SHSAT Sunset

Practice Test

7.EE.3 multi-step rational number problems

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Question 1 of 10

2026 Practice Test C Math, question 11

Question

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

  • A198
  • B612
  • C620
  • D632
Show the answer

Correct answer: B

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

5×576=2,8805 \times 576 = 2{,}880

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

6×378=2,2686 \times 378 = 2{,}268

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

2,880−2,268=6122{,}880 - 2{,}268 = 612

Question analysis

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

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

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

    612, 620 and 632 lie within 20 of one another, so rounding 576 and 378 to friendly numbers before multiplying lands on a wrong choice; only exact products separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    Two products of three-digit numbers, 5 × 576 = 2,880 and 6 × 378 = 2,268, must be done by hand and then subtracted with regrouping. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Allison has 5 albums with 576 stamps in each. She moves all her stamps into 6 new albums. Each new album holds 378 stamps. Some stamps will not fit. You need the number left over.

  2. Count all her stamps

    Use the equal-groups formula. Split 576 into 500 + 76 to make it easy:

    5×576=5×500+5×765 \times 576 = 5 \times 500 + 5 \times 76

    2,500+380=2,8802{,}500 + 380 = 2{,}880

    Allison has 2,880 stamps.

  3. Count what the new albums hold

    Do the same for the 6 new albums. Split 378 into 300 + 78:

    6×378=6×300+6×786 \times 378 = 6 \times 300 + 6 \times 78

    1,800+468=2,2681{,}800 + 468 = 2{,}268

    The new albums hold 2,268 stamps in all.

  4. Subtract to find the leftovers

    The stamps left over are the ones she has minus the ones that fit:

    2,880−2,268=6122{,}880 - 2{,}268 = 612

    Watch out: 576−378=198576 - 378 = 198, choice A, compares just one old album with one new one. Rounding to 580 and 380 first gives 620, choice C. Rounding only 576 gives 632, choice D. Use the exact numbers.

    So the answer is B, 612.

Check your answer

Add the leftovers to the stamps that fit: 2,268+612=2,8802{,}268 + 612 = 2{,}880. That is exactly the number Allison started with. So no stamps were lost or counted twice.

More questions like this

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

Question 2 of 10

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 3 of 10

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 4 of 10

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 5 of 10

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 6 of 10

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 7 of 10

2023 Test B Math, question 5

Question

Solve the equation for x:

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

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

Show the answer

Correct answer: 7

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

(2149)(1448)x=0.875\left(\frac{21}{49}\right)\left(\frac{14}{48}\right)x = 0.875

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

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

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

Question analysis

This equation puts x in a fraction with (21)(14) on top and (49)(48) below, set equal to the decimal 0.875. The grid-in asks for the value of x. Students do best to cancel common factors before multiplying anything out, and it helps to know the fraction that 0.875 equals.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.2c properties for multiplying and dividing
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Seventh graders solve equations like ax = b, usually with a simple coefficient. Here the coefficient is a product of four numbers that is best cancelled first, and a decimal has to be matched with a fraction.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 6 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-unknown equation dressed in large numbers, and the work is in reducing them without slipping. Students who cancel early find that the whole coefficient is one eighth and that 0.875 is seven eighths, which makes the last step mental. Those who multiply out face a four-digit product and a decimal division, and with nothing to check against, an early slip goes unnoticed.

  1. Error-prone arithmetic moderate Working it out

    Multiplied out, the coefficient is 294 over 2,352 and the right side is the decimal 0.875; both invite slips unless 21/49 and 14/48 are cancelled first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. No obvious method minor Working it out

    The quick route pairs 21 with 49 and 14 with 48 before multiplying, so that 3/7 × 7/24 collapses to 1/8; the layout of the fraction does not suggest that pairing. (Schoenfeld, 1985)

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

    The result is entered with no choices, so a coefficient’s denominator or a quotient taken the wrong way looks as complete as the right value. (Enright, Morley & Sheehan, 2002)

Strategy: Cancel before multiplying: reduce 21 over 49 and 14 over 48, multiply the simplified fractions, then undo that coefficient, writing 0.875 as a fraction if it helps.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying fractions
ab×cd=a×cb×d\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}
a and c are the tops, or numerators. b and d are the bottoms, or denominators. Multiply the tops together and the bottoms together. It also works backward: you can split one fraction into two smaller ones. That lets you simplify before the numbers get big.

  1. Sort out the question

    You need the value of x. On top, 21, 14 and x are multiplied. The bottom is 49 times 48. The right side is 0.875.

    Multiplying 49 by 48 makes a big number. It is easier to simplify first.

  2. Split the fraction into two

    Use the rule for multiplying fractions backward. Pair 21 with 49, and 14 with 48.

