Practice Test

6.RP.3b unit rate problems

Question 1 of 25

2026 Practice Test B Math, question 3

Question

At a diner, 18 dozen buns were used yesterday for 150 lunch customers. Today, there were 250 lunch customers. Assuming that the bun usage rate is the same for both days, which proportions can be used to determine how many dozen buns are needed for 250 lunch customers?

Select the three correct answers.

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

A $\frac{18}{150} = \frac{x}{250}$
B $\frac{18}{250} = \frac{x}{150}$
C $\frac{150}{18} = \frac{250}{x}$
D $\frac{x}{18} = \frac{250}{150}$
E $\frac{x}{150} = \frac{18}{250}$

Show the answer

Correct answer: A, C, and D

The three correct answers are:

$\frac{18}{150} = \frac{x}{250}$

$\frac{150}{18} = \frac{250}{x}$

$\frac{x}{18} = \frac{250}{150}$

Set up a proportion to compare the number of buns per customer. These are all equivalent:

$\frac{18\text{ dozen buns}}{150\text{ customers}} = \frac{x\text{ dozen buns}}{250\text{ customers}}$

To check for equivalency, solve each proportion for x.

Each of the three correct proportions simplifies to $150x = 18(250)$ .

Then solve for x to determine how many dozen buns are needed for 250 lunch customers: $150x = 4{,}500$

$x = \frac{4{,}500}{150}$

$x = 30$

Question analysis

A diner used 18 dozen buns for 150 lunch customers yesterday, and students select the three proportions, out of five, that can find how many dozen buns 250 customers need at the same rate. It tests setting up proportions and recognizing equivalent ones. All five equations use the same numbers, so each has to be checked for which quantities it pairs.

NYS standards: 7.RP.2c proportional relationship equations
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Setting up a proportion from a rate is everyday grade 7 work. Here nothing is solved: five setups built from the same numbers must each be judged, some arranged in unfamiliar ways, and all three usable ones found.
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: Every choice must be checked moderate
A very close wrong answer moderate
An answer in an unexpected form minor
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 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

Nothing needs solving here; the question is which pairings keep the usage rate fixed. Students who check only for buns over customers on both sides find exactly three such setups, two of them wrong, and pass over the two right ones written another way. The wrong pair gives 10.8 dozen, fewer buns for more customers, which a quick sense check would catch, and cross-multiplying every setup settles all five.

Every choice must be checked moderate Choosing the answer
All five setups are built from 18, 150, 250 and x, so none can be dismissed on sight; each has to be cross-multiplied and compared with 150x = 18 × 250. (Embretson & Wetzel, 1987)
A very close wrong answer moderate Choosing the answer
The first two setups differ only in where 150 and 250 sit, 18/150 = x/250 against 18/250 = x/150, and both show buns over customers on each side, so a units check passes both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
An answer in an unexpected form minor Choosing the answer
One right setup puts customers over dozens of buns, and another sets today’s x over yesterday’s 18 equal to 250 customers over 150, forms that look wrong to students who learned only rate equals rate. (Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Three of the five must be picked and all three must be right to score, so the verdict on each setup has to be kept until the last one is tested. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Write one proportion you trust by matching both units and days, cross-multiply it, then cross-multiply each choice and keep only those that give the same equation.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Checks only that both fractions show dozens of buns over customers, missing that 18 sits over today’s 250 customers; cross-multiplying gives 250x = 2,700, so x = 10.8.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Left out by students who expect each fraction to compare buns with customers, not seeing that today over yesterday for buns and for customers also gives 150x = 4,500.
E Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Accepts x over 150 because it shows buns over customers like the trusted setup, but today’s buns sit over yesterday’s customers, which yields only 10.8 dozen for more people.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A rate: $\text{rate} = \frac{\text{dozens of buns}}{\text{customers}}$; The rate is how many dozen buns are used for each customer. Yesterday it was 18 dozen over 150 customers. Today it is x dozen over 250 customers. The question says the rate is the same both days, so those two fractions are equal.
Cross-multiplying a proportion: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; A proportion says two fractions are equal. a and c are the tops, and b and d are the bottoms. Multiply each top by the other bottom, and the two products are equal. This turns any proportion into a plain equation, so you can compare setups that look different.

Sort out the question
Yesterday the diner used 18 dozen buns for 150 customers. Today it has 250 customers. Let x be the number of dozen buns needed today. The bun usage rate is the same both days.
A proportion is two equal fractions. You must pick the three proportions that give the right value of x.
Write one proportion you trust
Put each day’s buns over that same day’s customers. The rates are equal, so the fractions are equal:
$\frac{18}{150} = \frac{x}{250}$
This is choice A. Cross-multiply it:
$150 \times x = 18 \times 250$
$150x = 4{,}500$
Any choice that leads to 150x = 4,500 is a right one.
Solve for x
Divide both sides by 150 to get x by itself:
$x = 4{,}500 \div 150 = 30$
So the diner needs 30 dozen buns today. A wrong setup gives a different number.
Test choices C and D
Cross-multiply each one.
- C: $\frac{150}{18} = \frac{250}{x}$ gives $150 \times x = 18 \times 250$ . That is the same equation, so C works. It is choice A with both fractions flipped.
- D: $\frac{x}{18} = \frac{250}{150}$ gives $x \times 150 = 18 \times 250$ . The same equation again, so D works too.
D compares today with yesterday. Today’s buns over yesterday’s buns equals today’s customers over yesterday’s customers.
Watch out: D looks odd because it puts buns over buns. Do not skip it. It still keeps the rate the same.
Test choices B and E
- B: $\frac{18}{250} = \frac{x}{150}$ gives $250 \times x = 18 \times 150$ , so 250x = 2,700.
- E: $\frac{x}{150} = \frac{18}{250}$ gives $x \times 250 = 150 \times 18$ , the same 250x = 2,700.
Both give $x = 2{,}700 \div 250 = 10.8$ . That is fewer buns for more customers, which makes no sense.
Watch out: B and E show buns over customers, so they look right. But each one mixes up the days. B puts yesterday’s 18 dozen over today’s 250 customers. E puts today’s buns over yesterday’s 150 customers.
Name the three answers
Choices A, C and D all give 150x = 4,500, so x = 30. On the test, you check the box next to each one. So the answers are A, C and D.

Check your answer

Find the buns a second way. Today has $\frac{250}{150} = \frac{5}{3}$ times as many customers as yesterday. So it needs 5/3 times as many buns: $18 \times \frac{5}{3} = 30$ dozen. That matches the 30 from choices A, C and D.

More questions like this

The original DOE question

Question 3 as shown on the DOE’s digital test, page 75 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 75, question 3.

Question 2 of 25

2026 Practice Test C Math, question 9

Question

Maria rides her bicycle to school at a constant speed of 15 miles per hour. If the distance to school is 6 miles, how many minutes will it take Maria to get to school?

Show the answer

Correct answer: C

Maria’s rate of speed is given as 15 miles per hour. Since 1 hour is equivalent to 60 minutes, her rate of speed, in miles per minute, can be determined.

$\frac{15\text{ miles}}{1\text{ hour}} = \frac{15\text{ miles}}{60\text{ minutes}} = \frac{1\text{ mile}}{4\text{ minutes}}$ , or $\frac{1}{4}$ mile per minute.

Therefore, it takes Maria 4 minutes to ride 1 mile.

The distance to school is 6 miles, so it will take $6 \times 4 = 24$ minutes to ride to school.

Question analysis

Maria bikes 6 miles to school at a constant 15 miles per hour, and students find how many minutes the ride takes. It tests using a rate to find a time and converting hours to minutes. With the speed given per hour and the question asking, in bold, for minutes, keeping track of units is the main hurdle.

NYS standards: 6.RP.3b unit rate problems
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Distance, speed and time with whole numbers is grade 6 rate work. The only extra demand is an answer in minutes when the speed is per hour, a conversion grade 7 students make routinely.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The numbers are small, but the speed is per hour and the answer is wanted in minutes, so students have to choose a route through the units. Finding the minutes per mile first, 4 at 15 miles an hour, turns the rest into a single multiplication. The wrong choices come from using a number in the stem as the answer, from mixing the 60 minutes with the 6 miles, or from rounding 0.4 hour to a half hour.

Something you must already know minor Working it out
Nothing on the page says that time is distance divided by speed or that an hour has 60 minutes, and the bold word minutes requires both. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
6 ÷ 15 is less than 1, 0.4 hour, and turning it into minutes means 0.4 × 60; the cleaner route, 60 ÷ 15 = 4 minutes a mile, avoids decimals. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find how many minutes Maria needs to ride one mile at 15 miles per hour, then multiply by the number of miles to school.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the 60 minutes in an hour by the 6 miles, 60 ÷ 6 = 10, as if the ride took a full hour, and ignores the 15 mph speed.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the 15 in 15 miles per hour as the number of minutes, a number copied from the question without any calculation.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Sees that 6 miles is a little under half of 15, rounds the time of 0.4 hour up to half an hour, and gives 30 minutes.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Minutes for each mile: $\text{minutes per mile} = 60 \div \text{miles per hour}$; Miles per hour tells how many miles you ride in one hour, which is 60 minutes. Sharing the 60 minutes evenly among those miles gives the minutes for each mile. This works when the speed stays the same the whole time.
Distance, speed and time: $\text{time} = \text{distance} \div \text{speed}$; Distance is how far you go, here in miles. Speed is how far you go in one hour. Dividing the distance by the speed gives the time in hours.

Sort out the question
Maria rides at a constant speed of 15 miles per hour. Constant means it never changes. School is 6 miles away. You need the time in minutes, not hours. You must know that one hour has 60 minutes.
Find the minutes for one mile
15 miles per hour means Maria rides 15 miles in 60 minutes. Use the minutes-per-mile formula:
$60 \div 15 = 4$
So each mile takes Maria 4 minutes.
Find the minutes for 6 miles
Each of the 6 miles takes 4 minutes:
$6 \times 4 = 24$
So the ride takes 24 minutes.
Watch out: 15 is the speed, not a time, so choice B is a trap. And $60 \div 6 = 10$ , choice A, ignores the speed.
Choose the answer
Another way is the time formula. $6 \div 15 = 0.4$ , so the ride takes 0.4 of an hour. In minutes, that is $0.4 \times 60 = 24$ .
Watch out: rounding 0.4 hour up to half an hour gives choice D, 30.
So the answer is C, 24.

Check your answer

Check with the speed. 24 minutes is $\frac{24}{60} = \frac{2}{5}$ of an hour. In a full hour Maria rides 15 miles. So in 2/5 of an hour she rides $\frac{2}{5} \times 15 = 6$ miles. That is exactly the distance to school.

More questions like this

The original DOE question

Question 9 as shown on the DOE’s digital test, page 82 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 82, question 9.

Question 3 of 25

2026 Practice Test C Math, question 47

Question

The height of a plant, in millimeters, is proportional to the number of days the plant has grown. The plant has grown 20 millimeters in 8 days. What is the constant of proportionality, in millimeters per day?

A0.4
B2.5
C12
D20

Show the answer

Correct answer: B

The constant of proportionality is determined by dividing the number of millimeters by the number of days:

$\frac{20\text{ mm}}{8\text{ days}} = \frac{20}{8} = 2.5\text{ mm/day}$

Question analysis

A plant's height in millimeters is proportional to the number of days it has grown, and it grew 20 millimeters in 8 days. Students find the constant of proportionality in millimeters per day. It tests finding a unit rate, and the order of the division decides whether the rate comes out in the units asked for.

NYS standards: 7.RP.2b constant of proportionality
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a constant of proportionality from one pair of values is a basic grade 7 skill. With one division and the units named in the question, it asks less than most grade 7 problems.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 7, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One division gives the constant, and the question itself names the order by asking for millimeters per day. The tempting slip is dividing the smaller number by the larger, 8 by 20, out of habit, which gives a rate in days per millimeter. A plant that grows 20 millimeters in 8 days grows more than 2 millimeters a day, a quick sense check on any result.

Something you must already know minor Working it out
Students must know that the constant is the height divided by the days; the units in the question, millimeters per day, name the order of the division. (Anderson & Krathwohl, 2001)

Strategy: Use the units in the question, millimeters per day, as the instruction: divide the millimeters grown by the number of days.

