SHSAT Sunset

Practice Test

Misunderstands the concept

This test has the first 120 of the 123 questions that match.

Change this testNew TestYour practice tests

Question 1 of 120

2026 Practice Test A Math, question 10

Question

Three expressions are given:

−2(2.5n+4)-2(2.5n + 4)

5−n5 - n

3(2n+7)3(2n + 7)

Which of these are equivalent to the sum of these three expressions?

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.

  • A5−n+6n+21−5n−85 - n + 6n + 21 - 5n - 8
  • B5−n+6n+21−5n+85 - n + 6n + 21 - 5n + 8
  • C5−n+5n+21−5n−85 - n + 5n + 21 - 5n - 8
  • D18−n18 - n
  • E18
  • F34
Show the answer

Correct answer: A and E

Add the three expressions: 5−n+3(2n+7)+(−2)(2.5n+4)5 - n + 3(2n + 7) + (-2)(2.5n + 4).

Distribute 3 and −2: 5−n+6n+21−5n−85 - n + 6n + 21 - 5n - 8.

Combine the variable terms and combine the constants: (−n+6n−5n)+(5+21−8)=18(-n + 6n - 5n) + (5 + 21 - 8) = 18.

Question analysis

This algebra question gives three expressions, one with a decimal coefficient and a negative factor, and asks which two of six choices are equivalent to their sum. It tests distributing, combining like terms and recognizing equivalent expressions. Signs from the negative factor are easy to lose, and several choices are built from exactly those slips.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.4 identifying equivalent expressions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Expanding and combining linear expressions with a negative factor and a decimal coefficient is standard grade 7 work. Choosing the two equivalent ones out of six, where each common slip appears as a choice, takes more care than one simplification.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 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

The six choices form three consistent pairs, each an expansion with its simplified total: the right pair, a pair built on a sign slip with −2 × 4 that totals 34, and a pair built on 3 × 2n read as 5n that totals 18 − n. A student who makes either slip finds two choices that agree with each other and seem to confirm the work. The question rewards one careful simplification and the confidence to accept a sum with no n left.

  1. Error-prone arithmetic moderate Working it out

    The first expression puts a negative factor in front of a decimal term: −2 times 2.5n is −5n, and the −2 must reach the 4 as well, giving −8, not +8. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The first two expansions are identical except for their last sign, − 8 against + 8, and the third differs from the first only in 5n for 6n, so only a term-by-term check separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Many steps minor Working it out

    Two distributions come first, then the n terms and the constants are collected separately, and six choices are compared with the result; both right ones must be picked to score. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    One right answer is the expansion before any terms are combined and the other is the plain number 18, which has to be accepted as equal to a sum in which every expression held n. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute each factor across its parentheses, keeping the sign with every product, then combine the n terms and the constants separately and test each of the six choices against the result.

Traps

  • Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
  • 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

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, and b and c are the terms inside. You multiply a by every term inside, not only the first one. If a is negative, its minus sign goes into every product. It works because a groups of b + c is a groups of b plus a groups of c.
Multiplying signed numbers
(−a)×b=−(a×b)(-a) \times b = -(a \times b)
a and b are positive numbers, so −a is negative. A negative number times a positive number is negative. Use it whenever a minus sign sits in front of the parentheses: −2 times 4 is −8, not 8.
Combining like terms
an+bn=(a+b)nan + bn = (a + b)n
Like terms have the same letter part, such as 6n and −5n. a and b are their coefficients, the numbers in front of n. Add the coefficients and keep the n. Constants, the plain numbers, combine only with other constants.

  1. Sort out the question

    An expression is a math phrase made of numbers, letters and signs, with no equal sign. The sum of the three expressions means you add all of them:

    −2(2.5n+4)+(5−n)+3(2n+7)-2(2.5n + 4) + (5 - n) + 3(2n + 7)

    You need the two choices that are equivalent to this sum. Equivalent means equal for every value of n. Some choices show the sum before it is simplified. Others show the result.

  2. Distribute the 3

    To distribute means to multiply the number outside the parentheses by each term inside.

    3(2n+7)=3×2n+3×7=6n+213(2n + 7) = 3 \times 2n + 3 \times 7 = 6n + 21

    Watch out: 3 times 2n is 6n. Adding 3 and 2 instead gives 5n. That slip leads to choice C, and then to 18 − n, choice D.

  3. Distribute the −2

    Now multiply −2 by both terms. A negative times a positive is negative, so both products are negative.

    −2(2.5n+4)=−2×2.5n+(−2)×4=−5n−8-2(2.5n + 4) = -2 \times 2.5n + (-2) \times 4 = -5n - 8

    Watch out: the −2 multiplies the 4 too. Writing +8 instead of −8 leads to choice B. That sum works out to 34, choice F.

  4. Write the whole sum

    Put the three pieces together, starting with 5 − n:

    5−n+6n+21−5n−85 - n + 6n + 21 - 5n - 8

    This matches choice A exactly, term by term. So A is one of the answers.

  5. Combine like terms

    Gather the n terms first. Remember that −n means −1 times n.

    −n+6n−5n=(−1+6−5)n=0n=0-n + 6n - 5n = (-1 + 6 - 5)n = 0n = 0

    Then gather the constants:

    5+21−8=185 + 21 - 8 = 18

    The n terms cancel out, so the whole sum is just 18.

    Watch out: a sum with no n left is fine. Keeping an extra −n leads to choice D.

  6. Pick the two answers

    Choice A is the sum before you combine like terms. Choice E, 18, is the sum after you combine them. Both equal the sum for every value of n. So the answers are A and E.

Check your answer

Try a value, such as n = 2, in each expression. −2(2.5×2+4)=−2×9=−18-2(2.5 \times 2 + 4) = -2 \times 9 = -18. 5−2=35 - 2 = 3. 3(2×2+7)=3×11=333(2 \times 2 + 7) = 3 \times 11 = 33. The sum is −18+3+33=18-18 + 3 + 33 = 18, which matches E. Choice D would give 18 − 2 = 16 instead.

More questions like this

The original DOE question
Question 10 as shown on the DOE’s digital test, page 81 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test A Math, page 81, question 10.

Question 2 of 120

2026 Practice Test A Math, question 17

Question

Consider this expression:

|(−8)−(−3)+6|−|6−11||(-8) - (-3) + 6| - |6 - 11|

Create an equivalent expression.

Move the correct answer to each box. Not all answers will be used.

|| − ||

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: 1; blank 2: −5

To create an expression equivalent to |(−8)−(−3)+6|−|6−11||(-8) - (-3) + 6| - |6 - 11|:

Step 1: Simplify the first absolute value term.

(−8)−(−3)=−8+3=−5(-8) - (-3) = -8 + 3 = -5

−5+6=1-5 + 6 = 1

Simplify the second absolute value term.

6−11=−56 - 11 = -5

Therefore the equivalent expression is

|1|−|−5||1| - |-5|.

Question analysis

This absolute value question shows an expression with two pairs of bars and asks students to build an equivalent one by placing number tiles in two boxes. It tests subtracting negative integers and understanding what absolute value bars do. Not every tile is used, and several tiles match the results of common sign mistakes.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Simplifying signed integers inside absolute value bars, including subtracting a negative, is core grade 7 work. This asks only for the values inside the bars, not the final number, but one sign slip leads straight to a wrong tile.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 6, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every move here involves a sign, and the tiles are built from the usual misreadings of absolute value: 11 for making each number positive before combining, 17 for erasing every minus sign. A slip on − (−3) gives −5 for the first pair, but the second pair needs the only −5 tile, so that slip leaves a box with no tile to fit. Students who simplify each inside completely, and leave the bars untouched, finish quickly.

  1. Error-prone arithmetic moderate Working it out

    Inside the first bars, subtracting −3 must become adding 3, so −8 + 3 = −5, and adding 6 then crosses zero to reach 1; reading − (−3) as − 3 gives −5 instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Two pairs of bars are simplified separately, and both boxes need the right tile, kept in the original order, before the answer is right. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    Each box takes what is left inside a pair of bars, not its absolute value, so the second box needs the tile −5 rather than a 5, and the expression's value, −4, is never entered. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite subtracting −3 as adding 3, then simplify the inside of each pair of bars separately, left to right; each box replaces the inside of one pair, in the same order.

Traps

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

Subtracting a negative number
a−(−b)=a+ba - (-b) = a + b
a is any number, and b is a positive number, so −b is negative. Subtracting a negative number is the same as adding its opposite, the same number with the other sign. Think of money: taking away a debt of $3 leaves you $3 richer.
Absolute value
|a|=the distance from a to 0|a| = \text{the distance from } a \text{ to } 0
The bars around a number stand for its absolute value. a is any number. The absolute value is how far a is from 0 on a number line, so it is never negative. Both |5| and |−5| equal 5. Always simplify inside the bars first.

  1. Sort out the question

    The expression has two pairs of absolute value bars. The new expression keeps both pairs, with a box inside each one:

    |□|−|□||\square| - |\square|

    Each box gets the number you reach when you simplify inside that pair of bars. The bars stay, so you do not take the absolute value yet. Some tiles are not used.

  2. Simplify inside the first bars

    Inside the first bars is (−8)−(−3)+6(-8) - (-3) + 6. Subtracting −3 is the same as adding 3. Work from left to right:

    (−8)−(−3)=−8+3=−5(-8) - (-3) = -8 + 3 = -5

    −5+6=1-5 + 6 = 1

    The first box gets the tile 1.

    Watch out: the bars do not make each number positive before you combine. Changing −8 and −3 to 8 and 3 first gives 8 − 3 + 6 = 11. Dropping every minus sign gives 8 + 3 + 6 = 17. Both are tiles, and both are wrong.

  3. Simplify inside the second bars

    Inside the second bars is 6−116 - 11. You take a bigger number from a smaller one, so the result is negative:

    6−11=−56 - 11 = -5

    The second box gets the tile −5.

    Watch out: do not change −5 to 5. The bars are still there in the new expression, so the box holds what is inside them.

  4. Keep the boxes in order

    The first bars came first, so 1 goes in the first box and −5 in the second. The new expression has the same value as the old one:

    |1|−|−5|=1−5=−4|1| - |-5| = 1 - 5 = -4

    Swapping the tiles gives |−5|−|1|=5−1=4|-5| - |1| = 5 - 1 = 4, which is not equal. So the boxes get 1 and −5, making |1| − |−5|.

    On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Use a number line for the first pair. Start at −8 and move 3 to the right to reach −5. Then move 6 more to the right to reach 1. For the second pair, start at 6 and move 11 to the left. You land on −5. Both boxes check out.

More questions like this

The original DOE question
Question 17 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 A Math, page 88, question 17.

Question 3 of 120

2026 Practice Test A Math, question 37

Question

When a student buys 4 notebooks, the total cost is $12. Each notebook costs the same amount. Select each point on the graph that best represents y, the total cost of x notebooks.

Select all the correct answers.

A graph titled BUYING NOTEBOOKS, with Number of Notebooks on the x-axis from 0 to 6 and Total Cost (dollars) on the y-axis from 0 to 15, with grid lines every 1 unit. Seven points are plotted: (1, 1), (1, 3), (2, 5), (3, 7), (3, 9), (4, 12) and (5, 14).

On the test you click the points in the figure; here, choose them below. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

  • A(1, 1)
  • B(1, 3)
  • C(2, 5)
  • D(3, 7)
  • E(3, 9)
  • F(4, 12)
  • G(5, 14)
Show the answer

Correct answer: B, E, and F

Solution:

There are three correct points: (1, 3), (3, 9), and (4, 12).

Find the cost per notebook: 124=3\frac{12}{4} = 3.

The equation of the line is y=3xy = 3x, where y is the total cost of x notebooks.

Check each point by plugging the x-value into the equation y=3xy = 3x. If the y-value from the equation matches the y-value of the point, then the point is on the line y=3xy = 3x.

Check all seven points:

(1, 1) y=3(1)=3y = 3(1) = 3 y-values do not match; incorrect point

(1, 3) y=3(1)=3y = 3(1) = 3 y-values match; CORRECT point

(2, 5) y=3(2)=6y = 3(2) = 6 y-values do not match; incorrect point

(3, 7) y=3(3)=9y = 3(3) = 9 y-values do not match; incorrect point

(3, 9) y=3(3)=9y = 3(3) = 9 y-values match; CORRECT point

(4, 12) y=3(4)=12y = 3(4) = 12 y-values match; CORRECT point

(5, 14) y=3(5)=15y = 3(5) = 15 y-values do not match; incorrect point

Question analysis

A graph titled Buying Notebooks shows seven points, and students select every point that fits the total cost of buying notebooks, given the cost of a few notebooks and that each costs the same. It tests finding a unit rate and recognizing points of a proportional relationship. Some points line up in tempting patterns, and the item does not say how many to select.

NYS standards
  • 7.RP.2b constant of proportionality
  • 7.RP.2c proportional relationship equations
  • 6.RP.3a tables of equivalent ratios
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding a unit rate and deciding which points fit a proportional relationship are grade 7 skills, and the arithmetic here is easy. What adds demand is testing all seven points with no count given, since one miss or extra pick loses the item.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Finding the price is one division, 12 ÷ 4 = 3, so the question is really about checking: each chosen dot must fit y = 3x exactly, and the item does not say how many do. The graph punishes judging by eye, since a straight row of dots climbs from (1, 3) at $2 a notebook and two dots sit a single gridline off the true line. Multiplying each x by 3 and reading each y from the axis, not from the pattern, settles all seven.

  1. Every choice must be checked moderate Choosing the answer

    Seven dots must each be tested against $3 a notebook, and the item never says how many fit; one missed point or one extra loses the whole item, so no dot can be skipped. (Embretson & Wetzel, 1987)

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

    The graph stacks two dots above 1 notebook and two above 3, and each total cost has to be read off a grid ruled at every dollar up to 15. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. A very close wrong answer minor Choosing the answer

    (5, 14) sits one gridline below (5, 15), where 5 notebooks at $3 would be, and (2, 5) sits one below (2, 6), so judging by the line's direction cannot rule either out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out the cost of one notebook from the total given in the question, then multiply each point's x-coordinate by that price and keep only the points whose y-coordinate matches exactly.

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.
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit price
price of one=total cost÷number bought\text{price of one} = \text{total cost} \div \text{number bought}
The unit price is what one item costs. Divide the total cost by the number of items bought. This works because every notebook costs the same, so the total splits into equal parts.
A proportional relationship
y=kxy = kx
y is the total cost in dollars, and x is the number of notebooks. k is the price of one notebook, the number you always multiply by. A point fits only if its y is exactly k times its x. The graph is a straight line through (0, 0).

  1. Sort out the question

    Four notebooks cost $12, and each one costs the same. On the graph, x is the number of notebooks, and y is the total cost in dollars. The graph shows seven points. You must select every point that fits. The question does not say how many fit, so test all seven.

  2. Find the price of one notebook

    Split the $12 equally among the 4 notebooks:

    12÷4=312 \div 4 = 3

    Each notebook costs $3. So the total cost is 3 times the number of notebooks:

    y=3xy = 3x

  3. Test each point

    For each point, multiply its x by 3. Keep the point only if the result equals its y.

    ChoicePoint3 × xFits?
    A(1, 1)3No
    B(1, 3)3Yes
    C(2, 5)6No
    D(3, 7)9No
    E(3, 9)9Yes
    F(4, 12)12Yes
    G(5, 14)15No

  4. Watch for look-alike points

    Watch out: the dots (1, 3), (2, 5) and (3, 7) sit in a neat straight row. But that row climbs $2 per notebook, not $3. Lining up is not enough: (2, 5) and (3, 7), choices C and D, do not fit.

    Watch out: 5 notebooks cost 5×3=155 \times 3 = 15 dollars. The dot above 5 sits one gridline lower, at 14. Read the y-axis carefully, and you will skip choice G.

  5. Name the points

    Three points fit: (1, 3), (3, 9) and (4, 12). On the test, you click each of these dots on the graph. On this site, you check their boxes. So the answers are B, E and F.

Check your answer

All three points fit y=3xy = 3x: 3=3×13 = 3 \times 1, 9=3×39 = 3 \times 3 and 12=3×412 = 3 \times 4. They also line up with (0, 0), since 0 notebooks cost $0. And (4, 12) is the fact the question gave you.

More questions like this

The original DOE question
Question 37 as shown on the DOE’s digital test, page 108 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test A Math, page 108, question 37.

Question 4 of 120

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 5 of 120

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 6 of 120

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 7 of 120

2026 Practice Test B Math, question 46

Question

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

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

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

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

Show the answer

Correct answer: y = 20x + 50

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

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

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Amiya pays two kinds of charges:

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

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

  2. Work out a few totals

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

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

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

  3. Match each number to its job

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

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

  4. Write the equation

    Put the rate and the fee into the model:

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

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

  5. Enter it on the keypad

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

    So you would enter y = 20x + 50.

Check your answer

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

More questions like this

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

Question 8 of 120

2026 Practice Test C Math, question 1

Question

What is the prime factorization of 756?

  • A22⋅32⋅212^{2} \cdot 3^{2} \cdot 21
  • B22⋅3⋅7⋅92^{2} \cdot 3 \cdot 7 \cdot 9
  • C22⋅7⋅272^{2} \cdot 7 \cdot 27
  • D22⋅33⋅72^{2} \cdot 3^{3} \cdot 7
Show the answer

Correct answer: D

Determine the prime factors of 756 using a factor tree:

A factor tree for 756: 756 splits into 2 and 378; 378 into 2 and 189; 189 into 3 and 63; 63 into 3 and 21; and 21 into 3 and 7.

The prime factors are 2, 2, 3, 3, 3, and 7, or 2⋅2⋅3⋅3⋅3⋅72 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 7, which can be expressed using exponents as 22⋅33⋅72^{2} \cdot 3^{3} \cdot 7.

Question analysis

This number question asks for the prime factorization of 756, written with exponents. Students break the number into prime factors, with a factor tree or repeated division, and match the result to one of four products. Every choice multiplies back to 756, so the real test is whether each factor shown is actually prime.

NYS standards
  • 6.NS.4 common factors and multiples
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Factor trees and exponent notation are grade 6 skills, and 756 splits quickly by small divisors. What adds a little is that every choice multiplies out to 756, so checking the product alone settles nothing.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 11, 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

The factoring itself is quick: halving twice and dividing by 3 three times leaves 7. The question is built so that multiplying back proves nothing, since every choice equals 756, and three of them simply stop the factor tree at a composite number. Students who check only the product, or who take an odd factor such as 21 or 27 for a prime, pick a factorization that is unfinished.

  1. A very close wrong answer moderate Choosing the answer

    2² · 3² · 21 shares the right answer's 2² and differs only in 3² · 21 against 3³ · 7, and both equal 756, so neither an estimate nor a product check separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Students must know that a prime factorization may hold only primes, and must spot that 21, 9 and 27 are composite even though none of them is even. (Anderson & Krathwohl, 2001)

Strategy: Divide 756 by 2 while the result is even, then by 3 while it divides evenly, and keep going until every factor left is prime; then compare factor by factor, not by product.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Writing a repeated factor with an exponent
an=a×a×⋯×aa^{n} = a \times a \times \cdots \times a
a is the number being multiplied, called the base. n is the exponent, the small raised number. It tells how many times a appears in the product. So 2² means 2 × 2, and 3³ means 3 × 3 × 3. Exponents are a short way to write a factor that repeats.

  1. Sort out the question

    You need the prime factorization of 756. That means writing 756 as a product of prime numbers only. A product is the answer to a multiplication. A prime number has exactly two factors, 1 and itself. Examples are 2, 3, 5 and 7.

    All four choices multiply to 756. So multiplying a choice back does not pick the answer. You must check that every factor is prime.

  2. Divide by 2 while you can

    A factor tree splits a number into two factors, again and again. Start with the smallest prime, 2. 756 is even, so 2 goes into it:

    756=2×378756 = 2 \times 378

    378 is even too:

    378=2×189378 = 2 \times 189

    189 is odd, so 2 does not go into it. You have pulled out two 2s.

  3. Divide by 3 while you can

    Add the digits of 189: 1+8+9=181 + 8 + 9 = 18. A number can be divided by 3 when its digits add up to a number that 3 goes into. 3 goes into 18, so it goes into 189:

    189=3×63189 = 3 \times 63

    63=3×2163 = 3 \times 21

    21=3×721 = 3 \times 7

    7 is prime, so the tree stops. You have pulled out three 3s and one 7.

  4. Write the factors with exponents

    Collect every prime you pulled out:

    756=2×2×3×3×3×7756 = 2 \times 2 \times 3 \times 3 \times 3 \times 7

    Use the exponent rule. Two 2s make 222^{2}, and three 3s make 333^{3}. The choices use a raised dot, ·, for times. So 756 is 22⋅33⋅72^{2} \cdot 3^{3} \cdot 7.

  5. Choose the answer

    Watch out: each wrong choice is a factor tree that stopped too soon. A ends in 21, which is 3 × 7. B has 9, which is 3 × 3. C has 27, which is 3 × 3 × 3. Being odd does not make a number prime.

    Only D uses primes alone. So the answer is D, 22⋅33⋅72^{2} \cdot 3^{3} \cdot 7.

Check your answer

Multiply your answer back: 22×33=4×27=1082^{2} \times 3^{3} = 4 \times 27 = 108, and 108×7=756108 \times 7 = 756. Then look at each factor. 2, 3 and 7 are all prime. D passes both tests.

More questions like this

The original DOE question
Question 1 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 C Math, page 74, question 1.

Question 9 of 120

2026 Practice Test C Math, question 8

Question

Jordan is helping plan snacks for a school dance. He knows that 0.75 of the students voted for pizza, and he wants to record the same amount using a fraction.

Which fractions are equivalent to the decimal 0.75?

Select the four 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.

  • A34\frac{3}{4}
  • B1520\frac{15}{20}
  • C75100\frac{75}{100}
  • D25\frac{2}{5}
  • E68\frac{6}{8}
  • F810\frac{8}{10}
Show the answer

Correct answer: A, B, C, and E

Check each fraction to determine whether it is equal to 0.75 when simplified or converted to a decimal.

34→3÷4=0.75\frac{3}{4} \rightarrow 3 \div 4 = 0.75
This is the most common fraction equal to 0.75.

1520→15÷20=0.75\frac{15}{20} \rightarrow 15 \div 20 = 0.75
Simplifies to 3/4, because 15 and 20 can both be divided by 5.

75100→75÷100=0.75\frac{75}{100} \rightarrow 75 \div 100 = 0.75
This is a direct match to 0.75 when written as a decimal.

25→2÷5=0.4\frac{2}{5} \rightarrow 2 \div 5 = 0.4, which is less than 0.75.

68→6÷8=0.75\frac{6}{8} \rightarrow 6 \div 8 = 0.75
Simplifies to 34\frac{3}{4} because both 6 and 8 can be divided by 2.

810→8÷10=0.8\frac{8}{10} \rightarrow 8 \div 10 = 0.8, which is more than 0.75.

Question analysis

Jordan wants to record 0.75 of the students' pizza votes as a fraction, and students select the four fractions, out of six, that equal 0.75. It tests converting between decimals and fractions and recognizing equivalent fractions. Each choice has to be checked by its value, not by how its numbers look.

NYS standards
  • 7.NS.2d fractions to decimals by long division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Matching fractions to a decimal such as 0.75 builds on grade 4 to 6 work, and the denominators here divide easily. The added demand is the format: four of the six must be chosen, and one wrong or missing pick loses the item.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Three of the four right fractions, 3/4, 75/100 and 6/8, are familiar forms of 0.75, so the item comes down to the fourth. Students who stop after the familiar ones must decide between 15/20, which needs dividing by 5 to recognize, and 8/10, which sits close to 0.75 on a number line. Dividing each numerator by its denominator, or scaling 3/4 up by 5, settles it.

  1. Every choice must be checked moderate Choosing the answer

    Six fractions with six different denominators have to be tested one by one, by dividing or by simplifying, since 15/20 and 8/10 cannot be judged by how they look. (Embretson & Wetzel, 1987)

  2. A very close wrong answer minor Choosing the answer

    8/10 is 0.8, only 0.05 above 0.75, and it follows 6/8 by adding 2 to the top and the bottom, so it looks like one more member of the family. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Much to keep track of minor Working it out

    Four of the six must be chosen and all four must be right for the item to count, so each verdict has to be kept while the remaining fractions are tested. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Turn each choice into a decimal by dividing its numerator by its denominator, or reduce it to lowest terms, and keep only the ones whose value is exactly the decimal given.

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

Changing a fraction to a decimal
ab=a÷b\frac{a}{b} = a \div b
a is the numerator, the top. b is the denominator, the bottom. A fraction bar means divide. So dividing the top by the bottom gives the fraction’s value as a decimal.
Equivalent fractions
ab=a÷nb÷n\frac{a}{b} = \frac{a \div n}{b \div n}
a is the top and b is the bottom. n is a number that goes evenly into both. Dividing the top and the bottom by the same number gives an equal fraction. Equivalent fractions name the same amount in different ways.

  1. Sort out the question

    Six fractions are given, A to F. You need the four that equal the decimal 0.75. Equivalent means equal in value. Test every choice, because some fractions look alike but are not equal.

  2. Write 0.75 as a fraction

    0.75 means 75 hundredths:

    0.75=751000.75 = \frac{75}{100}

    So choice C works. Now simplify it. 25 goes into both 75 and 100:

    75÷25100÷25=34\frac{75 \div 25}{100 \div 25} = \frac{3}{4}

    So choice A, 3/4, works too.

  3. Test choices B and E

    Simplify each one to see if it becomes 3/4:

    • B: 5 goes into 15 and 20. 15÷520÷5=34\frac{15 \div 5}{20 \div 5} = \frac{3}{4}. It works.
    • E: 2 goes into 6 and 8. 6÷28÷2=34\frac{6 \div 2}{8 \div 2} = \frac{3}{4}. It works.

  4. Test choices D and F

    Divide the top by the bottom:

    • D: 2÷5=0.42 \div 5 = 0.4. That is less than 0.75.
    • F: 8÷10=0.88 \div 10 = 0.8. That is more than 0.75.

    Watch out: F looks like 6/8 with 2 added to the top and the bottom. Adding the same number to both changes the value. Only multiplying or dividing both keeps it.

  5. Name the four answers

    A, B, C and E all equal 0.75. On the test, you check the box beside each of them. On this site, you choose all four, then check your answer.

    So the answers are A, B, C and E.

Check your answer

Turn each answer into a decimal. 3÷43 \div 4, 15÷2015 \div 20, 75÷10075 \div 100 and 6÷86 \div 8 all give 0.75. Exactly four choices passed, and the question asks for four. So none were missed.

More questions like this

The original DOE question
Question 8 as shown on the DOE’s digital test, page 81 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 81, question 8.

Question 10 of 120

2026 Practice Test C Math, question 18

Question

I. s−ts - t
II. st2st^{2}
III. sts^{t}

If s is a positive integer and t is a negative integer, which of the above expressions must be a positive number?

  • AII only
  • BI and II only
  • CII and III only
  • DI, II, and III
Show the answer

Correct answer: D

Each of the three expressions must be considered:

Expression I: s−ts - t, where s is a positive integer and t is a negative integer:

Subtracting a negative integer is the equivalent of adding a positive integer, so s−ts - t is equivalent to finding the value of s+(−t)s + (-t), which must result in a positive number.

Expression II: st2st^{2}, where s is a positive integer and t is a negative integer:

Squaring a negative integer results in a positive integer, so s(−t)2s(-t)^{2} is equivalent to finding the value of st2st^{2}, which must result in a positive number.

Expression III: sts^{t}, where s is a positive integer and t is a negative integer:

A negative exponent results in the reciprocal of the same exponential expression, so sts^{t} is equivalent to finding the value of 1s(−t)\frac{1}{s^{(-t)}}, which must result in a positive number.

Therefore, Expressions I, II, and III must all result in a positive number.

Question analysis

Three expressions, s − t, s times t squared, and s to the power t, are listed, and students decide which must be positive when s is a positive integer and t is a negative integer. It tests sign rules for subtraction and multiplication and the meaning of a negative exponent. Each expression has to be judged for every possible pair, not just one example.

NYS standards
  • 8.EE.1 integer exponent properties
  • 7.NS.1c rational number subtraction
  • 7.NS.2a rational number multiplication
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students learn sign rules for subtracting and multiplying signed numbers. Here they reason about letters rather than numbers, decide what must always hold, and handle a negative exponent, a grade 8 idea.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 5, 42 words.
Corrected from the DOE handbook
  • Two steps of the explanation that treated t as a positive number were corrected.
  • A punctuation error in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The three expressions test three different sign facts: subtracting a negative, squaring a negative, and raising to a negative power. The last is the trap, since a negative exponent seems to promise a negative result when it really gives a fraction, and that belief alone leads to a listed choice. Trying s = 2 and t = −3 gives 5, 18 and 1/8, all positive, and the rules show why no other pair can change that.

  1. Variables instead of numbers moderate Understanding the question

    No values are given for s and t, only their signs, and must means each expression has to be positive for every such pair, not just for one example. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know moderate Working it out

    Expression III needs the meaning of a negative exponent, s to the power t equals 1 over s to the power −t, a grade 8 rule the question does not state. (Anderson & Krathwohl, 2001)

  3. Every choice must be checked minor Choosing the answer

    Each of the three expressions is judged on its own and the choices combine the verdicts, so one wrong judgment anywhere points to a wrong combination. (Embretson & Wetzel, 1987)

Strategy: Try a pair such as s = 2 and t = −3 in each expression, then explain why its sign could or could not change for any other positive s and negative t.

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.
  • 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

What was corrected from the DOE handbook, and why

  • Explanation: The question makes t the negative integer, but two steps of the DOE’s explanation write the negative integer as −t, as if t were positive. For Expression I it says “s − (−t) is equivalent to finding the value of s + t, which must result in a positive number”: s − (−t) is not Expression I, and with the question’s t, s + t need not be positive (s = 1 and t = −2 give −1). For Expression III it says “s^(−t) is equivalent to finding the value of 1/s^t”, about s^(−t) rather than Expression III. The site now says that s − t is equivalent to s + (−t), and that s^t is equivalent to 1/s^(−t). Since t is negative, −t is a positive integer, so s + (−t) is the sum of two positive integers and 1/s^(−t) is the reciprocal of a positive integer: both positive, and the answer, I, II and III, is unchanged. The step for Expression II, s(−t)² = st², is true for any t and is unchanged.
  • Explanation: The DOE’s explanation introduces Expression III as “s^t. where s is a positive integer and t is a negative integer”, with a period where Expressions I and II (and the 2022–23 guide, which first printed this explanation) have a comma. The site has the comma.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a number
s−t=s+(−t)s - t = s + (-t)
s and t are integers, and −t is the opposite of t. Subtracting a number is the same as adding its opposite. When t is negative, −t is positive. So subtracting t adds a positive amount.
A negative exponent
st=1s−ts^{t} = \frac{1}{s^{-t}}
s is the base and t is the exponent. When t is negative, −t is positive. So a negative exponent means 1 over the base to a positive exponent. For example, 2 to the power −3 is 1 over 2³, which is 1/8.

  1. Sort out the question

    An integer is a whole number or its opposite, such as −2, 0 or 5. Here s is a positive integer, like 1, 2 or 3. And t is a negative integer, like −1, −2 or −3. Must means always. An expression counts only if it is positive for every such pair.

  2. Try one pair of numbers

    Try s = 2 and t = −3 in each expression:

    • I: 2−(−3)=2+3=52 - (-3) = 2 + 3 = 5
    • II: 2×(−3)2=2×9=182 \times (-3)^{2} = 2 \times 9 = 18
    • III: 2−3=123=182^{-3} = \frac{1}{2^{3}} = \frac{1}{8}

    All three are positive this time. Now see why that always happens.

  3. Explain expression I

    Use the rule for subtracting:

    s−t=s+(−t)s - t = s + (-t)

    t is negative, so its opposite, −t, is positive. So s − t is a positive plus a positive. That is always positive.

    Watch out: subtracting does not always make a number smaller. Subtracting a negative makes it bigger. Missing this leads to choice C.

  4. Explain expression II

    t2t^{2} means t × t. A negative times a negative is positive, so t2t^{2} is always positive. Then s×t2s \times t^{2} is a positive times a positive. That is always positive too.

  5. Explain expression III

    Use the rule for a negative exponent:

    st=1s−ts^{t} = \frac{1}{s^{-t}}

    −t is a positive integer, so s−ts^{-t} is a positive integer. So sts^{t} is 1 over a positive number. That is a positive fraction.

    Watch out: a negative exponent does not make the answer negative. It makes a fraction. Thinking it does leads to choice B, I and II only.

  6. Choose the answer

    Each of the three expressions is positive for every positive s and negative t. None of them can ever be zero or negative.

    So the answer is D, I, II, and III.

Check your answer

Try the values closest to 0: s = 1 and t = −1. I gives 1−(−1)=21 - (-1) = 2. II gives 1×(−1)2=11 \times (-1)^{2} = 1. III gives 1−1=11=11^{-1} = \frac{1}{1} = 1. All three are still positive.

More questions like this

The original DOE question
Question 18 as shown on the DOE’s digital test, page 91 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 91, question 18. This is the DOE’s version, before the corrections listed in the analysis.

Question 11 of 120

2026 Practice Test C Math, question 21

Question

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

  • A2
  • B3
  • C18
  • D38
Show the answer

Correct answer: A

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

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

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

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

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

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

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

Solve for the value of a:

48a=9648a = 96

a=2a = 2

Question analysis

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

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

Why it is hard

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

  1. Turning words into math minor Understanding the question

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

  2. Much to keep track of minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Write the proportion

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

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

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

  3. Simplify the known ratio

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

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

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

  4. Find the apples

    Now the proportion is:

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

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

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

    So the answer is A, 2.

Check your answer

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

More questions like this

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

Question 12 of 120

2026 Practice Test C Math, question 26

Question

A coordinate grid with the x-axis and y-axis each from 0 to 10, grid lines every 1 unit, labeled at 2, 4, 6, 8 and 10, and the origin labeled O. Four straight lines are drawn: one through M (0, 8) and S (8, 0); one through V (2, 3) and P (3, 6); a horizontal one through V (2, 3) and Q (6, 3); and one drawn from the origin O through T (4, 1) and R (8, 2).

Which straight line is the graph of a proportional relationship?

  • AMS↔\overleftrightarrow{\mathrm{MS}}
  • BVP↔\overleftrightarrow{\mathrm{VP}}
  • CVQ↔\overleftrightarrow{\mathrm{VQ}}
  • DTR↔\overleftrightarrow{\mathrm{TR}}
Show the answer

Correct answer: D

The graph of any proportional line must intersect the origin, (0,0)(0, 0).

The only line on the coordinate plane that contains (0,0)(0, 0) is TR↔\overleftrightarrow{\mathrm{TR}}.

Both point T(4,1)\mathrm{T}\,(4, 1) and point R(8,2)\mathrm{R}\,(8, 2) have a y-value that is 14\frac{1}{4} of the x-value, and every point on the line fits the proportional equation y=14xy = \frac{1}{4}x.

When x=0, y=14(0)x = 0,\ y = \frac{1}{4}(0). So the line contains the origin, (0,0)(0, 0).

Question analysis

A coordinate grid shows four straight lines, each named by two labeled points such as M and S or V and Q, and students choose the one that is the graph of a proportional relationship. It tests recognizing proportional relationships on a graph. Every line is straight, so being straight is not enough, and the lines must be told apart by what defines proportionality.

NYS standards
  • 7.RP.2a testing for proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Deciding from a graph whether a relationship is proportional is a named grade 7 skill, and this is a typical case: four lines on a clean grid, each checked against the same rule, with no calculation needed.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 9, 14 words.
Corrected from the DOE handbook
  • Coordinates and an equation in the explanation were corrected to match the graph.
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 origin rule is recalled, the graph answers the question at a glance: only the line from O through T and R starts at (0, 0). Students who think of proportional as any steady pattern have three other candidates, from the falling line through M and S to the flat line at height 3. The steep line through V and P is the closest call, because its lower end disappears near the corner where the axes meet.

  1. Something you must already know minor Working it out

    The rule that a proportional graph is a straight line through the origin is not printed, and all four lines are straight, so straightness alone rules none of them out. (Anderson & Krathwohl, 2001)

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

    Line VP runs off the grid just right of O, crossing the x-axis at (1, 0), so whether it passes through the origin has to be read from the gridlines, not from its general direction. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Recall what every proportional graph must look like, then test each line: see where it would be when x is 0, and whether y divided by x stays the same at both labeled points.

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.
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation says “Both point T (1, 4) and point R (2, 8) have a y-value that is 4 times the x-value, and every point on the line fits the proportional equation y = 4x”, and then “When x = 0, y = 4(0).” On the graph, T is at (4, 1) and R at (8, 2): each y-value is 1/4 of its x-value, and the line through the origin, T and R is y = (1/4)x. (The line y = 4x would pass through (1, 4) and (2, 8), which are not marked.) The 2022–23 guide, which first printed this question with the same graph, has the same slip. The site now gives T (4, 1) and R (8, 2), “a y-value that is 1/4 of the x-value”, y = (1/4)x, and y = (1/4)(0) at x = 0. The answer is unchanged: TR is still the only line through the origin.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship
y=kxy = kx
y and x are two amounts that change together. k is the constant of proportionality: the number you always multiply x by to get y. When x is 0, y is k times 0, which is 0. So the graph is a straight line through the origin, the point (0, 0) where the axes meet.

  1. Sort out the question

    The grid shows four straight lines. Each one is named by two labeled points on it: line MS, line VP, line VQ and line TR.

    You need the line that shows a proportional relationship. In that kind of relationship, y is always the same number times x. For example, y might always be 3 times x.

  2. Recall what a proportional graph looks like

    Use the rule y=kxy = kx. Put in 0 for x:

    y=k×0=0y = k \times 0 = 0

    So every proportional graph goes through the origin. That is the corner point (0, 0), labeled O, where the two axes meet. The graph must also be a straight line.

    All four lines here are straight. So being straight does not settle it. Look for the line through O.

  3. Check where each line starts

    • Line MS meets the y-axis at M (0, 8). When x is 0, y is 8, not 0.
    • Line VP is steep. Its lower end crosses the x-axis at (1, 0), one unit to the right of O.
    • Line VQ is flat at a height of 3. It never comes down to 0.
    • Line TR starts at O and runs up through T and R.

    Watch out: the lower end of line VP looks close to O. Follow it along the grid lines. It misses O by one unit, so choice B is wrong.

  4. Test the ratio on line TR

    In a proportional relationship, the ratio y ÷ x is the same at every point. T is at (4, 1) and R is at (8, 2):

    T: 14R: 28=14\text{T: } \frac{1}{4} \qquad \text{R: } \frac{2}{8} = \frac{1}{4}

    Both give 14\frac{1}{4}. So each y-value is 14\frac{1}{4} of its x-value, and the line is y=14xy = \frac{1}{4}x.

    Watch out: line VQ keeps the same y-value, 3, but not the same ratio. At V it is 32\frac{3}{2}, and at Q it is 36=12\frac{3}{6} = \frac{1}{2}. So choice C is wrong.

  5. Give the answer

    Line MS starts at 8 on the y-axis, so choice A is out too. Only line TR goes through the origin and keeps y ÷ x the same.

    So the answer is D, line TR.

Check your answer

Put the origin into the rule for line TR: y=14×0=0y = \frac{1}{4} \times 0 = 0. So (0, 0) is on the line. Then test R: 14×8=2\frac{1}{4} \times 8 = 2, which matches R (8, 2). The rule fits every labeled point on the line.

More questions like this

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

Question 13 of 120

2026 Practice Test C Math, question 27

Question

A graph titled ICE COSTS, with Weight of Ice (lb) on the x-axis from 0 to 20, labeled every 2, and Cost (dollars) on the y-axis from 0 to 3.0, labeled every 0.3, with a grid line at each label. A straight line starts at the origin, O, and rises 0.3 dollars for every 2 pounds, passing through (4, 0.6) and (20, 3.0).

Some friends are buying bags of ice for a picnic. The graph shows the cost of the ice, in dollars, in relation to the weight, in pounds. Which statement about a point on the graph is true?

  • AThe point (20, 3) means that 3 lbs of ice costs $20.00.
  • BThe point (1, 0.15) means that 1 bag of ice costs $0.15.
  • CThe point (10, 1.5) means that 10 lbs of ice costs $1.50.
  • DThe point (0, 0) means that 0 lbs of ice will be needed if 0 people go to the picnic.
Show the answer

Correct answer: C

The x-axis represents the weight, in pounds (lbs), of ice purchased, and the y-axis represents the cost of the ice, in dollars.

The x-value in the point (10, 1.5) represents 10 lbs of ice, and the y-value represents 1.5 dollars, or $1.50. Therefore, the cost of 10 lbs of ice is $1.50.

Question analysis

Some friends buying ice for a picnic use a graph of the ice's cost in dollars against its weight in pounds, a straight line from the origin. Students pick the statement that correctly explains what a point on the graph means. Each statement names a point that really is on the line, so the units and the order of the coordinates decide.

NYS standards
  • 7.RP.2d points on proportional graphs
  • 7.RP.2b constant of proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Explaining what a point on a proportional graph means in its situation is a named grade 7 skill. The graph is clear and starts at the origin, but each statement has to be read closely against the axis labels.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 2, 77 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 computing, but each statement attaches words to two numbers, and three of them attach the wrong words. The point (1, 0.15) is the unit rate students are trained to look for, and the “bags” in the stem’s story make its statement sound right, though the graph measures pounds. Saying each point aloud as so many pounds for so many dollars exposes the swapped coordinates and the extra people as well.

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

    Each statement must be checked against the axis labels: x is the weight of ice in pounds and y the cost in dollars, on a grid ruled every 2 pounds and every $0.30. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Every choice must be checked minor Choosing the answer

    All four statements name a point that really lies on the line, (1, 0.15) included, so being on the graph rules nothing out; only the meaning given to each coordinate does. (Embretson & Wetzel, 1987)

Strategy: Read both axis labels and their units first, then for each choice say in words what its x-coordinate and its y-coordinate stand for, and compare that with the statement.

Traps

  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • 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

Unit rate
cost per pound=costweight\text{cost per pound} = \frac{\text{cost}}{\text{weight}}
The cost is in dollars, and the weight is in pounds. The unit rate is the cost of just 1 pound. On a straight line through the origin, it is the same at every point. So you can find it from any point you can read.

  1. Sort out the question

    The graph is called ICE COSTS. Read both axes first:

    • The x-axis, across the bottom, is the weight of the ice in pounds (lb).
    • The y-axis, up the side, is the cost in dollars.

    A point is written (x, y), weight first and cost second. So a point means: x pounds of ice cost y dollars. You need the statement that says this correctly.

  2. Find the cost of one pound

    The line reaches $3.00 at 20 pounds. Use the unit rate formula:

    cost per pound=320=0.15\text{cost per pound} = \frac{3}{20} = 0.15

    So 1 pound of ice costs $0.15. This helps you test each point.

  3. Test the point in choice C

    Find 10 on the weight axis and go straight up to the line. The line crosses there at 1.5 on the cost axis. The unit rate agrees:

    10×0.15=1.510 \times 0.15 = 1.5

    So the point (10, 1.5) means 10 pounds of ice cost $1.50. That is exactly what choice C says.

  4. Check the other statements

    Each one names a real point on the line. The problem is the meaning it gives the numbers.

    Watch out: choice A reads (20, 3) backward. It means 20 pounds cost $3, not 3 pounds cost $20.

    Watch out: choice B is tricky. The point (1, 0.15) is on the line, but the x-axis counts pounds, not bags. So it means 1 pound costs $0.15.

    Choice D talks about people, but the graph never counts people. The point (0, 0) means 0 pounds cost $0.

  5. Give the answer

    Only choice C gives the right meaning to both numbers.

    So the answer is C, 10 lbs of ice costs $1.50.

Check your answer

Use halves. The point (20, 3) says 20 pounds cost $3.00. Half as much ice, 10 pounds, should cost half as much. Half of $3.00 is $1.50. That matches choice C.

More questions like this

The original DOE question
Question 27 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 C Math, page 100, question 27.

Question 14 of 120

2026 Practice Test C Math, question 29

Question

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

SMOOTHIE RECIPE

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 0.25

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

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

1.5=6p1.5 = 6p

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

14=0.25=p\frac{1}{4} = 0.25 = p

The constant of proportionality is 0.25.

Question analysis

A smoothie shop's recipe table lists cups of fruit and cups of yogurt for four batches in a proportional relationship. Students type the constant of proportionality of yogurt to fruit. It tests finding a unit rate from a table, and since fruit is listed first while the question names yogurt first, the ratio is easy to take upside down.

NYS standards
  • 7.RP.2b constant of proportionality
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the constant of proportionality in a table is a core grade 7 skill, and these values are friendly decimals. Typing the answer with no choices, and taking the ratio in the order the question gives, are the added demands.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
1 step
Reading level
Question and choices: about grade 6, 75 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Any row gives the answer with one division, and the values divide evenly, so the points are lost on direction. Fruit fills the first column and comes first in the stem’s own ratio, which invites dividing it by the yogurt. The table also climbs by a steady half cup of yogurt, a number that looks like a constant but ignores the 2 cups of fruit added each time. Checking that 0.25 times each fruit value gives its yogurt value settles it.

  1. Turning words into math moderate Understanding the question

    The opening sentence speaks of a ratio of fruit to yogurt, but the question asks for the constant of yogurt to fruit, so the division is y ÷ x, against the table’s column order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The constant is typed into a box with no choices, so 4, the ratio taken the other way round, is a believable entry that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: Check which quantity the question names first, divide that column by the other in any row, and confirm the quotient is the same in a second row.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality
k=yxk = \frac{y}{x}
k is the constant of proportionality. y and x are two matching amounts from the same row of the table. In a proportional relationship, y = kx, so y is always k times x. That is why dividing y by x gives the same k in every row.

  1. Sort out the question

    The table lists cups of fruit (x) and cups of yogurt (y) for four batches. You need the constant of proportionality of yogurt to fruit.

    That is the number of cups of yogurt for each 1 cup of fruit. It is also the number you multiply x by to get y.

  2. Divide yogurt by fruit

    Use the first row: 6 cups of fruit and 1.5 cups of yogurt. Yogurt goes on top, because the question says yogurt to fruit:

    k=yx=1.56=0.25k = \frac{y}{x} = \frac{1.5}{6} = 0.25

    So there is 0.25 cup, or one quarter of a cup, of yogurt for each cup of fruit.

    Watch out: the first sentence says fruit to yogurt, but the question asks the other way. Dividing 6÷1.5=46 \div 1.5 = 4 turns the answer upside down.

  3. Check two more rows

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

    2.08=0.253.012=0.25\frac{2.0}{8} = 0.25 \qquad \frac{3.0}{12} = 0.25

    They match. Watch out: the yogurt goes up by 0.5 cup from row to row. That 0.5 is not the constant. The fruit goes up by 2 cups at the same time, and 0.5÷2=0.250.5 \div 2 = 0.25.

  4. Enter the number

    Type only the number in the box, with no units. You can type 0.25 or .25. On this site, 1/4 counts too.

    So you would enter 0.25.

Check your answer

Multiply each amount of fruit by 0.25. 6×0.25=1.56 \times 0.25 = 1.5, 8×0.25=28 \times 0.25 = 2, 10×0.25=2.510 \times 0.25 = 2.5 and 12×0.25=312 \times 0.25 = 3. Every result matches the yogurt column. Also, each batch has less yogurt than fruit, so the constant should be less than 1.

More questions like this

The original DOE question
Question 29 as shown on the DOE’s digital test, page 102 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 102, question 29.

Question 15 of 120

2026 Practice Test C Math, question 30

Question

Out of 2,000 high school seniors, 191 were enrolled in the Honor Society. What percentage of the entire group of 2,000 seniors were in the Honor Society?

  • A0.955%
  • B1.91%
  • C9.55%
  • D19.1%
Show the answer

Correct answer: C

The fraction 1912000\frac{191}{2000} represents the 191 high school seniors out of 2,000 enrolled in the Honor Society.

To determine the percentage, convert the fraction to a decimal. Then, rewrite the decimal as a percentage.

1912000=0.0955=9.55%\frac{191}{2000} = 0.0955 = 9.55\%

Question analysis

Out of 2,000 high school seniors, 191 belong to the Honor Society, and students find what percent of the whole group that is. It tests writing a part of a whole as a percent. Several choices share the same digits, so the decimal point has to be placed with care when the fraction becomes a percent.

NYS standards
  • 6.RP.3c percent as a rate per 100
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Writing a part of a whole as a percent is grade 6 work, done here with one division and one conversion. The awkward numbers, 191 out of 2,000, make placing the decimal point the only real demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 24 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is a single fraction, 191 over 2,000, so the question tests decimal places. The four choices share the digits 955 or 191 and differ only in where the point falls, which means a slip of one place still finds a choice waiting. A rough check settles it: 191 is a little under 200, and 200 is 10% of 2,000, so the share must be a little under 10%.

  1. Error-prone arithmetic moderate Working it out

    Dividing 191 by 2,000 gives 0.0955, with a zero right after the decimal point, and writing it as a percent moves the point two more places; each move is a chance to land one place off. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the Honor Society members as a fraction of all 2,000 seniors, divide, and move the decimal point two places to the right to turn the decimal into a percent.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a whole
percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100
The part is the smaller group you care about. The whole is the entire group. The fraction tells what share of the whole the part is. Multiplying by 100 turns that share into a percent, which means out of 100.

  1. Sort out the question

    There are 2,000 seniors in all. Of them, 191 are in the Honor Society. You need the percent of all 2,000 seniors who are in it.

    So 191 is the part, and 2,000 is the whole.

  2. Write the share as a fraction

    Put the part over the whole:

    partwhole=1912,000\frac{\text{part}}{\text{whole}} = \frac{191}{2{,}000}

    This fraction says 191 out of every 2,000 seniors. Next, turn it into a decimal by dividing.

  3. Divide to get a decimal

    Dividing by 2,000 is the same as dividing by 2 and then by 1,000:

    191÷2=95.5191 \div 2 = 95.5

    95.5÷1,000=0.095595.5 \div 1{,}000 = 0.0955

    Dividing by 1,000 moves the decimal point 3 places to the left. So the share is 0.0955.

  4. Change the decimal to a percent

    Multiply by 100. That moves the decimal point 2 places to the right:

    0.0955×100=9.550.0955 \times 100 = 9.55

    So the share is 9.55%.

    Watch out: moving the point only 1 place gives 0.955%, choice A. Dividing by 1,000 instead of 2,000 gives 19.1%, choice D. Dividing 191 by 100 gives 1.91%, choice B.

  5. Give the answer

    The 191 Honor Society members are 9.55% of all 2,000 seniors. That is a little less than one tenth of them.

    So the answer is C, 9.55%.

Check your answer

Estimate. 191 is a little less than 200. And 200 out of 2,000 is one tenth, which is 10%. So the answer should be a little less than 10%. Only 9.55% fits.

More questions like this

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

Question 16 of 120

2026 Practice Test C Math, question 36

Question

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

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

Correct answer: B

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

52−12=4052 - 12 = 40

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

13−3=1013 - 3 = 10

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

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

Question analysis

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

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

Why it is hard

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

  1. Much to keep track of minor Working it out

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

  2. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Count the cards left

    Take Sara’s 12 cards out of the deck:

    52−12=4052 - 12 = 40

    So 40 cards are left in the deck.

  3. Count the hearts left

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

    13−3=1013 - 3 = 10

    So 10 hearts are left in the deck.

  4. Find the probability

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

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

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

  5. Give the answer

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

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

Check your answer

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

More questions like this

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

Question 17 of 120

2026 Practice Test C Math, question 37

Question

The table shows the probability of randomly picking each flavor of candy from a bowl.

