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Question 1 of 99

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 99

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 99

2026 Practice Test A Math, question 27

Question

Complete the steps to show one way to solve the equation 24=−0.5(x+8)24 = -0.5(x + 8) for x.

Move the correct answer to each box. Each answer may be used more than once. Not all answers will be used.

24=−0.5(x+8)24 = -0.5(x + 8)

24=−0.5x+24 = -0.5x +

=−0.5x= -0.5x

=x= x

On the test you drag tiles into the boxes; here, choose each box’s tile from its menu. A tile can be used more than 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: −4; blank 2: 28; blank 3: −56

Solution:

Answer for second equation: −4

Answer for third equation: 28

Answer for fourth equation: −56

24=−0.5(x+8)24 = -0.5(x + 8)

Distribute the −0.5 on the right side of the equation: 24=−0.5x−424 = -0.5x - 4.

Add 4 to both sides of the equation to eliminate the −4 on the right side: 28=−0.5x28 = -0.5x.

Divide both sides by −0.5: −56=x-56 = x.

Question analysis

Students complete a step-by-step solution of the equation 24 = −0.5(x + 8) by dragging number tiles into three boxes: one after distributing, one after collecting the constants, and one for x. It tests solving a two-step linear equation with a negative decimal. Tiles may be reused, and many come in positive and negative pairs, so every sign counts.

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 Solving equations of the form p(x + q) = r with rational numbers is named in the grade 7 standards. Here the steps are laid out, but the negative decimal factor forces a sign decision at every one of the three boxes.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The layout settles the method, so the question comes down to one sign at each box. The printed plus before the first box nudges students toward the positive 4, and the pool then supplies every number that follows from it, down to −40. Students who distribute correctly can still lose the item at the last box by dividing 28 by 0.5. Substituting −56 back, −0.5 × (−48) = 24, confirms the final box.

  1. Error-prone arithmetic moderate Working it out

    Two of the three boxes turn on a negative decimal: −0.5 times 8 must give −4, and dividing 28 by −0.5 means multiplying by −2, which doubles the size and makes it negative. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer minor Choosing the answer

    The pool pairs −4 with 4, −40 with 40 and −56 with 56, so a box filled with the right size and the wrong sign looks just as finished; estimating cannot separate −56 from 56. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Many steps minor Working it out

    The three boxes form a chain and all three must be right, so a slip in the first carries through; the pool even holds the whole wrong chain 4, 20 and −40. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The second line prints a plus sign before its box, so the distributed −0.5x − 4 must be written as −0.5x + (−4), with the negative tile in the box. (Enright, Morley & Sheehan, 2002)

Strategy: Multiply −0.5 by both x and 8, undo the constant term by doing the opposite operation to both sides, then divide by −0.5, which is the same as multiplying by −2.

Traps

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

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. Multiply a by each term inside. If a is negative, the minus sign goes into every product. It works because a groups of b + c is the same as a groups of b plus a groups of c.
Signs when you multiply or divide
(−)×(+)=(−)(−)×(−)=(+)(-) \times (+) = (-) \qquad (-) \times (-) = (+)
A negative number times a positive number gives a negative result. A negative times a negative gives a positive result. The same rules hold when you divide. Use them at every step where a negative number appears.
Doing the same to both sides
if a=b, then a+c=b+c and ac=bc\text{if } a = b \text{, then } a + c = b + c \text{ and } \frac{a}{c} = \frac{b}{c}
a and b are the two sides of an equation, and c is any number (not zero when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it level. That is how you get x by itself.

  1. Sort out the question

    You solve 24=−0.5(x+8)24 = -0.5(x + 8) by filling three boxes:

    • Box 1 comes after −0.5x+-0.5x +, once the parentheses are gone.
    • Box 2 is the number equal to −0.5x-0.5x.
    • Box 3 is the value of x.

    A tile can be used more than once, and some tiles are not used.

  2. Distribute the −0.5

    Multiply −0.5 by each term inside the parentheses. A negative times a positive is negative:

    −0.5×x=−0.5x-0.5 \times x = -0.5x

    −0.5×8=−4-0.5 \times 8 = -4

    So 24=−0.5x+(−4)24 = -0.5x + (-4). The line already shows a plus sign, so box 1 gets −4.

    Watch out: the −0.5 multiplies the 8 too. Writing 4 instead starts a wrong chain: 4, then 20, then −40. Every number in that chain is a tile.

  3. Add 4 to both sides

    You want −0.5x alone on one side, and the −4 is in the way. Adding 4 undoes subtracting 4, so add 4 to both sides:

    24+4=−0.5x−4+424 + 4 = -0.5x - 4 + 4

    28=−0.5x28 = -0.5x

    Box 2 gets 28.

    Watch out: taking 4 away from 24 instead gives 20, another tile.

  4. Divide both sides by −0.5

    Now x is multiplied by −0.5. Dividing undoes multiplying, so divide both sides by −0.5:

    28÷(−0.5)=−5628 \div (-0.5) = -56

    Dividing by 0.5 is the same as multiplying by 2, since every whole holds two halves. A positive divided by a negative is negative. So x=−56x = -56.

    Watch out: dropping the minus sign gives 56, which is also a tile.

  5. Name the three tiles

    From top to bottom, the finished steps read:

    24=−0.5x+(−4)24 = -0.5x + (-4)

    28=−0.5x28 = -0.5x

    −56=x-56 = x

    So the boxes get −4, 28 and −56, in that order. On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Put x=−56x = -56 back into the first equation. −56+8=−48-56 + 8 = -48. Then −0.5×(−48)=24-0.5 \times (-48) = 24, since a negative times a negative is positive. That gives 24, the left side, so −56 is right.

More questions like this

The original DOE question
Question 27 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 A Math, page 98, question 27.

Question 4 of 99

2026 Practice Test A Math, question 43

Question

Graph the solution set for the inequality −3≤2x−5-3 \leq 2x - 5.

Select a ray. Then, move the endpoint of the ray to the correct position on the number line.

On the test you draw the ray on the number line; here, choose its endpoint, whether the endpoint is closed or open, and its direction. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

-5-4-3-2-1012345

Show the answer

Correct answer: a ray with a closed endpoint at 1, pointing right

Solution:

A number line from −5 to 5 with a tick at every integer, labeled at −5, 0 and 5. A solid dot at 1, with the line drawn thick from 1 to the right and ending in an arrow.

Solve the inequality for x, then graph the solution on the number line:

−3≤2x−5-3 \leq 2x - 5

Add 5 to both sides of the inequality: 2≤2x2 \leq 2x

Divide both sides by 2: 1≤x1 \leq x

Rewrite the inequality with x on the left side: x≥1x \geq 1

The graph of x≥1x \geq 1 is a closed ray (solid point) beginning at 1 and extending to the right on the number line.

Question analysis

Students graph the solution set of a two-step inequality on a number line by choosing a ray: where it starts, whether its endpoint is open or closed, and which way it points. It tests solving a linear inequality and reading its solution. The variable sits on the right of the inequality symbol, so reading the result the right way round is part of the challenge.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving a two-step inequality and graphing its solution set is a named grade 7 skill. This one puts the variable on the right of the symbol and asks students to build the graph themselves rather than pick it from four.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 26 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Solving takes two routine moves; the points are lost in turning 1 ≤ x into a picture. Because x sits on the right, the symbol seems to point left, and a student who shades the way it points draws the values below 1, which all fail. The dot must also be solid, since x = 1 makes both sides −3. Testing x = 2, where 2x − 5 = −1 and −3 ≤ −1 holds, confirms the direction.

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

    The solution comes out as 1 ≤ x, with x on the right, so the ≤ sign points away from the shaded side; it has to be read as x ≥ 1 before the ray is drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Two solving moves, adding 5 and then halving, come before three graph choices, the endpoint, a solid dot and a direction, and a mistake in any of the five loses the item. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The ray is built, by picking one of four ray shapes and dragging its endpoint, rather than chosen from four finished graphs, so nothing on screen flags a ray that starts at the right number but points the wrong way. (Enright, Morley & Sheehan, 2002)

Strategy: Undo the subtraction and then the multiplication on both sides, rewrite the result with x first, decide whether the endpoint is included, and test one value on your ray in the original inequality.

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides
if a≤b, then a+c≤b+c\text{if } a \leq b \text{, then } a + c \leq b + c
a and b are the two sides of an inequality, and c is any number. Adding the same number to both sides keeps the inequality true. So does dividing both sides by the same positive number. Dividing by a negative number would flip the symbol, but that does not happen here.
Turning an inequality around
a≤b means the same as b≥aa \leq b \text{ means the same as } b \geq a
a and b are any two numbers or terms. You may swap the two sides if you also turn the symbol around. The pointed end of the symbol always faces the smaller side. So 1 ≤ x and x ≥ 1 say the same thing.

  1. Sort out the question

    The symbol ≤ means is less than or equal to. The solution set is every value of x that makes −3≤2x−5-3 \leq 2x - 5 true. You show it as a ray. A ray is a dot, called the endpoint, with an arrow going one way forever. You need three things: where the dot goes, whether it is closed or open, and which way the ray points.

  2. Add 5 to both sides

    The right side is 2x − 5. Adding 5 undoes subtracting 5, so add 5 to both sides:

    −3+5≤2x−5+5-3 + 5 \leq 2x - 5 + 5

    2≤2x2 \leq 2x

    Watch out: taking 5 away from −3 instead gives −8 ≤ 2x. That wrong move puts the dot at −4.

  3. Divide both sides by 2

    Now x is multiplied by 2, so divide both sides by 2. You divide by a positive number, so the symbol stays the same:

    22≤2x2\frac{2}{2} \leq \frac{2x}{2}

    1≤x1 \leq x

  4. Read it with x first

    1≤x1 \leq x says 1 is less than or equal to x. So x is 1 or any number bigger than 1. With x first, that is:

    x≥1x \geq 1

    Bigger numbers are to the right on a number line, so the ray points right.

    Watch out: in 1 ≤ x, the symbol seems to point left. Do not follow it. A ray pointing left from 1 shows numbers that do not work.

  5. Choose a closed dot

    The words or equal to mean 1 itself is a solution. Check it: 2×1−5=−32 \times 1 - 5 = -3, and −3 ≤ −3 is true. So the dot at 1 is closed, which means filled in. An open dot, an empty circle, leaves the endpoint out. Use it only for < or >.

    Watch out: an open dot at 1 would leave out a number that works.

  6. Draw the ray

    On the test, choose the ray with a filled dot and an arrow pointing right. Then drag its dot to 1. On this site, choose 1 for the endpoint, then closed (filled dot) and pointing right.

    So the answer is a closed dot at 1, with the ray pointing right.

Check your answer

Test a number on your ray, such as 3. 2×3−5=12 \times 3 - 5 = 1, and −3 ≤ 1 is true. Now test a number off the ray, such as 0. 2×0−5=−52 \times 0 - 5 = -5, and −3 ≤ −5 is false. So the ray points the right way.

More questions like this

The original DOE question
Question 43 as shown on the DOE’s digital test, page 114 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test A Math, page 114, question 43.

Question 5 of 99

2026 Practice Test B Math, question 4

Question

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

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

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

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

How many minks will Maria receive from the exchange?

Enter your answer in the space.

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

Show the answer

Correct answer: 116

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

  2. Many steps minor Working it out

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

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

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

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

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

  2. Find what one zog is worth

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

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

  3. Turn the bart rule around

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

    4×14=14 \times \frac{1}{4} = 1

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

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

  4. Change the zogs into minks

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

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

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

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

  5. Change the barts into minks

    Each of the 24 barts is worth 4 minks:

    24×4=9624 \times 4 = 96

    So 24 barts trade for 96 minks.

  6. Add the minks

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

    20+96=11620 + 96 = 116

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

Check your answer

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

More questions like this

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

Question 6 of 99

2026 Practice Test C Math, question 13

Question

Start with the number 135.28 and perform the following steps in order:

  • Step 1: Multiply by 10.
  • Step 2: Add 0.5 to the result of Step 1.
  • Step 3: Drop the digits after the decimal point.
  • Step 4: Divide by 10.

What is the result?

  • A13.5
  • B135
  • C135.2
  • D135.3
Show the answer

Correct answer: D

Begin by multiplying 135.28 by 10:

135.28×10=1,352.8135.28 \times 10 = 1{,}352.8

Add 0.5:

1,352.8+0.5=1,353.31{,}352.8 + 0.5 = 1{,}353.3

Drop the digits after the decimal point, so 1,353.31{,}353.3 becomes 1,3531{,}353.

Divide by 10:

1,353÷10=135.31{,}353 \div 10 = 135.3

Question analysis

Students start with 135.28 and carry out four listed steps: multiply by 10, add 0.5, drop the digits after the decimal point, and divide by 10. It tests place value and operations with decimals. Skipping a step, or doing two of them in the wrong order, changes the result, and the choices are built from those slips.

NYS standards
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Multiplying and dividing a decimal by 10 and adding 0.5 are grade 5 and 6 skills. The four steps are spelled out, so the demand is following them exactly and in order, not choosing a method.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 3, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The steps spell out how to round 135.28 to the nearest tenth, but nothing says so, and they have to be followed exactly. The decisive moment is step 2, where adding 0.5 pushes 1,352.8 past 1,353; skip it and the digits are simply cut, not rounded. Students who recognize the routine can check the result directly, since 135.28 lies closer to 135.3 than to 135.2.

  1. A very close wrong answer moderate Choosing the answer

    135.3 and 135.2 differ only in the tenths digit, and 135 shares their whole-number part, so a rough sense of the result's size rules out nothing. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Many steps minor Working it out

    Four listed steps move the decimal point twice, and doing them in another order, or skipping one, still produces a believable number near 135. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    Adding 0.5 to 1,352.8 carries into the ones place, since 8 tenths and 5 tenths make 1.3, which turns 1,352 into 1,353 before the decimals are dropped. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write down the number after each step instead of doing them in your head, and carry out the steps exactly in the order listed.

Traps

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

Multiplying and dividing by 10
a×10anda÷10a \times 10 \quad \text{and} \quad a \div 10
a is any decimal number. Multiplying by 10 moves its decimal point one place to the right, so each digit is worth ten times as much. Dividing by 10 moves the point one place to the left. The two steps undo each other.

  1. Sort out the question

    You start with 135.28 and do four steps, in order. Write down the number after each step. Doing them in your head, or in another order, leads to a wrong choice. You need the number at the end.

  2. Multiply by 10

    Step 1. Multiplying by 10 moves the decimal point one place to the right:

    135.28×10=1,352.8135.28 \times 10 = 1{,}352.8

    The digits stay the same. Only the decimal point moves.

  3. Add 0.5

    Step 2. Line up the decimal points. 8 tenths plus 5 tenths is 13 tenths. That is 1 whole and 3 tenths, so carry the 1 into the ones place:

    1,352.8+0.5=1,353.31{,}352.8 + 0.5 = 1{,}353.3

    Watch out: skipping this step, or forgetting to carry, keeps 1,352. That leads to choice C, 135.2.

  4. Drop the digits after the point

    Step 3. Cut off the digits after the decimal point. Do not round. So 1,353.3 becomes 1,353.

    Watch out: do step 3 before step 4. Dividing first gives 135.33. Dropping its decimals then gives 135, choice B.

  5. Divide by 10

    Step 4. Dividing by 10 moves the decimal point one place to the left:

    1,353÷10=135.31{,}353 \div 10 = 135.3

    Skipping step 1 gives 13.5, choice A. So the answer is D, 135.3.

Check your answer

These four steps are a recipe for rounding to the nearest tenth. 135.28 is between 135.2 and 135.3. Its hundredths digit is 8, which is 5 or more, so it rounds up to 135.3. The recipe and the rounding agree.

More questions like this

The original DOE question
Question 13 as shown on the DOE’s digital test, page 86 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 86, question 13.

Question 7 of 99

2026 Practice Test C Math, question 14

Question

If x=−2x = -2, what is the value of |x+x2+x3|−|x+2x+3x||x + x^{2} + x^{3}| - |x + 2x + 3x|?

  • A−18
  • B−6
  • C−2
  • D2
Show the answer

Correct answer: B

Substitute −2-2 for each value of x in the expression:

|(−2)+(−2)2+(−2)3|\left|(-2) + (-2)^{2} + (-2)^{3}\right|
minus
|(−2)+2(−2)+3(−2)||(-2) + 2(-2) + 3(-2)|

Evaluate the terms involving exponents:

|(−2)+4+(−8)||(-2) + 4 + (-8)|
minus
|(−2)+2(−2)+3(−2)||(-2) + 2(-2) + 3(-2)|

Evaluate the terms involving multiplication:

|(−2)+4+(−8)||(-2) + 4 + (-8)|
minus
|(−2)+(−4)+(−6)||(-2) + (-4) + (-6)|

Simplify the expression:

|2+(−8)|−|(−2)+(−10)||2 + (-8)| - |(-2) + (-10)|

|−6|−|−12|\lvert -6 \rvert - \lvert -12 \rvert

The absolute value of (−6)(-6) is 6, and the absolute value of (−12)(-12) is 12, so |−6|−|−12|\lvert -6 \rvert - \lvert -12 \rvert is the equivalent of subtracting 12 from 6:

6−12=(−6)6 - 12 = (-6)

Question analysis

This substitution question sets x = −2 in an expression that subtracts one absolute value from another: the first holds x, x squared and x cubed, the second x, 2x and 3x. Students evaluate the expression. It tests powers of negative numbers, absolute value and integer subtraction, and the signs of the square and the cube are where answers go wrong.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.EE.2c evaluating expressions and order of operations
  • 6.NS.7c absolute value as distance
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Substituting a negative number and taking absolute values is typical grade 7 sign practice. This expression puts a square and a cube of a negative inside one pair of bars and then subtracts two absolute values, so there are several places to slip.
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 7, 8 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is routine, but the expression stacks sign decisions, and each wrong choice records a different one going wrong. The powers of −2 alternate in sign, which students who treat every power as positive, or the square as negative, miss. The last step catches students who know the rules: both absolute values are positive, yet 6 − 12 is still negative, so the answer keeps its minus sign.

  1. Error-prone arithmetic moderate Working it out

    With x = −2 the square is 4 but the cube is −8, so the first bars hold −2 + 4 − 8 = −6, and the sign changes from each term to the next. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Six terms are substituted, two pairs of bars are simplified and made positive, and only then is 12 subtracted from 6, four stages that must come in a fixed order. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out each power of −2 separately, keeping its sign, add the terms inside each pair of bars, take both absolute values, and subtract only at the end.

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.)
  • 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

Absolute value
|a|=the distance from a to 0|a| = \text{the distance from } a \text{ to } 0
a is any number. The bars mean absolute value: how far a is from 0 on the number line. A distance is never negative, so |−6| = 6 and |6| = 6. Work out what is inside the bars first, then take the absolute value.
Powers of a negative number
(−2)2=(−2)×(−2)(−2)3=(−2)×(−2)×(−2)(-2)^{2} = (-2) \times (-2) \quad (-2)^{3} = (-2) \times (-2) \times (-2)
The small raised number is the exponent. It tells how many times −2 is multiplied by itself. Two negatives multiply to a positive. So an even number of negative factors gives a positive answer, and an odd number gives a negative one.

  1. Sort out the question

    You need the value of |x+x2+x3|−|x+2x+3x||x + x^{2} + x^{3}| - |x + 2x + 3x| when x = −2. The bars | | mean absolute value. Work out each pair of bars on its own. Subtract only at the end.

  2. Find the powers of −2

    Put −2 in for x. Keep each negative sign:

    (−2)2=(−2)×(−2)=4(-2)^{2} = (-2) \times (-2) = 4

    (−2)3=4×(−2)=−8(-2)^{3} = 4 \times (-2) = -8

    The square is positive, but the cube is negative.

    Watch out: taking (−2)3(-2)^{3} as +8 makes the first bars hold 10. That leads to choice C, −2.

  3. Work out the first bars

    Add the three terms inside the first pair of bars:

    |(−2)+4+(−8)|=|−6|=6|(-2) + 4 + (-8)| = |-6| = 6

    −2 + 4 is 2, and 2 + (−8) is −6. The distance from −6 to 0 is 6.

  4. Work out the second bars

    Here 2x means 2 times x, so it is −4. And 3x is −6:

    |(−2)+(−4)+(−6)|=|−12|=12|(-2) + (-4) + (-6)| = |-12| = 12

    All three terms are negative, so they add to −12. Its absolute value is 12.

  5. Subtract the second value

    Now subtract the second value from the first:

    6−12=−66 - 12 = -6

    Watch out: the bars make each part positive, but the difference can still be negative. Keeping −6 inside the first bars gives −6 − 12 = −18, choice A.

    So the answer is B, −6.

Check your answer

Check the second bars a faster way. x+2x+3x=6xx + 2x + 3x = 6x, and 6×(−2)=−126 \times (-2) = -12. That matches. Also, the second value, 12, is bigger than the first, 6. So the difference must be negative, as −6 is.

More questions like this

The original DOE question
Question 14 as shown on the DOE’s digital test, page 87 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 87, question 14.

Question 8 of 99

2026 Practice Test C Math, question 15

Question

A grocery store sells 1-gallon containers of milk for $3.99. The store also sells orange juice in a 6-pack of 5.5-fluid-ounce bottles for $1.79. Suppose the store wants to sell its orange juice in gallon containers instead. To the nearest dollar, how much more would a gallon of orange juice cost than a gallon of milk?

Enter your answer in the space.

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

Show the answer

Correct answer: 3

To determine the cost of a gallon of orange juice, begin by finding the total number of ounces in the 6-pack of bottles:

5.5 oz×6=33 oz5.5\text{ oz} \times 6 = 33\text{ oz}

So, 33 oz of orange juice costs $1.79.

There are 128 oz in 1 gallon. Set up a proportion using the $1.79 cost per 33 oz of orange juice and the unknown cost (x) per 1 gallon of orange juice:

33 oz$1.79=128 ozx\frac{33\text{ oz}}{\$1.79} = \frac{128\text{ oz}}{x}

Cross multiply to solve for x in the proportion:

(33)×(x)=($1.79)×(128)(33) \times (x) = (\$1.79) \times (128)

33x=$229.1233x = \$229.12

x≈$6.94x \approx \$6.94

So, the orange juice costs $6.94 per gallon, and the milk costs $3.99 per gallon.

Subtract the cost per gallon of milk from the cost per gallon of orange juice.

$6.94−$3.99=$2.95\$6.94 - \$3.99 = \$2.95

Round the difference to the nearest dollar.

$2.95 rounds to $3.

Question analysis

A store sells milk by the gallon for $3.99 and orange juice in six-packs of 5.5-fluid-ounce bottles for $1.79. Students work out what a gallon of the juice would cost at the same price per ounce and how much more that is than the milk, to the nearest dollar. The ounces in a gallon must be known, and several steps involve decimals.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3d converting measurement units
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Unit prices are grade 6 and 7 work, but this chains five steps with decimals and needs a fact the question does not give, the number of fluid ounces in a gallon. Typing a rounded difference, with no choices, adds risk.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
5 steps
Reading level
Question and choices: about grade 5, 60 words.
Corrected from the DOE handbook
  • A rounded value in the explanation is now marked as approximate (≈ instead of =).
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 calculation is long, and its key fact is missing from the page. Students who know the 128 fluid ounces still face a division by 33 that does not come out evenly, which tempts them to round the price per ounce to 5 cents and lose about 50 cents a gallon. Keeping the exact quotient to the end, or scaling 33 ounces up by about 3.9, keeps the difference near $2.95, which rounds to 3.

  1. Something you must already know moderate Working it out

    The question never says that a gallon holds 128 fluid ounces, yet the whole comparison depends on it; the 6-pack holds only 6 × 5.5 = 33 ounces. (Anderson & Krathwohl, 2001)

  2. Many steps moderate Working it out

    Five moves follow one another: the ounces in a pack, the scale-up to 128 ounces, the division by 33, the subtraction of $3.99 and the rounding to a whole dollar. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    1.79 × 128 = 229.12 and 229.12 ÷ 33 are long decimal calculations without a calculator, and a slip in either moves the price of a gallon. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The rounded difference is typed with no choices, so an unrounded $2.95, or a figure rounded too soon, looks just as finished as the whole-dollar answer. (Enright, Morley & Sheehan, 2002)

Strategy: Find the ounces in one six-pack, scale its price up to the 128 fluid ounces in a gallon, subtract the milk's price, and round only at the end.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • 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 DOE’s explanation goes from 33x = $229.12 to “x = $6.94”. But 229.12 ÷ 33 = 6.9430…, so $6.94 is the cost per gallon rounded to the cent. The site writes x ≈ $6.94. Nothing else changes: $6.94 − $3.99 = $2.95, which rounds to $3, and the unrounded cost gives the same answer.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fluid ounces in a gallon
fluid ounces=128×gallons\text{fluid ounces} = 128 \times \text{gallons}
A gallon holds 128 fluid ounces. This is a fact you must know, because the question does not give it. A fluid ounce measures an amount of liquid. Multiply gallons by 128 to change them into fluid ounces.
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 ratios are equal. Here each ratio is ounces over dollars. 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. That turns the proportion into an equation.

  1. Sort out the question

    Milk costs $3.99 for 1 gallon. Orange juice comes in a 6-pack of 5.5-fluid-ounce bottles for $1.79. You need how much more a gallon of juice would cost than a gallon of milk. Round to the nearest dollar.

    The question doesn’t say how big a gallon is. You must know this fact: 1 gallon is 128 fluid ounces.

  2. Find the ounces in a 6-pack

    Each of the 6 bottles holds 5.5 fluid ounces:

    6×5.5=336 \times 5.5 = 33

    So $1.79 buys 33 fluid ounces of juice.

  3. Set up a proportion

    Let x be the cost in dollars of a gallon, 128 fluid ounces, of juice. The price per ounce stays the same. So ounces over dollars is the same for the pack and the gallon:

    331.79=128x\frac{33}{1.79} = \frac{128}{x}

    Cross-multiply:

    33×x=1.79×128=229.1233 \times x = 1.79 \times 128 = 229.12

  4. Find the cost of a gallon

    Divide both sides by 33:

    x=229.12÷33≈6.94x = 229.12 \div 33 \approx 6.94

    The sign ≈ means about equal. The exact value is 6.943…, so a gallon of juice costs about $6.94.

    Watch out: don’t round too early. One ounce costs about $0.054. Rounding that to $0.05 makes a gallon $6.40, and your answer becomes 2.

  5. Subtract the milk’s price

    Find how much more the juice costs:

    6.94−3.99=2.956.94 - 3.99 = 2.95

    A gallon of juice costs $2.95 more than a gallon of milk.

    Watch out: entering 7, the juice’s price, forgets to subtract the milk.

  6. Round and enter it

    The question asks for the nearest dollar. $2.95 is between $2 and $3. Its cents, 95, are 50 or more, so it rounds up to $3. Do not enter 2.95.

    Type only the number in the box. So you would enter 3.

Check your answer

Estimate. 128 ounces is almost 4 packs, since 4×33=1324 \times 33 = 132. Four packs cost 4×1.79=7.164 \times 1.79 = 7.16 dollars, so a gallon costs a bit less than that. Milk costs about $4. About $7 minus $4 is $3, which matches.

More questions like this

The original DOE question
Question 15 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 C Math, page 88, question 15. This is the DOE’s version, before the corrections listed in the analysis.

Question 9 of 99

2026 Practice Test C Math, question 17

Question

A number line from 0 to 8 with a tick and a label at every whole number. Four points are marked on it: M at 1, R at 3, S at 6 and N at 7.

What is the distance between the midpoints of MN\overline{MN} and RS\overline{RS}?

  • A12\frac{1}{2} unit
  • B1 unit
  • C1121\frac{1}{2} units
  • D2 units
Show the answer

Correct answer: A

MN\overline{MN} begins at 1 and ends at 7. Determine the length of MN\overline{MN}:

7−1=67 - 1 = 6

The midpoint of a line segment is located exactly halfway between the two endpoints. Since MN\overline{MN} is 6 units in length, the two endpoints are 6 units apart. Determine the location of the midpoint, in relation to each endpoint, by dividing 6 by 2:

6÷2=36 \div 2 = 3

The location of the midpoint of MN\overline{MN} is 3 units from either endpoint. Find the midpoint by either adding 3 to the location of M (1) or subtracting 3 from the location of N (7):

1+3=41 + 3 = 4

7−3=47 - 3 = 4

The midpoint of MN\overline{MN} is located at 4.

RS\overline{RS} begins at 3 and ends at 6. Determine the length of RS\overline{RS}:

6−3=36 - 3 = 3

Since RS\overline{RS} is 3 units in length, the two endpoints are 3 units apart. Determine the location of the midpoint, in relation to each endpoint, by dividing 3 by 2:

3÷2=1.53 \div 2 = 1.5

The location of the midpoint of RS\overline{RS} is 1.5 units from either endpoint. Find the midpoint by either adding 1.5 to the location of R (3) or subtracting 1.5 from the location of S (6):

3+1.5=4.53 + 1.5 = 4.5

6−1.5=4.56 - 1.5 = 4.5

The midpoint of RS\overline{RS} is located at 4.5.

Determine the distance between the two midpoints by subtracting:

4.5−4=0.54.5 - 4 = 0.5

0.5 is equivalent to 12\frac{1}{2}, making the distance between the two midpoints 12\frac{1}{2} unit.

Question analysis

A number line from 0 to 8 shows points M, R, S and N, and students find the distance between the midpoint of segment MN and the midpoint of segment RS. It tests locating points and finding distances on a number line. One segment sits inside the other, and several lengths in the figure can be mistaken for the distance wanted.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding distances between points on a number line is grade 7 work, here with whole-number positions. Two midpoints must be located before their distance can be found, so the task takes three short steps rather than one.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 17 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The positions are whole numbers, so the midpoints come from simple averages, (1 + 7) ÷ 2 and (3 + 6) ÷ 2. The difficulty is keeping several lengths apart: the half-lengths of the two segments, the gaps at their ends, and the distance actually asked for, half a unit. Marking both midpoints on the number line before subtracting makes the small gap visible.

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

    Segment RS lies inside MN, from 3 to 6, and its midpoint, 4 1/2, falls between two ticks of the number line rather than on a labeled point. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Two midpoints have to be found, at 4 and 4 1/2, before their distance is taken, and the half-length of RS met on the way, 1 1/2, is among the choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find each midpoint as the average of its segment's two endpoints, then subtract the smaller midpoint from the larger one.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint on a number line
midpoint=a+b2\text{midpoint} = \frac{a + b}{2}
a and b are the numbers at the two ends of a segment. The midpoint is the point exactly halfway between them. Adding the ends and dividing by 2 finds their average, which is the middle.
Distance on a number line
distance=larger−smaller\text{distance} = \text{larger} - \text{smaller}
The distance between two points on a number line is how many units apart they are. Subtract the smaller number from the larger one, so the distance is never negative.

  1. Sort out the question

    The number line shows M at 1, R at 3, S at 6 and N at 7. Segment MN goes from 1 to 7. Segment RS goes from 3 to 6. A midpoint is the point exactly halfway along a segment. You need the distance between the two midpoints.

  2. Find the midpoint of MN

    Use the midpoint formula with the ends 1 and 7:

    1+72=82=4\frac{1 + 7}{2} = \frac{8}{2} = 4

    The midpoint of MN is at 4, which is 3 units from each end.

  3. Find the midpoint of RS

    Use the same formula with the ends 3 and 6:

    3+62=92=4.5\frac{3 + 6}{2} = \frac{9}{2} = 4.5

    The midpoint of RS is at 4.5. That is halfway between the marks for 4 and 5, where no point is labeled.

  4. Subtract the two midpoints

    Use the distance formula:

    4.5−4=0.54.5 - 4 = 0.5

    0.5 is one half, so the midpoints are 1/2 unit apart.

    Watch out: half of MN is 3, and half of RS is 1 1/2. Subtracting those gives 1 1/2, choice C. But they are half-lengths, not positions. And 2, choice D, is only the distance from M to R.

