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2026 Practice Test B Math

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Question 1 of 50

2026 Practice Test B Math, question 1

Question

The decimal 0.06 can be written as the fraction x50\frac{x}{50} . What is the value of xx ? 

  • A3
  • B6
  • C12
  • D30
Show the answer

Correct answer: A

0.06=61000.06=\frac{6}{100}

Simplify the fraction to find the  answer:

6100=350\frac{6}{100}=\frac{3}{50}

x=3x=3

Question analysis

This question asks students to rewrite the decimal 0.06 as an equivalent fraction with a denominator of 50 and give the missing numerator. It tests converting between decimals and fractions. The step that trips students up is moving from hundredths to fiftieths, which means scaling the numerator in the right direction.

NYS standards
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Converting a decimal to an equivalent fraction is upper elementary and grade 6 work. This is a single routine conversion with no context and no traps beyond a place-value slip, well below grade 7 demand.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

A one-idea question: write 0.06 as a fraction and rename it with a denominator of 50. The only real risk is keeping the 6 when the denominator changes, or scaling the numerator in the wrong direction. Checking that 3 ÷ 50 really gives 0.06 settles it.

  1. Error-prone arithmetic minor Working it out

    The decimal is in hundredths but the target denominator is 50, so the numerator has to be halved along with the denominator rather than kept as 6. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the decimal as a fraction over 100, then find the equivalent fraction over 50 by dividing numerator and denominator by the same number.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a÷kb÷k\frac{a}{b} = \frac{a \div k}{b \div k}
a is the top number and b is the bottom number. k is any number that divides both. Dividing the top and bottom by the same number keeps the value the same. Use it to give a fraction a new bottom number.

  1. Sort out the question

    You need to write 0.06 as a fraction with 50 on the bottom. The top number of that fraction is xx. You need to find xx.

  2. Write the decimal as a fraction

    0.06 has two digits after the decimal point. That means it counts hundredths. So 0.06 is six hundredths.

    0.06=61000.06 = \frac{6}{100}

  3. Change the bottom to 50

    To turn 100 into 50, divide by 2. The rule for equivalent fractions says to do the same to the top.

    6100=6÷2100÷2=350\frac{6}{100} = \frac{6 \div 2}{100 \div 2} = \frac{3}{50}

    So x=3x = 3.

    Watch out: the 6 must change too. Keeping the 6 gives choice B. Doubling it to 12 gives choice C.

  4. Give the answer

    The fraction 350\frac{3}{50} is the same amount as 0.06, and its top number is 3. So the answer is A, 3.

Check your answer

Divide to turn the fraction back into a decimal: 3÷50=0.063 \div 50 = 0.06. You can also multiply: 0.06×50=30.06 \times 50 = 3. Both match. Choice D, 30, comes from reading 0.06 as 0.6.

More questions like this

The original DOE question
Question 104 as printed on page 162 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 104. Open page 162 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 2 of 50

2026 Practice Test B Math, question 2

Question

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

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

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

Move the correct answer to each box.

There were students surveyed, and students like only soccer.

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

Show the answer

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

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

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

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

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

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

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

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

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

Question analysis

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

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Overlapping groups like these are seldom taught on their own in grade 7, though the arithmetic is simple whole-number addition and subtraction. Two results come from the same four facts, and both boxes must be right to score.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 6, 78 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Much to keep track of moderate Working it out

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

  2. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The survey gives you four facts:

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

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

  2. See who is counted twice

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

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

  3. Count who likes at least one sport

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

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

    So 37 students like basketball, soccer or both.

  4. Add the students who like neither

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

    37+10=4737 + 10 = 47

    So 47 students were surveyed.

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

  5. Find who likes only soccer

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

    19−7=1219 - 7 = 12

    So 12 students like only soccer.

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

  6. Fill in both boxes

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

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

Check your answer

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

More questions like this

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

Question 3 of 50

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.

  • A18150=x250\frac{18}{150} = \frac{x}{250}
  • B18250=x150\frac{18}{250} = \frac{x}{150}
  • C15018=250x\frac{150}{18} = \frac{250}{x}
  • Dx18=250150\frac{x}{18} = \frac{250}{150}
  • Ex150=18250\frac{x}{150} = \frac{18}{250}
Show the answer

Correct answer: A, C, and D

The three correct answers are:

18150=x250\frac{18}{150} = \frac{x}{250}

15018=250x\frac{150}{18} = \frac{250}{x}

x18=250150\frac{x}{18} = \frac{250}{150}

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

18 dozen buns150 customers=x dozen buns250 customers\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)150x = 18(250).

Then solve for x to determine how many dozen buns are needed for 250 lunch customers: 150x=4,500150x = 4{,}500

x=4,500150x = \frac{4{,}500}{150}

x=30x = 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
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.
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.

  1. 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)

  2. 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)

  3. 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)

  4. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A rate
rate=dozens of bunscustomers\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
ab=cd means a×d=b×c\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.

  1. 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.

  2. 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:

    18150=x250\frac{18}{150} = \frac{x}{250}

    This is choice A. Cross-multiply it:

    150×x=18×250150 \times x = 18 \times 250

    150x=4,500150x = 4{,}500

    Any choice that leads to 150x = 4,500 is a right one.

  3. Solve for x

    Divide both sides by 150 to get x by itself:

    x=4,500÷150=30x = 4{,}500 \div 150 = 30

    So the diner needs 30 dozen buns today. A wrong setup gives a different number.

  4. Test choices C and D

    Cross-multiply each one.

    • C: 15018=250x\frac{150}{18} = \frac{250}{x} gives 150×x=18×250150 \times x = 18 \times 250. That is the same equation, so C works. It is choice A with both fractions flipped.
    • D: x18=250150\frac{x}{18} = \frac{250}{150} gives x×150=18×250x \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.

  5. Test choices B and E

    • B: 18250=x150\frac{18}{250} = \frac{x}{150} gives 250×x=18×150250 \times x = 18 \times 150, so 250x = 2,700.
    • E: x150=18250\frac{x}{150} = \frac{18}{250} gives x×250=150×18x \times 250 = 150 \times 18, the same 250x = 2,700.

    Both give x=2,700÷250=10.8x = 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.

  6. 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 250150=53\frac{250}{150} = \frac{5}{3} times as many customers as yesterday. So it needs 5/3 times as many buns: 18×53=3018 \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 4 of 50

2026 Practice Test B Math, question 4

Question

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

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

1 mink=14 bart1\text{ mink} = \frac{1}{4}\text{ bart}

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

How many minks will Maria receive from the exchange?

Enter your answer in the space.

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

Show the answer

Correct answer: 116

Step 1: Convert the zogs to minks.
Since 1 mink=3 zogs1\text{ mink} = 3\text{ zogs}, divide the number of zogs Maria has by 3: 60÷3=2060 \div 3 = 20.

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

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

So, 24 barts=96 minks24\text{ barts} = 96\text{ minks}.

Step 3: Add the mink conversions.
Add the results from Steps 1 and 2 to find the total number of minks Maria will receive: 20+96=11620 + 96 = 116.

Question analysis

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

NYS standards
  • 6.RP.3d converting measurement units
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Converting units with a given rate is grade 6 ratio reasoning. Invented units remove any real-world sense check, and the two conversions run in opposite directions, one through a fraction, which matches the care grade 7 problems need.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 2, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    Both rates put the mink on the left, so “1 mink = 1/4 bart” has to be turned around into 1 bart = 4 minks, and the invented units give no sense of which is worth more. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Two conversions run in opposite directions, dividing the zogs by 3 and multiplying the barts by 4, and their results are added, so a slip in either step changes the total. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The total is typed into a box with no choices to compare, so a conversion done the wrong way still produces a believable whole number that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

    • 3 zogs trade for 1 mink.
    • 1 mink is worth 14\frac{1}{4} of a bart.
    • Maria has 60 zogs and 24 barts.

    You need the total number of minks. Zogs and barts are different units, like dimes and dollars. So you cannot just add them.

    Watch out: 60 + 24 = 84 adds zogs to barts. That is not a number of minks.

  2. Find what one zog is worth

    It takes 3 zogs to make 1 mink. So one zog is only part of a mink. One zog is worth 13\frac{1}{3} of a mink.

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

  3. Turn the bart rule around

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

    4×14=14 \times \frac{1}{4} = 1

    So 4 minks make 1 bart. One bart is worth 4 minks. A bart is bigger than a mink.

    Watch out: the fraction tempts you to divide by 4. Then 24 barts become only 6 minks, and the total is 26.

  4. Change the zogs into minks

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

    60×13=60÷3=2060 \times \frac{1}{3} = 60 \div 3 = 20

    Multiplying by 1/3 is the same as dividing by 3. So 60 zogs trade for 20 minks.

    Watch out: multiplying 60 by 3 gives 180. That treats a zog as bigger than a mink, and the total becomes 276.

  5. Change the barts into minks

    Each of the 24 barts is worth 4 minks:

    24×4=9624 \times 4 = 96

    So 24 barts trade for 96 minks.

  6. Add the minks

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

    20+96=11620 + 96 = 116

    Maria receives 116 minks. Type only the number in the box, with no units. So you would enter 116.

Check your answer

Work backward. 20 minks are worth 20×3=6020 \times 3 = 60 zogs. 96 minks are worth 96×14=2496 \times \frac{1}{4} = 24 barts. Both match what Maria started with. Also, a bart is worth more than a mink, so 24 barts should give many more than 24 minks. They do.

More questions like this

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

Question 5 of 50

2026 Practice Test B Math, question 5

Question

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

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

Show the answer

Correct answer: 120

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

The total number of possibilities is

5 × 4 × 3 × 2 × 1 = 120

Question analysis

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

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 probability work builds sample spaces with lists and tree diagrams for two or three stages. This needs the multiplication principle with a shrinking number of choices across five places, which listing cannot reach.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 11, 19 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know moderate Working it out

    Nothing in the stem names the counting principle: students must already know to multiply the digits left for each place, 5 × 4 × 3 × 2 × 1, because listing 5-digit numbers by hand cannot finish. (Anderson & Krathwohl, 2001)

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

    As a grid-in there are no choices to compare with, so a count of 3,125 or 25 cannot be caught by noticing that it is missing from a list. (Enright, Morley & Sheehan, 2002)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    You have five digits: 2, 3, 5, 7 and 8. A digit is one of the symbols 0 to 9. You build 5-digit numbers from them.

    “Without repeating” means each digit is used only once in a number. So 23578 is allowed, but 22578 is not.

    You need to find how many different numbers you can make.

  2. Count the choices for each spot

    Think of the number as five empty spots. Fill them from left to right.

    • First spot: any of the 5 digits, so 5 choices.
    • Second spot: one digit is used up, so 4 are left.
    • Third spot: 3 are left.
    • Fourth spot: 2 are left.
    • Last spot: only 1 is left.

  3. Multiply the choices together

    Use the counting principle. Each choice for one spot works with every choice for the next spot. So you multiply. You do not add.

    total ways=5×4×3×2×1=120\text{total ways} = 5 \times 4 \times 3 \times 2 \times 1 = 120

    There are 120 different numbers.

    Watch out: Adding 5 + 4 + 3 + 2 + 1 gives 15, which is far too small. Using 5 choices for every spot gives 3,125. That count lets digits repeat, which the question does not allow.

    So you would grid 120.

Check your answer

Try a smaller case you can list. With the digits 2, 3 and 5, the rule gives 3×2×1=63 \times 2 \times 1 = 6 numbers. The list is 235, 253, 325, 352, 523 and 532. That is 6, so the rule works.

More questions like this

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

Question 6 of 50

2026 Practice Test B Math, question 6

Question

Parallelogram PQRS with Q and R along the top and P and S along the bottom. Point T is on the bottom side PS, and segment QT cuts off triangle PQT. Angle PQT, at Q, is marked 50°; angle PTQ, at T, is marked 70°; angle QRS, at R, is marked with an arc and no measure.

In the figure above, PQRS is a parallelogram. The measure of ∠PQT is 50°, and the measure of ∠PTQ is 70°. What is the measure of ∠QRS?

  • A60°
  • B70°
  • C80°
  • D120°
Show the answer

Correct answer: A

Find the missing angle, angle QPT, of triangle PQT:

180° − 70° − 50° = 60°

In parallelogram PQRS, angle QPT is congruent to angle QRS, so the measure of angle QRS is also 60°.

Question analysis

This geometry question uses parallelogram PQRS and a triangle PQT inside it, giving two of the triangle's angles and asking for angle QRS. Students need the triangle angle sum and the fact that opposite angles of a parallelogram are equal. The challenge is connecting the angles found in the triangle to the angles of the parallelogram.

NYS standards
  • 8.G.5 angle sums, transversals, angle-angle similarity
  • 7.G.5 angle relationships and equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 angle work uses vertical, adjacent and supplementary angles. This also needs the triangle angle sum and the opposite angles of a parallelogram, ideas New York places in grade 8, though the arithmetic is short.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 4, 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

Few steps are needed, but the figure misleads. It draws triangle PQT as isosceles, so the angle at P looks the same as the marked 70°, and the angle at R looks the same again; a student who trusts the drawing never computes. Those who do compute the angle at P must still choose the opposite angle of the parallelogram, not the angle next to it, which is where 120° comes from.

  1. Reading a figure, table or graph major Understanding the question

    The figure carries facts the stem leaves out: T lies on side PS, the 50° is only part of the angle at Q, and the 70° is at T, not at a corner of the parallelogram. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    Two separate facts link up: the angles of triangle PQT sum to 180°, and the angle at P, once known, equals the opposite angle at R of parallelogram PQRS. (Embretson, 1998; Daniel & Embretson, 2010)

  3. A very close wrong answer moderate Choosing the answer

    60° and 70° are close enough that judging the angle at R by eye cannot separate them, and the drawing even favors 70°; only computing the triangle's third angle does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  4. Something you must already know minor Working it out

    Neither the triangle angle sum nor the rule that opposite angles of a parallelogram are equal is given; students must bring both to the figure. (Anderson & Krathwohl, 2001)

Strategy: Find the third angle of triangle PQT from the angle sum, work out which angle of the parallelogram it matches, then use the parallelogram's angle facts to reach angle QRS.

Traps

  • Trusting the picture: The figure suggests something it does not state; work from the given facts.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles in a triangle
a+b+c=180a + b + c = 180
a, b and c are the three angles inside any triangle. They always add up to 180 degrees. If you know two of the angles, subtract them from 180 to find the third.
Opposite angles of a parallelogram
∠P=∠R\angle P = \angle R
A parallelogram is a four-sided shape whose opposite sides are parallel, meaning they never meet. In parallelogram PQRS, the corners at P and R are opposite each other. Opposite angles of a parallelogram are always equal.

  1. Sort out the figure

    PQRS is a parallelogram. That is a four-sided shape whose opposite sides are parallel, meaning they never meet.

    Point T is on the bottom side PS. Segment QT cuts off triangle PQT on the left. In that triangle, angle PQT is 50° and angle PTQ is 70°.

    You need angle QRS, the corner at R.

  2. Find the third angle of the triangle

    Triangle PQT has angles at Q, T and P. You know two of them. The angle at P is angle QPT. All three must add to 180°.

    a+b+c=180a + b + c = 180

    ∠QPT+50+70=180\angle QPT + 50 + 70 = 180

    ∠QPT=180−50−70=60\angle QPT = 180 - 50 - 70 = 60

    So angle QPT is 60°.

  3. Match angle P to angle R

    T lies on side PS. So angle QPT is also the corner of the parallelogram at P. Angle P and angle R are opposite corners. Opposite angles are equal, so angle QRS is also 60°.

    Watch out: Angle Q is next to angle P, not opposite it. Angle Q is 120°, choice D, but the question asks about angle R. Also, do not trust the drawing. Angle R may look like 70°, choice B.

    So the answer is A, 60°.

Check your answer

The four corners of the parallelogram are 60°, 120°, 60° and 120°. They add to 360°, as the angles of any four-sided shape must. Also, the 50° angle at Q fits inside the 120° corner at Q.

More questions like this

The original DOE question
Question 64 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 64. 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 7 of 50

2026 Practice Test B Math, question 7

Question

M=3N=P4=Q+5=R7>0M=3N=\frac{P}{4}=Q+5=\frac{R}{7}> 0

Based on the statement above, which variable has the greatest value?

  • AMM
  • BNN
  • CPP
  • DRR
Show the answer

Correct answer: D

Break the equations apart to each equal MM:

M=3NM=3N

M=P4M=\frac{P}{4}

M=Q+5M=Q+5

M=R7M=\frac{R}{7}

Pick a number to substitute into the equations, and solve the equations to find the values of MM, NN, PP, QQ, and RR.

Let M=2M=2. Since all the equations are equal to 2, substitute 2 to find each variable.

M=3NM=3N

2=3N2=3N

23=N\frac{2}{3}=N

M=P4M=\frac{P}{4}

2=P42=\frac{P}{4}

8=P8=P

M=Q+5M=Q+5

2=Q+52=Q+5

−3=Q-3=Q

M=R7M=\frac{R}{7}

2=R72=\frac{R}{7}

14=R14=R

Variable RR has the greatest value.

Question analysis

This algebra question sets five expressions equal to one another and to a positive number, then asks which variable has the greatest value. It tests whether students can rewrite each variable in terms of one shared value and compare them. What makes it tricky is that there are no numbers to compute, so each variable's size has to be reasoned out.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students rewrite expressions to see how quantities relate, usually one equation at a time. This chain of five linked expressions, with no numbers to compute and a positivity condition to use, is a reasoning task beyond routine grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 10, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards a strategy, not a calculation: once the shared value is named, each variable follows in one line and R is clearly largest. Without that step, students judge by how the expressions look, and the look is backwards, since the 3 in 3N makes N small and the 7 under R makes R large. The positivity condition at the end is easy to skip but is what fixes the order.

  1. Variables instead of numbers major Understanding the question

    There are no numbers to compute: five variables are tied together in the one chain M = 3N = P/4 = Q + 5 = R/7, and each must be compared through the shared value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. No obvious method moderate Working it out

    Nothing says how to compare; the efficient move is to name the common value, or pick one such as 12, and rewrite each variable, which students must think of themselves. (Schoenfeld, 1985)

  3. Much to keep track of minor Working it out

    Five relations and the condition > 0 must be held at once; the condition matters, since with a negative common value the order of M, N, P and R would reverse. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Call the shared value k, a positive number, and rewrite each variable as an expression in k; or substitute a convenient positive value such as 12 and compare the results.

Traps

  • Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing division
ab=c means a=b×c\frac{a}{b} = c \text{ means } a = b \times c
a is the top of the fraction, b is the bottom, and c is the result. Multiplying undoes dividing. So if a divided by b is c, then a is b times c.
Undoing multiplication
b×a=c means a=c÷bb \times a = c \text{ means } a = c \div b
b is the number multiplied by a, and c is the result. Dividing undoes multiplying. So to get a alone, divide c by b.

  1. Understand the chain

    The statement is a chain of equal signs. It says five expressions all have the same value. The last part says that value is greater than 0, so it is positive.

    You need to find which letter has the largest value.

  2. Pick a value to test

    All five expressions are equal, so give them one shared value. Pick a positive number that 3 divides evenly. That way N will be a whole number.

    Let the shared value be 12. Then M=12M = 12, and each of the other four expressions also equals 12.

  3. Solve for each letter

    Undo each operation to find each letter. Divide to undo times. Multiply to undo divide. Subtract to undo plus.

    3N=12, so N=12÷3=43N = 12 \text{, so } N = 12 \div 3 = 4

    P4=12, so P=12×4=48\frac{P}{4} = 12 \text{, so } P = 12 \times 4 = 48

    Q+5=12, so Q=12−5=7Q + 5 = 12 \text{, so } Q = 12 - 5 = 7

    R7=12, so R=12×7=84\frac{R}{7} = 12 \text{, so } R = 12 \times 7 = 84

  4. Compare the values

    The values are M = 12, N = 4, P = 48, Q = 7 and R = 84. R is the largest.

    Watch out: 3N does not make N big. It means 3 times N equals M, so N is only one third of M. That trap points to choice B. Also, P is big, but R is bigger. Check every letter before you pick choice C.

    So the answer is D, R.

Check your answer

Think without numbers. R divided by 7 is M, so R is 7 times M. P is only 4 times M. N is a third of M, and Q is 5 less than M. Since M is positive, 7 times M is the biggest.

More questions like this

The original DOE question
Question 65 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 65. Open page 154 of the handbook (PDF)

Question 8 of 50

2026 Practice Test B Math, question 8

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?

  • A5
  • B12
  • C13
  • D14
Show the answer

Correct answer: C

Set up a proportion:

x416=396\frac{x}{416}=\frac{3}{96}

96x=1,24896x=1{,}248

x=13 bundlesx=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
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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
area per bundle=area coveredbundles used\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
bundles=roof areaarea per bundle\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.

  1. 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.

  2. Find what one bundle covers

    Finding the amount for one bundle makes it easy to scale up to any roof.

    area per bundle=963=32\text{area per bundle} = \frac{96}{3} = 32

    One bundle covers 32 square feet.