PICKING CANDY FROM
A BOWL

Flavor Probability
Butterscotch 0.57
Lemon 0.095
Maple 0.125
Strawberry 0.21

Which flavor is most likely to be randomly picked?

  • Abutterscotch
  • Blemon
  • Cmaple
  • Dstrawberry
Show the answer

Correct answer: A

The flavor most likely to be picked at random is butterscotch, because its probability is the greatest of the probabilities in the list. More than half the candies in the bowl are butterscotch.

Question analysis

A table lists the probability of picking butterscotch, lemon, maple or strawberry candy at random from a bowl, each as a decimal. Students decide which flavor is most likely to be picked. It tests the meaning of probability and comparing decimals, which have different numbers of digits here and must be compared by place value.

NYS standards
  • 6.SP.6 probability from 0 to 1
  • 6.NS.7 ordering and absolute value
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Reading probabilities as likelihoods is grade 6 content, and this needs only a comparison of four decimals. The one demand is comparing decimals with different numbers of digits, a skill from earlier grades.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 9, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One idea answers the question: the most likely flavor has the greatest probability. The only obstacle is comparing decimals of different lengths, where 0.125 and 0.095 look bigger than 0.57 to students who count digits. Writing each as thousandths, 570, 95, 125 and 210, or noticing that 0.57 alone is more than one half, makes the order plain.

  1. Error-prone arithmetic minor Working it out

    The probabilities have two and three decimal places, 0.57 beside 0.095 and 0.125, so comparing them by length, or by their digits as whole numbers, ranks them wrongly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Line the four decimals up by place value, adding zeros so each has three decimal places, and compare them digit by digit starting from the tenths place.

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

Probability
P=candies of that flavorall candies in the bowlP = \frac{\text{candies of that flavor}}{\text{all candies in the bowl}}
P is the probability of picking a flavor. It is a number from 0 to 1. The more candies of a flavor there are, the bigger its probability. So the flavor with the greatest probability is the most likely to be picked.

  1. Sort out the question

    The table gives the probability of picking each of four flavors. You need the flavor that is most likely to be picked.

    The greater the probability, the more likely the pick. So you need the greatest of the four decimals.

  2. Line up the decimals

    The decimals have different numbers of digits. Add zeros at the end so each has three decimal places. This does not change their values:

    FlavorProbability
    Butterscotch0.570
    Lemon0.095
    Maple0.125
    Strawberry0.210

  3. Compare the tenths digits

    Start with the first digit after the decimal point, the tenths place. The tenths digits are 5, 0, 1 and 2. The biggest is 5, so 0.570 is the greatest.

    Watch out: 0.125 has more digits than 0.57, but it is smaller. Compare place by place, not by length. That mistake leads to choice C. Also, lemon is 0.095, not 0.95, so choice B is out.

  4. Give the answer

    Butterscotch has the greatest probability, 0.57. So it is the flavor most likely to be picked.

    So the answer is A, butterscotch.

Check your answer

0.57 is more than 0.5, which is one half. So more than half the candies are butterscotch. The other three flavors add up to 0.095+0.125+0.21=0.430.095 + 0.125 + 0.21 = 0.43, less than half. So butterscotch must be the most likely.

More questions like this

The original DOE question
Question 37 as shown on the DOE’s digital test, page 110 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 110, question 37.

Question 18 of 120

2026 Practice Test C Math, question 38

Question

This formula can be used to determine p, the number of pages that will be in a book that contains w words, with an average of r words per page, and f extra pages in the front for the title and publishing information.

p=1rw+fp = \frac{1}{r}w + f

If the formula is solved for w, it can be used to find the number of words in a book with p pages. What will the formula be if it is solved for w?

  • Aw=r(p−f)w = r(p - f)
  • Bw=1rp+fw = \frac{1}{r}p + f
  • Cw=f+rpw = f + rp
  • Dw=r+p+fw = r + p + f
Show the answer

Correct answer: A

Apply the properties of equality to solve for w.

p=1rw+fp = \frac{1}{r}w + f

Subtract f from both sides of the equation.

p−f=1rwp - f = \frac{1}{r}w

Multiply both sides by r.

r(p−f)=r(1rw)r(p - f) = r\left(\frac{1}{r}w\right)

r(p−f)=wr(p - f) = w

w=r(p−f)w = r(p - f)

Question analysis

A formula gives the number of pages p in a book from its number of words w, the average words per page r, and the extra front pages f. Students choose the formula solved for w. It tests rearranging a formula with inverse operations, and with only letters there are no numbers to check, so the order of the steps matters.

NYS standards
  • 8.EE.7 linear equations in one variable
  • 7.EE.4a two-step equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve two-step equations such as 5 = x/4 + 2 with numbers. Rearranging a formula with four letters to solve for a different one is grade 8 work, with no numbers to check against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 82 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 formula is short, but solving it for w means undoing two operations in reverse order, with letters only. The tempting shortcut multiplies p by r and leaves the f untouched, which yields a formula that looks as finished as the right one. Swapping p and w without undoing anything is the other quick trap. Substituting a small book, 5,000 words at 250 a page with 4 front pages, shows which formula gives back 5,000.

  1. Variables instead of numbers moderate Understanding the question

    Four letters and no numbers mean each choice has to be judged by its structure, unless students pick values such as r = 250, w = 5,000 and f = 4 to test it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Many steps minor Working it out

    Two inverse operations must come in order, subtracting f before multiplying by r; multiplying first means the f would have to be multiplied by r as well. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Treat it like a two-step equation with numbers: undo what was done to w last, then the step before, applying each operation to the whole of the other side.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying by the reciprocal
r×1r=1r \times \frac{1}{r} = 1
r is any number except 0. The reciprocal of r is 1/r, the number flipped upside down. A number times its reciprocal is always 1. So multiplying by r undoes the 1/r in front of w and leaves w by itself.

  1. Sort out the question

    The formula is p=1rw+fp = \frac{1}{r}w + f. Here p is the number of pages, w the number of words, r the words on each page, and f the extra front pages.

    You need the formula solved for w. That means w alone on one side of the equal sign.

  2. See what happens to w

    Read the formula like a recipe. Start with w. First it is multiplied by 1r\frac{1}{r}, which is the same as dividing by r. Then f is added.

    To get w back, undo the steps in reverse order. Take away f first. Then undo the 1r\frac{1}{r}.

  3. Subtract f from both sides

    Taking f from both sides keeps them equal:

    p−f=1rwp - f = \frac{1}{r}w

    Now the only thing left to undo is the 1r\frac{1}{r}.

  4. Multiply both sides by r

    Use the reciprocal rule. Multiply both whole sides by r:

    r(p−f)=r×1rw=wr(p - f) = r \times \frac{1}{r}w = w

    So w=r(p−f)w = r(p - f).

    Watch out: w=f+rpw = f + rp, choice C, multiplies only p by r. The f must be taken away before you multiply. Or, if you multiply first, every term gets multiplied by r.

  5. Give the answer

    Choice B only trades the places of p and w, and it undoes nothing. Choice D just adds the letters.

    So the answer is A, w=r(p−f)w = r(p - f).

Check your answer

Try a book with 5,000 words, 250 words a page and 4 front pages. The first formula gives p=5,000÷250+4=24p = 5{,}000 \div 250 + 4 = 24 pages. Choice A gives back the words: 250×(24−4)=5,000250 \times (24 - 4) = 5{,}000. Choice C gives 4+250×24=6,0044 + 250 \times 24 = 6{,}004, which is wrong.

More questions like this

The original DOE question
Question 38 as shown on the DOE’s digital test, page 111 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 111, question 38.

Question 19 of 120

2026 Practice Test C Math, question 43

Question

A softball player bought bottles of water for her team to have at practice. The total number of bottles she bought, b, is proportional to n, the number of players at practice. Which equation represents the relationship between b and n? The constant of proportionality is p.

  • Ab=pnb = pn
  • Bb=p+nb = p + n
  • Cb+n=pb + n = p
  • Db+n+p=1b + n + p = 1
Show the answer

Correct answer: A

In this context, the constant of proportionality, p, represents the relationship between b, the number of bottles of water, and n, the number of players on the team.

The number of bottles of water is the product of the constant of proportionality and the number of players.

The equation that represents this relationship is b=pnb = pn.

Question analysis

A softball player buys bottles of water for practice, and the number of bottles, b, is proportional to the number of players, n, with constant of proportionality p. Students choose the equation that represents the relationship. It tests writing an equation for a proportional relationship using letters only, so the meaning of proportional has to be clear.

NYS standards
  • 7.RP.2c proportional relationship equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 students write equations for proportional relationships from tables, graphs and descriptions, usually with a number to find. Here the constant is already given as a letter, so only the general form must be recalled, a short task.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 7, 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

This is a recall question dressed in a short story: the softball team adds nothing but names for the letters. The one idea needed is that proportional means multiplied by a constant, so doubling the players doubles the bottles. Students who think of a constant as something added, or as a fixed total, are drawn to the two sums that use the same letters.

  1. Something you must already know minor Working it out

    The general form of a proportional relationship, one amount equal to the constant times the other, has to be recalled; the stem names p but never says it multiplies. (Anderson & Krathwohl, 2001)

  2. Variables instead of numbers minor Understanding the question

    No numbers are given, only b, n and p, so a choice cannot be checked by computing unless students make up a case, such as 2 bottles per player for 10 players. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Recall how two proportional quantities are linked through the constant of proportionality, then test each choice with sample numbers: if the players double, the bottles should double too.

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

Proportional relationship
y=kxy = kx
y and x are two amounts that change together. k is the constant of proportionality, the number you always multiply x by to get y. Because it is a multiplication, doubling x doubles y.

  1. Sort out the question

    The letter b is the number of bottles, and n is the number of players. The bottles are proportional to the players. The constant of proportionality is p.

    You need the equation that links b and n. There are no numbers, only letters.

  2. Recall what proportional means

    Proportional means one amount is always a fixed number times the other. For example, say there are 2 bottles for each player. Then 10 players need 20 bottles, and 20 players need 40 bottles. Doubling the players doubles the bottles.

    The fixed number, here 2, is the constant of proportionality.

  3. Write it with these letters

    Start with the rule y=kxy = kx. The bottles take the place of y, the players take the place of x, and p is the constant:

    b=pnb = pn

    Here pn means p times n.

  4. Check the other choices

    Watch out: choice B, b=p+nb = p + n, adds instead of multiplying. With p = 2, 10 players give 12 bottles. Then 20 players give 22, which is not double.

    Choice C makes p a total of bottles and players, not a multiplier. Choice D makes all three letters add up to 1, which does not fit at all.

  5. Give the answer

    Only b=pnb = pn multiplies the players by a constant. So it is the only choice where doubling the players doubles the bottles.

    So the answer is A, b=pnb = pn.

Check your answer

Try p = 3 bottles per player. With 5 players, b=3×5=15b = 3 \times 5 = 15. With 10 players, b=3×10=30b = 3 \times 10 = 30. Twice the players, twice the bottles. Also, 15÷5=315 \div 5 = 3, so b ÷ n gives back the constant.

More questions like this

The original DOE question
Question 43 as shown on the DOE’s digital test, page 116 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 116, question 43.

Question 20 of 120

2026 Practice Test C Math, question 44

Question

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

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

The acceptable screen width is from to millimeters.

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

Show the answer

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

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

The least allowed measure is

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

The greatest allowed measure is

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

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

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding a percent of a number and adding or subtracting it is grade 7 percent work. The ± sign, the decimal result and two linked menus, which must both be right, add a little to the usual demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 59 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. Turning words into math minor Understanding the question

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

  3. A very close wrong answer minor Choosing the answer

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

  4. Many steps minor Working it out

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

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

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Find 4% of 72 mm

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

    0.04×72=2.880.04 \times 72 = 2.88

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

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

  3. Find the minimum width

    Take the change away from the design width:

    72−2.88=69.1272 - 2.88 = 69.12

    The narrowest allowed screen is 69.12 mm wide.

  4. Find the maximum width

    Add the change to the design width:

    72+2.88=74.8872 + 2.88 = 74.88

    The widest allowed screen is 74.88 mm wide.

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

  5. Choose from each menu

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

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

Check your answer

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

More questions like this

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

Question 21 of 120

2026 Practice Test C Math, question 45

Question

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

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

Correct answer: B

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

12−10=212 - 10 = 2

Divide the increase amount by the original hourly pay rate.

2÷10=0.22 \div 10 = 0.2

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

0.2×100=200.2 \times 100 = 20

Question analysis

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

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

Why it is hard

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

  1. Something you must already know minor Working it out

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

  2. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Find the increase in dollars

    Subtract the old pay from the new pay:

    12−10=212 - 10 = 2

    His pay goes up by $2 an hour.

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

  3. Compare the increase with the original

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

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

    So the pay goes up by 20%.

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

  4. Give the answer

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

    So the answer is B, 20%.

Check your answer

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

More questions like this

The original DOE question
Question 45 as shown on the DOE’s digital test, page 118 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 118, question 45.

Question 22 of 120

2026 Practice Test C Math, question 48

Question

Which table shows a proportional relationship?

  • A
    xy
    00
    432\frac{3}{2}
    83
  • B
    xy
    01
    472\frac{7}{2}
    86
  • C
    xy
    0−1
    4−12-\frac{1}{2}
    80
  • D
    xy
    00
    412
    816
Show the answer

Correct answer: A

The first table is the only one with a constant of proportionality, 38\frac{3}{8}.

In each ordered pair, 38x=y\frac{3}{8}x = y.

38(0)=0\frac{3}{8}(0) = 0

38(4)=128=32\frac{3}{8}(4) = \frac{12}{8} = \frac{3}{2}

38(8)=248=3\frac{3}{8}(8) = \frac{24}{8} = 3

Question analysis

Four tables each give y for x = 0, 4 and 8, with some values written as fractions or negatives. Students choose the table that shows a proportional relationship. It tests recognizing proportional relationships from a table, which means checking every row against one rule rather than judging by the pattern.

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 a relationship is proportional is a core grade 7 skill. The fractions and negative values in the tables make each ratio a careful check rather than a glance.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 9, 13 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 quick tests decide proportionality, and the tables are built so that each test alone lets a wrong table through. The last table starts at (0, 0) like the right one, while the middle two grow by equal steps like the right one. Only checking that y ÷ x is the same in every row with x not 0 separates the one table whose ratio stays at 3/8.

  1. Every choice must be checked moderate Choosing the answer

    Each table has to be tested row by row, and two of them start at (0, 0), so the opening row settles nothing; only y ÷ x in the other rows separates them. (Embretson & Wetzel, 1987)

  2. Error-prone arithmetic minor Working it out

    The y-values include 3/2, 7/2 and −1/2, so the ratios mean dividing fractions by 4, such as (3/2) ÷ 4 = 3/8, and comparing them with 3 ÷ 8. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: For each table, divide y by x in every row where x is not zero, and check that the quotient never changes and that y is 0 when x is 0.

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.
  • Checks only one number: Tries one value in each choice and takes the first that works, when that value happens to work for a wrong choice too; a second value would rule it out.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportional relationship
y=kxy = kx
y and x are the two amounts in each row. k is the constant of proportionality, the same number in every row. Two tests follow from it. When x is 0, y must be 0. And y ÷ x must equal k in every row where x is not 0.

  1. Sort out the question

    Each table gives y for x = 0, 4 and 8. You need the one that shows a proportional relationship. That means y is always the same number times x.

    Use two tests. First, y must be 0 when x is 0. Second, y ÷ x must be the same in the other rows.

  2. Test the row where x is 0

    Look at the top row of each table:

    • Table A: y = 0. It passes.
    • Table B: y = 1. It fails.
    • Table C: y = −1. It fails.
    • Table D: y = 0. It passes.

    Watch out: tables B and C go up by equal steps. Steady steps are not enough, because they do not start at 0. Table C does have a 0, but it is next to x = 8.

  3. Test the ratio in table A

    Divide y by x. Dividing by 4 is the same as multiplying by 14\frac{1}{4}:

    32÷4=32×14=38\frac{3}{2} \div 4 = \frac{3}{2} \times \frac{1}{4} = \frac{3}{8}

    3÷8=383 \div 8 = \frac{3}{8}

    Both rows give 38\frac{3}{8}. Table A passes both tests.

  4. Test the ratio in table D

    Divide y by x in the same way:

    12÷4=316÷8=212 \div 4 = 3 \qquad 16 \div 8 = 2

    The ratios are different, so table D is not proportional.

    Watch out: table D starts at (0, 0) like a proportional table. If you check only that row, or only the row with 4, you will pick choice D.

  5. Give the answer

    Only table A starts at (0, 0) and keeps y ÷ x the same.

    So the answer is A, the table with y = 0, 32\frac{3}{2} and 3.

Check your answer

Use the rule for table A, y=38xy = \frac{3}{8}x. When x = 4, y=38×4=128=32y = \frac{3}{8} \times 4 = \frac{12}{8} = \frac{3}{2}. When x = 8, y=38×8=3y = \frac{3}{8} \times 8 = 3. Both match. Doubling x from 4 to 8 also doubles y.

More questions like this

The original DOE question
Question 48 as shown on the DOE’s digital test, page 121 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 121, question 48.

Question 23 of 120

2026 Practice Test C Math, question 49

Question

The box plot illustrates the number of hours eighth-grade students spend on screens during a typical week. What is the difference between the range and the interquartile range (IQR) of the screen time data?

A box plot titled NUMBER OF SCREEN HOURS FOR 8th GRADE STUDENTS above a number line labeled Hours, from 5 to 20 with a labeled tick at every 1. The left whisker starts at 5, the box runs from 8 to 15 with the median line at 12, and the right whisker ends at 20.

Move the correct answer to each box. Not all answers will be used.

The range of screen time hours is hours.

The IQR is hours.

The difference between the range and the IQR is hours.

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: 15; blank 2: 7; blank 3: 8

Determine the range by subtracting the minimum value on the box plot from the maximum value on the box plot:

Maximum value =20 hr= 20\text{ hr}, Minimum value =5 hr= 5\text{ hr}

20 hr−5 hr=15 hr20\text{ hr} - 5\text{ hr} = 15\text{ hr}

Determine the interquartile range by subtracting the lower quartile value on the box plot from the upper quartile value on the box plot:

Upper Quartile Value =15 hr= 15\text{ hr}, Lower Quartile Value =8 hr= 8\text{ hr}

15 hr−8 hr=7 hr15\text{ hr} - 8\text{ hr} = 7\text{ hr}

Determine the difference between the range (15 hr) and the interquartile range (7 hr):

15 hr−7 hr=8 hr15\text{ hr} - 7\text{ hr} = 8\text{ hr}

Question analysis

A box plot shows how many hours a week eighth-grade students spend on screens, on a number line marked every hour. Students move number tiles to complete three sentences: the range, the interquartile range, and the difference between them. Each tile can be used only once, and keeping the range and the interquartile range apart is the tricky part.

NYS standards
  • 7.SP.1 box plots and interquartile range
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Reading a box plot and finding its range and interquartile range are grade 7 statistics skills. The values sit on labeled whole-number ticks and the three sentences lead through the steps, a typical grade 7 task.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 69 words.
Corrected from the DOE handbook
  • A missing comma in the explanation was restored.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The plot is drawn on whole-hour ticks, and the three sentences lay out the steps, so this is a direct check of two definitions. The main risk is taking a value off the plot, such as the lower quartile of 8, for a spread between two values. Because the four tiles allow only one use each, a student who confuses the measures can still fill every box, which hides the error.

  1. Something you must already know minor Working it out

    The sentences name the range and the IQR but never define them; students must know the range spans the whiskers and the IQR spans only the box. (Anderson & Krathwohl, 2001)

  2. Many steps minor Working it out

    Three boxes must all be right, the third depends on the first two, and each tile can be used once, so one mix-up spreads to a second box. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Read the five key values off the plot, subtract the least from the greatest for the range, subtract the lower quartile from the upper for the IQR, then subtract those results.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation runs two values together: “Maximum value = 20 hr Minimum value = 5 hr”. Its next line is “Upper Quartile Value = 15 hr, Lower Quartile Value = 8 hr”, and Question 32’s explanation has “Maximum value = 84° F, Minimum value = 39° F”. The site adds the comma.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range
range=maximum−minimum\text{range} = \text{maximum} - \text{minimum}
The maximum is the greatest value, at the right end of the right whisker. The minimum is the least value, at the left end of the left whisker. The range tells how far the whole set of data spreads.
Interquartile range (IQR)
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
Q3 is the upper quartile, the right edge of the box. Q1 is the lower quartile, the left edge of the box. The IQR is the width of the box. It tells how far the middle half of the data spreads.

  1. Sort out the question

    The box plot shows the hours of screen time for eighth graders in a week. You need three numbers: the range, the IQR, and the difference between them.

    There are four tiles: 5, 7, 8 and 15. Each tile is used at most once, and one tile is left over.

  2. Read the five values

    The number line has a tick at every hour, and every value sits on a tick:

    • The left whisker starts at 5. This is the minimum.
    • The box starts at 8. This is the lower quartile, the middle of the lower half of the data.
    • The line inside the box is at 12. This is the median, the middle value.
    • The box ends at 15. This is the upper quartile, the middle of the upper half.
    • The right whisker ends at 20. This is the maximum.

  3. Find the range

    Use the range formula with the two ends of the whiskers:

    range=20−5=15\text{range} = 20 - 5 = 15

    The screen times spread over 15 hours, from the least to the most.

  4. Find the IQR

    Use the IQR formula with the two edges of the box:

    IQR=15−8=7\text{IQR} = 15 - 8 = 7

    The middle half of the students spend between 8 and 15 hours, a spread of 7 hours.

    Watch out: the IQR is a width, not a point on the plot. The tile 8 is the lower quartile. Putting it in the IQR box also makes the last box 7, which is wrong.

  5. Find the difference

    Subtract the IQR from the range:

    15−7=815 - 7 = 8

    Watch out: do not swap the two measures. The range uses the whiskers’ ends, and the IQR uses only the box.

  6. Fill in the three boxes

    On the test, you drag a tile into each box. On this site, you choose each tile from its box’s menu. The range box gets 15, the IQR box gets 7, and the difference box gets 8. The tile 5 is not used.

    So the boxes get 15, 7 and 8.

Check your answer

The difference is just the two whiskers added together. The left whisker is 8−5=38 - 5 = 3 hours long. The right whisker is 20−15=520 - 15 = 5 hours long. 3+5=83 + 5 = 8, which matches the third box.

More questions like this

The original DOE question
Question 49 as shown on the DOE’s digital test, page 122 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 122, question 49. This is the DOE’s version, before the corrections listed in the analysis.

Question 24 of 120

2026 Practice Test D Math, question 1

Question

What is the prime factorization of 3,575?

  • A5⋅11⋅135 \cdot 11 \cdot 13
  • B52⋅11⋅135^{2} \cdot 11 \cdot 13
  • C5⋅7155 \cdot 715
  • D52⋅1435^{2} \cdot 143
Show the answer

Correct answer: B

By observation, 3,575 ends in 75; therefore, 3,575 is divisible by 25.

3,575÷25=1433{,}575 \div 25 = 143

Test 143 for factors and eliminate 3, 5, and 7, but 11 is a factor.

143÷11=13143 \div 11 = 13; both 11 and 13 are prime numbers.

The first factor, 25, is not prime but is the product of primes, 5×5=525 \times 5 = 5^{2}.

The prime factorization of 3,575 is 52×11×135^{2} \times 11 \times 13.

Question analysis

This number question asks for the prime factorization of 3,575, chosen from four products of factors. Students divide by primes, using divisibility tests, until only primes remain. The four products look much alike, so the work must be carried all the way through, keeping count of every prime divided out along the way.

NYS standards
  • 6.NS.4 common factors and multiples
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Breaking a number into primes is grade 6 number work that relies on divisibility tests. The number here is larger than in most class examples, but the work is still a few divisions, lighter than typical grade 7 problems.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 11, 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

Three of the four choices multiply back to 3,575, so checking a product rules out only the one that equals 715. The question turns on whether 143 is prime: it survives the quick tests for 2, 3 and 5, and the factor 11 is easy to miss. Students who keep dividing until every factor is prime, and who write a prime as often as it divides, finish in under a minute.

  1. A very close wrong answer moderate Choosing the answer

    5² · 143 matches the right factorization in its first factor and multiplies back to 3,575 just as well, so a product check cannot separate them; only splitting 143 into 11 × 13 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Nothing on the page defines a prime factorization or gives a test for 11, and 143 passes the familiar quick tests for 2, 3 and 5, so it looks prime. (Anderson & Krathwohl, 2001)

Strategy: Divide by the smallest prime that goes in evenly, again and again, writing down every prime you use until the quotient reaches 1, then compare that list with each choice.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisibility by 25
last two digits 00,25,50 or 75→divisible by 25\text{last two digits } 00, 25, 50 \text{ or } 75 \rightarrow \text{divisible by } 25
A number is divisible by 25 when 25 goes into it with no remainder. You only need to look at the last two digits. If they are 00, 25, 50 or 75, the number is divisible by 25. This works because 100 is 4 × 25, so the hundreds never leave a remainder.
Writing a repeated factor with an exponent
a2=a×aa^{2} = a \times a
a is any number. The small raised 2 is an exponent. It tells you how many times a is used as a factor, so 5² means 5 × 5 = 25. Use an exponent in a prime factorization when the same prime shows up more than once.

  1. Sort out the question

    A prime number is a whole number greater than 1 whose only factors are 1 and itself. Examples are 2, 3, 5, 7, 11 and 13. The prime factorization of 3,575 writes it as a product (the answer to a multiplication) of prime numbers only. So you must break 3,575 down until every factor is prime.

  2. Divide by 25 first

    3,575 ends in 75. So it is divisible by 25:

    3,575÷25=1433{,}575 \div 25 = 143

    That means 3,575 = 25 × 143. But 25 is not prime. It is 5 × 5, which you can write as 5².

  3. Test 143 for prime factors

    Now see if 143 splits. Try the primes in order. 143 is odd, so 2 does not work. Its digits add to 1 + 4 + 3 = 8, which is not a multiple of 3, so 3 does not work. It does not end in 0 or 5, so 5 does not work. Since 7 × 20 = 140, 143 ÷ 7 leaves a remainder of 3, so 7 does not work. Now try 11:

    143÷11=13143 \div 11 = 13

    So 143 = 11 × 13, and both 11 and 13 are prime.

    Watch out: 143 looks prime, but it is not. Stopping at 5² · 143 leads to choice D.

  4. Put all the primes together

    Collect every prime you found. The 5 appears twice, then 11 and 13 once each. The dot · means times, just like ×.

    3,575=5×5×11×13=52⋅11⋅133{,}575 = 5 \times 5 \times 11 \times 13 = 5^{2} \cdot 11 \cdot 13

    Watch out: write the 5 twice, since it divides 3,575 twice. Writing it once gives 5 · 11 · 13, choice A, which equals only 715. Choice C, 5 · 715, stops too early, because 715 ends in 5 and splits further.

    So the answer is B, 52⋅11⋅135^{2} \cdot 11 \cdot 13.

Check your answer

Multiply the primes back together: 25 × 11 = 275, and 275 × 13 = 3,575. That is the number you started with. Also, 5, 11 and 13 are all prime, so nothing can be split further.

More questions like this

The original DOE question
Question 1 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 D Math, page 74, question 1.

Question 25 of 120

2026 Practice Test D Math, question 9

Question

Let m be a negative integer and k be a positive integer.

Which of the expressions below must result in a positive value?

Select the three correct answers.

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

  • A−m+k-m + k
  • B−m⋅k-m \cdot k
  • Cm−km - k
  • D−k⋅m-k \cdot m
  • Emk\frac{m}{k}
Show the answer

Correct answer: A, B, and D

Each of the expressions must be considered:

−m+k-m + k, where m is a negative integer and k is a positive integer: Correct

The “−” in front of m acts like a −1-1 multiplier that flips the sign of m, so −m+k-m + k is positive when m is negative and k is positive.

−m⋅k-m \cdot k, where m is a negative integer and k is a positive integer: Correct

The “−” in front of m acts like a −1-1 multiplier that flips the sign of m, so −m⋅k-m \cdot k is positive when m is negative and k is positive.

m−km - k, where m is a negative integer and k is a positive integer:

A negative number minus a positive number gives a negative number.

−k⋅m-k \cdot m, where m is a negative integer and k is a positive integer: Correct

A positive times a negative is a negative. Then, taking the opposite of that makes it positive.

mk\frac{m}{k}, where m is a negative integer and k is a positive integer:

A negative divided by a positive remains negative.

Therefore, −m+k-m + k, −m⋅k-m \cdot k, and −k⋅m-k \cdot m must all result in a positive number.

Question analysis

Given that m is a negative integer and k a positive integer, students select the three expressions, out of five, that must have a positive value. It tests the sign rules for adding, subtracting, multiplying and dividing integers. Minus signs in front of letters, and letters in place of numbers, make each sign easy to misjudge, and all three must be found.

NYS standards
  • 7.NS.2a rational number multiplication
  • 7.NS.1c rational number subtraction
  • 7.NS.2b rational number division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Sign rules for adding, subtracting, multiplying and dividing integers are core grade 7 content. Here they are applied to letters with known signs instead of numbers, across five expressions, and all three right ones must be chosen with no wrong one.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 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

Each expression needs only one sign rule, but the minus signs are placed to mislead. In −k · m the leading minus seems to make the result negative until the sign of m is counted, and in m − k two negatives seem to cancel, though subtracting makes the result more negative. Substituting one negative and one positive value into all five gives three positive results and two negative ones, the count the item asks for.

  1. Variables instead of numbers moderate Understanding the question

    Only the signs of m and k are known, so the sign of each expression must be reasoned from the rules or tested with values such as m = −2 and k = 3. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    All five expressions mix the same two letters with minus signs, so none can be dismissed at a glance; each is judged on its own and exactly three must be picked. (Embretson & Wetzel, 1987)

  3. Much to keep track of minor Working it out

    The item scores only when all three positive expressions are chosen and both negative ones left out, so five separate verdicts must be kept and combined. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Substitute sample values, such as m = −2 and k = 3, into all five expressions and note each sign, then confirm with the sign rules that no other values could change it.

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.)
  • 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

The opposite of a number
−(−a)=a-(-a) = a
a is any positive number, so −a is negative. A minus sign in front of a number or letter means its opposite: the same distance from 0, on the other side. The opposite of a negative number is positive. So if m is negative, −m is positive.
Signs when you multiply or divide
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers, so −a and −b are negative. Two negatives multiply to a positive. A negative and a positive multiply or divide to a negative. These rules tell you the sign of an answer without knowing the numbers.

  1. Sort out the question

    m is a negative integer, a whole number below 0, such as −2. k is a positive integer, such as 3. You must pick the three expressions that are always positive. Check all five, and test m = −2 and k = 3 as you go.

  2. Understand the minus in front

    A minus sign in front of a letter means the opposite of that number. Since m is negative, −m is positive. Since k is positive, −k is negative. With the test values, −m = 2 and −k = −3.

  3. Check choices A and B

    A: −m+k-m + k is a positive plus a positive, so it is always positive. Test: 2 + 3 = 5.

    B: −m⋅k-m \cdot k is a positive times a positive, so it is always positive. The dot · means times. Test: 2 × 3 = 6.

  4. Check choices C and E

    C: m−km - k starts below 0 and moves further down, so it is always negative. Test: −2 − 3 = −5.

    E: mk\frac{m}{k} is a negative divided by a positive, so it is always negative. Test: −2 ÷ 3 = −2/3.

    Watch out: two negatives make a positive only when you multiply or divide them. Choice C subtracts, so it stays negative. Choice E shows no minus sign, but m itself is negative.

  5. Check choice D

    D: −k⋅m-k \cdot m is a negative times a negative, so it is always positive. Test: −3 × (−2) = 6.

    Watch out: the minus sign in front makes D look negative. But m is negative too, and the two negatives make a positive. Skipping D leaves you one answer short.

  6. Choose the three answers

    Choices C and E are always negative, so leave them out. The three positive ones are A, B and D. On the test, you check the box next to each one.

    So the answers are A, B and D.

Check your answer

Try different values, such as m = −10 and k = 1. A: 10 + 1 = 11. B: 10 × 1 = 10. D: −1 × (−10) = 10. All three are positive again. C gives −11 and E gives −10, both negative.

More questions like this

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

Question 26 of 120

2026 Practice Test D Math, question 11

Question

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

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

Correct answer: A

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Turning words into math minor Understanding the question

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

  2. Many steps minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the drawing’s perimeter

    Add the three sides of the drawing:

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

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

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

  3. Find the scale

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

    25÷20=1.2525 \div 20 = 1.25

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

  4. Scale the shortest side

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

    5×1.25=6.255 \times 1.25 = 6.25

    The real shortest side is 6.25 miles.

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

    So the answer is A, 6.25 mi.

Check your answer

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

More questions like this

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

Question 27 of 120

2026 Practice Test D Math, question 19

Question

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

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

Correct answer: C

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

456×1.25=570456 \times 1.25 = 570

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

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

The sales tax on the selling price is:

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

The customer pays the selling price plus the sales tax:

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

Question analysis

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

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

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

  2. Many steps moderate Working it out

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

  3. Error-prone arithmetic minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the increase

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

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

    The increase is $570.

  3. Find the selling price

    Increases by means the increase is added to the cost:

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

    The selling price is $1,026.

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

  4. Find the sales tax

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

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

    The tax is $82.08.

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

  5. Add the tax to the price

    The customer pays the selling price plus the tax:

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

    So the answer is C, $1,108.08.

Check your answer

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

More questions like this

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

Question 28 of 120

2026 Practice Test D Math, question 23

Question

Which situation involves quantities that combine to equal zero?

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

Correct answer: A

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

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

Question analysis

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

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

Why it is hard

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

  1. Turning words into math minor Understanding the question

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

  2. Every choice must be checked minor Choosing the answer

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Write each choice with signs

    Turn each situation into signed numbers and add them up:

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

  3. Look at the tricky choices

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

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

  4. Choose the zero total

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

    5+(−5)=05 + (-5) = 0

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

Check your answer

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

More questions like this

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

Question 29 of 120

2026 Practice Test D Math, question 25

Question

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

  • A8
  • B10
  • C12
  • D14
Show the answer

Correct answer: B

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

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

450÷45=10 hours450 \div 45 = 10\text{ hours}

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Speed, distance and time problems are grade 6 rate work, and this one uses whole numbers in two steps. Its only twist is how the time responds when the speed changes, which makes it easier than typical grade 7 rate problems.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 3, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. No obvious method moderate Working it out

    The question never mentions the trip's length, yet the 450-mile distance is what stays fixed; setting up a proportion between hours and speeds, the usual reflex, points the wrong way. (Schoenfeld, 1985)

  2. Something you must already know minor Working it out

    Students must bring distance = speed × time themselves, to turn 9 hours at 50 miles per hour into miles before dividing by the new speed. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    A bus trip takes 9 hours at a mean speed of 50 miles per hour. The mean speed is the average speed over the whole trip. You need the time for the same trip at 45 miles per hour. The length of the trip does not change, so find it first.

  2. Find the length of the trip

    Use distance equals speed times time, with a speed of 50 miles per hour and a time of 9 hours:

    d=50×9=450d = 50 \times 9 = 450

    The trip is 450 miles long.

  3. Find the new time

    Now the bus goes 45 miles per hour over the same 450 miles. Divide the distance by the speed:

    t=450÷45=10t = 450 \div 45 = 10

    The trip takes 10 hours.

  4. Check that it makes sense

    A slower bus needs more time for the same trip. 10 hours is more than 9 hours, so it fits.

    Watch out: time does not shrink with speed. Scaling 9 hours by 45/50 gives about 8 hours, choice A, which would make the slower bus arrive sooner. Adding the drop of 5 miles per hour to 9 hours gives 14, choice D, which has no reason behind it.

    So the answer is B, 10 hours.

Check your answer

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

More questions like this

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

Question 30 of 120

2026 Practice Test D Math, question 28

Question

A graph shows the relationship between the number of gallons of water, y, that has been added to a tank and the number of hours, x, that water has been added to the tank at a constant rate. What does the ordered pair (3, 24) represent?

  • AThree gallons of water was added in 24 hours.
  • BThree gallons of water was added per hour.
  • CTwenty-four gallons of water was added in 3 hours.
  • DTwenty-four gallons of water was added per hour.
Show the answer

Correct answer: C

It is given that x-values on the graph represent the numbers of hours for which water was added to the tank, and the y-values represent the numbers of gallons of water added in x hours. In the ordered pair (3, 24), 3 represents an x-value and 24 represents a y-value.

Therefore, (3, 24) represents a data point for 24 gallons of water being added to the tank in a 3-hour period.

Question analysis

A graph relates the gallons of water added to a tank, y, to the hours it has been filling at a constant rate, x. Students choose what the ordered pair (3, 24) means in the situation. It tests interpreting coordinates in context, and the four statements mix totals with rates and switch the two numbers around.

NYS standards
  • 7.RP.2d points on proportional graphs
  • 6.EE.9 dependent and independent variables
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Explaining what a point on the graph of a proportional relationship means is named in the grade 7 standards. This is a direct case with no graph to read, though the stem introduces the two quantities in an order that needs care.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
1 step
Reading level
Question and choices: about grade 5, 76 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 fact decides the question: an ordered pair lists x first. The choices then test two confusions, which number is the gallons and whether a point is a rate. Reading the pair aloud as “after 3 hours, 24 gallons”, and noting that 24 ÷ 3 gives a rate of 8 gallons an hour that no choice states, rules out all three wrong statements.

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

    No graph is drawn, and the stem names y, the gallons, before x, the hours, so the pair (3, 24) has to be matched to its letters, not to the order the quantities were mentioned. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. A very close wrong answer minor Choosing the answer

    The four statements reuse 3 and 24 in two frames, a total added “in” some hours and an amount “per hour”, so the right statement and its swapped twin differ only in which number counts gallons. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Recall that an ordered pair lists x first and y second, write down what x and y measure here, and read the pair as one sentence about hours and gallons.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

An ordered pair
(x,y)(x, y)
An ordered pair names one point on a graph. The first number is always the x-value. The second number is always the y-value. Here x counts hours and y counts gallons. So a pair tells how many gallons were added after that many hours.
Rate from a point
rate=yx\text{rate} = \frac{y}{x}
When water is added at a constant rate, divide the y-value by the x-value. The result is the rate: gallons added in 1 hour. A point itself shows a total, not a rate.

  1. Sort out the question

    The graph links two amounts:

    • x is the number of hours water has been added.
    • y is the number of gallons added.

    You need to say what the point (3, 24) means. No graph is drawn, so you work from the words.

  2. Match each number to its letter

    In an ordered pair, x always comes first and y second. So in (3, 24):

    • x = 3, which means 3 hours.
    • y = 24, which means 24 gallons.

    Watch out: the question names gallons first and hours second. Do not match the numbers to the words in that order. That mistake reads 3 as gallons and 24 as hours, choice A.

  3. Read the pair as a sentence

    Put the two meanings together: after 3 hours, 24 gallons have been added in all. The 24 is a total for the whole 3 hours.

    Watch out: 24 is not the amount for each hour. The rate per hour is 24÷3=824 \div 3 = 8 gallons. So choice D, 24 gallons per hour, is wrong. Choice B, 3 gallons per hour, is also not the rate.

    So the answer is C, twenty-four gallons of water was added in 3 hours.

Check your answer

Use the rate of 8 gallons each hour. In 3 hours, 3×8=243 \times 8 = 24 gallons go in. That matches choice C. No choice says 8 gallons per hour, so both “per hour” choices must be wrong.

More questions like this

The original DOE question
Question 28 as shown on the DOE’s digital test, page 101 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 101, question 28.

Question 31 of 120

2026 Practice Test D Math, question 31

Question

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

  • A80
  • B300
  • C400
  • D500
Show the answer

Correct answer: C

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

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

2x=8002x = 800

x=400x = 400

Question analysis

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

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

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

  2. Much to keep track of minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    You know:

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

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

  2. Set up the proportion

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

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

  3. Scale up the sample

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

    8×50=4008 \times 50 = 400

    So the box has about 400 green marbles.

  4. Avoid the tempting wrong totals

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

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

    So the answer is C, 400.

Check your answer

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

More questions like this

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

Question 32 of 120

2026 Practice Test D Math, question 34

Question

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

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

Correct answer: C

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

Question analysis

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

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 6.SP.8b probability models from observed frequencies
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Compound probability is grade 7 work, often done with a table of outcomes. Each chance here comes from one week's frequencies, and joining the morning and afternoon events in one step is about the usual grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 48 words.
Corrected from the DOE handbook
  • An inaccurate detail in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know moderate Working it out

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

  2. Turning words into math minor Understanding the question

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

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

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Last week Emily:

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

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

  2. Write each chance

    Use last week to find each chance:

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

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

  3. Multiply the two chances

    Both rides must happen, so multiply:

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

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

  4. Avoid the tempting wrong choices

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

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

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

Check your answer

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

More questions like this

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

Question 33 of 120

2026 Practice Test D Math, question 38

Question

Melissa has a bag of marbles that are all the same size. The bag contains three red (R) marbles, three green (G) marbles, and two white (W) marbles. For an experiment, she will take two marbles out of the bag without looking. Which list shows the sample space for Melissa’s experiment?

  • AR, G, W
  • BRG, RW, GW
  • CR, R, R, G, G, G, W, W
  • DRR, RG, RW, GG, GW, WW
Show the answer

Correct answer: D

The sample space of a random experiment is the collection of all possible outcomes. For this experiment, R represents a red marble, G represents a green marble, and W represents a white marble.

Since two marbles are being drawn, the possible outcomes for the first marble drawn are R, G, W.

List the possible results for each initial outcome combined with the second choice of marble. For example, if the first marble and second marble are both red, the sample space will include RR (red, red).

RR, RG, RW

GR, GG, GW

WR, WG, WW

Since order does not matter in this selection, remove any pairing that presents a duplicate choice, such as RG and GR—only one of the two pairings will be included in the sample space. Those are removed and leave a sample space of:

RR, RG, RW, GG, GW, WW

Question analysis

A bag holds three red, three green and two white marbles, and Melissa will take out two without looking. This probability question asks which list shows the sample space for her experiment. Students must know what a sample space is and what one outcome of drawing two marbles looks like, then decide which color pairs are possible.

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 Representing the sample space of a compound event is a named grade 7 skill. Nothing is computed here; the demand is knowing what one outcome of drawing two marbles looks like and which pairs the bag allows, much as in grade 7 classwork.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 3, 71 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question tests a definition more than a skill, and three choices describe something real: the colors of one marble, the contents of the bag, and the pairs of different colors. The strongest pull comes from treating the colors as three single objects, which leaves out the pairs of one color although the bag holds three reds, three greens and two whites. Asking what Melissa could be holding after the draw settles it.

  1. Something you must already know moderate Working it out

    The term sample space is not defined in the question: students must know it means every possible result of the whole experiment, here a pair of colors drawn together. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The three mixed pairs, RG, RW and GW, form a complete-looking list that is the right list minus RR, GG and WW, so only asking whether two marbles of one color can come out together separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Describe one possible result of taking out two marbles, then build the list systematically, starting from each color and pairing it with every color that could come out with it.

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.
  • Without replacement: In a second pick there is one fewer item, which changes the probability.
  • 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

Sample space
sample space=the list of every possible outcome\text{sample space} = \text{the list of every possible outcome}
An outcome is one possible result of an experiment. The sample space lists every outcome once. For this experiment, one outcome is the pair of colors Melissa holds after she takes out two marbles.

  1. Sort out the question

    The bag holds 3 red (R), 3 green (G) and 2 white (W) marbles. Melissa takes out 2 marbles without looking.

    You need the sample space: every possible result of taking out 2 marbles. So each outcome must name two colors, like RG for one red and one green.

  2. List pairs by the first color

    Start with each color the first marble could be. Then pair it with each color the second could be:

    • First red: RR, RG, RW
    • First green: GR, GG, GW
    • First white: WR, WG, WW

    That gives 9 pairs.

  3. Remove pairs that repeat

    Melissa ends up holding two marbles. One red and one green is the same result whichever came out first. So GR is the same as RG. Cross out GR, WR and WG. Six pairs are left:

    RR, RG, RW, GG, GW, WW

  4. Check the same-color pairs

    Can both marbles be the same color? Yes. The bag has 3 reds, 3 greens and 2 whites, so RR, GG and WW can all happen.

    Watch out: choice B leaves out RR, GG and WW. That would only be right if the bag had one marble of each color. Choice A is the colors of one marble. Choice C lists the marbles in the bag, not results.

    So the answer is D, RR, RG, RW, GG, GW, WW.

Check your answer

Ask what Melissa could be holding. Two reds? Yes, there are 3. Two whites? Yes, there are exactly 2. A green and a white? Yes. Every pair in D can happen, and no possible pair is missing. Choice B misses three pairs that can happen.

More questions like this

The original DOE question
Question 38 as shown on the DOE’s digital test, page 111 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 111, question 38.

Question 34 of 120

2026 Practice Test D Math, question 39

Question

The table shows the probabilities of a hockey team scoring different numbers of goals in a game.

HOCKEY GOAL PROBABILITIES

Number of Goals Probability
0 0.16
1 0.21
2 0.29
3 0.21
4 0.09
5 0.04

What is the probability that the team will score three or more goals in the next game?

  • A0.13
  • B0.21
  • C0.34
  • D0.50
Show the answer

Correct answer: C

To find the probability that any one of several mutually exclusive events occurs, use the addition rule, and add the probabilities of each event:

P(A or B)=P(A)+P(B)\mathrm{P}(\mathrm{A} \text{ or } \mathrm{B}) = \mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B})

Using the data in the table, the probability that the team will score 3 or more goals in the next game is represented by

P(3)+P(4)+P(5)=\mathrm{P}(3) + \mathrm{P}(4) + \mathrm{P}(5) =

0.21+0.09+0.04=0.34.0.21 + 0.09 + 0.04 = 0.34.

Question analysis

A table gives the probabilities of a hockey team scoring 0 to 5 goals in a game, and students find the chance that it scores three or more goals in the next one. This probability question tests adding the probabilities of separate outcomes. Reading exactly which rows the phrase covers is the main point of care.

NYS standards
  • 6.SP.8 probability models for simple events
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the probability of an event from a given probability model is grade 6 work in New York. Here the model is a table, and the task is choosing the right rows and adding decimals, shorter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 8, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Adding three decimals is easy, so the question turns on which rows the words cover. Including the 3-goal row, and not stopping at it, are the two decisions, and each misreading gives a listed choice. Checking the complement, 1 − (0.16 + 0.21 + 0.29), confirms both the sum and the choice of rows.

  1. Turning words into math minor Understanding the question

    The phrase “three or more goals” has to become the three rows for 3, 4 and 5 goals, starting at the 3-goal row itself. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table lists six probabilities, two of them both 0.21, so the rows for 3, 4 and 5 goals must be picked out by their goal counts, not by their values. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide which rows the words "three or more" cover, add their probabilities, and check by subtracting the total of the other rows from 1.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding probabilities
P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)
A and B are two events that cannot both happen at once, like scoring exactly 3 goals and exactly 4 goals. The chance that one or the other happens is the sum of their chances.
The chance it does not happen
P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)
The chances of all possible outcomes add up to 1. So the chance an event does not happen is 1 minus the chance it does. Use it to check an answer from the other side.

  1. Sort out the question

    The table gives the chance of each number of goals, from 0 to 5. You need the chance the team scores three or more goals in the next game.

  2. Pick the right rows

    Three or more means 3 goals, 4 goals or 5 goals. The row for 3 counts too. Read those three rows:

    • 3 goals: 0.21
    • 4 goals: 0.09
    • 5 goals: 0.04

    Watch out: leaving out the 3-goal row gives 0.09 + 0.04 = 0.13, choice A. Using only the 3-goal row gives 0.21, choice B.

  3. Add the chances

    A game can end with only one number of goals. So add the three chances:

    0.21+0.09+0.04=0.340.21 + 0.09 + 0.04 = 0.34

    Watch out: three of the six rows is half the rows, but that does not make the chance 0.50, choice D. The rows have different chances.

    So the answer is C, 0.34.

Check your answer

Find the chance of fewer than 3 goals, the other rows: 0.16+0.21+0.29=0.660.16 + 0.21 + 0.29 = 0.66. Then the chance of 3 or more is 1−0.66=0.341 - 0.66 = 0.34. Both ways give 0.34.

More questions like this

The original DOE question
Question 39 as shown on the DOE’s digital test, page 112 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 112, question 39.

Question 35 of 120

2026 Practice Test D Math, question 40

Question

A team is participating in a contest to collect canned goods for a food drive. The team has 53 points and must have a total of at least 100 points to win a prize. The team earns 5 points for each can of food a member collects. Which inequality can be used to find x, the number of cans of food the team needs to collect to win a prize?

  • A5+53x≤1005 + 53x \leq 100
  • B5+53x≥1005 + 53x \geq 100
  • C53+5x≤10053 + 5x \leq 100
  • D53+5x≥10053 + 5x \geq 100
Show the answer

Correct answer: D

Since each can is worth 5 points, the total number of points earned from all the cans, x, is 5x. The team started with 53 points. That is a constant value and doesn’t change. So the total number of points the team will earn is 53+5x53 + 5x. The team wants to earn at least (≥) 100 points, so they need 53+5x≥10053 + 5x \geq 100 total points.

Question analysis

A team has 53 points in a canned-food drive, earns 5 points for each can, and needs at least 100 points to win a prize. Students choose the inequality for x, the number of cans the team must collect. It tests writing an inequality from words, where the starting points and the points per can must each take the right place.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing an inequality for a word problem with a starting amount and a rate is grade 7 work. The four choices pair two expressions with two symbols, so students make one decision about the terms and one about the phrase "at least".
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 71 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The choices form a two-by-two grid: which number goes with x, and which way the symbol points. Each decision can be made separately, so a student who settles both gets the item, while one mistake lands on a listed choice. Testing two numbers of cans, 5, which leaves the team at 78 points, and 10, which lifts it to 103, rules out every wrong inequality.

  1. Turning words into math moderate Understanding the question

    The 53 points the team “has” are a fixed start, the “5 points for each can” go with x, and “at least 100” calls for ≥, though “least” can sound like less. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    The four choices are two expressions, 5 + 53x and 53 + 5x, each with ≤ and with ≥, so the right one and the strongest wrong one differ only in the symbol. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out the team's total points after 2 cans and then after x cans, translate "at least 100" into a symbol, and compare the result with each choice.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total with a start and a rate
total=start+rate×x\text{total} = \text{start} + \text{rate} \times x
The start is the amount you have now, here 53 points. The rate is what each item adds, here 5 points for each can. x is the number of items. The start is added once, and the rate is multiplied by x.
At least
total≥goal\text{total} \geq \text{goal}
The sign ≥ means is greater than or equal to. At least 100 means 100 or more. So the total must be greater than or equal to 100.

  1. Sort out the question

    You know:

    • The team already has 53 points.
    • It earns 5 points for each can.
    • It needs at least 100 points to win.

    x is the number of cans. You need the inequality that shows this.

  2. Write the total points

    Try 2 cans first. The team gets 5 points for each, so 53+5×2=6353 + 5 \times 2 = 63 points. For x cans, the total is:

    53+5x53 + 5x

    Watch out: the 53 points do not repeat for each can. 5+53x5 + 53x gives 53 points for each can, as in choices A and B.

  3. Choose the sign

    At least 100 means 100 or more, so use ≥:

    53+5x≥10053 + 5x \geq 100

    Watch out: “least” sounds like “less,” but at least means the total can be 100 or higher. Using ≤ gives choice C.

    So the answer is D, 53+5x≥10053 + 5x \geq 100.