  5. Choose the answer

    The midpoints, 4 and 4.5, are close together because RS sits near the middle of MN. So the answer is A, 1/2 unit.

Check your answer

Find each midpoint another way: walk half the length from the left end. MN is 7−1=67 - 1 = 6 units long, and 1+3=41 + 3 = 4. RS is 6−3=36 - 3 = 3 units long, and 3+1.5=4.53 + 1.5 = 4.5. Same midpoints, 1/2 unit apart.

More questions like this

The original DOE question
Question 17 as shown on the DOE’s digital test, page 90 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 90, question 17.

Question 10 of 99

2026 Practice Test C Math, question 31

Question

NUMBER OF VISITORS
TO BROOKSIDE ZOO

Male Female
Children 360 340
Adults 210 290

The table above shows the number of people who visited Brookside Zoo in one day. The largest of the four groups made up what percentage of all visitors on that day?

  • A30%
  • B28%
  • C24%
  • D17%
Show the answer

Correct answer: A

First, determine the total number of visitors.

360+340+210+290=1200360 + 340 + 210 + 290 = 1200

The largest group is 360 (Male Children).

Find the percent that represents 3601200\frac{360}{1200}.

3601200×1001%=36012×11%=30%\frac{360}{1200} \times \frac{100}{1}\% = \frac{360}{12} \times \frac{1}{1}\% = 30\%

Question analysis

A two-way table counts one day's visitors to Brookside Zoo by age, children or adults, and by male or female. Students find what percent of all visitors the largest of the four groups made up. They must total every cell first, then make sure the count they divide is really the largest one.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding what percent one count is of a total is grade 6 work. A two-way table adds two quick steps, totaling the four groups and spotting the largest, but the numbers divide evenly, so it stays below 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 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

The arithmetic is friendly, since 360 over 1,200 reduces to 3 over 10, so the work is in reading the table. Two of its cells are close, 340 just below 360, and every wrong choice is the share of one of the other groups. Adding all four counts first, and circling the largest before dividing, keeps the right cell in play.

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

    The two-way table gives four separate counts and no totals, so the 1,200 visitors must be added up from the cells and the largest group picked out of a grid. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Three short steps run in order: total the four cells, pick the largest, and divide; using the wrong cell in the last step still gives a believable percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add all four cells for the day's total, pick the biggest single cell, and divide it by the total; simplifying the fraction before dividing keeps the arithmetic short.

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.
  • 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

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

  1. Sort out the question

    The table splits one day’s visitors into four groups:

    MaleFemale
    Children360340
    Adults210290

    You need the largest group as a percent of all the visitors. The table gives no total, so you must find it.

  2. Add up all the visitors

    Add all four groups to get the whole:

    360+340+210+290=1,200360 + 340 + 210 + 290 = 1{,}200

    So 1,200 people visited the zoo that day.

  3. Find the largest group

    Compare the four numbers. The biggest is 360, the male children.

    Watch out: 340 is close to 360, but it is smaller. Using 340 gives about 28%, choice B. Using the 290 female adults gives about 24%, choice C. And 210 is the smallest group, which gives 17.5%, near choice D.

  4. Find the percent

    Put the part over the whole. Then simplify by dividing the top and bottom by 120:

    3601,200=310\frac{360}{1{,}200} = \frac{3}{10}

    Now multiply by 100 to get a percent:

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

    So the male children made up 30% of the visitors.

  5. Give the answer

    The largest group, 360 male children, is 30% of all 1,200 visitors.

    So the answer is A, 30%.

Check your answer

Work backward. 30% of 1,200 is 0.30×1,200=3600.30 \times 1{,}200 = 360, the male children. Also, with four groups an even split would be 25% each. The largest group should be a bit more than 25%, and 30% is.

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 C Math, page 104, question 31.

Question 11 of 99

2026 Practice Test C Math, question 32

Question

A box plot titled MONTHLY HIGH TEMPERATURES IN NEW YORK above a number line labeled Temperature (°F), from 36 to 88 with a labeled tick every 4. The left whisker starts at about 39, just left of the tick for 40. The box runs from 48 to about 78, between the ticks for 76 and 80, with the median line at 64. The right whisker ends at 84.

The box plot shows the average monthly high temperatures in New York City for 12 months. What is the difference between the range and the interquartile range of the temperature data?

Enter your answer in the space.

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

Show the answer

Correct answer: 15

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

Maximum value =84∘ F= 84^{\circ}\text{ F}, Minimum value =39∘ F= 39^{\circ}\text{ F}

84∘ F−39∘ F=45∘ F84^{\circ}\text{ F} - 39^{\circ}\text{ F} = 45^{\circ}\text{ F}

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 =78∘ F= 78^{\circ}\text{ F}, Lower Quartile Value =48∘ F= 48^{\circ}\text{ F}

78∘ F−48∘ F=30∘ F78^{\circ}\text{ F} - 48^{\circ}\text{ F} = 30^{\circ}\text{ F}

Determine the difference between the range (45° F) and the interquartile range (30° F):

45∘ F−30∘ F=15∘ F45^{\circ}\text{ F} - 30^{\circ}\text{ F} = 15^{\circ}\text{ F}

Question analysis

A box plot of New York City's average monthly high temperatures for 12 months sits above a number line labeled every 4 degrees. Students type the difference between the range and the interquartile range of the data. Two of the plot's values fall between labeled ticks, so careful reading of the scale decides the answer.

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
Above grade 7 Reading a box plot and finding its interquartile range are grade 7 skills. Here two of the five values fall between ticks spaced 4 apart, and the range, the IQR and their difference must all be found and typed with no choices.
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, 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 method is standard, but the key rests on reading the scale between its labels. The axis is ticked only every 4 degrees, and the minimum and the upper quartile land off the ticks, so rounding either one to the nearest label moves the result by a degree or two with nothing to flag it. Students who find the range correctly can also stop there, entering 45, the larger of the two spreads, instead of the difference the question asks for.

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

    Two of the five values fall between ticks spaced 4 degrees apart: the left whisker ends just short of the 40 tick, at 39, and the box’s right edge sits midway between 76 and 80, at 78. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Four values are read and combined in three subtractions, the range, the interquartile range and then their difference, so one misread value changes the final entry. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Something you must already know minor Working it out

    Neither measure is defined on the page: the range spans the whiskers, and the interquartile range is the width of the box, the upper quartile minus the lower. (Anderson & Krathwohl, 2001)

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

    The result is typed with no choices to compare, so an entry one or two degrees off, from rounding a value to the nearest tick, looks just as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Read the minimum, both quartiles and the maximum, judging where each mark falls between the ticks, then subtract for the range and for the IQR, and subtract those two results.

Traps

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 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

    A box plot shows data with five numbers. The two whiskers are the lines on each side of the box. Their ends are the minimum, the least value, and the maximum, the greatest. The box runs from the lower quartile to the upper quartile. These are the middles of the lower half and the upper half of the data. The line inside the box is the median, the middle value.

    You need the range minus the interquartile range (IQR), in degrees.

  2. Read the values between the ticks

    The ticks on the number line are 4 degrees apart. So each degree is one quarter of the space between two ticks. Three values sit right on ticks: the box’s left edge at 48, the median at 64, and the right whisker’s end at 84.

    • The left whisker starts just left of the 40 tick. It sits about one quarter of the way back toward 36. So the minimum is 40−1=3940 - 1 = 39 degrees.
    • The box’s right edge is halfway between 76 and 80. Halfway is 78 degrees.

  3. Find the range

    Use the range formula:

    range=84−39=45\text{range} = 84 - 39 = 45

    The data spread over 45 degrees.

    Watch out: rounding the minimum to the 40 tick gives a range of 44. Then the final answer comes out 14, which is wrong.

  4. Find the IQR

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

    IQR=78−48=30\text{IQR} = 78 - 48 = 30

    The middle half of the data spreads over 30 degrees.

    Watch out: reading the box’s right edge as 76 gives an IQR of 28. Then the final answer comes out 17.

  5. Subtract the IQR from the range

    The question asks for the difference between the two:

    45−30=1545 - 30 = 15

    Watch out: 45 is only the range. Do not stop there. Type only the number, with no degree sign.

    So you would enter 15.

Check your answer

The range covers the whole plot, and the IQR covers only the box. So their difference is the two whiskers added together. The left whisker is 48−39=948 - 39 = 9 degrees long. The right whisker is 84−78=684 - 78 = 6 degrees long. 9+6=159 + 6 = 15, which matches.

More questions like this

The original DOE question
Question 32 as shown on the DOE’s digital test, page 105 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 105, question 32.

Question 12 of 99

2026 Practice Test C Math, question 35

Question

If 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a, what is the value of a in terms of b?

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

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

Show the answer

Correct answer: a = 10b

To find the value of a in terms of b, solve 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a for a. Begin by distributing 4 on the left side of the equation:

(4×a)+(4×2b)=28b+2a(4 \times a) + (4 \times 2b) = 28b + 2a

4a+8b=28b+2a4a + 8b = 28b + 2a

Then, get the a terms on one side and the b terms on the other:

4a−2a=28b−8b4a - 2a = 28b - 8b

Solve for the value of a in terms of b by combining like terms and then dividing each side by 2:

2a=20b2a = 20b

a=10ba = 10b

Question analysis

An algebra question gives an equation with a and b on both sides, one side written as 4 times a sum in parentheses, and students type a in terms of b after a printed "a =". It tests the distributive property and solving linear equations. The answer is typed, so a slip in distributing or moving terms is not caught by any choices.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve equations like 4(x + 2) = 28 with numbers. This one has letters on both sides, needs the distributive property and like terms, and asks for a in terms of b, which is grade 8 algebra.
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, 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

Every step is routine, yet there are four of them, each with a classic slip, and the typed answer catches none. Multiplying the 4 into only the a, or carrying 8b across without changing its sign, gives another whole multiple of b that looks finished. Testing b = 1, where 4(a + 2) and 28 + 2a must come out equal, confirms an entry quickly.

  1. Variables instead of numbers moderate Understanding the question

    Both letters appear on both sides, and the answer must stay a multiple of b, so there is no number to check an entry against unless students invent one. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Many steps moderate Working it out

    Four moves run in a chain, distributing the 4, gathering the a terms, gathering the b terms and dividing by 2, and an error at any one carries into the typed result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The answer is typed after a printed “a =”, with no choices, so 20b, the line before dividing by 2, can be entered without anything on screen questioning it. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute the 4 to both terms in the parentheses, gather the a terms on one side and the b terms on the other, then divide by the coefficient of a.

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.
  • 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

Distributive property
k(x+y)=kx+kyk(x + y) = kx + ky
k is the number outside the parentheses. x and y are the terms inside. The number outside multiplies every term inside, not just the first one. It works because 4 groups of (x + y) hold 4 x’s and 4 y’s.
Doing the same thing to both sides
if A=B, then A−C=B−C and A2=B2\text{if } A = B, \text{ then } A - C = B - C \text{ and } \frac{A}{2} = \frac{B}{2}
A and B are the two equal sides of an equation. C is any amount. Taking the same amount from both sides, or dividing both by the same number, keeps them equal, like a balanced scale.

  1. Sort out the question

    The equation is 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a. You need a in terms of b. That means a alone on one side, with only b and numbers on the other.

    The answer box already shows a =, so you type only what comes after it.

  2. Multiply out the parentheses

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

    4×a+4×2b=28b+2a4 \times a + 4 \times 2b = 28b + 2a

    4a+8b=28b+2a4a + 8b = 28b + 2a

    Watch out: multiplying only the a by 4 gives 4a + 2b. That leads to the wrong answer 13b.

  3. Gather the a terms

    Like terms are terms with the same letter, such as 4a and 2a. Put the a terms on the left. Subtract 2a from both sides:

    4a−2a+8b=28b4a - 2a + 8b = 28b

    2a+8b=28b2a + 8b = 28b

  4. Gather the b terms

    Now subtract 8b from both sides:

    2a=28b−8b2a = 28b - 8b

    2a=20b2a = 20b

    Watch out: moving 8b without changing its sign gives 28b+8b=36b28b + 8b = 36b. That leads to 18b, which is wrong.

  5. Divide both sides by 2

    2a means 2 times a. Divide both sides by 2 to get a alone:

    a=20b2=10ba = \frac{20b}{2} = 10b

    Watch out: 20b is equal to 2a, not a. Do not stop there.

  6. Enter it after the printed a =

    On the test, the box already shows a =. Press 1, then 0, then the b key. You do not need a times sign. Do not type a = again. On this site, type 10b in the box.

    So you would enter 10b, which makes a = 10b.

Check your answer

Pick b = 1, so a = 10. The left side is 4(10+2)=4×12=484(10 + 2) = 4 \times 12 = 48. The right side is 28+2×10=4828 + 2 \times 10 = 48. The sides match, so 10b works.

More questions like this

The original DOE question
Question 35 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 C Math, page 108, question 35.

Question 13 of 99

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 14 of 99

2026 Practice Test C Math, question 39

Question

A shipment contains 170 small boxes of medical supplies. In a random sample of 20 of these boxes, 8 have damage and the others are undamaged. Based on this sample, what is the best prediction of the number of undamaged boxes in the shipment, not including the boxes in the random sample?

  • A60
  • B68
  • C90
  • D102
Show the answer

Correct answer: C

Determine the fraction of undamaged boxes in the sample, given 8 of the 20 boxes are damaged.

20−820=1220=35 or 0.6\frac{20-8}{20} = \frac{12}{20} = \frac{3}{5}\text{ or }0.6

Subtract the number of boxes in the sample from 170.

170−20=150170 - 20 = 150

Basing the prediction of undamaged boxes on the sample, multiply the number of boxes not in the sample by the fraction of undamaged boxes in the sample.

150×0.6=90150 \times 0.6 = 90

Question analysis

A shipment holds 170 boxes of medical supplies, and in a random sample of 20, 8 are damaged. Students predict how many undamaged boxes are in the shipment, leaving out the boxes in the sample. It tests using a sample to make a prediction, and the question's two conditions each lead to a different tempting wrong choice.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.SP.1b samples and populations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Predicting from a random sample is grade 7 proportional reasoning. This version adds two conditions, undamaged rather than damaged and the shipment without the sample, so it takes three steps and a close reading.
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 8, 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 proportion itself is routine; the question is built around two conditions that each change the numbers. The stem reports the damaged boxes, but the question asks about undamaged ones, and its bold not removes the 20 sampled boxes from the shipment. Each wrong choice matches one way of mishandling the two conditions, so checking the result against both words, undamaged and not, is what the question rewards.

  1. Much to keep track of moderate Working it out

    Two conditions point away from the obvious numbers: the question wants undamaged boxes while the stem counts the 8 damaged, and it leaves out the 20 sampled, so 150 boxes, not 170. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Many steps minor Working it out

    Three steps follow in order, the undamaged share 12/20, the 150 boxes outside the sample, and their product, and each wrong turn lands on a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the fraction of the sample that is undamaged, count the boxes that were not sampled, and multiply; then check that the answer describes the group of boxes asked about.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from a sample
prediction=part of the samplesample size×group size\text{prediction} = \frac{\text{part of the sample}}{\text{sample size}} \times \text{group size}
A sample is a small group chosen at random from a bigger group. The fraction is the share of the sample with some feature. A random sample is likely to look like the bigger group. So use the same share of the group you are predicting for.

  1. Sort out the question

    The shipment has 170 boxes. A random sample of 20 boxes was checked, and 8 of them were damaged.

    You need to predict the undamaged boxes in the shipment, not counting the 20 in the sample. Two words matter: undamaged and not.

  2. Find the undamaged share

    The sample has 20−8=1220 - 8 = 12 undamaged boxes. Write that as a share of the sample, and simplify:

    1220=35=0.6\frac{12}{20} = \frac{3}{5} = 0.6

    So 0.6, or 60%, of the sample is undamaged.

    Watch out: 820=0.4\frac{8}{20} = 0.4 is the damaged share. Using it leads to choice A or B.

  3. Count the boxes outside the sample

    The question leaves out the 20 boxes in the sample. Take them away from the shipment:

    170−20=150170 - 20 = 150

    So the prediction is about 150 boxes.

  4. Make the prediction

    Use the same undamaged share for the 150 boxes:

    0.6×150=900.6 \times 150 = 90

    About 90 of those boxes should be undamaged.

    Watch out: 0.6×170=1020.6 \times 170 = 102, choice D, forgets the bold not. It counts the sample boxes again.

  5. Give the answer

    The best prediction is 90 undamaged boxes outside the sample.

    So the answer is C, 90.

Check your answer

Predict the damaged boxes outside the sample too: 0.4×150=600.4 \times 150 = 60. Then 90+60=15090 + 60 = 150, every box outside the sample. Also, 12 out of 20 is a bit more than half. And 90 is a bit more than half of 150.

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 C Math, page 112, question 39.

Question 15 of 99

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 16 of 99

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 17 of 99

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 18 of 99

2026 Practice Test D Math, question 13

Question

A right triangle with the right angle marked at its bottom left vertex. The angle at the top vertex is labeled (2x + 9)°, and the angle at the bottom right vertex is labeled (3x − 14)°.

In the right triangle shown above, what is the measure of the smallest angle?

  • A43°
  • B44°
  • C46°
  • D47°
Show the answer

Correct answer: A

Since the three angles of a triangle have a sum of 180°, and since one angle is a right angle, the other two angles have a sum of 90°. By writing an equation and using inverse operations, the value of x can be found.

(2x+9)+(3x−14)=90(2x + 9) + (3x - 14) = 90

5x−5=905x - 5 = 90

5x=955x = 95

x=19x = 19

Now, use substitution to find the measure of the two angles to determine the smallest angle of the triangle.

2x+9=2(19)+9=472x + 9 = 2(19) + 9 = 47

3x−14=3(19)−14=433x - 14 = 3(19) - 14 = 43

The smallest angle is 43°.

Question analysis

A right triangle has its two other angles labeled (2x + 9)° and (3x − 14)°, and students find the measure of the smallest angle. It tests the angle sum of a triangle and solving a linear equation. Finding x is not the end: both angles must be worked out and compared to see which one is smallest.

NYS standards
  • 8.G.5 angle sums, transversals, angle-angle similarity
  • 7.G.5 angle relationships and equations
  • 7.EE.4a two-step equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students write equations for unknown angles using complementary and supplementary angles. This also needs the angle sum of a triangle, which New York places in grade 8, plus substituting back and comparing, though each step is short.
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, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Setting the sum of the two labeled angles to 90° is the decision; the algebra after it is routine. The trap waits at the end: x = 19 is not an angle, and the expression at the top vertex, 2x + 9, gives the larger acute angle, 47°. Evaluating both expressions and checking that they add to 90° takes seconds and shows which one is smaller.

  1. Combines ideas moderate Working it out

    The angle facts and the algebra meet in one equation: the right angle means the two labeled angles add to 90°, and only that sum makes (2x + 9) + (3x − 14) solvable. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Many steps minor Working it out

    After solving 5x − 5 = 90, both angle expressions must be evaluated and compared, because the question wants an angle, not x, and not the first angle found. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Something you must already know minor Working it out

    The 180° angle sum of a triangle is not printed, and only the small square at the bottom left shows that one angle is 90°. (Anderson & Krathwohl, 2001)

Strategy: Use the right angle to see that the other two angles add to 90°, solve that equation for x, then substitute into both expressions and compare 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.
  • 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

Angles in a triangle
a+b+c=180a + b + c = 180
a, b and c are the three angles of any triangle, in degrees. They always add up to 180 degrees. If you tear off the three corners and set them side by side, they make a straight line.
The two other angles of a right triangle
a+b=90a + b = 90
a and b are the two angles that are not the right angle. A right angle is 90 degrees, so the other two share what is left of the 180: 180 − 90 = 90.

  1. Sort out the question

    The small square at the bottom left corner marks a right angle, which is 90°. The top angle is (2x + 9)°, and the bottom right angle is (3x − 14)°. You need the smallest of the three angles. First find x, then find each angle.

  2. Write an equation

    The three angles add to 180°. The right angle uses 90° of that, so the other two angles add to 90°:

    (2x+9)+(3x−14)=90(2x + 9) + (3x - 14) = 90

  3. Solve for x

    Combine the x terms, 2x + 3x = 5x. Combine the numbers, 9 − 14 = −5:

    5x−5=905x - 5 = 90

    Add 5 to both sides, then divide both sides by 5:

    5x=955x = 95

    x=95÷5=19x = 95 \div 5 = 19

  4. Find both angles

    Put x = 19 into each expression:

    2x+9=2×19+9=38+9=472x + 9 = 2 \times 19 + 9 = 38 + 9 = 47

    3x−14=3×19−14=57−14=433x - 14 = 3 \times 19 - 14 = 57 - 14 = 43

    The three angles are 47°, 43° and 90°.

    Watch out: x = 19 is not an angle. And the first angle you find, 47°, is not the smallest one. Stopping there gives choice D.

  5. Pick the smallest angle

    Compare 47°, 43° and 90°. The smallest is 43°. The figure is not drawn to scale, so do not judge the angles by how they look. Use the numbers.

    So the answer is A, 43°.

Check your answer

Add the three angles: 47 + 43 + 90 = 180. That is the right total for a triangle. Also, 47 + 43 = 90, so the two smaller angles fill exactly what the right angle leaves.

More questions like this

The original DOE question
Question 13 as shown on the DOE’s digital test, page 86 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 86, question 13.

Question 19 of 99

2026 Practice Test D Math, question 14

Question

James must read 84 pages of a book for a class assignment. If he reads 14\frac{1}{4} of the pages in the morning and 13\frac{1}{3} of the remaining pages in the afternoon, how many pages will he still need to read?

  • A35
  • B42
  • C45
  • D56
Show the answer

Correct answer: B

Pages read in the morning: (14)(84)=21\left(\frac{1}{4}\right)(84) = 21

Pages remaining after the morning: 84−21=6384 - 21 = 63

Pages read in the afternoon: (13)(63)=21\left(\frac{1}{3}\right)(63) = 21

Pages he still needs to read: 63−21=4263 - 21 = 42

Question analysis

James must read 84 pages for a class assignment: 1/4 of them in the morning, then 1/3 of the remaining pages in the afternoon. Students find how many pages he still has left to read. It tests finding a fraction of a quantity in stages, and the bold word remaining decides which amount the second fraction is taken of.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Taking fractions of a quantity in stages is typical grade 7 work. The numbers divide evenly and the bold word flags the trap, so the four short steps match routine grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 7, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is simple, but the second fraction has a different whole from the first. Students who take 1/3 of the full 84 pages get a tidy 28 and a listed answer, despite the bold word that warns against it. Here the pages read and the pages left both come to 42, so confusing those two quantities costs nothing, and checking that 21 + 21 + 42 = 84 confirms the work.

  1. Many steps moderate Working it out

    Four operations in order, a quarter of 84, a subtraction, a third of the 63 left, and a second subtraction, each feed the next, so a slip early carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The bold “remaining” moves the base of the second fraction from all 84 pages to the 63 left after the morning, and “still need to read” asks for what is left, not what was read. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work in order: find the morning's pages, subtract to get what remains, take 1/3 of that remainder, and subtract again; a quick bar sketch of the 84 pages helps.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A fraction of a number
ab of n=ab×n\frac{a}{b} \text{ of } n = \frac{a}{b} \times n
n is the whole amount, and a/b is the fraction you take. The word of means multiply. When the top, a, is 1, this is just n divided by b. For example, 1/4 of 84 is 84 ÷ 4.

  1. Sort out the question

    James has 84 pages to read. In the morning he reads 1/4 of all the pages. In the afternoon he reads 1/3 of the remaining pages, the ones still left after the morning. You need the number of pages he still has to read after that.

  2. Find the morning pages

    Take 1/4 of 84:

    14×84=84÷4=21\frac{1}{4} \times 84 = 84 \div 4 = 21

    He reads 21 pages in the morning. That leaves:

    84−21=6384 - 21 = 63

    63 pages are left after the morning.

  3. Find the afternoon pages

    The afternoon fraction is 1/3 of the 63 pages left, not of all 84:

    13×63=63÷3=21\frac{1}{3} \times 63 = 63 \div 3 = 21

    He reads 21 pages in the afternoon.

    Watch out: taking 1/3 of all 84 pages gives 28. Then 84 − 21 − 28 = 35, choice A. The bold word remaining tells you to use 63.

  4. Find the pages still to read

    Take the afternoon pages away from the 63 pages left after the morning:

    63−21=4263 - 21 = 42

    He still needs to read 42 pages.

    Watch out: subtract both readings, morning and afternoon. Taking only 28 from 84 gives 56, choice D.

    So the answer is B, 42.

Check your answer

Add up all the parts: 21 pages in the morning, 21 in the afternoon and 42 still to read. 21 + 21 + 42 = 84, the whole assignment. Another way: after he reads 1/3 of the 63 pages, 2/3 of them are left, and 2/3 of 63 is 42.

More questions like this

The original DOE question
Question 14 as shown on the DOE’s digital test, page 87 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 87, question 14.

Question 20 of 99

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 21 of 99

2026 Practice Test D Math, question 20

Question

What is the value of −9(−4)2+36(−4)+304−4\frac{-9(-4)^{2} + 36(-4) + 304}{-4}?

  • A−76
  • B−4
  • C4
  • D148
Show the answer

Correct answer: B

Use the order of operations to evaluate the expression. It is important to recognize that the fraction bar is a grouping symbol and so the entire numerator should be evaluated first. Exponents are evaluated first.

−9(−4)2+36(−4)+304−4\frac{-9(-4)^{2} + 36(-4) + 304}{-4}

−9(16)+36(−4)+304−4\frac{-9(16) + 36(-4) + 304}{-4}

Then multiplication, recognizing that the product of a negative factor and a positive factor is negative.

−9(16)+36(−4)+304−4\frac{-9(16) + 36(-4) + 304}{-4}

−144−144+304−4\frac{-144 - 144 + 304}{-4}

Then addition and subtraction in order from left to right, recognizing the rules for adding and subtracting signed numbers.

−144−144+304−4\frac{-144 - 144 + 304}{-4}

−288+304−4=16−4\frac{-288 + 304}{-4} = \frac{16}{-4}

And finally the fraction can be evaluated by division, recognizing that the quotient of a positive and a negative number is negative.

16−4=−4\frac{16}{-4} = -4

Question analysis

This arithmetic question asks for the value of a fraction whose numerator, −9(−4)² + 36(−4) + 304, is divided by −4. Students follow the order of operations with negative numbers: a square, two products, a sum and a division. Every step carries a sign, and one dropped negative changes the result completely.

NYS standards
  • 7.NS.2a rational number multiplication
  • 7.NS.2b rational number division
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Order of operations with negative numbers is standard grade 7 practice. This expression stacks a squared negative, two products and a division by a negative into one fraction, so it takes more steps than usual, each with a sign to track.
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, 6 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Nothing here is conceptually new; the question measures care with signs under the order of operations. The parentheses in (−4)² decide whether −9 multiplies 16 or −16, and the two products of −144 then nearly cancel the 304, leaving a small numerator whose sign is easy to lose. Writing each intermediate value with its sign, line by line, prevents most errors.

  1. Error-prone arithmetic moderate Working it out

    Five sign decisions sit in one expression: squaring −4, two products with a negative factor, a sum that climbs from −288 past zero, and a division by −4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    The numerator alone takes an exponent, two multiplications and two additions before the single division, and all of it must be finished before dividing. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The right value and the choice 4 differ only in sign, so a student with the right size must still check the sign of a positive numerator over −4. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Treat the fraction bar as grouping: square −4 first, do the two multiplications, combine the numerator from left to right, and only then divide by −4, writing each sign as you go.

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations
grouping→exponents→×,÷→+,−\text{grouping} \rightarrow \text{exponents} \rightarrow \times, \div \rightarrow +, -
This is the order for working out an expression. Grouping means parentheses and the fraction bar. Exponents are small raised numbers, like the 2 in (−4)². Next, multiply and divide from left to right. Last, add and subtract from left to right.
Squaring a negative number
(−a)2=(−a)×(−a)=a2(-a)^{2} = (-a) \times (-a) = a^{2}
a is a positive number, so −a is negative. Squaring means multiplying a number by itself. A negative times a negative is positive, so a negative number squared is positive. For example, (−4)² = 16.
Signs when you multiply or divide
positivenegative=negative\frac{\text{positive}}{\text{negative}} = \text{negative}
When two numbers have different signs, their product or quotient is negative. When they have the same sign, it is positive. A quotient is the answer to a division.

  1. Sort out the question

    You need the value of one big fraction. The fraction bar is a grouping symbol, so finish the whole top first. Then divide by −4. Follow the order of operations, and track every minus sign.

  2. Do the exponent first

    The only exponent is (−4)². Squaring means multiplying a number by itself:

    (−4)2=(−4)×(−4)=16(-4)^{2} = (-4) \times (-4) = 16

    Watch out: (−4)² is 16, not −16. Using −16 makes the top 304, which gives −76, choice A.

  3. Do the multiplication

    Now multiply. A negative times a positive is negative:

    −9×16=−144-9 \times 16 = -144

    36×(−4)=−14436 \times (-4) = -144

    So the top becomes:

    −144+(−144)+304-144 + (-144) + 304

  4. Add from left to right

    Add the two negatives first:

    −144+(−144)=−288-144 + (-144) = -288

    Then add 304. Since 304 is more than 288, the result is positive:

    −288+304=16-288 + 304 = 16

    The whole top is 16.

  5. Divide by −4

    Now divide the top by the bottom. A positive divided by a negative is negative:

    16−4=−4\frac{16}{-4} = -4

    Watch out: positive 4, choice C, ignores the negative bottom. Dropping every minus sign gives 592 ÷ 4 = 148, choice D.

    So the answer is B, −4.

Check your answer

Multiply back. If the answer is −4, then −4 times the bottom, −4, should give the top. It does: (−4) × (−4) = 16. Also, the top is positive and the bottom is negative, so the answer must be negative.

More questions like this

The original DOE question
Question 20 as shown on the DOE’s digital test, page 93 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 93, question 20.

Question 22 of 99

2026 Practice Test D Math, question 27

Question

A birdseed mixture is 25\frac{2}{5} sunflower seeds. The cost of the sunflower seeds is $0.10 per ounce. At this rate, what is the cost of the sunflower seeds in 25 pounds of this birdseed? (Note: 1 lb = 16 oz.)

  • A$10.00
  • B$16.00
  • C$40.00
  • D$160.00
Show the answer

Correct answer: B

First, calculate the amount of sunflower seeds in 25 pounds of this birdseed by using ratio reasoning to find that there are 10 pounds of sunflower seeds in the birdseed.

25=1025\frac{2}{5} = \frac{10}{25}

Then, use ratio reasoning to find the cost of the sunflower seeds per pound.

($0.101 ounce)(16 ounces1 pound)=$1.60\left(\frac{\$0.10}{1 \text{ ounce}}\right)\left(\frac{16 \text{ ounces}}{1 \text{ pound}}\right) = \$1.60

The cost of the sunflower seeds is $1.60 per pound.

Finally, find the total cost using ratios by multiplying the unit cost per pound by 10 pounds.

$1.601 pound×10 pounds1=$16.00\frac{\$1.60}{1 \text{ pound}} \times \frac{10 \text{ pounds}}{1} = \$16.00

The first two steps above could be done in the other order; that is, find $1.60 per pound first and 10 pounds of sunflower seeds second.

Question analysis

A birdseed mixture is two-fifths sunflower seeds, which cost 10 cents an ounce, and students find the cost of the sunflower seeds in 25 pounds of the mix. It combines a fraction of a whole, a pounds-to-ounces conversion and a unit price. Each step's result looks like a possible answer, so keeping track of which quantity is in hand matters.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3d converting measurement units
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Multistep ratio problems with a unit conversion are typical grade 7 work. This one chains three short steps, a fraction of the mixture, pounds to ounces and a price per ounce, and the numbers are chosen to come out evenly.
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, 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

Each step is routine, but the steps can be done in more than one order, and every partial result looks like a price once a dollar sign is put in front of it. Each wrong choice leaves something out: the 2/5, the price per ounce, or the conversion and the price together. Labelling each number with its unit, pounds, ounces or dollars, as it is found keeps the chain straight and shows when it is finished.

