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Question 1 of 43

2026 Practice Test B Math, question 4

Question

Maria is playing a game in which players earn points called minks, zogs, and barts.

$1\text{ mink} = 3\text{ zogs}$

$1\text{ mink} = \frac{1}{4}\text{ bart}$

Maria has 60 zogs and 24 barts. She wants to exchange her zogs and barts for minks.

How many minks will Maria receive from the exchange?

Enter your answer in the space.

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

Show the answer

Correct answer: 116

Step 1: Convert the zogs to minks.
Since $1\text{ mink} = 3\text{ zogs}$ , divide the number of zogs Maria has by 3: $60 \div 3 = 20$ .

So, $60\text{ zogs} = 20\text{ minks}$ .

Step 2: Convert the barts to minks.
Since $1\text{ mink} = \frac{1}{4}\text{ bart}$ , multiply the number of barts by 4 (because 1 bart is equal to 4 minks): $24 \times 4 = 96$ .

So, $24\text{ barts} = 96\text{ minks}$ .

Step 3: Add the mink conversions.
Add the results from Steps 1 and 2 to find the total number of minks Maria will receive: $20 + 96 = 116$ .

Question analysis

In a game with made-up points called minks, zogs and barts, Maria trades 60 zogs and 24 barts for minks, and students type how many minks she receives. It tests unit conversion with invented units and a fraction. One rate says a mink equals a fraction of a bart, so deciding whether to multiply or divide is the tricky part.

NYS standards: 6.RP.3d converting measurement units
6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Converting units with a given rate is grade 6 ratio reasoning. Invented units remove any real-world sense check, and the two conversions run in opposite directions, one through a fraction, which matches the care grade 7 problems need.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 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

Each step is one multiplication, division or addition; the risk is direction. The two rates point opposite ways, three zogs to a mink but four minks to a bart, and the fraction in the second rate invites dividing by 4. Asking of each unit whether one of it is worth more or less than a mink fixes the direction before any arithmetic, and it also shows why the 60 and the 24 can never simply be added.

Turning words into math moderate Understanding the question
Both rates put the mink on the left, so “1 mink = 1/4 bart” has to be turned around into 1 bart = 4 minks, and the invented units give no sense of which is worth more. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Two conversions run in opposite directions, dividing the zogs by 3 and multiplying the barts by 4, and their results are added, so a slip in either step changes the total. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The total is typed into a box with no choices to compare, so a conversion done the wrong way still produces a believable whole number that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: For each kind of point, decide whether one of it is worth more or less than a mink, then multiply or divide to turn it into minks, and add the two results.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

26 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Gets 20 minks from the zogs but divides the barts by 4, 24 ÷ 4 = 6, as if a bart were worth a quarter of a mink, for 20 + 6 = 26.
276 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies the 60 zogs by 3, as if each zog were worth 3 minks, then adds the 96 minks from the barts correctly: 180 + 96 = 276.
84 Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the 60 zogs and 24 barts as they stand, 60 + 24 = 84, without converting either kind of point into minks first.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Converting with a rate: $\text{minks} = \text{number of points} \times \text{minks for one point}$; Number of points is how many zogs or barts Maria has. Minks for one point is what one zog, or one bart, is worth in minks. Find that worth first for each kind of point. Then multiply, because every point of the same kind trades for the same number of minks.

Sort out the question
Maria trades two kinds of points for minks. You know:
- 3 zogs trade for 1 mink.
- 1 mink is worth $\frac{1}{4}$ of a bart.
- Maria has 60 zogs and 24 barts.
You need the total number of minks. Zogs and barts are different units, like dimes and dollars. So you cannot just add them.
Watch out: 60 + 24 = 84 adds zogs to barts. That is not a number of minks.
Find what one zog is worth
It takes 3 zogs to make 1 mink. So one zog is only part of a mink. One zog is worth $\frac{1}{3}$ of a mink.
A zog is smaller than a mink. So Maria will get fewer minks than she has zogs.
Turn the bart rule around
The rule says 1 mink is $\frac{1}{4}$ of a bart. So a mink is a quarter of a bart. It takes 4 quarters to make a whole:
$4 \times \frac{1}{4} = 1$
So 4 minks make 1 bart. One bart is worth 4 minks. A bart is bigger than a mink.
Watch out: the fraction tempts you to divide by 4. Then 24 barts become only 6 minks, and the total is 26.
Change the zogs into minks
Use the rate. Each of the 60 zogs is worth 1/3 of a mink:
$60 \times \frac{1}{3} = 60 \div 3 = 20$
Multiplying by 1/3 is the same as dividing by 3. So 60 zogs trade for 20 minks.
Watch out: multiplying 60 by 3 gives 180. That treats a zog as bigger than a mink, and the total becomes 276.
Change the barts into minks
Each of the 24 barts is worth 4 minks:
$24 \times 4 = 96$
So 24 barts trade for 96 minks.
Add the minks
Both amounts are now in minks, so you can add them:
$20 + 96 = 116$
Maria receives 116 minks. Type only the number in the box, with no units. So you would enter 116.

Check your answer

Work backward. 20 minks are worth $20 \times 3 = 60$ zogs. 96 minks are worth $96 \times \frac{1}{4} = 24$ barts. Both match what Maria started with. Also, a bart is worth more than a mink, so 24 barts should give many more than 24 minks. They do.

More questions like this

The original DOE question

Question 4 as shown on the DOE’s digital test, page 76 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 76, question 4.

Question 2 of 43

2026 Practice Test B Math, question 17

Question

How many positive integers satisfy the inequality $x + 7 < 23$ ?

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

Solve the inequality for x.

$x + 7 < 23$

$x < 16$

There are 15 positive integers that are less than 16.

Question analysis

This inequality question asks how many positive integers make x + 7 < 23 true, and the count is typed into a box. Students solve a one-step inequality, then count its whole-number solutions. The strict less-than sign and the word positive decide where the count starts and stops, so both ends need checking.

NYS standards: 7.EE.4b two-step inequalities
6.EE.5 solutions and substitution
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 inequality problems usually take two steps, such as 2x − 5 > 9. This one needs a single subtraction; the care goes into counting the whole-number solutions and deciding whether the boundary itself counts.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: An answer in an unexpected form minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 12, 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

One subtraction solves the inequality, so the points are lost in the count that follows. The boundary 16 is the last number the solving produces and is easy to type in, or to count up to; checking that 16 + 7 is 23, which is not less than 23, rules it out. Listing 1 through 15, the positive integers below the boundary, settles the count.

An answer in an unexpected form minor Choosing the answer
Solving gives x < 16, but the box asks how many positive integers make the inequality true, a count that still has to be made after the solving is done. (Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Two conditions set the ends of the count at once: the word positive starts it at 1 rather than 0, and the strict < stops it below 16. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Undo the addition to find the boundary for x, then count the positive whole numbers below it, checking the largest one and the boundary in the original inequality.

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).
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

16 Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Solves to x < 16 and then enters 16, either the boundary itself or a count of 1 through 16, though x = 16 makes x + 7 equal 23, not less than it.
29 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves the 7 to the right side without changing its sign, 23 + 7 = 30, and counts the 29 positive integers below 30.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides: $\text{if } a < b \text{, then } a - c < b - c$; a and b are the two sides of an inequality, and c is any number. Taking the same number away from both sides keeps the inequality true. Think of two people who each walk down the same number of stairs: the lower one stays lower. Use it to get x by itself.

Sort out the question
An inequality compares two amounts. The sign < means is less than. A positive integer is a whole number bigger than 0: 1, 2, 3, and so on. You need to count how many of them make $x + 7 < 23$ true. The answer is a count, not a value of x.
Get x by itself
The 7 is added to x. Subtracting 7 undoes adding 7, so subtract 7 from both sides:
$x + 7 - 7 < 23 - 7$
$x < 16$
So x must be less than 16.
Watch out: adding 7 to 23 instead gives x < 30, and a count of 29.
Check the boundary number
The number 16 is the boundary, where the solutions stop. Does 16 itself work? Try it:
$16 + 7 = 23$
But 23 is not less than 23. It is equal. So 16 does not count. The largest number that works is 15, since $15 + 7 = 22$ , and 22 is less than 23.
Watch out: entering 16 counts the boundary, which fails.
Count the positive integers
The numbers that work start at 1, the smallest positive integer. Zero is not positive. They stop at 15:
$1, 2, 3, \ldots, 14, 15$
Counting from 1 up to 15 gives 15 numbers.
Enter the count
There are 15 positive integers that make the inequality true. Type the count in the box as a plain whole number. So you would enter 15.

Check your answer

Test both ends. For 1: $1 + 7 = 8$ , and 8 is less than 23, so 1 works. For 15: 22 is less than 23, so 15 works. For 16: 23 is not less than 23, so 16 fails. That leaves exactly the numbers 1 through 15.

More questions like this

The original DOE question

Question 17 as shown on the DOE’s digital test, page 89 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 89, question 17.

Question 3 of 43

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:

$5.5\text{ oz} \times 6 = 33\text{ oz}$

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:

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

Cross multiply to solve for x in the proportion:

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

$33x = \$229.12$

$x \approx \$6.94$

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.

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

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Many steps moderate
Error-prone arithmetic minor
An answer in an unexpected form minor
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.

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)
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)
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)
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

7 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the juice's cost per gallon, about $6.94, rounds it to $7, and enters that without subtracting the $3.99 price of the milk.
2 Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Rounds the price per ounce, 1.79 ÷ 33 ≈ 0.054, to $0.05, so a gallon costs $6.40 and the difference, $2.41, rounds to 2.
2.95 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Works out the difference of $2.95 correctly but enters it as it stands, though the question asks for it to the nearest dollar.

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: $\text{fluid ounces} = 128 \times \text{gallons}$; 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: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; 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.

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.
Find the ounces in a 6-pack
Each of the 6 bottles holds 5.5 fluid ounces:
$6 \times 5.5 = 33$
So $1.79 buys 33 fluid ounces of juice.
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:
$\frac{33}{1.79} = \frac{128}{x}$
Cross-multiply:
$33 \times x = 1.79 \times 128 = 229.12$
Find the cost of a gallon
Divide both sides by 33:
$x = 229.12 \div 33 \approx 6.94$
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.
Subtract the milk’s price
Find how much more the juice costs:
$6.94 - 3.99 = 2.95$
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.
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 $4 \times 33 = 132$ . Four packs cost $4 \times 1.79 = 7.16$ 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 4 of 43

2026 Practice Test C Math, question 22

Question

During a hiking trip, a group of students recorded the elevation (in feet) at several points along the trail:

$-120\text{ ft},\ -80\text{ ft},\ 0\text{ ft},\ 45\text{ ft},\ -30\text{ ft},\ 90\text{ ft}$

What is the difference between the highest and lowest elevations recorded on the trail?

Enter your answer in the space.

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

Show the answer

Correct answer: 210

Determine the difference between the highest and lowest elevations:

Determine the highest elevation:

Of −120, −80, 0, 45, −30, and 90, the greatest value is 90 because its location is farthest to the right on a number line.

Highest elevation $= 90\text{ ft}$

Determine the lowest elevation:

Of −120, −80, 0, 45, −30, and 90, the lowest value is −120 because its location is farthest to the left on a number line.

Lowest elevation $= -120\text{ ft}$

Determine the difference between the two elevations:

$90\text{ ft} - (-120)\text{ ft} = 90\text{ ft} + 120\text{ ft} = 210\text{ ft}$

Question analysis

A group of students on a hiking trip records six elevations in feet, some below zero and some above, and finds the difference between the highest and the lowest, typing the result. It tests ordering positive and negative numbers and subtracting integers. Spotting the lowest reading among several negatives and subtracting a negative number correctly are where errors creep in.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7b order statements in context
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the gap between a positive and a negative value by subtracting is core grade 7 work with integers. Picking the extremes from six readings is quick, so the demand is the subtraction of a negative number.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 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

Picking the highest and lowest of six readings is quick, though −120 must be recognized as lower than −80 and −30. The points are lost in the subtraction: the span runs 120 feet up to zero and 90 more above it, and students who treat the readings as plain sizes find only 30. Picturing the two readings on a vertical number line shows the whole span.

Error-prone arithmetic moderate Working it out
The difference is 90 − (−120), and subtracting a negative means adding 120, so the part of the trail below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The difference is typed with no choices, so 30 or −30, the results of mishandling the negative, gets no warning that the two readings lie on opposite sides of zero. (Enright, Morley & Sheehan, 2002)

Strategy: Pick the greatest and the least elevation by their places on a number line, then subtract the least from the greatest, turning the subtraction of a negative into an addition.

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.)
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).
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

30 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Subtracts the sizes, 120 − 90 = 30, as if both readings were on the same side of zero, leaving out the stretch below zero.
-30 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Combines the two extremes by adding, 90 + (−120) = −30, the net of the readings, instead of subtracting the lowest from the highest.
-210 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Subtracts in the reverse order, lowest minus highest, −120 − 90 = −210, and enters a negative difference.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Difference between highest and lowest: $\text{difference} = \text{highest} - \text{lowest}$; The difference between the highest and lowest values is how far apart they are. Subtract the lowest from the highest, so the answer is never negative.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Subtracting a negative is the same as adding a positive. On a number line, it means moving to the right.

Sort out the question
Elevation is height above or below a starting level, here 0 feet. Negative elevations are below 0. The six readings are −120, −80, 0, 45, −30 and 90 feet. You need the difference between the highest and the lowest.
Find the highest and the lowest
Picture the readings on a number line. The highest is farthest right: 90 feet. The lowest is farthest left: −120 feet.
For negative numbers, the bigger the number after the minus sign, the lower it is. So −120 is lower than −80.
Subtract the lowest from the highest
Use the difference formula. Subtracting a negative is the same as adding:
$90 - (-120) = 90 + 120 = 210$
So the two readings are 210 feet apart.
Watch out: $120 - 90 = 30$ treats both readings as if they were above 0. Adding them, $90 + (-120) = -30$ , is wrong too.
Enter the answer
A difference like this is a distance, so it is positive. Subtracting in the other order gives −210, which is wrong. Type only the number, with no units.
So you would enter 210.

