Practice Test

Percent of the wrong amount

Question 1 of 16

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: $0.80x$

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

End of second week: $0.80(0.80x) = 0.64x$ or 64%

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer major
Turning words into math moderate
Variables instead of numbers minor
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.

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)
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)
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

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: moderate
Adds the weekly losses, 20% + 20% = 40%, and gives that figure, which describes the amount lost rather than the percentage still left.
B Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Subtracts 20% of the original amount twice, 100 − 20 − 20 = 60, though the second week’s 20% comes from the 80% left, not the starting amount.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds that 80% remains after the first week and stops, never taking the second week’s 20% from that remainder.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of an 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: $\text{percent left} = 100\% - \text{percent lost}$; 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.

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.
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.
Work out the first week
Use the percent formula. The first week starts with 100 grams:
$\frac{20}{100} \times 100 = 20$
The chemical loses 20 grams, so $100 - 20 = 80$ grams are left. That is 80% after one week.
Watch out: 80% is only the first week. Stopping here gives choice D.
Work out the second week
The second week starts with 80 grams, not 100. So take 20% of 80:
$\frac{20}{100} \times 80 = 16$
This time the chemical loses only 16 grams, so $80 - 16 = 64$ 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.
Read the answer as a percent
Week Start (grams) Loss (grams) Left (grams)
1 100 20 80
2 80 16 64
Another way: each week keeps 80%, or 0.8, of its start. Two weeks gives $0.8 \times 0.8 = 0.64$ , which is 64%.
64 grams out of the original 100 grams is 64%. So the answer is C, 64%.

Week	Start (grams)	Loss (grams)	Left (grams)
1	100	20	80
2	80	16	64

Check your answer

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

More questions like this

The original DOE question

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

Question 2 of 16

2026 Practice Test D Math, question 50

Question

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

BIRDS AT BIRD FEEDER

Day	Number of Birds
Monday	30
Tuesday	24

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

A6%
B20%
C24%
D25%

Show the answer

Correct answer: B

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

$30 - 24 = 6$

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

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

0.20 represents 20%.

There was a 20% decrease from Monday to Tuesday.

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 3 of 16

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.

$\frac{\text{King}}{\text{Jones}} = \frac{x}{550}$

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

Then solve the proportion for x.

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

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

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

$550(1) = x(1.10)$

$550 = 1.10x$

$550 \div 1.10 = (1.10 \div 1.10)x$

$500 = x$

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

$550 - 500 = 50$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math major
A very close wrong answer moderate
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.

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)
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

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

B Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes 10% of 550, measuring the increase from Jones's herd instead of from King's smaller herd, which is the amount the 10% is based on.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: weak
Takes 10% off 550 to get King's herd, 550 − 55 = 495, using the wrong base, then gives that herd instead of the difference.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds King's herd correctly, 550 ÷ 1.1 = 500, then stops there instead of subtracting it from 550 to get how many more.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{new} = \text{original} \times \left(1 + \frac{p}{100}\right)$; 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.

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.
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.
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.
$\text{new} = \text{original} \times \left(1 + \frac{p}{100}\right)$
$550 = K \times 1.10$
$K = 550 \div 1.10 = 500$
Mr. King has 500 goats. Watch out: 500 is Mr. King’s herd, choice D. It is not the difference.
Find the difference
$550 - 500 = 50$
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 $0.10 \times 500 = 50$ , and $500 + 50 = 550$ , 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 4 of 16

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?

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.

$c = 40 + m$

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%):

$\frac{60}{80\%} = \frac{60}{0.80} = 75$

Use $75 as the total cost to find m.

$40 + m = 75$

$m = 35$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Combines ideas moderate
A very close wrong answer moderate
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.

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)
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)
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

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Takes 20% off the $60 again, 60 × 0.8 = 48, instead of undoing the discount, then subtracts the $40 fee to get 8 miles.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Ignores the coupon and treats the $60 as the full cost, so it subtracts the $40 fee and gets 20 miles.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Undoes the coupon by adding 20% of $60 to get $72, though the 20% was taken of the larger pre-coupon cost, then subtracts the $40 fee.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rental cost: $c = 40 + 1 \times m$; 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: $\text{price paid} = \text{full cost} \times \left(1 - \frac{p}{100}\right)$; 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.

