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Practice Test

Turning words into math

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Question 1 of 120

2026 Practice Test A Math, question 8

Question

Liam and Kevin each collect stamps. Liam has twice as many stamps as Kevin. Liam gives 8 of his stamps to Kevin. Liam now has 12 more stamps than Kevin.

How many stamps did Liam start with?

Select the correct answer from each drop-down to complete the sentences.

An equation that relates the number of stamps originally in Liam’s and Kevin’s collections is L=L = KK.

After Liam gives 8 stamps to Kevin, the equation becomes L−8=L - 8 = .

By solving the equation, it can be determined that Liam started with stamps.

Choose an answer from each menu, then check. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Show the answer

Correct answer: blank 1: 2; blank 2: K + 20; blank 3: 56

Liam originally has twice as many stamps as Kevin, so the equation is L=2KL = 2K.

After Liam gives 8 stamps to Kevin, the equation becomes L−8=(K+8)+12L - 8 = (K + 8) + 12, which simplifies to L−8=K+20L - 8 = K + 20.

Substitute L=2KL = 2K into the equation: 2K−8=K+202K - 8 = K + 20.

Solve for K:

2K−K=20+82K - K = 20 + 8

K=28K = 28

Substitute to find L: L=2K=2×28=56L = 2K = 2 \times 28 = 56.

So, Liam started with 56 stamps.

Question analysis

This word problem about two stamp collections is answered with three drop-down menus: an equation for the starting counts, an equation for after Liam gives Kevin 8 stamps, and the number Liam started with. Students turn "twice as many" and "12 more" into algebra and solve. Tracking both boys' counts before and after the gift is the tricky part, and every menu must be right.

NYS standards
  • 8.EE.8c systems in real-world problems
  • 7.EE.4 equations and inequalities from problems
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve one equation in one unknown. The menus lay out the equations, but the problem still ties two unknowns together before and after a gift, a grade 8 system, and all three menus must be right.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 5, 86 words.
Students, last 30 days
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 menus do much of the setting up, since each sentence already shows the shape of its equation. The second menu is the hard one: the 8 stamps leave Liam and join Kevin before the gap of 12 is measured. Solving gives Kevin's 28 first, and K + 20 then equals 48, Liam's count after the gift, which waits on the last menu beside his starting 56. Testing each starting count against the story is a quick check on the last menu.

  1. Turning words into math moderate Understanding the question

    The gift changes both counts, so "Liam now has 12 more stamps than Kevin" compares Liam's L − 8 with Kevin's K + 8, and the second menu's K + 20 holds a 20 that appears nowhere in the stem. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of moderate Working it out

    Each boy has a count before and after the gift, four numbers in all, and all three menus must be right; the last asks for Liam's start, not the 48 he has after giving 8 away. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let K be Kevin's starting count, write each boy's count after the gift in terms of K, and set Liam's 12 above Kevin's; testing each number in the last menu is a fast check.

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.
  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Twice as many as
L=2KL = 2K
L is the number of stamps Liam starts with, and K is the number Kevin starts with. Twice as many means 2 times as many, so Liam’s count is 2 times Kevin’s. Because L and 2K are equal, you can put 2K in place of L in any equation.
Doing the same to both sides
if a=b, then a+c=b+c\text{if } a = b \text{, then } a + c = b + c
a and b are the two sides of an equation, and c is any number or term. An equation is like a balanced scale. Add the same thing to both sides, or take the same thing away from both, and the scale stays level. This is how you move a term to the other side.

  1. Sort out the question

    Let L be Liam’s stamps at the start and K be Kevin’s. You know three facts:

    • Liam has twice as many stamps as Kevin.
    • Liam gives 8 of his stamps to Kevin.
    • After that, Liam has 12 more stamps than Kevin.

    The first menu finishes the starting equation. The second finishes the equation after the gift. The third is Liam’s count at the start.

  2. Fill in the first menu

    Twice as many means 2 times as many. Liam has twice as many as Kevin, so:

    L=2KL = 2K

    Pick 2 in the first menu.

    Watch out: the 0.5 in the menu turns the fact around. L=0.5KL = 0.5K would give Liam half as many as Kevin. Liam has more, so his count must be the bigger one.

  3. Write both counts after the gift

    The gift changes both counts. Liam loses 8 stamps, and Kevin gains those same 8.

    • Liam now has L−8L - 8 stamps.
    • Kevin now has K+8K + 8 stamps.

    Liam has 12 more than Kevin. So Liam’s count equals Kevin’s count plus 12:

    L−8=(K+8)+12=K+20L - 8 = (K + 8) + 12 = K + 20

    Pick K + 20 in the second menu.

    Watch out: the 12 goes on Kevin’s count, not Liam’s. Picking L + 12 adds it to the wrong boy.

  4. Solve for K

    Now use the first equation. Since L=2KL = 2K, put 2K in place of L:

    2K−8=K+202K - 8 = K + 20

    Subtract K from both sides. Then add 8 to both sides. Each move keeps the scale level.

    K−8=20K - 8 = 20

    K=28K = 28

    Kevin started with 28 stamps.

  5. Find Liam’s starting count

    Liam started with twice as many as Kevin:

    L=2K=2×28=56L = 2K = 2 \times 28 = 56

    Liam started with 56 stamps.

    Watch out: K+20=48K + 20 = 48 is Liam’s count after the gift. The menu offers 48, but the question asks about the start.

    So the menus read 2, K + 20 and 56.

Check your answer

At the start, Liam has 56 stamps and Kevin has 28, and 56 is twice 28. After the gift, Liam has 48 and Kevin has 36. 48−36=1248 - 36 = 12, so Liam has 12 more. Every fact fits.

More questions like this

The original DOE question
Question 8 as shown on the DOE’s digital test, page 79 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test A Math, page 79, question 8.

Question 2 of 120

2026 Practice Test B Math, question 2

Question

The list shows the results of a survey. A survey was taken to see whether students like basketball, soccer, both, or neither. These were the results:

  • 25 students like basketball.
  • 19 students like soccer.
  • 7 students like both basketball and soccer.
  • 10 students don’t like either basketball or soccer.

Use the information provided to determine how many students were surveyed and how many like only soccer.

Move the correct answer to each box.

There were students surveyed, and students like only soccer.

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

Show the answer

Correct answer: blank 1: 47; blank 2: 12

There were 47 students surveyed, and 12 students like only soccer.

To determine how many students were surveyed, start with the total number of students who like basketball and soccer separately and subtract the number of students who were counted twice.

There were 25 students who like basketball, 19 students who like soccer, and 7 students who like both basketball and soccer.

So, the total number of students who like basketball or soccer is (25+19)−7=37(25 + 19) - 7 = 37.

Then, add the students who don’t like either basketball or soccer: 10 students don’t like either sport.

Total students surveyed: 37+10=4737 + 10 = 47

To determine how many students like only soccer, subtract the number of students who like both basketball and soccer from the total number of students who like soccer.

Students who like only soccer: 19−7=1219 - 7 = 12

Question analysis

This survey problem lists how many students like basketball, soccer, both sports or neither, and students drag number tiles into two boxes: how many were surveyed and how many like only soccer. It is a Venn diagram counting question. Keeping track of the students who like both sports, who appear in two of the listed counts, is the tricky part.

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 Overlapping groups like these are seldom taught on their own in grade 7, though the arithmetic is simple whole-number addition and subtraction. Two results come from the same four facts, and both boxes must be right to score.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 6, 78 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light; the work is keeping track of one overlap that touches both answers. Students often remove the 7 for one box and not the other, and the pool holds 54 and 19, the tiles those half-finished methods produce. A sketch of two overlapping circles, with 7 in the middle, 18 and 12 beside it and 10 outside, turns both boxes into reading off the picture.

  1. Much to keep track of moderate Working it out

    The 7 students who like both sports sit inside the 25 and inside the 19, so they come out once for the total and again for only soccer, and both boxes must be right to score. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    The list never says that the 25 basketball fans already include the 7 who like both sports, and only the word “only” in the question separates the 12 from all 19 soccer fans. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Sketch two overlapping circles, write the students who like both sports in the middle, fill each outer part by subtraction, then add every region and the students who like neither.

Traps

  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting two groups that overlap
total=B+S−both+neither\text{total} = B + S - \text{both} + \text{neither}
B is the number who like basketball, and S is the number who like soccer. The students who like both are inside B and inside S, so B + S counts them twice. Subtract them once to fix that. Then add the students who like neither sport, since they are in neither group.
Only one group
only soccer=S−both\text{only soccer} = S - \text{both}
S is everyone who likes soccer, and both is the number who like both sports. Only soccer means the soccer fans who do not also like basketball. Take the both group out of the soccer group to find them.

  1. Sort out the question

    The survey gives you four facts:

    • 25 students like basketball.
    • 19 students like soccer.
    • 7 students like both sports.
    • 10 students like neither sport.

    You need two numbers. The first box is how many students were surveyed in all. The second box is how many like only soccer, which means soccer but not basketball. Each tile can be used once.

  2. See who is counted twice

    The 7 students who like both sports are part of the 25 basketball fans. They are also part of the 19 soccer fans. So if you add 25 and 19, those 7 students get counted two times.

    A Venn diagram shows this. It is a picture of two circles that overlap. The 7 go in the middle, where the circles share space.

  3. Count who likes at least one sport

    Use the overlap rule. Add the two groups, then take away the 7 who were counted twice:

    25+19−7=44−7=3725 + 19 - 7 = 44 - 7 = 37

    So 37 students like basketball, soccer or both.

  4. Add the students who like neither

    The 10 students who like neither sport are not in either group. They were still surveyed, so add them:

    37+10=4737 + 10 = 47

    So 47 students were surveyed.

    Watch out: adding 25 + 19 + 10 = 54 counts the 7 students twice. The tile 54 is waiting for that mistake.

  5. Find who likes only soccer

    The 19 soccer fans include the 7 who like basketball too. Take those 7 away:

    19−7=1219 - 7 = 12

    So 12 students like only soccer.

    Watch out: the word only matters. The tile 19 counts every soccer fan, not just the ones who like only soccer.

  6. Fill in both boxes

    On the test, you drag the tile 47 into the first box and the tile 12 into the second box. On this site, you choose each tile from its box’s menu.

    So the boxes get 47 and 12: there were 47 students surveyed, and 12 students like only soccer.

Check your answer

Split the students into four groups that do not overlap. Only basketball is 25−7=1825 - 7 = 18. Both sports is 7. Only soccer is 12. Neither is 10. Add the four groups: 18+7+12+10=4718 + 7 + 12 + 10 = 47. That matches the first box.

More questions like this

The original DOE question
Question 2 as shown on the DOE’s digital test, page 74 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B Math, page 74, question 2.

Question 3 of 120

2026 Practice Test B Math, question 4

Question

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

1 mink=3 zogs1\text{ mink} = 3\text{ zogs}

1 mink=14 bart1\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 mink=3 zogs1\text{ mink} = 3\text{ zogs}, divide the number of zogs Maria has by 3: 60÷3=2060 \div 3 = 20.

So, 60 zogs=20 minks60\text{ zogs} = 20\text{ minks}.

Step 2: Convert the barts to minks.
Since 1 mink=14 bart1\text{ mink} = \frac{1}{4}\text{ bart}, multiply the number of barts by 4 (because 1 bart is equal to 4 minks): 24×4=9624 \times 4 = 96.

So, 24 barts=96 minks24\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=11620 + 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.
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Converting with a rate
minks=number of points×minks for one point\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.

  1. Sort out the question

    Maria trades two kinds of points for minks. You know:

    • 3 zogs trade for 1 mink.
    • 1 mink is worth 14\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.

  2. 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 13\frac{1}{3} of a mink.

    A zog is smaller than a mink. So Maria will get fewer minks than she has zogs.

  3. Turn the bart rule around

    The rule says 1 mink is 14\frac{1}{4} of a bart. So a mink is a quarter of a bart. It takes 4 quarters to make a whole:

    4×14=14 \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.

  4. Change the zogs into minks

    Use the rate. Each of the 60 zogs is worth 1/3 of a mink:

    60×13=60÷3=2060 \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.

  5. Change the barts into minks

    Each of the 24 barts is worth 4 minks:

    24×4=9624 \times 4 = 96

    So 24 barts trade for 96 minks.

  6. Add the minks

    Both amounts are now in minks, so you can add them:

    20+96=11620 + 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 20×3=6020 \times 3 = 60 zogs. 96 minks are worth 96×14=2496 \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 4 of 120

2026 Practice Test B Math, question 24

Question

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

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

Correct answer: C

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

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

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

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 A typical grade 7 percent task applies one increase or decrease to one amount. This one applies a weekly decrease twice in a row, an early form of exponential decay that goes beyond typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
2 steps
Reading level
Question and choices: about grade 8, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. A very close wrong answer major Choosing the answer

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

  2. Turning words into math moderate Understanding the question

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

  3. Variables instead of numbers minor Understanding the question

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Pick an easy starting amount

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

  3. Work out the first week

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

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

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

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

  4. Work out the second week

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

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

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

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

  5. Read the answer as a percent

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

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

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

Check your answer

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

More questions like this

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

Question 5 of 120

2026 Practice Test B Math, question 46

Question

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

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

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

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

Show the answer

Correct answer: y = 20x + 50

Solution: y=20x+50y = 20x + 50

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

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

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

Question analysis

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

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.9 dependent and independent variables
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing an equation for a fixed fee plus a steady monthly charge is a familiar grade 7 task, and the numbers are whole. Typing the full equation, with both variables and no choices to compare, makes it less forgiving.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 51 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    Only the words “onetime setup fee” and “per month” tell which amount is multiplied by x and which is added once, and the $50 comes first in the sentence. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The whole equation, with y, an equals sign and x, is built on a keypad with no choices, so a reversed or incomplete equation looks as finished as the right one. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Amiya pays two kinds of charges:

    • a setup fee of $50, paid one time only;
    • a charge of $20 for every month of service.

    y is her total cost in dollars, and x is the number of months. You need an equation that links them. The answer is an equation, not a number.

  2. Work out a few totals

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

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

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

  3. Match each number to its job

    Use the fee-and-rate model, y=(rate)x+(fee)y = (\text{rate})x + (\text{fee}). The rate is the amount that repeats each month, $20. It multiplies the months, so it becomes 20x. The fee, $50, is paid once, so it is added once.

    Watch out: the $50 comes first in the question, but it is not the rate. y=50x+20y = 50x + 20 charges $50 a month. y=70xy = 70x charges the fee every month. y=20xy = 20x forgets the fee.

  4. Write the equation

    Put the rate and the fee into the model:

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

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

  5. Enter it on the keypad

    Type the whole equation, including y =. Without it, you have an expression, not an equation. On the test, the keypad has x, y and = keys. Press y, =, 2, 0, x, +, 5, 0. You do not need a times sign or dollar signs. On this site, type it the same way.

    So you would enter y = 20x + 50.

Check your answer

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

More questions like this

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

Question 6 of 120

2026 Practice Test C Math, question 5

Question

Maya read n pages for her English homework. Trevor read three times as many pages as Maya. Together, Maya and Trevor read 12 more pages than Ali read. Write an expression that shows how many pages Ali read, in terms of n.

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

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

Show the answer

Correct answer: 4n – 12

Let n represent the number of pages that Maya read.

Trevor read three times as many pages as Maya: Trevor’s pages =3n= 3n

To find how many pages they read together, add Maya’s and Trevor’s pages: n+3n=4nn + 3n = 4n

This means that together, they read 4n4n pages.

Ali read 12 fewer pages than Maya and Trevor read together. So 12 should be subtracted from the total:

Ali’s pages =4n−12= 4n - 12

Question analysis

Maya reads n pages, Trevor reads three times as many, and together they read 12 more pages than Ali. Students type an expression in n for the number of pages Ali read. It tests writing and simplifying expressions from words, and the comparison with Ali has to be translated with care.

NYS standards
  • 6.EE.6 variables representing unknown numbers
  • 6.EE.2a writing algebraic expressions
  • 6.EE.3 generating equivalent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Writing an expression from words is grade 6 content, but this chains three relationships, and the last compares the pair with Ali rather than describing Ali directly. Typing the answer with no choices to compare adds risk.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 52 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Building the pair's total, n + 3n = 4n, is quick; the question turns on the direction of the last comparison. The pair read more than Ali, so Ali's count is the smaller one, yet the word more pulls toward adding 12. Testing n = 10 settles it: Maya and Trevor read 40 pages together, so Ali must have read 28, not 52.

  1. Turning words into math moderate Understanding the question

    “Together, Maya and Trevor read 12 more pages than Ali read” compares the pair with Ali, so Ali read 12 fewer than 4n, though the word more invites adding 12. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    Every count is written in terms of n, so there is no page total to check an expression against unless the student picks a value for n and tests it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The expression is typed into a box with no choices to compare, so an entry such as 4n + 12 gets no warning that the comparison was read backward. (Enright, Morley & Sheehan, 2002)

Strategy: Write Trevor's pages and the pair's total in terms of n, then test your expression with a value such as n = 10 to check that the pair really read 12 more than Ali.

Traps

  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • 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.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding like terms
n+3n=4nn + 3n = 4n
n is the number of pages Maya read. Terms with the same letter are called like terms. Add them by adding the numbers in front: 1 n plus 3 n is 4 n. It works because 1 group of n pages and 3 more groups make 4 groups.
Turning a more-than fact around
if T=A+12, then A=T−12\text{if } T = A + 12 \text{, then } A = T - 12
T is the pages Maya and Trevor read together. A is the pages Ali read. The pair read 12 more than Ali, so Ali read 12 fewer than the pair. Subtracting 12 undoes adding 12.

  1. Sort out the question

    Maya read n pages. n is a variable, a letter that stands for a number you don’t know. Trevor read 3 times as many pages as Maya. Together they read 12 more pages than Ali. You need Ali’s pages, written with n.

  2. Write Trevor’s pages

    Three times as many as n is 3 times n:

    Trevor=3n\text{Trevor} = 3n

    A number written next to a letter means multiply. So 3n means 3 × n pages.

  3. Add Maya’s and Trevor’s pages

    Together means add. Maya read 1 group of n, and Trevor read 3 groups:

    n+3n=4nn + 3n = 4n

    Together they read 4n pages.

    Watch out: do not leave Maya out. Using only Trevor’s 3n gives 3n − 12, which is wrong.

  4. Decide who read more

    The pair read 12 more than Ali. So Ali read 12 fewer than the pair. Take 12 away from their total:

    Ali=4n−12\text{Ali} = 4n - 12

    Watch out: the word more tempts you to add 12. But 4n + 12 would mean Ali read more than the pair. And 12 − 4n subtracts in the wrong order.

  5. Enter the expression

    On the test, the keypad has an n key. Press 4, then n. Then press the minus key beside the plus key, then 1 and 2. You don’t need a times sign. On this site, type 4n-12.

    So you would enter 4n − 12. Any expression equal to it, such as −12 + 4n, is also right.

Check your answer

Pick a number for n, such as 10. Maya read 10 pages and Trevor read 30, so together they read 40. Your expression gives Ali 4×10−12=284 \times 10 - 12 = 28 pages. Is 40 twelve more than 28? Yes, 28+12=4028 + 12 = 40. So the expression works.

More questions like this

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

Question 7 of 120

2026 Practice Test C Math, question 21

Question

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

  • A2
  • B3
  • C18
  • D38
Show the answer

Correct answer: A

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

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

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

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

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

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

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

Solve for the value of a:

48a=9648a = 96

a=2a = 2

Question analysis

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

NYS standards
  • 6.RP.3 ratio and rate reasoning
  • 6.RP.1 ratio concepts and language
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Equal ratios with whole numbers are grade 6 work, and the numbers divide evenly. Grade 7 proportion problems are usually longer; here the only care is keeping apples with oranges in the same order as cherries with walnuts.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 32 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers divide evenly, since 16/48 is 1/3, so the work is in the setup. Writing the two ratios in the order the sentence gives them makes the apples a third of the oranges; writing either one upside down makes them three times the oranges. A quick sense check helps: cherries are fewer than walnuts, so apples must be fewer than oranges.

  1. Turning words into math minor Understanding the question

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

  2. Much to keep track of minor Working it out

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

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

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Stops one step early: A choice is a number from partway through the solution.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Write the proportion

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

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

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

  3. Simplify the known ratio

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

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

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

  4. Find the apples

    Now the proportion is:

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

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

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

    So the answer is A, 2.

Check your answer

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

More questions like this

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

Question 8 of 120

2026 Practice Test C Math, question 24

Question

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

  • A36
  • B63
  • C84
  • D108
Show the answer

Correct answer: B

The ratio of paintings to drawings to photographs is 3:5:43 : 5 : 4.

First, determine the number of actual paintings and photographs.

Since the number of actual drawings is 45, each ratio is multiplied by 9, since 5×9=455 \times 9 = 45.

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

Then, add the actual number of paintings and photographs.

27+36=6327 + 36 = 63

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.1 ratio concepts and language
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Scaling a ratio from one known part is grade 6 and 7 work. A three-part ratio, and a question about two of the three groups combined, make it a three-step task typical of grade 7 ratio problems.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 10, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Much to keep track of moderate Working it out

    The ratio has three parts, the known count belongs to the middle part, drawings, and the question wants the first and third parts combined, so three roles must be tracked. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    The word respectively assigns 3, 5 and 4 to paintings, drawings and photographs in that order, and the question then names two of them in the other order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The ratio of paintings to drawings to photographs is 3:5:4. Respectively means in that order: 3 for paintings, 5 for drawings and 4 for photographs. There were 45 drawings. You need the number of photographs and paintings together.

  2. Find the value of one part

    Drawings are 5 parts, and there were 45 drawings:

    45÷5=945 \div 5 = 9

    So each part of the ratio stands for 9 works.

  3. Find each group

    Multiply each group’s parts by 9:

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

    The drawings row gives 45, which matches the question.

  4. Add only the two groups asked for

    Add the paintings and the photographs:

    27+36=6327 + 36 = 63

    Watch out: 36 alone, choice A, leaves out the paintings. Adding the 45 drawings too gives 108, choice D, the whole show.

    So the answer is B, 63.

Check your answer

Paintings and photographs are 3+4=73 + 4 = 7 parts. Each part is 9 works, so 7×9=637 \times 9 = 63. That matches. Also, 63 to 45 is the same as 7 to 5, just as the ratio says.

More questions like this

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

Question 9 of 120

2026 Practice Test C Math, question 25

Question

For each 38\frac{3}{8} foot of curtain width, 56\frac{5}{6} yard of fabric is needed to make ruffles. How many yards of fabric are needed per foot of curtain width?

  • A516\frac{5}{16} yd
  • B2924\frac{29}{24} yd
  • C209\frac{20}{9} yd
  • D72\frac{7}{2} yd
Show the answer

Correct answer: C

It takes 56\frac{5}{6} yard of fabric to make 38\frac{3}{8} foot of curtain width.

To find the ratio of 1 foot of curtain width to 38\frac{3}{8} foot of curtain width (1:38)\left(1:\frac{3}{8}\right), divide.

1÷38=1×83=831 \div \frac{3}{8} = 1 \times \frac{8}{3} = \frac{8}{3}

So the amount of fabric needed for 1 foot of curtain width is 83\frac{8}{3} times the amount needed for 38\frac{3}{8} foot of curtain width.

83×56=4018\frac{8}{3} \times \frac{5}{6} = \frac{40}{18}

40÷218÷2=209\frac{40 \div 2}{18 \div 2} = \frac{20}{9}

The amount of fabric needed for 1 foot of curtain width is 209\frac{20}{9} yards.

Question analysis

Making ruffles takes 5/6 yard of fabric for every 3/8 foot of curtain width, and students find how many yards of fabric one foot of curtain width needs. It tests finding a unit rate when both quantities are fractions. The two units differ, and choosing whether to multiply, divide or add the fractions is the crucial decision.

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 the grade 7 rate standard itself. Two units, feet and yards, and an unfamiliar sewing setting make it harder to see which fraction divides which.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 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

Two fractions and two units, feet and yards, leave students to decide whether to multiply, divide or add, and each wrong choice records one of those decisions. A sense check rules out the most tempting one before any arithmetic: a whole foot is more than 3/8 foot, so it needs more than 5/6 yard, yet multiplying gives only 5/16. Dividing the yards by the feet then picks out the answer among the rest.

  1. Turning words into math moderate Understanding the question

    “For each 3/8 foot of curtain width, 5/6 yard of fabric” gives the rate as a pair of fractions, and “per foot” means dividing the yards by the feet. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Dividing by 3/8 means multiplying by 8/3, and the product 40/18 has to be reduced to 20/9, an improper fraction rather than a whole number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Divide the yards of fabric by the feet of curtain width, since a rate per foot puts feet in the denominator, then simplify the fraction.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
yards per foot=yards÷feet\text{yards per foot} = \text{yards} \div \text{feet}
A unit rate tells how much for exactly 1 of something. Here it is yards of fabric for 1 foot of curtain. Per foot means for each foot, so feet go on the bottom. Divide the yards by the feet.
Dividing by a fraction
a÷bc=a×cba \div \frac{b}{c} = a \times \frac{c}{b}
a is any number, and b/c is a fraction. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down. It works because b/c fits into 1 exactly c/b times.

  1. Sort out the question

    For every 3/8 foot of curtain width, you need 5/6 yard of fabric. You need the yards of fabric for 1 whole foot. That is a unit rate, an amount for exactly 1 foot.

    3/8 foot is less than a whole foot. So a whole foot needs more than 5/6 yard.

  2. Set up the division

    Per foot means divide the yards by the feet:

    56÷38\frac{5}{6} \div \frac{3}{8}

    Watch out: multiplying 5/6 by 3/8 gives 5/16, choice A. That is less fabric for more curtain, which makes no sense. Adding them gives 29/24, choice B, but you can’t add yards and feet.

  3. Multiply by the reciprocal

    Flip 3/8 to get 8/3, and multiply:

    56×83=5×86×3=4018\frac{5}{6} \times \frac{8}{3} = \frac{5 \times 8}{6 \times 3} = \frac{40}{18}

    The 8/3 says a whole foot is 8/3 times as long as 3/8 foot. So it needs 8/3 times as much fabric.

    Watch out: adding 8/3 and 5/6 gives 7/2, choice D. Multiply them instead.

  4. Simplify the fraction

    2 goes into both 40 and 18:

    40÷218÷2=209\frac{40 \div 2}{18 \div 2} = \frac{20}{9}

    So 1 foot of curtain width needs 20/9 yards of fabric, a little more than 2 yards. So the answer is C, 20/9 yd.

Check your answer

Multiply back. 3/8 foot should need 3/8 of the fabric for a whole foot: 209×38=6072=56\frac{20}{9} \times \frac{3}{8} = \frac{60}{72} = \frac{5}{6} yard. That is the amount the question gives. Also, 20/9 is more than 5/6, as it should be.

More questions like this

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

Question 10 of 120

2026 Practice Test C Math, question 29

Question

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

SMOOTHIE RECIPE

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 0.25

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

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

1.5=6p1.5 = 6p

1.56=6p6\frac{1.5}{6} = \frac{6p}{6}

14=0.25=p\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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality
k=yxk = \frac{y}{x}
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.

  1. 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.

  2. 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=yx=1.56=0.25k = \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÷1.5=46 \div 1.5 = 4 turns the answer upside down.

  3. Check two more rows

    The constant must be the same in every row. Test two more:

    2.08=0.253.012=0.25\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÷2=0.250.5 \div 2 = 0.25.

  4. 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. 6×0.25=1.56 \times 0.25 = 1.5, 8×0.25=28 \times 0.25 = 2, 10×0.25=2.510 \times 0.25 = 2.5 and 12×0.25=312 \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 11 of 120

2026 Practice Test C Math, question 33

Question

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

  • A6
  • B15
  • C18
  • D30
Show the answer

Correct answer: A

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

18÷6120÷6=320\frac{18 \div 6}{120 \div 6} = \frac{3}{20}

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

320(40)=12020=6\frac{3}{20}(40) = \frac{120}{20} = 6

Question analysis

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

NYS standards
  • 6.SP.7 long-run relative frequency
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Predicting from repeated trials is grade 6 probability, and grade 7 students make similar estimates from samples. The twist is scaling down, from 120 draws to a bag of 40 tiles, rather than up to a larger group.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 61 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

    The 18 red draws come from 120 draws with the tile put back each time, so they give a share, 3/20, to take of the 40 tiles, not a count of red tiles. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Students must know that the share of red in repeated draws estimates the share of red tiles in the bag; the stem never says to multiply 40 by that share. (Anderson & Krathwohl, 2001)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The bag has 40 tiles, each red, green or blue. A tile is drawn, its color is written down, and it goes back in the bag. This happens 120 times, and red comes up 18 times.

    You need the most likely number of red tiles in the bag.

  2. See why 18 is not the answer

    Each tile goes back in the bag after it is drawn. So the same red tile can be drawn again and again. That means 18 counts draws, not tiles.

    Watch out: choice C, 18, mixes up the number of red draws with the number of red tiles.

  3. Find the share of red draws

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

    18120=320\frac{18}{120} = \frac{3}{20}

    As a decimal, that is 0.15, or 15%.

    Watch out: 15 is a percent here, not a number of tiles. Stopping now gives choice B.

  4. Take that share of the bag

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

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

    So about 6 of the 40 tiles are red.

  5. Give the answer

    The best prediction is 6 red tiles. The other 34 tiles are probably green or blue.

    So the answer is A, 6.

Check your answer

There were 120 draws for 40 tiles. That is 3 draws for every tile. So divide the red draws by 3: 18÷3=618 \div 3 = 6. Same answer. Choice D, 30 red tiles, would be three quarters of the bag. Then red would come up far more than 18 times.

More questions like this

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

Question 12 of 120

2026 Practice Test C Math, question 36

Question

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

  • A110\frac{1}{10}
  • B14\frac{1}{4}
  • C313\frac{3}{13}
  • D13\frac{1}{3}
Show the answer

Correct answer: B

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

52−12=4052 - 12 = 40

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

13−3=1013 - 3 = 10

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

10 hearts remaining40 total cards remaining=1040=14\frac{10\text{ hearts remaining}}{40\text{ total cards remaining}} = \frac{10}{40} = \frac{1}{4}

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Finding a simple probability is grade 6 content. This one adds a step many grade 7 problems skip: the cards in Sara's hand must be taken out of both the hearts and the whole deck first.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 51 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Much to keep track of minor Working it out

    Four counts are in play, 52 cards, 13 hearts, Sara’s 12 cards and her 3 hearts, and both the hearts and the total must be reduced before the fraction is formed. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    “The other cards remain in the deck” and “the remainder of the deck” point to the 40 cards Sara is not holding, not to the full 52 or to her hand. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    A full deck has 52 cards, and 13 of them are hearts. Sara holds 12 cards, and 3 of those are hearts. The rest of the cards stay in the deck.

    You need the chance that a card drawn from the rest of the deck is a heart. Her hand changes both counts.

  2. Count the cards left

    Take Sara’s 12 cards out of the deck:

    52−12=4052 - 12 = 40

    So 40 cards are left in the deck.

  3. Count the hearts left

    Sara holds 3 of the 13 hearts. Take those out:

    13−3=1013 - 3 = 10

    So 10 hearts are left in the deck.

  4. Find the probability

    Use the probability formula. Put the hearts left over all the cards left, then simplify:

    P=1040=14P = \frac{10}{40} = \frac{1}{4}

    Watch out: comparing the 10 hearts with the 30 cards that are not hearts gives 13\frac{1}{3}, choice D. The bottom must be all 40 cards. Choice C, 313\frac{3}{13}, describes Sara’s hand, not the deck. Choice A, 110\frac{1}{10}, uses the 10 hearts but forgets the 40 cards.

  5. Give the answer

    The chance of drawing a heart from the rest of the deck is 1 in 4.

    So the answer is B, 14\frac{1}{4}.

Check your answer

10 is one quarter of 40, because 4×10=404 \times 10 = 40. Here is another check. Sara’s hand is 3 hearts out of 12, which is also one quarter. Taking a hand that is one quarter hearts out of a deck that is one quarter hearts leaves one quarter hearts.

More questions like this

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

Question 13 of 120

2026 Practice Test C Math, question 41

Question

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

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

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

Show the answer

Correct answer: 12.50 + 4.75n ≤ 35.00

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

Since notebooks cost $4.75 each and Elijah will buy n notebooks, the total cost just for the notebooks is found by multiplying $4.75 by n, or 4.75n4.75n.

Combining the cost of a water bottle ($12.50) with the cost of n notebooks, the inequality is 12.50+4.75n≤35.0012.50 + 4.75n \leq 35.00.

This may also be written as 4.75n+12.50≤35.004.75n + 12.50 \leq 35.00.

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing an inequality for a budget with a one-time cost and a cost per item is typical grade 7 work. The answer is typed, so both the symbol for "at most" and the role of each price must be right.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 50 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    The $12.50 bottle is bought once and the $4.75 is paid for each of n notebooks, and “at most $35.00” has to become ≤, which allows a total of exactly $35.00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The whole inequality, with n, its symbol and both prices, is built on a keypad offering ≤, <, ≥ and >, so a wrong symbol looks as complete as the right one. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Elijah buys one water bottle for $12.50. He also buys n notebooks at $4.75 each. He can spend at most $35.00.

    You need an inequality for n. An inequality compares two amounts with a sign such as ≤ instead of =.

  2. Price a few cases

    Work out the cost for a few numbers of notebooks:

    NotebooksTotal ($)
    112.50+4.75=17.2512.50 + 4.75 = 17.25
    212.50+9.50=22.0012.50 + 9.50 = 22.00
    312.50+14.25=26.7512.50 + 14.25 = 26.75

    The $12.50 shows up once. The $4.75 shows up once for each notebook.

  3. Write the total cost

    Use the total cost formula. The bottle is the one-time cost, and each notebook costs $4.75:

    12.50+4.75n12.50 + 4.75n

    Here 4.75n means 4.75 times n.

    Watch out: do not add the two prices first. 17.25n17.25n charges for a new bottle with every notebook.

  4. Choose the sign

    At most $35.00 means the total can be less than $35.00 or exactly $35.00. It cannot be more. So use ≤:

    12.50+4.75n≤35.0012.50 + 4.75n \leq 35.00

    Watch out: ≥ means at least, the opposite. And < leaves out a total of exactly $35.00, which Elijah is allowed to spend.

  5. Enter it on the keypad

    Type the whole inequality. Nothing is printed before the box. On the test, press 1, 2, ., 5, 0, +, 4, ., 7, 5, the n key, the ≤ key, then 3, 5, ., 0, 0. The ≤ key is in the fourth row. Skip the dollar signs. On this site, type <= for ≤.

    So you would enter 12.50 + 4.75n ≤ 35.00. An equal form, such as 4.75n + 12.50 ≤ 35.00, counts too.

Check your answer

Test some numbers. For 4 notebooks, 12.50+4.75×4=31.5012.50 + 4.75 \times 4 = 31.50, which is not over 35.00. For 5 notebooks, 12.50+4.75×5=36.2512.50 + 4.75 \times 5 = 36.25, which is too much. So Elijah can buy up to 4 notebooks. That makes sense for a budget of $35.00.

More questions like this

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

Question 14 of 120

2026 Practice Test C Math, question 44

Question

A smartphone screen is designed to be 72 millimeters wide. The quality control team allows for a variation of ±4%\pm 4\% in the width. Select the minimum and maximum screen widths that fall within the acceptable range of screen widths, in millimeters.

Select the correct answer from each drop-down to complete the sentence.

The acceptable screen width is from to millimeters.

Choose an answer from each menu, then check. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Show the answer

Correct answer: blank 1: 69.12; blank 2: 74.88

A variation of ±4%\pm 4\% in the width means that the width of the smartphone can be from 4% less than the standard of 72 mm to 4% greater than 72 mm.

The least allowed measure is

72−0.04(72)=72−2.88=69.12 mm72 - 0.04(72) = 72 - 2.88 = 69.12\text{ mm}

The greatest allowed measure is

72+0.04(72)=72+2.88=74.88 mm72 + 0.04(72) = 72 + 2.88 = 74.88\text{ mm}

Therefore, the allowable range is from 69.12 mm to 74.88 mm in width.

Question analysis

A smartphone screen is designed to be 72 millimeters wide, and quality control allows a variation of plus or minus 4%. Students choose the minimum and maximum acceptable widths from two drop-down menus. It tests finding a percent of a quantity, and the 4% has to be turned into millimeters before it is added or subtracted.

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 Finding a percent of a number and adding or subtracting it is grade 7 percent work. The ± sign, the decimal result and two linked menus, which must both be right, add a little to the usual demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 59 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards turning the percent into millimeters before doing anything else. Students who treat ±4% as ±4 millimeters find the two whole-number options at either end of the menus, 68.00 and 76.00, which look like a natural pair. Those who compute 2.88 correctly can still lose the item on the maximum by rounding to 3 and picking 75.00. Multiplying 72 by 0.96 and by 1.04 is a quick double-check.

  1. Error-prone arithmetic moderate Working it out

    Four percent of 72 is 0.04 × 72 = 2.88, a decimal product, and taking it from 72 means borrowing across the decimal point to reach 69.12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The phrase “a variation of ±4% in the width” has to become 4% of 72 millimeters, a length of 2.88 mm, before anything is added or subtracted. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. A very close wrong answer minor Choosing the answer

    The second menu lists 75.00 just 0.12 above 74.88, so a rounded estimate of 72 + 3 lands on the wrong option, and estimating cannot separate the two. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  4. Many steps minor Working it out

    Both menus must be right to score, so the minimum and the maximum each need their own subtraction or addition after the shared percent is found. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find 4% of the 72-millimeter design width first, then subtract that amount for the minimum and add it for the maximum, before looking at the menus.

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.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}
The percent is how many out of 100. Dividing it by 100 turns it into a decimal, so 4% becomes 0.04. The whole is the amount you take the percent of. The part is the amount that the percent stands for.

  1. Sort out the question

    The screen is designed to be 72 millimeters (mm) wide. The sign ±4% means plus or minus 4%. So a screen may be up to 4% narrower or up to 4% wider than 72 mm.

    You need the minimum, the smallest allowed width, and the maximum, the largest. You pick each one from a drop-down menu.

  2. Find 4% of 72 mm

    Use the percent formula. 4% is 4100=0.04\frac{4}{100} = 0.04:

    0.04×72=2.880.04 \times 72 = 2.88

    So the width may change by up to 2.88 mm either way.

    Watch out: 4% is not 4 mm. Taking 4 away and adding 4 gives 68.00 and 76.00. Both are in the menus, and both are wrong.

  3. Find the minimum width

    Take the change away from the design width:

    72−2.88=69.1272 - 2.88 = 69.12

    The narrowest allowed screen is 69.12 mm wide.

  4. Find the maximum width

    Add the change to the design width:

    72+2.88=74.8872 + 2.88 = 74.88

    The widest allowed screen is 74.88 mm wide.

    Watch out: rounding 2.88 up to 3 gives 75.00. That is in the second menu too, but it is too wide.

  5. Choose from each menu

    Open the first drop-down menu and choose 69.12. Open the second one and choose 74.88.

    So the sentence reads: the acceptable screen width is from 69.12 to 74.88 millimeters.

Check your answer

Use multipliers. A screen 4% narrower keeps 96% of the width: 0.96×72=69.120.96 \times 72 = 69.12. A screen 4% wider is 104% of it: 1.04×72=74.881.04 \times 72 = 74.88. Both match. Also, both ends are the same 2.88 mm away from 72.

More questions like this

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

Question 15 of 120

2026 Practice Test C Math, question 45

Question

Malik earns $10 per hour at his job. He wants to change to a job that will pay $12 per hour. What will be the percent increase in Malik’s hourly pay if he makes this job change?

  • A2%
  • B20%
  • C83%
  • D120%
Show the answer

Correct answer: B

Subtract to determine the amount of increase in Malik’s hourly pay if he changes jobs.

12−10=212 - 10 = 2

Divide the increase amount by the original hourly pay rate.

2÷10=0.22 \div 10 = 0.2

Multiply the quotient by 100 to express the decimal amount as a percentage.

0.2×100=200.2 \times 100 = 20

Question analysis

Malik earns $10 an hour and wants to switch to a job that pays $12 an hour. Students find the percent increase in his hourly pay. It tests finding a percent change. The numbers are easy, so the challenge is knowing which amounts to compare and how to turn that comparison into a percent.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Percent increase is grade 7 content, and this case uses whole-dollar wages that make the arithmetic easy. The demand is comparing the change with the right amount rather than reporting the change itself or a ratio of the wages.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 2, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers could hardly be friendlier, a $2 raise on $10, so the question checks what to divide by what. Dividing $12 by $10 gives 1.2, the new pay as a percent of the old, and stopping there is tempting because the division feels like the whole job. Reading the $2 itself as 2% is the other quick error, since dollars of change are not percent of change.

  1. Something you must already know minor Working it out

    The rule for percent change, the change divided by the starting amount, is not given, and each wrong choice comes from a different comparison of the $10 and the $12. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    “Percent increase in Malik’s hourly pay” has to be read as the $2 rise measured against the $10 he earns now, not as the new wage compared with the old. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find how many dollars the hourly pay goes up, then divide that change by the pay Malik earns now and write the result as a percent.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
percent increase=new−originaloriginal×100\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100
The original is the amount before the change. The new amount is the amount after it. The top is how much it went up. Dividing by the original compares the change with where you started. Multiplying by 100 makes it a percent.

  1. Sort out the question

    Malik earns $10 an hour now. The new job pays $12 an hour. You need the percent increase.

    That means: how much does his pay go up, compared with what he earns now? So $10 is the original amount, and $12 is the new amount.

  2. Find the increase in dollars

    Subtract the old pay from the new pay:

    12−10=212 - 10 = 2

    His pay goes up by $2 an hour.

    Watch out: $2 is an amount of money, not a percent. Calling it 2% gives choice A.

  3. Compare the increase with the original

    Use the percent increase formula. Divide the $2 by the original $10, then multiply by 100:

    210×100=0.2×100=20\frac{2}{10} \times 100 = 0.2 \times 100 = 20

    So the pay goes up by 20%.

    Watch out: 12÷10=1.212 \div 10 = 1.2 is 120%, choice D. That is the new pay as a percent of the old, not the increase. And 10÷1210 \div 12 is about 83%, choice C, which is upside down.

  4. Give the answer

    The $2 raise is 20% of Malik’s current pay.

    So the answer is B, 20%.

Check your answer

Work forward. 20% of $10 is 0.20×10=20.20 \times 10 = 2 dollars. Add that to $10, and you get $12, the new pay. It works. Note that 120% is the new pay compared with the old pay, which is the same as a 20% increase.

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 C Math, page 118, question 45.

Question 16 of 120

2026 Practice Test C Math, question 46

Question

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

  • A1
  • B2.5
  • C4
  • D10
Show the answer

Correct answer: C

Set up a proportion to solve:

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

2.5x=102.5x = 10

x=10÷2.5x = 10 \div 2.5

x=4x = 4

10 centimeters is approximately 4 inches.

Question analysis

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

NYS standards
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Converting units with a given conversion fact is grade 6 content, and this needs a single division with friendly numbers, lighter than grade 7 work with rates and percents.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 10, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The rain gauge shows 10 centimeters (cm) of rain. You know that 1 inch is about 2.5 cm. You need the rain in inches.

  2. Decide whether to multiply or divide

    An inch is longer than a centimeter. So it takes fewer inches than centimeters to measure the same rain. The answer must be less than 10. That tells you to divide, not multiply.

    Watch out: choice D, 10, keeps the same number, as if inches and centimeters were the same size.

  3. Divide by 2.5

    Use the conversion formula:

    10÷2.5=410 \div 2.5 = 4

    You can see it by adding: 2.5+2.5+2.5+2.5=102.5 + 2.5 + 2.5 + 2.5 = 10. Four groups of 2.5 cm fit into 10 cm. So 10 cm is about 4 inches.

  4. Give the answer

    Watch out: choices A and B, 1 and 2.5, are just the numbers in the conversion fact. They do not use the 10 cm at all.

    About 4 inches of rain fell. So the answer is C, 4.

Check your answer

Change back. Each inch is 2.5 cm, so 4 inches is 4×2.5=104 \times 2.5 = 10 cm. That is what the gauge showed. Also, 4 inches is less than 10, as it should be, because inches are longer.

More questions like this

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

Question 17 of 120

2026 Practice Test C Math, question 50

Question

Terry wants to pour cement around the edge of the circular patio in her backyard. The patio has a radius of 5 feet. What is the distance, in feet, around the edge of the patio? Use 3.14 for π\pi.

  • A15.7
  • B31.4
  • C49.3
  • D78.5
Show the answer

Correct answer: B

To find the distance around a circle if the radius is known, use the formula for circumference, C=2πrC = 2\pi r.

C=2(3.14)(5)=31.4C = 2(3.14)(5) = 31.4

Question analysis

Terry wants to pour cement around the edge of a circular patio with a radius of 5 feet. Students find the distance around the edge, using 3.14 for π. It tests the circumference of a circle, and because the formulas for a circle's area and circumference both use the radius, they are easy to mix up.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Applying a circle formula is a grade 7 standard, and this is a one-formula problem with a given value for π. The only demand beyond routine is recalling the right formula, which the test does not print.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One formula and one multiplication finish the problem once the right formula is chosen. The cement in the story pulls toward area, and the area, 78.5, is listed, as is half the circumference, from multiplying 3.14 by the radius alone. Remembering that the circumference is π times the diameter, here 10 feet, and that the answer is in feet, not square feet, settles it.

  1. Something you must already know moderate Working it out

    The circumference formula is not printed, and the area formula uses the same radius and π, so recalling the wrong one gives a number that looks just as finished. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    Pouring cement suggests covering a surface, but the question asks for “the distance, in feet, around the edge”, which is the circumference, not the area. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Decide which circle measurement the distance around the edge is, recall its formula, and substitute the radius and 3.14, doubling the radius wherever the formula calls for the diameter.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle
C=2πrC = 2 \pi r
C is the circumference, the distance around the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, about 3.14. 2r is the diameter, the distance all the way across. The distance around is always about 3.14 times that.
Area of a circle
A=πr2A = \pi r^2
A is the area, the space inside the circle, in square units. r² means r times r. It is here so you can tell the two formulas apart. This question does not use it.

  1. Sort out the question

    The patio is a circle with a radius of 5 feet. The radius is the distance from the center to the edge. Terry wants cement around the edge.

    You need the distance around the edge, in feet. The distance around a circle is called its circumference. Use 3.14 for π.

  2. Choose the right formula

    The distance around is a length, so use the circumference formula, C=2πrC = 2\pi r.

    Watch out: pouring cement may make you think of area. The area, 3.14×5×5=78.53.14 \times 5 \times 5 = 78.5 square feet, is choice D. That is the space inside the patio, not the distance around it.

  3. Put in the numbers

    Put 3.14 in for π and 5 in for r. Multiply 2 by 5 first, because that is easy:

    C=2×3.14×5=10×3.14=31.4C = 2 \times 3.14 \times 5 = 10 \times 3.14 = 31.4

    The distance around the patio is 31.4 feet.

    Watch out: 3.14×5=15.73.14 \times 5 = 15.7, choice A, forgets the 2. It goes only halfway around.

  4. Give the answer

    The distance around the edge of the patio is 31.4 feet.

    So the answer is B, 31.4.

Check your answer

Estimate. The diameter is 10 feet. The distance around a circle is a little more than 3 times its diameter. 3×10=303 \times 10 = 30, so about 31 feet makes sense. Choice D, 78.5, is far too big, and it is in square feet.

More questions like this

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

Question 18 of 120

2026 Practice Test D Math, question 6

Question

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

  • A12+2(p+3)12 + 2(p + 3)
  • B12+2(p−3)12 + 2(p - 3)
  • C14p+314p + 3
  • D14p−314p - 3
Show the answer

Correct answer: B

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

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

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

  • 4 pounds is

12+(4−3)(2)=1412 + (4 - 3)(2) = 14

  • 5 pounds is

12+(5−3)(2)=1612 + (5 - 3)(2) = 16

  • p pounds is

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

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

Question analysis

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

NYS standards
  • 6.EE.6 variables representing unknown numbers
  • 6.EE.2a writing algebraic expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Writing an expression from a cost description is grade 6 algebra. Because the per-pound charge starts only after the first 3 pounds, it is less routine than a fee plus a simple rate, in line with typical grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 6, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    “$2 for each additional pound over 3 pounds” must become 2(p − 3), a charge on the pounds beyond the first 3, and the word “over” invites writing p + 3 instead. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    12 + 2(p + 3) and 12 + 2(p − 3) differ only in the sign inside the parentheses, so their shape cannot separate them; only a test weight or a careful reading can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    The weight is the letter p, so there is no cost to work out and compare; choosing a weight such as 5 pounds, which costs $16, turns every choice into a number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Shipping costs $12 to start. Each pound over 3 pounds adds $2. The package weighs p pounds, and p is at least 3. You need an expression for the total cost in dollars.

  2. Count the extra pounds

    Only the pounds above 3 cost extra. To find them, take 3 away from the weight:

    extra pounds=p−3\text{extra pounds} = p - 3

    For example, a 5-pound package has 5 − 3 = 2 extra pounds.

    Watch out: over 3 pounds means more than 3. It does not tell you to add 3. Writing p + 3 leads to choice A.

  3. Build the expression

    Each extra pound costs $2, so the extra pounds cost 2(p − 3) dollars. Add the $12 base fee:

    total=12+2(p−3)\text{total} = 12 + 2(p - 3)

    Watch out: the $12 is paid once, not for every pound. Choices C and D charge $14 for every pound, which mixes up the fee and the rate.

  4. Test each choice with 5 pounds

    A 5-pound package costs $12 plus two $2 charges, which is $16. Put p = 5 into each choice:

    ChoiceValue when p = 5
    A12 + 2(8) = 28
    B12 + 2(2) = 16
    C14(5) + 3 = 73
    D14(5) − 3 = 67

    Only B gives $16. So the answer is B, 12+2(p−3)12 + 2(p - 3).

Check your answer

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

More questions like this

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

Question 19 of 120

2026 Practice Test D Math, question 11

Question

In a scale drawing of a triangular park, the sides measure 5 inches, 7 inches, and 8 inches. The perimeter of the actual park is 25 miles. What is the actual length of the shortest side of the park?

  • A6.25 mi
  • B8.75 mi
  • C10 mi
  • D15.625 mi
Show the answer

Correct answer: A

To find the perimeter of a triangle, add the measures of the three sides. The perimeter of the scale drawing is 5+7+8=205 + 7 + 8 = 20 inches.