Traps

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

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides the days by the millimeters, 8 ÷ 20 = 0.4, which gives days per millimeter, the rate turned upside down.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts the days from the millimeters, 20 − 8 = 12, combining the two numbers without thinking about a rate.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the 20 millimeters the plant grew over all 8 days, the total growth, rather than the growth for each single day.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality: $k = \frac{y}{x}$; Here y is the height in millimeters, and x is the number of days. k is the growth in one day, in millimeters per day. In a proportional relationship, y = kx, so dividing y by x gives k.

Sort out the question
The plant’s height is proportional to the number of days it grows. It grew 20 millimeters (mm) in 8 days.
You need the constant of proportionality in millimeters per day. Per day means for each day. So you need how much the plant grows in one day.
Divide millimeters by days
The units tell you the order: millimeters per day means millimeters divided by days. Use the formula:
$k = \frac{20}{8} = \frac{5}{2} = 2.5$
So the plant grows 2.5 mm each day.
Avoid the tempting slips
Watch out: $8 \div 20 = 0.4$ , choice A, is days per millimeter. That is the rate upside down.
Choice C, 12, comes from $20 - 8$ . Subtracting does not make a rate. Choice D, 20, is the growth over all 8 days, not one day.
Give the answer
The constant of proportionality is 2.5 mm per day.
So the answer is B, 2.5.

Check your answer

Multiply back. At 2.5 mm a day for 8 days, the plant grows $2.5 \times 8 = 20$ mm. That matches. Also, 2 mm a day would give only 16 mm, and 3 mm a day would give 24 mm. So the rate must be between 2 and 3.

More questions like this

The original DOE question

Question 47 as shown on the DOE’s digital test, page 120 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 120, question 47.

Question 4 of 25

2026 Practice Test D Math, question 25

Question

A bus trip takes 9 hours if the mean speed is 50 miles per hour. How many hours would this trip take if the mean speed was 45 miles per hour?

Show the answer

Correct answer: B

First, find the length of the bus trip in miles by multiplying the time, 9 hours, by the average (mean) speed, 50 miles per hour, to get 450 miles.

Then, divide the total distance, 450 miles, by the average (mean) speed, 45 miles per hour, to get the total time:

$450 \div 45 = 10\text{ hours}$

Question analysis

A bus trip takes 9 hours at a mean speed of 50 miles per hour, and students find how long the same trip takes at 45 miles per hour. It tests speed, distance and time. Working out the fixed distance first is the key, since guessing how the time should change leads many students astray.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Speed, distance and time problems are grade 6 rate work, and this one uses whole numbers in two steps. Its only twist is how the time responds when the speed changes, which makes it easier than typical grade 7 rate problems.
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: No obvious method moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two short calculations solve it, but only after noticing that speed and time move in opposite directions: a slower bus takes longer. Students who scale the 9 hours by 45/50, as in a direct proportion, get a little over 8 hours, a result that should feel wrong for a slower trip. Finding the distance first turns the question into one division with no direction to decide.

No obvious method moderate Working it out
The question never mentions the trip's length, yet the 450-mile distance is what stays fixed; setting up a proportion between hours and speeds, the usual reflex, points the wrong way. (Schoenfeld, 1985)
Something you must already know minor Working it out
Students must bring distance = speed × time themselves, to turn 9 hours at 50 miles per hour into miles before dividing by the new speed. (Anderson & Krathwohl, 2001)

Strategy: Find the trip's length in miles from the first speed and time, since that length stays the same, then divide it by the new mean speed.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Treats time as proportional to speed, scaling 9 hours by 45/50 to 8.1, about 8, though a slower bus needs more time, not less.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 12 hours; it needs a slip such as 540 miles for 9 × 50, so it mostly draws guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the 5-mile-per-hour drop in speed to the 9 hours, 9 + 5 = 14, combining numbers from the question without a model of the trip.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, speed and time: $d = r \times t$; d is the distance traveled. r is the rate, or speed, such as miles per hour. t is the time. It works because at 50 miles each hour, 9 hours covers 9 groups of 50 miles.
Finding the time: $t = d \div r$; This is the same rule turned around. d is the distance and r is the speed. Divide the distance by the speed to find t, the number of hours. Use it when you know how far and how fast.

Sort out the question
A bus trip takes 9 hours at a mean speed of 50 miles per hour. The mean speed is the average speed over the whole trip. You need the time for the same trip at 45 miles per hour. The length of the trip does not change, so find it first.
Find the length of the trip
Use distance equals speed times time, with a speed of 50 miles per hour and a time of 9 hours:
$d = 50 \times 9 = 450$
The trip is 450 miles long.
Find the new time
Now the bus goes 45 miles per hour over the same 450 miles. Divide the distance by the speed:
$t = 450 \div 45 = 10$
The trip takes 10 hours.
Check that it makes sense
A slower bus needs more time for the same trip. 10 hours is more than 9 hours, so it fits.
Watch out: time does not shrink with speed. Scaling 9 hours by 45/50 gives about 8 hours, choice A, which would make the slower bus arrive sooner. Adding the drop of 5 miles per hour to 9 hours gives 14, choice D, which has no reason behind it.
So the answer is B, 10 hours.

Check your answer

Check the distance both ways. At 50 miles per hour for 9 hours, the bus goes 50 × 9 = 450 miles. At 45 miles per hour for 10 hours, it goes 45 × 10 = 450 miles. Same trip, same distance.

More questions like this

The original DOE question

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

Question 5 of 25

2026 Practice Test D Math, question 43

Question

Two student clubs are selling snacks as part of a fundraiser. The Art Club sells a 24-pack of granola bars for $12.00. The Science Club sells a 10-pack of granola bars for $6.50. To the nearest cent, what is the difference in cost per bar between the two clubs?

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

The cost for the Art Club’s bar is per bar.

The cost for the Science Club’s bar is per bar.

The difference between the cost of the bars is .

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

Show the answer

Correct answer: blank 1: $0.50; blank 2: $0.65; blank 3: $0.15

To determine the cost per bar from the Art Club, divide the total cost by the total number of bars:

$\frac{\$12.00}{24} = \$0.50 \text{ per bar}$

To determine the cost per bar from the Science Club, divide the total cost by the total number of bars:

$\frac{\$6.50}{10} = \$0.65 \text{ per bar}$

To find the difference in price per bar, subtract the cost per bar for the Art Club from the cost per bar for the Science Club:

$\$0.65 \text{ per bar} - \$0.50 \text{ per bar} = \$0.15 \text{ per bar}$

Question analysis

Two clubs sell granola bars for a fundraiser: the Art Club a 24-pack for $12.00 and the Science Club a 10-pack for $6.50. Students complete three sentences with drop-down menus, giving each club's cost per bar and the difference between them. It tests unit prices with money, and all three menus must be right.

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Comparing unit prices is grade 6 rate work. The menus split it into three short steps, two divisions of money amounts and one subtraction, which is lighter than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Many steps minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 87 words.
Corrected from the DOE handbook: The order of the answers in the DOE’s key was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The menus guide the work, one sentence per step, and every option is a reasonable price, so the question rewards accurate division more than planning. No common misconception leads to the wrong options; they catch slips, such as misreading a pack size or subtracting carelessly. Checking that 24 bars at the first price cost $12.00 and 10 bars at the second cost $6.50 confirms both menus before the subtraction.

Many steps minor Working it out
Three menus must all be right to score: two divisions for the costs per bar, then a subtraction that uses both, so a slip in either division spoils the third menu. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Each division is with money, $12.00 ÷ 24 and $6.50 ÷ 10, so the decimal point must be placed carefully, though the menus offer only sensible amounts. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Divide each pack's price by the number of bars in it, then subtract the smaller cost per bar from the larger; checking that 24 bars at your first price make $12.00 catches slips.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

CBA Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Divides $12.00 by 20 instead of 24, misreading the size of the Art Club’s pack, so $0.60 a bar leaves a gap of only $0.05.
BBB An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Finds both costs per bar correctly, then slips in the subtraction, taking $0.65 − $0.50 as $0.10, the option beside the right one.

What was corrected from the DOE handbook, and why

Answer key: The DOE’s explanation opens with the key “($0.65; $0.50; $0.15)”, which puts $0.65 in the first drop-down and $0.50 in the second. Neither is possible: the first menu (the Art Club’s cost per bar) offers $0.40, $0.50, $0.60 and $0.75, and the second (the Science Club’s) offers $0.55, $0.65, $0.75 and $0.85. The explanation’s own working gives the Art Club $12.00 ÷ 24 = $0.50 and the Science Club $6.50 ÷ 10 = $0.65, so the key is $0.50, $0.65, $0.15. The site’s key was already in that order; the DOE’s key line is not shown on the site.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost of one item: $\text{cost per bar} = \text{price of the pack} \div \text{number of bars}$; The price of the pack is what you pay for all the bars together. Dividing shares that price equally among the bars. The result is the price of one bar, called the unit price.

Sort out the question
You know:
- The Art Club sells 24 bars for $12.00.
- The Science Club sells 10 bars for $6.50.
There are three menus. The first is the Art Club’s cost per bar. The second is the Science Club’s. The third is the difference, how much more one bar costs than the other.
Find the Art Club’s cost
Share $12.00 among 24 bars:
$12.00 \div 24 = 0.50$
Each bar costs $0.50. That makes sense: 24 half-dollars make $12. Pick $0.50 in the first menu.
Watch out: dividing by 20 instead of 24 gives $0.60, which is also in the menu.
Find the Science Club’s cost
Share $6.50 among 10 bars. Dividing by 10 moves the decimal point one place to the left:
$6.50 \div 10 = 0.65$
Each bar costs $0.65. Pick $0.65 in the second menu.
Find the difference
Subtract the smaller cost from the larger one:
$0.65 - 0.50 = 0.15$
The bars differ by $0.15. Pick $0.15 in the third menu.
Watch out: $0.10 sits right next to $0.15 in the menu. A slip in the subtraction lands there.
Fill in the three menus
On the test and on this site, you choose each answer from a drop-down menu.
So the menus read $0.50, $0.65 and $0.15.

Check your answer

Multiply back. $24 \times 0.50 = 12.00$ and $10 \times 0.65 = 6.50$ , so both costs per bar match the pack prices. And $0.50 + 0.15 = 0.65$ , so the difference fits too.

More questions like this

The original DOE question

Question 43 as shown on the DOE’s digital test, page 116 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 116, question 43. This is the DOE’s version, before the corrections listed in the analysis.

Question 6 of 25

2024 Test A Math, question 24

Question

In 1991, the total public debt of the United States was about $3,600,000,000,000. In that year, there were about 250,000,000 people in the United States. Which amount is the best estimate of the public debt per person for that year?

A$1,440
B$14,400
C$144,000
D$14,400,000,000

Show the answer

Correct answer: B

The question asks for the amount of the public debt per person, so divide the total public debt by the number of people.

$\frac{3{,}600{,}000{,}000{,}000}{250{,}000{,}000} =$

$\frac{3{,}600{,}00\cancel{0}{,}\cancel{0}\cancel{0}\cancel{0}{,}\cancel{0}\cancel{0}\cancel{0}}{25\cancel{0}{,}\cancel{0}\cancel{0}\cancel{0}{,}\cancel{0}\cancel{0}\cancel{0}} =$

$\frac{360{,}000}{25} = 14{,}400$

The amount of the public debt per person was about $14,400.

Question analysis

An estimation question with very large numbers: the 1991 United States public debt of about 3.6 trillion dollars and a population of about 250 million. Students find the debt per person. The choices differ by powers of ten, so cancelling zeros and keeping track of place value is the whole challenge.

NYS standards: 6.NS.2 multi-digit division
6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Dividing large whole numbers and using place value are grade 5 and 6 skills. Grade 7 work goes further with rational numbers; this asks only for one careful division, with counting zeros as the main risk.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 37 words.
Corrected from the DOE handbook: A misprinted number in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The division itself is easy once the zeros are handled, and the question is built to punish miscounting them. Three choices carry the same digits, 144, at different sizes, so a student who loses or gains one zero still finds a match. A quick estimate in powers of ten, 3.6 trillion over a quarter of a billion, confirms that the answer is in the tens of thousands.

Error-prone arithmetic moderate Working it out
The debt has 13 digits and the population 9; cancelling seven zeros from each leaves 360,000 ÷ 25, and losing or adding one zero shifts the answer tenfold. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
Three choices share the digits 144 and differ only by powers of ten, so getting the digits right does not settle it; only exact place value does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Cancel the same number of zeros from the debt and the population, then divide the smaller numbers that remain; check the size by estimating with powers of ten.