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.
Write what she paid
20% off means she paid what was left of the full cost.
$100\% - 20\% = 80\%$
As a decimal, 80% is 0.80.
$\text{price paid} = \text{full cost} \times \left(1 - \frac{p}{100}\right)$
$60 = \text{full cost} \times 0.80$
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.
$\text{full cost} = 60 \div 0.80 = 600 \div 8 = 75$
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.
Find the miles
$c = 40 + 1 \times m$
$75 = 40 + m$
$m = 75 - 40 = 35$
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 $0.20 \times 75 = 15$ dollars. Then $75 - 15 = 60$ . 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 5 of 16

2024 Test A Math, question 38

Question

By what percent did the price of a cup of coffee increase if its price was increased from $1.25 to $1.35?

A7%
B8%
C10%
D12%

Show the answer

Correct answer: B

Subtract to determine the amount of the increase.

$1.35 - 1.25 = 0.10$

Divide the increase by the original price.

$0.10 \div 1.25 = 0.08$

Multiply the quotient by 100 to express the decimal answer as a percent.

$0.08 \times 100 = 8$

Question analysis

This percent increase question gives the price of a cup of coffee before and after a raise, from $1.25 to $1.35, and asks by what percent the price went up. Students find the change in price and compare it with a price. Deciding which price to compare the change with is the tricky part.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Percent increase is grade 7 content, and this is a standard case with two prices. What makes it harder than routine is deciding which price the change is measured against and dividing small decimals accurately.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 6, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Most students see that the price rose by 10 cents; the question is what to compare that with. Percent change is measured from the original amount, and dividing by the new price gives a number close enough to the answer that rounding hides the mistake. Reading the 10 cents directly as 10 percent is the other quick trap, since 10 cents is only 10 percent of a dollar.

A very close wrong answer moderate Choosing the answer
Dividing the 10-cent rise by either price gives a single-digit percent, 8% from $1.25 and about 7.4% from $1.35, so only the choice of base separates the two listed answers. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic moderate Working it out
Without a calculator, 0.10 ÷ 1.25 is a division by a decimal; rewriting it in cents as 10/125, which simplifies to 8/100, is the clean way through. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract to get the increase, then divide it by the price before the raise and multiply by 100; working in cents keeps the decimals out of the way.

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.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Divides the 10-cent increase by the new price, $1.35, measuring from the wrong amount; 0.10 ÷ 1.35 is about 7.4%, so 7%.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads the 10-cent increase as 10 percent, as if every percent change were a percent of one dollar rather than of the $1.25 price.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Divides the wrong way round, $1.25 ÷ $0.10 = 12.5, and takes the nearest choice, 12%.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$; New is the price after the change, and original is the price before it. The top is the amount of the increase. Dividing by the original compares the increase with where you started. Multiplying by 100 turns that decimal into a percent.

Sort out the question
The coffee cost $1.25 before. That is the original price. Now it costs $1.35, the new price. You need the percent increase. That compares how much the price went up with the original price.
Find the amount of the increase
$1.35 - 1.25 = 0.10$
The price went up $0.10, or 10 cents. Watch out: 10 cents is not 10%. It would be 10% only if the price started at $1.00. That mistake gives choice C.
Divide by the original price
$\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$
$\frac{1.35 - 1.25}{1.25} \times 100 = \frac{0.10}{1.25} \times 100$
Dividing by a decimal is easier in cents: 10 cents out of 125 cents. Divide the top and bottom by 5. Then multiply both by 4 to get hundredths.
$\frac{10}{125} = \frac{2}{25} = \frac{8}{100} = 0.08$
Change to a percent
$0.08 \times 100 = 8$
The price went up 8%. Watch out: divide by the original price, $1.25, not the new one. Dividing by $1.35 gives about 7.4%, which rounds to choice A.
So the answer is B, 8%.

Check your answer

Work forward from the answer. 8% of $1.25 is $0.08 \times 1.25 = 0.10$ , or 10 cents. Then $1.25 + 0.10 = 1.35$ . That is the new price, so 8% is right.

More questions like this

The original DOE question

Question 95 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 86, question 95. Open page 86 of the handbook (PDF)

Question 6 of 16

2024 Test A Math, question 40

Question

The price of a sandwich was raised from $6.25 to $6.75. What was the percent increase in the price?