Now, use the perimeters to find the scale factor for the scale drawing to the actual perimeter.

The scale factor is 20 inches to 25 miles, which reduces to a scale factor of 4 inches to 5 miles. Note that the scale for the perimeter is the same as the scale for the side lengths.

Since the smallest side on the scale drawing is 5 inches, use the scale factor to find the actual side length of the smallest side:

5 inches(5 miles4 inches)=6.25 miles5\text{ inches}\left(\frac{5\text{ miles}}{4\text{ inches}}\right) = 6.25\text{ miles}

Question analysis

A scale drawing of a triangular park has sides of 5, 7 and 8 inches, and the real park's perimeter is 25 miles. Students find the actual length of the park's shortest side. It tests scale drawings and proportional reasoning; the scale is not given directly, so it must be worked out from the two perimeters first.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Scale drawings are a grade 7 topic, and this is a standard scale problem. The one twist is that the scale is not stated: it must come from comparing the two perimeters before any side can be converted.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

A scale drawing multiplies every length by one factor, and the perimeter shares that factor, so the work is short once the two perimeters are paired. The common loss is reasoning with differences: the perimeter grows by 5, so a side seems to grow by 5 as well. Converting to miles per inch first, then checking that the three real sides add back to 25 miles, guards against that error and against scaling the wrong side.

  1. Turning words into math minor Understanding the question

    The 25 miles is the park's perimeter, not a side, so it must be matched with the drawing's perimeter, 5 + 7 + 8 = 20 inches, before any side can be scaled. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three operations in a row, adding the sides, dividing 25 by 20 for 1.25 miles per inch, and multiplying the 5-inch side, each produce a number the next step needs. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the three drawn sides to get the drawing's perimeter, divide the real perimeter by it to find the miles per inch, then multiply the shortest drawn side by that scale.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a triangle
P=a+b+cP = a + b + c
P is the perimeter, the distance all the way around. a, b and c are the lengths of the three sides. You add them because going once around the triangle covers each side once.
Using a scale
actual length=drawing length×miles per inch\text{actual length} = \text{drawing length} \times \text{miles per inch}
The drawing length is measured on the drawing, in inches. Miles per inch is how many real miles each inch stands for. In a scale drawing, every length is multiplied by the same number, so the perimeter and each side use the same scale.

  1. Sort out the question

    A scale drawing is a smaller copy, with every length shrunk by the same amount. The drawing’s sides are 5, 7 and 8 inches. The real park’s perimeter, the distance all the way around, is 25 miles. You need the real length of the shortest side.

  2. Find the drawing’s perimeter

    Add the three sides of the drawing:

    P=5+7+8=20P = 5 + 7 + 8 = 20

    The drawing’s perimeter is 20 inches. It stands for the real perimeter of 25 miles.

    Watch out: 25 miles is the whole way around, not one side. Matching it with the 8-inch side gives 15.625 miles, choice D.

  3. Find the scale

    20 inches stand for 25 miles. Divide to find how many miles 1 inch stands for:

    25÷20=1.2525 \div 20 = 1.25

    So each inch on the drawing is 1.25 miles in the park.

  4. Scale the shortest side

    The shortest side on the drawing is 5 inches. Multiply it by the scale:

    5×1.25=6.255 \times 1.25 = 6.25

    The real shortest side is 6.25 miles.

    Watch out: the perimeter grew by 5, from 20 to 25, but you cannot just add 5 to a side. That gives 10 miles, choice C. Scaling multiplies. Also use the 5-inch side, not the 7-inch side, which gives 8.75 miles, choice B.

    So the answer is A, 6.25 mi.

Check your answer

Scale all three sides: 5 × 1.25 = 6.25, 7 × 1.25 = 8.75 and 8 × 1.25 = 10. Add them: 6.25 + 8.75 + 10 = 25 miles. That is the park’s perimeter, so the scale is right.

More questions like this

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

Question 20 of 120

2026 Practice Test D Math, question 14

Question

James must read 84 pages of a book for a class assignment. If he reads 14\frac{1}{4} of the pages in the morning and 13\frac{1}{3} of the remaining pages in the afternoon, how many pages will he still need to read?

  • A35
  • B42
  • C45
  • D56
Show the answer

Correct answer: B

Pages read in the morning: (14)(84)=21\left(\frac{1}{4}\right)(84) = 21

Pages remaining after the morning: 84−21=6384 - 21 = 63

Pages read in the afternoon: (13)(63)=21\left(\frac{1}{3}\right)(63) = 21

Pages he still needs to read: 63−21=4263 - 21 = 42

Question analysis

James must read 84 pages for a class assignment: 1/4 of them in the morning, then 1/3 of the remaining pages in the afternoon. Students find how many pages he still has left to read. It tests finding a fraction of a quantity in stages, and the bold word remaining decides which amount the second fraction is taken of.

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 Taking fractions of a quantity in stages is typical grade 7 work. The numbers divide evenly and the bold word flags the trap, so the four short steps match routine grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 7, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is simple, but the second fraction has a different whole from the first. Students who take 1/3 of the full 84 pages get a tidy 28 and a listed answer, despite the bold word that warns against it. Here the pages read and the pages left both come to 42, so confusing those two quantities costs nothing, and checking that 21 + 21 + 42 = 84 confirms the work.

  1. Many steps moderate Working it out

    Four operations in order, a quarter of 84, a subtraction, a third of the 63 left, and a second subtraction, each feed the next, so a slip early carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The bold “remaining” moves the base of the second fraction from all 84 pages to the 63 left after the morning, and “still need to read” asks for what is left, not what was read. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work in order: find the morning's pages, subtract to get what remains, take 1/3 of that remainder, and subtract again; a quick bar sketch of the 84 pages helps.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A fraction of a number
ab of n=ab×n\frac{a}{b} \text{ of } n = \frac{a}{b} \times n
n is the whole amount, and a/b is the fraction you take. The word of means multiply. When the top, a, is 1, this is just n divided by b. For example, 1/4 of 84 is 84 ÷ 4.

  1. Sort out the question

    James has 84 pages to read. In the morning he reads 1/4 of all the pages. In the afternoon he reads 1/3 of the remaining pages, the ones still left after the morning. You need the number of pages he still has to read after that.

  2. Find the morning pages

    Take 1/4 of 84:

    14×84=84÷4=21\frac{1}{4} \times 84 = 84 \div 4 = 21

    He reads 21 pages in the morning. That leaves:

    84−21=6384 - 21 = 63

    63 pages are left after the morning.

  3. Find the afternoon pages

    The afternoon fraction is 1/3 of the 63 pages left, not of all 84:

    13×63=63÷3=21\frac{1}{3} \times 63 = 63 \div 3 = 21

    He reads 21 pages in the afternoon.

    Watch out: taking 1/3 of all 84 pages gives 28. Then 84 − 21 − 28 = 35, choice A. The bold word remaining tells you to use 63.

  4. Find the pages still to read

    Take the afternoon pages away from the 63 pages left after the morning:

    63−21=4263 - 21 = 42

    He still needs to read 42 pages.

    Watch out: subtract both readings, morning and afternoon. Taking only 28 from 84 gives 56, choice D.

    So the answer is B, 42.

Check your answer

Add up all the parts: 21 pages in the morning, 21 in the afternoon and 42 still to read. 21 + 21 + 42 = 84, the whole assignment. Another way: after he reads 1/3 of the 63 pages, 2/3 of them are left, and 2/3 of 63 is 42.

More questions like this

The original DOE question
Question 14 as shown on the DOE’s digital test, page 87 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 87, question 14.

Question 21 of 120

2026 Practice Test D Math, question 16

Question

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

BREAD RECIPE

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 0.5

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

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

1=2p1 = 2p

12=2p2\frac{1}{2} = \frac{2p}{2}

0.5=p0.5 = 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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality
y=kxsok=yxy = 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.

  1. 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.

  2. 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=yx=1.02=0.5k = \frac{y}{x} = \frac{1.0}{2} = 0.5

    So there is 0.5 ounce of raisins for each cup of oatmeal.

  3. Check the other rows

    A constant must be the same in every row:

    1.53=0.53.06=0.5\frac{1.5}{3} = 0.5 \qquad \frac{3.0}{6} = 0.5

    Every row gives 0.5, so the rule is y=0.5xy = 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.

  4. 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 22 of 120

2026 Practice Test D Math, question 17

Question

At Midville High School, 64 students are members of either the orchestra or the choir, or both. Of these students, 38 are members of the orchestra, and 46 are members of the choir. What fraction of the total is the number of students who are members of both the orchestra and the choir?

  • A1621\frac{16}{21}
  • B1019\frac{10}{19}
  • C1023\frac{10}{23}
  • D516\frac{5}{16}
Show the answer

Correct answer: D

First, add the number of students that are members of the orchestra and the number of students that are members of the choir.

38+46=8438 + 46 = 84

Then, subtract to find the number of students that are members of both orchestra and choir.

84−64=2084 - 64 = 20

Then, find the fraction of the 64 students that are members of both orchestra and choir.

2064=516\frac{20}{64} = \frac{5}{16}

Question analysis

At Midville High School, 64 students belong to the orchestra, the choir or both; 38 are in the orchestra and 46 in the choir. Students find what fraction of the 64 belong to both groups. It is a Venn diagram counting question with a fraction, and the students in both groups are included in each membership number.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 6.RP.1 ratio concepts and language
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Overlapping groups are seldom taught directly in grade 7, though the arithmetic is whole-number addition and subtraction. Students must first find the overlap and then choose the right whole for the fraction, two decisions where errors are common.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 54 words.
Corrected from the DOE handbook
  • A misstatement in the explanation’s first step was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question hides its key number: the students in both groups appear only as the amount by which 38 + 46 overshoots 64. A sketch of two overlapping circles, with 20 in the middle, 18 in orchestra only and 26 in choir only, shows the whole group at once. After that, the remaining risk is the denominator, since the question asks for a fraction of all 64 students, not of one group.

  1. Turning words into math moderate Understanding the question

    “Either the orchestra or the choir, or both” means the 64 counts each student once, while 38 + 46 counts the students in both groups twice; the overlap itself is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method minor Working it out

    Nothing names the step that finds the overlap: the excess of 38 + 46 = 84 over 64 is the number in both groups, an idea students must bring themselves. (Schoenfeld, 1985)

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

    The fraction 20/64 appears only in lowest terms, so it must be simplified by 4 before it can be matched with a choice. (Enright, Morley & Sheehan, 2002)

Strategy: Add the orchestra and choir counts, subtract the 64 students to find how many were counted twice, then put that number over the total and simplify.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation begins “First, add the number of students that are members of both orchestra and choir. 38 + 46 = 84.” But 38 is the number of orchestra members and 46 the number of choir members; the number of students in both is what the next step finds (84 − 64 = 20). The site now says “First, add the number of students that are members of the orchestra and the number of students that are members of the choir.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting two groups that overlap
total=O+C−both\text{total} = O + C - \text{both}
O is the number in the orchestra, and C is the number in the choir. Students in both groups are counted in O and again in C. Subtracting them once makes each student count one time. Here every student is in at least one group, so this gives everyone.
Fraction of a total
fraction=partwhole\text{fraction} = \frac{\text{part}}{\text{whole}}
The part is the group you are asked about. The whole is the full group it comes from. Here the part is the students in both groups, and the whole is all 64 students. Simplify by dividing the top and bottom by the same number.

  1. Sort out the question

    There are 64 students in all. Each one is in the orchestra, the choir or both. 38 are in the orchestra, and 46 are in the choir. Some students are in both, but the question does not say how many. You need that number as a fraction of all 64 students.

  2. Add the two groups

    Add the number in the orchestra and the number in the choir:

    38+46=8438 + 46 = 84

    That is more than the 64 students there are. Why? A student in both groups is counted once with the orchestra and once more with the choir.

  3. Find the students in both

    The extra amount over 64 is the students who were counted twice:

    84−64=2084 - 64 = 20

    So 20 students are in both groups. This is the overlap rule turned around: both = O + C − total.

    Watch out: 84 is not the number of students. Putting 64 over 84 gives 16/21, choice A.

  4. Write the fraction and simplify

    The part is 20 students, and the whole is all 64 students. Both numbers divide by 4:

    2064=20÷464÷4=516\frac{20}{64} = \frac{20 \div 4}{64 \div 4} = \frac{5}{16}

    Watch out: the whole is all 64 students, not one group. Using 38 gives 10/19, choice B. Using 46 gives 10/23, choice C.

    So the answer is D, 516\frac{5}{16}.

Check your answer

Split the 64 students into three groups that do not overlap. Only orchestra: 38 − 20 = 18. Only choir: 46 − 20 = 26. Both: 20. Add them: 18 + 26 + 20 = 64. That matches the total, so 20 is right.

More questions like this

The original DOE question
Question 17 as shown on the DOE’s digital test, page 90 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 90, question 17. This is the DOE’s version, before the corrections listed in the analysis.

Question 23 of 120

2026 Practice Test D Math, question 19

Question

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

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

Correct answer: C

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

456×1.25=570456 \times 1.25 = 570

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

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

The sales tax on the selling price is:

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

The customer pays the selling price plus the sales tax:

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

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students find markups and sales tax, usually with increases under 100%. An increase of more than 100% is unusual and easy to misread, and two percent changes must be applied in order, which goes beyond routine grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 7, 41 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

  2. Many steps moderate Working it out

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

  3. Error-prone arithmetic minor Working it out

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

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

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the increase

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

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

    The increase is $570.

  3. Find the selling price

    Increases by means the increase is added to the cost:

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

    The selling price is $1,026.

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

  4. Find the sales tax

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

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

    The tax is $82.08.

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

  5. Add the tax to the price

    The customer pays the selling price plus the tax:

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

    So the answer is C, $1,108.08.

Check your answer

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

More questions like this

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

Question 24 of 120

2026 Practice Test D Math, question 21

Question

Maxie borrowed $7.75 from her mother, $11.00 from her father, and $4.50 from her brother to purchase a video game. Her grandmother gave her $25.00 as a gift. How much money will Maxie have left or still owe if she uses the money her grandmother gave her to pay back the money she borrowed?

  • AMaxie will still owe $2.25.
  • BMaxie will still owe $1.75.
  • CMaxie will have $1.75 left.
  • DMaxie will have $2.25 left.
Show the answer

Correct answer: C

First, determine the total amount Maxie borrowed, in dollars:

7.75+11.00+4.50=23.257.75 + 11.00 + 4.50 = 23.25

Then use subtraction to determine the difference between the amount her grandmother gave her and the amount she borrowed:

25.00−23.25=1.7525.00 - 23.25 = 1.75

The difference is positive, which means that Maxie will still have $1.75 left from the amount her grandmother gave her.

Question analysis

Maxie borrowed money from her mother, father and brother to buy a video game, then received $25.00 from her grandmother to pay them back. Students decide how much money she will have left over or still owe. It tests adding and subtracting decimals with money, and the choices also test which way the difference goes.

NYS standards
  • 6.NS.3 multi-digit decimal operations
  • 6.NS.5 positive and negative numbers
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Adding and subtracting money amounts is grade 6 decimal work. Here it takes one sum and one difference, and the only further demand is deciding whether the result is money left or money owed, well below typical grade 7 tasks.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 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

Few students miss this one, since it needs only one addition and one subtraction with money. The choices come in pairs, so the right amount must also be matched with the right outcome. Lining up the decimal points, and asking whether $25.00 is more or less than the $23.25 borrowed, settles both the amount and whether Maxie owes or keeps money.

  1. Error-prone arithmetic minor Working it out

    The loans add to $23.25, and subtracting that from $25.00 means borrowing across two zeros, a classic place for a digit to go wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    Each amount appears twice among the choices, once as money left and once as money owed, so the result must also be read for direction by asking which is larger, the gift or the loans. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add the three amounts borrowed, compare the total with the $25.00 gift, and decide from which is larger whether Maxie ends up with money left or still owes some.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Money left over
left=money she has−money she owes\text{left} = \text{money she has} - \text{money she owes}
Money she has is the $25.00 gift. Money she owes is the total she borrowed. If the result is positive, she has that much left. If it were negative, she would still owe that much. Comparing the two amounts first tells you which.

  1. Sort out the question

    Maxie borrowed from three people: $7.75, $11.00 and $4.50. Her grandmother gave her $25.00. She uses the $25.00 to pay everyone back. You need to know if money is left over or if she still owes some, and how much.

  2. Add up what she borrowed

    Line up the decimal points and add:

    7.75+11.00+4.50=23.257.75 + 11.00 + 4.50 = 23.25

    She owes $23.25 in all. That is the amount she must pay back.

  3. Compare the gift with the debt

    The gift, $25.00, is more than the $23.25 she owes. So she can pay everyone back and still have money left. She will not owe anything.

    Watch out: once you find $1.75, keep its meaning straight. Saying she still owes $1.75, choice B, gets the direction backwards.

  4. Subtract to find what is left

    Take the debt away from the gift:

    25.00−23.25=1.7525.00 - 23.25 = 1.75

    You need to regroup (borrow) across the zeros. Think of $25.00 as 24 dollars and 100 cents. Then 100 − 25 = 75 cents, and 24 − 23 = 1 dollar.

    Watch out: taking the smaller digit from the larger one in each column gives $2.25, choice D.

    So the answer is C: Maxie will have $1.75 left.

Check your answer

Add back. The $23.25 she pays plus the $1.75 she keeps is 23.25 + 1.75 = 25.00. That is the whole gift, so the subtraction is right.

More questions like this

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

Question 25 of 120

2026 Practice Test D Math, question 22

Question

Seth grows strawberries in his garden.

  • He picked 2122\frac{1}{2} cups of strawberries on Monday.
  • He picked 4 cups of strawberries on Tuesday.
  • He ate 34\frac{3}{4} cup of strawberries for breakfast each day on Wednesday, Thursday, and Friday.

How many cups of strawberries did Seth have left?

  • A3123\frac{1}{2}
  • B4144\frac{1}{4}
  • C6
  • D9
Show the answer

Correct answer: B

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

212+4+(3)(−34)2\frac{1}{2} + 4 + (3)\left(-\frac{3}{4}\right)

212+4−2142\frac{1}{2} + 4 - 2\frac{1}{4}

612−2146\frac{1}{2} - 2\frac{1}{4}

624−2146\frac{2}{4} - 2\frac{1}{4}

4144\frac{1}{4}

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

212+4+(−34)+(−34)+(−34)2\frac{1}{2} + 4 + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right)

4144\frac{1}{4}

Question analysis

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

NYS standards
  • 7.NS.3 four operations with rational numbers
  • 7.EE.3 multi-step rational number problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding mixed numbers and multiplying a fraction by a whole number are skills from before grade 7. The three steps are short and the numbers friendly, so the demand is below typical grade 7 work with signed rational numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    Three breakfasts of 3/4 cup make 9/4 = 2 1/4 cups, and taking that from 6 1/2 means renaming the half as 2/4, two fraction steps where a slip is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    Picking adds and eating subtracts, and “each day on Wednesday, Thursday, and Friday” means the 3/4 cup counts three times, not once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Seth picked 2 1/2 cups on Monday and 4 cups on Tuesday. Then he ate 3/4 cup each day on Wednesday, Thursday and Friday. That is 3 days of eating. Picking adds strawberries, and eating takes them away. You need the number of cups he has left.

  2. Add what he picked

    Add the two days of picking:

    212+4=6122\frac{1}{2} + 4 = 6\frac{1}{2}

    He picked 6 1/2 cups in all, before he ate any.

  3. Find how much he ate

    He ate 3/4 cup on each of 3 days. Multiply:

    3×34=94=2143 \times \frac{3}{4} = \frac{9}{4} = 2\frac{1}{4}

    To change 9/4 to a mixed number, note that 4 fits into 9 twice, with 1 left over. He ate 2 1/4 cups.

    Watch out: he ate 3/4 cup three times, not 3 cups. Subtracting 3 gives 3 1/2, choice A.

  4. Take away what he ate

    Write the half as fourths, so both fractions have the same bottom: 1/2 = 2/4. Then subtract the wholes and the fourths:

    612−214=624−214=4146\frac{1}{2} - 2\frac{1}{4} = 6\frac{2}{4} - 2\frac{1}{4} = 4\frac{1}{4}

    Watch out: eating takes away. Adding the 2 1/4 cups gives 8 3/4, and the nearest choice is 9, choice D.

    So the answer is B, 4144\frac{1}{4} cups.

Check your answer

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

More questions like this

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

Question 26 of 120

2026 Practice Test D Math, question 23

Question

Which situation involves quantities that combine to equal zero?

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

Correct answer: A

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

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

Question analysis

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

NYS standards
  • 7.NS.1a opposite quantities making 0
  • 6.NS.5 positive and negative numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Describing situations where opposite quantities make zero is named in the grade 7 standards, and this is its most direct form: one reading of four everyday situations, easier than typical grade 7 work with signed numbers.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 5, 60 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

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

  2. Every choice must be checked minor Choosing the answer

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Write each choice with signs

    Turn each situation into signed numbers and add them up:

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

  3. Look at the tricky choices

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

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

  4. Choose the zero total

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

    5+(−5)=05 + (-5) = 0

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

Check your answer

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

More questions like this

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

Question 27 of 120

2026 Practice Test D Math, question 24

Question

An athlete runs 18\frac{1}{8} kilometer in 34\frac{3}{4} minute. At this rate, how many kilometers would the athlete run in 1 minute?

  • A332\frac{3}{32}
  • B16\frac{1}{6}
  • C1121\frac{1}{2}
  • D6
Show the answer

Correct answer: B

To determine the number of kilometers in 1 minute, find the unit rate in kilometers per minute: 18 kilometer34 minute\frac{\frac{1}{8}\text{ kilometer}}{\frac{3}{4}\text{ minute}}.

To evaluate the fraction, use the reciprocal when dividing to get 16\frac{1}{6} kilometer per 1 minute.

18÷34=18×43=424=16\frac{1}{8} \div \frac{3}{4} = \frac{1}{8} \times \frac{4}{3} = \frac{4}{24} = \frac{1}{6}

Question analysis

An athlete runs 1/8 kilometer in 3/4 minute, and students find how many kilometers the athlete would run in 1 minute at the same rate. It tests unit rates with fractions. Both the distance and the time are fractions, so deciding which one to divide by which is the main challenge.

NYS standards
  • 7.RP.1 unit rates with fractions
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Computing a unit rate from a ratio of fractions is the grade 7 standard itself, and this is a standard example: one fraction divided by another in two short steps, with the order of division the only real decision.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 9, 23 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is a unit rate with fractions, and its wrong choices are the other ways to combine 1/8 and 3/4. Since 1 minute is only a third longer than 3/4 minute, the distance must be only a third more than 1/8 kilometer, which rules out the product and both large choices at once. Dividing with the units written in, kilometers over minutes, keeps the order right.

  1. Error-prone arithmetic moderate Working it out

    The rate is a fraction divided by a fraction, 1/8 ÷ 3/4, and only the 3/4 may be flipped; multiplying straight across or dividing the other way round both give listed choices. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    Kilometers in 1 minute means kilometers per minute, so the 1/8 goes on top even though the time, 3/4, is the larger number. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Divide the distance by the time, kilometers over minutes, by multiplying 1/8 by the reciprocal of 3/4, and check that the units of the result are kilometers per minute.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
rate=distance÷time\text{rate} = \text{distance} \div \text{time}
A unit rate tells how much happens in 1 unit of time. Here the distance is in kilometers and the time is in minutes, so the rate is kilometers per minute. You divide because you are sharing the distance equally over the minutes.
Dividing fractions
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). Flip only the second fraction, the one you divide by.

  1. Sort out the question

    The athlete runs 1/8 kilometer in 3/4 minute, which is less than a full minute. You need how far the athlete runs in 1 whole minute at the same speed. That is the unit rate, in kilometers per minute.

  2. Set up the division

    Kilometers per minute means kilometers divided by minutes. So the distance comes first and the time second:

    rate=18÷34\text{rate} = \frac{1}{8} \div \frac{3}{4}

    Watch out: dividing the other way, 3/4 ÷ 1/8, gives 6. That is minutes per kilometer, the rate turned upside down, choice D.

  3. Divide the fractions

    Flip 3/4 to get 4/3, then multiply. Multiply the tops and multiply the bottoms:

    18×43=424\frac{1}{8} \times \frac{4}{3} = \frac{4}{24}

    Simplify by dividing the top and bottom by 4:

    424=16\frac{4}{24} = \frac{1}{6}

    Watch out: multiplying 1/8 × 3/4 without flipping gives 3/32, choice A.

  4. Make sense of the answer

    In 1 minute, the athlete runs 1/6 kilometer. That is a bit more than 1/8 kilometer, which makes sense. A full minute is a bit longer than 3/4 minute, so the athlete goes a bit farther. Choice C, 1 1/2 kilometers, would be 12 times as far in barely more time.

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

Check your answer

Work in quarter minutes. In 3 quarter minutes, the athlete runs 1/8 km. So in 1 quarter minute, the athlete runs 1/8 ÷ 3 = 1/24 km. A full minute has 4 quarters: 4 × 1/24 = 4/24 = 1/6 km.

More questions like this

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

Question 28 of 120

2026 Practice Test D Math, question 26

Question

Jordan has a budget of $120 to rent a food truck. The rental company charges $18 per hour to rent the truck, plus a one-time cleaning fee of $30.

Write an inequality that represents this situation, in which x is the number of hours Jordan can rent the truck.

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

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

Show the answer

Correct answer: 30 + 18x ≤ 120

Let x represent the number of hours Jordan can rent the truck.

The hourly rental cost can be represented as $18⋅x\$18 \cdot x, written as 18x18x.

The one-time cleaning fee of $30 is added to the hourly cost to obtain the total cost.

The sum of the cleaning fee and the hourly rental cost is represented as 30+18x30 + 18x.

As Jordan has $120 to spend, the total cost must be less than or equal to $120.

This is represented as 30+18x≤12030 + 18x \leq 120 or 120≥30+18x120 \geq 30 + 18x.

Question analysis

Jordan has $120 to rent a food truck that costs $18 an hour plus a one-time $30 cleaning fee, and students type an inequality for x, the number of hours. It tests writing a two-step inequality from a word problem. Matching each dollar amount to its role and choosing the symbol that fits a spending limit are the tricky parts.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing a two-step inequality from a word problem is the grade 7 standard itself. Nothing is offered to choose from: students type the whole inequality, so the numbers, the variable term and the symbol are all theirs to get right.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 57 words.
Corrected from the DOE handbook
  • A spelling error in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is small, so the points ride on two decisions the keypad leaves open: which symbol, and whether to stop before solving. A budget means Jordan may spend all $120 but no more, so the total cost must stay at or under it, and a strict < leaves out the five-hour rental that costs exactly $120. Testing x = 5 and x = 6 in the typed inequality confirms both the symbol and the setup.

  1. Turning words into math moderate Understanding the question

    The stem never says “at most”: the limit comes only from “a budget of $120”, while “$18 per hour” must multiply x and the “one-time cleaning fee of $30” is added just once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The whole inequality is typed on a keypad that offers <, ≤, > and ≥ side by side, with no choices to compare, and the solved form x ≤ 5 is not accepted. (Enright, Morley & Sheehan, 2002)

Strategy: Work out what 1 hour and 2 hours would cost Jordan in all, see which amount repeats each hour, build the total for x hours, then compare it with the budget and test a few values.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation says “The onetime cleaning fee of $30 is added to the hourly cost …”. “Onetime” means “former”; a fee charged once is a “one-time” fee, as the question itself spells it. The site now says “one-time”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a fee and a rate
total cost=fee+rate×x\text{total cost} = \text{fee} + \text{rate} \times x
The fee is a charge paid only once, so you add it once. The rate is the charge for each hour. x is the number of hours. You multiply the rate by x because it is paid again for every hour.
Staying within a budget
total cost≤budget\text{total cost} \leq \text{budget}
The sign ≤ means is less than or equal to. A budget is the most money you can spend. You may spend all of it, or less, but never more. So the total cost must be less than or equal to the budget.

  1. Sort out the question

    You know three facts:

    • Jordan can spend at most $120. That is the budget.
    • The truck costs $18 for each hour.
    • A cleaning fee of $30 is paid one time only.

    x is the number of hours. You need an inequality. That is a math sentence that compares two amounts with a sign such as < or ≤. It must show all three facts.

  2. Work out a few totals

    Find the total cost for a few hours. Each hour adds one more $18:

    HoursHourly charges ($)Total cost ($)
    118×1=1818 \times 1 = 1830+18=4830 + 18 = 48
    218×2=3618 \times 2 = 3630+36=6630 + 36 = 66
    x18x18x30+18x30 + 18x

    The $30 shows up once in every total. The $18 shows up once for each hour. So x hours cost 30 + 18x dollars. Here 18x means 18 times x.

  3. Choose the right sign

    The total cost must fit inside the budget. Jordan may spend exactly $120, or less, but not more. That is less than or equal to, the sign ≤. So:

    30+18x≤12030 + 18x \leq 120

    Watch out: the sign < means less than, but not equal. 30+18x<12030 + 18x < 120 leaves out spending exactly $120. The sign ≥ is backward too. It says Jordan must spend at least $120. Also keep each number in its place. 30x+1830x + 18 charges $30 an hour.

  4. Stop before you solve

    You could solve the inequality. Subtract 30 from both sides to get 18x≤9018x \leq 90. Then divide both sides by 18 to get x≤5x \leq 5. So Jordan can rent the truck for at most 5 hours.

    Watch out: the question asks for an inequality that models the situation. To model means to show the facts of the story in math. The solved form, x ≤ 5, is true, but it hides the $18, the $30 and the $120. So x ≤ 5 is not accepted. Enter the inequality before you solve it.

  5. Enter it on the keypad

    On the test, build the inequality on the keypad. Press 3, 0, +, 1, 8 and the x key. Then press the ≤ key, in the fourth row of keys. Then press 1, 2, 0. You need no dollar signs and no times sign. On this site, type it, with <= for ≤: 30 + 18x <= 120.

    So you would enter 30 + 18x ≤ 120. The same inequality written the other way round, 120 ≥ 30 + 18x, is also right. The solved form, x ≤ 5, is not accepted, because it does not model the situation.

Check your answer

Test two numbers of hours. For 5 hours, 30+18×5=12030 + 18 \times 5 = 120. Is 120 ≤ 120? Yes, so 5 hours fit the budget. For 6 hours, 30+18×6=13830 + 18 \times 6 = 138. Is 138 ≤ 120? No, and $138 is over budget. The inequality gets both right.

More questions like this

The original DOE question
Question 26 as shown on the DOE’s digital test, page 99 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 99, question 26. This is the DOE’s version, before the corrections listed in the analysis.

Question 29 of 120

2026 Practice Test D Math, question 30

Question

A recipe uses 23\frac{2}{3} cup of sugar for every 1141\frac{1}{4} cups of flour. A cook wants to increase the recipe. How many cups of sugar are needed for each cup of flour?

  • A815\frac{8}{15}
  • B56\frac{5}{6}
  • C1781\frac{7}{8}
  • D3233\frac{2}{3}
Show the answer

Correct answer: A

To find the cups of sugar for each cup of flour, find the unit rate using this ratio:

23 cup of sugar114 cups of flour\frac{\frac{2}{3} \text{ cup of sugar}}{1\frac{1}{4} \text{ cups of flour}}

Then, convert the denominator to an improper fraction and divide to find the cups of sugar per cup of flour.

2354=23÷54=23×45=815\frac{\frac{2}{3}}{\frac{5}{4}} = \frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}

Question analysis

A recipe uses 2/3 cup of sugar for every 1 1/4 cups of flour, and this unit rate question asks how many cups of sugar go with each cup of flour. Students divide one fraction by a mixed number. Deciding which amount to divide by which, and converting the mixed number first, are where mistakes creep in.

NYS standards
  • 7.RP.1 unit rates with fractions
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Computing a unit rate from a ratio of fractions is exactly the grade 7 standard. This one divides by a mixed number, and the cook's plan to enlarge the recipe adds wording without adding any numbers to use.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 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 work is a single fraction division, but it involves a mixed number and an awkward result. Students unsure of the rule multiply the two amounts or divide them the other way round, and each gives a tidy listed fraction. A size check settles it: 1 1/4 cups of flour take only 2/3 cup of sugar, so one cup of flour needs less than 2/3 cup, which leaves a single choice.

  1. Error-prone arithmetic moderate Working it out

    The divisor is the mixed number 1 1/4, which must become 5/4 before dividing; then 2/3 × 4/5 gives 8/15, a fraction that does not simplify and looks unfinished. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The phrase “for each cup of flour” puts flour in the denominator, sugar ÷ flour, and the sentence about increasing the recipe adds nothing to the calculation. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Write the sugar over the flour, change 1 1/4 to an improper fraction, and divide by multiplying by its reciprocal, keeping sugar on top because the question asks for sugar per cup of flour.

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).
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
unit rate=cups of sugarcups of flour\text{unit rate} = \frac{\text{cups of sugar}}{\text{cups of flour}}
A unit rate tells how much of one thing goes with 1 of another. Here it is the cups of sugar for 1 cup of flour. The amount you want goes on top. The amount you want to make 1 goes on the bottom.
Mixed number to fraction
wnd=w×d+ndw\frac{n}{d} = \frac{w \times d + n}{d}
w is the whole number, and n/d is the fraction part. Multiply the whole number by the bottom, add the top, and keep the bottom. It works because each whole holds d pieces of size 1/d.
Dividing fractions
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Flip only the fraction you divide by.

  1. Sort out the question

    The recipe uses 23\frac{2}{3} cup of sugar for every 1141\frac{1}{4} cups of flour. You need the cups of sugar for each 1 cup of flour. The part about making more of the recipe gives no numbers, so you can set it aside.

  2. Set up the unit rate

    You want sugar for each cup of flour. So sugar goes on top and flour goes on the bottom:

    unit rate=23÷114\text{unit rate} = \frac{2}{3} \div 1\frac{1}{4}

    Watch out: dividing the other way, flour by sugar, gives 158=178\frac{15}{8} = 1\frac{7}{8}. That is cups of flour for each cup of sugar, choice C.

  3. Change the mixed number

    You cannot divide by a mixed number directly. Change it to a fraction first. Each whole has 4 fourths:

    114=1×4+14=541\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{5}{4}

  4. Divide the fractions

    Multiply by the reciprocal of 5/4, which is 4/5. Multiply the tops and multiply the bottoms:

    23÷54=23×45=815\frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}

    The fraction 8/15 cannot be simplified, since 8 and 15 share no factor but 1.

    Watch out: multiplying without flipping gives 23×54=1012=56\frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6}, choice B.

    So the answer is A, 815\frac{8}{15}.

Check your answer

More than 1 cup of flour takes only 2/3 cup of sugar. So 1 cup of flour needs a bit less than 2/3. Since 23=1015\frac{2}{3} = \frac{10}{15}, the answer 8/15 is a bit less. Also, 815×54=4060=23\frac{8}{15} \times \frac{5}{4} = \frac{40}{60} = \frac{2}{3}, which brings back the recipe.

More questions like this

The original DOE question
Question 30 as shown on the DOE’s digital test, page 103 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 103, question 30.

Question 30 of 120

2026 Practice Test D Math, question 31

Question

A box contains an unknown number of green marbles. An experimenter adds 100 red marbles to the box, mixes the marbles thoroughly, and then draws out 10 marbles at random. Assume that this sample is representative of the proportion in the box. If there are 2 red marbles and 8 green marbles in this sample, approximately how many green marbles are in the box?

  • A80
  • B300
  • C400
  • D500
Show the answer

Correct answer: C

In this context, the phrase “this sample is representative” means that the ratio of 2 red marbles to 8 green marbles is the same as the ratio of red to green marbles in the entire box. Use this ratio of 2 to 8 to set up a proportion and solve, where x represents the number of green marbles in the box.

28=100x\frac{2}{8} = \frac{100}{x}

2x=8002x = 800

x=400x = 400

Question analysis

An experimenter adds 100 red marbles to a box of green marbles, mixes them, and draws a sample of 10 with 2 red and 8 green. This sampling question asks students to estimate how many green marbles the box holds. It applies a ratio from a representative sample, worked from a count known for the box rather than for the sample.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a representative sample to reason about a population is grade 6 and 7 work. The twist is that the known count, the 100 red marbles added, belongs to the box rather than the sample, so the usual scaling runs in an unfamiliar direction.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 8, 60 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is the logic of estimating a population by tagging, and the twist is that the known count belongs to the marbles that were added, not to the ones being estimated. Scaling the sample up by 50, so that its 2 red become the box’s 100, turns its 8 green into the estimate in one step. Students who scale by way of the whole box often stop at the box’s total, and those who treat 100 as the whole box scale the wrong amount.

  1. Turning words into math moderate Understanding the question

    The 100 red marbles are counted in the box, while the 2 red and 8 green are counted in a sample of 10, so “representative of the proportion” must become 2 : 8 = 100 : x. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Three ratios can be read from the sample, red to green, red to all and green to all, and only the first two connect to the known 100 red marbles. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Use the sample's ratio of red to green marbles, then scale it up until the red part matches the 100 red marbles known to be in the box.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios (a proportion)
red in samplegreen in sample=red in boxgreen in box\frac{\text{red in sample}}{\text{green in sample}} = \frac{\text{red in box}}{\text{green in box}}
A ratio compares two amounts. A proportion says two ratios are equal. A sample that is representative has the same mix as the whole box. So the box is the sample scaled up.

  1. Sort out the question

    You know:

    • The box has an unknown number of green marbles.
    • 100 red marbles are added to the box.
    • A sample of 10 marbles has 2 red and 8 green.

    Representative means the sample has the same mix as the box. You need the number of green marbles in the box. Notice that the 100 red marbles are counted in the box, not in the sample.

  2. Set up the proportion

    In the sample, red to green is 2 to 8. In the box, red to green is 100 to x, where x is the number of green marbles. The two ratios are equal:

    28=100x\frac{2}{8} = \frac{100}{x}

  3. Scale up the sample

    How do you get from 2 red to 100 red? Multiply by 50, since 2×50=1002 \times 50 = 100. The green marbles grow by the same factor:

    8×50=4008 \times 50 = 400

    So the box has about 400 green marbles.

  4. Avoid the tempting wrong totals

    Watch out: there is another way that can trip you up. In the sample, 2 of 10 marbles are red. So the 100 red marbles are 2/10 of the box, and the box holds 500 marbles in all. But 100 of those are red. Stopping at 500 gives choice D.

    Taking away the red ones gives 500−100=400500 - 100 = 400 green marbles, the same answer. Treating 100 as the whole box gives 80 green, choice A. Reading 2 to 8 as red to all marbles gives 300, choice B.

    So the answer is C, 400.

Check your answer

Picture the finished box: 400 green and 100 red, so 500 marbles. The red part is 100500=15\frac{100}{500} = \frac{1}{5}. The sample had 210=15\frac{2}{10} = \frac{1}{5} red too. The mix matches, so 400 makes sense.

More questions like this

The original DOE question
Question 31 as shown on the DOE’s digital test, page 104 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 104, question 31.

Question 31 of 120

2026 Practice Test D Math, question 32

Question

There were 36 people at Emily’s party. If each person at the party drank 2 cups of juice, how many gallons of juice did they drink altogether?

Note: 1 gallon = 4 quarts; 1 quart = 4 cups.

  • A2142\frac{1}{4}
  • B4124\frac{1}{2}
  • C9
  • D72
Show the answer

Correct answer: B

Since 36 people each drank 2 cups of juice, the total amount of juice they drank was 36×2=7236 \times 2 = 72 cups.

Convert 72 cups to gallons:

72 cups×1 quart4 cups×1 gallon4 quarts=72 \text{ cups} \times \frac{1 \text{ quart}}{4 \text{ cups}} \times \frac{1 \text{ gallon}}{4 \text{ quarts}} =

7216 gallons=\frac{72}{16} \text{ gallons} =

92 gallons=\frac{9}{2} \text{ gallons} =

412 gallons4\frac{1}{2} \text{ gallons}

Question analysis

At Emily's party 36 people each drank 2 cups of juice, and this unit conversion question asks how many gallons they drank in all, given 4 quarts to a gallon and 4 cups to a quart. Students multiply to get the total in cups, then convert through quarts to gallons, and the bold word gallons sets the unit of the answer.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Converting units with given factors is grade 6 work. This adds a first step, the total number of cups, and a two-stage conversion through quarts, but the numbers are friendly and the path is routine for grade 7 students.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 32 words.
Corrected from the DOE handbook
  • A stray word at the end of the explanation was removed.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

With both conversion facts printed, the question is mainly about keeping the units straight. Students who skip the word each convert 36 cups instead of 72, and those who stop at cups find a total that is right in the wrong unit. Working out that a gallon is 4 × 4 = 16 cups before dividing turns two conversions into one and avoids adding the 4s.

  1. Many steps minor Working it out

    A multiplication, 36 × 2, comes before two divisions by 4, cups to quarts and quarts to gallons, and stopping after the first step leaves a believable 72. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The word “each” turns 36 people into 72 cups, and the bold word gallons sets the unit of the answer, a change the note supplies in two steps. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the total number of cups first, then divide by 4 to get quarts and by 4 again to get gallons, or work out how many cups make a gallon and divide once.

Traps

  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation ends with a line that reads only “lineExtend”, a formatting code left in the document by mistake. The site leaves it out.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total amount
total=number of people×amount each\text{total} = \text{number of people} \times \text{amount each}
When every person has the same amount, multiply the number of people by the amount each one has. The word each tells you to multiply.
Cups in a gallon
cups in 1 gallon=4×4=16\text{cups in 1 gallon} = 4 \times 4 = 16
A gallon holds 4 quarts, and each quart holds 4 cups. So a gallon holds 4 groups of 4 cups. You multiply, not add, because each of the 4 quarts has its own 4 cups.
Changing cups to gallons
gallons=cups÷16\text{gallons} = \text{cups} \div 16
It takes 16 cups to make 1 gallon. Dividing by 16 finds how many groups of 16 cups there are. A gallon is bigger than a cup, so the number of gallons is smaller.

  1. Sort out the question

    You know:

    • 36 people were at the party.
    • Each person drank 2 cups of juice.
    • 1 gallon = 4 quarts, and 1 quart = 4 cups.

    You need the total amount of juice in gallons. The bold word tells you the unit of the answer.

  2. Find the total in cups

    Each of the 36 people drank 2 cups:

    36×2=7236 \times 2 = 72

    They drank 72 cups in all.

    Watch out: skipping the word each uses only 36 cups. Then 36÷16=21436 \div 16 = 2\frac{1}{4} gallons, choice A.

  3. Find the cups in a gallon

    A gallon is 4 quarts, and each quart is 4 cups:

    4×4=164 \times 4 = 16

    So 1 gallon holds 16 cups.

    Watch out: adding the two 4s gives 8 cups in a gallon. Then 72÷8=972 \div 8 = 9, choice C.

  4. Change cups to gallons

    Divide the cups by 16. Simplify the fraction by dividing the top and bottom by 8:

    72÷16=7216=92=41272 \div 16 = \frac{72}{16} = \frac{9}{2} = 4\frac{1}{2}

    They drank 4124\frac{1}{2} gallons of juice.

    Watch out: 72 is the amount in cups. The question asks for gallons, so 72, choice D, is in the wrong unit.

    So the answer is B, 4124\frac{1}{2}.

Check your answer

Work backward. 4124\frac{1}{2} gallons is 412×4=184\frac{1}{2} \times 4 = 18 quarts. 18 quarts is 18×4=7218 \times 4 = 72 cups. Shared among 36 people, that is 72÷36=272 \div 36 = 2 cups each. It 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 D Math, page 105, question 32. This is the DOE’s version, before the corrections listed in the analysis.

Question 32 of 120

2026 Practice Test D Math, question 34

Question

Last week, Emily rode the bus to school on 3 of the 5 mornings and rode the bus home on 4 of the 5 afternoons. Based on last week’s events, what is the probability that Emily will ride the bus in the morning and the afternoon next Monday?

  • A15\frac{1}{5}
  • B725\frac{7}{25}
  • C1225\frac{12}{25}
  • D710\frac{7}{10}
Show the answer

Correct answer: C

Using the joint probability formula, find the product of the two independent events. The probability that Emily will ride the bus to school on any given morning is 35\frac{3}{5}. The probability that Emily will ride the bus home on any given afternoon is 45\frac{4}{5}. Therefore, the probability that Emily will ride the bus in the morning and the afternoon is 35×45=1225\frac{3}{5} \times \frac{4}{5} = \frac{12}{25}.

Question analysis

Last week Emily rode the bus to school on 3 of 5 mornings and home on 4 of 5 afternoons. This probability question asks how likely she is to ride the bus both ways next Monday, based on those frequencies. Students turn each count into a probability and combine the two, deciding how chances join when both events must happen.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 6.SP.8b probability models from observed frequencies
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Compound probability is grade 7 work, often done with a table of outcomes. Each chance here comes from one week's frequencies, and joining the morning and afternoon events in one step is about the usual grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 48 words.
Corrected from the DOE handbook
  • An inaccurate detail 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

Once the two chances are written, the question is how they combine. Pooling the week into 7 bus rides out of 10 trips, or adding fractions in some way, gives listed choices that look reasonable, since each lies between 0 and 1. Picturing the 25 pairings of a morning with an afternoon, 3 × 4 = 12 of which have both rides, shows why the chances multiply.

  1. Something you must already know moderate Working it out

    The rule that two independent events both happen with the product of their chances, 3/5 times 4/5, is not given, and “the morning and the afternoon” is the only signal for it. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The counts “3 of the 5 mornings” and “4 of the 5 afternoons” have to become two separate probabilities based on last week, not one pool of 10 trips. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Turn each count into a probability, then multiply them, since both rides must happen; picturing a table of 5 mornings by 5 afternoons shows why multiplying works.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • 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

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation says “The probability that Emily will ride the bus to school on any given afternoon is 4/5.” The question says she rode the bus home on 4 of the 5 afternoons; her rides to school are the morning ones. The site now says “ride the bus home on any given afternoon”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability from past results
P=times it happenedtimes it could have happenedP = \frac{\text{times it happened}}{\text{times it could have happened}}
P is the probability, the chance something happens. You use what happened before to predict the chance it happens again. Emily rode in 3 of 5 mornings, so the chance for one morning is 3/5.
Both of two independent events
P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)
A and B are two events. Independent means one does not change the chance of the other. To find the chance that both happen, multiply their chances. The result is smaller than either one, since both must happen.

  1. Sort out the question

    Last week Emily:

    • rode the bus to school on 3 of the 5 mornings;
    • rode the bus home on 4 of the 5 afternoons.

    You need the chance that she rides the bus in the morning and in the afternoon next Monday. The word and means both rides must happen.

  2. Write each chance

    Use last week to find each chance:

    • Bus to school on a morning: 35\frac{3}{5}
    • Bus home on an afternoon: 45\frac{4}{5}

    Keep them apart. The morning rides go to school, and the afternoon rides go home.

  3. Multiply the two chances

    Both rides must happen, so multiply:

    35×45=1225\frac{3}{5} \times \frac{4}{5} = \frac{12}{25}

    Why multiply? Pair each of the 5 mornings with each of the 5 afternoons. That makes 25 pairs. In 3×4=123 \times 4 = 12 of them, she rides both times. So the chance is 12 out of 25.

  4. Avoid the tempting wrong choices

    Watch out: pooling the week gives 7 rides out of 10 trips, or 7/10, choice D. That is the chance for one trip, not for two trips on one day.

    Adding the tops but multiplying the bottoms gives 7/25, choice B. Subtracting the chances gives 1/5, choice A.

    So the answer is C, 1225\frac{12}{25}.

Check your answer

The chance of both rides must be less than the chance of each ride alone. As decimals, 12/25 = 0.48. That is less than 3/5 = 0.6 and 4/5 = 0.8. Choice D, 7/10 = 0.7, is more than 3/5, so it cannot be right.

More questions like this

The original DOE question
Question 34 as shown on the DOE’s digital test, page 107 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 107, question 34. This is the DOE’s version, before the corrections listed in the analysis.

Question 33 of 120

2026 Practice Test D Math, question 35

Question

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

  • A4
  • B6
  • C12
  • D24
Show the answer

Correct answer: C

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

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

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

AB1, AB2, AC1, AC2, AD1, AD2

BC1, BC2, BD1, BD2

CD1, CD2

There are 12 possible groups with only 1 male.

Question analysis

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

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students list outcomes with organized lists and tree diagrams, usually ordered ones. Counting unordered groups of three under an "only one" condition, built from two kinds of people, goes beyond that typical work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 7, 41 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Combines ideas moderate Working it out

    Two counts are joined: the 6 ways to pick 2 of the 4 women, then the 2 ways to pick the man, and the total is their product. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Something you must already know minor Working it out

    Students must already know that a group is the same whichever order its members are named, so picking one woman and then another repeats the reverse pick. (Anderson & Krathwohl, 2001)

  3. Turning words into math minor Understanding the question

    The bold word only means exactly one man, so the other two places in each group of 3 must go to two of the 4 women, a step the stem leaves unsaid. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Six people work for the summer. Two are male, so the other 4 are female. Three will be hired in the fall.

    You need the number of groups of 3 with only 1 male. That means exactly 1 man and 2 women. In a group, order does not matter. Ann, Bea and Carl is the same group as Carl, Ann and Bea.

  2. Give each person a name

    Call the women A, B, C and D. Call the men 1 and 2. Every group you want has 2 letters and 1 number, like AB1.

  3. List the pairs of women

    List each pair once. Start with A, then pair B only with letters after it, and so on:

    • AB, AC, AD
    • BC, BD
    • CD

    That is 6 pairs. BA is not new, since it is the same pair as AB.

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

  4. Add a man to each pair

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

    6×2=126 \times 2 = 12

    Watch out: stopping at the 6 pairs of women leaves out the men. That gives 6, choice B.

    So the answer is C, 12.

Check your answer

Write out every group. With man 1: AB1, AC1, AD1, BC1, BD1, CD1. With man 2: AB2, AC2, AD2, BC2, BD2, CD2. Each list has 6 groups, and 6+6=126 + 6 = 12. No group repeats, and each has just one man.

More questions like this

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

Question 34 of 120

2026 Practice Test D Math, question 39

Question

The table shows the probabilities of a hockey team scoring different numbers of goals in a game.