Check your answer

Test two numbers of cans. With 10 cans, 53+5×10=10353 + 5 \times 10 = 103, and 103 ≥ 100 is true. The team wins. With 9 cans, 53+5×9=9853 + 5 \times 9 = 98, and 98 ≥ 100 is false. The team is short. Choice D gets both right.

More questions like this

The original DOE question
Question 40 as shown on the DOE’s digital test, page 113 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 113, question 40.

Question 36 of 120

2026 Practice Test D Math, question 46

Question

A company charges $3.00 to ship a 4-pound package. The cost includes a base fee of $1.50 plus $0.50 for each pound over the first pound. The table below illustrates how the cost depends on the total weight of the package. Complete the table by matching each total weight to its correct total shipping cost.

Move the correct answer to each box. Not all answers will be used.

SHIPPING COSTS

Total Weight (pounds) Total Cost to Ship (dollars)
2
2.50
6
5.50

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

Show the answer

Correct answer: blank 1: 2; blank 2: 3; blank 3: 4; blank 4: 9

Total Weight (pounds) Total Cost to Ship (dollars)
2 2
3 2.50
6 4
9 5.50

Let x represent the weight of the package in pounds. The first pound is included in the base fee of $1.50, so (x−1)(x - 1) represents the number of pounds that cost $0.50 each.

The cost equation can be written as Cost=$1.50+$0.50(x−1)\text{Cost} = \$1.50 + \$0.50\,(x - 1)

  • The cost for a 2-pound package is $1.50+$0.50(2−1)=$1.50+$0.50(1)=$2.00\$1.50 + \$0.50(2 - 1) = \$1.50 + \$0.50(1) = \$2.00
  • The cost for a 6-pound package is $1.50+$0.50(6−1)=$1.50+$0.50(5)=$1.50+$2.50=$4.00\$1.50 + \$0.50(6 - 1) = \$1.50 + \$0.50(5) = \$1.50 + \$2.50 = \$4.00

  • The weight of a package that costs $2.50 to ship can be found by solving

    $2.50=$1.50+$0.50(x−1)\$2.50 = \$1.50 + \$0.50(x - 1)

    $1.00=$0.50(x−1)\$1.00 = \$0.50(x - 1)

    2=x−12 = x - 1

    x=3 poundsx = 3 \text{ pounds}

  • The weight of a package that costs $5.50 to ship can be found by solving

    $5.50=$1.50+$0.50(x−1)\$5.50 = \$1.50 + \$0.50(x - 1)

    $4.00=$0.50(x−1)\$4.00 = \$0.50(x - 1)

    8=x−18 = x - 1

    x=9 poundsx = 9 \text{ pounds}

To check:

Given that the total cost is $3.00 for the initial package

$3.00=$1.50+$0.50(x−1)\$3.00 = \$1.50 + \$0.50(x - 1)

$1.50=$0.50(x−1)\$1.50 = \$0.50(x - 1)

3=x−13 = x - 1

x=4x = 4 pounds fits the linear pattern of the table

Question analysis

A shipping company charges a base fee plus 50 cents for each pound over the first, and a 4-pound package costs $3.00. Students drag number tiles into a table to complete two costs and two weights, with some tiles left unused. It tests building and using a linear rule, and the wording about the first pound must be applied in every row.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.9 dependent and independent variables
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve word problems with equations like p(x + q) = r. This table needs that rule four times, twice run backward from a cost to a weight, and the phrase about the first pound must be applied every time.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 4, 75 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The rule is stated in words and confirmed by the 4-pound example, so the work is applying it consistently in both directions. Charging every pound breaks the $3.00 example at once, since 4 pounds would cost $3.50, and models that ignore the wording leave costs or weights with no matching tile. A shortcut helps: each extra pound adds $0.50 to the $3.00, so the table can be filled by counting up or down from 4 pounds.

  1. Turning words into math moderate Understanding the question

    The phrase “$0.50 for each pound over the first pound” means the first pound is covered by the $1.50 fee, so a package of x pounds pays for only x − 1 of them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Four boxes must all be right, two found forward from a weight and two backward from a cost, and the backward rows need the first pound added back. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Write the rule as the base fee plus $0.50 times the pounds after the first, check it against the 4-pound example, then use it forward for the costs and backward for the weights.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Shipping cost
C=1.50+0.50×(w−1)C = 1.50 + 0.50 \times (w - 1)
C is the total cost in dollars, and w is the weight in pounds. The $1.50 base fee is paid once, and it covers the first pound. Each pound after the first costs $0.50. There are w − 1 of those pounds.

  1. Sort out the question

    The base fee of $1.50 covers the first pound. Each pound over the first adds $0.50. A 4-pound package costs $3.00.

    You fill four boxes: the costs for 2 and 6 pounds, and the weights that cost $2.50 and $5.50. Each tile is used once at most, and four tiles are not used.

  2. Test the rule on 4 pounds

    Check the rule with the example. A 4-pound package has 3 pounds over the first:

    C=1.50+0.50×(4−1)=1.50+1.50=3.00C = 1.50 + 0.50 \times (4 - 1) = 1.50 + 1.50 = 3.00

    That is $3.00, so the rule fits.

    Watch out: charging $0.50 for every pound, the first one too, gives $3.50 for 4 pounds. That does not match.

  3. Find the two costs

    For 2 pounds, 1 pound is over the first:

    1.50+0.50×1=2.001.50 + 0.50 \times 1 = 2.00

    For 6 pounds, 5 pounds are over the first:

    1.50+0.50×5=1.50+2.50=4.001.50 + 0.50 \times 5 = 1.50 + 2.50 = 4.00

    The costs are the tiles 2 and 4.

    Watch out: 6 pounds is 3 times 2 pounds, but the cost is not 3 times $2. The base fee is paid once. Tripling gives the tile 6. Leaving out the base fee gives the tile 2.5.

  4. Work back to the two weights

    For $2.50, take off the base fee: 2.50−1.50=1.002.50 - 1.50 = 1.00. At $0.50 a pound, $1.00 pays for 2 extra pounds. Add the first pound: 2+1=32 + 1 = 3 pounds.

    For $5.50: 5.50−1.50=4.005.50 - 1.50 = 4.00. $4.00 pays for 8 extra pounds. Add the first pound: 8+1=98 + 1 = 9 pounds.

    Watch out: forgetting to add back the first pound gives 2 and 8.

  5. Fill in the four boxes

    From top to bottom, the boxes get 2, 3, 4 and 9. The tiles 2.5, 3.5, 6 and 7 are not used. On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Each extra pound adds $0.50, so count from 4 pounds at $3.00. Going down: 3 pounds cost $2.50 and 2 pounds cost $2.00. Going up: 5 pounds $3.50, 6 pounds $4.00, 7 pounds $4.50, 8 pounds $5.00 and 9 pounds $5.50. All four boxes match.

More questions like this

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

Question 37 of 120

2026 Practice Test D Math, question 47

Question

A tall right rectangular prism standing on end. The bottom front edge is labeled 2 in., the bottom edge running back on the right is labeled 2 in., and the height is labeled 6 in. Right-angle marks are drawn at the lower left corner of the front face and at the front left corner of the top face.

What shape will result from a horizontal slice of the figure above?

  • Aa square that has 2-inch sides
  • Ba triangle that has 2-inch sides
  • Ca rectangle that has a 2-inch side and a 6-inch side
  • Da triangle that has one 2-inch side and two 6-inch sides
Show the answer

Correct answer: A

Based on the diagram, the figure shown is a right rectangular prism with a square base. A horizontal slice of the figure will be a plane parallel to the square base, as in this example.

The same prism, 2 in. by 2 in. by 6 in., drawn with its hidden edges dashed and a horizontal slice partway up: a square parallel to the base and the same size as the base.

Since the horizontal slice is parallel to the base, it will be congruent to the base, which is a square with 2-inch sides.

Question analysis

A figure shows a right rectangular prism measuring 2 inches by 2 inches by 6 inches, standing on end. This geometry question asks what two-dimensional shape a horizontal slice of the prism makes. It tests visualizing cross-sections of solids, using the labeled lengths to name both the shape and its size.

NYS standards
  • 7.G.3 slicing three-dimensional solids
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Describing the shape made by slicing a prism parallel to its base is exactly the grade 7 standard. This is its simplest case, a single slice of a labeled right prism, typical of a classroom example.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 2, 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

A horizontal cut runs parallel to the ground, and this prism stands on a square base, so the slice copies the base. The main danger is picturing the cut along the tall front face, which gives a 2-by-6 rectangle, the shape a vertical slice makes. The two triangles need a slanted cut through a corner and catch only students guessing.

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

    The prism is drawn standing on end, 6 inches tall on a 2-by-2 base, so the cut has to be pictured in a flat drawing where the tall front face is the most visible shape. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Students must know that a slice parallel to the base of a prism has the same shape and size as the base, which here is the square at the bottom. (Anderson & Krathwohl, 2001)

Strategy: Picture a flat cut across the prism parallel to the ground, ask which face of the prism that cut runs parallel to, and read that face's edge lengths from the labels.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A slice parallel to the base
slice parallel to the base≅base\text{slice parallel to the base} \cong \text{base}
The sign ≅ means congruent: the same shape and the same size. A prism has the same shape all the way up, like a stack of matching cards. So a cut parallel to its base, at any height, is a copy of the base.

  1. Sort out the question

    The figure is a right rectangular prism, a box shape whose faces are all rectangles. It stands on end. The two edges along the bottom are labeled 2 in. The tall edge is labeled 6 in.

    You need the shape you get from a horizontal slice. Horizontal means flat, like the floor or a tabletop.

  2. Find the base

    The base is the face the prism stands on. The two bottom edges you can see are both labeled 2 in. The right-angle mark on the top face shows its corners are square. The top and the bottom match. So the base is a square with 2-inch sides.

  3. Picture the slice

    A horizontal slice cuts straight across the prism, parallel to the base. The walls of the prism go straight up, so the slice is a copy of the base. It is a square with 2-inch sides, at whatever height you cut.

    The 6-inch height does not matter. It only tells how tall the stack of squares is.

  4. Rule out the other shapes

    Watch out: a 2-inch by 6-inch rectangle, choice C, comes from a vertical slice. That cut goes straight down, through the 6-inch height. A triangle, choice B or D, needs a slanted cut through a corner.

    So the answer is A, a square that has 2-inch sides.

Check your answer

Picture the slice sliding up and down. At the bottom, it matches the base, 2 in. by 2 in. At the top, it matches the top face, the same square. In between, the walls are straight, so its size never changes.

More questions like this

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

Question 38 of 120

2026 Practice Test D Math, question 48

Question

The perimeter of square W is twice the perimeter of equilateral triangle X. What is the ratio of the length of one side of square W to the length of one side of equilateral triangle X?

  • A3:8
  • B2:3
  • C3:2
  • D2:1
Show the answer

Correct answer: C

First, set up the ratio of the perimeter of the square to the triangle as 2:1 because the question states that the perimeter of the square is twice the perimeter of the triangle.

Recognizing that the square has 4 equal sides and that an equilateral triangle has 3 sides, use division to find the ratio of the sides.

Perimeter: 2:1=212 : 1 = \frac{2}{1}

Sides:

2413=24÷13=\frac{\frac{2}{4}}{\frac{1}{3}} = \frac{2}{4} \div \frac{1}{3} =

24×31=64=\frac{2}{4} \times \frac{3}{1} = \frac{6}{4} =

32=3:2\frac{3}{2} = 3 : 2

Question analysis

The perimeter of square W is twice the perimeter of equilateral triangle X, and students find the ratio of a side of the square to a side of the triangle. This ratio question uses perimeters of regular shapes with no lengths given. Turning a comparison of perimeters into one of sides, and keeping the ratio in the order asked, are the tricky parts.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4 equations and inequalities from problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 ratio problems usually supply numbers to work with. This one gives only a relationship between two perimeters, so students must bring in the number of sides of each shape themselves and keep the ratio in the order asked.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 9, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns a statement about perimeters into a ratio of sides, and the natural guess, that the sides keep the perimeters’ 2 to 1, ignores that the square shares its perimeter among 4 sides and the triangle among 3. Choosing a triangle perimeter such as 12 makes the comparison concrete. The order of the ratio still has to be checked at the end, because its reverse is listed too.

  1. Turning words into math moderate Understanding the question

    “The perimeter of square W is twice the perimeter of equilateral triangle X” must become 4s = 2 × 3t, with the 2 on the triangle’s side; written backward, it gives the triangle the larger perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers moderate Understanding the question

    No side length is given, so students must either work with letters for the two sides or choose a convenient perimeter themselves, such as 12 for the triangle. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The ratios 3:2 and 2:3 use the same two numbers, so only keeping the square’s side first, as the question orders it, separates the answer from its reverse. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Pick an easy side length for the triangle, find its perimeter, double that for the square, divide by the square's number of sides, and write square to triangle.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a square
P=4sP = 4s
P is the perimeter, the distance around the shape. s is the length of one side. A square has 4 equal sides, so its perimeter is 4 times one side.
Perimeter of an equilateral triangle
P=3tP = 3t
An equilateral triangle has 3 equal sides. t is the length of one side. So its perimeter is 3 times one side.
Ratio
a:ba : b
A ratio compares two amounts in order. a : b means a compared with b. The order matters, so write the amounts in the order the question names them.

  1. Sort out the question

    Square W has a perimeter twice as long as the perimeter of triangle X. Equilateral means all 3 sides of X are equal.

    You need the ratio of one side of the square to one side of the triangle, in that order. No lengths are given, so you can pick a size to try.

  2. Pick a size for the triangle

    Choose an easy side length for the triangle, such as 2 units. All 3 sides are 2, so its perimeter is:

    P=3×2=6P = 3 \times 2 = 6

  3. Find the square’s side

    The square’s perimeter is twice the triangle’s:

    2×6=122 \times 6 = 12

    The square has 4 equal sides, so divide by 4:

    s=12÷4=3s = 12 \div 4 = 3

    Each side of the square is 3 units.

  4. Write the ratio in order

    The square comes first, then the triangle. Their sides are 3 and 2, so the ratio is 3 : 2.

    Watch out: 2 : 1, choice D, compares the perimeters, not the sides. The square splits its perimeter into 4 sides, the triangle into only 3. Choice B, 2 : 3, puts the triangle first. Choice A, 3 : 8, comes from giving the triangle the bigger perimeter.

    So the answer is C, 3:2.

Check your answer

Try a different size. Give the triangle sides of 4. Its perimeter is 12, so the square’s is 24. Each side of the square is 24÷4=624 \div 4 = 6. The ratio 6 : 4 simplifies to 3 : 2. The size does not change the ratio.

More questions like this

The original DOE question
Question 48 as shown on the DOE’s digital test, page 121 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 121, question 48.

Question 39 of 120

2026 Practice Test D Math, question 50

Question

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

BIRDS AT BIRD FEEDER

Day Number of Birds
Monday 30
Tuesday 24

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

  • A6%
  • B20%
  • C24%
  • D25%
Show the answer

Correct answer: B

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

30−24=630 - 24 = 6

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

630=0.20\frac{6}{30} = 0.20

0.20 represents 20%.

There was a 20% decrease from Monday to Tuesday.

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Percent decrease is core grade 7 work, and this is a plain example: one subtraction and one division with friendly numbers. The only real demand is measuring the change against the right day's count.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 3, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. A very close wrong answer minor Choosing the answer

    Dividing the drop of 6 by Monday’s 30 gives 20% and by Tuesday’s 24 gives 25%, two round, listed percents only 5 points apart, so only the choice of base separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Students must know that a percent decrease is the change divided by the starting amount, here Monday’s count, a rule the question does not state. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The table shows 30 birds on Monday and 24 birds on Tuesday. You need the percent decrease from Monday to Tuesday. From Monday means Monday is the starting amount.

  2. Find the drop

    Subtract Tuesday’s count from Monday’s:

    30−24=630 - 24 = 6

    There were 6 fewer birds.

    Watch out: 6 is a number of birds, not a percent. Calling it 6% leads to choice A.

  3. Compare the drop with Monday

    Use the formula. Divide the drop by Monday’s count, then multiply by 100:

    630×100=0.20×100=20\frac{6}{30} \times 100 = 0.20 \times 100 = 20

    The number of birds went down by 20%.

    Watch out: dividing by Tuesday’s 24 gives 624=0.25\frac{6}{24} = 0.25, or 25%, choice D. Measure the change from the starting day. Choice C, 24%, just copies a number from the table.

    So the answer is B, 20%.

Check your answer

Take 20% of Monday’s count: 0.20×30=60.20 \times 30 = 6. Then 30−6=2430 - 6 = 24, which is Tuesday’s count. The answer checks out.

More questions like this

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

Question 40 of 120

2024 Test A Math, question 5

Question

RESULTS FROM SURVEY OF 110 FAMILIES

Number of Children in the Family Number of Families
0 45
1 32
2 19
3 8
4 6

The table above shows the number of children in each of 110 families. What is the median number of children in these families?

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

Show the answer

Correct answer: 1

Since 110 families were surveyed, the median will have 55 values less than or equal to it, and 55 values greater than or equal to it.

If the values in the table were listed in order, they would look like this:

0,…,0⏞45,1,…,1⏞32,2,…,2⏞19,3,…,3⏞8,4,…,4⏞6\overbrace{0,\ldots,0}^{45},\overbrace{1,\ldots,1}^{32},\overbrace{2,\ldots,2}^{19},\overbrace{3,\ldots,3}^{8},\overbrace{4,\ldots,4}^{6}

The first row of the table indicates that 45 families had 0 children.

The second row of the table indicates that 32 families had 1 child.

So 77 families had either 0 children or 1 child, and 45 families had 0 children.

Since 55 is between 45 and 77, the median must be a value in the second row of the table.

So the median is 1.

Question analysis

This statistics grid-in gives a frequency table from a survey of 110 families, showing how many families have each number of children from none to four. Students find the median number of children per family. The table lists counts rather than the 110 data values themselves, so locating the middle value takes a running total of the families.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.5a number of observations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 median problems usually list a dozen or so values. Here 110 values are hidden in a frequency table, so students must reason about positions with a running count, a less familiar and more error-prone task.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 7, 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

Students know a median as the middle of a list, but here the list is never written out. Treating the table's columns as the data gives quick, confident answers, 2 from the middle of the children column or 19 from the middle of the family counts, and neither can be checked against choices. The question rewards counting positions: 45 families have no children, so the 55th and 56th families both fall among the 32 with one child.

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

    The table hides 110 data values in five rows: the 0 to 4 column lists the possible values, and 45, 32, 19, 8 and 6 say how many families have each. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. No obvious method moderate Working it out

    The median of 110 values lies between the 55th and 56th; finding them takes a running count of families, 45 and then 77, a method seldom practiced with tables. (Schoenfeld, 1985)

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

    As a grid-in there are no choices to compare, so a student who takes the middle row's 2 has nothing on the page to suggest it is wrong. (Enright, Morley & Sheehan, 2002)

Strategy: With 110 values the median sits between the 55th and 56th; add the family counts row by row until the running total passes 55, and read that row's number of children.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Spot of the median
middle spot=n+12\text{middle spot} = \frac{n + 1}{2}
The median is the middle value when all the values are put in order. n is how many values there are. This gives the spot of the middle value, with the same number of values before and after it. If it ends in a half, the median is halfway between the two values on either side.

  1. Sort out the table

    The table covers 110 families. The left column is a number of children. The right column says how many families have that many children. For example, 45 families have 0 children. You need the median number of children. The median is the middle value when all the values are in order.

  2. Picture the values in order

    Think of a list with one number for each family: how many children it has. The list has 45 zeros, then 32 ones, then 19 twos, then 8 threes, then 6 fours. It is already in order, from least to most.

    45+32+19+8+6=11045 + 32 + 19 + 8 + 6 = 110

    There are 110 values. Watch out: the median is not the middle row, 2. Each row stands for a different number of families.

  3. Find the middle spot

    Use the median rule with n = 110.

    middle spot=n+12\text{middle spot} = \frac{n + 1}{2}

    110+12=55.5\frac{110 + 1}{2} = 55.5

    The middle is halfway between the 55th and 56th values. So you need those two values in the list.

  4. Count down the list

    Keep a running total of families, row by row. The zeros fill spots 1 to 45. Now add the 32 families with 1 child.

    45+32=7745 + 32 = 77

    So the ones fill spots 46 to 77. Spot 55 and spot 56 are both in that range. So both values are 1.

  5. Give the median

    The 55th and 56th values are both 1. Halfway between 1 and 1 is 1. So the median is 1 child. Watch out: the median is a number of children, not of families. Gridding 19, from the families column, is wrong. So is the mean, or average, which is about 1.07. So you would grid 1.

Check your answer

Half of 110 is 55. Families with 0 children: 45, which is less than half. So the median is not 0. Families with 0 or 1 child: 77, which is more than half. So the median is not more than 1. It must be 1.

More questions like this

The original DOE question
Question 62 as printed on page 76 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 76, question 62. Open page 76 of the handbook (PDF)

Question 41 of 120

2024 Test A Math, question 9

Question

A list of consecutive integers begins with m and ends with n. If n−m=66n - m = 66, how many integers are in the list?

  • A2
  • B33
  • C66
  • D67
Show the answer

Correct answer: D

Consider the case where m=1m = 1.

n−m=66n - m = 66

n−1=66n - 1 = 66

n−1+1=66+1n - 1 + 1 = 66 + 1

n=67n = 67

The list of integers from 1 to 67 contains 67 numbers.

Question analysis

A number question about a list of consecutive integers that starts at m and ends at n, where n minus m is 66. Students decide how many integers the list contains. The tricky part is relating the difference between the two ends to the number of integers in the list.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.EE.6 variables representing unknown numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Consecutive integers and differences are familiar grade 7 ideas, and the arithmetic is trivial. What keeps this at grade 7 rather than below it is the counting trap, which needs a moment of careful reasoning.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is trivial, so the question tests one idea: the number of integers from m to n is one more than their difference, because both ends belong to the list. The stem supplies 66 ready-made, and taking it as the count is the natural first reaction. A five-number list, 1 to 5, whose ends differ by 4, shows the extra one in seconds.

  1. A very close wrong answer moderate Choosing the answer

    The choices 66 and 67 differ by one, and 66 is the difference the stem hands over; only counting both ends of the list separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Variables instead of numbers moderate Understanding the question

    The list is given only by its ends, m and n, so the count has to be reasoned out in general, or checked on a small list such as 1 through 5. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Try a small list you can count, such as 1 through 5, and compare its count with the difference between its first and last numbers.

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

Counting consecutive integers
count=n−m+1\text{count} = n - m + 1
m is the first number in the list and n is the last. n − m counts the steps from each number to the next. A list always has one more number than it has steps, so you add 1. This works for any list that goes up by 1.

  1. Sort out the question

    Integers are whole numbers and their negatives, like −2, 0 and 5. Consecutive integers come one after another, going up by 1, like 4, 5, 6. The list starts at m and ends at n. The difference n−mn - m is 66. You need how many integers are in the list.

  2. Try a small list

    Letters are hard to count, so try a list you can count: 1, 2, 3, 4, 5. Here m = 1 and n = 5.

    n−m=5−1=4n - m = 5 - 1 = 4

    But the list has 5 numbers, not 4. The difference counts the steps: 1 to 2, 2 to 3, 3 to 4 and 4 to 5. Both ends are in the list, so there is one more number than steps.

  3. Use the counting rule

    count=n−m+1\text{count} = n - m + 1

    count=66+1=67\text{count} = 66 + 1 = 67

    Watch out: 66 is the difference, not the count. Choosing 66, choice C, leaves out one end of the list. Also, consecutive integers go up by 1, not 2, so halving 66 to get 33 is wrong. So the answer is D, 67.

Check your answer

Try m = 1. Then n=1+66=67n = 1 + 66 = 67. The list 1, 2, 3, and so on up to 67, plainly has 67 numbers. Any other start gives the same count, because the list just slides along.

More questions like this

The original DOE question
Question 66 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 66. Open page 78 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 42 of 120

2024 Test A Math, question 12

Question

If n is a whole number, and 0.01 is between 1n\frac{1}{n} and 1n+2\frac{1}{n + 2}, what is the value of n?

  • A0
  • B1
  • C2
  • D99
Show the answer

Correct answer: D

0.01=11000.01 = \frac{1}{100}, which is between 1101\frac{1}{101} and 199\frac{1}{99}.

Therefore, n+2=101n + 2 = 101 and n=99n = 99.

Question analysis

A number sense question: 0.01 lies between 1 over n and 1 over n + 2, where n is a whole number, and students find n. It tests comparing a decimal with unit fractions and knowing how a fraction changes as its denominator grows. The unusual form, with n in two denominators, is what slows students down.

NYS standards
  • 6.NS.7a inequality as relative position
  • 6.EE.5 solutions and substitution
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students compare fractions and decimals routinely, but not with an unknown in the denominators. The comparison is simple once 0.01 is written as a fraction; the unfamiliar setup is the main demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 9, 18 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The setup looks unfamiliar, with n in two denominators, but one rewrite makes it routine: 0.01 is 1/100, so the two denominators must sit on either side of 100. The three small choices fail a quick check, since every fraction from 1/1 to 1/4 is far above 0.01, and 0 cannot be a denominator at all. Students who test the choices instead of rewriting the decimal spend a little longer but reach the same place.

  1. Variables instead of numbers moderate Understanding the question

    The unknown sits in two denominators, n and n + 2, so students must reason about which unit fractions lie on either side of 0.01 rather than solve an equation. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know minor Working it out

    Two facts are assumed: 0.01 equals 1/100, and a unit fraction shrinks as its denominator grows, so 1/99 is just above 0.01 and 1/101 just below. (Anderson & Krathwohl, 2001)

Strategy: Rewrite 0.01 as a unit fraction, recall that a larger denominator gives a smaller fraction, and test each choice in both 1/n and 1/(n + 2).

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions
if a<b, then 1a>1b\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}
A unit fraction has 1 on top, like one fifth. a and b are positive whole numbers on the bottom. A bigger bottom cuts the whole into more pieces, so each piece is smaller. So the fraction with the bigger bottom is the smaller number.

  1. Sort out the question

    n is a whole number: 0, 1, 2, 3 and so on. The number 0.01 is between 1n\frac{1}{n} and 1n+2\frac{1}{n + 2}. Between means one fraction is bigger than 0.01 and the other is smaller. You need the value of n.

  2. Write 0.01 as a fraction

    0.01 means one hundredth.

    0.01=11000.01 = \frac{1}{100}

    Now all three numbers are fractions with 1 on top. That makes them easy to compare.

  3. Find fractions on each side

    With 1 on top, a bigger bottom makes a smaller fraction.

    if a<b, then 1a>1b\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}

    So 199\frac{1}{99} is a little more than 1100\frac{1}{100}, and 1101\frac{1}{101} is a little less.

    1101<1100<199\frac{1}{101} < \frac{1}{100} < \frac{1}{99}

    The bottoms 99 and 101 are 2 apart, just like n and n + 2.

  4. Match the fractions to n

    The smaller bottom is n, and the larger bottom is n + 2.

    n=99 and n+2=101n = 99 \text{ and } n + 2 = 101

    Both fit. Watch out: small choices do not work. For n = 1, the fractions are 1 and 13\frac{1}{3}. Both are far bigger than 0.01. For n = 0, 10\frac{1}{0} is not a number, because you can’t divide by zero. So the answer is D, 99.

Check your answer

Write the fractions as decimals. 199≈0.0101\frac{1}{99} \approx 0.0101, a bit more than 0.01. 1101≈0.0099\frac{1}{101} \approx 0.0099, a bit less. The sign ≈ means about. So 0.01 sits right between them.

More questions like this

The original DOE question
Question 69 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 69. Open page 78 of the handbook (PDF)

Question 43 of 120

2024 Test A Math, question 13

Question

When asked a certain question in a poll, 72% of the people polled answered yes. If 56 people did not answer yes to that question, what is the total number of people who were polled?

  • A78
  • B128
  • C144
  • D200
Show the answer

Correct answer: D

72% of the people polled answered yes, so 28% of the people did not answer yes.

Therefore, 56 people is 28% of the total, x.

56x=28100\frac{56}{x} = \frac{28}{100}

56x=725\frac{56}{x} = \frac{7}{25}

56(25)=7x56(25) = 7x

56(25)7=x\frac{56(25)}{7} = x

8(25)=x8(25) = x

200=x200 = x

Question analysis

A percent word problem about a poll: 72% of the people answered yes, and 56 people did not. Students find the total number of people polled. They must first match the 56 people with the right percent and then find the whole from a part, which is where the main slip happens.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding a whole from a part and a percent is grade 6 and 7 work. This adds one short step, finding the complementary percent first, which makes it a standard grade 7 multistep percent problem.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The only percent printed is 72 and the only count is 56, so the first instinct is to pair them, even though the bold not says they describe different groups. Once 56 is matched with the 28% who did not say yes, a single division finds the total. The listed wrong answers come from that mismatch, from adding the two numbers, and from stopping at the yes count.

  1. Turning words into math moderate Understanding the question

    The 56 people are the ones who did not answer yes, so they match 100% − 72% = 28%, not the 72% printed in the sentence before them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Finding the total means 56 ÷ 0.28, a division by a decimal; scaling 28% up to 100% at 2 people per percent avoids it. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract 72% from 100% to find the percent who did not answer yes, then find the total that makes 56 equal to that percent.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a whole
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
The whole is the total amount, here everyone polled. p is the percent. The part is the amount that p percent stands for. Percent means out of 100, so p% is p hundredths of the whole.

  1. Sort out the question

    In a poll, 72% of the people answered yes. 56 people did not answer yes. You need the total number of people polled. Notice the bold word not. The 56 people are the ones who did not say yes.

  2. Find the percent who did not say yes

    Everyone polled makes 100%. Take away the 72% who said yes.

    100−72=28100 - 72 = 28

    So 28% did not answer yes. That is the group of 56 people. Watch out: do not match the 56 people with 72%. That leads to about 78, choice A.

  3. Set up the percent equation

    Call the total number of people T. The 56 people are 28% of T.

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    56=28100×T56 = \frac{28}{100} \times T

  4. Solve for the total

    First find how many people make 1%. Divide 56 people by 28.

    56÷28=256 \div 28 = 2

    So each 1% is 2 people. All 100% is 100 times that.

    T=2×100=200T = 2 \times 100 = 200

    Watch out: do not multiply 2 by 72. That gives 144, the people who said yes, choice C. So the answer is D, 200.

Check your answer

Find 28% of 200: 0.28×200=560.28 \times 200 = 56. That matches the people who did not say yes. Also, 72% of 200 is 144 people who said yes, and 144+56=200144 + 56 = 200, everyone.

More questions like this

The original DOE question
Question 70 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 79, question 70. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 44 of 120

2024 Test A Math, question 15

Question

A program on a computer randomly generates a sequence of whole numbers from 1 to 9, inclusive. If the computer generates a sequence of 300 numbers, what is the best prediction of the number of odd numbers in the sequence?

  • A120
  • B133
  • C150
  • D167
Show the answer

Correct answer: D

There are five odd numbers in the range 1 through 9: 1, 3, 5, 7, and 9.

So the proportion of odd numbers in the range 1 through 9 is 59\frac{5}{9}.

The expected number of odd values in the sequence from the computer program is

300∙59=166.66.300 \bullet \frac{5}{9} = 166.\overline{66}.

So the best prediction for the number of odd numbers in the sequence is 167.

Question analysis

A probability question: a computer randomly generates 300 whole numbers from 1 to 9, and students predict how many will be odd. It asks students to find the probability of an odd number and use it to predict a count. Counting carefully which of the nine numbers are odd, and rounding the prediction sensibly, are the points to watch.

NYS standards
  • 6.SP.7 long-run relative frequency
  • 6.SP.8a uniform probability models
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a probability to predict how often an event happens is grade 6 and 7 work. This takes two short steps plus a rounding decision, about the demand of a typical grade 7 probability task.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 11, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is short, a probability times 300, so the question is decided by the probability. Odd and even feel like an even split, and half of 300 is a listed choice, but the numbers 1 to 9 hold five odd and four even. Listing the nine numbers before computing removes the trap, and the result, about 167, then needs only rounding.

  1. Turning words into math minor Understanding the question

    The phrase "whole numbers from 1 to 9, inclusive" means nine equally likely values, five of them odd, so the chance of odd is 5/9, not the one half that odd and even suggest. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    The stem never says that an expected count is the probability times the number of trials, here 5/9 of the 300 numbers generated. (Anderson & Krathwohl, 2001)

  3. Error-prone arithmetic minor Working it out

    The product 300 × 5/9 is 166.67, not a whole number, so the prediction has to be rounded to the nearest whole number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the odd numbers from 1 to 9, write the probability as that count over 9, and multiply by 300, rounding to the nearest whole number.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=wanted resultspossible resultsP = \frac{\text{wanted results}}{\text{possible results}}
P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of the results that you want.
Expected count
E=P×nE = P \times n
E is the expected count, your best guess of how many times something will happen. P is its probability, and n is the number of tries. If something happens 5 times out of 9, it should happen about 5 ninths of the time in many tries.

  1. Sort out the question

    The computer picks whole numbers from 1 to 9. Inclusive means 1 and 9 count too. Randomly means each number has the same chance. It picks 300 numbers. You need the best guess of how many of them are odd.

  2. Count the odd numbers

    An odd number does not split evenly by 2. From 1 to 9, the odd numbers are 1, 3, 5, 7 and 9. That is 5 of the 9 numbers. The even ones, 2, 4, 6 and 8, are only 4. Watch out: odd and even are not equally likely here, since there is one more odd number. Taking half of 300 gives 150, choice C.

  3. Find the chance of odd

    P=wanted resultspossible resultsP = \frac{\text{wanted results}}{\text{possible results}}

    P=59P = \frac{5}{9}

    Each number the computer picks has a 5 in 9 chance of being odd. Watch out: do not use the 4 even numbers. That gives about 133, choice B.

  4. Multiply by the 300 numbers

    E=P×nE = P \times n

    E=59×300=15009≈166.7E = \frac{5}{9} \times 300 = \frac{1500}{9} \approx 166.7

    The sign ≈ means about. You can’t have part of a number, so round to the nearest whole number. 166.7 is closer to 167 than to 166. So the answer is D, 167.

Check your answer

The odd numbers are a bit more than half of the nine. So the answer should be a bit more than half of 300, which is 150. Only 167 is a bit more than 150. Also, 167 odd plus about 133 even makes 300.

More questions like this

The original DOE question
Question 72 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 79, question 72. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 45 of 120

2024 Test A Math, question 28

Question

QUIZ SCORES IN
MRS. ARCH’S CLASS

Quiz Score Number of Students
60 9
70 7
80 4
90 5
100 3

In the table above, what is the mean quiz score?

  • A60
  • B70
  • C75
  • D80
Show the answer

Correct answer: C

First, find the total number of students who took the quiz.

9+7+4+5+3=289 + 7 + 4 + 5 + 3 = 28

Multiply each quiz score by the number of students who received that score. Add those products together to find the total number of points scored.

(60×9)+(70×7)+(80×4)+(90×5)+(100×3)=2,100(60 \times 9) + (70 \times 7) + (80 \times 4) + (90 \times 5) + (100 \times 3) = 2{,}100

To find the mean quiz score, divide the total number of points by the number of students who took the quiz.

2,100÷28=752{,}100 \div 28 = 75

Question analysis

A frequency table question: the table lists five quiz scores from 60 to 100 and how many students in Mrs. Arch's class earned each. Students find the mean quiz score for the whole class. Each score has to count once for every student who earned it, and the class size has to be added up from the table.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.5a number of observations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 A mean from a frequency table is grade 6 content in a three-step form that grade 7 students meet regularly. Five rows of products make it slightly longer than a typical exercise, but the arithmetic is friendly.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 2, 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 table invites a shortcut: average the five scores in the first column and get a round number that is a choice. That ignores the fact that more students earned the low scores than the high ones, which pulls the true mean down. Weighting each score by its count is a five-row calculation with friendly numbers, and the other choices catch students who confuse the mean with the most common or the middle score.

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

    Each row of the table stands for several students, 9 with 60, 7 with 70 and so on, so a score counts once per student and the class size, 28, must be added up. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Five products, 540, 490, 320, 450 and 300, must be found and added to 2,100 before dividing by the 28 students. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    The last step, 2,100 ÷ 28, is not a familiar fact; noticing that 28 × 75 = 2,100, or testing the choices, keeps it quick. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply each score by its number of students, add the products for the total points, add the student counts, and divide the points by the students.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean from a table
mean=total pointsnumber of students\text{mean} = \frac{\text{total points}}{\text{number of students}}
The mean is the average score. Total points is every student’s score added up. Each score in the table counts once for every student who got it. Number of students is how many took the quiz.

  1. Sort out the table

    The left column lists quiz scores. The right column says how many students got each score. For example, 9 students scored 60. You need the mean quiz score, which is the average.

  2. Count the students

    9+7+4+5+3=289 + 7 + 4 + 5 + 3 = 28

    28 students took the quiz. So there are 28 scores in all, not just 5.

  3. Find the total points

    Each score counts once for each student who got it. So multiply each score by its number of students.

    ScoreStudentsPoints
    60960×9=54060 \times 9 = 540
    70770×7=49070 \times 7 = 490
    80480×4=32080 \times 4 = 320
    90590×5=45090 \times 5 = 450
    1003100×3=300100 \times 3 = 300

    540+490+320+450+300=2,100540 + 490 + 320 + 450 + 300 = 2{,}100

    The class scored 2,100 points in all.

  4. Divide to find the mean

    mean=total pointsnumber of students\text{mean} = \frac{\text{total points}}{\text{number of students}}

    mean=2,10028=75\text{mean} = \frac{2{,}100}{28} = 75

    To divide, try 28×70=1,96028 \times 70 = 1{,}960. That leaves 2,100−1,960=1402{,}100 - 1{,}960 = 140, and 140=28×5140 = 28 \times 5. So the mean is 70+5=7570 + 5 = 75. Watch out: do not just average the five scores in the table. That gives 80, choice D. Also, 70 is the median and 60 is the most common score, choices B and A. So the answer is C, 75.

Check your answer

More students scored low, 60 or 70, than high. So the mean should be below 80, the middle of the listed scores. 75 is below 80. Also, 28×75=2,10028 \times 75 = 2{,}100, the total points.

More questions like this

The original DOE question
Question 85 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 85. Open page 83 of the handbook (PDF)

Question 46 of 120

2024 Test A Math, question 35

Question

Evaluate:

|(−8)−12+(−17)−(−31)|−|24|\left|(-8) - 12 + (-17) - (-31)\right| - \left|24\right|

  • A−30
  • B−18
  • C18
  • D44
Show the answer

Correct answer: B

|(−8)−12+(−17)−(−31)|−|24|\left|(-8) - 12 + (-17) - (-31)\right| - \left|24\right|

Simplify negatives inside the first absolute value brackets.

|−8−12−17+31|−|24|\left|-8 - 12 - 17 + 31\right| - \left|24\right|

Simplify the first absolute value.

|−6|−|24|\left|-6\right| - \left|24\right|

Calculate the absolute values.

6−246 - 24

Perform the final subtraction.

6−24=−186 - 24 = -18

Question analysis

This integer question asks students to evaluate an expression with two absolute values: the first holds four positive and negative numbers, and the second is the absolute value of 24. It tests subtracting negative numbers and understanding absolute value. Subtracting a negative number and handling each pair of bars separately are where most slips happen.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students add and subtract integers and know absolute value as distance from zero. This expression puts four signed terms inside one pair of bars and subtracts a second absolute value, a typical but sign-heavy grade 7 exercise.
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 16+, 2 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 the question stacks sign decisions, and one slip anywhere produces a clean wrong number. The last step is the subtle one: each pair of bars gives a positive number, but 6 minus 24 is still negative. Students who believe bars guarantee a positive result flip that subtraction, and the choices offer the flipped number beside the right one.

  1. Error-prone arithmetic moderate Working it out

    The first bars hold four signed terms, and two must be rewritten, + (−17) as − 17 and − (−31) as + 31, before they combine to −6. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The answer, −18, and the choice 18 differ only in sign, so estimating cannot separate them; only doing the last step as 6 − 24 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Many steps minor Working it out

    Each pair of bars is simplified inside and made positive before the outer subtraction, so three separate jobs must be done in the right order. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite each subtraction of a negative as addition, combine the four numbers inside the first bars, take both absolute values, and only then do the final subtraction.

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding and subtracting negatives
a+(−b)=a−banda−(−b)=a+ba + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b
a is any number, and b is a positive number, so −b is negative. Adding a negative moves you left on the number line, just like subtracting. Subtracting a negative is like taking away a debt, so you end up with more. It is the same as adding.
Absolute value
|a|=the distance from a to 0\left|a\right| = \text{the distance from } a \text{ to } 0
The bars around a number mean absolute value. a is any number. Its absolute value is how far a is from 0 on the number line. A distance is never negative, so the absolute value of −6 is 6.

  1. Sort out the expression

    The expression has two absolute values. The bars mean the distance from 0, so what comes out of them is never negative. Work in this order. First simplify inside the first bars. Then take both absolute values. Last, do the subtraction between them.

  2. Rewrite the signs inside

    Two signs inside the first bars are tricky. Adding −17 is the same as subtracting 17. Subtracting −31 is the same as adding 31.

    a+(−b)=a−banda−(−b)=a+ba + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b

    (−8)−12+(−17)−(−31)=−8−12−17+31(-8) - 12 + (-17) - (-31) = -8 - 12 - 17 + 31

    Watch out: minus negative 31 is plus 31. Treating it as minus 31 makes the inside −68. That leads to 44, choice D.

  3. Combine the four numbers

    Go from left to right, one number at a time.

    −8−12=−20-8 - 12 = -20

    −20−17=−37-20 - 17 = -37

    −37+31=−6-37 + 31 = -6

    The inside of the first bars is −6.

  4. Take the absolute values

    |a|=the distance from a to 0\left|a\right| = \text{the distance from } a \text{ to } 0

    |−6|=6and|24|=24\left|-6\right| = 6 \quad \text{and} \quad \left|24\right| = 24

    Now the expression is 6−246 - 24. Watch out: the bars turn −6 into 6. Keeping −6 gives −6−24=−30-6 - 24 = -30, choice A.

  5. Do the last subtraction

    6−24=−186 - 24 = -18

    You take 24 away from 6. Taking a bigger number from a smaller one gives a negative answer. Watch out: absolute values are never negative, but this answer can be. Doing 24−624 - 6 instead gives 18, choice C.

    So the answer is B, −18.

Check your answer

Check the inside a new way. The positive part is 31. The negative parts add to 8+12+17=378 + 12 + 17 = 37. Since 31−37=−631 - 37 = -6, the inside is right. Then, on a number line, start at 6 and move 24 to the left. Six steps reach 0, and 18 more land on −18.

More questions like this

The original DOE question
Question 92 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 85, question 92. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 47 of 120

2024 Test A Math, question 38

Question

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

  • A7%
  • B8%
  • C10%
  • D12%
Show the answer

Correct answer: B

Subtract to determine the amount of the increase.

1.35−1.25=0.101.35 - 1.25 = 0.10

Divide the increase by the original price.

0.10÷1.25=0.080.10 \div 1.25 = 0.08

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

0.08×100=80.08 \times 100 = 8

Question analysis

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

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

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

    Dividing the 10-cent rise by either price gives a single-digit percent, 8% from $1.25 and about 7.4% from $1.35, so only the choice of base separates the two listed answers. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic moderate Working it out

    Without a calculator, 0.10 ÷ 1.25 is a division by a decimal; rewriting it in cents as 10/125, which simplifies to 8/100, is the clean way through. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

  • Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The coffee cost $1.25 before. That is the original price. Now it costs $1.35, the new price. You need the percent increase. That compares how much the price went up with the original price.

  2. Find the amount of the increase

    1.35−1.25=0.101.35 - 1.25 = 0.10

    The price went up $0.10, or 10 cents. Watch out: 10 cents is not 10%. It would be 10% only if the price started at $1.00. That mistake gives choice C.

  3. Divide by the original price

    percent increase=new−originaloriginal×100\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100

    1.35−1.251.25×100=0.101.25×100\frac{1.35 - 1.25}{1.25} \times 100 = \frac{0.10}{1.25} \times 100

    Dividing by a decimal is easier in cents: 10 cents out of 125 cents. Divide the top and bottom by 5. Then multiply both by 4 to get hundredths.

    10125=225=8100=0.08\frac{10}{125} = \frac{2}{25} = \frac{8}{100} = 0.08

  4. Change to a percent

    0.08×100=80.08 \times 100 = 8

    The price went up 8%. Watch out: divide by the original price, $1.25, not the new one. Dividing by $1.35 gives about 7.4%, which rounds to choice A.

    So the answer is B, 8%.

Check your answer

Work forward from the answer. 8% of $1.25 is 0.08×1.25=0.100.08 \times 1.25 = 0.10, or 10 cents. Then 1.25+0.10=1.351.25 + 0.10 = 1.35. That is the new price, so 8% is right.

More questions like this

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

Question 48 of 120

2024 Test A Math, question 39

Question

Triangle with its two slanted sides each labeled 13 in. and its base labeled 10 in. Beside it, a vertical arrow as tall as the triangle, from its base to its top vertex, is labeled 12 in.

Raquel is cutting out pieces of cardboard to make a pyramid. She will use a square piece for the base and identical triangular pieces for the sides. The figure above shows the dimensions of the triangle for each side. What will be the total surface area, in square inches, of the pyramid, including the square base?

  • A280
  • B295
  • C340
  • D360
Show the answer

Correct answer: C

Surface area is found by calculating the area of each side, including the base, and adding them all together.

The base of the pyramid measures 10 inches on each side, so the area of the base is

10×10=10010 \times 10 = 100

Each triangular face has a base of 10 inches and a height of 12 inches, so the area of each face is

12×10×12=60\frac{1}{2} \times 10 \times 12 = 60

There are 4 triangular faces (because the square base has 4 edges, each of which will have a triangular face). So the total surface area of the pyramid is the sum of the areas of 4 triangles and 1 square.

(4×60)+100=240+100(4 \times 60) + 100 = 240 + 100

So the total surface area is 340 square inches.

Question analysis

This surface area word problem shows one triangular face, with sides of 13 inches, a base of 10 inches and a height of 12 inches, that Raquel will use to build a cardboard pyramid with a square base. Students find the total area of the base and all the triangular sides. Picturing the solid and choosing the right measurements from the figure are the tricky parts.

NYS standards
  • 7.G.6 area, surface area, and volume
  • 6.G.4 nets and surface area
  • 6.G.1 area of polygons
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students find the surface area of a pyramid from a drawn net or solid. Here only one triangular face is shown, so students must picture the whole pyramid, count its faces, and choose which labeled length belongs in the area formula.
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, 56 words.
Corrected from the DOE handbook
  • A wrong unit 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

Students have to build the pyramid in their heads from one flat triangle, then choose the right measurement from three labels. The 13-inch sides look like the natural height, and using them gives a listed total; picturing a pyramid with three sloping faces gives another. Sketching the square base with a triangle on each edge, and marking 12 as the height, settles both the face count and the formula.

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

    The figure gives one triangle with three length labels, 13, 13 and 10, and puts its height, 12, on a separate arrow beside it, where it is easy to overlook. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Combines ideas moderate Working it out

    The solid is never drawn: students must see that the square's side equals the triangle's 10-inch base, count four triangles, and add triangle and square areas. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Something you must already know minor Working it out

    A triangle's area uses its perpendicular height, not a slanted side, and a pyramid on a square base has four triangular faces; neither fact is given. (Anderson & Krathwohl, 2001)

Strategy: Sketch the pyramid: a square base whose side is the triangle's base, and one triangular face on each side of the square. Find each area with the matching formula, then add them all.

Traps

  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • Information you don't need: The question gives numbers or facts that play no part in the answer.
  • 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

What was corrected from the DOE handbook, and why

  • Explanation: The handbook ends “So the total surface area is 340 square feet.” The triangle’s dimensions are in inches and the question asks for square inches, so the site now reads “340 square inches.”

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, at a right angle to the base. You take half because a triangle is half of a rectangle with the same base and height.
Area of a square
A=s×sA = s \times s
A is the area and s is the length of one side. All four sides of a square are equal, so the area is side times side.
Surface area
surface area=sum of the areas of all the faces\text{surface area} = \text{sum of the areas of all the faces}
A face is one flat side of a solid shape. The surface area is the total area of the outside. Picture unfolding the solid into flat pieces. Then add up the area of every piece.

  1. Sort out the question

    Raquel’s pyramid has a square base. Each side is a triangle like the one in the figure. The triangle’s base is 10 in. Its two slanted sides are 13 in. Its height, the arrow beside it, is 12 in. You need the total surface area, including the square base, in square inches.

  2. Picture the pyramid

    Surface area is the total area of all the flat sides, called faces. Each triangle stands on one edge of the square. So each edge of the square is 10 inches, the same as the triangle’s base. A square has 4 edges, so the pyramid has 4 triangles.

    surface area=sum of the areas of all the faces\text{surface area} = \text{sum of the areas of all the faces}

    Here that is 4 triangles plus 1 square. Watch out: counting only 3 triangles gives 280, choice A.

  3. Find the area of one triangle

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×10×12=60A = \frac{1}{2} \times 10 \times 12 = 60

    Each triangle is 60 square inches. Watch out: the height is 12 in., the straight-up arrow. It is not 13 in., the slanted side. Using 13 gives 65 for each triangle and 360 in all, choice D.

  4. Find the area of the base

    The base is a square with sides of 10 inches, the same as each triangle’s base.

    A=s×sA = s \times s

    A=10×10=100A = 10 \times 10 = 100

    The square base is 100 square inches.

  5. Add all five faces

    surface area=4×60+100=240+100=340\text{surface area} = 4 \times 60 + 100 = 240 + 100 = 340

    The four triangles and the square together cover 340 square inches.

    So the answer is C, 340 square inches.

Check your answer

Each triangle is half of a 10 by 12 rectangle, which has an area of 120. So two triangles make 120, and four make 240. Add the base: 240+100=340240 + 100 = 340. The units fit too: inches times inches gives square inches.

More questions like this

The original DOE question
Question 96 as printed on page 86 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 86, question 96. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 49 of 120

2024 Test A Math, question 40

Question

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

  • A5%
  • B8%
  • C7%
  • D50%
Show the answer

Correct answer: B

Use the percent increase expression to find the answer:

new price−original priceoriginal price\frac{\text{new price} - \text{original price}}{\text{original price}}

6.75−6.256.25=0.506.25=0.08=8%\frac{6.75 - 6.25}{6.25} = \frac{0.50}{6.25} = 0.08 = 8\%

Question analysis

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

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

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

    Measured against $6.75 the 50-cent rise is about 7.4%, and against $6.25 it is 8%; both are listed side by side, so only the right base separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic moderate Working it out

    The division 0.50 ÷ 6.25 is easiest as 50/625, which reduces to 2/25 and then to 8/100; a three-digit decimal divisor invites a slip. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The sandwich cost $6.25 before. That is the original price. Now it costs $6.75, the new price. You need the percent increase. That compares the increase with the original price. Note that the choices are not in order: C, 7%, comes after B, 8%.

  2. Find the amount of the increase

    6.75−6.25=0.506.75 - 6.25 = 0.50

    The price went up $0.50, or 50 cents. Watch out: 50 cents is not 50%. It would be 50% only if the price started at $1.00. That mistake gives choice D.

  3. Divide by the original price

    percent increase=new−originaloriginal×100\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100

    6.75−6.256.25×100=0.506.25×100\frac{6.75 - 6.25}{6.25} \times 100 = \frac{0.50}{6.25} \times 100

    In cents, this is 50 out of 625. Divide the top and bottom by 25. Then multiply both by 4 to get hundredths.