A5%
B8%
C7%
D50%

Show the answer

Correct answer: B

Use the percent increase expression to find the answer:

$\frac{\text{new price} - \text{original price}}{\text{original price}}$

$\frac{6.75 - 6.25}{6.25} = \frac{0.50}{6.25} = 0.08 = 8\%$

Question analysis

A sandwich's price went up from $6.25 to $6.75, and this percent change question asks for the percent increase in the price. Students divide the 50-cent increase by one of the prices and write the result as a percent. Picking which price is the base, and doing the decimal division accurately, decide whether the answer comes out right.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a percent increase is a core grade 7 skill, and this is the plain version with no context beyond two prices. Its demand lies in measuring the change against the correct price and dividing by $6.25 without a calculator.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 2, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The 50-cent rise is easy to find; the percent depends on what it is compared with. Dividing by the new price, $6.75, gives a value that rounds to a listed choice one point below the answer, and reading 50 cents as 50 percent gives another. The choices are printed out of order, 8% before 7%, so students should read each one rather than scan the list.

A very close wrong answer moderate Choosing the answer
Measured against $6.75 the 50-cent rise is about 7.4%, and against $6.25 it is 8%; both are listed side by side, so only the right base separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic moderate Working it out
The division 0.50 ÷ 6.25 is easiest as 50/625, which reduces to 2/25 and then to 8/100; a three-digit decimal divisor invites a slip. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Put the 50-cent change over the original price, 0.50 over 6.25, then scale the fraction to hundredths or divide to read off the percent.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 5%: dividing the 50-cent rise by either price does not come near it, so it mostly draws guesses.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Divides the 50-cent increase by the new price, $6.75, instead of the original, which gives about 7.4%, so 7%.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads the 50-cent increase as 50 percent, treating cents as a percent of one dollar rather than of the $6.25 price.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$; New is the price after the change, and original is the price before it. The top is the amount of the increase. Dividing by the original compares the increase with where you started. Multiplying by 100 turns that decimal into a percent.

Sort out the question
The sandwich cost $6.25 before. That is the original price. Now it costs $6.75, the new price. You need the percent increase. That compares the increase with the original price. Note that the choices are not in order: C, 7%, comes after B, 8%.
Find the amount of the increase
$6.75 - 6.25 = 0.50$
The price went up $0.50, or 50 cents. Watch out: 50 cents is not 50%. It would be 50% only if the price started at $1.00. That mistake gives choice D.
Divide by the original price
$\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$
$\frac{6.75 - 6.25}{6.25} \times 100 = \frac{0.50}{6.25} \times 100$
In cents, this is 50 out of 625. Divide the top and bottom by 25. Then multiply both by 4 to get hundredths.
$\frac{50}{625} = \frac{2}{25} = \frac{8}{100} = 0.08$
Change to a percent
$0.08 \times 100 = 8$
The price went up 8%. Watch out: divide by the original price, $6.25, not the new one. Dividing by $6.75 gives about 7.4%, which is choice C.
So the answer is B, 8%.

Check your answer

Work forward from the answer. 8% of $6.25 is $0.08 \times 6.25 = 0.50$ , or 50 cents. Then $6.25 + 0.50 = 6.75$ . That is the new price, so 8% is right.

More questions like this

The original DOE question

Question 97 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 86, question 97. Open page 86 of the handbook (PDF)

Question 7 of 16

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:

$200\times 25\%=200\times 0.25=50$

So the number of employees in 2010 is $200+50=250$ .

Next, determine the number of employees in 2015:

$250\times 10\%=250\times 0.10=25$

So the total number of employees in 2015 is $250+25=275$ .

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Turning words into math moderate
Many steps minor
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.

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)
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)
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

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Adds the two percents to the employee count as if they were numbers of people, 200 + 25 + 10 = 235.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals partial knowledge Expected pull: weak
Finds the 250 employees in 2010 correctly, then adds 10 employees instead of 10% of 250, treating the second percent as a head count.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes both increases from the original 200, 50 + 20 = 70 more, missing that the second 10% applies to the 250 employees of 2010.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{new} = \text{old} + \frac{p}{100} \times \text{old}$; Old is the starting amount, and p is the percent it grows by. First find p percent of the old amount. That is the extra. Then add the extra to the old amount to get the new amount.

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.
Find the number in 2010
$\text{new} = \text{old} + \frac{p}{100} \times \text{old}$
$200 + \frac{25}{100} \times 200 = 200 + 50 = 250$
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.
Grow from the new number
The second increase starts in 2010. So take 10% of 250, not 10% of 200.
$250 + \frac{10}{100} \times 250 = 250 + 25 = 275$
In 2015 there were 275 employees.
Avoid the traps and answer
Watch out: taking both percents of the first 200 gives $200 + 50 + 20 = 270$ , 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. $200 \times 1.25 \times 1.1 = 250 \times 1.1 = 275$ . 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 8 of 16

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 $84=1.05x$ .