HOCKEY GOAL PROBABILITIES

Number of Goals Probability
0 0.16
1 0.21
2 0.29
3 0.21
4 0.09
5 0.04

What is the probability that the team will score three or more goals in the next game?

  • A0.13
  • B0.21
  • C0.34
  • D0.50
Show the answer

Correct answer: C

To find the probability that any one of several mutually exclusive events occurs, use the addition rule, and add the probabilities of each event:

P(A or B)=P(A)+P(B)\mathrm{P}(\mathrm{A} \text{ or } \mathrm{B}) = \mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B})

Using the data in the table, the probability that the team will score 3 or more goals in the next game is represented by

P(3)+P(4)+P(5)=\mathrm{P}(3) + \mathrm{P}(4) + \mathrm{P}(5) =

0.21+0.09+0.04=0.34.0.21 + 0.09 + 0.04 = 0.34.

Question analysis

A table gives the probabilities of a hockey team scoring 0 to 5 goals in a game, and students find the chance that it scores three or more goals in the next one. This probability question tests adding the probabilities of separate outcomes. Reading exactly which rows the phrase covers is the main point of care.

NYS standards
  • 6.SP.8 probability models for simple events
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the probability of an event from a given probability model is grade 6 work in New York. Here the model is a table, and the task is choosing the right rows and adding decimals, shorter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 8, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Adding three decimals is easy, so the question turns on which rows the words cover. Including the 3-goal row, and not stopping at it, are the two decisions, and each misreading gives a listed choice. Checking the complement, 1 − (0.16 + 0.21 + 0.29), confirms both the sum and the choice of rows.

  1. Turning words into math minor Understanding the question

    The phrase “three or more goals” has to become the three rows for 3, 4 and 5 goals, starting at the 3-goal row itself. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table lists six probabilities, two of them both 0.21, so the rows for 3, 4 and 5 goals must be picked out by their goal counts, not by their values. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide which rows the words "three or more" cover, add their probabilities, and check by subtracting the total of the other rows from 1.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding probabilities
P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)
A and B are two events that cannot both happen at once, like scoring exactly 3 goals and exactly 4 goals. The chance that one or the other happens is the sum of their chances.
The chance it does not happen
P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)
The chances of all possible outcomes add up to 1. So the chance an event does not happen is 1 minus the chance it does. Use it to check an answer from the other side.

  1. Sort out the question

    The table gives the chance of each number of goals, from 0 to 5. You need the chance the team scores three or more goals in the next game.

  2. Pick the right rows

    Three or more means 3 goals, 4 goals or 5 goals. The row for 3 counts too. Read those three rows:

    • 3 goals: 0.21
    • 4 goals: 0.09
    • 5 goals: 0.04

    Watch out: leaving out the 3-goal row gives 0.09 + 0.04 = 0.13, choice A. Using only the 3-goal row gives 0.21, choice B.

  3. Add the chances

    A game can end with only one number of goals. So add the three chances:

    0.21+0.09+0.04=0.340.21 + 0.09 + 0.04 = 0.34

    Watch out: three of the six rows is half the rows, but that does not make the chance 0.50, choice D. The rows have different chances.

    So the answer is C, 0.34.

Check your answer

Find the chance of fewer than 3 goals, the other rows: 0.16+0.21+0.29=0.660.16 + 0.21 + 0.29 = 0.66. Then the chance of 3 or more is 1−0.66=0.341 - 0.66 = 0.34. Both ways give 0.34.

More questions like this

The original DOE question
Question 39 as shown on the DOE’s digital test, page 112 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 112, question 39.

Question 35 of 120

2026 Practice Test D Math, question 40

Question

A team is participating in a contest to collect canned goods for a food drive. The team has 53 points and must have a total of at least 100 points to win a prize. The team earns 5 points for each can of food a member collects. Which inequality can be used to find x, the number of cans of food the team needs to collect to win a prize?

  • A5+53x≤1005 + 53x \leq 100
  • B5+53x≥1005 + 53x \geq 100
  • C53+5x≤10053 + 5x \leq 100
  • D53+5x≥10053 + 5x \geq 100
Show the answer

Correct answer: D

Since each can is worth 5 points, the total number of points earned from all the cans, x, is 5x. The team started with 53 points. That is a constant value and doesn’t change. So the total number of points the team will earn is 53+5x53 + 5x. The team wants to earn at least (≥) 100 points, so they need 53+5x≥10053 + 5x \geq 100 total points.

Question analysis

A team has 53 points in a canned-food drive, earns 5 points for each can, and needs at least 100 points to win a prize. Students choose the inequality for x, the number of cans the team must collect. It tests writing an inequality from words, where the starting points and the points per can must each take the right place.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing an inequality for a word problem with a starting amount and a rate is grade 7 work. The four choices pair two expressions with two symbols, so students make one decision about the terms and one about the phrase "at least".
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 71 words.
Students, last 30 days
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 choices form a two-by-two grid: which number goes with x, and which way the symbol points. Each decision can be made separately, so a student who settles both gets the item, while one mistake lands on a listed choice. Testing two numbers of cans, 5, which leaves the team at 78 points, and 10, which lifts it to 103, rules out every wrong inequality.

  1. Turning words into math moderate Understanding the question

    The 53 points the team “has” are a fixed start, the “5 points for each can” go with x, and “at least 100” calls for ≥, though “least” can sound like less. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    The four choices are two expressions, 5 + 53x and 53 + 5x, each with ≤ and with ≥, so the right one and the strongest wrong one differ only in the symbol. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out the team's total points after 2 cans and then after x cans, translate "at least 100" into a symbol, and compare the result with each choice.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total with a start and a rate
total=start+rate×x\text{total} = \text{start} + \text{rate} \times x
The start is the amount you have now, here 53 points. The rate is what each item adds, here 5 points for each can. x is the number of items. The start is added once, and the rate is multiplied by x.
At least
total≥goal\text{total} \geq \text{goal}
The sign ≥ means is greater than or equal to. At least 100 means 100 or more. So the total must be greater than or equal to 100.

  1. Sort out the question

    You know:

    • The team already has 53 points.
    • It earns 5 points for each can.
    • It needs at least 100 points to win.

    x is the number of cans. You need the inequality that shows this.

  2. Write the total points

    Try 2 cans first. The team gets 5 points for each, so 53+5×2=6353 + 5 \times 2 = 63 points. For x cans, the total is:

    53+5x53 + 5x

    Watch out: the 53 points do not repeat for each can. 5+53x5 + 53x gives 53 points for each can, as in choices A and B.

  3. Choose the sign

    At least 100 means 100 or more, so use ≥:

    53+5x≥10053 + 5x \geq 100

    Watch out: “least” sounds like “less,” but at least means the total can be 100 or higher. Using ≤ gives choice C.

    So the answer is D, 53+5x≥10053 + 5x \geq 100.

Check your answer

Test two numbers of cans. With 10 cans, 53+5×10=10353 + 5 \times 10 = 103, and 103 ≥ 100 is true. The team wins. With 9 cans, 53+5×9=9853 + 5 \times 9 = 98, and 98 ≥ 100 is false. The team is short. Choice D gets both right.

More questions like this

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

Question 36 of 120

2026 Practice Test D Math, question 42

Question

A student reads 112\frac{1}{12} of a book in 13\frac{1}{3} hour. What fraction of the book can the student read in 1 hour?

  • A136\frac{1}{36}
  • B14\frac{1}{4}
  • C512\frac{5}{12}
  • D123\frac{12}{3}
Show the answer

Correct answer: B

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

112 book13 hour\frac{\frac{1}{12} \text{ book}}{\frac{1}{3} \text{ hour}}

Then use division to find the unit rate.

11213=112÷13=112×31=312=14\frac{\frac{1}{12}}{\frac{1}{3}} = \frac{1}{12} \div \frac{1}{3} = \frac{1}{12} \times \frac{3}{1} = \frac{3}{12} = \frac{1}{4}

Another way to solve the problem is to set up the ratio and then multiply both the numerator and denominator by 3, since 13\frac{1}{3} hour times 3 equals 1 hour.

3×112 book3×13 hour=14 book1 hour\frac{3 \times \frac{1}{12} \text{ book}}{3 \times \frac{1}{3} \text{ hour}} = \frac{\frac{1}{4} \text{ book}}{1 \text{ hour}}

Question analysis

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

NYS standards
  • 7.RP.1 unit rates with fractions
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Computing a unit rate from a ratio of fractions is the grade 7 standard itself. This is a short example with unit fractions, where scaling the time up to a whole hour is as quick as dividing.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

    Both amounts are unit fractions, and 1/12 ÷ 1/3 means multiplying by 3, which gives 3/12 that still has to be simplified to match a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The phrase “fraction of the book … in 1 hour” asks for book per hour, so the fraction of the book is divided by the time, not the other way round. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The student reads 112\frac{1}{12} of a book in 13\frac{1}{3} of an hour. You need the part of the book read in 1 whole hour. That is a unit rate.

  2. Scale up to one hour

    One hour holds three thirds of an hour. So in 1 hour the student reads 3 times as much:

    3×112=312=143 \times \frac{1}{12} = \frac{3}{12} = \frac{1}{4}

    To simplify 3/12, divide the top and bottom by 3. The student reads 1/4 of the book in 1 hour.

  3. Check with the unit rate

    Divide the part of the book by the time. Multiply by the reciprocal of 1/3, which is 3/1:

    112÷13=112×31=312=14\frac{1}{12} \div \frac{1}{3} = \frac{1}{12} \times \frac{3}{1} = \frac{3}{12} = \frac{1}{4}

    Both ways give 1/4.

  4. Avoid the wrong operations

    Watch out: multiplying gives 112×13=136\frac{1}{12} \times \frac{1}{3} = \frac{1}{36}, choice A. That is less than 1/12. But in a full hour, the student must read more than in a third of an hour. Dividing the time by the book gives 12/3, choice D. That is 4 whole books.

    So the answer is B, 14\frac{1}{4}.

Check your answer

If the student reads 1/4 of the book each hour, the whole book takes 4 hours. In 1/3 of an hour, the student reads 14×13=112\frac{1}{4} \times \frac{1}{3} = \frac{1}{12} of the book. That matches the question.

More questions like this

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

Question 37 of 120

2026 Practice Test D Math, question 45

Question

Marcus and Isla are saving money to buy concert tickets.

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 6

Define the variable:

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

Construct an equation:

Start with $8.

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

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

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

Solve the equation:

8+3w=268 + 3w = 26

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

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

Marcus and Isla are saving $6 per week.

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an equation of the form px + q = r from a word problem is the grade 7 standard. This one uses small whole numbers and two steps, a typical classroom example, with no choices to compare.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 51 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

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

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

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

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

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    You know:

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

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

  2. Write an equation

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

    8+3w=268 + 3w = 26

  3. Subtract the starting money

    Take 8 away from both sides:

    3w=26−8=183w = 26 - 8 = 18

    They saved $18 over the 3 weeks.

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

  4. Divide by the number of weeks

    Divide both sides by 3:

    w=18÷3=6w = 18 \div 3 = 6

    They save $6 each week.

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

  5. Enter the answer

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

Check your answer

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

More questions like this

The original DOE question
Question 45 as shown on the DOE’s digital test, page 118 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 118, question 45.

Question 38 of 120

2026 Practice Test D Math, question 46

Question

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

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

SHIPPING COSTS

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

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

Show the answer

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

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

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

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

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

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

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

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

    2=x−12 = x - 1

    x=3 poundsx = 3 \text{ pounds}

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

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

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

    8=x−18 = x - 1

    x=9 poundsx = 9 \text{ pounds}

To check:

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

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

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

3=x−13 = x - 1

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

Question analysis

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

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.9 dependent and independent variables
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve word problems with equations like p(x + q) = r. This table needs that rule four times, twice run backward from a cost to a weight, and the phrase about the first pound must be applied every time.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 4, 75 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

  2. Many steps minor Working it out

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

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

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Test the rule on 4 pounds

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

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

    That is $3.00, so the rule fits.

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

  3. Find the two costs

    For 2 pounds, 1 pound is over the first:

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

    For 6 pounds, 5 pounds are over the first:

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

    The costs are the tiles 2 and 4.

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

  4. Work back to the two weights

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

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

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

  5. Fill in the four boxes

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

Check your answer

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

More questions like this

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

Question 39 of 120

2026 Practice Test D Math, question 48

Question

The perimeter of square W is twice the perimeter of equilateral triangle X. What is the ratio of the length of one side of square W to the length of one side of equilateral triangle X?

  • A3:8
  • B2:3
  • C3:2
  • D2:1
Show the answer

Correct answer: C

First, set up the ratio of the perimeter of the square to the triangle as 2:1 because the question states that the perimeter of the square is twice the perimeter of the triangle.

Recognizing that the square has 4 equal sides and that an equilateral triangle has 3 sides, use division to find the ratio of the sides.

Perimeter: 2:1=212 : 1 = \frac{2}{1}

Sides:

2413=24÷13=\frac{\frac{2}{4}}{\frac{1}{3}} = \frac{2}{4} \div \frac{1}{3} =

24×31=64=\frac{2}{4} \times \frac{3}{1} = \frac{6}{4} =

32=3:2\frac{3}{2} = 3 : 2

Question analysis

The perimeter of square W is twice the perimeter of equilateral triangle X, and students find the ratio of a side of the square to a side of the triangle. This ratio question uses perimeters of regular shapes with no lengths given. Turning a comparison of perimeters into one of sides, and keeping the ratio in the order asked, are the tricky parts.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4 equations and inequalities from problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 ratio problems usually supply numbers to work with. This one gives only a relationship between two perimeters, so students must bring in the number of sides of each shape themselves and keep the ratio in the order asked.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 9, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns a statement about perimeters into a ratio of sides, and the natural guess, that the sides keep the perimeters’ 2 to 1, ignores that the square shares its perimeter among 4 sides and the triangle among 3. Choosing a triangle perimeter such as 12 makes the comparison concrete. The order of the ratio still has to be checked at the end, because its reverse is listed too.

  1. Turning words into math moderate Understanding the question

    “The perimeter of square W is twice the perimeter of equilateral triangle X” must become 4s = 2 × 3t, with the 2 on the triangle’s side; written backward, it gives the triangle the larger perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers moderate Understanding the question

    No side length is given, so students must either work with letters for the two sides or choose a convenient perimeter themselves, such as 12 for the triangle. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The ratios 3:2 and 2:3 use the same two numbers, so only keeping the square’s side first, as the question orders it, separates the answer from its reverse. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Pick an easy side length for the triangle, find its perimeter, double that for the square, divide by the square's number of sides, and write square to triangle.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a square
P=4sP = 4s
P is the perimeter, the distance around the shape. s is the length of one side. A square has 4 equal sides, so its perimeter is 4 times one side.
Perimeter of an equilateral triangle
P=3tP = 3t
An equilateral triangle has 3 equal sides. t is the length of one side. So its perimeter is 3 times one side.
Ratio
a:ba : b
A ratio compares two amounts in order. a : b means a compared with b. The order matters, so write the amounts in the order the question names them.

  1. Sort out the question

    Square W has a perimeter twice as long as the perimeter of triangle X. Equilateral means all 3 sides of X are equal.

    You need the ratio of one side of the square to one side of the triangle, in that order. No lengths are given, so you can pick a size to try.

  2. Pick a size for the triangle

    Choose an easy side length for the triangle, such as 2 units. All 3 sides are 2, so its perimeter is:

    P=3×2=6P = 3 \times 2 = 6

  3. Find the square’s side

    The square’s perimeter is twice the triangle’s:

    2×6=122 \times 6 = 12

    The square has 4 equal sides, so divide by 4:

    s=12÷4=3s = 12 \div 4 = 3

    Each side of the square is 3 units.

  4. Write the ratio in order

    The square comes first, then the triangle. Their sides are 3 and 2, so the ratio is 3 : 2.

    Watch out: 2 : 1, choice D, compares the perimeters, not the sides. The square splits its perimeter into 4 sides, the triangle into only 3. Choice B, 2 : 3, puts the triangle first. Choice A, 3 : 8, comes from giving the triangle the bigger perimeter.

    So the answer is C, 3:2.

Check your answer

Try a different size. Give the triangle sides of 4. Its perimeter is 12, so the square’s is 24. Each side of the square is 24÷4=624 \div 4 = 6. The ratio 6 : 4 simplifies to 3 : 2. The size does not change the ratio.

More questions like this

The original DOE question
Question 48 as shown on the DOE’s digital test, page 121 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 121, question 48.

Question 40 of 120

2026 Practice Test D Math, question 49

Question

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

  • A3
  • B5
  • C15
  • D45
Show the answer

Correct answer: C

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

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

45=3×3×545 = 3 \times 3 \times 5

75=3×5×575 = 3 \times 5 \times 5

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

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

Question analysis

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

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Greatest common factors of two numbers up to 100 are grade 6 content. The cloth setting asks students to recognize that this is what is needed, but the work is one short factor search, below typical grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

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

  2. Something you must already know minor Working it out

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Break each number into primes

    Write each length as a product of prime numbers:

    45=3×3×545 = 3 \times 3 \times 5

    75=3×5×575 = 3 \times 5 \times 5

    The two lists share one 3 and one 5.

  3. Multiply the shared primes

    Multiply the primes the two numbers share:

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

    The squares can be 15 inches on a side.

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

  4. Rule out the full width

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

    So the answer is C, 15.

Check your answer

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

More questions like this

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

Question 41 of 120

2024 Test A Math, question 2

Question

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

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

Show the answer

Correct answer: 150

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

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

r+m=180r + m = 180

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

r=5mr = 5m

5m+m=1805m + m = 180

6m=1806m = 180

m=1806=30m = \frac{180}{6} = 30

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

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Supplementary angles
M+R=180M + 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.

  1. 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.

  2. Write the facts as equations

    Supplementary angles add up to 180°.

    M+R=180M + R = 180

    The sentence “The measure of angle R is 5 times the measure of angle M” becomes this:

    R=5×MR = 5 \times M

    Watch out: supplementary means 180°, not 90°. Using 90° gives R = 75, which is wrong.

  3. Find angle M

    Put 5×M5 \times M in place of R. Then the equation has only one unknown, M.

    M+5×M=180M + 5 \times M = 180

    6×M=1806 \times M = 180

    M=180÷6=30M = 180 \div 6 = 30

    Think of 6 equal parts. M is 1 part and R is 5 parts. So angle M is 30°.

  4. Find angle R

    Angle R is 5 times angle M.

    R=5×30=150R = 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=18030 + 150 = 180. So they are supplementary. Also, 150÷30=5150 \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 42 of 120

2024 Test A Math, question 3

Question

A juice mixture contains 316\frac{3}{16} gallon of apple juice and 340\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:

316 apple340 cranberry=x apple1 cranberry\frac{\frac{3}{16}\text{ apple}}{\frac{3}{40}\text{ cranberry}} = \frac{x\text{ apple}}{1\text{ cranberry}}

316×403=x\frac{3}{16} \times \frac{40}{3} = x

4016=x\frac{40}{16} = x

2.5=x2.5 = 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.
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.

  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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
r=a÷br = 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
ab÷cd=ab×dc\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.

  1. Sort out the question

    The mix has 316\frac{3}{16} gallon of apple juice and 340\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.

  2. 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÷br = a \div b

    r=316÷340r = \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.

  3. Multiply by the reciprocal

    To divide by 340\frac{3}{40}, multiply by its reciprocal, 403\frac{40}{3}.

    ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

    316×403=3×4016×3=4016\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÷3=13 \div 3 = 1. Watch out: do not just multiply the two amounts. That gives 9640\frac{9}{640}, about 0.014, which is far too small.

  4. Change the fraction to a decimal

    Make 4016\frac{40}{16} simpler by dividing the top and bottom by 8. Then divide.

    4016=52=5÷2=2.5\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, 316\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÷40=0.0753 \div 40 = 0.075 and 0.075×2.5=0.18750.075 \times 2.5 = 0.1875, which is 3÷163 \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 43 of 120

2024 Test A Math, question 4

Question

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

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

Show the answer

Correct answer: 5.5

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

8÷2=48 \div 2 = 4

4×2184 \times 2\frac{1}{8}

4×2.125=8.54 \times 2.125 = 8.5

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

8.5+4=12.58.5 + 4 = 12.5

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

8÷2=48 \div 2 = 4

4×1344 \times 1\frac{3}{4}

4×1.75=74 \times 1.75 = 7

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

12.5−7=5.512.5 - 7 = 5.5

Question analysis

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

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve multi-step problems with mixed numbers and convert between fractions and decimals. This one strings four such steps together, including counting the 2-week periods, which is typical grade 7 work done carefully.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 3, 49 words.
Corrected from the DOE handbook
  • A misspelled name in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Many steps moderate Working it out

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

  2. Error-prone arithmetic moderate Working it out

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

  3. Turning words into math minor Understanding the question

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

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

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

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

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Count the 2-week periods

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

    8÷2=48 \div 2 = 4

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

  3. Change the mixed numbers to decimals

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

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

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

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

  4. Add up the growing

    The lawn grows 2.125 inches, 4 times.

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

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

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

    4+8.5=12.54 + 8.5 = 12.5

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

  5. Add up the cutting

    Mr. Chan cuts off 1.75 inches, 4 times.

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

    He cuts off 7 inches in all.

  6. Subtract the cutting

    12.5−7=5.512.5 - 7 = 5.5

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

Check your answer

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

More questions like this

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

Question 44 of 120

2024 Test A Math, question 6

Question

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

  • A50
  • B55
  • C495
  • D500
Show the answer

Correct answer: A

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

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

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

Then solve the proportion for x.

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

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

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

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

550=1.10x550 = 1.10x

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

500=x500 = x

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

550−500=50550 - 500 = 50

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

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students usually find a percent increase from a known original amount. This runs the task backward: the increased amount is given, and the original must be recovered before the difference can be found, a step many do not expect.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 1, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is one division and one subtraction, so the question is really about the base of a percent. The sentence gives the bigger number and calls it 10% more, and the reflex is to take 10% of the number on the page. That shortcut lands on a listed choice five goats from the answer. Students who find King's herd first, 550 ÷ 1.1, still have to remember that the question wants the difference, not the herd.

  1. Turning words into math major Understanding the question

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

  2. A very close wrong answer moderate Choosing the answer

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Decide what the 10% is of

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

  3. Find Mr. King’s herd

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

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

    550=K×1.10550 = K \times 1.10

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

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

  4. Find the difference

    550−500=50550 - 500 = 50

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

Check your answer

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

More questions like this

The original DOE question
Question 63 as printed on page 77 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 77, question 63. Open page 77 of the handbook (PDF)

Question 45 of 120

2024 Test A Math, question 11

Question

Jar 1 and Jar 2 each contain 12\frac{1}{2} cup of water. If 14\frac{1}{4} of the water in Jar 1 is poured into Jar 2, how much water is now in Jar 2?

  • A18\frac{1}{8} cup
  • B14\frac{1}{4} cup
  • C58\frac{5}{8} cup
  • D34\frac{3}{4} cup
Show the answer

Correct answer: C

Jar 1 starts with 12\frac{1}{2} cup of water. If 14\frac{1}{4} of its water is poured into Jar 2, then the amount of water added to Jar 2 is 12×14=18\frac{1}{2} \times \frac{1}{4} = \frac{1}{8} cup.

Jar 2 starts with 12\frac{1}{2} cup of water. If 18\frac{1}{8} cup of water is added to it, then the amount of water now in Jar 2 is 12+18=48+18=58\frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8} cup.

Question analysis

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

NYS standards
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Multiplying and adding positive fractions is review by grade 7, and the problem takes two short steps. It is easier than typical grade 7 work with negative numbers or longer rational-number problems, apart from careful reading.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    The phrase "1/4 of the water in Jar 1" means a quarter of Jar 1's half cup, an eighth of a cup, not a quarter cup; the word of calls for multiplying. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Adding the 1/8 cup to Jar 2's 1/2 cup needs the common denominator 8, so 1/2 becomes 4/8 before the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Each jar starts with 12\frac{1}{2} cup of water. You pour “14\frac{1}{4} of the water in Jar 1” into Jar 2. You need how much water is in Jar 2 after that, in cups.

  2. Find how much water is poured

    Jar 1 has 12\frac{1}{2} cup. You pour 14\frac{1}{4} of it. Of means multiply.

    ab of N=ab×N\frac{a}{b} \text{ of } N = \frac{a}{b} \times N

    14×12=1×14×2=18\frac{1}{4} \times \frac{1}{2} = \frac{1 \times 1}{4 \times 2} = \frac{1}{8}

    You pour 18\frac{1}{8} cup. Watch out: a quarter of the water is not a quarter cup. It is a quarter of a half cup. Adding 14\frac{1}{4} cup gives 34\frac{3}{4} cup, choice D.

  3. Make the bottoms match

    Now add the poured water to the 12\frac{1}{2} cup already in Jar 2. The bottoms are 2 and 8, so rewrite the half with 8 on the bottom. Multiply its top and bottom by 4.

    12=1×42×4=48\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}

  4. Add the water in Jar 2

    ad+bd=a+bd\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}

    48+18=58\frac{4}{8} + \frac{1}{8} = \frac{5}{8}

    Jar 2 now holds 58\frac{5}{8} cup. Watch out: 18\frac{1}{8} cup is only the water poured in. Stopping there gives choice A. So the answer is C, 58\frac{5}{8} cup.

Check your answer

Jar 2 started with a half cup and got a little more. So it should hold just over a half cup, and 58\frac{5}{8} is. Also, the water in both jars still adds up to 1 cup: Jar 1 has 38\frac{3}{8} cup left, and 38+58=1\frac{3}{8} + \frac{5}{8} = 1.

More questions like this

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

Question 46 of 120

2024 Test A Math, question 13

Question

When asked a certain question in a poll, 72% of the people polled answered yes. If 56 people did not answer yes to that question, what is the total number of people who were polled?

  • A78
  • B128
  • C144
  • D200
Show the answer

Correct answer: D

72% of the people polled answered yes, so 28% of the people did not answer yes.

Therefore, 56 people is 28% of the total, x.

56x=28100\frac{56}{x} = \frac{28}{100}

56x=725\frac{56}{x} = \frac{7}{25}

56(25)=7x56(25) = 7x

56(25)7=x\frac{56(25)}{7} = x

8(25)=x8(25) = x

200=x200 = x

Question analysis

A percent word problem about a poll: 72% of the people answered yes, and 56 people did not. Students find the total number of people polled. They must first match the 56 people with the right percent and then find the whole from a part, which is where the main slip happens.

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 Finding a whole from a part and a percent is grade 6 and 7 work. This adds one short step, finding the complementary percent first, which makes it a standard grade 7 multistep percent problem.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The only percent printed is 72 and the only count is 56, so the first instinct is to pair them, even though the bold not says they describe different groups. Once 56 is matched with the 28% who did not say yes, a single division finds the total. The listed wrong answers come from that mismatch, from adding the two numbers, and from stopping at the yes count.

  1. Turning words into math moderate Understanding the question

    The 56 people are the ones who did not answer yes, so they match 100% − 72% = 28%, not the 72% printed in the sentence before them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Finding the total means 56 ÷ 0.28, a division by a decimal; scaling 28% up to 100% at 2 people per percent avoids it. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract 72% from 100% to find the percent who did not answer yes, then find the total that makes 56 equal to that percent.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a whole
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
The whole is the total amount, here everyone polled. p is the percent. The part is the amount that p percent stands for. Percent means out of 100, so p% is p hundredths of the whole.

  1. Sort out the question

    In a poll, 72% of the people answered yes. 56 people did not answer yes. You need the total number of people polled. Notice the bold word not. The 56 people are the ones who did not say yes.

  2. Find the percent who did not say yes

    Everyone polled makes 100%. Take away the 72% who said yes.

    100−72=28100 - 72 = 28

    So 28% did not answer yes. That is the group of 56 people. Watch out: do not match the 56 people with 72%. That leads to about 78, choice A.

  3. Set up the percent equation

    Call the total number of people T. The 56 people are 28% of T.

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    56=28100×T56 = \frac{28}{100} \times T

  4. Solve for the total

    First find how many people make 1%. Divide 56 people by 28.

    56÷28=256 \div 28 = 2

    So each 1% is 2 people. All 100% is 100 times that.

    T=2×100=200T = 2 \times 100 = 200

    Watch out: do not multiply 2 by 72. That gives 144, the people who said yes, choice C. So the answer is D, 200.

Check your answer

Find 28% of 200: 0.28×200=560.28 \times 200 = 56. That matches the people who did not say yes. Also, 72% of 200 is 144 people who said yes, and 144+56=200144 + 56 = 200, everyone.

More questions like this

The original DOE question
Question 70 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 79, question 70. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 47 of 120

2024 Test A Math, question 15

Question

A program on a computer randomly generates a sequence of whole numbers from 1 to 9, inclusive. If the computer generates a sequence of 300 numbers, what is the best prediction of the number of odd numbers in the sequence?

  • A120
  • B133
  • C150
  • D167
Show the answer

Correct answer: D

There are five odd numbers in the range 1 through 9: 1, 3, 5, 7, and 9.

So the proportion of odd numbers in the range 1 through 9 is 59\frac{5}{9}.

The expected number of odd values in the sequence from the computer program is

300∙59=166.66.300 \bullet \frac{5}{9} = 166.\overline{66}.

So the best prediction for the number of odd numbers in the sequence is 167.

Question analysis

A probability question: a computer randomly generates 300 whole numbers from 1 to 9, and students predict how many will be odd. It asks students to find the probability of an odd number and use it to predict a count. Counting carefully which of the nine numbers are odd, and rounding the prediction sensibly, are the points to watch.

NYS standards
  • 6.SP.7 long-run relative frequency
  • 6.SP.8a uniform probability models
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a probability to predict how often an event happens is grade 6 and 7 work. This takes two short steps plus a rounding decision, about the demand of a typical grade 7 probability task.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 11, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is short, a probability times 300, so the question is decided by the probability. Odd and even feel like an even split, and half of 300 is a listed choice, but the numbers 1 to 9 hold five odd and four even. Listing the nine numbers before computing removes the trap, and the result, about 167, then needs only rounding.

  1. Turning words into math minor Understanding the question

    The phrase "whole numbers from 1 to 9, inclusive" means nine equally likely values, five of them odd, so the chance of odd is 5/9, not the one half that odd and even suggest. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    The stem never says that an expected count is the probability times the number of trials, here 5/9 of the 300 numbers generated. (Anderson & Krathwohl, 2001)

  3. Error-prone arithmetic minor Working it out

    The product 300 × 5/9 is 166.67, not a whole number, so the prediction has to be rounded to the nearest whole number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the odd numbers from 1 to 9, write the probability as that count over 9, and multiply by 300, rounding to the nearest whole number.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=wanted resultspossible resultsP = \frac{\text{wanted results}}{\text{possible results}}
P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of the results that you want.
Expected count
E=P×nE = P \times n
E is the expected count, your best guess of how many times something will happen. P is its probability, and n is the number of tries. If something happens 5 times out of 9, it should happen about 5 ninths of the time in many tries.

  1. Sort out the question

    The computer picks whole numbers from 1 to 9. Inclusive means 1 and 9 count too. Randomly means each number has the same chance. It picks 300 numbers. You need the best guess of how many of them are odd.

  2. Count the odd numbers

    An odd number does not split evenly by 2. From 1 to 9, the odd numbers are 1, 3, 5, 7 and 9. That is 5 of the 9 numbers. The even ones, 2, 4, 6 and 8, are only 4. Watch out: odd and even are not equally likely here, since there is one more odd number. Taking half of 300 gives 150, choice C.

  3. Find the chance of odd

    P=wanted resultspossible resultsP = \frac{\text{wanted results}}{\text{possible results}}

    P=59P = \frac{5}{9}

    Each number the computer picks has a 5 in 9 chance of being odd. Watch out: do not use the 4 even numbers. That gives about 133, choice B.

  4. Multiply by the 300 numbers

    E=P×nE = P \times n

    E=59×300=15009≈166.7E = \frac{5}{9} \times 300 = \frac{1500}{9} \approx 166.7

    The sign ≈ means about. You can’t have part of a number, so round to the nearest whole number. 166.7 is closer to 167 than to 166. So the answer is D, 167.

Check your answer

The odd numbers are a bit more than half of the nine. So the answer should be a bit more than half of 300, which is 150. Only 167 is a bit more than 150. Also, 167 odd plus about 133 even makes 300.

More questions like this

The original DOE question
Question 72 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 79, question 72. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 48 of 120

2024 Test A Math, question 16

Question

A truck rental company charges a one-time fee of $40 plus $1 per mile driven. Dalia rented a truck and used a coupon for 20% off the total rental cost. After the coupon was applied, she spent a total of $60. How many miles did she drive?

  • A8
  • B20
  • C32
  • D35
Show the answer

Correct answer: D

Let m equal the miles driven during Dalia’s rental. The total cost, c, can be expressed as an equation.

c=40+mc = 40 + m

Dalia spent a total of $60, which included a 20% off coupon. Find the cost before the coupon by dividing the final cost ($60) by the percentage paid (80%):

6080%=600.80=75\frac{60}{80\%} = \frac{60}{0.80} = 75

Use $75 as the total cost to find m.

40+m=7540 + m = 75

m=35m = 35

Dalia drove 35 miles.

Question analysis

A multistep word problem about a truck rental that costs a $40 fee plus $1 per mile, with a 20% off coupon applied to the total. Given that Dalia paid $60 after the coupon, students find how many miles she drove. Undoing the discount correctly before removing the fixed fee is the hard part.

NYS standards
  • 7.EE.4a two-step equations
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve two-step equations and work with discounts, usually separately. This combines them and makes students undo a 20% discount starting from the price paid, a reversal many grade 7 students have not practised.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 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

Two ordinary skills, undoing a discount and solving a two-step equation, are stacked, and the first runs in reverse. The natural way to undo 20% off is to add 20% back, which gives $72 instead of $75 and a choice three miles from the answer. Students who check by applying the coupon to the full cost of 40 + 35 = 75 dollars see at once that 80% of it is 60.

  1. Turning words into math moderate Understanding the question

    The sentence "After the coupon was applied, she spent a total of $60" makes 60 the price after 20% off, so it is 80% of the full cost, 40 + m. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Combines ideas moderate Working it out

    A percent discount has to be undone before the fee-plus-mileage equation can be solved: 60 ÷ 0.8 = 75, then 75 − 40 = 35 miles. (Embretson, 1998; Daniel & Embretson, 2010)

  3. A very close wrong answer moderate Choosing the answer

    The choices 32 and 35 are three miles apart, too close to tell apart by estimate; only dividing 60 by 0.8, rather than adding 20% of 60, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Since $60 is 80% of the full cost, divide 60 by 0.80 to find the cost before the coupon, then subtract the $40 fee to get the miles.

Traps

  • Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rental cost
c=40+1×mc = 40 + 1 \times m
c is the full cost in dollars, before any coupon. 40 is the one-time fee, paid once no matter how far you drive. m is the number of miles, and each mile costs 1 dollar.
Price after a discount
price paid=full cost×(1−p100)\text{price paid} = \text{full cost} \times \left(1 - \frac{p}{100}\right)
The full cost is the price before the discount. p is the percent taken off. You pay what is left: 100% minus p%. For 20% off you pay 80%, so you multiply by 0.80.

  1. Sort out the question

    The truck costs a $40 fee, plus $1 for each mile. Dalia’s coupon took 20% off the total. After the coupon, she paid $60. You need how many miles she drove. The $60 is the price after the discount, not the full cost.

  2. Write what she paid

    20% off means she paid what was left of the full cost.

    100%−20%=80%100\% - 20\% = 80\%

    As a decimal, 80% is 0.80.

    price paid=full cost×(1−p100)\text{price paid} = \text{full cost} \times \left(1 - \frac{p}{100}\right)

    60=full cost×0.8060 = \text{full cost} \times 0.80

  3. Undo the discount

    To undo times 0.80, divide by 0.80. Move the decimal point one place in both numbers to make it easier.

    full cost=60÷0.80=600÷8=75\text{full cost} = 60 \div 0.80 = 600 \div 8 = 75

    The full cost was $75. Watch out: the 20% was taken off the full cost, not off $60. Adding 20% of $60 gives $72 and leads to 32 miles, choice C. Taking 20% off $60 again leads to 8 miles, choice A.

  4. Find the miles

    c=40+1×mc = 40 + 1 \times m

    75=40+m75 = 40 + m

    m=75−40=35m = 75 - 40 = 35

    She drove 35 miles. Watch out: do not skip the coupon. Using $60 as the full cost gives 20 miles, choice B. So the answer is D, 35.

Check your answer

Go forward. 35 miles cost $35. Add the $40 fee to get $75. The coupon takes off 20% of $75, which is 0.20×75=150.20 \times 75 = 15 dollars. Then 75−15=6075 - 15 = 60. That is exactly what she paid.

More questions like this

The original DOE question
Question 73 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 79, question 73. Open page 79 of the handbook (PDF)

Question 49 of 120

2024 Test A Math, question 17

Question

The probability of drawing a red candy at random from a bag of 25 candies is 25\frac{2}{5}. After 5 red candies are removed from the bag, what is the probability of randomly drawing a red candy from the bag?

  • A0
  • B110\frac{1}{10}
  • C15\frac{1}{5}
  • D14\frac{1}{4}
Show the answer

Correct answer: D

Start with a proportion to determine the total number of red candies in the bag:

25=r25\frac{2}{5} = \frac{r}{25}

25(25)=r\frac{2}{5}(25) = r

10=r10 = r

There are currently 10 red candies out of 25 candies in the bag. After 5 red candies are removed, there would be 5 red candies left in the bag and 20 total candies.

The probability of randomly drawing a red candy would then be 520=14\frac{5}{20} = \frac{1}{4}.

Question analysis

A probability question about a bag of 25 candies in which the chance of drawing red is 2/5; then 5 red candies are taken out. Students find the new probability of drawing a red candy. They must work out how many red candies there were, then rebuild the fraction from the counts left in the bag.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Simple probability as a fraction of outcomes is grade 6 and 7 content. This adds a step back from the probability to a count and an update after removal, which is typical grade 7 multi-step work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 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

The question gives a probability, not a count, so the first job is to rebuild the bag: 10 red candies among 25. Removing candies changes two numbers at once, and the strongest wrong choice comes from updating the red count while leaving the total at 25. A student who writes both new counts before forming the fraction avoids it.

  1. Turning words into math moderate Understanding the question

    The probability 2/5 has to be turned into a count, 10 red of 25, and "after 5 red candies are removed" lowers both the red count and the bag's total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three steps follow in order: 2/5 of 25 = 10, then 5 red among 20 candies left, then 5/20 reduced to lowest terms. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use 2/5 of 25 to find the number of red candies, subtract 5 from both the red count and the total, and write the new fraction in lowest terms.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=red candiesall candiesP = \frac{\text{red candies}}{\text{all candies}}
P is the probability, the chance of drawing a red candy. Red candies is how many red ones are in the bag. All candies is how many candies are in the bag in all. Each candy is equally likely, so the chance is the fraction of the bag that is red.

  1. Sort out the question

    The bag has 25 candies. The chance of drawing red is 25\frac{2}{5}. Then 5 red candies are taken out. You need the new chance of drawing red. Taking out candies changes both the red count and the total.

  2. Find the number of red candies

    A chance of 25\frac{2}{5} means 2 out of every 5 candies are red. Call the red count r.

    P=red candiesall candiesP = \frac{\text{red candies}}{\text{all candies}}

    25=r25\frac{2}{5} = \frac{r}{25}

    r=25×25=2×5=10r = \frac{2}{5} \times 25 = 2 \times 5 = 10

    There are 10 red candies. Watch out: do not stop at 25÷5=525 \div 5 = 5. You must also multiply by 2. Thinking there are 5 red candies leads to 0, choice A.

  3. Take out the 5 red candies

    red: 10−5=5\text{red: } 10 - 5 = 5

    all: 25−5=20\text{all: } 25 - 5 = 20

    Now 5 of the 20 candies are red. Watch out: the total goes down too. Keeping 25 gives 525=15\frac{5}{25} = \frac{1}{5}, choice C.

  4. Find the new probability

    P=520=14P = \frac{5}{20} = \frac{1}{4}

    To simplify, divide the top and bottom by 5. So the answer is D, 14\frac{1}{4}.

Check your answer

At first, 25\frac{2}{5} is 40% of the bag. Taking out only red candies should lower the chance. 14\frac{1}{4} is 25%, which is lower but not zero. That makes sense, because 5 red candies are still in the bag.

More questions like this

The original DOE question
Question 74 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 74. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 50 of 120

2024 Test A Math, question 19

Question

8.9, 8.2, 8.5, 9.0, 8.4, 8.6, 8.8

At a skating championship, there are seven judges who each award a score for each skater’s performance. The highest and lowest scores given to each skater are discarded, and the mean of the remaining scores is then calculated and reported as the skater’s final score. What is the final score for the skater who received the scores shown above from the judges?

  • A8.60
  • B8.62
  • C8.64
  • D8.70
Show the answer

Correct answer: C

The highest and lowest scores are discarded: 8.2 and 9.0.

To find the mean, add the remaining scores and then divide that sum by the total number of remaining scores.

8.9+8.5+8.4+8.6+8.8=43.28.9 + 8.5 + 8.4 + 8.6 + 8.8 = 43.2

43.2÷5=8.6443.2 \div 5 = 8.64

Question analysis

A statistics question about a skating championship with seven judges: the highest and lowest of a skater's seven scores are dropped, and the rest are averaged. Students apply that rule to the listed scores, decimals close together. Following the dropping rule exactly and adding the decimals carefully both matter, since the choices are very close.

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
Below grade 7 Finding a mean is grade 6 content, and this adds only a rule about which scores to drop. Grade 7 statistics usually compares whole data sets, so this is easier, though the close choices demand exact decimal arithmetic.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 62 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is short, but the choices are spaced so closely that a small slip or a skipped rule lands on a neighbor. Averaging all seven scores, or picking the middle score, gives values within a few hundredths of the answer. The question rewards following the stated rule exactly and adding the five remaining scores with care.

  1. A very close wrong answer moderate Choosing the answer

    The choices 8.60, 8.62 and 8.64 lie within 0.04 of one another, so no estimate separates them; only the exact sum of the five kept scores, 43.2, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    Five one-decimal scores must be added and divided by 5; subtracting 8 from each first leaves 0.9, 0.5, 0.4, 0.6 and 0.8, which are easier to add. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Turning words into math minor Understanding the question

    The scoring rule takes two sentences to state: discard the single highest and the single lowest score, then average the five that remain. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Cross out the single highest and the single lowest score, add the five that remain, and divide that sum by 5.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean, or average
mean=sum of the scoresnumber of scores\text{mean} = \frac{\text{sum of the scores}}{\text{number of scores}}
The mean is the average. The sum is what you get when you add all the scores. Divide it by how many scores you added. It shares the total out evenly, as if every score were the same.

  1. Sort out the rule

    Seven judges gave the seven scores shown. The rule has two parts. First, throw out the highest score and the lowest score. Then find the mean of the scores that are left. The mean is the average. That is the final score you need.

  2. Cross out the highest and lowest

    The highest score is 9.0. The lowest is 8.2. Cross both out. Five scores are left: 8.9, 8.5, 8.4, 8.6 and 8.8. Watch out: drop both. Dropping only 8.2 gives 8.70, choice D. Keeping all seven gives about 8.63, which is closest to 8.62, choice B.

  3. Add the five scores

    8.9+8.5+8.4+8.6+8.8=43.28.9 + 8.5 + 8.4 + 8.6 + 8.8 = 43.2

    A shortcut: each score is 8 plus a bit. The bits are 0.9, 0.5, 0.4, 0.6 and 0.8, and they add to 3.2. Five 8s make 40, and 40+3.2=43.240 + 3.2 = 43.2.

  4. Divide by the number of scores

    mean=sum of the scoresnumber of scores\text{mean} = \frac{\text{sum of the scores}}{\text{number of scores}}

    mean=43.25=8.64\text{mean} = \frac{43.2}{5} = 8.64

    Watch out: do not just take the middle score, 8.6. That is the median, choice A. The question asks for the mean. So the answer is C, 8.64.

Check your answer

Multiply back: 5×8.64=43.25 \times 8.64 = 43.2, the sum. Also, the mean should sit between the lowest kept score, 8.4, and the highest kept score, 8.9. 8.64 does.

More questions like this

The original DOE question
Question 76 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 76. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 51 of 120

2024 Test A Math, question 20

Question

A piece of wood that is 4124\frac{1}{2} feet long is cut into 2 pieces of different lengths. The shorter piece has a length of x feet. Which inequality expresses all possible values of x?

  • A0<x<2140 < x < 2\frac{1}{4}
  • B0≤x≤2140 \leq x \leq 2\frac{1}{4}
  • C0<x<4120 < x < 4\frac{1}{2}
  • D214<x<4122\frac{1}{4} < x < 4\frac{1}{2}
Show the answer

Correct answer: A

The original piece of wood is 4124\frac{1}{2} feet long.

The shorter piece is x feet long, and it must be less than half the length of the original piece of wood.

Since half of 4124\frac{1}{2} is 2142\frac{1}{4}, it follows that x<214x < 2\frac{1}{4}.

The shorter piece of wood cannot be 0 feet long, so it follows that 0<x0 < x.

This gives the compound inequality 0<x<2140 < x < 2\frac{1}{4}.

Question analysis

An inequality question: a piece of wood 4½ feet long is cut into two pieces of different lengths, and the shorter piece is x feet long. Students choose the inequality that gives every possible value of x. They must turn each condition in the words into a boundary and decide whether each end of the range is included.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.8 writing simple inequalities
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph simple inequalities. Writing a compound inequality from a description, where the words decide whether each end is strict, is less familiar and calls for careful reading of every condition.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 5, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Halving 4 1/2 feet is the only computation; the rest is deciding whether each end of the range can actually happen. A shorter piece of 2 1/4 feet would make the two pieces equal, and a piece of 0 feet is no piece at all, so both ends are excluded. Students who find the right bounds but never test the ends are drawn to the version with the inclusive symbols.

  1. Turning words into math moderate Understanding the question

    Two phrases set the two ends: "cut into 2 pieces" means x is more than 0, and "of different lengths" with "the shorter piece" means x is less than half of 4 1/2. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    The inequalities 0 < x < 2 1/4 and 0 ≤ x ≤ 2 1/4 share both bounds and differ only in their symbols; only checking whether each end can happen separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    The question asks for every possible length of the shorter piece, a range rather than one length, so students must picture the cut moving along the board. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Picture cutting the board in different places, find the largest and smallest values the shorter piece could take, and check whether each boundary value is actually possible.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Two pieces make the whole board
x+L=412x + L = 4\frac{1}{2}
x is the length of the shorter piece, and L is the length of the longer piece, both in feet. Together they make the whole board. Since x is smaller than L, x must be less than half of the board.
Inequality signs
a<b and a≤ba < b \text{ and } a \leq b
a < b says a is less than b. a ≤ b says a is less than or equal to b, so a can also equal b. A chain like 0 < x < 5 means x is more than 0 and less than 5.

  1. Sort out the question

    A board is 4124\frac{1}{2} feet long. It is cut into 2 pieces of different lengths. The shorter piece is x feet long. You need an inequality for every possible value of x. An inequality uses signs like <, less than, to show a range of numbers.

  2. Find the top end for x

    If you cut exactly in the middle, each piece is half the board.

    412÷2=92×12=94=2144\frac{1}{2} \div 2 = \frac{9}{2} \times \frac{1}{2} = \frac{9}{4} = 2\frac{1}{4}

    But the pieces have different lengths, so the cut is not in the middle. The shorter piece must be less than half.

    x<214x < 2\frac{1}{4}

  3. Find the bottom end for x

    The shorter piece can be very small, like 0.1 foot. But it can’t be 0 feet. Then nothing would be cut off, and there would not be 2 pieces.

    0<x0 < x

  4. Put the two ends together

    0<x<2140 < x < 2\frac{1}{4}

    This says x is more than 0 and less than 2142\frac{1}{4}. Watch out: choice B uses ≤, which allows x = 0 or x=214x = 2\frac{1}{4}. Neither can happen. Choice C lets x be more than half, so it would not be the shorter piece. Choice D describes the longer piece. So the answer is A, 0<x<2140 < x < 2\frac{1}{4}.

Check your answer

Test values. Cut off 1 foot: the pieces are 1 foot and 3123\frac{1}{2} feet. The shorter one is 1, and 1 is between 0 and 2142\frac{1}{4}. If x were 3, the other piece would be 1121\frac{1}{2} feet, so x would not be the shorter one. Choice A leaves 3 out, as it should.

More questions like this

The original DOE question
Question 77 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 81, question 77. Open page 81 of the handbook (PDF)

Question 52 of 120

2024 Test A Math, question 23

Question

Let x be an odd number. In terms of x, what is the sum of the two even numbers closest to x?

  • Axx
  • B2x2x
  • C2x−22x - 2
  • D2x−42x - 4
Show the answer

Correct answer: B

If x is odd, then the two even numbers closest to x are (x−1)(x - 1) and (x+1)(x + 1). Find the sum of these two expressions:

(x−1)+(x+1)=2x(x - 1) + (x + 1) = 2x

Question analysis

An algebra question about odd and even numbers: x is odd, and students write the sum of the two even numbers closest to x in terms of x. It tests representing neighboring integers with expressions and adding them. Deciding exactly which even numbers are closest to an odd number is where errors start.

NYS standards
  • 6.EE.2a writing algebraic expressions
  • 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 write and combine linear expressions; here the expressions describe neighboring integers, a familiar idea from consecutive-integer problems. Two short steps and a quick number check put it at typical grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
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 algebra is a single addition, so the work is in naming the right two numbers. An odd x sits between two even numbers, one step away on each side, and the minus one and plus one cancel. Wrong choices come from picking both neighbors below x, from treating x as even, or from giving the middle of the pair; a test value catches each of them.

  1. Variables instead of numbers moderate Understanding the question

    Here x stands for any odd number, so the two even neighbors must be written as x − 1 and x + 1; trying x = 7, with neighbors 6 and 8, makes it concrete. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The phrase "the two even numbers closest to x" means one just below and one just above x, not two in a row on the same side. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Substitute a real odd number such as 7 for x, find its two closest even numbers and their sum, then see which expression gives that sum.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Neighbors of a number
x−1 and x+1x - 1 \text{ and } x + 1
x is any whole number. The number just below it is x − 1, and the number just above it is x + 1. Whole numbers take turns being odd and even. So if x is odd, both of its neighbors are even.
Combining like terms
x+x=2xx + x = 2x
Like terms have the same letter, such as x and 2x. To add them, add the numbers in front. x by itself means 1x, so x + x is 2x, which is two x’s.