    21×1449×48=2149×1448\frac{21 \times 14}{49 \times 48} = \frac{21}{49} \times \frac{14}{48}

    Now the equation says these two fractions times x equal 0.875.

  3. Simplify each fraction

    To simplify, divide the top and bottom by the same number. Both 21 and 49 divide by 7. Both 14 and 48 divide by 2.

    2149=371448=724\frac{21}{49} = \frac{3}{7} \qquad \frac{14}{48} = \frac{7}{24}

    Watch out: 48 ÷ 2 is 24, not 12. Writing 712\frac{7}{12} leads to 3.5.

  4. Multiply the simple fractions

    37×724=3×77×24=324=18\frac{3}{7} \times \frac{7}{24} = \frac{3 \times 7}{7 \times 24} = \frac{3}{24} = \frac{1}{8}

    The 7 on top and the 7 on the bottom cancel, since 7 ÷ 7 = 1. Then divide 3 and 24 by 3 to get 18\frac{1}{8}. So the equation becomes:

    18×x=0.875\frac{1}{8} \times x = 0.875

  5. Solve for x

    One eighth of x is 0.875. To undo dividing by 8, multiply both sides by 8.

    x=0.875×8=7x = 0.875 \times 8 = 7

    Watch out: 8 is not the answer, and don’t divide 0.875 by 8. So you would grid 7.

Check your answer

0.875 is the same as 78\frac{7}{8}. One eighth of 7 is 78\frac{7}{8}, so x = 7 works. You can also multiply it all out: 21×14×7=2,05821 \times 14 \times 7 = 2{,}058 and 49×48=2,35249 \times 48 = 2{,}352. Then 2,058÷2,352=0.8752{,}058 \div 2{,}352 = 0.875.

More questions like this

The original DOE question
Question 62 as printed on page 192 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 62. Open page 192 of the handbook (PDF)

Question 8 of 10

2022 Test A Math, question 28

Question

The table shows the income and expenses in a bicycle shop on one day.

BICYCLE SHOP TRANSACTIONS

Description Income Expense
Paid employees $120×4\$120 \times 4
Sold 2 bicycles $400×2\$400 \times 2
Sold clothing $95
Bought new door $500
Sold bike parts p

The sum of all transactions was $70. What is the value of p?

  • A−$275
  • B−$85
  • C$70
  • D$155
Show the answer

Correct answer: D

The sum of all transactions ($70) includes the income data (positive values) and the expense data (negative values). Write an equation to determine the value of p by adding all the income data and subtracting all the expense data:

400∙2+95+p−120∙4−500=70400 \bullet 2 + 95 + p - 120 \bullet 4 - 500 = 70

Simplify the expression to determine the value of p:

800+95+p−480−500=70800 + 95 + p - 480 - 500 = 70

895+p−980=70895 + p - 980 = 70

p−85=70p - 85 = 70

p=155p = 155

Question analysis

A table lists one day of income and expenses at a bicycle shop, including two products to work out and an unknown amount, p, from selling bike parts. Given that all the transactions add up to $70, students find p. They must treat income as positive and expenses as negative, then solve an equation.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve multi-step money problems with positive and negative amounts. This table adds an unknown income entry, so students treat income as positive and expenses as negative before solving a one-step equation.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 5, 44 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The table turns a one-variable equation into bookkeeping: each entry must be multiplied out and given a sign by its column before anything is added. The known entries come to a loss of $85, so the bike parts must cover that loss and still leave $70, which is why p is larger than the day's total. Students who stop at the running total, or who copy the $70, never use the equation at all.

  1. Turning words into math moderate Understanding the question

    “The sum of all transactions was $70” has to be read with income positive and expenses negative, giving 800 + 95 + p − 480 − 500 = 70. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    Two cells hold products, $120 × 4 and $400 × 2, that must be worked out, and p sits in the Income column beside the sales, so it is added. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Error-prone arithmetic minor Working it out

    The known entries total 895 − 980 = −85, a negative running sum, so p must exceed 70 by 85 rather than fall short of it. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work out the two products in the table, add the income entries and subtract the expenses, set that total plus p equal to 70, and solve for p.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of all transactions
sum=income−expenses\text{sum} = \text{income} - \text{expenses}
A transaction is money coming in or going out. Income is money that comes in, so it counts as plus. Expenses are money that goes out, so they count as minus. The sum is what is left, like money added to and taken from a wallet.
Keeping an equation balanced
if a=b, then a+c=b+c, a−c=b−c and a÷c=b÷c\text{if } a = b \text{, then } a + c = b + c \text{, } a - c = b - c \text{ and } a \div c = b \div c
a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

  1. Sort out the question

    The table lists money that came into the shop, called income. It also lists money that went out, called expenses. The bike parts brought in p dollars, an amount you don’t know. All the income minus all the expenses is $70. You need p.