Check your answer

Count the distance in two parts. From −120 feet up to 0 is 120 feet. From 0 up to 90 is 90 more feet. Together, $120 + 90 = 210$ feet. That matches.

More questions like this

The original DOE question

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

Question 5 of 43

2026 Practice Test C Math, question 29

Question

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

SMOOTHIE RECIPE

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 0.25

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

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

$1.5 = 6p$

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

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

The constant of proportionality is 0.25.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 6 of 43

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

$84^{\circ}\text{ F} - 39^{\circ}\text{ F} = 45^{\circ}\text{ F}$

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

$78^{\circ}\text{ F} - 48^{\circ}\text{ F} = 30^{\circ}\text{ F}$

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

$45^{\circ}\text{ F} - 30^{\circ}\text{ F} = 15^{\circ}\text{ F}$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
Many steps minor
Something you must already know minor
An answer in an unexpected form minor
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.

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)
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)
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)
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

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.

The wrong answers, classified

14 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: strong
Reads the left whisker’s end as 40, the nearest labeled tick, instead of 39, so the range becomes 84 − 40 = 44 and 44 − 30 = 14.
45 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the range, 84 − 39 = 45 degrees, and enters it, stopping before the interquartile range of 30 is subtracted from it.
17 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads the box’s right edge as 76, the tick just left of it, instead of 78, so the IQR becomes 76 − 48 = 28 and 45 − 28 = 17.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range: $\text{range} = \text{maximum} - \text{minimum}$; 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): $\text{IQR} = Q_3 - Q_1$; 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.

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.
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 $40 - 1 = 39$ degrees.
- The box’s right edge is halfway between 76 and 80. Halfway is 78 degrees.
Find the range
Use the range formula:
$\text{range} = 84 - 39 = 45$
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.
Find the IQR
Use the IQR formula with the two edges of the box:
$\text{IQR} = 78 - 48 = 30$
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.
Subtract the IQR from the range
The question asks for the difference between the two:
$45 - 30 = 15$
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 $48 - 39 = 9$ degrees long. The right whisker is $84 - 78 = 6$ degrees long. $9 + 6 = 15$ , 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 7 of 43

2026 Practice Test C Math, question 42

Question

Tariq is making a clay block in the shape of a cube with a side length of $20\text{ cm}$ . He hollows out the center by removing clay to form a rectangular prism-shaped hole with a base of $5\text{ cm} \cdot 5\text{ cm}$ and a height of $20\text{ cm}$ .

$A cube labeled 20 cm along its front right edge. A tall rectangular prism-shaped hole runs from the top face of the cube down to the bottom face, the full height of the cube; its square opening on the top face is shaded gray and labeled 5 cm across. Hidden edges of the cube and the hole are dashed.$

What is the volume, in cubic centimeters, of the block after the clay is removed from the center?

Enter your answer in the space.

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

Show the answer

Correct answer: 7,500

First, find the volume of the cube using the formula $V = lwh$ .

The length, width, and height of the cube are each equal to 20 cm.

$V_c = 20\text{ cm} \times 20\text{ cm} \times 20\text{ cm} = 8{,}000\text{ cm}^3$

Next, find the volume of the cut-out square prism, with equal length and width of 5 cm and height of 20 cm.

$V_p = 5\text{ cm} \times 5\text{ cm} \times 20\text{ cm} = 500\text{ cm}^3$

Finally, subtract the volume of the prism from the volume of the cube.

$8{,}000\text{ cm}^3 - 500\text{ cm}^3 = 7{,}500\text{ cm}^3$

Question analysis

Tariq makes a clay cube 20 centimeters on each side and removes a rectangular prism from its center, with a 5 cm by 5 cm base and a height of 20 cm. Students type the volume of the block that is left, in cubic centimeters. It tests the volume of rectangular prisms, and the hole's size must be read carefully from the words and figure.

NYS standards: 7.G.6 area, surface area, and volume
6.G.2 volume with fractional edge lengths
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the volume of a solid made from rectangular prisms, here by taking one away, is named in the grade 7 standards. This is a direct case with whole numbers, two volumes and one subtraction.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Reading a figure, table or graph minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 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

This is a direct composite-volume problem with whole numbers: one cube, one hole, one subtraction. What it checks is that both volumes are real volumes, the hole as tall as the cube, and that the subtraction is done at the end. Students who stop at the hole’s volume, or who subtract only the square face, give amounts that are easy to type and hard to doubt without choices.

Something you must already know minor Working it out
The volume of a rectangular prism, length times width times height, is not given, and it is needed twice, once for the 20 cm cube and once for the hole. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The figure labels only the 5 cm opening and the cube’s 20 cm edge; the hole’s 20 cm height comes from the words, while the drawing shows it reaching from top to bottom. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
An answer in an unexpected form minor Choosing the answer
The volume is typed with no choices to compare, so the hole’s 500 cubic centimeters or the cube’s 8,000 can be entered without any warning. (Enright, Morley & Sheehan, 2002)

Strategy: Find the volume of the full cube, then the volume of the square hole that runs from the top face to the bottom, and subtract the hole from the cube.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
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

500 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the volume of the hole, 5 × 5 × 20 = 500 cubic centimeters, and gives the clay removed instead of the clay that is left.
7,875 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Treats the hole as a 5 cm cube, 5 × 5 × 5 = 125, though it runs the full 20 cm height, so 8,000 − 125 = 7,875.
7,975 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Subtracts only the area of the square opening, 5 × 5 = 25, from the cube’s 8,000, mixing a face’s area with the hole’s volume.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a rectangular prism: $V = l \times w \times h$; V is the volume, the space inside a solid. l is the length, w is the width and h is the height. A cube is a prism with all three the same. Volume is measured in cubic centimeters, tiny cubes 1 cm on each side.

Sort out the question
Tariq’s block starts as a cube 20 cm on each side. He removes a hole shaped like a rectangular prism, a box shape with flat rectangle sides. Its base is 5 cm by 5 cm, and its height is 20 cm. The figure shows it running from the top face to the bottom face.
You need the volume of the clay that is left, in cubic centimeters.
Find the volume of the cube
Use the volume formula. The length, width and height are all 20 cm:
$V = 20 \times 20 \times 20 = 8{,}000$
The full cube holds 8,000 cubic cm of clay.
Find the volume of the hole
The hole is a prism too. Its length and width are 5 cm, and its height is 20 cm:
$V = 5 \times 5 \times 20 = 500$
So 500 cubic cm of clay are removed.
Watch out: the hole is 20 cm tall, not 5 cm. Treating it as a small 5 cm cube gives 125 and the wrong answer 7,875. And 5 × 5 = 25 is only the area of the opening, not a volume. Subtracting it gives 7,975, also wrong.
Subtract the hole from the cube
The clay left is the cube minus the hole:
$8{,}000 - 500 = 7{,}500$
The block has 7,500 cubic cm of clay left.
Watch out: 500 is the clay taken out, not the clay that is left.
Enter the number
Type only the number, with no units. You can type 7500 or 7,500.
So you would enter 7,500.

Check your answer

Think of the cube as 20 flat layers, each 1 cm thick. Each layer is a 20 by 20 square, 400 square cm, with a 5 by 5 hole, 25 square cm. So each layer holds $400 - 25 = 375$ cubic cm of clay. Then $20 \times 375 = 7{,}500$ . It matches.

More questions like this

The original DOE question

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

Question 8 of 43

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 $\frac{1}{40} = 0.025$ .

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

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: 6.SP.8a uniform probability models
6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding 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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: An answer in an unexpected form moderate
Error-prone arithmetic minor
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.

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)
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

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".
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

0.025 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the probability as a decimal, 1 ÷ 40 = 0.025, and enters it without multiplying by 100, though the box asks for a percentage.
25 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Loses the zero after the decimal point when dividing, taking 1 ÷ 40 as 0.25, as 1 ÷ 4 would be, and so enters 25.
1/40 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Enters the probability as the fraction 1/40, the form it was found in, overlooking the bold word percentage in the question.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; 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: $\text{percent} = \frac{a}{b} \times 100$; 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%.

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.
Write the probability as a fraction
Only 1 book is number 223, and there are 40 books in all:
$P = \frac{1}{40}$
The numbers on the labels do not matter beyond this. They just give each book its own name.
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:
$1 \div 40 = 0.025$
Watch out: there is a 0 right after the decimal point. Losing it gives 0.25 and then 25, which is wrong.
Change the decimal to a percent
Multiply by 100 to get a percent. That moves the decimal point two places to the right:
$0.025 \times 100 = 2.5$
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.
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 9 of 43

2026 Practice Test D Math, question 8

Question

A square metal plate has a side length of 3 inches. A circular hole with a diameter of 2 inches is cut out from the center of the plate.

$A square, shaded gray, with a white circle cut out of its center. A side of the square is labeled 3 in., and the diameter of the circle is labeled 2 in. The shaded region is the part of the square outside the circle.$

What is the area of the shaded region, to the nearest hundredth of a square inch?

Enter your answer in the space.

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

Show the answer

Correct answer: 5.86

To determine the area of a square, use $A = s^{2}$ .

Begin by finding the length of the side of the square, which is 3 inches. The area of the square is $(3\text{ in})^{2} = 9\text{ square inches}$ .

To determine the area of the circle, use $A = \pi r^{2}$ .

The radius of the circle (the hole) is one-half the diameter or $2\text{ in.} \div 2 = 1\text{ in.}$ , so the area of the entire circle is $\pi(1\text{ in.})^{2} \approx 3.14\text{ sq in.}$

To determine the area of the shaded region, subtract the area of the hole in the square from the area of the square:

$9\text{ sq in.} - \pi\text{ sq in.} =$

$9\text{ sq in.} - (3.14)\text{ sq in.} =$

$5.86\text{ sq in.}$

The shaded area to the nearest hundredth of a square inch $= 5.86$ sq in.

Question analysis

A 3-inch square metal plate has a circular hole 2 inches across cut from its center, and students find the area of the metal that is left, to the nearest hundredth of a square inch. It tests the areas of a circle and a square. Using the radius, not the diameter, and subtracting the right areas are the keys.

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 a circle's area is a grade 7 standard, and subtracting a hole from a shape is common grade 7 practice. The radius must come from the diameter and the result must be rounded, but the steps are routine.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Combines ideas minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 49 words.
Corrected from the DOE handbook: A wrong symbol 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

Every number is small, so the work is choosing the right pieces: the area of the square, the area of the hole, and the difference. A radius of 1 makes the circle's area simply π, which is easy to confuse with the circumference, 2π, when the formulas are half remembered. The region asked for is the gray metal, not the hole, and the final rounding goes up, since the third decimal place is 8.

Something you must already know moderate Working it out
Neither area formula is given, and the hole is labeled by its 2-inch diameter, so students must recall A = πr² and halve the diameter to get the radius 1 first. (Anderson & Krathwohl, 2001)
Combines ideas minor Working it out
The shaded metal is a square with a circle removed, so two area formulas are combined by subtraction rather than either one used alone. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
The result is typed to the nearest hundredth with no choices to compare, so the long decimal 5.858… must be rounded up, not cut off after two places. (Enright, Morley & Sheehan, 2002)

Strategy: Halve the diameter to get the radius, find the area of the square and of the circle, subtract the circle from the square, and round only at the end.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

2.72 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses the circumference, 2π × 1 ≈ 6.28, in place of the circle's area and subtracts it from the square's 9 square inches, getting about 2.72.
3.14 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the area of the circular hole, π × 1² ≈ 3.14 square inches, and enters it instead of the area of the shaded metal left around it.
5.85 Rounds too early (King et al., 2004)
Reveals a slip Expected pull: weak
Works the area out correctly as about 5.858 but cuts off the extra digits instead of rounding up to the nearest hundredth.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation gives the area of the hole as “π(1 in.)² ≅ 3.14 sq in.” The symbol ≅ means “is congruent to”, which an area and a number cannot be; the line means “is approximately equal to”, written ≈. The site now says “π(1 in.)² ≈ 3.14 sq in.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s^{2}$; A is the area, the amount of flat space inside. s is the length of one side. s² means s × s. It works because a square with side s holds s rows of s little unit squares.
Area of a circle: $A = \pi r^{2}$; A is the area. r is the radius, the distance from the center to the edge. It is half the diameter, the distance straight across through the center. π (pi) is a number a little more than 3, about 3.14. r² means r × r.

Sort out the question
The plate is a square with 3-inch sides. A circle with a 2-inch diameter is cut out of the middle. The diameter is the distance straight across the circle through its center. The shaded region is the metal that is left. You need its area to the nearest hundredth of a square inch.
Find the area of the square
Use the formula for the area of a square, with s = 3 inches:
$A = s^{2} = 3^{2} = 3 \times 3 = 9$
The whole plate, before the hole is cut, is 9 square inches.
Find the radius of the hole
The circle formula needs the radius, not the diameter. The figure labels the 2 in. distance across the hole. The radius is half of it:
$r = 2 \div 2 = 1$
The radius is 1 inch.
Find the area of the hole
Use the formula for the area of a circle, with r = 1 inch:
$A = \pi r^{2} = \pi \times 1^{2} = \pi \approx 3.14$
The sign ≈ means is about equal to. The hole is about 3.14 square inches.
Watch out: 2πr is the circumference, the distance around the circle, not its area. Using it gives 9 − 6.28 = 2.72. And 3.14 is the hole, not the metal, so do not stop there.
Subtract the hole from the square
The shaded metal is the square with the circle taken away:
$9 - \pi \approx 9 - 3.14 = 5.86$
The shaded region is about 5.86 square inches.
Round and enter the answer
With more digits of π, 3.1416, you get 9 − 3.1416 = 5.8584. The digit after the hundredths place is 8, which is 5 or more, so round up to 5.86. Chopping it to 5.85 is wrong. Type only the number, with no units.
So you would enter 5.86.