  1. Sort out the question

    x is an odd number. You need the sum of the two even numbers closest to x. Sum means the result of adding. In terms of x means write the answer using the letter x.

  2. Try a real number first

    Pick an odd number, like 7. The even numbers closest to 7 are 6 and 8, one on each side.

    6+8=146 + 8 = 14

    Notice that 14 is 2×72 \times 7.

  3. Write the neighbors with x

    Odd and even numbers take turns. So both numbers right next to an odd x are even.

    x−1 and x+1x - 1 \text{ and } x + 1

    For x = 7, these are 6 and 8. Watch out: take one even number below x and one above. Using two even numbers below x, x−1x - 1 and x−3x - 3, gives 2x−42x - 4, choice D.

  4. Add the two neighbors

    (x−1)+(x+1)=x+x−1+1=2x(x - 1) + (x + 1) = x + x - 1 + 1 = 2x

    The −1 and the +1 cancel, and x+x=2xx + x = 2x. Watch out: x is the number in the middle, not the sum, so choice A is wrong. So the answer is B, 2x2x.

Check your answer

Try another odd number, 11. Its even neighbors are 10 and 12, and 10+12=2210 + 12 = 22. The answer 2x gives 2×11=222 \times 11 = 22. They match.

More questions like this

The original DOE question
Question 80 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 82, question 80. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 53 of 120

2024 Test A Math, question 27

Question

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

  • A60
  • B80
  • C120
  • D160
Show the answer

Correct answer: D

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

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

t+t+100+t−50=230t + t + 100 + t - 50 = 230

t+t+t+100−50=230t + t + t + 100 - 50 = 230

3t+50=2303t + 50 = 230

3t+(50−50)=230−503t + (50 - 50) = 230 - 50

3t=1803t = 180

(3÷3)t=180÷3(3 \div 3)t = 180 \div 3

t=60t = 60

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

t+100=60+100=160t + 100 = 60 + 100 = 160

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

Question analysis

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

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

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    Both comparisons are made with Tuesday, 100 more loaves on Wednesday and 50 fewer on Thursday, so the three days become t, t + 100 and t − 50. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    The equation's unknown is Tuesday's count, while the question asks for Wednesday's, so one more step remains after the equation is solved. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The baker made bread on Tuesday, Wednesday and Thursday. Wednesday had 100 more loaves than Tuesday. Thursday had 50 fewer loaves than Tuesday. All three days add up to 230 loaves. You need the number of loaves made on Wednesday.

  2. Write each day using Tuesday

    Both other days are compared with Tuesday. So call Tuesday’s loaves t.

    DayLoaves
    Tuesdaytt
    Wednesdayt+100t + 100
    Thursdayt−50t - 50

    100 more means add 100. 50 fewer means subtract 50.

  3. Write the equation and combine

    t+(t+100)+(t−50)=230t + (t + 100) + (t - 50) = 230

    Group the t’s together and the plain numbers together. Three t’s make 3t, and 100−50=50100 - 50 = 50.

    3t+50=2303t + 50 = 230

    Watch out: do not just subtract 100 and 50 from 230. That gives 80, choice B, and skips writing each day with t.

  4. Solve for Tuesday

    Subtract 50 from both sides. Then divide both sides by 3.

    if a=b, then a−c=b−c\text{if } a = b \text{, then } a - c = b - c

    3t+50−50=230−503t + 50 - 50 = 230 - 50

    3t=1803t = 180

    t=180÷3=60t = 180 \div 3 = 60

    Tuesday had 60 loaves. Watch out: 60 is Tuesday, choice A. The question asks about Wednesday.

  5. Find Wednesday’s loaves

    Wednesday had 100 more loaves than Tuesday, so add 100 to t.

    t+100=60+100=160t + 100 = 60 + 100 = 160

    Wednesday had 160 loaves. So the answer is D, 160.

Check your answer

Find all three days. Tuesday is 60, Wednesday is 160, and Thursday is 60−50=1060 - 50 = 10. Add them: 60+160+10=23060 + 160 + 10 = 230. That is the total in the question, so 160 is right.

More questions like this

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

Question 54 of 120

2024 Test A Math, question 30

Question

The reciprocal of 14\frac{1}{4} is added to the reciprocal of 3. What is the reciprocal of this sum?

  • A313\frac{3}{13}
  • B34\frac{3}{4}
  • C45\frac{4}{5}
  • D134\frac{13}{4}
Show the answer

Correct answer: A

The reciprocal of 14\frac{1}{4} is added to the reciprocal of 3.

41+13=123+13=133\frac{4}{1} + \frac{1}{3} = \frac{12}{3} + \frac{1}{3} = \frac{13}{3}

The reciprocal of this sum is 313\frac{3}{13}.

Question analysis

This fractions question asks for the reciprocal of a sum: the reciprocal of one fourth is added to the reciprocal of 3, and students must flip the total. It tests knowing what a reciprocal is and adding a whole number to a fraction. Keeping the three flips in the right order, without skipping one, is what makes it tricky.

NYS standards
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students use reciprocals when dividing fractions, usually one at a time. This item chains three reciprocals around one addition, with positive numbers only, so the demand is following the wording in order rather than new content.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is easy, so the question is about order: flip each number, add, then flip the total. The common failures are flipping at the wrong moment, term by term instead of the whole sum, and adding a whole number to a fraction straight across. Writing the full expression 1 ÷ (4 + 1/3) before doing any arithmetic keeps the three reciprocals in their places.

  1. Turning words into math moderate Understanding the question

    The two sentences hold three reciprocals and a sum, which must become 1 ÷ (4 + 1/3) before any arithmetic; "this sum" means flipping the total, not each part. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Adding 4 and 1/3 means writing 4 as 12/3 to get 13/3, a whole number plus a fraction, where adding tops and bottoms gives 5/4 instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work in the order of the words: flip each number to get its reciprocal, add the two using a common denominator, and then flip that sum.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reciprocal
reciprocal of ab=ba\text{reciprocal of } \frac{a}{b} = \frac{b}{a}
a is the top of a fraction and b is the bottom. The reciprocal is the fraction flipped upside down. A whole number such as 3 is the fraction 3 over 1, so its reciprocal is 1 over 3. A number times its reciprocal always makes 1.
Adding fractions with the same bottom
ac+bc=a+bc\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}
c is the shared bottom number, called the denominator. a and b are the tops, called numerators. When the pieces are the same size, you just add how many pieces you have. The size of the pieces, c, stays the same.

  1. Sort out the wording

    The question has three reciprocals in it. A reciprocal is a fraction flipped upside down. Do the jobs in the order of the words:

    1. Flip 14\frac{1}{4}.
    2. Flip 3.
    3. Add the two results to get the sum.
    4. Flip that sum.

    The last flip gives the answer.

  2. Flip each number

    reciprocal of ab=ba\text{reciprocal of } \frac{a}{b} = \frac{b}{a}

    Flip 14\frac{1}{4} to get 41\frac{4}{1}, which is 4. The number 3 is the same as 31\frac{3}{1}, so its reciprocal is 13\frac{1}{3}.

  3. Add the two reciprocals

    You need 4+134 + \frac{1}{3}. To add, both numbers need the same bottom. So write 4 in thirds. One whole is 3 thirds, so 4 wholes are 12 thirds.

    ac+bc=a+bc\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}

    123+13=133\frac{12}{3} + \frac{1}{3} = \frac{13}{3}

    The sum is 133\frac{13}{3}. Watch out: don’t add the tops and the bottoms straight across. That gives 54\frac{5}{4}, which flips to 45\frac{4}{5}, choice C.

  4. Flip the sum

    reciprocal of 133=313\text{reciprocal of } \frac{13}{3} = \frac{3}{13}

    Watch out: flip the whole sum, not each part. Flipping 4 and 13\frac{1}{3} one at a time takes you back to 14+3=134\frac{1}{4} + 3 = \frac{13}{4}, choice D.

    So the answer is A, 313\frac{3}{13}.

Check your answer

A number times its reciprocal is 1. Test it: 133×313=3939=1\frac{13}{3} \times \frac{3}{13} = \frac{39}{39} = 1. Also estimate. The sum is a bit more than 4, so its reciprocal should be a bit less than 14\frac{1}{4}. 313\frac{3}{13} is about 0.23, just under 0.25.

More questions like this

The original DOE question
Question 87 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 87. Open page 83 of the handbook (PDF)

Question 55 of 120

2024 Test A Math, question 32

Question

INGREDIENTS FOR 4 SERVINGS OF OATMEAL

Ingredient Cups
Oats 23\frac{2}{3}
Water 3143\frac{1}{4}

The table shows the cups of ingredients used to make 4 servings of oatmeal. What is the unit rate for cups of oats per cup of water?

  • A16\frac{1}{6}
  • B839\frac{8}{39}
  • C1316\frac{13}{16}
  • D138\frac{13}{8}
Show the answer

Correct answer: B

The unit rate for cups of oats per cup of water can be found by dividing the number of cups of oats by the number of cups of water:

23314\frac{\frac{2}{3}}{3\frac{1}{4}}

The denominator can be rewritten as 134\frac{13}{4}.

The quotient 23134\frac{\frac{2}{3}}{\frac{13}{4}} can be rewritten as:

23×413=839.\frac{2}{3} \times \frac{4}{13} = \frac{8}{39}.

Therefore, the unit rate for cups of oats per cup of water is 839\frac{8}{39}.

Question analysis

This unit rate question uses a recipe table for 4 servings of oatmeal, with 2/3 cup of oats and 3 1/4 cups of water, and asks for cups of oats per cup of water. Students divide one fraction by another, including a mixed number. Deciding which quantity goes on top, and whether the number of servings matters, is the tricky part.

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 Unit rates from fractions, such as 1/2 mile in 1/4 hour, are exactly grade 7 work. This one adds a mixed number, a fraction divided by a fraction, and a table titled for 4 servings, so students must decide which numbers the rate uses.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards reading the rate's wording exactly and then carrying out a fraction division without slips. The 4 servings in the title is the hook: dividing by 4 gives a per-serving amount that sounds like a unit rate and is listed. Students who write oats over water as one fraction, convert the mixed number and multiply by the reciprocal finish in a few lines.

  1. Error-prone arithmetic moderate Working it out

    The rate is a fraction over a mixed number, 2/3 ÷ 3 1/4, so 3 1/4 must become 13/4 and the division a multiplication by 4/13. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    In "cups of oats per cup of water" the quantity before "per" goes on top; reading the phrase loosely puts water over oats or brings in servings. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table's title, "for 4 servings", supplies the only other number on the page, and it plays no part in a rate of oats to water. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Write the rate as cups of oats over cups of water, change the mixed number to an improper fraction, and divide by multiplying by its reciprocal.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
unit rate=first amountsecond amount\text{unit rate} = \frac{\text{first amount}}{\text{second amount}}
A unit rate tells how much of the first amount goes with 1 of the second amount. The word per means for each. The amount named before per goes on top, and the amount after it goes on the bottom.
Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, b is the top of the fraction, and c is the bottom. Each whole holds c pieces, so a wholes hold a times c pieces. Then add the b extra pieces.
Dividing by a fraction
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. This works because dividing by a number undoes multiplying by it.

  1. Sort out the question

    The table shows 23\frac{2}{3} cup of oats and 3143\frac{1}{4} cups of water. You need the unit rate for “cups of oats per cup of water.” That is how many cups of oats go with 1 cup of water.

    Watch out: the 4 servings play no part here. Dividing the oats by 4 servings gives 16\frac{1}{6} cup for each serving, choice A.

  2. Set up the rate

    Oats comes before per, so oats goes on top. Water goes on the bottom.

    unit rate=first amountsecond amount\text{unit rate} = \frac{\text{first amount}}{\text{second amount}}

    unit rate=23÷314\text{unit rate} = \frac{2}{3} \div 3\frac{1}{4}

    Watch out: water over oats flips the rate. That gives a number bigger than 1, like choice D.

  3. Change the mixed number

    A mixed number is a whole number plus a fraction. Change 3143\frac{1}{4} to a fraction so you can divide.

    abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}

    314=3×4+14=1343\frac{1}{4} = \frac{3 \times 4 + 1}{4} = \frac{13}{4}

    So 3 and one quarter cups is 13 quarters of a cup.

  4. Divide by multiplying

    To divide by 134\frac{13}{4}, multiply by its reciprocal, 413\frac{4}{13}.

    ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

    23÷134=23×413=2×43×13=839\frac{2}{3} \div \frac{13}{4} = \frac{2}{3} \times \frac{4}{13} = \frac{2 \times 4}{3 \times 13} = \frac{8}{39}

    No whole number bigger than 1 divides both 8 and 39, so the fraction can’t be made simpler. The rate is 839\frac{8}{39} cup of oats per cup of water.

    So the answer is B, 839\frac{8}{39}.

Check your answer

There is much more water than oats. So the oats for 1 cup of water should be small, and 839\frac{8}{39} is about 0.2. Also multiply back: 839×134=104156=23\frac{8}{39} \times \frac{13}{4} = \frac{104}{156} = \frac{2}{3}. That is the cups of oats in the table.

More questions like this

The original DOE question
Question 89 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 84, question 89. Open page 84 of the handbook (PDF)

Question 56 of 120

2024 Test A Math, question 34

Question

In a certain state, the sales tax rate increased from 7.0% to 7.5%. What was the increase in the sales tax on a $200 item?

  • A$1
  • B$10
  • C$14
  • D$15
Show the answer

Correct answer: A

The increase in sales tax percentage is

7.5%−7.0%=0.5%7.5\% - 7.0\% = 0.5\%

0.5% of $200 is

(0.5%)(200)=(0.005)(200)=1(0.5\%)(200) = (0.005)(200) = 1

Question analysis

This percent word problem says a state's sales tax rate rose from 7.0% to 7.5% and asks how much more tax is charged on a $200 item. Students find the change in the rate and apply it to the price, or compute both taxes and compare. Converting a fraction of a percent to a decimal is where errors creep in.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Sales tax is a standard grade 7 percent topic, usually finding the tax or the total on one price. This one compares two tax rates whose difference is a fraction of a percent, where placing the decimal point is the main hazard.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 22 words.
Corrected from the DOE handbook
  • A dash printed in place of a minus sign in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is tiny once 0.5% is written correctly, so the question turns on the decimal point. A percent below 1 is unfamiliar, and moving the point one place instead of two gives a choice ten times too large. Students can sidestep it by finding both taxes, $14 and $15, and subtracting, though either tax on its own is also a listed choice.

  1. Error-prone arithmetic moderate Working it out

    The change in rate is half of one percent, and writing 0.5% as a decimal moves the point two places, to 0.005; one place gives 0.05 and a tenfold error. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The question asks for "the increase in the sales tax", a difference in dollars, while the stem's first sentence is about the two rates themselves. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Subtract the old rate from the new rate, then take that percent of $200, writing it as a decimal carefully; or find both taxes and subtract.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes the subtraction with an em dash, “7.5% — 7.0% = 0.5%”. The site uses a minus sign: 7.5% − 7.0% = 0.5%. Nothing else changes; 0.5% of $200 is $1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. Percent means out of 100, so p percent is p over 100. The whole is the amount you take the percent of, here the $200 price. The part is the result. To write a percent as a decimal, move the decimal point two places left.

  1. Sort out the question

    Sales tax is extra money added to a price. The tax rate went from 7.0% to 7.5%. The item costs $200. You need “the increase in the sales tax” in dollars. That is how much more tax you pay now.

    Watch out: the tax at the old rate is $14, choice C. The tax at the new rate is $15, choice D. Neither one is the increase.

  2. Find how much the rate went up

    Subtract the old rate from the new rate.

    7.5−7.0=0.57.5 - 7.0 = 0.5

    The rate went up 0.5%. That is half of one percent.

  3. Find 0.5% of $200

    The extra tax is 0.5% of the price. Use the percent formula with p = 0.5 and the whole = 200.

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    part=0.5100×200=0.005×200=1\text{part} = \frac{0.5}{100} \times 200 = 0.005 \times 200 = 1

    The tax goes up by $1. Watch out: 0.5% is 0.005, not 0.05. Moving the point only one place gives $10, choice B.

    So the answer is A, $1.

Check your answer

Find both taxes and subtract. Old tax: 0.07×200=140.07 \times 200 = 14, so $14. New tax: 0.075×200=150.075 \times 200 = 15, so $15. The difference is 15−14=115 - 14 = 1, or $1. Both ways give the same answer.

More questions like this

The original DOE question
Question 91 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 85, question 91. Open page 85 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 57 of 120

2024 Test A Math, question 41

Question

Terrell played 5 computer games and earned a mean score of 8 points per game. If his mean score for the first 4 games was 7 points per game, how many points was his score in the fifth game?

  • A9
  • B11
  • C12
  • D14
Show the answer

Correct answer: C

If the mean number of points for the first four games is 7, the total number of points earned in those games is 28.

4×7=284 \times 7 = 28

If the mean number of points for all five games is 8, the total number of points earned is 40.

5×8=405 \times 8 = 40

To find the number of points in the fifth game, find the difference between 40 and 28.

40−28=1240 - 28 = 12

Question analysis

This mean, or average, word problem says Terrell averaged 8 points over 5 computer games and 7 points over the first 4, and asks for his score in the fifth game. Students turn each average into a total and compare the totals. Combining the two averages directly, without going through totals, is the tempting shortcut that fails.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.3 measures of center and variation
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Computing a mean is grade 6 content. Working backward from two means to the totals they stand for, then to one missing score, is a common grade 7 test task, and it comes with a tempting shortcut.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 4, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards turning both means into totals, 40 points for five games and 28 for four, after which one subtraction finishes it. Averaging 7 with the unknown score is a quick shortcut that gives a listed choice, but it weighs four games as one. Checking that 28 plus the answer, divided by 5, gives 8 catches that error and the near miss next to the answer.

  1. Turning words into math moderate Understanding the question

    Each mean has to become a total before the fifth game can be found: "a mean score of 8 points per game" over 5 games means 40 points in all. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Students must know that a total is the mean times the number of games, and that means over different numbers of games cannot simply be averaged. (Anderson & Krathwohl, 2001)

Strategy: Turn each mean into a total by multiplying by its number of games, then subtract the four-game total from the five-game total.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total from a mean
total=mean×number of games\text{total} = \text{mean} \times \text{number of games}
The mean, or average, is what each game would score if the total were shared out equally. So the total is the mean times the number of games. Use it to turn each mean back into a total. Totals can be subtracted, but means over different numbers of games can’t.

  1. Sort out the question

    Terrell played 5 games. His mean score for all 5 was 8 points per game. His mean for the first 4 games was 7 points per game. The mean, or average, is the total split equally among the games. You need his score in game 5.

  2. Find the total for all 5 games

    Turn the mean for all 5 games into a total. Multiply the mean by the number of games.

    total=mean×number of games\text{total} = \text{mean} \times \text{number of games}

    total=8×5=40\text{total} = 8 \times 5 = 40

    He scored 40 points in all 5 games together.

  3. Find the total for the first 4 games

    Use the same formula with the mean of 7 and 4 games.

    total=7×4=28\text{total} = 7 \times 4 = 28

    He scored 28 points in the first 4 games.

  4. Subtract to find game 5

    The 40 points include the first 4 games and game 5. Take away the first 4 games to leave game 5.

    40−28=1240 - 28 = 12

    He scored 12 points in game 5. Watch out: don’t average 7 and the fifth score as if they were two games. That gives 9, choice A. The 7 stands for 4 games, not 1. Watch out: adding 4 to 7 gives 11, choice B. Each of the first 4 games was 1 point below 8. So game 5 needs 8 plus 4 more points.

    So the answer is C, 12 points.

Check your answer

Try scores of 7, 7, 7, 7 and 12. They add up to 28+12=4028 + 12 = 40, and 40÷5=840 \div 5 = 8. That is the mean for all 5 games, so 12 is right.

More questions like this

The original DOE question
Question 98 as printed on page 87 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 87, question 98. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 58 of 120

2024 Test A Math, question 47

Question

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

  • A5,600
  • B9,000
  • C11,300
  • D24,800
Show the answer

Correct answer: A

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

45200=x25,000\frac{45}{200} = \frac{x}{25{,}000}

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

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

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

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a sample to estimate a count for a whole population is typical grade 7 statistics and ratio work. This version takes one proportion with large numbers and a rounding step, about the usual grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 13, 50 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

    The survey's 45 out of 200 is a rate to scale up to the town's 25,000 adults, a fraction of the sample rather than a percent. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    The scale factor 25,000 ÷ 200 = 125 and the product 45 × 125 = 5,625 involve large numbers, and the result must then be rounded to the nearest hundred. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    A survey asked 200 adults, and 45 of them read the online paper. The town has 25,000 adults. You need a good estimate of how many adults in the whole town read it. Assume the town is like the survey. Then the same fraction of adults read it.

  2. Set up a proportion

    A proportion says two fractions are equal. Let x be the number of readers in the town. The fraction of readers is the same in both groups.

    ab=cd\frac{a}{b} = \frac{c}{d}

    45200=x25,000\frac{45}{200} = \frac{x}{25{,}000}

    Watch out: multiplying 45 by 200 gives 9,000, choice B. That mixes the numbers without a plan.

  3. Find the scale factor

    The scale factor is how many times bigger the town is than the survey group.

    25,000÷200=12525{,}000 \div 200 = 125

    The town is 125 times as big. So multiply the top, 45, by 125 too.

  4. Scale up the readers

    x=45×125x = 45 \times 125

    Break 125 into 100 and 25.

    45×100=4,50045 \times 100 = 4{,}500

    45×25=1,12545 \times 25 = 1{,}125

    4,500+1,125=5,6254{,}500 + 1{,}125 = 5{,}625

    About 5,625 adults read it. Watch out: 45 out of 200 is not 45%. A percent is out of 100. Using 45% gives about 11,300, choice C.

  5. Round to the nearest hundred

    The choices are rounded, since this is an estimate. 5,625 is between 5,600 and 5,700. Its last two digits, 25, are less than 50, so round down to 5,600.

    So the answer is A, 5,600.

Check your answer

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

More questions like this

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

Question 59 of 120

2024 Test A Math, question 48

Question

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

  • Ay=317xy = \frac{3}{17}x
  • By=3xy = 3x
  • Cy=173xy = \frac{17}{3}x
  • Dy=17xy = 17x
Show the answer

Correct answer: C

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

yx=173\frac{y}{x} = \frac{17}{3}

y=173xy = \frac{17}{3}x

So the equation y=173xy = \frac{17}{3}x describes the relationship.

Question analysis

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

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

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

    The coefficients 17/3 and 3/17 use the same two numbers, one the other inverted, so only checking that 3 days gives 17 miles separates the two equations. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Turning words into math minor Understanding the question

    With x as days and y as miles, "17 miles in 3 days" becomes miles per day, 17 ÷ 3, the number that multiplies x. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The hiker plans 17 miles in 3 days. x is the number of days, and y is the number of miles. You need an equation that links them. Assume the hiker goes the same distance each day. Then y is always the same number times x.

  2. Find the miles for each day

    That number is the constant of proportionality, k. It is miles divided by days.

    k=yxk = \frac{y}{x}

    k=173k = \frac{17}{3}

    That is 5235\frac{2}{3} miles each day. Watch out: miles go on top, because y is miles. 317\frac{3}{17} is days per mile, upside down. That is choice A.

  3. Write the equation

    y=kxy = kx

    y=173xy = \frac{17}{3}x

    Watch out: 17 is the miles for all 3 days, not for 1 day. So y=17xy = 17x, choice D, is too fast. And y=3xy = 3x, choice B, uses the days as the rate.

    So the answer is C, y=173xy = \frac{17}{3}x.

Check your answer

Put in x = 3, for 3 days: y=173×3=17y = \frac{17}{3} \times 3 = 17. That is 17 miles, which matches the plan. Try choice A the same way: 317×3=917\frac{3}{17} \times 3 = \frac{9}{17}. That is less than 1 mile, far too short.

More questions like this

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

Question 60 of 120

2024 Test A Math, question 49

Question

Carolyn walked 3 miles from her house to the library and then 2122\frac{1}{2} miles farther to the grocery store. Returning home by the same route, she walked 1231\frac{2}{3} miles before stopping at a friend’s house. How many miles did Carolyn have left to walk home?

  • A3563\frac{5}{6}
  • B4164\frac{1}{6}
  • C4234\frac{2}{3}
  • D7167\frac{1}{6}
Show the answer

Correct answer: A

The distance from Carolyn’s house to the grocery store is 3+212=5123 + 2\frac{1}{2} = 5\frac{1}{2} miles.

Carolyn walked 1231\frac{2}{3} miles back, so she still had 512−1235\frac{1}{2} - 1\frac{2}{3} miles left to walk home.

512−123=112−535\frac{1}{2} - 1\frac{2}{3} = \frac{11}{2} - \frac{5}{3}

336−106=236=356\frac{33}{6} - \frac{10}{6} = \frac{23}{6} = 3\frac{5}{6}

So Carolyn had 3563\frac{5}{6} miles left to walk home.

Question analysis

Carolyn walks 3 miles from home to the library and 2 1/2 miles more to the grocery store, then heads back the same way and stops at a friend's house after 1 2/3 miles. This fractions word problem asks how far she still has to walk home. Students add and subtract mixed numbers, and the subtraction needs regrouping across unlike denominators.

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 word problems with mixed numbers. This one has only positive numbers, but it needs a clear picture of the route and a subtraction that requires regrouping across unlike denominators.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The story needs a quick sketch, but the real test is one mixed-number subtraction in which the second fraction is larger than the first. Many students subtract the fractions the easy way round, which gives a choice close to the right one. Converting both numbers to sixths, 33/6 − 10/6, sidesteps the regrouping, and adding the result to 1 2/3 to get back 5 1/2 is a quick check.

  1. Error-prone arithmetic moderate Working it out

    The subtraction 5 1/2 − 1 2/3 needs sixths, 5 3/6 − 1 4/6, and the fraction on top is the smaller one, so a whole must be regrouped. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    Subtracting the fraction parts the wrong way round gives 4 1/6, only 1/3 away from the answer, so a rough estimate cannot tell the two apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Turning words into math minor Understanding the question

    The route has to be pictured: the walk back starts at the grocery store, 5 1/2 miles from home, not at the library. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Sketch the route as a line from home to the library to the store, find the total distance out, then subtract the distance already walked back, using sixths.

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.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, b is the top of the fraction, and c is the bottom. Each whole holds c pieces, so a wholes hold a times c pieces. Then add the b extra pieces.
Equivalent fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a is the top of a fraction and b is the bottom. n is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, cut into smaller pieces. Use it to give two fractions the same bottom before you add or subtract.
Subtracting fractions with the same bottom
ac−bc=a−bc\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}
c is the shared bottom number. a and b are the tops. When the pieces are the same size, you just take away how many pieces you have. The size of the pieces, c, stays the same.

  1. Picture the route

    Draw a line: home, then the library, then the store. Home to the library is 3 miles. The library to the store is 2122\frac{1}{2} miles more. Carolyn walks back the same way. After 1231\frac{2}{3} miles she stops at a friend’s house. You need the miles still left to get home.

  2. Find the distance from home to the store

    3+212=5123 + 2\frac{1}{2} = 5\frac{1}{2}

    The store is 5125\frac{1}{2} miles from home. She starts back from the store, so this is the full trip home. Watch out: adding all three distances gives 7167\frac{1}{6}, choice D. That is how far she has walked, not how far is left.

  3. Change to improper fractions

    The miles left are 512−1235\frac{1}{2} - 1\frac{2}{3}. An improper fraction has a top bigger than its bottom. Change each mixed number to one.

    abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}

    512=5×2+12=1125\frac{1}{2} = \frac{5 \times 2 + 1}{2} = \frac{11}{2}

    123=1×3+23=531\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3}

  4. Use sixths for both

    To subtract, the fractions need the same bottom. 2 and 3 both go into 6, so change both to sixths.

    ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}

    112=11×32×3=336\frac{11}{2} = \frac{11 \times 3}{2 \times 3} = \frac{33}{6}

    53=5×23×2=106\frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6}

  5. Subtract and simplify

    ac−bc=a−bc\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}

    336−106=236\frac{33}{6} - \frac{10}{6} = \frac{23}{6}

    23 sixths is 3 wholes (18 sixths) and 5 sixths more. So 236=356\frac{23}{6} = 3\frac{5}{6}. Watch out: 12\frac{1}{2} is less than 23\frac{2}{3}. Subtracting the fraction parts the wrong way round gives 4164\frac{1}{6}, choice B.

    So the answer is A, 3563\frac{5}{6} miles.

Check your answer

Add back: 356+123=356+146=496=536=5123\frac{5}{6} + 1\frac{2}{3} = 3\frac{5}{6} + 1\frac{4}{6} = 4\frac{9}{6} = 5\frac{3}{6} = 5\frac{1}{2}. That is the full distance from the store to home. Also estimate: 5.5 minus about 1.7 is about 3.8, and 3563\frac{5}{6} is about 3.83.

More questions like this

The original DOE question
Question 106 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 89, question 106. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 61 of 120

2024 Test B Math, question 4

Question

At one location, the temperature increased from −7° F to 4° F between 5:00 a.m. and 9:00 a.m. By 11:00 a.m., the temperature was 3.5 times the 9:00 a.m. temperature. What was the total increase in temperature between 5:00 a.m. and 11:00 a.m., in degrees Fahrenheit?

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

Show the answer

Correct answer: 21

The 11:00 a.m. temperature is 3.5 times the 9:00 a.m. temperature. The 9:00 a.m. temperature was 4° F.

4×3.5=144\times 3.5=14

The temperature increased from −7° F at 5:00 a.m. to 14° F at 11:00 a.m., which is an increase of 21° F.

Question analysis

A temperature word problem with negative numbers: the temperature rises from −7° F to 4° F by 9:00 a.m., and by 11:00 a.m. it is 3.5 times the 9:00 a.m. reading. Students find the total rise from 5:00 to 11:00 a.m. Subtracting a negative starting temperature is where most slips happen.

NYS standards
  • 7.NS.3 four operations with rational numbers
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Using signed numbers for temperature change is standard grade 7 work. This adds a decimal multiplier and asks for a change rather than a temperature, two short steps, which keeps it at a typical grade 7 level.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 5, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Finding the 11:00 a.m. temperature is a single multiplication; the points are lost in the subtraction that follows. A rise that starts below zero has two parts, 7 degrees up to zero and 14 more above it, and students who treat −7 as 7 lose the first part. The times in the stem also invite measuring from 9:00 a.m. instead of 5:00 a.m., which covers only the last two hours.

  1. Error-prone arithmetic moderate Working it out

    The last step is 14 − (−7): subtracting the negative starting temperature means adding 7, and the part of the rise below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    “3.5 times the 9:00 a.m. temperature” multiplies the reading of 4°, not the 11-degree rise, and “total increase” runs from the 5:00 a.m. reading, not from 9:00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Much to keep track of minor Working it out

    Three times of day carry three temperatures, −7°, 4° and 14°, and the answer compares the first with the last, not either of them with the middle one. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find the 11:00 a.m. temperature from the 9:00 a.m. one, then subtract the 5:00 a.m. temperature from it, remembering that subtracting a negative number adds its size.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Stops one step early: A choice is a number from partway through the solution.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Change in temperature
change=end−start\text{change} = \text{end} - \text{start}
End is the last temperature, and start is the first one. Subtract start from end to find how much it went up. It works with negative numbers too. A positive change means the temperature rose.
Subtracting a negative number
a−(−b)=a+ba - (-b) = a + b
a is any number, and −b is a negative number. Taking away a negative is the same as adding its size. On a thermometer, going from −7 up to 0 is 7 degrees, so those 7 degrees get added.

  1. Sort out what you know

    There are three times and three temperatures, in degrees Fahrenheit (°F).

    • 5:00 a.m.: −7°F, which is 7 degrees below zero
    • 9:00 a.m.: 4°F
    • 11:00 a.m.: 3.5 times the 9:00 a.m. temperature

    You need the total increase from 5:00 a.m. to 11:00 a.m. That is how much warmer it got from the first time to the last.

  2. Find the 11:00 temperature

    “3.5 times the 9:00 a.m. temperature” means 3.5 times 4°F.

    3.5×4=143.5 \times 4 = 14

    At 11:00 a.m. it was 14°F. Watch out: 14 is a temperature, not the increase. Gridding 14 stops too early.

  3. Subtract the start from the end

    The increase runs from 5:00 a.m. to 11:00 a.m. So it starts at −7°F and ends at 14°F.

    change=end−start\text{change} = \text{end} - \text{start}

    change=14−(−7)\text{change} = 14 - (-7)

    Watch out: start at 5:00 a.m., not at 9:00 a.m. Starting from 4°F gives only 10.

  4. Subtract the negative number

    Taking away a negative number is the same as adding its size.

    a−(−b)=a+ba - (-b) = a + b

    14−(−7)=14+7=2114 - (-7) = 14 + 7 = 21

    The temperature rose 21°F. Watch out: writing 14 − 7 = 7 loses the minus sign. It skips the 7 degrees the air warmed just to reach zero.

    So you would grid 21.

Check your answer

Count up on a thermometer. From −7°F up to 0°F is 7 degrees. From 0°F up to 14°F is 14 more. 7+14=217 + 14 = 21. Or add the two parts of the morning: it rose 11 degrees by 9:00 and 10 more by 11:00, and 11+10=2111 + 10 = 21.

More questions like this

The original DOE question
Question 61 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 199, question 61. Open page 199 of the handbook (PDF)

Question 62 of 120

2024 Test B Math, question 9

Question

A car dealership is having a sale where all cars have a 12% discount. What is the discounted price of a car that originally cost $15,600?

  • A$13,728
  • B$14,300
  • C$14,400
  • D$15,588
Show the answer

Correct answer: A

15,600×12%=15{,}600\times 12\%=

15,600×0.12=1,87215{,}600\times 0.12=1{,}872

15,600−1,872=13,72815{,}600-1{,}872=13{,}728

Question analysis

A percent word problem about a car sale: every car is 12% off, and one car originally cost $15,600. Students find the sale price of that car. The discount has to be worked out as a dollar amount and then taken off, with care over the decimal point in 12%.

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 Discounts are one of the percent applications in grade 7, and this is a routine two-step example with whole-dollar amounts. It matches typical grade 7 work, with no reverse percent or second change.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 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

This is a standard two-step discount, and a quick estimate does most of the work: 10% of $15,600 is $1,560, so 12% is a little under $1,900, and only one choice sits about that far below the price. The wrong choices take off $1,300, $1,200 or just $12, so the question mainly catches students unsure what 12% of a price means.

  1. Error-prone arithmetic minor Working it out

    Finding 12% of $15,600 by hand, 0.12 × 15,600 = 1,872, means placing a decimal point in a product of large numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The phrase “discounted price” asks for what the buyer pays, so the $1,872 discount has to be taken off the price rather than reported. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Write 12% as 0.12 and multiply by the original price to get the discount, then subtract it; or multiply the price by 0.88 in one step.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
x=p100×Wx = \frac{p}{100} \times W
W is the whole amount, here the first price. p is the percent, and x is the part you want. Percent means out of 100, so p percent is the fraction p over 100. Multiplying the whole by that fraction gives the part.
Sale price
sale price=first price−discount\text{sale price} = \text{first price} - \text{discount}
The first price is what the car cost before the sale. The discount is the amount taken off. The sale price is what the buyer pays, so you subtract.

  1. Sort out the question

    The car first cost $15,600. A discount is an amount taken off the price. Here it is 12% of the price. You need the discounted price, the amount the buyer pays after the discount.

  2. Find 12% of the price

    x=p100×Wx = \frac{p}{100} \times W

    x=12100×15,600=0.12×15,600x = \frac{12}{100} \times 15{,}600 = 0.12 \times 15{,}600

    Break it into easy parts. 10% is 15,600÷10=1,56015{,}600 \div 10 = 1{,}560. 1% is 15,600÷100=15615{,}600 \div 100 = 156, so 2% is 2×156=3122 \times 156 = 312.

    1,560+312=1,8721{,}560 + 312 = 1{,}872

    The discount is $1,872. Watch out: 12% does not mean $12. Taking off $12 gives $15,588, choice D.

  3. Take the discount off

    sale price=first price−discount\text{sale price} = \text{first price} - \text{discount}

    15,600−1,872=13,72815{,}600 - 1{,}872 = 13{,}728

    The buyer pays $13,728. Watch out: 12% is not one twelfth. Dividing by 12 takes off only $1,300 and gives $14,300, choice B.

    So the answer is A, $13,728.

Check your answer

After a 12% discount, the buyer pays 88% of the price, because 100−12=88100 - 12 = 88. Then 0.88×15,600=13,7280.88 \times 15{,}600 = 13{,}728. That matches. Also, $13,728 is less than $15,600, as a sale price should be.

More questions like this

The original DOE question
Question 66 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 201, question 66. Open page 201 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 63 of 120

2024 Test B Math, question 11

Question

In a sports league, each team has 36 players and 3 coaches. There are also a number of team assistants. The ratio of team assistants to players is 16\frac{1}{6}. What is the ratio of coaches to assistants?

  • A14\frac{1}{4}
  • B12\frac{1}{2}
  • C23\frac{2}{3}
  • D56\frac{5}{6}
Show the answer

Correct answer: B

Let x be the number of assistants.

Since the ratio of assistants to players is 16\frac{1}{6}, then:

16=x36\frac{1}{6}=\frac{x}{36}

366=x\frac{36}{6}=x

6=x6=x

There are a total of 6 assistants.

So the ratio of coaches to assistants is 36=12\frac{3}{6}=\frac{1}{2}.

Question analysis

A ratio word problem about a sports team with 36 players, 3 coaches and an unknown number of assistants, where the ratio of assistants to players is 1 to 6. Students find how many assistants there are, then form a new ratio of coaches to assistants. Keeping the order of each ratio straight matters.

NYS standards
  • 6.RP.3 ratio and rate reasoning
  • 6.RP.1 ratio concepts and language
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a missing amount from a ratio and then writing a second ratio is grade 6 ratio reasoning with whole numbers. Typical grade 7 work adds fractions, rates or percents, which this two-step problem does not.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two short steps, finding 6 assistants and then writing 3 to 6, make this one of the easier ratio questions. The care needed is in keeping each ratio's order and its two groups straight, since the given ratio compares assistants with players and the asked one compares coaches with assistants. None of the wrong choices matches a common slip closely, so students who find the 6 assistants seldom go wrong.

  1. Turning words into math minor Understanding the question

    “The ratio of team assistants to players is 1/6” must become 6 assistants for 36 players, and the question then asks for a ratio in a new order, coaches first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Three groups are named, 36 players, 3 coaches and an unknown number of assistants, and each ratio pairs a different two of them. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Use the 1 to 6 ratio to find the number of assistants from the 36 players, then write coaches over assistants, in that order, and simplify.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Ratio as a fraction
ratio=first groupsecond group\text{ratio} = \frac{\text{first group}}{\text{second group}}
A ratio compares two amounts. The group named first goes on top, and the group named second goes on the bottom. Order matters. Assistants to players is not the same as players to assistants.
Equal ratios
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a and b are the two numbers in a ratio, written as a fraction. n is any number except zero. Multiplying both numbers by the same n gives an equal ratio. This is how you scale a small ratio up to real counts.

  1. Sort out what you know

    Each team has 36 players and 3 coaches. It also has some assistants, but you are not told how many. The ratio of assistants to players is 1/6. That means 1 assistant for every 6 players. You need the ratio of coaches to assistants.

  2. Find the number of assistants

    Write the ratio as a fraction, assistants on top and players on the bottom. To get from 6 players to 36, multiply by 6. Do the same to the top.

    ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}

    16=1×66×6=636\frac{1}{6} = \frac{1 \times 6}{6 \times 6} = \frac{6}{36}

    So there are 6 assistants for the 36 players.

  3. Write the new ratio

    Now the order changes. Coaches come first, so coaches go on top.

    ratio=first groupsecond group\text{ratio} = \frac{\text{first group}}{\text{second group}}

    coachesassistants=36=12\frac{\text{coaches}}{\text{assistants}} = \frac{3}{6} = \frac{1}{2}

    Divide the top and the bottom by 3 to simplify.

  4. Avoid the traps and answer

    Watch out: use only coaches and assistants. Putting all 9 staff over the 36 players gives 1/4, choice A. Putting the 6 assistants over all 9 staff gives 2/3, choice C. And 1/6 is not a share of everyone, so 5/6, choice D, has no meaning here.

    So the answer is B, 12\frac{1}{2}.

Check your answer

A ratio of 1/2 means 1 coach for every 2 assistants. With 3 coaches, that makes 6 assistants. Then 6 assistants to 36 players is 636=16\frac{6}{36} = \frac{1}{6}, the ratio the question gave.

More questions like this

The original DOE question
Question 68 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 201, question 68. Open page 201 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 64 of 120

2024 Test B Math, question 14

Question

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

  • A4
  • B5
  • C7
  • D15
Show the answer

Correct answer: B

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

60+60∙25100=60+60\bullet\frac{25}{100}=

60+60∙0.25=60+60\bullet 0.25=

60+15=7560+15=75

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

Karen: 1,500 pagesk days=60 pages1 day\frac{1{,}500\text{ pages}}{k\text{ days}}=\frac{60\text{ pages}}{1\text{ day}}

60k=1,50060k=1{,}500

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

k=25 daysk=25\text{ days}

Martina: 1,500 pagesm days=75 pages1 day\frac{1{,}500\text{ pages}}{m\text{ days}}=\frac{75\text{ pages}}{1\text{ day}}

75m=1,50075m=1{,}500

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

m=20 daysm=20\text{ days}

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

25−20=525-20=5

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

Question analysis

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

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

Why it is hard

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

  1. Many steps moderate Working it out

    Four operations chain together: 25% of 60, Martina's 75 pages, the two finishing times 1,500 ÷ 60 and 1,500 ÷ 75, and finally their difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Combines ideas moderate Working it out

    A percent increase in a daily rate has to be turned into a change in days, and 25% more pages a day does not mean 25% fewer days. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Turning words into math minor Understanding the question

    “25% more pages per day than Karen” makes Martina's rate 1.25 × 60, and “how many days sooner” asks for a difference of two times, not of two rates. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}
Old is the starting amount, and p is the percent it grows by. First find p percent of the old amount. That is the extra. Then add the extra to the old amount to get the new amount.
Time from a rate
days=total pages÷pages per day\text{days} = \text{total pages} \div \text{pages per day}
Total pages is the whole assignment. Pages per day is how fast someone reads. Dividing tells how many days of reading fit into the whole job.

  1. Sort out what you know

    Karen reads 60 pages per day. Martina reads 25% more pages per day than Karen. Each girl must read 1,500 pages. You need how many days sooner Martina finishes. That is the difference between their numbers of days.

  2. Find Martina’s pages per day

    25% more means Karen’s 60 pages plus 25% of 60.

    new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}

    60+25100×60=60+15=7560 + \frac{25}{100} \times 60 = 60 + 15 = 75

    Martina reads 75 pages per day. Watch out: 15 is extra pages, not days. Stopping here gives 15, choice D. Adding 25 pages instead of 25% gives 85 pages and leads to 7, choice C.

  3. Find each girl’s number of days

    Divide the whole assignment by each girl’s pages per day.

    days=total pages÷pages per day\text{days} = \text{total pages} \div \text{pages per day}

    Karen: 1,500÷60=25\text{Karen: } 1{,}500 \div 60 = 25

    Martina: 1,500÷75=20\text{Martina: } 1{,}500 \div 75 = 20

    Karen needs 25 days, and Martina needs 20 days.

  4. Subtract the days

    “How many days sooner” asks for the difference.

    25−20=525 - 20 = 5

    Martina finishes 5 days sooner. Watch out: dividing 60 by 15 mixes two rates and gives 4, choice A.

    So the answer is B, 5.

Check your answer

Multiply back. Karen: 25×60=1,50025 \times 60 = 1{,}500 pages. Martina: 20×75=1,50020 \times 75 = 1{,}500 pages. Both read all 1,500 pages, and 25 days is 5 days more than 20 days.

More questions like this

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

Question 65 of 120

2024 Test B Math, question 15

Question

Claire is running on a circular track. She has already run 650 meters. Each lap around the track is a distance of 120 meters. Her goal is to run at least 2,500 meters in all. How many more laps does she need to complete in order to reach her goal?

  • A5
  • B6
  • C15
  • D16
Show the answer

Correct answer: D

Claire’s goal is to run at least 2,500 meters, and she has already run 650 meters.

2,500−650=1,8502{,}500-650=1{,}850

Claire has at least 1,850 meters left to run. Divide by the distance around the track to find the number of laps she needs to complete:

1,850÷120≈15.4171{,}850\div 120\approx 15.417

Since she wants to run at least 2,500 meters, round 15.417 up to 16. Claire needs to run 16 laps to meet her goal.

Question analysis

A word problem about running laps: Claire has already run 650 meters on a 120-meter track and wants to run at least 2,500 meters in all. Students find how many more laps she must complete. Subtracting and dividing are straightforward; deciding what to do with the leftover part of a lap is the real test.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve word problems that lead to inequalities like px + q ≥ r and interpret the solution in context. This is a typical three-step example, where the solution must be turned into a whole number of laps.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic leads to about 15.4 laps, and the question is entirely about what to do with the fraction of a lap. Rounding to the nearest whole number is a habit that fails here, because the goal must be met or passed. Multiplying back, 15 × 120 = 1,800 and 16 × 120 = 1,920, shows which count actually covers the 1,850 meters still to run.

  1. Turning words into math moderate Understanding the question

    “At least 2,500 meters in all” means a part lap must still be run in full, so 1,850 ÷ 120 ≈ 15.4 has to be rounded up, against the usual rounding rule. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    15 and 16 laps are one apart and both sit next to the quotient 15.4, so estimating cannot decide between them; only the condition that the goal be reached can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Error-prone arithmetic minor Working it out

    Dividing 1,850 by 120 by hand does not come out even, and long division by a three-digit number invites slips in placing the quotient. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract what she has run from 2,500, divide by 120, and check your whole number by multiplying it back: those laps must cover all of the remaining distance.

Traps

  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance left
left=goal−done\text{left} = \text{goal} - \text{done}
Goal is the distance she wants to reach in all. Done is the distance she has already run. Subtract to find how far she still has to go.
Number of laps
laps=distance÷one lap\text{laps} = \text{distance} \div \text{one lap}
Distance is how far she still has to run. One lap is the length of the track, 120 meters. Dividing tells how many laps fit into that distance. If it does not come out even, part of a lap is left over.

  1. Sort out what you know

    Claire has run 650 meters. One lap is 120 meters. Her goal is “at least 2,500 meters in all”. At least means 2,500 or more, so she must not fall short. You need how many more laps she must run.

  2. Find the distance left

    left=goal−done\text{left} = \text{goal} - \text{done}

    left=2,500−650=1,850\text{left} = 2{,}500 - 650 = 1{,}850

    She still needs 1,850 meters. Watch out: divide the distance left, not the 650 meters she already ran. Dividing 650 by 120 leads to 5 or 6, choices A and B.

  3. Divide by the length of a lap

    laps=distance÷one lap\text{laps} = \text{distance} \div \text{one lap}

    1,850÷120≈15.41{,}850 \div 120 \approx 15.4

    The sign ≈ means about. Check with whole laps: 15×120=1,80015 \times 120 = 1{,}800. That leaves 1,850−1,800=501{,}850 - 1{,}800 = 50 meters still to go. So 15 laps are not quite enough.

  4. Round up to reach the goal

    She needs part of one more lap. She has to complete whole laps, so she runs a 16th lap. Here you round up, even though 15.4 would usually round down. Watch out: 15 laps leave her 50 meters short, so choice C is wrong.

    So the answer is D, 16.

Check your answer

Add up her total. With 15 more laps: 650+1,800=2,450650 + 1{,}800 = 2{,}450 meters, which is short of 2,500. With 16 more laps: 650+16×120=650+1,920=2,570650 + 16 \times 120 = 650 + 1{,}920 = 2{,}570 meters. That reaches the goal.

More questions like this

The original DOE question
Question 72 as printed on page 202 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 202, question 72. Open page 202 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 66 of 120

2024 Test B Math, question 16

Question

A company with 200 employees grew in size by 25% between 2005 and 2010. The company grew again by 10% between 2010 and 2015. What was the total number of employees in 2015?

  • A235
  • B260
  • C270
  • D275
Show the answer

Correct answer: D

First, determine the number of employees in 2010:

200×25%=200×0.25=50200\times 25\%=200\times 0.25=50

So the number of employees in 2010 is 200+50=250200+50=250.

Next, determine the number of employees in 2015:

250×10%=250×0.10=25250\times 10\%=250\times 0.10=25

So the total number of employees in 2015 is 250+25=275250+25=275.

Question analysis

A percent change word problem: a company of 200 employees grows by 25% from 2005 to 2010 and then by another 10% from 2010 to 2015. Students find the number of employees in 2015. Two percent changes in a row are the tricky part, since each percent must be taken of the right amount.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Percent increase is a grade 7 topic, and two increases in a row are a common grade 7 multistep problem. The numbers are friendly, so the demand matches typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 5, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each increase alone is easy, and the whole question is about the base of the second one. Adding the percents into a single 35% feels like a shortcut and lands five employees short, a gap estimation cannot reveal. Working period by period, or multiplying 200 by 1.25 and then by 1.1, keeps each percent attached to the right amount.

  1. A very close wrong answer moderate Choosing the answer

    270 and 275 employees are five apart, too close to separate by estimate; only taking the second 10% of 250 rather than of 200 decides between them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Turning words into math moderate Understanding the question

    “Grew again by 10%” refers to the company's size in 2010, so the base for the second percent is the new 250 employees, not the original 200. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Many steps minor Working it out

    Two percent increases make four operations, a percent and an addition for each period, and the second period starts from the first one's result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work one period at a time: find the 2010 total by increasing 200 by 25%, then increase that new total by 10%.

Traps

  • Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}
Old is the starting amount, and p is the percent it grows by. First find p percent of the old amount. That is the extra. Then add the extra to the old amount to get the new amount.

  1. Sort out what you know

    The company has 200 employees at the start, in 2005. From 2005 to 2010 it grows by 25%. From 2010 to 2015 it grows again, by 10%. You need the number of employees in 2015.