Dividing both sides of this equation by 1.05 gives $x=80$ .

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 $(10\%)(80)=8.0^{\circ}\text{ F}$ .

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Much to keep track of minor
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.

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)
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)
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

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds Monday's 80° F and then gives the 4-degree rise from Monday to Tuesday, 84 − 80, instead of Wednesday's 10% drop.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes 10% of Tuesday's 84° F, getting 8.4, instead of first working back to Monday's temperature, the base for Wednesday's drop.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Undoes Tuesday's increase in the wrong direction, adding 5% to 84 to get 88.2° F for Monday, then takes 10% of it, about 8.8.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{new} = \text{old} \times \left(1 + \frac{p}{100}\right)$; 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: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; 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.

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.
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.
$84 = x \times \left(1 + \frac{5}{100}\right) = 1.05x$
Solve for Monday
Divide both sides by 1.05 to undo the multiplying.
$x = 84 \div 1.05 = 80$
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.
Find Wednesday’s drop
Wednesday was 10% lower than Monday. So the drop is 10% of Monday’s 80°F.
$\text{part} = \frac{10}{100} \times 80 = 8$
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 9 of 16

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% $(100 - 15)$ , so multiply the original price by 85% (0.85) to find the discounted price:

$\$85.75 \times 0.85 \approx \$72.89$

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% $(100 - 25)$ , so multiply the price found in the first step by 75% (0.75) to find the second discounted price:

$\$72.89 \times 0.75 \approx \$54.67$

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% $(100 + 8.5)$ , so multiply the second discounted price by 108.5% (1.085) to find the final cost:

$\$54.67 \times 1.085 \approx \$59.32$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic major
Turning words into math moderate
Slow to do carefully moderate
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.

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)
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)
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

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes the 25% coupon off the original $85.75 instead of the sale price, so the discounts add to 40%: 85.75 × 0.60 × 1.085 ≈ $55.82.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives $59.08; it sits 24 cents below the right total and matches no discount or tax mistake, so it mostly catches guesses.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Applies both discounts correctly, reaching $54.67, but adds $8.50 as the tax, reading 8.5% as a dollar amount instead of a percent of the price.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Price after a discount: $\text{new price} = \text{price} \times \left(1 - \frac{p}{100}\right)$; 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: $\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)$; 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.

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.
Take off the 15% discount
You save 15%, so you pay 85%. That is 0.85 of the price.
$\text{new price} = \text{price} \times \left(1 - \frac{p}{100}\right)$
$\text{new price} = 85.75 \times \left(1 - \frac{15}{100}\right) = 85.75 \times 0.85 = 72.8875$
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.
$72.8875 \approx 72.89$
The sign ≈ means about. The sale price is $72.89.
Take 25% off the sale price
The coupon takes 25% off $72.89, so you pay 75% of it.
$72.89 \times 0.75 = 54.6675 \approx 54.67$
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.
Add the 8.5% tax
You pay the price plus 8.5% more, which is 108.5% of it.
$\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)$
$\text{total} = 54.67 \times 1.085 = 59.31695 \approx 59.32$
Watch out: 8.5% is a percent of the price, not $8.50. Adding $8.50 gives $63.17, choice D.
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:
$85.75 \times 0.85 \times 0.75 \times 1.085 = 59.3122\ldots \approx 59.31$
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 $0.085 \times 54.67 = 4.64695$ , about $4.65. Then $54.67 + 4.65 = 59.32$ . 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 10 of 16

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:

$\$805.95 \div 1.08 = \$746.25$

Then, divide that quotient by 15 to find the price of 1 pair of cleats:

$\$746.25 \div 15 = \$49.75$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Error-prone arithmetic moderate
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.

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)
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)
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

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes 8% of the $805.95 total and subtracts it, 805.95 × 0.92 ÷ 15 ≈ $49.43, though the tax was 8% of the smaller pre-tax cost.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides the $805.95 total by 15 pairs without removing the tax, which gives the price of a pair with tax rather than before it.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides the total by 15 and then multiplies by 1.08 instead of dividing, adding the tax a second time: 53.73 × 1.08 ≈ $58.03.

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: $\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)$; 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: $\text{if } a \times b = c \text{, then } a = c \div b$; 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.