    50625=225=8100=0.08\frac{50}{625} = \frac{2}{25} = \frac{8}{100} = 0.08

  4. Change to a percent

    0.08×100=80.08 \times 100 = 8

    The price went up 8%. Watch out: divide by the original price, $6.25, not the new one. Dividing by $6.75 gives about 7.4%, which is choice C.

    So the answer is B, 8%.

Check your answer

Work forward from the answer. 8% of $6.25 is 0.08×6.25=0.500.08 \times 6.25 = 0.50, or 50 cents. Then 6.25+0.50=6.756.25 + 0.50 = 6.75. That is the new price, so 8% is right.

More questions like this

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

Question 50 of 120

2024 Test A Math, question 41

Question

Terrell played 5 computer games and earned a mean score of 8 points per game. If his mean score for the first 4 games was 7 points per game, how many points was his score in the fifth game?

  • A9
  • B11
  • C12
  • D14
Show the answer

Correct answer: C

If the mean number of points for the first four games is 7, the total number of points earned in those games is 28.

4×7=284 \times 7 = 28

If the mean number of points for all five games is 8, the total number of points earned is 40.

5×8=405 \times 8 = 40

To find the number of points in the fifth game, find the difference between 40 and 28.

40−28=1240 - 28 = 12

Question analysis

This mean, or average, word problem says Terrell averaged 8 points over 5 computer games and 7 points over the first 4, and asks for his score in the fifth game. Students turn each average into a total and compare the totals. Combining the two averages directly, without going through totals, is the tempting shortcut that fails.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.3 measures of center and variation
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Computing a mean is grade 6 content. Working backward from two means to the totals they stand for, then to one missing score, is a common grade 7 test task, and it comes with a tempting shortcut.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 4, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards turning both means into totals, 40 points for five games and 28 for four, after which one subtraction finishes it. Averaging 7 with the unknown score is a quick shortcut that gives a listed choice, but it weighs four games as one. Checking that 28 plus the answer, divided by 5, gives 8 catches that error and the near miss next to the answer.

  1. Turning words into math moderate Understanding the question

    Each mean has to become a total before the fifth game can be found: "a mean score of 8 points per game" over 5 games means 40 points in all. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Students must know that a total is the mean times the number of games, and that means over different numbers of games cannot simply be averaged. (Anderson & Krathwohl, 2001)

Strategy: Turn each mean into a total by multiplying by its number of games, then subtract the four-game total from the five-game total.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total from a mean
total=mean×number of games\text{total} = \text{mean} \times \text{number of games}
The mean, or average, is what each game would score if the total were shared out equally. So the total is the mean times the number of games. Use it to turn each mean back into a total. Totals can be subtracted, but means over different numbers of games can’t.

  1. Sort out the question

    Terrell played 5 games. His mean score for all 5 was 8 points per game. His mean for the first 4 games was 7 points per game. The mean, or average, is the total split equally among the games. You need his score in game 5.

  2. Find the total for all 5 games

    Turn the mean for all 5 games into a total. Multiply the mean by the number of games.

    total=mean×number of games\text{total} = \text{mean} \times \text{number of games}

    total=8×5=40\text{total} = 8 \times 5 = 40

    He scored 40 points in all 5 games together.

  3. Find the total for the first 4 games

    Use the same formula with the mean of 7 and 4 games.

    total=7×4=28\text{total} = 7 \times 4 = 28

    He scored 28 points in the first 4 games.

  4. Subtract to find game 5

    The 40 points include the first 4 games and game 5. Take away the first 4 games to leave game 5.

    40−28=1240 - 28 = 12

    He scored 12 points in game 5. Watch out: don’t average 7 and the fifth score as if they were two games. That gives 9, choice A. The 7 stands for 4 games, not 1. Watch out: adding 4 to 7 gives 11, choice B. Each of the first 4 games was 1 point below 8. So game 5 needs 8 plus 4 more points.

    So the answer is C, 12 points.

Check your answer

Try scores of 7, 7, 7, 7 and 12. They add up to 28+12=4028 + 12 = 40, and 40÷5=840 \div 5 = 8. That is the mean for all 5 games, so 12 is right.

More questions like this

The original DOE question
Question 98 as printed on page 87 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 87, question 98. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 51 of 120

2024 Test A Math, question 47

Question

In a survey of 200 adults in the town of Waskegon, 45 reported reading the online version of the Waskegon Bulletin the previous day. If 25,000 adults live in Waskegon, which number is the best estimate of the number of adults who read the online version of the Waskegon Bulletin the previous day?

  • A5,600
  • B9,000
  • C11,300
  • D24,800
Show the answer

Correct answer: A

In the survey, 45 out of 200 adults reported reading the online version of the Waskegon Bulletin, and there are 25,000 adults living in Waskegon. Set up a proportion to find the number of adults who read the online version.

45200=x25,000\frac{45}{200} = \frac{x}{25{,}000}

x=(45)(25,000)200=x = \frac{(45)(25{,}000)}{200} =

(45)(125)=5,625(45)(125) = 5{,}625

Rounded to the nearest 100, this number is 5,600.

Question analysis

In a survey of 200 adults in Waskegon, 45 said they read the online Waskegon Bulletin the day before, and the town has 25,000 adults. This sampling question asks for the best estimate of how many adults in town read it. Students scale the sample's rate up to the population with a proportion and round.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a sample to estimate a count for a whole population is typical grade 7 statistics and ratio work. This version takes one proportion with large numbers and a rounding step, about the usual grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 13, 50 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The idea is a single proportion, and the choices are far apart, so a reasonable estimate is enough. The main trap is treating 45 out of 200 as 45 percent, which doubles the estimate; the other wrong choices combine the survey numbers without a proportion. Noting that 45 is a bit under a quarter of 200, so the town figure must be a bit under a quarter of 25,000, confirms the choice.

  1. Turning words into math minor Understanding the question

    The survey's 45 out of 200 is a rate to scale up to the town's 25,000 adults, a fraction of the sample rather than a percent. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    The scale factor 25,000 ÷ 200 = 125 and the product 45 × 125 = 5,625 involve large numbers, and the result must then be rounded to the nearest hundred. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Set the survey's fraction, 45 out of 200, equal to the unknown number out of 25,000, then scale up: 25,000 is 125 times 200.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportion
ab=cd\frac{a}{b} = \frac{c}{d}
A proportion says two ratios are equal. A ratio compares two numbers by dividing. Here a out of b in the survey matches c out of d in the whole town. Use it when a small group stands for a bigger group.

  1. Sort out the question

    A survey asked 200 adults, and 45 of them read the online paper. The town has 25,000 adults. You need a good estimate of how many adults in the whole town read it. Assume the town is like the survey. Then the same fraction of adults read it.

  2. Set up a proportion

    A proportion says two fractions are equal. Let x be the number of readers in the town. The fraction of readers is the same in both groups.

    ab=cd\frac{a}{b} = \frac{c}{d}

    45200=x25,000\frac{45}{200} = \frac{x}{25{,}000}

    Watch out: multiplying 45 by 200 gives 9,000, choice B. That mixes the numbers without a plan.

  3. Find the scale factor

    The scale factor is how many times bigger the town is than the survey group.

    25,000÷200=12525{,}000 \div 200 = 125

    The town is 125 times as big. So multiply the top, 45, by 125 too.

  4. Scale up the readers

    x=45×125x = 45 \times 125

    Break 125 into 100 and 25.

    45×100=4,50045 \times 100 = 4{,}500

    45×25=1,12545 \times 25 = 1{,}125

    4,500+1,125=5,6254{,}500 + 1{,}125 = 5{,}625

    About 5,625 adults read it. Watch out: 45 out of 200 is not 45%. A percent is out of 100. Using 45% gives about 11,300, choice C.

  5. Round to the nearest hundred

    The choices are rounded, since this is an estimate. 5,625 is between 5,600 and 5,700. Its last two digits, 25, are less than 50, so round down to 5,600.

    So the answer is A, 5,600.

Check your answer

45 out of 200 is a bit less than 50 out of 200, which is one fourth. One fourth of 25,000 is 6,250. So the answer should be a bit less than 6,250. 5,600 fits, and the other choices are all much bigger.

More questions like this

The original DOE question
Question 104 as printed on page 88 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 88, question 104. Open page 88 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 52 of 120

2024 Test A Math, question 51

Question

3(0.01)−3(0.1)=3(0.01) - 3(0.1) =

  • A−0.33
  • B−0.27
  • C0
  • D0.33
Show the answer

Correct answer: B

3(0.01)−3(0.1)=3(0.01) - 3(0.1) =

0.03−0.3=−0.270.03 - 0.3 = -0.27

Question analysis

This decimal question asks for the value of 3 times 0.01 minus 3 times 0.1. Students multiply each decimal by 3 and subtract, keeping the order of the subtraction as written. The two decimals look alike, so place value and the sign of the result both need care.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students subtract rational numbers and get negative results. This short computation adds place-value care, since 0.01 and 0.1 look alike, but it is a routine two-step grade 7 exercise.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 16+, 1 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 quick multiplications leave 0.03 − 0.3, a subtraction whose result is negative. The usual errors are adding the two products, with or without a minus sign, and treating 0.01 and 0.1 as the same number. Rewriting the subtraction as −(0.3 − 0.03) keeps the place values lined up and makes the sign plain.

  1. Error-prone arithmetic moderate Working it out

    The products 0.03 and 0.3 have their digits in different places, and taking the larger from the smaller crosses zero, so both alignment and sign need care. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer minor Choosing the answer

    The choice −0.33 has the right sign and is only 0.06 from the answer, so a sense that the result is a little below zero cannot separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Multiply each decimal by 3, then subtract with the place values lined up; or factor out the 3 and compute 3 times (0.01 − 0.1).

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a bigger number
a−b=−(b−a)a - b = -(b - a)
a and b are numbers, and b is the bigger one. When you take a bigger number away from a smaller one, the answer is negative. Its size is the gap between them, b minus a. So subtract the other way, then put a minus sign in front.

  1. Sort out the problem

    3(0.01) means 3 times 0.01. A number right next to parentheses means multiply. Then subtract 3 times 0.1. Note that 0.01 is one hundredth, and 0.1 is one tenth. They are not equal. Watch out: treating them as equal makes the two parts cancel to 0, choice C.

  2. Multiply each part

    3×0.01=0.033 \times 0.01 = 0.03

    3×0.1=0.33 \times 0.1 = 0.3

    So the problem is 0.03−0.30.03 - 0.3. Write 0.3 as 0.30 so both have two decimal places. Now you can see that 0.30 is bigger than 0.03.

  3. Subtract the bigger number

    You take a bigger number from a smaller one, so the answer is negative.

    a−b=−(b−a)a - b = -(b - a)

    0.03−0.30=−(0.30−0.03)=−0.270.03 - 0.30 = -(0.30 - 0.03) = -0.27

    Watch out: find the gap between the numbers. Don’t add them. Adding gives 0.33, choice D, or −0.33, choice A.

    So the answer is B, −0.27.

Check your answer

Pull out the 3 first: 3×(0.01−0.1)=3×(−0.09)=−0.273 \times (0.01 - 0.1) = 3 \times (-0.09) = -0.27. Same answer. Also think of a number line. Start at 0.03, just right of 0. Move 0.30 to the left. You land a little left of 0, at −0.27.

More questions like this

The original DOE question
Question 108 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 89, question 108. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 53 of 120

2024 Test A Math, question 54

Question

The probability of an event occurring is 0.05. What is the chance that the event will occur?

  • Alikely
  • Bunlikely
  • Cimpossible
  • Dneither likely nor unlikely
Show the answer

Correct answer: B

The probability of the event is 0.05, which equals 5%, which is less than 50%.

Option A:

An event that is likely has a probability over 50%.

Option B:

An event that is unlikely has a probability less than 50% but greater than 0%.

Option C:

An event that is impossible has a 0% probability.

Option D:

An event that is neither likely nor unlikely has a 50% probability.

Question analysis

This probability question gives an event a probability of 0.05 and asks how likely the event is to occur, described in words. Students place the decimal on the scale from 0, impossible, to 1, certain. Reading the place value of the decimal correctly is the only real hurdle.

NYS standards
  • 6.SP.6 probability from 0 to 1
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Describing in words how likely an event is from its probability between 0 and 1 is a grade 6 standard. This is one quick judgment with no computation, well below grade 7 probability work with compound events.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 10, 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

One judgment answers the question: 0.05 is a small probability, but it is not zero. The two tempting errors are treating a small chance as no chance at all and misreading 0.05 as 0.5, an even chance. Writing the probability as 5% or as 1 in 20 makes its size plain.

  1. Something you must already know minor Working it out

    The probability scale, 0 for impossible, 1/2 for an even chance and 1 for certain, is assumed, and 0.05 has to be placed near its low end. (Anderson & Krathwohl, 2001)

Strategy: Picture the probability scale from 0 to 1, with 0 for impossible, 1/2 for an even chance and 1 for certain, and see where 0.05 falls.

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.
  • 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

Probability scale
0≤P≤10 \leq P \leq 1
P is the probability of an event, a number that tells how likely it is to happen. It is always from 0 to 1. 0 means impossible, and 1 means certain. One half, or 0.5, means an even chance. The closer P is to 0, the less likely the event.

  1. Sort out the question

    Probability is a number that tells how likely an event is. An event is something that might happen. Here the probability is 0.05. You need the word that best describes that chance.

  2. Change 0.05 to a percent

    0.05=51000.05 = \frac{5}{100}

    So 0.05 is 5 out of 100, or 5%. That is the same as 1 chance in 20. Watch out: 0.05 is not 0.5. A probability of 0.5 would be an even chance, choice D.

  3. Place it on the scale

    0≤P≤10 \leq P \leq 1

    ProbabilityChance
    0impossible
    more than 0, less than 0.5unlikely
    0.5neither likely nor unlikely
    more than 0.5, less than 1likely
    1certain

    0.05 is more than 0 but much less than 0.5. So the event is unlikely.

  4. Rule out the other choices

    Watch out: a small chance is still a chance. 0.05 is more than 0, so the event can happen. It is not impossible, choice C. Likely, choice A, would need more than 0.5.

    So the answer is B, unlikely.

Check your answer

Picture 20 tries. With a 5% chance, the event would happen about once, since 5% of 20 is 1. It would not happen about 19 times. That is rare but possible, so unlikely fits.

More questions like this

The original DOE question
Question 111 as printed on page 90 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 90, question 111. Open page 90 of the handbook (PDF)

Question 54 of 120

2024 Test A Math, question 55

Question

The table below shows the number of cups of red paint and blue paint used to make a purple paint mixture.

PURPLE PAINT

Cups of Red Paint Cups of Blue Paint
1 1.5
4 6
11 y

Based on the relationship between the number of cups of red paint and the number of cups of blue paint, what is the value of y?

  • A10.5
  • B13
  • C16.5
  • D24
Show the answer

Correct answer: C

Divide the number of cups of blue paint by the number of cups of red paint to find the constant of proportionality.

1.51=64=1.5\frac{1.5}{1} = \frac{6}{4} = 1.5

Multiply to find the value of y.

11×1.5=16.511 \times 1.5 = 16.5

Question analysis

A table gives cups of red and blue paint for a purple mixture: 1 and 1.5, 4 and 6, and 11 and an unknown y. This ratio table question asks for y based on the relationship between the two colors. The tricky part is noticing whether the relationship works by multiplying or by adding.

NYS standards
  • 7.RP.2b constant of proportionality
  • 6.RP.3a tables of equivalent ratios
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a constant of proportionality from a table is grade 7 work, but here one row already pairs 1 cup of red with its cups of blue, so a single multiplication finishes it, lighter than typical grade 7 tasks.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 57 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 relationship is multiplicative, 1.5 cups of blue for every cup of red, and one multiplication finishes the question. The trap is additive thinking: red grows by 7 cups from the second row to the third, and adding the same 7 to the blue column gives a listed answer. Testing a rule on every row exposes the additive one, since blue is 0.5 more than red in the first row but 2 more in the second.

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

    The table's first row, 1 cup of red to 1.5 cups of blue, hands over the constant of proportionality; the other two rows only confirm it. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the row with 1 cup of red paint to read the cups of blue per cup of red, multiply that rate by 11, and check the rate against the row with 4 cups.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Constant of proportionality
k=yxk = \frac{y}{x}
x and y are two amounts that change together. k is the constant of proportionality: how much y goes with 1 of x. In a proportional relationship, y divided by x gives the same k every time.
Proportional relationship
y=k×xy = k \times x
x is the cups of red paint, and y is the cups of blue paint. k is the cups of blue for each cup of red. In a proportional relationship, y is always k times x. Use it to find y for any amount of red.

  1. Sort out the table

    The table shows cups of red and blue paint for the same purple. 1 cup of red goes with 1.5 cups of blue. 4 cups of red go with 6 cups of blue. You need y, the cups of blue for 11 cups of red. The same color means the paints mix at the same rate in every row.

  2. Find the blue for each cup of red

    k=yxk = \frac{y}{x}

    k=1.51=1.5andk=64=1.5k = \frac{1.5}{1} = 1.5 \quad \text{and} \quad k = \frac{6}{4} = 1.5

    Both rows give 1.5. So there are always 1.5 cups of blue for each cup of red. This number is the constant of proportionality.

  3. Use the rate for 11 cups

    y=k×xy = k \times x

    y=1.5×11=16.5y = 1.5 \times 11 = 16.5

    Think of it as 11 cups plus half of 11 cups: 11+5.5=16.511 + 5.5 = 16.5. So y is 16.5 cups of blue.

  4. Avoid the traps and answer

    Watch out: red goes up 7 cups from 4 to 11. Adding the same 7 cups to the 6 cups of blue gives 13, choice B. Paint mixes by multiplying, not by adding. Watch out: 7×1.5=10.57 \times 1.5 = 10.5, choice A, is only the extra blue for the 7 extra cups of red.

    So the answer is C, 16.5.

Check your answer

Divide to check: 16.5÷11=1.516.5 \div 11 = 1.5. That matches the other rows. Also, every row has more blue than red, so y must be more than 11. That rules out 10.5.

More questions like this

The original DOE question
Question 112 as printed on page 90 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 90, question 112. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 55 of 120

2024 Test B Math, question 2

Question

Two number cubes have sides labeled 1 through 6. Both number cubes are rolled. How many different outcomes will have a sum of 6?

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

Show the answer

Correct answer: 5

There will be 5 different outcomes with a sum of 6:

1+5=61+5=6

2+4=62+4=6

3+3=63+3=6

4+2=64+2=6

5+1=65+1=6

Question analysis

A counting grid-in with two number cubes, each numbered 1 to 6, rolled together. Students count how many different outcomes give a sum of 6. The catch is deciding whether a 2 on the first cube with a 4 on the second counts as a different outcome from the reverse.

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students list sample spaces for two number cubes when finding compound probabilities. This asks for one count from that list, a short task, though as a grid-in it offers no choices to expose an incomplete list.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 3, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is trivial; the question turns on what counts as a different outcome. Treating the cubes as a first and a second settles it: the first cube runs from 1 to 5 and the second is forced each time. Students who think in pairs of numbers rather than in rolls of two cubes undercount, and those who double every pair overcount by one at the double 3 and 3.

  1. Something you must already know moderate Working it out

    Nothing in the stem says that a 2 on the first cube with a 4 on the second differs from the reverse; students must already know that two cubes give 36 ordered outcomes. (Anderson & Krathwohl, 2001)

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

    As a grid-in it offers no list of counts to compare, so a student who counts unordered pairs gets no sign that ordered outcomes were wanted. (Enright, Morley & Sheehan, 2002)

Strategy: Treat the cubes as a first and a second cube, run the first cube's value from 1 to 6, and for each value check whether the second cube can complete the sum.

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.
  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle
total outcomes=a×b\text{total outcomes} = a \times b
a is the number of ways the first cube can land, and b is the number of ways the second cube can land. Multiply them to count every pair. It works because each face of the first cube can go with each face of the second.

  1. Sort out the question

    You roll two number cubes. Each one shows 1, 2, 3, 4, 5 or 6. An outcome is one way the two cubes can land. You need the number of outcomes whose sum (the two numbers added) is 6.

    The question does not say if order matters. The answer key counts the cubes as a first cube and a second cube. So 1 on the first and 5 on the second is a different outcome from 5 on the first and 1 on the second.

  2. See why order counts

    Picture a red cube and a blue cube. Red 1 with blue 5 is not the same roll as red 5 with blue 1. Both can happen, so both count.

    This is how all the outcomes of two cubes are counted. Use the counting principle.

    total outcomes=6×6=36\text{total outcomes} = 6 \times 6 = 36

    Watch out: 36 is every outcome, not just the ones that add up to 6.

  3. List the pairs that make 6

    Go through the first cube one number at a time. The second cube has to make up the rest of the 6.

    First cubeSecond cubeSum
    156
    246
    336
    426
    516

    The first cube cannot show 6. Then the second cube would need 0, and no face shows 0.

  4. Count the outcomes

    The table has 5 rows, so there are 5 outcomes. Watch out: counting 1 and 5 as the same as 5 and 1 gives only 3. Watch out: 3 and 3 is one outcome, not two. Doubling 3 pairs to get 6 counts it twice.

    So you would grid 5.

Check your answer

Each number from 1 to 5 on the first cube goes with exactly one number on the second cube. A 6 on the first cube goes with nothing. So there are exactly 5 outcomes. The table shows the same pattern: as the first number goes up by 1, the second goes down by 1.

More questions like this

The original DOE question
Question 59 as printed on page 198 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 198, question 59. Open page 198 of the handbook (PDF)

Question 56 of 120

2024 Test B Math, question 6

Question

This graph shows y, the weight of a pole of length x.

Graph titled WEIGHT OF POLE. The x-axis, Length (feet), is marked 1, 2, 3 and 4, and the y-axis, Weight (ounces), is marked 4, 8, 12 and 16, with grid lines at each mark; the origin is labeled O. A straight line starts at O and rises through the grid corners (1, 4), (2, 8), (3, 12) and (4, 16), ending in an arrow, and the point (1, 4) is marked with a dot and labeled (1, 4).

What is represented by the point with coordinates (1, 4)?

  • AThe unit rate is 4 ounces per foot.
  • BThe y-intercept is 4.
  • CA pole 4 feet long weighs 1 ounce.
  • DThe length increases 4 feet for each 1 ounce of weight.
Show the answer

Correct answer: A

The x-axis represents the length of the pole, in feet, and the y-axis represents the weight of the pole, in ounces.

The point (1, 4) means that a 1-foot pole would weigh 4 ounces.

Since the graph is linear, the pole weighs 4 ounces for every additional 1 foot of length.

So the unit rate is 4 ounces per foot.

Question analysis

A graph of a pole's weight in ounces against its length in feet is a straight line through the origin, with the point (1, 4) marked. Students decide what that point says about poles, choosing among statements about rates, intercepts and particular poles, so the meaning of each coordinate has to be pinned down.

NYS standards
  • 7.RP.2d points on proportional graphs
  • 7.RP.2b constant of proportionality
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Explaining what a point on a proportional graph means in the situation is a named grade 7 skill. This is a direct, one-step example with a clear graph, typical of grade 7 classwork.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
1 step
Reading level
Question and choices: about grade 2, 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

Nothing needs computing; the question tests whether a point on a graph can be put into words. Each wrong statement misuses the same two numbers, as an intercept, with the coordinates swapped, or as a rate turned upside down. Saying the point aloud with its units, a 1-foot pole weighs 4 ounces, and comparing that sentence with each statement settles it quickly.

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

    The axes carry the meaning: length in feet runs along x and weight in ounces up y, so (1, 4) describes a 1-foot pole, and the line through O makes that a rate. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. A very close wrong answer minor Choosing the answer

    Both rate statements pair 4 with 1, one as ounces per foot and the other as feet per ounce; only the axis labels say which way the rate runs. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Something you must already know minor Working it out

    The choices use the terms unit rate and y-intercept, which the question does not define, and two statements hang on knowing what each one means. (Anderson & Krathwohl, 2001)

Strategy: Note which quantity each axis measures, say in words what the x-coordinate and the y-coordinate of the marked point stand for, then compare that sentence with each choice.

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).
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • 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

Unit rate
unit rate=yx\text{unit rate} = \frac{y}{x}
For a point on the line, y is the weight in ounces and x is the length in feet. The unit rate is the weight of 1 foot of pole. This works for a straight line that starts at the origin, the point (0, 0).

  1. Read the axes

    The graph has two axes (number lines). The x-axis, across the bottom, shows length in feet. The y-axis, going up, shows weight in ounces.

    A point is written (x, y). The across number comes first, then the up number. So in (1, 4), x is 1 and y is 4.

  2. Say what the point means

    x = 1 means a length of 1 foot. y = 4 means a weight of 4 ounces. So a pole 1 foot long weighs 4 ounces.

    Watch out: do not swap the numbers. “A pole 4 feet long weighs 1 ounce” reads them the wrong way round. That is choice C.

  3. See that this is a unit rate

    A unit rate is an amount for 1 unit, here the ounces in 1 foot. The line starts at O, the origin (0, 0), and it is straight. So each extra foot adds the same weight.

    unit rate=yx=41=4\text{unit rate} = \frac{y}{x} = \frac{4}{1} = 4

    The unit rate is 4 ounces per foot. Watch out: choice D turns this upside down. It gives feet for each ounce, not ounces for each foot.

  4. Rule out the y-intercept

    The y-intercept is where the line crosses the y-axis. This line crosses it at O, where y is 0. So the y-intercept is 0, not 4, and choice B is wrong.

    So the answer is A, the unit rate is 4 ounces per foot.

Check your answer

Try another point on the line, such as (3, 12). A pole 3 feet long weighs 12 ounces. 12÷3=412 \div 3 = 4, so it is still 4 ounces per foot. The rate stays the same all along the line.

More questions like this

The original DOE question
Question 63 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 200, question 63. Open page 200 of the handbook (PDF)

Question 57 of 120

2024 Test B Math, question 9

Question

A car dealership is having a sale where all cars have a 12% discount. What is the discounted price of a car that originally cost $15,600?

  • A$13,728
  • B$14,300
  • C$14,400
  • D$15,588
Show the answer

Correct answer: A

15,600×12%=15{,}600\times 12\%=

15,600×0.12=1,87215{,}600\times 0.12=1{,}872

15,600−1,872=13,72815{,}600-1{,}872=13{,}728

Question analysis

A percent word problem about a car sale: every car is 12% off, and one car originally cost $15,600. Students find the sale price of that car. The discount has to be worked out as a dollar amount and then taken off, with care over the decimal point in 12%.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Discounts are one of the percent applications in grade 7, and this is a routine two-step example with whole-dollar amounts. It matches typical grade 7 work, with no reverse percent or second change.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 24 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a standard two-step discount, and a quick estimate does most of the work: 10% of $15,600 is $1,560, so 12% is a little under $1,900, and only one choice sits about that far below the price. The wrong choices take off $1,300, $1,200 or just $12, so the question mainly catches students unsure what 12% of a price means.

  1. Error-prone arithmetic minor Working it out

    Finding 12% of $15,600 by hand, 0.12 × 15,600 = 1,872, means placing a decimal point in a product of large numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The phrase “discounted price” asks for what the buyer pays, so the $1,872 discount has to be taken off the price rather than reported. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Write 12% as 0.12 and multiply by the original price to get the discount, then subtract it; or multiply the price by 0.88 in one step.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
x=p100×Wx = \frac{p}{100} \times W
W is the whole amount, here the first price. p is the percent, and x is the part you want. Percent means out of 100, so p percent is the fraction p over 100. Multiplying the whole by that fraction gives the part.
Sale price
sale price=first price−discount\text{sale price} = \text{first price} - \text{discount}
The first price is what the car cost before the sale. The discount is the amount taken off. The sale price is what the buyer pays, so you subtract.

  1. Sort out the question

    The car first cost $15,600. A discount is an amount taken off the price. Here it is 12% of the price. You need the discounted price, the amount the buyer pays after the discount.

  2. Find 12% of the price

    x=p100×Wx = \frac{p}{100} \times W

    x=12100×15,600=0.12×15,600x = \frac{12}{100} \times 15{,}600 = 0.12 \times 15{,}600

    Break it into easy parts. 10% is 15,600÷10=1,56015{,}600 \div 10 = 1{,}560. 1% is 15,600÷100=15615{,}600 \div 100 = 156, so 2% is 2×156=3122 \times 156 = 312.

    1,560+312=1,8721{,}560 + 312 = 1{,}872

    The discount is $1,872. Watch out: 12% does not mean $12. Taking off $12 gives $15,588, choice D.

  3. Take the discount off

    sale price=first price−discount\text{sale price} = \text{first price} - \text{discount}

    15,600−1,872=13,72815{,}600 - 1{,}872 = 13{,}728

    The buyer pays $13,728. Watch out: 12% is not one twelfth. Dividing by 12 takes off only $1,300 and gives $14,300, choice B.

    So the answer is A, $13,728.

Check your answer

After a 12% discount, the buyer pays 88% of the price, because 100−12=88100 - 12 = 88. Then 0.88×15,600=13,7280.88 \times 15{,}600 = 13{,}728. That matches. Also, $13,728 is less than $15,600, as a sale price should be.

More questions like this

The original DOE question
Question 66 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 201, question 66. Open page 201 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 58 of 120

2024 Test B Math, question 11

Question

In a sports league, each team has 36 players and 3 coaches. There are also a number of team assistants. The ratio of team assistants to players is 16\frac{1}{6}. What is the ratio of coaches to assistants?

  • A14\frac{1}{4}
  • B12\frac{1}{2}
  • C23\frac{2}{3}
  • D56\frac{5}{6}
Show the answer

Correct answer: B

Let x be the number of assistants.

Since the ratio of assistants to players is 16\frac{1}{6}, then:

16=x36\frac{1}{6}=\frac{x}{36}

366=x\frac{36}{6}=x

6=x6=x

There are a total of 6 assistants.

So the ratio of coaches to assistants is 36=12\frac{3}{6}=\frac{1}{2}.

Question analysis

A ratio word problem about a sports team with 36 players, 3 coaches and an unknown number of assistants, where the ratio of assistants to players is 1 to 6. Students find how many assistants there are, then form a new ratio of coaches to assistants. Keeping the order of each ratio straight matters.

NYS standards
  • 6.RP.3 ratio and rate reasoning
  • 6.RP.1 ratio concepts and language
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a missing amount from a ratio and then writing a second ratio is grade 6 ratio reasoning with whole numbers. Typical grade 7 work adds fractions, rates or percents, which this two-step problem does not.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two short steps, finding 6 assistants and then writing 3 to 6, make this one of the easier ratio questions. The care needed is in keeping each ratio's order and its two groups straight, since the given ratio compares assistants with players and the asked one compares coaches with assistants. None of the wrong choices matches a common slip closely, so students who find the 6 assistants seldom go wrong.

  1. Turning words into math minor Understanding the question

    “The ratio of team assistants to players is 1/6” must become 6 assistants for 36 players, and the question then asks for a ratio in a new order, coaches first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Three groups are named, 36 players, 3 coaches and an unknown number of assistants, and each ratio pairs a different two of them. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Use the 1 to 6 ratio to find the number of assistants from the 36 players, then write coaches over assistants, in that order, and simplify.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Ratio as a fraction
ratio=first groupsecond group\text{ratio} = \frac{\text{first group}}{\text{second group}}
A ratio compares two amounts. The group named first goes on top, and the group named second goes on the bottom. Order matters. Assistants to players is not the same as players to assistants.
Equal ratios
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a and b are the two numbers in a ratio, written as a fraction. n is any number except zero. Multiplying both numbers by the same n gives an equal ratio. This is how you scale a small ratio up to real counts.

  1. Sort out what you know

    Each team has 36 players and 3 coaches. It also has some assistants, but you are not told how many. The ratio of assistants to players is 1/6. That means 1 assistant for every 6 players. You need the ratio of coaches to assistants.

  2. Find the number of assistants

    Write the ratio as a fraction, assistants on top and players on the bottom. To get from 6 players to 36, multiply by 6. Do the same to the top.

    ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}

    16=1×66×6=636\frac{1}{6} = \frac{1 \times 6}{6 \times 6} = \frac{6}{36}

    So there are 6 assistants for the 36 players.

  3. Write the new ratio

    Now the order changes. Coaches come first, so coaches go on top.

    ratio=first groupsecond group\text{ratio} = \frac{\text{first group}}{\text{second group}}

    coachesassistants=36=12\frac{\text{coaches}}{\text{assistants}} = \frac{3}{6} = \frac{1}{2}

    Divide the top and the bottom by 3 to simplify.

  4. Avoid the traps and answer

    Watch out: use only coaches and assistants. Putting all 9 staff over the 36 players gives 1/4, choice A. Putting the 6 assistants over all 9 staff gives 2/3, choice C. And 1/6 is not a share of everyone, so 5/6, choice D, has no meaning here.

    So the answer is B, 12\frac{1}{2}.

Check your answer

A ratio of 1/2 means 1 coach for every 2 assistants. With 3 coaches, that makes 6 assistants. Then 6 assistants to 36 players is 636=16\frac{6}{36} = \frac{1}{6}, the ratio the question gave.

More questions like this

The original DOE question
Question 68 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 201, question 68. Open page 201 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 59 of 120

2024 Test B Math, question 14

Question

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

  • A4
  • B5
  • C7
  • D15
Show the answer

Correct answer: B

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

60+60∙25100=60+60\bullet\frac{25}{100}=

60+60∙0.25=60+60\bullet 0.25=

60+15=7560+15=75

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

Karen: 1,500 pagesk days=60 pages1 day\frac{1{,}500\text{ pages}}{k\text{ days}}=\frac{60\text{ pages}}{1\text{ day}}

60k=1,50060k=1{,}500

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

k=25 daysk=25\text{ days}

Martina: 1,500 pagesm days=75 pages1 day\frac{1{,}500\text{ pages}}{m\text{ days}}=\frac{75\text{ pages}}{1\text{ day}}

75m=1,50075m=1{,}500

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

m=20 daysm=20\text{ days}

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

25−20=525-20=5

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

Question analysis

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

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

Why it is hard

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

  1. Many steps moderate Working it out

    Four operations chain together: 25% of 60, Martina's 75 pages, the two finishing times 1,500 ÷ 60 and 1,500 ÷ 75, and finally their difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Combines ideas moderate Working it out

    A percent increase in a daily rate has to be turned into a change in days, and 25% more pages a day does not mean 25% fewer days. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Turning words into math minor Understanding the question

    “25% more pages per day than Karen” makes Martina's rate 1.25 × 60, and “how many days sooner” asks for a difference of two times, not of two rates. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}
Old is the starting amount, and p is the percent it grows by. First find p percent of the old amount. That is the extra. Then add the extra to the old amount to get the new amount.
Time from a rate
days=total pages÷pages per day\text{days} = \text{total pages} \div \text{pages per day}
Total pages is the whole assignment. Pages per day is how fast someone reads. Dividing tells how many days of reading fit into the whole job.

  1. Sort out what you know

    Karen reads 60 pages per day. Martina reads 25% more pages per day than Karen. Each girl must read 1,500 pages. You need how many days sooner Martina finishes. That is the difference between their numbers of days.

  2. Find Martina’s pages per day

    25% more means Karen’s 60 pages plus 25% of 60.

    new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}

    60+25100×60=60+15=7560 + \frac{25}{100} \times 60 = 60 + 15 = 75

    Martina reads 75 pages per day. Watch out: 15 is extra pages, not days. Stopping here gives 15, choice D. Adding 25 pages instead of 25% gives 85 pages and leads to 7, choice C.

  3. Find each girl’s number of days

    Divide the whole assignment by each girl’s pages per day.

    days=total pages÷pages per day\text{days} = \text{total pages} \div \text{pages per day}

    Karen: 1,500÷60=25\text{Karen: } 1{,}500 \div 60 = 25

    Martina: 1,500÷75=20\text{Martina: } 1{,}500 \div 75 = 20

    Karen needs 25 days, and Martina needs 20 days.

  4. Subtract the days

    “How many days sooner” asks for the difference.

    25−20=525 - 20 = 5

    Martina finishes 5 days sooner. Watch out: dividing 60 by 15 mixes two rates and gives 4, choice A.

    So the answer is B, 5.

Check your answer

Multiply back. Karen: 25×60=1,50025 \times 60 = 1{,}500 pages. Martina: 20×75=1,50020 \times 75 = 1{,}500 pages. Both read all 1,500 pages, and 25 days is 5 days more than 20 days.

More questions like this

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

Question 60 of 120

2024 Test B Math, question 16

Question

A company with 200 employees grew in size by 25% between 2005 and 2010. The company grew again by 10% between 2010 and 2015. What was the total number of employees in 2015?

  • A235
  • B260
  • C270
  • D275
Show the answer

Correct answer: D

First, determine the number of employees in 2010:

200×25%=200×0.25=50200\times 25\%=200\times 0.25=50

So the number of employees in 2010 is 200+50=250200+50=250.

Next, determine the number of employees in 2015:

250×10%=250×0.10=25250\times 10\%=250\times 0.10=25

So the total number of employees in 2015 is 250+25=275250+25=275.

Question analysis

A percent change word problem: a company of 200 employees grows by 25% from 2005 to 2010 and then by another 10% from 2010 to 2015. Students find the number of employees in 2015. Two percent changes in a row are the tricky part, since each percent must be taken of the right amount.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Percent increase is a grade 7 topic, and two increases in a row are a common grade 7 multistep problem. The numbers are friendly, so the demand matches typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 5, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each increase alone is easy, and the whole question is about the base of the second one. Adding the percents into a single 35% feels like a shortcut and lands five employees short, a gap estimation cannot reveal. Working period by period, or multiplying 200 by 1.25 and then by 1.1, keeps each percent attached to the right amount.

  1. A very close wrong answer moderate Choosing the answer

    270 and 275 employees are five apart, too close to separate by estimate; only taking the second 10% of 250 rather than of 200 decides between them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Turning words into math moderate Understanding the question

    “Grew again by 10%” refers to the company's size in 2010, so the base for the second percent is the new 250 employees, not the original 200. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Many steps minor Working it out

    Two percent increases make four operations, a percent and an addition for each period, and the second period starts from the first one's result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work one period at a time: find the 2010 total by increasing 200 by 25%, then increase that new total by 10%.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}
Old is the starting amount, and p is the percent it grows by. First find p percent of the old amount. That is the extra. Then add the extra to the old amount to get the new amount.

  1. Sort out what you know

    The company has 200 employees at the start, in 2005. From 2005 to 2010 it grows by 25%. From 2010 to 2015 it grows again, by 10%. You need the number of employees in 2015.

  2. Find the number in 2010

    new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}

    200+25100×200=200+50=250200 + \frac{25}{100} \times 200 = 200 + 50 = 250

    In 2010 there were 250 employees. Watch out: 25% does not mean 25 people. Adding 25 people and then 10 people gives 235, choice A.

  3. Grow from the new number

    The second increase starts in 2010. So take 10% of 250, not 10% of 200.

    250+10100×250=250+25=275250 + \frac{10}{100} \times 250 = 250 + 25 = 275

    In 2015 there were 275 employees.

  4. Avoid the traps and answer

    Watch out: taking both percents of the first 200 gives 200+50+20=270200 + 50 + 20 = 270, choice C. That misses the growth of the 50 new workers. Adding 10 people instead of 10% gives 260, choice B.

    So the answer is D, 275.

Check your answer

Growing by 25% is the same as multiplying by 1.25. Growing by 10% is the same as multiplying by 1.1. 200×1.25×1.1=250×1.1=275200 \times 1.25 \times 1.1 = 250 \times 1.1 = 275. The answer should also be a bit more than 270, since the second 10% is taken of a bigger number.

More questions like this

The original DOE question
Question 73 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 73. Open page 203 of the handbook (PDF)

Question 61 of 120

2024 Test B Math, question 17

Question

A bowl contains 12 green candies, 4 yellow candies, and some red candies. The probability of choosing a green candy at random is twice as great as the probability of choosing a red candy at random. What is the probability of choosing a yellow candy at random?

  • A211\frac{2}{11}
  • B29\frac{2}{9}
  • C14\frac{1}{4}
  • D311\frac{3}{11}
Show the answer

Correct answer: A

The probability of choosing a green candy is twice as great as the probability of choosing a red candy. There are 12 green candies. Let r = the number of red candies.

12=2r12=2r

r=6r=6

So there are 6 red candies. 12 green candies + 4 yellow candies + 6 red candies = 22 total candies.

There are 4 yellow candies out of a total of 22 candies. So the probability of choosing a yellow candy is

422=211\frac{4}{22}=\frac{2}{11}.

Question analysis

A probability question about a bowl of 12 green candies, 4 yellow candies and an unknown number of red ones. A green candy is twice as likely to be picked as a red one, and students find the chance of picking a yellow candy. The missing red count has to be worked out before any probability can be found.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Simple probability is grade 6 content in New York, and grade 7 problems usually give every count. Working backward from a comparison of two probabilities to a missing count, then forward to a third probability, is more than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 49 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question withholds the red count, and everything depends on recovering it from the comparison of probabilities. Students who cannot make that step tend to leave red out altogether, while those who find 6 red candies can still slip at the end by using the wrong total or the wrong color. Writing all three counts and the total of 22 before forming any fraction prevents both endings.

  1. Turning words into math moderate Understanding the question

    “Twice as great as the probability of choosing a red candy” must become 12 = 2 × the red count, which works only because both probabilities share one total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Combines ideas moderate Working it out

    An unknown count has to be found from a ratio before any probability can be written, joining proportional reasoning with probability as a part over the total. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Turn twice as likely into a statement about numbers of candies to find the red count, add all three colors for the total, then put the yellow count over that total and simplify.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of the kind you wanttotal numberP = \frac{\text{number of the kind you want}}{\text{total number}}
P is the probability, the chance that something happens. The top is how many candies are the kind you want. The bottom is how many candies there are in all. Each candy is just as likely to be picked, so this fraction gives the chance.

  1. Sort out what you know

    The bowl has 12 green candies, 4 yellow candies and some red candies. You are not told how many are red. Let r be the number of red candies.

    The chance of picking green is twice the chance of picking red. You need the probability (the chance) of picking yellow.

  2. Compare green and red

    Write each chance with the probability formula.

    P=number of the kind you wanttotal numberP = \frac{\text{number of the kind you want}}{\text{total number}}

    Green and red have the same bottom number, the total number of candies. So if the chance of green is twice the chance of red, there are twice as many green candies as red ones.

    12=2×r12 = 2 \times r

  3. Find the red candies and the total

    r=12÷2=6r = 12 \div 2 = 6

    There are 6 red candies. Now add up all the candies.

    12+4+6=2212 + 4 + 6 = 22

    There are 22 candies in all. Watch out: do not leave out the red candies just because their number was not given. Using 16 as the total gives 1/4, choice C.

  4. Find the chance of yellow

    There are 4 yellow candies out of 22.

    P=422=211P = \frac{4}{22} = \frac{2}{11}

    Divide the top and the bottom by 2 to simplify. Watch out: the question asks for yellow, not red. 6/22 = 3/11 is the chance of red, choice D. The bottom must be all 22 candies. Using 18 gives 2/9, choice B.

    So the answer is A, 211\frac{2}{11}.

Check your answer

Check the clue. The chance of green is 1222\frac{12}{22}, and the chance of red is 622\frac{6}{22}. Twelve is twice 6, so green is twice as likely. Also, the three chances add up to 1: 1222+422+622=2222=1\frac{12}{22} + \frac{4}{22} + \frac{6}{22} = \frac{22}{22} = 1.

More questions like this

The original DOE question
Question 74 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 74. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 62 of 120

2024 Test B Math, question 18

Question

A taxi driver recorded the number of fares each day for one month. The box plot displays the data.

Box plot titled DAILY FARES above a number line labeled Number of Fares, with arrows at both ends and a tick labeled at every even number from 0 to 32. The left whisker starts at 2, the box runs from 12 to 24 with the median line at 16, and the right whisker ends at 32.

Based on the box plot, which statement is a correct interpretation of the data?

  • AThe data point 2 is an outlier.
  • BThe mean of the data is 16.
  • CThe interquartile range (IQR) of the data is 30.
  • DOn about 25% of the days, there were between 12 and 16 fares.
Show the answer

Correct answer: D

The amount of data in a box plot is broken into 4 parts, or quartiles, which each represent 25% of the data. So the amount of data represented between Quartile 1 (12) and the median (16) is about 25% of the data.

Question analysis

A statistics question on a box plot of the number of fares a taxi driver recorded each day for one month. Students decide which of four statements is a correct interpretation of the data, using what a box plot does and does not show. The statements draw on several different box plot ideas, so each one must be checked.

NYS standards
  • 7.SP.1 box plots and interquartile range
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Reading box plots, finding the interquartile range and testing for outliers are all grade 7 standards in New York. This item gathers several of them in one question, but each check is short, typical of grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 4, 63 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

No calculation here is long; the challenge is knowing what a box plot does and does not show. Two of the wrong statements sound reasonable because they borrow real features of the plot, its long left whisker and its middle line, and give them the wrong meaning. Remembering that the plot splits the days into four groups of about equal size makes the true statement easy to confirm.

  1. Something you must already know moderate Working it out

    Each statement draws on a box plot fact the test does not supply: what the middle line marks, how the IQR is found, and when a value counts as an outlier. (Anderson & Krathwohl, 2001)

  2. Every choice must be checked moderate Choosing the answer

    The four statements test four separate ideas, an outlier, the mean, the IQR and a quartile's share, so ruling out one says nothing about the next. (Embretson & Wetzel, 1987)

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

    The scale is labeled every 2 fares with the numbers crowded together, so the five values 2, 12, 16, 24 and 32 must be read off with care. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the minimum, quartiles, median and maximum off the plot, then test each statement in turn, computing the IQR and the outlier cutoffs rather than judging by eye.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range (IQR)
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
Q1 is the first quartile, the left edge of the box. Q3 is the third quartile, the right edge. The IQR is the length of the box. It tells how spread out the middle half of the data is.
Low cutoff for an outlier
low cutoff=Q1−1.5×IQR\text{low cutoff} = Q_1 - 1.5 \times \text{IQR}
Q1 and the IQR are found from the box. An outlier is a value far away from the rest. A value below this cutoff counts as an outlier on the low side. A value at or above it does not.

  1. Read the five numbers

    A box plot sums up a set of data with five numbers. Here each piece of data is the number of fares on one day.

    • Minimum, the smallest value: 2
    • First quartile, Q1, the left edge of the box: 12
    • Median, the middle value, the line inside the box: 16
    • Third quartile, Q3, the right edge of the box: 24
    • Maximum, the largest value: 32

  2. Learn what each part holds

    These five numbers split the days into four parts. Each part holds about one fourth of the days, or about 25%. For example, about 25% of the days fall between the minimum and Q1. Another 25% fall between Q1 and the median.

  3. Check choice D

    Choice D talks about days with between 12 and 16 fares. 12 is Q1, and 16 is the median. That stretch is one of the four parts. So on about 25% of the days, there were between 12 and 16 fares. Choice D is correct.

  4. Rule out choices B and C

    Choice B: a box plot shows the median, 16, not the mean (the average). You cannot find the mean from it. Watch out: do not mix up the median and the mean.

    IQR=Q3−Q1=24−12=12\text{IQR} = Q_3 - Q_1 = 24 - 12 = 12

    Choice C says the IQR is 30. But 30 is the range, 32−232 - 2. Watch out: the IQR uses the edges of the box, not the ends of the whiskers.

  5. Rule out choice A and answer

    A long whisker does not make 2 an outlier. Use the cutoff rule.

    low cutoff=Q1−1.5×IQR\text{low cutoff} = Q_1 - 1.5 \times \text{IQR}

    low cutoff=12−1.5×12=12−18=−6\text{low cutoff} = 12 - 1.5 \times 12 = 12 - 18 = -6

    2 is above −6, so 2 is not an outlier. The long whisker just shows that the lowest fourth of the days is spread out.

    So the answer is D.

Check your answer

Go back over the four choices. Only D uses what a box plot really shows: four parts of about 25% each. Choice B needs the mean, which the plot does not show. Choice C gives the range, not the IQR. And 2 is not far enough out to be an outlier, so A is wrong.

More questions like this

The original DOE question
Question 75 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 75. Open page 203 of the handbook (PDF)

Question 63 of 120

2024 Test B Math, question 21

Question

−149, −49, −17, −1-\frac{1}{49},\ -49,\ -\frac{1}{7},\ -1

Which number shown above is the greatest?

  • A−149-\frac{1}{49}
  • B−49-49
  • C−17-\frac{1}{7}
  • D−1-1
Show the answer

Correct answer: A

All the numbers are negative, so they are all less than zero. Consider the absolute value of each number and where it falls on a number line. The negative number closest to 0 will be the greatest.

The negative number with the greatest absolute value is −49, so it has the least value.

The negative number with the next greatest absolute value is −1, so it would fall next in order from least to greatest.

−49, −1-49,\ -1

Compare the absolute values of −149-\frac{1}{49} and −17-\frac{1}{7}.

149<17\frac{1}{49}<\frac{1}{7}

Therefore, the numbers listed from least to greatest are:

−49, −1, −17, −149-49,\ -1,\ -\frac{1}{7},\ -\frac{1}{49}

So the number −149-\frac{1}{49} is the greatest.

Question analysis

Four negative numbers are listed: negative 1/49, negative 49, negative 1/7 and negative 1. Students decide which of them is the greatest. Comparing negative fractions with negative whole numbers tests whether students order numbers by their position on the number line or by their size without the sign.

NYS standards
  • 6.NS.7a inequality as relative position
  • 6.NS.7d absolute value versus order
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Ordering positive and negative rational numbers on a number line is grade 6 content in New York. This is a one-step comparison of four numbers, lighter than typical grade 7 work with signed numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 10, 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

With no arithmetic to do, the question tests how firmly students picture negative numbers on a line. Comparing sizes without the sign picks the number farthest from 0, and comparing the two fractions by habit picks the wrong one of the pair. Placing all four as points left of 0, with −1 between −49 and the fractions, makes the order plain.

  1. A very close wrong answer moderate Choosing the answer

    −1/49 and −1/7 differ only in the denominator, and for negative numbers the larger denominator gives the greater value, the reverse of what their sizes suggest. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Students must bring the number-line fact that a negative number's value falls as its distance from 0 grows, so −49 is the least of the four, not the greatest. (Anderson & Krathwohl, 2001)

Strategy: Picture each number's place on the number line: among negative numbers, the one closest to 0 is the greatest, so compare their distances from 0.

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.)
  • 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

Comparing negative numbers
if a<b, then −a>−b\text{if } a \lt b \text{, then } -a \gt -b
a and b are positive numbers. Putting a minus sign on both flips their order. The one with the smaller size is closer to 0, so it is greater. For example, 2 is less than 5, but −2 is greater than −5.

  1. Sort out the question

    All four numbers are negative, which means less than 0. You need the greatest one, the one with the highest value. On a number line, greater numbers are farther to the right.

  2. Think about distance from 0

    For negative numbers, the one closest to 0 is the greatest. The size of a number without its sign is its absolute value. It is the number’s distance from 0. So look for the smallest size.

    if a<b, then −a>−b\text{if } a \lt b \text{, then } -a \gt -b

  3. Compare the sizes

    The sizes are 1/49, 49, 1/7 and 1. Both 1/49 and 1/7 have 1 on top. Cutting 1 into 49 pieces gives much smaller pieces than cutting it into 7. So 1/49 is less than 1/7.

    149<17<1<49\frac{1}{49} \lt \frac{1}{7} \lt 1 \lt 49

    So 1/49 is the smallest size. That puts −149-\frac{1}{49} closest to 0.

  4. Put the numbers in order

    Flip the order of the sizes to order the negative numbers, from least to greatest.