  2. Find the number in 2010

    new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}

    200+25100×200=200+50=250200 + \frac{25}{100} \times 200 = 200 + 50 = 250

    In 2010 there were 250 employees. Watch out: 25% does not mean 25 people. Adding 25 people and then 10 people gives 235, choice A.

  3. Grow from the new number

    The second increase starts in 2010. So take 10% of 250, not 10% of 200.

    250+10100×250=250+25=275250 + \frac{10}{100} \times 250 = 250 + 25 = 275

    In 2015 there were 275 employees.

  4. Avoid the traps and answer

    Watch out: taking both percents of the first 200 gives 200+50+20=270200 + 50 + 20 = 270, choice C. That misses the growth of the 50 new workers. Adding 10 people instead of 10% gives 260, choice B.

    So the answer is D, 275.

Check your answer

Growing by 25% is the same as multiplying by 1.25. Growing by 10% is the same as multiplying by 1.1. 200×1.25×1.1=250×1.1=275200 \times 1.25 \times 1.1 = 250 \times 1.1 = 275. The answer should also be a bit more than 270, since the second 10% is taken of a bigger number.

More questions like this

The original DOE question
Question 73 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 73. Open page 203 of the handbook (PDF)

Question 67 of 120

2024 Test B Math, question 17

Question

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

  • A211\frac{2}{11}
  • B29\frac{2}{9}
  • C14\frac{1}{4}
  • D311\frac{3}{11}
Show the answer

Correct answer: A

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

12=2r12=2r

r=6r=6

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

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

422=211\frac{4}{22}=\frac{2}{11}.

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Simple probability is grade 6 content in New York, and grade 7 problems usually give every count. Working backward from a comparison of two probabilities to a missing count, then forward to a third probability, is more than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 49 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    “Twice as great as the probability of choosing a red candy” must become 12 = 2 × the red count, which works only because both probabilities share one total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Combines ideas moderate Working it out

    An unknown count has to be found from a ratio before any probability can be written, joining proportional reasoning with probability as a part over the total. (Embretson, 1998; Daniel & Embretson, 2010)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out what you know

    The bowl has 12 green candies, 4 yellow candies and some red candies. You are not told how many are red. Let r be the number of red candies.

    The chance of picking green is twice the chance of picking red. You need the probability (the chance) of picking yellow.

  2. Compare green and red

    Write each chance with the probability formula.

    P=number of the kind you wanttotal numberP = \frac{\text{number of the kind you want}}{\text{total number}}

    Green and red have the same bottom number, the total number of candies. So if the chance of green is twice the chance of red, there are twice as many green candies as red ones.

    12=2×r12 = 2 \times r

  3. Find the red candies and the total

    r=12÷2=6r = 12 \div 2 = 6

    There are 6 red candies. Now add up all the candies.

    12+4+6=2212 + 4 + 6 = 22

    There are 22 candies in all. Watch out: do not leave out the red candies just because their number was not given. Using 16 as the total gives 1/4, choice C.

  4. Find the chance of yellow

    There are 4 yellow candies out of 22.

    P=422=211P = \frac{4}{22} = \frac{2}{11}

    Divide the top and the bottom by 2 to simplify. Watch out: the question asks for yellow, not red. 6/22 = 3/11 is the chance of red, choice D. The bottom must be all 22 candies. Using 18 gives 2/9, choice B.

    So the answer is A, 211\frac{2}{11}.

Check your answer

Check the clue. The chance of green is 1222\frac{12}{22}, and the chance of red is 622\frac{6}{22}. Twelve is twice 6, so green is twice as likely. Also, the three chances add up to 1: 1222+422+622=2222=1\frac{12}{22} + \frac{4}{22} + \frac{6}{22} = \frac{22}{22} = 1.

More questions like this

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

Question 68 of 120

2024 Test B Math, question 20

Question

Centerville will hold a vote on whether to build a new park. The town council surveyed a random sample of 200 residents to determine whether they would be in favor of building this park. The table shows the results of the survey.

SURVEY RESULTS

Opinion Number of Residents
Yes 142
No 38
Undecided 20

The mayor estimates that half the undecided voters will vote in favor of building the park. Based on the survey results, if the mayor is correct, how many of the 24,100 voters will be in favor of building the park?

  • A17,111
  • B18,316
  • C19,401
  • D19,521
Show the answer

Correct answer: B

If half the undecided voters (half of 20) vote in favor of building the park, the total number of “yes” votes from the survey will be 152.

142+10=152142+10=152

There are 24,100 voters, and 200 residents were surveyed. Set up a proportion to solve:

152200=x24,100\frac{152}{200}=\frac{x}{24{,}100}

1925=x24,100\frac{19}{25}=\frac{x}{24{,}100}

(24,100)(1925)=x(24{,}100)\left(\frac{19}{25}\right)=x

18,316=x18{,}316=x

Question analysis

A survey question: a random sample of 200 Centerville residents answered yes, no or undecided about building a park, and the mayor expects half the undecided to vote yes. Students use the sample to predict how many of 24,100 voters will favor the park. It joins a proportion from a sample with large-number arithmetic.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.SP.1b samples and populations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Using a random sample to estimate a count in a whole population is a typical grade 7 proportion task. The extra step for the undecided group and the large numbers add care without going beyond grade 7.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 5, 88 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The setup is a sample-to-population proportion, and the trap is the mayor's assumption in the middle of the stem. Students who scale the 142 yes answers directly, or who count every undecided resident, reach believable totals that sit right among the choices. Once the count of 152 is settled, a rough estimate, about three quarters of 24,000, is enough to confirm the exact product.

  1. Error-prone arithmetic moderate Working it out

    Scaling 152 out of 200 up to 24,100 voters means computing 24,100 × 0.76, or 120.5 × 152, by hand with five-digit numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    The mayor's sentence adds a step: “half the undecided voters” turns the 20 undecided residents into 10 more yes votes, giving 152 before any scaling. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table gives Yes, No and Undecided counts, and the half must be taken of the Undecided row, not of the No row just above it. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Count the sample's expected yes votes, including half of the undecided, write them as a fraction of 200, and multiply that fraction by 24,100; estimate first to check the choice.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scaling up a sample
yes in samplesample size=yes in townvoters in town\frac{\text{yes in sample}}{\text{sample size}} = \frac{\text{yes in town}}{\text{voters in town}}
A random sample is a smaller group picked by chance. Its share of yes votes should be about the same as the whole town’s share. So the two fractions are set equal. Use it to scale a survey up to a whole group.

  1. Sort out what you know

    The town surveyed 200 people. 142 said Yes, 38 said No, and 20 were Undecided (not sure yet). The mayor thinks half of the undecided people will vote yes. There are 24,100 voters in all. You need how many of them will vote yes.

  2. Count the yes votes in the survey

    Half of the 20 undecided people is 20÷2=1020 \div 2 = 10. Add them to the yes votes.

    142+10=152142 + 10 = 152

    So 152 of the 200 people surveyed would vote yes. Watch out: take half of the Undecided row, not the No row. Half of 38 leads to 19,401, choice C.

  3. Find the share who vote yes

    152200=0.76\frac{152}{200} = 0.76

    That is 76% of the people surveyed. The sample was random, so about 76% of all the voters should vote yes too. Watch out: using only the 142 yes votes gives 17,111, choice A. Counting all 20 undecided people gives 19,521, choice D.

  4. Scale up to all the voters

    yes in samplesample size=yes in townvoters in town\frac{\text{yes in sample}}{\text{sample size}} = \frac{\text{yes in town}}{\text{voters in town}}

    152200=x24,100\frac{152}{200} = \frac{x}{24{,}100}

    x=0.76×24,100x = 0.76 \times 24{,}100

    Break it into easy parts. 75% is 34×24,100=18,075\frac{3}{4} \times 24{,}100 = 18{,}075, and 1% is 24,100÷100=24124{,}100 \div 100 = 241.

    x=18,075+241=18,316x = 18{,}075 + 241 = 18{,}316

    So the answer is B, 18,316.

Check your answer

Estimate. 76% is a little more than three fourths. Three fourths of 24,000 is 18,000. So the answer should be a little more than 18,000, and 18,316 fits. Also, 18,316÷24,100=0.7618{,}316 \div 24{,}100 = 0.76, the same share as the survey.

More questions like this

The original DOE question
Question 77 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 204, question 77. Open page 204 of the handbook (PDF)

Question 69 of 120

2024 Test B Math, question 23

Question

Each loaf of bread a baker makes requires 15\frac{1}{5} ounce of salt. How many loaves can be made if 40 ounces of salt are used?

  • A8
  • B20
  • C45
  • D200
Show the answer

Correct answer: D

Each loaf of bread requires 15\frac{1}{5} ounce of salt. Therefore, the baker can make 5 loaves of bread with 1 ounce of salt. Multiply to find the number of loaves that can be made with 40 ounces of salt:

5×40=2005\times 40=200

Question analysis

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

NYS standards
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Dividing a whole number by a fraction in a word problem is a grade 6 standard. This takes one step with a unit fraction, simpler than typical grade 7 rational number problems.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 3, 24 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out what you know

    Each loaf uses 15\frac{1}{5} ounce of salt. The baker uses 40 ounces of salt. You need how many loaves that makes. That means how many fifths of an ounce fit into 40 ounces.

  2. Decide to divide

    You are splitting the salt into equal groups of 1/5 ounce. Each group makes one loaf. So divide the total salt by the salt for one loaf.

    groups=total÷size of one group\text{groups} = \text{total} \div \text{size of one group}

    loaves=40÷15\text{loaves} = 40 \div \frac{1}{5}

    Watch out: do not multiply by 1/5. 40×15=840 \times \frac{1}{5} = 8, choice A, is a fifth of the salt, not a number of loaves.

  3. Divide by the fraction

    Each ounce holds 5 fifths, so 1 ounce makes 5 loaves. Dividing by 1/5 is the same as multiplying by 5.

    a÷1n=a×na \div \frac{1}{n} = a \times n

    40÷15=40×5=20040 \div \frac{1}{5} = 40 \times 5 = 200

    The salt makes 200 loaves. Watch out: keep both zeros. 20, choice B, drops one.

    So the answer is D, 200.

Check your answer

Multiply back. 200 loaves at 15\frac{1}{5} ounce each use 200×15=40200 \times \frac{1}{5} = 40 ounces. That is all the salt. Also, each loaf uses less than 1 ounce, so there should be more loaves than ounces. 200 is more than 40.

More questions like this

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

Question 70 of 120

2024 Test B Math, question 26

Question

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

  • A1.33
  • B6.00
  • C7.25
  • D10.67
Show the answer

Correct answer: D

Set up a proportion:

1 sind0.75 plunk=x sinds8 plunks\frac{1\text{ sind}}{0.75\text{ plunk}}=\frac{x\text{ sinds}}{8\text{ plunks}}

1(8)=0.75x1(8)=0.75x

8=75100x8=\frac{75}{100}x

8=34x8=\frac{3}{4}x

8(43)=x8\left(\frac{4}{3}\right)=x

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

1023=x10\frac{2}{3}=x

10.67≈x10.67\approx x

Question analysis

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

NYS standards
  • 6.RP.3d converting measurement units
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Converting units with ratio reasoning is grade 6 content, and made-up unit names are a familiar SHSAT twist. Working with a decimal rate and rounding to hundredths put it at about the level of typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    “1 sind = 0.75 plunk” says a sind is smaller than a plunk, so 8 plunks hold more than 8 sinds; seeing that decides between dividing and multiplying by 0.75. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    8 ÷ 0.75 is easiest as 8 × 4/3 = 32/3, and the repeating decimal 10.666… must then be rounded to the nearest hundredth. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
a, b, c and d are numbers or letters, and b and d are not zero. When two fractions are equal, multiply each top by the other bottom. The two products are equal. It works because it is the same as multiplying both sides by b and by d.
Dividing by a fraction
a÷bc=a×cba \div \frac{b}{c} = a \times \frac{c}{b}
a is any number, and b over c is the fraction you divide by. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down. It asks how many of that fraction fit into a.

  1. Sort out what you know

    Sinds and plunks are made-up units, like inches and feet. One sind equals 0.75 plunk, so a sind is smaller than a plunk. You need how many sinds equal 8 plunks. Round to the nearest hundredth, two places after the decimal point.

  2. Decide if the answer is more than 8

    A sind is only 3/4 of a plunk. So it takes more than 8 sinds to make 8 plunks. The answer must be more than 8. Watch out: 8×0.75=68 \times 0.75 = 6, choice B, goes the wrong way. That is the number of plunks in 8 sinds.

  3. Set up a proportion

    A proportion says that two ratios are equal. Put sinds on top and plunks on the bottom on both sides. Let x be the number of sinds.

    10.75=x8\frac{1}{0.75} = \frac{x}{8}

    ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c

    1×8=0.75×x1 \times 8 = 0.75 \times x

    So 0.75x=80.75x = 8.

  4. Divide by 0.75

    0.75 is the same as 34\frac{3}{4}. Divide both sides by it. To divide by a fraction, multiply by its reciprocal.

    a÷bc=a×cba \div \frac{b}{c} = a \times \frac{c}{b}

    x=8÷34=8×43=323=10.666…x = 8 \div \frac{3}{4} = 8 \times \frac{4}{3} = \frac{32}{3} = 10.666\ldots

    Watch out: 1.33, choice A, is the number of sinds in just 1 plunk. Keep going to 8 plunks.

  5. Round to the nearest hundredth

    Look at the third digit after the point, the thousandths digit. It is 6, which is 5 or more. So round the hundredths digit up, from 6 to 7.

    10.666…≈10.6710.666\ldots \approx 10.67

    So the answer is D, 10.67.

Check your answer

Multiply back. Each sind is 0.75 plunk, so 10.67 sinds are 10.67×0.75≈8.0010.67 \times 0.75 \approx 8.00 plunks. That matches the 8 plunks. The answer is also more than 8, as it must be.

More questions like this

The original DOE question
Question 83 as printed on page 206 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 206, question 83. Open page 206 of the handbook (PDF)

Question 71 of 120

2024 Test B Math, question 29

Question

Jar Q contains 12 balls, and Jar R contains 8 balls. The probability of drawing a yellow ball at random from Jar Q is 13\frac{1}{3}. The probability of drawing a yellow ball at random from Jar R is 34\frac{3}{4}. All the balls are poured into an empty jar, Jar S. What is the probability of drawing a yellow ball at random from Jar S?

  • A14\frac{1}{4}
  • B13\frac{1}{3}
  • C12\frac{1}{2}
  • D47\frac{4}{7}
Show the answer

Correct answer: C

Jar Q contains 12 balls, and the probability of drawing a yellow ball at random is 13\frac{1}{3}.

13×44=412\frac{1}{3}\times\frac{4}{4}=\frac{4}{12}

There are 4 yellow balls in Jar Q.

Jar R contains 8 balls, and the probability of drawing a yellow ball at random is 34\frac{3}{4}.

34×22=68\frac{3}{4}\times\frac{2}{2}=\frac{6}{8}

There are 6 yellow balls in Jar R.

Jar S contains all the balls from Jar Q and Jar R.

12+8=2012+8=20

There are 20 balls in Jar S.

4+6=104+6=10

There are 10 yellow balls in Jar S.

1020÷1010=12\frac{10}{20}\div\frac{10}{10}=\frac{1}{2}

The probability of drawing a yellow ball at random from Jar S is 12\frac{1}{2}.

Question analysis

Two jars hold different numbers of balls and give different chances of drawing a yellow ball, and then everything is poured into a third jar. Students find the probability of drawing yellow from the combined jar. The tricky part is that the jars are different sizes, so their probabilities cannot be merged without first counting the yellow balls.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 probability tasks usually give the counts and ask for a fraction. Here each jar's probability must be turned back into a count of yellow balls before the jars are pooled, a reverse step that typical classwork rarely asks for.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 6, 66 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light; the hurdle is seeing that chances from jars of different sizes cannot be combined as fractions. Every wrong choice works on 1/3 and 3/4 directly, by multiplying them, by keeping one, or by adding tops and bottoms, while the right path runs through counts of balls. Asking how many yellow balls each jar holds before anything is poured avoids all three.

  1. Turning words into math moderate Understanding the question

    Each jar's probability, 1/3 of 12 balls and 3/4 of 8 balls, has to be turned back into a count of yellow balls, 4 and 6, before the two jars can be combined. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    The straight-across 4/7, about 0.57, sits just above the right 1/2, and both lie between the jars' chances of 1/3 and 3/4, so a rough estimate does not rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Turn each jar's probability into a number of yellow balls by multiplying it by that jar's total, then divide all the yellow balls by all the balls.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P(yellow)=yellow ballsall ballsP(\text{yellow}) = \frac{\text{yellow balls}}{\text{all balls}}
P(yellow) is the probability, or chance, of drawing a yellow ball. It is a number from 0 to 1. The top counts the yellow balls. The bottom counts all the balls. It works because every ball is equally likely to be drawn.

  1. Sort out what you know

    Jar Q holds 12 balls. The probability, or chance, of drawing yellow from it is 13\frac{1}{3}.

    Jar R holds 8 balls. The chance of drawing yellow from it is 34\frac{3}{4}.

    All the balls are poured into Jar S. You need the chance of drawing yellow from Jar S.

  2. Count the yellow balls in Jar Q

    The jars are different sizes, so you can’t mix their chances directly. Count the yellow balls instead. The chance tells what fraction of the balls are yellow. So multiply it by the number of balls.

    yellow balls=13×12=4\text{yellow balls} = \frac{1}{3} \times 12 = 4

    Jar Q has 4 yellow balls.

  3. Count the yellow balls in Jar R

    Do the same for Jar R. Three fourths of its 8 balls are yellow.

    yellow balls=34×8=6\text{yellow balls} = \frac{3}{4} \times 8 = 6

    Jar R has 6 yellow balls.

  4. Count the balls in Jar S

    Jar S gets every ball from both jars. Add all the balls. Then add the yellow balls.

    12+8=2012 + 8 = 20

    4+6=104 + 6 = 10

    Jar S has 20 balls, and 10 of them are yellow.

  5. Find the chance for Jar S

    Use the probability formula with the Jar S counts.

    P(yellow)=1020=12P(\text{yellow}) = \frac{10}{20} = \frac{1}{2}

    Watch out: Adding the tops and the bottoms, 1 + 3 over 3 + 4, gives 47\frac{4}{7}, choice D. Those numbers are parts of fractions, not balls. Multiplying 13×34\frac{1}{3} \times \frac{3}{4} gives 14\frac{1}{4}, choice A. That is the chance of two yellow draws, one from each jar.

    So the answer is C, 12\frac{1}{2}.

Check your answer

Half of the 20 balls is 10, and 4 + 6 = 10 yellow balls. That matches. Also, the mixed jar’s chance should land between the two jars’ chances, 13\frac{1}{3} and 34\frac{3}{4}. One half does.

More questions like this

The original DOE question
Question 86 as printed on page 207 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 207, question 86. Open page 207 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 72 of 120

2024 Test B Math, question 35

Question

The high temperature on Monday was x° F. On Tuesday, the high temperature was 84° F, which was a 5% increase from Monday’s high temperature. On Wednesday, the high temperature was 10% lower than it was on Monday. How much lower was the high temperature on Wednesday than the high temperature on Monday?

  • A4.0° F
  • B8.0° F
  • C8.4° F
  • D8.8° F
Show the answer

Correct answer: B

The temperature on Tuesday was 84° F, which was 5% higher than the temperature on Monday.

This can be represented by the equation 84=1.05x84=1.05x.

Dividing both sides of this equation by 1.05 gives x=80x=80.

Therefore, the temperature on Monday was 80° F.

Since the temperature on Wednesday was 10% lower than the temperature on Monday, the difference between the two temperatures would equal (10%)(80)=8.0∘ F(10\%)(80)=8.0^{\circ}\text{ F}.

Question analysis

Tuesday's high temperature is given along with its percent increase over Monday's, and Wednesday's high is a percent below Monday's. Students find how many degrees cooler Wednesday was than Monday. It tests percent increase and decrease, and the difficulty is working back to Monday, the base for both percents, rather than starting from Tuesday's number.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually apply a percent to a known amount. Here Monday's temperature must first be recovered from a percent increase and then a second percent applied to it, a reverse step most grade 7 students find unfamiliar.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 54 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns on noticing that both percents are measured from Monday, a temperature the stem never states. Students who take 10% of the only temperature given, 84, land on a listed choice just 0.4 degree off. Working back from 84 is forgiving: the exact 84 ÷ 1.05 = 80 and even the rougher 84 × 0.95 = 79.8 lead to the same choice, so the real test is not skipping Monday.

  1. Turning words into math moderate Understanding the question

    "84° F, which was a 5% increase from Monday’s high temperature" makes 84 the result of the increase, so Monday comes from solving 1.05x = 84, not from 5% of 84. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    The choices 8.0, 8.4 and 8.8 lie within a degree of each other, so only the base of the 10%, Monday's 80 rather than Tuesday's 84, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Much to keep track of minor Working it out

    Three days and two percents have to be kept apart: the 5% compares Tuesday with Monday, and the 10% compares Wednesday with Monday. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let Monday's temperature be x, write Tuesday's 84 degrees as 1.05 times x and solve, then take 10 percent of Monday's value.

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.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=old×(1+p100)\text{new} = \text{old} \times \left(1 + \frac{p}{100}\right)
old is the starting amount. p is the percent it goes up. new is the amount after. The 1 keeps all of the old amount, and p over 100 adds the extra part. So a 5% rise means multiply by 1.05.
Percent of a number
part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}
The whole is the amount the percent is taken of. The percent tells how many out of every 100. The part is the piece you get, here a number of degrees.

  1. Sort out the question

    Monday’s high is unknown. Call it x degrees. Tuesday’s high was 84°F, which is 5% more than Monday. Wednesday’s high was 10% less than Monday.

    You need how many degrees lower Wednesday was than Monday. Both percents compare with Monday. So find Monday first.

  2. Write an equation for Tuesday

    Tuesday is Monday plus 5% of Monday. That is 105% of Monday. Use the percent increase formula with p = 5.

    84=x×(1+5100)=1.05x84 = x \times \left(1 + \frac{5}{100}\right) = 1.05x

  3. Solve for Monday

    Divide both sides by 1.05 to undo the multiplying.

    x=84÷1.05=80x = 84 \div 1.05 = 80

    Monday’s high was 80°F. Watch out: 84 is the result of the increase, not the start. Adding 5% to 84 gives about 88.2°F. Then 10% of that is about 8.8, choice D.

  4. Find Wednesday’s drop

    Wednesday was 10% lower than Monday. So the drop is 10% of Monday’s 80°F.

    part=10100×80=8\text{part} = \frac{10}{100} \times 80 = 8

    Wednesday was 8 degrees lower. Watch out: Taking 10% of Tuesday’s 84 gives 8.4, choice C. The 10% is of Monday’s high. Also, 84 − 80 = 4, choice A, is Tuesday’s rise, not Wednesday’s drop.

    So the answer is B, 8.0°F.

Check your answer

Work forward from 80°F. 5% of 80 is 4, and 80 + 4 = 84. That matches Tuesday. Then 80 − 8 = 72, so Wednesday’s high was 72°F. That is 8 degrees below Monday.

More questions like this

The original DOE question
Question 92 as printed on page 209 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 92. Open page 209 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 73 of 120

2024 Test B Math, question 37

Question

Raoul gave two musical performances. His second performance occurred 86 days after his first performance. If Raoul’s second performance was on a Tuesday, on what day of the week was his first performance?

  • ASunday
  • BMonday
  • CWednesday
  • DThursday
Show the answer

Correct answer: A

The total number of days (86) is divided by the number of days in a week (7).

86÷7=12, with a remainder of 286\div 7=12\text{, with a remainder of }2

The 2 days are counted back from Tuesday, so the first performance was on a Sunday.

Question analysis

Raoul's second musical performance came 86 days after his first and fell on a Tuesday. Students work out the day of the week of the first performance. It tests division with remainders in a repeating weekly cycle, and the traps are counting in the wrong direction or miscounting the leftover days.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Dividing with a remainder is grade 6 arithmetic, but using the remainder to step through a seven-day cycle, and counting backward, is less routine than grade 7 classwork. It needs care with direction rather than new content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 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

Counting 86 days one at a time is hopeless in the time allowed, so the question rewards seeing that whole weeks return to the same weekday. Two decisions remain after that, and each has a listed wrong answer: which way to count, and whether Tuesday itself is one of the 2 days. Saying the days aloud, Monday is 1 back and Sunday is 2 back, avoids both.

  1. No obvious method moderate Working it out

    Nothing in the stem says to divide by 7; the key is that 84 of the 86 days are exactly 12 weeks, which leaves only 2 days to count. (Schoenfeld, 1985)

  2. Turning words into math minor Understanding the question

    "His second performance occurred 86 days after his first" puts the unknown day before the Tuesday, so the leftover days are counted backward. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Remove whole weeks from the 86 days, since each full week lands on the same weekday, then count the leftover days backward from Tuesday.

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.)
  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Weeks and leftover days
days=7×weeks+leftover\text{days} = 7 \times \text{weeks} + \text{leftover}
days is the total number of days. weeks is the number of full weeks inside it. leftover is the days that remain, from 0 to 6. The day of the week repeats every 7 days, so only the leftover days change it.

  1. Sort out the question

    The second show was on a Tuesday. It came 86 days after the first show. So the first show was 86 days before that Tuesday.

    You need the day of the week of the first show. So you count back 86 days from Tuesday.

  2. Take out the full weeks

    Counting back 86 days one at a time is slow. Here is a shortcut. Every 7 days, the same weekday comes back. So divide 86 by 7.

    86÷7=12 remainder 286 \div 7 = 12 \text{ remainder } 2

    86=7×12+286 = 7 \times 12 + 2

    That is 12 full weeks and 2 extra days. Going back 12 weeks from a Tuesday lands on a Tuesday again.

  3. Count back the 2 extra days

    Start on Tuesday and go back one day at a time. One day back is Monday. Two days back is Sunday.

    Watch out: Do not count Tuesday itself as one of the 2 days. That lands on Monday, choice B. And count backward, not forward. Counting forward 2 days gives Thursday, choice D.

    So the answer is A, Sunday.

Check your answer

Count forward to be sure. Start on a Sunday. After 84 days, which is 12 weeks, it is Sunday again. Two more days are Monday and Tuesday. That is day 86, a Tuesday. It matches.

More questions like this

The original DOE question
Question 94 as printed on page 210 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 94. Open page 210 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 74 of 120

2024 Test B Math, question 38

Question

Michael received a check for $213 for working 20 hours at his part-time job. The amount on the check was his total pay after $32 in taxes was deducted. What was Michael’s hourly pay rate before taxes were deducted?

  • A$9.05
  • B$10.65
  • C$12.25
  • D$21.35
Show the answer

Correct answer: C

Add the taxes to the amount of the check to find Michael’s total pay before taxes were deducted:

213+32=245213+32=245

Divide the total pay by the number of hours to find the hourly pay rate:

245÷20=12.25245\div 20=12.25

Question analysis

Michael's paycheck shows what he earned for 20 hours after taxes were taken out, and the tax amount is given. Students find his hourly pay rate before taxes. It tests unit rates in a money word problem, where students must decide what to do with the tax that was deducted.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Unit rate problems are grade 6 content, and this one needs one addition and one division with a decimal result. It is easier than typical grade 7 work, apart from deciding which amount is the pay before taxes.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 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

Two operations in the right order, adding the tax back and dividing by the hours, are all this takes. The danger is the word deducted, which pulls students toward subtracting the $32 a second time, and the phrase before taxes, which rules out the take-home rate from dividing the check alone. Checking that $12.25 × 20 − $32 gives back $213 confirms the answer.

  1. Turning words into math minor Understanding the question

    "The amount on the check was his total pay after $32 in taxes was deducted" makes $213 what remained, so the $32 is added back, not taken away. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Rebuild Michael's pay before taxes from the check and the tax that was taken out, then divide that total by the 20 hours worked.

Traps

  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Hourly pay rate
rate=total payhours\text{rate} = \frac{\text{total pay}}{\text{hours}}
rate is the pay for one hour. total pay is the money earned for all the hours. hours is the time worked. Dividing shares the total pay equally among the hours. Use it when the pay for each hour is the same.

  1. Sort out the question

    Michael’s check was for $213. That is what was left after $32 in taxes was deducted, which means taken away. He worked 20 hours.

    You need his hourly pay before taxes. So first find his full pay before the $32 was taken out.

  2. Add the taxes back

    The $32 was taken away from his full pay to leave $213. To undo taking away, add it back.

    213+32=245213 + 32 = 245

    His pay before taxes was $245. Watch out: The word “deducted” tempts you to subtract again. That gives (213−32)÷20=9.05(213 - 32) \div 20 = 9.05, choice A.

  3. Divide by the hours

    rate=24520=12.25\text{rate} = \frac{245}{20} = 12.25

    He earned $12.25 for each hour before taxes. Watch out: Dividing the $213 check by 20 gives $10.65, choice B. That is his pay for each hour after taxes.

    So the answer is C, $12.25.

Check your answer

Work forward. 12.25×20=24512.25 \times 20 = 245, so 20 hours pay $245. Take away $32 in taxes: 245−32=213245 - 32 = 213. That is the amount on the check.

More questions like this

The original DOE question
Question 95 as printed on page 210 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 95. Open page 210 of the handbook (PDF)

Question 75 of 120

2024 Test B Math, question 40

Question

The value of a particular car decreases at a constant rate. If the car is worth $15,000 three years after its original purchase date and $10,000 five years after its original purchase date, what was the value of the car on its original purchase date?

  • A$18,000
  • B$20,000
  • C$22,500
  • D$25,000
Show the answer

Correct answer: C

The value of the car decreased $5,000 in two years: $15,000 to $10,000.

5,000÷2=2,5005{,}000\div 2=2{,}500

Therefore, the value of the car decreases $2,500 each year. After the first 3 years, the car’s value had decreased $7,500.

So 7,500+15,000=22,5007{,}500+15{,}000=22{,}500.

Question analysis

A car loses value at a constant rate, and its worth three years and five years after purchase is given. Students find what the car was worth on the day it was bought. It tests rate of change and initial value in a linear relationship, and the tricky part is counting the right number of years when working back.

NYS standards
  • 8.F.4 modeling with linear functions
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Finding a starting value from two later values of a steady change is grade 8 linear-function work. A grade 7 student can reason with a yearly rate, but extending it back to the purchase date goes beyond typical grade 7 problems.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 10, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is a straight-line pattern read backward, and its numbers are friendly; the difficulty is counting years correctly. The $5,000 drop happens over 2 years, not 5, and the purchase lies 3 years before the $15,000 value, not 2. Each miscount gives a listed answer, so sketching a timeline from year 0 to year 5 before computing pays off.

  1. Turning words into math moderate Understanding the question

    The phrases "three years after its original purchase date" and "five years after" must become points at year 3 and year 5 on a timeline whose start, year 0, holds the unknown value. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three moves are chained: the $5,000 drop over 2 years, the $2,500 yearly rate, and three years of that rate added back to $15,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the yearly drop from the change between the two given years, then add that drop back once for each year between the purchase and year 3.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rate of change
rate=change in valuechange in time\text{rate} = \frac{\text{change in value}}{\text{change in time}}
change in value is how many dollars the value went up or down. change in time is how many years passed. rate is the change for one year. A constant rate means the change is the same every year.

  1. Sort out the question

    The car’s value “decreases at a constant rate.” That means the car loses the same number of dollars each year. Year 0 is the day it was bought.

    YearValue
    0?
    3$15,000
    5$10,000

    You need the value at year 0.

  2. Find the drop for one year

    From year 3 to year 5 is 2 years. In that time, the value fell by 15,000−10,000=5,00015{,}000 - 10{,}000 = 5{,}000 dollars.

    rate=5,0002=2,500\text{rate} = \frac{5{,}000}{2} = 2{,}500

    The car loses $2,500 each year. Watch out: Divide by the 2 years between the values, not by 5. Dividing by 5 leads to $18,000, choice A.

  3. Go back to year 0

    From year 0 to year 3 is 3 years. The car lost $2,500 in each of them. So add that amount back 3 times.

    15,000+3×2,500=15,000+7,500=22,50015{,}000 + 3 \times 2{,}500 = 15{,}000 + 7{,}500 = 22{,}500

    Watch out: Adding back only one $5,000 gives $20,000, choice B. That is the value at year 1, not year 0.

    So the answer is C, $22,500.

Check your answer

Count forward from $22,500, taking off $2,500 each year. Year 1 is $20,000. Year 2 is $17,500. Year 3 is $15,000. Year 4 is $12,500. Year 5 is $10,000. Both given values match.

More questions like this

The original DOE question
Question 97 as printed on page 210 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 97. Open page 210 of the handbook (PDF)

Question 76 of 120

2024 Test B Math, question 43

Question

A community garden had 50 members in 2001. In each of the next three years, the number of members increased by 40% over the number of members from the previous year. How many members were there in 2004? (Round to the nearest whole number.)

  • A70
  • B98
  • C110
  • D137
Show the answer

Correct answer: D

To find out how many members there were in 2002, calculate 40% of 50 and then add it to 50.

50×0.40=2050\times 0.40=20

20+50=70 members20+50=70\text{ members}

To find out how many members there were in 2003, calculate 40% of 70 and then add it to 70.

70×0.40=2870\times 0.40=28

28+70=98 members28+70=98\text{ members}

To find out how many members there were in 2004, calculate 40% of 98 and then add it to 98.

98×0.40=39.298\times 0.40=39.2

The result is 39.2, which is rounded to 39 to keep members to the nearest whole number.

39+98=137 members39+98=137\text{ members}

Question analysis

A community garden has 50 members in 2001, and membership grows by 40 percent each year over the year before. Students find the number of members in 2004, rounded to a whole number. It tests repeated percent increase, where each of the three rounds of growth must start from the previous year's total.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Percent increase is grade 7 content, and each step here is a routine one. Repeating it three times, each time on the new total, adds length and bookkeeping but no new idea, so it matches grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Compound growth is the whole question: each year's 40% is taken of a larger number, so the increases grow from 20 to 28 to about 39. Students who add 20 three times treat the growth as simple, and those who lose count of the years stop at an earlier total that is also listed. Multiplying by 1.4 three times, or checking that the third increase is the largest, separates the right count from the rest.

  1. Many steps moderate Working it out

    Three yearly increases have to be applied in turn, 50 to 70 to 98 to 137.2, each taking 40% of the new total rather than of the original 50. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The phrase "increased by 40% over the number of members from the previous year" sets a new base each year, the wording that separates compound from simple growth. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    The third year's 40% of 98 is 39.2, the first decimal in the chain, and the rounding instruction applies only to the final count. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply the membership by 1.4 once for each year from 2002 through 2004, carrying the unrounded value forward, and round only at the end.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}
old is last year’s number of members. p is the percent increase, here 40. new is this year’s number. You add p percent of the old amount to the old amount. Each year, the new number becomes next year’s old number.

  1. Sort out the question

    In 2001 the garden had 50 members. In each of the next three years, 2002, 2003 and 2004, the members grew by 40%.

    The 40% is of last year’s members, so the base changes every year. You need the number for 2004, rounded to the nearest whole number.

  2. Find the members in 2002

    40% of 50 is 20. Add it to 50.

    new=50+40100×50=50+20=70\text{new} = 50 + \frac{40}{100} \times 50 = 50 + 20 = 70

    There were 70 members in 2002.

  3. Find the members in 2003

    Now 70 is the old number. Take 40% of 70, not of 50.

    new=70+40100×70=70+28=98\text{new} = 70 + \frac{40}{100} \times 70 = 70 + 28 = 98

    There were 98 members in 2003.

  4. Find the members in 2004

    Now 98 is the old number.

    new=98+40100×98=98+39.2=137.2\text{new} = 98 + \frac{40}{100} \times 98 = 98 + 39.2 = 137.2

    Round to the nearest whole number. 137.2 is closer to 137 than to 138, so there were 137 members.

    Watch out: Adding 20 members every year uses 40% of 50 each time. That gives 110, choice C. Stopping after one or two years gives 70 or 98, choices A and B.

    So the answer is D, 137.

Check your answer

A 40% rise is the same as multiplying by 1.4. So 50×1.4×1.4×1.4=50×2.744=137.250 \times 1.4 \times 1.4 \times 1.4 = 50 \times 2.744 = 137.2, which rounds to 137. It matches.

More questions like this

The original DOE question
Question 100 as printed on page 211 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 100. Open page 211 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 77 of 120

2024 Test B Math, question 45

Question

A box contains a total of 30 paper clips with colors in the ratio red:white:blue=2:5:3\text{red}:\text{white}:\text{blue}=2:5:3. Suppose that one paper clip of each color is removed from the box and not replaced. What is the probability that the next paper clip chosen will be blue?

  • A527\frac{5}{27}
  • B27\frac{2}{7}
  • C827\frac{8}{27}
  • D310\frac{3}{10}
Show the answer

Correct answer: C

The ratio is red:white:blue=2:5:3\text{red}:\text{white}:\text{blue}=2:5:3. Find the number of each color in a box of 30 paper clips.

2x+3x+5x=302x+3x+5x=30

10x=3010x=30

x=3x=3

Red: 2x=2(3)=6\text{Red: }2x=2(3)=6

White: 5x=5(3)=15\text{White: }5x=5(3)=15

Blue: 3x=3(3)=9\text{Blue: }3x=3(3)=9

If one paper clip of each color is removed (3 total), there will be 8 blue paper clips out of 27 in the box. The probability that the next paper clip chosen will be blue is 827\frac{8}{27}.

Question analysis

A box of 30 paper clips has red, white and blue clips in the ratio 2 to 5 to 3, and one clip of each color is taken out. Students find the probability that the next clip drawn is blue. It tests ratios and simple probability; the ratio gives shares rather than counts, and the removals change both the blue count and the total.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Splitting a total by a three-part ratio and finding a simple probability are both familiar skills. Combining them, then adjusting the counts after the removals, adds steps but stays within typical grade 7 multi-step work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 5, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains a ratio problem into a probability problem, with a small change in between. Students who skip the removal get 3/10, a choice that is almost the right value and looks natural because it matches the ratio. The cleanest route lists the actual counts, 6, 15 and 9, takes one from each, and writes blue's 8 over the remaining 27.

  1. Many steps moderate Working it out

    The ratio 2:5:3 must become counts of 6, 15 and 9 clips, one clip of each color must come out, and only then can blue's 8 be put over 27. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    3/10 = 0.3 and 8/27, about 0.296, are almost equal, so an estimate cannot separate them; only subtracting the three removed clips does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Turning words into math minor Understanding the question

    The removal of one clip of each color, not replaced, happens before the draw, so it changes both the blue count and the total in the box. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Divide 30 by the sum of the ratio parts to find each color's count, take one clip away from each color, then put the blue count over the new total.

Traps

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Size of one ratio part
one part=totalsum of the ratio numbers\text{one part} = \frac{\text{total}}{\text{sum of the ratio numbers}}
The ratio numbers tell how many parts each color gets. Their sum is the number of parts in all. total is the number of clips. Dividing gives the size of one part. Each color is then its ratio number times one part.
Probability
P(blue)=blue clipsall clipsP(\text{blue}) = \frac{\text{blue clips}}{\text{all clips}}
P(blue) is the chance of picking a blue clip. The top counts the blue clips in the box. The bottom counts all the clips in the box. It works because each clip is equally likely to be picked.

  1. Sort out the question

    The box has 30 clips. The ratio red : white : blue = 2 : 5 : 3 means that for every 2 red clips, there are 5 white clips and 3 blue clips.

    Then one clip of each color is taken out and not put back. You need the chance that the next clip picked is blue.

  2. Find the size of one part

    The ratio has 2+5+3=102 + 5 + 3 = 10 parts in all. The 30 clips are shared among those 10 parts.

    one part=3010=3\text{one part} = \frac{30}{10} = 3

    Each part is 3 clips.

  3. Count the clips of each color

    Multiply each ratio number by the size of one part.

    red=2×3=6\text{red} = 2 \times 3 = 6

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

    blue=3×3=9\text{blue} = 3 \times 3 = 9

    Check: 6 + 15 + 9 = 30 clips. That matches the total.

  4. Take out one of each color

    One red, one white and one blue clip come out. Each color drops by 1, and the total drops by 3.

    ColorBeforeAfter
    Red65
    White1514
    Blue98
    All3027

  5. Find the chance of blue

    P(blue)=827P(\text{blue}) = \frac{8}{27}

    Watch out: 930=310\frac{9}{30} = \frac{3}{10}, choice D, ignores the 3 clips taken out. 527\frac{5}{27}, choice A, is the chance of red. Taking 1 from each ratio number, as if the box held only 10 clips, gives 27\frac{2}{7}, choice B.

    So the answer is C, 827\frac{8}{27}.

Check your answer

Add the clips left: 5 + 14 + 8 = 27. That matches. Also, 827\frac{8}{27} is about 0.296, a bit less than blue’s share of the full box, 0.3. That makes sense, since one blue clip came out.

More questions like this

The original DOE question
Question 102 as printed on page 212 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 102. Open page 212 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 78 of 120

2024 Test B Math, question 47

Question

Three different colors of paint are available to paint the window frame, the door, and the wall of a house. If no two parts of the house may be painted the same color, how many different ways are there to paint the three parts of the house?

  • A3
  • B6
  • C9
  • D27
Show the answer

Correct answer: B

One method of solving the problem is to use a table to represent the sample space. Let the three colors be red, blue, and green.

PAINT COLORS

Window Frame Door Wall
Red Blue Green
Red Green Blue
Blue Red Green
Blue Green Red
Green Red Blue
Green Blue Red

There are 6 possible combinations of colors for the window frame, the door, and the wall where no two parts are painted the same color.

Question analysis

Three paint colors are available for a house's window frame, door and wall, and no two of the three parts may be painted the same color. Students count the different ways to paint them. It tests counting arrangements with a sample space, and the rule against repeating a color is what shapes the count.

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Counting arrangements with an organized list is grade 7 work, and this one has only three parts and three colors. It is typical practice, with the no-repeat rule as the one condition to keep in mind.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 8, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

With three colors and three parts, the six arrangements can be listed in well under a minute, so the question rewards either a quick organized list or the counting principle. The main loss comes from ignoring the rule that no two parts match, which turns 3 × 2 × 1 into 3 × 3 × 3. The other wrong choices simply combine the two 3s in the stem.

  1. Something you must already know minor Working it out

    The counting principle is assumed: 3 color choices for the window frame, 2 left for the door, then 1 for the wall, multiplied together. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The rule that "no two parts of the house may be painted the same color" means each color is used exactly once, which removes one choice at every step. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Count the color choices for the window frame, then for the door with one color already used, then for the wall, and multiply; or list the arrangements.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    There are 3 colors and 3 parts to paint: the window frame, the door and the wall. No two parts may be the same color. So each color is used exactly once.

    You need the number of different ways to paint the three parts.

  2. Count the choices for each part

    Paint the parts in order. Each color you use is gone for the next part.

    • Window frame: any of the 3 colors, so 3 choices.
    • Door: one color is used, so 2 choices are left.
    • Wall: only 1 color is left, so 1 choice.

  3. Multiply the choices

    Use the counting principle. Each choice for the window frame goes with each choice for the door, and so on. So you multiply.

    total ways=3×2×1=6\text{total ways} = 3 \times 2 \times 1 = 6

    Watch out: 3×3×3=273 \times 3 \times 3 = 27, choice D, lets a color be used twice. 3×3=93 \times 3 = 9, choice C, just multiplies the numbers in the question.

    So the answer is B, 6.

Check your answer

List them. Use R for red, B for blue and G for green, in the order frame, door, wall: RBG, RGB, BRG, BGR, GRB and GBR. That is 6 ways, and each one uses all three colors once.

More questions like this

The original DOE question
Question 104 as printed on page 212 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 104. Open page 212 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 79 of 120

2024 Test B Math, question 50

Question

Chris has 1 red shirt, 2 white shirts, and 3 black shirts. Chris picks a shirt at random. Without putting the first shirt back, he picks a second shirt at random. What is the probability that the first shirt is black and the second shirt is red?

  • A112\frac{1}{12}
  • B110\frac{1}{10}
  • C15\frac{1}{5}
  • D710\frac{7}{10}
Show the answer

Correct answer: B

Use the joint probability formula: P (black shirt first) • P (red shirt second, given black shirt first).

Initially there are 3 black shirts out of 6 total shirts, so the probability of picking a black shirt first is 36\frac{3}{6}.

Once a black shirt is picked first, there is 1 red shirt out of 5 total shirts remaining.

So the probability of picking a red shirt second, given that a black shirt was picked first, is 15\frac{1}{5}.

Then P (black shirt first) • P (red shirt second, given black shirt first)
=36∙15=330=110=\frac{3}{6}\bullet\frac{1}{5}=\frac{3}{30}=\frac{1}{10}.

Question analysis

Chris has red, white and black shirts and picks two at random, one after the other, without putting the first back. Students find the probability that the first shirt is black and the second is red. It tests the probability of two dependent events, where the second pick has one fewer shirt to choose from.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 compound probability is usually found with a list or tree of equally likely outcomes. Only two picks are involved, but adjusting the second pick for the shirt already taken goes beyond typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two dependent picks with small numbers make this a textbook question, but the choices punish each half-step. Students who leave the first shirt in the drawer get a fraction just below the right one, students who find only the second pick's chance stop at a listed value, and students who add the two chances get a probability far too large. Writing both fractions, 3/6 and 1/5, and multiplying is the whole method.

  1. Something you must already know moderate Working it out

    The rule for two dependent picks, the first pick's chance times the second's given the first, is assumed, along with the drop to 5 shirts for the second pick. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    1/10 and 1/12 differ only in the second pick's denominator, 5 shirts or 6, so estimating cannot separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Turning words into math minor Understanding the question

    "Without putting the first shirt back" sets the second pick's total at 5 shirts, and the order, black first then red, fixes which fraction comes first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the chance of black on the first pick, then the chance of red on the second pick with one shirt fewer, and multiply the two.

Traps

  • Without replacement: In a second pick there is one fewer item, which changes the probability.
  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of wanted outcomestotal number of outcomesP = \frac{\text{number of wanted outcomes}}{\text{total number of outcomes}}
P is the probability, the chance that something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row
P(A then B)=P(A)×P(B after A)P(A \text{ then } B) = P(A) \times P(B \text{ after } A)
A and B are two events that happen one after the other. P(A) is the chance of the first. P(B after A) is the chance of the second, counting only what is left after the first. You multiply because the second must also happen inside the share of times the first did.

  1. Sort out the question

    Chris has 1 red, 2 white and 3 black shirts. That is 6 shirts. He picks one and keeps it out. Then he picks a second shirt.

    You need the chance that the first shirt is black and the second is red. Both must happen, in that order.

  2. Find the chance of black first

    3 of the 6 shirts are black. Each shirt is equally likely to be picked.

    P(black first)=36P(\text{black first}) = \frac{3}{6}

    That is one half.

  3. Find the chance of red second

    A black shirt is now out and not put back. That leaves 5 shirts: 1 red, 2 white and 2 black.

    P(red second)=15P(\text{red second}) = \frac{1}{5}

    Watch out: The bottom is 5, not 6. Using 6 gives 112\frac{1}{12}, choice A, as if the shirt went back.

  4. Multiply the two chances

    Use the rule for events in a row. The second pick must go right too, inside the half of the time the first one did.

    P(black, then red)=36×15=330=110P(\text{black, then red}) = \frac{3}{6} \times \frac{1}{5} = \frac{3}{30} = \frac{1}{10}

    Watch out: Adding the chances gives 710\frac{7}{10}, choice D. Stopping at 15\frac{1}{5} gives choice C.

    So the answer is B, 110\frac{1}{10}.

Check your answer

Count every ordered pair of picks. There are 6 choices for the first and 5 for the second, so 6×5=306 \times 5 = 30 pairs. Pairs with black then red: 3 black shirts times 1 red shirt is 3. So the chance is 330=110\frac{3}{30} = \frac{1}{10}.

More questions like this

The original DOE question
Question 107 as printed on page 213 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 107. Open page 213 of the handbook (PDF)

Question 80 of 120

2024 Test B Math, question 54

Question

A rectangular kitchen floor is 12 feet long and 7.5 feet wide. It will be covered completely with new square tiles measuring 0.75 foot on each side. How many tiles will cover the floor?

  • A68
  • B78
  • C120
  • D160
Show the answer

Correct answer: D

First, determine the area of the kitchen floor in square feet by multiplying the length and width:

12×7.5=90 square feet12\times 7.5=90\text{ square feet}

Next, determine the area of each square tile:

0.75×0.75=34×34=9160.75\times 0.75=\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}

The number of tiles needed is found by dividing the area of the floor by the area of the tiles:

90÷916=90×169=16090\div\frac{9}{16}=90\times\frac{16}{9}=160

Question analysis

A rectangular kitchen floor measuring 12 feet by 7.5 feet will be covered with square tiles 0.75 foot on a side. Students find how many tiles are needed. It tests area and division with decimals, and working with a decimal side length while keeping lengths and areas apart is what makes it tricky.

NYS standards
  • 6.G.1 area of polygons
  • 6.NS.3 multi-digit decimal operations
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Area of a rectangle is grade 6 content, but here it combines with dividing by a decimal tile area, or by the tile's side along each wall. Three decimal steps and a tempting shortcut make it typical grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The floor's 90 square feet comes quickly; the trap is what to divide it by. A tile is described by its side, and dividing by 0.75 instead of 0.5625 gives a listed answer that looks reasonable. Counting tiles along each wall, 16 along the length and 10 along the width, avoids both the decimal division and the side-versus-area mistake.

  1. Error-prone arithmetic moderate Working it out

    The tile's area, 0.75 × 0.75 = 0.5625 square foot, must divide 90, a decimal division that is easier as 90 × 16/9 or as 16 tiles by 10 tiles. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The tiles are "square tiles measuring 0.75 foot on each side", a side length, which has to be squared before it can be compared with the floor's area in square feet. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find how many tiles fit along the length and how many along the width by dividing each by 0.75, then multiply the two counts.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle
A=l×wA = l \times w
A is the area, the flat space inside. l is the length and w is the width. Multiplying counts how many unit squares fit, row by row.
Area of a square
A=s2A = s^2
A is the area. s is the length of one side. s² means s × s. A square is a rectangle whose length and width are both s.

  1. Sort out the question

    The floor is a rectangle 12 feet long and 7.5 feet wide. Each tile is a square, 0.75 foot on each side. You need the number of tiles that cover the whole floor.