  3. Divide the roof by 32

    Now see how many 32-square-foot pieces fit in the roof.

    bundles=41632=13\text{bundles} = \frac{416}{32} = 13

    To check the division, 32×13=320+96=41632 \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 4×96+32=384+32=4164 \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 9 of 50

2026 Practice Test B Math, question 9

Question

To make party invitations, Macie could buy a package of paper for $10.50, or she could buy x individual sheets of the same paper for $0.15 each. What is the largest value of x that would make buying the individual sheets less expensive than buying the package?

  • A60
  • B65
  • C69
  • D70
Show the answer

Correct answer: C

Set up an inequality to compare the costs:

0.15x < 10.50
x < 70

Therefore, 70 individual sheets of paper would cost $10.50, so 69 is the greatest number of individual sheets of paper that Macie can buy that would be less expensive than the package.

Question analysis

This inequality word problem compares buying a package of paper with buying single sheets at 15 cents each, and asks for the largest number of sheets for which the single sheets cost less. Students write and solve an inequality with decimals. The tricky part is interpreting the solution as a whole number of sheets.

NYS standards
  • 7.EE.4b two-step inequalities
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an inequality from a real-world situation is a grade 7 standard. This one is short, but it asks students to interpret a strict inequality as a count of sheets, the kind of interpretation grade 7 tests expect.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 10, 45 words.
Corrected from the DOE handbook
  • An inequality sign in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The computation leads straight to 70, and the whole question is about what to do with that number. Students who treat the boundary as the answer forget that equal cost is not less cost, while the other choices only catch those who misdivide or ignore the word largest. The question rewards checking the boundary case in the story: 70 sheets cost exactly the package price.

  1. A very close wrong answer moderate Choosing the answer

    70 and 69 sheets differ by one, and 70 is exactly where the two costs meet at $10.50; only the strict “less expensive” separates them, ruling out 70. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Turning words into math minor Understanding the question

    The words “less expensive than buying the package” must become a strict inequality, 0.15x < 10.50, and “largest value” must then pick the biggest whole number below the boundary. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Dividing 10.50 by 0.15 means shifting the decimal point in both numbers, a step where a misplaced digit gives a wrong boundary. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Divide the package price by the price of one sheet to see where the two costs meet, then find the largest whole number of sheets that still costs less than $10.50.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook sets up 0.15x ≤ 10.50 and x ≤ 70, which would allow 70 sheets, and then says that 70 sheets cost exactly $10.50. Buying the sheets has to be less expensive than the package, so the site uses strict inequalities, 0.15x < 10.50 and x < 70, which make 69 the largest number of sheets.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost
cost=price of one×number bought\text{cost} = \text{price of one} \times \text{number bought}
Price of one is what a single sheet costs, here $0.15. Number bought is how many sheets, here x. Multiplying gives the total you pay, because you pay the same price for every sheet.

  1. Sort out the question

    The package costs $10.50. Single sheets cost $0.15 each. Macie buys x sheets.

    You need the largest x that makes the sheets cost less than the package. Costing the same is not enough.

  2. Write the inequality

    An inequality is a math sentence that compares two amounts. The sign < means is less than.

    The cost of x sheets is 0.15 times x. It must be less than 10.50.

    cost=price of one×number bought\text{cost} = \text{price of one} \times \text{number bought}

    0.15×x<10.500.15 \times x < 10.50

  3. Find where the costs are equal

    First find how many sheets cost exactly $10.50. Divide the package price by the price of one sheet. Multiply both numbers by 100 to clear the decimals.

    10.50÷0.15=1050÷15=7010.50 \div 0.15 = 1050 \div 15 = 70

    So 70 sheets cost exactly $10.50.

  4. Pick the largest number below 70

    x must be less than 70. You can only buy whole sheets. The largest whole number below 70 is 69.

    Watch out: 70 sheets cost the same as the package, not less. That is why choice D is wrong. Choices A and B are cheaper too, but they are not the largest.

    So the answer is C, 69.

Check your answer

Test both numbers. 69×0.15=10.3569 \times 0.15 = 10.35, and $10.35 is less than $10.50. But 70×0.15=10.5070 \times 0.15 = 10.50, which is not less. So 69 is the largest number that works.

More questions like this

The original DOE question
Question 67 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 67. Open page 154 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 10 of 50

2026 Practice Test B Math, question 10

Question

At 1:00 p.m. one day, the temperature was 8 degrees above zero. During the rest of the day, the temperature fell 3 degrees per hour. What was the temperature at 7:00 p.m. that day?

  • A−13°
  • B−10°
  • C−7°
  • D5°
Show the answer

Correct answer: B

7:00 p.m. is 6 hours after 1:00 p.m. Calculate the number of degrees the temperature dropped in 6 hours:

3 × 6 = 18  degrees

Subtract that from the starting point (8 degrees) to find the solution:

8 − 18 = −10 degrees

Question analysis

This integer word problem starts with a temperature 8 degrees above zero at 1:00 p.m. and has it fall 3 degrees every hour. Students find the temperature at 7:00 p.m. by counting hours and subtracting into negative numbers. Getting the number of hours between the two times right is the main hurdle.

NYS standards
  • 7.NS.3 four operations with rational numbers
  • 7.NS.1b rational number addition
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students use positive and negative numbers for real changes such as temperature. This is a standard three-step example of that skill, with a counting slip as its only real trap.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 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 arithmetic is light; the question turns on counting the hours from 1:00 to 7:00 correctly. Counting the clock times instead of the gaps between them gives seven hours, starting the drop an hour late gives five, and both errors are built into the choices. Subtracting the times, 7 − 1, is safer than counting on fingers.

  1. Error-prone arithmetic minor Working it out

    The total drop of 18 degrees is larger than the starting 8 degrees, so the subtraction crosses zero and the result must be negative. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the hours that pass from 1:00 to 7:00 by subtracting the times, multiply by the 3-degree drop, and subtract that total from 8.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total change
change=rate×time\text{change} = \text{rate} \times \text{time}
The rate is how much the temperature changes each hour, here 3 degrees down. The time is the number of hours. Multiply them to get the total change, because the same drop happens every hour.

  1. Write down what you know

    At 1:00 p.m. it was 8 degrees above zero. Each hour after that, it dropped 3 degrees.

    You need the temperature at 7:00 p.m.

  2. Count the hours

    Subtract the clock times: 7 − 1 = 6. So 6 hours pass.

    Watch out: Do not count the clock numbers 1, 2, 3, 4, 5, 6, 7. That is 7 numbers but only 6 hours. Using 7 hours gives −13°, choice A. Starting the count at 2:00 gives 5 hours and −7°, choice C.

  3. Find the total drop

    The temperature drops the same amount each hour, so multiply.

    change=rate×time=3×6=18\text{change} = \text{rate} \times \text{time} = 3 \times 6 = 18

    The temperature fell 18 degrees in all.

  4. Subtract from the start

    Start at 8 degrees and take away the 18-degree drop.

    8−18=−108 - 18 = -10

    The drop is bigger than 8, so the temperature goes below zero. A temperature below zero is a negative number. It is 10 degrees below zero.

    So the answer is B, −10°.

Check your answer

Count down hour by hour: 8 at 1:00, 5 at 2:00, 2 at 3:00, −1 at 4:00, −4 at 5:00, −7 at 6:00 and −10 at 7:00. You land on −10.

More questions like this

The original DOE question
Question 68 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 68. 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).

Question 11 of 50

2026 Practice Test B Math, question 11

Question

A bag contains 75 marbles that are red, blue, or green. The ratio of red to blue marbles is 15:715:7, and the ratio of blue to green marbles is 7:37:3. If 2 blue marbles are removed and replaced with 2 green marbles, what will be the new ratio of red to green marbles?

  • A3:13:1
  • B5:15:1
  • C15:315:3
  • D45:1145:11
Show the answer

Correct answer: D

The ratio of red to blue to green is 15 : 7 : 3.

Find the proportion of blue marbles. Add the numbers of the ratio and use the total sum as the denominator:

715+7+3=725\frac{7}{15+7+3}=\frac{7}{25}

Find the proportion of green marbles:

325\frac{3}{25}

Since there are a total of 75 marbles, the number of blue marbles is

725×75=21\frac{7}{25}\times 75=21

The number of green marbles is

325×75=9\frac{3}{25}\times 75=9

 The number of red marbles is

75−21−9=4575-21-9=45

If 2 blue marbles are removed and replaced with 2 green marbles, the number of blue marbles is now 19 and the number of green marbles is now 11. The ratio of red marbles to green marbles is 45 : 11.

Question analysis

This ratio problem describes a bag of 75 marbles with a red-to-blue ratio and a blue-to-green ratio that share the blue term, then swaps 2 blue marbles for 2 green ones. Students combine the ratios, find the actual number of each color, and form a new ratio. The swap at the end is what turns a routine ratio exercise into a long one.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 ratio problems usually involve one ratio and one scaling step. This links two ratios through a shared term, converts to counts and then changes the counts, a longer chain than typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
5 steps
Reading level
Question and choices: about grade 5, 54 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is easy because both ratios share the 7 for blue, so the work is about keeping the chain going without stopping early. The main danger is doing the swap on the ratio numbers or forgetting it altogether. Two of the wrong choices, 5:1 and 15:3, are the same ratio, so a student who notices that can rule out both before doing any arithmetic.

  1. Many steps moderate Working it out

    The chain is long: merge 15:7 and 7:3 into 15:7:3, find that 25 parts make 75 marbles, scale to 45, 21 and 9, then add 2 to the green count. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    Three colors' counts must be tracked through the swap, which lowers blue and raises green while red stays where it was. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Turning words into math minor Understanding the question

    The phrase “2 blue marbles are removed and replaced with 2 green marbles” must be read as changing the actual counts, not the ratio numbers 7 and 3. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Merge the two ratios into one red-to-blue-to-green ratio, scale it so the parts total 75, make the swap on those counts, then write red to green.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Value of one part
one part=totalsum of the ratio numbers\text{one part} = \frac{\text{total}}{\text{sum of the ratio numbers}}
A ratio like 15:7:3 splits a total into equal parts. The sum of the ratio numbers is how many parts there are. Dividing the total by that sum tells how many marbles are in one part.

  1. Write down what you know

    There are 75 marbles: red, blue and green. A ratio compares two amounts. Red to blue is 15:7. Blue to green is 7:3.

    Then 2 blue marbles are taken out, and 2 green ones are put in. You need the new ratio of red to green.

  2. Join the two ratios

    Blue is 7 in both ratios. So you can join them into one ratio.

    red:blue:green=15:7:3\text{red} : \text{blue} : \text{green} = 15 : 7 : 3

    This means for every 15 red marbles, there are 7 blue and 3 green.

  3. Find the size of one part

    The ratio has 15+7+3=2515 + 7 + 3 = 25 parts in all. The 75 marbles fill those 25 parts.

    one part=totalsum of the ratio numbers=7525=3\text{one part} = \frac{\text{total}}{\text{sum of the ratio numbers}} = \frac{75}{25} = 3

    Each part is 3 marbles.

  4. Count each color

    Multiply each ratio number by 3 to get the real counts.

    15×3=457×3=213×3=915 \times 3 = 45 \qquad 7 \times 3 = 21 \qquad 3 \times 3 = 9

    So there are 45 red, 21 blue and 9 green. They add to 75, so the counts are right.

  5. Make the swap

    Take out 2 blue: 21−2=1921 - 2 = 19 blue. Put in 2 green: 9+2=119 + 2 = 11 green. Red does not change. It is still 45.

    The new ratio of red to green is 45:11. It cannot be simplified, because 45 and 11 share no factor except 1.

    Watch out: 45:9 is the ratio before the swap. It simplifies to 5:1, choice B. Choice C, 15:3, is that same old ratio. Make the swap with the real counts.

    So the answer is D, 45:11.

Check your answer

Count the marbles after the swap: 45+19+11=7545 + 19 + 11 = 75. The total is still 75, as it should be. A swap moves marbles in and out but does not change the total.

More questions like this

The original DOE question
Question 69 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 69. Open page 155 of the handbook (PDF)

Question 12 of 50

2026 Practice Test B Math, question 12

Question

DESSERT CHOICES

Dessert Number of Times Ordered
Cookies 42
Pie 23
Cake 47
Ice Cream 48

The table above shows the number of times that different desserts were ordered at a restaurant. Based on this information, what is the probability of a customer ordering ice cream as a dessert?

  • A25%
  • B30%
  • C40%
  • D48%
Show the answer

Correct answer: B

The total number of desserts ordered is

42 + 23 + 47 + 48 = 160

The probability that ice cream was chosen is

48160=310=30%\frac{48}{160}=\frac{3}{10}=30\%.

Question analysis

This probability question gives a table of how many times four desserts were ordered at a restaurant and asks how likely a customer is to order ice cream, as a percent. Students total the orders and divide. The trap is using a raw count, or the wrong total, in place of the full number of desserts ordered.

NYS standards
  • 6.SP.8b probability models from observed frequencies
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a probability from observed frequencies is grade 6 content in New York. Here it takes one sum, one division and a conversion to a percent with friendly numbers, lighter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
3 steps
Reading level
Question and choices: about grade 5, 45 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a direct relative-frequency question with friendly numbers, and most students finish it quickly. The risk is using a number from the table as if it were already a percent, or treating the four desserts as equally likely. Remembering that a probability compares one count with the total of all counts prevents both mistakes.

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

    The four counts sit in a table, and the probability needs all of them: the ice cream row gives the numerator and the sum of the column gives the denominator. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Error-prone arithmetic minor Working it out

    Adding 42 + 23 + 47 + 48 by hand and then simplifying 48/160 to 3/10 are small steps where a careless total spoils the percent. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add the four counts in the table to get all the orders, divide the ice cream count by that total, and convert the fraction to a percent.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=outcomes you wantall outcomesP = \frac{\text{outcomes you want}}{\text{all outcomes}}
P is the probability, the chance that something happens. The top counts the outcomes you want, here ice cream orders. The bottom counts every outcome, here all dessert orders. It works because each order counts the same.
Fraction to percent
percent=fraction×100\text{percent} = \text{fraction} \times 100
Percent means out of 100. Multiply a fraction or a decimal by 100 to write it as a percent. For example, 3/10 times 100 is 30, so 3/10 is 30%.

  1. Sort out the table

    The table shows how many times each dessert was ordered. Ice cream was ordered 48 times.

    You need the chance that a customer orders ice cream. The orders in the table show how often each dessert is picked.

  2. Find the total orders

    Add all four rows of the table.

    42+23+47+48=16042 + 23 + 47 + 48 = 160

    There were 160 dessert orders in all. This is the bottom of the fraction, because it counts every order.

  3. Write the probability

    Put the counts into the probability formula. The top is the 48 ice cream orders. The bottom is all 160 orders.

    P=outcomes you wantall outcomes=48160P = \frac{\text{outcomes you want}}{\text{all outcomes}} = \frac{48}{160}

    Simplify the fraction. Both numbers can be divided by 16.

    48÷16160÷16=310\frac{48 \div 16}{160 \div 16} = \frac{3}{10}

  4. Change to a percent

    310×100=30\frac{3}{10} \times 100 = 30

    The probability is 30%.

    Watch out: 48 orders is not 48%, choice D. You must divide by the total. Also, the four desserts are not equally popular. So the chance is not 1 out of 4, or 25%, choice A.

    So the answer is B, 30%.

Check your answer

Work backward: 0.30×160=480.30 \times 160 = 48. That matches the ice cream row. Also, 48 is a bit less than a third of 160, and 30% is a bit less than a third.

More questions like this

The original DOE question
Question 70 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 70. 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).

Question 13 of 50

2026 Practice Test B Math, question 13

Question

A basket contains 3 apples, 4 oranges, 5 bananas, and 8 pears.

If one piece of fruit is randomly pulled from the basket, what is the probability that it will be a pear?

Select the place on the number line to plot the point.

On the test you click a point on the number line; here, choose it from the menu.

00.10.20.30.40.50.60.70.80.91

Show the answer

Correct answer: the point 0.4

The total number of fruits is:

3(apples)+4(oranges)+5(bananas)+8(pears)=20 fruits3(\text{apples}) + 4(\text{oranges}) + 5(\text{bananas}) + 8(\text{pears}) = 20\text{ fruits}

The probability of picking a pear is:

8 (pears)20 (total)=820 or 0.4\frac{8\ (\text{pears})}{20\ (\text{total})} = \frac{8}{20} \text{ or } 0.4

Question analysis

A basket holds apples, oranges, bananas and pears, and students find the probability that a piece of fruit pulled at random is a pear, then plot it as a point on a number line from 0 to 1. It tests simple probability as a part of a whole. With no choices to pick from, the fraction has to be turned into tenths to find its place.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the probability of a simple event from counts is grade 6 work in New York. This one adds four whole numbers, forms one fraction and plots it among marked tenths, a shorter and more routine task than grade 7 probability.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
3 steps
Reading level
Question and choices: about grade 6, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Adding the four counts to 20 and writing 8 over 20 is routine, so most students succeed. The losses come from the number line: its decimal labels make it easy to read the 8 pears as 0.8, and a part-to-part fraction such as 8/12 falls between marks and has to be rounded, which should prompt a second look. Simplifying 8/20 to 2/5 and then to 4/10 makes the mark plain.

  1. Something you must already know minor Working it out

    The question does not say that a probability is the favorable count over the whole count, here the 8 pears over all 20 pieces of fruit rather than over the other 12. (Anderson & Krathwohl, 2001)

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

    The number line is labeled only in tenths from 0 to 1, so 8/20 has to be rewritten as a decimal before the right mark can be chosen. (Enright, Morley & Sheehan, 2002)

Strategy: Add up all the fruit for the total, write the number of pears over that total, rewrite the fraction in tenths, and select that mark on the line.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event
P=number of results you wanttotal number of resultsP = \frac{\text{number of results you want}}{\text{total number of results}}
P is the probability, the chance that something happens. It goes from 0 (never) to 1 (always). The top counts the results you want, here pears. The bottom counts every possible result, here every piece of fruit. It works when each result is equally likely, like grabbing fruit without looking.
Equal fractions
ab=a÷nb÷n\frac{a}{b} = \frac{a \div n}{b \div n}
a is the top of a fraction, and b is the bottom. n is any number except 0. Dividing the top and the bottom by the same number gives an equal fraction. Use it to get 10 on the bottom, which is easy to write as a decimal.

  1. Sort out the question

    The basket has 3 apples, 4 oranges, 5 bananas and 8 pears. You pull out one piece without looking. You need the probability that it is a pear. Then you plot that number as a point. The number line runs from 0 to 1, with a mark at every tenth.

  2. Count all the fruit

    The bottom of the fraction is every piece of fruit you could pull out. Add all four kinds:

    3+4+5+8=203 + 4 + 5 + 8 = 20

    There are 20 pieces of fruit in all.

  3. Write the probability as a fraction

    Use the probability formula. There are 8 pears out of 20 pieces of fruit:

    P=820P = \frac{8}{20}

    Watch out: the bottom is all the fruit, not just the other fruit. 812\frac{8}{12} compares the pears with the 12 fruits that are not pears. It is about 0.67, and it leads you to the mark at 0.7. And 12 over 20 is the chance of not getting a pear.

  4. Change it to tenths

    The number line is marked in tenths. So rewrite the fraction with 10 on the bottom. Divide the top and the bottom by 2:

    820=8÷220÷2=410=0.4\frac{8}{20} = \frac{8 \div 2}{20 \div 2} = \frac{4}{10} = 0.4

    Four tenths is written 0.4.

    Watch out: the 8 pears do not make 0.8. That skips dividing by the 20 pieces of fruit.

  5. Plot the point

    Start at 0 and count four marks to the right: 0.1, 0.2, 0.3, 0.4. On the test, you click the mark labeled 0.4. On this site, you choose 0.4 from the menu.

    So the answer is the point at 0.4.

Check your answer

The chance of not getting a pear is 1220=0.6\frac{12}{20} = 0.6. A pear and not a pear cover every result, so the two must add to 1: 0.4+0.6=10.4 + 0.6 = 1. Also, pears are less than half of the fruit, so the point should be left of 0.5. It is.

More questions like this

The original DOE question
Question 13 as shown on the DOE’s digital test, page 85 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B Math, page 85, question 13.

Question 14 of 50

2026 Practice Test B Math, question 14

Question

A cooler contains three types of beverages: 5 bottles of apple juice, 3 bottles of grape juice, and 6 bottles of orange juice. What is the probability that a bottle chosen at random from this cooler is not apple juice?

  • A19\frac{1}{9}
  • B514\frac{5}{14}
  • C914\frac{9}{14}
  • D23\frac{2}{3}
Show the answer

Correct answer: C

The total number of bottles of juice in the cooler is

5 + 3 + 6 = 14

The number of bottles of juice that are not apple juice (grape juice and orange juice) is

3 + 6 = 9

So the probability is 914\frac{9}{14} .

Question analysis

This probability question describes a cooler holding apple, grape and orange juice bottles in different numbers and asks for the chance that a randomly chosen bottle is not apple juice. Students count favorable bottles over all bottles. The word not is the trap, along with counting kinds of drink instead of bottles.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Simple probability with equally likely outcomes is grade 6 content in New York. This needs one count and one fraction, and the only extra demand is reading the word not, so it sits below grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 41 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The counting is easy, 14 bottles in all, so the question mainly tests careful reading. The word “not” turns the event around, and the probability of the opposite event is waiting among the choices. The other wrong choices catch students who count kinds of juice instead of bottles, or who build a fraction with the wrong total.