Check your answer

The hole is a bit more than a third of the plate, since 3.14 is a bit more than 9 ÷ 3 = 3. So the metal left should be a bit less than two thirds of 9, which is 6. The answer 5.86 is just under 6, so it makes sense.

More questions like this

The original DOE question

Question 8 as shown on the DOE’s digital test, page 81 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 81, question 8. This is the DOE’s version, before the corrections listed in the analysis.

Question 10 of 43

2026 Practice Test D Math, question 16

Question

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

BREAD RECIPE

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 0.5

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

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

$1 = 2p$

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

$0.5 = p$

The constant of proportionality is 0.5.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 11 of 43

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: $(90 - 45) + 1 = 46$ pages

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

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

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

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

$\frac{128}{500} = \frac{128 \div 4}{500 \div 4} = \frac{32}{125}$

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

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Many steps moderate
An answer in an unexpected form minor
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.

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)
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)
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

25 Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Subtracts the first page from the last for each day, 45 + 35 + 45 = 125 pages, forgetting that an inclusive range counts both end pages, so 125 ÷ 500 = 25%.
0.256 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Counts the 128 pages and divides by 500 correctly, then enters the decimal 0.256, one step short of multiplying by 100 to make a percent.
26 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds 25.6% correctly but rounds it to the nearest whole percent, overlooking the instruction to round to the nearest tenth.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting pages in an inclusive range: $\text{pages} = \text{last page} - \text{first page} + 1$; Inclusive means the first and last pages are both read. Subtracting counts the steps from the first page to the last. That misses one page, so add 1. Test it: pages 1 to 3 are 3 pages, and 3 − 1 + 1 = 3.
Percent of a total: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the amount you have, here the pages 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.

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.
Count the pages each day
Use the rule for an inclusive range: last page minus first page, plus 1.
$90 - 45 + 1 = 46$
$135 - 100 + 1 = 36$
$190 - 145 + 1 = 46$
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.
Add the three days
Add the pages from all three days to get her total:
$46 + 36 + 46 = 128$
Leah read 128 pages in all. This is the part you compare with the goal.
Change to a percent
Divide the pages read by the goal, then multiply by 100:
$\frac{128}{500} \times 100 = 0.256 \times 100 = 25.6$
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.
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: $0.256 \times 500 = 128$ 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 12 of 43

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:

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

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:

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

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

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Much to keep track of minor
Something you must already know minor
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.

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)
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

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

The wrong answers, classified

632 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Takes the first multiple of 8 after 630, 8 × 79 = 632, and stops without checking divisibility by 5, which a number ending in 2 fails.
645 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Picks a number ending in 5 so that 5 divides it, then slips on the check for 8, which leaves a remainder of 5.
40 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Finds the least common multiple of 5 and 8, 40, and enters it, one step before finding its multiple between 630 and 650.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisible by 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: $\text{LCM}(5, 8) = 5 \times 8 = 40$; 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.

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.
Use the rule for 5
A number divisible by 5 ends in 0 or 5. Between 630 and 650, only three numbers do:
$635, \; 640, \; 645$
Only these need checking for 8.
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 $8 \times 79$ , passes for 8 but ends in 2, so it fails for 5.
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. $40 \times 16 = 640$ . 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 13 of 43

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 $3 \cdot w$ or $3w$ .

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

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

Solve the equation:

$8 + 3w = 26$

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

$w = 6$ (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: 7.EE.4a two-step equations
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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
An answer in an unexpected form minor
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.

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)
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

18 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Subtracts the $8 already saved from the $26 total and stops, giving the amount saved over all 3 weeks rather than in each week.
26/3 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides the $26 total by the 3 weeks, 26 ÷ 3 = 8 2/3, about $8.67, forgetting that $8 of it was saved before the weekly saving began.
3 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds the weekly amount and then halves it for each person, though the question asks what Marcus and Isla save together each week.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total savings: $\text{total} = \text{start} + \text{weeks} \times \text{amount each week}$; 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: $\text{if } a = b \text{, then } a - c = b - c \text{ and } \frac{a}{c} = \frac{b}{c}$; 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.

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.
Write an equation
Let w be the dollars saved each week. Three weeks add 3w dollars to the $8 they started with:
$8 + 3w = 26$
Subtract the starting money
Take 8 away from both sides:
$3w = 26 - 8 = 18$
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.
Divide by the number of weeks
Divide both sides by 3:
$w = 18 \div 3 = 6$
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.
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: $8 + 6 = 14$ . After week 2: $14 + 6 = 20$ . After week 3: $20 + 6 = 26$ . 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 14 of 43

2024 Test A Math, question 1

Question

Simplify:

$-3.8 + 2.3 - (-1.1)$

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

Show the answer

Correct answer: -0.4

Subtracting a negative number is the same as adding a positive number. Rewrite the expression to simplify:

$-3.8 + 2.3 - (-1.1)$

$-3.8 + 2.3 + 1.1$

$-3.8 + 3.4$

$-0.4$

Question analysis

This grid-in asks students to simplify a short expression that mixes negative and positive decimals and ends by subtracting a negative number. It tests adding and subtracting rational numbers in order. The double negative, and keeping track of the sign of the final result, are where errors creep in.

NYS standards: 7.NS.1c rational number subtraction
7.NS.1b rational number addition
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding and subtracting negative decimals is core grade 7 content, and this is a routine three-term version. It matches a typical classroom exercise; the only extra demand is gridding a signed decimal correctly.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 16+, 2 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 three-term expression with small decimals, so the method is never in doubt; the points are lost on signs. The double negative at the end is the classic place to slip; the other is the result itself, which sits just below zero and is easy to grid without its minus sign. Grouping the positive terms first, 2.3 + 1.1, and comparing that sum with 3.8 settles both the size and the sign.

Error-prone arithmetic moderate Working it out
The last term, − (−1.1), has to turn into + 1.1, and the total then crosses zero, since 2.3 + 1.1 = 3.4 falls just short of the 3.8 below zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in it shows no choices, so a double negative read as a single one, or a result gridded without its minus sign, gets no warning from the page. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite subtracting −1.1 as adding 1.1, combine the two positive decimals first, then compare their total with 3.8 to decide the sign and 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.)

The wrong answers, classified

-2.6 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Treats subtracting −1.1 as subtracting 1.1, so it works out −3.8 + 2.3 − 1.1 = −1.5 − 1.1 and moves further below zero.
0.4 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Rewrites the double negative correctly and finds the gap 3.8 − 3.4 = 0.4, but drops the minus sign that the larger negative term gives the result.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number: $a - (-b) = a + b$; a is any number, and b is a positive number, so −b is negative. Taking away a negative number is the same as adding the positive number. Think of money: if someone takes away a debt of b dollars, you are b dollars better off.
Adding a negative and a positive: $-a + b = -(a - b)$; a and b are positive numbers, and a is the larger one. A negative and a positive partly cancel out. So find the difference of their sizes, a − b. The result keeps the sign of the number farther from zero, which here is the negative one.

Sort out the expression
You need to simplify $-3.8 + 2.3 - (-1.1)$ . Simplify means work it out to one number. Two of the numbers are negative: −3.8 and −1.1. One is positive: 2.3. The tricky part is at the end. It says minus negative 1.1. The answer can be negative, so keep track of every sign.
Change minus a negative to plus
Taking away a negative is the same as adding a positive. Use the rule with b = 1.1.
$a - (-b) = a + b$
$-3.8 + 2.3 - (-1.1) = -3.8 + 2.3 + 1.1$
Now only one negative number is left. Watch out: do not just subtract 1.1. That gives $-3.8 + 2.3 - 1.1 = -2.6$ , which is wrong.
Add the two positive numbers
Group the positive numbers first. This makes the rest of the work short.
$2.3 + 1.1 = 3.4$
Now the problem is $-3.8 + 3.4$ . One number is negative and one is positive.
Combine the negative and the positive
They partly cancel. Use the rule with a = 3.8 and b = 3.4.
$-a + b = -(a - b)$
$-3.8 + 3.4 = -(3.8 - 3.4) = -0.4$
The negative number, 3.8, is farther from zero than 3.4 is. So the result is negative. Watch out: keep the minus sign. Dropping it gives 0.4, which is wrong. So you would grid -0.4, with the minus sign in its own column.

Check your answer

Use a number line. Start at −3.8. Adding 2.3 moves you right, to −1.5. Subtracting −1.1 also moves you right, by 1.1, to −0.4. You end just to the left of zero. So a small negative answer makes sense.

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 2025 Admissions, page 75, question 58. Open page 75 of the handbook (PDF)

Question 15 of 43

2024 Test A Math, question 2

Question

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

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

Show the answer

Correct answer: 150

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

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

$r + m = 180$

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

$r = 5m$

$5m + m = 180$

$6m = 180$

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

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

Therefore, the measure of angle R is 150°.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 16 of 43

2024 Test A Math, question 3

Question

A juice mixture contains $\frac{3}{16}$ gallon of apple juice and $\frac{3}{40}$ gallon of cranberry juice. How many gallons of apple juice per gallon of cranberry juice does the mixture contain? (Express your answer as a decimal.)

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

Show the answer

Correct answer: 2.5

Set up a proportion to solve:

$\frac{\frac{3}{16}\text{ apple}}{\frac{3}{40}\text{ cranberry}} = \frac{x\text{ apple}}{1\text{ cranberry}}$

$\frac{3}{16} \times \frac{40}{3} = x$

$\frac{40}{16} = x$

$2.5 = x$

Question analysis

This ratio grid-in describes a juice mixture with a fraction of a gallon of apple juice and a smaller fraction of a gallon of cranberry juice. Students find the unit rate, gallons of apple juice per gallon of cranberry juice, by dividing one fraction by the other and writing it as a decimal. Dividing in the wrong order is the main risk.

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of two fractions is exactly the grade 7 standard, and this is a textbook example of it. The only added demands are dividing in the right order and answering as a decimal.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 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 one division of fractions, but the question tests whether the division runs the right way. The word per tells which juice goes on top, and a student who divides the smaller amount by the larger gets a tidy decimal for the reverse rate. A sense check helps: the mixture holds more apple juice than cranberry, so the rate must be greater than 1.

Turning words into math moderate Understanding the question
The phrase "gallons of apple juice per gallon of cranberry juice" fixes the order: apple's 3/16 goes on top and cranberry's 3/40 below, though the two fractions invite either order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic moderate Working it out
Dividing 3/16 by 3/40 means multiplying by 40/3; the 3s cancel to leave 40/16, which still has to be reduced and written as a decimal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The answer must be gridded as a decimal, and with no choices shown, the upside-down 0.4 gives no sign that it answers the reverse question. (Enright, Morley & Sheehan, 2002)

Strategy: Put the apple juice over the cranberry juice and divide the fractions by multiplying by the reciprocal; the matching 3s cancel, leaving a simple fraction to turn into a decimal.

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).

The wrong answers, classified

0.4 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides cranberry by apple, 3/40 ÷ 3/16 = 16/40 = 0.4, giving gallons of cranberry per gallon of apple, the rate turned upside down.
0.014 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Multiplies the two amounts instead of dividing them, 3/16 × 3/40 = 9/640 ≈ 0.014, as if comparing the juices called for a product.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $r = a \div b$; r is the rate: how much of one thing there is for each 1 of another. a is the amount of the first thing. b is the amount of the second thing, the one named after the word per. Dividing shares a out evenly over each 1 of b.
Dividing by a fraction: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a/b and c/d are fractions, and b, c and d are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. It works because flipping a fraction gives the number that undoes it.

Sort out the question
The mix has $\frac{3}{16}$ gallon of apple juice and $\frac{3}{40}$ gallon of cranberry juice. You need the “gallons of apple juice per gallon of cranberry juice”. Per means for each. So the question asks: for each 1 gallon of cranberry juice, how many gallons of apple juice are there? The answer must be a decimal.
Set up the division
To find an amount for each 1 gallon, divide. Apple juice goes first, because it comes before per. Cranberry juice comes second.
$r = a \div b$
$r = \frac{3}{16} \div \frac{3}{40}$
Watch out: do not flip the order. Cranberry divided by apple gives 0.4. That is gallons of cranberry juice for each gallon of apple juice, the reverse question.
Multiply by the reciprocal
To divide by $\frac{3}{40}$ , multiply by its reciprocal, $\frac{40}{3}$ .
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
$\frac{3}{16} \times \frac{40}{3} = \frac{3 \times 40}{16 \times 3} = \frac{40}{16}$
The 3 on top and the 3 on the bottom cancel, since $3 \div 3 = 1$ . Watch out: do not just multiply the two amounts. That gives $\frac{9}{640}$ , about 0.014, which is far too small.
Change the fraction to a decimal
Make $\frac{40}{16}$ simpler by dividing the top and bottom by 8. Then divide.
$\frac{40}{16} = \frac{5}{2} = 5 \div 2 = 2.5$
There are 2.5 gallons of apple juice for each gallon of cranberry juice. So you would grid 2.5.

Check your answer

Both fractions have 3 on top. The one with the smaller bottom, $\frac{3}{16}$ , is bigger. So there is more apple juice than cranberry juice, and the answer must be more than 1. That rules out 0.4. Also, $3 \div 40 = 0.075$ and $0.075 \times 2.5 = 0.1875$ , which is $3 \div 16$ .

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 2025 Admissions, page 75, question 60. Open page 75 of the handbook (PDF)

Question 17 of 43

2024 Test A Math, question 4

Question

Mr. Chan’s lawn grows $2\frac{1}{8}$ inches every 2 weeks. He mows his lawn every 2 weeks and cuts off the top $1\frac{3}{4}$ 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 $2\frac{1}{8}$ inches every 2 weeks. Find the total number of inches it grows in 8 weeks.