Traps

An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Cancels one zero too many while dividing and ends with $1,440, a per-person amount ten times too small.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Cancels one zero too few from the debt and ends with $144,000, a per-person amount ten times too large.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Divides the debt by 250 alone, dropping the six zeros that follow 250 in the population without removing any from the debt.

What was corrected from the DOE handbook, and why

Explanation: The handbook’s first division writes the debt as “3,6000,000,000,000”, with an extra zero. The question gives $3,600,000,000,000, and the handbook’s next line (where the zeros are cancelled) uses 3,600,000,000,000, so the site now reads 3,600,000,000,000 ÷ 250,000,000. The answer, $14,400, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Amount per person: $\text{per person} = \text{total} \div \text{number of people}$; The total is the whole amount, here the debt in dollars. Dividing by the number of people shares it out equally. The result is how much goes with each one person.
Canceling zeros: $\frac{a \times 10}{b \times 10} = \frac{a}{b}$; a and b are any numbers, and b is not zero. Dividing the top and bottom of a fraction by 10 does not change its value. So you can cross off a zero at the end of the top and one at the end of the bottom, as a pair.

Sort out the question
The debt was about $3,600,000,000,000, which is 3.6 trillion dollars. There were about 250,000,000 people, which is 250 million. You need the debt per person. Per person means the amount for each one person.
Set up the division
Share the debt equally by dividing it by the number of people. The debt goes on top, and the people go on the bottom.
$\text{per person} = \text{total} \div \text{number of people}$
$\text{per person} = \frac{3{,}600{,}000{,}000{,}000}{250{,}000{,}000}$
Cancel zeros in pairs
Count the zeros at the end of each number. The debt ends in 11 zeros. The population ends in 7. Cross off 7 zeros from each. Each pair you cross off divides the top and bottom by 10.
$\frac{360{,}000}{25}$
Watch out: cross off the same number from each. One too many from the debt gives $1,440, choice A. One too few gives $144,000, choice C.
Divide what is left
Dividing by 25 is the same as dividing by 100 and then multiplying by 4. That is because 4 groups of 25 make 100.
$360{,}000 \div 100 = 3{,}600$
$3{,}600 \times 4 = 14{,}400$
The debt was about $14,400 per person. Watch out: choice D divides by 250 only and forgets the zeros after it. So the answer is B, $14,400.

Check your answer

Multiply back. First, $14{,}400 \times 25 = 360{,}000$ . Then put back the 7 zeros you crossed off: 3,600,000,000,000. That is the debt, so $14,400 is right.

More questions like this

The original DOE question

Question 81 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 82, question 81. Open page 82 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 7 of 25

2024 Test A Math, question 42

Question

Lian bought enough oranges to fill 4 bags. Each bag contains 8 oranges. The total cost was $11.52. At that rate, how much would Lian pay for 42 oranges?

A$17.28
B$15.12
C$15.02
D$12.52

Show the answer

Correct answer: B

First, find the amount, in dollars, that Lian paid for each orange. There are 4 bags with 8 oranges each.

$4 \times 8 = 32$

Divide the total cost by the total number of oranges to get the cost per orange:

$11.52 \div 32 = 0.36$

Multiply to find the cost of 42 oranges:

$0.36 \times 42 = 15.12$

Question analysis

This rate word problem says Lian paid $11.52 for 4 bags of 8 oranges each and asks how much 42 oranges would cost at the same rate. Students find the price of one orange and multiply. Since 42 oranges is not a whole number of bags, working by the bag instead of by the orange invites mistakes.

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a unit price and scaling it up is grade 6 rate work. The decimals are multi-digit, but the plan is routine, with less to manage than grade 7 problems built on percents or negative numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The plan is familiar, total oranges, price per orange, then the new cost, so the work lies in the decimals. Because 42 oranges is five bags and two oranges, students who price by the bag either round up to six bags or must handle the leftover. Estimating, since 42 is about a third more than 32, gives roughly $15 and rules out the far choices but not the near one.

Error-prone arithmetic moderate Working it out
Two decimal operations are done by hand: 11.52 ÷ 32 = 0.36 for one orange, then 0.36 × 42 = 15.12, where a dropped carry is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The choices $15.12 and $15.02 differ by 10 cents, so a rough estimate of about $15 cannot choose between them; only the exact product can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Multiply to find how many oranges were bought, divide the total cost by that number to get the price of one orange, then multiply by 42.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Prices whole bags, rounding 42 oranges up to 6 bags of 8, and pays 6 × $2.88 = $17.28, missing that "at that rate" means per orange.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Uses the right method but slips one digit in multiplying 0.36 by 42, a tens-place error that lands 10 cents short of the total.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives $12.52, exactly one dollar more than $11.52; no price per orange or per bag leads there, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit price: $\text{unit price} = \frac{\text{total cost}}{\text{number of items}}$; The unit price is what one item costs. Divide the total cost by how many items you got. It works because the total is shared equally among the items.
Cost: $\text{cost} = \text{unit price} \times \text{number of items}$; The unit price is the cost of one item, here one orange. Multiply it by how many items you buy to get the total cost. This works when every item costs the same.

Sort out the question
Lian bought 4 bags with 8 oranges in each bag. She paid $11.52 in all. You need the cost of 42 oranges. “At that rate” means each orange costs the same as before. So find the price of one orange first.
Count the oranges
Each of the 4 bags holds 8 oranges, so multiply to find how many she bought.
$4 \times 8 = 32$
Lian bought 32 oranges for $11.52.
Find the price of one orange
$\text{unit price} = \frac{\text{total cost}}{\text{number of items}}$
$\text{unit price} = \frac{11.52}{32} = 0.36$
This is easier in cents: $1{,}152 \div 32 = 36$ . So one orange costs $0.36, or 36 cents.
Find the cost of 42 oranges
$\text{cost} = \text{unit price} \times \text{number of items}$
$\text{cost} = 0.36 \times 42$
Break 42 into 40 and 2 to make it easier.
$0.36 \times 40 = 14.40$
$0.36 \times 2 = 0.72$
$14.40 + 0.72 = 15.12$
The 42 oranges cost $15.12.
Avoid the traps and answer
Watch out: $15.02, choice C, is only 10 cents away. Multiply and add with care. Watch out: Lian pays for each orange, not for full bags. Buying 6 whole bags would cost $17.28, choice A.
So the answer is B, $15.12.

Check your answer

42 oranges is 10 more than 32. Those 10 oranges cost $3.60, since $10 \times 0.36 = 3.60$ . Then $11.52 + 3.60 = 15.12$ . Same answer, so $15.12 is right.

More questions like this

The original DOE question

Question 99 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 87, question 99. Open page 87 of the handbook (PDF)

Question 8 of 25

2024 Test A Math, question 45

Question

Misha wants to use ribbon to make 2 straps for a backpack. The ribbon costs $5.00 a yard. If each strap requires $1\frac{1}{4}$ yards of ribbon, how much will Misha pay for the ribbon (not including tax)?

A$4.00
B$6.25
C$11.25
D$12.50

Show the answer

Correct answer: D

Each strap requires $1\frac{1}{4}$ yards of ribbon, and ribbon costs $5.00 a yard.

$1\frac{1}{4} \times 5 = \frac{5}{4} \times \frac{5}{1} = \frac{25}{4} = 6.25$ , so the ribbon for each strap costs $6.25.

Misha wants to make 2 straps.

$6.25 \times 2 = 12.50$ , so the cost of the ribbon for 2 straps is $12.50.

Question analysis

Misha needs ribbon for 2 backpack straps, each using 1 1/4 yards, and ribbon costs $5.00 a yard. This cost word problem asks how much she will pay for the ribbon before tax. Students multiply a mixed number by a price per yard and by the number of straps; it is short, but both the mixed number and the two straps need care.

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Multiplying a mixed number by a whole number and finding a total cost are grade 5 and 6 skills. With two short steps and friendly numbers, this is easier than typical grade 7 problems with percents or negatives.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is short and the numbers are friendly; points are lost by stopping at one strap or by mishandling the mixed number. Pricing a single strap gives a listed choice, as does doubling only the whole-number part of 1 1/4. Finding the total ribbon first, 2 1/2 yards, before multiplying by the price keeps both errors away.

Error-prone arithmetic minor Working it out
The length per strap is a mixed number, 1 1/4 yards, so the arithmetic is easiest as 2 × 1.25 = 2.5 yards, then 2.5 × $5.00. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Much to keep track of minor Working it out
The number of straps comes in the first sentence and the length of each in the last, so the 2 has to be carried through to the cost. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find the total ribbon first by doubling the length for one strap, then multiply the yards by $5.00; writing 1 1/4 as 1.25 keeps the arithmetic simple.

Traps

Stops one step early: A choice is a number from partway through the solution.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides the price by the length, $5.00 ÷ 1.25 = 4, instead of multiplying yards by dollars per yard.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Prices the ribbon for one strap, 1 1/4 × $5.00 = $6.25, and forgets that Misha is making 2 straps.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Doubles only the whole-number part of 1 1/4, getting 2 1/4 yards instead of 2 1/2, and pays 2.25 × $5.00 = $11.25.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost: $\text{cost} = \text{price per unit} \times \text{number of units}$; Price per unit is the cost of one unit, here $5.00 for one yard. Number of units is how many you buy, here yards of ribbon. Multiply them to get the total before tax.

Sort out the question
Misha makes 2 straps. Each strap needs $1\frac{1}{4}$ yards of ribbon. A yard is a length of 3 feet. Ribbon costs $5.00 a yard. You need the total cost for both straps, with no tax.
Find the total ribbon
First change the mixed number to a decimal. One quarter is 0.25, so $1\frac{1}{4}$ is 1.25. Then double it for 2 straps.
$2 \times 1.25 = 2.5$
Misha needs 2.5 yards of ribbon. Watch out: double the whole mixed number, the 1 and the quarter. Doubling only the 1 gives $2\frac{1}{4}$ yards and $11.25, choice C.
Find the cost
$\text{cost} = \text{price per unit} \times \text{number of units}$
$\text{cost} = 5 \times 2.5 = 12.5$
The ribbon costs $12.50. Watch out: this is for both straps. One strap’s ribbon costs $6.25, choice B. Don’t stop there.
So the answer is D, $12.50.

Check your answer

Estimate. Each strap needs a bit more than 1 yard, so both need a bit more than 2 yards. Two yards cost $10, so the answer should be a bit more than $10. $12.50 fits. Also, 2 yards cost $10 and the extra half yard costs $2.50, which makes $12.50.

More questions like this

The original DOE question

Question 102 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 87, question 102. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 9 of 25

2024 Test B Math, question 14

Question

Karen reads at the rate of 60 pages per day. Martina reads 25% more pages per day than Karen. If the two students begin 1,500-page reading assignments on the same day, how many days sooner will Martina complete the assignment than Karen?

Show the answer

Correct answer: B

First, determine the rate at which Martina reads, in pages per day, using the rate at which Karen reads, 60 pages per day.

$60+60\bullet\frac{25}{100}=$

$60+60\bullet 0.25=$

$60+15=75$

Next, determine the number of days it will take each person to complete the 1,500-page reading assignment.

Karen: $\frac{1{,}500\text{ pages}}{k\text{ days}}=\frac{60\text{ pages}}{1\text{ day}}$

$60k=1{,}500$

$(60\div 60)k=1{,}500\div 60$

$k=25\text{ days}$

Martina: $\frac{1{,}500\text{ pages}}{m\text{ days}}=\frac{75\text{ pages}}{1\text{ day}}$

$75m=1{,}500$

$(75\div 75)m=1{,}500\div 75$

$m=20\text{ days}$

Finally, determine the difference between the two numbers of days.

$25-20=5$

So Martina will complete the assignment 5 days sooner than Karen.

Question analysis

A reading-rate word problem: Karen reads 60 pages a day, and Martina reads 25% more pages a day. Both start a 1,500-page assignment on the same day, and students find how many days sooner Martina finishes. It combines a percent increase with rates and times, and asks for a difference in days.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 percent problems usually stop after one increase or decrease. This one turns the percent into a new rate, finds two finishing times and compares them, four steps that go beyond a typical grade 7 task.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Combines ideas moderate
Turning words into math minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is easy on its own; the difficulty is in carrying the chain to the end. Three of the choices are numbers a student meets or makes along the way: the extra pages per day, a quotient of two rates, or a rate built on 25 pages instead of 25%. Writing both finishing times down before subtracting keeps the target, a number of days, in view.