  1. Turning words into math minor Understanding the question

    The event “not apple juice” must be turned into grape plus orange, 3 + 6 bottles over all 14, so the word “not” decides which counts go in the numerator. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add all the bottles for the denominator, then count only the grape and orange bottles for the numerator, or subtract the chance of apple juice from 1.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=outcomes you wantall outcomesP = \frac{\text{outcomes you want}}{\text{all outcomes}}
P is the probability, the chance that something happens. The top counts the outcomes you want, here bottles that are not apple. The bottom counts all outcomes, here every bottle. It works when every bottle has the same chance.
Probability of “not”
P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)
A is an event, such as picking apple juice. P(A) is its chance. Every bottle either is apple or is not, so the two chances add to 1. Subtract from 1 to get the chance that A does not happen.

  1. Sort out the question

    The cooler has 5 apple, 3 grape and 6 orange bottles. One bottle is picked at random. That means every bottle has the same chance.

    You need the chance that it is not apple juice.

  2. Count all the bottles

    5+3+6=145 + 3 + 6 = 14

    There are 14 bottles. This is the bottom of the probability fraction, since any of them could be picked.

  3. Count the bottles that are not apple

    The bottles that are not apple are the grape and orange ones.

    3+6=93 + 6 = 9

    There are 9 such bottles. This is the top of the fraction.

  4. Write the probability

    P=outcomes you wantall outcomes=914P = \frac{\text{outcomes you want}}{\text{all outcomes}} = \frac{9}{14}

    This fraction cannot be simplified. 9 and 14 share no factor except 1.

    Watch out: 514\frac{5}{14} is the chance of picking apple juice. That is choice B. Read the word “not” carefully.

    So the answer is C, 914\frac{9}{14}.

Check your answer

Use the “not” rule. The chance of apple is 514\frac{5}{14}. So the chance of not apple is 1−514=9141 - \frac{5}{14} = \frac{9}{14}. Both ways give the same answer.

More questions like this

The original DOE question
Question 73 as printed on page 156 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 73. Open page 156 of the handbook (PDF)

Question 15 of 50

2026 Practice Test B Math, question 15

Question

A large circular dinner plate has a radius of 20 centimeters. A smaller circular plate with a circumference of 20π20\pi centimeters is placed in the center of the larger dinner plate. What is the area of the part of the larger dinner plate that is not covered by the smaller plate?

  • A20π20\pi sq cm
  • B100π100\pi sq cm
  • C200π200\pi sq cm
  • D300π300\pi sq cm
Show the answer

Correct answer: D

The radius of the large plate is 20 cm. Use that to find the area of the large plate:

A=πr2=π(202)=400π sq cmA=\pi r^2=\pi(20^2)=400\pi\text{ sq cm}

The circumference of the smaller plate is 20π cm. Use that to find the radius, and then the area, of the smaller plate:

C=2πrC=2\pi r

20π=2πr20\pi=2\pi r

r=10r=10

A=πr2=π(102)=100π sq cmA=\pi r^2=\pi(10^2)=100\pi\text{ sq cm}

Subtract the area of the small plate from the area of the large plate:

400π − 100π = 300π sq cm

Question analysis

This circle question places a smaller plate, described by its circumference, on a larger dinner plate described by its radius, and asks for the area of the large plate left uncovered. Students recall the circumference and area formulas, find the small radius and subtract areas. Because the two plates are described in different ways, circumference and area have to be kept apart.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Applying circle area and circumference formulas is the grade 7 standard, and subtracting one area from another is a common grade 7 exercise. Working backward from a circumference adds one step but no new content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 6, 62 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question describes the two plates in different measures, one by radius and one by circumference, so students must convert before they can compare. The trap is a halving argument: the small circumference is half the large one, but the area it covers is only a quarter. Students who keep both areas in terms of π and subtract finish quickly; those who reason from the circumferences alone go wrong.

  1. Many steps moderate Working it out

    The chain runs through four stages: the large area 400π, the small radius 10 from the circumference 20π, the small area 100π, then the uncovered difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Something you must already know moderate Working it out

    Neither formula is given: students need C = 2πr to recover the small plate's radius and A = πr² for both areas. (Anderson & Krathwohl, 2001)

Strategy: Use C = 2πr to get the small plate's radius, find both areas with A = πr², keep everything in terms of π, and subtract.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle
C=2×π×rC = 2 \times \pi \times r
C is the circumference, the distance around the circle. r is the radius, the distance from the center to the edge. π, said “pie,” is a number a little more than 3. The distance around is always about 6.28 times the radius.
Area of a circle
A=π×r2A = \pi \times r^2
A is the area, the space inside the circle. r is the radius. r² means r times r. π is about 3.14. The radius is squared because area spreads out in two directions.

  1. Write down what you know

    The big plate has a radius of 20 cm. The radius is the distance from the center to the edge.

    The small plate has a circumference of 20π cm. Circumference is the distance around a circle.

    You need the area of the big plate that the small plate does not cover. Area is the space inside a shape.

  2. Find the big plate's area

    A=π×r2A = \pi \times r^2

    A=π×202=π×400=400πA = \pi \times 20^2 = \pi \times 400 = 400\pi

    The big plate is 400π sq cm. Keep π as a symbol, because the choices use it.

  3. Find the small plate's radius

    You know the small plate’s circumference, not its radius. Use the circumference formula to work back.

    C=2×π×rC = 2 \times \pi \times r

    20π=2×π×r20\pi = 2 \times \pi \times r

    Divide both sides by 2π. That leaves r=10r = 10. The small radius is 10 cm.

  4. Find the small plate's area

    Use the area formula again, with the small radius.

    A=π×102=100πA = \pi \times 10^2 = 100\pi

    The small plate covers 100π sq cm.

    Watch out: 100π is the covered part. The question asks for the part not covered, so do not stop at choice B.

  5. Subtract to find the uncovered part

    400π−100π=300π400\pi - 100\pi = 300\pi

    The uncovered area is 300π sq cm.

    Watch out: Do not subtract the distances around the plates. That gives 40π − 20π = 20π, choice A.

    So the answer is D, 300π sq cm.

Check your answer

The small radius is half the big radius. So the small area is 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} of the big area. That leaves three quarters uncovered: 34×400π=300π\frac{3}{4} \times 400\pi = 300\pi. It matches.

More questions like this

The original DOE question
Question 74 as printed on page 156 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 74. Open page 156 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 16 of 50

2026 Practice Test B Math, question 16

Question

Which tables represent a proportional relationship between the number of books purchased and the total cost?

Select the two 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
    Number of Books Purchased10152025
    Total Cost ($)406080100
  • B
    Number of Books Purchased10152025
    Total Cost ($)70110150250
  • C
    Number of Books Purchased10152025
    Total Cost ($)80120160200
  • D
    Number of Books Purchased10152025
    Total Cost ($)90130170210
  • E
    Number of Books Purchased10152025
    Total Cost ($)100100100100
Show the answer

Correct answer: A and C

The two tables that represent a proportional relationship are tables A and C.

In table A, the total costs divided by the numbers of books for each set of quantities are:

4010=4; 6015=4; 8020=4; 10025=4\frac{40}{10} = 4;\ \frac{60}{15} = 4;\ \frac{80}{20} = 4;\ \frac{100}{25} = 4

These ratios are equivalent; therefore, these quantities are in a proportional relationship.

In table C, the total costs divided by the numbers of books for each set of quantities are:

8010=8; 12015=8; 16020=8; 20025=8\frac{80}{10} = 8;\ \frac{120}{15} = 8;\ \frac{160}{20} = 8;\ \frac{200}{25} = 8

These ratios are equivalent; therefore, these quantities are in a proportional relationship.

Question analysis

Five tables list the total cost of buying 10, 15, 20 and 25 books, and students select the two tables that show a proportional relationship between books and cost. It tests recognizing proportional relationships from a constant unit rate. A steady rise in cost is not enough on its own, so each pair of values has to be checked.

NYS standards
  • 7.RP.2a testing for proportionality
  • 7.RP.2b constant of proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Deciding from a table whether two quantities are proportional is a core grade 7 skill, and the divisions here are easy. The demand lies in checking every column of five tables rather than judging by how the costs change.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 51 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The divisions are easy, so the question tests what proportional means. Three tables grow by the same amount for every 5 books, and the one with a $10 head start looks just as regular as the right one it copies; only its cost per book, drifting from $9 down to $8.40, gives it away. The flat $100 table and the one that jumps to $250 catch fewer students, and dividing the first and last column of each table is a quick, reliable check.

  1. A very close wrong answer moderate Choosing the answer

    The table running $90, $130, $170, $210 is the $80-to-$200 table with $10 added to every cost; both climb $40 for each 5 books, and only the cost per book separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Every choice must be checked minor Choosing the answer

    Five tables of four columns each have to be tested by dividing cost by books, and both right tables must be picked to score, so no table can be judged from one column. (Embretson & Wetzel, 1987)

  3. Something you must already know minor Working it out

    Students must know that proportional means the same cost per book in every column, a line through zero, not merely a cost that grows by the same amount each time. (Anderson & Krathwohl, 2001)

Strategy: Divide the total cost by the number of books in every column of each table; a table is proportional only when that cost per book stays the same all the way across.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost per book
cost per book=total cost÷number of books\text{cost per book} = \text{total cost} \div \text{number of books}
This is the unit rate, the price of one book. Divide the total cost in dollars by the number of books bought. If every book has the same price, this number comes out the same in every column.
A proportional relationship
y=kxy = kx
x is the number of books, and y is the total cost in dollars. k is the cost per book, a number that never changes. In a proportional relationship, y is always k times x. So 0 books cost $0, and twice as many books cost twice as much.

  1. Sort out the question

    Each table shows the total cost of 10, 15, 20 and 25 books. Proportional means the cost per book is the same in every column. In other words, each book has one fixed price. You must pick the two tables where that is true.

  2. Test the first table

    Divide each total cost by its number of books. For table A:

    40÷10=460÷15=480÷20=4100÷25=440 \div 10 = 4 \qquad 60 \div 15 = 4 \qquad 80 \div 20 = 4 \qquad 100 \div 25 = 4

    Every column gives $4 per book. So table A is proportional.

  3. Test the other four tables

    Do the same division for every table. The numbers are dollars per book:

    Table10 books15 books20 books25 booksSame?
    A4444Yes
    B7about 7.337.5010No
    C8888Yes
    D9about 8.678.508.40No
    E10about 6.6754No

    Only tables A and C keep the same price all the way across.

  4. Watch for steady growth

    Watch out: table D climbs $40 for every 5 more books, just like table C. That steady climb looks proportional, but it is not enough. D starts $10 higher than C, so its price per book slides from $9 down to $8.40.

    Watch out: table E never changes. But $100 for 10 books is $10 each, and $100 for 25 books is $4 each. A total that stays the same is not a price that stays the same.

  5. Name the two answers

    Table A charges $4 per book and table C charges $8 per book, in every column. On the test, you check the box next to each of those tables. So the answers are A and C.

Check your answer

In a proportional table, twice the books cost twice as much. In table A, 10 books cost $40 and 20 books cost $80, which is double. In table C, $80 doubles to $160. In table D, $90 would have to double to $180, but the table shows $170. So D fails.

More questions like this

The original DOE question
Question 16 as shown on the DOE’s digital test, page 88 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B Math, page 88, question 16.

Question 17 of 50

2026 Practice Test B Math, question 17

Question

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

Enter your answer in the space.

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

Show the answer

Correct answer: 15

Solve the inequality for x.

x+7<23x + 7 < 23

x<16x < 16

There are 15 positive integers that are less than 16.

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 inequality problems usually take two steps, such as 2x − 5 > 9. This one needs a single subtraction; the care goes into counting the whole-number solutions and deciding whether the boundary itself counts.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 12, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

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

    Solving gives x < 16, but the box asks how many positive integers make the inequality true, a count that still has to be made after the solving is done. (Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    Two conditions set the ends of the count at once: the word positive starts it at 1 rather than 0, and the strict < stops it below 16. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    An inequality compares two amounts. The sign < means is less than. A positive integer is a whole number bigger than 0: 1, 2, 3, and so on. You need to count how many of them make x+7<23x + 7 < 23 true. The answer is a count, not a value of x.

  2. Get x by itself

    The 7 is added to x. Subtracting 7 undoes adding 7, so subtract 7 from both sides:

    x+7−7<23−7x + 7 - 7 < 23 - 7

    x<16x < 16

    So x must be less than 16.

    Watch out: adding 7 to 23 instead gives x < 30, and a count of 29.

  3. Check the boundary number

    The number 16 is the boundary, where the solutions stop. Does 16 itself work? Try it:

    16+7=2316 + 7 = 23

    But 23 is not less than 23. It is equal. So 16 does not count. The largest number that works is 15, since 15+7=2215 + 7 = 22, and 22 is less than 23.

    Watch out: entering 16 counts the boundary, which fails.

  4. Count the positive integers

    The numbers that work start at 1, the smallest positive integer. Zero is not positive. They stop at 15:

    1,2,3,…,14,151, 2, 3, \ldots, 14, 15

    Counting from 1 up to 15 gives 15 numbers.

  5. Enter the count

    There are 15 positive integers that make the inequality true. Type the count in the box as a plain whole number. So you would enter 15.

Check your answer

Test both ends. For 1: 1+7=81 + 7 = 8, and 8 is less than 23, so 1 works. For 15: 22 is less than 23, so 15 works. For 16: 23 is not less than 23, so 16 fails. That leaves exactly the numbers 1 through 15.

More questions like this

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

Question 18 of 50

2026 Practice Test B Math, question 18

Question

If 36y=4x\frac{36}{y}=4x, what is the value of xx when y=3y=3?

  • A3
  • B4
  • C9
  • D12
Show the answer

Correct answer: A

Substitute 3 for yy and solve for xx:

36y=4x\frac{36}{y}=4x

363=4x\frac{36}{3}=4x

12=4x12=4x

3=x3=x

Question analysis

This equation question gives 36 divided by y equal to 4x and asks for the value of x at one given value of y. Students substitute, simplify the fraction and solve a one-step equation. It is short, so the risk lies in stopping one step early or dividing by the wrong number.

NYS standards
  • 6.EE.7 one-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Substituting a value and solving a one-step equation are grade 6 skills. With two quick steps and whole numbers, this is lighter than the multistep equations typical of grade 7.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 8, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is substitution followed by a one-step equation, with small whole numbers throughout, so most students answer correctly. The only real risk is stopping at 12, which is the value of 4x rather than of x. Writing 12 = 4x as its own line before reading the choices makes the last division hard to skip.

  1. Variables instead of numbers minor Understanding the question

    The variable y sits in a denominator, 36/y, which looks less familiar than an equation in one unknown, though substituting y = 3 removes it at once. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Replace y with its value, simplify the left side to a single number, and then divide by the coefficient of x.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing multiplication
a×x=b means x=b÷aa \times x = b \text{ means } x = b \div a
a is the number multiplied by x, and b is the result. Division undoes multiplication. So to get x alone, divide both sides by a.

  1. Write down what you know

    The equation is 36y=4x\frac{36}{y} = 4x. You are told y = 3. You need the value of x.

    The letter y is on the bottom of a fraction. Putting in its value turns the fraction into a plain number.

  2. Put in the value of y

    Replace y with 3.

    363=4x\frac{36}{3} = 4x

    The fraction means 36 divided by 3, which is 12.

    12=4x12 = 4x

    Watch out: 12 is the value of 4x, not x. Stopping here gives choice D.

  3. Divide both sides by 4

    4x means 4 times x. To undo multiplying by 4, divide both sides by 4.

    x=12÷4=3x = 12 \div 4 = 3

    So the answer is A, 3.

Check your answer

Put x = 3 and y = 3 back in. The left side is 36÷3=1236 \div 3 = 12. The right side is 4×3=124 \times 3 = 12. Both sides match, so x = 3 is right.

More questions like this

The original DOE question
Question 77 as printed on page 157 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 77. Open page 157 of the handbook (PDF)

Question 19 of 50

2026 Practice Test B Math, question 19

Question

 Points XX , YY, and ZZ are on a straight line, and YY is between XX and ZZ . Length YZ=35XY\overline{YZ}=\frac{3}{5}\overline{XY}, and length XY=20\overline{XY}=20 centimeters. What is the length of XZ\overline{XZ}?

  • A12 cm
  • B24 cm
  • C30 cm
  • D32 cm
Show the answer

Correct answer: D

Since XY=20\overline{XY}=20 cm, use that to find YZ\overline{YZ} :

YZ=35XY=35(20)=12 cm\overline{YZ}=\frac{3}{5}\overline{XY}=\frac{3}{5}(20)=12\text{ cm}

XZ=XY+YZ=20+12=32 cm\overline{XZ}=\overline{XY}+\overline{YZ}=20+12=32\text{ cm}

Question analysis

This segment question places points X, Y and Z on a line with Y between the others, gives XY as 20 centimeters, and says YZ is three fifths of XY. Students find the length of the whole segment XZ. Keeping track of which segment the question asks for matters more than the arithmetic.

NYS standards
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a fraction of a length and adding segments takes only two short steps with friendly numbers. Grade 7 rational-number problems are usually longer or involve negatives, so this sits below grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

There is no figure, so students who sketch the three points see at once that XZ is made of the two pieces XY and YZ. The arithmetic, three fifths of 20, is friendly. Most errors come from answering with the piece that was just calculated instead of the whole segment the question asks for.

  1. Turning words into math minor Understanding the question

    The condition “Y is between X and Z” has to become XZ = XY + YZ, and the overbar notation for segment lengths may be unfamiliar to some students. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find three fifths of 20 to get YZ, then add the two parts, since Y lies between X and Z.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a number
ab of n=n÷b×a\frac{a}{b} \text{ of } n = n \div b \times a
n is the starting amount. b is the bottom of the fraction: split n into b equal pieces. a is the top: take a of those pieces. Here n is 20, b is 5 and a is 3.
Adding segment lengths
XZ=XY+YZ\overline{XZ} = \overline{XY} + \overline{YZ}
X, Y and Z are points on one line, with Y between X and Z. XY and YZ are the lengths of the two short pieces. XZ is the whole length. The two pieces join end to end to make the whole.

  1. Picture the line

    Points X, Y and Z are on one straight line, with Y between X and Z. So the order is X, then Y, then Z.

    A segment is the straight piece between two points. The bar over two letters means the length of that segment.

    XY is 20 cm. You need XZ, the whole length.

  2. Find the length of YZ

    YZ is three fifths of XY. To find three fifths of 20, first find one fifth. Then take 3 of those.

    35 of 20=20÷5×3=4×3=12\frac{3}{5} \text{ of } 20 = 20 \div 5 \times 3 = 4 \times 3 = 12

    So YZ is 12 cm.

    Watch out: 12 cm is only one piece of the line. Stopping here gives choice A.

  3. Add the two pieces

    Y is between X and Z, so the two pieces join to make the whole.

    XZ=XY+YZ=20+12=32\overline{XZ} = \overline{XY} + \overline{YZ} = 20 + 12 = 32

    XZ is 32 cm.

    So the answer is D, 32 cm.

Check your answer

Check the fraction: 1220=35\frac{12}{20} = \frac{3}{5}, so YZ really is three fifths of XY. Also, XZ must be longer than XY, since it holds XY and more. 32 is longer than 20.

More questions like this

The original DOE question
Question 78 as printed on page 157 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 78. Open page 157 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 50

2026 Practice Test B Math, question 20

Question

Bryana bought 1341\frac{3}{4} yards of cloth at $8.00 per yard. If there was an 8% sales tax, what was the total cost of the cloth?

  • A$12.96
  • B$14.08
  • C$15.12
  • D$16.08
Show the answer

Correct answer: C

Calculate the cost of the cloth before tax:

134×8=74×8=$141\frac{3}{4}\times 8=\frac{7}{4}\times 8=\$14

Now find the tax for $14 worth of cloth:

14×8%=14×8100=$1.1214\times 8\%=14\times\frac{8}{100}=\$1.12

Finally, add the cost of the fabric and the tax:

$14 + $1.12 = $15.12

Question analysis

This percent word problem has Bryana buy 1 3/4 yards of cloth at $8.00 a yard and pay 8% sales tax. Students multiply a mixed number by a price, find the tax and add it to get the total cost. Handling the mixed number and taking the tax as a percent of the price both need care.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Sales tax on a purchase is a standard grade 7 percent problem. This one adds a mixed-number multiplication first, but the numbers are friendly and the steps are routine for grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 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 method is standard for a sales-tax question, so the work lies in the number handling. Converting 1 3/4 yards correctly and taking 8% of the cost, rather than adding 8 cents, are the two places students slip. The choices punish each slip but not stopping early, since neither $14 nor $1.12 appears, which helps students who check their result against the list.