    −49<−1<−17<−149-49 \lt -1 \lt -\frac{1}{7} \lt -\frac{1}{49}

    Watch out: −49 has the biggest digits, but it is the least, so choice B is wrong. Watch out: 1/7 is more than 1/49, but −1/7 is less than −1/49. So choice C is wrong.

    So the answer is A, −149-\frac{1}{49}.

Check your answer

Change the fractions to decimals: −149≈−0.02-\frac{1}{49} \approx -0.02 and −17≈−0.14-\frac{1}{7} \approx -0.14. Think of money you owe. Owing 2 cents is better than owing 14 cents, $1 or $49. So −1/49 is the greatest.

More questions like this

The original DOE question
Question 78 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 205, question 78. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 64 of 120

2024 Test B Math, question 28

Question

A community center has two 30-member exercise groups. The director recorded the ages of each group’s members in the box plots.

Two box plots titled AGES OF GROUPS over one number line from 15 to 33, with a tick at every whole number and labels at 15, 18, 21, 24, 27, 30 and 33. Group P: whiskers from 17 to 33, box from 19 to 27, median line at 24. Group Q: whiskers from 15 to 29, box from 17 to 26, median line at 21.

Which statement about the two groups can be verified based on the box plots?

  • ABoth groups have at least one member who is exactly 29 years old.
  • BGroup P has more members who are at least 17 years old than Group Q has.
  • CBoth groups have an equal number of members who are between 21 and 24 years old.
  • DGroup Q has fewer members who are between 19 and 27 years old than Group P has.
Show the answer

Correct answer: B

Groups P and Q have an equal number of members. All the members of Group P are 17 or older, but only about 34\frac{3}{4} of the members of Group Q are 17 or older.

Question analysis

Box plots show the ages of the members of two 30-member exercise groups at a community center, Group P and Group Q. Students decide which statement comparing the groups can be verified from the plots alone. The challenge is knowing what a box plot guarantees about individual ages and counts, and what it leaves unknown.

NYS standards
  • 7.SP.1 box plots and interquartile range
  • 7.SP.4 comparative inferences about populations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students read box plots and compare medians and spreads. Deciding which claims two box plots can prove, from what the five key values guarantee about counts, is a harder reasoning task than typical 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 5, 91 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 question about evidence: the right statement is the only one the box plots guarantee, while the others may or may not be true. Students who treat the plots as pictures of every member, assuming ages fill each whisker or spread evenly through each box, find support for claims that cannot be checked. Asking what must hold for any 30 ages with these summaries separates the verifiable claim from the merely plausible ones.

  1. Every choice must be checked major Choosing the answer

    Three of the statements are not false but unverifiable, so each must be tested against what the five-number summaries guarantee, and none can be dismissed at a glance. (Embretson & Wetzel, 1987)

  2. Something you must already know moderate Working it out

    Students must know that the minimum and maximum are actual ages, that each quarter of a plot holds about a quarter of the 30 members, and that a whisker shows nothing in between. (Anderson & Krathwohl, 2001)

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

    The scale is labeled only every 3 years, so values such as 17, 19, 26 and 29 have to be read from unlabeled ticks on two separate plots. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: List each group's minimum, first quartile, median, third quartile and maximum, then test every statement against only what those five values guarantee about the 30 ages.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Members in each part of a box plot
members in each part≈14×n\text{members in each part} \approx \frac{1}{4} \times n
n is the number of members in a group, here 30. A box plot splits the data into four parts at the minimum, Q1, the median, Q3 and the maximum. Each part holds about one fourth of the data. It does not show how the ages spread out inside a part.

  1. Read the two box plots

    A box plot shows five numbers. The minimum is the youngest age, and the maximum is the oldest. Q1 is the left edge of the box, and Q3 is the right edge. The median is the line inside the box.

    GroupMinQ1MedianQ3Max
    P1719242733
    Q1517212629

    Each group has 30 members.

  2. Know what can be verified

    To verify a statement means to know for sure that it is true. A box plot shows only five numbers, not every age. The minimum and maximum are real ages of real members. The other ages are hidden inside the box and whiskers.

  3. Check choice B

    In Group P, the youngest member is 17. So all 30 members are at least 17. In Group Q, the youngest member is 15, and 15 is under 17. So at most 29 members of Group Q are at least 17.

    Group P has 30 such members, and Group Q has 29 or fewer. So choice B is true for sure.

  4. Rule out choices A and C

    A: Group Q’s oldest member is 29, so Group Q has a 29-year-old. But in Group P, 29 is inside the right whisker. A whisker does not show single ages. Watch out: 29 is in Group P’s range, but maybe no one in Group P is exactly 29.

    C: The plots do not show how many members are between 21 and 24 in either group.

  5. Rule out choice D and answer

    Group P’s box, from 19 to 27, is two of its four parts.

    members in each part≈14×30=7.5\text{members in each part} \approx \frac{1}{4} \times 30 = 7.5

    So about 15 members of Group P are between 19 and 27. But 19 and 27 fall in the middle of Group Q’s parts, and ages inside a part can bunch up in any way. Group Q could have more or fewer. Watch out: D assumes the ages are spread out evenly, and a box plot never tells you that.

    So the answer is B.

Check your answer

Choice B uses only the two minimums, which are real ages. Group P’s youngest member is 17, and Group Q’s is 15. That alone proves it. Each other choice needs ages inside a box or whisker, which a box plot does not show.

More questions like this

The original DOE question
Question 85 as printed on page 207 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 207, question 85. Open page 207 of the handbook (PDF)

Question 65 of 120

2024 Test B Math, question 29

Question

Jar Q contains 12 balls, and Jar R contains 8 balls. The probability of drawing a yellow ball at random from Jar Q is 13\frac{1}{3}. The probability of drawing a yellow ball at random from Jar R is 34\frac{3}{4}. All the balls are poured into an empty jar, Jar S. What is the probability of drawing a yellow ball at random from Jar S?

  • A14\frac{1}{4}
  • B13\frac{1}{3}
  • C12\frac{1}{2}
  • D47\frac{4}{7}
Show the answer

Correct answer: C

Jar Q contains 12 balls, and the probability of drawing a yellow ball at random is 13\frac{1}{3}.

13×44=412\frac{1}{3}\times\frac{4}{4}=\frac{4}{12}

There are 4 yellow balls in Jar Q.

Jar R contains 8 balls, and the probability of drawing a yellow ball at random is 34\frac{3}{4}.

34×22=68\frac{3}{4}\times\frac{2}{2}=\frac{6}{8}

There are 6 yellow balls in Jar R.

Jar S contains all the balls from Jar Q and Jar R.

12+8=2012+8=20

There are 20 balls in Jar S.

4+6=104+6=10

There are 10 yellow balls in Jar S.

1020÷1010=12\frac{10}{20}\div\frac{10}{10}=\frac{1}{2}

The probability of drawing a yellow ball at random from Jar S is 12\frac{1}{2}.

Question analysis

Two jars hold different numbers of balls and give different chances of drawing a yellow ball, and then everything is poured into a third jar. Students find the probability of drawing yellow from the combined jar. The tricky part is that the jars are different sizes, so their probabilities cannot be merged without first counting the yellow balls.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 probability tasks usually give the counts and ask for a fraction. Here each jar's probability must be turned back into a count of yellow balls before the jars are pooled, a reverse step that typical classwork rarely asks for.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 6, 66 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light; the hurdle is seeing that chances from jars of different sizes cannot be combined as fractions. Every wrong choice works on 1/3 and 3/4 directly, by multiplying them, by keeping one, or by adding tops and bottoms, while the right path runs through counts of balls. Asking how many yellow balls each jar holds before anything is poured avoids all three.

  1. Turning words into math moderate Understanding the question

    Each jar's probability, 1/3 of 12 balls and 3/4 of 8 balls, has to be turned back into a count of yellow balls, 4 and 6, before the two jars can be combined. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    The straight-across 4/7, about 0.57, sits just above the right 1/2, and both lie between the jars' chances of 1/3 and 3/4, so a rough estimate does not rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Turn each jar's probability into a number of yellow balls by multiplying it by that jar's total, then divide all the yellow balls by all the balls.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P(yellow)=yellow ballsall ballsP(\text{yellow}) = \frac{\text{yellow balls}}{\text{all balls}}
P(yellow) is the probability, or chance, of drawing a yellow ball. It is a number from 0 to 1. The top counts the yellow balls. The bottom counts all the balls. It works because every ball is equally likely to be drawn.

  1. Sort out what you know

    Jar Q holds 12 balls. The probability, or chance, of drawing yellow from it is 13\frac{1}{3}.

    Jar R holds 8 balls. The chance of drawing yellow from it is 34\frac{3}{4}.

    All the balls are poured into Jar S. You need the chance of drawing yellow from Jar S.

  2. Count the yellow balls in Jar Q

    The jars are different sizes, so you can’t mix their chances directly. Count the yellow balls instead. The chance tells what fraction of the balls are yellow. So multiply it by the number of balls.

    yellow balls=13×12=4\text{yellow balls} = \frac{1}{3} \times 12 = 4

    Jar Q has 4 yellow balls.

  3. Count the yellow balls in Jar R

    Do the same for Jar R. Three fourths of its 8 balls are yellow.

    yellow balls=34×8=6\text{yellow balls} = \frac{3}{4} \times 8 = 6

    Jar R has 6 yellow balls.

  4. Count the balls in Jar S

    Jar S gets every ball from both jars. Add all the balls. Then add the yellow balls.

    12+8=2012 + 8 = 20

    4+6=104 + 6 = 10

    Jar S has 20 balls, and 10 of them are yellow.

  5. Find the chance for Jar S

    Use the probability formula with the Jar S counts.

    P(yellow)=1020=12P(\text{yellow}) = \frac{10}{20} = \frac{1}{2}

    Watch out: Adding the tops and the bottoms, 1 + 3 over 3 + 4, gives 47\frac{4}{7}, choice D. Those numbers are parts of fractions, not balls. Multiplying 13×34\frac{1}{3} \times \frac{3}{4} gives 14\frac{1}{4}, choice A. That is the chance of two yellow draws, one from each jar.

    So the answer is C, 12\frac{1}{2}.

Check your answer

Half of the 20 balls is 10, and 4 + 6 = 10 yellow balls. That matches. Also, the mixed jar’s chance should land between the two jars’ chances, 13\frac{1}{3} and 34\frac{3}{4}. One half does.

More questions like this

The original DOE question
Question 86 as printed on page 207 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 207, question 86. Open page 207 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 66 of 120

2024 Test B Math, question 30

Question

GIA’S MIX

Snack Number in Bag
Pretzels 12
Raisins 9

Gia and her friends counted the numbers of pretzels and raisins in their snack mix. The table above shows the numbers in Gia’s mix. Which other mix has a ratio of pretzels to raisins in the same proportional relationship as Gia’s mix?

  • A

    ADELE’S MIX

    SnackNumber in Bag
    Pretzels6
    Raisins18
  • B

    CARL’S MIX

    SnackNumber in Bag
    Pretzels15
    Raisins12
  • C

    ISABEL’S MIX

    SnackNumber in Bag
    Pretzels18
    Raisins15
  • D

    TREVON’S MIX

    SnackNumber in Bag
    Pretzels16
    Raisins12
Show the answer

Correct answer: D

The ratio of pretzels to raisins in Gia’s bag of snack mix, 12:912:9, is equivalent to the ratio of pretzels to raisins in Trevon’s bag, 16:1216:12.

129=1612=43\frac{12}{9}=\frac{16}{12}=\frac{4}{3}

Question analysis

Gia's snack mix table lists how many pretzels and raisins are in her bag, and four friends' tables list theirs. Students pick the mix whose ratio of pretzels to raisins matches Gia's. It tests equivalent ratios and proportional relationships, and some mixes look close at a glance, so each ratio has to be tested exactly.

NYS standards
  • 7.RP.2a testing for proportionality
  • 6.RP.3a tables of equivalent ratios
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Testing whether two ratios are equivalent is core grade 7 proportional reasoning. The numbers are small whole numbers in four short tables, so the work is routine, though every table has to be checked against Gia's.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 77 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 short ratio check made harder only by an additive pull: Carl's and Isabel's mixes keep the gap of 3 between pretzels and raisins that Gia's mix has, so students who compare differences instead of ratios see two matches. Reducing every table to lowest terms, Gia's 12:9 to 4:3, shows that exactly one mix keeps the ratio.

  1. Every choice must be checked minor Choosing the answer

    Each of the four tables has to be reduced in turn, to 1:3, 5:4, 6:5 and 4:3, because none of them can be ruled out at a glance. (Embretson & Wetzel, 1987)

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

    The counts sit in five separate tables, and pretzels must stay first in every ratio, in the same order as Gia's 12 pretzels to 9 raisins. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Simplify the pretzels-to-raisins ratio in Gia's table to lowest terms, then simplify each other mix's ratio in the same order and compare.

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

Equivalent ratios
a:b=(a÷k):(b÷k)a : b = (a \div k) : (b \div k)
a and b are the two amounts being compared, in order. k is any number that divides both evenly. Dividing both amounts by the same number keeps the ratio the same. When no number but 1 divides both, the ratio is in simplest form.

  1. Sort out the question

    A ratio compares two amounts. Gia’s mix has 12 pretzels and 9 raisins. So her ratio of pretzels to raisins is 12 : 9.

    The “same proportional relationship” means the ratios are equal once you simplify them. For example, 2 : 1 and 6 : 3 are equal. Always keep pretzels first and raisins second.

  2. Simplify Gia’s ratio

    Both 12 and 9 can be divided by 3. So divide both by 3.

    12:9=(12÷3):(9÷3)=4:312 : 9 = (12 \div 3) : (9 \div 3) = 4 : 3

    Gia’s mix has 4 pretzels for every 3 raisins.

  3. Simplify each other mix

    Do the same for each choice. Divide both numbers by the biggest number that goes into both.

    ChoicePretzels : raisinsDivide bySimplest form
    A, Adele6 : 1861 : 3
    B, Carl15 : 1235 : 4
    C, Isabel18 : 1536 : 5
    D, Trevon16 : 1244 : 3

  4. Pick the matching mix

    Only Trevon’s mix simplifies to 4 : 3, the same as Gia’s.

    Watch out: Carl’s and Isabel’s mixes each have 3 more pretzels than raisins, just like Gia’s. But adding the same number to both amounts changes the ratio. Equal ratios come from multiplying or dividing. That trap leads to choice B or C.

    So the answer is D, Trevon’s mix.

Check your answer

Go the other way. Multiply both parts of 4 : 3 by 4. You get 16 : 12, which are Trevon’s exact numbers. You can also divide pretzels by raisins. Both 12÷912 \div 9 and 16÷1216 \div 12 equal 43\frac{4}{3}.

More questions like this

The original DOE question
Question 87 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 208, question 87. Open page 208 of the handbook (PDF)

Question 67 of 120

2024 Test B Math, question 42

Question

Jevon earns 12%\frac{1}{2}\% per year on the money in his savings account. Which expression is not another way to write 12%\frac{1}{2}\%?

  • A0.005100\frac{0.005}{100}
  • B12100\frac{\frac{1}{2}}{100}
  • C51,000\frac{5}{1{,}000}
  • D1200\frac{1}{200}
Show the answer

Correct answer: A

Rewrite the percent as a decimal:

12%=0.5%=0.005\frac{1}{2}\%=0.5\%=0.005

So the correct answer must not equal 0.005:

Option A

0.005100=0.00005\frac{0.005}{100}=0.00005

Option B

12100=0.5÷100=0.005\frac{\frac{1}{2}}{100}=0.5\div 100=0.005

Option C

51,000=0.005\frac{5}{1{,}000}=0.005

Option D

1200=0.005\frac{1}{200}=0.005

All answer options except Option A equal 0.005. Therefore, Option A is the correct answer.

Question analysis

Jevon's savings account earns one-half percent a year, and students must find the expression that is not another way to write one-half percent. It tests converting a percent less than 1 into fractions and decimals. Reading the word not carefully, and evaluating each stacked or unfamiliar fraction correctly, are what make it tricky.

NYS standards
  • 6.RP.3c percent as a rate per 100
  • 7.NS.2d fractions to decimals by long division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Converting between percents, fractions and decimals is grade 6 and 7 work. A percent smaller than 1 is less familiar, and the question asks which form is not equal, so all four must be checked, which keeps it at typical grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 7, 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

Every expression is built from the same idea, so the question tests whether students can convert a percent smaller than 1 without losing track of decimal places. The odd one out, 0.005 over 100, looks like a correct form, because 0.005 is one-half percent and dividing by 100 seems to turn it into a percent. Converting one-half percent to 0.005 once and comparing each expression with it beats reasoning about each form.

  1. A question in the negative moderate Understanding the question

    The bold word not asks for the one expression unequal to one-half percent, so the three expressions that check out are exactly the ones to reject. (Haladyna, Downing & Rodriguez, 2002)

  2. Every choice must be checked moderate Choosing the answer

    Three of the four forms are equal, so each has to be converted to a decimal and compared with 0.005 before the odd one out can be named. (Embretson & Wetzel, 1987)

  3. Error-prone arithmetic minor Working it out

    One-half percent is 0.5 with the point moved two places, 0.005, and dividing that by 100 moves it two more places, to 0.00005. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write one-half percent as a single decimal by dividing one half by 100, then evaluate each expression as a decimal and compare it with that value.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction
p%=p100p\% = \frac{p}{100}
p is any number. The % sign means out of 100. So p percent is p divided by 100. Dividing by 100 moves the decimal point two places to the left.

  1. Sort out the question

    The question asks which expression is not equal to 12%\frac{1}{2}\%. So three choices are equal to it, and one is different. You need the odd one out.

  2. Write the percent as a decimal

    One half is 0.5. A percent means out of 100, so divide by 100.

    12%=0.5%=0.5100=0.005\frac{1}{2}\% = 0.5\% = \frac{0.5}{100} = 0.005

    So 12%\frac{1}{2}\% is 0.005. It is half of one percent, a very small amount.

  3. Test each choice

    Turn each choice into a decimal. Then compare it with 0.005.

    ChoiceWork it outEqual to 0.005?
    A0.005÷100=0.000050.005 \div 100 = 0.00005No
    B0.5÷100=0.0050.5 \div 100 = 0.005Yes
    C5÷1,000=0.0055 \div 1{,}000 = 0.005Yes
    D1÷200=0.0051 \div 200 = 0.005Yes

  4. Pick the one that is different

    Choice A divides by 100 twice. The 0.005 was already divided by 100 once. Dividing again makes it 100 times too small.

    Watch out: Choice B is the plainest way to write 12%\frac{1}{2}\%: one half over 100. Picking it means missing the word “not.” Also, 1 ÷ 200 is 0.005, not 0.05, so D is equal too.

    So the answer is A, 0.005100\frac{0.005}{100}.

Check your answer

Use fractions instead. Half of one hundredth is 12×1100=1200\frac{1}{2} \times \frac{1}{100} = \frac{1}{200}, which is choice D. And 51,000\frac{5}{1{,}000} simplifies to 1200\frac{1}{200} too. Only A is a different amount.

More questions like this

The original DOE question
Question 99 as printed on page 211 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 99. Open page 211 of the handbook (PDF)

Question 68 of 120

2024 Test B Math, question 43

Question

A community garden had 50 members in 2001. In each of the next three years, the number of members increased by 40% over the number of members from the previous year. How many members were there in 2004? (Round to the nearest whole number.)

  • A70
  • B98
  • C110
  • D137
Show the answer

Correct answer: D

To find out how many members there were in 2002, calculate 40% of 50 and then add it to 50.

50×0.40=2050\times 0.40=20

20+50=70 members20+50=70\text{ members}

To find out how many members there were in 2003, calculate 40% of 70 and then add it to 70.

70×0.40=2870\times 0.40=28

28+70=98 members28+70=98\text{ members}

To find out how many members there were in 2004, calculate 40% of 98 and then add it to 98.

98×0.40=39.298\times 0.40=39.2

The result is 39.2, which is rounded to 39 to keep members to the nearest whole number.

39+98=137 members39+98=137\text{ members}

Question analysis

A community garden has 50 members in 2001, and membership grows by 40 percent each year over the year before. Students find the number of members in 2004, rounded to a whole number. It tests repeated percent increase, where each of the three rounds of growth must start from the previous year's total.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Percent increase is grade 7 content, and each step here is a routine one. Repeating it three times, each time on the new total, adds length and bookkeeping but no new idea, so it matches grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Compound growth is the whole question: each year's 40% is taken of a larger number, so the increases grow from 20 to 28 to about 39. Students who add 20 three times treat the growth as simple, and those who lose count of the years stop at an earlier total that is also listed. Multiplying by 1.4 three times, or checking that the third increase is the largest, separates the right count from the rest.

  1. Many steps moderate Working it out

    Three yearly increases have to be applied in turn, 50 to 70 to 98 to 137.2, each taking 40% of the new total rather than of the original 50. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The phrase "increased by 40% over the number of members from the previous year" sets a new base each year, the wording that separates compound from simple growth. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    The third year's 40% of 98 is 39.2, the first decimal in the chain, and the rounding instruction applies only to the final count. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply the membership by 1.4 once for each year from 2002 through 2004, carrying the unrounded value forward, and round only at the end.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}
old is last year’s number of members. p is the percent increase, here 40. new is this year’s number. You add p percent of the old amount to the old amount. Each year, the new number becomes next year’s old number.

  1. Sort out the question

    In 2001 the garden had 50 members. In each of the next three years, 2002, 2003 and 2004, the members grew by 40%.

    The 40% is of last year’s members, so the base changes every year. You need the number for 2004, rounded to the nearest whole number.

  2. Find the members in 2002

    40% of 50 is 20. Add it to 50.

    new=50+40100×50=50+20=70\text{new} = 50 + \frac{40}{100} \times 50 = 50 + 20 = 70

    There were 70 members in 2002.

  3. Find the members in 2003

    Now 70 is the old number. Take 40% of 70, not of 50.

    new=70+40100×70=70+28=98\text{new} = 70 + \frac{40}{100} \times 70 = 70 + 28 = 98

    There were 98 members in 2003.

  4. Find the members in 2004

    Now 98 is the old number.

    new=98+40100×98=98+39.2=137.2\text{new} = 98 + \frac{40}{100} \times 98 = 98 + 39.2 = 137.2

    Round to the nearest whole number. 137.2 is closer to 137 than to 138, so there were 137 members.

    Watch out: Adding 20 members every year uses 40% of 50 each time. That gives 110, choice C. Stopping after one or two years gives 70 or 98, choices A and B.

    So the answer is D, 137.

Check your answer

A 40% rise is the same as multiplying by 1.4. So 50×1.4×1.4×1.4=50×2.744=137.250 \times 1.4 \times 1.4 \times 1.4 = 50 \times 2.744 = 137.2, which rounds to 137. It matches.

More questions like this

The original DOE question
Question 100 as printed on page 211 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 100. Open page 211 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 69 of 120

2024 Test B Math, question 45

Question

A box contains a total of 30 paper clips with colors in the ratio red:white:blue=2:5:3\text{red}:\text{white}:\text{blue}=2:5:3. Suppose that one paper clip of each color is removed from the box and not replaced. What is the probability that the next paper clip chosen will be blue?

  • A527\frac{5}{27}
  • B27\frac{2}{7}
  • C827\frac{8}{27}
  • D310\frac{3}{10}
Show the answer

Correct answer: C

The ratio is red:white:blue=2:5:3\text{red}:\text{white}:\text{blue}=2:5:3. Find the number of each color in a box of 30 paper clips.

2x+3x+5x=302x+3x+5x=30

10x=3010x=30

x=3x=3

Red: 2x=2(3)=6\text{Red: }2x=2(3)=6

White: 5x=5(3)=15\text{White: }5x=5(3)=15

Blue: 3x=3(3)=9\text{Blue: }3x=3(3)=9

If one paper clip of each color is removed (3 total), there will be 8 blue paper clips out of 27 in the box. The probability that the next paper clip chosen will be blue is 827\frac{8}{27}.

Question analysis

A box of 30 paper clips has red, white and blue clips in the ratio 2 to 5 to 3, and one clip of each color is taken out. Students find the probability that the next clip drawn is blue. It tests ratios and simple probability; the ratio gives shares rather than counts, and the removals change both the blue count and the total.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Splitting a total by a three-part ratio and finding a simple probability are both familiar skills. Combining them, then adjusting the counts after the removals, adds steps but stays within typical grade 7 multi-step work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 5, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains a ratio problem into a probability problem, with a small change in between. Students who skip the removal get 3/10, a choice that is almost the right value and looks natural because it matches the ratio. The cleanest route lists the actual counts, 6, 15 and 9, takes one from each, and writes blue's 8 over the remaining 27.

  1. Many steps moderate Working it out

    The ratio 2:5:3 must become counts of 6, 15 and 9 clips, one clip of each color must come out, and only then can blue's 8 be put over 27. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    3/10 = 0.3 and 8/27, about 0.296, are almost equal, so an estimate cannot separate them; only subtracting the three removed clips does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Turning words into math minor Understanding the question

    The removal of one clip of each color, not replaced, happens before the draw, so it changes both the blue count and the total in the box. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Divide 30 by the sum of the ratio parts to find each color's count, take one clip away from each color, then put the blue count over the new total.

Traps

  • Without replacement: In a second pick there is one fewer item, which changes the probability.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Size of one ratio part
one part=totalsum of the ratio numbers\text{one part} = \frac{\text{total}}{\text{sum of the ratio numbers}}
The ratio numbers tell how many parts each color gets. Their sum is the number of parts in all. total is the number of clips. Dividing gives the size of one part. Each color is then its ratio number times one part.
Probability
P(blue)=blue clipsall clipsP(\text{blue}) = \frac{\text{blue clips}}{\text{all clips}}
P(blue) is the chance of picking a blue clip. The top counts the blue clips in the box. The bottom counts all the clips in the box. It works because each clip is equally likely to be picked.

  1. Sort out the question

    The box has 30 clips. The ratio red : white : blue = 2 : 5 : 3 means that for every 2 red clips, there are 5 white clips and 3 blue clips.

    Then one clip of each color is taken out and not put back. You need the chance that the next clip picked is blue.

  2. Find the size of one part

    The ratio has 2+5+3=102 + 5 + 3 = 10 parts in all. The 30 clips are shared among those 10 parts.

    one part=3010=3\text{one part} = \frac{30}{10} = 3

    Each part is 3 clips.

  3. Count the clips of each color

    Multiply each ratio number by the size of one part.

    red=2×3=6\text{red} = 2 \times 3 = 6

    white=5×3=15\text{white} = 5 \times 3 = 15

    blue=3×3=9\text{blue} = 3 \times 3 = 9

    Check: 6 + 15 + 9 = 30 clips. That matches the total.

  4. Take out one of each color

    One red, one white and one blue clip come out. Each color drops by 1, and the total drops by 3.

    ColorBeforeAfter
    Red65
    White1514
    Blue98
    All3027

  5. Find the chance of blue

    P(blue)=827P(\text{blue}) = \frac{8}{27}

    Watch out: 930=310\frac{9}{30} = \frac{3}{10}, choice D, ignores the 3 clips taken out. 527\frac{5}{27}, choice A, is the chance of red. Taking 1 from each ratio number, as if the box held only 10 clips, gives 27\frac{2}{7}, choice B.

    So the answer is C, 827\frac{8}{27}.

Check your answer

Add the clips left: 5 + 14 + 8 = 27. That matches. Also, 827\frac{8}{27} is about 0.296, a bit less than blue’s share of the full box, 0.3. That makes sense, since one blue clip came out.

More questions like this

The original DOE question
Question 102 as printed on page 212 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 102. Open page 212 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 70 of 120

2024 Test B Math, question 48

Question

Two triangles. Triangle MNP has N at the top and base MP labeled 10 cm; sides MN and NP each have a tick mark, showing they are equal. The smaller triangle XYZ has Y at the top, base XZ labeled 6 cm, and sides XY and YZ each labeled 9 cm.

Triangle MNP is similar to triangle XYZ, and MN=NP\mathrm{MN}=\mathrm{NP}. What is the perimeter of triangle MNP?

  • A25 cm
  • B30 cm
  • C36 cm
  • D40 cm
Show the answer

Correct answer: D

Similar geometric figures have sides that are proportional.

106=x9\frac{10}{6}=\frac{x}{9}

53=x9\frac{5}{3}=\frac{x}{9}

x=(5)(9)3=15x=\frac{(5)(9)}{3}=15

Because MN=NP\mathrm{MN}=\mathrm{NP}, both unknown sides are equal to 15. Therefore, the three sides of MNP are 15, 15, and 10. The perimeter of a triangle is the sum of its three side lengths.

15+15+10=4015+15+10=40

Question analysis

Two isosceles triangles are drawn, one with all its side lengths labeled and the other with only its base labeled, and the triangles are similar. Students find the perimeter of triangle MNP in centimeters. It tests similar figures and proportional sides, and the catch is that two sides must be found from the scale and the equal-sides marks.

NYS standards
  • 8.G.4 similarity and dilations
  • 7.G.1 scale drawings
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Similar triangles are grade 8 geometry, though grade 7 scale drawings use the same proportional reasoning. Students also need the equal-sides marks to find a missing side, a combined step beyond routine grade 7 scale work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 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 chains a scale factor to a perimeter, and both links have a trap. Some students scale by adding, since the base grew by 4, and others find the 15 cm leg correctly but then add only some of the sides. Writing all three sides of the large triangle, 15, 15 and 10, before adding is what the question rewards.

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

    The figure carries every number: tick marks show MN = NP, the base MP is 10 cm, and triangle XYZ has 9 cm legs on a 6 cm base. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Students must know that corresponding sides of similar triangles share one scale factor, here 10/6 = 5/3, though the stem never says so. (Anderson & Krathwohl, 2001)

  3. Error-prone arithmetic minor Working it out

    The scale factor 5/3 is a fraction, and 9 × 5/3 = 15 is easier to get right by dividing 9 by 3 first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Compare the two bases to get the scale factor, multiply a leg of the small triangle by it, then use the equal sides to add up the perimeter.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Similar triangles
MNXY=MPXZ\frac{\mathrm{MN}}{\mathrm{XY}} = \frac{\mathrm{MP}}{\mathrm{XZ}}
Similar triangles have the same shape but can differ in size. MN matches XY, and MP matches XZ. Each side of the big triangle is the same number of times as long as its matching side. So these two fractions are equal.
Perimeter of a triangle
P=a+b+cP = a + b + c
P is the perimeter, the distance all the way around. a, b and c are the lengths of the three sides. Add them to go once around the shape.

  1. Sort out the question

    In triangle MNP, the base MP is 10 cm. The tick marks on MN and NP show those two sides are equal. In triangle XYZ, the base XZ is 6 cm, and XY and YZ are each 9 cm.

    Similar means the same shape, with every side scaled by the same amount. You need the perimeter of MNP, the sum of its three sides.

  2. Match the sides

    The letters line up in order: M with X, N with Y, and P with Z. So the base MP matches the base XZ. The side MN matches XY, and NP matches YZ.

  3. Find the scale factor

    The scale factor is how many times bigger MNP is. Compare the two bases.

    MPXZ=106=53\frac{\mathrm{MP}}{\mathrm{XZ}} = \frac{10}{6} = \frac{5}{3}

    Every side of MNP is 53\frac{5}{3} times as long as its matching side in XYZ.

  4. Find the equal sides

    Use the similar triangles formula with XY = 9.

    MN9=53\frac{\mathrm{MN}}{9} = \frac{5}{3}

    MN=9×53=3×5=15\mathrm{MN} = 9 \times \frac{5}{3} = 3 \times 5 = 15

    Divide 9 by 3 first to keep the numbers small. MN is 15 cm. Since MN = NP, NP is 15 cm too.

    Watch out: The base grew by 4 cm, but adding 4 cm to each side is wrong. That gives 13 cm sides and 36, choice C. Similar shapes grow by multiplying.

  5. Add the three sides

    P=15+15+10=40P = 15 + 15 + 10 = 40

    The perimeter is 40 cm. Watch out: Leaving out one 15 cm side gives 25, choice A. Leaving out the base gives 30, choice B.

    So the answer is D, 40 cm.

Check your answer

The small triangle’s perimeter is 9 + 9 + 6 = 24 cm. Perimeters grow by the same scale factor as the sides. So 24×53=4024 \times \frac{5}{3} = 40. It matches.

More questions like this

The original DOE question
Question 105 as printed on page 212 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 105. Open page 212 of the handbook (PDF)

Question 71 of 120

2024 Test B Math, question 49

Question

Suppose that w, x, y, and z are integers, none of which are zero. Under which set of conditions would the product x2yz3|w|x^{2}yz^{3}|w| always be positive?

  • Ax is positive and w is positive.
  • Bx is negative and w is negative.
  • Cy is negative and z is negative.
  • Dx and y are negative, and z and w are positive.
Show the answer

Correct answer: C

First, analyze the role of each variable in the expression x2yz3|w|x^{2}yz^{3}|w|.

x2x^{2} will always be a positive number.

y can be a positive or a negative number.

z3z^{3} can be a positive or a negative number.

|w||w| will always be a positive number.

The expressions x2x^{2} and |w||w| will always be positive. Therefore, making the values of y and z negative would result in the expression yz3yz^{3} being positive, since a negative multiplied by a negative is positive. Thus the given product would always be positive.

Question analysis

Four nonzero integers appear in a product that includes a square, a cube and an absolute value. Students choose the set of conditions that makes the product positive every time. It tests the sign rules for multiplication together with exponents and absolute value, and the trap is judging the sign by counting negative variables without looking at what is done to each.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.EE.1 whole-number exponent expressions
  • 6.NS.7c absolute value as distance
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students learn sign rules for products of signed numbers. Here they must reason about four unknown integers under an even power, an odd power and an absolute value, and decide what is always true, which is more abstract than typical 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 6, 58 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One insight decides the question: x² and |w| are positive whatever x and w are, so only the signs of y and z³ matter. Students who count negative letters instead of negative factors pick conditions that look balanced but leave y as the only negative factor. Students who test a single set of values can be fooled by a choice that leaves y and z unspecified.

  1. Variables instead of numbers moderate Understanding the question

    There are four variables and no numbers, and always positive must hold for every allowed value, so one test case can rule a choice out but never confirm it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    Each set of conditions has to be checked against all four factors; the first two fix only x and w and leave y and z free to make the product negative. (Embretson & Wetzel, 1987)

  3. Something you must already know minor Working it out

    Students must know that a nonzero square and an absolute value are always positive, while an odd power such as z³ keeps the sign of z. (Anderson & Krathwohl, 2001)

Strategy: Work out, factor by factor, which signs each part of the product can take, then test each set of conditions against every factor rather than counting negatives.

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.)
  • 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

Signs in multiplication
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.

  1. Sort out the question

    w, x, y and z are integers, meaning whole numbers or their negatives. None of them is 0. The product x2yz3|w|x^2 y z^3 |w| means x2×y×z3×|w|x^2 \times y \times z^3 \times |w|.

    You need the conditions that make the product positive every time, for every number that fits them.

  2. Look at x squared and |w|

    x2x^2 means x × x. Two positives or two negatives multiply to a positive. So x2x^2 is always positive.

    |w||w| is the absolute value of w, its distance from 0. It is always positive, since w is not 0.

    So x and w never change the sign of the product. Their signs do not matter.

  3. Look at y and z cubed

    y can be positive or negative. z3z^3 means z × z × z, and it has the same sign as z. For example, (−2)3=−8(-2)^3 = -8.

    So the sign of the whole product is the sign of y×z3y \times z^3. It depends only on y and z.

  4. Test each choice

    • A and B say nothing about y and z. If y = −1 and z = 1, the product is negative. So they do not always work.
    • C: y is negative, and z3z^3 is negative. A negative times a negative is positive. This always works.
    • D: y is negative and z is positive. So the product is always negative.

    Watch out: D has two negative letters, x and y. But x2x^2 is positive anyway, so y is the only negative factor.

    So the answer is C, y is negative and z is negative.

Check your answer

Try numbers for C. Let x = −2, y = −3, z = −1 and w = −5. Then x2×y×z3×|w|=4×(−3)×(−1)×5=60x^2 \times y \times z^3 \times |w| = 4 \times (-3) \times (-1) \times 5 = 60, which is positive. Even with x and w negative, the product stays positive.

More questions like this

The original DOE question
Question 106 as printed on page 213 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 106. Open page 213 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 72 of 120

2024 Test B Math, question 53

Question

Which graph represents the solution to |x+1|≥2|x+1|\geq 2?

  • ANumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1. The line is shaded from −3 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.
  • BNumber line from −3 to 3 with a tick at every integer. Closed dots at −1 and 1. The line is shaded from −1 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.
  • CNumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1, with the segment between them shaded.
  • DNumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 3. The line is shaded from −3 to the left, ending in an arrow, and from 3 to the right, ending in an arrow.
Show the answer

Correct answer: A

Since |x+1|≥2|x+1|\geq 2, either x+1≤−2x+1\leq -2 or x+1≥2x+1\geq 2.

Solving x+1≤−2x+1\leq -2 for x gives x≤−3x\leq -3.

Solving x+1≥2x+1\geq 2 for x gives x≥1x\geq 1.

A filled circle on a number line represents “or equal to.” So the correct number line representation has filled circles at −3 and 1, with the number line filled to the left of −3 for “less than” and to the right of 1 for “greater than.”

Question analysis

Students choose the number line graph that shows every solution of an inequality in which x + 1 sits inside absolute value bars. It tests absolute value as distance and graphing inequalities. The tricky part is turning one absolute value statement into two pieces and placing each boundary point and shaded ray correctly.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph two-step inequalities and know absolute value as distance. An inequality with an absolute value has to be split into two cases, which is outside the grade 6 to 8 standards and well above typical 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 9, 7 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

Reading |x + 1| as the distance from x to −1 makes this quick: the points 2 units away are −3 and 1, and greater than or equal means outside them. Students who solve only one case and mirror it across 0 land on graphs centered on 0, and students who find both endpoints may still shade between them. The graphs differ in just one dot or one shading choice, so each must be read carefully.

  1. Something you must already know moderate Working it out

    Solving |x + 1| ≥ 2 needs two cases, x + 1 ≤ −2 or x + 1 ≥ 2, joined by or; absolute value inequalities go beyond grade 7, and the stem gives no hint. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The right graph and the segment from −3 to 1 share both filled endpoints, so only deciding between outside and between the boundaries separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    Four number lines must be read for filled dots and arrow directions, and two of them differ from the right one only in where a single dot sits. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the absolute value as distance from −1 on the number line, find the points exactly 2 units away, and decide whether the solutions lie between or outside them.

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.
  • 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

Absolute value as distance
|x−a|=distance from x to a|x - a| = \text{distance from } x \text{ to } a
x and a are points on the number line. The bars mean absolute value. |x − a| is how far apart the two points are, which is never negative. Since x + 1 = x − (−1), |x + 1| is the distance from x to −1.

  1. Sort out the question

    You need the graph of every x that makes |x+1|≥2|x + 1| \geq 2 true. The sign ≥ means greater than or equal to. The bars mean absolute value.

    On a number line, a filled dot means that point counts. A thick line with an arrow means all the numbers that way count too.

  2. Read it as a distance

    Adding 1 is the same as subtracting −1, so x+1=x−(−1)x + 1 = x - (-1). By the formula, |x+1||x + 1| is the distance from x to −1.

    So the inequality says that x is at least 2 units away from −1.

  3. Find the two boundary points

    Start at −1 and go 2 units each way.

    −1−2=−3-1 - 2 = -3

    −1+2=1-1 + 2 = 1

    At −3 and at 1, the distance is exactly 2. The words or equal to let them count, so they get filled dots.

  4. Decide which side to shade

    Being at least 2 away means being farther out. Numbers between −3 and 1 are too close to −1. For example, 0 is only 1 unit away. So shade to the left of −3 and to the right of 1.

    In algebra, that is x+1≤−2x + 1 \leq -2 or x+1≥2x + 1 \geq 2. These give x≤−3x \leq -3 or x≥1x \geq 1.

  5. Match the graph

    Choice A has filled dots at −3 and 1, with arrows going outward. That matches.

    Watch out: Choice C shades between −3 and 1. Those are the points within 2 of −1, the answer to |x+1|≤2|x + 1| \leq 2. Choices B and D are centered on 0, but the center here is −1.

    So the answer is A, the graph shaded outward from −3 and 1.

Check your answer

Test a point from each part. For x = 2, |2+1|=3|2 + 1| = 3, and 3 is at least 2. For x = 0, |0+1|=1|0 + 1| = 1, which is less than 2. For x = −3, |−3+1|=2|-3 + 1| = 2, which counts. Graph A agrees with all three.

More questions like this

The original DOE question
Question 110 as printed on page 214 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 110. Open page 214 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 73 of 120

2024 Test B Math, question 56

Question

A rectangular field had a length of 80 meters and a width of 50 meters. The field was recently enlarged, with the length increased by 15% and the width increased by 10 meters. By what percentage was the area of the field increased?

  • A26.5%
  • B35%
  • C38%
  • D42.5%
Show the answer

Correct answer: C

The original area of the field was 4,000 square meters.

80×50=4,00080\times 50=4{,}000

When the field was enlarged, the length increased by 15%, so the new length is 92 meters.

80×15%80\times 15\%

80×0.15=1280\times 0.15=12

80+12=9280+12=92

The width increased by 10 meters, so the new width is 60 meters.

50+10=6050+10=60

The new area of the field is 5,520 square meters.

92×60=5,52092\times 60=5{,}520

The percent increase is found by finding the difference between the two areas and dividing the increase by the original area.

5,520−4,000=1,5205{,}520-4{,}000=1{,}520

1,520÷4,000=0.38=38%1{,}520\div 4{,}000=0.38=38\%

Question analysis

A rectangular field 80 meters long and 50 meters wide has its length increased by 15 percent and its width by 10 meters. Students find the percent by which the field's area grew. It tests percent increase with rectangle area, and the tricky part is that one side changes by a percent and the other by a fixed length.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.G.1 area of polygons
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually apply one percent change to one quantity. This changes a length by a percent and a width by a fixed amount, then asks for the percent change in area, a five-step chain with two kinds of change.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
5 steps
Reading level
Question and choices: about grade 6, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is ordinary, but the two changes are stated in different units, one a percent and one in meters, which tempts students to force them into the same form. Adding the two percents is the most natural shortcut and lands close to the right value, because it misses only the small extra area where both changes overlap. Computing the old and new areas directly is slower but leaves nothing to chance.

  1. Combines ideas moderate Working it out

    Percent change has to be combined with area: the length grows by a percent and the width by meters, and the question asks for the percent change of their product. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Many steps moderate Working it out

    Five computations are chained: the new length 92, the new width 60, the areas 4,000 and 5,520, and the increase 1,520 as a share of 4,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The 35% from adding 15% and 20% sits only 3 points below 38%, so a rough estimate of the new area cannot rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Compute the new length and the new width, find the old and new areas, and divide the increase in area by the original area.

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.
  • 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

Area of a rectangle
A=l×wA = l \times w
A is the area, the flat space inside. l is the length and w is the width. Multiplying counts how many unit squares fit, row by row.
Percent increase
percent increase=new−oldold×100\text{percent increase} = \frac{\text{new} - \text{old}}{\text{old}} \times 100
old is the starting amount, and new is the amount after. new − old is how much it grew. Dividing by old compares the growth with the start. Multiplying by 100 turns that into a percent.

  1. Sort out the question

    The field was 80 meters long and 50 meters wide. The length grew by 15%, a percent. The width grew by 10 meters, a number of meters. These are two different kinds of increase.

    You need the percent by which the area grew.

  2. Find the old area

    You need the old area to compare with the new one.

    A=80×50=4,000A = 80 \times 50 = 4{,}000

    The old field covered 4,000 square meters.

  3. Find the new length and width

    15% of 80 is 0.15×80=120.15 \times 80 = 12. So the new length is 80+12=9280 + 12 = 92 meters.

    The new width is 50+10=6050 + 10 = 60 meters.

    Watch out: 10 meters is not 10%. Treating it as 10% leads to 26.5%, choice A. Treating 15% as 15 meters leads to 42.5%, choice D.

  4. Find the new area

    Multiply the new length by the new width.

    A=92×60=5,520A = 92 \times 60 = 5{,}520

    The new field covers 5,520 square meters. That is 5,520−4,000=1,5205{,}520 - 4{,}000 = 1{,}520 square meters more.

  5. Find the percent increase

    percent increase=5,520−4,0004,000×100=1,5204,000×100=38\text{percent increase} = \frac{5{,}520 - 4{,}000}{4{,}000} \times 100 = \frac{1{,}520}{4{,}000} \times 100 = 38

    The area grew by 38%. Watch out: The width grew by 20%, since 10 is 20% of 50. Adding 15% + 20% = 35% gives choice B. That misses the extra corner piece where both new strips meet.

    So the answer is C, 38%.

Check your answer

Use growth factors. The length is multiplied by 1.15. The width goes from 50 to 60, so it is multiplied by 1.2. The area is multiplied by 1.15×1.2=1.381.15 \times 1.2 = 1.38, which is 38% more.

More questions like this

The original DOE question
Question 113 as printed on page 215 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 113. Open page 215 of the handbook (PDF)

Question 74 of 120

2024 Test B Math, question 57

Question

If x and y are both negative integers with x<yx<y, which expression has the least value?

  • Ax2yx^{2}y
  • Bxy2xy^{2}
  • C−x2y-x^{2}y
  • D−xy2-xy^{2}
Show the answer

Correct answer: A

Since x and y are both negative integers, both x2x^{2} and y2y^{2} are positive.

Therefore, both x2yx^{2}y and xy2xy^{2} are negative, and both −x2y-x^{2}y and −xy2-xy^{2} are positive.

Because x<yx<y and both x and y are negative, |y|<|x||y|<|x| and y2<x2y^{2}<x^{2}. So |y2|<|x2||y^{2}|<|x^{2}|.

Since |y|<|x||y|<|x| and x and y are integers, |xy2|<|x2y||xy^{2}|<|x^{2}y|.

So x2y<xy2x^{2}y<xy^{2}.

−xy2=|xy2| and −x2y=|x2y|-xy^{2}=|xy^{2}|\text{ and }-x^{2}y=|x^{2}y|

Therefore, x2y<xy2<−xy2<−x2yx^{2}y<xy^{2}<-xy^{2}<-x^{2}y.

Question analysis

Two negative integers x and y, with x less than y, appear in four expressions built from squares and minus signs. Students find the expression with the least value. It tests multiplying negative numbers and comparing their sizes, and the difficulty is tracking how each minus sign and each square changes both the sign and the size of the result.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.NS.7d absolute value versus order
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students multiply negative numbers, but this asks them to compare four variable expressions for any pair of negative integers. Deciding how the squares compare when x is less than y goes beyond typical grade 7 sign practice.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 8, 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

Two facts settle the question: squares of negatives are positive, and the more negative number has the larger square. Students who track the minus signs carelessly think a leading minus makes an expression negative, and students who assume the larger y has the larger square choose the wrong negative expression. Substituting x = −3 and y = −2 gives −18, −12, 18 and 12, which orders all four at once.

  1. Variables instead of numbers moderate Understanding the question

    No values are given, only that x and y are negative integers with x < y, so students must reason about signs or invent a pair such as −3 and −2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    x²y and xy² use the same letters with the exponent moved; both are negative, and only comparing x² with y² separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Every choice must be checked minor Choosing the answer

    All four expressions have to be placed in order, because the minus signs in front of the last two turn those expressions positive. (Embretson & Wetzel, 1987)

Strategy: Pick two negative integers with x less than y, such as −3 and −2, evaluate all four expressions, and confirm the order with a second pair.

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.)
  • 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

Signs in multiplication
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.
Squaring a number
x2=x×xx^2 = x \times x
x is any number. x² (x squared) means x times itself. The square of a number other than 0 is always positive, because two negatives or two positives multiply to a positive.

  1. Sort out the question

    x and y are negative integers, such as −3 and −2. Since x < y, x is to the left of y on the number line. So x is more negative, farther from 0.

    You need the expression with the least value. That is the most negative one.

  2. Find the sign of each expression

    Squares are positive, so x2x^2 and y2y^2 are positive.

    • x2yx^2 y is a positive times a negative, so it is negative.
    • xy2x y^2 is a negative times a positive, so it is negative.
    • −x2y-x^2 y and −xy2-x y^2 are the opposites of these, so they are positive.

    So the least value must be A or B. Watch out: The minus sign in front of C and D does not make them negative. It turns a negative into a positive.

  3. Compare A and B with numbers

    Try x = −3 and y = −2, since −3 < −2.

    x2y=(−3)2×(−2)=9×(−2)=−18x^2 y = (-3)^2 \times (-2) = 9 \times (-2) = -18

    xy2=(−3)×(−2)2=(−3)×4=−12x y^2 = (-3) \times (-2)^2 = (-3) \times 4 = -12

    Since −18 < −12, A is smaller.

  4. See why it always works

    Both expressions contain x × y, which is positive. x2y=(xy)×xx^2 y = (xy) \times x and xy2=(xy)×yx y^2 = (xy) \times y. Multiplying the same positive number by the more negative x gives the more negative result.

    Watch out: y is bigger than x, but y2y^2 is smaller than x2x^2: 4 is less than 9. Thinking y² is bigger leads to choice B.

    So the answer is A, x2yx^2 y.

Check your answer

Try another pair, x = −5 and y = −1. Then A is 25×(−1)=−2525 \times (-1) = -25, B is (−5)×1=−5(-5) \times 1 = -5, C is 25 and D is 5. A is still the least.

More questions like this

The original DOE question
Question 114 as printed on page 215 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 114. Open page 215 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 75 of 120

2023 Test A Math, question 4

Question

Line l is vertical and line n is horizontal, and a small square marks the right angle where they meet. Line m rises from the lower left to the upper right; it crosses line l below line n and crosses line n to the right of line l. Where m crosses l, the angle between the lower part of l and the lower-left part of m is marked 35°. Where m crosses n, the angle between the right part of n and the upper-right part of m is marked x°.

In the figure above, line l is perpendicular to line n. What is the value of x?

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

Show the answer

Correct answer: 55

Since vertical angles are congruent, the right triangle in the figure has acute angles with measures of 35° and x°. The interior angles of a triangle sum to 180°, so set up an equation to solve for x:

90+35+x=18090 + 35 + x = 180

125+x=180125 + x = 180

x=55x = 55

Question analysis

Three lines cross in a figure: line l is perpendicular to line n, and a slanted line m meets them at a marked 35° angle and an unknown angle x. This geometry grid-in asks for the value of x. Students must see the right triangle the lines form and carry the marked angles into it using vertical angles.

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 problems use vertical, supplementary and complementary angles. This one also needs the angle sum of a triangle, which New York places in grade 8, and the triangle is not drawn as a separate shape, so students must spot it.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 4, 17 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The figure hides a small right triangle between the three lines, and the question is easy once students see it. Its angles are the right angle, the vertical partner of the 35° angle and the vertical partner of x, so x is what the right angle and 35° leave of 180°. Students who never spot the triangle tend to guess a direct relationship between the two marked angles, which no angle fact supports.

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

    Neither marked angle is inside the right triangle that lines l, m and n form: the 35° opens below the point where m crosses l, and x opens above line n, so each must be carried in as a vertical angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know moderate Working it out

    The stem states only that l is perpendicular to n; students must supply that vertical angles are equal and that the three angles of a triangle add to 180°. (Anderson & Krathwohl, 2001)

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

    As a grid-in it offers no choices, so 125, the right angle plus the 35° angle, can be gridded with nothing to signal the missing subtraction. (Enright, Morley & Sheehan, 2002)

Strategy: Find the triangle formed by the three lines, use vertical angles to move the 35° and x° angles inside it, and apply the triangle's angle sum with the right angle.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Vertical angles
angle a=angle b\text{angle } a = \text{angle } b
When two lines cross, they make an X. The angles straight across from each other in the X are called vertical angles. a and b are one such pair. They are always equal, because each one plus the angle between them makes a straight line.
Angles of a triangle
A+B+C=180A + B + C = 180
A, B and C are the three angles inside any triangle, in degrees. They always add up to 180°. Use it when you know two of the angles and need the third.