  2. Find the floor’s area

    Area tells how much flat space the tiles must cover. Use the rectangle formula.

    A=12×7.5=90A = 12 \times 7.5 = 90

    The floor is 90 square feet.

  3. Find one tile’s area

    The tile’s side is a length. You need its area to compare it with the floor’s area. Use the square formula.

    A=0.752=0.75×0.75=0.5625A = 0.75^2 = 0.75 \times 0.75 = 0.5625

    As a fraction, 0.75=340.75 = \frac{3}{4}, so the area is 34×34=916\frac{3}{4} \times \frac{3}{4} = \frac{9}{16} square foot.

  4. Divide the floor by one tile

    The number of tiles is the floor’s area divided by one tile’s area. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down.

    90÷916=90×169=10×16=16090 \div \frac{9}{16} = 90 \times \frac{16}{9} = 10 \times 16 = 160

    Watch out: Dividing 90 by 0.75, the side, gives 120, choice C. You must divide by the tile’s area. Multiplying 90 by 0.75 gives about 68, choice A.

    So the answer is D, 160.

Check your answer

Count the tiles in rows. Along the 12-foot length, 12÷0.75=1612 \div 0.75 = 16 tiles fit. Along the 7.5-foot width, 7.5÷0.75=107.5 \div 0.75 = 10 tiles fit. That makes 16×10=16016 \times 10 = 160 tiles.

More questions like this

The original DOE question
Question 111 as printed on page 214 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 111. Open page 214 of the handbook (PDF)

Question 81 of 120

2024 Test B Math, question 55

Question

Typically, an amusement park sells 3 times as many tickets on Sunday as on any weekday, and 4 times as many tickets on Saturday as on Sunday. If the number of tickets sold on Saturday and Sunday totaled 30,000, how many tickets are expected to be sold on the following Monday?

  • A2,000
  • B2,500
  • C3,000
  • D4,285
Show the answer

Correct answer: A

First, determine the number of tickets sold on Sunday, x, using the total number of tickets sold on Saturday and Sunday, 30,000, and the number of tickets sold on Saturday, 4x4x.

30,000=4x+x30{,}000=4x+x

30,000=5x30{,}000=5x

6,000=x6{,}000=x

Then determine the number of tickets expected to be sold on Monday, y, using the number of tickets sold on Sunday, 6,000.

6,000=3y6{,}000=3y

2,000=y2{,}000=y

The number of tickets expected to be sold on Monday is 2,000.

Question analysis

An amusement park sells 3 times as many tickets on Sunday as on a weekday and 4 times as many on Saturday as on Sunday. Given the weekend total, students find the expected ticket sales for the following Monday. It tests writing an equation from multiplicative comparisons, and the trap is comparing Saturday with the wrong day.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving a one-variable equation from a word problem is grade 7 work. Two linked multiplicative comparisons, with Saturday measured against Sunday rather than a weekday, add a layer of translation, but the equation itself stays simple.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 12, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is two comparisons chained through Sunday, and most wrong answers come from linking them incorrectly. Students who read Saturday's 4 times as 4 weekdays, or who leave Sunday out of the 30,000, find listed answers, one of them from a division that does not come out even. Letting a weekday be one share, so Sunday is 3 and Saturday 12, makes the weekend 15 shares and the check easy.

  1. Turning words into math moderate Understanding the question

    "4 times as many tickets on Saturday as on Sunday" compares Saturday with Sunday, not with a weekday, so Saturday is 4 × 3 = 12 weekdays' worth. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Three days have to be held in relation at once, a weekday, Sunday and Saturday, with the 30,000 covering only the last two. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let Sunday's sales be the unknown, write Saturday as 4 times it, solve the weekend equation for Sunday, then divide Sunday's figure by 3 for a weekday.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing multiplication
a×x=b means x=b÷aa \times x = b \text{ means } x = b \div a
x is the unknown number. a and b are known numbers. Division undoes multiplication. So if a times x is b, you find x by dividing b by a. This is the same as dividing both sides by a.

  1. Sort out the question

    Sunday sells 3 times as many tickets as a weekday. Saturday sells 4 times as many as Sunday. Saturday and Sunday together sell 30,000.

    Monday is a weekday. You need Monday’s tickets. Notice that Saturday is compared with Sunday, not with a weekday.

  2. Write an equation for the weekend

    Let x be Sunday’s tickets. Then Saturday’s tickets are 4x. Together they make 30,000.

    4x+x=30,0004x + x = 30{,}000

    5x=30,0005x = 30{,}000

  3. Solve for Sunday

    Use the rule for undoing multiplication. Divide both sides by 5.

    x=30,000÷5=6,000x = 30{,}000 \div 5 = 6{,}000

    Sunday sells 6,000 tickets. Saturday sells 4×6,000=24,0004 \times 6{,}000 = 24{,}000.

  4. Find Monday’s tickets

    Sunday sells 3 times as many as a weekday. Let y be Monday’s tickets.

    3×y=6,000 means y=6,000÷3=2,0003 \times y = 6{,}000 \text{ means } y = 6{,}000 \div 3 = 2{,}000

    Watch out: If you read Saturday as 4 times a weekday, you split 30,000 into 7 shares. That gives about 4,285, choice D. Dividing 30,000 by 12 leaves out Sunday and gives 2,500, choice B.

    So the answer is A, 2,000.

Check your answer

Think in shares. A weekday is 1 share. Sunday is 3 shares, and Saturday is 4×3=124 \times 3 = 12 shares. The weekend is 15 shares, and 30,000÷15=2,00030{,}000 \div 15 = 2{,}000. It matches.

More questions like this

The original DOE question
Question 112 as printed on page 215 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 112. Open page 215 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 82 of 120

2023 Test A Math, question 1

Question

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

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

Show the answer

Correct answer: 5

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

prt=yprt = y

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

100t=500100t = 500

t=5t = 5

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Simple interest is a grade 7 percent application, usually asked as how much interest builds up over a given time. This one gives the interest and asks for the time, a short reverse step with friendly numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know minor Working it out

    The stem says 4% simple interest annually but never says that each year adds the same 4% of the $2,500 deposit, $100, so the yearly amount has to come from memory. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The question gives the interest, $500, and asks for the time, so the familiar formula runs backward: 2,500 × 0.04 × t = 500 has to be solved for t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    As a grid-in it shows no choices, so 20, the interest as a percent of the deposit, can be gridded with nothing on the page to say it is not a number of years. (Enright, Morley & Sheehan, 2002)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest
I=P×r×tI = P \times r \times t
I is the interest, the extra money the account pays you. P is the deposit, the money you put in. r is the yearly rate as a decimal. t is the time in years. Simple means each year’s interest is figured on the deposit only, so every year adds the same amount.
Percent to decimal
p%=p100p\% = \frac{p}{100}
p is the percent. Percent means out of 100. So divide by 100 to write it as a decimal. For example, 4% is 4 ÷ 100, which is 0.04.

  1. Sort out the question

    Ms. Li puts $2,500 into an account. This first amount is called the deposit. Each year the account pays 4% simple interest. Interest is extra money the bank pays you for keeping your money there. You need the number of years it takes to earn $500 of interest. The answer is a number of years, not dollars and not a percent.

  2. Find one year’s interest

    First write 4% as a decimal. 4% means 4 out of 100.

    4%=4100=0.044\% = \frac{4}{100} = 0.04

    Now use the formula for 1 year, so t = 1.

    I=P×r×tI = P \times r \times t

    I=2,500×0.04×1=100I = 2{,}500 \times 0.04 \times 1 = 100

    The account earns $100 each year. Watch out: 4% is 0.04, not 0.4. Using 0.4 makes one year’s interest $1,000, and you would grid 0.5 years.

  3. Find how many years make $500

    Simple interest adds the same $100 every year. So the time t must make the interest $500.

    2,500×0.04×t=5002{,}500 \times 0.04 \times t = 500

    100×t=500100 \times t = 500

    t=500÷100=5t = 500 \div 100 = 5

    It takes 5 years. Watch out: $500 is 20% of $2,500. But 20 is a percent, not a number of years. Don’t grid 20. So you would grid 5.

Check your answer

Count up one year at a time: $100, $200, $300, $400, $500. That is 5 years. Or put t = 5 back into the formula: 2,500×0.04×5=5002{,}500 \times 0.04 \times 5 = 500. That matches the $500 in the question.

More questions like this

The original DOE question
Question 58 as printed on page 75 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 75, question 58. Open page 75 of the handbook (PDF)

Question 83 of 120

2023 Test A Math, question 8

Question

Jamel works at a computer store. He is paid an hourly rate plus a 15% commission on all computer products he sells. Last week, Jamel was paid $802.50 for working 30 hours and selling $1,250.00 worth of computer products. What is Jamel’s hourly rate?

  • A$20.50/hr
  • B$26.75/hr
  • C$33.00/hr
  • D$37.65/hr
Show the answer

Correct answer: A

First, determine the amount of Jamel’s weekly pay that is from commission:

$1,250.00×15%=$187.50\$1{,}250.00 \times 15\% = \$187.50

Next, subtract Jamel’s commission amount from his weekly pay:

$802.50−$187.50=$615.00\$802.50 - \$187.50 = \$615.00

Last, divide the result by 30 hours to find Jamel’s hourly rate:

$615.00÷30=$20.50\$615.00 \div 30 = \$20.50

Question analysis

Jamel earns an hourly wage plus a 15% commission on his sales, and one week he is paid $802.50 for 30 hours of work and $1,250 in sales. Students find his hourly rate. The question mixes a percent with a rate, and the commission has to be taken out of the total pay before dividing by the hours.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Commission is one of the percent applications named in the grade 7 standards, and the remaining work is a two-step equation with decimals. That matches a typical grade 7 multistep problem, with the order of the steps as the main demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 44 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Jamel's pay has two parts, and the question asks for only one of them, so the commission has to be removed before dividing by the hours. Students who divide the whole $802.50 by 30 get a believable hourly figure that still contains the commission. Checking the answer against the story, 30 × $20.50 + $187.50 = $802.50, confirms both parts.

  1. Turning words into math moderate Understanding the question

    The phrase "paid an hourly rate plus a 15% commission on all computer products he sells" makes $802.50 the sum of two parts, 30 hours of pay and 15% of $1,250. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps moderate Working it out

    Three operations in order: 15% of $1,250 is $187.50, $802.50 − $187.50 = $615.00, and $615.00 ÷ 30 = $20.50; dividing before subtracting gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    Every amount carries cents, and 15% of $1,250 and $615 ÷ 30 are worked by hand, so a misplaced decimal point shifts the hourly rate. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Take 15% of the $1,250 in sales to find the commission, remove that from the week's pay, and divide what is left by the 30 hours worked.

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.
  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Pay with commission
pay=rate×hours+commission\text{pay} = \text{rate} \times \text{hours} + \text{commission}
The rate is the money Jamel earns for each hour. Hours is how long he worked. The commission is extra money that is a percent of what he sells. His pay is these two amounts added together.

  1. Sort out the question

    Jamel’s pay has two parts. One part is an hourly rate, the money he earns for each hour. The other part is a commission. A commission is extra pay that is a percent of what he sells. He worked 30 hours and sold $1,250.00 of products. In all he got $802.50. You need his hourly rate.

  2. Find the commission

    His commission is 15% of what he sold. 15% means 15 out of 100, or 0.15.

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    part=0.15×1,250=187.50\text{part} = 0.15 \times 1{,}250 = 187.50

    His commission was $187.50. Watch out: take 15% of the $1,250 he sold, not of his $802.50 pay. Mixing them up leads to $37.65, choice D.

  3. Take the commission out of his pay

    The rest of his pay came from his hours.

    802.50−187.50=615.00802.50 - 187.50 = 615.00

    He earned $615.00 for 30 hours of work. Watch out: the word plus does not mean add here. Adding the $187.50 to $802.50 counts the commission twice. That leads to $33.00, choice C.

  4. Divide by the hours he worked

    Share the $615.00 equally over the 30 hours.

    615÷30=20.50615 \div 30 = 20.50

    Watch out: don’t divide all $802.50 by 30. That gives $26.75, choice B. That amount still includes the commission. So the answer is A, $20.50 per hour.

Check your answer

Work forward from the answer. 30 hours at $20.50 is 30×20.50=61530 \times 20.50 = 615 dollars. Add the commission: 615+187.50=802.50615 + 187.50 = 802.50. That is exactly Jamel’s pay for the week.

More questions like this

The original DOE question
Question 65 as printed on page 77 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 65. Open page 77 of the handbook (PDF)

Question 84 of 120

2023 Test A Math, question 17

Question

A customer wants to buy a pair of hiking boots.

  • The original price of the boots is $85.75.
  • The store is offering a 15% discount on all boot purchases.
  • The customer has a coupon for an additional 25% off the sale price.
  • The tax rate is 8.5%.

What is the final cost of the boots, including the tax, to the nearest cent?

  • A$55.82
  • B$59.08
  • C$59.32
  • D$63.17
Show the answer

Correct answer: C

First, determine the sale price after the 15% discount is applied.

Since the customer is saving 15%, this means they are paying 85% (100−15)(100 - 15), so multiply the original price by 85% (0.85) to find the discounted price:

$85.75×0.85≈$72.89\$85.75 \times 0.85 \approx \$72.89

Next, find the price after the 25% coupon has been applied:

Since the customer is saving 25% off the sale price, this means they are paying 75% (100−25)(100 - 25), so multiply the price found in the first step by 75% (0.75) to find the second discounted price:

$72.89×0.75≈$54.67\$72.89 \times 0.75 \approx \$54.67

Last, find the final cost after an 8.5% tax is added to the second discounted price.

Since the customer is paying 8.5% tax, this means they are paying 108.5% (100+8.5)(100 + 8.5), so multiply the second discounted price by 108.5% (1.085) to find the final cost:

$54.67×1.085≈$59.32\$54.67 \times 1.085 \approx \$59.32

Question analysis

A pair of hiking boots costs $85.75, goes on sale at 15% off, gets an extra 25% off with a coupon, and is then taxed at 8.5%. Students find the final cost to the nearest cent. Each percent applies to the price left after the step before, so the discounts cannot simply be added, and the decimals call for careful multiplication.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 A discount followed by sales tax is standard grade 7 percent work. This item stacks two discounts and a tax, with cents to keep at every stage, so it takes about twice the steps and decimal multiplication of a typical problem.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 4, 58 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question tests whether students treat successive percents as successive multipliers, and then holds them to three long multiplications. Adding the 15% and 25% into one discount is the tempting shortcut, and a tax read as a dollar amount is the other. The key rounds to the cent at each stage, as a register would; carried exactly, the result is $59.31, so the nearest choice, $59.32, still stands.

  1. Error-prone arithmetic major Working it out

    Three products of cents and decimals are worked by hand: 85.75 × 0.85 = 72.8875, then 72.89 × 0.75 = 54.6675, then 54.67 × 1.085 ≈ 59.32, each rounded to the cent. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    The phrase "an additional 25% off the sale price" means 25% of the already discounted $72.89, not of the $85.75 original, so the two discounts multiply rather than add. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Slow to do carefully moderate Choosing the answer

    A reliable method needs three long decimal multiplications without a calculator, well past the minute and a half a question allows, which tempts students to combine the percents. (Lu & Sireci, 2007)

Strategy: Multiply by 0.85 for the sale, by 0.75 for the coupon on the sale price, and by 1.085 for the tax, rounding to the cent at each stage as a register would.

Traps

  • Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Price after a discount
new price=price×(1−p100)\text{new price} = \text{price} \times \left(1 - \frac{p}{100}\right)
p is the percent off. A discount is money taken off the price. If you save p percent, you pay the rest, 100 − p percent. For 15% off you pay 85%, so multiply by 0.85.
Price with tax
total=price×(1+t100)\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)
The price is the cost before tax. t is the tax rate, a percent. Tax is extra money added on top. You pay the whole price, 100%, plus t percent more. So for 8.5% tax you multiply by 1.085.

  1. Sort out the question

    The boots cost $85.75 at first. The store takes 15% off. Then a coupon takes 25% off the sale price. The sale price is the price after the first discount. Then 8.5% tax is added. You need the final cost to the nearest cent. A cent is $0.01, the second place after the decimal point.

  2. Take off the 15% discount

    You save 15%, so you pay 85%. That is 0.85 of the price.

    new price=price×(1−p100)\text{new price} = \text{price} \times \left(1 - \frac{p}{100}\right)

    new price=85.75×(1−15100)=85.75×0.85=72.8875\text{new price} = 85.75 \times \left(1 - \frac{15}{100}\right) = 85.75 \times 0.85 = 72.8875

    A price can’t have part of a cent, so round to two decimal places. The third digit is 7, which is 5 or more, so round up.

    72.8875≈72.8972.8875 \approx 72.89

    The sign ≈ means about. The sale price is $72.89.

  3. Take 25% off the sale price

    The coupon takes 25% off $72.89, so you pay 75% of it.

    72.89×0.75=54.6675≈54.6772.89 \times 0.75 = 54.6675 \approx 54.67

    The price is now $54.67. Watch out: take the coupon off the sale price, not the first price. Adding 15% and 25% to get 40% off $85.75 leads to $55.82, choice A.

  4. Add the 8.5% tax

    You pay the price plus 8.5% more, which is 108.5% of it.

    total=price×(1+t100)\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)

    total=54.67×1.085=59.31695≈59.32\text{total} = 54.67 \times 1.085 = 59.31695 \approx 59.32

    Watch out: 8.5% is a percent of the price, not $8.50. Adding $8.50 gives $63.17, choice D.

  5. See why rounding each time matters

    A store rounds to the cent after each step, as you just did. The test’s answer is worked the same way. Keep every digit and round only at the end, and you get this:

    85.75×0.85×0.75×1.085=59.3122…≈59.3185.75 \times 0.85 \times 0.75 \times 1.085 = 59.3122\ldots \approx 59.31

    That is $59.31, which is not a choice. The first two roundings each went up by a tiny part of a cent. That was just enough to push the total from $59.31 to $59.32. So here, round at each step, as a store would. So the answer is C, $59.32.

Check your answer

Check the tax a second way. 8.5% of $54.67 is 0.085×54.67=4.646950.085 \times 54.67 = 4.64695, about $4.65. Then 54.67+4.65=59.3254.67 + 4.65 = 59.32. Also estimate: two discounts leave a bit less than two-thirds of $86, near $55. Tax adds under $5, so about $59 or $60 makes sense.

More questions like this

The original DOE question
Question 74 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 74. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 85 of 120

2023 Test A Math, question 18

Question

Sheila is saving money for her summer vacation. She starts the summer with a balance of $90.00 and plans to save 15% of her earnings each week. She earns the same amount each week. After 12 weeks, Sheila has saved a total of $472.59. How much money does Sheila earn each week?

  • A$212.55
  • B$262.55
  • C$302.55
  • D$312.55
Show the answer

Correct answer: A

Let x be the amount that Sheila needs to earn each week.

$90.00+0.15×12x=$472.59\$90.00 + 0.15 \times 12x = \$472.59

First, subtract 90 from both sides:

0.15×12x=$382.590.15 \times 12x = \$382.59

1.8x=$382.591.8x = \$382.59

Last, divide both sides by 1.8:

x=$212.55x = \$212.55

Question analysis

Sheila starts the summer with $90 and saves 15% of what she earns each week, earning the same amount every week, and after 12 weeks she has $472.59. Students find her weekly earnings. They must write an equation that separates the starting balance from the savings and then work with a percent and long decimals.

NYS standards
  • 7.EE.4a two-step equations
  • 7.RP.3 multistep ratio and percent problems
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 core grade 7 work. Here the coefficient combines a percent with 12 weeks, and the decimals are long, which adds steps but no new idea.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The story has three layers, a starting balance, a weekly saving rate and a 12-week total, and the unknown sits under the percent. Students lose the point by forgetting the $90 was already there, or by moving it the wrong way. Testing a choice is a dependable alternative: 15% of $212.55 is about $31.88, and 12 of those plus $90 gives $472.59.

  1. Turning words into math moderate Understanding the question

    The total of $472.59 includes the $90.00 she started with, and only 15% of each week's pay is saved, so the equation is 90 + 12 × 0.15x = 472.59. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic moderate Working it out

    Dividing $382.59 by 1.8 by hand means shifting the decimal point to get 3,825.9 ÷ 18, which comes out to exactly $212.55 only if every digit is placed correctly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Many steps minor Working it out

    Subtract the $90, then divide by 12 weeks and by 0.15, or by their product 1.8; doing the division before the subtraction gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Subtract the starting $90 from the final total, then divide by 12 weeks and by 0.15, or test each choice by taking 15% of it, multiplying by 12 and adding $90.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Savings over time
total=start+weekly savings×weeks\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}
The start is the money she had at the beginning. Weekly savings is what she adds each week. Weeks is how many weeks pass. Each week adds the same amount, so multiply. Then add what she started with.

  1. Sort out the question

    Sheila starts with $90.00. Each week she earns the same amount and saves 15% of it. After 12 weeks she has $472.59 in all. That total includes the $90 she started with. You need how much she earns each week, not how much she saves.

  2. Write an equation

    Call her weekly earnings x. She saves 15% of x each week. That is 0.15×x0.15 \times x.

    total=start+weekly savings×weeks\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}

    472.59=90+0.15x×12472.59 = 90 + 0.15x \times 12

  3. Take away the starting $90

    Only the money added during the 12 weeks came from her pay. Subtract 90 from both sides.

    0.15x×12=472.59−90=382.590.15x \times 12 = 472.59 - 90 = 382.59

    She saved $382.59 from her pay. Watch out: subtract the $90, don’t add it. Adding leads to $312.55, choice D. Forgetting the $90 leads to $262.55, choice B.

  4. Combine the numbers next to x

    0.15×12=1.80.15 \times 12 = 1.8

    1.8x=382.591.8x = 382.59

    Each week she saves 0.15 of one week’s pay. Over 12 weeks, that adds up to 1.8 times one week’s pay.

  5. Divide by 1.8

    Divide both sides by 1.8. First move the decimal point one place right in both numbers. Then you divide by the whole number 18.

    x=382.59÷1.8=3,825.9÷18=212.55x = 382.59 \div 1.8 = 3{,}825.9 \div 18 = 212.55

    Sheila earns $212.55 each week. So the answer is A, $212.55.

Check your answer

Work forward. 12 weeks of saving 15% is 1.8 weeks of pay: 1.8×212.55=382.591.8 \times 212.55 = 382.59. Add the $90 she started with: 382.59+90=472.59382.59 + 90 = 472.59. That is the total in the question.

More questions like this

The original DOE question
Question 75 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 75. Open page 80 of the handbook (PDF)

Question 86 of 120

2023 Test A Math, question 21

Question

A soccer coach purchased 15 pairs of cleats for team members and spent a total of $805.95, including tax. If the tax rate is 8%, what is the price of each pair of cleats before tax?

  • A$49.43
  • B$49.75
  • C$53.73
  • D$58.03
Show the answer

Correct answer: B

Since the total cost of $805.95 includes an 8% tax, this means that the cost of 15 pairs of cleats before tax was multiplied by 1.08 to find the total cost of the cleats including the tax. So, first, use division to find the cost of the 15 pairs of cleats before tax:

$805.95÷1.08=$746.25\$805.95 \div 1.08 = \$746.25

Then, divide that quotient by 15 to find the price of 1 pair of cleats:

$746.25÷15=$49.75\$746.25 \div 15 = \$49.75

Question analysis

A soccer coach pays $805.95, tax included, for 15 pairs of cleats, and the tax rate is 8%. Students find the price of one pair before tax. The total already includes the tax, so students must work backward from a percent increase, and a common shortcut lands very close to the right price.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3b unit rate problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Typical grade 7 tax problems add tax to a price. This one works backward from a total that already includes the tax, a reverse percent step many students find unfamiliar, and then shares the cost among 15 pairs.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
2 steps
Reading level
Question and choices: about grade 5, 33 words.
Corrected from the DOE handbook
  • A repeated word in the question 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

Removing a tax is harder than adding one, because the 8% was taken of an amount the problem never states. Taking 8% off the $805.95 total feels natural and lands within 32 cents of the answer, which is why this question is hard. A quick check exposes it: 15 pairs at $49.75 cost $746.25, and adding 8% of that gives exactly $805.95.

  1. Turning words into math moderate Understanding the question

    The phrase "spent a total of $805.95, including tax" makes $805.95 equal to 108% of the pre-tax cost, so the tax is removed by dividing by 1.08, not by taking 8% off. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    $49.43 and $49.75 are 32 cents apart, and taking 8% off the total lands on the lower one; no estimate separates them, only dividing by 1.08. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Error-prone arithmetic moderate Working it out

    805.95 ÷ 1.08 = 746.25 and 746.25 ÷ 15 = 49.75 are long decimal divisions by hand; dividing by 15 first, to 53.73, and then by 1.08 is no easier. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Treat the total as 108% of the pre-tax cost: divide $805.95 by 1.08, then divide by 15 pairs, and check by adding 8% back.

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.
  • 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

What was corrected from the DOE handbook, and why

  • Stem: The handbook’s question reads “what is the the price of each pair of cleats before tax?”, with “the” printed twice (p. 80, question 78). The site now reads “what is the price of each pair of cleats before tax?”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Price with tax
total=price×(1+t100)\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)
The price is the cost before tax. t is the tax rate, a percent. Tax is extra money added on top. You pay the whole price, 100%, plus t percent more. So for 8% tax you multiply by 1.08.
Undoing a multiplication
if a×b=c, then a=c÷b\text{if } a \times b = c \text{, then } a = c \div b
a is the unknown, b is the number it was multiplied by, and c is the result. Dividing by b undoes multiplying by b. Use it to work backward to the amount before tax.

  1. Sort out the question

    The coach bought 15 pairs of cleats, which are sports shoes. He paid $805.95 in all, and that includes 8% tax. Tax is extra money added to the price. You need the price of one pair before tax.

  2. Write the total with tax

    The tax is 8% of the cost before tax. So the coach paid 100% + 8% = 108% of that cost. 108% is 1.08. Call the pre-tax cost of all 15 pairs c.

    total=price×(1+t100)\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)

    c×1.08=805.95c \times 1.08 = 805.95

  3. Divide to remove the tax

    To undo multiplying by 1.08, divide by 1.08. Move both decimal points two places right first.

    c=805.95÷1.08=80,595÷108=746.25c = 805.95 \div 1.08 = 80{,}595 \div 108 = 746.25

    All 15 pairs cost $746.25 before tax. Watch out: don’t take 8% of $805.95 and subtract it. The tax was 8% of the smaller pre-tax cost. That way leads to $49.43, choice A.

  4. Find the price of one pair

    All 15 pairs cost the same, so share the cost equally.

    746.25÷15=49.75746.25 \div 15 = 49.75

    Watch out: dividing $805.95 by 15 without removing the tax gives $53.73, choice C. That price still includes the tax. So the answer is B, $49.75.

Check your answer

Work forward. 15 pairs at $49.75 cost 15×49.75=746.2515 \times 49.75 = 746.25 dollars. The tax is 0.08×746.25=59.700.08 \times 746.25 = 59.70 dollars. Then 746.25+59.70=805.95746.25 + 59.70 = 805.95, the total in the question.

More questions like this

The original DOE question
Question 78 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 78. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 87 of 120

2023 Test A Math, question 27

Question

A basket contains red balls, green balls, and white balls. There are 12 red balls in the basket. The probability of randomly choosing a red ball is 1 in 3. If the probability of randomly choosing a green ball is 1 in 4, how many green balls are in the basket?

  • A3
  • B8
  • C9
  • D16
Show the answer

Correct answer: C

First, use the information about the red balls to calculate the total number of balls in the basket:

13=12x\frac{1}{3} = \frac{12}{x} Cross multiply.
x=36x = 36

There are a total of 36 balls in the basket. The probability of selecting a green ball is 1 in 4. Use that information to calculate how many green balls are in the basket:

g36=14\frac{g}{36} = \frac{1}{4} Cross multiply.
4g=364g = 36 Apply the multiplicative inverse property; divide both sides of the equation by 4.
g=9g = 9

There are 9 green balls in the basket.

Question analysis

A basket holds red, green and white balls, including 12 red ones, and the chances of drawing red and green are 1 in 3 and 1 in 4. Students find how many green balls are in the basket. Getting the total from the red count is the step students skip when they apply a probability to the wrong number.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Simple probability is grade 6 content, but here it runs backward: a count and a probability give the total, which then gives another count. Two linked steps like this are typical of grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 46 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The basket's total is never stated, and that missing number is the whole problem: it has to be recovered from the red balls before the green probability means anything. Applying the green chance straight to the 12 red balls is the quick wrong move. Checking that 12 red and 9 green fit in 36 balls with 15 white left over confirms the setup.

  1. Turning words into math minor Understanding the question

    The sentence "The probability of randomly choosing a red ball is 1 in 3" has to become 12 = one third of the total, which makes the basket hold 36 balls. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Two linked steps: the red balls give the total, 12 × 3 = 36, and only then does the green probability give a count, 36 ÷ 4 = 9. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use the red balls to find the total, since 12 is one third of it, then take one fourth of that total for the green balls.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=balls of one colorall the ballsP = \frac{\text{balls of one color}}{\text{all the balls}}
P is the probability, the chance of picking that color. Each ball is equally likely to be picked. So the chance is the fraction of all the balls that have that color. A chance of 1 in 3 means one third of the balls.

  1. Sort out the question

    The basket has red, green and white balls. There are 12 red balls. The chance of picking red is 1 in 3. The chance of picking green is 1 in 4. You need the number of green balls. You don’t know the total number of balls yet, so find it first.

  2. Find the total number of balls

    1 in 3 means one third of all the balls are red. So 12 is one third of the total. Call the total number of balls b.

    12b=13\frac{12}{b} = \frac{1}{3}

    b=12×3=36b = 12 \times 3 = 36

    There are 36 balls in all.

  3. Find the number of green balls

    1 in 4 means one fourth of all 36 balls are green. Call the number of green balls g.

    g36=14\frac{g}{36} = \frac{1}{4}

    g=36÷4=9g = 36 \div 4 = 9

    Watch out: take one fourth of all 36 balls, not of the 12 red ones. 12÷4=312 \div 4 = 3 gives choice A. Multiplying 12 by 43\frac{4}{3} gives 16, choice D. So the answer is C, 9.

Check your answer

A 1 in 4 chance is smaller than a 1 in 3 chance. So there should be fewer green balls than red ones, and 9 is less than 12. Check: 936=14\frac{9}{36} = \frac{1}{4}. The rest, 36−12−9=1536 - 12 - 9 = 15, are white.

More questions like this

The original DOE question
Question 84 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 84. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 88 of 120

2023 Test A Math, question 30

Question

Lamel has a jar containing 6 red chips, 10 blue chips, and 4 yellow chips. If he removes one chip at random, what is the probability that it will not be red?

  • A45\frac{4}{5}
  • B710\frac{7}{10}
  • C310\frac{3}{10}
  • D15\frac{1}{5}
Show the answer

Correct answer: B

There are currently 20 chips in the jar (6+10+46 + 10 + 4). Of those chips, 14 are not red. So the probability of choosing a chip that is not red is 1420=710\frac{14}{20} = \frac{7}{10}.

Question analysis

In this probability word problem, Lamel's jar holds red, blue and yellow chips, and students find the chance that one chip drawn at random is not red. It tests counting the favorable outcomes against the total and writing the probability as a fraction. Missing the bold word not, or counting the wrong colors, leads to tempting wrong fractions.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Probability of a single event with equally likely outcomes is grade 6 content in New York. Students add three counts and form one fraction, and the only extra care is the bold word not, so it sits below grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-event probability with small whole numbers, so the bold not carries the question. A student who skims past it computes the chance of red, a tidy fraction that is waiting among the choices. The other wrong fractions come from removing the wrong color or keeping only one of the two colors that are not red. Adding blue and yellow first, then dividing by all 20 chips, avoids each of them.

  1. Turning words into math minor Understanding the question

    The bold not turns the question around: the chips that count are the 10 blue and 4 yellow ones, 14 in all, or equivalently 1 minus the red share of 6/20. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The count of 14 out of 20 matches no choice until it is reduced to 7/10, so a correct fraction left unsimplified can look missing from the list. (Enright, Morley & Sheehan, 2002)

Strategy: Add all the chips for the total, then count the chips that are blue or yellow, or subtract the chance of red from 1, and simplify the fraction.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=outcomes you wantall outcomesP = \frac{\text{outcomes you want}}{\text{all outcomes}}
P is the probability, the chance that something happens. The top counts the outcomes you want. The bottom counts every outcome that could happen. It works when each outcome is equally likely, like pulling one chip without looking.

  1. Sort out the question

    Lamel’s jar holds 6 red chips, 10 blue chips and 4 yellow chips. He takes one chip without looking. You need the probability that it is not red. That means the chip is blue or yellow.

  2. Count all the chips

    6+10+4=206 + 10 + 4 = 20

    There are 20 chips in all. That is the bottom of the fraction, because any of the 20 could come out.

  3. Count the chips that are not red

    The blue and yellow chips are the ones that are not red.

    10+4=1410 + 4 = 14

    So 14 chips are not red. Watch out: the question asks for not red. Counting the 6 red chips gives 620=310\frac{6}{20} = \frac{3}{10}, choice C. Counting only the 4 yellow chips gives 15\frac{1}{5}, choice D.

  4. Find and simplify the probability

    P=outcomes you wantall outcomesP = \frac{\text{outcomes you want}}{\text{all outcomes}}

    P=1420=14÷220÷2=710P = \frac{14}{20} = \frac{14 \div 2}{20 \div 2} = \frac{7}{10}

    Dividing the top and bottom by 2 puts the fraction in lowest terms. So the answer is B, 710\frac{7}{10}.

Check your answer

A chip is either red or not red, so the two chances must add up to 1. The chance of red is 620=310\frac{6}{20} = \frac{3}{10}. Then 310+710=1\frac{3}{10} + \frac{7}{10} = 1. It works.

More questions like this

The original DOE question
Question 87 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 87. Open page 83 of the handbook (PDF)

Question 89 of 120

2023 Test A Math, question 32

Question

Mei-Ling is one of 6 members of a committee. If 2 members of that committee are selected to go to a conference, how many of the possible pairs of members would include Mei-Ling?

  • A5
  • B6
  • C10
  • D12
Show the answer

Correct answer: A

Call the 6 members A, B, C, D, E, and M (for Mei-Ling). The possible combinations are AB, AC, AD, AE, AM, BC, BD, BE, BM, CD, CE, CM, DE, DM, and EM. Of those pairs, 5 include Mei-Ling.

Question analysis

This counting question describes a six-member committee sending two members to a conference and asks how many of the possible pairs include Mei-Ling. Students count combinations in which order does not matter. The traps are counting a pair twice, pairing Mei-Ling with herself, or multiplying the numbers in the question instead of listing the pairs.

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students count outcomes with organized lists and tree diagrams. Here the list of pairs is short and a single condition, that one named person belongs to the pair, filters it, which matches routine grade 7 counting work.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 8, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The count is small enough to list, so the question rewards a clear picture of what a pair is: two different people, in no particular order. The tempting errors count Mei-Ling as her own partner, count each pair twice, or multiply the two numbers in the stem. Writing the five pairs out, M with each of A through E, takes seconds and rules out all three.

  1. Turning words into math minor Understanding the question

    Pairs that include Mei-Ling are simply Mei-Ling with each of the other 5 members, a count of her possible partners, though the phrase “possible pairs” invites counting all 15 pairs. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Listing pairs by letter, as the explanation does with A to E and M, means tracking which pairs are already written, so that AM and MA are not both counted. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Label the members with letters and list every pair that contains Mei-Ling, remembering that a pair is two different people and that order does not matter.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of pairs in a group
pairs=n×(n−1)2\text{pairs} = \frac{n \times (n - 1)}{2}
n is the number of people in the group. Each person can pair up with the n − 1 others. Multiplying counts every pair twice, once from each partner’s side. So you divide by 2.

  1. Sort out the question

    The committee has 6 members, and Mei-Ling is one of them. Two members go to the conference together. That makes a pair. You don’t need every possible pair. You only need the pairs that include Mei-Ling.

  2. Find Mei-Ling’s possible partners

    A pair with Mei-Ling is Mei-Ling plus one other member. Call Mei-Ling M, and call the other members A, B, C, D and E. Her pairs are M with A, M with B, M with C, M with D and M with E. That is 5 pairs.

    Watch out: she can’t be paired with herself. Counting all 6 members as partners gives 6, choice B.

  3. Don’t count a pair twice

    Mei-Ling with A is the same pair as A with Mei-Ling. The same two people go either way. Watch out: counting both orders doubles the count to 10, choice C. Multiplying 6 × 2 = 12 gives choice D, but that does not count any pairs.

  4. Check against all the pairs

    Count every pair in the committee.

    pairs=n×(n−1)2\text{pairs} = \frac{n \times (n - 1)}{2}

    pairs=6×52=15\text{pairs} = \frac{6 \times 5}{2} = 15

    Each pair has 2 people, so the 15 pairs have 30 name spots. All 6 members get the same share: 30÷6=530 \div 6 = 5. So the answer is A, 5.

Check your answer

List all 15 pairs: AB, AC, AD, AE, AM, BC, BD, BE, BM, CD, CE, CM, DE, DM, EM. Now circle the ones with M in them: AM, BM, CM, DM, EM. There are exactly 5.

More questions like this

The original DOE question
Question 89 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 89. Open page 83 of the handbook (PDF)

Question 90 of 120

2023 Test A Math, question 35

Question

A number line with tick marks every 2 units, labeled −6, −4, −2, 0, 2, 4 and 6. Point J is the dot at −4 and point K is the dot at 6.

On the number line above, point L (not shown) is located on line segment JK so that JL=23LK\mathrm{JL} = \frac{2}{3}\mathrm{LK}. What is the position of point L?

  • A−2
  • B0
  • C2
  • D4
Show the answer

Correct answer: B

JK=6−(−4)=10\mathrm{JK} = 6 - (-4) = 10 units. Let x=JLx = \mathrm{JL}. Then LK=10−x\mathrm{LK} = 10 - x. Substitute these expressions in the given equation:

x=23(10−x)x = \frac{2}{3}(10 - x)

Solve for x:

x=23(10−x)x = \frac{2}{3}(10 - x) Apply the multiplicative inverse property; multiply both sides of the equation by 3.
3x=2(10−x)3x = 2(10 - x) Apply the distributive property; multiply the 2 by 10 and x.
3x=20−2x3x = 20 - 2x Apply the additive inverse property; add 2x2x to both sides of the equation.
5x=205x = 20 Apply the multiplicative inverse property; divide both sides of the equation by 5.
x=4x = 4

So, point L is 4 units to the right of point J: −4+4=0-4 + 4 = 0. Point L is located at 0 on the number line.

Question analysis

This number line question puts J at −4 and K at 6 and asks for the position of a point L on segment JK where JL is two thirds of LK. Students find the segment's length, divide it in a 2-to-3 ratio, and turn a distance into a coordinate. Reading the ratio backwards, or giving a distance instead of a position, leads to wrong answers.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Distances on a number line that crosses zero are grade 7 rational-number work, and splitting a length in a given ratio is grade 7 proportional reasoning. Turning a fraction of one part into a split, then into a position, makes a typical multistep task.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The ratio is stated between two unknown pieces, which is less familiar than a fraction of a whole, and the final step turns a length into a position. Students who stop at JL's length, read 2/3 the wrong way round, or use its 2 as a distance all reach listed points. Sketching five equal shares of 2 units from −4 to 6, two for JL and three for LK, makes the position plain.

  1. Turning words into math moderate Understanding the question

    JL = 2/3 LK compares the two pieces with each other, not a piece with the whole JK, so the 10 units must be split in the ratio 2 : 3, giving JL 4 and LK 6. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    J's position, −4, and K's, 6, are read from the number line, whose labeled ticks run every 2 units, and L itself is not drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    Three steps follow in order: the length JK = 6 − (−4), the length JL from the ratio, and finally L's position counted from J rather than from 0. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length of JK, split it into five equal shares with two for JL and three for LK, then count JL's length to the right from J.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
Subtract the position on the left from the position on the right. The answer is how many units apart the two points are. A distance is never negative.
Splitting a length in a ratio
one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}
A ratio a : b compares two pieces. If a length is split in the ratio a : b, think of it as a + b equal parts. The first piece gets a of them and the second gets b. Divide the whole length by a + b to find one part.

  1. Sort out the question

    Read the number line. Point J is at −4 and point K is at 6. Point L sits somewhere between them. JL is the length from J to L, and LK is the length from L to K. The rule says JL is 23\frac{2}{3} of LK. You need the position of L.

  2. Find the length of JK

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    K is on the right, so subtract J’s position from K’s.

    JK=6−(−4)=6+4=10\mathrm{JK} = 6 - (-4) = 6 + 4 = 10

    Taking away −4 is the same as adding 4. J and K are 10 units apart.

  3. Turn the rule into equal parts

    JL is 23\frac{2}{3} of LK. So for every 2 units in JL, LK has 3 units. That means JK is made of 2 + 3 = 5 equal parts. JL gets 2 parts and LK gets 3.

    one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}

    one part=102+3=2\text{one part} = \frac{10}{2 + 3} = 2

    Each part is 2 units long.

  4. Find the two lengths

    JL=2×2=4\mathrm{JL} = 2 \times 2 = 4

    LK=3×2=6\mathrm{LK} = 3 \times 2 = 6

    Watch out: JL is 23\frac{2}{3} of LK, not the other way round. Flipping the ratio makes JL 6 and puts L at 2, choice C.

  5. Find the position of L

    L is 4 units to the right of J. Start at J’s position, −4, and move 4 to the right.

    −4+4=0-4 + 4 = 0

    Watch out: 4 is the length of JL, not where L sits. Answering 4 is choice D. So the answer is B, 0.

Check your answer

Put L at 0 and test the rule. JL = 0 − (−4) = 4, and LK = 6 − 0 = 6. Then 23×6=4\frac{2}{3} \times 6 = 4, which is JL. Also 4 + 6 = 10, the whole length JK.

More questions like this

The original DOE question
Question 92 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 92. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 91 of 120

2023 Test A Math, question 36

Question

Josef and Mai divided some stamps between themselves. Josef got 60% of the stamps. If Josef received 500 more stamps than Mai, how many stamps did Josef receive?

  • A300
  • B800
  • C1,500
  • D2,500
Show the answer

Correct answer: C

Let j be the number of stamps Josef has. Then Mai has j−500j - 500 stamps. The total number of stamps is j+(j−500)=2j−500j + (j - 500) = 2j - 500. Since Josef has 60% of the stamps, set up an equation to solve for j:

j2j−500=60100\frac{j}{2j - 500} = \frac{60}{100} Cross multiply.
100j=60(2j−500)100j = 60(2j - 500) Apply the distributive property; multiply the 60 by 2j2j and 500.
100j=120j−30,000100j = 120j - 30{,}000 Apply the additive inverse property; subtract 120j120j from both sides of the equation.
−20j=−30,000-20j = -30{,}000 Apply the multiplicative inverse property; divide both sides of the equation by −20.
j=1,500j = 1{,}500

Question analysis

This percent word problem says Josef and Mai split some stamps, Josef got 60% of them, and he received 500 more than Mai; it asks how many stamps Josef received. Students must connect a difference in counts to a difference in percents. Stopping at the total, or taking a percent of 500, gives tempting wrong answers.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here neither share nor the total is known; students must link the 500-stamp gap to the gap between two percents before finding the total and then Josef's part.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 26 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is easy once the 500 is seen as the gap between two percents of one total; seeing that is the hard part. The tempting move is to apply the 60% to the only number given, the 500, which treats a difference as a whole. Students who do find the total must still take Josef's share of it, since the total waits among the choices. Checking that 1,500 and 1,000 differ by 500 confirms the answer.

  1. Turning words into math moderate Understanding the question

    The stem describes one split in two ways, Josef's 60% of the total and his 500-stamp lead over Mai, and the link, that the lead is 60% − 40% = 20% of the total, is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method moderate Working it out

    No equation is suggested: the quick route treats the 500 as 20 percentage points and scales up to 100%, while the handbook's route, j over 2j − 500, needs cross-multiplying and a negative coefficient. (Schoenfeld, 1985)

  3. Many steps minor Working it out

    Even the short route takes three moves, the 20% gap, the 2,500 total and Josef's 60% of it, and the total is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: If Josef has 60%, Mai has the other 40%, so the 500-stamp difference is a known percent of the total; use it to find the total, then take Josef's share.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the total amount, and the part is the piece you want. Percent means “out of 100,” so 60% is 60 out of every 100. Use it to find a part, or work backward to find the whole.

  1. Sort out the question

    Josef and Mai split some stamps. Josef got 60% of them. He also got 500 more stamps than Mai. You need how many stamps Josef got. You don’t know the total yet, so find it first.

  2. Find Mai’s percent

    All the stamps make 100%. Josef has 60%, so Mai has the rest.

    100%−60%=40%100\% - 60\% = 40\%

    Mai has 40% of the stamps.

  3. Find what the 500 stamps are

    Josef’s extra 500 stamps are the gap between his share and Mai’s share.

    60%−40%=20%60\% - 40\% = 20\%

    So 500 stamps are 20% of all the stamps.

  4. Find the total number of stamps

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    500=20100×whole500 = \frac{20}{100} \times \text{whole}

    whole=500÷20100=500×5=2,500\text{whole} = 500 \div \frac{20}{100} = 500 \times 5 = 2{,}500

    There are 2,500 stamps in all. Watch out: that is everyone’s stamps, choice D. Don’t stop here.

  5. Find Josef’s share

    part=60100×2,500=1,500\text{part} = \frac{60}{100} \times 2{,}500 = 1{,}500

    Watch out: taking 60% of 500 gives 300, choice A. The 500 is the gap between them, not the total. So the answer is C, 1,500 stamps.

Check your answer

Mai gets 40% of 2,500, which is 1,000 stamps. Josef’s 1,500 minus Mai’s 1,000 is 500, just as the question says. And 1,500 + 1,000 = 2,500, the whole amount.

More questions like this

The original DOE question
Question 93 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 93. Open page 84 of the handbook (PDF)

Question 92 of 120

2023 Test A Math, question 39

Question

A scientist mixed three chemicals, R, S, and T, in a glass container. The amount of R is 3 times the amount of S, and the amount of T is 16\frac{1}{6} the amount of S. What is the ratio of the amount of R to the amount of T?

  • A1:18
  • B2:1
  • C3:1
  • D18:1
Show the answer

Correct answer: D

First, set up the equations based on the information in the question.

R=3S\mathrm{R} = 3\mathrm{S}

T=16S\mathrm{T} = \frac{1}{6}\mathrm{S}

So, the ratio of R:T is

3S:16S3\mathrm{S}:\frac{1}{6}\mathrm{S}

Divide both sides by S to eliminate the variable. Then, multiply both sides by 6 to eliminate the fraction. The resulting ratio is 18:1.

Question analysis

This ratio question describes three chemicals mixed in a container: R is 3 times the amount of S, and T is one sixth of S. Students must combine the two relationships through S to find the ratio of R to T. Writing the ratio in the wrong order, or using only one of the relationships, produces the tempting wrong choices.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.1 ratio concepts and language
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 ratio problems often link two quantities through a third. Here R and T are each described relative to S, one as a multiple and one as a fraction, and combining them into one ratio is typical grade 7 work with an order trap.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains two comparisons through S, and the 1/6 makes the chain awkward to do with letters. Choosing S = 6 turns it into whole numbers at once. The listed wrong ratios come from writing the right numbers in the wrong order, combining 3 and 6 by division, or stopping at the first comparison. Rereading which amount the question names first settles the order.

  1. Turning words into math moderate Understanding the question

    Both sentences compare a chemical with S, R as 3 times S and T as 1/6 of S, so S must serve as the link before R can be compared with T. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    The ratios 1:18 and 18:1 use the same two numbers in opposite order, so only checking that R, the larger amount, comes first separates the right ratio from its reverse. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    No amounts are given, only relationships, so students either reason with S as a letter or choose a number for it, such as 6, which makes R 18 and T 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Pick a convenient amount for S, such as 6, so that both R and T come out as whole numbers, then write R to T in that order and simplify.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios
a:b=(a×n):(b×n)a : b = (a \times n) : (b \times n)
a and b are two amounts being compared, and n is any number except 0. Multiplying or dividing both parts of a ratio by the same number gives an equal ratio. For example, 2 : 1 is the same as 4 : 2.

  1. Sort out the question

    There are three chemicals, R, S and T. R is 3 times S. T is 16\frac{1}{6} of S. You need the ratio of R to T, written R : T. Both facts compare a chemical with S, so S links R and T.

  2. Pick an easy amount for S

    No amounts are given, so you may choose one. Pick 6 units for S, because 6 divides evenly by 6.

    R=3×6=18\mathrm{R} = 3 \times 6 = 18

    T=16×6=1\mathrm{T} = \frac{1}{6} \times 6 = 1

    So R is 18 units and T is 1 unit.

  3. Write the ratio of R to T

    R:T=18:1\mathrm{R} : \mathrm{T} = 18 : 1

    R comes first because the question asks for R to T. Watch out: writing it the other way round gives 1:18, choice A. Using only the first fact gives R : S = 3 : 1, choice C.

  4. Check with letters instead

    Without picking a number, R = 3S and T is 16S\frac{1}{6}\mathrm{S}. Divide both parts by S, then multiply both parts by 6.

    a:b=(a×n):(b×n)a : b = (a \times n) : (b \times n)

    3S:16S=3:16=18:13\mathrm{S} : \frac{1}{6}\mathrm{S} = 3 : \frac{1}{6} = 18 : 1

    So the answer is D, 18:1.

Check your answer

R should be much bigger than T. R is 3 times S, while T is only a small piece of S. Try another amount, S = 12. Then R = 36 and T = 2, and 36 : 2 is also 18 : 1.

More questions like this

The original DOE question
Question 96 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 96. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 93 of 120

2023 Test A Math, question 40

Question

Ken has k video games, and Jeff has j video games. If Ken gives 6 video games to Jeff, Ken will have twice as many video games as Jeff. Which equation shows the relationship between k and j?