$8 \div 2 = 4$

$4 \times 2\frac{1}{8}$

$4 \times 2.125 = 8.5$

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

$8.5 + 4 = 12.5$

Mr. Chan mows his lawn every 2 weeks and cuts off the top $1\frac{3}{4}$ inches. Find the total number of inches that are cut off in 8 weeks.

$8 \div 2 = 4$

$4 \times 1\frac{3}{4}$

$4 \times 1.75 = 7$

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

$12.5 - 7 = 5.5$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Error-prone arithmetic moderate
Turning words into math minor
An answer in an unexpected form minor
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.

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)
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)
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)
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

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

1.5 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the net gain of 3/8 inch per period and multiplies by the 4 periods, 1.5 inches, but stops before adding the lawn's starting height of 4 inches.
12.5 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Adds four periods of growth, 8.5 inches, to the starting 4 inches and stops there, never subtracting the 1 3/4 inches cut at each of the four mowings.
7 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Applies the net gain of 3/8 inch to each of the 8 weeks instead of each 2-week period, 8 × 3/8 = 3, then adds the starting 4.

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: $a\frac{b}{c} = a + b \div c$; 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: $\text{total} = n \times a$; 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.

Sort out the question
The lawn starts at 4 inches tall. Every 2 weeks it grows $2\frac{1}{8}$ inches. Every 2 weeks Mr. Chan also cuts off $1\frac{3}{4}$ inches. You need the height after 8 weeks, as a decimal. Three things matter: the start, the growing and the cutting.
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.
$8 \div 2 = 4$
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.
Change the mixed numbers to decimals
The answer must be a decimal, so change both amounts now.
$2\frac{1}{8} = 2 + 1 \div 8 = 2.125$
$1\frac{3}{4} = 1 + 3 \div 4 = 1.75$
Each period, the lawn grows 2.125 inches and loses 1.75 inches.
Add up the growing
The lawn grows 2.125 inches, 4 times.
$\text{total} = n \times a$
$\text{total} = 4 \times 2.125 = 8.5$
It grows 8.5 inches in all. Add the starting height of 4 inches.
$4 + 8.5 = 12.5$
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.
Add up the cutting
Mr. Chan cuts off 1.75 inches, 4 times.
$\text{total} = 4 \times 1.75 = 7$
He cuts off 7 inches in all.
Subtract the cutting
$12.5 - 7 = 5.5$
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 $2.125 - 1.75 = 0.375$ inch. In 4 periods that is $4 \times 0.375 = 1.5$ inches. Add the starting 4 inches: $4 + 1.5 = 5.5$ . 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 18 of 43

2024 Test A Math, question 5

Question

RESULTS FROM SURVEY OF 110 FAMILIES

Number of Children in the Family	Number of Families
0	45
1	32
2	19
3	8
4	6

The table above shows the number of children in each of 110 families. What is the median number of children in these families?

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

Show the answer

Correct answer: 1

Since 110 families were surveyed, the median will have 55 values less than or equal to it, and 55 values greater than or equal to it.

If the values in the table were listed in order, they would look like this:

$\overbrace{0,\ldots,0}^{45},\overbrace{1,\ldots,1}^{32},\overbrace{2,\ldots,2}^{19},\overbrace{3,\ldots,3}^{8},\overbrace{4,\ldots,4}^{6}$

The first row of the table indicates that 45 families had 0 children.

The second row of the table indicates that 32 families had 1 child.

So 77 families had either 0 children or 1 child, and 45 families had 0 children.

Since 55 is between 45 and 77, the median must be a value in the second row of the table.

So the median is 1.

Question analysis

This statistics grid-in gives a frequency table from a survey of 110 families, showing how many families have each number of children from none to four. Students find the median number of children per family. The table lists counts rather than the 110 data values themselves, so locating the middle value takes a running total of the families.

NYS standards: 6.SP.5c range, center, and overall pattern
6.SP.5a number of observations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 median problems usually list a dozen or so values. Here 110 values are hidden in a frequency table, so students must reason about positions with a running count, a less familiar and more error-prone task.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
No obvious method moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Students know a median as the middle of a list, but here the list is never written out. Treating the table's columns as the data gives quick, confident answers, 2 from the middle of the children column or 19 from the middle of the family counts, and neither can be checked against choices. The question rewards counting positions: 45 families have no children, so the 55th and 56th families both fall among the 32 with one child.

Reading a figure, table or graph major Understanding the question
The table hides 110 data values in five rows: the 0 to 4 column lists the possible values, and 45, 32, 19, 8 and 6 say how many families have each. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
No obvious method moderate Working it out
The median of 110 values lies between the 55th and 56th; finding them takes a running count of families, 45 and then 77, a method seldom practiced with tables. (Schoenfeld, 1985)
An answer in an unexpected form minor Choosing the answer
As a grid-in there are no choices to compare, so a student who takes the middle row's 2 has nothing on the page to suggest it is wrong. (Enright, Morley & Sheehan, 2002)

Strategy: With 110 values the median sits between the 55th and 56th; add the family counts row by row until the running total passes 55, and read that row's number of children.

Traps

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

The wrong answers, classified

2 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Takes the middle entry of the children column, 0 to 4, ignoring that each number of children belongs to a different number of families.
19 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: moderate
Finds the median of the Number of Families column, 6, 8, 19, 32 and 45, a middle family count rather than a number of children.
1.07 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: weak
Computes the mean instead, 118 children among 110 families, about 1.07, a different measure of center from the median asked for.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Spot of the median: $\text{middle spot} = \frac{n + 1}{2}$; The median is the middle value when all the values are put in order. n is how many values there are. This gives the spot of the middle value, with the same number of values before and after it. If it ends in a half, the median is halfway between the two values on either side.

Sort out the table
The table covers 110 families. The left column is a number of children. The right column says how many families have that many children. For example, 45 families have 0 children. You need the median number of children. The median is the middle value when all the values are in order.
Picture the values in order
Think of a list with one number for each family: how many children it has. The list has 45 zeros, then 32 ones, then 19 twos, then 8 threes, then 6 fours. It is already in order, from least to most.
$45 + 32 + 19 + 8 + 6 = 110$
There are 110 values. Watch out: the median is not the middle row, 2. Each row stands for a different number of families.
Find the middle spot
Use the median rule with n = 110.
$\text{middle spot} = \frac{n + 1}{2}$
$\frac{110 + 1}{2} = 55.5$
The middle is halfway between the 55th and 56th values. So you need those two values in the list.
Count down the list
Keep a running total of families, row by row. The zeros fill spots 1 to 45. Now add the 32 families with 1 child.
$45 + 32 = 77$
So the ones fill spots 46 to 77. Spot 55 and spot 56 are both in that range. So both values are 1.
Give the median
The 55th and 56th values are both 1. Halfway between 1 and 1 is 1. So the median is 1 child. Watch out: the median is a number of children, not of families. Gridding 19, from the families column, is wrong. So is the mean, or average, which is about 1.07. So you would grid 1.

Check your answer

Half of 110 is 55. Families with 0 children: 45, which is less than half. So the median is not 0. Families with 0 or 1 child: 77, which is more than half. So the median is not more than 1. It must be 1.

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 2025 Admissions, page 76, question 62. Open page 76 of the handbook (PDF)

Question 19 of 43

2024 Test B Math, question 1

Question

$\frac{3}{5}(2x+5)-2x$

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 $\frac{3}{5}$ over $(2x+5)$ to get the expression $\frac{6}{5}x+3-2x$ .

Express $-2x$ as $-\frac{10}{5}x$ and then add like terms:

$\frac{6}{5}x+3+\left(-\frac{10}{5}x\right)=-\frac{4}{5}x+3$ .

The coefficient of x is $-\frac{4}{5}$ .

When expressed as a decimal,

$-\frac{4}{5}=-0.8$ .

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
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.

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)
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

0.8 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Combines 6/5x and −2x to a coefficient of the right size, 4/5, but drops the minus sign, or subtracts 6/5 from 2, and grids 0.8.
1.2 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Distributes 3/5 to get 6/5x + 3 and grids 6/5 = 1.2, stopping before the −2x outside the parentheses is combined with it.
-1.4 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Multiplies 3/5 by 2 as 6/10, doubling the denominator along with the numerator, so the coefficient comes out 0.6 − 2 = −1.4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property: $a(b + c) = a \times b + a \times c$; 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: $\frac{a}{b} \times c = \frac{a \times c}{b}$; 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.

Sort out the question
You start with this expression. An expression is a math phrase with numbers, letters and signs, but no equals sign.
$\frac{3}{5}(2x + 5) - 2x$
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.
Multiply out the parentheses
The 3/5 multiplies everything inside the parentheses. Use the distributive property.
$a(b + c) = a \times b + a \times c$
$\frac{3}{5}(2x + 5) = \frac{3}{5} \times 2x + \frac{3}{5} \times 5$
Multiply only the top of 3/5 by 2: $\frac{3}{5} \times 2 = \frac{6}{5}$ . Also, $\frac{3}{5} \times 5 = 3$ . Now the whole expression is:
$\frac{6}{5}x + 3 - 2x$
Watch out: do not double the bottom number too. Writing 6/10 leads to −1.4.
Write 2x in fifths
There are now two x terms: $\frac{6}{5}x$ and $-2x$ . To combine them, they need the same denominator (bottom number). One whole is 5/5, so 2 wholes are 10/5.
$2x = \frac{10}{5}x$
Watch out: the −2x is outside the parentheses, but it still counts. Stopping at 6/5 and gridding 1.2 is a common mistake.
Combine the x terms
Six fifths take away ten fifths leaves negative four fifths.
$\frac{6}{5}x - \frac{10}{5}x = -\frac{4}{5}x$
The 3 has no x, so it stays as it is. The simplified expression is $-\frac{4}{5}x + 3$ . So the coefficient of x is $-\frac{4}{5}$ . Watch out: 6 − 10 is negative. Dropping the minus sign gives 0.8.
Change it to a decimal
A fraction means the top divided by the bottom: $4 \div 5 = 0.8$ . Keep the minus sign.
$-\frac{4}{5} = -0.8$
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: $\frac{3}{5} \times 7 - 2 = 4.2 - 2 = 2.2$ . The simplified form: $-0.8 + 3 = 2.2$ . 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 20 of 43

2024 Test B Math, question 2

Question

Two number cubes have sides labeled 1 through 6. Both number cubes are rolled. How many different outcomes will have a sum of 6?

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

Show the answer

Correct answer: 5

There will be 5 different outcomes with a sum of 6:

$1+5=6$

$2+4=6$

$3+3=6$

$4+2=6$

$5+1=6$

Question analysis

A counting grid-in with two number cubes, each numbered 1 to 6, rolled together. Students count how many different outcomes give a sum of 6. The catch is deciding whether a 2 on the first cube with a 4 on the second counts as a different outcome from the reverse.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students list sample spaces for two number cubes when finding compound probabilities. This asks for one count from that list, a short task, though as a grid-in it offers no choices to expose an incomplete list.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 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

The arithmetic is trivial; the question turns on what counts as a different outcome. Treating the cubes as a first and a second settles it: the first cube runs from 1 to 5 and the second is forced each time. Students who think in pairs of numbers rather than in rolls of two cubes undercount, and those who double every pair overcount by one at the double 3 and 3.

Something you must already know moderate Working it out
Nothing in the stem says that a 2 on the first cube with a 4 on the second differs from the reverse; students must already know that two cubes give 36 ordered outcomes. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in it offers no list of counts to compare, so a student who counts unordered pairs gets no sign that ordered outcomes were wanted. (Enright, Morley & Sheehan, 2002)

Strategy: Treat the cubes as a first and a second cube, run the first cube's value from 1 to 6, and for each value check whether the second cube can complete the sum.

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.
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

3 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Counts unordered pairs, 1 and 5, 2 and 4, 3 and 3, treating a 1 on the first cube with a 5 on the second as the same outcome as the reverse.
6 Overlapping groups (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Knows order matters and doubles the 3 pairs to 6, but 3 and 3 reads the same both ways, so that outcome is counted twice.
36 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Grids the total number of outcomes for two cubes, 6 × 6 = 36, instead of only the outcomes whose sum is 6.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle: $\text{total outcomes} = a \times b$; a is the number of ways the first cube can land, and b is the number of ways the second cube can land. Multiply them to count every pair. It works because each face of the first cube can go with each face of the second.

Sort out the question
You roll two number cubes. Each one shows 1, 2, 3, 4, 5 or 6. An outcome is one way the two cubes can land. You need the number of outcomes whose sum (the two numbers added) is 6.
The question does not say if order matters. The answer key counts the cubes as a first cube and a second cube. So 1 on the first and 5 on the second is a different outcome from 5 on the first and 1 on the second.
See why order counts
Picture a red cube and a blue cube. Red 1 with blue 5 is not the same roll as red 5 with blue 1. Both can happen, so both count.
This is how all the outcomes of two cubes are counted. Use the counting principle.
$\text{total outcomes} = 6 \times 6 = 36$
Watch out: 36 is every outcome, not just the ones that add up to 6.
List the pairs that make 6
Go through the first cube one number at a time. The second cube has to make up the rest of the 6.
First cube Second cube Sum
1 5 6
2 4 6
3 3 6
4 2 6
5 1 6
The first cube cannot show 6. Then the second cube would need 0, and no face shows 0.
Count the outcomes
The table has 5 rows, so there are 5 outcomes. Watch out: counting 1 and 5 as the same as 5 and 1 gives only 3. Watch out: 3 and 3 is one outcome, not two. Doubling 3 pairs to get 6 counts it twice.
So you would grid 5.