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.
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.
$\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)$
$c \times 1.08 = 805.95$
Divide to remove the tax
To undo multiplying by 1.08, divide by 1.08. Move both decimal points two places right first.
$c = 805.95 \div 1.08 = 80{,}595 \div 108 = 746.25$
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.
Find the price of one pair
All 15 pairs cost the same, so share the cost equally.
$746.25 \div 15 = 49.75$
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 $15 \times 49.75 = 746.25$ dollars. The tax is $0.08 \times 746.25 = 59.70$ dollars. Then $746.25 + 59.70 = 805.95$ , 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 11 of 16

2023 Test A Math, question 36

Question

Josef and Mai divided some stamps between themselves. Josef got 60% of the stamps. If Josef received 500 more stamps than Mai, how many stamps did Josef receive?

A300
B800
C1,500
D2,500

Show the answer

Correct answer: C

Let j be the number of stamps Josef has. Then Mai has $j - 500$ stamps. The total number of stamps is $j + (j - 500) = 2j - 500$ . Since Josef has 60% of the stamps, set up an equation to solve for j:

$\frac{j}{2j - 500} = \frac{60}{100}$	Cross multiply.
$100j = 60(2j - 500)$	Apply the distributive property; multiply the 60 by $2j$ and 500.
$100j = 120j - 30{,}000$	Apply the additive inverse property; subtract $120j$ from both sides of the equation.
$-20j = -30{,}000$	Apply the multiplicative inverse property; divide both sides of the equation by −20.
$j = 1{,}500$

Question analysis

This percent word problem says Josef and Mai split some stamps, Josef got 60% of them, and he received 500 more than Mai; it asks how many stamps Josef received. Students must connect a difference in counts to a difference in percents. Stopping at the total, or taking a percent of 500, gives tempting wrong answers.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here neither share nor the total is known; students must link the 500-stamp gap to the gap between two percents before finding the total and then Josef's part.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
No obvious method moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 26 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is easy once the 500 is seen as the gap between two percents of one total; seeing that is the hard part. The tempting move is to apply the 60% to the only number given, the 500, which treats a difference as a whole. Students who do find the total must still take Josef's share of it, since the total waits among the choices. Checking that 1,500 and 1,000 differ by 500 confirms the answer.

Turning words into math moderate Understanding the question
The stem describes one split in two ways, Josef's 60% of the total and his 500-stamp lead over Mai, and the link, that the lead is 60% − 40% = 20% of the total, is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
No obvious method moderate Working it out
No equation is suggested: the quick route treats the 500 as 20 percentage points and scales up to 100%, while the handbook's route, j over 2j − 500, needs cross-multiplying and a negative coefficient. (Schoenfeld, 1985)
Many steps minor Working it out
Even the short route takes three moves, the 20% gap, the 2,500 total and Josef's 60% of it, and the total is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: If Josef has 60%, Mai has the other 40%, so the 500-stamp difference is a known percent of the total; use it to find the total, then take Josef's share.

Traps

Stops one step early: A choice is a number from partway through the solution.
Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: moderate
Takes 60% of 500, applying the percent to the difference between the two shares as if the 500 stamps were the whole collection.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes 60% of 500 to get 300 and then adds the 500 extra stamps, combining the stem's numbers without a model of the two shares.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Sees that the 500-stamp gap is 20% of the total, finds the total of 2,500, and stops before taking Josef's 60% of it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{p}{100} \times \text{whole}$; p is the percent. The whole is the total amount, and the part is the piece you want. Percent means “out of 100,” so 60% is 60 out of every 100. Use it to find a part, or work backward to find the whole.

Sort out the question
Josef and Mai split some stamps. Josef got 60% of them. He also got 500 more stamps than Mai. You need how many stamps Josef got. You don’t know the total yet, so find it first.
Find Mai’s percent
All the stamps make 100%. Josef has 60%, so Mai has the rest.
$100\% - 60\% = 40\%$
Mai has 40% of the stamps.
Find what the 500 stamps are
Josef’s extra 500 stamps are the gap between his share and Mai’s share.
$60\% - 40\% = 20\%$
So 500 stamps are 20% of all the stamps.
Find the total number of stamps
$\text{part} = \frac{p}{100} \times \text{whole}$
$500 = \frac{20}{100} \times \text{whole}$
$\text{whole} = 500 \div \frac{20}{100} = 500 \times 5 = 2{,}500$
There are 2,500 stamps in all. Watch out: that is everyone’s stamps, choice D. Don’t stop here.
Find Josef’s share
$\text{part} = \frac{60}{100} \times 2{,}500 = 1{,}500$
Watch out: taking 60% of 500 gives 300, choice A. The 500 is the gap between them, not the total. So the answer is C, 1,500 stamps.