Many steps moderate Working it out
Four operations chain together: 25% of 60, Martina's 75 pages, the two finishing times 1,500 ÷ 60 and 1,500 ÷ 75, and finally their difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Combines ideas moderate Working it out
A percent increase in a daily rate has to be turned into a change in days, and 25% more pages a day does not mean 25% fewer days. (Embretson, 1998; Daniel & Embretson, 2010)
Turning words into math minor Understanding the question
“25% more pages per day than Karen” makes Martina's rate 1.25 × 60, and “how many days sooner” asks for a difference of two times, not of two rates. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Increase 60 by 25% to get Martina's daily pages, divide 1,500 by each girl's daily rate to get her number of days, then subtract the two numbers of days.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides Karen's 60 pages a day by Martina's 15 extra pages a day, combining two rates without finding how long either girl takes.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Adds 25 pages instead of 25%, giving Martina 85 pages a day, so she needs 1,500 ÷ 85 ≈ 17.6, or 18 days, and 25 − 18 = 7.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that Martina reads 25% of 60 = 15 more pages each day and stops there, giving a number of pages instead of days.

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.
Time from a rate: $\text{days} = \text{total pages} \div \text{pages per day}$; Total pages is the whole assignment. Pages per day is how fast someone reads. Dividing tells how many days of reading fit into the whole job.

Sort out what you know
Karen reads 60 pages per day. Martina reads 25% more pages per day than Karen. Each girl must read 1,500 pages. You need how many days sooner Martina finishes. That is the difference between their numbers of days.
Find Martina’s pages per day
25% more means Karen’s 60 pages plus 25% of 60.
$\text{new} = \text{old} + \frac{p}{100} \times \text{old}$
$60 + \frac{25}{100} \times 60 = 60 + 15 = 75$
Martina reads 75 pages per day. Watch out: 15 is extra pages, not days. Stopping here gives 15, choice D. Adding 25 pages instead of 25% gives 85 pages and leads to 7, choice C.
Find each girl’s number of days
Divide the whole assignment by each girl’s pages per day.
$\text{days} = \text{total pages} \div \text{pages per day}$
$\text{Karen: } 1{,}500 \div 60 = 25$
$\text{Martina: } 1{,}500 \div 75 = 20$
Karen needs 25 days, and Martina needs 20 days.
Subtract the days
“How many days sooner” asks for the difference.
$25 - 20 = 5$
Martina finishes 5 days sooner. Watch out: dividing 60 by 15 mixes two rates and gives 4, choice A.
So the answer is B, 5.

Check your answer

Multiply back. Karen: $25 \times 60 = 1{,}500$ pages. Martina: $20 \times 75 = 1{,}500$ pages. Both read all 1,500 pages, and 25 days is 5 days more than 20 days.

More questions like this

The original DOE question

Question 71 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 202, question 71. Open page 202 of the handbook (PDF)

Question 10 of 25

2024 Test B Math, question 38

Question

Michael received a check for $213 for working 20 hours at his part-time job. The amount on the check was his total pay after $32 in taxes was deducted. What was Michael’s hourly pay rate before taxes were deducted?

A$9.05
B$10.65
C$12.25
D$21.35

Show the answer

Correct answer: C

Add the taxes to the amount of the check to find Michael’s total pay before taxes were deducted:

$213+32=245$

Divide the total pay by the number of hours to find the hourly pay rate:

$245\div 20=12.25$

Question analysis

Michael's paycheck shows what he earned for 20 hours after taxes were taken out, and the tax amount is given. Students find his hourly pay rate before taxes. It tests unit rates in a money word problem, where students must decide what to do with the tax that was deducted.

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Unit rate problems are grade 6 content, and this one needs one addition and one division with a decimal result. It is easier than typical grade 7 work, apart from deciding which amount is the pay before taxes.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two operations in the right order, adding the tax back and dividing by the hours, are all this takes. The danger is the word deducted, which pulls students toward subtracting the $32 a second time, and the phrase before taxes, which rules out the take-home rate from dividing the check alone. Checking that $12.25 × 20 − $32 gives back $213 confirms the answer.

Turning words into math minor Understanding the question
"The amount on the check was his total pay after $32 in taxes was deducted" makes $213 what remained, so the $32 is added back, not taken away. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Rebuild Michael's pay before taxes from the check and the tax that was taken out, then divide that total by the 20 hours worked.

Traps

Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Follows the word deducted and subtracts the $32 from the $213 check again, (213 − 32) ÷ 20 = 9.05, instead of adding it back.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides the $213 check by 20 hours, which gives Michael's hourly pay after taxes rather than before.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard working gives $21.35, which equals 32 − 10.65; dividing $213 by 10 hours instead of 20, $21.30, comes closest, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Hourly pay rate: $\text{rate} = \frac{\text{total pay}}{\text{hours}}$; rate is the pay for one hour. total pay is the money earned for all the hours. hours is the time worked. Dividing shares the total pay equally among the hours. Use it when the pay for each hour is the same.

Sort out the question
Michael’s check was for $213. That is what was left after $32 in taxes was deducted, which means taken away. He worked 20 hours.
You need his hourly pay before taxes. So first find his full pay before the $32 was taken out.
Add the taxes back
The $32 was taken away from his full pay to leave $213. To undo taking away, add it back.
$213 + 32 = 245$
His pay before taxes was $245. Watch out: The word “deducted” tempts you to subtract again. That gives $(213 - 32) \div 20 = 9.05$ , choice A.
Divide by the hours
$\text{rate} = \frac{245}{20} = 12.25$
He earned $12.25 for each hour before taxes. Watch out: Dividing the $213 check by 20 gives $10.65, choice B. That is his pay for each hour after taxes.
So the answer is C, $12.25.

Check your answer

Work forward. $12.25 \times 20 = 245$ , so 20 hours pay $245. Take away $32 in taxes: $245 - 32 = 213$ . That is the amount on the check.

More questions like this

The original DOE question

Question 95 as printed on page 210 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 95. Open page 210 of the handbook (PDF)

Question 11 of 25

2023 Test A Math, question 9

Question

A revolving sign makes 1 complete revolution every 90 minutes. If the sign starts moving at 2:30 p.m., at what time will the sign complete 8 revolutions?

A1:00 a.m.
B2:30 a.m.
C4:00 p.m.
D12:00 midnight

Show the answer

Correct answer: B

First, use a proportion to find out how many minutes it will take the sign to complete 8 revolutions. Let x equal the total number of minutes:

$\frac{x}{8} = \frac{90}{1}$

$x = 720$

Next, convert 720 minutes into hours and minutes:

$\frac{720}{60} = 12\text{ hours}$

Finally, 12 hours after 2:30 p.m. is 2:30 a.m.

Question analysis

A revolving sign turns once every 90 minutes and starts at 2:30 in the afternoon. Students find the clock time when it finishes 8 full revolutions. The question combines a rate with converting minutes to hours, and then adding the elapsed time to the starting time on a 12-hour clock.

NYS standards: 6.RP.3b unit rate problems
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 A constant rate, a multiplication and a conversion from minutes to hours are grade 6 skills. The only step beyond routine work is adding the elapsed time to a starting clock time.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is routine, so most points are lost in keeping elapsed time and clock time apart. The 12 hours is a duration, and the clock reading that ends it is 12 hours after 2:30 in the afternoon, not 12:00. A sense check helps: 12 hours after any time shows the same digits on the clock with a.m. and p.m. swapped.

Many steps minor Working it out
Three moves: 8 × 90 = 720 minutes, 720 ÷ 60 = 12 hours, then 12 hours counted on from the 2:30 start, which crosses midnight into the morning. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Counting hours past midnight means switching from p.m. to a.m. on a 12-hour clock, which the stem leaves to the student; the choices mix both halves of the day. (Anderson & Krathwohl, 2001)

Strategy: Multiply 90 minutes by 8 to get the total time, change it to hours, and then count that many hours forward from 2:30 p.m.

Traps

Stops one step early: A choice is a number from partway through the solution.
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Counts only 7 revolutions, 630 minutes or 10½ hours, as if the first turn were already done at the start, and lands at 1:00 a.m.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds a single revolution's 90 minutes to the 2:30 start, giving 4:00 in the afternoon, without multiplying by the 8 revolutions.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds that 8 revolutions take 12 hours and picks 12:00 midnight, reading the elapsed time as a clock time instead of counting on from 2:30.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total time: $\text{total time} = \text{time for one} \times \text{how many}$; Time for one is how long one revolution takes. How many is the number of revolutions. Each one takes the same time, so multiply to get the total time.
Minutes to hours: $\text{hours} = \text{minutes} \div 60$; There are 60 minutes in 1 hour. So to change minutes to hours, divide by 60. Each group of 60 minutes makes 1 hour.

Sort out the question
The sign turns all the way around once every 90 minutes. One full turn is called a revolution. The sign starts at 2:30 p.m. You need the clock time when it finishes 8 revolutions.
Find the total time in minutes
Each revolution takes 90 minutes, and there are 8 of them.
$\text{total time} = 90 \times 8 = 720$
The 8 revolutions take 720 minutes. Watch out: use all 8 revolutions. Counting only 7 gives 630 minutes and 1:00 a.m., choice A. Adding just one 90 minutes gives 4:00 p.m., choice C.
Change the minutes to hours
Clock times are easier to count in hours. There are 60 minutes in an hour.
$\text{hours} = 720 \div 60 = 12$
The sign turns for 12 hours.
Count forward from 2:30 p.m.
Start at 2:30 p.m. and add 12 hours. When you pass midnight, p.m. changes to a.m. So 12 hours after 2:30 p.m. is 2:30 a.m. the next morning. Watch out: 12 hours is an amount of time, not a clock time. Reading it as 12:00 midnight gives choice D. So the answer is B, 2:30 a.m.

Check your answer

From 2:30 p.m. to 2:30 a.m. is one full trip around the clock, 12 hours. 12 hours is $12 \times 60 = 720$ minutes. And $720 \div 90 = 8$ , the number of revolutions in the question.

More questions like this

The original DOE question

Question 66 as printed on page 77 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 66. Open page 77 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 12 of 25

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 13 of 25

2023 Test A Math, question 28

Question

It took Lars 2 hours to ride his bicycle 48 kilometers. What was his average speed in miles per hour? (Use the approximation 1 mile = 1.6 kilometers.)

A1.5
B15.0
C30.0
D38.4

Show the answer

Correct answer: B

First, calculate Lars’s average speed in kilometers per hour:

$\frac{48}{2} = 24$

Then, convert kilometers to miles:

$\frac{m}{24} = \frac{1}{1.6}$	Cross multiply.
$m(1.6) = 24(1)$	Apply the multiplicative inverse property; divide both sides of the equation by 1.6.
$m = \frac{24}{1.6} = 15$

Question analysis

Lars rides 48 kilometers in 2 hours, and this rate question asks for his average speed in miles per hour, using 1 mile = 1.6 kilometers. Students find a unit rate and convert it to other units. Deciding whether to multiply or divide by 1.6, and remembering the 2 hours, are where students go wrong.

NYS standards: 6.RP.3d converting measurement units
6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a speed and converting units with a given rate are grade 6 ratio skills. The question takes two short steps with friendly numbers, lighter than typical grade 7 rate problems with fractional quantities.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question tests the direction of a conversion more than the arithmetic. A mile is longer than a kilometer, so a distance has fewer miles than kilometers, and multiplying by 1.6 runs the conversion backward. The other trap is converting the 48 kilometers to 30 miles and forgetting that the ride took 2 hours, which gives a distance instead of a speed.

Combines ideas minor Working it out
A rate and a unit conversion are combined: 48 ÷ 2 = 24 kilometers per hour, then 24 ÷ 1.6 = 15 miles per hour, and the two steps can come in either order. (Embretson, 1998; Daniel & Embretson, 2010)
Error-prone arithmetic minor Working it out
Dividing by 1.6 means dividing 240 by 16; a decimal point in the wrong place makes the speed ten times too small, and that value is offered. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the speed in kilometers per hour first, then divide by 1.6, since a mile is longer than a kilometer and the number of miles must be smaller.