First cube	Second cube	Sum
1	5	6
2	4	6
3	3	6
4	2	6
5	1	6

Check your answer

Each number from 1 to 5 on the first cube goes with exactly one number on the second cube. A 6 on the first cube goes with nothing. So there are exactly 5 outcomes. The table shows the same pattern: as the first number goes up by 1, the second goes down by 1.

More questions like this

The original DOE question

Question 59 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 59. Open page 198 of the handbook (PDF)

Question 21 of 43

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 $(13-8)$ .

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 $(6-3)$ .

So the total perimeter of the scale drawing is

$13+3+5+3+8+6=38\text{ cm}$

Use the scale to find the perimeter of the actual garden:

$\frac{1\text{ cm}}{2.5\text{ m}}=\frac{38\text{ cm}}{p\text{ m}}$

$p=2.5\times 38=95$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Combines ideas minor
An answer in an unexpected form minor
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.

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)
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)
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

75 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: strong
Adds only the four labeled sides, 13 + 3 + 8 + 6 = 30 cm, and scales that to 75 m, missing the two unlabeled sides of the notch.
38 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the drawing's perimeter, 38 cm, correctly and grids it without applying the scale of 2.5 meters per centimeter.
15.2 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides the 38 cm perimeter by 2.5 instead of multiplying, shrinking the garden though each centimeter of the drawing stands for 2.5 meters.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter: $P = \text{sum of all the side lengths}$; 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: $\text{actual length} = \text{drawing length} \times s$; 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.

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.
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.
$13 - 8 = 5$
The missing side across is 5 cm.
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.
$6 - 3 = 3$
The missing side going down is 3 cm.
Add all six sides
$P = \text{sum of all the side lengths}$
$P = 13 + 3 + 5 + 3 + 8 + 6 = 38$
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.
Change to real meters
Each centimeter stands for 2.5 meters, so multiply by 2.5.
$\text{actual length} = \text{drawing length} \times s$
$\text{actual length} = 38 \times 2.5 = 76 + 19 = 95$
Here $38 \times 2 = 76$ and $38 \times 0.5 = 19$ . 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: $2 \times (13 + 6) = 38$ . 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 22 of 43

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.

$4\times 3.5=14$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Much to keep track of minor
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.

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)
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)
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

7 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds 14° at 11:00 a.m. but subtracts 7 instead of −7, so 14 − 7 = 7, leaving out the part of the rise below zero.
14 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the 11:00 a.m. temperature, 4 × 3.5 = 14, and grids it without subtracting the 5:00 a.m. temperature to get the increase.
10 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Measures the rise from 9:00 a.m. instead of 5:00 a.m., 14 − 4 = 10, which is only the last two hours of the total increase.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Change in temperature: $\text{change} = \text{end} - \text{start}$; 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: $a - (-b) = a + b$; 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.

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.
Find the 11:00 temperature
“3.5 times the 9:00 a.m. temperature” means 3.5 times 4°F.
$3.5 \times 4 = 14$
At 11:00 a.m. it was 14°F. Watch out: 14 is a temperature, not the increase. Gridding 14 stops too early.
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.
$\text{change} = \text{end} - \text{start}$
$\text{change} = 14 - (-7)$
Watch out: start at 5:00 a.m., not at 9:00 a.m. Starting from 4°F gives only 10.
Subtract the negative number
Taking away a negative number is the same as adding its size.
$a - (-b) = a + b$
$14 - (-7) = 14 + 7 = 21$
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. $7 + 14 = 21$ . Or add the two parts of the morning: it rose 11 degrees by 9:00 and 10 more by 11:00, and $11 + 10 = 21$ .

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 23 of 43

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 $\frac{1}{2}bh=\frac{1}{2}(4)(8)=16$ square inches.

The height of the triangular prism is 2 inches.

So the volume of the triangular prism, in cubic inches, is $16\bullet 2=32$ .

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Reading a figure, table or graph minor
An answer in an unexpected form minor
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.

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)
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)
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

64 Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 4 × 8 × 2 = 64 as for a rectangular box, leaving out the one half in the area of the triangular base.
16 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the triangle's area, 1/2 × 4 × 8 = 16 square inches, and stops before multiplying by the prism's 2-inch height.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, 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: $V = B \times H$; 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.

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.
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.
Find the area of the triangle
$A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times 4 \times 8 = 16$
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.
Multiply by the prism’s height
The height of the prism is the distance between the two triangles. That is the 2 in. edge.
$V = B \times H$
$V = 16 \times 2 = 32$
The volume is 32 cubic inches. Watch out: do not forget the one half. $4 \times 8 \times 2 = 64$ 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 $4 \times 8 \times 2 = 64$ 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 24 of 43

2023 Test A Math, question 1

Question

Ms. Li opened a retirement account with a deposit of $2,500. This account earns 4% simple interest annually. How many years will it take her to earn $500 on her $2,500 deposit?

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

Show the answer

Correct answer: 5

Simple interest (y) is calculated by multiplying the initial deposit (p), the interest rate (r), and time (t):

$prt = y$

$(2{,}500)(0.04)t = 500$

$100t = 500$

$t = 5$

Question analysis

Ms. Li deposits $2,500 in a retirement account that pays 4% simple interest each year, and this grid-in asks how many years it takes to earn $500 in interest. Students find one year's interest and work out how many years are needed. It turns the usual simple interest question around, asking for the time rather than the interest.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Simple interest is a grade 7 percent application, usually asked as how much interest builds up over a given time. This one gives the interest and asks for the time, a short reverse step with friendly numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Turning words into math minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once one year's interest is known, $100, the rest is counting how many years make $500, so the arithmetic is light. The question turns the usual simple interest problem around, giving the interest and asking for the time, and that reversal is where students stall or grid a percent instead of a number of years. Writing 4% as 0.04, not 0.4, keeps the yearly interest from coming out ten times too big.

Something you must already know minor Working it out
The stem says 4% simple interest annually but never says that each year adds the same 4% of the $2,500 deposit, $100, so the yearly amount has to come from memory. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The question gives the interest, $500, and asks for the time, so the familiar formula runs backward: 2,500 × 0.04 × t = 500 has to be solved for t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
As a grid-in it shows no choices, so 20, the interest as a percent of the deposit, can be gridded with nothing on the page to say it is not a number of years. (Enright, Morley & Sheehan, 2002)

Strategy: Work out one year's interest as 4% of $2,500, then find how many of those yearly amounts add up to the $500 of interest.

Traps

Stops one step early: A choice is a number from partway through the solution.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

20 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that the $500 of interest is 20% of the $2,500 deposit and grids 20, stopping before dividing by the 4% earned each year.
0.5 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Writes 4% as 0.4 instead of 0.04, so one year's interest looks like $1,000 and the $500 seems to take only half a year.
125 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the $500 of interest by the 4 in 4%, combining two numbers from the stem without first finding 4% of the deposit.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest: $I = P \times r \times t$; I is the interest, the extra money the account pays you. P is the deposit, the money you put in. r is the yearly rate as a decimal. t is the time in years. Simple means each year’s interest is figured on the deposit only, so every year adds the same amount.
Percent to decimal: $p\% = \frac{p}{100}$; p is the percent. Percent means out of 100. So divide by 100 to write it as a decimal. For example, 4% is 4 ÷ 100, which is 0.04.

Sort out the question
Ms. Li puts $2,500 into an account. This first amount is called the deposit. Each year the account pays 4% simple interest. Interest is extra money the bank pays you for keeping your money there. You need the number of years it takes to earn $500 of interest. The answer is a number of years, not dollars and not a percent.
Find one year’s interest
First write 4% as a decimal. 4% means 4 out of 100.
$4\% = \frac{4}{100} = 0.04$
Now use the formula for 1 year, so t = 1.
$I = P \times r \times t$
$I = 2{,}500 \times 0.04 \times 1 = 100$
The account earns $100 each year. Watch out: 4% is 0.04, not 0.4. Using 0.4 makes one year’s interest $1,000, and you would grid 0.5 years.
Find how many years make $500
Simple interest adds the same $100 every year. So the time t must make the interest $500.
$2{,}500 \times 0.04 \times t = 500$
$100 \times t = 500$
$t = 500 \div 100 = 5$
It takes 5 years. Watch out: $500 is 20% of $2,500. But 20 is a percent, not a number of years. Don’t grid 20. So you would grid 5.

Check your answer

Count up one year at a time: $100, $200, $300, $400, $500. That is 5 years. Or put t = 5 back into the formula: $2{,}500 \times 0.04 \times 5 = 500$ . That matches the $500 in the question.

More questions like this

The original DOE question

Question 25 of 43

2023 Test A Math, question 2

Question

$6 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}$

What is the value of the expression shown above?

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

Show the answer

Correct answer: -9

$6 - 9 \div \left\|-3\right\| + (-2)^{3} \times 1\frac{1}{2}$	Calculate the exponent and convert the mixed number to an improper fraction.
$6 - 9 \div \left\|-3\right\| + (-8) \times \frac{3}{2}$	Next, calculate the absolute value.
$6 - 9 \div 3 + (-8) \times \frac{3}{2}$	Perform the indicated multiplication and division.
$6 - 3 + (-12)$	Perform the addition and subtraction.
$3 + (-12) = -9$

Question analysis

This grid-in asks for the value of an arithmetic expression that mixes an absolute value, a negative number raised to the third power, a division and a mixed number. Students apply the order of operations with signed numbers. Cubing a negative base, and doing the division and multiplication before any adding, are where most slips happen.

NYS standards: 7.NS.2a rational number multiplication
6.EE.2c evaluating expressions and order of operations
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students routinely evaluate expressions with negative numbers in the right order. This one combines an absolute value, a cubed negative and a mixed number in a single line, so it takes more steps than usual, each with a sign to watch.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 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 expression packs four kinds of notation into one line: a subtraction, a division by an absolute value, a negative base raised to the third power and a mixed number. None of the operations is hard alone, but each has a rule that is easy to break under time pressure, and the slips lead to very different totals. Rewriting the power and the mixed number first, then dividing and multiplying before any addition, avoids them.

Error-prone arithmetic moderate Working it out
(−2)³ must stay negative, −8, and the mixed number 1½ must become 3/2 before multiplying, so the last product is −12; a sign or fraction slip here moves the total far from −9. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Six numbers and six operations, including a power and an absolute value, have to be worked in a planned order, and every addition waits until the division and multiplication are done. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The value is negative and the grid-in offers no choices, so the minus sign has to be gridded and nothing on the page confirms that the total falls below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the absolute value and the power first, change the mixed number to an improper fraction, do the division and multiplication, and only then add and subtract from left to right.

Traps

Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

15 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes (−2)³ as positive 8, as if every power of a negative were positive, so the last product is +12 and 6 − 3 + 12 = 15.
-13.5 Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Works strictly from left to right: 6 − 9 = −3, divided by 3 is −1, plus −8 is −9, and times 1½ gives −13.5.
-6 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Multiplies the base by the exponent, taking (−2)³ as −6, so the product with 3/2 is −9 and the total is 6 − 3 − 9 = −6.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\text{grouping} \rightarrow \text{powers} \rightarrow \times \text{ and } \div \rightarrow + \text{ and } -$; This is the order for working out an expression. First work inside grouping symbols, such as parentheses and absolute value bars. Next do powers. Then multiply and divide from left to right. Last, add and subtract from left to right. Using one order means everyone gets the same answer.
A power: $a^{3} = a \times a \times a$; a is the base, the number being multiplied. The small 3 is the exponent. It says how many times the base is used as a factor, a number being multiplied. It does not mean a times 3.
Absolute value: $\left|a\right| = \text{the distance from } a \text{ to } 0$; The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so the absolute value of −3 is 3.
Mixed number to fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; a is the whole number part, and b over c is the fraction part. Each whole is c pieces, so a wholes are a × c pieces. Add the b extra pieces. The bottom number stays c.

Sort out the expression
You need the value of $6 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}$ . It has many operations, so the order matters. It has absolute value bars, a power of a negative number and a mixed number. The answer may be negative, so keep track of every sign.
Do the absolute value and the power
Grouping symbols and powers come first.
$\left|-3\right| = 3$
$(-2)^{3} = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$
Two negatives multiply to a positive, 4. One more negative makes it negative again. Watch out: (−2)³ is −8, not +8. Using +8 leads to 15. It is also not −2 × 3. That mistake leads to a total of −6.
Change the mixed number to a fraction
A fraction is easier to multiply than a mixed number.
$1\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}$
Now the expression looks like this:
$6 - 9 \div 3 + (-8) \times \frac{3}{2}$
Multiply and divide next
Multiplying and dividing come before adding and subtracting.
$9 \div 3 = 3$
$(-8) \times \frac{3}{2} = \frac{-24}{2} = -12$
A negative times a positive is negative. Watch out: don’t just work from left to right. Doing 6 − 9 first, then dividing by 3, leads to −13.5.
Add and subtract last
Now only adding and subtracting are left. Work from left to right.
$6 - 3 + (-12) = 3 + (-12) = -9$
Adding −12 is the same as taking away 12. So you would grid −9, with the minus sign in its own column.

Check your answer

Look at the size of each part. The part 6 − 3 is a small positive number, 3. The last part, −8 times 1½, is a bigger negative number, −12. A big negative plus a small positive is negative. So an answer below zero, −9, makes sense.

More questions like this

The original DOE question

Question 26 of 43

2023 Test A Math, question 3

Question

Solve for x:

$7x + 3 - 2(2x + 1) = 13$

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

Show the answer

Correct answer: 4

$7x + 3 - 2(2x + 1) = 13$	Apply the distributive property; multiply the −2 by $2x$ and 1.
$7x + 3 - 4x - 2 = 13$	Combine like terms.
$3x + 1 = 13$	Apply the additive inverse property; subtract 1 from both sides of the equation.
$3x = 12$	Apply the multiplicative inverse property; divide both sides of the equation by 3.
$x = 4$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
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.