  1. Many steps moderate Working it out

    Three operations are chained: 2/5 of 25 pounds gives 10 pounds of sunflower seeds, the note turns them into 160 ounces, and $0.10 an ounce prices them; stopping at 10 pounds or 160 ounces, or skipping the 2/5, gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Combines ideas minor Working it out

    A fraction of a mixture, a pounds-to-ounces conversion and a price per ounce meet in one question, and the price is per ounce while the amount is given in pounds. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Find the weight of the sunflower seeds first, change it to ounces with the note, then multiply by the price per ounce; or turn the price into dollars per pound first.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A fraction of an amount
part=fraction×whole\text{part} = \text{fraction} \times \text{whole}
The whole is the full amount, here 25 pounds of birdseed. The fraction tells how much of it you want. To find 2/5 of something, split it into 5 equal parts and take 2 of them. Multiplying does both at once.
Changing pounds to ounces
ounces=pounds×16\text{ounces} = \text{pounds} \times 16
The note says 1 pound is 16 ounces. So each pound turns into 16 ounces. An ounce is smaller than a pound, so you get a bigger number.
Total cost
cost=number of units×price of one unit\text{cost} = \text{number of units} \times \text{price of one unit}
The units here are ounces. The price of one unit is $0.10 for one ounce. Every ounce costs the same, so you multiply the number of ounces by the price of one.

  1. Sort out the question

    You know:

    • The birdseed is 25\frac{2}{5} sunflower seeds.
    • There are 25 pounds of birdseed.
    • Sunflower seeds cost $0.10 per ounce.
    • 1 pound = 16 ounces.

    You need the cost, in dollars, of just the sunflower seeds. The price is per ounce, but the weight is in pounds. So you will need to change pounds to ounces.

  2. Find the weight of sunflower seeds

    Only 2/5 of the birdseed is sunflower seeds. Find 2/5 of 25 pounds:

    25×25=505=10\frac{2}{5} \times 25 = \frac{50}{5} = 10

    Another way: 25 ÷ 5 = 5, and 2 fifths is 2 × 5 = 10. So there are 10 pounds of sunflower seeds.

    Watch out: 10 is a number of pounds, not dollars. Stopping here and calling it $10.00 leads to choice A.

  3. Change pounds to ounces

    The price is per ounce, so turn the 10 pounds into ounces. Each pound is 16 ounces:

    10×16=16010 \times 16 = 160

    There are 160 ounces of sunflower seeds.

    Watch out: 160 is a number of ounces, not dollars. Calling it $160.00 leads to choice D.

  4. Find the cost

    Each ounce costs $0.10. Multiply:

    160×0.10=16.00160 \times 0.10 = 16.00

    The sunflower seeds cost $16.00.

    Watch out: skipping the 2/5 prices all 25 pounds. That is 400 ounces, and 400×0.10=40.00400 \times 0.10 = 40.00, choice C.

    So the answer is B, $16.00.

Check your answer

Do it in another order. One pound of sunflower seeds costs 16×0.10=1.6016 \times 0.10 = 1.60 dollars. Ten pounds cost 10×1.60=16.0010 \times 1.60 = 16.00 dollars. You get $16.00 again.

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 D Math, page 100, question 27.

Question 23 of 99

2026 Practice Test D Math, question 32

Question

There were 36 people at Emily’s party. If each person at the party drank 2 cups of juice, how many gallons of juice did they drink altogether?

Note: 1 gallon = 4 quarts; 1 quart = 4 cups.

  • A2142\frac{1}{4}
  • B4124\frac{1}{2}
  • C9
  • D72
Show the answer

Correct answer: B

Since 36 people each drank 2 cups of juice, the total amount of juice they drank was 36×2=7236 \times 2 = 72 cups.

Convert 72 cups to gallons:

72 cups×1 quart4 cups×1 gallon4 quarts=72 \text{ cups} \times \frac{1 \text{ quart}}{4 \text{ cups}} \times \frac{1 \text{ gallon}}{4 \text{ quarts}} =

7216 gallons=\frac{72}{16} \text{ gallons} =

92 gallons=\frac{9}{2} \text{ gallons} =

412 gallons4\frac{1}{2} \text{ gallons}

Question analysis

At Emily's party 36 people each drank 2 cups of juice, and this unit conversion question asks how many gallons they drank in all, given 4 quarts to a gallon and 4 cups to a quart. Students multiply to get the total in cups, then convert through quarts to gallons, and the bold word gallons sets the unit of the answer.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Converting units with given factors is grade 6 work. This adds a first step, the total number of cups, and a two-stage conversion through quarts, but the numbers are friendly and the path is routine for grade 7 students.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 32 words.
Corrected from the DOE handbook
  • A stray word at the end of the explanation 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

With both conversion facts printed, the question is mainly about keeping the units straight. Students who skip the word each convert 36 cups instead of 72, and those who stop at cups find a total that is right in the wrong unit. Working out that a gallon is 4 × 4 = 16 cups before dividing turns two conversions into one and avoids adding the 4s.

  1. Many steps minor Working it out

    A multiplication, 36 × 2, comes before two divisions by 4, cups to quarts and quarts to gallons, and stopping after the first step leaves a believable 72. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The word “each” turns 36 people into 72 cups, and the bold word gallons sets the unit of the answer, a change the note supplies in two steps. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the total number of cups first, then divide by 4 to get quarts and by 4 again to get gallons, or work out how many cups make a gallon and divide once.

Traps

  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • 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 DOE’s explanation ends with a line that reads only “lineExtend”, a formatting code left in the document by mistake. The site leaves it out.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total amount
total=number of people×amount each\text{total} = \text{number of people} \times \text{amount each}
When every person has the same amount, multiply the number of people by the amount each one has. The word each tells you to multiply.
Cups in a gallon
cups in 1 gallon=4×4=16\text{cups in 1 gallon} = 4 \times 4 = 16
A gallon holds 4 quarts, and each quart holds 4 cups. So a gallon holds 4 groups of 4 cups. You multiply, not add, because each of the 4 quarts has its own 4 cups.
Changing cups to gallons
gallons=cups÷16\text{gallons} = \text{cups} \div 16
It takes 16 cups to make 1 gallon. Dividing by 16 finds how many groups of 16 cups there are. A gallon is bigger than a cup, so the number of gallons is smaller.

  1. Sort out the question

    You know:

    • 36 people were at the party.
    • Each person drank 2 cups of juice.
    • 1 gallon = 4 quarts, and 1 quart = 4 cups.

    You need the total amount of juice in gallons. The bold word tells you the unit of the answer.

  2. Find the total in cups

    Each of the 36 people drank 2 cups:

    36×2=7236 \times 2 = 72

    They drank 72 cups in all.

    Watch out: skipping the word each uses only 36 cups. Then 36÷16=21436 \div 16 = 2\frac{1}{4} gallons, choice A.

  3. Find the cups in a gallon

    A gallon is 4 quarts, and each quart is 4 cups:

    4×4=164 \times 4 = 16

    So 1 gallon holds 16 cups.

    Watch out: adding the two 4s gives 8 cups in a gallon. Then 72÷8=972 \div 8 = 9, choice C.

  4. Change cups to gallons

    Divide the cups by 16. Simplify the fraction by dividing the top and bottom by 8:

    72÷16=7216=92=41272 \div 16 = \frac{72}{16} = \frac{9}{2} = 4\frac{1}{2}

    They drank 4124\frac{1}{2} gallons of juice.

    Watch out: 72 is the amount in cups. The question asks for gallons, so 72, choice D, is in the wrong unit.

    So the answer is B, 4124\frac{1}{2}.

Check your answer

Work backward. 4124\frac{1}{2} gallons is 412×4=184\frac{1}{2} \times 4 = 18 quarts. 18 quarts is 18×4=7218 \times 4 = 72 cups. Shared among 36 people, that is 72÷36=272 \div 36 = 2 cups each. It matches.

More questions like this

The original DOE question
Question 32 as shown on the DOE’s digital test, page 105 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 105, question 32. This is the DOE’s version, before the corrections listed in the analysis.

Question 24 of 99

2026 Practice Test D Math, question 33

Question

A car costing $24,000 is subject to a sales tax of 8%. If Bindu made a $10,000 down payment on this car, what is the total amount she has left to pay?

  • A$14,000
  • B$15,620
  • C$15,920
  • D$25,920
Show the answer

Correct answer: C

First, find the sales tax. Then, add the sales tax to the cost of the car. Finally, subtract the down payment to find out how much she has left to pay.

(0.08)(24,000)=1,920(0.08)(24{,}000) = 1{,}920

$24,000+$1,920=$25,920\$24{,}000 + \$1{,}920 = \$25{,}920

$25,920−$10,000=$15,920\$25{,}920 - \$10{,}000 = \$15{,}920

Question analysis

Bindu buys a $24,000 car with an 8% sales tax and makes a $10,000 down payment, and students find how much she still has to pay. This percent word problem combines finding a tax, adding it to the price and subtracting a payment. Every step must be done in order, since leaving one out gives an amount that looks just as reasonable.

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 Adding sales tax to a price is a standard grade 7 percent task. This one also subtracts a down payment at the end, three routine steps with round numbers, much like a typical classroom problem.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 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 method is the standard one for tax on a purchase, so the question checks order and completeness. Forgetting the tax or forgetting the down payment each produces an amount as reasonable-looking as the answer, and a careless 8 × 24 produces one just $300 below it. Multiplying the price by 1.08 in one step, which gives $25,920, is a second route that catches a slip in the tax.

  1. Many steps minor Working it out

    Three operations come in a fixed order, 8% of $24,000, then adding the tax, then subtracting the $10,000, and leaving out the first or the last gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Finding 8% of $24,000 comes down to 8 × 24 = 192 with two zeros added, and a lost carry in that product turns $1,920 into $1,620. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find 8% of the price and add it, or multiply the price by 1.08, to get the full cost with tax, then subtract what Bindu has already paid.

Traps

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}
Percent means out of 100. So 8% is 8/100, or 0.08. To find a percent of a number, change the percent to a decimal and multiply it by the whole.
Amount left to pay
amount left=price+tax−down payment\text{amount left} = \text{price} + \text{tax} - \text{down payment}
The price plus the tax is the full cost. A down payment is money paid at the start. Subtract it, because that part is already paid.

  1. Sort out the question

    You know:

    • The car costs $24,000.
    • The sales tax is 8%. Sales tax is extra money added to a price.
    • Bindu paid $10,000 at the start. That is the down payment.

    You need the amount Bindu still has to pay. Work in order: tax, then full cost, then subtract.

  2. Find the sales tax

    Change 8% to the decimal 0.08. Multiply it by the price:

    0.08×24,000=1,9200.08 \times 24{,}000 = 1{,}920

    A quick way: 1% of 24,000 is 240, so 8% is 8×240=1,9208 \times 240 = 1{,}920. The tax is $1,920.

    Watch out: in 8 × 24 you carry a 3 from 8 × 4 = 32. Dropping it gives 162, so the tax looks like $1,620. That leads to $15,620, choice B.

  3. Find the full cost

    Add the tax to the price:

    24,000+1,920=25,92024{,}000 + 1{,}920 = 25{,}920

    The car costs $25,920 with tax.

    Watch out: this is not the answer yet. Stopping here gives $25,920, choice D. Bindu already paid part of it.

  4. Subtract the down payment

    Bindu already paid $10,000. Take it away:

    25,920−10,000=15,92025{,}920 - 10{,}000 = 15{,}920

    She has $15,920 left to pay.

    Watch out: leaving out the tax gives 24,000−10,000=14,00024{,}000 - 10{,}000 = 14{,}000, choice A.

    So the answer is C, $15,920.

Check your answer

Use a shortcut for the full cost. The price plus 8% more is 108% of the price, or 1.08 times it: 24,000×1.08=25,92024{,}000 \times 1.08 = 25{,}920. Then 25,920−10,000=15,92025{,}920 - 10{,}000 = 15{,}920. Same answer.

More questions like this

The original DOE question
Question 33 as shown on the DOE’s digital test, page 106 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 106, question 33.

Question 25 of 99

2026 Practice Test D Math, question 36

Question

Leah has set a goal to read a total of 500 pages of a textbook. The table shows the book page numbers for the reading she did during three days.

DAILY READING LOG

Day Pages Read
1 pages 45–90, inclusive
2 pages 100–135, inclusive
3 pages 145–190, inclusive

What percentage of her 500-page goal did Leah read on these three days? Round your answer to the nearest tenth.

Enter your answer in the space.

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

Show the answer

Correct answer: 25.6%

First, determine the number of pages read each day. Since the pages per assignment are inclusive, subtract the first page from the last page assigned and then add 1 to determine the pages read for each day.

Day 1: (90−45)+1=46(90 - 45) + 1 = 46 pages

Day 2: (135−100)+1=36(135 - 100) + 1 = 36 pages

Day 3: (190−145)+1=46(190 - 145) + 1 = 46 pages

Leah read a total of 128 pages, since 46+36+46=12846 + 36 + 46 = 128.

To determine the percent, express 128 pages out of 500 pages as a fraction.

128500=128÷4500÷4=32125\frac{128}{500} = \frac{128 \div 4}{500 \div 4} = \frac{32}{125}

32÷125=0.256=25.6%32 \div 125 = 0.256 = 25.6\%

Question analysis

Leah's daily reading log lists three ranges of textbook pages, each marked inclusive, and her goal is 500 pages. This percent question asks what percentage of the goal she read over the three days, rounded to the nearest tenth and typed in. Counting the pages in each inclusive range correctly is the tricky step before the percent.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Finding what percent one number is of another is grade 6 and 7 work. This adds three inclusive page ranges to count, which takes more care than subtracting, then a rounding step, all typed with no choices to check against.
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 3, 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

Counting pages in a range is the whole difficulty; the percent at the end is routine once the total is right. Subtracting first from last feels natural, and here it even rewards the student with a round 25%, so nothing looks wrong. Checking the rule on a tiny range, pages 1 to 3, which is 3 pages and not 2, fixes the count, and the instruction to round to the nearest tenth hints that the answer is not a whole number.

  1. Something you must already know moderate Working it out

    Every row says inclusive, but nothing says that pages 45 to 90 hold 90 − 45 + 1 = 46 pages; subtracting alone loses one page on each of the three days. (Anderson & Krathwohl, 2001)

  2. Many steps moderate Working it out

    Three page counts, a sum, a division by 500 and a change to a percent come in sequence, and a one-page error in each count carries through to the end. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The answer is typed with no choices, and the miscounted total, 125 pages, gives a tidy 25% that looks more finished than the right value with its tenth. (Enright, Morley & Sheehan, 2002)

Strategy: Count each day's pages carefully, testing your method on a short range such as pages 1–3, add the three days, divide by 500, and write the result as a percent to the nearest tenth.

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).
  • Stops one step early: A choice is a number from partway through the solution.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting pages in an inclusive range
pages=last page−first page+1\text{pages} = \text{last page} - \text{first page} + 1
Inclusive means the first and last pages are both read. Subtracting counts the steps from the first page to the last. That misses one page, so add 1. Test it: pages 1 to 3 are 3 pages, and 3 − 1 + 1 = 3.
Percent of a total
percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100
The part is the amount you have, here the pages read. The whole is the amount you compare with, here the 500-page goal. Dividing gives a decimal. Multiplying by 100 turns it into a percent, a number out of 100.

  1. Sort out the question

    Leah’s goal is 500 pages. The log shows three days:

    • Day 1: pages 45 to 90
    • Day 2: pages 100 to 135
    • Day 3: pages 145 to 190

    Each range is inclusive. You need the percent of 500 pages she read, rounded to the nearest tenth.

  2. Count the pages each day

    Use the rule for an inclusive range: last page minus first page, plus 1.

    90−45+1=4690 - 45 + 1 = 46

    135−100+1=36135 - 100 + 1 = 36

    190−145+1=46190 - 145 + 1 = 46

    Watch out: subtracting alone gives 45, 35 and 45 pages. That is 125 pages, which is exactly 25%. A round answer like that can hide the mistake.

  3. Add the three days

    Add the pages from all three days to get her total:

    46+36+46=12846 + 36 + 46 = 128

    Leah read 128 pages in all. This is the part you compare with the goal.

  4. Change to a percent

    Divide the pages read by the goal, then multiply by 100:

    128500×100=0.256×100=25.6\frac{128}{500} \times 100 = 0.256 \times 100 = 25.6

    Leah read 25.6% of her goal. That already ends in tenths, so no rounding is needed.

    Watch out: 0.256 is a decimal, not a percent. And 26 rounds to a whole number, not to the nearest tenth.

  5. Enter the answer

    The box has no % sign, but the question asks for a percentage, so the number in the box is the percent. Type it with its decimal point. Typing 25.6% also counts on this site. So you would enter 25.6.

Check your answer

Estimate. 10% of 500 pages is 50 pages. 128 pages is a bit more than 2 × 50 = 100, so the answer is between 20% and 30%. Then check exactly: 0.256×500=1280.256 \times 500 = 128 pages. It matches.

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 D Math, page 109, question 36.

Question 26 of 99

2026 Practice Test D Math, question 43

Question

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

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

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

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

The difference between the cost of the bars is .

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

Show the answer

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

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

$12.0024=$0.50 per bar\frac{\$12.00}{24} = \$0.50 \text{ per bar}

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

$6.5010=$0.65 per bar\frac{\$6.50}{10} = \$0.65 \text{ per bar}

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

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

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Comparing unit prices is grade 6 rate work. The menus split it into three short steps, two divisions of money amounts and one subtraction, which is lighter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 87 words.
Corrected from the DOE handbook
  • The order of the answers in the DOE’s key was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Many steps minor Working it out

    Three menus must all be right to score: two divisions for the costs per bar, then a subtraction that uses both, so a slip in either division spoils the third menu. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Each division is with money, $12.00 ÷ 24 and $6.50 ÷ 10, so the decimal point must be placed carefully, though the menus offer only sensible amounts. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    You know:

    • The Art Club sells 24 bars for $12.00.
    • The Science Club sells 10 bars for $6.50.

    There are three menus. The first is the Art Club’s cost per bar. The second is the Science Club’s. The third is the difference, how much more one bar costs than the other.

  2. Find the Art Club’s cost

    Share $12.00 among 24 bars:

    12.00÷24=0.5012.00 \div 24 = 0.50

    Each bar costs $0.50. That makes sense: 24 half-dollars make $12. Pick $0.50 in the first menu.

    Watch out: dividing by 20 instead of 24 gives $0.60, which is also in the menu.

  3. Find the Science Club’s cost

    Share $6.50 among 10 bars. Dividing by 10 moves the decimal point one place to the left:

    6.50÷10=0.656.50 \div 10 = 0.65

    Each bar costs $0.65. Pick $0.65 in the second menu.

  4. Find the difference

    Subtract the smaller cost from the larger one:

    0.65−0.50=0.150.65 - 0.50 = 0.15

    The bars differ by $0.15. Pick $0.15 in the third menu.

    Watch out: $0.10 sits right next to $0.15 in the menu. A slip in the subtraction lands there.

  5. Fill in the three menus

    On the test and on this site, you choose each answer from a drop-down menu.

    So the menus read $0.50, $0.65 and $0.15.

Check your answer

Multiply back. 24×0.50=12.0024 \times 0.50 = 12.00 and 10×0.65=6.5010 \times 0.65 = 6.50, so both costs per bar match the pack prices. And 0.50+0.15=0.650.50 + 0.15 = 0.65, so the difference fits too.

More questions like this

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

Question 27 of 99

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 28 of 99

2024 Test A Math, question 4

Question

Mr. Chan’s lawn grows 2182\frac{1}{8} inches every 2 weeks. He mows his lawn every 2 weeks and cuts off the top 1341\frac{3}{4} inches of lawn. If Mr. Chan’s lawn was 4 inches tall at the beginning of the season, how many inches tall, in decimal form, is Mr. Chan’s lawn after 8 weeks?

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

Show the answer

Correct answer: 5.5

The lawn grows 2182\frac{1}{8} inches every 2 weeks. Find the total number of inches it grows in 8 weeks.

8÷2=48 \div 2 = 4

4×2184 \times 2\frac{1}{8}

4×2.125=8.54 \times 2.125 = 8.5

The lawn was 4 inches tall at the beginning of the season.

8.5+4=12.58.5 + 4 = 12.5

Mr. Chan mows his lawn every 2 weeks and cuts off the top 1341\frac{3}{4} inches. Find the total number of inches that are cut off in 8 weeks.

8÷2=48 \div 2 = 4

4×1344 \times 1\frac{3}{4}

4×1.75=74 \times 1.75 = 7

The height after 8 weeks is the total number of inches grown minus the total number of inches cut off.

12.5−7=5.512.5 - 7 = 5.5

Question analysis

This multistep word problem follows Mr. Chan's lawn over 8 weeks: it grows a mixed number of inches every 2 weeks, and he mows a mixed number of inches off on the same schedule. Students combine the growth, the cuts and the starting height and give the final height as a decimal. Leaving out one of those three pieces is the usual error.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve multi-step problems with mixed numbers and convert between fractions and decimals. This one strings four such steps together, including counting the 2-week periods, which is typical grade 7 work done carefully.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 3, 49 words.
Corrected from the DOE handbook
  • A misspelled name in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

No single step is hard, but three parts must all reach the final height: the starting 4 inches, four rounds of growth and four rounds of mowing. Leaving out any one of them gives a clean decimal that can be gridded without a second thought. Working one 2-week period at a time, in which the lawn gains 3/8 inch net, keeps the bookkeeping short.

  1. Many steps moderate Working it out

    Four pieces have to be combined: count the four 2-week periods in 8 weeks, total the growth, total the mowing, and start from the lawn's first 4 inches. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic moderate Working it out

    The amounts are mixed numbers with unlike denominators, 2 1/8 and 1 3/4, which must be multiplied by 4 or subtracted and then gridded as a decimal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Turning words into math minor Understanding the question

    The phrase "every 2 weeks" governs both the growing and the mowing, so 8 weeks means four rounds of each, not eight. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    As a grid-in it gives no choices, so a height that leaves out the starting 4 inches, or leaves out the mowing, looks as finished as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Find the net change for one 2-week period by subtracting the cut from the growth, multiply by the number of such periods in 8 weeks, and add the starting height.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook’s explanation says “Mr. Chen mows his lawn every 2 weeks.” The question calls him Mr. Chan throughout, so the site now reads “Mr. Chan.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mixed number to decimal
abc=a+b÷ca\frac{b}{c} = a + b \div c
A mixed number has a whole part, a, and a fraction part, b over c. To write it as a decimal, divide b by c to get the fraction part as a decimal. Then add the whole part. For example, one and a half is 1 + 0.5 = 1.5.
Repeated amount
total=n×a\text{total} = n \times a
n is how many times something happens. a is the amount each time. Adding the same amount n times is the same as multiplying it by n.

  1. Sort out the question

    The lawn starts at 4 inches tall. Every 2 weeks it grows 2182\frac{1}{8} inches. Every 2 weeks Mr. Chan also cuts off 1341\frac{3}{4} inches. You need the height after 8 weeks, as a decimal. Three things matter: the start, the growing and the cutting.

  2. Count the 2-week periods

    Both the growing and the cutting happen every 2 weeks. So count how many 2-week periods fit in 8 weeks.

    8÷2=48 \div 2 = 4

    There are 4 periods. The lawn grows 4 times and is cut 4 times. Watch out: do not count 8 rounds, one for each week. That leads to the wrong answer 7.

  3. Change the mixed numbers to decimals

    The answer must be a decimal, so change both amounts now.

    218=2+1÷8=2.1252\frac{1}{8} = 2 + 1 \div 8 = 2.125

    134=1+3÷4=1.751\frac{3}{4} = 1 + 3 \div 4 = 1.75

    Each period, the lawn grows 2.125 inches and loses 1.75 inches.

  4. Add up the growing

    The lawn grows 2.125 inches, 4 times.

    total=n×a\text{total} = n \times a

    total=4×2.125=8.5\text{total} = 4 \times 2.125 = 8.5

    It grows 8.5 inches in all. Add the starting height of 4 inches.

    4+8.5=12.54 + 8.5 = 12.5

    Without mowing, the lawn would be 12.5 inches tall. Watch out: do not forget the starting 4 inches. Leaving them out leads to 1.5.

  5. Add up the cutting

    Mr. Chan cuts off 1.75 inches, 4 times.

    total=4×1.75=7\text{total} = 4 \times 1.75 = 7

    He cuts off 7 inches in all.

  6. Subtract the cutting

    12.5−7=5.512.5 - 7 = 5.5

    The lawn is 5.5 inches tall after 8 weeks. Watch out: 12.5 is not the answer. It leaves out the mowing. So you would grid 5.5.

Check your answer

Work one period at a time. Each period the lawn gains 2.125−1.75=0.3752.125 - 1.75 = 0.375 inch. In 4 periods that is 4×0.375=1.54 \times 0.375 = 1.5 inches. Add the starting 4 inches: 4+1.5=5.54 + 1.5 = 5.5. Same answer.

More questions like this

The original DOE question
Question 61 as printed on page 76 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 76, question 61. Open page 76 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 29 of 99

2024 Test A Math, question 10

Question

Simplify:

(39(x−3)3)+3913\frac{\left(\frac{39(x - 3)}{3}\right) + 39}{13}

  • Axx
  • Bx−2x - 2
  • C13x−3613x - 36
  • Dx+3613x + \frac{36}{13}
Show the answer

Correct answer: A

(39(x−3)3)+3913\frac{\left(\frac{39(x - 3)}{3}\right) + 39}{13}

Divide the first 39 by 3.

13(x−3)+3913\frac{13(x - 3) + 39}{13}

Distribute 13 through (x−3)(x - 3).

13x−39+3913\frac{13x - 39 + 39}{13}

In the numerator, the terms −39 and 39 cancel out.

13x13\frac{13x}{13}

Finally, divide the numerator and denominator by 13 to get x.

Question analysis

An algebra question that asks students to simplify a stacked fraction: 39 times (x − 3) over 3, plus 39, all divided by 13. It tests dividing, distributing and combining like terms in the right order. The nested fraction bars make it easy to divide only some of the terms or to distribute to only one.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.3 generating equivalent expressions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students expand and simplify linear expressions such as 13(x − 3) + 39. This wraps that work inside two fraction bars, which adds reading load and chances to misstep but no new mathematics.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 16+, 6 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The expression looks heavier than it is: every number is a multiple of 13, and the constants cancel to leave a single term. What makes it risky is scope, knowing which terms each fraction bar and each multiplier covers. Each wrong choice comes from stretching or shrinking that scope by one term, so working outward from the innermost parentheses, one line per step, is the safe way.

  1. Error-prone arithmetic moderate Working it out

    Two fraction bars are nested: 39 ÷ 3 applies only to the first term, the 13 must reach both x and −3, and only then does the whole numerator divide by 13. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Four moves come in a fixed order, divide 39 by 3, distribute 13, cancel −39 with +39, divide by 13, and each is a chance to apply a step to the wrong terms. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work from the inside out: divide 39 by 3, distribute through the parentheses, combine the constants in the numerator, and only then divide everything by 13.

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
a(b−c)=a×b−a×ca(b - c) = a \times b - a \times c
a is the number outside the parentheses. b and c are the terms inside. Multiply a by each term inside, and keep the minus sign between them. It works because a groups of (b − c) are a groups of b, take away a groups of c.

  1. Sort out the expression

    You need to simplify the big fraction, which means write it in its shortest form. It has a small fraction inside its top. The top is 39(x−3)3+39\frac{39(x - 3)}{3} + 39, and the bottom is 13. Work from the inside out, one step at a time.

  2. Divide the first 39 by 3

    The small fraction bar is only under 39(x−3)39(x - 3). So divide just that 39 by 3.

    39(x−3)3=13(x−3)\frac{39(x - 3)}{3} = 13(x - 3)

    Now the top is 13(x−3)+3913(x - 3) + 39. Watch out: the small bar does not reach the added 39. Dividing that 39 by 3 as well leads to x−2x - 2, choice B.

  3. Multiply out the parentheses

    a(b−c)=a×b−a×ca(b - c) = a \times b - a \times c

    13(x−3)=13×x−13×3=13x−3913(x - 3) = 13 \times x - 13 \times 3 = 13x - 39

    Watch out: multiply the 3 by 13 too, not just the x. Leaving −3 as it is leads to choice D.

  4. Combine the plain numbers on top

    13x−39+39=13x13x - 39 + 39 = 13x

    −39 and 39 add up to 0, so they cancel. The top is now just 13x, which means 13 times x.

  5. Divide by the bottom

    Now divide the whole top by 13. Dividing by 13 undoes times 13.

    13x13=x\frac{13x}{13} = x

    Watch out: the long bar is under the whole top. Dividing only the last 39 by 13 gives 13x−3613x - 36, choice C. So the answer is A, xx.

Check your answer

Try x = 5. Then x−3=2x - 3 = 2 and 39×2=7839 \times 2 = 78. Next, 78÷3=2678 \div 3 = 26 and 26+39=6526 + 39 = 65. Last, 65÷13=565 \div 13 = 5. You get back 5, the same as x. So x is right.

More questions like this

The original DOE question
Question 67 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 67. Open page 78 of the handbook (PDF)

Question 30 of 99

2024 Test A Math, question 17

Question

The probability of drawing a red candy at random from a bag of 25 candies is 25\frac{2}{5}. After 5 red candies are removed from the bag, what is the probability of randomly drawing a red candy from the bag?

  • A0
  • B110\frac{1}{10}
  • C15\frac{1}{5}
  • D14\frac{1}{4}
Show the answer

Correct answer: D

Start with a proportion to determine the total number of red candies in the bag:

25=r25\frac{2}{5} = \frac{r}{25}

25(25)=r\frac{2}{5}(25) = r

10=r10 = r

There are currently 10 red candies out of 25 candies in the bag. After 5 red candies are removed, there would be 5 red candies left in the bag and 20 total candies.

The probability of randomly drawing a red candy would then be 520=14\frac{5}{20} = \frac{1}{4}.

Question analysis

A probability question about a bag of 25 candies in which the chance of drawing red is 2/5; then 5 red candies are taken out. Students find the new probability of drawing a red candy. They must work out how many red candies there were, then rebuild the fraction from the counts left in the bag.

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 Simple probability as a fraction of outcomes is grade 6 and 7 content. This adds a step back from the probability to a count and an update after removal, which is typical grade 7 multi-step work.
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, 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 question gives a probability, not a count, so the first job is to rebuild the bag: 10 red candies among 25. Removing candies changes two numbers at once, and the strongest wrong choice comes from updating the red count while leaving the total at 25. A student who writes both new counts before forming the fraction avoids it.

  1. Turning words into math moderate Understanding the question

    The probability 2/5 has to be turned into a count, 10 red of 25, and "after 5 red candies are removed" lowers both the red count and the bag's total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three steps follow in order: 2/5 of 25 = 10, then 5 red among 20 candies left, then 5/20 reduced to lowest terms. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use 2/5 of 25 to find the number of red candies, subtract 5 from both the red count and the total, and write the new fraction in lowest terms.

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=red candiesall candiesP = \frac{\text{red candies}}{\text{all candies}}
P is the probability, the chance of drawing a red candy. Red candies is how many red ones are in the bag. All candies is how many candies are in the bag in all. Each candy is equally likely, so the chance is the fraction of the bag that is red.

  1. Sort out the question

    The bag has 25 candies. The chance of drawing red is 25\frac{2}{5}. Then 5 red candies are taken out. You need the new chance of drawing red. Taking out candies changes both the red count and the total.

  2. Find the number of red candies

    A chance of 25\frac{2}{5} means 2 out of every 5 candies are red. Call the red count r.

    P=red candiesall candiesP = \frac{\text{red candies}}{\text{all candies}}

    25=r25\frac{2}{5} = \frac{r}{25}

    r=25×25=2×5=10r = \frac{2}{5} \times 25 = 2 \times 5 = 10

    There are 10 red candies. Watch out: do not stop at 25÷5=525 \div 5 = 5. You must also multiply by 2. Thinking there are 5 red candies leads to 0, choice A.