  1. Sort out the figure

    Line l is perpendicular to line n. Perpendicular means they meet at a right angle, 90°. The small square in the figure marks it. Line m crosses both lines. The three lines make a triangle with a right angle. You need x, the angle marked x° where m crosses n.

  2. Move the 35° angle into the triangle

    The 35° angle is below the point where m crosses l. So it is outside the triangle. Lines l and m make an X there. The triangle’s angle is straight across from the 35° angle. They are vertical angles, so they are equal. The triangle’s angle at that corner is 35°.

  3. Move the x° angle into the triangle

    The x° angle is above line n, so it is outside the triangle too. Lines m and n make another X. The triangle’s angle at that crossing is straight across from x°. So the third angle of the triangle is also x°. Watch out: line m does not cross l and n at the same angle. Lines l and n are not parallel, so x is not 35.

  4. Use the triangle’s angle sum

    The triangle’s three angles are 90°, 35° and x°.

    A+B+C=180A + B + C = 180

    90+35+x=18090 + 35 + x = 180

    125+x=180125 + x = 180

    x=180−125=55x = 180 - 125 = 55

    Watch out: 125 is only the two known angles added together. You still subtract it from 180. Gridding 125 is wrong. So you would grid 55.

Check your answer

In a right triangle, the two smaller angles add up to 90°. That is because the right angle uses 90° of the 180°. Check: 35+55=9035 + 55 = 90. Also, in the figure the x° angle looks a bit bigger than the 35° angle, and 55 is.

More questions like this

The original DOE question
Question 61 as printed on page 76 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 76, question 61. Open page 76 of the handbook (PDF)

Question 76 of 120

2023 Test A Math, question 12

Question

If n is an integer and 3n+33n + 3 is an even number, which expression must also represent an even number?

  • A5n+15n + 1
  • B4n+54n + 5
  • C2n+32n + 3
  • Dn+2n + 2
Show the answer

Correct answer: A

The expression 3n+33n + 3 is equal to 3(n+1)3(n + 1), and it is given that this is an even number. The expression (n+1)(n + 1) must also represent an even number, because 3 is an odd number: if (n+1)(n + 1) were odd, 3(n+1)3(n + 1) would be the product of two odd numbers, which is an odd number. If (n+1)(n + 1) is an even number, then n must be an odd number. Choose an odd number to substitute for n in the answer options to find which expression represents an even number:

Let n=3n = 3.

Option A:

5n+1=5(3)+1=165n + 1 = 5(3) + 1 = 16

Option B:

4n+5=4(3)+5=174n + 5 = 4(3) + 5 = 17

Option C:

2n+3=2(3)+3=92n + 3 = 2(3) + 3 = 9

Option D:

n+2=3+2=5n + 2 = 3 + 2 = 5

Since 16 is the only even number, the correct answer is option A.

Question analysis

This number properties question says that 3n + 3 is even for an integer n and asks which of four expressions must also be even. Students reason about odd and even numbers written with a variable. The hidden step is deciding what the given fact says about n itself before testing the expressions.

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
Above grade 7 Grade 7 students rewrite and evaluate expressions, but rarely reason about whether an expression is odd or even for every value of a variable. Here that reasoning comes in two layers: first about n, then about each choice.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 23 words.
Corrected from the DOE handbook
  • The explanation gave the wrong reason why n + 1 must be even; the reason was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question hides its key step: 3n + 3 = 3(n + 1), and since 3 is odd, n + 1 must be even, so n is odd. A student who skips that step and tests a convenient even value never checks the condition and lands on a wrong expression. Once n is known to be odd, any odd test value, such as 1, picks out 5n + 1 at once.

  1. Variables instead of numbers moderate Understanding the question

    No value of n is given; the only fact is that 3n + 3 is even, and it has to be turned into a fact about n itself, that n is odd, before any choice is judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know minor Working it out

    The parity rules are assumed, not stated: an odd number times an odd number is odd, and an even number plus an odd number is odd. (Anderson & Krathwohl, 2001)

  3. Every choice must be checked minor Choosing the answer

    Each expression has to be tested with an odd n or by parity rules; three of the four come out odd whenever n is odd, so none can be dismissed on sight. (Embretson & Wetzel, 1987)

Strategy: First work out what 3n + 3 being even says about n, by factoring out 3 or testing small integers, then check each expression with a value of n that fits.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation said that n + 1 must be even “because 3 is an odd number and the product of an odd number and an even number is an even number.” That fact does not show that n + 1 is even; the reason is that if n + 1 were odd, 3(n + 1) would be the product of two odd numbers, which is odd, and it is given as even. The site now gives that reason. The answer, A, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Pulling out a common factor
a×b+a×c=a(b+c)a \times b + a \times c = a(b + c)
a is a factor, a number being multiplied, that both terms share. b and c are what is left. This is the distributive property run backward. It turns a sum into a product, which is easier to reason about.
Multiplying odd numbers
odd×odd=odd\text{odd} \times \text{odd} = \text{odd}
An even number splits into pairs, like 2, 4 or 6. An odd number has one left over, like 1, 3 or 5. An odd number of odd groups always leaves one extra. So the product is odd.
Adding odd and even numbers
odd+odd=even,even+odd=odd\text{odd} + \text{odd} = \text{even}, \quad \text{even} + \text{odd} = \text{odd}
Two odd numbers each have one left over. Together the two extras make a pair, so the sum is even. An even number plus an odd number still has one left over, so the sum is odd.

  1. Sort out the question

    n is an integer, a whole number that can be positive, negative or 0. You are told that 3n+33n + 3 is even. You need the expression that must be even too. Must means it is even every time. First find out what the fact tells you about n.

  2. Pull out the 3

    Both terms in 3n+33n + 3 have a factor of 3. Pull it out.

    3n+3=3(n+1)3n + 3 = 3(n + 1)

    So 3 times (n+1)(n + 1) is an even number.

  3. Decide whether n is odd or even

    3 is odd. If n+1n + 1 were odd, 3(n+1)3(n + 1) would be odd times odd. That is always odd. But we are told it is even. So n+1n + 1 must be even. One less than an even number is odd. So n is odd. Watch out: don’t test n = 2. Then 3n+3=93n + 3 = 9, which is odd. With n = 2, only n+2n + 2 comes out even, choice D.

  4. Test each choice with an odd n

    Pick an odd number, n = 3. It fits, since 3×3+3=123 \times 3 + 3 = 12 is even.

    • A: 5n+1=5×3+1=165n + 1 = 5 \times 3 + 1 = 16, even
    • B: 4n+5=4×3+5=174n + 5 = 4 \times 3 + 5 = 17, odd
    • C: 2n+3=2×3+3=92n + 3 = 2 \times 3 + 3 = 9, odd
    • D: n+2=3+2=5n + 2 = 3 + 2 = 5, odd

    Only A is even.

  5. Make sure A always works

    Must means every odd n, not just 3. If n is odd, 5n is odd times odd, so it is odd. Odd plus 1 is even. So 5n+15n + 1 is always even. Watch out: 4n is even, but adding the odd 5 makes 4n+54n + 5 odd, choice B. In the same way, 2n+32n + 3 is always odd, choice C. So the answer is A, 5n+15n + 1.

Check your answer

Try another odd number, n = 1. Then 3n+3=63n + 3 = 6, which is even, so n = 1 fits. And 5n+1=65n + 1 = 6 is even too. The other three choices give 9, 5 and 3, all odd. Choice A wins again.

More questions like this

The original DOE question
Question 69 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 69. Open page 78 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 77 of 120

2023 Test A Math, question 15

Question

A swim instructor used the box plot below to display the distribution of the ages of students who signed up for swim lessons.

Box plot titled STUDENTS’ AGES above a number line from 5 to 13 with a tick and a label at every whole number and arrows at both ends. The left whisker starts at 7, the box runs from 8 to 10 with the median line at 9, and the right whisker ends at 12.

Which statement about the distribution of ages is true?

  • AThe data contain an outlier.
  • BApproximately half the students are exactly 9 years old.
  • CApproximately one-fourth of the students are at least 10 years old.
  • DThe interquartile range is 3 greater than the range of the data.
Show the answer

Correct answer: C

The data points in the box plot are as follows:

  • minimum: 7
  • first quartile: 8
  • median: 9
  • third quartile: 10
  • maximum: 12

Since the value of the third quartile is 10, approximately one-fourth of the students are 10–12 years old.

Question analysis

A swim instructor's box plot shows the ages of students who signed up for lessons. Students decide which of four statements about the distribution is true, using the minimum, quartiles, median and maximum. The statements test different box plot ideas, including outliers, the median, the quartiles and the range, so each needs its own check.

NYS standards
  • 7.SP.1 box plots and interquartile range
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 New York places box plots, the interquartile range and outliers in grade 7. Here four statements test those ideas one at a time; each check is short, so the item matches typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 66 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Reading the five values off the plot is quick; the work is knowing what a box plot can and cannot say. It shows where quarters of the data begin and end, not how many students share any one age, and a long whisker is not an outlier unless it passes the 1.5 × IQR cutoff, here 13. The true statement uses the third quartile: about a quarter of the students are 10 or older.

  1. Something you must already know moderate Working it out

    The statements rely on box plot facts the question does not give: each section holds about a quarter of the data, the middle line is the median, and outliers lie beyond 1.5 IQRs. (Anderson & Krathwohl, 2001)

  2. Every choice must be checked moderate Choosing the answer

    Four statements test four separate ideas, an outlier, the share at one age, a quarter of the data and two spreads, so ruling one out says nothing about the rest. (Embretson & Wetzel, 1987)

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

    The five values, 7, 8, 9, 10 and 12, are read off a 5-to-13 scale, where the longer right whisker draws the eye more than the box does. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the five key values off the plot, then test each statement: compute the range, the interquartile range and the outlier cutoffs, and recall what share of the data each part holds.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range
range=maximum−minimum\text{range} = \text{maximum} - \text{minimum}
The maximum is the largest value, the right end of a box plot. The minimum is the smallest, the left end. The range is how far apart they are. It shows how spread out all the data are.
Interquartile range
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
IQR is short for interquartile range. Q1, the first quartile, is the left edge of the box. Q3, the third quartile, is the right edge. The IQR is the width of the box. It is the spread of the middle half of the data. The four parts of a box plot each hold about one-fourth of the data.
Outlier test
below Q1−1.5×IQR or above Q3+1.5×IQR\text{below } Q_1 - 1.5 \times \text{IQR} \text{ or above } Q_3 + 1.5 \times \text{IQR}
An outlier is a value far from the rest of the data. A common rule says a value is an outlier if it lies more than 1.5 IQRs below Q1 or above Q3.

  1. Read the five values off the plot

    A box plot shows five numbers. The median is the middle value when the ages are in order.

    PartAge
    Minimum, the left whisker’s end7
    First quartile, Q1, the box’s left edge8
    Median, the line in the box9
    Third quartile, Q3, the box’s right edge10
    Maximum, the right whisker’s end12

  2. Check choice A, an outlier

    IQR=10−8=2\text{IQR} = 10 - 8 = 2

    Q3+1.5×2=13 and Q1−1.5×2=5Q_3 + 1.5 \times 2 = 13 \text{ and } Q_1 - 1.5 \times 2 = 5

    An outlier would be above 13 or below 5. The ages only run from 7 to 12. So there is no outlier. Watch out: a long whisker is not an outlier. It just means the top ages are spread out. Choice A is false.

  3. Check choice B, the median

    The median, 9, is the middle age. About half the students are 9 or younger. About half are 9 or older. It does not tell how many are exactly 9. Watch out: the median line does not mean half the students are 9. Choice B is false.

  4. Check choice C, the top fourth

    The box plot splits the students into four groups of about the same size. Each group holds about one-fourth of them. The last group runs from Q3, 10, to the maximum, 12. At least 10 means 10 or more. So about one-fourth of the students are at least 10 years old. Choice C is true.

  5. Check choice D, the two spreads

    range=12−7=5\text{range} = 12 - 7 = 5

    IQR=2\text{IQR} = 2

    The IQR is 3 less than the range, not 3 greater. The box always fits inside the whiskers, so the IQR can never be bigger. Watch out: swapping the two measures gives choice D. So the answer is C: about one-fourth of the students are at least 10 years old.

Check your answer

Picture 20 students. About 5 would be in each fourth of the plot. The top 5 would be ages 10 to 12. That is about one-fourth, just as choice C says. Each other choice fails a fact you can read straight off the plot.

More questions like this

The original DOE question
Question 72 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 72. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 78 of 120

2023 Test A Math, question 17

Question

A customer wants to buy a pair of hiking boots.

  • The original price of the boots is $85.75.
  • The store is offering a 15% discount on all boot purchases.
  • The customer has a coupon for an additional 25% off the sale price.
  • The tax rate is 8.5%.

What is the final cost of the boots, including the tax, to the nearest cent?

  • A$55.82
  • B$59.08
  • C$59.32
  • D$63.17
Show the answer

Correct answer: C

First, determine the sale price after the 15% discount is applied.

Since the customer is saving 15%, this means they are paying 85% (100−15)(100 - 15), so multiply the original price by 85% (0.85) to find the discounted price:

$85.75×0.85≈$72.89\$85.75 \times 0.85 \approx \$72.89

Next, find the price after the 25% coupon has been applied:

Since the customer is saving 25% off the sale price, this means they are paying 75% (100−25)(100 - 25), so multiply the price found in the first step by 75% (0.75) to find the second discounted price:

$72.89×0.75≈$54.67\$72.89 \times 0.75 \approx \$54.67

Last, find the final cost after an 8.5% tax is added to the second discounted price.

Since the customer is paying 8.5% tax, this means they are paying 108.5% (100+8.5)(100 + 8.5), so multiply the second discounted price by 108.5% (1.085) to find the final cost:

$54.67×1.085≈$59.32\$54.67 \times 1.085 \approx \$59.32

Question analysis

A pair of hiking boots costs $85.75, goes on sale at 15% off, gets an extra 25% off with a coupon, and is then taxed at 8.5%. Students find the final cost to the nearest cent. Each percent applies to the price left after the step before, so the discounts cannot simply be added, and the decimals call for careful multiplication.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 A discount followed by sales tax is standard grade 7 percent work. This item stacks two discounts and a tax, with cents to keep at every stage, so it takes about twice the steps and decimal multiplication of a typical problem.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 4, 58 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question tests whether students treat successive percents as successive multipliers, and then holds them to three long multiplications. Adding the 15% and 25% into one discount is the tempting shortcut, and a tax read as a dollar amount is the other. The key rounds to the cent at each stage, as a register would; carried exactly, the result is $59.31, so the nearest choice, $59.32, still stands.

  1. Error-prone arithmetic major Working it out

    Three products of cents and decimals are worked by hand: 85.75 × 0.85 = 72.8875, then 72.89 × 0.75 = 54.6675, then 54.67 × 1.085 ≈ 59.32, each rounded to the cent. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    The phrase "an additional 25% off the sale price" means 25% of the already discounted $72.89, not of the $85.75 original, so the two discounts multiply rather than add. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Slow to do carefully moderate Choosing the answer

    A reliable method needs three long decimal multiplications without a calculator, well past the minute and a half a question allows, which tempts students to combine the percents. (Lu & Sireci, 2007)

Strategy: Multiply by 0.85 for the sale, by 0.75 for the coupon on the sale price, and by 1.085 for the tax, rounding to the cent at each stage as a register would.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Price after a discount
new price=price×(1−p100)\text{new price} = \text{price} \times \left(1 - \frac{p}{100}\right)
p is the percent off. A discount is money taken off the price. If you save p percent, you pay the rest, 100 − p percent. For 15% off you pay 85%, so multiply by 0.85.
Price with tax
total=price×(1+t100)\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)
The price is the cost before tax. t is the tax rate, a percent. Tax is extra money added on top. You pay the whole price, 100%, plus t percent more. So for 8.5% tax you multiply by 1.085.

  1. Sort out the question

    The boots cost $85.75 at first. The store takes 15% off. Then a coupon takes 25% off the sale price. The sale price is the price after the first discount. Then 8.5% tax is added. You need the final cost to the nearest cent. A cent is $0.01, the second place after the decimal point.

  2. Take off the 15% discount

    You save 15%, so you pay 85%. That is 0.85 of the price.

    new price=price×(1−p100)\text{new price} = \text{price} \times \left(1 - \frac{p}{100}\right)

    new price=85.75×(1−15100)=85.75×0.85=72.8875\text{new price} = 85.75 \times \left(1 - \frac{15}{100}\right) = 85.75 \times 0.85 = 72.8875

    A price can’t have part of a cent, so round to two decimal places. The third digit is 7, which is 5 or more, so round up.

    72.8875≈72.8972.8875 \approx 72.89

    The sign ≈ means about. The sale price is $72.89.

  3. Take 25% off the sale price

    The coupon takes 25% off $72.89, so you pay 75% of it.

    72.89×0.75=54.6675≈54.6772.89 \times 0.75 = 54.6675 \approx 54.67

    The price is now $54.67. Watch out: take the coupon off the sale price, not the first price. Adding 15% and 25% to get 40% off $85.75 leads to $55.82, choice A.

  4. Add the 8.5% tax

    You pay the price plus 8.5% more, which is 108.5% of it.

    total=price×(1+t100)\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)

    total=54.67×1.085=59.31695≈59.32\text{total} = 54.67 \times 1.085 = 59.31695 \approx 59.32

    Watch out: 8.5% is a percent of the price, not $8.50. Adding $8.50 gives $63.17, choice D.

  5. See why rounding each time matters

    A store rounds to the cent after each step, as you just did. The test’s answer is worked the same way. Keep every digit and round only at the end, and you get this:

    85.75×0.85×0.75×1.085=59.3122…≈59.3185.75 \times 0.85 \times 0.75 \times 1.085 = 59.3122\ldots \approx 59.31

    That is $59.31, which is not a choice. The first two roundings each went up by a tiny part of a cent. That was just enough to push the total from $59.31 to $59.32. So here, round at each step, as a store would. So the answer is C, $59.32.

Check your answer

Check the tax a second way. 8.5% of $54.67 is 0.085×54.67=4.646950.085 \times 54.67 = 4.64695, about $4.65. Then 54.67+4.65=59.3254.67 + 4.65 = 59.32. Also estimate: two discounts leave a bit less than two-thirds of $86, near $55. Tax adds under $5, so about $59 or $60 makes sense.

More questions like this

The original DOE question
Question 74 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 74. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 79 of 120

2023 Test A Math, question 19

Question

x:35=20:28x:35 = 20:28

For what value of x is the proportion shown above true?

  • A27
  • B25
  • C16
  • D13
Show the answer

Correct answer: B

Write the ratios in fraction form and solve for x:

x35=2028\frac{x}{35} = \frac{20}{28} Cross multiply.
28x=70028x = 700 Apply the multiplicative inverse property; divide both sides by 28.
x=70028=25x = \frac{700}{28} = 25

Question analysis

This question gives the proportion x:35 = 20:28 and asks for the value of x that makes it true. Students use equivalent ratios or cross-multiplication. The trap is reasoning by adding or subtracting the difference between terms rather than multiplying, which gives numbers that look reasonable but break the ratio.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a missing term in a proportion is grade 6 ratio reasoning, and here the given ratio simplifies to small numbers. It is easier than typical grade 7 proportional reasoning, which usually involves fractions or a real context.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 5, 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 single proportion with whole numbers, and the only real trap is reasoning by differences instead of by a common factor. Since 35 is 7 more than 28, adding or subtracting 7 to 20 is tempting, but equal ratios scale; they do not shift. Reducing 20:28 to 5:7 shows at once that the scale factor to 35 is 5.

  1. Error-prone arithmetic minor Working it out

    Cross-multiplying gives 28x = 700, and 700 ÷ 28 is not a table fact; simplifying 20:28 to 5:7 first turns it into 5 × 5. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer minor Choosing the answer

    27 is only 2 more than 25, so a rough sense that x is a little more than 20 fits both; only the exact scale factor, 35/28 = 5/4, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Simplify 20:28 to lowest terms first, then scale it up so its second term becomes 35, or cross-multiply and divide by 28.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Ratio as a fraction
a:b=aba:b = \frac{a}{b}
a:b is a ratio, read a to b. It compares two amounts by division. You can write it as the fraction a over b. Two equal ratios make a proportion.
Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a over b is a fraction, and k is any number except 0. Multiplying or dividing the top and the bottom by the same number gives an equal fraction.

  1. Sort out the question

    x:35=20:28x:35 = 20:28 is a proportion. A proportion says two ratios are equal. A ratio compares two numbers, like 20 to 28. You need the value of x that makes the two ratios equal.

  2. Write the ratios as fractions

    Write each ratio as a fraction. The first number of the ratio goes on top. The second goes on the bottom.

    x35=2028\frac{x}{35} = \frac{20}{28}

  3. Simplify the known fraction

    20 and 28 both divide by 4. Smaller numbers are easier to work with.

    2028=20÷428÷4=57\frac{20}{28} = \frac{20 \div 4}{28 \div 4} = \frac{5}{7}

    So now x35=57\frac{x}{35} = \frac{5}{7}.

  4. Scale up so the bottom is 35

    7 times 5 is 35. To keep the fraction equal, multiply the top by 5 too.

    57=5×57×5=2535\frac{5}{7} = \frac{5 \times 5}{7 \times 5} = \frac{25}{35}

    So x = 25. Watch out: a ratio is about multiplying, not adding. 35 is 7 more than 28, but adding 7 to 20 gives 27, choice A. Also, x must be more than 20, because 35 is more than 28. So 16 and 13 are too small. So the answer is B, 25.

Check your answer

Cross-multiply, as the test’s own explanation does. 28×25=70028 \times 25 = 700 and 35×20=70035 \times 20 = 700. The two products are equal, so the ratios 25:35 and 20:28 are equal.

More questions like this

The original DOE question
Question 76 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 76. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 80 of 120

2023 Test A Math, question 23

Question

{0.1,0.01,0.001,0.0001,0.00001}\{0.1, 0.01, 0.001, 0.0001, 0.00001\}

If a person chooses a number at random from the set above, what is the probability that the number is less than 0.005?

  • A15\frac{1}{5}
  • B25\frac{2}{5}
  • C35\frac{3}{5}
  • D23\frac{2}{3}
Show the answer

Correct answer: C

Three of the five numbers listed (0.001, 0.0001, and 0.00001) are less than 0.005. So, the probability of a person selecting a number less than 0.005 is 35\frac{3}{5}.

Question analysis

A set holds five decimals, from one tenth down to one hundred-thousandth, and one number is chosen at random. Students find the probability that the number is less than 0.005. Comparing decimals with many zeros is the real task here, along with putting the count over the right total.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.NS.7a inequality as relative position
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Probability with equally likely outcomes is grade 6 content in New York, and this needs one comparison per number and one fraction. The only real demand is comparing decimals with many zeros correctly.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 27 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each number in the set is a tenth of the one before, so only the three smallest fall below 0.005, which lies between 0.01 and 0.001. Students lose the point either by counting the numbers above 0.005 instead of below it or by writing the count over the wrong total. Lining the decimals up with the same number of places makes every comparison easy.

  1. Error-prone arithmetic moderate Working it out

    0.005 has to be compared with decimals from 0.1 down to 0.00001; written to five places, 0.00500 against 0.01000 and 0.00100, the order becomes a comparison of whole numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    The probability is the count of numbers below 0.005 over all 5 numbers in the set; the stem leaves that part-to-whole rule to the student. (Anderson & Krathwohl, 2001)

Strategy: Write each number with five decimal places, compare it with 0.00500, count those that are smaller, and divide by the five numbers in the set.

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
P=wanted resultspossible resultsP = \frac{\text{wanted results}}{\text{possible results}}
P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.
Adding zeros to a decimal
0.1=0.10=0.1000.1 = 0.10 = 0.100
Zeros added to the right end of a decimal don’t change its value. One tenth is the same as ten hundredths. Giving every number the same number of decimal places lets you compare them like whole numbers.

  1. Sort out the question

    The set has five numbers. A person picks one at random, so each number has the same chance. You need the probability, or chance, that the number picked is less than 0.005.

  2. Give every number five decimal places

    Decimals are easier to compare with the same number of places. Add zeros to the end of each.

    NumberWith five places
    0.10.10000
    0.010.01000
    0.0010.00100
    0.00010.00010
    0.000010.00001
    0.0050.00500

  3. Count the numbers below 0.005

    Compare each number with 0.00500 like whole numbers: 10000, 1000, 100, 10 and 1 against 500. Two numbers are bigger: 0.1 and 0.01. Three are smaller: 0.001, 0.0001 and 0.00001. Watch out: 0.01 has fewer digits than 0.005, but it is bigger. Counting the 2 bigger numbers gives 25\frac{2}{5}, choice B.

  4. Write the probability

    P=wanted resultspossible results=35P = \frac{\text{wanted results}}{\text{possible results}} = \frac{3}{5}

    Watch out: divide by all 5 numbers in the set. Comparing the 2 bigger numbers with the 3 smaller ones gives 23\frac{2}{3}, choice D. So the answer is C, 35\frac{3}{5}.

Check your answer

The numbers below 0.005 and the numbers above it make up the whole set: 3 + 2 = 5. So the two chances, 35\frac{3}{5} and 25\frac{2}{5}, add up to 1, as they should.

More questions like this

The original DOE question
Question 80 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 80. Open page 81 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 81 of 120

2023 Test A Math, question 24

Question

A student uses Spinner R and Spinner T to generate a list of two-digit numbers.

Two spinners, each with an arrow pointing out from its center. Spinner R is a circle divided into 4 equal sections, labeled 1 and 2 across the top and 3 and 4 across the bottom. Spinner T is a circle divided into 6 equal sections, labeled, clockwise from the top left, 1, 1, 2, 3, 4 and 4.

Spinner R determines the digit in the tens place, and Spinner T determines the digit in the ones place. What is the probability that the two-digit number determined by spinning each spinner one time is a prime number?

  • A38\frac{3}{8}
  • B512\frac{5}{12}
  • C35\frac{3}{5}
  • D58\frac{5}{8}
Show the answer

Correct answer: A

Spinner R has 4 equal sections and Spinner T has 6 equal sections, so the number of possible two-digit numbers that can be created is found by multiplying 4 and 6, which is 24. List the possible outcomes:

11 11 12 13 14 14
21 21 22 23 24 24
31 31 32 33 34 34
41 41 42 43 44 44

There are 9 prime numbers out of 24 possible outcomes: 11, 11, 13, 23, 31, 31, 41, 41, and 43.

So the probability that the two-digit number created is a prime number is 924\frac{9}{24}, which is equivalent to 38\frac{3}{8}.

Question analysis

Spinner R, with four equal sections, gives the tens digit and Spinner T, with six equal sections and some labels repeated, gives the ones digit of a two-digit number. Students find the probability that the number is prime. Building the full table of outcomes, counting repeated sections separately, and checking whether each number is prime all take care.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.SP.8a compound event as fraction of outcomes
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Listing a sample space for two spinners is grade 7 work, usually with distinct labels and a simple event. Here one spinner repeats two numbers, the space has 24 outcomes, and each two-digit result must be tested for being prime.
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 7, 57 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 combines a compound sample space with a test for primes, and its figure adds a twist: two labels repeat on Spinner T. Counting the 24 section pairs rather than the 16 different numbers is the careful route, though here both happen to give the same fraction. The points are lost in the prime test, where 33 is easy to count as prime, and in writing the count as a share of all 24 outcomes.

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

    Spinner T has six equal sections but only four labels, with 1 and 4 each appearing twice, so the 24 equally likely outcomes include repeats such as 11 and 11. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Much to keep track of moderate Working it out

    Twenty-four two-digit numbers have to be built and each tested for primality while the repeated sections are kept track of; a 4-by-6 table is the reliable way to hold them. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Something you must already know minor Working it out

    Which two-digit numbers are prime is not given; students must test 21, 33, 43 and the rest themselves, and 21 = 3 × 7 and 33 = 3 × 11 look prime at a glance. (Anderson & Krathwohl, 2001)

Strategy: Make a 4-by-6 table with Spinner R's digits as tens and Spinner T's six sections as ones, mark each prime, and count every section, including repeated labels.

Traps

  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting outcomes
outcomes=a×b\text{outcomes} = a \times b
a is the number of equal sections on the first spinner, and b is the number on the second. Each section of the first can go with each section of the second. So the number of pairs is a times b.
Probability
P=wanted resultspossible resultsP = \frac{\text{wanted results}}{\text{possible results}}
P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.

  1. Sort out the question

    Spinner R gives the tens digit. Spinner T gives the ones digit. Together they make a two-digit number, like 23. You need the chance that the number is prime. A prime number has exactly two factors, 1 and itself. For example, 13 is prime, but 21 = 3 × 7 is not.

  2. Count all the outcomes

    Spinner R has 4 equal sections. Spinner T has 6 equal sections. Look closely: T has two 1s and two 4s. Each section counts as its own outcome.

    outcomes=4×6=24\text{outcomes} = 4 \times 6 = 24

    There are 24 outcomes, all equally likely.

  3. Make a table of the numbers

    Each row is a tens digit from R. Each column is a section of T. Prime numbers are in bold.

    R112344
    1111112131414
    2212122232424
    3313132333434
    4414142434444

  4. Count the prime numbers

    Numbers ending in 2 or 4 are even, so 2 divides them. They are not prime. Check the rest. 11, 13, 23, 31, 41 and 43 are prime. But 21 = 3 × 7 and 33 = 3 × 11 are not. Count every section, so 11, 31 and 41 each count twice. That makes 3 + 1 + 2 + 3 = 9 primes. Watch out: counting 33 as prime gives 1024=512\frac{10}{24} = \frac{5}{12}, choice B.

  5. Find the probability

    P=924=9÷324÷3=38P = \frac{9}{24} = \frac{9 \div 3}{24 \div 3} = \frac{3}{8}

    Watch out: divide by all 24 outcomes. Comparing the 9 primes with the 15 others gives 915=35\frac{9}{15} = \frac{3}{5}, choice C. So the answer is A, 38\frac{3}{8}.

Check your answer

There are 24−9=1524 - 9 = 15 numbers that are not prime. Their chance is 1524=58\frac{15}{24} = \frac{5}{8}. And 38+58=1\frac{3}{8} + \frac{5}{8} = 1, as it should be. Notice that 58\frac{5}{8}, choice D, is the chance of not prime.

More questions like this

The original DOE question
Question 81 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 81. Open page 81 of the handbook (PDF)

Question 82 of 120

2023 Test A Math, question 26

Question

Which statement must be true if x is a whole number greater than or equal to 1?

  • A1x+1>1x+2\frac{1}{x + 1} > \frac{1}{x + 2}
  • B1x+1<1x+2\frac{1}{x + 1} < \frac{1}{x + 2}
  • C1x+1−1x+2>1\frac{1}{x + 1} - \frac{1}{x + 2} > 1
  • D1x+1−1x+2>1x\frac{1}{x + 1} - \frac{1}{x + 2} > \frac{1}{x}
Show the answer

Correct answer: A

Let x=2x = 2, and substitute that value for x in each inequality to determine which one is correct:

Option A:

13>14\frac{1}{3} > \frac{1}{4}. This is true.

Option B:

13<14\frac{1}{3} < \frac{1}{4}. This is false.

Option C:

13−14>1\frac{1}{3} - \frac{1}{4} > 1, or 112>1\frac{1}{12} > 1. This is false.

Option D:

13−14>12\frac{1}{3} - \frac{1}{4} > \frac{1}{2}, or 112>12\frac{1}{12} > \frac{1}{2}. This is false.

Option A is the only true inequality.

Question analysis

This question asks which of four inequalities must be true when x is a whole number of at least 1. The inequalities compare fractions with denominators x + 1 and x + 2, or their difference. Students reason about how the size of a fraction changes as its denominator grows, and must be sure a statement holds for every allowed value.

NYS standards
  • 6.NS.7a inequality as relative position
  • 6.EE.5 solutions and substitution
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Comparing fractions is familiar, but these fractions have variable denominators and the question asks what must be true for every whole number from 1 up. That abstraction makes it as demanding as typical grade 7 work, though the arithmetic is light.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 6, 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

Choosing a small value, such as x = 1 or 2, and testing every statement settles this quickly, because three of the four fail for any value tried. The trap is the belief that a bigger denominator makes a bigger fraction, which turns the true comparison around. Students unsure of fraction subtraction lose time on the two difference statements, though both differences are tiny, never more than 1/6.

  1. Variables instead of numbers moderate Understanding the question

    All four statements are about the unknown x, and the bold must means each has to hold for every whole number from 1 up, not just for a single test value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    The statements are four separate inequalities, two comparisons and two differences, so each has to be tested; one counterexample is enough to rule a statement out. (Embretson & Wetzel, 1987)

  3. Error-prone arithmetic minor Working it out

    Testing the differences means subtracting unlike fractions, such as 1/3 − 1/4 = 1/12 when x = 2, and comparing the result with 1 and with 1/x. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute a small whole number such as 1 or 2 for x in all four statements, rule out any that fail, and confirm the survivor by thinking about larger denominators.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions
if 0<a<b, then 1a>1b\text{if } 0 < a < b \text{, then } \frac{1}{a} > \frac{1}{b}
a and b are positive numbers, and b is bigger. One over a means one whole cut into a equal pieces. Cutting the same whole into more pieces makes each piece smaller. So the fraction with the bigger bottom is smaller.
Subtracting fractions
1a−1b=b−aa×b\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}
a and b are the bottoms of the two fractions. Rewrite both fractions over the common bottom a × b. Then subtract the tops. The bottom stays a × b.

  1. Sort out the question

    x is a whole number that is 1 or more, so x can be 1, 2, 3 and so on. You need the statement that must be true. That means it is true for every such x. One x that makes a statement false rules it out.

  2. Test x = 1 in each statement

    With x = 1, x + 1 is 2 and x + 2 is 3.

    • A: 12>13\frac{1}{2} > \frac{1}{3}. True.
    • B: 12<13\frac{1}{2} < \frac{1}{3}. False.
    • C: 12−13=36−26=16\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}, which is not more than 1. False.
    • D: 16\frac{1}{6} is not more than 11\frac{1}{1}. False.

    Only A is true when x = 1. So B, C and D are out.

  3. Make sure A is always true

    Must means A has to work for every x, not just 1. In 1x+1\frac{1}{x + 1} and 1x+2\frac{1}{x + 2}, both tops are 1. The bottom x + 2 is always 1 more than x + 1. Cutting a whole into more pieces makes each piece smaller. So A is always true. Watch out: a bigger bottom makes a smaller fraction, not a bigger one. Thinking the other way gives choice B.

  4. See why C and D always fail

    Use the subtracting rule with a = x + 1 and b = x + 2.

    1a−1b=b−aa×b\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}

    1x+1−1x+2=(x+2)−(x+1)(x+1)(x+2)=1(x+1)(x+2)\frac{1}{x + 1} - \frac{1}{x + 2} = \frac{(x + 2) - (x + 1)}{(x + 1)(x + 2)} = \frac{1}{(x + 1)(x + 2)}

    The bottom is at least 2 × 3 = 6, so the difference is at most 16\frac{1}{6}. That is less than 1. Its bottom is also bigger than x, so it is less than 1x\frac{1}{x}. Watch out: keep the bottom when you subtract. Keeping only the top, 1, leads to choice D. So the answer is A.

Check your answer

Try a bigger number, x = 9. A tenth of a pizza is more than an eleventh, so 110>111\frac{1}{10} > \frac{1}{11}. The difference, 1110\frac{1}{110}, is tiny. It is less than 1 and less than 19\frac{1}{9}. A still holds, and C and D still fail.

More questions like this

The original DOE question
Question 83 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 83. Open page 82 of the handbook (PDF)

Question 83 of 120

2023 Test A Math, question 42

Question

{1,2,3,4,5,…,198,199,200}\{1,2,3,4,5,\ldots,198,199,200\}

How many members of the set shown above are multiples of 6 but not multiples of 9?

  • A11
  • B13
  • C20
  • D22
Show the answer

Correct answer: D

First, list the first several multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54 . . .

Every third multiple (i.e., 18, 36, 54, . . .) is also a multiple of 9. Thus, 23\frac{2}{3} of the multiples of 6 are not multiples of 9.

Next, calculate how many multiples of 6 are in the given number range. The last multiple of 6 would be 198. So, there are a total of 1986=33\frac{198}{6} = 33 multiples of 6. Since 23\frac{2}{3} of those are not multiples of 9, the answer is 23(33)=22\frac{2}{3}(33) = 22.

Question analysis

This number properties question shows the set of whole numbers from 1 to 200 and asks how many of them are multiples of 6 but not multiples of 9. Students count multiples in a range and remove the overlap, the numbers divisible by both. Finding the right common multiple, and removing only the overlap, is where it gets tricky.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Multiples and common multiples are grade 6 content, but counting the members of a set that meet one condition and fail another is rare in grade 7 work. The overlap between the two kinds of multiples must be found and removed exactly.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 6, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Counting multiples of 6 is routine; the harder decision is how to remove the ones that are also multiples of 9. Those are the multiples of 18, the least common multiple, not of 54 or 15, and students who count them correctly must still subtract them rather than report them. Listing 6, 12, 18, 24, 30, 36 shows the pattern, two kept for every one removed, which also gives the answer as two thirds of 33.

  1. No obvious method moderate Working it out

    The stem never mentions a common multiple: students must see that the multiples of 6 that are also multiples of 9 are the multiples of 18, every third multiple of 6. (Schoenfeld, 1985)

  2. Much to keep track of minor Working it out

    Three counts are in play at once, 33 multiples of 6 up to 198, 11 multiples of 18 up to 198, and their difference, and two of them are listed choices. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Turning words into math minor Understanding the question

    The bold not in “multiples of 6 but not multiples of 9” asks for what is left after a removal, a set difference rather than a count of the numbers removed. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Count the multiples of 6 up to 200, find the smallest number that is a multiple of both 6 and 9, count its multiples in the set, and subtract.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting multiples
last multiple=n×k\text{last multiple} = n \times k
n is the number whose multiples you count, such as 6. The first multiple is n × 1, the second is n × 2, and so on. So if the largest multiple in the set is n × k, there are k multiples in all.
Least common multiple
LCM(6,9)=18\text{LCM}(6, 9) = 18
The least common multiple, or LCM, is the smallest number that two numbers both divide evenly. A number is a multiple of both 6 and 9 exactly when it is a multiple of 18.

  1. Sort out the question

    The set holds the whole numbers from 1 to 200. A multiple of 6 is a number that 6 divides evenly, like 6, 12 or 18. You need the multiples of 6 that are not multiples of 9. So count all the multiples of 6. Then take away those that are also multiples of 9.

  2. Count the multiples of 6

    200 is not a multiple of 6, but 198 is. So 198 is the last multiple of 6 in the set.

    last multiple=n×k\text{last multiple} = n \times k

    198=6×33198 = 6 \times 33

    There are 33 multiples of 6.

  3. Find the numbers to take away

    List the multiples of 6: 6, 12, 18, 24, 30, 36. The first one 9 also divides is 18, then 36. So every third multiple of 6 is also a multiple of 9.

    LCM(6,9)=18\text{LCM}(6, 9) = 18

    Watch out: the shared multiples are multiples of 18, not of 6 + 9 = 15. Using 15 leads to 13 or 20, choices B and C.

  4. Count the multiples of 18

    The last multiple of 18 in the set is also 198.

    198=18×11198 = 18 \times 11

    There are 11 of them. Watch out: these are the numbers you take away, not the answer. Stopping here gives 11, choice A.

  5. Take away the multiples of 18

    33−11=2233 - 11 = 22

    There are 22 multiples of 6 left over. So the answer is D, 22.

Check your answer

Every third multiple of 6 is a multiple of 18, so two out of every three multiples of 6 stay. Two thirds of 33 is 23×33=22\frac{2}{3} \times 33 = 22. That matches. Test one: 30 is a multiple of 6, and 30 ÷ 9 is not a whole number, so 30 counts.

More questions like this

The original DOE question
Question 99 as printed on page 86 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 99. Open page 86 of the handbook (PDF)

Question 84 of 120

2023 Test A Math, question 45

Question

A rectangle. Point R is on the top side, near its right end. Line segment RS runs down and to the right from R, crosses the right side of the rectangle, and ends at point S outside it. At R, the angle between the top side, to the left of R, and segment RS is marked 130°. Where RS crosses the right side, the angle outside the rectangle between the right side, above the crossing, and the part of RS going on to S is marked y.

The diagram above shows line segment RS intersecting a rectangle. What is the measure of angle y?

  • A140°
  • B130°
  • C50°
  • D40°
Show the answer

Correct answer: A

First, find the measures of the interior angles of the triangle formed by the line segment and the rectangle. The top angle on the right is a vertex of the rectangle, so the measure of that angle is 90°. The top angle on the left is supplementary to the given angle of 130°. So, the measure of that angle is 180−130=50180 - 130 = 50. The measure of the lower angle is 40°, since 180−90−50=40180 - 90 - 50 = 40. Angle y is supplementary to the lower angle of 40°, so the measure of angle y is 180−40=140180 - 40 = 140.

Question analysis

This geometry question shows line segment RS cutting across a corner of a rectangle, with a 130° angle marked where RS meets the top side, and asks for angle y where it crosses the right side. Students combine supplementary angles, the right angle of a rectangle and the angle sum of a triangle. Stopping at an angle inside the triangle is the main trap.

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 problems use supplementary, vertical and adjacent angles. This one adds the angle sum of a triangle, which New York places in grade 8, and chains three facts around a small triangle that students must first spot in the figure.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 7, 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 angle facts are basic, but the route runs through a triangle the figure does not point out, and the answer lies outside it. Students who stop inside the triangle give its 40° angle or its 50° angle, and those who assume y matches the marked 130° skip the work entirely. Checking that y and the triangle's angle beside it form a straight line, 140 + 40 = 180, confirms the result.

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

    The small triangle cut off at the rectangle's top right corner is never outlined or labeled, its 90° corner is unmarked, and y sits outside it, between RS and the rectangle's right side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Three steps follow in order, a straight angle at R, the triangle's angle sum, and a straight angle at the right side, and the first two results are both listed choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Something you must already know minor Working it out

    The rectangle's corner must be taken as 90° although no right-angle mark is drawn, and the 180° totals for a straight line and a triangle are assumed. (Anderson & Krathwohl, 2001)

Strategy: Look at the small triangle cut off at the rectangle's corner: get its angle at R from the straight top side, use the 90° corner and the triangle's angle sum, then use the straight right side.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line
a+b=180a + b = 180
a and b are two angles side by side that together make a straight line. A straight line makes an angle of 180 degrees. So if you know one of them, subtract it from 180 to get the other.
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 them, subtract both from 180 to find the third.
Outside angle of a triangle
outside angle=a+b\text{outside angle} = a + b
An outside angle sits next to a triangle’s angle, along a straight line. a and b are the two inside angles at the other corners. The outside angle equals their sum.

  1. Sort out the figure

    Segment RS starts at R on the top side of the rectangle. It goes down to the right, crosses the right side, and ends at S. Call the rectangle’s top right corner P. Call the point where RS crosses the right side Q. Then R, P and Q make a small triangle. Angle y is at Q, outside the rectangle.

  2. Find the triangle’s angle at R

    The 130° angle and the triangle’s angle at R sit side by side along the top side, a straight line.

    a+b=180a + b = 180

    130+∠R=180130 + \angle \mathrm{R} = 180

    ∠R=180−130=50\angle \mathrm{R} = 180 - 130 = 50

    Watch out: 50° is only the first step. Stopping here gives choice C.

  3. Find the triangle’s angle at Q

    P is a corner of the rectangle, so its angle is 90°.

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

    50+90+∠Q=18050 + 90 + \angle \mathrm{Q} = 180

    ∠Q=180−140=40\angle \mathrm{Q} = 180 - 140 = 40

    Watch out: y is outside the triangle, so this 40° angle, choice D, is not y.

  4. Find angle y

    Angle y and the triangle’s 40° angle at Q sit side by side along RS, a straight line.

    a+b=180a + b = 180

    y+40=180y + 40 = 180

    y=140y = 140

    So the answer is A, 140°.

Check your answer

Use the outside angle rule. Angle y is outside the triangle at Q. The two inside angles at the other corners are 50° and 90°, so y=50+90=140y = 50 + 90 = 140. Also, y looks wider than a right angle in the figure, so it must be more than 90°.

More questions like this

The original DOE question
Question 102 as printed on page 87 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 87, question 102. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 85 of 120

2023 Test A Math, question 47

Question

Which percentage is closest in value to 0.0099?

  • A0%
  • B0.1%
  • C1%
  • D100%
Show the answer

Correct answer: C

Since 0.0099 is equal to 0.99%, the answer option closest to that value is 1%.

Question analysis

This percent question asks which of four percentages is closest in value to the decimal 0.0099. Students convert a small decimal to a percent by multiplying by 100, then compare it with choices that differ by factors of ten. Moving the decimal point the wrong number of places, or reading the digits 99 as a percent, leads to wrong answers.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Converting a decimal to a percent is grade 6 work. This decimal has two zeros before its first digit, so the point must move carefully, but it is a single conversion followed by picking the nearest choice, below typical grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 7 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 conversion answers the question, and the choices are spaced by powers of ten, so the work is all in the decimal places. A decimal this small looks like nothing, and students who forget to multiply by 100 see it as nearly 0%. Others move the point the wrong number of places or read the digits 99 as a percent. Seeing that 0.01 is exactly 1% puts 0.0099 just below it.

  1. Error-prone arithmetic minor Working it out

    The decimal 0.0099 has two zeros before its digits, so moving the point two places gives 0.99, and miscounting the places lands on 0.1% or 99%. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    Students must know that a percent is a number of hundredths, so 0.0099 is multiplied by 100 to become 0.99%, a rule the question takes for granted. (Anderson & Krathwohl, 2001)

Strategy: Multiply the decimal by 100 to write it as a percent, moving the point two places to the right, then choose the percent nearest to that value.

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.
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimal to percent
percent=decimal×100\text{percent} = \text{decimal} \times 100
A percent is a number of hundredths. To change a decimal into a percent, multiply by 100. That moves the decimal point two places to the right. For example, 0.25 × 100 = 25, so 0.25 is 25%.

  1. Sort out the question

    The number 0.0099 is a decimal, but the choices are percents. To compare them, change 0.0099 into a percent first. Then find the choice nearest to it.

  2. Change the decimal to a percent

    percent=decimal×100\text{percent} = \text{decimal} \times 100

    0.0099×100=0.990.0099 \times 100 = 0.99

    So 0.0099 is 0.99%. The decimal point moved two places to the right. Watch out: 0.0099 is not already a percent. Reading it as 0.0099% points to 0%, choice A.

  3. Find the nearest choice

    Find how far 0.99% is from each choice.

    ChoiceDistance from 0.99%
    A, 0%0.99
    B, 0.1%0.89
    C, 1%0.01
    D, 100%99.01

    Watch out: moving the point only one place gives about 0.1, choice B. So the answer is C, 1%.

Check your answer

Work the other way. 1% is 1 out of 100, or 0.01. And 0.0099 is just a tiny bit less than 0.0100. So 0.0099 is very close to 1%, just a little less.

More questions like this

The original DOE question
Question 104 as printed on page 87 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 87, question 104. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 86 of 120

2023 Test A Math, question 48

Question

The manager of a shipping company recorded the weights, in pounds, of the last 9 packages the company shipped. The manager displayed the distribution of the weights in a box plot with the five-number summary shown below:

  • minimum: 29
  • first quartile: 31
  • median: 42
  • third quartile: 73
  • maximum: 98

Which statement about the distribution of weights is supported by the box plot the manager created?

  • AThe minimum, 29, is an outlier.
  • BThe maximum, 98, is an outlier.
  • CThe interquartile range of the data is 11.
  • DThe interquartile range and the median are equal.
Show the answer

Correct answer: D

First, identify the first quartile and the third quartile:

  • first quartile: 31
  • third quartile: 73

Then, find the difference between the first quartile and the third quartile:

73−31=4273 - 31 = 42

The median is given: 42.

So, the interquartile range and the median are equal.

Question analysis

This statistics question gives the five-number summary of nine package weights shown in a box plot and asks which statement about the distribution is supported. Students compute the interquartile range and decide whether the minimum or the maximum is an outlier. Judging outliers by eye, or subtracting the wrong pair of values for the IQR, makes wrong statements look true.

NYS standards
  • 7.SP.1 box plots and interquartile range
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the interquartile range and testing for outliers are grade 7 standards in New York. There is no plot to read here, only five numbers, but every statement has to be checked by calculation, which makes it a careful 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 7, 84 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 given outright, so the work is knowing what each statement claims and testing it. The outlier statements sound right because 98 looks far from 73 and 29 is the smallest value, but the 1.5 × IQR rule clears both. The true statement needs only one subtraction, 73 − 31, and a look at the median, which happens to have the same value.

  1. Something you must already know moderate Working it out

    Neither the interquartile range, the third quartile minus the first, nor the outlier rule, more than 1.5 × IQR beyond a quartile, is defined on the page. (Anderson & Krathwohl, 2001)

  2. Every choice must be checked moderate Choosing the answer

    The four statements test different ideas, two possible outliers, a value for the IQR and a comparison of the IQR with the median, so ruling out one says nothing about the next. (Embretson & Wetzel, 1987)

Strategy: Compute the interquartile range from the two quartiles, use it to set the outlier cutoffs 1.5 IQR beyond each quartile, and test every statement against those numbers.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
Q1, the first quartile, is the middle of the lower half of the data. Q3, the third quartile, is the middle of the upper half. The interquartile range, or IQR, is the distance between them. It is the width of the box in a box plot. It shows how spread out the middle half of the data is.
Outlier rule
outlier: below Q1−1.5×IQR or above Q3+1.5×IQR\text{outlier: below } Q_1 - 1.5 \times \text{IQR} \text{ or above } Q_3 + 1.5 \times \text{IQR}
An outlier is a value far away from the rest of the data. Q1 and Q3 are the quartiles, and IQR is the interquartile range. A value is an outlier if it lies more than 1.5 times the IQR below Q1 or above Q3.

  1. Sort out the question

    The manager’s box plot is drawn from the five numbers in the list, so you can work from the list. The weights, in pounds, are: minimum 29, first quartile 31, median 42, third quartile 73, maximum 98. You need the one statement that is true. So check each choice.

  2. Find the interquartile range

    IQR=Q3−Q1\text{IQR} = Q_3 - Q_1

    IQR=73−31=42\text{IQR} = 73 - 31 = 42

    The interquartile range is 42 pounds. That makes choice C, which says 11, false.

  3. Compare the IQR with the median

    The median is 42 pounds. The interquartile range is also 42 pounds. They are equal, so statement D is true.

    Watch out: 42 − 31 = 11 subtracts Q1 from the median, not from Q3. That mistake makes choice C look right.

  4. Test the ends for outliers

    outlier: below Q1−1.5×IQR or above Q3+1.5×IQR\text{outlier: below } Q_1 - 1.5 \times \text{IQR} \text{ or above } Q_3 + 1.5 \times \text{IQR}

    1.5×42=631.5 \times 42 = 63

    31−63=−3273+63=13631 - 63 = -32 \qquad 73 + 63 = 136

    An outlier would be below −32 or above 136. The minimum, 29, is not below −32. The maximum, 98, is not above 136. So A and B are false. Watch out: 98 looks far away, but not far enough to be an outlier.

  5. Choose the answer

    Only statement D is supported by the five numbers. So the answer is D, the interquartile range and the median are equal.