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)
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

3 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Multiplies only the 2x by −2 and leaves the + 1 untouched, so the equation becomes 3x + 4 = 13 and x = 3.
8/3 Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Gives the minus sign to 4x but not to the 2, writing −4x + 2, so the equation becomes 3x + 5 = 13 and x = 8/3.
12 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Simplifies correctly to 3x + 1 = 13, subtracts 1 to reach 3x = 12, and grids 12 without dividing by 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property: $a(b + c) = a \times b + a \times c$; 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: $\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c$; 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.

Sort out the equation
You need the value of x that makes $7x + 3 - 2(2x + 1) = 13$ true. x is the unknown number. The tricky part is the −2 in front of the parentheses. It multiplies everything inside them.
Multiply out the parentheses
Use the distributive property with a = −2, b = 2x and c = 1.
$-2(2x + 1) = (-2) \times 2x + (-2) \times 1 = -4x - 2$
So the equation becomes this:
$7x + 3 - 4x - 2 = 13$
Watch out: the −2 must multiply the 1 too. Leaving + 1 gives x = 3. Writing + 2 instead of − 2 gives $x = \frac{8}{3}$ .
Combine like terms
Like terms are terms of the same kind. Put the x terms together. Put the plain numbers together.
$7x - 4x = 3x$
$3 - 2 = 1$
$3x + 1 = 13$
Get x by itself
Undo the + 1 first. Subtract 1 from both sides, so the equation stays balanced.
$3x + 1 - 1 = 13 - 1$
$3x = 12$
3x means 3 times x. So divide both sides by 3.
$x = 12 \div 3 = 4$
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: $7 \times 4 + 3 - 2(2 \times 4 + 1) = 28 + 3 - 2 \times 9 = 31 - 18 = 13$ . That matches the 13 on the right side, so x = 4 is right.

More questions like this

The original DOE question

Question 27 of 43

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:

$90 + 35 + x = 180$

$125 + x = 180$

$x = 55$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know moderate
An answer in an unexpected form minor
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.

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)
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)
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

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

125 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Adds the right angle and the 35° angle, 90 + 35 = 125, and grids that sum without subtracting it from the triangle's 180°.
35 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Assumes line m crosses l and n at equal angles, as it would cross two parallel lines, though l and n are perpendicular.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Vertical angles: $\text{angle } a = \text{angle } b$; 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: $A + B + C = 180$; 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.

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.
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°.
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.
Use the triangle’s angle sum
The triangle’s three angles are 90°, 35° and x°.
$A + B + C = 180$
$90 + 35 + x = 180$
$125 + x = 180$
$x = 180 - 125 = 55$
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: $35 + 55 = 90$ . 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 28 of 43

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:

$\frac{x}{8} = 17$

$x = 136$

Next, subtract the 3 discarded numbers from the sum:

$136 - 9 - 11 - 20 = 96$

Finally, calculate the mean of the remaining 5 numbers by dividing the new sum, 96, by 5:

$\frac{96}{5} = 19.2$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Much to keep track of minor
An answer in an unexpected form minor
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.

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)
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)
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

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

The wrong answers, classified

12 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds the remaining total, 136 − 40 = 96, correctly but divides it by the original 8 numbers instead of the 5 that are left.
96 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Subtracts the three discarded numbers from the total of 136 and grids the new sum, 96, without dividing by the 5 numbers left.
27.2 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Turns the mean into the total, 136, and divides it by 5 without first subtracting the three discarded numbers, 9, 11 and 20.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean, or average: $\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}$; 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: $\text{sum} = \text{mean} \times \text{how many numbers}$; 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.

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.
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.
$\text{sum} = \text{mean} \times \text{how many numbers}$
$\text{sum} = 17 \times 8 = 136$
The 8 numbers add up to 136.
Take away the three discarded numbers
Subtract the three numbers that were thrown away.
$136 - 9 - 11 - 20 = 136 - 40 = 96$
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.
Divide by the 5 numbers left
Now use the mean formula for the numbers that are left.
$\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}$
$\text{mean} = \frac{96}{5} = 19.2$
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, $19.2 \times 5 = 96$ , the sum of the 5 numbers left.

More questions like this

The original DOE question

Question 29 of 43

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:

$\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000}$

$x = \frac{(2,250)(1,000)}{9{,}000} = \frac{2{,}250}{9} = 250$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Reading a figure, table or graph minor
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.

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)
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

Stops one step early: A choice is a number from partway through the solution.
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

25 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds that 2,250 of 9,000 sales is 25 percent and enters that percent, stopping before taking a quarter of next month’s 1,000 vehicles.
650 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads the four-door row instead of the two-door row and scales it correctly: 5,850 of 9,000 is 650 of 1,000.
187.5 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Divides last year’s 2,250 two-door sales by 12 months, a monthly average that ignores the stated 1,000 vehicles for next month.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportion: $\frac{\text{part}}{\text{whole}} = \frac{\text{part}}{\text{whole}}$; 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.

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.
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.
$\frac{2{,}250}{9{,}000} = \frac{1}{4}$
This works because $2{,}250 \times 4 = 9{,}000$ . So 1 out of every 4 vehicles sold was a two-door car.
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.
$\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000} = \frac{1}{4}$
To get x alone, multiply both sides by 1,000.
$x = \frac{1}{4} \times 1{,}000 = 250$
So about 250 two-door cars will be sold.
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 $\frac{1}{9}$ of last year’s 9,000. So take $\frac{1}{9}$ of the 2,250 two-door cars: $2{,}250 \div 9 = 250$ . 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 30 of 43

2023 Test B Math, question 2

Question

In a scale drawing of a rectangular garden, the length is 15 inches and the width is 9 inches. In the drawing, 2 inches represents 3 yards. What is the width of the actual garden, in yards?

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

Show the answer

Correct answer: 13.5

Let w represent the width of the actual garden. Set up a proportion based on the information given:

$\frac{w}{9} = \frac{3}{2}$

$w = \frac{3(9)}{2} = \frac{27}{2} = 13.5$

Question analysis

In a scale drawing of a rectangular garden, 2 inches stands for 3 yards, and the drawing measures 15 inches by 9 inches. This grid-in asks for the real width of the garden in yards. Students apply the scale to one side only: the length is given but plays no part, and the scale has to be used the right way round.

NYS standards: 7.G.1 scale drawings
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings are grade 7 content, and this is a typical one: a single length converted with a stated scale. The only extra demand is using the width rather than the length and entering the result with no choices to check.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 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 drawing gives two lengths and only the width matters, so the first decision is which number to scale. The second is direction: the real garden must measure more in yards than the drawing’s 9 in inches, which a student can check before entering anything. Errors come from scaling the length, turning the scale over, or treating 2 to 3 as an increase of one.

Turning words into math minor Understanding the question
“2 inches represents 3 yards” must become 1.5 yards for every inch of drawing, so the 9-inch width is multiplied by 3/2, not by 2/3. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The answer is a decimal entered with no choices to compare, so a result from the 15-inch length or from an upside-down scale looks just as finished. (Enright, Morley & Sheehan, 2002)

Strategy: Find how many yards one inch of the drawing stands for, then multiply the drawing's width, not its length, by that number.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

22.5 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Applies the scale to the 15-inch length instead of the 9-inch width: 15 × 3 ÷ 2 = 22.5 yards.
6 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Turns the scale upside down, multiplying the 9 inches by 2/3 instead of 3/2, so the real garden comes out smaller than the drawing’s number.
10 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Reads 2 inches to 3 yards as adding 1, keeping a constant difference instead of a constant ratio, so 9 inches becomes 9 + 1 = 10 yards.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scale proportion: $\frac{\text{real size}}{\text{drawing size}} = \frac{\text{real size}}{\text{drawing size}}$; A scale drawing shrinks everything by the same amount. So real size over drawing size makes the same fraction for every part of the drawing. Keep the real sizes on top on both sides, and the drawing sizes on the bottom.

Sort out the question
The drawing shows a rectangle, a four-sided shape with square corners. On the drawing, the length is 15 inches and the width is 9 inches. The scale says 2 inches on the drawing stand for 3 yards in real life. You need the real width, in yards.
Only the width matters. The 15-inch length is extra.
Find what one inch stands for
If 2 inches stand for 3 yards, then 1 inch stands for half as much.
$3 \div 2 = 1.5$
So each inch on the drawing is 1.5 yards in real life. That means the number of yards will be bigger than the number of inches.
Set up the proportion and solve
Let w be the real width, in yards. Put yards on top and inches on the bottom, on both sides.
$\frac{w}{9} = \frac{3}{2}$
To get w alone, multiply both sides by 9.
$w = \frac{3 \times 9}{2} = \frac{27}{2} = 13.5$
The real width is 13.5 yards.
Give the answer
Watch out: don’t flip the scale. Using $9 \times \frac{2}{3}$ gives 6, which is smaller than the drawing’s number. And don’t use the length: 15 inches would give 22.5 yards. So you would grid 13.5.

Check your answer

Count groups of 2 inches. The 9-inch width holds 4.5 groups of 2 inches. Each group stands for 3 yards, so the real width is $4.5 \times 3 = 13.5$ yards. That matches.

More questions like this

The original DOE question

Question 59 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 59. Open page 191 of the handbook (PDF)

Question 31 of 43

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:

$72 = 2^{3} \times 3^{2}$

$60 = 2^{2} \times 3 \times 5$

$84 = 2^{2} \times 3 \times 7$

Then, find the factors that all three have in common, which is $2^{2} \times 3 = 12$ .

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 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.

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)
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

6 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Spots that 6 divides 72, 60 and 84 and stops there, a common factor not yet multiplied by the extra 2 that all three share.
2520 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Finds the least common multiple of 72, 60 and 84, 2,520, confusing a number all three counts divide into with one that divides all three.
4 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Takes only the shared 2 × 2 from the prime factorizations and drops the factor 3 that 72, 60 and 84 also share.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor: $\text{GCF} = \text{product of the shared prime factors}$; 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.

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.
Break each number into primes
Write each number as a product of primes. A product is the answer to a multiplication.
$72 = 2 \times 2 \times 2 \times 3 \times 3$
$60 = 2 \times 2 \times 3 \times 5$
$84 = 2 \times 2 \times 3 \times 7$
You can check each one by multiplying it back out.
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.
$\text{GCF} = 2 \times 2 \times 3 = 12$
Watch out: keep every shared prime. Using only one 2 gives 6. Leaving out the 3 gives 4.
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 $72 \div 12 = 6$ crayons, $60 \div 12 = 5$ pencils and $84 \div 12 = 7$ 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 32 of 43

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):

$prt = y$

$900(0.05)t = 360$

$45t = 360$

$t = 8$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Turning words into math minor
Error-prone arithmetic minor
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.

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)
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)
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

40 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds that $360 is 40 percent of the $900 deposit and enters 40, stopping before dividing by the 5 percent earned each year.
72 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads 5% as $5 a year, a fixed amount instead of a share of the $900, and divides 360 by 5.
80 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Writes 5% as 0.005, so a year’s interest seems to be $4.50, and 360 ÷ 4.5 = 80 years.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest: $I = P \times r \times t$; I is the interest, the 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.

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.
Write 5% as a decimal
Percent means “out of 100.” So 5% is 5 out of 100.
$\frac{5}{100} = 0.05$
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.
Find the interest for one year
Each year the account earns 5% of $900.
$900 \times 0.05 = 45$
So it earns $45 every year. Watch out: 5% is not $5 a year. Dividing 360 by 5 gives 72, which is wrong.
Use the formula to find the years
Put the numbers into the formula. I is 360, P is 900 and r is 0.05.
$I = P \times r \times t$
$360 = 900 \times 0.05 \times t$
$360 = 45 \times t$
To undo “times 45,” divide both sides by 45.
$t = 360 \div 45 = 8$
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 $45 \times 8 = 360$ 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 33 of 43

2023 Test B Math, question 5

Question

Solve the equation for x:

$\frac{(21)(14)x}{(49)(48)} = 0.875$

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

Show the answer

Correct answer: 7

$\frac{(21)(14)x}{(49)(48)} = 0.875$

$\left(\frac{21}{49}\right)\left(\frac{14}{48}\right)x = 0.875$

$\left(\frac{3}{7}\right)\left(\frac{7}{24}\right)x = 0.875$

$\left(\frac{1}{8}\right)x = 0.875$

$x = (0.875)(8) = 7$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
No obvious method minor
An answer in an unexpected form minor
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.

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)
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)
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

8 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Simplifies the coefficient correctly to 1/8 but enters the 8 from its denominator, never multiplying it by 0.875.
0.109375 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides 0.875 by 8 instead of multiplying, undoing the coefficient 1/8 the wrong way, so x comes out as 0.109375.
3.5 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Slips in reducing 14/48, taking it as 7/12 instead of 7/24, so the coefficient becomes 1/4 and x = 0.875 × 4 = 3.5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; 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.

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.
Split the fraction into two
Use the rule for multiplying fractions backward. Pair 21 with 49, and 14 with 48.
$\frac{21 \times 14}{49 \times 48} = \frac{21}{49} \times \frac{14}{48}$
Now the equation says these two fractions times x equal 0.875.
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.
$\frac{21}{49} = \frac{3}{7} \qquad \frac{14}{48} = \frac{7}{24}$
Watch out: 48 ÷ 2 is 24, not 12. Writing $\frac{7}{12}$ leads to 3.5.
Multiply the simple fractions
$\frac{3}{7} \times \frac{7}{24} = \frac{3 \times 7}{7 \times 24} = \frac{3}{24} = \frac{1}{8}$
The 7 on top and the 7 on the bottom cancel, since 7 ÷ 7 = 1. Then divide 3 and 24 by 3 to get $\frac{1}{8}$ . So the equation becomes:
$\frac{1}{8} \times x = 0.875$
Solve for x
One eighth of x is 0.875. To undo dividing by 8, multiply both sides by 8.
$x = 0.875 \times 8 = 7$
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 $\frac{7}{8}$ . One eighth of 7 is $\frac{7}{8}$ , so x = 7 works. You can also multiply it all out: $21 \times 14 \times 7 = 2{,}058$ and $49 \times 48 = 2{,}352$ . Then $2{,}058 \div 2{,}352 = 0.875$ .