  3. Take out the 5 red candies

    red: 10−5=5\text{red: } 10 - 5 = 5

    all: 25−5=20\text{all: } 25 - 5 = 20

    Now 5 of the 20 candies are red. Watch out: the total goes down too. Keeping 25 gives 525=15\frac{5}{25} = \frac{1}{5}, choice C.

  4. Find the new probability

    P=520=14P = \frac{5}{20} = \frac{1}{4}

    To simplify, divide the top and bottom by 5. So the answer is D, 14\frac{1}{4}.

Check your answer

At first, 25\frac{2}{5} is 40% of the bag. Taking out only red candies should lower the chance. 14\frac{1}{4} is 25%, which is lower but not zero. That makes sense, because 5 red candies are still in the bag.

More questions like this

The original DOE question
Question 74 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 2025 Admissions, page 80, question 74. 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 31 of 99

2024 Test A Math, question 18

Question

Each number in a sequence is formed by doubling the previous number and then adding 1. If the 9th number in the sequence is 63, what is the 10th number minus the 7th number?

  • A96
  • B111
  • C112
  • D127
Show the answer

Correct answer: C

First, write an equation based on the rule described in the item. Let x be the previous number in the sequence and y be the number in the sequence. The ninth number in the sequence is 63.

2x+1=y2x + 1 = y

Determine the 10th number in the sequence, y, by making x=63x = 63 and solving:

2(63)+1=y2(63) + 1 = y

126+1=y126 + 1 = y

127=y127 = y

Determine the 8th number in the sequence, x, by making y=63y = 63 and solving:

2x+1=632x + 1 = 63

Subtract one from both sides of the equation.

2x+(1−1)=63−12x + (1 - 1) = 63 - 1

2x=622x = 62

2x2=622\frac{2x}{2} = \frac{62}{2}

x=31x = 31

Determine the 7th number in the sequence, x, by making y=31y = 31 and solving:

2x+1=312x + 1 = 31

2x+(1−1)=31−12x + (1 - 1) = 31 - 1

2x=302x = 30

2x2=302\frac{2x}{2} = \frac{30}{2}

x=15x = 15

Subtract the 7th number in the sequence from the 10th number:

127−15=112127 - 15 = 112

The difference is 112.

Question analysis

A sequence question: each number is found by doubling the one before it and adding 1, and the 9th number is 63. Students find the 10th number minus the 7th. That means running the rule forward once and backward twice, while keeping track of which position each number holds.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve two-step equations but rarely have to run a rule backward through a sequence. This mixes forward and backward steps with position counting, so it asks for more planning than 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 7, 32 words.
Corrected from the DOE handbook
  • A misplaced parenthesis in the explanation’s working 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

Applying the rule forward is routine; running it backward is not, and the question needs both. Students must also count positions carefully, since the 9th number is given and the 7th is two steps back, not one. Each listed wrong answer stops or slips at a different stage, so writing the four numbers in a row, 15, 31, 63 and 127, before subtracting is the reliable way.

  1. No obvious method moderate Working it out

    The rule is stated only forward, double and add 1, but the 7th and 8th numbers come before the given 9th, so the rule must be undone: subtract 1, then halve. (Schoenfeld, 1985)

  2. Many steps moderate Working it out

    One forward step reaches the 10th number, 127, two backward steps reach the 8th and 7th, 31 and 15, and a final subtraction finishes. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Much to keep track of minor Working it out

    Four positions and their values, 7th to 10th, have to be kept straight, and the 8th number is easy to mistake for the 7th. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Apply the rule to 63 to get the 10th number; to go back a position, subtract 1 and halve, doing it twice to reach the 7th; then subtract.

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).
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: Finding the 7th number, the handbook writes “2x + (1 − 1 = 31 − 1)”, with the closing parenthesis after the whole equation, so the line reads as an equation inside an expression (and 1 − 1 = 31 − 1 is false). The site now reads 2x + (1 − 1) = 31 − 1, the same step the handbook writes correctly for the 8th number (2x + (1 − 1) = 63 − 1). It leads to 2x = 30 and x = 15 as before.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule
y=2x+1y = 2x + 1
A sequence is a list of numbers that follows a rule. x is a number in the list, and y is the next number. To get the next number, double x and then add 1.
Going back one number
x=(y−1)÷2x = (y - 1) \div 2
y is a number in the list, and x is the one just before it. To go back, undo the rule in reverse order. The rule doubles and then adds 1. So first subtract 1, then divide by 2.

  1. Sort out the question

    Each number is “formed by doubling the previous number and then adding 1”. Previous means the one just before. The 9th number is 63. You need the 10th number minus the 7th number. So you go forward one number and back two numbers.

  2. Find the 10th number

    Use the rule with x = 63.

    y=2x+1y = 2x + 1

    y=2×63+1=126+1=127y = 2 \times 63 + 1 = 126 + 1 = 127

    The 10th number is 127. Watch out: remember to add 1. Doubling only gives 126 and leads to 111, choice B.

  3. Plan how to go back

    The 7th number comes before the 9th, so you must undo the rule. The rule doubled, then added 1. Undo the last step first: subtract 1. Then undo the doubling: divide by 2.

    x=(y−1)÷2x = (y - 1) \div 2

  4. Find the 8th number

    x=(63−1)÷2=62÷2=31x = (63 - 1) \div 2 = 62 \div 2 = 31

    The 8th number is 31. Watch out: 31 is the 8th number, not the 7th. Using it gives 127−31=96127 - 31 = 96, choice A.

  5. Find the 7th number

    Step back once more, from the 8th number. Use the same rule with y = 31.

    x=(31−1)÷2=30÷2=15x = (31 - 1) \div 2 = 30 \div 2 = 15

    The 7th number is 15.

  6. Subtract the 7th from the 10th

    127−15=112127 - 15 = 112

    Watch out: 127 alone is choice D. You still have to subtract the 7th number. So the answer is C, 112.

Check your answer

Go forward from 15 using the rule. 2×15+1=312 \times 15 + 1 = 31, then 2×31+1=632 \times 31 + 1 = 63, then 2×63+1=1272 \times 63 + 1 = 127. You land on 63 as the 9th number, so the list is right. And 127−15=112127 - 15 = 112.

More questions like this

The original DOE question
Question 75 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 2025 Admissions, page 80, question 75. Open page 80 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 32 of 99

2024 Test A Math, question 22

Question

F=95C+32F = \frac{9}{5}C + 32

Yesterday in Centerville, the highest Fahrenheit temperature, F, was 86°, and the lowest was 68°. What was the difference between these temperatures, in degrees Celsius, C?

  • A10.0° C
  • B15.0° C
  • C20.0° C
  • D32.4° C
Show the answer

Correct answer: A

The highest temperature was 86° F.

F=95C+32F = \frac{9}{5}C + 32

86=95C+3286 = \frac{9}{5}C + 32

Subtract 32 from both sides of the equation:

86−32=95C86 - 32 = \frac{9}{5}C

54=95C54 = \frac{9}{5}C

Multiply both sides of the equation by 59\frac{5}{9}:

54(59)=(59)(95C)54\left(\frac{5}{9}\right) = \left(\frac{5}{9}\right)\left(\frac{9}{5}C\right)

30=C30 = C

The lowest temperature was 68° F.

F=95C+32F = \frac{9}{5}C + 32

68=95C+3268 = \frac{9}{5}C + 32

68−32=95C68 - 32 = \frac{9}{5}C

36=95C36 = \frac{9}{5}C

(59)(36)=C\left(\frac{5}{9}\right)(36) = C

20=C20 = C

The difference, in degrees Celsius, is 30 minus 20.

30−20=1030 - 20 = 10

Question analysis

A formula question: given F = 9/5 C + 32 and a day's high and low of 86 and 68 degrees Fahrenheit, students find the difference between the two temperatures in degrees Celsius. It tests solving a two-step equation with a fraction coefficient. Answering with the difference, in the right scale, is what to watch.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving equations of the form px + q = r with fraction coefficients is grade 7 work. Doing it twice and comparing the results is a typical multistep grade 7 task, with extra care over which scale the answer is in.
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 7, 29 words.
Corrected from the DOE handbook
  • A stray term in the explanation’s working 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

The formula is written to give Fahrenheit from Celsius, and the question runs it the other way, twice. Plugging 86 and 68 straight into the right-hand side is the tempting misuse, and it produces a listed choice. A shortcut avoids most of the work: the 18-degree Fahrenheit gap times 5/9 is the Celsius gap, because the 32 cancels when two readings are subtracted.

  1. Many steps moderate Working it out

    Each Fahrenheit reading needs its own two-step solve, subtracting 32 and multiplying by 5/9, and then the two Celsius values must be subtracted. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Multiplying 54 and 36 by 5/9 is easy only if the division by 9 comes first; multiplying by 5 first leaves 270/9 and 180/9 to reduce. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Solve the formula for C at 86° and at 68° by subtracting 32 and multiplying by 5/9, then subtract; or convert the 18-degree gap directly with 5/9.

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: After “Subtract 32 from both sides of the equation,” the handbook writes “86 − 32 = (9/5)C = 32”. The extra “= 32” would mean (9/5)C = 32, which is false: 86 − 32 = 54, and the next line is 54 = (9/5)C. The site now reads 86 − 32 = (9/5)C, as the handbook writes the same step for 68°.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing Celsius to Fahrenheit
F=95C+32F = \frac{9}{5}C + 32
F is a temperature in degrees Fahrenheit. C is the same temperature in degrees Celsius. Multiply the Celsius number by 9 fifths, then add 32. To go the other way, undo each step: subtract 32, then multiply by 5 ninths.

  1. Sort out the question

    Fahrenheit and Celsius are two ways to measure temperature. The formula changes Celsius, C, into Fahrenheit, F. Yesterday’s high was 86° F and the low was 68° F. You need the difference between them in degrees Celsius. So change each one to Celsius, then subtract.

  2. Undo the + 32 for the high

    Put 86 in for F. Then subtract 32 from both sides. That keeps the equation balanced.

    F=95C+32F = \frac{9}{5}C + 32

    86=95C+3286 = \frac{9}{5}C + 32

    86−32=95C86 - 32 = \frac{9}{5}C

    54=95C54 = \frac{9}{5}C

  3. Solve for the high in Celsius

    To undo times 95\frac{9}{5}, multiply both sides by 59\frac{5}{9}. That is its reciprocal, the fraction flipped upside down.

    C=54×59=6×5=30C = 54 \times \frac{5}{9} = 6 \times 5 = 30

    Dividing 54 by 9 first gives 6, which keeps the numbers small. The high was 30° C.

  4. Change the low to Celsius

    Do the same with 68.

    68=95C+3268 = \frac{9}{5}C + 32

    68−32=95C68 - 32 = \frac{9}{5}C

    36=95C36 = \frac{9}{5}C

    C=36×59=4×5=20C = 36 \times \frac{5}{9} = 4 \times 5 = 20

    The low was 20° C. Watch out: 20° C is the low, not the difference. That is choice C.

  5. Subtract to find the difference

    30−20=1030 - 20 = 10

    The difference is 10 degrees Celsius. Watch out: do not multiply the 18-degree Fahrenheit gap by 95\frac{9}{5}. That goes the wrong way and gives 32.4, choice D. So the answer is A, 10.0° C.

Check your answer

A shortcut: the gap in Fahrenheit is 86−68=1886 - 68 = 18 degrees. The + 32 moves both temperatures the same, so it does not change the gap. Multiply the gap by 59\frac{5}{9}: 18×59=1018 \times \frac{5}{9} = 10. Same answer.

More questions like this

The original DOE question
Question 79 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 2025 Admissions, page 81, question 79. Open page 81 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 33 of 99

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 34 of 99

2024 Test A Math, question 33

Question

If (35−12)x=14+23\left(\frac{3}{5} - \frac{1}{2}\right)x = \frac{1}{4} + \frac{2}{3}, what is the value of x?

  • A11120\frac{11}{120}
  • B27\frac{2}{7}
  • C56\frac{5}{6}
  • D556\frac{55}{6}
Show the answer

Correct answer: D

(35−12)x=14+23\left(\frac{3}{5} - \frac{1}{2}\right)x = \frac{1}{4} + \frac{2}{3}

Rewrite with equivalent fractions that have like denominators.

(3×25×2−1×52×5)x=1×34×3+2×43×4\left(\frac{3 \times 2}{5 \times 2} - \frac{1 \times 5}{2 \times 5}\right)x = \frac{1 \times 3}{4 \times 3} + \frac{2 \times 4}{3 \times 4}

(610−510)x=312+812\left(\frac{6}{10} - \frac{5}{10}\right)x = \frac{3}{12} + \frac{8}{12}

(110)x=1112\left(\frac{1}{10}\right)x = \frac{11}{12}

Multiply both sides of the equation by 10.

10(110)x=10(1112)10\left(\frac{1}{10}\right)x = 10\left(\frac{11}{12}\right)

x=11012=556x = \frac{110}{12} = \frac{55}{6}

Question analysis

This equation has fractions on both sides: the coefficient of x is 3/5 minus 1/2, and the right side is 1/4 plus 2/3. Students simplify each side using common denominators, then solve for x. Each step is routine, but four fraction operations in a row leave many places to slip, especially when undoing the coefficient.

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 Grade 7 students solve equations such as (2/3)x = 4/5 with fractional coefficients. Here both sides must first be simplified over unlike denominators, so four fraction operations come before x is found: on level, but long.
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 8, 12 words.
Corrected from the DOE handbook
  • A minus sign printed in place of an equals sign in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Nothing here is new, but the question strings together four fraction operations, and the last one runs against habit. Having found that one tenth of x is 11/12, students must multiply by 10; multiplying by 1/10 instead gives a tiny listed value. The size of the answer is a useful check: if a tenth of x is almost 1, then x must be almost 10.

  1. Error-prone arithmetic moderate Working it out

    Both sides need common denominators, tenths on the left and twelfths on the right, and the coefficient comes out as 1/10, a fraction to divide by. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps moderate Working it out

    Four operations come before x appears: 3/5 − 1/2, then 1/4 + 2/3, then 11/12 divided by 1/10, then reducing 110/12 to 55/6. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Simplify the parentheses on the left and the sum on the right over common denominators, then undo the coefficient by multiplying both sides by its reciprocal.

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes “(1/10)x − 11/12”, an expression, where the working needs the equation (1/10)x = 11/12: the line before is (6/10 − 5/10)x = 3/12 + 8/12, and the next step multiplies both sides of the equation by 10. The site now reads (1/10)x = 11/12, which gives x = 110/12 = 55/6.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a is the top of a fraction and b is the bottom. n is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, cut into smaller pieces. Use it to give two fractions the same bottom before you add or subtract.
Keeping an equation balanced
if a=b, then n×a=n×b\text{if } a = b \text{, then } n \times a = n \times b
a and b are the two sides of an equation, and n is any number. An equation is like a balanced scale. If you multiply both sides by the same number, it stays balanced. Use it to undo a number that multiplies x.

  1. Sort out the equation

    An equation says two things are equal. On the left, the subtraction in the parentheses multiplies x. On the right is a sum. You need the value of x. First make each side as simple as you can. Then get x by itself.

  2. Simplify the left side

    To subtract, the fractions need the same bottom. Both 5 and 2 go into 10, so use tenths.

    ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}

    35−12=3×25×2−1×52×5=610−510=110\frac{3}{5} - \frac{1}{2} = \frac{3 \times 2}{5 \times 2} - \frac{1 \times 5}{2 \times 5} = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

    Watch out: this is a minus sign. Adding instead gives 1110\frac{11}{10}, and that leads to 56\frac{5}{6}, choice C.

  3. Simplify the right side

    Both 4 and 3 go into 12, so use twelfths.

    14+23=1×34×3+2×43×4=312+812=1112\frac{1}{4} + \frac{2}{3} = \frac{1 \times 3}{4 \times 3} + \frac{2 \times 4}{3 \times 4} = \frac{3}{12} + \frac{8}{12} = \frac{11}{12}

    Now the equation is much simpler:

    110x=1112\frac{1}{10}x = \frac{11}{12}

  4. Get x by itself

    x is multiplied by 110\frac{1}{10}. To undo that, multiply both sides by 10. That is the reciprocal of 110\frac{1}{10}, the fraction flipped upside down.

    if a=b, then n×a=n×b\text{if } a = b \text{, then } n \times a = n \times b

    10×110x=10×111210 \times \frac{1}{10}x = 10 \times \frac{11}{12}

    x=11012x = \frac{110}{12}

    Watch out: multiplying 1112\frac{11}{12} by 110\frac{1}{10} does not undo anything. It gives 11120\frac{11}{120}, choice A.

  5. Simplify the fraction

    Both 110 and 12 are even, so divide the top and the bottom by 2.

    x=110÷212÷2=556x = \frac{110 \div 2}{12 \div 2} = \frac{55}{6}

    So the answer is D, 556\frac{55}{6}.

Check your answer

Put 556\frac{55}{6} back in for x. 110×556=5560=1112\frac{1}{10} \times \frac{55}{6} = \frac{55}{60} = \frac{11}{12}. That matches the right side. Also estimate. A tenth of x is almost 1, so x should be almost 10. 556\frac{55}{6} is about 9.2.

More questions like this

The original DOE question
Question 90 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 90. 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). This is the DOE’s version, before the corrections listed in the analysis.

Question 35 of 99

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 36 of 99

2024 Test A Math, question 36

Question

CELL PHONE SALES BY COLOR

Color Percentage of Cell Phones Sold
White 8k8k
Black 30
Blue 30−2k30 - 2k
Red k+5k + 5
Total 100

The table above shows cell phone sales by color. What percentage of the cell phones sold were blue?

  • A18%
  • B20%
  • C22%
  • D28%
Show the answer

Correct answer: B

First, use the information given in the table and properties of equations to determine the value of k:

8k+30+30−2k+k+5=1008k + 30 + 30 - 2k + k + 5 = 100

8k−2k+k+30+30+5=1008k - 2k + k + 30 + 30 + 5 = 100

7k+65=1007k + 65 = 100

7k+(65−65)=100−657k + (65 - 65) = 100 - 65

7k=357k = 35

k=5k = 5

Then, determine the percentage of cell phones sold that were blue by finding the value of the expression 30−2k30 - 2k (given in the table), when k=5k = 5:

30−2k=30−2(5)=30 - 2k = 30 - 2(5) =

30−10=2030 - 10 = 20

The percentage of cell phones sold that were blue is 20%.

Question analysis

A table of cell phone sales by color gives each color's percentage as an expression in k, such as 8k for white and 30 − 2k for blue, with a total of 100. Students write and solve an equation for k, then find the percentage for blue. Combining terms with a negative coefficient, and remembering that k itself is not the answer, both need care.

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 Grade 7 students solve two-step equations that are written out for them. Here the equation must be built from a table whose percentages total 100, with like terms to combine, and the value found must then be substituted back.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 4, 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 hides an equation in a table, and the key insight is that the four percentages must add to 100. After that the algebra is routine, and no listed choice rewards stopping early, since neither k nor any other color's percentage appears. The wrong choices sit near the answer or use the blue expression with k ignored, so students who cannot set up the equation are left guessing.

  1. Many steps moderate Working it out

    The answer needs a chain: add the four expressions, set the sum to 100, collect 8k − 2k + k into 7k, solve for k, then substitute into 30 − 2k. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    No equation is printed; the table's Total row of 100 is what turns the four percentages written in k into one equation to solve. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Error-prone arithmetic minor Working it out

    Collecting like terms includes the negative coefficient in the blue row, −2k, and the constants 30, 30 and 5 have to be gathered too. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add the four color expressions and set the sum equal to 100, combine like terms to solve for k, then substitute k into the blue row's expression.

Traps

  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Combining like terms
ak+bk=(a+b)kak + bk = (a + b)k
k is the unknown number. a and b are the numbers in front of k, called coefficients. Terms with the same letter are called like terms. You can add their coefficients, because a groups of k plus b groups of k make a + b groups of k.
Keeping an equation balanced
if a=b, then a−c=b−c and a÷c=b÷c\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c
a and b are the two sides of an equation, and c is any number (not 0 when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it balanced. Use it to undo each step and get k alone.

  1. Sort out the table

    The table gives the percent of phones sold in each color. Some entries use k, a letter that stands for an unknown number. 8k means 8 times k. You need the percent that were blue, 30−2k30 - 2k. To find it, you first need k.

  2. Write an equation for the total

    Every phone sold is one of the four colors. So the four percents add up to the total, 100, as the last row of the table shows.

    8k+30+(30−2k)+(k+5)=1008k + 30 + (30 - 2k) + (k + 5) = 100

  3. Combine like terms

    Terms with k in them are called like terms, so they can be combined. Group the k terms together and the plain numbers together. A k on its own means 1k.

    ak+bk=(a+b)kak + bk = (a + b)k

    8k−2k+k=(8−2+1)k=7k8k - 2k + k = (8 - 2 + 1)k = 7k

    30+30+5=6530 + 30 + 5 = 65

    So the equation becomes 7k+65=1007k + 65 = 100. Watch out: the blue row takes away 2k. Keep its minus sign, or the k terms will be wrong.

  4. Solve for k

    Undo the adding first, then the multiplying.

    if a=b, then a−c=b−c and a÷c=b÷c\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c

    7k+65−65=100−657k + 65 - 65 = 100 - 65

    7k=357k = 35

    k=35÷7=5k = 35 \div 7 = 5

    So k is 5. But k is not the answer. You still need the blue row.

  5. Find the blue percent

    Put 5 in for k in the blue row.

    30−2k=30−2×5=30−10=2030 - 2k = 30 - 2 \times 5 = 30 - 10 = 20

    So 20% of the phones sold were blue. Watch out: 2k means 2 times k, not 2. Reading 30−2k30 - 2k as 30−230 - 2 gives 28%, choice D.

    So the answer is B, 20%.

Check your answer

Put k = 5 into every row. White: 8×5=408 \times 5 = 40. Black: 30. Blue: 20. Red: 5+5=105 + 5 = 10. Now add them: 40+30+20+10=10040 + 30 + 20 + 10 = 100. The total matches the table, so k = 5 is right.

More questions like this

The original DOE question
Question 93 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 93. Open page 85 of the handbook (PDF)

Question 37 of 99

2024 Test A Math, question 37

Question

Number line with points X, Y and Z marked from left to right; Y is closer to X than to Z. Above the line, a dimension arrow from X to Y is labeled 1 and one from Y to Z is labeled 2.

On the number line above, the distance between X and Y is 1 unit, and the distance between Y and Z is 2 units. What is the distance, in units, between Y and the midpoint of X and Z?

  • A12\frac{1}{2}
  • B1
  • C1121\frac{1}{2}
  • D3
Show the answer

Correct answer: A

First, find the distance between X and Z by adding the distance between X and Y to the distance between Y and Z.

1+2=31 + 2 = 3

Since the distance between X and Z is 3, the midpoint between X and Z is 1121\frac{1}{2} units from X.

The midpoint between X and Z is 1121\frac{1}{2} units from X, and the distance between X and Y is 1. The distance between Y and the midpoint of X and Z is 112−1=121\frac{1}{2} - 1 = \frac{1}{2}.

Question analysis

This number line question shows points X, Y and Z, with 1 unit from X to Y and 2 units from Y to Z, and asks how far Y is from the midpoint of X and Z. Students find the length of XZ, locate its midpoint, and compare it with Y. Each step produces a tempting number, so stopping early is the usual mistake.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 number line problems usually involve negative numbers and fractions as coordinates. This one has no coordinates, only two whole-number distances, and needs three short steps: a total length, a midpoint and a difference.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 39 words.
Corrected from the DOE handbook
  • An imprecise statement about the midpoint 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 arithmetic is light, but every intermediate result is waiting among the choices, so the risk is stopping before the last subtraction. Students also have to keep two midpoints apart: the one asked about lies between X and Z, not halfway along the 2-unit stretch from Y. Marking X at 0 and the midpoint at 1.5 on the drawing makes the half-unit gap visible.

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

    The number line has no coordinates, only dimension arrows labeled 1 and 2, so students must assign positions themselves, such as X at 0, Y at 1 and Z at 3. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Three short results come in sequence, the length XZ, the midpoint's distance from X, then its distance from Y, and each is also a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Place X at 0 so that Y and Z get positions from the given distances, find the midpoint of X and Z as half their distance, then compare it with Y's position.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says “the midpoint between X and Z is 1½” (twice). A midpoint is a point, and this number line has no numbers on it, so the 1½ is the midpoint’s distance from X (half of XZ = 3). Both sentences now say the midpoint is “1½ units from X.” The rest is unchanged: Y is 1 unit from X, so the distance between Y and the midpoint is 1½ − 1 = ½.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding lengths on a line
XZ=XY+YZXZ = XY + YZ
X, Y and Z are points on one line, with Y between X and Z. XY and YZ are the lengths of the two short pieces. XZ is the length of the whole. The two pieces join end to end to make the whole.
Midpoint of a segment
distance from an end to the midpoint=length of the segment2\text{distance from an end to the midpoint} = \frac{\text{length of the segment}}{2}
A segment is the straight piece of a line between two points, its ends. The midpoint is the point exactly halfway between the ends. It splits the segment into two equal halves. So it is half the length away from each end.

  1. Sort out the picture

    The points go X, then Y, then Z, from left to right. X to Y is 1 unit, and Y to Z is 2 units. You need the midpoint of X and Z, the point halfway between them. Then you need how far Y is from that midpoint. The line has no numbers, so measure everything from X.

  2. Find the length from X to Z

    XZ=XY+YZXZ = XY + YZ

    XZ=1+2=3XZ = 1 + 2 = 3

    X to Z is 3 units. Watch out: 3 is the whole length, choice D. It is not a distance to the midpoint.

  3. Place the midpoint

    The midpoint of X and Z is half of 3 units from X.

    distance from an end to the midpoint=length of the segment2\text{distance from an end to the midpoint} = \frac{\text{length of the segment}}{2}

    32=112\frac{3}{2} = 1\frac{1}{2}

    So the midpoint is 1121\frac{1}{2} units to the right of X. Watch out: this is its distance from X, not from Y. Stopping here gives choice C.

  4. Measure from Y to the midpoint

    Y is 1 unit to the right of X. The midpoint is 1121\frac{1}{2} units to the right of X. Subtract to find the gap between them.

    112−1=121\frac{1}{2} - 1 = \frac{1}{2}

    Watch out: use the midpoint of X and Z, not of Y and Z. Halving the 2 units from Y to Z gives 1, choice B.

    So the answer is A, 12\frac{1}{2} unit.

Check your answer

Give each point a number, with X at 0. Then Y is at 1 and Z is at 3. Halfway between 0 and 3 is 1121\frac{1}{2}. From Y at 1 to 1121\frac{1}{2} is 12\frac{1}{2}. This fits the picture too: Y sits a little left of the middle of XZ.

More questions like this

The original DOE question
Question 94 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 94. 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 38 of 99

2024 Test A Math, question 44

Question

A tank with a 500-gallon capacity currently contains 75 gallons of water. Additional water is poured into this tank at a rate of 5 gallons per minute. After 45 minutes of adding water, what percentage of the tank’s total capacity will be filled? (Assume that there is no loss of water from the tank.)

  • A45%
  • B55%
  • C60%
  • D70%
Show the answer

Correct answer: C

If 5 gallons of water is added to the tank per minute for 45 minutes, then 5×45=2255 \times 45 = 225 gallons is added to the tank.

The tank started with 75 gallons of water in it. After the 45 minutes, it contains 75+225=30075 + 225 = 300 gallons, assuming no loss of water.

The tank has a capacity of 500 gallons, and 300500=0.6\frac{300}{500} = 0.6.

Rewrite decimal 0.6 as a percentage:

0.6=610=60100=60%.0.6 = \frac{6}{10} = \frac{60}{100} = 60\%.

Question analysis

A 500-gallon tank holds 75 gallons, and water flows in at 5 gallons per minute for 45 minutes. This rate and percent word problem asks what percentage of the tank's capacity will then be filled. Students find the water added, the new amount in the tank and its percent of 500, keeping track of every quantity of water.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve multistep rate and percent problems. This one combines a pouring rate, a starting amount and a percent of capacity in three short steps with friendly numbers, which 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 8, 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 friendly, so the question tests whether students track every quantity of water. Forgetting the 75 gallons that were there from the start is the costly slip, because the share of the tank that was poured in is itself a listed choice. A quick check, that 300 gallons is more than half of 500 but well under three quarters, confirms the answer.

  1. Much to keep track of minor Working it out

    The 75 gallons already in the tank, given in the first sentence, must be held aside through the rate calculation and added back before dividing. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Many steps minor Working it out

    The chain runs from 5 gallons a minute for 45 minutes to 225 gallons, then to 300 gallons in the tank, then to 300 out of 500 as a percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply the rate by the time to get the water added, account for the water already in the tank, and divide the total by the 500-gallon capacity.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Amount from a rate
amount=rate×time\text{amount} = \text{rate} \times \text{time}
The rate is how much water goes in each minute, here 5 gallons. The time is how many minutes it runs. Multiply them to get the amount added. It works because the same amount goes in every minute.
Percent of a whole
percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100
The part is the amount you have, and the whole is the full amount. Dividing gives what fraction of the whole you have. Multiplying by 100 turns it into a percent, which means out of 100.

  1. Sort out the question

    The tank holds at most 500 gallons. That is its capacity. It already has 75 gallons in it. Water is added at 5 gallons per minute for 45 minutes. You need the percent of the tank that will be full.

  2. Find the water added

    The same amount of water goes in every minute, so multiply the rate by the time.

    amount=rate×time\text{amount} = \text{rate} \times \text{time}

    amount=5×45=225\text{amount} = 5 \times 45 = 225

    So 225 gallons are poured in during the 45 minutes.

  3. Find the total water in the tank

    Add the water that was already there.

    75+225=30075 + 225 = 300

    The tank holds 300 gallons after 45 minutes. Watch out: don’t forget the first 75 gallons. The 225 gallons alone fill 45% of the tank, choice A.

  4. Find the percent full

    percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100

    percent=300500×100=35×100=60\text{percent} = \frac{300}{500} \times 100 = \frac{3}{5} \times 100 = 60

    The tank is 60% full. Watch out: find the part that is full, not the part that is empty. Choice B, 55%, is the empty part if you also forget the first 75 gallons.

    So the answer is C, 60%.

Check your answer

10% of 500 gallons is 50 gallons. The tank has 300 gallons, which is 6 groups of 50. So it is 60% full. Also, 300 gallons is less than 500, so the tank never overflows.

More questions like this

The original DOE question
Question 101 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 101. Open page 87 of the handbook (PDF)

Question 39 of 99

2024 Test A Math, question 52

Question

What is the value of 1012+(−513)−(−234)10\frac{1}{2} + \left(-5\frac{1}{3}\right) - \left(-2\frac{3}{4}\right)?

  • A25122\frac{5}{12}
  • B711127\frac{11}{12}
  • C1311213\frac{1}{12}
  • D1871218\frac{7}{12}
Show the answer

Correct answer: B

Rewrite the mixed numbers so that they all have equivalent denominators.

1012+(−513)−(−234)10\frac{1}{2} + \left(-5\frac{1}{3}\right) - \left(-2\frac{3}{4}\right)

10612+(−5412)−(−2912)10\frac{6}{12} + \left(-5\frac{4}{12}\right) - \left(-2\frac{9}{12}\right)

Subtracting a negative is the same as adding.

10612−5412+291210\frac{6}{12} - 5\frac{4}{12} + 2\frac{9}{12}

Group the whole numbers together and the fractions together.

10612−5412+291210\frac{6}{12} - 5\frac{4}{12} + 2\frac{9}{12}

(10−5+2)+(612−412+912)(10 - 5 + 2) + \left(\frac{6}{12} - \frac{4}{12} + \frac{9}{12}\right)

711127\frac{11}{12}

Question analysis

This signed-number question asks for the value of 10 1/2 plus negative 5 1/3 minus negative 2 3/4. It tests adding and subtracting negative mixed numbers with unlike denominators. Turning each sign combination into a plain addition or subtraction, and then working in twelfths, is where students most often go wrong.