Traps

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

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Divides 24 by 16 instead of 1.6, a decimal point in the wrong place that makes the speed ten times too small, 1.5.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Converts the 48 kilometers to 30 miles correctly and stops, forgetting to divide by the 2 hours, so it gives a distance rather than a speed.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 24 kilometers per hour by 1.6 instead of dividing, converting in the wrong direction and getting more miles than kilometers.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average speed: $\text{speed} = \frac{\text{distance}}{\text{time}}$; Distance is how far you go. Time is how long it takes. Speed is how far you go in each unit of time, such as each hour. Average speed shares the whole trip out evenly over the time.
Kilometers to miles: $\text{miles} = \text{kilometers} \div 1.6$; A kilometer is a unit of length used in most countries. 1 mile is about 1.6 kilometers, so a mile is longer. To find how many miles, find how many groups of 1.6 kilometers fit. So divide by 1.6.

Sort out the question
Lars rode 48 kilometers in 2 hours. You need his average speed in miles per hour. That means miles, not kilometers. Use the fact that 1 mile is about 1.6 kilometers.
Find the speed in kilometers per hour
Divide the distance by the time.
$\text{speed} = \frac{48}{2} = 24$
Lars went 24 kilometers each hour. This speed is still in kilometers, so you are not done yet.
Change kilometers to miles
Divide by 1.6. To make it easier, multiply both numbers by 10 first. Then you divide by the whole number 16.
$24 \div 1.6 = 240 \div 16 = 15$
Watch out: divide by 1.6, don’t multiply. A mile is longer, so there are fewer miles than kilometers. $24 \times 1.6 = 38.4$ is choice D. Dividing 24 by 16 gives 1.5, choice A, with the decimal point in the wrong place.
Name the answer
Lars rode 15 miles each hour. Watch out: changing 48 kilometers to 30 miles is only part of the work. 30 is a distance, choice C. You still have to divide by the 2 hours. So the answer is B, 15.0 miles per hour.

Check your answer

Work backward. At 15 miles per hour for 2 hours, Lars rides 30 miles. And $30 \times 1.6 = 48$ kilometers, the distance in the question. Also, 15 is smaller than 24, as it should be, since miles are longer than kilometers.

More questions like this

The original DOE question

Question 85 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 85. Open page 82 of the handbook (PDF)

Question 14 of 25

2023 Test A Math, question 43

Question

Kim jogs 8 kilometers in 1 hour 40 minutes. At that rate, how many meters does she jog per minute?

A0.08
B80
C800
D8,000

Show the answer

Correct answer: B

First, calculate Kim’s jogging time in minutes:

1 hour 40 minutes = 100 minutes

Then, convert kilometers to meters:

8 kilometers = 8,000 meters

Now simplify the ratio to find the answer:

$\frac{8{,}000}{100} = 80$

Question analysis

This rate question says Kim jogs 8 kilometers in 1 hour 40 minutes and asks how many meters she jogs per minute. Students convert a mixed time to minutes and kilometers to meters before finding a unit rate. Skipping a conversion, or stopping before dividing, gives answers that are off by factors of ten or more.

NYS standards: 6.RP.3d converting measurement units
6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Unit rates and metric conversions are grade 6 topics. This needs two conversions, hours and minutes to minutes and kilometers to meters, then one division with whole numbers, simpler than grade 7 rate problems built on fractions.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The rate itself is one division; what makes the question work is that the distance and the time both arrive in the wrong units. Converting only the time gives a speed in kilometers per minute, and converting only the distance and stopping gives the whole trip in meters. Writing the units beside every number, meters over minutes, shows when the answer is ready.

Something you must already know minor Working it out
The conversions are not given: 1 kilometer is 1,000 meters and 1 hour is 60 minutes, so 8 kilometers becomes 8,000 meters and 1 hour 40 minutes becomes 100 minutes. (Anderson & Krathwohl, 2001)
Many steps minor Working it out
Two conversions, one for distance and one for time, come before the single division, and skipping the distance conversion leaves an answer a thousand times too small. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the time entirely in minutes and the distance entirely in meters, then divide distance by time; check that the answer's units are meters per minute.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Stops one step early: A choice is a number from partway through the solution.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides the 8 kilometers by the 100 minutes but never changes kilometers to meters, giving a speed of 0.08 kilometer per minute.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Converts both units correctly but cancels only one zero when dividing 8,000 by 100, getting a rate ten times too large.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Converts 8 kilometers to 8,000 meters and stops, giving the whole distance before dividing by the 100 minutes it took.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Kilometers to meters: $\text{meters} = \text{kilometers} \times 1{,}000$; Each kilometer is 1,000 meters, so multiply the number of kilometers by 1,000 to get meters. The word part kilo means a thousand.
Rate: $\text{rate} = \frac{\text{distance}}{\text{time}}$; The rate, or speed, is how far you go in one unit of time. Distance is how far, and time is how long. Dividing shares the distance equally among the minutes, so you get the distance for one minute.

Sort out the question
Kim jogs 8 kilometers in 1 hour 40 minutes. You need her speed in meters per minute: how many meters she jogs in one minute. The question gives kilometers and hours, so change both units first.
Change the time to minutes
One hour is 60 minutes. Add the extra 40 minutes.
$60 + 40 = 100$
Kim jogs for 100 minutes.
Change the distance to meters
$\text{meters} = \text{kilometers} \times 1{,}000$
$8 \times 1{,}000 = 8{,}000$
Kim jogs 8,000 meters. Watch out: 8,000 is the whole distance, not meters per minute. Stopping here gives choice D.
Divide the distance by the time
$\text{rate} = \frac{\text{distance}}{\text{time}}$
$\text{rate} = \frac{8{,}000}{100} = 80$
She jogs 80 meters each minute. Watch out: if you skip the change to meters, you get 8 ÷ 100 = 0.08 kilometer per minute, choice A. So the answer is B, 80 meters per minute.

Check your answer

Work backward. At 80 meters a minute, 100 minutes gives $80 \times 100 = 8{,}000$ meters. That is 8 kilometers, just as the question says.

More questions like this

The original DOE question

Question 100 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 100. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 15 of 25

2023 Test A Math, question 54

Question

A train travels 2,200 miles from Phoenix to New York City. The train covers the first 240 miles in 5 hours. If the train continues to travel at this rate, how many more hours will it take to reach New York City? Round your answer to the nearest whole hour.

Show the answer

Correct answer: D

First, find the speed of the train in miles per hour: $240 \div 5 = 48$ .

The number of miles left to travel is $2{,}200 - 240 = 1{,}960$ .

To find the number of hours left, use the equation $rt = d$ :

$48t = 1{,}960$	Apply the multiplicative inverse property; divide both sides of the equation by 48.
$t = \frac{1{,}960}{48} = 40\frac{5}{6}$

That number rounded to the nearest whole hour is 41 hours.

Question analysis

This rate word problem describes a 2,200-mile train trip from Phoenix to New York City in which the first 240 miles take 5 hours, and asks how many more hours the rest will take, rounded to the nearest hour. Students find a unit rate and the remaining distance. The choices are close together, so the time for the whole trip is a tempting trap.

NYS standards: 6.RP.3b unit rate problems
6.NS.2 multi-digit division
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Unit rates are grade 6 content, but this chains four steps: a speed, the distance left, a long division that does not come out even, and rounding. The word more must be read closely, which puts it at typical grade 7 demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Turning words into math minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is routine, but the question has a long chain and asks for only part of the trip. Dividing the whole 2,200 miles by the speed is the natural shortcut and gives a listed choice, and the rounding instruction adds a second place to go wrong. Checking that 41 hours at 48 miles per hour covers about 1,970 miles, close to the 1,960 left, confirms the answer.

Many steps moderate Working it out
Four steps follow in order, the speed 240 ÷ 5 = 48, the miles left 2,200 − 240 = 1,960, the time 1,960 ÷ 48, and the rounding, and an early slip carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The bold more in “how many more hours” limits the question to the 1,960 miles still ahead, though 2,200, the whole trip, is the distance the stem names first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Dividing 1,960 by 48 leaves 40 with a remainder of 40, so the time, 40 5/6 hours, must be rounded up to the nearest hour rather than cut short. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the train's speed from the first 240 miles, subtract those miles from the total to get the distance left, divide by the speed, and round only at the end.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the whole 2,200 miles by 48 miles per hour, about 45.8, or 46 hours, the time for the entire trip rather than the hours still to go.
B Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: weak
Divides the whole 2,200 miles by 48 and then cuts 45.8 down to 45 instead of rounding to the nearest hour, two errors at once.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Subtracts 240 from 2,200 without reducing the hundreds after borrowing, getting 2,060 instead of 1,960, and 2,060 ÷ 48 ≈ 42.9 rounds to 43.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, rate and time: $d = r \times t$; d is the distance, r is the rate or speed, and t is the time. If you know two of them, you can find the third. To find the speed, divide the distance by the time. To find the time, divide the distance by the speed.

Sort out the question
The whole trip is 2,200 miles. The train goes the first 240 miles in 5 hours, and it keeps that speed. You need how many more hours it takes, rounded to the nearest whole hour. “More” means the time after the first 240 miles.
Find the train’s speed
$d = r \times t$
$240 = r \times 5$
$r = 240 \div 5 = 48$
Divide the 240 miles by the 5 hours. The train goes 48 miles per hour.
Find the miles still to go
$2{,}200 - 240 = 1{,}960$
There are 1,960 miles left. Watch out: use the miles left, not the whole trip. Dividing 2,200 by 48 gives about 45.8 hours, which leads to 46 or 45, choices A and B.
Find the hours still to go
$1{,}960 = 48 \times t$
$t = 1{,}960 \div 48 = 40\frac{40}{48} = 40\frac{5}{6}$
Here 48 × 40 = 1,920, and 40 miles are left over. That leftover is $\frac{40}{48} = \frac{5}{6}$ of an hour.
Round to the nearest hour
$\frac{5}{6}$ of an hour is more than half an hour, so round up to 41. Watch out: don’t just drop the fraction.
So the answer is D, 41 hours.

Check your answer

At 48 miles per hour, 41 hours covers $48 \times 41 = 1{,}968$ miles. That is just a bit more than 1,960, as it should be after rounding up. In 40 hours, the train would cover only 1,920 miles, which is not enough.

More questions like this

The original DOE question

Question 111 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 111. Open page 90 of the handbook (PDF)

Question 16 of 25

2023 Test B Math, question 25

Question

The town of Waterville covers an area of 50 square miles. In 2005, its population was 84,000. In 2006, its population had decreased so that the average population per square mile was equal to 1,500. What was the total decrease in population from 2005 to 2006?

A7,500
B8,400
C9,000
D9,200

Show the answer

Correct answer: C

Let p represent the population of Waterville in 2006. The population per square mile is:

$\frac{p}{50} = 1{,}500$

Solve for p:

$p = 1{,}500(50) = 75{,}000$

The population decrease from 2005 to 2006 is $84{,}000 - 75{,}000 = 9{,}000$ .

Question analysis

The town of Waterville covers 50 square miles and had 84,000 people in 2005. By 2006 the population had dropped so that there were 1,500 people per square mile on average. Students find how much the population decreased, turning the population density into a total population before subtracting.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Using a rate such as people per square mile is grade 6 content. Here it is one step inside a two-step problem with large round numbers, which matches typical grade 7 multi-step rate work.
Estimated difficulty: Straightforward: two steps or one careful reading.
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: 2 steps
Reading level: Question and choices: about grade 8, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question gives a total for 2005 but only a rate for 2006, so the two years cannot be compared until the rate becomes a population. After that, one subtraction finishes it. Students who stop at the 2006 population, or who reach for a number shaped like 84,000, land on listed choices, while the near miss above the answer comes from no sensible working.

Turning words into math moderate Understanding the question
“The average population per square mile was equal to 1,500” is a density, which has to become a total, 1,500 × 50 = 75,000, before it can be compared with 84,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
The 2006 population is only a midpoint; subtracting it from 84,000 is a second step, and the question asks for that decrease, not either population. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply 1,500 people per square mile by 50 square miles to get the 2006 population, then subtract it from the 2005 population.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the 2006 population, 1,500 × 50 = 75,000, and stops; 7,500, one zero short, is the choice whose digits match it.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes a tenth of the 2005 population, 84,000 ÷ 10, a number built from the question rather than from the density.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 9,200; it lies just above the right decrease and needs an arithmetic slip with no reason behind it, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Population per square mile: $\text{people per square mile} = \frac{\text{population}}{\text{area}}$; The population is the number of people who live in a place. The area is its size, here in square miles. Dividing gives the average number of people for each square mile. To get back the population, multiply people per square mile by the area.