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 34 of 43

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 $x$ ?

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
Combines ideas moderate
An answer in an unexpected form minor
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.

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)
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)
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

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

The wrong answers, classified

108 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds angle PSR = 180 − 72 = 108 and carries it to angle PQR, then grids that interior angle instead of the angle x outside it.
198 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: moderate
Reads the 72° as the parallelogram's interior angle at S, so takes angle PQR as 72 and grids 360 − 90 − 72 = 198.
252 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Overlooks the right-angle mark at Q and subtracts only the 108° angle PQR from 360, giving 252.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line: $a + b = 180$; 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: $\angle PQR = \angle PSR$; 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: $a + b + c = 360$; 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.

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.
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°.
$a + b = 180$
$\angle PSR + 72 = 180$
$\angle PSR = 180 - 72 = 108$
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.
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.
$\angle PQR = \angle PSR = 108$
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.
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.
$a + b + c = 360$
$90 + 108 + x = 360$
$198 + x = 360$
Watch out: don’t skip the right angle at Q. Leaving out the 90 gives 252.
Solve for x
Subtract 198 from 360 to find what is left for x.
$x = 360 - 198 = 162$
So x is 162°. So you would grid 162.

Check your answer

Add the three angles at Q: $90 + 108 + 162 = 360$ . 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 35 of 43

2018 Test A Math, question 2

Question

The owner of a tree farm plants pine trees and oak trees in a ratio of $8:3$ . 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 $x$ be the number of oak trees when 264 pine trees are planted. Set up a proportion and solve for $x$ :

$\frac{x}{264}=\frac{3}{8}$

$8x=792$

$x=99$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
An answer in an unexpected form minor
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.

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)
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

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

The wrong answers, classified

704 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: strong
Turns the ratio upside down, treating the 264 pines as 3 parts: 264 ÷ 3 = 88, then 88 × 8 = 704.
33 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Divides 264 by 8 to get the size of one part, 33 trees, and stops before multiplying by the 3 parts that are oak.
363 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds the 99 oak trees correctly and then adds the 264 pines, gridding the total number of trees planted.

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: $\frac{a}{b} = \frac{a \times n}{b \times n}$; 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.

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.
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.
$264 \div 8 = 33$
Each part is 33 trees. Watch out: 33 is one part, not the number of oak trees. Don’t stop here.
Find the number of oak trees
Oak trees fill 3 parts. Multiply the part size by 3.
$3 \times 33 = 99$
You can see the same thing with equal ratios. Multiply both numbers of 8:3 by 33.
$\frac{8}{3} = \frac{8 \times 33}{3 \times 33} = \frac{264}{99}$
So 264 pine trees go with 99 oak trees.
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: $264 \div 33 = 8$ and $99 \div 33 = 3$ . 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 36 of 43

2018 Test A Math, question 3

Question

For what value of $w$ is $4w=2w-8$ ?

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

Show the answer

Correct answer: -4

$4w=2w-8$

$2w=-8$

$w=-4$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
An answer in an unexpected form minor
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.

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)
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

4 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Gets 2w = −8 correctly but drops the sign when dividing by 2, gridding 4 instead of a negative value.
-8 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Subtracts 2w from both sides to reach 2w = −8 and grids −8 without dividing by the coefficient 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c$; a and b are the two sides of an equation, and c is any number 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.

Sort out the equation
An equation says two things are equal. You need the value of w that makes $4w$ equal to $2w - 8$ . 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.
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.
$4w - 2w = 2w - 8 - 2w$
$2w = -8$
Now 2 times w is −8.
Divide both sides by 2
To undo times 2, divide both sides by 2. A negative number divided by a positive number is negative.
$w = -8 \div 2 = -4$
Watch out: keep the minus sign. Dropping it gives 4. Also, don’t stop at $2w = -8$ 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: $4 \times (-4) = -16$ . Right side: $2 \times (-4) - 8 = -8 - 8 = -16$ . 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 37 of 43

2018 Test A Math, question 4

Question

A survey asked students what pets they have. Based on the results, the following statements are all true.

20 students have cats.
23 students have dogs.
3 students have both dogs and cats.
5 students have no dogs or cats.

How many students were surveyed?

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

Show the answer

Correct answer: 45

Let x = number of students with only cats as pets.

Let y = number of students with only dogs as pets.

Calculate x and y using the given information:

There are 20 students who have cats, and of those 20 students, 3 have both cats and dogs.

Thus,

$x=20-3=17$

There are 23 students who have dogs, and of those 23 students, 3 have both cats and dogs.

Thus,

$y=23-3=20$

To find the total number of students surveyed, add the number of students who only have cats (x), the number of students who only have dogs (y), the number of students who have both (3), and the number of students who have neither (5):

$3+5+x+y=8+17+20=45$

Question analysis

A grid-in counting question about a survey of students' pets: some have cats, some dogs, some both, and some neither. Students find how many were surveyed by sorting everyone into groups without counting anyone twice. The students who own both pets are the trap in this classic Venn diagram problem.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 classes organize outcomes in lists and tables, but overlapping groups like these are rarely taught directly. The arithmetic is simple whole-number addition and subtraction; the demand lies in spotting the double count.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Much to keep track of minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is small; the trap is in the wording. Because each bullet reads like a separate group, students tend to add the numbers as listed and count the students with both pets twice. A Venn diagram with the overlap filled in first solves the problem, and the remaining risk is forgetting that students with no pets were surveyed too.

Turning words into math moderate Understanding the question
The statement "20 students have cats" already includes the 3 who have both pets, and so does "23 students have dogs", which the separate bulleted list hides. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Four facts, two overlapping groups, their overlap and an outside group of 5 with neither pet, all have to be placed before anything is added. (Sweller, 1988; Carpenter, Just & Shell, 1990)
An answer in an unexpected form minor Choosing the answer
As a grid-in it gives no choices, so a total that counts the 3 twice or leaves out the 5 looks just as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Draw two overlapping circles: put the both-pets group in the middle, subtract it from each pet total to fill the outer parts, then add every region and the students with neither.

Traps

Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

48 Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: strong
Adds 20 + 23 + 5 = 48, counting the 3 students with both pets once among the cat owners and again among the dog owners.
40 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Removes the overlap correctly, 20 + 23 − 3 = 40, but counts only the pet owners, leaving out the 5 surveyed students with no dogs or cats.
51 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds all four numbers in the list, 20 + 23 + 3 + 5 = 51, treating the both-pets group as an extra group of students.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting two groups that overlap: $T = C + D - B + N$; T is the total number of students. C is the number with cats, and D is the number with dogs. B is the number with both pets, and N is the number with neither. Students with both are counted in C and again in D, so you subtract B once. Then add N.

Sort out the four facts
You need the number of students who were surveyed. Here is what you know:
- 20 students have cats.
- 23 students have dogs.
- 3 students have both dogs and cats.
- 5 students have no dogs or cats.
The tricky part: the 3 students with both pets are already inside the 20 and inside the 23.
Split the pet owners into groups
Picture two circles that overlap, one for cats and one for dogs. The overlap holds the 3 students with both pets. Take them out of each pet total to find who has only one kind of pet.
$20 - 3 = 17$
So 17 students have only cats.
$23 - 3 = 20$
So 20 students have only dogs.
Add every group once
Now every student is in exactly one group.
Group Students
Only cats 17
Only dogs 20
Both pets 3
Neither pet 5
$17 + 20 + 3 + 5 = 45$
The formula gives the same total.
$T = 20 + 23 - 3 + 5 = 45$
Avoid the traps and answer
Watch out: adding $20 + 23 + 5$ counts the 3 students with both pets twice. That gives 48. Watch out: the 5 students with no pets were surveyed too. Leaving them out gives 40. Adding all four numbers in the list gives 51, which is also wrong. So you would grid 45.

Group	Students
Only cats	17
Only dogs	20
Both pets	3
Neither pet	5

Check your answer

Check each fact against your groups. Cat owners: $17 + 3 = 20$ . Dog owners: $20 + 3 = 23$ . Both: 3. Neither: 5. Every fact matches, so 45 students were surveyed.

More questions like this

The original DOE question

Question 61 as printed on page 81 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 61. Open page 81 of the handbook (PDF)

Question 38 of 43

2018 Test A Math, question 5

Question

The sum of two consecutive integers is −15. If 1 is added to the smaller integer and 2 is subtracted from the larger integer, what is the product of the two resulting integers?

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

Show the answer

Correct answer: 63

If x is the smaller consecutive integer, then

$x+1$

is the larger consecutive integer. Use their sum (−15) to find x:

$x+(x+1)=-15$

$2x+1=-15$

$2x=-16$

$x=-8$

The two consecutive integers are −8 and −7. One is added to the smaller integer:

$-8+1=-7$

and 2 is subtracted from the larger integer:

$-7-2=-9$

Find the product:

$(-7)(-9)=63$

Question analysis

A grid-in problem about two consecutive integers whose sum is −15. Students find the two integers, add 1 to the smaller and subtract 2 from the larger, then multiply the results. Deciding which negative integer is the smaller one, and keeping signs straight through every step, are what make it tricky.

NYS standards: 7.EE.4a two-step equations
7.NS.2a rational number multiplication
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 A typical grade 7 task solves one two-step equation or multiplies signed numbers. This links both: set up a consecutive-integer equation, solve it with negatives, adjust each integer, and multiply, with no choices to check against.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math moderate
Many steps moderate
Solution: 4 steps
Reading level: Question and choices: about grade 9, 30 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every step is routine, yet the question collects most of the ways students lose points with negatives. The crucial judgment is which of −8 and −7 is the smaller; a student who picks −7 still produces a tidy, wrong product. Gridding the product of the original pair, or giving a negative product, are the other usual losses. Writing each intermediate pair down, rather than holding them in mind, protects against all three.

Error-prone arithmetic moderate Working it out
Every number in play is negative: −8 and −7, then −7 and −9, and a product of two negatives that must come out positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math moderate Understanding the question
"1 is added to the smaller integer and 2 is subtracted from the larger" depends on knowing that −8, not −7, is the smaller, since it lies farther below zero. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps moderate Working it out
Four linked steps: set up x + (x + 1) = −15, solve for x, adjust each integer, then multiply; an error early in the chain carries through to the product. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Halve the sum to see which two integers sit on either side of it, remember that the smaller negative is farther from zero, then adjust each and multiply.

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

The wrong answers, classified

60 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes −7 as the smaller integer because 7 is less than 8, so adjusts to −7 + 1 = −6 and −8 − 2 = −10, and multiplies to 60.
56 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds −8 and −7 and multiplies them to get 56, skipping the sentence that adds 1 to one integer and subtracts 2 from the other.
-63 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Finds the adjusted integers −7 and −9 correctly but gives their product a negative sign, gridding −63.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive integers: $n, \; n + 1$; Integers are whole numbers and their opposites, like −2, −1, 0, 1 and 2. Consecutive integers come one right after another. If n is the smaller one, the next one is n + 1, because each step up the number line adds 1.
Multiplying two negative numbers: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. When you multiply two negative numbers, the answer is positive. The two minus signs cancel out.

Sort out the question
Two consecutive integers add up to −15. Integers are whole numbers and their negatives. Consecutive means one comes right after the other, like 4 and 5. You need to do three things:
1. Find the two integers.
2. Add 1 to the smaller one. Subtract 2 from the larger one.
3. Find the product of the two new numbers. The product is the answer when you multiply.
Find the two integers
Call the smaller integer n. Then the larger one is n + 1. Their sum is −15.
$n + (n + 1) = -15$
$2n + 1 = -15$
Subtract 1 from both sides. Then divide both sides by 2.
$2n = -16$
$n = -8$
So the integers are −8 and −7.
Decide which one is smaller
On a number line, −8 is to the left of −7. Numbers to the left are smaller. So −8 is the smaller integer, and −7 is the larger one. Watch out: 8 is bigger than 7, but −8 is smaller than −7. Mixing them up gives −6 and −10, and the wrong answer 60.
Change each integer
Add 1 to the smaller integer.
$-8 + 1 = -7$
Subtract 2 from the larger integer.
$-7 - 2 = -9$
The two new numbers are −7 and −9.
Multiply the new numbers
Both numbers are negative, so their product is positive.
$(-7) \times (-9) = 63$
Watch out: the answer is not −63, because two negatives make a positive. Also, don’t multiply the first pair, −8 and −7. That gives 56 and skips the changes. So you would grid 63.

Check your answer

Check the pair: $-8 + (-7) = -15$ , and they are one apart. Half of −15 is −7.5, and −8 and −7 sit on either side of it. Then $7 \times 9 = 63$ , and the sign is positive.

More questions like this

The original DOE question

Question 62 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 62. Open page 81 of the handbook (PDF)

Question 39 of 43

2018 Test B Math, question 1

Question

How many 5-digit numbers can be created using the digits 2, 3, 5, 7, and 8 without repeating any digits within that 5-digit number?

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

Show the answer

Correct answer: 120

There are 5 choices for the first digit, 4 choices for the second digit, 3 choices for the third digit, 2 choices for the fourth digit, and 1 choice for the final digit.

The total number of possibilities is

5 × 4 × 3 × 2 × 1 = 120

Question analysis

This grid-in counting question asks how many different 5-digit numbers can be made from five given digits when no digit may repeat. It tests the counting principle for arrangements rather than arithmetic. The catch is the no-repeat rule: each place has one fewer digit to choose from, and ignoring that gives a far larger count.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 probability work builds sample spaces with lists and tree diagrams for two or three stages. This needs the multiplication principle with a shrinking number of choices across five places, which listing cannot reach.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 11, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is tiny; the difficulty is knowing that arrangements are counted by multiplying, and that the no-repeat rule removes one digit for each later place. Students who have met the counting principle finish in seconds, while those who start listing numbers run out of time or guess a product they half remember. With no choices to check against, the 3,125 that comes from ignoring the rule looks as convincing as the right count.