NYS standards
  • 7.NS.1c rational number subtraction
  • 7.NS.1d properties for adding and subtracting
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Adding and subtracting negative mixed numbers is central grade 7 work. This three-term expression is typical in length, but it needs a common denominator of 12 and two different sign rules: adding a negative and subtracting one.
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 10, 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 expression is short, but it packs both sign rules for negative numbers into one line, and each misapplication has its own choice. Subtracting a negative is the rule most often broken, and the result is a reasonable-looking mixed number. Estimating with decimals, 10.5 − 5.3 + 2.8 is about 8, separates the answer from every wrong choice.

  1. Error-prone arithmetic moderate Working it out

    Two different sign rules meet: adding −5 1/3 means subtracting 5 1/3, while subtracting −2 3/4 means adding 2 3/4, and mixing them up gives a listed value. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    After the signs, halves, thirds and quarters become twelfths, and the wholes, 10 − 5 + 2, are combined apart from the fractions, 6/12 − 4/12 + 9/12. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite adding a negative as subtraction and subtracting a negative as addition, convert the fractions to twelfths, then combine the whole numbers and the fractions separately.

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

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.
Equivalent fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a is the top of a fraction and b is the bottom. n is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, cut into smaller pieces. Use it to give two fractions the same bottom before you add or subtract.

  1. Sort out the expression

    You need 1012+(−513)−(−234)10\frac{1}{2} + \left(-5\frac{1}{3}\right) - \left(-2\frac{3}{4}\right). There are three mixed numbers, and two of them are negative. A mixed number is a whole number plus a fraction. Plan: fix the signs, give the fractions the same bottom, then work with the whole numbers and the fractions one group at a time.

  2. Fix the signs

    Adding −513-5\frac{1}{3} means subtracting 5135\frac{1}{3}. Subtracting −234-2\frac{3}{4} means adding 2342\frac{3}{4}.

    a+(−b)=a−banda−(−b)=a+ba + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b

    1012−513+23410\frac{1}{2} - 5\frac{1}{3} + 2\frac{3}{4}

    Watch out: subtracting a negative is adding. Subtracting 2342\frac{3}{4} instead gives 25122\frac{5}{12}, choice A. Dropping the minus signs gives C or D.

  3. Use twelfths for the fractions

    The bottoms are 2, 3 and 4. All three go into 12, so change each fraction to twelfths.

    ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}

    12=612,13=412,34=912\frac{1}{2} = \frac{6}{12}, \quad \frac{1}{3} = \frac{4}{12}, \quad \frac{3}{4} = \frac{9}{12}

    Now the problem is 10612−5412+291210\frac{6}{12} - 5\frac{4}{12} + 2\frac{9}{12}.

  4. Combine whole numbers, then fractions

    Work with the whole numbers and the fractions one group at a time. Keep each sign.

    10−5+2=710 - 5 + 2 = 7

    612−412+912=1112\frac{6}{12} - \frac{4}{12} + \frac{9}{12} = \frac{11}{12}

    The fraction part is less than 1, so just put them together: 711127\frac{11}{12}.

    So the answer is B, 711127\frac{11}{12}.

Check your answer

Estimate with decimals. 10.5−5.33+2.7510.5 - 5.33 + 2.75 is about 7.92. And 711127\frac{11}{12} is also about 7.92, since 1112\frac{11}{12} is almost 1. They match.

More questions like this

The original DOE question
Question 109 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 109. Open page 89 of the handbook (PDF)

Question 40 of 99

2024 Test A Math, question 56

Question

Number line from −3 to 8 with a tick at every integer. Point P is at −3, point Q is at 1 and point R is at 7, each marked with a dot.

On the number line above, what is the distance, in units, between the midpoint of PQ\overline{PQ} and the midpoint of QR\overline{QR}?

  • A3
  • B4
  • C5
  • D6
Show the answer

Correct answer: C

To find the midpoint of a segment, subtract the value on the left from the value on the right and then divide by 2. Then add the result to the value on the left of the segment.

Find the midpoint of PQ\overline{PQ}:

1−(−3)=41 - (-3) = 4

4÷2=24 \div 2 = 2

2+(−3)=−12 + (-3) = -1

Find the midpoint of QR\overline{QR}:

7−1=67 - 1 = 6

6÷2=36 \div 2 = 3

3+1=43 + 1 = 4

The distance between −1 and 4 on the number line is 5 units.

Question analysis

A number line marks points P, Q and R at labeled integer positions, with P to the left of zero. This midpoint and distance question asks how far apart the midpoints of segments PQ and QR are. Students locate both midpoints and then measure the distance between them, which takes care with negative positions.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students measure distances on a number line as the difference of two positions, including across zero. Here two midpoints must be located first, but all three points sit on labeled integers, which keeps it at grade 7 level.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 11, 21 words.
Corrected from the DOE handbook
  • A reversed instruction 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 points sit on labeled integers, so the midpoints can be found by counting, and the question is mostly about finishing carefully. Dropping the negative in 4 − (−1), stopping at the second midpoint, and counting tick marks instead of spaces each give a listed choice. Marking −1 and 4 on the figure and counting the unit spaces between them confirms the distance.

  1. Error-prone arithmetic minor Working it out

    P sits at −3, so the midpoint of PQ is (−3 + 1) ÷ 2 = −1, and the final distance, 4 − (−1), subtracts a negative number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Three results come in sequence, the midpoint of PQ, the midpoint of QR, then the gap between them, and the second of them, 4, is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Read the positions of P, Q and R, average each pair of endpoints to place the two midpoints, then subtract the left midpoint from the right one.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Stops one step early: A choice is a number from partway through the solution.
  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook’s rule says to “subtract the value on the right from the value on the left and then divide by 2,” but its own working does the opposite: 1 − (−3) = 4 for PQ and 7 − 1 = 6 for QR, the right value minus the left. Following the rule as printed gives (−3 − 1) ÷ 2 = −2 and (1 − 7) ÷ 2 = −3, which would put the midpoints at −5 and −2, outside the segments, and lead to a distance of 3, a wrong choice. The site now says “subtract the value on the left from the value on the right.” The midpoints, −1 and 4, and the distance between them, 5 units, are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint on a number line
midpoint=left end+right end−left end2\text{midpoint} = \text{left end} + \frac{\text{right end} - \text{left end}}{2}
The left end and right end are the numbers at the two ends of a segment. Right end minus left end is the segment’s length. The midpoint is halfway, so it is half the length to the right of the left end.
Distance on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
On a number line, numbers grow as you move right. The distance between two points is the larger number minus the smaller one. That is the right point minus the left point, so a distance is never negative.

  1. Read the number line

    P is at −3, Q is at 1 and R is at 7. A bar over two letters, such as PQ\overline{PQ}, means the segment between them. A segment is the straight piece of line between two points. You need the midpoint of each segment, the point exactly halfway between its ends. Then you need the distance between those two midpoints.

  2. Find the midpoint of PQ

    The left end is −3 and the right end is 1.

    midpoint=left end+right end−left end2\text{midpoint} = \text{left end} + \frac{\text{right end} - \text{left end}}{2}

    midpoint=−3+1−(−3)2=−3+42=−3+2=−1\text{midpoint} = -3 + \frac{1 - (-3)}{2} = -3 + \frac{4}{2} = -3 + 2 = -1

    So the midpoint of PQ is at −1. Note that 1−(−3)1 - (-3) is 1+31 + 3, which is 4.

  3. Find the midpoint of QR

    The left end is 1 and the right end is 7.

    midpoint=1+7−12=1+62=1+3=4\text{midpoint} = 1 + \frac{7 - 1}{2} = 1 + \frac{6}{2} = 1 + 3 = 4

    So the midpoint of QR is at 4. Watch out: 4 is where a midpoint is, not the answer. Stopping here gives choice B.

  4. Measure between the midpoints

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

    distance=4−(−1)=4+1=5\text{distance} = 4 - (-1) = 4 + 1 = 5

    Watch out: keep the minus sign on −1. Using 4−1=34 - 1 = 3 gives choice A. Also count the spaces, not the tick marks. From −1 to 4 there are 5 spaces but 6 marks, and 6 is choice D.

    So the answer is C, 5 units.

Check your answer

Each midpoint should sit in the middle of its segment. −1 is 2 units from both −3 and 1. 4 is 3 units from both 1 and 7. There is also a shortcut. The gap between the midpoints is half of PR: 7−(−3)2=102=5\frac{7 - (-3)}{2} = \frac{10}{2} = 5.

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 2025 Admissions, page 91, question 113. Open page 91 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 41 of 99

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 42 of 99

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 43 of 99

2024 Test B Math, question 24

Question

What is the value of (157−167)(347−367)\frac{\left(1\frac{5}{7}-1\frac{6}{7}\right)}{\left(3\frac{4}{7}-3\frac{6}{7}\right)}?

  • A−12-\frac{1}{2}
  • B−17-\frac{1}{7}
  • C17\frac{1}{7}
  • D12\frac{1}{2}
Show the answer

Correct answer: D

First, subtract to simplify.

157−167347−367=−17−27=−17÷−27\frac{1\frac{5}{7}-1\frac{6}{7}}{3\frac{4}{7}-3\frac{6}{7}}=\frac{-\frac{1}{7}}{-\frac{2}{7}}=-\frac{1}{7}\div -\frac{2}{7}

Follow the process for dividing fractions:

(−17)×(−72)=714=12\left(-\frac{1}{7}\right)\times\left(-\frac{7}{2}\right)=\frac{7}{14}=\frac{1}{2}

Question analysis

A fractions question: find the value of a quotient whose top is 1 5/7 minus 1 6/7 and whose bottom is 3 4/7 minus 3 6/7. Students subtract mixed numbers that share a denominator and then divide one difference by the other. Keeping track of positive and negative signs through both steps is the main hurdle.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students subtract and divide rational numbers, including negatives, and use the sign rules for quotients. This item chains three such operations with easy sevenths, a typical grade 7 exercise.
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, 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 mixed numbers look heavy, but the whole parts cancel and only sevenths remain. What decides the question is sign handling: each difference is negative, and one negative divided by another is positive. Students who stop after the top, subtract instead of divide, or keep a minus sign each find their result among the choices.

  1. Error-prone arithmetic moderate Working it out

    Both differences are negative, 1 5/7 − 1 6/7 = −1/7 and 3 4/7 − 3 6/7 = −2/7, and dividing them needs both the sign rule and a reciprocal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Two subtractions must be finished before the division, and a sign lost in either one carries into the final quotient. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out the top and the bottom separately, writing each difference as a signed fraction in sevenths, then divide by multiplying the top by the reciprocal of the bottom and apply the sign rule.

Traps

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

Dividing fractions
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Only the second fraction flips.
Dividing two negative numbers
(−a)÷(−b)=a÷b(-a) \div (-b) = a \div b
a and b are positive numbers, and b is not zero. A negative number divided by a negative number gives a positive answer. The two minus signs cancel. Multiplying two negatives works the same way.

  1. Sort out the problem

    The long fraction bar means divide. So you need the top difference divided by the bottom difference. A difference is the answer to a subtraction. Work out the top and the bottom first, then divide.

  2. Work out the top

    Both numbers on top have the same whole number, 1. So the wholes cancel, and you only need 57−67\frac{5}{7} - \frac{6}{7}. You take away more than you have, so the answer is negative.

    157−167=57−67=−171\frac{5}{7} - 1\frac{6}{7} = \frac{5}{7} - \frac{6}{7} = -\frac{1}{7}

  3. Work out the bottom

    Both numbers on the bottom have the whole number 3, so those cancel too.

    347−367=47−67=−273\frac{4}{7} - 3\frac{6}{7} = \frac{4}{7} - \frac{6}{7} = -\frac{2}{7}

    Watch out: do not stop after the top. −17-\frac{1}{7}, choice B, is only part of the problem.

  4. Divide the top by the bottom

    Both results are negative. The two minus signs cancel.

    (−a)÷(−b)=a÷b(-a) \div (-b) = a \div b

    −17÷(−27)=17÷27-\frac{1}{7} \div \left(-\frac{2}{7}\right) = \frac{1}{7} \div \frac{2}{7}

    Now flip the second fraction and multiply.

    ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

    17×72=714=12\frac{1}{7} \times \frac{7}{2} = \frac{7}{14} = \frac{1}{2}

  5. Avoid the traps and answer

    Watch out: keeping a minus sign gives −12-\frac{1}{2}, choice A. Subtracting the bottom from the top, instead of dividing, gives 17\frac{1}{7}, choice C.

    So the answer is D, 12\frac{1}{2}.

Check your answer

Think in sevenths. The top is −1 seventh and the bottom is −2 sevenths. That is like (−1)÷(−2)=12(-1) \div (-2) = \frac{1}{2}. Or multiply back: 12×(−27)=−17\frac{1}{2} \times \left(-\frac{2}{7}\right) = -\frac{1}{7}, which is the top.

More questions like this

The original DOE question
Question 81 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 81. Open page 205 of the handbook (PDF)

Question 44 of 99

2024 Test B Math, question 40

Question

The value of a particular car decreases at a constant rate. If the car is worth $15,000 three years after its original purchase date and $10,000 five years after its original purchase date, what was the value of the car on its original purchase date?

  • A$18,000
  • B$20,000
  • C$22,500
  • D$25,000
Show the answer

Correct answer: C

The value of the car decreased $5,000 in two years: $15,000 to $10,000.

5,000÷2=2,5005{,}000\div 2=2{,}500

Therefore, the value of the car decreases $2,500 each year. After the first 3 years, the car’s value had decreased $7,500.

So 7,500+15,000=22,5007{,}500+15{,}000=22{,}500.

Question analysis

A car loses value at a constant rate, and its worth three years and five years after purchase is given. Students find what the car was worth on the day it was bought. It tests rate of change and initial value in a linear relationship, and the tricky part is counting the right number of years when working back.

NYS standards
  • 8.F.4 modeling with linear functions
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Finding a starting value from two later values of a steady change is grade 8 linear-function work. A grade 7 student can reason with a yearly rate, but extending it back to the purchase date goes beyond typical grade 7 problems.
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, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is a straight-line pattern read backward, and its numbers are friendly; the difficulty is counting years correctly. The $5,000 drop happens over 2 years, not 5, and the purchase lies 3 years before the $15,000 value, not 2. Each miscount gives a listed answer, so sketching a timeline from year 0 to year 5 before computing pays off.

  1. Turning words into math moderate Understanding the question

    The phrases "three years after its original purchase date" and "five years after" must become points at year 3 and year 5 on a timeline whose start, year 0, holds the unknown value. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three moves are chained: the $5,000 drop over 2 years, the $2,500 yearly rate, and three years of that rate added back to $15,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the yearly drop from the change between the two given years, then add that drop back once for each year between the purchase and year 3.

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".
  • 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

Rate of change
rate=change in valuechange in time\text{rate} = \frac{\text{change in value}}{\text{change in time}}
change in value is how many dollars the value went up or down. change in time is how many years passed. rate is the change for one year. A constant rate means the change is the same every year.

  1. Sort out the question

    The car’s value “decreases at a constant rate.” That means the car loses the same number of dollars each year. Year 0 is the day it was bought.

    YearValue
    0?
    3$15,000
    5$10,000

    You need the value at year 0.

  2. Find the drop for one year

    From year 3 to year 5 is 2 years. In that time, the value fell by 15,000−10,000=5,00015{,}000 - 10{,}000 = 5{,}000 dollars.

    rate=5,0002=2,500\text{rate} = \frac{5{,}000}{2} = 2{,}500

    The car loses $2,500 each year. Watch out: Divide by the 2 years between the values, not by 5. Dividing by 5 leads to $18,000, choice A.

  3. Go back to year 0

    From year 0 to year 3 is 3 years. The car lost $2,500 in each of them. So add that amount back 3 times.

    15,000+3×2,500=15,000+7,500=22,50015{,}000 + 3 \times 2{,}500 = 15{,}000 + 7{,}500 = 22{,}500

    Watch out: Adding back only one $5,000 gives $20,000, choice B. That is the value at year 1, not year 0.

    So the answer is C, $22,500.

Check your answer

Count forward from $22,500, taking off $2,500 each year. Year 1 is $20,000. Year 2 is $17,500. Year 3 is $15,000. Year 4 is $12,500. Year 5 is $10,000. Both given values match.

More questions like this

The original DOE question
Question 97 as printed on page 210 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 97. Open page 210 of the handbook (PDF)

Question 45 of 99

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 46 of 99

2024 Test B Math, question 44

Question

Which inequality is equivalent to 6x−20<5x−2(5−3x)6x-20<5x-2(5-3x)?

  • Ax<16x<16
  • Bx<52x<\frac{5}{2}
  • Cx<−2x<-2
  • Dx>−2x>-2
Show the answer

Correct answer: D

6x−20<5x−2(5−3x)6x-20<5x-2(5-3x)

Distribute the coefficient −2 on the right-hand side:

6x−20<5x+(−2)(5)+(−2)(−3x)6x-20<5x+(-2)(5)+(-2)(-3x)

6x−20<5x−10+6x6x-20<5x-10+6x

Subtract 6x6x from both sides:

−20<5x−10-20<5x-10

Add 10 to both sides:

−10<5x-10<5x

Divide both sides by 5:

−2<x-2<x

This is equivalent to

x>−2x>-2

Question analysis

Students choose the simplest equivalent form of a linear inequality that has x on both sides and a negative number multiplied into parentheses. It tests distributing, combining like terms and solving inequalities. Sign errors in the distribution, and reading the direction of the final inequality, are where most mistakes happen.

NYS standards
  • 7.EE.4b two-step inequalities
  • 8.EE.7b linear equations with rational coefficients
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 inequalities have the form px + q < r. This one has x on both sides and a negative factor to distribute across a subtraction, steps that belong to grade 8 equation work, so it is harder than typical grade 7 practice.
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 16+, 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

Every move is standard algebra, but the inequality is built so that small slips change the answer completely. The −2 has to reach both terms in the parentheses, and once the boundary −2 is found its direction still has to be read correctly, since the variable ends up on the right as −2 < x. Rewriting that as x > −2, or substituting x = 0 into the original, catches the trap.

  1. Error-prone arithmetic moderate Working it out

    Distributing −2 over (5 − 3x) gives −10 + 6x, two sign decisions in one move, and the x terms then sit on both sides of the inequality. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    x < −2 and x > −2 share the boundary and differ only in direction, which depends on reading −2 < x correctly or flipping the sign when dividing by a negative. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Many steps minor Working it out

    Four moves follow one another, distributing, collecting 5x + 6x, moving the x terms and dividing by 5, and a slip at any of them changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Distribute the −2 to both terms in the parentheses, combine the x terms on the right, then gather the x terms on the side that keeps their coefficient positive.

Traps

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

Distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. Multiply a by each one. A minus sign inside counts as a negative term, so 5 − 3x is 5 + (−3x). A negative times a negative is positive.
Keeping an inequality true
if a<b, then a+c<b+c\text{if } a < b \text{, then } a + c < b + c
a and b are the two sides, and c is any number. Adding or subtracting the same number on both sides keeps the inequality true. Dividing both sides by a positive number also keeps it true. Only multiplying or dividing by a negative number flips the sign.

  1. Sort out the question

    An inequality compares two sides with a sign like <, which means is less than. To solve it, you get x alone on one side.

    You need the short inequality that has the same answers as the long one.

  2. Distribute the −2

    Multiply −2 by both terms inside the parentheses, 5 and −3x.

    −2(5−3x)=(−2)×5+(−2)×(−3x)=−10+6x-2(5 - 3x) = (-2) \times 5 + (-2) \times (-3x) = -10 + 6x

    A negative times a negative is positive, so the second part is +6x. Watch out: If the −2 multiplies only the 5, you end up with x<52x < \frac{5}{2}, choice B.

  3. Combine the x terms

    The right side is now 5x − 10 + 6x. Put the x terms together: 5x + 6x = 11x.

    6x−20<11x−106x - 20 < 11x - 10

  4. Gather the x terms on one side

    Subtract 6x from both sides. That leaves the x terms on the right, where they stay positive.

    6x−20−6x<11x−10−6x6x - 20 - 6x < 11x - 10 - 6x

    −20<5x−10-20 < 5x - 10

  5. Get x by itself

    Add 10 to both sides. Then divide both sides by 5.

    −10<5x-10 < 5x

    −2<x-2 < x

    You divided by a positive 5, so the sign does not flip.

  6. Read it with x first

    −2<x-2 < x says −2 is less than x. That is the same as saying x is greater than −2, just as 3 < 5 means 5 > 3.

    x>−2x > -2

    Watch out: x<−2x < -2, choice C, has the same number but points the wrong way.

    So the answer is D, x>−2x > -2.

Check your answer

Test x = 0, which is greater than −2. The left side is 6(0)−20=−206(0) - 20 = -20. The right side is 5(0)−2(5−0)=−105(0) - 2(5 - 0) = -10. And −20 < −10 is true. Now test x = −3. The left side is −38 and the right side is −43. That is false, as it should be.

More questions like this

The original DOE question
Question 101 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 101. Open page 211 of the handbook (PDF)

Question 47 of 99

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 48 of 99

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 49 of 99

2023 Test A Math, question 2

Question

6−9÷|−3|+(−2)3×1126 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}

What is the value of the expression shown above?

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

Show the answer

Correct answer: -9

6−9÷|−3|+(−2)3×1126 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2} Calculate the exponent and convert the mixed number to an improper fraction.
6−9÷|−3|+(−8)×326 - 9 \div \left|-3\right| + (-8) \times \frac{3}{2} Next, calculate the absolute value.
6−9÷3+(−8)×326 - 9 \div 3 + (-8) \times \frac{3}{2} Perform the indicated multiplication and division.
6−3+(−12)6 - 3 + (-12) Perform the addition and subtraction.
3+(−12)=−93 + (-12) = -9

Question analysis

This grid-in asks for the value of an arithmetic expression that mixes an absolute value, a negative number raised to the third power, a division and a mixed number. Students apply the order of operations with signed numbers. Cubing a negative base, and doing the division and multiplication before any adding, are where most slips happen.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.EE.2c evaluating expressions and order of operations
  • 6.NS.7c absolute value as distance
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students routinely evaluate expressions with negative numbers in the right order. This one combines an absolute value, a cubed negative and a mixed number in a single line, so it takes more steps than usual, each with a sign to watch.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 5, 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 expression packs four kinds of notation into one line: a subtraction, a division by an absolute value, a negative base raised to the third power and a mixed number. None of the operations is hard alone, but each has a rule that is easy to break under time pressure, and the slips lead to very different totals. Rewriting the power and the mixed number first, then dividing and multiplying before any addition, avoids them.

  1. Error-prone arithmetic moderate Working it out

    (−2)³ must stay negative, −8, and the mixed number 1½ must become 3/2 before multiplying, so the last product is −12; a sign or fraction slip here moves the total far from −9. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Six numbers and six operations, including a power and an absolute value, have to be worked in a planned order, and every addition waits until the division and multiplication are done. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    The value is negative and the grid-in offers no choices, so the minus sign has to be gridded and nothing on the page confirms that the total falls below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the absolute value and the power first, change the mixed number to an improper fraction, do the division and multiplication, and only then add and subtract from left to right.

Traps

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

Order of operations
grouping→powers→× and ÷→+ and −\text{grouping} \rightarrow \text{powers} \rightarrow \times \text{ and } \div \rightarrow + \text{ and } -
This is the order for working out an expression. First work inside grouping symbols, such as parentheses and absolute value bars. Next do powers. Then multiply and divide from left to right. Last, add and subtract from left to right. Using one order means everyone gets the same answer.
A power
a3=a×a×aa^{3} = a \times a \times a
a is the base, the number being multiplied. The small 3 is the exponent. It says how many times the base is used as a factor, a number being multiplied. It does not mean a times 3.
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. It is how far the number is from 0 on a number line. A distance is never negative, so the absolute value of −3 is 3.
Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, and b over c is the fraction part. Each whole is c pieces, so a wholes are a × c pieces. Add the b extra pieces. The bottom number stays c.

  1. Sort out the expression

    You need the value of 6−9÷|−3|+(−2)3×1126 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}. It has many operations, so the order matters. It has absolute value bars, a power of a negative number and a mixed number. The answer may be negative, so keep track of every sign.

  2. Do the absolute value and the power

    Grouping symbols and powers come first.

    |−3|=3\left|-3\right| = 3

    (−2)3=(−2)×(−2)×(−2)=4×(−2)=−8(-2)^{3} = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8

    Two negatives multiply to a positive, 4. One more negative makes it negative again. Watch out: (−2)³ is −8, not +8. Using +8 leads to 15. It is also not −2 × 3. That mistake leads to a total of −6.

  3. Change the mixed number to a fraction

    A fraction is easier to multiply than a mixed number.

    112=1×2+12=321\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}

    Now the expression looks like this:

    6−9÷3+(−8)×326 - 9 \div 3 + (-8) \times \frac{3}{2}

  4. Multiply and divide next

    Multiplying and dividing come before adding and subtracting.

    9÷3=39 \div 3 = 3

    (−8)×32=−242=−12(-8) \times \frac{3}{2} = \frac{-24}{2} = -12

    A negative times a positive is negative. Watch out: don’t just work from left to right. Doing 6 − 9 first, then dividing by 3, leads to −13.5.

  5. Add and subtract last

    Now only adding and subtracting are left. Work from left to right.

    6−3+(−12)=3+(−12)=−96 - 3 + (-12) = 3 + (-12) = -9

    Adding −12 is the same as taking away 12. So you would grid −9, with the minus sign in its own column.

Check your answer

Look at the size of each part. The part 6 − 3 is a small positive number, 3. The last part, −8 times 1½, is a bigger negative number, −12. A big negative plus a small positive is negative. So an answer below zero, −9, makes sense.

More questions like this

The original DOE question
Question 59 as printed on page 75 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 75, question 59. Open page 75 of the handbook (PDF)

Question 50 of 99

2023 Test A Math, question 5

Question

The mean value of 8 numbers is 17. Three of these numbers (9, 11, and 20) are discarded. What is the mean of the 5 remaining numbers?

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

Show the answer

Correct answer: 19.2

First, find the sum of the original 8 numbers. Let x equal the sum of those numbers and use the mean formula to solve for x:

x8=17\frac{x}{8} = 17

x=136x = 136

Next, subtract the 3 discarded numbers from the sum:

136−9−11−20=96136 - 9 - 11 - 20 = 96

Finally, calculate the mean of the remaining 5 numbers by dividing the new sum, 96, by 5:

965=19.2\frac{96}{5} = 19.2

Question analysis

Eight numbers have a mean of 17, and three of them, 9, 11 and 20, are removed. This statistics grid-in asks for the mean of the five numbers that are left. Students have to rebuild the total from the mean before removing anything, and then divide by the new count rather than the old one.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.NS.3 multi-digit decimal operations
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, removing values and averaging again is a typical grade 7 multistep task. As a grid-in, it offers no choices to check a wrong total against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 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

A mean cannot be adjusted directly when numbers are removed, so the question rewards switching from the mean to the total and back. The removed numbers, 9, 11 and 20, average less than 17, so the new mean must rise above 17; that quick check rules out an answer such as 12. Points are lost by dividing by the wrong count or by stopping at the new total.

  1. Many steps moderate Working it out

    Three operations come in a fixed order: multiply the mean by 8 to get 136, subtract 9, 11 and 20 to leave 96, then divide by the numbers that remain. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    The count changes partway through, from 8 numbers to 5, and the divisor at the end must be the new count, not the 8 from the first sentence. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

    The answer, 19.2, is a decimal that must be gridded with its point, and with no choices a total of 96 or a mean over 8 numbers looks complete. (Enright, Morley & Sheehan, 2002)

Strategy: Turn the mean into a total by multiplying by 8, subtract the three discarded numbers, and divide what is left by the count of numbers that remain.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean, or average
mean=sum of the numbershow many numbers\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}
The mean is the average. The sum is what you get when you add all the numbers. Divide it by how many numbers there are. The mean is the value each number would have if the total were shared out equally.
Sum from a mean
sum=mean×how many numbers\text{sum} = \text{mean} \times \text{how many numbers}
This is the mean formula turned around. If you know the mean and how many numbers there are, multiply them to get the sum. It works even when you don’t know the numbers themselves.

  1. Sort out the question

    There are 8 numbers. Their mean, or average, is 17. You don’t know what the 8 numbers are. Three of them, 9, 11 and 20, are discarded, or thrown away. That leaves 5 numbers. You need the mean of those 5 numbers.

  2. Find the sum of all 8 numbers

    You can’t add the 8 numbers, because you don’t know them. But the mean tells you their sum.

    sum=mean×how many numbers\text{sum} = \text{mean} \times \text{how many numbers}

    sum=17×8=136\text{sum} = 17 \times 8 = 136

    The 8 numbers add up to 136.

  3. Take away the three discarded numbers

    Subtract the three numbers that were thrown away.

    136−9−11−20=136−40=96136 - 9 - 11 - 20 = 136 - 40 = 96

    The 5 numbers left add up to 96. Watch out: 96 is a sum, not a mean, so gridding 96 is wrong. Also, dividing 136 by 5 without subtracting first gives 27.2, which is wrong.

  4. Divide by the 5 numbers left

    Now use the mean formula for the numbers that are left.

    mean=sum of the numbershow many numbers\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}

    mean=965=19.2\text{mean} = \frac{96}{5} = 19.2

    Watch out: divide by 5, not by 8. Only 5 numbers are left. Dividing by 8 gives 12. So you would grid 19.2, with the decimal point in its own column.

Check your answer

The three numbers thrown away add to 40, so their mean is about 13. That is less than 17. Removing small numbers should make the mean go up, and 19.2 is more than 17. Also, 19.2×5=9619.2 \times 5 = 96, the sum of the 5 numbers left.

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 2024 Admissions, page 76, question 62. Open page 76 of the handbook (PDF)

Question 51 of 99

2023 Test A Math, question 7

Question

In one week, 1341\frac{3}{4} inches of rain fell on Monday, 2232\frac{2}{3} inches fell on Tuesday, and 78\frac{7}{8} inch fell on Wednesday. How many inches of rain fell during those three days?

  • A57245\frac{7}{24}
  • B51245\frac{1}{24}
  • C3453\frac{4}{5}
  • D3123\frac{1}{2}
Show the answer

Correct answer: A

To solve, add the three given fractions:

134+223+781\frac{3}{4} + 2\frac{2}{3} + \frac{7}{8}

The common denominator is 24:

=11824+21624+2124= 1\frac{18}{24} + 2\frac{16}{24} + \frac{21}{24}

=41024+2124= 4\frac{10}{24} + \frac{21}{24}

=5724= 5\frac{7}{24}

Question analysis

A rainfall word problem gives the inches of rain on Monday, Tuesday and Wednesday as two mixed numbers and a fraction, and asks for the total over the three days. Students add fractions with unlike denominators. Finding a common denominator for fourths, thirds and eighths, and regrouping the fractional part into wholes, is where errors come in.

NYS standards
  • 7.NS.3 four operations with rational numbers
  • 7.NS.1d properties for adding and subtracting
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding fractions and mixed numbers with unlike denominators is taught before grade 7, and this is a direct word problem. It takes more care than a typical exercise only because three different denominators must be rewritten over a common one.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The setup is plain addition, so the difficulty is all in the fractions. Students who know to rename over a common denominator rarely miss it, while those who add the fraction parts straight across, or put the numerators over 24 without scaling them, reach choices with a whole part of 3. Since 1 3/4 + 2 2/3 alone is more than 4, any total below 4 can be ruled out at a glance.

