right person. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Students must hold Carmen's future age 17, her present age 15, Nicole's present age 45 and the 5-year shift while deciding which to report. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find Carmen's age now by going back two years from 17, triple it for Nicole's current age, then subtract five.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • 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.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Age at another time
age in n years=age now+n\text{age in } n \text{ years} = \text{age now} + n
n is a number of years. To find an age in the future, add n to the age now. To find an age in the past, subtract n instead. Everyone gets older at the same speed, one year per year.

  1. Sort out the times

    The question talks about three times: now, two years from now, and 5 years ago. Nicole is 3 times as old as Carmen now.

    Carmen will be 17 in two years. You need Nicole’s age 5 years ago.

  2. Find Carmen's age now

    In two years Carmen will be 17. So right now she is 2 years younger.

    17−2=1517 - 2 = 15

    Carmen is 15 now.

  3. Find Nicole's age now

    Nicole’s age now is three times Carmen’s age now.

    3×15=453 \times 15 = 45

    Nicole is 45 now.

    Watch out: 45 is her age now, choice C. The question asks about 5 years ago.

  4. Go back 5 years

    Going back in time means subtracting.

    45−5=4045 - 5 = 40

    Nicole was 40 five years ago.

    Watch out: Adding 5 gives 50, choice D. Also, the 2 years only apply to Carmen. Taking 2 more years off Nicole gives 38, choice A.

    So the answer is B, 40 yr.

Check your answer

Work forward from your answer. Nicole was 40 five years ago, so she is 45 now. Carmen is a third of that, 15. In two years Carmen will be 17. That matches the question.

More questions like this

The original DOE question
Question 84 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 84. Open page 158 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 28

Question

A chemical decays in such a way that the amount left at the end of each week is 20% less than the amount at the beginning of that same week. What percent of the original amount is left after two weeks?

  • A40%
  • B60%
  • C64%
  • D80%
Show the answer

Correct answer: C

Let xx be the original amount of the chemical. It loses 20% after each week, which means 80% of the chemical remains at the end of each week. End of first week:

0.80x0.80x

At the end of the second week, 80% of the amount left at the end of the first week remains. End of second week:

0.80(0.80x)=0.64x or 64%0.80(0.80x)=0.64x\text{ or }64\%

Question analysis

This percent problem describes a chemical that loses 20% of what is present each week and asks what percent of the original amount remains after two weeks. Students apply a percent decrease twice in a row. Deciding which amount each week's 20 percent is taken from is the heart of the problem.

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 percent problems usually apply a single increase or decrease. Applying a decrease twice, each time to the new amount, is an early form of exponential decay that goes beyond typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
2 steps
Reading level
Question and choices: about grade 8, 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

This is a two-step percent decrease, and the intuitive answer, 20% plus 20%, is wrong for a reason students rarely check: the second loss is taken from what is left after the first. The stem states this rule in its own words, but quickly. Starting with 100 units and working week by week makes the difference visible, since the second week removes 16 units, not 20.

  1. A very close wrong answer major Choosing the answer

    60% and 64% are four points apart, too close for a rough estimate; only seeing that week two's 20% comes from 80, not 100, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Turning words into math moderate Understanding the question

    The stem measures each week's loss against “the amount at the beginning of that same week”, and students must see that this resets the base to 80% for the second week. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Variables instead of numbers minor Understanding the question

    No starting amount is given, so students must supply one, such as 100 units, or reason directly with a percent of a percent. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Start with 100 units, take away 20 percent of the current amount for the first week, then take 20 percent of what is left for the second week.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent decrease
amount left=amount×(1−p100)\text{amount left} = \text{amount} \times \left(1 - \frac{p}{100}\right)
p is the percent taken away, here 20. Taking away 20% leaves 80% of the amount, so you multiply by 0.80. Each week starts from the amount left at the end of the week before.

  1. Understand the decay

    Each week, the chemical loses 20% of what it had at the start of that week. The amount at the start of each week keeps changing.

    So the second week’s 20% is taken from a smaller amount. You need the percent left after two weeks.

  2. Start with 100 units

    No amount is given, so pick an easy one. Start with 100 units. Then the number of units left is also the percent left.

  3. Work out week one

    20% of 100 is 20, so 20 units are lost.

    100−20=80100 - 20 = 80

    After week one, 80 units are left. That is 80% of the start.