  2. Work out the products in the table

    Two cells in the table show multiplication.

    120×4=480120 \times 4 = 480

    400×2=800400 \times 2 = 800

    So paying the employees cost $480. The 2 bicycles brought in $800.

  3. Write the equation

    The income is $800, $95 and p. The expenses are $480 and $500. p sits in the Income column, so it is added.

    income−expenses=70\text{income} - \text{expenses} = 70

    800+95+p−480−500=70800 + 95 + p - 480 - 500 = 70

  4. Combine the known numbers

    800+95=895480+500=980800 + 95 = 895 \qquad 480 + 500 = 980

    895−980=−85895 - 980 = -85

    The known entries come to −85 dollars. That is a loss of $85. Now the equation is shorter:

    p−85=70p - 85 = 70

    Watch out: −85 is choice B, but it is not p. You still have to solve for p.

  5. Solve for p

    Add 85 to both sides. This keeps the equation balanced.

    p−85+85=70+85p - 85 + 85 = 70 + 85

    p=155p = 155

    The bike parts brought in $155. Watch out: $70 is the total for the day, not the bike parts. Don’t just copy it, choice C. So the answer is D, $155.

Check your answer

Put 155 back in. Income: 800+95+155=1,050800 + 95 + 155 = 1{,}050 dollars. Expenses: 480+500=980480 + 500 = 980 dollars. 1,050−980=701{,}050 - 980 = 70. That matches. And p should be more than 70, because the bike parts had to cover the $85 loss first.

More questions like this

The original DOE question
Question 85 as printed on page 86 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 86, question 85. Open page 86 of the handbook (PDF)

Question 9 of 10

2018 Test A Math, question 28

Question

215+3310+425+5122\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}

What is the value of the expression shown above?

  • A1472014\frac{7}{20}
  • B142514\frac{2}{5}
  • C1572015\frac{7}{20}
  • D152515\frac{2}{5}
Show the answer

Correct answer: D

215+3310+425+5122\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}

Convert all the fractions to a common denominator (10):

2210+3310+4410+55102\frac{2}{10}+3\frac{3}{10}+4\frac{4}{10}+5\frac{5}{10}

=(2+3+4+5)+(2+3+4+510)=(2+3+4+5)+\left(\frac{2+3+4+5}{10}\right)

=14+1410=1525=14+1\frac{4}{10}=15\frac{2}{5}

Question analysis

A fractions question: add four mixed numbers whose fractions have denominators of 5, 10, 5 and 2. Students rewrite the fractions over a common denominator, add the whole numbers and the fractions, regroup any extra whole, and simplify. Regrouping and lowest terms both matter, since the choices differ in exactly those places.

NYS standards
  • 7.NS.1d properties for adding and subtracting
  • 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 with unlike denominators is grade 5 content. With four terms and a regrouping step it takes care, but it asks less than typical grade 7 work with negative rational numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are friendly: every fraction converts to tenths, and the whole parts add to 14. The risk is in the bookkeeping, either forgetting to carry the extra whole from the fractions or adding numerators over a made-up denominator. Converting to decimals, 2.2 + 3.3 + 4.4 + 5.5, is a fast check that avoids both.

  1. Error-prone arithmetic moderate Working it out

    Four mixed numbers with denominators 5, 10, 5 and 2 must be rewritten as tenths, 2 + 3 + 4 + 5 tenths, and the 14/10 regrouped into a whole and a fraction. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    15 2/5 and 14 2/5 are written alike and share the fraction 2/5; only the whole carried from 14/10 separates them, so a missed carry is invisible at a glance. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    The fractions total 1 4/10, but every choice is in lowest terms, so the 4/10 has to be reduced to 2/5 before it matches. (Enright, Morley & Sheehan, 2002)

Strategy: Convert each mixed number to a decimal, since these fractions all convert exactly, add them, and turn the decimal part back into a fraction in lowest terms.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a is the top number, called the numerator. b is the bottom number, called the denominator. n is any whole number except zero. Multiplying the top and bottom by the same n gives an equal fraction. Use it to give fractions the same denominator so you can add them.

  1. Sort out the question

    You need to add four mixed numbers. A mixed number is a whole number plus a fraction, like 2152\frac{1}{5}. The fractions have different denominators, or bottom numbers: 5, 10, 5 and 2. To add fractions, they need the same denominator.

  2. Add the whole numbers

    Add the whole numbers first. You will add the fractions on their own after that.

    2+3+4+5=142 + 3 + 4 + 5 = 14

    The whole numbers make 14.

  3. Rewrite each fraction in tenths

    Use 10 as the denominator, since 5 and 2 both go into 10.

    15=1×25×2=210\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10}

    25=2×25×2=410\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}

    12=1×52×5=510\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}

    310\frac{3}{10} is already in tenths. Watch out: when you change a bottom number, change the top too. Adding the old tops, 1 + 3 + 2 + 1 = 7, and putting them over 20 gives 7/20. That is in choices A and C.