  1. Error-prone arithmetic moderate Working it out

    Two number forms meet: the mixed number 1 3/4 times $8 must become $14, and 8% of $14 must become $1.12, a decimal with two places. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    The steps, price before tax, then tax, then total, must be done in order, and stopping after any of them gives a number that is not among the choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Turn 1 3/4 into 7/4 or 1.75 and multiply by $8, then multiply that cost by 1.08 to include the tax in one step.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost
cost=price per unit×number of units\text{cost} = \text{price per unit} \times \text{number of units}
Price per unit is the cost of one unit, here $8 for one yard. Number of units is how many you buy, here yards of cloth. Multiply them to get the total before tax.
Sales tax
tax=tax rate100×price\text{tax} = \frac{\text{tax rate}}{100} \times \text{price}
The tax rate is the percent, here 8. Percent means out of 100, so 8% is 8/100, or 0.08. The price is the cost before tax. The tax is added on top of the price.

  1. Write down what you know

    Bryana bought 1341\frac{3}{4} yards of cloth. Each yard costs $8.00. There is an 8% sales tax. Sales tax is an extra amount added to the price.

    You need the total cost, with the tax.

  2. Find the cost before tax

    First change the mixed number to a fraction. A mixed number is a whole number plus a fraction. 1341\frac{3}{4} is the same as 74\frac{7}{4}, because 1 whole is 4 quarters.

    cost=price per unit×number of units\text{cost} = \text{price per unit} \times \text{number of units}

    cost=8×74=564=14\text{cost} = 8 \times \frac{7}{4} = \frac{56}{4} = 14

    The cloth costs $14 before tax.

    Watch out: 1341\frac{3}{4} is 1.75, not 1.5. Using 1.5 gives $12.96, choice A.

  3. Find the tax

    tax=tax rate100×price\text{tax} = \frac{\text{tax rate}}{100} \times \text{price}

    tax=8100×14=0.08×14=1.12\text{tax} = \frac{8}{100} \times 14 = 0.08 \times 14 = 1.12

    The tax is $1.12.

    Watch out: 8% of $14 is not 8 cents. Adding only $0.08 gives $14.08, choice B.

  4. Add the tax to the price

    The total is the price before tax plus the tax.

    14+1.12=15.1214 + 1.12 = 15.12

    The total is $15.12.

    So the answer is C, $15.12.

Check your answer

Estimate. 10% of $14 is $1.40, so 8% is a little less, about $1.10. Then $14 plus about $1.10 is about $15.10. Only $15.12 is that close.

More questions like this

The original DOE question
Question 79 as printed on page 157 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 79. Open page 157 of the handbook (PDF)

Question 21 of 50

2026 Practice Test B Math, question 21

Question

A number line with point M on the left and point N on the right. N is labeled 1 1/3; no other coordinates are marked.

On the number line above, MN=556MN=5\frac{5}{6}. What is the position of point MM?

  • A−716-7\frac{1}{6}
  • B−412-4\frac{1}{2}
  • C4124\frac{1}{2}
  • D7167\frac{1}{6}
Show the answer

Correct answer: B

To find MM, subtract N−MN-M and set it equal to the length:

113−M=5561\frac{1}{3}-M=5\frac{5}{6}

−M=556−113-M=5\frac{5}{6}-1\frac{1}{3}

−M=556−126-M=5\frac{5}{6}-1\frac{2}{6}

−M=436-M=4\frac{3}{6}

M=−412M=-4\frac{1}{2}

Question analysis

This number line question gives the distance between points M and N as a mixed number and asks for the position of M. Students subtract fractions with unlike denominators, moving past zero into negative numbers. Direction along the line and the sign of the result are where it gets tricky.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students find distances on a number line as the absolute value of a difference. Doing it with mixed numbers that have unlike denominators, and crossing zero, is a typical and slightly demanding grade 7 task.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 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 distance is given and one endpoint is known, so the question is really about direction and sign. Because the drawing shows no zero, students must reason that moving 5 5/6 to the left from 1 1/3 carries M past zero. The choices pair each size with each sign, so getting the size right is only half the work; the sign decides between two choices.

  1. Reading a figure, table or graph moderate Understanding the question

    The number line labels only N, at 1 1/3, and shows no zero, so students must see from the picture that M lies left of N and work out whether it passes zero. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Error-prone arithmetic moderate Working it out

    The subtraction 1 1/3 − 5 5/6 mixes unlike denominators and crosses zero: in sixths it is 8/6 − 35/6 = −27/6, which then reduces to a mixed number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. A very close wrong answer moderate Choosing the answer

    4 1/2 and −4 1/2 are written the same except for the sign; only noticing that M lies left of N, so below 1 1/3, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Read N's coordinate from the number line, convert everything to sixths, and move the given distance from N toward M's side of the line.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
On a number line, numbers grow as you move right. The distance between two points is the larger number minus the smaller one. That is the right point minus the left point, so a distance is never negative.
Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, b is the top of the fraction, and c is the bottom. Each whole holds c pieces, so a wholes hold a times c pieces. Then add the b extra pieces.

  1. Read the number line

    Point N is at 1131\frac{1}{3}. Point M is to the left of N. MN means the distance from M to N, and it is 5565\frac{5}{6}.

    You need the number at point M.

  2. Decide which way to move

    On a number line, numbers get smaller as you move left. M is left of N, so N is the right point.

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    556=113−M5\frac{5}{6} = 1\frac{1}{3} - M

    So M is what you get when you take the distance away from N.

    M=113−556M = 1\frac{1}{3} - 5\frac{5}{6}

    The distance is bigger than 1131\frac{1}{3}, so you will pass 0. M will be negative.

  3. Change both numbers to sixths

    To subtract, the fractions need the same bottom number. Use sixths, since 3 goes into 6. Change each mixed number to an improper fraction, a fraction whose top is bigger than its bottom.

    113=1×3+13=43=861\frac{1}{3} = \frac{1 \times 3 + 1}{3} = \frac{4}{3} = \frac{8}{6}

    556=5×6+56=3565\frac{5}{6} = \frac{5 \times 6 + 5}{6} = \frac{35}{6}

  4. Subtract and simplify

    M=86−356=−276M = \frac{8}{6} - \frac{35}{6} = -\frac{27}{6}

    27 sixths is 4 wholes (24 sixths) and 3 sixths more. And 36=12\frac{3}{6} = \frac{1}{2}. So M is −412-4\frac{1}{2}.

    Watch out: 4124\frac{1}{2} without the minus sign is choice C. That point would be to the right of N, not the left.

    So the answer is B, −412-4\frac{1}{2}.

Check your answer

Find the distance from your M to N. From −412-4\frac{1}{2} to 0 is 4124\frac{1}{2}, or 276\frac{27}{6}. From 0 to N is 86\frac{8}{6}. Together, 276+86=356=556\frac{27}{6} + \frac{8}{6} = \frac{35}{6} = 5\frac{5}{6}. That matches MN.

More questions like this

The original DOE question
Question 80 as printed on page 157 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 80. Open page 157 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 22 of 50

2026 Practice Test B Math, question 22

Question

A United States presidential coin is made from an alloy of four metals—copper, zinc, manganese, and nickel—with weights in the ratio of 177:12:7:4177:12:7:4, respectively. The coin weighs a total of 8 grams. What is the weight of the zinc in this coin?

  • A0.28 g
  • B0.48 g
  • C0.96 g
  • D48 g
Show the answer

Correct answer: B

Add the four values in the ratio

(177 + 12 + 7 + 4 = 200)

and use the sum as the denominator. Use that to find the fraction of zinc in one of the coins. Then reduce the fraction:

12200=350\frac{12}{200}=\frac{3}{50}

Multiply this fraction by 8 to find the number of grams of zinc in decimal form:

350×8=2450=0.48 g\frac{3}{50}\times 8=\frac{24}{50}=0.48\text{ g}

Question analysis

This ratio word problem describes a United States presidential coin made of four metals in the ratio 177:12:7:4 by weight, and asks how many grams of zinc are in an 8-gram coin. Students turn a part-to-whole ratio into a fraction and multiply. Matching each metal to its number and handling the decimal result need care.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Using a part-to-whole ratio to split a total is ratio work grade 7 students practice often. The four-term ratio and a decimal answer add a little care, but the demand matches typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method, adding the parts and taking zinc's share of the total, is standard, and the parts conveniently add to 200. Students lose points by picking the wrong term of the four-part ratio or by assuming the parts are out of 100. The last check is size: no part of an 8-gram coin can weigh 48 grams.

  1. Hard words minor Understanding the question

    Construct-irrelevant reading: the words “alloy” and “respectively” carry the setup, and only “respectively” tells students that zinc is the second term, 12, of 177:12:7:4. (Perfetti, 2007; Freedle & Kostin, 1993)

  2. Error-prone arithmetic minor Working it out

    Taking 12 parts out of 200 of an 8-gram coin produces 96/200, a decimal under one gram, where a misplaced point gives 4.8 or 48. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add the four ratio numbers to get the total parts, write zinc's share as a fraction of that total, and multiply by the coin's 8 grams.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Share of a ratio
part=its ratio numbersum of the ratio numbers×total\text{part} = \frac{\text{its ratio number}}{\text{sum of the ratio numbers}} \times \text{total}
The ratio numbers tell how many equal shares each metal gets. Their sum is the number of shares in the whole coin. One metal's fraction is its shares over all the shares. Multiply by the total weight to get that metal's weight.

  1. Match each metal to its number

    The ratio is 177:12:7:4. “Respectively” means in the same order as the list of metals. So copper is 177, zinc is 12, manganese is 7 and nickel is 4.

    The whole coin weighs 8 grams. You need the weight of the zinc.

    Watch out: Using manganese’s 7 instead of 12 gives 0.28 g, choice A.

  2. Count the total parts

    177+12+7+4=200177 + 12 + 7 + 4 = 200

    The coin is made of 200 equal parts by weight. Zinc has 12 of those parts.

  3. Find zinc's fraction

    Zinc’s fraction is its ratio number over the sum of the ratio numbers.

    12200=12÷4200÷4=350\frac{12}{200} = \frac{12 \div 4}{200 \div 4} = \frac{3}{50}

    Zinc is 350\frac{3}{50} of the coin’s weight.

    Watch out: Do not divide by 100. The ratio adds to 200, not 100. Treating zinc as 12% gives 0.96 g, choice C.

  4. Multiply by the coin's weight

    Multiply zinc’s fraction by the weight of the whole coin, 8 grams.

    part=350×8=2450=48100=0.48\text{part} = \frac{3}{50} \times 8 = \frac{24}{50} = \frac{48}{100} = 0.48

    The zinc weighs 0.48 grams.

    So the answer is B, 0.48 g.

Check your answer

Zinc is a small share, 12 out of 200, so it must weigh well under 1 gram. 0.48 g fits. Choice D, 48 g, is heavier than the whole 8-gram coin, so it cannot be right.

More questions like this

The original DOE question
Question 81 as printed on page 157 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 81. Open page 157 of the handbook (PDF)

Question 23 of 50

2026 Practice Test B Math, question 23

Question

Nicole’s age now is three times Carmen’s age. If Carmen will be 17 in two years, how old was Nicole 5 years ago?

  • A38 yr
  • B40 yr
  • C45 yr
  • D50 yr
Show the answer

Correct answer: B

First, find out how old Nicole and Carmen are now.

Let NN = Nicole’s age now. Let CC = Carmen’s age now.

C+2=17C+2=17

C=15C=15 (Carmen’s age now)

N=3CN=3C

N=3(15)=45N=3(15)=45 (Nicole’s age now)

N−5=45−5=40N-5=45-5=40 (Nicole’s age 5 years ago)

Question analysis

This age word problem says Nicole is three times as old as Carmen now, and that Carmen will be 17 in two years. Students work out both present ages and then go back five years for Nicole. Keeping the times straight is the tricky part: two years ahead for Carmen, five years back for Nicole.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students write simple equations from word problems. Age problems like this one take three short steps, and the shifts in time are the only real hurdle, so the demand is typical of grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 1, 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 arithmetic is short, but the question moves through three points in time, and only one of them carries the three-times relationship. Students who find Nicole's present age often stop there, since it is a choice. Labeling each age with its time, now or then, keeps the two-year and five-year shifts with the right person and in the right direction.

  1. Turning words into math moderate Understanding the question

    Three points in time appear, now, in two years and 5 years ago, and “three times Carmen's age” holds only for now, so each shift must be applied to the right person. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Students must hold Carmen's future age 17, her present age 15, Nicole's present age 45 and the 5-year shift while deciding which to report. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find Carmen's age now by going back two years from 17, triple it for Nicole's current age, then subtract five.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Age at another time
age in n years=age now+n\text{age in } n \text{ years} = \text{age now} + n
n is a number of years. To find an age in the future, add n to the age now. To find an age in the past, subtract n instead. Everyone gets older at the same speed, one year per year.

  1. Sort out the times

    The question talks about three times: now, two years from now, and 5 years ago. Nicole is 3 times as old as Carmen now.

    Carmen will be 17 in two years. You need Nicole’s age 5 years ago.

  2. Find Carmen's age now

    In two years Carmen will be 17. So right now she is 2 years younger.

    17−2=1517 - 2 = 15

    Carmen is 15 now.

  3. Find Nicole's age now

    Nicole’s age now is three times Carmen’s age now.

    3×15=453 \times 15 = 45

    Nicole is 45 now.

    Watch out: 45 is her age now, choice C. The question asks about 5 years ago.

  4. Go back 5 years

    Going back in time means subtracting.

    45−5=4045 - 5 = 40

    Nicole was 40 five years ago.

    Watch out: Adding 5 gives 50, choice D. Also, the 2 years only apply to Carmen. Taking 2 more years off Nicole gives 38, choice A.

    So the answer is B, 40 yr.

Check your answer

Work forward from your answer. Nicole was 40 five years ago, so she is 45 now. Carmen is a third of that, 15. In two years Carmen will be 17. That matches the question.

More questions like this

The original DOE question
Question 84 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 84. Open page 158 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 50

2026 Practice Test B Math, question 24

Question

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

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

Correct answer: C

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

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

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

Question analysis

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

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

Why it is hard

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

  1. A very close wrong answer major Choosing the answer

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

  2. Turning words into math moderate Understanding the question

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

  3. Variables instead of numbers minor Understanding the question

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Pick an easy starting amount

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

  3. Work out the first week

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

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

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

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

  4. Work out the second week

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

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

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

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

  5. Read the answer as a percent

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

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

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

Check your answer

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

More questions like this

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

Question 25 of 50

2026 Practice Test B Math, question 25

Question

If w−1w-1 is an odd integer, which one of the following must be an even integer?

  • Aw+1w+1
  • B2w−12w-1
  • C2w−22w-2
  • D2w+12w+1
Show the answer

Correct answer: C

One more than an odd integer must be even. One more than w−1w-1 is ww, therefore ww must be even.

Two times an even integer must be even, therefore 2w2w is even. An even integer decreased by 2 must be even.

Therefore, 2w−22w-2 must be even.

Question analysis

This algebra question says that w − 1 is an odd integer and asks which of four expressions must be even. It tests reasoning about odd and even numbers written with a variable. The difficulty is that "must be" means even for every allowed value of w, so one example that happens to work is not enough.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.4 common factors and multiples
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 work evaluates and rewrites expressions but rarely asks whether one is odd or even for every value. The algebra here is light; the demand is reasoning about parity and reading "must be" carefully.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 8, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns on one early inference: since w − 1 is odd, w itself is even. Students who skip that step, or quietly treat w as the odd number, carry the wrong parity into every choice. After that the work is mechanical, because each expression adds an even or an odd amount to an even number. Trying one value, such as w = 4, confirms the choice in seconds.

  1. Variables instead of numbers moderate Understanding the question

    The stem gives no number at all: w − 1 is only known to be odd, so the parity of w itself has to be worked out before any choice can be judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    All four choices are built from w or 2w plus or minus 1 or 2, so each one has to be tested for evenness; three of them come out odd for every allowed w. (Embretson & Wetzel, 1987)

  3. Something you must already know minor Working it out

    The parity rules are assumed, not stated: one more than an odd number is even, any multiple of 2 is even, and an even number plus or minus 1 is odd. (Anderson & Krathwohl, 2001)

Strategy: Choose a value of w that makes w − 1 odd, evaluate all four expressions, and keep the one that comes out even; then confirm it with a second value.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding 1 to an odd number
odd+1=even\text{odd} + 1 = \text{even}
Odd and even numbers take turns on the number line: 1, 2, 3, 4 and so on. Adding 1 moves you to the next whole number. So the number right after an odd number is always even, and the one after an even number is always odd.
Doubling makes an even number
2×n=even2 \times n = \text{even}
n is any whole number. An even number is one you can split into two equal whole groups. Two times n is always two equal groups of n, so it is always even.

  1. Sort out what you know

    You know that w−1w-1 is an odd integer. An integer is a whole number, like 3, 0 or −5. An even number can be split into two equal groups, like 4 or 10. An odd number is 1 more than an even number, like 3 or 11.

    You need the choice that must be even. Must means it is even for every possible value of ww, not just one.

  2. Decide whether w is odd or even

    The number ww is 1 more than w−1w-1. One more than an odd number is always even. For example, if w−1w-1 is 3, then ww is 4. So ww is even.

    Watch out: do not treat ww itself as the odd number. That mistake makes w+1w+1 look even and leads to choice A.

  3. Show that 2w − 2 is even

    Two times any whole number is even, so 2w2w is even. Taking 2 away from an even number moves you back two spots. You land on another even number. So 2w−22w-2 is even.

    Here is another way to see it. 2w−22w-2 is the same as 2×(w−1)2 \times (w-1). That is 2 times a whole number, so it must be even.

  4. Check the other three choices

    Each one adds or takes away 1 from an even number. That always gives an odd number.

    • A: w+1w+1 is even plus 1, so it is odd.
    • B: 2w−12w-1 is even minus 1, so it is odd.
    • D: 2w+12w+1 is even plus 1, so it is odd.

    Watch out: starting with 2w2w does not make a number even. The +1 in choice D makes it odd.

  5. Give the answer

    Only one choice is even for every possible ww. So the answer is C, 2w−22w-2.

Check your answer

Pick a number. Let w=4w = 4, so w−1=3w-1 = 3, which is odd. Then A is 5, B is 7, C is 6 and D is 9. Only C is even. Try w=10w = 10 too. You get 11, 19, 18 and 21. C is even again.

More questions like this

The original DOE question
Question 86 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 86. Open page 158 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 26 of 50

2026 Practice Test B Math, question 26

Question

Simplify this expression:

4(7−3x)−(5−x)4\left ( 7-3x \right )-\left ( 5-x \right )

  • A23−4x23-4x
  • B23−11x23-11x
  • C28−4x28-4x
  • D28−12x28-12x
Show the answer

Correct answer: B

4( 7 − 3x ) −( 5 − x) = 28 − 12x − 5 + x = 23 − 11x

Question analysis

This question asks students to simplify an algebraic expression by distributing a number into one set of parentheses, subtracting another set, and combining like terms. It checks the distributive property with a minus sign in front of parentheses. The usual slip is subtracting only the first term inside the second parentheses, which changes the sign of an x term.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Expanding and combining linear expressions is core grade 7 work. This is a standard example, two distributions and one subtraction, with nothing beyond what a grade 7 lesson on the distributive property would cover.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 16+, 8 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is routine, but subtracting a whole group is where students go wrong: the minus sign has to reach the −x as well as the 5. Each choice pairs a constant with an x coefficient, so getting the constant right only narrows the field to two. The question rewards writing out both distributions in full rather than doing them mentally.

  1. Error-prone arithmetic moderate Working it out

    The minus sign in front of (5 − x) must change both terms, so −x becomes +x and then combines with −12x from the first group. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Two distributions, then collecting the constants 28 and −5 and the x terms separately: four small operations, each a place where a sign can slip. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    23 − 4x shares the right constant with 23 − 11x, so getting 23 does not settle it; only the x coefficient, −12x + x rather than −3x − x, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Rewrite the subtraction as adding −1 times the second group, distribute both multipliers, then collect the constant terms and the x terms separately.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The distributive property
a(b−c)=a×b−a×ca(b - c) = a \times b - a \times c
a is the number in front of the parentheses. b and c are the terms inside. You multiply a by each term inside, not only the first one. This works because a groups of (b − c) is a groups of b take away a groups of c.
Subtracting a group
−(b−c)=−b+c-(b - c) = -b + c
b and c are the terms inside the parentheses. A minus sign in front means take away the whole group. So you change the sign of each term inside. Taking away a minus c is the same as adding c.

  1. Sort out the question

    You need to simplify 4(7−3x)−(5−x)4(7-3x)-(5-x). Simplify means write it in the shortest way. A term is one piece of the expression, like 7 or −3x-3x. Like terms have the same letter part, so −12x-12x and xx are like terms. Plain numbers are like terms too.

  2. Multiply the 4 into the first group

    Use the distributive property. The 4 multiplies both terms inside.

    4(7−3x)=4×7−4×3x=28−12x4(7-3x) = 4 \times 7 - 4 \times 3x = 28 - 12x

    Watch out: the 4 must multiply the 3x3x too. Writing only 28−3x28-3x is the first slip that leads to choice A.

  3. Take away the second group

    The minus sign in front of (5−x)(5-x) means subtract the whole group. So change the sign of each term inside.