    Watch out: 80% is only one week. Stopping here gives choice D.

  4. Work out week two

    Now the week starts with 80 units, not 100. Take 20% of 80.

    20100×80=16\frac{20}{100} \times 80 = 16

    80−16=6480 - 16 = 64

    After two weeks, 64 units are left. That is 64% of the 100 units you started with.

    Watch out: Taking 20 away twice from 100 gives 60%, choice B. That uses the old amount in week two.

    So the answer is C, 64%.

Check your answer

Use the percent decrease formula. Each week keeps 80%, or 0.80, of the amount. Two weeks gives 0.80×0.80=0.640.80 \times 0.80 = 0.64, which is 64%. The two methods agree.

More questions like this

The original DOE question
Question 85 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 85. Open page 158 of the handbook (PDF)

Question 120 of 120

2018 Test B Math, question 30

Question

Three students stand at the starting line of a running track and begin running laps at the same time. Ann completes 1 lap every 2 minutes, Jack completes 1 lap every 3 minutes, and Lee completes 1 lap every 4 minutes. How many laps does Ann complete before all three runners are once again at the starting line at the same time?

  • A4
  • B6
  • C12
  • D20
Show the answer

Correct answer: B

Find the least common multiple of 2, 3, and 4 — which is 12. So, it takes 12 minutes before all three are back at the starting line. Ann completes 1 lap every 2 minutes, so in 12 minutes she has completed 6 laps.

Question analysis

Three runners start together and finish laps every 2, 3 and 4 minutes, and the question asks how many laps Ann, the fastest, completes before all three are at the starting line together again. It tests least common multiples in a word problem. The catch is keeping track of which quantity the question wants once the common time is found.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Sixth and seventh graders find least common multiples of two numbers in isolation; this applies the idea to three runners in a story and adds a conversion from minutes to laps.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 8, 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

Two quantities are in play, minutes and laps, and the question asks for the second while the natural first result is the first. Students who find the 12-minute meeting time often stop there, and students who do count laps can count Jack's or Lee's by mistake. The question rewards tracking whose laps are asked for and converting the time into laps at the very end.

  1. Turning words into math moderate Understanding the question

    "All three runners are once again at the starting line at the same time" has to be recognized as the first common multiple of 2, 3 and 4 minutes; the wording never says multiple. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Finding a least common multiple is assumed; multiplying the three lap times instead gives 24 minutes, which is a common time but not the first one. (Anderson & Krathwohl, 2001)

Strategy: Find the first time that is a multiple of all three lap times, then divide that time by Ann's lap time to count her laps.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of laps
laps=total time÷time for one lap\text{laps} = \text{total time} \div \text{time for one lap}
Total time is how long the runner runs. Time for one lap is how long one trip around the track takes. Dividing tells you how many of those trips fit into the total time.

  1. Sort out what you know

    Ann runs 1 lap every 2 minutes. Jack runs 1 lap every 3 minutes. Lee runs 1 lap every 4 minutes. They all start together.

    A runner is back at the start line at the end of every lap. First find when all three are “once again at the starting line at the same time”. Then count Ann’s laps up to that moment.

  2. List when each runner is back

    Write the minutes when each runner reaches the line.

    • Ann: 2, 4, 6, 8, 10, 12 …
    • Jack: 3, 6, 9, 12 …
    • Lee: 4, 8, 12 …

    The first number in all three lists is 12. This is the least common multiple of 2, 3 and 4. It is the smallest number that all three divide into with nothing left over. So they meet after 12 minutes.

  3. Count Ann’s laps

    Use the laps formula. Ann runs for 12 minutes, and each lap takes her 2 minutes.

    laps=12÷2=6\text{laps} = 12 \div 2 = 6

    Ann runs 6 laps.

    Watch out: 12 is the number of minutes, not laps. Choosing it gives C. Dividing by Jack’s 3 minutes gives his 4 laps, which is A.

  4. Give the answer

    The question asks only about Ann, and she runs 6 laps in 12 minutes. So the answer is B, 6.

Check your answer

Check that everyone is back at 12 minutes. Ann: 6×2=126 \times 2 = 12. Jack: 4×3=124 \times 3 = 12. Lee: 3×4=123 \times 4 = 12. No earlier time works. At 6 minutes, Lee is only halfway through his second lap.

More questions like this

The original DOE question
Question 87 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 87. Open page 158 of the handbook (PDF)