Check your answer

Mai gets 40% of 2,500, which is 1,000 stamps. Josef’s 1,500 minus Mai’s 1,000 is 500, just as the question says. And 1,500 + 1,000 = 2,500, the whole amount.

More questions like this

The original DOE question

Question 93 as printed on page 84 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 93. Open page 84 of the handbook (PDF)

Question 12 of 16

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.

$0.30d=15$

$d=\frac{15}{0.30}=50$

So, the distance from W to X, in miles, is $50-15=35$ .

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 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.

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)
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

Stops one step early: A choice is a number from partway through the solution.
Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: moderate
Takes 70% of the 15 miles, 0.7 × 15 = 10.5, applying the percent to XY instead of to the whole distance from W to Y.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 45; it would make XY a quarter of WY instead of 30%, so it mostly draws guesses.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the whole distance from W to Y, 15 ÷ 0.30 = 50 miles, and stops before subtracting the 15 miles from X to Y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; The whole is the full amount. 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.

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.
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.
$100 - 70 = 30$
So the 15 miles from X to Y is 30% of the whole distance.
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.
$15 = \frac{30}{100} \times d = 0.30 \times d$
Divide both sides by 0.30 to get d by itself.
$d = 15 \div 0.30 = 50$
The distance from W to Y is 50 miles.
Find the distance from W to X
Take the 15 miles from X to Y away from the whole.
$50 - 15 = 35$
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? $0.70 \times 50 = 35$ . 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 13 of 16

2023 Test B Math, question 57

Question

A box contains 14 red marbles, 10 black marbles, and 16 white marbles. How many more white marbles need to be added to the box so that the probability of randomly drawing a white marble is $\frac{3}{5}$ ?

Show the answer

Correct answer: D

Let w represent the number of additional white marbles.

Set the probability of drawing a white marble equal to $\frac{3}{5}$ and solve the equation for w:

$\frac{16+w}{14+10+16+w}=\frac{3}{5}$

$\frac{16+w}{40+w}=\frac{3}{5}$

$5(16+w)=3(40+w)$

$80+5w=120+3w$

$2w=40$

$w=20$

Question analysis

A box holds 14 red, 10 black and 16 white marbles, and students find how many white marbles must be added so that the chance of drawing a white one is 3/5. Working backward from a target probability to a number of marbles is the unusual step.

NYS standards: 6.SP.8a uniform probability models
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 probability questions usually give the counts and ask for a probability. This runs the other way: the target probability is given, and the number to add changes both the white count and the total.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The tempting shortcut is to take 3/5 of the 40 marbles already in the box, which forgets that the new white marbles join the total. Students who set up the probability with the unknown in both the top and the bottom solve it in a few lines, and testing choices works too: adding 20 gives 36 white of 60, exactly 3/5. The other two choices are counts copied from the stem.

Turning words into math moderate Understanding the question
“How many more white marbles need to be added” puts the unknown in two places, (16 + w)/(40 + w), because each white marble added also raises the total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
The unknown number of added marbles has to be carried as a letter through a proportion, 5(16 + w) = 3(40 + w), before it can be found. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Totaling the marbles, setting up the proportion, cross-multiplying and solving 80 + 5w = 120 + 3w make four steps, unless the choices are tested one by one. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Let w be the number of white marbles added, write the new probability with w in both the top and the bottom, set it equal to 3/5 and solve; or test each choice.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals partial knowledge Expected pull: strong
Takes 3/5 of the 40 marbles now in the box, 24, and subtracts the 16 white ones, forgetting that added marbles raise the total.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 10, the number of black marbles in the stem, without setting up the new probability or testing it.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 14, the number of red marbles in the stem, without checking that adding 14 white marbles gives 30 of 54, not 3/5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P(\text{white}) = \frac{\text{white marbles}}{\text{all marbles}}$; P(white) is the probability, or chance, of drawing a white marble. The top counts the white marbles. The bottom counts every marble in the box. It works because each marble is equally likely to be drawn.
Cross multiplication: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a and c are the tops. b and d are the bottoms. When two fractions are equal, each top times the other bottom gives the same product. It works because you multiply both sides by both bottoms. Use it to clear fractions from an equation.