Check your answer

Check the subtraction by adding back: 31+42=7331 + 42 = 73, which is Q3. The median, 42, sits inside the box, between 31 and 73, as it must. The two measures mean different things, but here they happen to be the same number.

More questions like this

The original DOE question
Question 105 as printed on page 88 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 88, question 105. Open page 88 of the handbook (PDF)

Question 87 of 120

2023 Test A Math, question 49

Question

A shaded rectangle inside a larger, unshaded outer rectangle. The bottom side of the outer rectangle is labeled 4 ft. The left side of the shaded rectangle is labeled 2 ft. The shaded rectangle is wider than it is tall.

In the figure above, the shaded rectangle is similar to the outer rectangle. The length of the outer rectangle is 4 feet, and the perimeter of the outer rectangle is 14 feet. If the width of the shaded rectangle is 2 feet, what is the area of the shaded rectangle?

  • A5135\frac{1}{3} sq ft
  • B6 sq ft
  • C9139\frac{1}{3} sq ft
  • D12 sq ft
Show the answer

Correct answer: A

For the outer rectangle, the length is 4 feet and the perimeter is 14 feet. Write and solve an equation to find the width:

2(4)+2w=142(4) + 2w = 14 Apply the additive inverse property; subtract 8 from both sides of the equation.
2w=62w = 6 Apply the multiplicative inverse property; divide both sides of the equation by 2.
w=3w = 3

Use that information in a proportion to find the length of the shaded rectangle:

x4=23\frac{x}{4} = \frac{2}{3} Cross multiply.
3x=4(2)3x = 4(2) Apply the multiplicative inverse property; divide both sides of the equation by 3.
x=83x = \frac{8}{3}

Now find the area of the shaded rectangle:

2×83=163=5132 \times \frac{8}{3} = \frac{16}{3} = 5\frac{1}{3}

Question analysis

This geometry question shows a shaded rectangle similar to an outer rectangle that is 4 feet long with a perimeter of 14 feet; the shaded rectangle is 2 feet wide. Students find the shaded area. They must find a missing side, match corresponding sides, and scale by a factor rather than a fixed amount, which makes it a multistep similar-figures problem.

NYS standards
  • 8.G.4 similarity and dilations
  • 7.G.1 scale drawings
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Similar figures are grade 8 geometry, though grade 7 scale drawings use the same proportions. Students must first find a missing side from the perimeter, match corresponding sides, solve a proportion with a fractional result, and then find an area.
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 5, 57 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 familiar, but the three together, perimeter to width, width to scale factor, and sides to area, demand a plan. The strongest pull is scaling by subtraction, shrinking each side by 1 foot because the width drops from 3 to 2, which gives a neat whole-number area. Others stop at the outer rectangle or add the sides instead of multiplying. Checking that 8/3 to 2 matches 4 to 3 confirms the shape is similar.

  1. Combines ideas moderate Working it out

    Perimeter, similarity and area are chained: 14 feet gives the outer width of 3 feet, the 2-foot side sets a scale factor of 2/3, and only then can the area be found. (Embretson, 1998; Daniel & Embretson, 2010)

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

    The figure marks the shaded rectangle's 2-foot side running up and down and the outer 4-foot length along the bottom, so the 2 feet must be paired with the outer 3-foot side, which is not labeled. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Error-prone arithmetic minor Working it out

    The scale factor 2/3 makes the shaded length 8/3 feet, and the area 2 × 8/3 = 16/3 has to be written as the mixed number 5 1/3. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Use the perimeter to find the outer rectangle's width, match the shaded width of 2 feet to it, scale the 4-foot length by the same factor, then multiply length by width.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a rectangle
P=2l+2wP = 2l + 2w
P is the perimeter, the distance all the way around. l is the length and w is the width. A rectangle has two lengths and two widths, so you add each one twice.
Scale factor of similar shapes
scale factor=small sidematching large side\text{scale factor} = \frac{\text{small side}}{\text{matching large side}}
Similar shapes have the same shape but different sizes. Each side of the small shape equals its matching side on the large shape times one number, the scale factor. So one pair of matching sides tells you the scale factor.
Area of a rectangle
A=l×wA = l \times w
A is the area, the space inside. l is the length and w is the width. You multiply because the rectangle is made of rows of unit squares: w rows with l squares in each.

  1. Sort out the question

    The outer rectangle is 4 feet long, and its perimeter is 14 feet. The shaded rectangle is similar to it, so it has the same shape. The shaded rectangle’s width, its left side, is 2 feet. You need the shaded area, so you also need the shaded length.

  2. Find the outer rectangle’s width

    P=2l+2wP = 2l + 2w

    14=2×4+2w14 = 2 \times 4 + 2w

    14=8+2w14 = 8 + 2w

    2w=6,w=32w = 6, \quad w = 3

    The outer rectangle is 3 feet wide. That is its up-and-down side.

  3. Find the scale factor

    The shaded width, 2 feet, matches the outer width, 3 feet.

    scale factor=small sidematching large side\text{scale factor} = \frac{\text{small side}}{\text{matching large side}}

    scale factor=23\text{scale factor} = \frac{2}{3}

    Every shaded side is 23\frac{2}{3} of the matching outer side. Watch out: similar shapes shrink by the same factor, not by the same number of feet. Taking 1 foot off each side gives 2 by 3, an area of 6, choice B.

  4. Find the shaded length

    Multiply the outer length, 4 feet, by the scale factor.

    l=23×4=83=223l = \frac{2}{3} \times 4 = \frac{8}{3} = 2\frac{2}{3}

    The shaded rectangle is 2232\frac{2}{3} feet long.

  5. Find the shaded area

    A=l×wA = l \times w

    A=83×2=163=513A = \frac{8}{3} \times 2 = \frac{16}{3} = 5\frac{1}{3}

    Watch out: 4 × 3 = 12 is the outer area, choice D. So the answer is A, 5135\frac{1}{3} sq ft.

Check your answer

Check the shape. For the outer rectangle, length ÷ width is 4÷3=434 \div 3 = \frac{4}{3}. For the shaded one, it is 83÷2=43\frac{8}{3} \div 2 = \frac{4}{3}. They match, so the shapes are similar. Also, 5135\frac{1}{3} square feet is less than the outer 12, as it should be.

More questions like this

The original DOE question
Question 106 as printed on page 88 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 88, question 106. Open page 88 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 88 of 120

2023 Test A Math, question 51

Question

At a carnival, visitors can win a prize if an employee cannot correctly guess their age. The employee records the first 15 ages that he guesses one Saturday. The box plot shown below displays data about the ages he recorded.

A box plot above a number line from 7 to 18 with a tick at every integer. The left whisker starts at 7, the box runs from 8 to 17 with the median line at 10, and the right whisker ends at 18.

What is the interquartile range of the box plot?

  • A7
  • B9
  • C10
  • D11
Show the answer

Correct answer: B

Quartile 1 is the median of the lower half of the data, 8.

Quartile 3 is the median of the upper half of the data, 17.

IQR=Q3−Q1=17−8=9\text{IQR} = \text{Q3} - \text{Q1} = 17 - 8 = 9

The interquartile range is 9.

Question analysis

This statistics question shows a box plot of 15 ages that a carnival employee guessed and asks for its interquartile range. Students read the quartiles from the edges of the box and subtract. Confusing the interquartile range with the full range, or using the median line in place of a quartile, gives the wrong choices.

NYS standards
  • 7.SP.1 box plots and interquartile range
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Reading a box plot and finding its interquartile range is a grade 7 standard in New York. The values sit on labeled whole-number ticks and one subtraction is needed, a routine grade 7 task once the box's edges are known as the quartiles.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is one subtraction, so the question tests whether students know which parts of a box plot to use. The whiskers' ends give the range, the line inside the box gives the median, and both are easy to grab in place of the quartiles. The story about the carnival and the 15 ages plays no part in the answer.

  1. Something you must already know minor Working it out

    The plot shows five values with no labels, so students must know that the box's ends mark the quartiles and that the interquartile range is the box's width, not the whiskers'. (Anderson & Krathwohl, 2001)

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

    The values 7, 8, 10, 17 and 18 are read off a scale ticked at every whole number, with the box's ends at 8 and 17 and the median line at 10. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the first and third quartiles on the plot by recalling which parts of a box plot mark them, then subtract the smaller quartile from the larger.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • 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.
  • Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range
IQR=Q3−Q1\text{IQR} = Q_3 - Q_1
Q1, the first quartile, is the middle of the lower half of the data. Q3, the third quartile, is the middle of the upper half. In a box plot, Q1 is the left end of the box and Q3 is the right end. So the IQR is the width of the box.

  1. Sort out the question

    A box plot shows five numbers. The whiskers, the lines sticking out, end at the smallest and largest values. The box covers the middle half of the data. The line inside the box is the median, the middle value. You need the interquartile range of the ages.

  2. Read the five numbers

    Read each one off the number line below the plot.

    • Smallest age, the left whisker: 7
    • Q1, the left end of the box: 8
    • Median, the line inside the box: 10
    • Q3, the right end of the box: 17
    • Largest age, the right whisker: 18

  3. Find the interquartile range

    IQR=Q3−Q1\text{IQR} = Q_3 - Q_1

    IQR=17−8=9\text{IQR} = 17 - 8 = 9

    The interquartile range is 9 years. Watch out: 18 − 7 = 11 is the range, the whole spread from whisker to whisker. That is choice D. The median, 10, is choice C.

    So the answer is B, 9.

Check your answer

Count the spaces across the box. From 8 to 17 there are 9 spaces of 1 year each, so the box is 9 years wide. The answer is also less than the whole range, 11, as it must be.

More questions like this

The original DOE question
Question 108 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 89, question 108. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 89 of 120

2023 Test A Math, question 56

Question

Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. The line is shaded from the unlabeled tick halfway between 0 and 1 to the unlabeled tick halfway between 1 and 2, with closed dots at both ends.

If all possible values of x are indicated by the shaded part of the number line above, which number line best shows all possible values of 1x\frac{1}{x}?

  • ANumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 1 and 2 to the tick halfway between 2 and 3, with closed dots at both ends.
  • BNumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from 1 to 2, with closed dots at both ends.
  • CNumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 0 and 1 to 2, with closed dots at both ends.
  • DNumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from a closed dot between the tick halfway between 0 and 1 and the tick at 1, about two-thirds of the way from 0 to 1, to a closed dot at 2.
Show the answer

Correct answer: D

The shaded region goes from 12\frac{1}{2} to 32\frac{3}{2}. So, the possible values of 1x\frac{1}{x} would be between the reciprocal of 12\frac{1}{2} and the reciprocal of 32\frac{3}{2}, which are 2 and 23\frac{2}{3}, respectively. Find the number line in the answer options that has a shaded region that ranges from 23\frac{2}{3} to 2, which is option D.

Question analysis

This number line question shades all possible values of x from one half to three halves and asks which number line shows all possible values of 1/x. Students find reciprocals of fractions and work out how a whole range of values changes. Keeping track of what happens to each end of the range, and placing fractions accurately on the line, makes it tricky.

NYS standards
  • 6.NS.1 dividing fractions by fractions
  • 6.NS.7a inequality as relative position
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Reciprocals of single fractions are grade 6 work. Applying them to every value in a shaded range at once, and working out where the new range begins and ends, is reasoning about a whole set of numbers that grade 7 classes rarely practice.
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 12, 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

Few students have flipped an interval before, so the question rewards testing its two ends and noticing that the order reverses. The harder part comes at the end: 2/3 falls between the tick marks, and a graph ending at the nearby 1/2 tick looks right at a glance. Students who believe a reciprocal is always at least 1, or who shift the interval instead of flipping it, reach the other graphs.

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

    The shaded segment's ends sit on unlabeled half-unit ticks, 1/2 and 3/2, and each choice must be read the same way, with 2/3 falling between ticks rather than on one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. A very close wrong answer moderate Choosing the answer

    The graph from 1/2 to 2 matches the right graph's right end and differs at the left end by only a sixth of a unit, 1/2 against 2/3. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    The question asks about 1/x for every x in a range, so a whole interval must be transformed, and its ends swap: the smallest x gives the largest 1/x. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Take the reciprocal of each end of the shaded part, ask whether a bigger x gives a bigger or a smaller 1/x, and look for the graph whose endpoints are exactly those values.

Traps

  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reciprocal
reciprocal of ab=ba\text{reciprocal of } \frac{a}{b} = \frac{b}{a}
a and b are the top and bottom of a fraction. The reciprocal flips the fraction upside down. The number 1 over x is the reciprocal of x. A number times its reciprocal always makes 1.

  1. Sort out the question

    x can be any number in the shaded part of the first number line. You need the number line that shows every possible value of 1x\frac{1}{x}, the reciprocal of x. The reciprocal is the number flipped upside down, like 12\frac{1}{2} for 2.

  2. Read the shaded part

    The ticks come every half unit. The shading starts at the tick halfway between 0 and 1, which is 12\frac{1}{2}. It ends halfway between 1 and 2, which is 32\frac{3}{2}, or 1121\frac{1}{2}. Both dots are filled in, so both ends count.

  3. Find the reciprocals of the ends

    Flip each end of the shaded part.

    reciprocal of ab=ba\text{reciprocal of } \frac{a}{b} = \frac{b}{a}

    reciprocal of 12=21=2\text{reciprocal of } \frac{1}{2} = \frac{2}{1} = 2

    reciprocal of 32=23\text{reciprocal of } \frac{3}{2} = \frac{2}{3}

    So the ends of the new shading are 2 and 23\frac{2}{3}.

  4. Notice that the ends swap

    When x gets bigger, 1x\frac{1}{x} gets smaller. So the smallest x, 12\frac{1}{2}, gives the biggest value, 2. The biggest x, 32\frac{3}{2}, gives the smallest, 23\frac{2}{3}. So 1x\frac{1}{x} runs from 23\frac{2}{3} to 2, ends included.

  5. Find the matching number line

    23\frac{2}{3} lies between the half tick and 1, a third of the way from 12\frac{1}{2} to 1. Line D starts there and ends at 2. Watch out: line C starts at the half tick, 12\frac{1}{2}, not at 23\frac{2}{3}. Line B starts at 1.

    So the answer is D, the line shaded from 23\frac{2}{3} to 2.

Check your answer

Test a value in the middle. x = 1 is shaded, and 11=1\frac{1}{1} = 1. On line D, 1 is shaded too. Test an end: 32×23=1\frac{3}{2} \times \frac{2}{3} = 1, so 23\frac{2}{3} really is the reciprocal of 32\frac{3}{2}.

More questions like this

The original DOE question
Question 113 as printed on page 91 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 91, question 113. Open page 91 of the handbook (PDF)

Question 90 of 120

2023 Test B Math, question 2

Question

In a scale drawing of a rectangular garden, the length is 15 inches and the width is 9 inches. In the drawing, 2 inches represents 3 yards. What is the width of the actual garden, in yards?

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

Show the answer

Correct answer: 13.5

Let w represent the width of the actual garden. Set up a proportion based on the information given:

w9=32\frac{w}{9} = \frac{3}{2}

w=3(9)2=272=13.5w = \frac{3(9)}{2} = \frac{27}{2} = 13.5

Question analysis

In a scale drawing of a rectangular garden, 2 inches stands for 3 yards, and the drawing measures 15 inches by 9 inches. This grid-in asks for the real width of the garden in yards. Students apply the scale to one side only: the length is given but plays no part, and the scale has to be used the right way round.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Scale drawings are grade 7 content, and this is a typical one: a single length converted with a stated scale. The only extra demand is using the width rather than the length and entering the result with no choices to check.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The drawing gives two lengths and only the width matters, so the first decision is which number to scale. The second is direction: the real garden must measure more in yards than the drawing’s 9 in inches, which a student can check before entering anything. Errors come from scaling the length, turning the scale over, or treating 2 to 3 as an increase of one.

  1. Turning words into math minor Understanding the question

    “2 inches represents 3 yards” must become 1.5 yards for every inch of drawing, so the 9-inch width is multiplied by 3/2, not by 2/3. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The answer is a decimal entered with no choices to compare, so a result from the 15-inch length or from an upside-down scale looks just as finished. (Enright, Morley & Sheehan, 2002)

Strategy: Find how many yards one inch of the drawing stands for, then multiply the drawing's width, not its length, by that number.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scale proportion
real sizedrawing size=real sizedrawing size\frac{\text{real size}}{\text{drawing size}} = \frac{\text{real size}}{\text{drawing size}}
A scale drawing shrinks everything by the same amount. So real size over drawing size makes the same fraction for every part of the drawing. Keep the real sizes on top on both sides, and the drawing sizes on the bottom.

  1. Sort out the question

    The drawing shows a rectangle, a four-sided shape with square corners. On the drawing, the length is 15 inches and the width is 9 inches. The scale says 2 inches on the drawing stand for 3 yards in real life. You need the real width, in yards.

    Only the width matters. The 15-inch length is extra.

  2. Find what one inch stands for

    If 2 inches stand for 3 yards, then 1 inch stands for half as much.

    3÷2=1.53 \div 2 = 1.5

    So each inch on the drawing is 1.5 yards in real life. That means the number of yards will be bigger than the number of inches.

  3. Set up the proportion and solve

    Let w be the real width, in yards. Put yards on top and inches on the bottom, on both sides.

    w9=32\frac{w}{9} = \frac{3}{2}

    To get w alone, multiply both sides by 9.

    w=3×92=272=13.5w = \frac{3 \times 9}{2} = \frac{27}{2} = 13.5

    The real width is 13.5 yards.

  4. Give the answer

    Watch out: don’t flip the scale. Using 9×239 \times \frac{2}{3} gives 6, which is smaller than the drawing’s number. And don’t use the length: 15 inches would give 22.5 yards. So you would grid 13.5.

Check your answer

Count groups of 2 inches. The 9-inch width holds 4.5 groups of 2 inches. Each group stands for 3 yards, so the real width is 4.5×3=13.54.5 \times 3 = 13.5 yards. That matches.

More questions like this

The original DOE question
Question 59 as printed on page 191 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 59. Open page 191 of the handbook (PDF)

Question 91 of 120

2023 Test B Math, question 3

Question

A kindergarten teacher has 72 crayons, 60 pencils, and 84 sheets of paper to distribute to the children in her class. If each child receives an equal number of each item and there are no items remaining, what is the greatest possible number of children in the class?

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

Show the answer

Correct answer: 12

The greatest possible number of children in the class is the greatest common factor of 72, 60, and 84. Write the prime factorization of each number:

72=23×3272 = 2^{3} \times 3^{2}

60=22×3×560 = 2^{2} \times 3 \times 5

84=22×3×784 = 2^{2} \times 3 \times 7

Then, find the factors that all three have in common, which is 22×3=122^{2} \times 3 = 12.

Question analysis

A kindergarten teacher shares 72 crayons, 60 pencils and 84 sheets of paper so that every child gets the same number of each item, with nothing left over. This grid-in asks for the greatest possible number of children in the class. Students must recognize a greatest common factor problem and not confuse it with a least common multiple.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Greatest common factors are grade 6 content, normally of two numbers. This one uses three numbers and a word setting that must be translated, but the factoring is short, so the demand stays below typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 45 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The story has to be recognized as a sharing problem before any arithmetic starts, and the word greatest points to the largest divisor shared by all three counts. Students who spot a small shared factor can stop too soon, while those who think of a common multiple head the wrong way entirely. Checking that 72, 60 and 84 all divide by the entry, and that no larger number does, settles it.

  1. Turning words into math moderate Understanding the question

    Nothing names a greatest common factor; “each child receives an equal number of each item and there are no items remaining” means the class size must divide 72, 60 and 84 exactly. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Finding the greatest common factor of three numbers, by prime factorization or by listing factors, is a grade 6 method the question assumes rather than explains. (Anderson & Krathwohl, 2001)

Strategy: Write 72, 60 and 84 as products of primes, multiply the primes all three share, and check that each amount divides evenly by the result.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor
GCF=product of the shared prime factors\text{GCF} = \text{product of the shared prime factors}
A factor of a number divides it with nothing left over. The GCF is the greatest number that is a factor of every number in the group. A prime number has only 1 and itself as factors, like 2, 3, 5 and 7. Multiply the primes that all the numbers share, as many times as each number has them.

  1. Sort out the question

    The teacher has 72 crayons, 60 pencils and 84 sheets of paper. Every child gets the same number of each item, with none left over.

    So the number of children must divide 72, 60 and 84 with no remainder. A number like that is a common factor. You need the greatest one.

  2. Break each number into primes

    Write each number as a product of primes. A product is the answer to a multiplication.

    72=2×2×2×3×372 = 2 \times 2 \times 2 \times 3 \times 3

    60=2×2×3×560 = 2 \times 2 \times 3 \times 5

    84=2×2×3×784 = 2 \times 2 \times 3 \times 7

    You can check each one by multiplying it back out.

  3. Pick out the shared primes

    Look for the primes that appear in all three lists. Each list has at least two 2s and one 3. Only 72 has a third 2 and a second 3. The 5 and the 7 appear in one list each.

    GCF=2×2×3=12\text{GCF} = 2 \times 2 \times 3 = 12

    Watch out: keep every shared prime. Using only one 2 gives 6. Leaving out the 3 gives 4.

  4. Give the answer

    Watch out: 2,520 is the smallest number that all three numbers divide into. That is the least common multiple, a different idea. The class can’t have 2,520 children when there are only 60 pencils. So you would grid 12.

Check your answer

Share the items among 12 children. Each child gets 72÷12=672 \div 12 = 6 crayons, 60÷12=560 \div 12 = 5 pencils and 84÷12=784 \div 12 = 7 sheets. Nothing is left over. The numbers 6, 5 and 7 share no factor but 1, so no bigger class works.

More questions like this

The original DOE question
Question 60 as printed on page 191 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 60. Open page 191 of the handbook (PDF)

Question 92 of 120

2023 Test B Math, question 4

Question

Mr. Smith opened a retirement account with a deposit of $900. This account earns 5% simple interest annually. How many years will it take for his $900 deposit to earn $360 interest?

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

Show the answer

Correct answer: 8

Calculate simple interest (y) by multiplying the initial deposit (p) by the interest rate (r) by the number of years (t):

prt=yprt = y

900(0.05)t=360900(0.05)t = 360

45t=36045t = 360

t=8t = 8

Question analysis

Mr. Smith deposits $900 in a retirement account that earns 5% simple interest each year. This grid-in asks how many years it takes for the deposit to earn $360 in interest. Students find one year's interest and then how many years add up to the total, and the percent must become the right decimal.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Simple interest is grade 7 percent work, and a typical problem asks for the interest earned over a given time. This one runs the formula backward to find the time, one extra step, with friendly numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 7, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are friendly once one year’s interest is found, so most of the difficulty is setting up the right relationship. Students who divide $360 by 5 treat the rate as a dollar amount, and students who compare $360 with $900 stop at a percent. Multiplying back, eight years of $45, confirms the $360.

  1. Something you must already know minor Working it out

    Simple interest is the deposit times the rate times the years; the stem does not give this rule, and it must be known to see that each year adds the same $45. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The question gives the total interest, $360, and asks for the time, so the usual interest calculation has to be run backward, as 45t = 360. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Writing 5% as 0.05, not 0.5 or 0.005, is the one place-value decision, and a slip there changes the yearly interest by a power of ten. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work out the interest for one year, 5 percent of $900, then divide the $360 target by that yearly amount.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest
I=P×r×tI = P \times r \times t
I is the interest, the money the account earns. P is the deposit you start with. r is the yearly rate, written as a decimal. t is the number of years. With simple interest, every year earns the same amount, because it is always based on the first deposit.

  1. Sort out the question

    Mr. Smith puts $900 into an account. It earns 5% simple interest each year. Interest is extra money the bank pays you for keeping your money there. You need the number of years until the interest adds up to $360.

  2. Write 5% as a decimal

    Percent means “out of 100.” So 5% is 5 out of 100.

    5100=0.05\frac{5}{100} = 0.05

    Watch out: 5% is not 0.5 or 0.005. Using 0.005 makes a year’s interest $4.50, and that leads to 80 years.

  3. Find the interest for one year

    Each year the account earns 5% of $900.

    900×0.05=45900 \times 0.05 = 45

    So it earns $45 every year. Watch out: 5% is not $5 a year. Dividing 360 by 5 gives 72, which is wrong.

  4. Use the formula to find the years

    Put the numbers into the formula. I is 360, P is 900 and r is 0.05.

    I=P×r×tI = P \times r \times t

    360=900×0.05×t360 = 900 \times 0.05 \times t

    360=45×t360 = 45 \times t

    To undo “times 45,” divide both sides by 45.

    t=360÷45=8t = 360 \div 45 = 8

  5. Give the answer

    It takes 8 years. Watch out: $360 is 40% of $900. Gridding 40 stops too soon, because each year adds only 5%. So you would grid 8.

Check your answer

Add up the interest year by year. $45 a year for 8 years is 45×8=36045 \times 8 = 360 dollars. Also, 8 years at 5% a year is 40%, and 40% of $900 is $360.

More questions like this

The original DOE question
Question 61 as printed on page 192 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 61. Open page 192 of the handbook (PDF)

Question 93 of 120

2023 Test B Math, question 9

Question

Bryan completed a 100-meter race in 11.74 seconds. Luis completed the same race in 11.69 seconds. What was the difference between their times, expressed as a fraction of a second?

  • A120\frac{1}{20}
  • B110\frac{1}{10}
  • C320\frac{3}{20}
  • D15\frac{1}{5}
Show the answer

Correct answer: A

First, subtract Luis’s time from Bryan’s time: 11.74−11.69=0.0511.74 - 11.69 = 0.05.

Then, convert to a fraction:

0.05=5100=1200.05 = \frac{5}{100} = \frac{1}{20}.

Question analysis

Bryan ran a 100-meter race in 11.74 seconds and Luis ran the same race in 11.69 seconds. Students find the difference between their times and express it as a fraction of a second. It tests subtracting decimals and rewriting a decimal as a fraction in lowest terms.

NYS standards
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Subtracting decimals and writing a decimal as a fraction are skills from before grade 7. The two times are close, so the subtraction is quick, and the only care needed is with place value and simplifying.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 5, 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

Two quick skills settle this question: lining up the decimals to subtract, and naming the result as a fraction in lowest terms. The two times differ only in their last two digits, so the subtraction is short, but it needs a borrow that some students skip. The fraction step is where a decimal’s digit gets mistaken for a denominator.

  1. Error-prone arithmetic minor Working it out

    Subtracting 11.69 from 11.74 needs a borrow in the hundredths, 14 − 9, after which the tenths cancel to 0, the step where place value goes wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    The difference, 0.05, must be read as five hundredths and reduced as a fraction; the question gives no help with converting a decimal. (Anderson & Krathwohl, 2001)

Strategy: Line up the decimal points, subtract, write the difference as a number of hundredths over 100, and reduce the fraction.

Traps

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

Decimals as hundredths
a decimal with two places=its digits100\text{a decimal with two places} = \frac{\text{its digits}}{100}
The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths, and you can write its digits over 100. For example, 0.37 is 37/100. Then simplify by dividing the top and bottom by the same number.

  1. Sort out the question

    Bryan ran the race in 11.74 seconds. Luis ran it in 11.69 seconds. The difference is how much longer one time is than the other. You need it as a fraction of a second.

  2. Subtract the two times

    Subtract the smaller time from the larger one. Line up the decimal points.

    11.74−11.69=0.0511.74 - 11.69 = 0.05

    In the hundredths place, 4 is less than 9. So regroup: trade 1 tenth for 10 hundredths. Then 14 − 9 = 5 hundredths. The tenths are now 6 − 6 = 0.

    Watch out: don’t just take the smaller digit from the larger one in each place. That gives 0.15, which is 320\frac{3}{20}, choice C.

  3. Write the difference as a fraction

    The 5 is in the hundredths place, so 0.05 is 5 hundredths.

    0.05=51000.05 = \frac{5}{100}

    Now simplify. Divide the top and bottom by 5.

    5÷5100÷5=120\frac{5 \div 5}{100 \div 5} = \frac{1}{20}

  4. Give the answer

    Watch out: 0.05 is not 15\frac{1}{5}. One fifth is 0.2, which is four times as big. So the answer is A, 120\frac{1}{20}.

Check your answer

Turn 120\frac{1}{20} back into a decimal: 1÷20=0.051 \div 20 = 0.05. Then add it to Luis’s time: 11.69+0.05=11.7411.69 + 0.05 = 11.74. That is Bryan’s time, so the difference is right.

More questions like this

The original DOE question
Question 66 as printed on page 193 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 66. Open page 193 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 94 of 120

2023 Test B Math, question 10

Question

A total of 24 students in a math class took a 100-point test:

  • None of the students scored 100 points.
  • 75% of the students scored less than 82 points.
  • The lowest score in the class was 54 points.

Which box plot best represents the test scores from this math class?

  • ABox plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 54, the box runs from 60 to about 72 with the median line at about 63, and the right whisker ends at about 82.5, halfway between 80 and 85.
  • BBox plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 54, the box runs from about 62.5 to about 82.5 with the median line at about 71, and the right whisker ends at about 98.
  • CBox plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 51, the box runs from 60 to about 83 with the median line at about 74, and the right whisker ends at about 99.
  • DBox plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 54, the box runs from about 63 to about 83.5 with the median line at 70, and the right whisker ends at 100.
Show the answer

Correct answer: B

Use the information in the bullet points to reason which box plot could represent the data set of test scores.

Since none of the students scored 100, the maximum value on the box plot must be less than 100.

If 75% of the students scored less than 82, then 25% scored 82 points or more. Since the third quartile (Q3) indicates 34\frac{3}{4} (75%) of the data is to the left and 14\frac{1}{4} (25%) of the data is to the right, it is reasonable that Q3 would be at or about 82.

Given that the lowest score in the class was 54, the minimum point on the box plot must be 54.

The box plot graph in option B is the only one that meets all three criteria.

Question analysis

Twenty-four students took a 100-point test: nobody scored 100, 75% scored below 82, and the lowest score was 54. Students choose the box plot that fits all three facts. The skill is linking each statement to a part of the plot, the whiskers, the box and the quartiles, and checking every plot closely.

NYS standards
  • 7.SP.1 box plots and interquartile range
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students usually read values off a box plot. This item runs the other way, turning a percent statement into a quartile and checking the whiskers against two more facts, but the content stays at grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 2, 45 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns three sentences into three features of a box plot, and a plot that matches two of them is easy to accept. The upper-quartile fact is the conceptual one: students who do not know that the box’s right edge marks 75 percent put 82 at the end of the whisker. The other two facts are quick visual checks, but only if every plot is actually checked.

  1. Every choice must be checked moderate Choosing the answer

    Each wrong plot fails a different fact, so all four must be tested against all three bullets: the left whisker against 54, the right whisker against 100, and the box’s right edge against 82. (Embretson & Wetzel, 1987)

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

    The axis is labeled every 5 points, and the deciding features fall between labels: one left whisker starts near 51 instead of 54, and box edges sit near 82.5. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Something you must already know minor Working it out

    Nothing explains that “75% of the students scored less than 82” locates the upper quartile, the right edge of the box, rather than the highest score. (Anderson & Krathwohl, 2001)

Strategy: Turn each fact into a feature to check: where the left whisker starts, where the right whisker ends, and where the box ends. Rule out any plot that breaks one.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts of a box plot
minimum, Q1, median, Q3, maximum\text{minimum},\ Q_1,\ \text{median},\ Q_3,\ \text{maximum}
A box plot shows five numbers. The left whisker starts at the minimum, the lowest value. The box starts at Q1, the first quartile, and ends at Q3, the third quartile. The line inside is the median, the middle value. The right whisker ends at the maximum. About a quarter of the data falls in each of the four sections.

  1. Sort out the question

    24 students took a test. You know three facts. Nobody scored 100. 75% scored less than 82. The lowest score was 54. You need the box plot that fits all three facts.

  2. Turn each fact into a feature

    Match each fact to a part of the box plot.

    • The lowest score is 54, so the left whisker starts at 54.
    • Nobody scored 100, so the right whisker ends before 100.
    • 75% is three quarters. Three quarters of the scores are below Q3. So the box should end at about 82.

  3. Read the number line

    The ticks are 5 points apart: 50, 55, 60 and so on. A score of 54 sits just left of the 55 tick. A score of 82 sits a little less than halfway from 80 to 85. Look closely, because these points fall between labels.

  4. Check plot A

    In plot A, the right whisker ends near 82, and the box ends near 72. That would make 82 the highest score. Watch out: 75% below 82 does not make 82 the top score. It tells you where the box ends. So A is out.

  5. Check plot C

    In plot C, the left whisker starts near 51, just right of the 50 tick. But the lowest score was 54. So C is out.

  6. Check plot D

    Plot D starts at 54, and its box ends near 82. But its right whisker reaches 100. That means someone scored 100, and nobody did. Watch out: D fits two facts, so it is easy to stop early. Check all three. So D is out.

  7. Check plot B and answer

    Plot B starts at 54. Its box ends near 82. Its right whisker stops near 98, below 100. It fits all three facts. So the answer is B.

Check your answer

Go over B once more with all three facts. Its lowest point is 54. Its box ends near 82. Its highest point is under 100. Each other plot breaks one fact: A’s box, C’s left whisker and D’s right whisker.

More questions like this

The original DOE question
Question 67 as printed on page 194 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 67. Open page 194 of the handbook (PDF)

Question 95 of 120

2023 Test B Math, question 13

Question

Shiow-Chen has a box that contains 4 green cards, 7 yellow cards, and 6 red cards. She randomly draws one card at a time from the box and does not return the cards to the box after they have been drawn. The first card she draws is yellow. The second card she draws is green. What is the probability that the third card she draws will also be green?

  • A15\frac{1}{5}
  • B417\frac{4}{17}
  • C14\frac{1}{4}
  • D13\frac{1}{3}
Show the answer

Correct answer: A

After Shiow-Chen draws a yellow card and a green card, the remaining cards are 3 green, 6 yellow, and 6 red. The probability of drawing a green card from that group is:

33+6+6=315=15\frac{3}{3 + 6 + 6} = \frac{3}{15} = \frac{1}{5}

Question analysis

Shiow-Chen's box holds 4 green, 7 yellow and 6 red cards, and she draws cards one at a time without putting them back. Her first card is yellow and her second is green. Students find the probability that the third card is also green, which means updating both the green count and the total first.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Simple probability is grade 6 content. The extra demand is tracking two earlier draws that were not put back, which changes the counts for the third draw, a step typical grade 7 problems often leave out.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 70 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The probability itself is one fraction, but it has to be built from a box that two draws have already changed. Students who use the original 17 cards, or who compare green with the other colors instead of with all the cards, reach listed choices that look reasonable. Listing the counts after each draw, then putting the green cards over all the cards left, avoids both.

  1. Much to keep track of moderate Working it out

    Three color counts and the total must each be updated after two draws, 4 green, 7 yellow and 6 red becoming 3, 6 and 6 out of 15, before the third draw is considered. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    “Does not return the cards” and the two named draws are the conditions that change the box, and the stem states them in passing before asking about the third card. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Take the two drawn cards out of the box: lower the yellow and green counts by one each, recount the total, then put the green cards left over the cards left.

Traps

  • Without replacement: In a second pick there is one fewer item, which changes the probability.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of results you wanttotal number of resultsP = \frac{\text{number of results you want}}{\text{total number of results}}
P is the probability, how likely something is, from 0 (never) to 1 (always). The top counts the results you want, here green cards. The bottom counts every possible result, here every card in the box. It works when each result is equally likely, as in a random draw.

  1. Sort out the question

    The box starts with 4 green, 7 yellow and 6 red cards. Shiow-Chen draws cards one at a time, and she does not put them back. The first card is yellow, and the second is green. You need the probability that the third card is green.

  2. Take out the two cards

    The cards are not put back, so the box changes. Lower the yellow count by 1 and the green count by 1.

    ColorAt the startAfter two draws
    Green43
    Yellow76
    Red66
    Total1715

  3. Use the probability formula

    For the third draw, the results you want are the 3 green cards left. The total is the 15 cards left.

    P=number of results you wanttotal number of resultsP = \frac{\text{number of results you want}}{\text{total number of results}}

    P=33+6+6=315P = \frac{3}{3 + 6 + 6} = \frac{3}{15}

    Watch out: using the full box gives 417\frac{4}{17}, choice B. But two cards are gone.

  4. Simplify and answer

    Divide the top and bottom by 3.

    315=15\frac{3}{15} = \frac{1}{5}

    Watch out: compare the green cards with all 15 cards, not with the 12 cards of other colors. That gives 14\frac{1}{4}, choice C. And the three colors are not equally likely, so 13\frac{1}{3} is wrong too. So the answer is A, 15\frac{1}{5}.

Check your answer

17 cards minus the 2 drawn leaves 15. One fifth of 15 is 3, the number of green cards left. Also, the chance went down from 417\frac{4}{17}. That makes sense, because a green card was taken out.

More questions like this

The original DOE question
Question 70 as printed on page 195 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 70. Open page 195 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 96 of 120

2023 Test B Math, question 17

Question

m∙t=72m \bullet t = 72

For the equation above, what is the sum of the values of m when t=1t = 1, when t=2t = 2, and when t=3t = 3?

  • A12
  • B72
  • C132
  • D135
Show the answer

Correct answer: C

First, find the value of m for each value of t:

For t=1t = 1:

1m=721m = 72

m=72m = 72

For t=2t = 2:

2m=722m = 72

m=36m = 36

For t=3t = 3:

3m=723m = 72

m=24m = 24

Then, add those three values together:

72+36+24=13272 + 36 + 24 = 132

Question analysis

The equation m • t = 72 links two variables, and the raised dot means multiplication. Students find m when t is 1, 2 and 3, and then add those three values of m. Each value is a one-step division, and the trap is combining the t values before dividing.

NYS standards
  • 6.EE.7 one-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Solving equations like 3m = 72 is grade 6 work. This item repeats that step three times and adds the results, more bookkeeping than difficulty, so it is below typical grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 7, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each value of m comes from one division, so the work is bookkeeping rather than algebra. The trap is structural: adding 1, 2 and 3 first and dividing once feels efficient but answers a different question, since m does not change in step with t. Writing the three values of m in a row before adding them keeps the steps separate.

  1. Turning words into math minor Understanding the question

    “The sum of the values of m when t = 1, when t = 2, and when t = 3” asks for three separate solutions added together, not one solution at a combined t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three divisions, 72 ÷ 1, 72 ÷ 2 and 72 ÷ 3, and an addition make four small steps, each easy but each a chance to drop a value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Divide 72 by each value of t in turn to get three separate values of m, then add the three quotients.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division undoes multiplication
if a×b=c, then a=c÷b\text{if } a \times b = c \text{, then } a = c \div b
a, b and c are numbers, and b is not 0. If you know a product and one of the numbers that made it, divide to find the other one. For example, 4 × 5 = 20, so 4 = 20 ÷ 5.

  1. Sort out the question

    The equation is m∙t=72m \bullet t = 72. The dot means times. You need m three times: when t is 1, when t is 2 and when t is 3. Then add the three values of m.

  2. Get m by itself

    Since m times t is 72, m is 72 divided by t.

    m=72÷tm = 72 \div t

    Now put in each value of t, one at a time.

  3. Find m for each value of t

    Each time, divide 72 by the value of t. A table keeps the three answers in order.

    Value of tValue of m
    1m=72÷1=72m = 72 \div 1 = 72
    2m=72÷2=36m = 72 \div 2 = 36
    3m=72÷3=24m = 72 \div 3 = 24

  4. Add the three values

    72+36+24=13272 + 36 + 24 = 132

    Watch out: don’t add the t values first. 72÷(1+2+3)=1272 \div (1 + 2 + 3) = 12 is choice A, and it is not the same thing. Also, 72 alone is only the first value, choice B. So the answer is C, 132.

Check your answer

Check each pair by multiplying: 72×1=7272 \times 1 = 72, 36×2=7236 \times 2 = 72 and 24×3=7224 \times 3 = 72. All three pairs make 72. Then add again: 72 + 36 = 108, and 108 + 24 = 132.

More questions like this

The original DOE question
Question 74 as printed on page 196 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 74. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 97 of 120

2023 Test B Math, question 20

Question

Five consecutive numbers have a sum of 110. What is the least of these consecutive numbers?

  • A11
  • B18
  • C20
  • D22
Show the answer

Correct answer: C

Let n represent the least of the five numbers. Then the next four consecutive numbers are (n+1)(n + 1), (n+2)(n + 2), (n+3)(n + 3), (n+4)(n + 4). Write an equation to solve for n:

n+(n+1)+(n+2)+(n+3)+(n+4)=110n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 110

5n+10=1105n + 10 = 110

5n=1005n = 100

n=20n = 20

Question analysis

Five consecutive numbers add up to 110, and students find the least of them. This number-properties question can be solved with an equation in one unknown, or by noting that the middle of an odd number of consecutive numbers equals their average. The bold word least matters, because the middle number is easier to find.

NYS standards
  • 7.EE.4a two-step equations
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 A sum of consecutive numbers leads to a two-step equation, standard grade 7 work. The average shortcut makes it quicker, and the only extra demand is answering for the least number rather than the middle one.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Dividing 110 by 5 is the natural first move, and it gives the middle number, which is a listed choice. The bold word least signals that one more step is needed, counting down two from the middle. The equation route, 5n + 10 = 110, goes straight to the least number but takes longer to set up.

  1. Turning words into math minor Understanding the question

    “Consecutive numbers” means each is 1 more than the last, and the bold “least” asks for the first of the five, not the middle one that dividing by 5 gives. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    None of the five numbers is given, so they must be written as n through n + 4, or pictured as two on each side of a middle value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Divide 110 by 5 to find the middle of the five numbers, then count down to the least of them.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive numbers
n, n+1, n+2, n+3, n+4n,\ n + 1,\ n + 2,\ n + 3,\ n + 4
Consecutive numbers come one right after another when you count, like 7, 8, 9. Each is 1 more than the one before. If n is the least number, the next ones are n + 1, n + 2, and so on.

  1. Sort out the question

    Five consecutive numbers add up to 110. Consecutive means each number is 1 more than the one before. You need the least, or smallest, of the five.

  2. Name the five numbers

    Let n be the least number. Then the five numbers are:

    n, n+1, n+2, n+3, n+4n,\ n + 1,\ n + 2,\ n + 3,\ n + 4

    Each one is 1 more than the one before it.

  3. Write the sum as an equation

    Add all five and set the sum equal to 110.

    n+(n+1)+(n+2)+(n+3)+(n+4)=110n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 110

    There are five n’s, which make 5n. The extra numbers add up to 1 + 2 + 3 + 4 = 10.

    5n+10=1105n + 10 = 110

  4. Solve for n

    Subtract 10 from both sides. Then divide both sides by 5.

    5n=1005n = 100

    n=100÷5=20n = 100 \div 5 = 20

    So the least number is 20.

  5. Give the answer

    Watch out: 110÷5=22110 \div 5 = 22 is the middle number, not the least. Picking 22 is choice D. The question asks for the least of the five. So the answer is C, 20.

Check your answer

Write out the five numbers and add them: 20+21+22+23+24=11020 + 21 + 22 + 23 + 24 = 110. The sum matches. Also, the middle number, 22, is 110 ÷ 5, and the least is 2 less than the middle.

More questions like this

The original DOE question
Question 77 as printed on page 197 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 77. Open page 197 of the handbook (PDF)

Question 98 of 120

2023 Test B Math, question 21

Question

−mn-mn

(−m)(−n)(-m)(-n)

|mn||mn|

|−mn||-mn|

|(−m)(−n)||(-m)(-n)|

If m and n are positive integers, how many of the five expressions above are equivalent to mnmn?

  • A1
  • B2
  • C3
  • D4
Show the answer

Correct answer: D

Since m and n are positive, the value of mnmn is also positive. Of the five options given:

−mn-mn is negative.

(−m)(−n)(-m)(-n) is positive.

|mn||mn| is positive.

|−mn||-mn| is positive.

|(−m)(−n)||(-m)(-n)| is positive.

There are 4 expressions that are positive, and each one is equal to mnmn.

Question analysis

Five expressions are listed: −mn, (−m)(−n), |mn|, |−mn| and |(−m)(−n)|. If m and n are positive integers, students count how many are equivalent to mn. It tests the sign rules for multiplying negatives together with what absolute value does, and each expression has to be judged on its own.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.NS.7c absolute value as distance
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Multiplying signed numbers and absolute value are grade 6 and 7 content. Checking five expressions one by one takes longer than a typical single computation, but each check uses a familiar rule.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
5 steps
Reading level
Question and choices: about grade 10, 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

Each expression takes only a moment, but the answer is a count, so one misjudged expression changes it. The minus signs are designed to unsettle: four of the five expressions show at least one, and only one of those actually stays negative. Substituting small positive numbers and comparing each value with mn turns the question into arithmetic.

  1. Variables instead of numbers moderate Understanding the question

    The five expressions use only m and n, so each sign rule has to be applied in general unless the student substitutes values such as m = 2 and n = 3. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    Five expressions are judged one at a time and a count kept, and two of them combine the product rule for negatives with absolute value bars. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Something you must already know minor Working it out

    The rules that two negative factors give a positive product and that absolute value removes a minus sign are assumed, not stated. (Anderson & Krathwohl, 2001)

Strategy: Pick small values such as m = 2 and n = 3, evaluate each of the five expressions, and count the ones equal to 6.

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.)
  • 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

Multiplying two negatives
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers. A negative number times a negative number gives a positive number. The two minus signs cancel out. One negative and one positive give a negative.
Absolute value
|a|=the distance from a to 0|a| = \text{the distance from } a \text{ to } 0
The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so |−6| = 6 and |6| = 6. The bars remove any minus sign.

  1. Sort out the question

    m and n are positive integers. That means whole numbers greater than 0. Writing mn means m times n, so mn is positive. You need to count how many of the five expressions always equal mn.

  2. Pick numbers to test

    Pick easy values: m=2m = 2 and n=3n = 3. Then mn=2×3=6mn = 2 \times 3 = 6. Any expression that equals mn must come out to 6.

  3. Test the first two expressions

    −mn=−(2×3)=−6-mn = -(2 \times 3) = -6

    This is negative, so it does not equal 6.

    (−m)(−n)=(−2)(−3)=6(-m)(-n) = (-2)(-3) = 6

    A negative times a negative is positive, so this one equals 6. Watch out: if you think a negative times a negative stays negative, you reject this one and get 3, choice C.

  4. Test the three absolute values

    |mn|=|6|=6|mn| = |6| = 6

    |−mn|=|−6|=6|-mn| = |-6| = 6

    |(−m)(−n)|=|6|=6|(-m)(-n)| = |6| = 6

    All three equal 6, because absolute value bars remove a minus sign. Watch out: the bars work even when the minus sign is inside them. Rejecting the last two gives 2, choice B.

  5. Count the matches

    Four expressions equal mn: (−m)(−n)(-m)(-n), |mn||mn|, |−mn||-mn| and |(−m)(−n)||(-m)(-n)|. Only −mn-mn does not. The sign rules work the same for any positive m and n, not just 2 and 3. So the answer is D, 4.

Check your answer

Try other numbers, such as m=1m = 1 and n=5n = 5. Then mn=5mn = 5. The five expressions give −5, 5, 5, 5 and 5. Again, four of them match mn.

More questions like this

The original DOE question
Question 78 as printed on page 197 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 78. Open page 197 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 99 of 120

2023 Test B Math, question 29

Question

Which number could be the sum of 4 consecutive integers?

  • A−12
  • B−10
  • C−8
  • D−4
Show the answer

Correct answer: B

The sum of 4 consecutive integers can be expressed as:

n+(n+1)+(n+2)+(n+3)n+(n+1)+(n+2)+(n+3)

=4n+6=4n+6

Set that expression equal to each of the options and solve for n:

Option A:

4n+6=−124n+6=-12

4n=−184n=-18

n=−184n=-\frac{18}{4} is not an integer.

Option B:

4n+6=−104n+6=-10

4n=−164n=-16

n=−4n=-4 is an integer.

Option C:

4n+6=−84n+6=-8

4n=−144n=-14

n=−144n=-\frac{14}{4} is not an integer.

Option D:

4n+6=−44n+6=-4

4n=−104n=-10

n=−104n=-\frac{10}{4} is not an integer.

The only option for which n is an integer is −10.

Question analysis

A number properties question about consecutive integers: which of four negative numbers could be the sum of 4 consecutive integers? Students write the sum with a variable, or add runs of negative integers, and test each choice. Keeping signs straight with negative numbers, and remembering that the integers must be whole numbers, is the tricky part.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 consecutive-integer problems usually give the sum and ask for the integers. This one reverses the task: students test which negative total is possible at all, using the structure of the sum rather than one equation.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 9 words.
Corrected from the DOE handbook
  • A stray word in the explanation’s last sentence was removed.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Three of the four choices are multiples of 4, and that is the whole trap: a student who pictures the sum as four times some integer finds three choices that seem to work. The sum of four consecutive integers is 4n + 6, so it is always even but never a multiple of 4. Trying the most natural run, −4, −3, −2 and −1, lands on −10 at once and is the fastest check.

  1. Every choice must be checked moderate Choosing the answer

    The stem asks which number “could be” the sum, so no single equation gives the answer; each of −12, −10, −8 and −4 has to be tested against the form a four-integer sum must take. (Embretson & Wetzel, 1987)

  2. No obvious method minor Working it out

    Nothing points to the shortcut that n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 is always 2 more than a multiple of 4, or to simply adding a run such as −1, −2, −3 and −4. (Schoenfeld, 1985)

  3. Error-prone arithmetic minor Working it out

    Every choice is negative, so solving 4n + 6 = −12 or adding a run of integers means subtracting 6 from a negative number and dividing a negative by 4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the integers as n, n + 1, n + 2 and n + 3, add them, and set the sum equal to each choice to see which one gives a whole-number n.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook ends with “The only option for which n is an integer answer is −10.” The word “answer” does not belong in that sentence; the site reads “The only option for which n is an integer is −10.” The working above it is unchanged: only 4n + 6 = −10 gives an integer, n = −4, and −4 + (−3) + (−2) + (−1) = −10.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of 4 consecutive integers
n+(n+1)+(n+2)+(n+3)=4n+6n + (n + 1) + (n + 2) + (n + 3) = 4n + 6
n is the smallest of the four integers. Each next integer is 1 more than the one before, so they are n, n + 1, n + 2 and n + 3. Adding them gives four n’s, plus 1 + 2 + 3, which is 6. Use it to add any four integers in a row.

  1. Sort out the question

    Integers are whole numbers and their negatives, like −2, 0 and 5. They have no fractions or decimals. Consecutive integers come one right after another, like 7, 8, 9, 10 or −3, −2, −1, 0.

    You need the choice that can be the sum of four consecutive integers. The sum is what you get when you add them.

  2. Write the sum with a letter

    You don’t know the integers yet. So call the smallest one n. Each next one is 1 more: n + 1, n + 2 and n + 3. Use the formula. Collect the four n’s, then add the numbers.

    n+n+n+n+1+2+3=4n+6n + n + n + n + 1 + 2 + 3 = 4n + 6

    Any four consecutive integers add up to 4n + 6.

  3. Test each choice

    Set 4n + 6 equal to each choice and solve for n. First subtract 6 from both sides. Then divide both sides by 4.

    Choice4n + 64nn
    A−12−18−4.5
    B−10−16−4
    C−8−14−3.5
    D−4−10−2.5

    n must be an integer. Only choice B gives one: n = −4.

  4. Name the four integers

    Start at n = −4 and count up by 1. The integers are −4, −3, −2 and −1.

    Watch out: Choices A, C and D are multiples of 4. A multiple of 4 is 4 times an integer, like 4 × (−3) = −12. It is tempting to think four integers must add to one of those. But the sum is 4n + 6, and the extra 6 gets in the way. Choice D works only with n = −2.5, and that is not an integer.