    −(5−x)=−5+x-(5-x) = -5 + x

    Taking away a minus xx gives plus xx.

    Watch out: if you stop before this step, you get 28−12x28-12x. That is choice D.

  4. Combine the like terms

    Now you have 28−12x−5+x28-12x-5+x. Put the plain numbers together. Then put the xx terms together.

    28−5=2328 - 5 = 23

    −12x+x=−11x-12x + x = -11x

    So the expression simplifies to 23−11x23-11x.

  5. Give the answer

    The plain number is 23, and the xx term is −11x-11x. So the answer is B, 23−11x23-11x.

Check your answer

Put in an easy number, like x=1x = 1. The original gives 4(7−3)−(5−1)=16−4=124(7-3)-(5-1) = 16-4 = 12. Choice B gives 23−11=1223-11 = 12. They match. Choice A would give 23−4=1923-4 = 19, which does not match.

More questions like this

The original DOE question
Question 88 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 88. Open page 158 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 27 of 50

2026 Practice Test B Math, question 27

Question

PET SURVEY

Number of Pets Number of Students
0 12
1 16
2 7
3 or more 5

Amy surveyed students at her school about the number of pets they have. What is the probability that a student who participated in the survey has at least 2 pets?

  • A740\frac{7}{40}
  • B112\frac{1}{12}
  • C18\frac{1}{8}
  • D310\frac{3}{10}
Show the answer

Correct answer: D

First, add the number of students for each category to find out how many total students were in the survey:

12 + 16 + 7 + 5 = 40

The number of students who had at least 2 pets are the ones who have 2 pets (7) plus the ones who have 3 or more (5). The total number of students with at least 2 pets is

7 + 5 = 12

The probability of a student in the survey having at least two pets is:

1240=310\frac{12}{40}=\frac{3}{10}

Question analysis

Amy's pet survey table shows how many pets each student has, and the question asks for the probability that a randomly chosen student has at least 2 pets. It tests finding a probability from survey data. The tricky part is reading "at least 2" correctly and using the right total as the denominator.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Simple probability from a frequency table is grade 6 work in New York. The only extra demand is interpreting "at least" and adding a column, two short steps a grade 7 student should handle easily.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light, so the question is decided by the phrase "at least 2" and by remembering that the denominator is everyone surveyed. Students who include only one of the two qualifying rows reach an answer with the right denominator, which makes the mistake hard to notice. Rereading the condition before choosing is the main safeguard.

  1. Turning words into math minor Understanding the question

    "At least 2 pets" has to be turned into two rows of the table, the 2-pet row and the 3-or-more row, not one. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The denominator is not printed anywhere; it has to be built by adding all four rows of the table, 12 + 16 + 7 + 5 = 40. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Total the students in every row to get the denominator, then add the counts for each row that satisfies "at least 2" and simplify the fraction.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number you wanttotal numberP = \frac{\text{number you want}}{\text{total number}}
P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

  1. Sort out the question

    The table shows how many students have each number of pets. You pick one student who took the survey. You need the chance that the student has “at least 2 pets”.

    At least 2 means 2 or more. So two rows count: the 2 pets row and the 3 or more row.

  2. Find the total number of students

    The bottom of the fraction is everyone in the survey. Add all four rows.

    12+16+7+5=4012 + 16 + 7 + 5 = 40

    There are 40 students in all.

  3. Count the students you want

    Add the 2 pets row and the 3 or more row.

    7+5=127 + 5 = 12

    So 12 students have at least 2 pets.

    Watch out: using only the 7 students with exactly 2 pets gives 740\frac{7}{40}, choice A. Using only the 5 with 3 or more gives 540=18\frac{5}{40} = \frac{1}{8}, choice C.

  4. Write and simplify the probability

    Use the probability formula.

    P=1240P = \frac{12}{40}

    Divide the top and bottom by 4, since 4 goes into both.

    1240=12÷440÷4=310\frac{12}{40} = \frac{12 \div 4}{40 \div 4} = \frac{3}{10}

    So the answer is D, 310\frac{3}{10}.

Check your answer

The students with fewer than 2 pets are 12+16=2812 + 16 = 28. Then 28+12=4028 + 12 = 40, so every student is counted once. Also, 12 out of 40 is a bit less than a third. So is 310\frac{3}{10}.

More questions like this

The original DOE question
Question 89 as printed on page 159 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 89. Open page 159 of the handbook (PDF)

Question 28 of 50

2026 Practice Test B Math, question 28

Question

A circle has a radius of 5 inches. What is the area of the circle in terms of π\pi?

Use π\pi to express your answer.

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

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

Show the answer

Correct answer: 25π

Solution:

25π25\pi

The area of a circle is its radius squared times π\pi.

A=πr2A = \pi r^2

r=5r = 5

A=π(5)2A = \pi(5)^2

A=25πA = 25\pi

Question analysis

This circle question gives a radius of 5 inches and asks for the area in terms of π, typed into a box rather than chosen. Students recall and apply the formula for the area of a circle. The formula is not printed on the test, and the area and circumference formulas are easy to mix up.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the area of a circle from its radius is the grade 7 circle standard applied directly, in one routine calculation. Typing the answer in terms of π, with no choices to compare, adds a little risk.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is one formula and one square, so the question comes down to recall and to the form of the answer. The area and circumference formulas both pair π with the radius, and 2πr gives a tidy value with π in it that looks just as right. The other losses are about form: replacing π with 3.14, which the instructions rule out twice, or dropping π after squaring 5.

  1. Something you must already know moderate Working it out

    The formula A = πr² is not printed with the question, and the circumference formula 2πr uses the same r and π, so recalling the wrong one gives a value that looks just as finished. (Anderson & Krathwohl, 2001)

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

    The answer is built on a keypad with a π key and must keep π as a symbol; a decimal worked out with 3.14 is not accepted, and no choices show the expected form. (Enright, Morley & Sheehan, 2002)

Strategy: Recall the area formula rather than the circumference formula, square the radius before multiplying, and leave π as a symbol instead of replacing it with 3.14.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a circle
A=πr2A = \pi r^2
A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, a little more than 3. r² means r times r. Square the radius first, then multiply by π. The area is in square units.
Circumference of a circle
C=2πrC = 2 \pi r
C is the circumference, the distance around the circle. r is the radius, and π is the same number as above. This formula measures a length, not the space inside. It is here so you can tell the two formulas apart.

  1. Sort out the question

    The circle has a radius of 5 inches. The radius is the distance from the center to the edge. You need the area, the space inside the circle. In terms of π means π stays in the answer as a symbol. Do not change it to 3.14.

  2. Choose the area formula

    The area formula is A=πr2A = \pi r^2. It uses the radius squared.

    Watch out: the circumference formula, C=2πrC = 2\pi r, also uses π and the radius. It gives 2×π×5=10π2 \times \pi \times 5 = 10\pi, the distance around the circle in inches. That is not the area.

  3. Square the radius

    Squaring a number means multiplying it by itself:

    r2=52=5×5=25r^2 = 5^2 = 5 \times 5 = 25

    Watch out: squaring is not doubling. 5×2=105 \times 2 = 10 would lead to 10π again.

  4. Multiply by π

    Put 25 into the formula in place of r²:

    A=π×25=25πA = \pi \times 25 = 25\pi

    The area is 25π square inches. Write the number first and then π, the way you write 3x for 3 times x.

    Watch out: leave π as π. Using 3.14 gives 78.5, which is not accepted. Dropping π leaves 25, which is not the area.

  5. Enter it with the π key

    On the test, the keypad has a π key in its bottom left corner. Press 2, then 5, then the π key. You do not need a times sign. Enter only the answer, with no A = and no units. On this site, type 25π or 25pi.

    So you would enter 25π.

Check your answer

Estimate. π is a little more than 3, so 25π is a little more than 75 square inches. A square drawn around the circle is 10 inches on each side, so its area is 100 square inches. The circle fits inside that square, so its area must be less than 100. It is.

More questions like this

The original DOE question
Question 28 as shown on the DOE’s digital test, page 100 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B Math, page 100, question 28.

Question 29 of 50

2026 Practice Test B Math, question 29

Question

Square MNPQ with N at the top left, P at the top right, Q at the bottom right and M at the bottom left; side PQ is labeled 6 cm. S is on the top side NP, T is on the bottom side MQ, and R is on the left side MN. Segments ST, RS and RT are drawn, and the two triangles RNS (top left corner) and TMR (bottom left corner) are shaded.

R, S, and T are midpoints of the sides of square MNPQ, as shown above. What is the sum of the areas of the shaded triangles?

  • A9 sq cm
  • B12 sq cm
  • C18 sq cm
  • D36 sq cm
Show the answer

Correct answer: A

The length of one side of the square is 6 cm. Since RR, SS, and TT are midpoints, then TM\overline{TM}, MR\overline{MR}, RN\overline{RN}, and NS\overline{NS} are all equal to 3 cm. Triangles TMRTMR and RNSRNS are both right triangles, so the area of one of the triangles is

12×3×3=92\frac{1}{2}\times 3\times 3=\frac{9}{2}

The triangles are congruent, so the sum of the areas is

92+92=9 sq cm\frac{9}{2}+\frac{9}{2}=9\text{ sq cm}

Question analysis

Points R, S and T are midpoints of the sides of square MNPQ, whose sides are 6 cm long, and students find the total area of the two shaded triangles in its corners. It tests the area of a triangle. The common slips are forgetting the one-half or measuring the wrong region.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Triangle area is grade 6 content, and the corner triangles here are right triangles with easy legs. Typical grade 7 area problems are similar or harder; the only extra step is using midpoints to get the lengths.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 1, 34 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The shaded triangles are right triangles tucked into the corners, so their legs lie along the square's sides and are easy to measure once the midpoints are used. Most lost points come from the one half in the area formula, or from reporting a larger region of the figure, such as the left half or the whole square. Labeling each 3 cm leg on the figure before computing pays off here.

  1. Reading a figure, table or graph moderate Understanding the question

    Only one length is labeled, the 6 cm side PQ; the legs NR, NS, MR and MT must each be read as half a side, 3 cm, from the word midpoints. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    The area of a triangle, one half times base times height, is not given, and leaving out the one half doubles each 4.5 sq cm triangle. (Anderson & Krathwohl, 2001)

Strategy: Use the midpoints to get each triangle's legs as half the square's side, apply one-half base times height, and remember there are two shaded triangles.

Traps

  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height, the straight-up distance from the base to the top corner. You take half because a triangle is half of a rectangle with the same base and height.

  1. Read the figure

    MNPQ is a square, and side PQ is 6 cm. All sides of a square are equal, so every side is 6 cm. A midpoint is the point exactly halfway along a side. R is the midpoint of side NM. S is the midpoint of NP, and T is the midpoint of MQ.

    The shaded triangles are RNS at the top and RMT at the bottom.

  2. Find the short sides

    Each midpoint cuts a 6 cm side into two equal parts.

    6÷2=36 \div 2 = 3

    So NR, NS, RM and MT are each 3 cm long.

  3. Find the area of one triangle

    Triangle RNS has a square corner at N. A square corner is a right angle. So NS can be the base and NR the height.

    A=12×3×3=4.5A = \frac{1}{2} \times 3 \times 3 = 4.5

    One triangle is 4.5 sq cm. Triangle RMT has the same sides, so it is also 4.5 sq cm.

    Watch out: do not forget the one half. Without it, each triangle is 9 sq cm and the total is 18. That is choice C.

  4. Add the two triangles

    There are two shaded triangles, so add their areas to get the total.

    4.5+4.5=94.5 + 4.5 = 9

    So the answer is A, 9 sq cm.

Check your answer

The left half of the square is rectangle NSTM, 3 cm by 6 cm. Its area is 3×6=183 \times 6 = 18 sq cm. The white triangle RST fills half of it. So the shaded part is the other half, 9 sq cm. Choice D, 36, is the whole square.

More questions like this

The original DOE question
Question 92 as printed on page 159 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 92. Open page 159 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 30 of 50

2026 Practice Test B Math, question 30

Question

The Chens spend $5 of every $8 they earn on planned expenses. If the family earns $29,600 in one year, how much will they spend on planned expenses that year?

  • A$1,850
  • B$3,700
  • C$5,920
  • D$18,500
Show the answer

Correct answer: D

Let xx be the amount spent on planned expense in one year:

x29,600=58\frac{x}{29{,}600}=\frac{5}{8}

x=58(29,600)=18,500x=\frac{5}{8}(29{,}600)=18{,}500

Question analysis

The Chens spend $5 out of every $8 they earn on planned expenses, and students apply that ratio to a yearly income of $29,600. It tests using a part-to-whole ratio with large numbers. The trap is using only part of the ratio, or using its two numbers the wrong way round.

NYS standards
  • 6.RP.3 ratio and rate reasoning
  • 6.RP.1 ratio concepts and language
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Applying a ratio to a total is grade 6 ratio reasoning. This is a two-step calculation with friendly numbers, easier than the multistep ratio and percent problems typical of grade 7.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 27 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a direct fraction-of-a-quantity problem, and most students who read the ratio as five-eighths will finish it. The common loss is stopping after the division by 8, since that one-eighth share is also offered. Checking that the result is a little more than half the income catches both that error and a misplaced zero.

  1. Turning words into math minor Understanding the question

    "$5 of every $8" has to be read as the fraction 5/8 of the income, not as an instruction to divide by 5 or by 8 alone. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Dividing $29,600 by 8 and multiplying by 5 works with five-digit numbers, where a dropped zero gives an answer ten times too small. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Treat 5 of every 8 dollars as the fraction five-eighths of the income: divide the yearly earnings by 8, then multiply by 5.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a number
part=ab×whole\text{part} = \frac{a}{b} \times \text{whole}
The whole is the full amount, here the money earned. The fraction a over b tells what share you want: a dollars out of every b dollars. Dividing the whole by b gives one share. Multiplying by a gives a of those shares.

  1. Sort out what you know

    The Chens earn $29,600 in one year. They spend “$5 of every $8” on planned expenses. That means 5 out of every 8 dollars. As a fraction, that is 58\frac{5}{8} of what they earn. You need to find how much they spend.

  2. Find one eighth of the money

    Split the money into 8 equal parts.

    29,600÷8=3,70029{,}600 \div 8 = 3{,}700

    One eighth is $3,700.

    Watch out: this is not the answer yet. Stopping here gives choice B.

  3. Take five of those parts

    They spend 5 of the 8 parts. Use the fraction formula.

    part=58×29,600=5×3,700=18,500\text{part} = \frac{5}{8} \times 29{,}600 = 5 \times 3{,}700 = 18{,}500

    They spend $18,500.

    Watch out: dividing only by 5 gives $5,920, choice C. Dropping a zero gives $1,850, choice A.

  4. Give the answer

    The planned expenses are 5 of the 8 equal parts, or $18,500. So the answer is D, $18,500.

Check your answer

58\frac{5}{8} is a little more than half. Half of $29,600 is $14,800. So the answer should be a bit more than $14,800. Only $18,500 fits. The other three choices are all less than half.

More questions like this

The original DOE question
Question 93 as printed on page 159 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 93. Open page 159 of the handbook (PDF)

Question 31 of 50

2026 Practice Test B Math, question 31

Question

A pizza shop offers a choice of 3 sizes (small, medium, and large) and 7 different toppings. Different pizzas can be created by changing the size and the choice of toppings. If Cody wants to order a pizza with exactly 2 different toppings, how many different pizzas can he create?

  • A6
  • B21
  • C63
  • D126
Show the answer

Correct answer: C

Figure out how many different topping pairs are possible. Use 1, 2, 3, 4, 5, 6, 7 to represent the toppings, and create a list of possible pairs:

1,2; 1,3; 1,4; 1,5; 1,6; 1,7
2,3; 2,4; 2,5; 2,6; 2,7
3,4; 3,5; 3,6; 3,7
4,5; 4,6; 4,7
5,6; 5,7
6,7

So there are 21 different topping combinations for one pizza.

Since there are 3 pizza sizes, multiply the total number of combinations by 3 to get the total number of different pizzas Cody can create: 3×21=633 \times 21 = 63.

Question analysis

Cody orders a pizza with exactly 2 of 7 toppings in one of 3 sizes, and students count how many different pizzas he could create. It tests counting combinations, where the order of the toppings does not matter, combined with a second choice. The challenge is to count every possible pizza exactly once.

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 In grade 7, compound events are usually ordered pairs, such as a coin with a number cube. Here an unordered pair of toppings is chosen from seven and then matched with three sizes, a combination count beyond typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question packs two counts into one sentence, each with its own trap. Order does not matter among the toppings, while the size multiplies every pair, and students tend to remember one of these and forget the other. Listing pairs systematically, each topping matched only with later ones, keeps the first count honest, and the stem’s “changing the size and the choice of toppings” is the reminder that size counts too.

  1. Combines ideas moderate Working it out

    Two counts are joined: the unordered pairs of toppings from 7, then the 3 sizes each pair can come in, and a mistake in either count changes the total. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Something you must already know minor Working it out

    Students must know that a pizza with mushroom and onion is the same as one with onion and mushroom, so the ordered count 7 × 6 has to be halved. (Anderson & Krathwohl, 2001)

Strategy: Number the toppings 1 to 7, list each pair once by matching every topping only with higher-numbered ones, count the list, then bring in the choice of size.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The counting principle
total=m×n\text{total} = m \times n
m is the number of ways to make one choice, and n is the number of ways to make a second choice. Every first choice can go with every second choice, so you multiply. Here, every pair of toppings can come in every size.
Pairs when order does not matter
pairs=n×(n−1)2\text{pairs} = \frac{n \times (n - 1)}{2}
n is the number of items to choose from. There are n ways to pick a first item and n − 1 ways to pick a different second one. That counts each pair twice, once in each order. Dividing by 2 counts each pair once.

  1. Sort out the question

    Each of Cody’s pizzas has one size and exactly 2 different toppings. There are 3 sizes and 7 toppings. Pizzas differ by “changing the size and the choice of toppings.” So a small and a large with the same toppings are two different pizzas. You need to count every pizza once.

  2. List the topping pairs

    Order does not matter: onion with peppers is the same pizza as peppers with onion. So number the toppings 1 to 7, and pair each one only with bigger numbers. Then no pair shows up twice.

    ToppingPairs withPairs
    12, 3, 4, 5, 6, 76
    23, 4, 5, 6, 75
    34, 5, 6, 74
    45, 6, 73
    56, 72
    671

    6+5+4+3+2+1=216 + 5 + 4 + 3 + 2 + 1 = 21

    There are 21 different pairs of toppings.

  3. Check the pairs with the formula

    Use the pairs formula with 7 toppings:

    7×62=422=21\frac{7 \times 6}{2} = \frac{42}{2} = 21

    Watch out: 7 × 6 = 42 counts every pair twice, once as 1 then 2 and once as 2 then 1. Keeping 42 leads to 42×3=12642 \times 3 = 126, choice D.

  4. Bring in the sizes

    Each of the 21 topping pairs can come in any of the 3 sizes. Use the counting principle:

    21×3=6321 \times 3 = 63

    Watch out: stopping at 21 leaves out the sizes, which gives choice B. Choice A, 6, multiplies the 3 sizes by the 2 toppings on one pizza. That does not count pizzas.

  5. Give the answer

    There are 21 topping pairs, and each one comes in 3 sizes. Cody can create 63 different pizzas. So the answer is C, 63.

Check your answer

Count one size at a time. A small pizza can have any of the 21 pairs, so there are 21 small pizzas. There are also 21 medium and 21 large ones: 21+21+21=6321 + 21 + 21 = 63. That is more than the 21 pairs but fewer than 126, so it is reasonable.

More questions like this

The original DOE question
Question 31 as shown on the DOE’s digital test, page 103 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B Math, page 103, question 31.

Question 32 of 50

2026 Practice Test B Math, question 32

Question

At North High School, a survey asked two questions, Question A and Question B. For each question, students could answer either “yes” or “no.” Of the 800 students who responded to the survey, 720 answered “yes” to Question A, and 640 answered “yes” to Question B. What is the least possible number of these students who could have answered “yes” to both questions?

  • A80
  • B160
  • C560
  • D640
Show the answer

Correct answer: C

None of the 80 students (800 − 720) who answered “no” to Question A could have answered “yes” to both questions.

Therefore, the least possible number of students who could have answered “yes” to both questions can be found by subtracting the 80 who answered “no” to Question A from the 640 who answered “yes” to Question B or 640 − 80 = 560.

Question analysis

In a survey of 800 students at North High School, 720 said yes to the first question and 640 said yes to the second. Students must find the smallest possible number who said yes to both. It tests reasoning about overlapping groups, and asking for the least possible overlap rather than an exact count makes it unusual.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.EE.4 equations and inequalities from problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students work with Venn diagrams and two-way tables when all counts are given. Finding the smallest possible overlap requires reasoning about extremes rather than computing, which is rare in grade 7 and new to most students.
Estimated difficulty
Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Solution
2 steps
Reading level
Question and choices: about grade 8, 60 words.
Corrected from the DOE handbook
  • A repeated phrase and a stray comma in the explanation were removed.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Most students have never been asked for a smallest overlap, so they fall back on a picture of nested groups and give the largest one. The key move is to push the two yes groups apart as far as possible: only the 80 students who said no to Question A can absorb some of Question B's yes answers. Once that is seen, the arithmetic is one subtraction, so the question rewards reasoning far more than computation.