  • Ak−6=2(j+6)k - 6 = 2(j + 6)
  • Bk−6=2j+6k - 6 = 2j + 6
  • C2(k−6)=j2(k - 6) = j
  • D2(k−6)=j+62(k - 6) = j + 6
Show the answer

Correct answer: A

If Ken gives 6 video games to Jeff, the number Ken has is k−6k - 6, and the number Jeff has is j+6j + 6. After the exchange, Ken has twice as many video games as Jeff. The equation is:

k−6=2(j+6)k - 6 = 2(j + 6)

Question analysis

This algebra word problem says Ken has k video games and Jeff has j, and if Ken gives Jeff 6 games, Ken will have twice as many as Jeff. Students choose the equation that models the situation. Getting the comparison the right way round, and grouping each new total correctly, is where most errors happen.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.2a writing algebraic expressions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students write equations from word problems. This one tracks two people's counts after a transfer and states a times-as-many comparison, so both counts change and the factor of 2 must go on the right side: typical but error-prone.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 4, 41 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is all setup: no solving is needed, just an equation that says what the story says. The trade moves games out of one count and into the other, and the comparison has to be written with the 2 on the smaller side. The wrong equations reverse the comparison, forget Jeff's gain, or drop the parentheses, and all four look alike. Plugging in sample counts is the surest check.

  1. Turning words into math moderate Understanding the question

    After the gift both counts change, Ken's to k − 6 and Jeff's to j + 6, and “twice as many video games as Jeff” must double Jeff's whole new count, putting the 2 on his side. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    All four equations share the pieces k − 6, j + 6 and a 2; the right one differs from 2(k − 6) = j + 6 only in where the 2 sits, and from k − 6 = 2j + 6 by parentheses. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    There are no numbers to check against, only k and j, so testing an equation means inventing counts that fit the story, such as Jeff with 4 games and Ken with 26. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Write each boy's number of games after the exchange as an expression, then translate "Ken will have twice as many as Jeff" into an equation and test it with sample numbers.

Traps

  • Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.
  • Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Twice as many
A=2×BA = 2 \times B
A and B are two amounts. “A has twice as many as B” means A equals 2 times B. The 2 goes with the smaller amount, because that is the one you double to match the bigger one.
The distributive property
a(b+c)=ab+aca(b + c) = ab + ac
a, b and c are numbers. Multiplying a by a sum means multiplying a by each part, then adding. For example, 2(4 + 6) = 8 + 12 = 20.

  1. Sort out the question

    Ken has k video games, and Jeff has j. Then Ken gives 6 of his games to Jeff. After that, Ken has twice as many games as Jeff. You need the equation that links k and j.

  2. Write each new amount

    Ken gives away 6 games, so he has k − 6 left. Jeff gets those 6 games, so he has j + 6. Both amounts change, so use both new amounts.

  3. Write “twice as many”

    After the gift, Ken has more games. So Jeff’s new amount is the one that gets doubled.

    A=2×BA = 2 \times B

    k−6=2(j+6)k - 6 = 2(j + 6)

    Watch out: putting the 2 on Ken’s side, as in 2(k−6)=j+62(k - 6) = j + 6, would mean Jeff has twice as many. That is choice D.

  4. Keep the parentheses

    a(b+c)=ab+aca(b + c) = ab + ac

    2(j+6)=2j+122(j + 6) = 2j + 12

    Twice Jeff’s new amount is 2j + 12, not 2j + 6. Watch out: dropping the parentheses gives choice B. So the answer is A, k−6=2(j+6)k - 6 = 2(j + 6).

Check your answer

Make up numbers that fit the story. Say Jeff has 4 games and Ken has 26. After the gift, Ken has 20 and Jeff has 10, and 20 is twice 10. Choice A gives 26−6=2(4+6)26 - 6 = 2(4 + 6), or 20 = 20. Choice B gives 20 = 14, which is false.

More questions like this

The original DOE question
Question 97 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 97. Open page 85 of the handbook (PDF)

Question 94 of 120

2023 Test A Math, question 41

Question

Yesterday Sarah read 15% of her entire book. Today she read another 17% of the entire book. In lowest terms, what fraction of the book is left for her to read?

  • A725\frac{7}{25}
  • B310\frac{3}{10}
  • C1725\frac{17}{25}
  • D710\frac{7}{10}
Show the answer

Correct answer: C

So far, Sarah has read 15%+17%=32%15\% + 17\% = 32\% of the entire book. That means she has 100%−32%=68%100\% - 32\% = 68\% left to read.

68%=68100=172568\% = \frac{68}{100} = \frac{17}{25}

Question analysis

This percent word problem says Sarah read 15% of her book yesterday and 17% today, and asks what fraction of the book she still has to read, in lowest terms. Students combine percents of the same whole, find the part remaining, and convert a percent to a simplified fraction. Rounding, or giving the part already read, leads to wrong choices.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 percent work often takes a percent of a percent or finds a percent change. Here both percents are of the same whole book, so they combine directly; the rest is a subtraction and simplifying a fraction, lighter than typical.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 33 words.
Corrected from the DOE handbook
  • Missing percent signs in the explanation were restored.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The percent work is a sum and a subtraction from 100, so the question mostly tests the last step, writing 68% as a fraction in lowest terms. Students who round to 30% read and 70% left to get a friendlier fraction land on a listed choice, as do those who give the part read. Dividing 68 and 100 by their common factor 4 keeps the answer exact.

  1. Turning words into math minor Understanding the question

    Both percents are “of the entire book,” so they simply add to 32%, and “left for her to read” asks for the rest, 100% − 32%, not the part already read. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The result has to be written as a fraction in lowest terms, so 68/100 must be divided by 4 to reach a form that appears among the choices. (Enright, Morley & Sheehan, 2002)

Strategy: Add the two percents, since both are of the whole book, subtract the total from 100%, then write the result over 100 and simplify it fully.

Traps

  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • 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

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes “So far, Sarah has read 15 + 17 = 32 of the entire book. That means she has 100 − 32 = 68 left to read.” (p. 143, question 98), with no percent signs, so 32 and 68 read as amounts of the book (pages, say) that the question never gives. The question gives 15% and 17%, and the handbook’s next line begins “68% =”, so the site now reads 15% + 17% = 32% and 100% − 32% = 68%. The answer, 17/25, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction
p%=p100p\% = \frac{p}{100}
p is the percent. Percent means “out of 100,” so p percent is p hundredths. For example, 50% is 50 out of 100, which is one half.
Simplifying a fraction
ab=a÷nb÷n\frac{a}{b} = \frac{a \div n}{b \div n}
a is the top number, b is the bottom, and n is a number that divides both evenly. Dividing the top and bottom by the same number keeps the value the same. A fraction is in lowest terms when no number bigger than 1 divides both.

  1. Sort out the question

    Sarah read 15% of her book yesterday and 17% today. Both are percents of the whole book. You need the part still left to read, as a fraction in lowest terms. That means a fraction that can’t be simplified any more.

  2. Find how much she has read

    15%+17%=32%15\% + 17\% = 32\%

    Both percents are of the same whole book, so you can add them. She has read 32% of it.

  3. Find how much is left

    The whole book is 100%.

    100%−32%=68%100\% - 32\% = 68\%

    Watch out: 32% is the part she has read, not the part left. Rounding 32% to 30% gives 310\frac{3}{10}, choice B.

  4. Write 68% as a fraction

    p%=p100p\% = \frac{p}{100}

    68%=6810068\% = \frac{68}{100}

    This fraction is not in lowest terms yet, because 68 and 100 are both even.

  5. Write it in lowest terms

    The number 4 divides both 68 and 100 evenly.

    ab=a÷nb÷n\frac{a}{b} = \frac{a \div n}{b \div n}

    68100=68÷4100÷4=1725\frac{68}{100} = \frac{68 \div 4}{100 \div 4} = \frac{17}{25}

    Watch out: don’t round 68% to 70%. That gives 710\frac{7}{10}, choice D. So the answer is C, 1725\frac{17}{25}.

Check your answer

Turn the answer back into a percent: 1725=68100\frac{17}{25} = \frac{68}{100}, which is 68%. Then 68% + 32% = 100%, the whole book. Also, 17 and 25 share no factor except 1, so the fraction is in lowest terms.

More questions like this

The original DOE question
Question 98 as printed on page 86 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 98. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 95 of 120

2023 Test A Math, question 42

Question

{1,2,3,4,5,…,198,199,200}\{1,2,3,4,5,\ldots,198,199,200\}

How many members of the set shown above are multiples of 6 but not multiples of 9?

  • A11
  • B13
  • C20
  • D22
Show the answer

Correct answer: D

First, list the first several multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54 . . .

Every third multiple (i.e., 18, 36, 54, . . .) is also a multiple of 9. Thus, 23\frac{2}{3} of the multiples of 6 are not multiples of 9.

Next, calculate how many multiples of 6 are in the given number range. The last multiple of 6 would be 198. So, there are a total of 1986=33\frac{198}{6} = 33 multiples of 6. Since 23\frac{2}{3} of those are not multiples of 9, the answer is 23(33)=22\frac{2}{3}(33) = 22.

Question analysis

This number properties question shows the set of whole numbers from 1 to 200 and asks how many of them are multiples of 6 but not multiples of 9. Students count multiples in a range and remove the overlap, the numbers divisible by both. Finding the right common multiple, and removing only the overlap, is where it gets tricky.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Multiples and common multiples are grade 6 content, but counting the members of a set that meet one condition and fail another is rare in grade 7 work. The overlap between the two kinds of multiples must be found and removed exactly.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 6, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Counting multiples of 6 is routine; the harder decision is how to remove the ones that are also multiples of 9. Those are the multiples of 18, the least common multiple, not of 54 or 15, and students who count them correctly must still subtract them rather than report them. Listing 6, 12, 18, 24, 30, 36 shows the pattern, two kept for every one removed, which also gives the answer as two thirds of 33.

  1. No obvious method moderate Working it out

    The stem never mentions a common multiple: students must see that the multiples of 6 that are also multiples of 9 are the multiples of 18, every third multiple of 6. (Schoenfeld, 1985)

  2. Much to keep track of minor Working it out

    Three counts are in play at once, 33 multiples of 6 up to 198, 11 multiples of 18 up to 198, and their difference, and two of them are listed choices. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Turning words into math minor Understanding the question

    The bold not in “multiples of 6 but not multiples of 9” asks for what is left after a removal, a set difference rather than a count of the numbers removed. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Count the multiples of 6 up to 200, find the smallest number that is a multiple of both 6 and 9, count its multiples in the set, and subtract.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting multiples
last multiple=n×k\text{last multiple} = n \times k
n is the number whose multiples you count, such as 6. The first multiple is n × 1, the second is n × 2, and so on. So if the largest multiple in the set is n × k, there are k multiples in all.
Least common multiple
LCM(6,9)=18\text{LCM}(6, 9) = 18
The least common multiple, or LCM, is the smallest number that two numbers both divide evenly. A number is a multiple of both 6 and 9 exactly when it is a multiple of 18.

  1. Sort out the question

    The set holds the whole numbers from 1 to 200. A multiple of 6 is a number that 6 divides evenly, like 6, 12 or 18. You need the multiples of 6 that are not multiples of 9. So count all the multiples of 6. Then take away those that are also multiples of 9.

  2. Count the multiples of 6

    200 is not a multiple of 6, but 198 is. So 198 is the last multiple of 6 in the set.

    last multiple=n×k\text{last multiple} = n \times k

    198=6×33198 = 6 \times 33

    There are 33 multiples of 6.

  3. Find the numbers to take away

    List the multiples of 6: 6, 12, 18, 24, 30, 36. The first one 9 also divides is 18, then 36. So every third multiple of 6 is also a multiple of 9.

    LCM(6,9)=18\text{LCM}(6, 9) = 18

    Watch out: the shared multiples are multiples of 18, not of 6 + 9 = 15. Using 15 leads to 13 or 20, choices B and C.

  4. Count the multiples of 18

    The last multiple of 18 in the set is also 198.

    198=18×11198 = 18 \times 11

    There are 11 of them. Watch out: these are the numbers you take away, not the answer. Stopping here gives 11, choice A.

  5. Take away the multiples of 18

    33−11=2233 - 11 = 22

    There are 22 multiples of 6 left over. So the answer is D, 22.

Check your answer

Every third multiple of 6 is a multiple of 18, so two out of every three multiples of 6 stay. Two thirds of 33 is 23×33=22\frac{2}{3} \times 33 = 22. That matches. Test one: 30 is a multiple of 6, and 30 ÷ 9 is not a whole number, so 30 counts.

More questions like this

The original DOE question
Question 99 as printed on page 86 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 99. Open page 86 of the handbook (PDF)

Question 96 of 120

2023 Test A Math, question 46

Question

ESSAY LENGTH

Number of Words Number of Essays
<100 6
100–250 4
251–500 11
>500 9

All 150 students in Grade 8 at a school are assigned to write an essay on the same topic. A teacher records the number of words in a random sample of the essays, as shown in the table above. Based on this sample, how many students in the entire grade would be expected to write essays with at least 100 words?

  • A20
  • B30
  • C100
  • D120
Show the answer

Correct answer: D

The number of students in the sample who wrote an essay with at least 100 words is 24 out of the 30 students. Write it as a fraction: 2430=45\frac{24}{30} = \frac{4}{5}. To predict the number of students in the entire grade who would write an essay with at least 100 words, find 45\frac{4}{5} of 150, which is 45(150)=120\frac{4}{5}(150) = 120.

Question analysis

This statistics question gives a table of word counts for a random sample of Grade 8 essays and asks how many of all 150 students would be expected to write at least 100 words. Students find the sample size, the share meeting the condition, and scale it to the whole grade. Matching the words at least to the table's ranges is the tricky part.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a random sample to predict a count for a whole population is typical grade 7 statistics and ratio work. Students total the sample from a table, read at least 100 against the table's ranges, and scale by a whole-number factor.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 65 words.
Students, last 30 days
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 proportion at the end is routine; the question is decided at the table. Students must turn “at least 100 words” into the right rows and also find the sample size, which is printed nowhere. Each listed wrong answer scales up a different group of rows, so a careless reading still produces a believable number. Noting that only 6 of 30 essays fall short, one fifth, gives a quick check.

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

    The table never prints the sample size; it has to be built by adding all four rows, 6 + 4 + 11 + 9 = 30, and the qualifying essays sit in three rows, not one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Turning words into math minor Understanding the question

    The bold “at least 100 words” must be matched to the rows 100–250, 251–500 and >500, keeping the row that begins at 100 and leaving out only the <100 row. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Many steps minor Working it out

    Four steps follow in order: the sample total, the qualifying count of 24, the fraction 24/30 = 4/5, and that fraction of all 150 students. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the table's counts to find the sample size, add the rows with 100 or more words, and scale that fraction of the sample up to all 150 students.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from a sample
expected number=count in samplesample size×whole group\text{expected number} = \frac{\text{count in sample}}{\text{sample size}} \times \text{whole group}
A sample is a smaller group picked at random from the whole group. The fraction of the sample with some feature is your best guess for the fraction of the whole group. So multiply that fraction by the size of the whole group.

  1. Sort out the question

    The table shows a random sample of the essays from a grade of 150 students. You need to predict how many of the 150 students wrote at least 100 words. At least 100 means 100 or more.

  2. Find the sample size

    The table doesn’t give the total, so add all four rows.

    6+4+11+9=306 + 4 + 11 + 9 = 30

    The sample has 30 essays.

  3. Count the essays with enough words

    Three rows count: 100–250, 251–500 and >500. Only the <100 row is left out.

    4+11+9=244 + 11 + 9 = 24

    Watch out: the 100–250 row counts, because 100 is at least 100. Leaving it out gives 100, choice C. Using the 6 short essays instead gives 30, choice B.

  4. Scale up to the whole grade

    expected number=count in samplesample size×whole group\text{expected number} = \frac{\text{count in sample}}{\text{sample size}} \times \text{whole group}

    2430×150=45×150=120\frac{24}{30} \times 150 = \frac{4}{5} \times 150 = 120

    About 4 out of every 5 students write at least 100 words. So the answer is D, 120 students.

Check your answer

The whole grade, 150, is 5 times the sample of 30. So multiply the sample count by 5: 24×5=12024 \times 5 = 120. The short essays scale to 6×5=306 \times 5 = 30, and 120 + 30 = 150, the whole grade.

More questions like this

The original DOE question
Question 103 as printed on page 87 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 87, question 103. Open page 87 of the handbook (PDF)

Question 97 of 120

2023 Test A Math, question 52

Question

Vicente and Carla each ran 8 laps around a track. They started at the same time and place. If Vicente ran 1.5 times as fast as Carla, how many laps did Carla have left to finish when Vicente finished his 8th lap?

  • A2122\frac{1}{2}
  • B2232\frac{2}{3}
  • C3133\frac{1}{3}
  • D5135\frac{1}{3}
Show the answer

Correct answer: B

Vicente ran 1.5 times as fast as Carla, so that means for every lap Carla finished, Vicente finished 1.5 laps. Set up a proportion to calculate c, the number of laps Carla had completed when Vicente finished 8 laps:

c1=81.5\frac{c}{1} = \frac{8}{1.5} Cross multiply.
1.5c=8(1)1.5c = 8(1) Apply the multiplicative inverse property; divide both sides of the equation by 1.5.
c=513c = 5\frac{1}{3}

So, the number of laps Carla still had to finish is 8−513=2238 - 5\frac{1}{3} = 2\frac{2}{3}.

Question analysis

This ratio word problem says Vicente and Carla each ran 8 laps, starting at the same time and place, and Vicente ran 1.5 times as fast. It asks how many laps Carla still had to run when Vicente finished. Students reason with a speed ratio and mixed numbers, and giving the laps run rather than the laps remaining is an easy slip.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 rate problems usually give a speed and a time. Here only the ratio of two speeds is given, so students must reason that distances run in the same time share that ratio, then work with a fractional number of laps.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 4, 44 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

No time is given, so students must reason that in equal time distance scales with speed, which turns the race into one division and one subtraction. Both steps hide a trap: the division by 1.5 leaves a remainder that is easy to misread, and subtracting a mixed number from 8 needs regrouping. The result of the first step, Carla's laps run, is itself a choice, so the question also rewards rereading what is asked.

  1. Turning words into math moderate Understanding the question

    The phrase “1.5 times as fast” has to become distances in the same time: while Vicente runs 8 laps, Carla runs 8 ÷ 1.5, not 8 − 1.5 or 8 × 1.5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic moderate Working it out

    Dividing 8 by 1.5 gives 16/3, or 5 1/3, a division by a decimal with a remainder, and 8 − 5 1/3 then needs a whole regrouped into thirds. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. A very close wrong answer minor Choosing the answer

    The choice 2 1/2 is only a sixth of a lap from 2 2/3, so a rough estimate of how far Carla trails cannot separate them; only the exact quotient does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: In the same amount of time the faster runner covers 1.5 times the distance, so divide Vicente's laps by 1.5 to get Carla's, then subtract that from 8.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, rate and time
d=r×td = r \times t
d is the distance, r is the rate or speed, and t is the time. When two runners run for the same time, the faster one goes farther by the same factor as the speed. So 1.5 times as fast means 1.5 times as far.
Dividing by a fraction
a÷mn=a×nma \div \frac{m}{n} = a \times \frac{n}{m}
a is any number and m over n is a fraction. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down.

  1. Sort out the question

    Vicente and Carla each run 8 laps, starting together. Vicente runs 1.5 times as fast as Carla. You need how many laps Carla still has to run when Vicente finishes. First find how far Carla has run at that moment.

  2. Compare their distances

    d=r×td = r \times t

    They have run for the same time. So Vicente’s distance is 1.5 times Carla’s. Let c be the laps Carla has run.

    8=1.5×c8 = 1.5 \times c

    To find c, divide 8 by 1.5.

  3. Divide 8 by 1.5

    Write 1.5 as the fraction 32\frac{3}{2}.

    a÷mn=a×nma \div \frac{m}{n} = a \times \frac{n}{m}

    c=8÷32=8×23=163=513c = 8 \div \frac{3}{2} = 8 \times \frac{2}{3} = \frac{16}{3} = 5\frac{1}{3}

    Carla has run 5135\frac{1}{3} laps. Watch out: that is how far she has run, not what is left. Stopping here gives choice D.

  4. Find the laps she has left

    8−513=243−163=83=2238 - 5\frac{1}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3} = 2\frac{2}{3}

    Watch out: 8 − 5135\frac{1}{3} is not 3133\frac{1}{3}. You must break 1 whole into thirds first. Answering 3133\frac{1}{3} is choice C.

    So the answer is B, 2232\frac{2}{3} laps.

Check your answer

Multiply back: 1.5×513=32×163=81.5 \times 5\frac{1}{3} = \frac{3}{2} \times \frac{16}{3} = 8. So when Carla has run 5135\frac{1}{3} laps, Vicente has run 8. And 513+223=85\frac{1}{3} + 2\frac{2}{3} = 8, her full race.

More questions like this

The original DOE question
Question 109 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 89, question 109. Open page 89 of the handbook (PDF)

Question 98 of 120

2023 Test A Math, question 53

Question

At West College, students are randomly assigned to one of 20 dormitories and one of 6 dining rooms. Kharleen likes 6 of the dormitories and 2 of the dining rooms. What is the probability that she is assigned to both a dormitory and a dining room that she likes?

  • A10%
  • B12%
  • C19%
  • D38%
Show the answer

Correct answer: A

Kharleen likes 620=310\frac{6}{20} = \frac{3}{10} of the dormitories. She likes 26=13\frac{2}{6} = \frac{1}{3} of the dining rooms. Multiply those fractions together to calculate the probability of Kharleen being assigned to both a dormitory and a dining room that she likes: 310×13=110=10%\frac{3}{10} \times \frac{1}{3} = \frac{1}{10} = 10\%.

Question analysis

This probability word problem says students at West College are randomly assigned one of 20 dormitories and one of 6 dining rooms, and Kharleen likes 6 dormitories and 2 dining rooms. It asks for the chance that she gets both a dormitory and a dining room she likes. Students combine two independent events and write the result as a percent.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the chance that two independent events both happen is grade 7 compound probability. Each event's chance comes straight from the counts given, and joining them and writing the result as a percent is a typical grade 7 task.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 45 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns on one rule, multiplying the chances of two independent events. Students who know it reduce the fractions first and finish in seconds; students who don't tend to combine the counts, and 6 × 2 = 12 reads naturally as 12%. The other two choices match no common working, so the real contest is between the product of the counts and the product of the chances.

  1. Something you must already know moderate Working it out

    The rule that the chance of two independent events both happening is the product of their chances, 3/10 × 1/3, is assumed; nothing in the stem says to multiply. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The phrase “both a dormitory and a dining room that she likes” describes two separate random assignments that must both go her way, so the two chances are combined, not compared. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. A very close wrong answer minor Choosing the answer

    The choice 12% sits only 2 points above 10%, so a rough sense that the answer is small cannot separate them; only computing 3/10 × 1/3 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the chance of a liked dormitory and the chance of a liked dining room as fractions, multiply them because both must happen, and convert the product to a percent.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=outcomes you wantall outcomesP = \frac{\text{outcomes you want}}{\text{all outcomes}}
P is the probability, the chance that something happens. The top counts the outcomes you want. The bottom counts every outcome that could happen. Each outcome must be equally likely.
Probability of two separate events
P(A and B)=P(A)×P(B)P(\text{A and B}) = P(\text{A}) \times P(\text{B})
A and B are two events that don’t affect each other, like two separate random picks. P(A) is the chance of A, and P(B) is the chance of B. The chance that both happen is the two chances multiplied.

  1. Sort out the question

    Kharleen gets one of 20 dorms, or dormitories, where students live. She also gets one of 6 dining rooms. Each is picked at random. She likes 6 dorms and 2 dining rooms. You need the chance that she gets a dorm she likes and a dining room she likes.

  2. Find the chance of a liked dorm

    P=outcomes you wantall outcomesP = \frac{\text{outcomes you want}}{\text{all outcomes}}

    P(dorm)=620=310P(\text{dorm}) = \frac{6}{20} = \frac{3}{10}

    She likes 6 of the 20 dorms. Divide the top and bottom by 2 to simplify.

  3. Find the chance of a liked dining room

    Use the same formula for the dining rooms.

    P(dining room)=26=13P(\text{dining room}) = \frac{2}{6} = \frac{1}{3}

    She likes 2 of the 6 dining rooms. Divide the top and bottom by 2.

  4. Multiply the two chances

    The two picks are separate. One doesn’t change the other.

    P(A and B)=P(A)×P(B)P(\text{A and B}) = P(\text{A}) \times P(\text{B})

    310×13=330=110\frac{3}{10} \times \frac{1}{3} = \frac{3}{30} = \frac{1}{10}

    Both must happen, so the chance is smaller than either one alone.

  5. Change to a percent

    110=10100=10%\frac{1}{10} = \frac{10}{100} = 10\%

    Watch out: multiplying 6 × 2 = 12 and calling it 12% skips dividing by all the possible assignments. That is choice B. So the answer is A, 10%.

Check your answer

Count every possible assignment. There are 20×6=12020 \times 6 = 120 dorm and dining room pairs, all equally likely. She likes 6×2=126 \times 2 = 12 of them. Then 12120=110\frac{12}{120} = \frac{1}{10}, which is 10%, the same answer.

More questions like this

The original DOE question
Question 110 as printed on page 90 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 110. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 99 of 120

2023 Test A Math, question 54

Question

A train travels 2,200 miles from Phoenix to New York City. The train covers the first 240 miles in 5 hours. If the train continues to travel at this rate, how many more hours will it take to reach New York City? Round your answer to the nearest whole hour.

  • A46
  • B45
  • C43
  • D41
Show the answer

Correct answer: D

First, find the speed of the train in miles per hour: 240÷5=48240 \div 5 = 48.

The number of miles left to travel is 2,200−240=1,9602{,}200 - 240 = 1{,}960.

To find the number of hours left, use the equation rt=drt = d:

48t=1,96048t = 1{,}960 Apply the multiplicative inverse property; divide both sides of the equation by 48.
t=1,96048=4056t = \frac{1{,}960}{48} = 40\frac{5}{6}

That number rounded to the nearest whole hour is 41 hours.

Question analysis

This rate word problem describes a 2,200-mile train trip from Phoenix to New York City in which the first 240 miles take 5 hours, and asks how many more hours the rest will take, rounded to the nearest hour. Students find a unit rate and the remaining distance. The choices are close together, so the time for the whole trip is a tempting trap.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Unit rates are grade 6 content, but this chains four steps: a speed, the distance left, a long division that does not come out even, and rounding. The word more must be read closely, which puts it at typical grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 4, 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 routine, but the question has a long chain and asks for only part of the trip. Dividing the whole 2,200 miles by the speed is the natural shortcut and gives a listed choice, and the rounding instruction adds a second place to go wrong. Checking that 41 hours at 48 miles per hour covers about 1,970 miles, close to the 1,960 left, confirms the answer.

  1. Many steps moderate Working it out

    Four steps follow in order, the speed 240 ÷ 5 = 48, the miles left 2,200 − 240 = 1,960, the time 1,960 ÷ 48, and the rounding, and an early slip carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The bold more in “how many more hours” limits the question to the 1,960 miles still ahead, though 2,200, the whole trip, is the distance the stem names first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Dividing 1,960 by 48 leaves 40 with a remainder of 40, so the time, 40 5/6 hours, must be rounded up to the nearest hour rather than cut short. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the train's speed from the first 240 miles, subtract those miles from the total to get the distance left, divide by the speed, 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.
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, rate and time
d=r×td = r \times t
d is the distance, r is the rate or speed, and t is the time. If you know two of them, you can find the third. To find the speed, divide the distance by the time. To find the time, divide the distance by the speed.

  1. Sort out the question

    The whole trip is 2,200 miles. The train goes the first 240 miles in 5 hours, and it keeps that speed. You need how many more hours it takes, rounded to the nearest whole hour. “More” means the time after the first 240 miles.

  2. Find the train’s speed

    d=r×td = r \times t

    240=r×5240 = r \times 5

    r=240÷5=48r = 240 \div 5 = 48

    Divide the 240 miles by the 5 hours. The train goes 48 miles per hour.

  3. Find the miles still to go

    2,200−240=1,9602{,}200 - 240 = 1{,}960

    There are 1,960 miles left. Watch out: use the miles left, not the whole trip. Dividing 2,200 by 48 gives about 45.8 hours, which leads to 46 or 45, choices A and B.

  4. Find the hours still to go

    1,960=48×t1{,}960 = 48 \times t

    t=1,960÷48=404048=4056t = 1{,}960 \div 48 = 40\frac{40}{48} = 40\frac{5}{6}

    Here 48 × 40 = 1,920, and 40 miles are left over. That leftover is 4048=56\frac{40}{48} = \frac{5}{6} of an hour.

  5. Round to the nearest hour

    56\frac{5}{6} of an hour is more than half an hour, so round up to 41. Watch out: don’t just drop the fraction.

    So the answer is D, 41 hours.

Check your answer

At 48 miles per hour, 41 hours covers 48×41=1,96848 \times 41 = 1{,}968 miles. That is just a bit more than 1,960, as it should be after rounding up. In 40 hours, the train would cover only 1,920 miles, which is not enough.

More questions like this

The original DOE question
Question 111 as printed on page 90 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 111. Open page 90 of the handbook (PDF)

Question 100 of 120

2023 Test A Math, question 55

Question

What is the least of four consecutive integers whose sum is 58?

  • A1
  • B2
  • C12
  • D13
Show the answer

Correct answer: D

Let n represent the least of the four integers. The other three integers are (n+1)(n + 1), (n+2)(n + 2), and (n+3)(n + 3). Write an equation to solve for n:

n+(n+1)+(n+2)+(n+3)=58n + (n + 1) + (n + 2) + (n + 3) = 58

4n+6=584n + 6 = 58

4n=524n = 52

n=13n = 13

Question analysis

This number properties question asks for the least of four consecutive integers that add up to 58. Students write the integers in terms of one unknown and solve a two-step equation, or reason from the average of the four. Starting the list at n + 1 instead of n, which shifts every integer by one, is the main trap.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.6 variables representing unknown numbers
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 is core grade 7 work. Expressing four consecutive integers in terms of the least one is a familiar setup, so this is a routine three-step grade 7 problem.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 6, 11 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One equation, 4n + 6 = 58, solves the question, and dividing 58 by 4 to find the middle of the four, 14.5, is quicker still. The only real risk is the count: writing the integers as n + 1 through n + 4 and reporting n gives a least integer one too small. The two smallest choices match no reasonable working, so they draw only guesses. Adding the four integers back up to 58 is a ten-second check.

  1. Turning words into math minor Understanding the question

    The phrase “four consecutive integers” has to become n, n + 1, n + 2 and n + 3, with n the least, so the sum is 4n + 6, not 4n + 10. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    The choices 12 and 13 differ by one, the size of a fence-post error, so only checking that 13 + 14 + 15 + 16 = 58 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Call the least integer n, write the next three as n + 1, n + 2 and n + 3, add them and solve; or divide 58 by 4 to find the middle of the four.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive integers
n,n+1,n+2,n+3n,\; n + 1,\; n + 2,\; n + 3
Integers are whole numbers and their negatives, like −2, 0 and 7. Consecutive integers come one right after another, like 4, 5, 6, 7. If n is the least one, each next one is 1 more than the one before.

  1. Sort out the question

    You need four consecutive integers, numbers in a row like 7, 8, 9, 10. Together they add up to 58. You need the least, or smallest, of the four.

  2. Name the four numbers

    Let n be the least number. Then the other three are 1, 2 and 3 more than n.

    n,n+1,n+2,n+3n,\; n + 1,\; n + 2,\; n + 3

  3. Write and simplify the equation

    Add the four numbers and set the sum equal to 58.

    n+(n+1)+(n+2)+(n+3)=58n + (n + 1) + (n + 2) + (n + 3) = 58

    4n+6=584n + 6 = 58

    To combine, count the n’s: there are four of them. The plain numbers make 1 + 2 + 3 = 6.

  4. Solve for n

    Subtract 6 from both sides. Then divide both sides by 4.

    4n=58−6=524n = 58 - 6 = 52

    n=52÷4=13n = 52 \div 4 = 13

    Watch out: if you call the numbers n + 1 to n + 4, you get n = 12. Then 12 is not one of the four numbers; the least one is 13. Answering 12 is choice C.

    So the answer is D, 13.

Check your answer

Add them up: 13+14+15+16=5813 + 14 + 15 + 16 = 58. It works. Another way: 58÷4=14.558 \div 4 = 14.5 is their average. It falls halfway between the two middle numbers, 14 and 15. So the four numbers are 13, 14, 15 and 16.

More questions like this

The original DOE question
Question 112 as printed on page 90 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 112. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 101 of 120

2023 Test B Math, question 1

Question

LAST YEAR’S TOTAL SALES

Vehicles Number
Four-door cars 5,850
Two-door cars 2,250
Trucks 900
Total 9,000

The table above shows the number of each type of vehicle sold at a dealership last year. If 1,000 vehicles will be sold next month, what is the best estimate (based on last year’s sales) of the number of two-door cars that will be sold?

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

Show the answer

Correct answer: 250

Let x represent the number of two-door cars that will be sold next month. Use the information in the table to set up a proportion:

x1,000=2,2509,000\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000}

x=(2,250)(1,000)9,000=2,2509=250x = \frac{(2,250)(1,000)}{9{,}000} = \frac{2{,}250}{9} = 250

Question analysis

A table shows last year's sales at a car dealership: four-door cars, two-door cars and trucks, 9,000 vehicles in all. This grid-in asks for the best estimate of how many two-door cars will be sold next month if 1,000 vehicles are sold. Students set up a proportion from the table; most of its rows are not needed.

NYS standards
  • 6.RP.3 ratio and rate reasoning
  • 6.SP.7 long-run relative frequency
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 proportion problems often take several steps or awkward numbers. This one scales a single share from a table to a new total, with numbers that divide cleanly, which is closer to grade 6 ratio work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 55 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the share is seen as 2,250 out of 9,000, a quarter, the arithmetic is light, which makes this one of the easier grid-ins. Points are lost by entering the share itself as a percent, by reading the wrong row of the table, or by building a monthly average when the question supplies next month’s total. With no choices to compare, checking that the entry is a quarter of 1,000 is the only safeguard.

  1. Turning words into math minor Understanding the question

    The phrase “best estimate (based on last year’s sales)” has to become a proportion: two-door cars keep the same share of next month’s 1,000 vehicles as of last year’s 9,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table lists three vehicle types and a total; only the two-door row, 2,250, and the 9,000 total are needed, so the four-door and truck rows are distractions. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find what fraction of last year's 9,000 vehicles were two-door cars, simplify it, then take that fraction of the 1,000 vehicles expected next month.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportion
partwhole=partwhole\frac{\text{part}}{\text{whole}} = \frac{\text{part}}{\text{whole}}
A proportion says two fractions are equal. Each fraction here is a part over its whole: two-door cars over all vehicles sold. Use it when a share stays the same while the total changes. Then you can find the missing part.

  1. Sort out the question

    The table shows last year’s sales. In all, 9,000 vehicles were sold, and 2,250 of them were two-door cars. Next month, 1,000 vehicles will be sold. You need about how many of them will be two-door cars.

    “Based on last year’s sales” means two-door cars keep the same share of all sales. A share is a part out of a whole. The four-door and truck rows don’t matter here.

  2. Find last year’s share of two-door cars

    Last year, 2,250 out of 9,000 vehicles were two-door cars. Write that as a fraction and simplify it.

    2,2509,000=14\frac{2{,}250}{9{,}000} = \frac{1}{4}

    This works because 2,250×4=9,0002{,}250 \times 4 = 9{,}000. So 1 out of every 4 vehicles sold was a two-door car.

  3. Set up the proportion and solve

    Let x be the number of two-door cars next month. Next month’s share must match last year’s share.

    x1,000=2,2509,000=14\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000} = \frac{1}{4}

    To get x alone, multiply both sides by 1,000.

    x=14×1,000=250x = \frac{1}{4} \times 1{,}000 = 250

    So about 250 two-door cars will be sold.

  4. Give the answer

    Watch out: one fourth is the same as 25%. Gridding 25 gives the percent, not the number of cars. Also be sure to use the two-door row. The four-door row would give 650. So you would grid 250.

Check your answer

Try it another way. Next month’s 1,000 vehicles are 19\frac{1}{9} of last year’s 9,000. So take 19\frac{1}{9} of the 2,250 two-door cars: 2,250÷9=2502{,}250 \div 9 = 250. Both ways give 250 cars.

More questions like this

The original DOE question
Question 58 as printed on page 191 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 58. Open page 191 of the handbook (PDF)

Question 102 of 120

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:

w9=32\frac{w}{9} = \frac{3}{2}

w=3(9)2=272=13.5w = \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
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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scale proportion
real sizedrawing size=real sizedrawing size\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.

  1. 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.

  2. Find what one inch stands for

    If 2 inches stand for 3 yards, then 1 inch stands for half as much.

    3÷2=1.53 \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.

  3. 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.

    w9=32\frac{w}{9} = \frac{3}{2}

    To get w alone, multiply both sides by 9.

    w=3×92=272=13.5w = \frac{3 \times 9}{2} = \frac{27}{2} = 13.5

    The real width is 13.5 yards.

  4. Give the answer

    Watch out: don’t flip the scale. Using 9×239 \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 4.5×3=13.54.5 \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 103 of 120

2023 Test B Math, question 3

Question

A kindergarten teacher has 72 crayons, 60 pencils, and 84 sheets of paper to distribute to the children in her class. If each child receives an equal number of each item and there are no items remaining, what is the greatest possible number of children in the class?

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

Show the answer

Correct answer: 12

The greatest possible number of children in the class is the greatest common factor of 72, 60, and 84. Write the prime factorization of each number:

72=23×3272 = 2^{3} \times 3^{2}

60=22×3×560 = 2^{2} \times 3 \times 5

84=22×3×784 = 2^{2} \times 3 \times 7

Then, find the factors that all three have in common, which is 22×3=122^{2} \times 3 = 12.

Question analysis

A kindergarten teacher shares 72 crayons, 60 pencils and 84 sheets of paper so that every child gets the same number of each item, with nothing left over. This grid-in asks for the greatest possible number of children in the class. Students must recognize a greatest common factor problem and not confuse it with a least common multiple.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Greatest common factors are grade 6 content, normally of two numbers. This one uses three numbers and a word setting that must be translated, but the factoring is short, so the demand stays below typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 45 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The story has to be recognized as a sharing problem before any arithmetic starts, and the word greatest points to the largest divisor shared by all three counts. Students who spot a small shared factor can stop too soon, while those who think of a common multiple head the wrong way entirely. Checking that 72, 60 and 84 all divide by the entry, and that no larger number does, settles it.

  1. Turning words into math moderate Understanding the question

    Nothing names a greatest common factor; “each child receives an equal number of each item and there are no items remaining” means the class size must divide 72, 60 and 84 exactly. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Finding the greatest common factor of three numbers, by prime factorization or by listing factors, is a grade 6 method the question assumes rather than explains. (Anderson & Krathwohl, 2001)

Strategy: Write 72, 60 and 84 as products of primes, multiply the primes all three share, and check that each amount divides evenly by the result.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor
GCF=product of the shared prime factors\text{GCF} = \text{product of the shared prime factors}
A factor of a number divides it with nothing left over. The GCF is the greatest number that is a factor of every number in the group. A prime number has only 1 and itself as factors, like 2, 3, 5 and 7. Multiply the primes that all the numbers share, as many times as each number has them.

  1. Sort out the question

    The teacher has 72 crayons, 60 pencils and 84 sheets of paper. Every child gets the same number of each item, with none left over.

    So the number of children must divide 72, 60 and 84 with no remainder. A number like that is a common factor. You need the greatest one.

  2. Break each number into primes

    Write each number as a product of primes. A product is the answer to a multiplication.

    72=2×2×2×3×372 = 2 \times 2 \times 2 \times 3 \times 3

    60=2×2×3×560 = 2 \times 2 \times 3 \times 5

    84=2×2×3×784 = 2 \times 2 \times 3 \times 7

    You can check each one by multiplying it back out.

  3. Pick out the shared primes

    Look for the primes that appear in all three lists. Each list has at least two 2s and one 3. Only 72 has a third 2 and a second 3. The 5 and the 7 appear in one list each.

    GCF=2×2×3=12\text{GCF} = 2 \times 2 \times 3 = 12

    Watch out: keep every shared prime. Using only one 2 gives 6. Leaving out the 3 gives 4.

  4. Give the answer

    Watch out: 2,520 is the smallest number that all three numbers divide into. That is the least common multiple, a different idea. The class can’t have 2,520 children when there are only 60 pencils. So you would grid 12.

Check your answer

Share the items among 12 children. Each child gets 72÷12=672 \div 12 = 6 crayons, 60÷12=560 \div 12 = 5 pencils and 84÷12=784 \div 12 = 7 sheets. Nothing is left over. The numbers 6, 5 and 7 share no factor but 1, so no bigger class works.

More questions like this

The original DOE question
Question 60 as printed on page 191 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 60. Open page 191 of the handbook (PDF)

Question 104 of 120

2023 Test B Math, question 4

Question

Mr. Smith opened a retirement account with a deposit of $900. This account earns 5% simple interest annually. How many years will it take for his $900 deposit to earn $360 interest?

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

Show the answer

Correct answer: 8

Calculate simple interest (y) by multiplying the initial deposit (p) by the interest rate (r) by the number of years (t):

prt=yprt = y

900(0.05)t=360900(0.05)t = 360

45t=36045t = 360

t=8t = 8

Question analysis

Mr. Smith deposits $900 in a retirement account that earns 5% simple interest each year. This grid-in asks how many years it takes for the deposit to earn $360 in interest. Students find one year's interest and then how many years add up to the total, and the percent must become the right decimal.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Simple interest is grade 7 percent work, and a typical problem asks for the interest earned over a given time. This one runs the formula backward to find the time, one extra step, with friendly numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 7, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are friendly once one year’s interest is found, so most of the difficulty is setting up the right relationship. Students who divide $360 by 5 treat the rate as a dollar amount, and students who compare $360 with $900 stop at a percent. Multiplying back, eight years of $45, confirms the $360.

  1. Something you must already know minor Working it out

    Simple interest is the deposit times the rate times the years; the stem does not give this rule, and it must be known to see that each year adds the same $45. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The question gives the total interest, $360, and asks for the time, so the usual interest calculation has to be run backward, as 45t = 360. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Writing 5% as 0.05, not 0.5 or 0.005, is the one place-value decision, and a slip there changes the yearly interest by a power of ten. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work out the interest for one year, 5 percent of $900, then divide the $360 target by that yearly amount.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest
I=P×r×tI = P \times r \times t
I is the interest, the money the account earns. P is the deposit you start with. r is the yearly rate, written as a decimal. t is the number of years. With simple interest, every year earns the same amount, because it is always based on the first deposit.

  1. Sort out the question

    Mr. Smith puts $900 into an account. It earns 5% simple interest each year. Interest is extra money the bank pays you for keeping your money there. You need the number of years until the interest adds up to $360.

  2. Write 5% as a decimal

    Percent means “out of 100.” So 5% is 5 out of 100.

    5100=0.05\frac{5}{100} = 0.05

    Watch out: 5% is not 0.5 or 0.005. Using 0.005 makes a year’s interest $4.50, and that leads to 80 years.

  3. Find the interest for one year

    Each year the account earns 5% of $900.

    900×0.05=45900 \times 0.05 = 45

    So it earns $45 every year. Watch out: 5% is not $5 a year. Dividing 360 by 5 gives 72, which is wrong.

  4. Use the formula to find the years

    Put the numbers into the formula. I is 360, P is 900 and r is 0.05.

    I=P×r×tI = P \times r \times t

    360=900×0.05×t360 = 900 \times 0.05 \times t

    360=45×t360 = 45 \times t

    To undo “times 45,” divide both sides by 45.

    t=360÷45=8t = 360 \div 45 = 8

  5. Give the answer

    It takes 8 years. Watch out: $360 is 40% of $900. Gridding 40 stops too soon, because each year adds only 5%. So you would grid 8.

Check your answer

Add up the interest year by year. $45 a year for 8 years is 45×8=36045 \times 8 = 360 dollars. Also, 8 years at 5% a year is 40%, and 40% of $900 is $360.

More questions like this

The original DOE question
Question 61 as printed on page 192 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 61. Open page 192 of the handbook (PDF)

Question 105 of 120

2023 Test B Math, question 7

Question

On Fran’s map, 2 centimeters represents 3 kilometers. The points on this map that represent Fran’s home and Bryant State Park are 4.5 centimeters apart. How many kilometers apart are Fran’s home and Bryant State Park, to the nearest kilometer?

  • A3
  • B5
  • C7
  • D9
Show the answer

Correct answer: C

Let x represent the distance, in kilometers, between Fran’s house and Bryant State Park. Set up a proportion to solve for x:

x4.5=32\frac{x}{4.5} = \frac{3}{2} Cross multiply.
2x=3(4.5)2x = 3(4.5) Apply the multiplicative inverse property; divide both sides of the equation by 2.
x=13.52=634≈7x = \frac{13.5}{2} = 6\frac{3}{4} \approx 7

Question analysis

On Fran's map, 2 centimeters represent 3 kilometers, and her home and Bryant State Park are 4.5 centimeters apart on the map. Students find the real distance to the nearest kilometer. It is a map-scale proportion problem: the scale must be applied in the right direction, and the result rounded.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Map scales are standard grade 7 proportional reasoning, and this problem is typical: one distance, one scale, then rounding. Nothing beyond a single proportion is needed.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-proportion map question, and the only judgments are the direction of the scale and when to round. The phrase to the nearest kilometer tempts some students to round the scale itself before multiplying, which lands on a listed answer two kilometers too far. Rounding only the final 6.75 keeps the result right.

  1. Turning words into math minor Understanding the question

    “2 centimeters represents 3 kilometers” gives 1.5 kilometers per centimeter; set up the other way round, the scale shrinks the 4.5 centimeters instead of enlarging them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    The map distance is a decimal, 4.5 × 3 ÷ 2 = 6.75, and the result must then be rounded to the nearest kilometer, two places where a quick estimate drifts. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work out how many kilometers 1 centimeter on the map stands for, multiply by 4.5, and round only at the very end.

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).
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scale proportion
real distancemap distance=real distancemap distance\frac{\text{real distance}}{\text{map distance}} = \frac{\text{real distance}}{\text{map distance}}
A map shrinks every distance by the same amount. So real distance over map distance makes the same fraction everywhere on the map. Put kilometers on top on both sides and centimeters on the bottom. Then solve for the missing number.

  1. Sort out the question

    On Fran’s map, 2 centimeters stand for 3 kilometers in real life. Her home and the park are 4.5 centimeters apart on the map. You need the real distance in kilometers, rounded to the nearest whole kilometer.

  2. Set up the proportion

    Let x be the real distance, in kilometers. Put kilometers on top and centimeters on the bottom, on both sides.

    x4.5=32\frac{x}{4.5} = \frac{3}{2}

  3. Solve for x

    To get x alone, multiply both sides by 4.5.

    x=3×4.52=13.52=6.75x = \frac{3 \times 4.5}{2} = \frac{13.5}{2} = 6.75

    So the real distance is 6.75 kilometers. Watch out: keep the scale exact until the end. Rounding 1.5 kilometers per centimeter up to 2 gives 4.5×2=94.5 \times 2 = 9, choice D.

  4. Round to the nearest kilometer

    6.75 is between 6 and 7. Its decimal part, 0.75, is more than 0.5, so round up to 7.

    Watch out: don’t flip the scale. Multiplying 4.5 by 23\frac{2}{3} gives 3, choice A. So the answer is C, 7.

Check your answer

Each centimeter stands for 3÷2=1.53 \div 2 = 1.5 kilometers. So 4 centimeters is 6 kilometers, and half a centimeter is 0.75 kilometer. That adds to 6.75, which rounds to 7. The real number should be bigger than 4.5, and it is.

More questions like this

The original DOE question
Question 64 as printed on page 193 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 64. Open page 193 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 106 of 120

2023 Test B Math, question 12

Question

The perimeter of a 12-sided polygon is 100 centimeters. The length of one side is 25 centimeters. The length of another side is 35 centimeters. The remaining sides are equal in length to one another. What is the length of each of the remaining sides, in centimeters?

  • A4
  • B5
  • C20
  • D40
Show the answer

Correct answer: A

Let x represent the length of one of the 10 remaining sides:

25+35+10x=10025 + 35 + 10x = 100

60+10x=10060 + 10x = 100

10x=4010x = 40

x=4x = 4

Question analysis

A 12-sided polygon has a perimeter of 100 centimeters: one side is 25 cm long, another is 35 cm, and the rest are all equal to one another. Students find the length of each remaining side. This perimeter problem turns into a two-step equation, and the tricky part is counting the sides that are left.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving a two-step equation from a short description is standard grade 7 work. The perimeter setting is simple; the only extra care is counting the sides that remain after the two given ones.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 44 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The perimeter and the two named sides make the subtraction obvious, so the real decision comes after it: what to divide the leftover 40 centimeters by. Students who stop at the leftover length, or who lose track of how many sides remain, reach a listed choice. Writing 25 + 35 + 10x = 100 keeps the count of 10 in view.

  1. Turning words into math minor Understanding the question

    “The remaining sides are equal in length to one another” hides the count: 12 sides minus the two given ones leaves 10 equal sides to share the rest of the perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Subtracting 25 and 35 from 100, counting the sides left, and dividing are three steps, and the first result, 40, is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Subtract the two known sides from the perimeter, count how many sides are left, and share the remaining length equally among them.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a polygon
P=the sum of all the side lengthsP = \text{the sum of all the side lengths}
P is the perimeter, the total distance around a shape. A polygon is a flat shape with straight sides. A sum is the answer to an addition. So add the lengths of every side to get P.

  1. Sort out the question

    The shape is a polygon with 12 sides. Its perimeter is 100 centimeters. One side is 25 centimeters, and another is 35 centimeters. The other sides all have the same length. You need that length.

  2. Count the equal sides

    Two of the 12 sides are already known. The rest are the equal sides.

    12−2=1012 - 2 = 10

    So there are 10 equal sides. Call the length of each one x.

  3. Write an equation for the perimeter

    Use the perimeter formula. All 12 sides add up to 100. Ten sides of length x add up to 10x, which means 10 times x.