  1. Error-prone arithmetic moderate Working it out

    The fractional parts have denominators 4, 3 and 8, so they must be renamed as 18, 16 and 21 twenty-fourths, and their sum, 55/24, regrouped as 2 7/24 before the wholes are added. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Renaming, adding the numerators, regrouping the improper fraction and adding the whole numbers are four separate moves, and losing the 2 carried from 55/24 drops the total by two inches. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the three fractional parts over a common denominator of 24, add the whole numbers and the fractions separately, and regroup any extra twenty-fourths as a whole.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a over b is a fraction, and k is any whole number except 0. Multiplying the top and the bottom by the same number gives an equal fraction. You cut each piece into k smaller pieces. So you have k times as many pieces, each k times smaller.
Adding fractions with the same bottom
ad+bd+cd=a+b+cd\frac{a}{d} + \frac{b}{d} + \frac{c}{d} = \frac{a + b + c}{d}
d is the denominator, the bottom number. It tells the size of the pieces. a, b and c are the numerators, the top numbers. They count the pieces. When the pieces are the same size, add the counts. The bottom stays the same.
Fraction to mixed number
ab=qrb\frac{a}{b} = q\frac{r}{b}
a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.

  1. Sort out the question

    Rain fell on three days: 1341\frac{3}{4} inches, 2232\frac{2}{3} inches and 78\frac{7}{8} inch. You need the total rain, in inches. So add the three amounts. Add the whole numbers and the fractions separately. The whole numbers are 1 and 2, so they add to 3 inches.

  2. Find a common denominator

    The fractions have denominators, or bottom numbers, 4, 3 and 8. You can only add pieces of the same size. So find the smallest number that 4, 3 and 8 all divide into. Count by 8s: 8, 16, 24. 24 works, since 24 = 4 × 6 and 24 = 3 × 8. So use 24ths.

  3. Rename each fraction in 24ths

    Multiply the top and the bottom by the same number.

    34=3×64×6=1824\frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}

    23=2×83×8=1624\frac{2}{3} = \frac{2 \times 8}{3 \times 8} = \frac{16}{24}

    78=7×38×3=2124\frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24}

    Watch out: change the tops too. Keeping 3, 2 and 7 over 24 gives 1224=12\frac{12}{24} = \frac{1}{2} and a total of 3123\frac{1}{2}, choice D.

  4. Add the fractions

    Now the pieces are the same size, so add the tops.

    1824+1624+2124=5524\frac{18}{24} + \frac{16}{24} + \frac{21}{24} = \frac{55}{24}

    That is more than 1 whole. 24 goes into 55 twice, with 7 left over.

    5524=2724\frac{55}{24} = 2\frac{7}{24}

    Watch out: never add the bottoms. Adding straight across gives 1215=45\frac{12}{15} = \frac{4}{5} and a total of 3453\frac{4}{5}, choice C.

  5. Put the parts together

    Add the 3 whole inches from before.

    3+2724=57243 + 2\frac{7}{24} = 5\frac{7}{24}

    So the answer is A, 57245\frac{7}{24} inches.

Check your answer

Use decimals. 1341\frac{3}{4} is 1.75, 2232\frac{2}{3} is about 2.67, and 78\frac{7}{8} is 0.875. They add to about 5.29. And 724\frac{7}{24} is about 0.29. So 57245\frac{7}{24} fits, and 51245\frac{1}{24} is too small.

More questions like this

The original DOE question
Question 64 as printed on page 77 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 64. Open page 77 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 52 of 99

2023 Test A Math, question 8

Question

Jamel works at a computer store. He is paid an hourly rate plus a 15% commission on all computer products he sells. Last week, Jamel was paid $802.50 for working 30 hours and selling $1,250.00 worth of computer products. What is Jamel’s hourly rate?

  • A$20.50/hr
  • B$26.75/hr
  • C$33.00/hr
  • D$37.65/hr
Show the answer

Correct answer: A

First, determine the amount of Jamel’s weekly pay that is from commission:

$1,250.00×15%=$187.50\$1{,}250.00 \times 15\% = \$187.50

Next, subtract Jamel’s commission amount from his weekly pay:

$802.50−$187.50=$615.00\$802.50 - \$187.50 = \$615.00

Last, divide the result by 30 hours to find Jamel’s hourly rate:

$615.00÷30=$20.50\$615.00 \div 30 = \$20.50

Question analysis

Jamel earns an hourly wage plus a 15% commission on his sales, and one week he is paid $802.50 for 30 hours of work and $1,250 in sales. Students find his hourly rate. The question mixes a percent with a rate, and the commission has to be taken out of the total pay before dividing by the hours.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Commission is one of the percent applications named in the grade 7 standards, and the remaining work is a two-step equation with decimals. That matches a typical grade 7 multistep problem, with the order of the steps as the main 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 3, 44 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Jamel's pay has two parts, and the question asks for only one of them, so the commission has to be removed before dividing by the hours. Students who divide the whole $802.50 by 30 get a believable hourly figure that still contains the commission. Checking the answer against the story, 30 × $20.50 + $187.50 = $802.50, confirms both parts.

  1. Turning words into math moderate Understanding the question

    The phrase "paid an hourly rate plus a 15% commission on all computer products he sells" makes $802.50 the sum of two parts, 30 hours of pay and 15% of $1,250. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps moderate Working it out

    Three operations in order: 15% of $1,250 is $187.50, $802.50 − $187.50 = $615.00, and $615.00 ÷ 30 = $20.50; dividing before subtracting gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    Every amount carries cents, and 15% of $1,250 and $615 ÷ 30 are worked by hand, so a misplaced decimal point shifts the hourly rate. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Take 15% of the $1,250 in sales to find the commission, remove that from the week's pay, and divide what is left by the 30 hours worked.

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.
  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • 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

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Pay with commission
pay=rate×hours+commission\text{pay} = \text{rate} \times \text{hours} + \text{commission}
The rate is the money Jamel earns for each hour. Hours is how long he worked. The commission is extra money that is a percent of what he sells. His pay is these two amounts added together.

  1. Sort out the question

    Jamel’s pay has two parts. One part is an hourly rate, the money he earns for each hour. The other part is a commission. A commission is extra pay that is a percent of what he sells. He worked 30 hours and sold $1,250.00 of products. In all he got $802.50. You need his hourly rate.

  2. Find the commission

    His commission is 15% of what he sold. 15% means 15 out of 100, or 0.15.

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    part=0.15×1,250=187.50\text{part} = 0.15 \times 1{,}250 = 187.50

    His commission was $187.50. Watch out: take 15% of the $1,250 he sold, not of his $802.50 pay. Mixing them up leads to $37.65, choice D.

  3. Take the commission out of his pay

    The rest of his pay came from his hours.

    802.50−187.50=615.00802.50 - 187.50 = 615.00

    He earned $615.00 for 30 hours of work. Watch out: the word plus does not mean add here. Adding the $187.50 to $802.50 counts the commission twice. That leads to $33.00, choice C.

  4. Divide by the hours he worked

    Share the $615.00 equally over the 30 hours.

    615÷30=20.50615 \div 30 = 20.50

    Watch out: don’t divide all $802.50 by 30. That gives $26.75, choice B. That amount still includes the commission. So the answer is A, $20.50 per hour.

Check your answer

Work forward from the answer. 30 hours at $20.50 is 30×20.50=61530 \times 20.50 = 615 dollars. Add the commission: 615+187.50=802.50615 + 187.50 = 802.50. That is exactly Jamel’s pay for the week.

More questions like this

The original DOE question
Question 65 as printed on page 77 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 65. Open page 77 of the handbook (PDF)

Question 53 of 99

2023 Test A Math, question 9

Question

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

  • A1:00 a.m.
  • B2:30 a.m.
  • C4:00 p.m.
  • D12:00 midnight
Show the answer

Correct answer: B

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

x8=901\frac{x}{8} = \frac{90}{1}

x=720x = 720

Next, convert 720 minutes into hours and minutes:

72060=12 hours\frac{720}{60} = 12\text{ hours}

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

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 A constant rate, a multiplication and a conversion from minutes to hours are grade 6 skills. The only step beyond routine work is adding the elapsed time to a starting clock time.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Many steps minor Working it out

    Three moves: 8 × 90 = 720 minutes, 720 ÷ 60 = 12 hours, then 12 hours counted on from the 2:30 start, which crosses midnight into the morning. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Something you must already know minor Working it out

    Counting hours past midnight means switching from p.m. to a.m. on a 12-hour clock, which the stem leaves to the student; the choices mix both halves of the day. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The sign turns all the way around once every 90 minutes. One full turn is called a revolution. The sign starts at 2:30 p.m. You need the clock time when it finishes 8 revolutions.

  2. Find the total time in minutes

    Each revolution takes 90 minutes, and there are 8 of them.

    total time=90×8=720\text{total time} = 90 \times 8 = 720

    The 8 revolutions take 720 minutes. Watch out: use all 8 revolutions. Counting only 7 gives 630 minutes and 1:00 a.m., choice A. Adding just one 90 minutes gives 4:00 p.m., choice C.

  3. Change the minutes to hours

    Clock times are easier to count in hours. There are 60 minutes in an hour.

    hours=720÷60=12\text{hours} = 720 \div 60 = 12

    The sign turns for 12 hours.

  4. Count forward from 2:30 p.m.

    Start at 2:30 p.m. and add 12 hours. When you pass midnight, p.m. changes to a.m. So 12 hours after 2:30 p.m. is 2:30 a.m. the next morning. Watch out: 12 hours is an amount of time, not a clock time. Reading it as 12:00 midnight gives choice D. So the answer is B, 2:30 a.m.

Check your answer

From 2:30 p.m. to 2:30 a.m. is one full trip around the clock, 12 hours. 12 hours is 12×60=72012 \times 60 = 720 minutes. And 720÷90=8720 \div 90 = 8, the number of revolutions in the question.

More questions like this

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

Question 54 of 99

2023 Test A Math, question 10

Question

A number line from 0 to 10 with a tick and a label at every whole number and arrows at both ends. Point W is marked with a dot at 0, and point X is marked with a dot at 2.

Points Y and Z are not shown on the number line above. If X is the midpoint of WY\overline{WY}, and Y is the midpoint of WZ\overline{WZ}, where on the number line would point Z be located?

  • A2
  • B4
  • C6
  • D8
Show the answer

Correct answer: D

First, find where Y is located on the number line. Point X is the midpoint of WY\overline{WY}. Since there are 2 units between W and X, there must also be 2 units between X and Y. Thus, Y is located at 4 on the number line.

Point Y is the midpoint of WZ\overline{WZ}. Use the same reasoning as above. Since there are 4 units between W and Y, there must be 4 units between Y and Z. So, Z is located at 8 on the number line.

Question analysis

A number line shows point W at 0 and point X at 2, and two more points, Y and Z, are described but not drawn. Students use two midpoint facts, one after the other, to find where Z lies. Each midpoint fact is about a different segment, so keeping track of which point is the middle one matters.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Midpoints on a number line rest on distance, which grade 7 students find as a difference. Here every point lies at a whole number of zero or more, so the work is two short steps with no negative numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 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

Each midpoint statement doubles a distance from W, so the problem is two short doublings, 2 to 4 and then 4 to 8. The risk is stopping halfway, since Y's position is found first and appears among the choices, or reusing X as an endpoint in the second statement. Rereading which point the question asks about, and checking that Y is halfway between W and Z, catches both.

  1. Many steps minor Working it out

    The two midpoint statements must be used in order: the first places Y at 4, and only then can the second, about segment WZ, place Z. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    Only W at 0 and X at 2 are marked on the 0-to-10 line, and the overbars name WY and WZ as segments, so Y and Z have to be placed from the statements alone. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the first midpoint to place Y as far beyond X as X is from W, then use the second midpoint the same way to place Z beyond Y.

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

Midpoint of a segment
AM=MBAM = MB
A and B are the ends of a segment, a straight piece of a line. M is its midpoint, the point exactly halfway. AM is the distance from A to M, and MB is the distance from M to B. A midpoint splits a segment into two equal parts, so these are equal.

  1. Sort out the question

    The number line shows W at 0 and X at 2. Points Y and Z are not shown. X is the midpoint of segment WY. Y is the midpoint of segment WZ. A midpoint is the point exactly halfway along a segment. You need the number where Z sits.

  2. Place point Y

    X is halfway from W to Y. So X to Y is as long as W to X.

    WX=2−0=2WX = 2 - 0 = 2

    XY=WX=2XY = WX = 2

    So Y is 2 units past X.

    2+2=42 + 2 = 4

    Point Y is at 4. Watch out: this is Y, not Z. Stopping here gives 4, choice B.

  3. Place point Z

    Now Y is halfway from W to Z. W to Y is 4 units, so Y to Z is also 4 units.

    YZ=WY=4−0=4YZ = WY = 4 - 0 = 4

    4+4=84 + 4 = 8

    Watch out: the second midpoint is for segment WZ, not XZ. Taking just 2 more units from Y puts Z at 6, choice C. So the answer is D, 8.

Check your answer

Check both midpoints. Halfway from 0 to 4 is (0+4)÷2=2(0 + 4) \div 2 = 2, which is X. Halfway from 0 to 8 is (0+8)÷2=4(0 + 8) \div 2 = 4, which is Y. Both facts in the question are true.

More questions like this

The original DOE question
Question 67 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 67. Open page 78 of the handbook (PDF)

Question 55 of 99

2023 Test A Math, question 18

Question

Sheila is saving money for her summer vacation. She starts the summer with a balance of $90.00 and plans to save 15% of her earnings each week. She earns the same amount each week. After 12 weeks, Sheila has saved a total of $472.59. How much money does Sheila earn each week?

  • A$212.55
  • B$262.55
  • C$302.55
  • D$312.55
Show the answer

Correct answer: A

Let x be the amount that Sheila needs to earn each week.

$90.00+0.15×12x=$472.59\$90.00 + 0.15 \times 12x = \$472.59

First, subtract 90 from both sides:

0.15×12x=$382.590.15 \times 12x = \$382.59

1.8x=$382.591.8x = \$382.59

Last, divide both sides by 1.8:

x=$212.55x = \$212.55

Question analysis

Sheila starts the summer with $90 and saves 15% of what she earns each week, earning the same amount every week, and after 12 weeks she has $472.59. Students find her weekly earnings. They must write an equation that separates the starting balance from the savings and then work with a percent and long decimals.

NYS standards
  • 7.EE.4a two-step equations
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an equation of the form px + q = r from a word problem is core grade 7 work. Here the coefficient combines a percent with 12 weeks, and the decimals are long, which adds steps but no new idea.
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 4, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The story has three layers, a starting balance, a weekly saving rate and a 12-week total, and the unknown sits under the percent. Students lose the point by forgetting the $90 was already there, or by moving it the wrong way. Testing a choice is a dependable alternative: 15% of $212.55 is about $31.88, and 12 of those plus $90 gives $472.59.

  1. Turning words into math moderate Understanding the question

    The total of $472.59 includes the $90.00 she started with, and only 15% of each week's pay is saved, so the equation is 90 + 12 × 0.15x = 472.59. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic moderate Working it out

    Dividing $382.59 by 1.8 by hand means shifting the decimal point to get 3,825.9 ÷ 18, which comes out to exactly $212.55 only if every digit is placed correctly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Many steps minor Working it out

    Subtract the $90, then divide by 12 weeks and by 0.15, or by their product 1.8; doing the division before the subtraction gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Subtract the starting $90 from the final total, then divide by 12 weeks and by 0.15, or test each choice by taking 15% of it, multiplying by 12 and adding $90.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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 of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Savings over time
total=start+weekly savings×weeks\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}
The start is the money she had at the beginning. Weekly savings is what she adds each week. Weeks is how many weeks pass. Each week adds the same amount, so multiply. Then add what she started with.

  1. Sort out the question

    Sheila starts with $90.00. Each week she earns the same amount and saves 15% of it. After 12 weeks she has $472.59 in all. That total includes the $90 she started with. You need how much she earns each week, not how much she saves.

  2. Write an equation

    Call her weekly earnings x. She saves 15% of x each week. That is 0.15×x0.15 \times x.

    total=start+weekly savings×weeks\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}

    472.59=90+0.15x×12472.59 = 90 + 0.15x \times 12

  3. Take away the starting $90

    Only the money added during the 12 weeks came from her pay. Subtract 90 from both sides.

    0.15x×12=472.59−90=382.590.15x \times 12 = 472.59 - 90 = 382.59

    She saved $382.59 from her pay. Watch out: subtract the $90, don’t add it. Adding leads to $312.55, choice D. Forgetting the $90 leads to $262.55, choice B.

  4. Combine the numbers next to x

    0.15×12=1.80.15 \times 12 = 1.8

    1.8x=382.591.8x = 382.59

    Each week she saves 0.15 of one week’s pay. Over 12 weeks, that adds up to 1.8 times one week’s pay.

  5. Divide by 1.8

    Divide both sides by 1.8. First move the decimal point one place right in both numbers. Then you divide by the whole number 18.

    x=382.59÷1.8=3,825.9÷18=212.55x = 382.59 \div 1.8 = 3{,}825.9 \div 18 = 212.55

    Sheila earns $212.55 each week. So the answer is A, $212.55.

Check your answer

Work forward. 12 weeks of saving 15% is 1.8 weeks of pay: 1.8×212.55=382.591.8 \times 212.55 = 382.59. Add the $90 she started with: 382.59+90=472.59382.59 + 90 = 472.59. That is the total in the question.

More questions like this

The original DOE question
Question 75 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 75. Open page 80 of the handbook (PDF)

Question 56 of 99

2023 Test A Math, question 20

Question

32+(−8)2+22(3−8+2)2=\frac{3^{2} + (-8)^{2} + 2^{2}}{(3 - 8 + 2)^{2}} =

  • A−60
  • B−173-\frac{17}{3}
  • C779\frac{77}{9}
  • D68
Show the answer

Correct answer: C

32+(−8)2+22(3−8+2)2=\frac{3^{2} + (-8)^{2} + 2^{2}}{(3 - 8 + 2)^{2}} =

9+64+4(−3)2=779\frac{9 + 64 + 4}{(-3)^{2}} = \frac{77}{9}

Question analysis

This arithmetic question asks for the value of a fraction whose numerator is the sum of the squares of 3, −8 and 2, and whose denominator is the square of 3 − 8 + 2. Students apply the order of operations with negative numbers. Squaring a negative number and adding inside parentheses before squaring are the points to watch.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.EE.2c evaluating expressions and order of operations
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Squaring negative numbers and simplifying with the order of operations is standard grade 7 practice. Here the fraction bar groups a numerator and a denominator that each involve a negative sign, but the work stays short.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 16+, 3 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 expression is built so that signs decide everything: the top's 3², (−8)² and 2² are all positive, and so is the square of the bottom's −3. Students who keep a negative after squaring, or who cancel the 9 on top against the 9 below, reach the other choices. The result, 77/9, does not reduce, which can make careful students doubt a correct answer.

  1. Error-prone arithmetic moderate Working it out

    Two negatives sit inside squares: (−8)² must be +64, and the denominator's 3 − 8 + 2 = −3 must be squared to +9, so every term ends up positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    The numerator needs three squares and a sum, the denominator a sum and then a square, and the fraction bar groups each side, so neither part can be split or cancelled. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Simplify the top and the bottom separately: square each number in the numerator, add inside the parentheses before squaring in the denominator, then write the fraction in lowest terms.

Traps

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

Square of a negative
(−a)2=(−a)×(−a)=a2(-a)^{2} = (-a) \times (-a) = a^{2}
a is any number. To square means to multiply a number by itself. A negative times a negative is positive. So the square of a negative number is always positive.
The fraction bar groups
topbottom=(top)÷(bottom)\frac{\text{top}}{\text{bottom}} = (\text{top}) \div (\text{bottom})
A fraction bar means divide. It also groups: work out the whole top and the whole bottom first, then divide. You can only cancel a number that multiplies the whole top, not one part of a sum.

  1. Sort out the expression

    The top, the numerator, is three squares added together. The bottom, the denominator, is a sum in parentheses that is then squared. To square means to multiply a number by itself. Work out the top and the bottom on their own. Then divide.

  2. Work out the top

    Square each number first. Then add.

    32=93^{2} = 9

    (−8)2=(−8)×(−8)=64(-8)^{2} = (-8) \times (-8) = 64

    22=42^{2} = 4

    9+64+4=779 + 64 + 4 = 77

    Watch out: (−8)² is +64, not −64. The minus sign is inside the parentheses, so it is squared too. Using −64 makes the top −51 and gives −173-\frac{17}{3}, choice B.

  3. Work out the bottom

    Work inside the parentheses first. Add and subtract before you square.

    3−8+2=−5+2=−33 - 8 + 2 = -5 + 2 = -3

    (−3)2=(−3)×(−3)=9(-3)^{2} = (-3) \times (-3) = 9

    The bottom is 9.

  4. Divide and simplify

    779\frac{77}{9}

    77 is 7 × 11 and 9 is 3 × 3. They share no factor, so the fraction is in lowest terms. Watch out: don’t cancel the 9 on top with the 9 below. That 9 is only one part of a sum. Canceling it leaves 64 + 4 = 68, choice D. So the answer is C, 779\frac{77}{9}.

Check your answer

Every square is positive or zero. So the top and the bottom are both positive, and the answer must be positive. That rules out −60 and −173-\frac{17}{3}. Also, 779\frac{77}{9} is about 8.6, which is far from 68.

More questions like this

The original DOE question
Question 77 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 77. Open page 80 of the handbook (PDF)

Question 57 of 99

2023 Test A Math, question 27

Question

A basket contains red balls, green balls, and white balls. There are 12 red balls in the basket. The probability of randomly choosing a red ball is 1 in 3. If the probability of randomly choosing a green ball is 1 in 4, how many green balls are in the basket?

  • A3
  • B8
  • C9
  • D16
Show the answer

Correct answer: C

First, use the information about the red balls to calculate the total number of balls in the basket:

13=12x\frac{1}{3} = \frac{12}{x} Cross multiply.
x=36x = 36

There are a total of 36 balls in the basket. The probability of selecting a green ball is 1 in 4. Use that information to calculate how many green balls are in the basket:

g36=14\frac{g}{36} = \frac{1}{4} Cross multiply.
4g=364g = 36 Apply the multiplicative inverse property; divide both sides of the equation by 4.
g=9g = 9

There are 9 green balls in the basket.

Question analysis

A basket holds red, green and white balls, including 12 red ones, and the chances of drawing red and green are 1 in 3 and 1 in 4. Students find how many green balls are in the basket. Getting the total from the red count is the step students skip when they apply a probability to the wrong number.

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 Simple probability is grade 6 content, but here it runs backward: a count and a probability give the total, which then gives another count. Two linked steps like this are typical of grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 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

The basket's total is never stated, and that missing number is the whole problem: it has to be recovered from the red balls before the green probability means anything. Applying the green chance straight to the 12 red balls is the quick wrong move. Checking that 12 red and 9 green fit in 36 balls with 15 white left over confirms the setup.

  1. Turning words into math minor Understanding the question

    The sentence "The probability of randomly choosing a red ball is 1 in 3" has to become 12 = one third of the total, which makes the basket hold 36 balls. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Two linked steps: the red balls give the total, 12 × 3 = 36, and only then does the green probability give a count, 36 ÷ 4 = 9. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use the red balls to find the total, since 12 is one third of it, then take one fourth of that total for the green balls.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=balls of one colorall the ballsP = \frac{\text{balls of one color}}{\text{all the balls}}
P is the probability, the chance of picking that color. Each ball is equally likely to be picked. So the chance is the fraction of all the balls that have that color. A chance of 1 in 3 means one third of the balls.

  1. Sort out the question

    The basket has red, green and white balls. There are 12 red balls. The chance of picking red is 1 in 3. The chance of picking green is 1 in 4. You need the number of green balls. You don’t know the total number of balls yet, so find it first.

  2. Find the total number of balls

    1 in 3 means one third of all the balls are red. So 12 is one third of the total. Call the total number of balls b.

    12b=13\frac{12}{b} = \frac{1}{3}

    b=12×3=36b = 12 \times 3 = 36

    There are 36 balls in all.

  3. Find the number of green balls

    1 in 4 means one fourth of all 36 balls are green. Call the number of green balls g.

    g36=14\frac{g}{36} = \frac{1}{4}

    g=36÷4=9g = 36 \div 4 = 9

    Watch out: take one fourth of all 36 balls, not of the 12 red ones. 12÷4=312 \div 4 = 3 gives choice A. Multiplying 12 by 43\frac{4}{3} gives 16, choice D. So the answer is C, 9.

Check your answer

A 1 in 4 chance is smaller than a 1 in 3 chance. So there should be fewer green balls than red ones, and 9 is less than 12. Check: 936=14\frac{9}{36} = \frac{1}{4}. The rest, 36−12−9=1536 - 12 - 9 = 15, are white.

More questions like this

The original DOE question
Question 84 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 84. Open page 82 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 99

2023 Test A Math, question 33

Question

If m=5m = 5 and t=−1t = -1, what is the value of 6−8(2−t)2m+4(3−m)\frac{6 - 8(2 - t)}{2m + 4(3 - m)}?

  • A−9
  • B−1
  • C1
  • D9
Show the answer

Correct answer: A

6−8(2−t)2m+4(3−m)=\frac{6 - 8(2 - t)}{2m + 4(3 - m)} =

6−8[2−(−1)]2(5)+4(3−5)=\frac{6 - 8[2 - (-1)]}{2(5) + 4(3 - 5)} =

6−8(3)10+4(−2)=\frac{6 - 8(3)}{10 + 4(-2)} =

6−2410+(−8)=\frac{6 - 24}{10 + (-8)} =

−182=−9\frac{-18}{2} = -9

Question analysis

This algebra question gives m = 5 and t = −1 and asks for the value of a fraction whose numerator and denominator both contain parentheses. Students substitute the values and follow the order of operations. Subtracting a negative number and multiplying by a negative difference make sign mistakes likely, and the four choices differ only in sign and size.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Substituting values into expressions with negative numbers is everyday grade 7 work. This fraction has a numerator and a denominator to evaluate separately, with a minus sign before a negative and a negative difference, so it is a longer version of a standard 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 8, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a plain evaluation with no story, and every point is lost on signs. Subtracting the negative t, then handling the two negative intermediate results, each invite a dropped minus, and the four choices are built so that each such slip lands on a listed value. Writing −1 in parentheses when substituting, and finishing the top and the bottom before dividing, keeps the signs in view.

  1. Error-prone arithmetic moderate Working it out

    The value t = −1 sits inside 2 − t, so the numerator needs 2 − (−1) = 3, and the work then passes through two negatives, 6 − 24 = −18 and 3 − 5 = −2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    The numerator and the denominator each take a substitution, a parenthesis, a product and a sum, worked separately before one final division, so a slip in either half changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The choices pair each size with each sign, −9 with 9 and −1 with 1, so getting 18 and 2 right still leaves the sign to decide between two choices. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Substitute with parentheses around −1, simplify the numerator and the denominator separately, innermost parentheses first, and divide only at the end, checking each sign as you go.

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

Subtracting a negative number
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. Think of a debt of 1 dollar being taken away: you are 1 dollar better off.
Dividing a negative by a positive
−ab=−ab\frac{-a}{b} = -\frac{a}{b}
a and b are positive numbers. A negative number divided by a positive number gives a negative answer. The answer is positive only when both signs are the same.

  1. Sort out the question

    You know m = 5 and t = −1. You need the value of the big fraction. Work out the top, called the numerator, and the bottom, called the denominator, one at a time. Then divide the top by the bottom.

  2. Put in the numbers

    6−8(2−t)2m+4(3−m)=6−8(2−(−1))2(5)+4(3−5)\frac{6 - 8(2 - t)}{2m + 4(3 - m)} = \frac{6 - 8(2 - (-1))}{2(5) + 4(3 - 5)}

    Put each number in parentheses, so the minus signs stay clear. Work inside the parentheses first, then multiply, then add or subtract.

  3. Work out the top

    a−(−b)=a+ba - (-b) = a + b

    2−(−1)=2+1=32 - (-1) = 2 + 1 = 3

    6−8(3)=6−24=−186 - 8(3) = 6 - 24 = -18

    The top is −18. It is negative because 24 is more than 6. Watch out: 2 − (−1) is 3, not 1. Using 1 makes the top −2 and the answer −1, choice B.

  4. Work out the bottom

    2(5)+4(3−5)=10+4(−2)=10−8=22(5) + 4(3 - 5) = 10 + 4(-2) = 10 - 8 = 2

    Here 3 − 5 = −2, because 5 is more than 3. Then 4 times −2 is −8. The bottom is 2.

  5. Divide the top by the bottom

    −ab=−ab\frac{-a}{b} = -\frac{a}{b}

    −182=−9\frac{-18}{2} = -9

    Watch out: if you lost the minus sign in 6 − 24, you would get 9, choice D. So the answer is A, −9.

Check your answer

Check the sign first. The top, 6 − 24, is clearly negative. The bottom, 10 − 8, is positive. Negative over positive is negative, so only −1 or −9 can work. Without the signs, the top is 18 and the bottom is 2, and 18 ÷ 2 = 9. So −9 fits.

More questions like this

The original DOE question
Question 90 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 90. Open page 84 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 99

2023 Test A Math, question 35

Question

A number line with tick marks every 2 units, labeled −6, −4, −2, 0, 2, 4 and 6. Point J is the dot at −4 and point K is the dot at 6.

On the number line above, point L (not shown) is located on line segment JK so that JL=23LK\mathrm{JL} = \frac{2}{3}\mathrm{LK}. What is the position of point L?

  • A−2
  • B0
  • C2
  • D4
Show the answer

Correct answer: B

JK=6−(−4)=10\mathrm{JK} = 6 - (-4) = 10 units. Let x=JLx = \mathrm{JL}. Then LK=10−x\mathrm{LK} = 10 - x. Substitute these expressions in the given equation:

x=23(10−x)x = \frac{2}{3}(10 - x)

Solve for x:

x=23(10−x)x = \frac{2}{3}(10 - x) Apply the multiplicative inverse property; multiply both sides of the equation by 3.
3x=2(10−x)3x = 2(10 - x) Apply the distributive property; multiply the 2 by 10 and x.
3x=20−2x3x = 20 - 2x Apply the additive inverse property; add 2x2x to both sides of the equation.
5x=205x = 20 Apply the multiplicative inverse property; divide both sides of the equation by 5.
x=4x = 4

So, point L is 4 units to the right of point J: −4+4=0-4 + 4 = 0. Point L is located at 0 on the number line.

Question analysis

This number line question puts J at −4 and K at 6 and asks for the position of a point L on segment JK where JL is two thirds of LK. Students find the segment's length, divide it in a 2-to-3 ratio, and turn a distance into a coordinate. Reading the ratio backwards, or giving a distance instead of a position, leads to wrong answers.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Distances on a number line that crosses zero are grade 7 rational-number work, and splitting a length in a given ratio is grade 7 proportional reasoning. Turning a fraction of one part into a split, then into a position, makes a typical multistep task.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 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

The ratio is stated between two unknown pieces, which is less familiar than a fraction of a whole, and the final step turns a length into a position. Students who stop at JL's length, read 2/3 the wrong way round, or use its 2 as a distance all reach listed points. Sketching five equal shares of 2 units from −4 to 6, two for JL and three for LK, makes the position plain.

  1. Turning words into math moderate Understanding the question

    JL = 2/3 LK compares the two pieces with each other, not a piece with the whole JK, so the 10 units must be split in the ratio 2 : 3, giving JL 4 and LK 6. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    J's position, −4, and K's, 6, are read from the number line, whose labeled ticks run every 2 units, and L itself is not drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    Three steps follow in order: the length JK = 6 − (−4), the length JL from the ratio, and finally L's position counted from J rather than from 0. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length of JK, split it into five equal shares with two for JL and three for LK, then count JL's length to the right from J.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • 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 on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
Subtract the position on the left from the position on the right. The answer is how many units apart the two points are. A distance is never negative.
Splitting a length in a ratio
one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}
A ratio a : b compares two pieces. If a length is split in the ratio a : b, think of it as a + b equal parts. The first piece gets a of them and the second gets b. Divide the whole length by a + b to find one part.

  1. Sort out the question

    Read the number line. Point J is at −4 and point K is at 6. Point L sits somewhere between them. JL is the length from J to L, and LK is the length from L to K. The rule says JL is 23\frac{2}{3} of LK. You need the position of L.