Sort out the box
The box has 14 red, 10 black and 16 white marbles.
$14 + 10 + 16 = 40$
That is 40 marbles. You will add some white marbles. Call that number w. You need w so that the chance of drawing white is $\frac{3}{5}$ .
Write the new probability
Adding w white marbles changes both counts. There will be 16 + w white marbles and 40 + w marbles in all. Use the probability formula.
$\frac{16 + w}{40 + w} = \frac{3}{5}$
Watch out: Taking $\frac{3}{5}$ of 40 gives 24, and 24 − 16 = 8, choice A. That forgets that the new marbles also make the box bigger.
Cross multiply
Fractions are hard to solve with. Clear them with cross multiplication.
$5 \times (16 + w) = 3 \times (40 + w)$
Then multiply each outside number by both terms inside the parentheses.
$80 + 5w = 120 + 3w$
Solve for w
Subtract 3w from both sides. Then subtract 80 from both sides.
$2w = 40$
Divide both sides by 2.
$w = 20$
You need to add 20 white marbles.
So the answer is D, 20.

Check your answer

Add 20 white marbles. Then there are 16 + 20 = 36 white marbles and 40 + 20 = 60 marbles in all. Divide the top and bottom by 12: $\frac{36}{60} = \frac{3}{5}$ . It works.

More questions like this

The original DOE question

Question 114 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 114. Open page 208 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 14 of 16

2018 Test B Math, question 4

Question

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

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

Show the answer

Correct answer: 300

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

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

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

$20x=6{,}000$

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

Question analysis

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

NYS standards: 6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the whole from a part and a percent is a grade 6 skill. It takes a single division here, so it sits below grade 7, although students who reflexively take 20 percent of 60 will go wrong.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Solution: 1 step
Reading level: Question and choices: about grade 3, 26 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is short and the arithmetic is one division, so it rests on reading the second sentence the right way round. Students who see 20% next to 60 and compute 20% of 60 answer a different problem, and a grid-in gives no list of choices to make them doubt it. A quick check that 60 is one fifth of the result catches every one of these errors.

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

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

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 61 as printed on page 153 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 61. Open page 153 of the handbook (PDF)

Question 15 of 16

2018 Test B Math, question 28

Question

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

A40%
B60%
C64%
D80%

Show the answer

Correct answer: C

Let $x$ 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:

$0.80x$

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

$0.80(0.80x)=0.64x\text{ or }64\%$

Question analysis

This percent problem describes a chemical that loses 20% of what is present each week and asks what percent of the original amount remains after two weeks. Students apply a percent decrease twice in a row. Deciding which amount each week's 20 percent is taken from is the heart of the problem.

NYS standards: 7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually apply a single increase or decrease. Applying a decrease twice, each time to the new amount, is an early form of exponential decay that goes beyond typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer major
Turning words into math moderate
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a two-step percent decrease, and the intuitive answer, 20% plus 20%, is wrong for a reason students rarely check: the second loss is taken from what is left after the first. The stem states this rule in its own words, but quickly. Starting with 100 units and working week by week makes the difference visible, since the second week removes 16 units, not 20.

A very close wrong answer major Choosing the answer
60% and 64% are four points apart, too close for a rough estimate; only seeing that week two's 20% comes from 80, not 100, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math moderate Understanding the question
The stem measures each week's loss against “the amount at the beginning of that same week”, and students must see that this resets the base to 80% for the second week. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
No starting amount is given, so students must supply one, such as 100 units, or reason directly with a percent of a percent. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Start with 100 units, take away 20 percent of the current amount for the first week, then take 20 percent of what is left for the second week.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: moderate
Adds the two 20% losses to get 40% and reports that amount lost, when the question asks how much is left.
B Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes both weeks' 20% from the original amount, 100 − 20 − 20, missing that the second week's loss comes from the smaller 80%.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the 80% left after one week and stops before applying the second week's decrease.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent decrease: $\text{amount left} = \text{amount} \times \left(1 - \frac{p}{100}\right)$; p is the percent taken away, here 20. Taking away 20% leaves 80% of the amount, so you multiply by 0.80. Each week starts from the amount left at the end of the week before.