  1. No obvious method major Working it out

    Nothing in the wording points to the method: the smallest overlap appears only when all 80 students who said no to Question A are imagined saying yes to Question B. (Schoenfeld, 1985)

  2. Turning words into math moderate Understanding the question

    "The least possible number of these students who could have answered “yes” to both questions" asks for an extreme case, which has to be turned into an arrangement of the 800 students. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. A very close wrong answer minor Choosing the answer

    640 is only 80 above the right 560, so a rough sense of size cannot choose between them; only asking whether the overlap is the least or the greatest separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Add the two yes counts and compare the sum with the number of students; any amount above the total must come from students counted in both groups.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook prints “(800 − 720)” twice in its first sentence, after “80 students” and again after “Question A,” and puts a comma between subject and verb in “the least possible number of students who could have answered “yes” to both questions, can be found.” The site keeps the first “(800 − 720)” and drops the comma.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Least overlap of two groups
least overlap=a+b−t\text{least overlap} = a + b - t
a is the size of the first group and b is the size of the second. t is the total number of people. If a + b is more than t, some people must be in both groups. The extra amount is the smallest number that can be in both.

  1. Sort out what you know

    There were 800 students. Of them, 720 said yes to Question A and 640 said yes to Question B. You need the least number who could have said yes to both. The overlap is the group in both yes groups. You want the smallest overlap.

  2. Find who said no to A

    Students who said no to Question A cannot be in the overlap.

    800−720=80800 - 720 = 80

    So 80 students said no to Question A.

    Watch out: 80 is not the answer. It is only a step, and it is choice A.

  3. Keep the overlap small

    To make the overlap as small as you can, put as many B yes answers outside it as possible. The only students outside it are the 80 who said no to A. So pretend all 80 of them said yes to B.

  4. Count the rest of the yes group

    The other yes answers to B must come from students who also said yes to A.

    640−80=560640 - 80 = 560

    So at least 560 students said yes to both.

    Watch out: 640 is the most that could say yes to both, not the least. That is choice D.

  5. Give the answer

    The fewest students who could say yes to both questions is 560. So the answer is C, 560.

Check your answer

Use the overlap formula: 720+640−800=560720 + 640 - 800 = 560. There are 1,360 yes answers but only 800 students, so 560 students must be counted twice. You can also check the groups. 560 said yes to both, 160 to A only and 80 to B only. That makes 560+160+80=800560 + 160 + 80 = 800.

More questions like this

The original DOE question
Question 110 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 110. Open page 163 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 33 of 50

2026 Practice Test B Math, question 33

Question

A wooden box has a square base. The height of this box is 3 times the length of one side of the base. If one side of the base is 3 feet long, what is the volume of this box?

  • A9 cu ft
  • B27 cu ft
  • C36 cu ft
  • D81 cu ft
Show the answer

Correct answer: D

One side of the square base is 3 ft long. Since the height of the box is 3 times the length, then the height is

3 × 3 = 9 ft.

The volume of a rectangular prism is length × height × width. The volume of the wooden box is

V= 3 × 3 × 9 = 81 cu ft.

Question analysis

A wooden box has a square base, and its height is 3 times the length of a side of the base; students find its volume in cubic feet. It tests the volume of a rectangular prism with dimensions described in words. The work is short, but the height must be built from the side length before multiplying.

NYS standards
  • 6.G.2 volume with fractional edge lengths
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Volume of a box is grade 5 and 6 content, and this is a two-step version with whole numbers. Grade 7 volume problems usually involve composite or triangular prisms, so this sits below.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 1, 46 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a routine volume question once all three dimensions are written down. Errors come from treating the box as a 3-foot cube, from stopping at the base area, or from mixing perimeter into the product. Listing the dimensions 3, 3 and 9 before multiplying prevents each of them.

  1. Turning words into math minor Understanding the question

    "The height of this box is 3 times the length of one side of the base" must become a height of 9 feet before the volume formula can be used. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    The volume of a box, length times width times height, is not given, and the square base means the length and width are both 3 feet. (Anderson & Krathwohl, 2001)

Strategy: Write down all three dimensions first, computing the height from the base side, then multiply length, width and height.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a box
V=l×w×hV = l \times w \times h
V is the volume, the space inside the box. l is the length, w is the width and h is the height. l times w is the area of the bottom. Multiplying by h stacks that many layers of cubes.

  1. Sort out what you know

    The box has a square base with sides of 3 feet. A square has equal sides, so the length and width are both 3 feet. The height is “3 times the length of one side of the base”. You need the volume, the space inside the box.

  2. Find the height

    The height is 3 times the 3-foot side.

    h=3×3=9h = 3 \times 3 = 9

    The height is 9 feet.

    Watch out: the height is not 3 feet. If you use 3 for the height, you get 3×3×3=273 \times 3 \times 3 = 27. That is choice B.

  3. Find the volume

    Use the volume formula with all three sizes.

    V=3×3×9=81V = 3 \times 3 \times 9 = 81

    The volume is 81 cubic feet.

    Watch out: 3×3=93 \times 3 = 9 is only the area of the bottom. Stopping there gives choice A.

  4. Give the answer

    The box is 3 feet long, 3 feet wide and 9 feet tall. So the answer is D, 81 cu ft.

Check your answer

Think in layers. The bottom layer holds 3×3=93 \times 3 = 9 cubes, each 1 foot on a side. The box is 9 feet tall, so it holds 9 layers. Then 9×9=819 \times 9 = 81 cubic feet. The units are cubic feet because the answer is a volume.

More questions like this

The original DOE question
Question 96 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 96. Open page 160 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 34 of 50

2026 Practice Test B Math, question 34

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?

  • A1
  • B5
  • C7
  • D9
Show the answer

Correct answer: B

Calculate each mean speed:

R=655=13 kphR=\frac{65}{5}=13\text{ kph}

S=724=18 kphS=\frac{72}{4}=18\text{ kph}

Then calculate the difference of both mean speeds:

S−R=18−13=5 kphS-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
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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Speed
speed=distance÷time\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.

  1. 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.

  2. Find Rajiv’s speed

    He went 65 kilometers in 5 hours. Use the speed formula.

    speed=65÷5=13\text{speed} = 65 \div 5 = 13

    Rajiv’s mean speed is 13 kph.

  3. Find Shaina’s speed

    She went 72 kilometers in 4 hours. Use the same formula.

    speed=72÷4=18\text{speed} = 72 \div 4 = 18

    Shaina’s mean speed is 18 kph.

  4. Subtract the two speeds

    Find how much less 13 is than 18.

    18−13=518 - 13 = 5

    Rajiv was 5 kph slower.

    Watch out: do not subtract the distances or the times. 72−65=772 - 65 = 7 gives choice C, and 5−4=15 - 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 13×5=6513 \times 5 = 65 kilometers. At 18 kph for 4 hours, Shaina goes 18×4=7218 \times 4 = 72 kilometers. Both match the question, so the speeds are right.

More questions like this

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)

Question 35 of 50

2026 Practice Test B Math, question 35

Question

Points PP, QQ, RR, and SS represent −3, −1, 0, and 2, respectively, on a number line. How many units is the midpoint of PQ\overline{PQ} from the midpoint of RS\overline{RS}?

  • A1
  • B2
  • C3
  • D4
Show the answer

Correct answer: C

Number line from −3 to 3 with a tick at each integer. P is at −3, Q is at −1, R is at 0 and S is at 2.

Find the midpoint of PQPQ and RSRS:

Half the length of PQPQ is −1−(−3)2=22=1\frac{-1-(-3)}{2}=\frac{2}{2}=1 unit.

The midpoint of PQPQ is located 1 unit from each endpoint, so the midpoint is at −2.

Half the length of RSRS is 2−02=22=1\frac{2-0}{2}=\frac{2}{2}=1 unit.

The midpoint of RS is located 1 unit from each endpoint, so the midpoint is at 1. The distance between the two midpoints is 1 −(−2) = 3 units.

Question analysis

Four points on a number line are given by their coordinates, some of them negative, and students find the distance between the midpoints of two segments. It tests locating midpoints and measuring distance with negative numbers. The care needed is in placing each midpoint correctly and in measuring a distance that crosses zero.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Distance on a number line with negative numbers is grade 7 content. Finding two midpoints first adds a step, but the numbers are small integers, so the demand matches typical grade 7 practice.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 12, 26 words.
Corrected from the DOE handbook
  • Two mislabeled steps in the explanation were 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

Each step is simple, but the negative coordinates make the last subtraction easy to get wrong, and there is no picture to check against. Students who sketch the line and mark −2 and 1 can count the three units directly. The wrong choices come from confusing a segment's length with the gap between midpoints, or from counting tick marks instead of spaces.

  1. Error-prone arithmetic moderate Working it out

    The midpoint of PQ is at −2, so the final distance is 1 − (−2); dropping the negative sign at that step turns 3 into 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Much to keep track of minor Working it out

    Four labeled points and two midpoints have to be held at once, and the question draws no number line, so sketching one is the safe move. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Average the coordinates of each segment's endpoints to place its midpoint, then find the distance between the two midpoints as the absolute value of their difference.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook labels (−1 − (−3))/2 = 1 unit as the “Midpoint of PQ” and (2 − 0)/2 = 1 unit as the “Midpoint of RS.” Those are half the lengths of the segments, not their midpoints: as the next sentences say, the midpoint of PQ is at −2 and the midpoint of RS is at 1. The site labels them “Half the length of PQ” and “Half the length of RS.” The distance between the midpoints is still 3 units.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint on a number line
midpoint=a+b2\text{midpoint} = \frac{a + b}{2}
a and b are the numbers at the two ends of a segment. The midpoint is the point exactly halfway between them. Adding the ends and dividing by 2 finds their average, which is the middle.
Distance on a number line
distance=right number−left number\text{distance} = \text{right number} - \text{left number}
The right number is the point farther to the right on the number line. The left number is the other point. Subtracting tells how many units apart they are. Taking the left from the right keeps the answer positive.

  1. Sort out what you know

    P is at −3, Q is at −1, R is at 0 and S is at 2. A segment is the piece of a line between two points. So PQ\overline{PQ} runs from −3 to −1. You need the distance between the midpoint of PQ and the midpoint of RS. A quick sketch of a number line helps.

  2. Find the midpoint of PQ

    Use the midpoint formula with −3 and −1.

    −3+(−1)2=−42=−2\frac{-3 + (-1)}{2} = \frac{-4}{2} = -2

    The midpoint of PQ is at −2.

  3. Find the midpoint of RS

    Use the midpoint formula with 0 and 2.

    0+22=22=1\frac{0 + 2}{2} = \frac{2}{2} = 1

    The midpoint of RS is at 1.

  4. Find the distance between them

    The midpoint at 1 is to the right of the one at −2. Use the distance formula.

    distance=1−(−2)=1+2=3\text{distance} = 1 - (-2) = 1 + 2 = 3

    Subtracting a negative number is the same as adding.

    Watch out: dropping the minus sign gives 2−1=12 - 1 = 1, choice A. Counting the tick marks −2, −1, 0 and 1 gives 4, choice D. Count the spaces between marks instead.

  5. Give the answer

    The midpoints are at −2 and 1, which are 3 units apart. So the answer is C, 3.

Check your answer

Count the jumps on a number line. Going from −2 to −1 is one jump. Going from −1 to 0 is two, and going from 0 to 1 is three. So the midpoints are 3 units apart.

More questions like this

The original DOE question
Question 98 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 98. Open page 160 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 36 of 50

2026 Practice Test B Math, question 36

Question

A quarter circle whose center, marked with a dot, is at the bottom left corner. A segment labeled x + 1 runs from the center to a point on the arc, and from that point a vertical segment drops to the bottom edge, meeting it at a right angle. Along the bottom edge, the part from the center to the foot of that segment is labeled y, and the part from the foot to the end of the arc is labeled 2.

In the quarter circle above, what is yy in terms of xx ?

  • Ax−1x-1
  • Bx+1x+1
  • Cx+12\frac{x+1}{2}
  • D(x+1)22\sqrt{\frac{\left ( x+1 \right )^{2}}{2}}
Show the answer

Correct answer: A

Both x+1x+1 and y+2y+2 are radii.

So, set them equal to each other and solve for yy.

y+2=x+1y+2=x+1

y=x−1y=x-1

Question analysis

In a quarter circle, one segment is labeled x + 1 and the bottom edge is split into lengths y and 2, and students must write y in terms of x. It tests basic circle facts together with solving a simple equation. The challenge is seeing which marked lengths must be equal.

NYS standards
  • 7.G.4 circle area and circumference
  • 7.EE.4 equations and inequalities from problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 circle work centers on area and circumference formulas. This needs only the definition of a radius and a one-step equation, but spotting the relationship in a figure is less routine than plugging into a formula.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 9, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The figure looks like a right-triangle problem, and that is the trap: the right angle and the slanted radius pull students toward the Pythagorean theorem, which cannot work without the vertical leg. The solution needs only the fact that all radii of a circle are equal, applied to the bottom edge. Students who mark every segment that starts at the center see it at once.

  1. Reading a figure, table or graph major Understanding the question

    The labels sit on two different radii: x + 1 on the slanted one, and y + 2 along the bottom edge, which runs from the center all the way to the arc. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Variables instead of numbers minor Understanding the question

    The answer is an expression in x, so there is no number to check; trying x = 5, which makes the radius 6 and y = 4, is the fallback. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    x + 1 differs from the right x − 1 by a single sign; only seeing that y plus the 2-unit piece makes a radius, so the 2 is subtracted, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Look for two lengths in the figure that are both radii of the quarter circle, set their expressions equal, and solve for y.

Traps

  • Trusting the picture: The figure suggests something it does not state; work from the given facts.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radii of a circle are equal
r1=r2r_1 = r_2
A radius is a segment from the center of a circle to its curved edge. r₁ and r₂ stand for any two radii of the same circle. They are always equal, because every point on the curve is the same distance from the center.

  1. Read the figure

    The shape is a quarter circle, one fourth of a whole circle. Its center is the corner with the dot. A radius is a segment from the center to the curved edge. The slanted segment, labeled x+1x+1, goes from the center to the curve. The bottom edge is split into two parts, yy and 2.

  2. Spot the two radii

    The slanted segment is a radius, so the radius is x+1x+1. The whole bottom edge also runs from the center to the curve. Its length is y+2y+2, so it is a radius too.

    Watch out: the slanted radius only looks like it is at 45 degrees. Do not assume the triangle’s two short sides are equal. That idea leads to choice D.

  3. Set the radii equal

    Use the rule that all radii of a circle are equal.

    y+2=x+1y + 2 = x + 1

    Now you have one equation with xx and yy.

  4. Solve for y

    Get yy alone by taking 2 from both sides.

    y=x+1−2=x−1y = x + 1 - 2 = x - 1

    Watch out: yy is only part of the bottom radius. Setting yy equal to x+1x+1 gives choice B. Dividing by 2 instead of subtracting 2 gives choice C.

  5. Give the answer

    yy is 2 less than the radius, and the radius is x+1x+1. So the answer is A, x−1x-1.

Check your answer

Pick a number. If x=5x = 5, the radius is 6. Then y=5−1=4y = 5 - 1 = 4, and 4+2=64 + 2 = 6, which matches the radius. Also, yy is a short side of the right triangle. It should be shorter than the slanted side, and 4 is less than 6.

More questions like this

The original DOE question
Question 100 as printed on page 161 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 100. Open page 161 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 37 of 50

2026 Practice Test B Math, question 37

Question

A number line with eight evenly spaced hash marks. Point M is on the second mark, labeled −1/4, and point T is on the seventh mark, labeled 5/8. Point R, with no label, is on the fifth mark, three marks to the right of M.

The hash marks on the number line above are evenly spaced. What is the coordinate of point RR?

  • A740\frac{7}{40}
  • B940\frac{9}{40}
  • C1140\frac{11}{40}
  • D2140\frac{21}{40}
Show the answer

Correct answer: C

There are 5 sections between M and T. To find the length of one of these sections, find the distance between M and T and divide by 5:

(58−(−14))÷51=\left(\frac{5}{8}-\left(-\frac{1}{4}\right)\right)\div\frac{5}{1}=

(58+28)÷51=\left(\frac{5}{8}+\frac{2}{8}\right)\div\frac{5}{1}=

78÷51=78×15=740\frac{7}{8}\div\frac{5}{1}=\frac{7}{8}\times\frac{1}{5}=\frac{7}{40}

R is 3 sections away from M, so add:

−14+3(740)=−1040+2140=1140-\frac{1}{4}+3(\frac{7}{40})=-\frac{10}{40}+\frac{21}{40}=\frac{11}{40}

R is located at 1140\frac{11}{40} .

Question analysis

Students find the fraction at point R on a number line with evenly spaced hash marks, where M is at −1/4 and T is at 5/8. It tests fraction arithmetic with negative numbers. Working out the size of one gap and counting gaps from M to R takes several careful steps.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students place fractions and negatives on number lines, but here they must subtract a negative fraction, divide by five, and then add back. Four fraction steps with unlike denominators make it harder than typical classwork.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 6, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Fractions, a negative starting point and a count of spaces all have to come out right together. Students lose points by counting tick marks instead of spaces, by subtracting −1/4 as if it were positive, or by measuring R from 0 instead of from M. Converting both labels to fortieths first, −10/40 and 25/40, makes each gap a whole number of fortieths.

  1. Error-prone arithmetic major Working it out

    The gap length needs 5/8 − (−1/4) = 7/8, then 7/8 ÷ 5 = 7/40, then −10/40 + 21/40: negatives and three different denominators meet in one problem. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The hash marks have to be counted as spaces: five from M to T, with R three spaces past M, while the unlabeled marks outside M and T play no part. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the distance from M to T by subtracting their coordinates, divide it by the number of equal gaps, then step from M toward T by the right number of gaps.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
distance=right number−left number\text{distance} = \text{right number} - \text{left number}
The right number is the point farther to the right on the number line. The left number is the other point. Subtracting tells how many units apart they are. Taking the left from the right keeps the answer positive.
Dividing a fraction by a whole number
ab÷c=ab×c\frac{a}{b} \div c = \frac{a}{b \times c}
a over b is the fraction, and c is the whole number you divide by. Splitting a fraction into c equal parts makes each part c times smaller. So you multiply the bottom number, b, by c.

  1. Read the number line

    M is at −14-\frac{1}{4} and T is at 58\frac{5}{8}. The hash marks are evenly spaced, so every space is the same length. Count the spaces. From M to T there are 5 spaces. From M to R there are 3 spaces. You need the number at point R.

  2. Find the distance from M to T

    Use the distance formula. First write −14-\frac{1}{4} in eighths, as −28-\frac{2}{8}.

    58−(−28)=58+28=78\frac{5}{8} - \left(-\frac{2}{8}\right) = \frac{5}{8} + \frac{2}{8} = \frac{7}{8}

    Taking away a negative is the same as adding. M and T are 78\frac{7}{8} apart.

  3. Find the length of one space

    The 5 spaces are equal, so divide by 5.

    78÷5=78×5=740\frac{7}{8} \div 5 = \frac{7}{8 \times 5} = \frac{7}{40}

    Each space is 740\frac{7}{40} long.

    Watch out: this is the length of one space, not where R is. Stopping here gives choice A.

  4. Step from M to R

    R is 3 spaces to the right of M.

    3×740=21403 \times \frac{7}{40} = \frac{21}{40}

    Now add that to M. Write −14-\frac{1}{4} as −1040-\frac{10}{40}.

    −1040+2140=1140-\frac{10}{40} + \frac{21}{40} = \frac{11}{40}

    Watch out: 2140\frac{21}{40} is how far R is from M, not where R is. Choosing it gives choice D.

  5. Give the answer

    Point R is at 1140\frac{11}{40}, a little to the right of 0. So the answer is C, 1140\frac{11}{40}.

Check your answer

Go backward from T instead. T is 2 spaces past R, and 58=2540\frac{5}{8} = \frac{25}{40}. So R is at 2540−1440=1140\frac{25}{40} - \frac{14}{40} = \frac{11}{40}. Both ways give the same point.

More questions like this

The original DOE question
Question 101 as printed on page 161 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 101. Open page 161 of the handbook (PDF)

Question 38 of 50

2026 Practice Test B Math, question 38

Question

Phan chose an Internet service that charges $18.00 per month plus $0.024 per minute. Deion chose an Internet service that charges $30.00 per month for unlimited usage. At the end of the month, Phan’s and Deion’s charges were identical. For how many minutes did Phan use the Internet service that month?

  • A50
  • B60
  • C100
  • D500
Show the answer

Correct answer: D

Let x be the number of minutes Phan used his internet service in the month. Phan’s monthly charges were

18 + 0.024x

Since Deion’s charges were the same as Phan’s, set the expression equal to 30 and solve for x:

18 + 0.024x = 30

0.024x = 12

x = 500

Phan used his service for 500 minutes.