Sort out the question
Waterville covers 50 square miles. In 2005, 84,000 people lived there. In 2006, there were 1,500 people per square mile, on average. You need how many fewer people lived there in 2006 than in 2005.
Find the 2006 population
Let p be the 2006 population. Use the formula.
$\text{people per square mile} = \frac{\text{population}}{\text{area}}$
$1{,}500 = \frac{p}{50}$
To undo dividing by 50, multiply both sides by 50.
$p = 1{,}500 \times 50 = 75{,}000$
So 75,000 people lived there in 2006.
Find the decrease
A decrease is how much a number went down. Subtract the 2006 population from the 2005 population.
$84{,}000 - 75{,}000 = 9{,}000$
The population went down by 9,000 people.
Give the answer
Watch out: 75,000 is the 2006 population, not the decrease. Choice A, 7,500, looks like it but has one zero fewer. The question asks how much the population dropped. So the answer is C, 9,000.

Check your answer

Work with people per square mile instead. In 2005, there were $84{,}000 \div 50 = 1{,}680$ people per square mile. In 2006, there were 1,500. That is 180 fewer per square mile, and $180 \times 50 = 9{,}000$ fewer people.

More questions like this

The original DOE question

Question 82 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 82. Open page 199 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 17 of 25

2022 Test B Math, question 30

Question

A robot moves forward at the rate of 9 steps every 6 seconds. If each step is x feet long, what must x be in order for the robot to travel exactly 270 feet per minute?

Show the answer

Correct answer: A

First, convert steps per second to steps per minute.

$\frac{9\text{ steps}}{6\text{ seconds}}=\frac{90\text{ steps}}{60\text{ seconds}}=\frac{90\text{ steps}}{1\text{ minute}}$

Then, use ratio reasoning to determine that each step must be 3 feet since 90 times 3 is 270 (or 270 divided by 90 is 3).

$\frac{90\text{ steps}}{1\text{ minute}}\times\frac{3\text{ feet}}{1\text{ step}}=\frac{270\text{ feet}}{1\text{ minute}}$

$\frac{270\text{ feet}}{1\text{ minute}}\div\frac{90\text{ steps}}{1\text{ minute}}=$

$\frac{270\text{ feet}}{1\text{ minute}}\times\frac{1\text{ minute}}{90\text{ steps}}=$

$\frac{3\text{ feet}}{1\text{ step}}$

Question analysis

A robot moves forward 9 steps every 6 seconds, and each step is x feet long. Students find the step length that makes the robot travel exactly 270 feet per minute. It tests rates and unit conversion: the rate is given per 6 seconds but the target is per minute, so the units must be matched before solving.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3d converting measurement units
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 rate problems often give one rate and ask for a total. This one links steps per time, feet per step and feet per minute, with seconds to convert, a typical but full multistep ratio problem.
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: Much to keep track of moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 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 numbers are friendly, but the rate is given per 6 seconds and the target per minute, so the units have to meet first. A minute holds ten 6-second stretches, so the robot takes 90 steps in it, and 270 feet over 90 steps settles the length. Students who divide 270 by the 9 steps forget that those steps take only 6 seconds, and those who scale 270 down to 6 seconds stop at 27 feet, a distance rather than a step length.

Much to keep track of moderate Working it out
Three rates are in play, 9 steps per 6 seconds, x feet per step and 270 feet per minute, and their units must be lined up before any two are combined. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Many steps minor Working it out
The rate is first scaled from 6 seconds to 60, giving 90 steps a minute, and only then compared with the 270 feet covered in that minute. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Change the rate to steps per minute first, since 6 seconds goes into a minute evenly, then ask what step length makes that many steps cover 270 feet.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Stops one step early: A choice is a number from partway through the solution.
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

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 9 steps by the 6 seconds, 54, and divides 270 by it, combining the stem’s numbers without a model of the rate.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Scales 270 feet per minute down to 27 feet per 6 seconds but stops before dividing by the 9 steps taken in that time.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides 270 feet by 9 steps, as if the robot took 9 steps a minute, never converting the 6-second rate to a minute.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rate: $\text{rate} = \frac{\text{amount}}{\text{time}}$; A rate compares an amount with the time it takes. Here the robot’s rate is 9 steps for every 6 seconds. To compare two rates, first write them for the same length of time, such as one minute, which is 60 seconds.
Distance from steps: $\text{distance} = \text{number of steps} \times \text{length of each step}$; Every step covers the same length. So the distance is the number of steps times the length of one step. Here the distance and the step length are both in feet.

Sort out the question
The robot takes 9 steps every 6 seconds. Each step is x feet long. You need the x that makes the robot go exactly 270 feet per minute.
The two rates use different times: 6 seconds and 1 minute. Match the times before you compare them.
Find the steps in one minute
One minute is 60 seconds. Find how many 6-second stretches fit in a minute.
$60 \div 6 = 10$
In each stretch, the robot takes 9 steps. So in one minute it takes 10 times as many.
$9 \times 10 = 90$
The robot takes 90 steps per minute.
Find the length of each step
In one minute, those 90 steps must cover 270 feet. Use the distance formula.
$\text{distance} = \text{number of steps} \times \text{length of each step}$
$270 = 90 \times x$
To get x alone, divide both sides by 90.
$x = 270 \div 90 = 3$
Each step must be 3 feet long.
Give the answer
Watch out: dividing 270 by 9 gives 30, choice D. That treats 9 steps as a whole minute’s steps, but they take only 6 seconds. Cutting 270 feet per minute down to 27 feet per 6 seconds and stopping gives 27, choice C. That is a distance, not a step length. So the answer is A, 3.

Check your answer

Check the units. 90 steps per minute times 3 feet per step gives $90 \times 3 = 270$ feet per minute. That is the goal. Or think in 6 seconds: 9 steps of 3 feet is 27 feet, and 10 of those stretches make 270 feet.

More questions like this

The original DOE question

Question 87 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 205, question 87. Open page 205 of the handbook (PDF)

Question 18 of 25

2018 Test A Math, question 10

Question

In a scale diagram, $0.125$ inch represents $125$ feet. How many inches represent $1$ foot?

A $0.001$
B $0.01$
C $0.1$
D $0.12$

Show the answer

Correct answer: A

Let $x$ be the number of inches representing 1 foot. Set up a proportion and solve for $x$ :

$\frac{x}{1}=\frac{0.125}{125}$

$x=0.001$ in.

Question analysis

This scale drawing question uses decimals: on a diagram, 0.125 inch stands for 125 feet. Students find how many inches represent just 1 foot, which is the unit rate of the scale. Dividing a decimal by a large whole number makes the decimal point easy to misplace.

NYS standards: 7.G.1 scale drawings
6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings and unit rates are core grade 7 topics, and this is a single division. The demand matches typical grade 7 work, with decimal place value as the only real hurdle.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Solution: 1 step
Reading level: Question and choices: about grade 11, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

A single division settles this question, and the choices are built to punish a misplaced decimal point, since three of them are each a power of ten apart. Students who see that 0.125 is one-thousandth of 125 can answer at once; students who divide by hand often lose a zero. Multiplying the chosen answer back by 125 is a quick check.

Error-prone arithmetic moderate Working it out
The one step is 0.125 ÷ 125, a decimal divided by a large whole number, where the place of the decimal point decides between 0.001, 0.01 and 0.1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
0.001 and 0.01 look almost the same, differing by one zero after the decimal point; only exact placement of the point in 0.125 ÷ 125 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Divide the inches by the feet to get inches per foot, then check by multiplying your result by 125 to see whether you get back 0.125.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Divides 0.125 by 125 but loses one zero in placing the decimal point, giving 0.01, ten times too large.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Sees that 125 ÷ 125 = 1 and puts the 1 in the tenths place, where 0.125 begins, instead of the thousandths place.
D Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Treats the 0.125 inch as if it already stood for 1 foot, never dividing by the 125 feet it represents, and picks the choice closest to it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $r = \frac{a}{b}$; r is the unit rate, the amount that goes with just 1 unit. a is the first amount, here 0.125 inch. b is the matching second amount, here 125 feet. Dividing shares the inches out equally, one share for each foot.

Sort out the scale
A scale diagram is a drawing where small lengths stand for real lengths. Here 0.125 inch on the drawing stands for 125 real feet. You need how many inches stand for just 1 foot. That is a unit rate, an amount for 1 unit.
Set up the division
Divide the inches by the feet. That splits the 0.125 inch into 125 equal shares, one for each foot.
$r = \frac{a}{b}$
$r = \frac{0.125}{125}$
Divide carefully
Write 0.125 as a fraction. It is 125 thousandths.
$0.125 = \frac{125}{1000}$
Now divide by 125. Split 125 thousandths into 125 shares, and each share is 1 thousandth.
$\frac{125}{1000} \div 125 = \frac{1}{1000} = 0.001$
So 1 foot is shown by 0.001 inch.
Match the decimal places
0.001 has the 1 in the third place after the decimal point. That place is thousandths. Watch out: 0.01 and 0.1 put the decimal point in the wrong place. They are choices B and C, and each is too big. So the answer is A, 0.001.

Check your answer

Multiply back. $0.001 \times 125 = 0.125$ inch, which matches the scale. Choice B would give $0.01 \times 125 = 1.25$ inches. That is ten times too long.

More questions like this

The original DOE question

Question 67 as printed on page 82 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 82, question 67. Open page 82 of the handbook (PDF)

Question 19 of 25

2018 Test A Math, question 29

Question

A car is traveling 55 miles per hour, and 1 mile = 5,280 feet. Which of the following calculations would give the car’s speed in feet per second?

A $\frac{55\cdot 5280}{1}$
B $\frac{55\cdot 5280}{3600}$
C $\frac{55\cdot 3600}{5280}$
D $\frac{55\cdot 5280}{60}$

Show the answer

Correct answer: B

Divide the rate by the number of seconds in an hour. (Since there are 60 minutes in an hour and 60 seconds in a minute, multiply 60 × 60 = 3,600 seconds in an hour):

$\frac{55}{3600}$ miles per second

Multiply by the number of feet in a mile (5,280):

$\frac{55\cdot 5280}{3600}$ feet per second

Question analysis

This unit conversion question gives a car's speed of 55 miles per hour and asks which calculation expresses it in feet per second. Students have to change miles to feet and hours to seconds in one expression. The tricky part is converting both units at once and keeping each conversion factor on the correct side of the fraction bar.

NYS standards: 6.RP.3d converting measurement units
6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students usually convert one unit at a time, such as feet to inches. This item converts both parts of a rate at once, miles per hour into feet per second, and asks students to recognize the right setup.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 28 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Nothing has to be calculated; the work is judging which setup turns miles per hour into feet per second. Students who convert only part of the way find a choice waiting, whether they stop at feet per hour or go from hours to minutes instead of seconds. Tracking which unit each factor cancels, rather than computing, is what the question rewards.

A very close wrong answer moderate Choosing the answer
The answer, 55 · 5,280/3,600, and the choice 55 · 5,280/60 are written identically except for the denominator, so only knowing that an hour has 3,600 seconds, not 60, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
The stem gives 5,280 feet per mile but not the 3,600 seconds in an hour, which students must build from 60 × 60 themselves. (Anderson & Krathwohl, 2001)

Strategy: Write the speed as 55 miles over 1 hour and multiply by unit fractions such as 5,280 feet over 1 mile, checking that miles and hours cancel and only feet per second remain.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Converts miles to feet with 55 × 5,280 but never changes hours to seconds, leaving the speed in feet per hour.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Turns both conversion factors upside down, multiplying by 3,600 and dividing by 5,280, so miles and hours never cancel.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides by 60, converting hours to minutes and stopping there, which gives the speed in feet per minute rather than feet per second.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Speed: $r = \frac{d}{t}$; r is the speed, or rate. d is the distance traveled. t is the time it took. Speed tells how far something goes in one unit of time, so you divide the distance by the time.