    So the answer is B, −10.

Check your answer

Add the integers: −4+(−3)+(−2)+(−1)=−10-4 + (-3) + (-2) + (-1) = -10. That is choice B. Also, 4n + 6 is always 2 more than a multiple of 4. And −10 is 2 more than −12, which is 4 × (−3).

More questions like this

The original DOE question
Question 86 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 86. Open page 200 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 100 of 120

2023 Test B Math, question 31

Question

If xy=30xy=30, what is the value of 10÷x10\div x in terms of y?

  • A110y\frac{1}{10}y
  • B13y\frac{1}{3}y
  • C3y3y
  • D10y10y
Show the answer

Correct answer: B

Solve xy=30xy=30, the given equation, for x:

xy=30xy=30

x=30yx=\frac{30}{y}

Then:

10÷x=10÷30y=10×y30=13y10\div x=10\div\frac{30}{y}=10\times\frac{y}{30}=\frac{1}{3}y

Question analysis

An algebra question: given that x times y equals 30, students express 10 divided by x using only y. It tests solving for one variable and dividing by an algebraic fraction. Working with letters instead of numbers, and dividing in the right order, are what trip students up.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.1 dividing fractions by fractions
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students rewrite expressions that have numbers and one variable. This one links two variables through their product and asks for a quotient, with no numbers given to check an answer against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 7, 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 algebra is three short moves, but each is a place to turn a fraction upside down: solving xy = 30 for x, dividing 10 by that fraction, and reducing 10/30. Students who reach for the bigger-over-smaller division they know from whole numbers land on 3y. Choosing two numbers with a product of 30, such as 3 and 10, and testing every choice settles the question in seconds.

  1. Variables instead of numbers moderate Understanding the question

    x and y are tied only by their product, 30, and the answer must be an expression in y, so there is no number to check unless students pick a pair such as x = 3 and y = 10. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Substituting x = 30/y turns 10 ÷ x into 10 divided by a fraction, which must be flipped to 10 × y/30 and then reduced. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. A very close wrong answer minor Choosing the answer

    The answer, (1/3)y, and the choice 3y use the same 3, one dividing y and one multiplying it, so a student unsure whether 10 ÷ 30 or 30 ÷ 10 applies sees both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Solve xy = 30 for x, substitute that fraction into 10 ÷ x, and divide by multiplying by its reciprocal; or choose values of x and y with a product of 30 and test each choice.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction
a÷bc=a×cba \div \frac{b}{c} = a \times \frac{c}{b}
a is the number you divide. b over c is the fraction you divide by: b is its top and c is its bottom. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. For example, 6 ÷ 1/2 = 6 × 2 = 12, because 12 halves fit in 6.

  1. Sort out the question

    xyxy means x times y. So x times y is 30. You need 10÷x10 \div x “in terms of y.” That means your answer can use y and numbers, but no x.

  2. Solve for x

    To swap x out, first get x by itself. Divide both sides of xy=30xy = 30 by y.

    x=30yx = \frac{30}{y}

    This says x is 30 divided by y.

  3. Put the fraction in place of x

    Replace x in 10÷x10 \div x with this fraction.

    10÷x=10÷30y10 \div x = 10 \div \frac{30}{y}

    Now use the rule for dividing by a fraction. Flip the fraction and multiply.

    10÷30y=10×y30=10y3010 \div \frac{30}{y} = 10 \times \frac{y}{30} = \frac{10y}{30}

  4. Simplify the fraction

    Both 10 and 30 can be divided by 10.

    10y30=y3=13y\frac{10y}{30} = \frac{y}{3} = \frac{1}{3}y

    Watch out: 10 ÷ 30 is not 3. Dividing the bigger number by the smaller one gives 3y, choice C. Also, dividing by x is not the same as multiplying by y. That only works when xy is 1, not 30. It leads to choice D.

    So the answer is B, 13y\frac{1}{3}y.

Check your answer

Pick numbers that multiply to 30, like x = 3 and y = 10. Then 10÷x=10÷3=10310 \div x = 10 \div 3 = \frac{10}{3}. Choice B gives 13×10=103\frac{1}{3} \times 10 = \frac{10}{3}. They match. Choice C would give 30, and choice D would give 100.

More questions like this

The original DOE question
Question 88 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 88. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 101 of 120

2023 Test B Math, question 39

Question

In a certain competition, 1 person finished in first place, 5 people tied for second place, 10 people tied for third place, and the other 4,984 people did not place. What percentage of the people in the competition placed first, second, or third altogether?

  • A0.0032%
  • B0.032%
  • C0.32%
  • D3.2%
Show the answer

Correct answer: C

The total number of people who participated in the competition was 1+5+10+4,984=5,0001+5+10+4{,}984=5{,}000. The total number who placed first, second, or third was 1+5+10=161+5+10=16.

165,000=3210,000=0.0032=0.32%\frac{16}{5{,}000}=\frac{32}{10{,}000}=0.0032=0.32\%

Question analysis

A percent question about a competition with 1 first-place finisher, 5 tied for second, 10 tied for third and 4,984 who did not place. Students find what percentage of everyone placed. All four choices use the same digits, so the challenge is placing the decimal point when a decimal becomes a percent.

NYS standards
  • 6.RP.3c percent as a rate per 100
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Writing part of a whole as a percent is grade 6 content. This one first totals four groups, then turns a fraction of a large total into a percent, where a slip of one place value is easy to make.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 10, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is short, so the question tests decimal places. All four choices share the digits 32 and differ only in where the point falls, so a slip of one place, or forgetting to multiply by 100, still finds a choice waiting. A rough check helps: 16 is a bit more than 15, and 15 of 5,000 is 0.3%, so the answer must sit just above 0.3%.

  1. Error-prone arithmetic moderate Working it out

    16 ÷ 5,000 = 0.0032 has two zeros after the decimal point, and writing it as a percent moves the point two more places; each move can land one place off. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    The total of 5,000 is not given; it has to be built from 1 + 5 + 10 + 4,984 before the 16 people who placed can be compared with it. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the people who placed and the people who did not to get the total, write the placers as a fraction of it, then change that fraction to a decimal and multiply by 100.

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.
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a whole
percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100
The part is the group you care about. The whole is everyone. The fraction tells what share of the whole the part is. Multiplying by 100 turns that share into a percent, which means out of 100.

  1. Sort out the groups

    Some people placed: 1 in first, 5 in second and 10 in third. The other 4,984 people did not place. You need the percent of all the people who placed.

  2. Count the people who placed

    These are the people who finished first, second or third. Add them.

    1+5+10=161 + 5 + 10 = 16

    So 16 people placed.

  3. Count everyone

    The whole is everyone in the competition: the people who placed plus the people who did not.

    16+4,984=5,00016 + 4{,}984 = 5{,}000

    So 5,000 people took part. Notice that 4,984 is not the whole. It leaves out the 16 who placed.

  4. Turn the share into a percent

    Use the percent formula.

    percent=165,000×100\text{percent} = \frac{16}{5{,}000} \times 100

    First divide: 16÷5,000=0.003216 \div 5{,}000 = 0.0032. Then multiply by 100. That moves the decimal point two places to the right.

    0.0032×100=0.320.0032 \times 100 = 0.32

    So 0.32% of the people placed.

  5. Avoid the decimal traps

    Watch out: 0.0032 is not a percent yet. Putting a % sign on it gives choice A, which is 100 times too small. Moving the point only one place gives choice B. Moving it three places gives choice D.

    So the answer is C, 0.32%.

Check your answer

Estimate. 1% of 5,000 is 50. The 16 people are about a third of 50. So the answer should be about a third of 1%, or about 0.3%. Choice C, 0.32%, fits.

More questions like this

The original DOE question
Question 96 as printed on page 202 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 96. Open page 202 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 102 of 120

2023 Test B Math, question 45

Question

The Barnes family and the Ramirez family each have a pizza of equal size. The first pizza is cut into 18 equal slices, and the second pizza is cut into 15 equal slices. If the Barnes family eats 11 slices from the first pizza, what is the greatest number of whole slices from the second pizza that the Ramirez family can eat without eating a greater percentage of a pizza than the Barnes family ate?

  • A6
  • B7
  • C8
  • D9
Show the answer

Correct answer: D

The fraction of the pizza that the Barnes family eats is 1118\frac{11}{18}. Let x represent the number of slices of pizza that the Ramirez family eats. Then the fraction of the pizza that the Ramirez family eats is x15\frac{x}{15}. Write an inequality and solve for x:

x15≤1118\frac{x}{15}\leq\frac{11}{18}

x≤11×1518x\leq\frac{11\times 15}{18}

x≤16518x\leq\frac{165}{18}

x≤916x\leq 9\frac{1}{6}

So, the greatest number of whole slices of pizza the Ramirez family can eat is 9.

Question analysis

Two pizzas of the same size are cut into 18 and 15 equal slices, and one family eats 11 slices of the first. Students find the most whole slices of the second pizza the other family can eat without eating a larger share. It tests comparing fractions with different denominators and rounding to a whole number in the right direction.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 inequality problems usually ask for the greatest whole number that fits a budget. This one wraps that idea in a long stem about shares of two pizzas sliced differently, with several conditions to keep straight at once.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 11, 72 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The math is one inequality between fractions, but the long final sentence hides it, so students who skim fall back on comparing slice counts. Reasoning that the second pizza has 3 fewer slices, so the Ramirez family should eat 3 fewer, is the strongest pull. Comparing 11/18 with x/15, or turning 11/18 into about 61% and taking 61% of 15, keeps the comparison about shares of a pizza.

  1. Complex sentences moderate Understanding the question

    The question is one 42-word sentence ending in a nested condition, “without eating a greater percentage of a pizza than the Barnes family ate”, a reading hurdle rather than more math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

  2. Turning words into math moderate Understanding the question

    The limit of no “greater percentage of a pizza” must become the inequality x/15 ≤ 11/18, comparing fractions of two pizzas cut into different numbers of slices, not slice counts. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Solving gives x ≤ 11 × 15 ÷ 18 = 165/18, a division by 18 that leaves 9 1/6, which must then be cut back to whole slices. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the Barnes family's share as a fraction of their pizza, then find the largest whole number of fifteenths that does not exceed it, by cross-multiplying or by converting to decimals.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a pizza
fraction eaten=slices eatenslices in the pizza\text{fraction eaten} = \frac{\text{slices eaten}}{\text{slices in the pizza}}
The top counts the slices eaten. The bottom counts all the equal slices the pizza was cut into. The fraction tells how much of the whole pizza was eaten. Use fractions to compare pizzas cut into different numbers of slices.

  1. Sort out the question

    The two pizzas are the same size. The Barnes pizza has 18 slices, and they eat 11. The Ramirez pizza has 15 slices, so its slices are bigger. You need the most whole slices the Ramirez family can eat without eating a bigger share of their pizza.

  2. Find the Barnes family’s share

    Their share is the slices they ate out of all the slices in their pizza. Use the fraction formula.

    fraction eaten=1118\text{fraction eaten} = \frac{11}{18}

    The Barnes family ate 1118\frac{11}{18} of their pizza.

  3. Write an inequality

    Let x be the number of Ramirez slices. Their share is x15\frac{x}{15}. It must be no more than the Barnes share. The sign ≤ means less than or equal to.

    x15≤1118\frac{x}{15} \leq \frac{11}{18}

  4. Solve for x

    Multiply both sides by 15 to get x by itself.

    x≤11×1518=16518x \leq \frac{11 \times 15}{18} = \frac{165}{18}

    Now divide. 18 goes into 165 nine times, since 18 × 9 = 162. There are 3 left over.

    x≤9318=916x \leq 9\frac{3}{18} = 9\frac{1}{6}

  5. Pick the whole number

    x must be a whole number no bigger than 9169\frac{1}{6}. So the most is 9 slices. 10 slices would be too many.

    Watch out: The Ramirez pizza has 3 fewer slices, but that does not mean they eat 3 fewer. Their slices are bigger. Eating 11 − 3 = 8 slices is choice C, which is not the most they can eat.

    So the answer is D, 9.

Check your answer

Use decimals. The Barnes share is 11÷18≈0.61111 \div 18 \approx 0.611. Nine Ramirez slices are 9÷15=0.69 \div 15 = 0.6, just under it. Ten slices are 10÷15≈0.66710 \div 15 \approx 0.667, too much. So 9 is the most.

More questions like this

The original DOE question
Question 102 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 102. Open page 204 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 103 of 120

2023 Test B Math, question 50

Question

{3,4,5,6,7}\{3, 4, 5, 6, 7\}

A student made a list of all possible products of 2 different numbers in the set above. What fraction of the products are odd?

  • A320\frac{3}{20}
  • B310\frac{3}{10}
  • C12\frac{1}{2}
  • D35\frac{3}{5}
Show the answer

Correct answer: B

First, list the products:

3×4=123\times 4=12

3×5=153\times 5=15

3×6=183\times 6=18

3×7=213\times 7=21

4×5=204\times 5=20

4×6=244\times 6=24

4×7=284\times 7=28

5×6=305\times 6=30

5×7=355\times 7=35

6×7=426\times 7=42

Of the 10 products, 3 are odd. The fraction is 310\frac{3}{10}.

Question analysis

A student lists every product of two different numbers from the set 3, 4, 5, 6, 7. Students find what fraction of those products are odd. It tests listing outcomes systematically without counting a pair twice, and reasoning about odd and even numbers.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.SP.8a compound event as fraction of outcomes
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Organized lists of outcomes are grade 7 work. Here students list the products of pairs from five numbers, without repeating a pair or pairing a number with itself, then count the ones that fit.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The count of odd products is easy, since only 3, 5 and 7 can make one; the total is where students go wrong. Counting ordered pairs, 5 × 4 = 20, while listing odd products once mixes two kinds of counting. Writing the 10 pairs in order, 3 with each larger number, then 4, and so on, keeps the total honest.

  1. Much to keep track of moderate Working it out

    Listing the products of 2 different numbers from five means tracking 10 pairs, with no pair repeated in reverse order and no number paired with itself. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Something you must already know minor Working it out

    The shortcut, that a product is odd only when both factors are odd, cuts the work to counting pairs from 3, 5 and 7, but it must be known. (Anderson & Krathwohl, 2001)

Strategy: List each pair of different numbers once, then find each product, or use the fact that a product is odd only when both factors are odd, and count.

Traps

  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a list
fraction=odd productsall products\text{fraction} = \frac{\text{odd products}}{\text{all products}}
The top counts the products that are odd. The bottom counts all the products in the list. A product is the answer when you multiply. The fraction tells what share of the list is odd.
Odd times odd
odd×odd=odd\text{odd} \times \text{odd} = \text{odd}
An odd number can’t be split into pairs, like 3 or 7. An even number can, like 4. A product is odd only when both numbers are odd. If either number is even, the product is even, because it splits into pairs.

  1. Sort out the question

    The set has 3, 4, 5, 6 and 7. The student multiplies 2 different numbers, every possible way. A product is the answer to a multiplication. Each pair counts once, since 3 × 4 and 4 × 3 give the same product. You need the fraction of the products that are odd.

  2. List every pair once

    Pair 3 with each bigger number. Then pair 4 with each bigger number, and so on. This way no pair is counted twice.

    3 × 4 = 123 × 5 = 153 × 6 = 183 × 7 = 21
    4 × 5 = 204 × 6 = 244 × 7 = 28
    5 × 6 = 305 × 7 = 35
    6 × 7 = 42

    There are 10 products.

  3. Count the odd products

    The odd products are 15, 21 and 35. That makes 3. They come from the only pairs where both numbers are odd: 3 and 5, 3 and 7, and 5 and 7.

  4. Write the fraction

    Use the fraction formula.

    fraction=310\text{fraction} = \frac{3}{10}

    Watch out: Counting 4 × 3 as well as 3 × 4 makes 20 pairs. Then 320\frac{3}{20}, choice A, mixes two ways of counting. And 35\frac{3}{5}, choice D, is the share of the numbers that are odd, not the products.

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

Check your answer

Count without listing. Each of the 5 numbers can pair with 4 others, which makes 20. But that counts every pair twice, so there are 20 ÷ 2 = 10 pairs. Only the odd numbers 3, 5 and 7 make odd products, and they form 3 pairs.

More questions like this

The original DOE question
Question 107 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 107. Open page 205 of the handbook (PDF)

Question 104 of 120

2023 Test B Math, question 51

Question

A number, n, is divided by 25. If the quotient is 10 with a remainder of 5, what is the quotient expressed as a decimal?

  • A10.20
  • B10.25
  • C10.50
  • D10.75
Show the answer

Correct answer: A

A quotient of 10 with a remainder of 5 means the quotient is

10+525=1015=10.2010+\frac{5}{25}=10\frac{1}{5}=10.20

Question analysis

A number is divided by 25, giving a quotient of 10 with a remainder of 5, and students write the quotient as a decimal. It tests what a remainder means and how it becomes part of a decimal quotient. Connecting the remainder to the decimal part of the answer is the key idea.

NYS standards
  • 6.NS.2 multi-digit division
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Division with remainders and decimal quotients are grade 6 skills. This takes two short steps, with no word problem to translate, though students must know what a remainder means once a quotient is written as a decimal.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question checks one idea: a remainder is a fraction of the divisor, not a string of digits to place after the point. Students who write the remainder 5 as tenths, or set the divisor 25 after the point, land on listed choices. Rebuilding n = 25 × 10 + 5 = 255 and dividing by 25 confirms the decimal.

  1. Something you must already know moderate Working it out

    A remainder of 5 means 5 out of the 25 needed for another whole, so it becomes 5/25 = 0.2, a link between remainders and decimals the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Rebuild n from the quotient and the remainder, then divide it by 25 to a decimal; or write the remainder as a fraction of the divisor.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Quotient with a remainder
quotient=q+rd\text{quotient} = q + \frac{r}{d}
d is the divisor, the number you divide by. q is the whole-number part of the quotient, the answer to the division. r is the remainder, what is left over. The leftover still has to be divided by d, so it becomes the fraction r over d.

  1. Sort out the words

    n is divided by 25. So 25 is the divisor, the number you divide by. The quotient is the answer to a division. Here it is 10 with a remainder of 5. The remainder is the amount left over. You need the exact quotient as a decimal.

  2. Turn the remainder into a fraction

    The 5 left over still has to be divided by 25. So it becomes the fraction 525\frac{5}{25}. Use the formula.

    quotient=10+525\text{quotient} = 10 + \frac{5}{25}

  3. Change the fraction to a decimal

    Simplify first. Divide the top and bottom by 5. Then write the fraction in hundredths.

    525=15=20100=0.20\frac{5}{25} = \frac{1}{5} = \frac{20}{100} = 0.20

    So the quotient is 10 + 0.20 = 10.20.

  4. Give the answer

    Watch out: A remainder is not the digits after the decimal point. Writing the 5 after the point gives 10.50, choice C. Writing the 25 after it gives 10.25, choice B.

    So the answer is A, 10.20.

Check your answer

Rebuild n. Ten 25s plus the 5 left over is 25×10+5=25525 \times 10 + 5 = 255. Now divide: 255÷25=10.2255 \div 25 = 10.2. That matches 10.20.

More questions like this

The original DOE question
Question 108 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 108. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 105 of 120

2018 Test A Math, question 8

Question

In a certain school, course grades range from 0 to 100. Adrianna took 4 courses and her mean course grade was 90. Roberto took 5 courses. If both students have the same sum of course grades, what was Roberto’s mean?

  • A72
  • B80
  • C90
  • D92
Show the answer

Correct answer: A

The sum of Adrianna’s course grades equals 4 times the mean (average) of her grades:

90 × 4 = 360

Roberto has the same sum (360) as Adrianna. Find the mean of his course grades:

360 ÷ 5 = 72

Question analysis

This averages word problem compares two students' course grades. One student's number of courses and mean grade are given, and a second student has the same total over a different number of courses. Students work back from a mean to a sum, then find a new mean, while the stated grade range plays no part.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.3 measures of center and variation
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a mean is grade 6 content, and this needs only the step back from a mean to a total and one division. Typical grade 7 statistics work compares two data sets, which asks more.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 2, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns on one idea, that a mean times the number of items gives the total, and the numbers are easy. Students who skim "the same sum" as "the same mean" carry Adrianna's 90 straight across, and the 0-to-100 range in the first sentence offers a second number to misuse. Converting to the shared total of 360 first makes the rest a single division.

  1. Turning words into math moderate Understanding the question

    "Both students have the same sum of course grades" is the bridge: Adrianna's mean of 90 must be turned into a total, 4 × 90, before it says anything about Roberto. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    The link between a mean and a total, sum = mean × number of courses, is not given and has to be used in both directions. (Anderson & Krathwohl, 2001)

Strategy: Multiply the first student's mean by her number of courses to get the shared total, then divide that total by Roberto's number of courses.

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.
  • Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean (average)
M=SnM = \frac{S}{n}
M is the mean, also called the average. S is the sum, all the values added together. n is how many values there are. Dividing shares the total out equally, so the mean is what each value would be if they were all the same.
Sum from the mean
S=M×nS = M \times n
This is the mean formula turned around. S is the sum, M is the mean, and n is how many values there are. If n values were each equal to the mean, they would add up to M times n. That is the same as the real sum.

  1. Sort out the question

    Adrianna took 4 courses, and her mean grade was 90. The mean is the average. Roberto took 5 courses. The key fact is that both have “the same sum of course grades”. The sum is all the grades added together. You need Roberto’s mean. The 0 to 100 range is not needed.

  2. Find Adrianna’s total

    You know her mean and her number of courses. Use the sum formula.

    S=M×nS = M \times n

    S=90×4=360S = 90 \times 4 = 360

    Her grades add up to 360 points.

  3. Find Roberto’s mean

    Roberto’s grades also add up to 360 points. But he spreads them over 5 courses. Use the mean formula.

    M=SnM = \frac{S}{n}

    M=3605=72M = \frac{360}{5} = 72

    His mean is 72.

  4. Avoid the traps and answer

    Watch out: the same sum does not mean the same mean. Roberto has more courses sharing the points, so his mean is lower. Choosing 90 is choice C. Watch out: don’t use 100, the top grade, in place of 90. That leads to 80, choice B. So the answer is A, 72.

Check your answer

More courses share the same 360 points, so Roberto’s mean must be lower than 90. 72 is lower. Multiply back: 5×72=3605 \times 72 = 360, which matches Adrianna’s total.

More questions like this

The original DOE question
Question 65 as printed on page 82 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 82, question 65. Open page 82 of the handbook (PDF)

Question 106 of 120

2018 Test A Math, question 23

Question

SCORES ON MATH QUIZ

Score Number of Students
85 4
75 4
65 2

What is the mean score of the 10 students in the table above?

  • A22.5
  • B75
  • C77
  • D85
Show the answer

Correct answer: C

For each row, multiply the number of students by the score. Then add those together and divide by the total number of students to find the mean (average) of the 10 students.

85(4)+75(4)+65(2)10=340+300+13010=77010=77\frac{85(4)+75(4)+65(2)}{10}=\frac{340+300+130}{10}=\frac{770}{10}=77

Question analysis

This statistics question gives a frequency table of quiz scores: three scores and how many of the 10 students earned each. Students find the mean score of all 10 students. Each score has to count as many times as students earned it, which is easy to overlook when only three scores are listed.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.5a number of observations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Computing a mean is grade 6 work, and this weighted version from a table takes two short steps. It is a little easier than typical grade 7 statistics, which compares whole data sets.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 2, 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 procedure is routine once students see that the table lists counts, not separate students. The tempting shortcut is to average the three scores in the first column, which gives a listed choice close to the right one. Multiplying each score by its count, then dividing by 10, takes well under a minute.

  1. A very close wrong answer moderate Choosing the answer

    77 and 75 are only 2 points apart, so a rough sense of the middle score cannot choose between them; only weighting each score by its 4, 4 or 2 students does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    The table pairs each score with a number of students, so the 10 students are 4, 4 and 2, and each score must be weighted by its count. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Multiply each score by its number of students, add the three products, and divide by the 10 students.

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

Mean (average)
M=SnM = \frac{S}{n}
M is the mean, also called the average. S is the sum, all the values added together. n is how many values there are. Dividing shares the total out equally, so the mean is what each value would be if they were all the same.

  1. Sort out the table

    The table lists 3 scores and how many students got each one. 4 students scored 85. 4 students scored 75. 2 students scored 65. That is 4+4+2=104 + 4 + 2 = 10 students. You need the mean, or average, of all 10 scores.

  2. Find the sum of all 10 scores

    Each score counts once for every student who got it. So multiply each score by its number of students.

    85×4=34085 \times 4 = 340

    75×4=30075 \times 4 = 300

    65×2=13065 \times 2 = 130

    Add the three results.

    340+300+130=770340 + 300 + 130 = 770

    All 10 scores add up to 770 points.

  3. Divide by the number of students

    M=SnM = \frac{S}{n}

    M=77010=77M = \frac{770}{10} = 77

    Watch out: averaging just the three scores gives (85+75+65)÷3=75(85 + 75 + 65) \div 3 = 75, choice B. That acts as if only one student got each score. So the answer is C, 77.

Check your answer

More students scored 85 than 65. So the mean should be a little above the middle score, 75. 77 fits. A mean must also be between the lowest and highest scores, 65 and 85. That rules out 22.5.

More questions like this

The original DOE question
Question 80 as printed on page 85 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 80. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 107 of 120

2018 Test A Math, question 26

Question

In a scale drawing of a triangular banner, one side measures 16 centimeters and the other two sides each measure 12 centimeters. On the actual banner, these two sides each measure 36 feet. What is the length of the remaining side of the actual banner?

  • A16 ft
  • B32 ft
  • C40 ft
  • D48 ft
Show the answer

Correct answer: D

Let xx be the remaining side of the actual banner.

Set up a proportion:

x16=3612\frac{x}{16}=\frac{36}{12}

12x=57612x=576

x=48x=48 ft

Question analysis

A scale drawing word problem about a triangular banner: the drawing's sides are 16 cm, 12 cm and 12 cm, and the two equal sides are 36 feet on the real banner. Students find the actual length of the remaining side. The key is using a scale factor, not adding the difference between lengths.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Scale drawings are a named grade 7 topic, and this is a direct one-factor problem. It matches typical grade 7 work; the only pull away is reasoning by differences instead of by multiplication.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 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

A scale drawing multiplies every length by the same factor, here 3 feet per centimeter, and the question checks whether students use that factor or a difference. The 16 cm side is 4 cm longer than the others, and adding 4 to 36 feels natural but treats the scale as an addition. Students who find the factor from the matching sides first have one multiplication left.

  1. Turning words into math minor Understanding the question

    "These two sides each measure 36 feet" refers back to the 12 cm sides, so the scale pairs 12 cm with 36 ft, centimeters on the drawing against feet on the banner. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the scale from the matching sides, feet per centimeter, then multiply the 16 cm side by that same factor.

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scale factor
k=actual lengthdrawing lengthk = \frac{\text{actual length}}{\text{drawing length}}
k is the scale factor: how many real units match 1 unit on the drawing. Divide a real length by its matching length on the drawing to find it. Every length on the drawing gets multiplied by the same k. That keeps the real shape the same as the drawing.

  1. Sort out the question

    A scale drawing is a smaller copy with the same shape. The drawing is a triangle with sides of 16 cm, 12 cm and 12 cm. On the real banner, the two 12 cm sides are 36 feet each. You need the real length of the 16 cm side.

  2. Find the scale factor

    Use the matching sides: 12 cm on the drawing is 36 feet on the banner.

    k=actual lengthdrawing lengthk = \frac{\text{actual length}}{\text{drawing length}}

    k=3612=3k = \frac{36}{12} = 3

    So every 1 cm on the drawing stands for 3 feet on the banner.

  3. Scale up the third side

    Multiply the 16 cm side by the same scale factor.

    16×3=4816 \times 3 = 48

    The third side is 48 feet. Watch out: a scale multiplies lengths. It does not add to them. 16 is 4 more than 12, but adding 4 to 36 gives 40, choice C. So the answer is D, 48 ft.

Check your answer

The 16 cm side is the longest side on the drawing. So it must be the longest on the banner too, longer than 36 feet. 48 fits. Also, 48÷16=348 \div 16 = 3, the same scale as 36÷12=336 \div 12 = 3.

More questions like this

The original DOE question
Question 83 as printed on page 86 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 83. Open page 86 of the handbook (PDF)

Question 108 of 120

2018 Test A Math, question 45

Question

A metal square used in an electronic device must have a thickness of 0.02 inch, with an allowable error of 1 percent. What is the greatest allowable thickness of the metal square?

  • A0.0002 in.
  • B0.02 in.
  • C0.0202 in.
  • D0.03 in.
Show the answer

Correct answer: C

Find 1% of 0.02:

(0.02)(1100)=0.0002(0.02)(\frac{1}{100})=0.0002

The greatest allowable thickness would be

0.02+0.0002=0.02020.02+0.0002=0.0202 inch.

Question analysis

This percent question says a metal square must be 0.02 inch thick with an allowable error of 1 percent, and asks for the greatest allowable thickness. Students must find 1 percent of a small decimal and add it to the target thickness. Keeping track of decimal places is what makes it tricky.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Percent error and tolerance problems appear in grade 7, so the idea is on level. The demand comes from taking a percent of a tiny decimal and adding the result, where a slipped decimal place is easy to miss.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 4, 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 idea of tolerance is simple, but the numbers are tiny, and the question turns on what the 1 percent is a percent of. Students who read it as 0.01 inch land on a choice that looks sensible, and those who find the error correctly may stop before adding it. Writing 0.02 × 1.01 in one step avoids both.

  1. Error-prone arithmetic moderate Working it out

    One percent of 0.02 is 0.0002, two more decimal places, and adding it to 0.02 needs the places lined up to get 0.0202. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    The phrase "allowable error of 1 percent" must be read as 1 percent of 0.02 inch, added for the greatest thickness, not 1 percent as a fixed 0.01 inch. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. A very close wrong answer minor Choosing the answer

    The choices 0.0202, 0.02 and 0.0002 are built from the same digits, and the first two differ by only 1 percent, so only careful counting of decimal places tells them apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Find 1 percent of 0.02 by dividing by 100, counting decimal places carefully, then add that error to 0.02 to get the upper limit.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
x=p100×nx = \frac{p}{100} \times n
x is the part you want. p is the percent. n is the whole amount. Percent means out of 100, so p percent of a number is p hundredths of it.

  1. Sort out what you know

    The square should be 0.02 inch thick. An allowable error of 1 percent means it may be off by up to 1 percent of 0.02 inch. It may be that much thinner or thicker.

    You need the greatest allowed thickness, so look at the thicker side.

  2. Find 1 percent of 0.02

    x=1100×0.02=0.0002x = \frac{1}{100} \times 0.02 = 0.0002

    Dividing by 100 moves the decimal point two places to the left. So the error is 0.0002 inch.

    Watch out: 1 percent here is not 0.01 inch. It is 1 percent of 0.02 inch. Adding 0.01 gives 0.03, which is choice D.

  3. Add the error to the thickness

    The thickest square is the right size plus the error. Line up the decimal points:

    0.0200+0.0002=0.02020.0200 + 0.0002 = 0.0202

    Watch out: 0.0002 is only the error, choice A. You must add it to 0.02. So the answer is C, 0.0202 in.

Check your answer

0.0202 is just a tiny bit more than 0.02, which makes sense for a small error. Check it: 0.0202 − 0.02 = 0.0002. And 0.0002 ÷ 0.02 = 0.01, which is 1 percent.

More questions like this

The original DOE question
Question 102 as printed on page 90 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 102. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 109 of 120

2018 Test A Math, question 46

Question

SCORES ON BIOLOGY TEST

Section Lowest Score Range
I 65 28
II 62 25
III 67 22

Mr. Blake’s biology class is divided into three sections. The same test was given to each section. The table above shows both the lowest score and the range of scores on this test for each section. What is the overall range of all scores in all three sections?

  • A25
  • B27
  • C28
  • D31
Show the answer

Correct answer: D

First, calculate the highest score for each section by adding the lowest score to the range:

Section I: 65 + 28 = 93

Section II: 62 + 25 = 87

Section III: 67 + 22 = 89

The overall highest score is 93, and the overall lowest score is 62. Thus the overall range is 93 − 62 = 31.

Question analysis

This statistics question uses a table of the lowest score and the range for three sections of a biology class and asks for the range of all the scores combined. Students must rebuild each section's highest score from its lowest score and range, then compare across sections. Reversing the usual range calculation is the tricky part.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.3 measures of center and variation
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students compute a range from a list of data. Here they must work backward from range to the highest score in each section and then combine three groups, one step beyond routine.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 59 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Range is usually computed from a list, so working backward from it is the unfamiliar part. The key fact, that the overall lowest and highest scores come from different sections, is invisible until all three highest scores are written down. Every wrong choice comes from skipping that step and reasoning with the ranges as given.

  1. No obvious method moderate Working it out

    The table gives each section's lowest score and range but no highest score, so students must reverse the range formula, high = low + range, before comparing sections. (Schoenfeld, 1985)

  2. A very close wrong answer moderate Choosing the answer

    The choices 28 and 27 sit within 4 of 31, so a sense of the spread cannot separate them; only rebuilding the highest scores, which puts the top at 93, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Much to keep track of minor Working it out

    The overall lowest, 62, comes from Section II and the overall highest, 93, from Section I, so values from two rows must be combined. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Add each section's range to its lowest score to get its highest score, then subtract the lowest of all scores from the highest of all scores.

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

Range
R=H−LR = H - L
R is the range, the spread of the scores. H is the highest score and L is the lowest score. The range tells how far apart the top and bottom scores are. If you know L and R, then H = L + R.

  1. Sort out what you know

    The table gives each section’s lowest score and its range. The range is the highest score minus the lowest score.

    The bold overall means one range for all three sections together. For that you need the highest and lowest scores of all.

  2. Find each section’s highest score

    Turn the range formula around: highest = lowest + range.

    SectionLowestRangeHighest
    I652865 + 28 = 93
    II622562 + 25 = 87
    III672267 + 22 = 89

  3. Pick the overall highest and lowest

    The highest score of all is 93, from Section I. The lowest score of all is 62, from Section II. They come from different sections, and that is fine.

  4. Subtract to find the range

    R=93−62=31R = 93 - 62 = 31

    Watch out: 28 is just Section I’s range, choice C. Averaging the three ranges gives 25, choice A. Neither covers all the scores at once.

    So the answer is D, 31.

Check your answer

The overall range must be at least as big as each section’s range, because it covers all their scores. 31 is bigger than 28, 25 and 22, so it is reasonable. Also, 62 + 31 = 93, the top score.

More questions like this

The original DOE question
Question 103 as printed on page 90 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 103. Open page 90 of the handbook (PDF)

Question 110 of 120

2018 Test A Math, question 47

Question

If 3n3n is a positive even number, how many odd numbers are in the range from 3n3n up to and including 3n+53n+5?

  • A2
  • B3
  • C4
  • D5
Show the answer

Correct answer: B

Since 3n3n is even, then 3n+13n+1 must be odd.

Thus 3n+33n+3 and 3n+53n+5 are also odd.

So there are a total of 3 numbers in this range that are odd.

Question analysis

This number properties question says 3n is a positive even number and asks how many odd numbers lie from 3n up to and including 3n + 5. Students must decide which integers in that span are odd. The coefficient in 3n is a distraction; only the fact that 3n is even matters.

NYS standards
  • 6.EE.2b parts of an expression
  • 6.EE.6 variables representing unknown numbers
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students work with expressions like 3n + 5 but rarely reason about whether they are even or odd. The idea is simple once the integers are written out, so the demand is about typical for grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 12, 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

Writing the six integers from 3n to 3n + 5 and marking them even, odd, even, odd, even, odd answers the question in seconds. The traps come from not writing them: dropping the last one, which "up to and including" keeps in, or reading the 5 in 3n + 5 as a count. Picking any even value for 3n, such as 6, makes the check concrete.

  1. Variables instead of numbers moderate Understanding the question

    No value of n is given, so students must reason about parity: since 3n is even, 3n + 1, 3n + 3 and 3n + 5 are the odd ones. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know minor Working it out

    The parity rules, even plus odd is odd and even plus even is even, are assumed; the coefficient 3 is a distraction once 3n is said to be even. (Anderson & Krathwohl, 2001)

  3. A very close wrong answer minor Choosing the answer

    The choices 2, 3, 4 and 5 are consecutive counts, so 2 and 3 cannot be told apart by sizing up; only checking whether 3n + 5 is inside the range separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write out every integer from 3n to 3n + 5, mark 3n as even, alternate even and odd from there, and count the odd ones.

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

Even plus odd
even+odd=odd\text{even} + \text{odd} = \text{odd}
An even number splits into pairs. An odd number has one left over. Adding them keeps that one left over, so the sum is odd. So adding 1, 3 or 5 to an even number gives an odd number.
Even plus even
even+even=even\text{even} + \text{even} = \text{even}
Two numbers that each split into pairs still split into pairs when you add them. So adding 2 or 4 to an even number gives an even number.

  1. Sort out what you know

    You are told that 3n is even. You do not need to know n itself. You need to count the odd numbers from 3n to 3n + 5.

    “Up to and including” means 3n + 5 counts too.

  2. List all the numbers

    Go up by 1 each time, from 3n to 3n + 5. That gives six numbers:

    3n, 3n+1, 3n+2, 3n+3, 3n+4, 3n+53n,\ 3n+1,\ 3n+2,\ 3n+3,\ 3n+4,\ 3n+5

  3. Mark each as even or odd

    3n is even. Adding 1, 3 or 5 makes it odd. Adding 2 or 4 keeps it even.

    NumberEven or odd
    3n3neven
    3n+13n+1odd
    3n+23n+2even
    3n+33n+3odd
    3n+43n+4even
    3n+53n+5odd

    Watch out: 3n is even because the question says so. The 3 being odd does not matter. Counting 3n as odd gives choice C, 4.

  4. Count the odd numbers

    The odd ones are 3n + 1, 3n + 3 and 3n + 5. That is 3 numbers.

    Watch out: leaving out 3n + 5 gives 2, choice A. It is the last number, and it is included. So the answer is B, 3.

Check your answer

Try a real number. If n = 2, then 3n = 6, which is even. The numbers from 6 to 11 are 6, 7, 8, 9, 10 and 11. The odd ones are 7, 9 and 11. That is 3.

More questions like this

The original DOE question
Question 104 as printed on page 90 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 104. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 111 of 120

2018 Test A Math, question 51

Question

A sports store has a container of handballs: 4 blue, 5 red, 8 yellow, 9 white, and 11 green. If one ball is picked from the container at random, what is the probability that it will be yellow?

  • A137\frac{1}{37}
  • B18\frac{1}{8}
  • C837\frac{8}{37}
  • D829\frac{8}{29}
Show the answer

Correct answer: C

The total number of handballs in the container is

4+5+8+9+11=374+5+8+9+11=37

Since there are 8 yellow handballs, the probability of selecting a yellow handball is

837\frac{8}{37}

Question analysis

This probability question describes a container of handballs in five colors and asks the chance that one ball picked at random is yellow. Students must add all the balls to get the total and compare the yellow count to it. Using the wrong total, such as only the balls that are not yellow, is the easy mistake.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the probability of a single event as favorable outcomes over total outcomes is grade 6 content. This item takes one addition and one fraction, so it is easier than typical grade 7 compound probability work.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-event probability: add the five colors to get 37 balls and put the 8 yellow ones over that total. The only real decision is the denominator, and the wrong choices offer the non-yellow count, the yellow count, and a numerator of 1. Students who say "favorable over total" to themselves avoid all three.

  1. A very close wrong answer minor Choosing the answer

    The choices 8/37 and 8/29 share the numerator 8 and differ only in the denominator, all 37 balls against the 29 that are not yellow. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Add up every ball in the container to get the total number of outcomes, then put the number of yellow balls over that total.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • 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=ftP = \frac{f}{t}
P is the probability, the chance that something happens. f is the number of favorable outcomes, the ones you want. t is the total number of outcomes. It works when every outcome is equally likely, like picking one item at random.

  1. Sort out what you know

    The container has 4 blue, 5 red, 8 yellow, 9 white and 11 green balls. One ball is picked at random, so every ball has the same chance. You need the chance it is yellow.

  2. Count all the balls

    Add up every color, because any ball could be picked:

    4+5+8+9+11=374 + 5 + 8 + 9 + 11 = 37

    There are 37 balls in all. That is the total number of outcomes.

  3. Put yellow over the total

    8 of the balls are yellow. Those are the favorable outcomes:

    P=837P = \frac{8}{37}

    Watch out: 829\frac{8}{29} compares yellow with the 29 balls that are not yellow. Probability compares the part with all the balls, so choice D is wrong. So the answer is C, 837\frac{8}{37}.

Check your answer

8 out of 37 is about 0.22, a little more than 1 in 5. That makes sense, since yellow is a fair share of the balls. Choice A, 1/37, is the chance of picking one certain ball, not any yellow one.

More questions like this

The original DOE question
Question 108 as printed on page 91 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 91, question 108. Open page 91 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 112 of 120

2018 Test A Math, question 56

Question

SERVINGS OF FRUITS AND VEGETABLES

Number of Servings of Fruits and Vegetables Number of Students
0 5
1 7
2 3
3 4
4 0
5 1

There are 20 students in a class. The frequency table above shows the number of students in this class who ate 0, 1, 2, 3, 4, or 5 servings of fruits and vegetables yesterday. What is the mean number of servings of fruits and vegetables eaten yesterday per student in this class?

  • A1121\frac{1}{2}
  • B33
  • C3133\frac{1}{3}
  • D44
Show the answer

Correct answer: A

First, determine the total number of servings of fruits and vegetables that the students ate by multiplying the number of servings by the number of students in each row of the table.

Then add that column to get the total number of servings:

Number of Servings of Fruits and Vegetables Number of Students Number of Servings × Number of Students
0 5 0
1 7 7
2 3 6
3 4 12
4 0 0
5 1 5

Total: 30

Calculate the mean by dividing the total number of servings of fruits and vegetables by the total number of students:

3020=112\frac{30}{20}=1\frac{1}{2}

Question analysis

This statistics question uses a frequency table showing how many of 20 students ate 0, 1, 2, 3, 4 or 5 servings of fruits and vegetables, and asks for the mean servings per student. Students must weight each serving amount by its number of students, add, and divide. Averaging the wrong column of the table is the common mistake.

NYS standards
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Finding a mean is grade 6 content; doing it from a frequency table, where each value must be weighted by its count, is the multi-step version grade 7 students meet regularly.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 64 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 answer is smaller than every wrong choice, which surprises students who average the table's columns directly. The question rewards seeing that each row stands for several students, so the total servings come from weighting each value by its count. The stem's reminder that there are 20 students points to the right divisor.

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

    The frequency table has a value column and a count column; the mean needs each value multiplied by its count, so the two columns play different roles. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Six products, 0, 7, 6, 12, 0 and 5, must be found and added to 30 before dividing by the 20 students. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply each serving amount by its number of students, add those products to get the total servings, then divide by the 20 students.

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

Mean
m=Snm = \frac{S}{n}
m is the mean, or average. S is the sum of all the values, here the total servings eaten. n is the number of values, here the number of students. The mean shares the total equally, as if every student ate the same amount.

  1. Sort out the table

    This is a frequency table. Each row says how many students ate a certain number of servings. For example, the second row says 7 students ate 1 serving each.

    There are 20 students. You need the mean, or average, number of servings per student.

  2. Find the servings in each row

    In each row, multiply the servings by the number of students:

    ServingsStudentsServings × Students
    050
    177
    236
    3412
    400
    515

  3. Add up all the servings

    0+7+6+12+0+5=300 + 7 + 6 + 12 + 0 + 5 = 30

    The class ate 30 servings in all. The students add up to 20, as the question says: 5 + 7 + 3 + 4 + 0 + 1 = 20.

  4. Divide by the number of students

    m=3020=11020=112m = \frac{30}{20} = 1\frac{10}{20} = 1\frac{1}{2}

    Watch out: do not just average the serving numbers 1 to 5. That gives 3, choice B, and ignores how many students ate each amount. Dividing 20 students by 6 rows gives choice C.

    So the answer is A, 1121\frac{1}{2}.

Check your answer

12 of the 20 students ate 0 or 1 serving. Only 5 ate 3 or more. So the mean should be small, between 1 and 2. 1121\frac{1}{2} fits. Choices B, C and D are all 3 or more, which is too big.

More questions like this

The original DOE question
Question 113 as printed on page 92 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 92, question 113. Open page 92 of the handbook (PDF)

Question 113 of 120

2018 Test B Math, question 3

Question

|(−6)−(−5)+4.2|−|3−9.6|\left | \left ( -6 \right )-\left ( -5 \right )+4.2 \right |-\left | 3-9.6 \right |

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

Show the answer

Correct answer: -3.4

|(−6)−(−5)+4.2|−|3−9.6|=|3.2|−|−6.6|\left|(-6)-(-5)+4.2\right|-\left|3-9.6\right|=\left|3.2\right|-\left|-6.6\right|

=3.2−6.6=−3.4=3.2-6.6=-3.4

Question analysis

This grid-in asks students to evaluate an expression that mixes negative integers, decimals and two absolute values. The hard parts are subtracting a negative number inside the first pair of bars and keeping each absolute value separate until the final subtraction, with no answer choices to check against.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students add and subtract signed numbers, and grade 6 students meet absolute value. This combines both in one expression with decimals and several sign decisions, a typical grade 7 exercise with no context to check against.
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 16+, 1 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 familiar, but the expression packs four of them with negatives and decimals, and a single sign slip anywhere gives a clean-looking wrong number. The subtlest point comes last: absolute value bars make each part positive, yet the difference of two absolute values can still be negative. Students who believe bars always produce a positive answer reverse the final subtraction, and the grid-in format offers no choices to expose it.

  1. Error-prone arithmetic moderate Working it out

    Three sign decisions stack up: (−6) − (−5) must become −6 + 5, then −1 + 4.2 crosses zero, and 3 − 9.6 is negative before its bars make it positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps moderate Working it out

    Each pair of bars must be simplified inside and then made positive, and only after both does the outer subtraction happen; doing the bars and the subtraction in the wrong order changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The result is a negative decimal to be gridded with its sign, and with no choices on screen nothing warns a student who has dropped the minus or flipped the subtraction. (Enright, Morley & Sheehan, 2002)

Strategy: Simplify inside each pair of absolute value bars first, rewriting minus a negative as plus, then take each absolute value, and only then subtract the second from the first.

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

Absolute value
|a|=distance from a to 0|a| = \text{distance from } a \text{ to } 0
The two bars around a number mean absolute value. a is any number. Its absolute value is how far it is from 0 on the number line. A distance is never negative. So |−6.6| = 6.6 and |3.2| = 3.2.
Subtracting a negative
a−(−b)=a+ba - (-b) = a + b
a and b are numbers. Taking away a negative number is the same as adding the positive number. For example, 5 − (−2) = 7. Think of it as taking away a debt: you end up richer.

  1. Plan the order of work

    The expression has two pairs of bars. The bars stand for absolute value, a number’s distance from 0.

    Work inside each pair of bars first. Then take each absolute value. Subtract the two results last.

  2. Work inside the first bars

    Start with (−6)−(−5)+4.2(-6) - (-5) + 4.2. Subtracting −5 is the same as adding 5.

    (−6)−(−5)+4.2=−6+5+4.2=−1+4.2=3.2(-6) - (-5) + 4.2 = -6 + 5 + 4.2 = -1 + 4.2 = 3.2

    3.2 is already positive, so |3.2|=3.2|3.2| = 3.2.

    Watch out: If you read the start as −6 − 5, you get −11. That path ends at the wrong answer 0.2.

  3. Work inside the second bars

    Next, 3−9.6=−6.63 - 9.6 = -6.6. This result is negative. Its distance from 0 is 6.6, so the absolute value is positive.

    |−6.6|=6.6|-6.6| = 6.6

  4. Subtract the two results

    Now take the second result away from the first.

    3.2−6.6=−3.43.2 - 6.6 = -3.4

    You take a bigger number away from a smaller one, so the result is negative.

    Watch out: The bars made each part positive. They do not make the final answer positive, so 3.4 is wrong. Also, bars are not parentheses. Working out 3.2 − (−6.6) = 9.8 is wrong too.

    So you would grid -3.4, with the minus sign.

Check your answer

Estimate. The first part is about 3. The second part is about 7. And 3 − 7 is about −4. The answer −3.4 is close to that. It is also negative, as the estimate says it should be.

More questions like this

The original DOE question
Question 60 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 60. Open page 153 of the handbook (PDF)

Question 114 of 120

2018 Test B Math, question 12

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 115 of 120

2018 Test B Math, question 13

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 116 of 120

2018 Test B Math, question 16

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 117 of 120

2018 Test B Math, question 17

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 118 of 120

2018 Test B Math, question 22

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 119 of 120

2018 Test B Math, question 24

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 120 of 120

2018 Test B Math, question 25

Question

Jack scored a mean of 15 points per game in his first 3 basketball games. In his 4th game, he scored 27 points. What is his mean score for the first 4 games?

  • A15
  • B17
  • C18
  • D21
Show the answer

Correct answer: C

Jack scored a mean of 15 points per game in each of the first 3 games, so he earned a total of 45 points for the first 3 games. Use that information to calculate the mean over the four games:

45+274=724=18\frac{45+27}{4}=\frac{72}{4}=18

Question analysis

This statistics question gives Jack's mean score for his first three basketball games and his score in the fourth, and asks for his mean over all four games. Students recover a total from a mean, add the new score and divide again. The earlier mean stands for three games, not one, which is easy to forget.

NYS standards
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Finding a mean is grade 6 content, but working back from a mean to a total and then updating it is a common grade 7 test item, with a tempting shortcut that gives the wrong value.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 1, 29 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The tempting shortcut, averaging 15 and 27, is quick, feels reasonable and gives a choice, so the question rewards students who slow down. Its error is weighting: the 15 stands for three games and the 27 for one. Rebuilding the total from the mean and dividing by four games takes only a moment longer.

  1. Turning words into math moderate Understanding the question

    The phrase “a mean of 15 points per game in his first 3 games” has to be turned into a total, 45 points, before the fourth game's 27 can be added. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Students must know that a mean is a total divided by a count, so a mean cannot be averaged with a single score as if both had equal weight. (Anderson & Krathwohl, 2001)

Strategy: Multiply the first mean by 3 to get the points so far, add the fourth game's points, and divide the new total by 4.

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

Mean (average)
mean=sum of the valuesnumber of values\text{mean} = \frac{\text{sum of the values}}{\text{number of values}}
The mean is the average. Add all the values, here the points in each game. Then divide by how many values there are, here the number of games. Turned around, it also says: sum = mean × number of values.

  1. Write down what you know

    Jack’s mean for his first 3 games was 15 points. The mean is the average: the total shared out equally among the games.

    In game 4 he scored 27 points. You need his mean for all 4 games.

  2. Find the total for 3 games

    A mean of 15 over 3 games means the total is 3 times 15. You need the total, because a new mean is built from the total.

    15×3=4515 \times 3 = 45

    He scored 45 points in the first 3 games.

  3. Add the fourth game

    Now add the 27 points from game 4 to get the total for all 4 games.

    45+27=7245 + 27 = 72

    He scored 72 points in 4 games.

  4. Divide by 4 games

    mean=sum of the valuesnumber of values=724=18\text{mean} = \frac{\text{sum of the values}}{\text{number of values}} = \frac{72}{4} = 18

    Watch out: Do not average 15 and 27. That gives 21, choice D. It treats one game as if it counted as much as three.

    So the answer is C, 18.

Check your answer

The mean should go up a little, since 27 is above 15. It went from 15 to 18, so that makes sense. Also, 18×4=7218 \times 4 = 72, which matches the total points.

More questions like this

The original DOE question
Question 82 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 82. 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).