  2. Find the length of JK

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

    K is on the right, so subtract J’s position from K’s.

    JK=6−(−4)=6+4=10\mathrm{JK} = 6 - (-4) = 6 + 4 = 10

    Taking away −4 is the same as adding 4. J and K are 10 units apart.

  3. Turn the rule into equal parts

    JL is 23\frac{2}{3} of LK. So for every 2 units in JL, LK has 3 units. That means JK is made of 2 + 3 = 5 equal parts. JL gets 2 parts and LK gets 3.

    one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}

    one part=102+3=2\text{one part} = \frac{10}{2 + 3} = 2

    Each part is 2 units long.

  4. Find the two lengths

    JL=2×2=4\mathrm{JL} = 2 \times 2 = 4

    LK=3×2=6\mathrm{LK} = 3 \times 2 = 6

    Watch out: JL is 23\frac{2}{3} of LK, not the other way round. Flipping the ratio makes JL 6 and puts L at 2, choice C.

  5. Find the position of L

    L is 4 units to the right of J. Start at J’s position, −4, and move 4 to the right.

    −4+4=0-4 + 4 = 0

    Watch out: 4 is the length of JL, not where L sits. Answering 4 is choice D. So the answer is B, 0.

Check your answer

Put L at 0 and test the rule. JL = 0 − (−4) = 4, and LK = 6 − 0 = 6. Then 23×6=4\frac{2}{3} \times 6 = 4, which is JL. Also 4 + 6 = 10, the whole length JK.

More questions like this

The original DOE question
Question 92 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 92. Open page 84 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 60 of 99

2023 Test A Math, question 36

Question

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

  • A300
  • B800
  • C1,500
  • D2,500
Show the answer

Correct answer: C

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

j2j−500=60100\frac{j}{2j - 500} = \frac{60}{100} Cross multiply.
100j=60(2j−500)100j = 60(2j - 500) Apply the distributive property; multiply the 60 by 2j2j and 500.
100j=120j−30,000100j = 120j - 30{,}000 Apply the additive inverse property; subtract 120j120j from both sides of the equation.
−20j=−30,000-20j = -30{,}000 Apply the multiplicative inverse property; divide both sides of the equation by −20.
j=1,500j = 1{,}500

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here neither share nor the total is known; students must link the 500-stamp gap to the gap between two percents before finding the total and then Josef's part.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 26 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    The stem describes one split in two ways, Josef's 60% of the total and his 500-stamp lead over Mai, and the link, that the lead is 60% − 40% = 20% of the total, is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method moderate Working it out

    No equation is suggested: the quick route treats the 500 as 20 percentage points and scales up to 100%, while the handbook's route, j over 2j − 500, needs cross-multiplying and a negative coefficient. (Schoenfeld, 1985)

  3. Many steps minor Working it out

    Even the short route takes three moves, the 20% gap, the 2,500 total and Josef's 60% of it, and the total is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the total amount, and the part is the piece you want. Percent means “out of 100,” so 60% is 60 out of every 100. Use it to find a part, or work backward to find the whole.

  1. Sort out the question

    Josef and Mai split some stamps. Josef got 60% of them. He also got 500 more stamps than Mai. You need how many stamps Josef got. You don’t know the total yet, so find it first.

  2. Find Mai’s percent

    All the stamps make 100%. Josef has 60%, so Mai has the rest.

    100%−60%=40%100\% - 60\% = 40\%

    Mai has 40% of the stamps.

  3. Find what the 500 stamps are

    Josef’s extra 500 stamps are the gap between his share and Mai’s share.

    60%−40%=20%60\% - 40\% = 20\%

    So 500 stamps are 20% of all the stamps.

  4. Find the total number of stamps

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    500=20100×whole500 = \frac{20}{100} \times \text{whole}

    whole=500÷20100=500×5=2,500\text{whole} = 500 \div \frac{20}{100} = 500 \times 5 = 2{,}500

    There are 2,500 stamps in all. Watch out: that is everyone’s stamps, choice D. Don’t stop here.

  5. Find Josef’s share

    part=60100×2,500=1,500\text{part} = \frac{60}{100} \times 2{,}500 = 1{,}500

    Watch out: taking 60% of 500 gives 300, choice A. The 500 is the gap between them, not the total. So the answer is C, 1,500 stamps.

Check your answer

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

More questions like this

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

Question 61 of 99

2023 Test A Math, question 43

Question

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

  • A0.08
  • B80
  • C800
  • D8,000
Show the answer

Correct answer: B

First, calculate Kim’s jogging time in minutes:

1 hour 40 minutes = 100 minutes

Then, convert kilometers to meters:

8 kilometers = 8,000 meters

Now simplify the ratio to find the answer:

8,000100=80\frac{8{,}000}{100} = 80

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Unit rates and metric conversions are grade 6 topics. This needs two conversions, hours and minutes to minutes and kilometers to meters, then one division with whole numbers, simpler than grade 7 rate problems built on fractions.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 17 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know minor Working it out

    The conversions are not given: 1 kilometer is 1,000 meters and 1 hour is 60 minutes, so 8 kilometers becomes 8,000 meters and 1 hour 40 minutes becomes 100 minutes. (Anderson & Krathwohl, 2001)

  2. Many steps minor Working it out

    Two conversions, one for distance and one for time, come before the single division, and skipping the distance conversion leaves an answer a thousand times too small. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Kim jogs 8 kilometers in 1 hour 40 minutes. You need her speed in meters per minute: how many meters she jogs in one minute. The question gives kilometers and hours, so change both units first.

  2. Change the time to minutes

    One hour is 60 minutes. Add the extra 40 minutes.

    60+40=10060 + 40 = 100

    Kim jogs for 100 minutes.

  3. Change the distance to meters

    meters=kilometers×1,000\text{meters} = \text{kilometers} \times 1{,}000

    8×1,000=8,0008 \times 1{,}000 = 8{,}000

    Kim jogs 8,000 meters. Watch out: 8,000 is the whole distance, not meters per minute. Stopping here gives choice D.

  4. Divide the distance by the time

    rate=distancetime\text{rate} = \frac{\text{distance}}{\text{time}}

    rate=8,000100=80\text{rate} = \frac{8{,}000}{100} = 80

    She jogs 80 meters each minute. Watch out: if you skip the change to meters, you get 8 ÷ 100 = 0.08 kilometer per minute, choice A. So the answer is B, 80 meters per minute.

Check your answer

Work backward. At 80 meters a minute, 100 minutes gives 80×100=8,00080 \times 100 = 8{,}000 meters. That is 8 kilometers, just as the question says.

More questions like this

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

Question 62 of 99

2023 Test A Math, question 44

Question

For what value of x is the equation x5−4=3(4−2x)−1\frac{x}{5} - 4 = 3(4 - 2x) - 1 true?

  • A7511\frac{75}{11}
  • B7531\frac{75}{31}
  • C157\frac{15}{7}
  • D6531\frac{65}{31}
Show the answer

Correct answer: B

x5−4=3(4−2x)−1\frac{x}{5} - 4 = 3(4 - 2x) - 1 Apply the distributive property; multiply the 3 by 4 and −2x-2x.
x5−4=12−6x−1\frac{x}{5} - 4 = 12 - 6x - 1 Combine like terms.
x5−4=11−6x\frac{x}{5} - 4 = 11 - 6x Apply the additive inverse property; add 6x6x and 4 to both sides of the equation.
x5+6x=15\frac{x}{5} + 6x = 15 Get a common denominator.
x+30x5=15\frac{x + 30x}{5} = 15 Apply the multiplicative inverse property; multiply both sides of the equation by 5.
31x=7531x = 75 Apply the multiplicative inverse property; divide both sides of the equation by 31.
x=7531x = \frac{75}{31}

Question analysis

This equation question asks for the value of x that makes x/5 − 4 = 3(4 − 2x) − 1 true. Students use the distributive property, collect variable terms on one side, and clear a fraction before solving. Each wrong choice matches a common slip with the parentheses or with the fraction, so careful bookkeeping matters more than speed.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 equations have the form px + q = r. This one has x on both sides, a fractional coefficient and a product to expand, which are grade 8 equation skills, and every wrong choice matches a plausible slip along the way.
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 8, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is a routine move in solving a linear equation, but there are five of them and the fraction raises the cost of each. The listed wrong answers come from distributing the 3 to only one term, clearing the fraction from one term only, and stretching the parentheses over the −1. Substituting the result back is slow with a denominator of 31, so careful step-by-step work matters more than checking.

  1. Error-prone arithmetic moderate Working it out

    The term x/5 sits beside whole-number terms and a negative −6x, so clearing the fraction means multiplying every term by 5, and the answer is the unfriendly 75/31. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps moderate Working it out

    Five moves in a row, distributing the 3, combining 12 and −1, collecting the x terms, clearing the fifth and dividing by 31, each give a slip a place to enter. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The fraction 75/11 shares its numerator with 75/31 and differs by one digit of the denominator, so only the coefficient of x, 6 1/5 rather than 2 1/5, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Distribute the 3 to both terms in the parentheses, combine constants, collect the x terms on one side, then multiply every term by 5 to clear the fraction; check by substituting.

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

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)=ab−aca(b - c) = ab - ac
a, b and c are numbers or terms. Multiplying a by a difference means multiplying a by each part. The number outside must reach every term inside the parentheses.
Doing the same to both sides
if a=b, then a+c=b+c and a×c=b×c\text{if } a = b\text{, then } a + c = b + c \text{ and } a \times c = b \times c
a and b are the two sides of an equation, and c is any number. An equation is like a balanced scale. Adding the same amount to both sides, or multiplying both sides by the same number, keeps it balanced.

  1. Sort out the question

    You need the value of x that makes both sides of the equation equal. The plan: clear the parentheses, gather the x terms on one side, and get rid of the fraction. Then divide.

  2. Clear the parentheses

    a(b−c)=ab−aca(b - c) = ab - ac

    3(4−2x)=12−6x3(4 - 2x) = 12 - 6x

    x5−4=12−6x−1=11−6x\frac{x}{5} - 4 = 12 - 6x - 1 = 11 - 6x

    Watch out: the 3 multiplies both 4 and 2x. Writing 12 − 2x leads to 7511\frac{75}{11}, choice A. The −1 is outside the parentheses, so the 3 does not touch it.

  3. Gather the x terms

    Add 6x to both sides to move the x terms to the left. Add 4 to both sides to move the plain numbers to the right.

    if a=b, then a+c=b+c and a×c=b×c\text{if } a = b\text{, then } a + c = b + c \text{ and } a \times c = b \times c

    x5+6x=11+4=15\frac{x}{5} + 6x = 11 + 4 = 15

  4. Clear the fraction

    Multiply both sides by 5. Every term gets multiplied, not only the fraction.

    5×x5+5×6x=5×155 \times \frac{x}{5} + 5 \times 6x = 5 \times 15

    x+30x=75x + 30x = 75

    31x=7531x = 75

    Watch out: multiplying only the fraction by 5 gives x + 6x = 15. That leads to 157\frac{15}{7}, choice C.

  5. Divide by 31

    x=7531x = \frac{75}{31}

    75 and 31 share no factor except 1, so the fraction is already simplified. So the answer is B, 7531\frac{75}{31}.

Check your answer

Put x=7531x = \frac{75}{31} back in. Then x5=1531\frac{x}{5} = \frac{15}{31} and 2x=150312x = \frac{150}{31}. Write 4 as 12431\frac{124}{31}. The left side is 1531−12431=−10931\frac{15}{31} - \frac{124}{31} = -\frac{109}{31}. On the right, 4−15031=−26314 - \frac{150}{31} = -\frac{26}{31}, and 3×(−2631)−1=−7831−3131=−109313 \times \left(-\frac{26}{31}\right) - 1 = -\frac{78}{31} - \frac{31}{31} = -\frac{109}{31}. Both sides match.

More questions like this

The original DOE question
Question 101 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 101. Open page 86 of the handbook (PDF)

Question 63 of 99

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 64 of 99

2023 Test A Math, question 46

Question

ESSAY LENGTH

Number of Words Number of Essays
<100 6
100–250 4
251–500 11
>500 9

All 150 students in Grade 8 at a school are assigned to write an essay on the same topic. A teacher records the number of words in a random sample of the essays, as shown in the table above. Based on this sample, how many students in the entire grade would be expected to write essays with at least 100 words?

  • A20
  • B30
  • C100
  • D120
Show the answer

Correct answer: D

The number of students in the sample who wrote an essay with at least 100 words is 24 out of the 30 students. Write it as a fraction: 2430=45\frac{24}{30} = \frac{4}{5}. To predict the number of students in the entire grade who would write an essay with at least 100 words, find 45\frac{4}{5} of 150, which is 45(150)=120\frac{4}{5}(150) = 120.

Question analysis

This statistics question gives a table of word counts for a random sample of Grade 8 essays and asks how many of all 150 students would be expected to write at least 100 words. Students find the sample size, the share meeting the condition, and scale it to the whole grade. Matching the words at least to the table's ranges is the tricky part.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a random sample to predict a count for a whole population is typical grade 7 statistics and ratio work. Students total the sample from a table, read at least 100 against the table's ranges, and scale by a whole-number factor.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 65 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 proportion at the end is routine; the question is decided at the table. Students must turn “at least 100 words” into the right rows and also find the sample size, which is printed nowhere. Each listed wrong answer scales up a different group of rows, so a careless reading still produces a believable number. Noting that only 6 of 30 essays fall short, one fifth, gives a quick check.

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

    The table never prints the sample size; it has to be built by adding all four rows, 6 + 4 + 11 + 9 = 30, and the qualifying essays sit in three rows, not one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Turning words into math minor Understanding the question

    The bold “at least 100 words” must be matched to the rows 100–250, 251–500 and >500, keeping the row that begins at 100 and leaving out only the <100 row. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Many steps minor Working it out

    Four steps follow in order: the sample total, the qualifying count of 24, the fraction 24/30 = 4/5, and that fraction of all 150 students. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the table's counts to find the sample size, add the rows with 100 or more words, and scale that fraction of the sample up to all 150 students.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from a sample
expected number=count in samplesample size×whole group\text{expected number} = \frac{\text{count in sample}}{\text{sample size}} \times \text{whole group}
A sample is a smaller group picked at random from the whole group. The fraction of the sample with some feature is your best guess for the fraction of the whole group. So multiply that fraction by the size of the whole group.

  1. Sort out the question

    The table shows a random sample of the essays from a grade of 150 students. You need to predict how many of the 150 students wrote at least 100 words. At least 100 means 100 or more.

  2. Find the sample size

    The table doesn’t give the total, so add all four rows.

    6+4+11+9=306 + 4 + 11 + 9 = 30

    The sample has 30 essays.

  3. Count the essays with enough words

    Three rows count: 100–250, 251–500 and >500. Only the <100 row is left out.

    4+11+9=244 + 11 + 9 = 24

    Watch out: the 100–250 row counts, because 100 is at least 100. Leaving it out gives 100, choice C. Using the 6 short essays instead gives 30, choice B.

  4. Scale up to the whole grade

    expected number=count in samplesample size×whole group\text{expected number} = \frac{\text{count in sample}}{\text{sample size}} \times \text{whole group}

    2430×150=45×150=120\frac{24}{30} \times 150 = \frac{4}{5} \times 150 = 120

    About 4 out of every 5 students write at least 100 words. So the answer is D, 120 students.

Check your answer

The whole grade, 150, is 5 times the sample of 30. So multiply the sample count by 5: 24×5=12024 \times 5 = 120. The short essays scale to 6×5=306 \times 5 = 30, and 120 + 30 = 150, the whole grade.

More questions like this

The original DOE question
Question 103 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 103. Open page 87 of the handbook (PDF)

Question 65 of 99

2023 Test A Math, question 54

Question

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

  • A46
  • B45
  • C43
  • D41
Show the answer

Correct answer: D

First, find the speed of the train in miles per hour: 240÷5=48240 \div 5 = 48.

The number of miles left to travel is 2,200−240=1,9602{,}200 - 240 = 1{,}960.

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

48t=1,96048t = 1{,}960 Apply the multiplicative inverse property; divide both sides of the equation by 48.
t=1,96048=4056t = \frac{1{,}960}{48} = 40\frac{5}{6}

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

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Unit rates are grade 6 content, but this chains four steps: a speed, the distance left, a long division that does not come out even, and rounding. The word more must be read closely, which puts it at typical grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 4, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Many steps moderate Working it out

    Four steps follow in order, the speed 240 ÷ 5 = 48, the miles left 2,200 − 240 = 1,960, the time 1,960 ÷ 48, and the rounding, and an early slip carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The bold more in “how many more hours” limits the question to the 1,960 miles still ahead, though 2,200, the whole trip, is the distance the stem names first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Dividing 1,960 by 48 leaves 40 with a remainder of 40, so the time, 40 5/6 hours, must be rounded up to the nearest hour rather than cut short. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The whole trip is 2,200 miles. The train goes the first 240 miles in 5 hours, and it keeps that speed. You need how many more hours it takes, rounded to the nearest whole hour. “More” means the time after the first 240 miles.

  2. Find the train’s speed

    d=r×td = r \times t

    240=r×5240 = r \times 5

    r=240÷5=48r = 240 \div 5 = 48

    Divide the 240 miles by the 5 hours. The train goes 48 miles per hour.

  3. Find the miles still to go

    2,200−240=1,9602{,}200 - 240 = 1{,}960

    There are 1,960 miles left. Watch out: use the miles left, not the whole trip. Dividing 2,200 by 48 gives about 45.8 hours, which leads to 46 or 45, choices A and B.

  4. Find the hours still to go

    1,960=48×t1{,}960 = 48 \times t

    t=1,960÷48=404048=4056t = 1{,}960 \div 48 = 40\frac{40}{48} = 40\frac{5}{6}

    Here 48 × 40 = 1,920, and 40 miles are left over. That leftover is 4048=56\frac{40}{48} = \frac{5}{6} of an hour.

  5. Round to the nearest hour

    56\frac{5}{6} of an hour is more than half an hour, so round up to 41. Watch out: don’t just drop the fraction.

    So the answer is D, 41 hours.

Check your answer

At 48 miles per hour, 41 hours covers 48×41=1,96848 \times 41 = 1{,}968 miles. That is just a bit more than 1,960, as it should be after rounding up. In 40 hours, the train would cover only 1,920 miles, which is not enough.

More questions like this

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

Question 66 of 99

2023 Test B Math, question 6

Question

What is the value of 314−123−(−412)3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)?

  • A−3112-3\frac{1}{12}
  • B−21112-2\frac{11}{12}
  • C61126\frac{1}{12}
  • D95129\frac{5}{12}
Show the answer

Correct answer: C

The first step is to rewrite the problem replacing all subtraction with adding the opposite:

314−123−(−412)3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)

314+(−123)+4123\frac{1}{4} + \left(-1\frac{2}{3}\right) + 4\frac{1}{2}

Then, rewrite all the fractions using a common denominator of 12, which is the least common multiple of the fractions’ denominators, 4, 3, and 2:

3312+(−1812)+46123\frac{3}{12} + \left(-1\frac{8}{12}\right) + 4\frac{6}{12}

Next, combine the positive numbers and, finally, add the negative number to the sum by subtracting the number with the least absolute value from the number with the greatest absolute value and using the sign of the number with the greatest absolute value:

7912+(−1812)7\frac{9}{12} + \left(-1\frac{8}{12}\right)

61126\frac{1}{12}

Question analysis

What is 3 1/4 − 1 2/3 − (−4 1/2)? This question on rational-number arithmetic combines mixed numbers with denominators of 4, 3 and 2 and ends by subtracting a negative number. Students must turn that last subtraction into an addition and find a common denominator before combining.

NYS standards
  • 7.NS.1c rational number subtraction
  • 7.NS.1d properties for adding and subtracting
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Subtracting negative mixed numbers is core grade 7 content. This expression is typical in length, but it needs a common denominator of 12 and careful handling of the rule for subtracting a negative.
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 10, 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 expression is short, but it mixes a subtraction that stays a subtraction with one that becomes an addition, and that difference is what most wrong answers miss. The fractions add a second load, since twelfths are needed before anything combines. Estimating first, 3 − 1.7 + 4.5 is about 6, shows the result is positive, which rules out both negative values.

  1. Error-prone arithmetic moderate Working it out

    Three mixed numbers with denominators 4, 3 and 2 must become twelfths, and the last term, −(−4 1/2), changes sign while the middle subtraction does not. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Rewriting the double negative, renaming each fraction, adding 3 1/4 and 4 1/2, then taking away 1 2/3 are four steps, each able to carry a slip forward. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Change subtracting −4 1/2 into adding 4 1/2, rename every fraction in twelfths, add the two positive numbers, then take away 1 2/3.

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

Subtracting a negative number
a−(−b)=a+ba - (-b) = a + b
a and b stand for any numbers. Taking away a negative number is the same as adding its opposite, the same number with the other sign. Think of it this way: taking away a debt leaves you richer.
Equivalent fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a is the top and b is the bottom of a fraction. Multiplying both by the same number n does not change its value. Use it to give fractions the same bottom number, so you can add or subtract them.

  1. Sort out the question

    You need the value of 314−123−(−412)3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right). Each number is a mixed number, a whole number plus a fraction. There are two subtractions. The second one takes away a negative number.

  2. Change the double negative

    Use the rule for subtracting a negative. Taking away −412-4\frac{1}{2} is the same as adding 4124\frac{1}{2}.

    314−123+4123\frac{1}{4} - 1\frac{2}{3} + 4\frac{1}{2}

    The first minus sign stays as it is. Watch out: if you subtract 4124\frac{1}{2} instead, you get −21112-2\frac{11}{12}, choice B.

  3. Give the fractions one denominator

    The denominator is the bottom number of a fraction. Here they are 4, 3 and 2. The smallest number all three divide into is 12. So change each fraction to twelfths.

    14=31223=81212=612\frac{1}{4} = \frac{3}{12} \qquad \frac{2}{3} = \frac{8}{12} \qquad \frac{1}{2} = \frac{6}{12}

    The problem is now:

    3312−1812+46123\frac{3}{12} - 1\frac{8}{12} + 4\frac{6}{12}

  4. Add the two positive numbers

    You may add and subtract in any order, as long as each number keeps its sign. Add the two positive numbers first.

    3312+4612=79123\frac{3}{12} + 4\frac{6}{12} = 7\frac{9}{12}

    That is 3 + 4 = 7 wholes and 3 + 6 = 9 twelfths.

  5. Take away the last number

    7912−1812=61127\frac{9}{12} - 1\frac{8}{12} = 6\frac{1}{12}

    That is 7 − 1 = 6 wholes and 9 − 8 = 1 twelfth. So the answer is C, 61126\frac{1}{12}.

Check your answer

Estimate with decimals: 3.25−1.67+4.5≈6.083.25 - 1.67 + 4.5 \approx 6.08. And 61126\frac{1}{12} is about 6.08 too. The answer must be positive, so A and B are out. Choice D comes from adding all three numbers.

More questions like this

The original DOE question
Question 63 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 63. Open page 193 of the handbook (PDF)

Question 67 of 99

2023 Test B Math, question 12

Question

The perimeter of a 12-sided polygon is 100 centimeters. The length of one side is 25 centimeters. The length of another side is 35 centimeters. The remaining sides are equal in length to one another. What is the length of each of the remaining sides, in centimeters?

  • A4
  • B5
  • C20
  • D40
Show the answer

Correct answer: A

Let x represent the length of one of the 10 remaining sides:

25+35+10x=10025 + 35 + 10x = 100

60+10x=10060 + 10x = 100

10x=4010x = 40

x=4x = 4

Question analysis

A 12-sided polygon has a perimeter of 100 centimeters: one side is 25 cm long, another is 35 cm, and the rest are all equal to one another. Students find the length of each remaining side. This perimeter problem turns into a two-step equation, and the tricky part is counting the sides that are left.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving a two-step equation from a short description is standard grade 7 work. The perimeter setting is simple; the only extra care is counting the sides that remain after the two given ones.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 44 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The perimeter and the two named sides make the subtraction obvious, so the real decision comes after it: what to divide the leftover 40 centimeters by. Students who stop at the leftover length, or who lose track of how many sides remain, reach a listed choice. Writing 25 + 35 + 10x = 100 keeps the count of 10 in view.

  1. Turning words into math minor Understanding the question

    “The remaining sides are equal in length to one another” hides the count: 12 sides minus the two given ones leaves 10 equal sides to share the rest of the perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Subtracting 25 and 35 from 100, counting the sides left, and dividing are three steps, and the first result, 40, is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Subtract the two known sides from the perimeter, count how many sides are left, and share the remaining length equally among them.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a polygon
P=the sum of all the side lengthsP = \text{the sum of all the side lengths}
P is the perimeter, the total distance around a shape. A polygon is a flat shape with straight sides. A sum is the answer to an addition. So add the lengths of every side to get P.

  1. Sort out the question

    The shape is a polygon with 12 sides. Its perimeter is 100 centimeters. One side is 25 centimeters, and another is 35 centimeters. The other sides all have the same length. You need that length.

  2. Count the equal sides

    Two of the 12 sides are already known. The rest are the equal sides.

    12−2=1012 - 2 = 10

    So there are 10 equal sides. Call the length of each one x.

  3. Write an equation for the perimeter

    Use the perimeter formula. All 12 sides add up to 100. Ten sides of length x add up to 10x, which means 10 times x.

    P=the sum of all the side lengthsP = \text{the sum of all the side lengths}

    25+35+10x=10025 + 35 + 10x = 100

  4. Solve for x

    60+10x=10060 + 10x = 100

    10x=100−60=4010x = 100 - 60 = 40

    The 10 equal sides add up to 40 centimeters. Watch out: 40 is all ten sides together, not one side. Stopping here gives choice D. Now share the 40 among the 10 sides.

    x=40÷10=4x = 40 \div 10 = 4

  5. Give the answer

    Each remaining side is 4 centimeters long. Watch out: sharing the 40 centimeters between only 2 sides gives 20, choice C. That treats the shape as if it had 4 sides, but it has 12. So the answer is A, 4.

Check your answer

Add all 12 sides: 25+35+10×4=25+35+40=10025 + 35 + 10 \times 4 = 25 + 35 + 40 = 100. That is the perimeter the question gives, so 4 centimeters is right.

More questions like this

The original DOE question
Question 69 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 69. Open page 194 of the handbook (PDF)

Question 68 of 99

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 69 of 99

2023 Test B Math, question 18

Question

A rectangle whose bottom side is labeled 16 ft and whose left side is labeled 8 ft. It is shaded gray except for a white triangle inside it: the triangle’s vertex is at the middle of the left side, and its base lies along the right side and is labeled 4 ft.

The diagram above shows a rectangle with an inscribed triangle. The triangle’s vertex and base touch the sides of the rectangle as shown. What is the total area of the shaded portion?

  • A32 sq ft
  • B64 sq ft
  • C96 sq ft
  • D128 sq ft
Show the answer

Correct answer: C

First, find the area of the rectangle, in square feet:

16×8=12816 \times 8 = 128

Then, find the area of the triangle:

12×4×16=32\frac{1}{2} \times 4 \times 16 = 32

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle:

128−32=96 sq ft128 - 32 = 96\text{ sq ft}

Question analysis

A 16-foot by 8-foot rectangle has a white triangle inside it, with one vertex at the middle of the left side and a 4-foot base along the right side. Students find the total area of the shaded part. They subtract the triangle's area, whose height is not labeled and must be read from the figure.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Composite areas built from rectangles and triangles appear in grades 6 and 7. This one is typical, but the triangle lies on its side, so students must see which length is its height rather than read it from a label.
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, 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 shaded region is a rectangle with a triangle cut out, so the plan is standard, but the triangle is turned so that its height runs across the figure. Students who see that the height is the 16-foot side have two short areas to find and one subtraction. The listed wrong answers stop at either area or leave out the one half.

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

    The triangle lies on its side: its 4-foot base runs along the right edge, so its height is the rectangle’s full 16-foot length, which the figure labels only as the bottom side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    The rectangle’s area, the triangle’s area and their difference are three separate results, and the first two, 128 and 32, are themselves listed choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the rectangle's area, find the triangle's area from its base on the right side and its height across the rectangle, then subtract.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • 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

Area of a rectangle
A=l×wA = l \times w
A is the area, the space inside a flat shape, in square units. l is the length and w is the width. It works because the rectangle is filled by w rows of unit squares, with l squares in each row.
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 you measure from. h is the height, the distance from the base to the opposite 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.

  1. Sort out the figure

    The figure is a rectangle 16 feet long and 8 feet tall. Inside it is a white triangle. Its tip touches the middle of the left side. Its base lies along the right side and is 4 feet long. The shaded part is the gray area around the triangle. You need its area.

    Plan: find the rectangle’s area, then take away the triangle’s area.

  2. Find the area of the rectangle

    A=l×wA = l \times w

    A=16×8=128A = 16 \times 8 = 128

    The rectangle’s area is 128 square feet. Watch out: this includes the white triangle. Stopping here gives choice D.

  3. Find the triangle’s height

    The triangle lies on its side. Its base is the 4-foot piece on the right side, which runs up and down. The height must meet the base at a right angle, a square corner. So the height runs straight across, from the base to the tip on the left side. That is the full length of the rectangle, 16 feet.

  4. Find the area of the triangle

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×4×16=32A = \frac{1}{2} \times 4 \times 16 = 32

    The triangle’s area is 32 square feet. Watch out: don’t forget the 12\frac{1}{2}. Using 4×16=644 \times 16 = 64 leaves 64 for the shaded part, choice B.

  5. Subtract to find the shaded area

    128−32=96128 - 32 = 96

    The shaded part is 96 square feet. Watch out: 32 is the white triangle, not the shaded part, so choice A is wrong. So the answer is C, 96 sq ft.

Check your answer

The triangle’s base, 4 feet, is half the rectangle’s height of 8 feet. Its height is the full 16 feet. So the triangle is half of half the rectangle, or one fourth of it. One fourth of 128 is 32, and 128−32=96128 - 32 = 96.

More questions like this

The original DOE question
Question 75 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 75. Open page 196 of the handbook (PDF)

Question 70 of 99

2023 Test B Math, question 22

Question

A vendor is selling 10 packages of homemade fudge at a bake sale. The vendor weighed and labeled each package. The package weights are given in ounces:

4.6 6.0 3.8 3.5 4.5
5.4 3.9 6.5 4.8 6.2

The vendor created a box plot of the package weights. Which box plot best represents the data?

  • ABox plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at about 4.1, the box runs from 5 to 6 with the median line at 5.5, and the right whisker ends at about 6.3.
  • BBox plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from 4 to 6 with the median line at 4.5, and the right whisker ends at 6.5.
  • CBox plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from about 3.9, just left of 4, to 6, with the median line halfway between 4.5 and 5, and the right whisker ends at 6.5.
  • DBox plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from 4 to 6 with the median line at 5, and the right whisker ends at 6.5.
Show the answer

Correct answer: C

A set of 10 data points is given. The first step for determining the quartiles and minimum and maximum values of the box plot is to order the data from smallest to greatest value. For this data set the order is:

3.5 3.8 3.9 4.5 4.6
4.8 5.4 6.0 6.2 6.5

The minimum value is the smallest decimal, 3.5, and the maximum value is the greatest decimal, 6.5.

To determine the median in a set with an even number of values, locate the two middle values (5th and 6th in this case): 4.6 and 4.8. Determine the mean, or average, of the two middle numbers by adding them together and dividing the sum by 2:

4.6+4.82=9.42=4.7\frac{4.6 + 4.8}{2} = \frac{9.4}{2} = 4.7

To determine the value of the first quartile (Q1), locate the middle value of the 5 terms to the left of the median:

3.5 3.8 3.9 4.5 4.6

To determine the value of the third quartile (Q3), locate the middle value of the 5 terms to the right of the median:

4.8 5.4 6.0 6.2 6.5

Box plot C identifies the correct minimum, maximum, median, and quartiles of the data set.