Something you must already know moderate Working it out
Nothing in the stem names the counting principle: students must already know to multiply the digits left for each place, 5 × 4 × 3 × 2 × 1, because listing 5-digit numbers by hand cannot finish. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in there are no choices to compare with, so a count of 3,125 or 25 cannot be caught by noticing that it is missing from a list. (Enright, Morley & Sheehan, 2002)

Strategy: Fill the five places from left to right, write how many digits are still available for each place, and multiply those five counts together.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.

The wrong answers, classified

3125 Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Uses the counting principle but lets digits repeat, multiplying 5 × 5 × 5 × 5 × 5 instead of using one fewer digit for each later place.
25 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Multiplies the five digits by the five places, 5 × 5, as if each digit simply took each position once.
15 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Sees that the choices shrink from 5 to 1 but adds them, 5 + 4 + 3 + 2 + 1, instead of multiplying.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle: $\text{total ways} = a \times b \times c \times \ldots$; a is the number of choices for the first spot, b for the second, c for the third, and so on. Multiply them to get every possible way. It works because each choice for one spot can go with each choice for the next spot.

Sort out the question
You have five digits: 2, 3, 5, 7 and 8. A digit is one of the symbols 0 to 9. You build 5-digit numbers from them.
“Without repeating” means each digit is used only once in a number. So 23578 is allowed, but 22578 is not.
You need to find how many different numbers you can make.
Count the choices for each spot
Think of the number as five empty spots. Fill them from left to right.
- First spot: any of the 5 digits, so 5 choices.
- Second spot: one digit is used up, so 4 are left.
- Third spot: 3 are left.
- Fourth spot: 2 are left.
- Last spot: only 1 is left.
Multiply the choices together
Use the counting principle. Each choice for one spot works with every choice for the next spot. So you multiply. You do not add.
$\text{total ways} = 5 \times 4 \times 3 \times 2 \times 1 = 120$
There are 120 different numbers.
Watch out: Adding 5 + 4 + 3 + 2 + 1 gives 15, which is far too small. Using 5 choices for every spot gives 3,125. That count lets digits repeat, which the question does not allow.
So you would grid 120.

Check your answer

Try a smaller case you can list. With the digits 2, 3 and 5, the rule gives $3 \times 2 \times 1 = 6$ numbers. The list is 235, 253, 325, 352, 523 and 532. That is 6, so the rule works.

More questions like this

The original DOE question

Question 58 as printed on page 153 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 58. Open page 153 of the handbook (PDF)

Question 40 of 43

2018 Test B Math, question 2

Question

$\frac{147-x}{12}=12$

What is the value of $x$ in the equation shown above?

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

Show the answer

Correct answer: 3

$\frac{147-x}{12}=12$

$147-x=144$

$x=3$

Question analysis

This grid-in asks students to solve a two-step linear equation in which an expression containing x is divided by 12 and set equal to 12. The algebra is short, but x is being subtracted from 147, so undoing the operations in the right order and keeping the minus sign straight is where students slip.

NYS standards: 7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 This is exactly the two-step equation practice of grade 7: undo a division, then undo a subtraction. The only twist is that the unknown is being subtracted, so the last step needs care with its sign.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a two-move equation that most students finish quickly: multiply both sides by 12, then find what must be taken from 147 to leave 144. Points are lost only at the last move, where the minus sign in front of x invites 144 − 147 or 147 + 144. Substituting the result back into the fraction catches either slip in a few seconds.

Error-prone arithmetic minor Working it out
After multiplying by 12, the unknown sits behind a minus sign in 147 − x = 144, so students must ask what is taken from 147 rather than subtract in the order the numbers appear. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply both sides by 12 to clear the fraction, then ask what number must be taken away from 147 to leave that product.

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.)

The wrong answers, classified

-3 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Reaches 147 − x = 144 but subtracts in the order the numbers appear, writing x = 144 − 147.
146 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: moderate
Divides the right side by 12 instead of multiplying, so sets 147 − x equal to 1 and gets 146.
291 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves 147 across the equals sign without changing its sign and drops the minus on x, so x = 144 + 147.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing division: $\frac{a}{b} = c \text{ means } a = b \times c$; a is the top of the fraction, b is the bottom, and c is the result. Division and multiplication undo each other. So if a divided by b gives c, then b times c gives back a. This is the same as multiplying both sides by b.

Read the equation in words
The equation says: start with 147, take away a mystery number x, then divide by 12. The result is 12.
You need to find x. To get x alone, undo the steps in reverse order. Dividing by 12 was done last, so undo it first.
Undo the division by 12
Multiply both sides by 12. Whatever you do to one side, you must do to the other. That keeps the equation balanced.
$\frac{147 - x}{12} = 12$
$147 - x = 12 \times 12 = 144$
Watch out: Do not divide 12 by 12. That gives 147 − x = 1 and the wrong answer 146.
Find what was taken away
Now ask: 147 minus what number leaves 144? The gap between the two numbers is x.
$x = 147 - 144 = 3$
Watch out: Do not work out 144 − 147. That gives −3. Since 147 is bigger than 144, x must be positive.
So you would grid 3.

Check your answer

Put 3 back in for x. First, $147 - 3 = 144$ . Then $144 \div 12 = 12$ . That matches the right side of the equation, so x = 3 is correct.

More questions like this

The original DOE question

Question 59 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 59. Open page 153 of the handbook (PDF)

Question 41 of 43

2018 Test B Math, question 3

Question

$\left | \left ( -6 \right )-\left ( -5 \right )+4.2 \right |-\left | 3-9.6 \right |$

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

Show the answer

Correct answer: -3.4

$=3.2-6.6=-3.4$

Question analysis

This grid-in asks students to evaluate an expression that mixes negative integers, decimals and two absolute values. The hard parts are subtracting a negative number inside the first pair of bars and keeping each absolute value separate until the final subtraction, with no answer choices to check against.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students add and subtract signed numbers, and grade 6 students meet absolute value. This combines both in one expression with decimals and several sign decisions, a typical grade 7 exercise with no context to check against.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps moderate
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 16+, 1 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 familiar, but the expression packs four of them with negatives and decimals, and a single sign slip anywhere gives a clean-looking wrong number. The subtlest point comes last: absolute value bars make each part positive, yet the difference of two absolute values can still be negative. Students who believe bars always produce a positive answer reverse the final subtraction, and the grid-in format offers no choices to expose it.

Error-prone arithmetic moderate Working it out
Three sign decisions stack up: (−6) − (−5) must become −6 + 5, then −1 + 4.2 crosses zero, and 3 − 9.6 is negative before its bars make it positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps moderate Working it out
Each pair of bars must be simplified inside and then made positive, and only after both does the outer subtraction happen; doing the bars and the subtraction in the wrong order changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The result is a negative decimal to be gridded with its sign, and with no choices on screen nothing warns a student who has dropped the minus or flipped the subtraction. (Enright, Morley & Sheehan, 2002)

Strategy: Simplify inside each pair of absolute value bars first, rewriting minus a negative as plus, then take each absolute value, and only then subtract the second from the first.

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.)

The wrong answers, classified

3.4 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Believes an expression built from absolute values cannot be negative, so reports 6.6 − 3.2 instead of 3.2 − 6.6.
9.8 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Treats the bars like parentheses, leaving 3 − 9.6 as −6.6, then subtracts that negative: 3.2 + 6.6.
0.2 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Reads (−6) − (−5) as −6 − 5 = −11, so the first bars give |−6.8| = 6.8, and 6.8 − 6.6 = 0.2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{distance from } a \text{ to } 0$; The two bars around a number 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 |−6.6| = 6.6 and |3.2| = 3.2.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Taking away a negative number is the same as adding the positive number. For example, 5 − (−2) = 7. Think of it as taking away a debt: you end up richer.

Plan the order of work
The expression has two pairs of bars. The bars stand for absolute value, a number’s distance from 0.
Work inside each pair of bars first. Then take each absolute value. Subtract the two results last.
Work inside the first bars
Start with $(-6) - (-5) + 4.2$ . Subtracting −5 is the same as adding 5.
$(-6) - (-5) + 4.2 = -6 + 5 + 4.2 = -1 + 4.2 = 3.2$
3.2 is already positive, so $|3.2| = 3.2$ .
Watch out: If you read the start as −6 − 5, you get −11. That path ends at the wrong answer 0.2.
Work inside the second bars
Next, $3 - 9.6 = -6.6$ . This result is negative. Its distance from 0 is 6.6, so the absolute value is positive.
$|-6.6| = 6.6$
Subtract the two results
Now take the second result away from the first.
$3.2 - 6.6 = -3.4$
You take a bigger number away from a smaller one, so the result is negative.
Watch out: The bars made each part positive. They do not make the final answer positive, so 3.4 is wrong. Also, bars are not parentheses. Working out 3.2 − (−6.6) = 9.8 is wrong too.
So you would grid -3.4, with the minus sign.

Check your answer

Estimate. The first part is about 3. The second part is about 7. And 3 − 7 is about −4. The answer −3.4 is close to that. It is also negative, as the estimate says it should be.

More questions like this

The original DOE question

Question 60 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 60. Open page 153 of the handbook (PDF)

Question 42 of 43

2018 Test B Math, question 4

Question

Tyler has completed 60 pages in his French workbook. This is 20% of the total number of pages in the workbook. How many pages are in the workbook?

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

Show the answer

Correct answer: 300

Let $x$ be the total number of pages in the workbook.

Then, 20% of $x$ is 60. Set up a proportion and solve for $x$ :

$\frac{20}{100}=\frac{60}{x}$

$20x=6{,}000$

$x=\frac{6{,}000}{20}=300\text{ pages}$

Question analysis

This grid-in percent word problem gives the number of French workbook pages Tyler has finished and says what percent of the whole book that is. Students must find the total number of pages by working backward from a part to the whole, which is where many take a percent of the wrong number.

NYS standards: 6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the whole from a part and a percent is a grade 6 skill. It takes a single division here, so it sits below grade 7, although students who reflexively take 20 percent of 60 will go wrong.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Solution: 1 step
Reading level: Question and choices: about grade 3, 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 question is short and the arithmetic is one division, so it rests on reading the second sentence the right way round. Students who see 20% next to 60 and compute 20% of 60 answer a different problem, and a grid-in gives no list of choices to make them doubt it. A quick check that 60 is one fifth of the result catches every one of these errors.

Turning words into math moderate Understanding the question
The sentence “This is 20% of the total number of pages” makes 60 the part and the total the unknown, the reverse of the familiar “find 20% of a number” pattern. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Write the finished pages as 20 percent of an unknown total and solve for the total, or think about what fraction of the whole book 20 percent is.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
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

12 Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes 20% of 60 instead of finding the number that 60 is 20% of, applying the percent to the wrong amount.
240 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Sees that 60 pages is one fifth of the book but multiplies 60 by 4, giving the pages Tyler still has left.
3 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals partial knowledge Expected pull: weak
Divides 60 by 20 instead of by 0.20, treating 20% as the whole number 20.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; The whole is the full amount, here all the pages in the book. The part is a piece of it, here the pages done. Percent means how many out of every 100. Use this when you know two of the three numbers and need the third.

Sort out the part and whole
Tyler has done 60 pages. That is 20% of the whole book. So 60 is the part. The total number of pages is the whole.
The whole is what you need to find. Call it x.
Write the percent equation
Put the numbers into the percent formula.
$\text{part} = \frac{\text{percent}}{100} \times \text{whole}$
$60 = \frac{20}{100} \times x$
The fraction $\frac{20}{100}$ simplifies to $\frac{1}{5}$ . So 60 pages is one fifth of the book.
Solve for the whole book
If one fifth of the book is 60 pages, the whole book is 5 times as big.
$x = 60 \times 5 = 300$
The book has 300 pages.
Watch out: Finding 20% of 60 gives 12. That treats 60 as the whole, but 60 is the part. Also, 300 − 60 = 240 is the pages still left to do, not the total.
So you would grid 300.

Check your answer

Find 20% of your answer: $\frac{20}{100} \times 300 = 60$ . That matches the 60 pages Tyler finished. Also, 300 is bigger than 60, as a whole should be.

More questions like this

The original DOE question

Question 61 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 61. Open page 153 of the handbook (PDF)

Question 43 of 43

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 $y$ ?

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 $x$ .

The sum of the four angles on the top of the figure is equal to 180°.

$x+y+30+60=180$

Since $x$ is a vertical angle with the 25° angle, then $x$ is also 25°. Use that to solve for $y$ .

$25+y+30+60=180$

$y+115=180$

$y=65$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
Combines ideas moderate
An answer in an unexpected form minor
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.

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)
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)
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

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.
Trusting the picture: The figure suggests something it does not state; work from the given facts.

The wrong answers, classified

90 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: strong
Misses the unlabeled angle between the 30° angle and y, so subtracts only 30 and 60 from 180.
115 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Adds the three known angles beside y, 25 + 30 + 60, and grids that sum without subtracting it from 180.
60 Trusting the picture (Lesh, Post & Behr, 1987)
Reveals a surface strategy Expected pull: moderate
Judges from the drawing, where y looks like a mirror image of the 60° angle across the vertical line, instead of computing it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line: $a + b + c + d = 180$; 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: $\angle 1 = \angle 2$; 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.

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.
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°.
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°.
$a + b + c + d = 180$
$y + 25 + 30 + 60 = 180$
$y + 115 = 180$
Solve for y
Take 115 away from 180 to get y alone.
$y = 180 - 115 = 65$
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: $65 + 25 + 30 + 60 = 180$ . 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)