Question analysis

Phan pays $18 a month plus 2.4 cents per minute for Internet service, Deion pays a flat $30, and their bills come out the same. Students find how many minutes Phan used. It tests writing and solving a two-step linear equation, and the small decimal rate makes the division error-prone.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving equations of the form px + q = r from a word problem is central grade 7 work. This is a typical example; only the three-decimal-place rate adds some care to the division.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Setting up the equation is routine, so the question is mostly a test of decimal division. Dividing 12 by 0.024 invites a misplaced decimal point, and one such slip gives a choice exactly ten times too small. A quick check, 500 × 0.024 = 12, confirms the answer in seconds.

  1. Error-prone arithmetic moderate Working it out

    The last step divides 12 by 0.024, a three-place decimal, and moving the decimal point one place too few gives 50 instead of 500. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    "Phan's and Deion's charges were identical" has to become 18 + 0.024m = 30, with the flat $30 on one side and Phan's fee plus per-minute cost on the other. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Set Phan's cost, the monthly fee plus the per-minute charge times the minutes, equal to Deion's flat fee, then subtract and divide, checking the decimal places.

Traps

  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost with a monthly fee
C=f+r×mC = f + r \times m
C is the total cost for the month. f is the fixed fee you pay no matter what. r is the price for each minute, and m is the number of minutes. Use it when a bill has a set fee plus a charge for each unit.

  1. Sort out what you know

    Phan pays $18.00 a month plus $0.024 for each minute. Deion pays $30.00 a month, however much he uses. Their bills were the same. You need the number of minutes Phan used. Call it mm.

  2. Write an equation

    Use the cost formula for Phan. His bill equals Deion’s $30.

    18+0.024×m=3018 + 0.024 \times m = 30

    An equation says two amounts are equal. Here it says the two bills are equal.

  3. Take away the monthly fee

    Subtract 18 from both sides. That leaves the minutes part alone.

    0.024×m=30−18=120.024 \times m = 30 - 18 = 12

    So Phan paid $12 for his minutes.

  4. Divide to find the minutes

    Divide 12 by 0.024. To get rid of the decimal, move the point 3 places right in both numbers.

    m=12÷0.024=12,000÷24=500m = 12 \div 0.024 = 12{,}000 \div 24 = 500

    Phan used 500 minutes.

    Watch out: 0.024 is not 0.24. Using 0.24 gives 12÷0.24=5012 \div 0.24 = 50, choice A. Move the point the same number of places in both numbers.

  5. Give the answer

    Phan used 500 minutes that month. That is the only choice that makes the two bills equal. So the answer is D, 500.

Check your answer

Put 500 back in. The minutes cost 500×0.024=12500 \times 0.024 = 12 dollars. Add the fee: 18+12=3018 + 12 = 30 dollars. That is Deion’s bill, so the bills match.

More questions like this

The original DOE question
Question 102 as printed on page 161 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 102. Open page 161 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 39 of 50

2026 Practice Test B Math, question 39

Question

In a sample of 50 cars at a local dealership, there are 12 red cars and 10 cars with backup cameras. Of the 12 red cars, 4 have backup cameras. If a car is selected at random from the given sample, what is the probability that both of the following are true: the car is not red and does not have a backup camera?

  • A35\frac{3}{5}
  • B1625\frac{16}{25}
  • C1925\frac{19}{25}
  • D45\frac{4}{5}
Show the answer

Correct answer: B

Create a chart using the given information and use subtraction to figure out how many cars are not red and do not have a back-up camera:

Red Not Red TOTAL
Back-up Camera 4 6
(10-4)		 10
No back-up Camera 32
(38-6)
TOTAL 12 38
(50-12)		 50

The probability of selecting a car that meets both conditions from the total of 50 cars at the dealership is 3250=1625\frac{32}{50}=\frac{16}{25} .

Question analysis

In a sample of 50 cars, 12 are red, 10 have backup cameras, and 4 of the red cars have cameras. Students find the probability that a randomly chosen car is neither red nor equipped with a camera. It tests compound probability with overlapping groups, and the overlap between red cars and camera cars is what makes it tricky.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 compound probability usually uses a ready-made table or tree. Here students must build the two-way table from words, handle an overlap, and use a "not this and not that" condition, several steps beyond typical work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 6, 63 words.
Corrected from the DOE handbook
  • A grammar slip in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is an overlap problem dressed as probability: the hard part is counting the cars that avoid both features, not the division by 50. A two-way table makes it routine, but students working in their heads often keep only one of the two conditions. The question rewards placing the 4 red cars with cameras in the table before doing any subtraction.

  1. Much to keep track of major Working it out

    Three overlapping counts, 12 red, 10 with cameras and 4 that are both, must be sorted into four groups before the wanted one, neither red nor with a camera, can be found. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Many steps moderate Working it out

    Find the non-red cars with cameras, 10 − 4 = 6, and the non-red cars, 50 − 12 = 38, subtract to get 32, then simplify 32/50. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer moderate Choosing the answer

    19/25, or 38/50, has the same denominator as the right 16/25 and sits just above it; only removing the 6 non-red cars that have cameras from the 38 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  4. Complex sentences minor Understanding the question

    The last sentence stacks "both of the following are true", a colon and two negatives, "not red" and "does not have"; untangling it is reading work, not extra math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Draw a two-way table with red and not red across, camera and no camera down, fill in the given counts, and complete the missing cells by subtraction.

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says “a car that meet both conditions”; the site says “a car that meets both conditions.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number you wanttotal numberP = \frac{\text{number you want}}{\text{total number}}
P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

  1. Sort out what you know

    There are 50 cars. Of these, 12 are red and 10 have backup cameras. Of the red cars, 4 have cameras. You pick one car at random. You need the chance that “both of the following are true”: the car is not red, and it has no camera.

  2. Set up a two-way table

    A two-way table sorts each car by two things at once: its color and its camera. Put in the numbers you know.

    RedNot redTotal
    Camera4?10
    No camera?
    Total12?50

  3. Fill in the missing numbers

    Not red with a camera: 10−4=610 - 4 = 6 cars.

    All cars that are not red: 50−12=3850 - 12 = 38 cars.

    Not red with no camera: 38−6=3238 - 6 = 32 cars.

    Watch out: the 38 cars that are not red include 6 with cameras. Using 38 gives 3850=1925\frac{38}{50} = \frac{19}{25}, choice C. Using all 40 cars with no camera gives 4050=45\frac{40}{50} = \frac{4}{5}, choice D.

  4. Write the probability

    Use the probability formula. There are 32 cars you want out of 50 cars in all.

    P=3250P = \frac{32}{50}

    Divide the top and bottom by 2.

    3250=1625\frac{32}{50} = \frac{16}{25}

    So the answer is B, 1625\frac{16}{25}.

Check your answer

Finish the table. Red cars with no camera: 12−4=812 - 4 = 8. Now add the four inside boxes: 4+6+8+32=504 + 6 + 8 + 32 = 50. That is every car, so the numbers fit. Also, the cars with no camera are 8+32=408 + 32 = 40, and 50−10=4050 - 10 = 40 too.

More questions like this

The original DOE question
Question 103 as printed on page 162 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 103. Open page 162 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 40 of 50

2026 Practice Test B Math, question 40

Question

147−x12=12\frac{147-x}{12}=12

What is the value of xx in the equation shown above?

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

Show the answer

Correct answer: 3

147−x12=12\frac{147-x}{12}=12

147−x=144147-x=144

x=3x=3

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 This is exactly the two-step equation practice of grade 7: undo a division, then undo a subtraction. The only twist is that the unknown is being subtracted, so the last step needs care with its sign.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 12 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

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

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

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Read the equation in words

    The equation says: start with 147, take away a mystery number x, then divide by 12. The result is 12.

    You need to find x. To get x alone, undo the steps in reverse order. Dividing by 12 was done last, so undo it first.

  2. Undo the division by 12

    Multiply both sides by 12. Whatever you do to one side, you must do to the other. That keeps the equation balanced.

    147−x12=12\frac{147 - x}{12} = 12

    147−x=12×12=144147 - x = 12 \times 12 = 144

    Watch out: Do not divide 12 by 12. That gives 147 − x = 1 and the wrong answer 146.

  3. Find what was taken away

    Now ask: 147 minus what number leaves 144? The gap between the two numbers is x.

    x=147−144=3x = 147 - 144 = 3

    Watch out: Do not work out 144 − 147. That gives −3. Since 147 is bigger than 144, x must be positive.

    So you would grid 3.

Check your answer

Put 3 back in for x. First, 147−3=144147 - 3 = 144. Then 144÷12=12144 \div 12 = 12. That matches the right side of the equation, so x = 3 is correct.

More questions like this

The original DOE question
Question 59 as printed on page 153 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 59. Open page 153 of the handbook (PDF)

Question 41 of 50

2026 Practice Test B Math, question 41

Question

Coordinate axes with origin O. A shaded triangle stands on the x-axis: its base runs from the point marked m to the point marked n, both to the right of O, and its top vertex touches a dashed horizontal line drawn at 4 on the y-axis.

What is the area of the shaded triangle shown above?

  • Am+nm+n
  • Bn−mn-m
  • C2(n−m)2\left ( n-m \right )
  • D4(n−m)4\left ( n-m \right )
Show the answer

Correct answer: C

The height of the triangle is 4 units. The length of the base is n−mn-m. So the area is

A=12(n−m)(4)=2(n−m)A=\frac{1}{2}(n-m)(4)=2(n-m)

Question analysis

Students find the area of a shaded triangle drawn on coordinate axes, with its base between the points m and n on the x-axis and its top vertex at a height of 4. It tests the triangle area formula with an algebraic side length. Writing the base as an expression in m and n is the step that needs care.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Triangle area is grade 6 content, but a base written with two variables is less familiar. With only two steps, the demand matches typical grade 7 work that mixes geometry with algebraic expressions.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 7, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is trivial, but students must treat a length as a difference of coordinates and keep a formula's one half when the answer is a letter expression. Those who forget the half, or who stop at the base length, find their result among the choices. Trying m = 1 and n = 4 gives a triangle of area 6 that checks the expressions quickly.

  1. Variables instead of numbers moderate Understanding the question

    The base runs from m to n on the x-axis, so its length is the expression n − m, and the area stays in letters with no number to check. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The height, 4, is read from the dashed line through the apex rather than labeled on the triangle, and the base endpoints appear only as tick labels. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. A very close wrong answer moderate Choosing the answer

    4(n − m) is the right 2(n − m) doubled, written the same way except for the coefficient; only remembering the one half in a triangle's area separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the base as the difference between its endpoint values, read the height from the figure, and apply one-half times base times height.

Traps

  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height, the straight-up distance from the base to the top corner. You take half because a triangle is half of a rectangle with the same base and height.
Length along the x-axis
length=right end−left end\text{length} = \text{right end} - \text{left end}
The right end and the left end are the numbers where a segment starts and stops on the x-axis. Subtracting the smaller from the larger tells how long the segment is.

  1. Read the figure

    The shaded triangle sits on the x-axis, the flat number line. Its base runs from mm to nn. Its top corner touches the dashed line at height 4. You need the area, written with the letters mm and nn.

  2. Find the length of the base

    The base starts at mm and ends at nn. Use the length rule: right end minus left end.

    base=n−m\text{base} = n - m

    Watch out: adding the ends, m+nm+n, does not give a length. That is choice A.

  3. Find the height

    The height is the straight-up distance from the base to the top corner. The top corner is on the dashed line, which is at 4 on the y-axis. So the height is 4.

  4. Find the area

    Use the triangle area formula.

    A=12×(n−m)×4=2(n−m)A = \frac{1}{2} \times (n - m) \times 4 = 2(n - m)

    Half of 4 is 2, so the area is 2(n−m)2(n-m).

    Watch out: forgetting the one half gives 4(n−m)4(n-m), choice D. Stopping at the base gives n−mn-m, choice B.

  5. Give the answer

    The area is 2 times the length of the base. So the answer is C, 2(n−m)2(n-m).

Check your answer

Try numbers. Say m=1m = 1 and n=4n = 4. The base is 3 and the height is 4, so the area is 12×3×4=6\frac{1}{2} \times 3 \times 4 = 6. Choice C gives 2(4−1)=62(4-1) = 6. It matches. Choice D would give 12.

More questions like this

The original DOE question
Question 105 as printed on page 162 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 105. Open page 162 of the handbook (PDF)

Question 42 of 50

2026 Practice Test B Math, question 42

Question

ANIMAL CARDS

Number of Cards Picture on Card
8 cat
6 dog
5 bird
4 fish
1 horse

The cards in the table above are mixed in a box. Which animal pictured on a card has exactly a 1 in 4 chance of being picked at random from the box?

  • Acat
  • Bdog
  • Cfish
  • Dhorse
Show the answer

Correct answer: B

The total number of cards in the box is

8 + 6 + 5 + 4 + 1 = 24

Set up a proportion to figure out which card has exactly a 1 in 4 chance of being picked at random.

x24=14\frac{x}{24}=\frac{1}{4}

x=6x=6

The dog card has a 1 in 4 chance of being randomly selected.

Question analysis

A table lists picture cards for five animals, from 8 cat cards down to 1 horse card, all mixed in one box. Students decide which animal has exactly a 1 in 4 chance of being drawn at random. It tests simple probability as a fraction of the total, and the numbers in the table invite matching the wrong count.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Simple probability from a list of counts is grade 6 work. This takes one sum and one division with small whole numbers, a shorter and more routine task than typical grade 7 probability problems.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 1, 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

Once the 24 cards are totaled, one fourth of them is 6 and the answer is a lookup. The wrong choices mostly attract students who match the numbers 1 and 4 in the phrase to the horse and fish rows, or who assume the most common card is the one asked about. Little time is needed, so the main risk is skipping the total.

  1. Turning words into math minor Understanding the question

    "Exactly a 1 in 4 chance" has to be turned into a count, one fourth of the 24 cards, rather than matched to a 1 or a 4 in the table. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The counts sit in the table's first column with the animals in the second, and the total of 24 has to be added up from all five rows. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add the cards to find the total in the box, find one-fourth of that total, and look for the animal with exactly that many cards.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number you wanttotal numberP = \frac{\text{number you want}}{\text{total number}}
P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

  1. Sort out the question

    The table shows 8 cat cards, 6 dog, 5 bird, 4 fish and 1 horse. All the cards are mixed in a box. You need the animal with “exactly a 1 in 4 chance” of being picked. A 1 in 4 chance means a probability of 14\frac{1}{4}.

  2. Find the total number of cards

    Add every row of the table. This total is the bottom of the probability fraction.

    8+6+5+4+1=248 + 6 + 5 + 4 + 1 = 24

    There are 24 cards in the box.

  3. Find one fourth of the cards

    The animal you want must have one fourth of all the cards.

    24÷4=624 \div 4 = 6

    So you need the animal with exactly 6 cards. The dog has 6 cards.

    P=624=14P = \frac{6}{24} = \frac{1}{4}

  4. Rule out the other choices

    Watch out: do not match the numbers 1 and 4 to the table. Fish has 4 cards, but its chance is 424=16\frac{4}{24} = \frac{1}{6}. So C is wrong. Horse has 1 card, a chance of 124\frac{1}{24}, so D is wrong. Cat has 824=13\frac{8}{24} = \frac{1}{3}, so A is wrong.

  5. Give the answer

    Only the dog has exactly one fourth of the cards. So the answer is B, dog.

Check your answer

Four groups of 6 cards make 24, the whole box. So 6 cards is one of four equal shares. That is a 1 in 4 chance. The dog cards are exactly one of those shares.

More questions like this

The original DOE question
Question 106 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 106. Open page 163 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 43 of 50

2026 Practice Test B Math, question 43

Question

Which number line below shows the solution set for 2x−2≤y≤4x+102x-2\leq y\leq 4x+10 when y=1y=1?

  • AA number line from −3 to 3 marked in quarter units, with a solid segment from −2 1/4 to −1 1/2; both ends are closed dots.
  • BA number line from −3 to 3 marked in quarter units, with a solid segment from −1 1/2 to 2 1/4; both ends are closed dots.
  • CA number line from −3 to 3 marked in quarter units, with a solid segment from −2 1/4 to 1 1/2; both ends are closed dots.
  • DA number line from −3 to 3 marked in quarter units, with a closed dot at 1 1/2 and a solid ray from it running left past −2 1/2 to an arrowhead.
Show the answer

Correct answer: C

Separate the compound inequality into two pieces:

2x−2≤y2x-2\leq y and y≤4x+10y\leq 4x+10

Substitute y=1y=1 into each inequality and solve for xx:

2x−2≤12x-2\leq1

2x≤32x\leq 3

x≤32x\leq\frac{3}{2}

1≤4x+101\leq 4x+10

−9≤4x-9\leq4x

−94≤x-\frac{9}{4}\leq x

The solution is the number line that shows that xx is greater than or equal to −214-2\frac{1}{4} and less than or equal to 1121\frac{1}{2} .

Question analysis

Students substitute y = 1 into the compound inequality 2x − 2 ≤ y ≤ 4x + 10 and pick the number line that shows the values of x that work. It tests solving two-step inequalities and graphing the solution. The four graphs differ in their endpoints and in whether the shading is closed off on both sides, so each endpoint has to be found exactly.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph one two-step inequality. This compound inequality in two variables requires substituting, solving two inequalities and combining them into one interval, a step toward algebra 1.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 9, 12 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Substituting y = 1 is easy; the work is solving two inequalities and matching a fractional interval to a graph. Students who solve only the left inequality find a matching ray, and sign slips on either endpoint produce intervals that shift or flip across zero. The graphs differ by quarter-unit ticks, so reading them carefully matters as much as the algebra.

  1. Many steps moderate Working it out

    The compound inequality must be split into 2x − 2 ≤ 1 and 1 ≤ 4x + 10, each solved for x, and the two results joined into one interval. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic moderate Working it out

    Both endpoints are fractions, 3/2 and −9/4, and the second comes from dividing −9 by 4, where a lost sign moves the endpoint across zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Reading a figure, table or graph moderate Understanding the question

    The graphs are marked in quarter units, so −2 1/4 and 1 1/2 have to be found by counting small ticks between the labeled integers. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  4. A very close wrong answer minor Choosing the answer

    The ray ending at 1 1/2 looks almost like the right segment, sharing its right end; only the left end, an arrow instead of a closed dot at −2 1/4, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Split the compound inequality into two separate inequalities, solve each for x after putting 1 in for y, then combine the two results into a single interval and match it to a graph.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Solving an inequality
if a≤b, then a+c≤b+c\text{if } a \leq b, \text{ then } a + c \leq b + c
a, b and c are numbers. The sign ≤ means is less than or equal to. Adding or subtracting the same number c on both sides keeps it true. Dividing both sides by the same positive number also keeps it true.

  1. Sort out the question

    The sign ≤\leq means is less than or equal to. An inequality compares two amounts that may not be equal. The statement 2x−2≤y≤4x+102x-2 \leq y \leq 4x+10 is two inequalities joined together. You put 1 in place of yy. Then you need every xx that makes both parts true.

  2. Split it into two parts

    With y=1y = 1, the statement becomes two inequalities.

    2x−2≤12x - 2 \leq 1

    1≤4x+101 \leq 4x + 10

    The answer must make both of them true at the same time.

  3. Solve the first part

    Add 2 to both sides. Then divide both sides by 2.

    2x≤32x \leq 3

    x≤32x \leq \frac{3}{2}

    So xx is at most 1121\frac{1}{2}. This is the right end of the answer.

    Watch out: the right end is a positive number. Choice A ends its bar at −112-1\frac{1}{2} instead.

  4. Solve the second part

    Subtract 10 from both sides. Then divide both sides by 4.

    −9≤4x-9 \leq 4x

    −94≤x-\frac{9}{4} \leq x

    So xx is at least −214-2\frac{1}{4}. This is the left end.

    Watch out: keep the minus sign. Choice B uses the opposite signs and ends at 2142\frac{1}{4}.

  5. Put the two parts together

    The answer is every number from −214-2\frac{1}{4} to 1121\frac{1}{2}, with both ends included. On the graph, that is a solid bar with a dot at each end. The small ticks are quarters. So count 1 tick left of −2 and 2 ticks right of 1.

    Watch out: choice D uses only the first part. Its arrow goes left forever.

  6. Give the answer

    Choice C shows a bar from −214-2\frac{1}{4} to 1121\frac{1}{2}, with dots at both ends. So the answer is C.

Check your answer

Test a number on the bar, like x=0x = 0. Then 2(0)−2=−22(0)-2 = -2 and 4(0)+10=104(0)+10 = 10, and 1 is between them. Now test x=−3x = -3, which is off the bar. Then 4(−3)+10=−24(-3)+10 = -2, and 1 is not less than −2. So the bar must stop on the left, which rules out D.

More questions like this

The original DOE question
Question 107 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 107. Open page 163 of the handbook (PDF)

Question 44 of 50

2026 Practice Test B Math, question 44

Question

1421=p7\frac{14}{21}=\frac{p}{7}

In the equation above, what is the value of pp?