Sort out what you know
The car goes 55 miles in 1 hour. One mile is 5,280 feet. You need the speed in feet per second. That means how many feet the car goes in 1 second.
The choices are setups, not numbers. So you only need to pick the right setup. You do not have to work out the final number.
Change the miles to feet
Each mile is 5,280 feet. So 55 miles is 55 groups of 5,280 feet. Multiply to get the distance in feet:
$55 \times 5{,}280$
That is how many feet the car goes in 1 hour. Watch out: if you stop here, you have feet per hour, not feet per second. That is choice A.
Change the hour to seconds
The question gives you miles to feet. It does not give you hours to seconds, so you build that yourself. An hour has 60 minutes, and each minute has 60 seconds:
$60 \times 60 = 3{,}600$
So 1 hour is 3,600 seconds. Watch out: 60 is the number of minutes in an hour, not seconds. Dividing by 60 gives feet per minute. That is choice D.
Divide the distance by the time
Use the speed formula. The distance is 55 × 5,280 feet. The time is 3,600 seconds:
$r = \frac{55 \times 5{,}280}{3{,}600}$
Feet on top and seconds on the bottom give feet per second. Watch out: choice C turns both changes upside down, so the miles and hours never cancel. So the answer is B, $\frac{55 \cdot 5{,}280}{3{,}600}$ .

Check your answer

Estimate. 55 × 5,280 is about 290,000 feet in an hour. Divide by 3,600 seconds and you get about 80 feet each second. A car on a highway covering 80 feet in one second sounds right.

More questions like this

The original DOE question

Question 86 as printed on page 87 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 86. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 20 of 25

2018 Test A Math, question 35

Question

For a presentation, Deion can create 5 slides in 20 minutes, working at a constant rate. Kyra can create 3 slides in 10 minutes, working at her own constant rate. What is the total number of slides the two of them can create in one hour?

Show the answer

Correct answer: C

Set up proportions to figure out how many slides Deion and Kyra can create in 1 hour:

Deion

$\frac{5}{20}=\frac{x}{60}$

$20x=300$

$x=15$

Deion can create 15 slides in 1 hour.

Kyra

$\frac{3}{10}=\frac{x}{60}$

$10x=180$

$x=18$

Kyra can create 18 slides in 1 hour.

Add Deion and Kyra to figure out how many slides they can create together in 1 hour:

15 + 18 = 33.

Question analysis

This rate word problem says how many slides Deion makes in 20 minutes and Kyra makes in 10 minutes, then asks how many slides they make together in one hour. Students must scale each person's rate to an hour before combining. Because their time periods differ, the two rates cannot simply be added as given.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 rate problems often ask for a single unit rate. This one needs two rates over different time periods, each scaled to an hour and then combined, which is typical grade 7 work built on grade 6 content.
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: Much to keep track of moderate
Many steps minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 42 words.
Students, last 30 days: 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 each person is scaled to an hour on their own; the trouble is the urge to combine first. Adding the slides and the minutes into one rate treats two different rates like fractions added across the top and bottom, and it gives a listed answer. Keeping the two workers separate until the last step avoids every trap.

Much to keep track of moderate Working it out
Deion's rate is given per 20 minutes and Kyra's per 10, so each needs its own scale factor to an hour, 3 and 6, before the totals can be added. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Many steps minor Working it out
Two scalings and an addition: 5 × 3 = 15 for Deion, 3 × 6 = 18 for Kyra, then the sum, each a simple step. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The choices 33 and 30 differ by only 3 slides, so a rough sense of the total cannot separate them; only scaling Kyra's own rate to 18 an hour, not 15, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Find how many 20-minute blocks and how many 10-minute blocks fit in 60 minutes, scale each person's slides by that number, and then add the two totals.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the rates straight across, 5 + 3 slides in 20 + 10 minutes, then doubles 8 slides per 30 minutes to get 16 in an hour.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds Deion's 15 slides an hour and doubles it for the two of them, missing that Kyra works at her own, faster rate.
D Units or scale (King et al., 2004)
Reveals a slip Expected pull: weak
Adds the per-minute rates, 0.25 + 0.3 = 0.55 slides, then scales by 100 instead of 60 minutes, getting 55.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Amount of work: $a = r \times t$; a is the amount of work done, here slides. r is the rate: how much gets done in one block of time. t is the number of those blocks. Use it when someone works at a steady speed.

Sort out what you know
Deion makes 5 slides every 20 minutes. Kyra makes 3 slides every 10 minutes. Each works at a constant rate, which means a steady speed that does not change.
One hour is 60 minutes. You need how many slides they make together in 60 minutes.
Find Deion’s slides in an hour
Deion’s rate is 5 slides per 20-minute block. An hour holds 3 of those blocks, since 60 ÷ 20 = 3:
$a = 5 \times 3 = 15$
Deion makes 15 slides in an hour.
Find Kyra’s slides in an hour
Kyra’s rate is 3 slides per 10-minute block. An hour holds 6 of those blocks, since 60 ÷ 10 = 6:
$a = 3 \times 6 = 18$
Kyra makes 18 slides in an hour. Watch out: Kyra has her own rate. Doubling Deion’s 15 to get 30, choice B, treats her as if she worked at his speed.
Add the two amounts
$15 + 18 = 33$
Together they make 33 slides. Watch out: do not add the slides and the minutes first. Mixing 5 + 3 slides with 20 + 10 minutes blends two different rates and leads to choice A, 16.
So the answer is C, 33.

Check your answer

Compare per minute. Deion makes 5 ÷ 20 = 0.25 slide a minute, and Kyra makes 3 ÷ 10 = 0.3. Together that is 0.55 slide a minute. Times 60 minutes gives 33. Choice D, 55, comes from multiplying by 100 instead of 60.

More questions like this

The original DOE question

Question 92 as printed on page 87 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 92. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 21 of 25

2018 Test A Math, question 38

Question

Ryan must read 150 pages for school this weekend. It took him 30 minutes to read the first 20 pages. At this rate, how much additional time will it take him to finish the reading?

A $2\frac{1}{6}\text{ hr}$
B $3\frac{1}{4}\text{ hr}$
C $3\frac{3}{4}\text{ hr}$
D $7\frac{1}{2}\text{ hr}$

Show the answer

Correct answer: B

Ryan has 130 pages left to read (150 − 20). He read 20 pages in 30 minutes, which means he read at a rate of 40 pages per 1 hour. To find out how much longer it will take him to finish the assignment, divide the total number of pages remaining (130) by the number of pages he is able to read per hour (40):

$\frac{130}{40}=3\frac{1}{4}$

Question analysis

This reading-rate word problem says Ryan read 20 of his 150 pages in 30 minutes and asks how much additional time he needs to finish at that rate. Students must find the pages left, use the rate, and convert minutes to hours. The word 'additional' is easy to overlook.

NYS standards: 6.RP.3b unit rate problems
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Unit rate problems like this are grade 6 content and common in grade 7. It adds a subtraction step and a conversion to hours, and the word 'additional' must be read closely, which keeps it at grade 7 level.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
A very close wrong answer moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The bold word "additional" carries the question, and missing it costs the point even when the rate work is perfect. The rate itself mixes pages, minutes and hours, so students who skip writing units drift into answers that count half-hour blocks or minutes as hours. Subtracting the pages already read before anything else keeps the work on the right quantity.

Many steps moderate Working it out
The solution chains four moves: subtract the 20 pages read, double 20 pages per half hour to 40 per hour, divide 130 by 40, and write the result in hours. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
The choices 3 1/4 and 3 3/4 hours differ by exactly the half hour already spent on the first 20 pages, so only subtracting those pages first separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
An answer in an unexpected form minor Choosing the answer
A student who works in minutes gets 195 minutes and must convert it to the mixed number 3 1/4 hours to find it among the choices. (Enright, Morley & Sheehan, 2002)

Strategy: Subtract the pages already read, work out how many pages he reads per hour, and divide the remaining pages by that hourly rate.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Divides the 130 remaining pages by 60, treating the rate as one page a minute, which gives 2 1/6 hours.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Times all 150 pages at 40 pages an hour, ignoring the bold "additional", and gets the whole assignment's 3 3/4 hours instead of the time left.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Divides 150 pages by 20 to get 7.5 blocks of reading and reports them as hours, though each 20-page block takes half an hour.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Time from a rate: $t = \frac{p}{r}$; t is the time needed. p is the number of pages to read. r is the rate: the pages read in one hour. Dividing tells you how many hours of reading the pages need.

Sort out what you know
Ryan must read 150 pages in all. He read the first 20 pages in 30 minutes. The word additional is in bold. It means extra time, after the 30 minutes he already spent.
So you need the time for the pages he has not read yet.
Find the pages left
$150 - 20 = 130$
He still has 130 pages to read. Watch out: timing all 150 pages gives $3\frac{3}{4}$ hours, which is choice C. That counts pages he already read.
Find his rate per hour
The choices are in hours, so find how many pages he reads in an hour. 30 minutes is half an hour. In a full hour he reads twice as much:
$20 \times 2 = 40$
His rate is 40 pages per hour.
Divide the pages by the rate
Dividing tells you how many hours the 130 pages take at 40 pages an hour:
$t = \frac{130}{40} = 3\frac{10}{40} = 3\frac{1}{4}$
It will take $3\frac{1}{4}$ more hours. So the answer is B, $3\frac{1}{4}$ hr.

Check your answer

Work in minutes. 130 pages is 6.5 groups of 20 pages. Each group takes 30 minutes, so 6.5 × 30 = 195 minutes. That is 3 hours and 15 minutes. And 15 minutes is a quarter of an hour, so it matches.

More questions like this

The original DOE question

Question 95 as printed on page 88 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 88, question 95. Open page 88 of the handbook (PDF)

Question 22 of 25

2018 Test A Math, question 41

Question

$Bar graph titled "Number of Schools in Five Cities". The vertical axis, Number of Schools, runs from 0 to 10 in steps of 1; the horizontal axis is City. The bars show M 8, N 5, P 9, Q 6 and R 4 schools.$

The graph above shows the number of schools per city for five small cities. Cities M and N each have 500 students per school. City P has 400 students per school. Cities Q and R each have 700 students per school. Which of the five cities has the greatest number of students?

ACity M
BCity P
CCity Q
DCity R

Show the answer

Correct answer: C

Take each city’s number of schools and multiply by the number of students. It is not necessary to calculate all 5 of these.

Cities M and N have the same number of students per school, so just calculate the number of students in City M because it has more schools than City N.

The same goes for Q and R — only Q needs to be calculated because it has more schools than R.

M= 8 × 500 = 4,000

P = 9 × 400 = 3,600

Q = 6 × 700 = 4,200

City Q has the greatest number of students.

Question analysis

This graph question shows the number of schools in each of five small cities and gives the number of students per school for each city. Students multiply schools by students per school to compare the total number of students. Comparing only the bars, or only the rates, can mislead.

NYS standards: 6.RP.3b unit rate problems
6.SP.5 summarizing data sets
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Reading a bar graph and multiplying is familiar from grade 6. Grade 7 data tasks often add percents or compare distributions; this one needs only a few whole-number products, so it is easier than typical grade 7 work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Much to keep track of minor
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 57 words.
Corrected from the DOE handbook: A misstatement in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Neither the tallest bar nor the highest rate wins, which is the point of the question: total students depend on both. The explanation's shortcut, comparing only M, P and Q because N and R each share a rate with a taller bar, cuts the work to three products. The top two totals are close, so the products must actually be computed rather than estimated.

A very close wrong answer moderate Choosing the answer
City M's 4,000 students and City Q's 4,200 differ by only 200, so judging from bar heights and rates cannot separate them; only computing 8 × 500 and 6 × 700 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Much to keep track of minor Working it out
The rates come in three groups in the stem, 500 for M and N, 400 for P, 700 for Q and R, and each must be matched to its bar. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Reading a figure, table or graph minor Understanding the question
The number of schools comes only from the bar heights, read against gridlines from 0 to 10; all five bars end exactly on a gridline. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: For each city, multiply the number of schools read from the graph by its students per school and compare the products; cities that share a rate need only their school counts compared.

Traps

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 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Computes M's 8 × 500 = 4,000 and P's 9 × 400 = 3,600 and stops, never working out Q's 6 × 700 = 4,200.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Picks the tallest bar, P's nine schools, which answers which city has the most schools rather than the most students.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks a city with the stem's highest rate, 700 students per school, without noticing that R has only 4 schools to Q's 6.