    P=the sum of all the side lengthsP = \text{the sum of all the side lengths}

    25+35+10x=10025 + 35 + 10x = 100

  4. Solve for x

    60+10x=10060 + 10x = 100

    10x=100−60=4010x = 100 - 60 = 40

    The 10 equal sides add up to 40 centimeters. Watch out: 40 is all ten sides together, not one side. Stopping here gives choice D. Now share the 40 among the 10 sides.

    x=40÷10=4x = 40 \div 10 = 4

  5. Give the answer

    Each remaining side is 4 centimeters long. Watch out: sharing the 40 centimeters between only 2 sides gives 20, choice C. That treats the shape as if it had 4 sides, but it has 12. So the answer is A, 4.

Check your answer

Add all 12 sides: 25+35+10×4=25+35+40=10025 + 35 + 10 \times 4 = 25 + 35 + 40 = 100. That is the perimeter the question gives, so 4 centimeters is right.

More questions like this

The original DOE question
Question 69 as printed on page 194 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 69. Open page 194 of the handbook (PDF)

Question 107 of 120

2023 Test B Math, question 13

Question

Shiow-Chen has a box that contains 4 green cards, 7 yellow cards, and 6 red cards. She randomly draws one card at a time from the box and does not return the cards to the box after they have been drawn. The first card she draws is yellow. The second card she draws is green. What is the probability that the third card she draws will also be green?

  • A15\frac{1}{5}
  • B417\frac{4}{17}
  • C14\frac{1}{4}
  • D13\frac{1}{3}
Show the answer

Correct answer: A

After Shiow-Chen draws a yellow card and a green card, the remaining cards are 3 green, 6 yellow, and 6 red. The probability of drawing a green card from that group is:

33+6+6=315=15\frac{3}{3 + 6 + 6} = \frac{3}{15} = \frac{1}{5}

Question analysis

Shiow-Chen's box holds 4 green, 7 yellow and 6 red cards, and she draws cards one at a time without putting them back. Her first card is yellow and her second is green. Students find the probability that the third card is also green, which means updating both the green count and the total first.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Simple probability is grade 6 content. The extra demand is tracking two earlier draws that were not put back, which changes the counts for the third draw, a step typical grade 7 problems often leave out.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 70 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The probability itself is one fraction, but it has to be built from a box that two draws have already changed. Students who use the original 17 cards, or who compare green with the other colors instead of with all the cards, reach listed choices that look reasonable. Listing the counts after each draw, then putting the green cards over all the cards left, avoids both.

  1. Much to keep track of moderate Working it out

    Three color counts and the total must each be updated after two draws, 4 green, 7 yellow and 6 red becoming 3, 6 and 6 out of 15, before the third draw is considered. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    “Does not return the cards” and the two named draws are the conditions that change the box, and the stem states them in passing before asking about the third card. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Take the two drawn cards out of the box: lower the yellow and green counts by one each, recount the total, then put the green cards left over the cards left.

Traps

  • Without replacement: In a second pick there is one fewer item, which changes the probability.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of results you wanttotal number of resultsP = \frac{\text{number of results you want}}{\text{total number of results}}
P is the probability, how likely something is, from 0 (never) to 1 (always). The top counts the results you want, here green cards. The bottom counts every possible result, here every card in the box. It works when each result is equally likely, as in a random draw.

  1. Sort out the question

    The box starts with 4 green, 7 yellow and 6 red cards. Shiow-Chen draws cards one at a time, and she does not put them back. The first card is yellow, and the second is green. You need the probability that the third card is green.

  2. Take out the two cards

    The cards are not put back, so the box changes. Lower the yellow count by 1 and the green count by 1.

    ColorAt the startAfter two draws
    Green43
    Yellow76
    Red66
    Total1715

  3. Use the probability formula

    For the third draw, the results you want are the 3 green cards left. The total is the 15 cards left.

    P=number of results you wanttotal number of resultsP = \frac{\text{number of results you want}}{\text{total number of results}}

    P=33+6+6=315P = \frac{3}{3 + 6 + 6} = \frac{3}{15}

    Watch out: using the full box gives 417\frac{4}{17}, choice B. But two cards are gone.

  4. Simplify and answer

    Divide the top and bottom by 3.

    315=15\frac{3}{15} = \frac{1}{5}

    Watch out: compare the green cards with all 15 cards, not with the 12 cards of other colors. That gives 14\frac{1}{4}, choice C. And the three colors are not equally likely, so 13\frac{1}{3} is wrong too. So the answer is A, 15\frac{1}{5}.

Check your answer

17 cards minus the 2 drawn leaves 15. One fifth of 15 is 3, the number of green cards left. Also, the chance went down from 417\frac{4}{17}. That makes sense, because a green card was taken out.

More questions like this

The original DOE question
Question 70 as printed on page 195 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 70. Open page 195 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 108 of 120

2023 Test B Math, question 14

Question

Each number in a sequence is 3 more than twice the number that comes just before it. If 93 is a number in the sequence, what number comes just before it?

  • A92
  • B90
  • C48
  • D45
Show the answer

Correct answer: D

Let n represent a number in the sequence, and let x represent the number that comes just before n:

n=3+2xn = 3 + 2x

93=3+2x93 = 3 + 2x

90=2x90 = 2x

45=x45 = x

Question analysis

In a sequence, each number is 3 more than twice the number just before it, and 93 is one of the numbers. Students find the number that comes just before 93. That means running the rule backward, undoing the adding and the doubling in reverse order.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students routinely solve two-step equations. The demand here is translating the wordy sequence rule into one, or running the rule backward in the right order, rather than any harder content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 29 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is two short steps, so the question rests on reading a wordy rule correctly and running it in reverse. Each wrong choice marks a different place to go wrong: ignoring the rule, stopping after the subtraction, or undoing the adding with more adding. Checking the answer forward, twice the number plus 3, confirms it reaches 93.

  1. Turning words into math moderate Understanding the question

    “Each number … is 3 more than twice the number that comes just before it” must become 2x + 3 = 93, with x the earlier number, not the later one. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method minor Working it out

    The rule is stated forward, from a number to the next, but the question runs backward, so each operation has to be undone in reverse order: subtract 3, then halve. (Schoenfeld, 1985)

Strategy: Call the earlier number x, write the rule as an expression in x set equal to 93, and solve it by undoing each operation in reverse order.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Stops one step early: A choice is a number from partway through the solution.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule
n=2x+3n = 2x + 3
x is any number in the sequence, a list that follows a rule. n is the number right after it. Twice means times 2, so twice x is 2x. Then 3 more means add 3. The same rule links every number to the one after it.

  1. Sort out the question

    A sequence is a list of numbers that follows a rule. Here, each number is “3 more than twice the number that comes just before it.” 93 is in the list. You need the number right before 93. Call it x.

  2. Write the rule as an equation

    Twice x is 2x. Three more than that is 2x + 3. The rule says this equals the next number, which is 93.

    n=2x+3n = 2x + 3

    93=2x+393 = 2x + 3

  3. Undo the adding of 3

    The rule doubles x first, then adds 3. To go backward, undo the steps in reverse order. First, subtract 3 from both sides.

    93−3=2x93 - 3 = 2x

    90=2x90 = 2x

    Watch out: 90 is twice the number, not the number itself. Stopping here gives choice B. And don’t add 3, which leads to 48, choice C.

  4. Undo the doubling

    Now divide both sides by 2.

    x=90÷2=45x = 90 \div 2 = 45

    Watch out: “just before” means just before in this list. It does not mean one less than 93, so 92 is wrong. So the answer is D, 45.

Check your answer

Run the rule forward from 45. Twice 45 is 90, and 3 more is 93: 2×45+3=932 \times 45 + 3 = 93. So 45 really does come just before 93.

More questions like this

The original DOE question
Question 71 as printed on page 195 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 71. Open page 195 of the handbook (PDF)

Question 109 of 120

2023 Test B Math, question 17

Question

m∙t=72m \bullet t = 72

For the equation above, what is the sum of the values of m when t=1t = 1, when t=2t = 2, and when t=3t = 3?

  • A12
  • B72
  • C132
  • D135
Show the answer

Correct answer: C

First, find the value of m for each value of t:

For t=1t = 1:

1m=721m = 72

m=72m = 72

For t=2t = 2:

2m=722m = 72

m=36m = 36

For t=3t = 3:

3m=723m = 72

m=24m = 24

Then, add those three values together:

72+36+24=13272 + 36 + 24 = 132

Question analysis

The equation m • t = 72 links two variables, and the raised dot means multiplication. Students find m when t is 1, 2 and 3, and then add those three values of m. Each value is a one-step division, and the trap is combining the t values before dividing.

NYS standards
  • 6.EE.7 one-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Solving equations like 3m = 72 is grade 6 work. This item repeats that step three times and adds the results, more bookkeeping than difficulty, so it is below typical grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 7, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each value of m comes from one division, so the work is bookkeeping rather than algebra. The trap is structural: adding 1, 2 and 3 first and dividing once feels efficient but answers a different question, since m does not change in step with t. Writing the three values of m in a row before adding them keeps the steps separate.

  1. Turning words into math minor Understanding the question

    “The sum of the values of m when t = 1, when t = 2, and when t = 3” asks for three separate solutions added together, not one solution at a combined t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three divisions, 72 ÷ 1, 72 ÷ 2 and 72 ÷ 3, and an addition make four small steps, each easy but each a chance to drop a value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Divide 72 by each value of t in turn to get three separate values of m, then add the three quotients.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division undoes multiplication
if a×b=c, then a=c÷b\text{if } a \times b = c \text{, then } a = c \div b
a, b and c are numbers, and b is not 0. If you know a product and one of the numbers that made it, divide to find the other one. For example, 4 × 5 = 20, so 4 = 20 ÷ 5.

  1. Sort out the question

    The equation is m∙t=72m \bullet t = 72. The dot means times. You need m three times: when t is 1, when t is 2 and when t is 3. Then add the three values of m.

  2. Get m by itself

    Since m times t is 72, m is 72 divided by t.

    m=72÷tm = 72 \div t

    Now put in each value of t, one at a time.

  3. Find m for each value of t

    Each time, divide 72 by the value of t. A table keeps the three answers in order.

    Value of tValue of m
    1m=72÷1=72m = 72 \div 1 = 72
    2m=72÷2=36m = 72 \div 2 = 36
    3m=72÷3=24m = 72 \div 3 = 24

  4. Add the three values

    72+36+24=13272 + 36 + 24 = 132

    Watch out: don’t add the t values first. 72÷(1+2+3)=1272 \div (1 + 2 + 3) = 12 is choice A, and it is not the same thing. Also, 72 alone is only the first value, choice B. So the answer is C, 132.

Check your answer

Check each pair by multiplying: 72×1=7272 \times 1 = 72, 36×2=7236 \times 2 = 72 and 24×3=7224 \times 3 = 72. All three pairs make 72. Then add again: 72 + 36 = 108, and 108 + 24 = 132.

More questions like this

The original DOE question
Question 74 as printed on page 196 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 74. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 110 of 120

2023 Test B Math, question 19

Question

Orange juice costs $2 per gallon, and grapefruit juice costs $3 per gallon. A recent shipment of equal amounts of the two types of juice costs a total of $250. How many gallons of orange juice are in the shipment?

  • A50
  • B62.5
  • C100
  • D125
Show the answer

Correct answer: A

Let x represent the number of gallons of orange juice. Since the shipment contains equal amounts of the two types of juice, x also represents the number of gallons of grapefruit juice. Write an equation to solve for x:

2x+3x=2502x + 3x = 250

5x=2505x = 250

x=50x = 50

Question analysis

Orange juice costs $2 a gallon and grapefruit juice costs $3 a gallon, and a shipment with equal amounts of each costs $250. Students find how many gallons of orange juice are in the shipment. The word problem leads to a simple equation; the key phrase is equal amounts of juice, not equal amounts of money.

NYS standards
  • 7.EE.4a two-step equations
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an equation from a short word problem is standard grade 7 work. This one has a single unknown, since both juices come in the same number of gallons, and only two short steps.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The equation is short once the key phrase is read as equal gallons, since one gallon of each juice costs $5 together and $250 buys 50 such pairs. Read as equal spending, the same phrase leads to a listed answer that is not a whole number of gallons. The other trap is giving the gallons of both juices together when only orange juice is asked for.

  1. Turning words into math moderate Understanding the question

    “Equal amounts of the two types of juice” means the same number of gallons of each, x, so the total cost is 2x + 3x, not two equal halves of $250. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Price one gallon of each juice together, then see how many such pairs $250 buys; that is the number of gallons of each kind.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost
cost=price per gallon×number of gallons\text{cost} = \text{price per gallon} \times \text{number of gallons}
The price per gallon is what one gallon costs. Multiply it by the number of gallons to get what you pay in all. For example, 3 gallons at $2 a gallon cost $6.

  1. Sort out the question

    Orange juice costs $2 a gallon. Grapefruit juice costs $3 a gallon. The shipment has equal amounts of each, so it has the same number of gallons of each juice. It costs $250 in all. You need the number of gallons of orange juice.

  2. Write the cost of each juice

    Let x be the number of gallons of each juice. Use the cost formula for each one.

    cost=price per gallon×number of gallons\text{cost} = \text{price per gallon} \times \text{number of gallons}

    orange juice cost=2×x=2x\text{orange juice cost} = 2 \times x = 2x

    grapefruit juice cost=3×x=3x\text{grapefruit juice cost} = 3 \times x = 3x

  3. Add the costs and solve

    The two costs add up to $250. Two x’s plus three x’s make five x’s.

    2x+3x=2502x + 3x = 250

    5x=2505x = 250

    x=250÷5=50x = 250 \div 5 = 50

    There are 50 gallons of each juice.

  4. Give the answer

    Watch out: equal amounts means equal gallons, not equal dollars. Spending $125 on each juice gives 62.5 gallons, choice B. Also, 100 gallons is both juices together, choice C. So the answer is A, 50.

Check your answer

Find each cost. 50 gallons of orange juice cost 50×2=10050 \times 2 = 100 dollars. 50 gallons of grapefruit juice cost 50×3=15050 \times 3 = 150 dollars. Together, 100+150=250100 + 150 = 250 dollars, the total in the question.

More questions like this

The original DOE question
Question 76 as printed on page 196 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 76. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 111 of 120

2023 Test B Math, question 20

Question

Five consecutive numbers have a sum of 110. What is the least of these consecutive numbers?

  • A11
  • B18
  • C20
  • D22
Show the answer

Correct answer: C

Let n represent the least of the five numbers. Then the next four consecutive numbers are (n+1)(n + 1), (n+2)(n + 2), (n+3)(n + 3), (n+4)(n + 4). Write an equation to solve for n:

n+(n+1)+(n+2)+(n+3)+(n+4)=110n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 110

5n+10=1105n + 10 = 110

5n=1005n = 100

n=20n = 20

Question analysis

Five consecutive numbers add up to 110, and students find the least of them. This number-properties question can be solved with an equation in one unknown, or by noting that the middle of an odd number of consecutive numbers equals their average. The bold word least matters, because the middle number is easier to find.

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 A sum of consecutive numbers leads to a two-step equation, standard grade 7 work. The average shortcut makes it quicker, and the only extra demand is answering for the least number rather than the middle one.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Dividing 110 by 5 is the natural first move, and it gives the middle number, which is a listed choice. The bold word least signals that one more step is needed, counting down two from the middle. The equation route, 5n + 10 = 110, goes straight to the least number but takes longer to set up.

  1. Turning words into math minor Understanding the question

    “Consecutive numbers” means each is 1 more than the last, and the bold “least” asks for the first of the five, not the middle one that dividing by 5 gives. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    None of the five numbers is given, so they must be written as n through n + 4, or pictured as two on each side of a middle value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Divide 110 by 5 to find the middle of the five numbers, then count down to the least of them.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive numbers
n, n+1, n+2, n+3, n+4n,\ n + 1,\ n + 2,\ n + 3,\ n + 4
Consecutive numbers come one right after another when you count, like 7, 8, 9. Each is 1 more than the one before. If n is the least number, the next ones are n + 1, n + 2, and so on.

  1. Sort out the question

    Five consecutive numbers add up to 110. Consecutive means each number is 1 more than the one before. You need the least, or smallest, of the five.

  2. Name the five numbers

    Let n be the least number. Then the five numbers are:

    n, n+1, n+2, n+3, n+4n,\ n + 1,\ n + 2,\ n + 3,\ n + 4

    Each one is 1 more than the one before it.

  3. Write the sum as an equation

    Add all five and set the sum equal to 110.

    n+(n+1)+(n+2)+(n+3)+(n+4)=110n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 110

    There are five n’s, which make 5n. The extra numbers add up to 1 + 2 + 3 + 4 = 10.

    5n+10=1105n + 10 = 110

  4. Solve for n

    Subtract 10 from both sides. Then divide both sides by 5.

    5n=1005n = 100

    n=100÷5=20n = 100 \div 5 = 20

    So the least number is 20.

  5. Give the answer

    Watch out: 110÷5=22110 \div 5 = 22 is the middle number, not the least. Picking 22 is choice D. The question asks for the least of the five. So the answer is C, 20.

Check your answer

Write out the five numbers and add them: 20+21+22+23+24=11020 + 21 + 22 + 23 + 24 = 110. The sum matches. Also, the middle number, 22, is 110 ÷ 5, and the least is 2 less than the middle.

More questions like this

The original DOE question
Question 77 as printed on page 197 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 77. Open page 197 of the handbook (PDF)

Question 112 of 120

2023 Test B Math, question 25

Question

The town of Waterville covers an area of 50 square miles. In 2005, its population was 84,000. In 2006, its population had decreased so that the average population per square mile was equal to 1,500. What was the total decrease in population from 2005 to 2006?

  • A7,500
  • B8,400
  • C9,000
  • D9,200
Show the answer

Correct answer: C

Let p represent the population of Waterville in 2006. The population per square mile is:

p50=1,500\frac{p}{50} = 1{,}500

Solve for p:

p=1,500(50)=75,000p = 1{,}500(50) = 75{,}000

The population decrease from 2005 to 2006 is 84,000−75,000=9,00084{,}000 - 75{,}000 = 9{,}000.

Question analysis

The town of Waterville covers 50 square miles and had 84,000 people in 2005. By 2006 the population had dropped so that there were 1,500 people per square mile on average. Students find how much the population decreased, turning the population density into a total population before subtracting.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a rate such as people per square mile is grade 6 content. Here it is one step inside a two-step problem with large round numbers, which matches typical grade 7 multi-step rate work.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 8, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question gives a total for 2005 but only a rate for 2006, so the two years cannot be compared until the rate becomes a population. After that, one subtraction finishes it. Students who stop at the 2006 population, or who reach for a number shaped like 84,000, land on listed choices, while the near miss above the answer comes from no sensible working.

  1. Turning words into math moderate Understanding the question

    “The average population per square mile was equal to 1,500” is a density, which has to become a total, 1,500 × 50 = 75,000, before it can be compared with 84,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    The 2006 population is only a midpoint; subtracting it from 84,000 is a second step, and the question asks for that decrease, not either population. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply 1,500 people per square mile by 50 square miles to get the 2006 population, then subtract it from the 2005 population.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Population per square mile
people per square mile=populationarea\text{people per square mile} = \frac{\text{population}}{\text{area}}
The population is the number of people who live in a place. The area is its size, here in square miles. Dividing gives the average number of people for each square mile. To get back the population, multiply people per square mile by the area.

  1. Sort out the question

    Waterville covers 50 square miles. In 2005, 84,000 people lived there. In 2006, there were 1,500 people per square mile, on average. You need how many fewer people lived there in 2006 than in 2005.

  2. Find the 2006 population

    Let p be the 2006 population. Use the formula.

    people per square mile=populationarea\text{people per square mile} = \frac{\text{population}}{\text{area}}

    1,500=p501{,}500 = \frac{p}{50}

    To undo dividing by 50, multiply both sides by 50.

    p=1,500×50=75,000p = 1{,}500 \times 50 = 75{,}000

    So 75,000 people lived there in 2006.

  3. Find the decrease

    A decrease is how much a number went down. Subtract the 2006 population from the 2005 population.

    84,000−75,000=9,00084{,}000 - 75{,}000 = 9{,}000

    The population went down by 9,000 people.

  4. Give the answer

    Watch out: 75,000 is the 2006 population, not the decrease. Choice A, 7,500, looks like it but has one zero fewer. The question asks how much the population dropped. So the answer is C, 9,000.

Check your answer

Work with people per square mile instead. In 2005, there were 84,000÷50=1,68084{,}000 \div 50 = 1{,}680 people per square mile. In 2006, there were 1,500. That is 180 fewer per square mile, and 180×50=9,000180 \times 50 = 9{,}000 fewer people.

More questions like this

The original DOE question
Question 82 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 82. Open page 199 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 113 of 120

2023 Test B Math, question 33

Question

A line segment with point W at its left end and point Y at its right end. Point X lies on the segment between W and Y, closer to Y.

The distance from W to X is 70% of the distance from W to Y. If the distance from X to Y is 15 miles, what is the distance, in miles, from W to X?

  • A10.5
  • B35
  • C45
  • D50
Show the answer

Correct answer: B

Since the distance from W to X is 70% of the distance from W to Y, the distance from X to Y is 30% of the distance from W to Y. Let d represent the distance from W to Y. Then 15 miles is 30% of d.

0.30d=150.30d=15

d=150.30=50d=\frac{15}{0.30}=50

So, the distance from W to X, in miles, is 50−15=3550-15=35.

Question analysis

A percent question with a line segment: point X lies between W and Y, the distance from W to X is 70% of the whole, and the distance from X to Y is 15 miles. Students find the distance from W to X. Matching the 15 miles to the right percent before finding the whole is the step students most often skip.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 percent problems often give a part and its percent and ask for the whole. This one gives the distance for the part the percent does not describe and asks for the other part, so the whole has to be found on the way.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The percent and the given distance describe different pieces of the road, which is the whole difficulty. Students who apply the 70% straight to the 15 miles get a small number fast, and students who solve correctly for the whole of WY must still remember to take XY away. Checking that 35 is 70% of 50 confirms the answer and catches both slips.

  1. Turning words into math moderate Understanding the question

    The 70% describes WX, but the 15 miles given is XY, so the sentence has to be recast as “XY is 30% of WY” before the 15 miles can be used. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Finding the complement, 30%, then the whole, 15 ÷ 0.30 = 50, then the part, 50 − 15, gives three results, and the middle one is itself a choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out what percent of WY the 15-mile stretch from X to Y must be, use it to find the whole of WY, then find the part from W to X.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}
The whole is the full amount. The part is a piece of it. The percent tells how many hundredths of the whole the part is. For example, 30% means 30 out of every 100, or 0.30. Use it whenever a percent describes a piece of a whole.

  1. Sort out the distances

    Point X sits between W and Y. So the trip from W to Y has two pieces: W to X, then X to Y.

    W to X is 70% of the whole distance from W to Y. X to Y is 15 miles. You need the distance from W to X, in miles.

  2. Find the percent for X to Y

    The whole distance from W to Y is 100%. W to X uses 70% of it. The rest is X to Y.

    100−70=30100 - 70 = 30

    So the 15 miles from X to Y is 30% of the whole distance.

  3. Find the whole distance

    Let d be the distance from W to Y. Use the percent formula. The part is 15, and the percent is 30.

    15=30100×d=0.30×d15 = \frac{30}{100} \times d = 0.30 \times d

    Divide both sides by 0.30 to get d by itself.

    d=15÷0.30=50d = 15 \div 0.30 = 50

    The distance from W to Y is 50 miles.

  4. Find the distance from W to X

    Take the 15 miles from X to Y away from the whole.

    50−15=3550 - 15 = 35

    W to X is 35 miles.

    Watch out: 50 miles is the whole distance from W to Y, choice D. You still have to subtract. Also, the 70% is of the whole distance, not of the 15 miles. Taking 70% of 15 gives 10.5, choice A.

    So the answer is B, 35 miles.

Check your answer

Is 35 really 70% of 50? 0.70×50=350.70 \times 50 = 35. Yes. And 35 + 15 = 50, the whole distance. The figure also shows X closer to Y. So W to X should be longer than 15 miles, and it is.

More questions like this

The original DOE question
Question 90 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 90. Open page 201 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 114 of 120

2023 Test B Math, question 36

Question

The side length of square S is shorter than the side length of square T by 1 centimeter. The area of square S is 9 square centimeters. What is the difference, in square centimeters, between the area of square T and the area of square S?

  • A3
  • B7
  • C8
  • D10
Show the answer

Correct answer: B

Let x represent the side length of square T, in centimeters. Then the side length of square S is (x−1)(x-1).

The area of square S is:

(x−1)(x−1)=9(x-1)(x-1)=9

(x−1)2=9(x-1)^{2}=9

x−1=3x-1=3

x=4x=4

So the side length of square T is 4 cm, and the area of square T is 16 sq cm.

The difference between the areas, in square centimeters, is 16−9=716-9=7.

Question analysis

Two squares differ in side length by 1 centimeter, and the smaller one has an area of 9 square centimeters. Students find how much greater the larger square's area is. It tests the link between a square's side and its area, and the extra area is easy to miscount when picturing the border that the longer sides add.

NYS standards
  • 6.G.5 perfect squares and perfect cubes
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 The area of a square and recognizing 9 as a perfect square are grade 6 ideas. This takes three short steps with small whole numbers, lighter than grade 7 work on composite figures, though the first step runs backward from area to side.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 44 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every number is small, so the work is keeping track of which quantity the question wants: a difference of areas, not a side or a single area. The added area forms an L-shaped strip around square S, and students who picture it as two strips 4 centimeters long count the corner square twice. Finding both areas, 9 and 16, and subtracting avoids reasoning about the strip at all.

  1. Many steps minor Working it out

    The work runs from area to side and back: the square root of 9 gives S’s side, adding 1 gives T’s, squaring gives 16, and subtracting gives the difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The phrase “shorter than the side length of square T by 1 centimeter” describes S from T’s point of view, so T’s side is S’s side plus 1, not minus 1. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the side of square S from its area, add 1 centimeter to get the side of square T, find that square's area, and subtract the smaller area from the larger.

Traps

  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square
A=s×sA = s \times s
A is the area, the space inside the square, in square units. s is the length of one side. All four sides of a square are equal, so the area is the side times itself. To go backward from an area, find the number that times itself gives the area.

  1. Sort out the two squares

    Square S has an area of 9 square centimeters. Its side is 1 centimeter shorter than the side of square T. You need the area of square T minus the area of square S, in square centimeters.

  2. Find the side of square S

    Use the area formula backward. Which number times itself makes 9?

    3×3=93 \times 3 = 9

    So each side of square S is 3 centimeters. This number is called the square root of 9.

  3. Find the side of square T

    Square T’s side is 1 centimeter longer than square S’s side.

    3+1=43 + 1 = 4

    Each side of square T is 4 centimeters.

  4. Find the area of square T

    Use the area formula with a side of 4 centimeters.

    A=4×4=16A = 4 \times 4 = 16

    The area of square T is 16 square centimeters.

  5. Subtract the areas

    Take the smaller area away from the larger one.

    16−9=716 - 9 = 7

    The difference is 7 square centimeters.

    Watch out: Picturing the new area as two strips, each 4 by 1, gives 8, choice C. But those strips share a 1-by-1 corner, so it gets counted twice. Also, 3 is a side length, not a difference of areas. That is choice A.

    So the answer is B, 7 square centimeters.

Check your answer

Draw square T and mark square S in one corner. The extra space is an L shape. It is a 3-by-1 strip, another 3-by-1 strip and a 1-by-1 corner. That is 3 + 3 + 1 = 7 square centimeters. It matches.

More questions like this

The original DOE question
Question 93 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 93. Open page 201 of the handbook (PDF)

Question 115 of 120

2023 Test B Math, question 40

Question

Angle A and angle B are complementary. If the measure of angle B is three times as great as the measure of angle A, what is the measure of angle B?

  • A45°
  • B67.5°
  • C75°
  • D135°
Show the answer

Correct answer: B

Let A represent the measure of angle A, and let B represent the measure of angle B, in degrees. Since the angles are complementary, A+B=90A+B=90. The question states that B=3AB=3A, or A=B3A=\frac{B}{3}. Substitute that expression for A in the first equation and solve for B:

A+B=90A+B=90

B3+B=90\frac{B}{3}+B=90

4B3=90\frac{4B}{3}=90

4B=2704B=270

B=67.5B=67.5

Question analysis

A geometry question about two complementary angles, where angle B is three times as large as angle A. Students find the measure of angle B. They must recall what complementary means, since the question does not say, and set up an equation that splits the total in the right proportion.

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 Writing and solving an equation from complementary angles is a grade 7 standard. This is the familiar version without a figure: one equation from the definition, then a division and a multiplication, typical of grade 7 classwork.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 7, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once complementary is read as a 90° total, the problem is one equation with a decimal answer, 4A = 90. The 22.5° for angle A is not listed, so stopping at the smaller angle is not rewarded here, but using 180° is: it gives 45° and 135°, and both appear among the choices. Checking that the chosen angle and one third of it add to 90° settles it.

  1. Something you must already know moderate Working it out

    The stem never says what complementary means; the 90° total has to be recalled, and a student who confuses it with the supplementary 180° doubles both angles. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The clause “the measure of angle B is three times as great as the measure of angle A” has to become B = 3A, so the 90° splits into four equal parts, not three. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Recall the total that complementary angles share, think of angle A as one part and angle B as three parts, find one part, then multiply for angle B.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Complementary angles
A+B=90A + B = 90
A and B are the measures of the two angles, in degrees. Complementary angles are two angles that add up to 90 degrees, the size of a square corner. Use this whenever a problem says two angles are complementary.

  1. Sort out the question

    Complementary angles are two angles that add up to 90 degrees. Angle B is three times as big as angle A. You need the measure of angle B, in degrees.

  2. Write the facts as equations

    Let A and B stand for the two measures. The angles are complementary, so use the formula.

    A+B=90A + B = 90

    Angle B is three times angle A, so:

    B=3AB = 3A

  3. Put 3A in place of B

    Replace B in the sum with 3A. Then there is only one letter to find.

    A+3A=90A + 3A = 90

    4A=904A = 90

    Think of it as 4 equal parts. Angle A is 1 part, and angle B is 3 parts.

  4. Find angle A

    Divide both sides by 4.

    A=90÷4=22.5A = 90 \div 4 = 22.5

    Angle A measures 22.5 degrees. That is not the answer yet, because the question asks for angle B.

  5. Find angle B

    Angle B is three times angle A.

    B=3×22.5=67.5B = 3 \times 22.5 = 67.5

    Angle B measures 67.5 degrees.

    Watch out: Angles that add to 180 degrees are called supplementary. Using 180 by mistake gives 45 and 135, choice D. Splitting 90 in half gives 45, choice A. That ignores the fact that B is three times A.

    So the answer is B, 67.5°.

Check your answer

Add the two angles: 22.5 + 67.5 = 90. So they are complementary. Now divide: 67.5 ÷ 22.5 = 3. So angle B is three times angle A. Both facts check out.

More questions like this

The original DOE question
Question 97 as printed on page 202 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 97. Open page 202 of the handbook (PDF)

Question 116 of 120

2023 Test B Math, question 41

Question

The height of a tree in 2013 is 300% of the height it was in 2003. By what percentage did its height increase over this ten-year period?

  • A3313%33\frac{1}{3}\%
  • B100%
  • C200%
  • D300%
Show the answer

Correct answer: C

Let h represent the height of the tree in 2003. Then the height of the tree in 2013 is 3h3h. Calculate the percent increase:

3h−hh=2hh=2=200%\frac{3h-h}{h}=\frac{2h}{h}=2=200\%

Question analysis

A percent question about a tree whose height in 2013 is 300% of its height in 2003. Students find the percent increase over the ten years. No heights are given, so the key is telling the difference between a percent of an amount and a percent change.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent-increase problems give two amounts to compare. This gives only a percent of the earlier height and no heights at all, so students must work out what that percent means for the change with nothing concrete to calculate with.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
2 steps
Reading level
Question and choices: about grade 3, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The stem hands students a percent and asks for a different percent, and the given 300% looks like a finished answer. The real step is separating “300% of” from “increased by”: a height that triples has grown by twice its original size. Picking a 2003 height of 100 units makes it concrete, since 300 − 100 = 200 units of growth on a base of 100.

  1. Turning words into math moderate Understanding the question

    “300% of the height it was in 2003” describes the new height, three times the old, not the increase; the question then asks for the change, which is a different percent. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers moderate Understanding the question

    No heights are given, only a percent, so students must invent a height such as 100 units or work with h and 3h to measure the change. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Something you must already know minor Working it out

    A percent increase divides the change by the starting amount, (3h − h) ÷ h, a rule the question assumes without stating. (Anderson & Krathwohl, 2001)

Strategy: Call the 2003 height 100 units, find the 2013 height from the percent given, then divide the change by the 2003 height and write the result as a percent.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
percent increase=new−originaloriginal×100\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100
Original is the starting amount. New is the amount after the change. New minus original is how much it grew. Dividing by the original compares the growth with the start. Multiplying by 100 turns that into a percent.

  1. Sort out the wording

    The 2013 height is 300% of the 2003 height. That means it is 3 times the 2003 height. You need the percent increase. That is how much the tree grew, compared with its 2003 height.

  2. Pick an easy height

    The question gives no real height, so choose one. Say the tree was 100 feet tall in 2003. Then the 2013 height is 300% of 100 feet.

    300100×100=300\frac{300}{100} \times 100 = 300

    In 2013, the tree was 300 feet tall.

  3. Find how much it grew

    The growth is how much taller the tree got. Subtract the old height from the new one.

    300−100=200300 - 100 = 200

    The tree grew 200 feet in ten years.

  4. Find the percent increase

    Use the formula. The original height is the 2003 height.

    percent increase=300−100100×100=2×100=200\text{percent increase} = \frac{300 - 100}{100} \times 100 = 2 \times 100 = 200

    The height went up by 200%.

    Watch out: 300% compares the new height with the old one. It is not the increase. That trap is choice D. Dividing the old height by the new one gives 3313%33\frac{1}{3}\%, choice A.

    So the answer is C, 200%.

Check your answer

Think about doubling. A tree that doubles adds 100% of its height. A tree that triples adds 200% of its height. So 100 feet plus 200% of 100 feet is 100 + 200 = 300 feet. That is 300% of 100 feet.

More questions like this

The original DOE question
Question 98 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 98. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 117 of 120

2023 Test B Math, question 45

Question

The Barnes family and the Ramirez family each have a pizza of equal size. The first pizza is cut into 18 equal slices, and the second pizza is cut into 15 equal slices. If the Barnes family eats 11 slices from the first pizza, what is the greatest number of whole slices from the second pizza that the Ramirez family can eat without eating a greater percentage of a pizza than the Barnes family ate?

  • A6
  • B7
  • C8
  • D9
Show the answer

Correct answer: D

The fraction of the pizza that the Barnes family eats is 1118\frac{11}{18}. Let x represent the number of slices of pizza that the Ramirez family eats. Then the fraction of the pizza that the Ramirez family eats is x15\frac{x}{15}. Write an inequality and solve for x:

x15≤1118\frac{x}{15}\leq\frac{11}{18}

x≤11×1518x\leq\frac{11\times 15}{18}

x≤16518x\leq\frac{165}{18}

x≤916x\leq 9\frac{1}{6}

So, the greatest number of whole slices of pizza the Ramirez family can eat is 9.

Question analysis

Two pizzas of the same size are cut into 18 and 15 equal slices, and one family eats 11 slices of the first. Students find the most whole slices of the second pizza the other family can eat without eating a larger share. It tests comparing fractions with different denominators and rounding to a whole number in the right direction.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 inequality problems usually ask for the greatest whole number that fits a budget. This one wraps that idea in a long stem about shares of two pizzas sliced differently, with several conditions to keep straight at once.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 11, 72 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The math is one inequality between fractions, but the long final sentence hides it, so students who skim fall back on comparing slice counts. Reasoning that the second pizza has 3 fewer slices, so the Ramirez family should eat 3 fewer, is the strongest pull. Comparing 11/18 with x/15, or turning 11/18 into about 61% and taking 61% of 15, keeps the comparison about shares of a pizza.

  1. Complex sentences moderate Understanding the question

    The question is one 42-word sentence ending in a nested condition, “without eating a greater percentage of a pizza than the Barnes family ate”, a reading hurdle rather than more math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

  2. Turning words into math moderate Understanding the question

    The limit of no “greater percentage of a pizza” must become the inequality x/15 ≤ 11/18, comparing fractions of two pizzas cut into different numbers of slices, not slice counts. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Solving gives x ≤ 11 × 15 ÷ 18 = 165/18, a division by 18 that leaves 9 1/6, which must then be cut back to whole slices. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the Barnes family's share as a fraction of their pizza, then find the largest whole number of fifteenths that does not exceed it, by cross-multiplying or by converting to decimals.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a pizza
fraction eaten=slices eatenslices in the pizza\text{fraction eaten} = \frac{\text{slices eaten}}{\text{slices in the pizza}}
The top counts the slices eaten. The bottom counts all the equal slices the pizza was cut into. The fraction tells how much of the whole pizza was eaten. Use fractions to compare pizzas cut into different numbers of slices.

  1. Sort out the question

    The two pizzas are the same size. The Barnes pizza has 18 slices, and they eat 11. The Ramirez pizza has 15 slices, so its slices are bigger. You need the most whole slices the Ramirez family can eat without eating a bigger share of their pizza.

  2. Find the Barnes family’s share

    Their share is the slices they ate out of all the slices in their pizza. Use the fraction formula.

    fraction eaten=1118\text{fraction eaten} = \frac{11}{18}

    The Barnes family ate 1118\frac{11}{18} of their pizza.

  3. Write an inequality

    Let x be the number of Ramirez slices. Their share is x15\frac{x}{15}. It must be no more than the Barnes share. The sign ≤ means less than or equal to.

    x15≤1118\frac{x}{15} \leq \frac{11}{18}

  4. Solve for x

    Multiply both sides by 15 to get x by itself.

    x≤11×1518=16518x \leq \frac{11 \times 15}{18} = \frac{165}{18}

    Now divide. 18 goes into 165 nine times, since 18 × 9 = 162. There are 3 left over.

    x≤9318=916x \leq 9\frac{3}{18} = 9\frac{1}{6}

  5. Pick the whole number

    x must be a whole number no bigger than 9169\frac{1}{6}. So the most is 9 slices. 10 slices would be too many.

    Watch out: The Ramirez pizza has 3 fewer slices, but that does not mean they eat 3 fewer. Their slices are bigger. Eating 11 − 3 = 8 slices is choice C, which is not the most they can eat.

    So the answer is D, 9.

Check your answer

Use decimals. The Barnes share is 11÷18≈0.61111 \div 18 \approx 0.611. Nine Ramirez slices are 9÷15=0.69 \div 15 = 0.6, just under it. Ten slices are 10÷15≈0.66710 \div 15 \approx 0.667, too much. So 9 is the most.

More questions like this

The original DOE question
Question 102 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 102. Open page 204 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 118 of 120

2023 Test B Math, question 48

Question

The outdoor temperature was 4.2° F at 10 p.m. After 3123\frac{1}{2} hours, the outdoor temperature was −0.7° F. Which value best represents the average change in temperature per hour?

  • A−1.4° F
  • B−1.0° F
  • C1.0° F
  • D1.4° F
Show the answer

Correct answer: A

First, recognize that the temperature decreased. This means that the order of subtraction matters and that the result is a negative number:

−0.7−4.2=−4.9-0.7-4.2=-4.9

The temperature decreased 4.9 degrees. Divide by 3.5 to get the average change per hour:

−4.9÷3.5=−1.4-4.9\div 3.5=-1.4

Question analysis

The temperature was 4.2° F at 10 p.m. and −0.7° F three and a half hours later. Students find the average change in temperature per hour. It tests subtracting and dividing with negative decimals, and the choices pair two sizes of change with both signs, so the sign has to be right as well as the size.

NYS standards
  • 7.NS.3 four operations with rational numbers
  • 7.NS.1c rational number subtraction
  • 7.NS.2b rational number division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Signed-number problems about temperature are standard grade 7 work. This one asks for an average rate, so a change that crosses zero must be divided by a mixed-number time, with the sign kept through both steps.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 5, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is short, but signs decide every choice: two choices come from treating −0.7 as 0.7, and one comes from getting the size right and the direction wrong. Students who think of the drop in two parts, 4.2 degrees down to zero and 0.7 more, find the 4.9-degree fall; dividing by 3.5 hours and remembering the temperature fell finishes it.

  1. Error-prone arithmetic moderate Working it out

    The change crosses zero, −0.7 − 4.2 = −4.9, and then a negative decimal is divided by 3.5, from the 3 1/2 hours, with the sign kept through both steps. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The phrase “average change in temperature per hour” means the change from the 4.2° start to the −0.7° end divided by the time, final minus initial, so a drop is negative. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. A very close wrong answer minor Choosing the answer

    The answer, −1.4° F, and the choice 1.4° F share every digit and differ only in sign, so a correct division followed by a careless sign cannot be caught by estimating. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Subtract the starting temperature from the ending one, keeping track of the sign, then divide that change by the 3.5 hours.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average change per hour
average change per hour=end−startnumber of hours\text{average change per hour} = \frac{\text{end} - \text{start}}{\text{number of hours}}
Start is the first temperature, and end is the last one. End minus start is the total change. It is negative when the temperature drops. Dividing by the number of hours shares that change equally over each hour.

  1. Sort out the question

    At 10 p.m. it was 4.2° F. After 3123\frac{1}{2} hours, it was −0.7° F. That is below zero. The temperature went down, so the change is negative. You need the average change for each hour.

  2. Find the total change

    Subtract the start from the end.

    −0.7−4.2=−4.9-0.7 - 4.2 = -4.9

    Picture a thermometer. Going from 4.2 down to 0 is a drop of 4.2 degrees. Going on to −0.7 is 0.7 more. In all, it fell 4.9 degrees.

    Watch out: 4.2 − 0.7 = 3.5 treats −0.7 as if it were above zero. That leads to choice B or C.

  3. Divide by the hours

    Write 3123\frac{1}{2} hours as 3.5 hours. Use the formula.

    average change per hour=−0.7−4.23.5=−4.93.5=−1.4\text{average change per hour} = \frac{-0.7 - 4.2}{3.5} = \frac{-4.9}{3.5} = -1.4

    To divide, think 49 ÷ 35 = 1.4, since 35 × 1.4 = 49. The change is negative, so it is −1.4 degrees per hour.

  4. Keep the sign

    The temperature fell, so the answer must be negative.

    Watch out: 1.4 has the right size, but it says the temperature went up. That is choice D.

    So the answer is A, −1.4° F.

Check your answer

Work forward. Dropping 1.4 degrees each hour for 3.5 hours is 1.4×3.5=4.91.4 \times 3.5 = 4.9 degrees. Then 4.2−4.9=−0.74.2 - 4.9 = -0.7. That is the temperature in the problem.

More questions like this

The original DOE question
Question 105 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 105. Open page 205 of the handbook (PDF)

Question 119 of 120

2023 Test B Math, question 57

Question

A box contains 14 red marbles, 10 black marbles, and 16 white marbles. How many more white marbles need to be added to the box so that the probability of randomly drawing a white marble is 35\frac{3}{5}?

  • A8
  • B10
  • C14
  • D20
Show the answer

Correct answer: D

Let w represent the number of additional white marbles.

Set the probability of drawing a white marble equal to 35\frac{3}{5} and solve the equation for w:

16+w14+10+16+w=35\frac{16+w}{14+10+16+w}=\frac{3}{5}

16+w40+w=35\frac{16+w}{40+w}=\frac{3}{5}

5(16+w)=3(40+w)5(16+w)=3(40+w)

80+5w=120+3w80+5w=120+3w

2w=402w=40

w=20w=20

Question analysis

A box holds 14 red, 10 black and 16 white marbles, and students find how many white marbles must be added so that the chance of drawing a white one is 3/5. Working backward from a target probability to a number of marbles is the unusual step.

NYS standards
  • 6.SP.8a uniform probability models
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 probability questions usually give the counts and ask for a probability. This runs the other way: the target probability is given, and the number to add changes both the white count and the total.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 8, 34 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The tempting shortcut is to take 3/5 of the 40 marbles already in the box, which forgets that the new white marbles join the total. Students who set up the probability with the unknown in both the top and the bottom solve it in a few lines, and testing choices works too: adding 20 gives 36 white of 60, exactly 3/5. The other two choices are counts copied from the stem.

  1. Turning words into math moderate Understanding the question

    “How many more white marbles need to be added” puts the unknown in two places, (16 + w)/(40 + w), because each white marble added also raises the total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    The unknown number of added marbles has to be carried as a letter through a proportion, 5(16 + w) = 3(40 + w), before it can be found. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Many steps minor Working it out

    Totaling the marbles, setting up the proportion, cross-multiplying and solving 80 + 5w = 120 + 3w make four steps, unless the choices are tested one by one. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Let w be the number of white marbles added, write the new probability with w in both the top and the bottom, set it equal to 3/5 and solve; or test each choice.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P(white)=white marblesall marblesP(\text{white}) = \frac{\text{white marbles}}{\text{all marbles}}
P(white) is the probability, or chance, of drawing a white marble. The top counts the white marbles. The bottom counts every marble in the box. It works because each marble is equally likely to be drawn.
Cross multiplication
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
a and c are the tops. b and d are the bottoms. When two fractions are equal, each top times the other bottom gives the same product. It works because you multiply both sides by both bottoms. Use it to clear fractions from an equation.

  1. Sort out the box

    The box has 14 red, 10 black and 16 white marbles.

    14+10+16=4014 + 10 + 16 = 40

    That is 40 marbles. You will add some white marbles. Call that number w. You need w so that the chance of drawing white is 35\frac{3}{5}.

  2. Write the new probability

    Adding w white marbles changes both counts. There will be 16 + w white marbles and 40 + w marbles in all. Use the probability formula.

    16+w40+w=35\frac{16 + w}{40 + w} = \frac{3}{5}

    Watch out: Taking 35\frac{3}{5} of 40 gives 24, and 24 − 16 = 8, choice A. That forgets that the new marbles also make the box bigger.

  3. Cross multiply

    Fractions are hard to solve with. Clear them with cross multiplication.

    5×(16+w)=3×(40+w)5 \times (16 + w) = 3 \times (40 + w)

    Then multiply each outside number by both terms inside the parentheses.

    80+5w=120+3w80 + 5w = 120 + 3w

  4. Solve for w

    Subtract 3w from both sides. Then subtract 80 from both sides.

    2w=402w = 40

    Divide both sides by 2.

    w=20w = 20

    You need to add 20 white marbles.

    So the answer is D, 20.

Check your answer

Add 20 white marbles. Then there are 16 + 20 = 36 white marbles and 40 + 20 = 60 marbles in all. Divide the top and bottom by 12: 3660=35\frac{36}{60} = \frac{3}{5}. It works.

More questions like this

The original DOE question
Question 114 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 114. Open page 208 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 120 of 120

2018 Test A Math, question 2

Question

The owner of a tree farm plants pine trees and oak trees in a ratio of 8:38:3. How many oak trees are planted if 264 pine trees are planted?

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

Show the answer

Correct answer: 99

Let xx be the number of oak trees when 264 pine trees are planted. Set up a proportion and solve for xx:

x264=38\frac{x}{264}=\frac{3}{8}

8x=7928x=792

x=99x=99

Question analysis

A ratio word problem about a tree farm that plants pine and oak trees in a fixed ratio of 8 to 3. Given how many pine trees are planted, students find the matching number of oak trees and grid it in. The main risk is pairing each number with the wrong kind of tree.

NYS standards
  • 6.RP.3a tables of equivalent ratios
  • 6.RP.1 ratio concepts and language
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 This is a one-ratio proportion with whole numbers, the kind of problem grade 6 students solve with equivalent ratios. A typical grade 7 task adds rates with fractions or a multistep percent; this has neither.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 4, 28 words.
Corrected from the DOE handbook
  • An arithmetic slip in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-ratio scaling problem with friendly numbers: 264 splits into 8 equal parts of 33. The only real risk is matching 264 to the wrong term of 8:3, and a quick sense check catches it, since oak trees are the smaller group and must number fewer than 264. Students who grid the one-part value or the total of both kinds of tree have answered a question that was not asked.

  1. Turning words into math minor Understanding the question

    The ratio 8:3 lists pine first, so the 264 pine trees match the 8 and the oak count matches the 3; reading the terms in the other order flips the setup. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    As a grid-in it offers no choices, so an upside-down proportion that gives 704 oak trees, far more than the 264 pines, has to be caught by the student's own check. (Enright, Morley & Sheehan, 2002)

Strategy: Divide the number of pine trees by the pine part of the ratio to get the size of one part, then multiply by the oak part.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook goes from x/264 = 3/8 to “8x = 762”. Cross-multiplying gives 8x = 3 × 264 = 792, so that line now reads 8x = 792. The handbook’s final answer, x = 99, was already right (792 ÷ 8 = 99).

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a and b are the two numbers in a ratio, written as a fraction. n is any number you choose, except zero. Multiplying both numbers by the same n gives an equal ratio. This is how you scale a small ratio up to real counts.

  1. Sort out the ratio

    A ratio compares two amounts. Here the ratio of pine trees to oak trees is 8:3. Pine comes first, so pine goes with the 8. Oak goes with the 3. So for every 8 pine trees, there are 3 oak trees.

    The farm planted 264 pine trees. You need the number of oak trees.

  2. Find the size of one part

    Think of the trees in equal groups, called parts. Pine trees fill 8 parts, and oak trees fill 3 parts. The 264 pine trees are the 8 parts, so divide by 8.

    264÷8=33264 \div 8 = 33

    Each part is 33 trees. Watch out: 33 is one part, not the number of oak trees. Don’t stop here.

  3. Find the number of oak trees

    Oak trees fill 3 parts. Multiply the part size by 3.

    3×33=993 \times 33 = 99

    You can see the same thing with equal ratios. Multiply both numbers of 8:3 by 33.

    83=8×333×33=26499\frac{8}{3} = \frac{8 \times 33}{3 \times 33} = \frac{264}{99}

    So 264 pine trees go with 99 oak trees.

  4. Avoid the traps and answer

    Watch out: if you match 264 to the 3 instead of the 8, you flip the ratio. That gives 704 oak trees. Watch out: the question asks only for oak trees. Adding the 264 pines gives 363. So you would grid 99.

Check your answer

The ratio 8:3 says there are fewer oak trees than pine trees. 99 is less than 264, so that fits. Also, divide both counts by 33: 264÷33=8264 \div 33 = 8 and 99÷33=399 \div 33 = 3. That gives back 8:3.

More questions like this

The original DOE question
Question 59 as printed on page 81 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 59. Open page 81 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.