  • A23\frac{2}{3}
  • B33
  • C143\frac{14}{3}
  • D1414
Show the answer

Correct answer: C

1421=p7\frac{14}{21}=\frac{p}{7}

21p=7(14)21p=7(14)

21p=9821p=98

p=9821=143p=\frac{98}{21}=\frac{14}{3}

Question analysis

Students solve a proportion that sets 14/21 equal to a fraction with an unknown numerator p over 7. It tests solving a proportion, either by cross-multiplying or by dividing numerator and denominator by the same factor. It is quick, and the main risk is losing track of what p stands for.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Solving a one-step proportion is grade 6 and 7 ratio work. This one is purely numerical with one or two steps, shorter than the multistep ratio word problems typical of grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 10, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is standard, but the result is not a whole number, and that surprises students who expect the numbers in a proportion to scale evenly. Two intermediate values, the ratio 2/3 and the scale factor 3, appear among the choices. Students who finish the computation and trust a fractional answer get it right.

  1. An answer in an unexpected form moderate Choosing the answer

    The answer is the improper fraction 14/3, not a whole number, so students who expect a tidy proportion distrust their result or reach for the factor 3. (Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Cross-multiplying gives 21p = 98, and 98/21 must be reduced by 7; simplifying 14/21 to 2/3 first and multiplying by 7 is quicker. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Cross-multiply to form an equation in p, divide both sides by p's coefficient, and simplify the fraction you get.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross multiplication
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
a and c are the top numbers. b and d are the bottom numbers. When two fractions are equal, the top of each times the bottom of the other gives the same answer. It works because you multiply both sides by b and by d.

  1. Sort out the question

    Two fractions are equal: 1421\frac{14}{21} and p7\frac{p}{7}. An equation like this, with two equal fractions, is called a proportion. You need the value of pp.

  2. Cross multiply

    Use cross multiplication. This gets rid of the fractions. Multiply 21 by pp, and multiply 14 by 7.

    21×p=14×7=9821 \times p = 14 \times 7 = 98

  3. Divide to find p

    Divide both sides by 21 to get pp alone.

    p=9821p = \frac{98}{21}

    Both 98 and 21 can be divided by 7.

    98÷721÷7=143\frac{98 \div 7}{21 \div 7} = \frac{14}{3}

  4. Make sense of the fraction

    The answer is 143\frac{14}{3}, which is 4234\frac{2}{3}. It is not a whole number, and that is fine.

    Watch out: 23\frac{2}{3} is the value of 1421\frac{14}{21}, not pp. That is choice A. The 3 is what 21 was divided by to get 7. That gives choice B.

  5. Give the answer

    The top number over 7 must be 143\frac{14}{3} to keep the two fractions equal. So the answer is C, 143\frac{14}{3}.

Check your answer

Think about the bottoms. 7 is one third of 21. So pp must be one third of 14, which is 143\frac{14}{3}. Check it: 143÷7=1421\frac{14}{3} \div 7 = \frac{14}{21}. Both ways agree.

More questions like this

The original DOE question
Question 108 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 108. Open page 163 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 45 of 50

2026 Practice Test B Math, question 45

Question

A ball is selected at random from a box that contains 7 black balls, 14 green balls, and 21 red balls. What is the probability that the ball selected is black?

  • A16\frac{1}{6}
  • B15\frac{1}{5}
  • C13\frac{1}{3}
  • D56\frac{5}{6}
Show the answer

Correct answer: A

The total number of balls in the box is

7+14+21=427+14+21=42

The probability that the ball is black is

742=16\frac{7}{42}=\frac{1}{6}

Question analysis

Students find the probability of drawing a black ball at random from a box with 7 black, 14 green and 21 red balls. It tests simple probability as a part of a whole. It is a short question, but the denominator must be chosen with care.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a simple probability from given counts is grade 6 work in New York. This one is a single fraction to simplify, a routine task well below the compound-event problems of grade 7.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 4, 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

This is a one-formula probability question with friendly numbers, since 7 divides 42 evenly. The wrong answers come from comparing black balls with the other balls only, from treating the three colors as equally likely, or from finding the chance of not drawing black. Students who write the total first rarely miss it.

  1. Something you must already know minor Working it out

    The probability of an outcome, favorable balls over all balls, is assumed; nothing in the stem says to add 7 + 14 + 21 for the denominator. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer minor Choosing the answer

    1/5 and the right 1/6 are neighboring unit fractions; only counting the 7 black balls in the denominator, 7/42 rather than 7/35, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Add all the balls to get the total, put the number of black balls over that total, and simplify the fraction.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number you wanttotal numberP = \frac{\text{number you want}}{\text{total number}}
P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

  1. Sort out what you know

    The box has 7 black balls, 14 green balls and 21 red balls. One ball is picked at random. That means every ball has the same chance. You need the probability, or chance, that it is black.

  2. Find the total number of balls

    The bottom of the fraction is every ball in the box.

    7+14+21=427 + 14 + 21 = 42

    There are 42 balls in all.

  3. Write and simplify the probability

    Use the probability formula. There are 7 black balls out of 42.

    P=742=7÷742÷7=16P = \frac{7}{42} = \frac{7 \div 7}{42 \div 7} = \frac{1}{6}

    Watch out: comparing black with the other 35 balls gives 735=15\frac{7}{35} = \frac{1}{5}, choice B. Saying 1 of 3 colors gives 13\frac{1}{3}, choice C. But the colors have different numbers of balls.

  4. Give the answer

    The chance of picking a black ball is 1 out of 6. So the answer is A, 16\frac{1}{6}.

Check your answer

Notice that 14 is 2 sevens and 21 is 3 sevens. So for every 1 black ball there are 2 green and 3 red. That is 1 black out of every 6 balls. Choice D, 56\frac{5}{6}, is the chance of not black.

More questions like this

The original DOE question
Question 109 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 109. Open page 163 of the handbook (PDF)

Question 46 of 50

2026 Practice Test B Math, question 46

Question

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

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

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

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

Show the answer

Correct answer: y = 20x + 50

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

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

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Amiya pays two kinds of charges:

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

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

  2. Work out a few totals

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

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

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

  3. Match each number to its job

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

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

  4. Write the equation

    Put the rate and the fee into the model:

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

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

  5. Enter it on the keypad

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

    So you would enter y = 20x + 50.

Check your answer

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

More questions like this

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

Question 47 of 50

2026 Practice Test B Math, question 47

Question

Raoul is at least 3 years older than Vahn. Which of the following inequalities gives the relationship between Raoul’s age (rr ) and Vahn’s age ( vv )?

  • Ar−v≥3r-v\geq 3
  • Br−v≤3r-v\leq 3
  • C3−v≤r3-v\leq r
  • D3−r≤v3-r\leq v
Show the answer

Correct answer: A

Raoul is at least 3 years older than Vahn, which can be written as

r ≥ v + 3

Rewrite this inequality to match the answer options:

r − v ≥ 3

Question analysis

Raoul is at least 3 years older than Vahn, and students choose the inequality that links their ages, r and v. It tests translating a phrase like "at least" into an inequality. The correct form may not match the one students write first, so they need to rearrange carefully and keep the direction of the inequality.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.8 writing simple inequalities
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing inequalities from words is grade 6 and 7 work. The extra demand is that "at least" must become the right symbol and the choices are rearranged forms, so students must manipulate inequalities rather than just recognize one.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 8, 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 skills are tested at once: translating "at least … older than" into an inequality, and recognizing that inequality after a rearrangement. The two choices that begin with 3 − v and 3 − r look like legitimate moves but say only that the two ages add up to at least 3. Testing one pair of ages that fits the sentence and one that does not, such as 20 and 12 against 13 and 12, sorts the choices without algebra.

  1. Turning words into math major Understanding the question

    "Raoul is at least 3 years older than Vahn" has to become r ≥ v + 3; "at least" calls for ≥, though the word "least" seems to point the other way. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    No choice is written as r ≥ v + 3, so the natural translation has to be rearranged to r − v ≥ 3 before it can be matched. (Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer moderate Choosing the answer

    r − v ≤ 3 is the right r − v ≥ 3 with the symbol reversed; only deciding that "at least" means greater than or equal to separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the sentence directly as Raoul's age compared with Vahn's age plus 3, then rearrange each choice into that form to see which one matches.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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.
  • Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greater than or equal to
a≥ba \geq b
a and b are two amounts. The sign ≥ means a is greater than or equal to b, so a is b or more. The words at least mean the same thing. The sign ≤ means less than or equal to, or at most.

  1. Sort out the question

    The letter rr is Raoul’s age, and vv is Vahn’s age. Raoul is “at least 3 years older than Vahn”. That means the gap between their ages is 3 years or more. An inequality is a math sentence that uses a sign like ≥\geq instead of =. You need the one that says this.

  2. Write the age gap

    Raoul is older, so the gap is his age minus Vahn’s age. The gap is r−vr - v. For example, ages 15 and 10 have a gap of 5.

  3. Turn at least into a sign

    At least 3 means 3 or more. So use ≥\geq, which means greater than or equal to.

    r−v≥3r - v \geq 3

    This is choice A.

    Watch out: the word least does not mean less. Choice B, r−v≤3r - v \leq 3, says the gap is 3 or less. Then Raoul could even be younger.

  4. Check choices C and D

    Add vv to both sides of choice C. You get 3≤r+v3 \leq r + v. That is about the sum of their ages, not the gap. Choice D works the same way. Neither one says Raoul is older.

  5. Give the answer

    Only one choice says the gap is 3 years or more. So the answer is A, r−v≥3r-v \geq 3.

Check your answer

Test ages that do not fit: Raoul 11 and Vahn 10. He is only 1 year older. Choice A gives 11−10=111 - 10 = 1, and 1 is not 3 or more. So A rightly says no. Choice B says yes, so B is wrong.

More questions like this

The original DOE question
Question 111 as printed on page 164 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 111. Open page 164 of the handbook (PDF)

Question 48 of 50

2026 Practice Test B Math, question 48

Question

There are now xx cans stacked on a shelf that holds 36 cans when full. If 4 of these cans were removed, the shelf would be half full. What is the value of xx ?

  • A14
  • B16
  • C18
  • D22
Show the answer

Correct answer: D

The shelf, when full, holds 36 cans. When the shelf is half full, it holds 18 cans.

x − 4 = 18

x = 22

Question analysis

A shelf holds 36 cans when full, and removing 4 of the cans now on it would leave it half full. Students find how many cans are on the shelf now. It tests writing and solving a one-step equation from words, and the risk is undoing the removal in the wrong direction.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 6 students solve equations like x − p = q, and this one needs only that and halving 36. It is a quick, routine word problem, easier than the two-step equations typical of grade 7.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 3, 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

Short and routine, this question still punishes a quick reading. Students must keep three amounts apart, the 36-can capacity, the 18 cans on a half-full shelf and the x cans there now, and then undo the removal by adding. Writing x − 4 = 18 before computing prevents every wrong choice.

  1. Turning words into math minor Understanding the question

    The 36 cans are the shelf's capacity, not its contents, and "if 4 of these cans were removed" describes a hypothetical, so the equation is x − 4 = 18. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    18 sits just below the right 22, and both are near half of 36; only remembering that x counts the cans before the 4 were removed separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out how many cans make a half-full shelf, then undo the removal of the 4 cans; write it as an equation in x if that helps.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing subtraction
x−a=b means x=b+ax - a = b \text{ means } x = b + a
x is the number you want. a is the amount taken away, and b is what is left. To undo taking away a, add a back. Adding the same number to both sides keeps the equation balanced.

  1. Sort out what you know

    The shelf holds 36 cans when it is full. That is its size, not how many cans are on it now. Right now it has xx cans. If 4 cans were taken off, the shelf would be half full. You need xx.

  2. Find half of a full shelf

    A full shelf holds 36 cans, so half full is half of that.

    36÷2=1836 \div 2 = 18

    A half-full shelf has 18 cans.

  3. Write an equation

    Taking 4 cans away from xx leaves 18 cans.

    x−4=18x - 4 = 18

    Watch out: 18 is the number after the cans are taken off. Picking 18 gives choice C.

  4. Add the 4 cans back

    Undo the subtraction by adding 4 to both sides.

    x=18+4=22x = 18 + 4 = 22

    Watch out: subtracting the 4 again gives 18−4=1418 - 4 = 14, choice A. The cans were taken from xx, so xx is more than 18.

  5. Give the answer

    There are 22 cans on the shelf now, before any are taken off. So the answer is D, 22.

Check your answer

Start with 22 cans and take off 4. That leaves 22−4=1822 - 4 = 18 cans. Half of 36 is 18, so the shelf is half full. The answer works.

More questions like this

The original DOE question
Question 113 as printed on page 164 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 113. Open page 164 of the handbook (PDF)

Question 49 of 50

2026 Practice Test B Math, question 49

Question

Carlos tossed a paper cup in the air 50 times and found that the probability of it landing on its side was 72%. If he tosses the cup in the air 150 more times, what is the total number of times he can expect the cup to land on its side?

  • A72
  • B108
  • C144
  • D158
Show the answer

Correct answer: C

The probability of the cup landing on its side is 72%. Carlos tossed the cup a total of 200 times (50 + 150). The number of times the cup lands on its side is 72% of 200:

0.72 × 200 = 144.

Question analysis

Carlos tossed a paper cup 50 times and found it landed on its side 72% of the time. If he tosses it 150 more times, students predict the total number of side landings he can expect. It tests using a probability to predict outcomes, and the word total decides which tosses to count.

NYS standards
  • 6.SP.7 long-run relative frequency
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Predicting outcomes from a probability is grade 6 content, and the arithmetic is a single percent. What lifts it to grade 7 level is deciding exactly which tosses the word "total" refers to.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 8, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The computation is one multiplication, so the question is decided by what "total" covers. Students who focus on the 150 new tosses answer for part of the experiment, and students who add the original 50 tosses as if every one landed on its side overshoot. Reading the last sentence twice, and noticing the word total, is the safeguard.

  1. Turning words into math moderate Understanding the question

    "150 more times" and "the total number of times" together mean all 200 tosses, so the expected count is 72% of 200, not of 150. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Students must know that an expected count is the probability times the number of trials; the stem never says to multiply. (Anderson & Krathwohl, 2001)

Strategy: Decide which tosses the word "total" covers, then multiply that number of tosses by 0.72 to get the expected count.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Expected count
expected count=P×n\text{expected count} = P \times n
P is the probability, the chance of the result on one try, written as a decimal. n is the number of tries. Multiplying tells how many times you can expect the result. It is a good guess, not a promise.

  1. Sort out what you know

    Carlos tossed the cup 50 times. It landed on its side 72% of the time. He will toss it 150 more times. You need “the total number of times” it should land on its side. Total means all the tosses, the first 50 and the next 150.

  2. Count all the tosses

    Add the first tosses and the new ones.

    50+150=20050 + 150 = 200

    There are 200 tosses in all.

    Watch out: using only the 150 new tosses gives 0.72×150=1080.72 \times 150 = 108, choice B.

  3. Change the percent to a decimal

    Percent means out of 100. So 72% is 72 out of 100. Move the decimal point two places left.

    72%=72100=0.7272\% = \frac{72}{100} = 0.72

  4. Multiply to find the expected count

    Use the expected count formula with 0.72 and 200 tosses.

    0.72×200=1440.72 \times 200 = 144

    He can expect the cup to land on its side 144 times.

    Watch out: 72 is the count for 100 tosses, not 200. That is choice A.

  5. Give the answer

    All 200 tosses count, so the expected total is 144. So the answer is C, 144.

Check your answer

Split it up. In the first 50 tosses, 0.72×50=360.72 \times 50 = 36. In the next 150, 0.72×150=1080.72 \times 150 = 108. Then 36+108=14436 + 108 = 144. Also, 144 is a bit less than three quarters of 200, which is 150. That fits 72%.

More questions like this

The original DOE question
Question 114 as printed on page 164 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 114. Open page 164 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 50 of 50

2026 Practice Test B Math, question 50

Question

Let b=3b = 3, c=4c = 4, and d=5d = 5. Determine the value of cdcb+cd\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}.

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

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

Show the answer

Correct answer: 7/5

Solution:

75\frac{7}{5}

Substitute b=3b = 3, c=4c = 4, and d=5d = 5 in cdcb+cd\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}: 4543+45\frac{\frac{4}{5}}{\frac{4}{3}} + \frac{4}{5}.

Then simplify the complex fraction by multiplying the complex fraction by a form of 1, (3434)\left(\frac{\frac{3}{4}}{\frac{3}{4}}\right): 4543+45=4543(3434)+45\frac{\frac{4}{5}}{\frac{4}{3}} + \frac{4}{5} = \frac{\frac{4}{5}}{\frac{4}{3}}\left(\frac{\frac{3}{4}}{\frac{3}{4}}\right) + \frac{4}{5}.

Simplify after multiplying by a form of 1: (45)(34)(43)(34)+45=351+45\frac{\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)}{\left(\frac{4}{3}\right)\left(\frac{3}{4}\right)} + \frac{4}{5} = \frac{\frac{3}{5}}{1} + \frac{4}{5}.

Then simplify again and add the two fractions: 351+45=35+45=75\frac{\frac{3}{5}}{1} + \frac{4}{5} = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}.

Question analysis

This substitution question gives values for b, c and d and asks for the value of an expression made of a fraction divided by a fraction, plus another fraction. Students substitute, simplify the complex fraction and add, then type the result. Dividing fractions in the stacked form is the main hurdle, and the item wants a fraction, not a decimal.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.NS.1 dividing fractions by fractions
  • 7.NS.2c properties for multiplying and dividing
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students evaluate expressions and divide and add fractions. A fraction stacked over another fraction looks unfamiliar and there are no choices to check, but with these values the work is three short steps.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the values are in, the work is one fraction division and one addition, but the division is printed as a fraction over a fraction, in small type. Multiplying the two fractions, or flipping the top one, gives a sum in fifteenths that looks as finished as the right one. Noticing that the 4s cancel, so the stacked part is simply 3/5, turns the last step into adding fifths.

  1. Error-prone arithmetic moderate Working it out

    Dividing 4/5 by 4/3 means multiplying by 3/4, giving 3/5, which is then added to 4/5; multiplying straight across or flipping the top fraction instead both lead to fifteenths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    Nothing on the page says that a fraction over a fraction is a division, or that dividing by 4/3 means multiplying by its reciprocal, and the stacked form is rare in grade 7 work. (Anderson & Krathwohl, 2001)

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

    The value is typed into a fraction box with no choices to check against, and the item asks for a fraction, so the equal decimal 1.4 is not accepted. (Enright, Morley & Sheehan, 2002)

Strategy: Substitute the values, rewrite the stacked fraction as the top fraction times the reciprocal of the bottom one, where the 4s cancel, then add the last fraction and leave the result as a fraction.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • 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.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Flip only the fraction you divide by. It works because a fraction times its reciprocal is 1, so the reciprocal undoes the fraction, just as dividing does.
Adding fractions with the same bottom
ad+bd=a+bd\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}
d is the denominator, the bottom number, which tells the size of each piece. a and b are the numerators, the top numbers, which count the pieces. When the pieces are the same size, add the tops and keep the bottom.

  1. Sort out the question

    You know b=3b = 3, c=4c = 4 and d=5d = 5. You need the value of:

    cdcb+cd\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}

    The first part is a fraction on top of a fraction. A fraction bar means divide, so this part is cd÷cb\frac{c}{d} \div \frac{c}{b}. The answer must be typed as a fraction.

  2. Put in the numbers

    Replace each letter with its value. Look closely: the bottom fraction has b under the c, not d.

    cd=45cb=43\frac{c}{d} = \frac{4}{5} \qquad \frac{c}{b} = \frac{4}{3}

    So the expression becomes:

    45÷43+45\frac{4}{5} \div \frac{4}{3} + \frac{4}{5}

  3. Divide the two fractions

    Use the rule for dividing fractions. The reciprocal of 4/3 is 3/4. To multiply fractions, multiply the tops and multiply the bottoms:

    45÷43=45×34=1220=35\frac{4}{5} \div \frac{4}{3} = \frac{4}{5} \times \frac{3}{4} = \frac{12}{20} = \frac{3}{5}

    To simplify 12/20, divide its top and bottom by 4.

    Watch out: flip only the bottom fraction. Multiplying without flipping gives 16/15 and a total of 28/15. Flipping the top one gives 5/3 and a total of 37/15.

  4. Add the last fraction

    Both fractions are now fifths, so add the tops and keep the bottom:

    35+45=3+45=75\frac{3}{5} + \frac{4}{5} = \frac{3 + 4}{5} = \frac{7}{5}

    The value is 7/5.

  5. Enter it as a fraction

    The item asks for a fraction. 7/5 equals the decimal 1.4, but 1.4 is not accepted. On the test, the answer box already holds a fraction template, a small box over a small box. Type 7 in the top box. Then click the bottom box and type 5. On this site, type 7/5.

    So you would enter 7/5.

Check your answer

Check with decimals. 45=0.8\frac{4}{5} = 0.8, and 4/3 is about 1.33. Then 0.8÷1.330.8 \div 1.33 is about 0.6, which is 3/5. Last, 0.6+0.8=1.40.6 + 0.8 = 1.4, and 1.4 is the same as 7/5.

More questions like this

The original DOE question
Question 50 as shown on the DOE’s digital test, page 121 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test A Math, page 121, question 50.