What was corrected from the DOE handbook, and why

Explanation: The handbook says “Cities M and N have the same number of students”. They have the same number of students per school (500), not the same total: M has 8 × 500 = 4,000 and N has 5 × 500 = 2,500, which is why only City M, with more schools, needs to be calculated. The sentence now says “the same number of students per school”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total students in a city: $S = n \times s$; S is the total number of students in a city. n is the number of schools, read from the bar. s is the number of students in each school. You multiply because every school in that city has the same number of students.

Read the graph
The bar graph shows how many schools each city has. Read the top of each bar against the numbers on the left. M has 8, N has 5, P has 9, Q has 6 and R has 4.
The question gives the students per school. You need the city with the most students in all.
Multiply for each city
Use the formula for each city:
City Schools Students per school Total students
M 8 500 4,000
N 5 500 2,500
P 9 400 3,600
Q 6 700 4,200
R 4 700 2,800
Compare the totals
The biggest total is 4,200, for City Q.
Watch out: City P has the tallest bar, but that means the most schools, not the most students. That leads to choice B. City M’s 4,000 is close, so do not stop before you work out Q. That is choice A.
So the answer is C, City Q.

City	Schools	Students per school	Total students
M	8	500	4,000
N	5	500	2,500
P	9	400	3,600
Q	6	700	4,200
R	4	700	2,800

Check your answer

A shortcut: M and N both have 500 per school, and M has more schools, so M beats N. Q and R both have 700, and Q beats R. So only M, P and Q matter: 4,000, 3,600 and 4,200. Q still wins.

More questions like this

The original DOE question

Question 98 as printed on page 89 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 89, question 98. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 23 of 25

2018 Test B Math, question 9

Question

A roofing contractor uses shingles at a rate of 3 bundles for every 96 square feet of roof covered. At this rate, how many bundles of shingles will he need in order to cover a roof that is 416 square feet?

Show the answer

Correct answer: C

Set up a proportion:

$\frac{x}{416}=\frac{3}{96}$

$96x=1{,}248$

$x=13\text{ bundles}$

Question analysis

This rate word problem says a roofing contractor uses 3 bundles of shingles for every 96 square feet of roof and asks how many bundles a 416-square-foot roof needs. Students set up a proportion or find a unit rate. Care is needed to answer in bundles rather than in 96-square-foot sections of roof.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Using a rate to scale up a quantity is grade 6 ratio reasoning. Here it takes two short steps with whole numbers, easier than the grade 7 proportion problems that involve fractions or percents.
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
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 38 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The rate is simple once it is turned into 32 square feet per bundle, and 416 ÷ 32 comes out exactly. Trouble starts for students who divide 416 by 96 first: 4.33 sections of roof invites rounding, and rounding before multiplying by 3 gives a neighbor of the right answer. Keeping the fraction exact, or using the unit rate, avoids the problem.

A very close wrong answer minor Choosing the answer
12 and 14 bundles sit on either side of the right 13, so a rough estimate of 416 ÷ 32 cannot pick among them; only the exact division, 32 × 13 = 416, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
The division 416 ÷ 96 does not come out even, about 4.33, so students who use it rather than 32 square feet per bundle must carry a repeating decimal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find how many square feet one bundle covers, then divide the roof's 416 square feet by that amount.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides 416 by 96 and rounds 4.33 up to 5, giving the number of 96-square-foot sections instead of the number of bundles.
B Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Rounds 416 ÷ 96 down to 4 sections before multiplying by 3 bundles, leaving the last 32 square feet uncovered.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Slips in dividing 1,248 by 96, or adds a spare bundle to be safe, landing one bundle above the exact count.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{area per bundle} = \frac{\text{area covered}}{\text{bundles used}}$; The area covered is in square feet. Bundles used is how many bundles covered that area. Dividing gives how much just one bundle covers. An amount for just one is called a unit rate.
Bundles needed: $\text{bundles} = \frac{\text{roof area}}{\text{area per bundle}}$; Roof area is the whole roof in square feet. Area per bundle is what one bundle covers. Dividing tells how many bundles it takes to cover the whole roof.

Write down what you know
3 bundles cover 96 square feet. A bundle is a pack of shingles. The new roof is 416 square feet.
You need the number of bundles for that roof, using the same rate.
Find what one bundle covers
Finding the amount for one bundle makes it easy to scale up to any roof.
$\text{area per bundle} = \frac{96}{3} = 32$
One bundle covers 32 square feet.
Divide the roof by 32
Now see how many 32-square-foot pieces fit in the roof.
$\text{bundles} = \frac{416}{32} = 13$
To check the division, $32 \times 13 = 320 + 96 = 416$ . So exactly 13 bundles cover the roof.
Watch out: 416 ÷ 96 is about 4.33. That counts groups of 96 square feet, not bundles. Rounding it to 5 gives choice A. Rounding down to 4 groups gives 12 bundles, choice B, and leaves part of the roof bare.
So the answer is C, 13.

Check your answer

Use the rate from the question. 13 bundles is 4 groups of 3 bundles plus 1 more bundle. That covers $4 \times 96 + 32 = 384 + 32 = 416$ square feet, the whole roof.

More questions like this

The original DOE question

Question 66 as printed on page 154 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 66. Open page 154 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 24 of 25

2018 Test B Math, question 15

Question

One day, the Early Bird Restaurant used 15 dozen eggs for 200 breakfast customers. At this rate, approximately how many dozen eggs are needed for 300 breakfast customers?

Show the answer

Correct answer: B

Let $x$ be the number of dozens of eggs for 300 customers. Set up a proportion:

$\frac{x}{300}=\frac{15}{200}$

$200x=4500$

$x=22.5\text{ dozen eggs}$

Round up to 23, because 22 dozen eggs would not be enough for 300 customers.

Question analysis

This proportion word problem says a restaurant used 15 dozen eggs for 200 breakfast customers and asks approximately how many dozen are needed for 300 customers. Students scale a rate up and round sensibly. Scaling by the right factor, rather than doubling or dividing the wrong pair of numbers, is the key step.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Scaling a rate from one quantity to another is grade 6 ratio reasoning. This needs one multiplication and a rounding decision, so it is lighter than the multistep proportion problems typical of grade 7.
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: An answer in an unexpected form minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 25 words.
Corrected from the DOE handbook: The explanation’s reason for rounding was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The proportion is simple, since 300 customers is one and a half times 200, and the only judgment is what to do with 22.5 dozen. The choices are spread wide enough that 23 is clearly nearest once 22.5 is found. Errors come mostly from students who skip the proportion and combine the numbers 300, 200 and 15 in some quicker way.

An answer in an unexpected form minor Choosing the answer
The scaled amount is 22.5 dozen, which is not a choice; the word “approximately” signals that the answer is the nearest whole number of dozens. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Scaling 15 dozen by 300/200 needs the factor 1.5, and 15 × 1.5 = 22.5 brings in a half that students must handle before matching a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Compare 300 customers with 200 to get the scale factor, multiply the 15 dozen eggs by it, and choose the closest answer.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Divides 300 customers by 15 dozen, combining two numbers from the question without a model of eggs per customer.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard route gives 25; at most a student who finds 22.5 rounds it to the nearest five instead of the nearest whole dozen.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Rounds the scale factor 300 ÷ 200 = 1.5 up to 2 before multiplying, doubling 15 dozen to 30.

What was corrected from the DOE handbook, and why

Explanation: The handbook says to round 22.5 up to 23 “because you can't have half an egg.” But 22.5 dozen is 270 eggs, a whole number; the half is half a dozen, or 6 eggs. The site gives the real reason: 22 dozen eggs would not be enough for 300 customers, so the amount rounds up to 23 dozen.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scaling a rate: $\text{new amount} = \text{old amount} \times \frac{\text{new count}}{\text{old count}}$; The old amount is the eggs used before, 15 dozen. The old count is the customers then, 200. The new count is the new number of customers, 300. The fraction tells how many times bigger the new group is. The eggs must grow by that same number of times.

Write down what you know
15 dozen eggs fed 200 customers. A dozen is 12, but you can keep everything in dozens.
You need the dozens of eggs for 300 customers, at the same rate.
Compare the customer counts
How many times bigger is 300 than 200? Divide to find out. This number is called the scale factor.
$\frac{300}{200} = 1.5$
There are 1.5 times as many customers. So you need 1.5 times as many eggs.
Scale up the eggs
$\text{new amount} = \text{old amount} \times \frac{\text{new count}}{\text{old count}}$
$15 \times 1.5 = 22.5$
You need 22.5 dozen eggs.
Watch out: Do not round 1.5 up to 2 first. That doubles 15 and gives 30, choice D.
Round to a choice
22.5 is not a choice. The question says “approximately,” so pick the closest whole number. 22.5 rounds up to 23.
Rounding up also makes sense, because 22 dozen would be too few.
So the answer is B, 23.

Check your answer

Use another route. 15 dozen for 200 customers is 7.5 dozen for every 100 customers. For 300 customers, that is $3 \times 7.5 = 22.5$ dozen. That also rounds to 23.

More questions like this

The original DOE question

Question 72 as printed on page 155 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 155, question 72. Open page 155 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 25 of 25

2018 Test B Math, question 40

Question

On a bike trip, Rajiv traveled 65 kilometers in 5 hours, while Shaina traveled 72 kilometers in 4 hours. How much less was Rajiv’s mean speed, in kilometers per hour (kph), than Shaina’s?

Show the answer

Correct answer: B

Calculate each mean speed:

$R=\frac{65}{5}=13\text{ kph}$

$S=\frac{72}{4}=18\text{ kph}$

Then calculate the difference of both mean speeds:

$S-R=18-13=5\text{ kph}$

Question analysis

Rajiv and Shaina ride different distances in different numbers of hours on a bike trip, and students find how much lower Rajiv's mean speed is, in kilometers per hour. It tests unit rates: speed as distance divided by time. The trap is comparing the raw distances or times instead of the speeds.

NYS standards: 6.RP.3b unit rate problems
6.RP.2 unit rate concept
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Unit rates and speed are grade 6 topics, and this needs only two divisions and a subtraction with whole-number results. It is easier than typical grade 7 rate problems, which often involve fractions.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Many steps minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 29 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Few students miss this one. The risk is subtracting raw numbers from the stem, the distances or the hours, instead of first turning each trip into a speed. Students who write each speed in kilometers per hour before comparing them avoid every wrong choice, and the arithmetic divides evenly.

Many steps minor Working it out
Two divisions, 65 ÷ 5 and 72 ÷ 4, have to be done before the subtraction, so no single operation on the given numbers answers the question. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
"How much less was Rajiv's mean speed" asks for a difference of two rates, while the listed distances and times invite subtracting them directly. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Divide each rider's distance by their time to get two speeds in kilometers per hour, then subtract the smaller speed from the larger.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts the travel times, 5 − 4 = 1 hour, comparing how long they rode rather than how fast.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Subtracts the distances, 72 − 65 = 7 kilometers, which says how much farther Shaina rode, not how much faster.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two travel times, 5 + 4 = 9, a number built from the stem that has nothing to do with either speed.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Speed: $\text{speed} = \text{distance} \div \text{time}$; Distance is how far someone went. Time is how long it took. Speed tells how far they go in one unit of time, like kilometers in one hour. Mean speed is the average speed for the whole trip.

Sort out what you know
Rajiv went 65 kilometers in 5 hours. Shaina went 72 kilometers in 4 hours. Mean speed means average speed. It is measured in kilometers per hour (kph), the distance in one hour. You need how much slower Rajiv was than Shaina.
Find Rajiv’s speed
He went 65 kilometers in 5 hours. Use the speed formula.
$\text{speed} = 65 \div 5 = 13$
Rajiv’s mean speed is 13 kph.
Find Shaina’s speed
She went 72 kilometers in 4 hours. Use the same formula.
$\text{speed} = 72 \div 4 = 18$
Shaina’s mean speed is 18 kph.
Subtract the two speeds
Find how much less 13 is than 18.
$18 - 13 = 5$
Rajiv was 5 kph slower.
Watch out: do not subtract the distances or the times. $72 - 65 = 7$ gives choice C, and $5 - 4 = 1$ gives choice A. Neither one is a speed.
So the answer is B, 5.

Check your answer

Work backward. At 13 kph for 5 hours, Rajiv goes $13 \times 5 = 65$ kilometers. At 18 kph for 4 hours, Shaina goes $18 \times 4 = 72$ kilometers. Both match the question, so the speeds are right.

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The original DOE question

Question 97 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 97. Open page 160 of the handbook (PDF)

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