  4. Add the fractions

    210+310+410+510=1410\frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} = \frac{14}{10}

    14 tenths is more than 1 whole. Ten tenths make 1, so 14 tenths is 1 whole and 4 tenths.

    1410=1410=125\frac{14}{10} = 1\frac{4}{10} = 1\frac{2}{5}

    To simplify 4/10, divide the top and bottom by 2.

  5. Put it all together

    Add the 14 wholes to the sum of the fractions.

    14+125=152514 + 1\frac{2}{5} = 15\frac{2}{5}

    Watch out: carry the extra whole into the 14. Missing it gives 142514\frac{2}{5}, choice B. So the answer is D, 152515\frac{2}{5}.

Check your answer

Change each mixed number to a decimal: 2.2+3.3+4.4+5.5=15.42.2 + 3.3 + 4.4 + 5.5 = 15.4. The 0.4 is 4 tenths, which is 2/5. So the sum is 152515\frac{2}{5}.

More questions like this

The original DOE question
Question 85 as printed on page 86 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 85. Open page 86 of the handbook (PDF)

Question 10 of 10

2018 Test A Math, question 52

Question

Each week, Leon has fixed expenses of $1,250 at his furniture shop. It costs him $150 to make a chair in his shop, and he sells each chair for $275. What is Leon’s profit if he makes and sells 25 chairs in 1 week?

  • A$1,875
  • B$2,500
  • C$3,125
  • D$4,375
Show the answer

Correct answer: A

Each chair costs Leon $150 to make, and he sells the chair for $275.

His profit is found by subtracting the cost from the price:

$275 − $150 = $125 per chair

If Leon makes and sells 25 chairs in a week, his initial profit is

25 × $125 = $3,125

However, Leon has additional fixed expenses of $1,250 per week, so this cost must also be subtracted to arrive at the profit. His final profit is

$3,125 − $1,250 = $1,875

Question analysis

This profit word problem gives Leon's weekly fixed expenses at his furniture shop, the cost to make each chair and its selling price, then asks his profit for 25 chairs in one week. Students must find the profit on each chair, multiply, and account for the fixed expenses. Forgetting the fixed expenses, or adding them, are the main slips.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Multi-step money problems are standard grade 7 work. This one has three whole-number steps and a familiar profit idea, so it matches typical grade 7 demand, with the fixed cost as the detail to remember.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is whole-number and friendly, so the question is about bookkeeping: which amounts are income, which are costs, and which happen once a week. The fixed expenses are named first in the stem and are easy to forget by the time the per-chair profit is multiplied. A final check that revenue minus all costs, 6,875 − 3,750 − 1,250, gives the same result confirms it.

  1. Much to keep track of moderate Working it out

    Three dollar amounts play different roles, a weekly fixed expense, a cost per chair and a price per chair, and each must enter the profit with the right sign. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Something you must already know minor Working it out

    The bold "profit" is not defined; students must know it means money taken in minus every cost, including costs that do not depend on the number of chairs. (Anderson & Krathwohl, 2001)

Strategy: Subtract the cost from the price to get the profit per chair, multiply by the number of chairs, then subtract the week's fixed expenses.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Profit
P=I−CP = I - C
P is the profit, the money Leon keeps. I is the income, the money he takes in from sales. C is the total cost: every dollar he spends, including fixed expenses. Profit is what is left after all costs are paid.

  1. Sort out what you know

    Leon pays $1,250 each week no matter what. These are fixed expenses, like rent. Each chair costs $150 to make, and he sells each one for $275.

    He makes and sells 25 chairs. Profit is the money left after paying all the costs.

  2. Find the profit on one chair

    Each chair sells for $275 but costs $150 to make. Subtract to see what he keeps:

    275−150=125275 - 150 = 125

    He keeps $125 on each chair he sells.

  3. Find the profit on 25 chairs

    25×125=3,12525 \times 125 = 3{,}125

    The chairs bring in $3,125 more than they cost to make. Watch out: $3,125 is choice C, but it is not the final profit. The fixed expenses still have to be paid.

  4. Subtract the fixed expenses

    P=3,125−1,250=1,875P = 3{,}125 - 1{,}250 = 1{,}875

    Watch out: the $1,250 is money spent, so subtract it. Adding it gives $4,375, choice D. So the answer is A, $1,875.

Check your answer

Work it the long way. Sales: 25 × $275 = $6,875. Costs: 25 × $150 = $3,750, plus $1,250, which makes $5,000. Profit: $6,875 − $5,000 = $1,875. Same answer.

More questions like this

The original DOE question
Question 109 as printed on page 91 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 91, question 109. Open page 91 of the handbook (PDF)