Understand the decay
Each week, the chemical loses 20% of what it had at the start of that week. The amount at the start of each week keeps changing.
So the second week’s 20% is taken from a smaller amount. You need the percent left after two weeks.
Start with 100 units
No amount is given, so pick an easy one. Start with 100 units. Then the number of units left is also the percent left.
Work out week one
20% of 100 is 20, so 20 units are lost.
$100 - 20 = 80$
After week one, 80 units are left. That is 80% of the start.
Watch out: 80% is only one week. Stopping here gives choice D.
Work out week two
Now the week starts with 80 units, not 100. Take 20% of 80.
$\frac{20}{100} \times 80 = 16$
$80 - 16 = 64$
After two weeks, 64 units are left. That is 64% of the 100 units you started with.
Watch out: Taking 20 away twice from 100 gives 60%, choice B. That uses the old amount in week two.
So the answer is C, 64%.

Check your answer

Use the percent decrease formula. Each week keeps 80%, or 0.80, of the amount. Two weeks gives $0.80 \times 0.80 = 0.64$ , which is 64%. The two methods agree.

More questions like this

The original DOE question

Question 85 as printed on page 158 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 85. Open page 158 of the handbook (PDF)

Question 16 of 16

2018 Test B Math, question 38

Question

SURVEY OF CATS PER FAMILY

Number of Cats	Number of Families
0	15
1	42
2	35
3 or more	8

The table above shows the number of cats per family in 100 households in the Blaine neighborhood. By what percentage is the number of families with 1 cat greater than the number of families with 2 cats?

A7%
B10%
C17%
D20%

Show the answer

Correct answer: D

To find by what percent the number of families with 1 cat is greater than the number of families with 2 cats, calculate the difference between the two numbers and divide by the number of families with 2 cats:

$\frac{42-35}{35}=\frac{7}{35}=\frac{1}{5}=0.20\text{ or }20\%$

Question analysis

A table lists how many cats live in each of 100 households in the Blaine neighborhood, and the question asks by what percent the number of one-cat families exceeds the number of two-cat families. It tests percent difference. The tricky part is choosing the right base for the comparison when the table's total is a convenient 100.

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 and decrease is grade 7 content, and this is one standard comparison. What makes it harder than a routine exercise is choosing the base and ignoring the tempting 100-family total.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 2 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 arithmetic is a single subtraction and division, so the question is really about choosing the base of a percent comparison. The 100-household total makes the count difference look like a percent already, and dividing by the larger count is the other classic slip. Students who say "greater than the 2-cat count" to themselves pick 35 as the base.

Turning words into math moderate Understanding the question
"Greater than the number of families with 2 cats" fixes the base of the percent as 35, the count being compared against, not 42 or the 100 households. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
17% and the right 20% both come from dividing the 7-family gap by a count, 42 or 35; only choosing 35, the count being compared against, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
Only two rows of the four-row table matter, the 42 families with 1 cat and the 35 with 2; the 0-cat and 3-or-more rows go unused. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the difference between the two counts, decide which count the comparison is measured against, divide by it, and convert to a percent.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Subtracts 42 − 35 = 7 and, since there are 100 households, reads that 7-point gap in shares as a 7% increase.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No division of the 7-family gap by a count in the table gives 10%, so this mostly attracts students estimating a round percent.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: moderate
Divides the 7-family difference by 42, the larger count, measuring the change from the wrong amount: 7/42 is about 16.7%, so 17%.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent greater than: $\text{percent} = \frac{\text{difference}}{\text{amount you compare to}} \times 100$; The difference is how much bigger one amount is than the other. The amount you compare to is the one after the word than. Dividing shows what share of that amount the difference is. Times 100 turns that share into a percent.

Sort out the question
From the table, 42 families have 1 cat and 35 families have 2 cats. You need the percent by which 42 is “greater than the number of families with 2 cats”. The rows for 0 cats and 3 or more cats do not matter here.
Find the difference
Subtract the two counts.
$42 - 35 = 7$
There are 7 more families with 1 cat.
Watch out: 7 is a number of families, not a percent. There are 100 households, so 7% looks tempting. It is choice A, and it is wrong.
Divide by the number you compare to
The question compares with the families that have 2 cats. So divide by 35, not 42.
$\frac{7}{35} = \frac{1}{5}$
Watch out: dividing by 42 gives about 17%. That is choice C.
Change the fraction to a percent
Use the percent formula. One fifth of 100 is 20.
$\text{percent} = \frac{7}{35} \times 100 = \frac{1}{5} \times 100 = 20$
So the answer is D, 20%.

Check your answer

Find 20% of 35. That is $0.20 \times 35 = 7$ . Add it on: $35 + 7 = 42$ . That is the number of families with 1 cat, so 20% is right.

More questions like this

The original DOE question

Question 95 as printed on page 160 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 160, question 95. Open page 160 of the handbook (PDF)

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