Question analysis

A vendor weighs 10 packages of homemade fudge, from 3.5 to 6.5 ounces, and makes a box plot of the weights. Students choose the box plot that matches the data. They must sort the decimals, find the median of an even number of values and both quartiles, then compare plots that differ only slightly.

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
Above grade 7 Building a box plot from data is grade 7 content. This version needs all five summary values from ten unsorted decimals, including a median between two values, then a close comparison of plots, longer and finer than typical.
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 4, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Most of the work happens before the plots are even read: the weights must be sorted, and the median of ten values is the average of two of them. After that, three plots look almost identical, so a careless median, taken from the unsorted list or placed halfway between the extremes, settles on a wrong one. The careful student checks the median line and the box’s left edge against values that fall between the tick marks.

  1. Many steps moderate Working it out

    Ten unsorted decimals must be ordered, then five values found: the median as the mean of the fifth and sixth, 4.6 and 4.8, and each quartile as the middle of its half. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    Three plots share whiskers at 3.5 and 6.5 and a box ending at 6; they differ only in the median line, at 4.5, near 4.75 or at 5, and in where the box starts. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    The axis is ticked every half ounce, and the correct plot’s median, 4.7, and lower quartile, 3.9, both fall between ticks, so each line’s position must be judged by eye. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  4. Slow to do carefully minor Choosing the answer

    Sorting ten decimals, computing a five-number summary and comparing four plots takes well over a minute and a half when done carefully. (Lu & Sireci, 2007)

Strategy: Sort the ten weights, then find the least and greatest, the median as the mean of the fifth and sixth values, and each quartile as the middle of its half.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • 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

Median of an even number of values
median=first middle value+second middle value2\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}
First put the values in order. With an even number of values, two sit in the middle. The median is their mean, or average: add them and divide by 2. The median splits the data in half.
Quartiles
Q1=middle of the lower halfQ3=middle of the upper halfQ_1 = \text{middle of the lower half} \qquad Q_3 = \text{middle of the upper half}
Q1 is the first quartile and Q3 is the third quartile. Split the ordered data in half at the median. Q1 is the middle of the lower half, and Q3 is the middle of the upper half. A box plot’s box runs from Q1 to Q3.

  1. Sort out the question

    The vendor weighed 10 packages of fudge, in ounces. You need the box plot that matches the weights. A box plot is built from five numbers. They are the minimum (least), the first quartile Q1, the median, the third quartile Q3 and the maximum (greatest). So find those five numbers first.

  2. Put the weights in order

    List the 10 weights from least to greatest.

    3.53.83.94.54.6
    4.85.46.06.26.5

    Watch out: you must sort first. The fifth weight as printed is 4.5. Using it as the median points to choice B.

  3. Find the least and greatest

    The minimum is 3.5 ounces and the maximum is 6.5 ounces. So the whiskers must run from 3.5 to 6.5. Plot A’s whiskers run from about 4.1 to about 6.3, so A is out.

  4. Find the median

    With 10 weights, the two middle ones are the 5th and 6th: 4.6 and 4.8.

    median=first middle value+second middle value2\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}

    median=4.6+4.82=9.42=4.7\text{median} = \frac{4.6 + 4.8}{2} = \frac{9.4}{2} = 4.7

    The median is 4.7 ounces. Watch out: the median is not halfway between the least and greatest. That gives (3.5+6.5)÷2=5(3.5 + 6.5) \div 2 = 5 and points to choice D.

  5. Find the two quartiles

    The lower half is the 5 weights below the median: 3.5, 3.8, 3.9, 4.5 and 4.6. Its middle value is Q1=3.9Q_1 = 3.9.

    The upper half is the 5 weights above the median: 4.8, 5.4, 6.0, 6.2 and 6.5. Its middle value is Q3=6.0Q_3 = 6.0.

  6. Match the summary to a plot

    The five numbers are 3.5, 3.9, 4.7, 6.0 and 6.5. Plots B, C and D all run from 3.5 to 6.5, and their boxes end at 6. They differ in the median line and where the box starts.

    • B: the box starts at 4, and the median line is at 4.5.
    • D: the box starts at 4, and the median line is at 5.
    • C: the box starts just left of 4, and the median line is between 4.5 and 5.

    C is the closest match to a Q1 of 3.9 and a median of 4.7. So the answer is C.

Check your answer

A median splits the data in half. Five weights, 3.5 to 4.6, are below 4.7. Five, 4.8 to 6.5, are above it. A median line at 5, as in D, would have six weights below it. A line at 4.5, as in B, would have only three below it.

More questions like this

The original DOE question
Question 79 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 2024 Admissions, page 198, question 79. Open page 198 of the handbook (PDF)

Question 71 of 99

2023 Test B Math, question 23

Question

FRUIT SOLD
AT STORE XYZ

Fruit Number Sold
Apples 8
Bananas 6
Cherries 12
Oranges 6
Peaches ?
Total 50

Peaches made up what percentage of the total number of pieces of fruit sold at Store XYZ, according to the table above?

  • A18%
  • B28%
  • C36%
  • D64%
Show the answer

Correct answer: C

First, find the number of peaches sold:

50−(8+6+12+6)=50 - (8 + 6 + 12 + 6) =

50−32=1850 - 32 = 18

Then, find the percentage:

1850=0.36=36%\frac{18}{50} = 0.36 = 36\%

Question analysis

A table lists the fruit sold at Store XYZ, apples, bananas, cherries and oranges, with the number of peaches shown as a question mark and a total of 50. Students find what percentage of the fruit sold were peaches. They first work out the missing count, then turn it into a percent of the total.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a missing value and writing a part as a percent of a whole are grade 6 skills. With a total of 50, the percent step is quick, so the task is lighter than typical grade 7 percent problems.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 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 total of 50 makes the percent step easy, since each piece of fruit is 2 percent, but the question first hides the count it needs. Students who find the 18 peaches and stop, or who give the share of the four listed fruits, reach listed choices. Checking that the peaches’ percent and the other fruits’ percent add to 100 confirms the answer.

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

    The table gives four counts, a question mark for peaches and a total of 50, so the missing count must be found by subtraction before any percent. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Adding 8, 6, 12 and 6, subtracting from 50, and turning 18 out of 50 into a percent are three steps, and the count of 18 is itself a listed percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the four known counts, subtract from 50 to get the peaches, then double that count, since each piece out of 50 is 2 percent.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a total
percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100
The part is the amount you are asking about, here the peaches. The whole is the total, here all 50 pieces of fruit. Dividing gives the fraction of the whole. Multiplying by 100 turns it into a percent, which means out of 100.

  1. Sort out the question

    The table shows how many of each fruit Store XYZ sold. The number of peaches is missing, shown by a question mark. The total is 50 pieces of fruit. You need the percent of the total that were peaches.

  2. Add the fruits you know

    8+6+12+6=328 + 6 + 12 + 6 = 32

    Apples, bananas, cherries and oranges add up to 32 pieces of fruit. That is every fruit except the peaches.

  3. Find the number of peaches

    The peaches are whatever is left of the 50.

    50−32=1850 - 32 = 18

    So 18 peaches were sold. Watch out: 18 is a count, not a percent. Choosing 18% is choice A.

  4. Turn the count into a percent

    percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100

    1850×100=0.36×100=36\frac{18}{50} \times 100 = 0.36 \times 100 = 36

    Peaches were 36% of the fruit sold. Watch out: 32 out of 50 is 64%, choice D. That is the other four fruits, not the peaches. So the answer is C, 36%.

Check your answer

Out of 50 pieces, each piece is 2% of the total, because 100÷50=2100 \div 50 = 2. So 18 peaches are 18×2=3618 \times 2 = 36 percent. Also, 36% for peaches plus 64% for the rest makes 100%.

More questions like this

The original DOE question
Question 80 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 2024 Admissions, page 198, question 80. Open page 198 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 99

2023 Test B Math, question 25

Question

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

  • A7,500
  • B8,400
  • C9,000
  • D9,200
Show the answer

Correct answer: C

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

p50=1,500\frac{p}{50} = 1{,}500

Solve for p:

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

The population decrease from 2005 to 2006 is 84,000−75,000=9,00084{,}000 - 75{,}000 = 9{,}000.

Question analysis

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

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

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    “The average population per square mile was equal to 1,500” is a density, which has to become a total, 1,500 × 50 = 75,000, before it can be compared with 84,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    The 2006 population is only a midpoint; subtracting it from 84,000 is a second step, and the question asks for that decrease, not either population. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Waterville covers 50 square miles. In 2005, 84,000 people lived there. In 2006, there were 1,500 people per square mile, on average. You need how many fewer people lived there in 2006 than in 2005.

  2. Find the 2006 population

    Let p be the 2006 population. Use the formula.

    people per square mile=populationarea\text{people per square mile} = \frac{\text{population}}{\text{area}}

    1,500=p501{,}500 = \frac{p}{50}

    To undo dividing by 50, multiply both sides by 50.

    p=1,500×50=75,000p = 1{,}500 \times 50 = 75{,}000

    So 75,000 people lived there in 2006.

  3. Find the decrease

    A decrease is how much a number went down. Subtract the 2006 population from the 2005 population.

    84,000−75,000=9,00084{,}000 - 75{,}000 = 9{,}000

    The population went down by 9,000 people.

  4. Give the answer

    Watch out: 75,000 is the 2006 population, not the decrease. Choice A, 7,500, looks like it but has one zero fewer. The question asks how much the population dropped. So the answer is C, 9,000.

Check your answer

Work with people per square mile instead. In 2005, there were 84,000÷50=1,68084{,}000 \div 50 = 1{,}680 people per square mile. In 2006, there were 1,500. That is 180 fewer per square mile, and 180×50=9,000180 \times 50 = 9{,}000 fewer people.

More questions like this

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

Question 73 of 99

2023 Test B Math, question 28

Question

The sum of the numbers x, y, and z is 40. The ratio of x to y is 1:2, and the ratio of y to z is 2:5. What is the value of y?

  • A2
  • B4
  • C8
  • D10
Show the answer

Correct answer: D

The ratio of x to y is 1:2, so xy=12\frac{x}{y} = \frac{1}{2} or x=12yx = \frac{1}{2}y. Similarly, the ratio of y to z is 2:5, so yz=25\frac{y}{z} = \frac{2}{5}. Solve for z in terms of y:

5y=2z5y = 2z

52y=z\frac{5}{2}y = z

Use the given sum of x, y, and z to find the value of y:

x+y+z=40x + y + z = 40

12y+y+52y=40\frac{1}{2}y + y + \frac{5}{2}y = 40

4y=404y = 40

y=10y = 10

Question analysis

This ratio question gives two ratios that share y: x to y is 1:2 and y to z is 2:5, and x, y and z total 40. Students find the value of y. The numbers in a ratio count parts, not amounts, so the work is to combine the ratios and find what one part is worth.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Most grade 7 ratio tasks split a total using one ratio. Here two ratios linked through y must first become a single three-part ratio; since y's term already matches, that step is short and the problem stays at grade 7.
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, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards seeing that the two ratios share y’s term, which makes one three-part ratio. From there, one part is worth 5 and y is two parts. The wrong choices are numbers that live inside the ratios, and adding all four ratio numbers, which counts y twice, also leads to the part count, so that choice draws students along two paths.

  1. No obvious method moderate Working it out

    Nothing says to merge the ratios, but y is 2 in both, so x : y : z becomes 1 : 2 : 5 at once; the explanation’s route through fractions of y is slower. (Schoenfeld, 1985)

  2. Variables instead of numbers minor Understanding the question

    x, y and z appear only in two ratios and a sum of 40, so the ratio numbers 1, 2 and 5 are easy to mistake for the values themselves. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Many steps minor Working it out

    Adding the 8 parts, dividing 40 by 8 to get 5 per part, and taking y’s 2 parts are three steps, and the part count is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Write x : y : z as one ratio using y's matching term, add the parts, divide 40 by that sum to get one part, and multiply by y's parts.

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.
  • 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

Splitting a total by a ratio
one part=total÷number of parts\text{one part} = \text{total} \div \text{number of parts}
Each number in a ratio counts parts that are all the same size. Add the numbers in the ratio to find how many parts make up the total. Dividing the total by that number gives the size of one part. Each amount is its number of parts times that size.

  1. Sort out the question

    x, y and z add up to 40. A ratio compares two amounts. The ratio of x to y is 1:2, so y is twice x. The ratio of y to z is 2:5. You need the actual value of y, not its number in a ratio.

  2. Combine the two ratios

    y is 2 in both ratios. So both ratios count the same size of part, and they join into one.

    x:y:z=1:2:5x : y : z = 1 : 2 : 5

    So x is 1 part, y is 2 parts and z is 5 parts.

  3. Find the total number of parts

    1+2+5=81 + 2 + 5 = 8

    The 40 is split into 8 equal parts. Watch out: 8 counts parts. It is not y, so choice C is wrong.

  4. Find the size of one part

    one part=total÷number of parts\text{one part} = \text{total} \div \text{number of parts}

    one part=40÷8=5\text{one part} = 40 \div 8 = 5

    The 40 is shared equally among the 8 parts. So each part is worth 5.

  5. Find y

    y is 2 parts, so multiply the size of one part by 2.

    y=2×5=10y = 2 \times 5 = 10

    Watch out: the 2 in the ratio counts parts, so y is not 2, choice A. Adding the two 2s to get 4 is wrong too, choice B. So the answer is D, 10.

Check your answer

Find all three: x = 5, y = 10 and z = 25. They add to 5+10+25=405 + 10 + 25 = 40. x to y is 5 to 10, which is 1:2. y to z is 10 to 25, which is 2:5.

More questions like this

The original DOE question
Question 85 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 85. Open page 199 of the handbook (PDF)

Question 74 of 99

2023 Test B Math, question 33

Question

A line segment with point W at its left end and point Y at its right end. Point X lies on the segment between W and Y, closer to Y.

The distance from W to X is 70% of the distance from W to Y. If the distance from X to Y is 15 miles, what is the distance, in miles, from W to X?

  • A10.5
  • B35
  • C45
  • D50
Show the answer

Correct answer: B

Since the distance from W to X is 70% of the distance from W to Y, the distance from X to Y is 30% of the distance from W to Y. Let d represent the distance from W to Y. Then 15 miles is 30% of d.

0.30d=150.30d=15

d=150.30=50d=\frac{15}{0.30}=50

So, the distance from W to X, in miles, is 50−15=3550-15=35.

Question analysis

A percent question with a line segment: point X lies between W and Y, the distance from W to X is 70% of the whole, and the distance from X to Y is 15 miles. Students find the distance from W to X. Matching the 15 miles to the right percent before finding the whole is the step students most often skip.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 percent problems often give a part and its percent and ask for the whole. This one gives the distance for the part the percent does not describe and asks for the other part, so the whole has to be found on the way.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The percent and the given distance describe different pieces of the road, which is the whole difficulty. Students who apply the 70% straight to the 15 miles get a small number fast, and students who solve correctly for the whole of WY must still remember to take XY away. Checking that 35 is 70% of 50 confirms the answer and catches both slips.

  1. Turning words into math moderate Understanding the question

    The 70% describes WX, but the 15 miles given is XY, so the sentence has to be recast as “XY is 30% of WY” before the 15 miles can be used. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Finding the complement, 30%, then the whole, 15 ÷ 0.30 = 50, then the part, 50 − 15, gives three results, and the middle one is itself a choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out what percent of WY the 15-mile stretch from X to Y must be, use it to find the whole of WY, then find the part from W to X.

Traps

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 whole is the full amount. The part is a piece of it. The percent tells how many hundredths of the whole the part is. For example, 30% means 30 out of every 100, or 0.30. Use it whenever a percent describes a piece of a whole.

  1. Sort out the distances

    Point X sits between W and Y. So the trip from W to Y has two pieces: W to X, then X to Y.

    W to X is 70% of the whole distance from W to Y. X to Y is 15 miles. You need the distance from W to X, in miles.

  2. Find the percent for X to Y

    The whole distance from W to Y is 100%. W to X uses 70% of it. The rest is X to Y.

    100−70=30100 - 70 = 30

    So the 15 miles from X to Y is 30% of the whole distance.

  3. Find the whole distance

    Let d be the distance from W to Y. Use the percent formula. The part is 15, and the percent is 30.

    15=30100×d=0.30×d15 = \frac{30}{100} \times d = 0.30 \times d

    Divide both sides by 0.30 to get d by itself.

    d=15÷0.30=50d = 15 \div 0.30 = 50

    The distance from W to Y is 50 miles.

  4. Find the distance from W to X

    Take the 15 miles from X to Y away from the whole.

    50−15=3550 - 15 = 35

    W to X is 35 miles.

    Watch out: 50 miles is the whole distance from W to Y, choice D. You still have to subtract. Also, the 70% is of the whole distance, not of the 15 miles. Taking 70% of 15 gives 10.5, choice A.

    So the answer is B, 35 miles.

Check your answer

Is 35 really 70% of 50? 0.70×50=350.70 \times 50 = 35. Yes. And 35 + 15 = 50, the whole distance. The figure also shows X closer to Y. So W to X should be longer than 15 miles, and it is.

More questions like this

The original DOE question
Question 90 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 90. Open page 201 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 75 of 99

2023 Test B Math, question 36

Question

The side length of square S is shorter than the side length of square T by 1 centimeter. The area of square S is 9 square centimeters. What is the difference, in square centimeters, between the area of square T and the area of square S?

  • A3
  • B7
  • C8
  • D10
Show the answer

Correct answer: B

Let x represent the side length of square T, in centimeters. Then the side length of square S is (x−1)(x-1).

The area of square S is:

(x−1)(x−1)=9(x-1)(x-1)=9

(x−1)2=9(x-1)^{2}=9

x−1=3x-1=3

x=4x=4

So the side length of square T is 4 cm, and the area of square T is 16 sq cm.

The difference between the areas, in square centimeters, is 16−9=716-9=7.

Question analysis

Two squares differ in side length by 1 centimeter, and the smaller one has an area of 9 square centimeters. Students find how much greater the larger square's area is. It tests the link between a square's side and its area, and the extra area is easy to miscount when picturing the border that the longer sides add.

NYS standards
  • 6.G.5 perfect squares and perfect cubes
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 The area of a square and recognizing 9 as a perfect square are grade 6 ideas. This takes three short steps with small whole numbers, lighter than grade 7 work on composite figures, though the first step runs backward from area to side.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 44 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every number is small, so the work is keeping track of which quantity the question wants: a difference of areas, not a side or a single area. The added area forms an L-shaped strip around square S, and students who picture it as two strips 4 centimeters long count the corner square twice. Finding both areas, 9 and 16, and subtracting avoids reasoning about the strip at all.

  1. Many steps minor Working it out

    The work runs from area to side and back: the square root of 9 gives S’s side, adding 1 gives T’s, squaring gives 16, and subtracting gives the difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The phrase “shorter than the side length of square T by 1 centimeter” describes S from T’s point of view, so T’s side is S’s side plus 1, not minus 1. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the side of square S from its area, add 1 centimeter to get the side of square T, find that square's area, and subtract the smaller area from the larger.

Traps

  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
  • 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

Area of a square
A=s×sA = s \times s
A is the area, the space inside the square, in square units. s is the length of one side. All four sides of a square are equal, so the area is the side times itself. To go backward from an area, find the number that times itself gives the area.

  1. Sort out the two squares

    Square S has an area of 9 square centimeters. Its side is 1 centimeter shorter than the side of square T. You need the area of square T minus the area of square S, in square centimeters.

  2. Find the side of square S

    Use the area formula backward. Which number times itself makes 9?

    3×3=93 \times 3 = 9

    So each side of square S is 3 centimeters. This number is called the square root of 9.

  3. Find the side of square T

    Square T’s side is 1 centimeter longer than square S’s side.

    3+1=43 + 1 = 4

    Each side of square T is 4 centimeters.

  4. Find the area of square T

    Use the area formula with a side of 4 centimeters.

    A=4×4=16A = 4 \times 4 = 16

    The area of square T is 16 square centimeters.

  5. Subtract the areas

    Take the smaller area away from the larger one.

    16−9=716 - 9 = 7

    The difference is 7 square centimeters.

    Watch out: Picturing the new area as two strips, each 4 by 1, gives 8, choice C. But those strips share a 1-by-1 corner, so it gets counted twice. Also, 3 is a side length, not a difference of areas. That is choice A.

    So the answer is B, 7 square centimeters.

Check your answer

Draw square T and mark square S in one corner. The extra space is an L shape. It is a 3-by-1 strip, another 3-by-1 strip and a 1-by-1 corner. That is 3 + 3 + 1 = 7 square centimeters. It matches.

More questions like this

The original DOE question
Question 93 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 93. Open page 201 of the handbook (PDF)

Question 76 of 99

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 77 of 99

2023 Test B Math, question 44

Question

In a certain class, Rosa will take a total of 4 tests. She has already taken 2 of the tests and earned scores of 81 and 83. What is the least possible score Rosa can earn on the third test and still be able to finish the class with an average score of 85 on all 4 tests? (Assume that test scores can range from 0 to 100.)

  • A76
  • B82
  • C88
  • D90
Show the answer

Correct answer: A

Let x represent Rosa’s score on the third test, and let y represent her score on the fourth test.

81+83+x+y4=85\frac{81+83+x+y}{4}=85

164+x+y=340164+x+y=340

x+y=176x+y=176

If Rosa scores 100 on the fourth test, then the least score she can earn on the third test is 76.

Question analysis

Rosa has scores of 81 and 83 on the first two of four tests and wants a mean of 85 on all four. Students find the lowest score she can earn on the third test and still reach that average, with scores capped at 100. Two tests are still unknown, so the problem needs a plan, not a single equation.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.EE.8 writing simple inequalities
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 mean problems usually find one missing score. Here two tests remain, and the least possible score depends on how high the other unknown score is allowed to go, a planning step beyond typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 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

Most students can find that Rosa needs 176 more points, but the question is about the least third score, which only comes from pushing the fourth score to its limit. Splitting the 176 evenly gives a natural-looking 88 that ignores the word least. The parenthesis about the 0 to 100 range is the key information, so it pays to ask what each stated limit is for.

  1. No obvious method moderate Working it out

    The least third score depends on how high the fourth can go; only the parenthesis “test scores can range from 0 to 100” tells students to set the fourth test at 100. (Schoenfeld, 1985)

  2. Many steps minor Working it out

    The 340 points needed, the 164 already earned, the 176 still needed and the 100 allowed on the last test make four steps, each a chance to lose track. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply the target mean by 4 for the total points needed, subtract the two scores so far, then decide how the remaining points can be split so the third score is as low as possible.

Traps

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

Average
average=sum of the scoresnumber of scores\text{average} = \frac{\text{sum of the scores}}{\text{number of scores}}
The sum is all the scores added together. The number of scores is how many there are. The average, or mean, shares the total equally among them. Turned around, the sum equals the average times the number of scores.

  1. Sort out the question

    Rosa takes 4 tests. She scored 81 and 83 on the first two. She wants an average of 85 on all 4. You need the lowest score she can get on test 3 and still reach that average. Each score is from 0 to 100.

  2. Find the total points needed

    Turn the average formula around. The sum is the average times the number of tests.

    sum of the scores=85×4=340\text{sum of the scores} = 85 \times 4 = 340

    Rosa needs 340 points in all.

  3. Find the points still needed

    Subtract the points she already has.

    81+83=16481 + 83 = 164

    340−164=176340 - 164 = 176

    Tests 3 and 4 must add up to 176 points.

  4. Make test 4 as high as possible

    To make test 3 as low as it can be, test 4 must carry as many points as it can. The highest score is 100.

    176−100=76176 - 100 = 76

    So the lowest score on test 3 is 76.

    Watch out: Splitting 176 evenly gives 88 on each test, choice C. That works, but it is not the least score. And 82 is just the average of her first two tests, choice B.

    So the answer is A, 76.

Check your answer

Add all four scores: 81 + 83 + 76 + 100 = 340. Then 340 ÷ 4 = 85, the average she wants. A score of 75 on test 3 would need 101 on test 4, which is not possible.

More questions like this

The original DOE question
Question 101 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 2024 Admissions, page 203, question 101. Open page 203 of the handbook (PDF)

Question 78 of 99

2023 Test B Math, question 54

Question

The speed 10 meters per second is equivalent to how many feet per hour? (Use the approximation 1 foot = 0.3 meter.)

  • A10,800
  • B12,000
  • C36,000
  • D120,000
Show the answer

Correct answer: D

10 m1 sec×3,600 sec1 h×1 ft0.3 m=\frac{10\text{ m}}{1\text{ sec}}\times\frac{3{,}600\text{ sec}}{1\text{ h}}\times\frac{1\text{ ft}}{0.3\text{ m}}=

36,000 ft0.3 h=120,000fthr\frac{36{,}000\text{ ft}}{0.3\text{ h}}=120{,}000\,\frac{\text{ft}}{\text{hr}}

Question analysis

This unit conversion question gives a speed of 10 meters per second and asks for the same speed in feet per hour, using 1 foot = 0.3 meter. Students convert both the distance unit and the time unit. Setting up each conversion factor so that the unwanted units cancel is what makes it tricky.

NYS standards
  • 6.RP.3d converting measurement units
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students usually convert one unit with a given factor. This converts both parts of a rate, seconds to hours and meters to feet, with a decimal conversion factor and large numbers to keep track of.
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, 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

Every wrong choice is a conversion handled badly: multiplying by 0.3 when the fact 1 foot = 0.3 meter calls for dividing, dividing by 3 instead of 0.3, or stopping in meters per hour. Writing the units as fractions, so that seconds and meters cancel, shows which way each factor goes. A size check also helps: a foot is shorter than a meter, so the number of feet must be larger than the 36,000 meters.

  1. Many steps moderate Working it out

    Both parts of the rate change: seconds become hours by multiplying by 3,600, and meters become feet by dividing by 0.3, two conversions that each have a wrong direction. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic moderate Working it out

    Dividing 36,000 by 0.3 means moving the decimal point in the divisor, and a slip of one place gives 12,000 instead of 120,000. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Something you must already know minor Working it out

    The stem gives 1 foot = 0.3 meter but not the 3,600 seconds in an hour, which students must build from 60 × 60. (Anderson & Krathwohl, 2001)

Strategy: Write the speed as a fraction, multiply by 3,600 seconds per hour, then by 1 foot over 0.3 meter, checking that seconds and meters cancel.

Traps

  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Seconds in an hour
seconds in 1 hour=60×60=3,600\text{seconds in 1 hour} = 60 \times 60 = 3{,}600
An hour has 60 minutes, and each minute has 60 seconds. So 1 hour holds 60 groups of 60 seconds. To change a speed per second into a speed per hour, multiply by 3,600, because an hour is 3,600 seconds long.
Meters to feet
feet=meters÷0.3\text{feet} = \text{meters} \div 0.3
The problem says 1 foot is about 0.3 meter. So each foot uses up 0.3 meter of the distance. To count the feet, divide the meters by 0.3. A foot is shorter than a meter, so you get more feet than meters.

  1. Sort out the question

    The speed is 10 meters per second. That means it moves 10 meters every second. You need the speed in feet per hour. Two units change: seconds become hours, and meters become feet. Change one at a time. Use 1 foot = 0.3 meter.

  2. Find the seconds in an hour

    An hour is much longer than a second. So in one hour, the object moves much farther. First find how many seconds are in an hour.

    60×60=3,60060 \times 60 = 3{,}600

    One hour is 3,600 seconds.

  3. Find the meters per hour

    Each second it moves 10 meters. In 3,600 seconds it moves 3,600 times as far.

    10×3,600=36,00010 \times 3{,}600 = 36{,}000

    The speed is 36,000 meters per hour.

    Watch out: Stopping here gives 36,000, choice C. That number is still in meters, not feet.

  4. Change meters to feet

    Each foot is 0.3 meter. To find how many feet fit in 36,000 meters, divide by 0.3. Multiplying both numbers by 10 makes the division easier.

    36,000÷0.3=360,000÷3=120,00036{,}000 \div 0.3 = 360{,}000 \div 3 = 120{,}000

    So 36,000 meters is 120,000 feet. The speed is 120,000 feet per hour.

  5. Give the answer

    Watch out: Multiplying by 0.3 instead of dividing gives 10,800, choice A. Dividing by 3 instead of 0.3 gives 12,000, choice B.

    So the answer is D, 120,000 feet per hour.

Check your answer

A foot is shorter than a meter. So any distance is more feet than meters. The answer must be bigger than 36,000, and only choice D is. Also, 120,000 feet at 0.3 meter each is 120,000 × 0.3 = 36,000 meters. That matches.

More questions like this

The original DOE question
Question 111 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 2024 Admissions, page 207, question 111. Open page 207 of the handbook (PDF)

Question 79 of 99

2023 Test B Math, question 56

Question

A coordinate grid with the origin O, the x-axis marked 1, 2, and 3, and the y-axis marked 1 through 5. Line t passes through the point (0, 1) and line s passes through the point (1, 0); they intersect at P (3, 5), and dashed lines run from P across to 5 on the y-axis and down to 3 on the x-axis. The shaded region is the four-sided figure with vertices O, (0, 1), P (3, 5), and (1, 0).

Lines s and t intersect at point P (3, 5). What is the area, in square units, of the shaded region in the graph above?

  • A2
  • B3
  • C4
  • D7
Show the answer

Correct answer: C

First, find the area of the rectangle in the graph:

3×5=153\times 5=15

Then, find the areas of the two right triangles that are not shaded:

Upper triangle:

12×4×3=6\frac{1}{2}\times 4\times 3=6

Lower triangle:

12×2×5=5\frac{1}{2}\times 2\times 5=5

The area of the shaded region, in square units, is calculated by subtracting the total area of the two right triangles from the area of the rectangle:

15−(6+5)=15−11=415-(6+5)=15-11=4

Question analysis

Two lines meet at the point (3, 5) on a coordinate grid, and the shaded region between them and the axes is a four-sided shape. Students find its area in square units. The shape is not a standard figure, so no single area formula applies to it directly.

NYS standards
  • 6.G.1 area of polygons
  • 6.G.3 polygons in the coordinate plane
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Areas of triangles and rectangles are grade 6 content, and grade 7 students often find composite areas. This region is an irregular four-sided shape with two slanted sides, so students need a plan, and the lengths come from coordinates rather than labels.
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, 23 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards a plan more than computation: the region has no name, so students must see it inside the dashed rectangle from O to P, with two right triangles cut away. Reading each triangle’s legs from the axes is the second hurdle, since the 1-unit intercepts shorten one leg of each, to 4 instead of 5 and to 2 instead of 3. Only one wrong choice, from using one coordinate as both heights, has a clear working behind it, so most errors here are guesses after a failed plan.

  1. No obvious method moderate Working it out

    The shaded shape has four sides, two of them slanted, and no formula of its own; it must be boxed in the 3-by-5 rectangle with two triangles removed, or cut along OP. (Schoenfeld, 1985)

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

    No lengths are labeled; every one comes from the coordinates of P (3, 5), the y-intercept 1 of line s and the x-intercept 1 of line t. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    The rectangle, two triangle areas and a subtraction make four computations, 15, 6 and 5, then 15 − 11, each built from different coordinates. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Draw the rectangle from O to P, find the areas of the two unshaded right triangles inside it from the coordinates, and subtract both from the rectangle's area.

Traps

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, in square units. l is the length and w is the width. Multiplying them counts the unit squares that fill the rectangle.
Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, one side. h is the height: the distance from the base straight up to the far corner. A right triangle has a square corner, and the two sides that meet there are the base and height. A triangle is half of a rectangle with the same base and height.

  1. Sort out the shaded region

    The shaded region has four corners: O at (0, 0), (0, 1) on the y-axis, P at (3, 5), and (1, 0) on the x-axis. Line t meets the y-axis at (0, 1), and line s meets the x-axis at (1, 0). There is no easy formula for this shape. But the dashed lines make a rectangle around it. So find the rectangle’s area and take away the white parts.

  2. Find the rectangle’s area

    The rectangle has corners O, (3, 0), P and (0, 5). It is 3 units wide and 5 units tall.

    A=3×5=15A = 3 \times 5 = 15

    The rectangle’s area is 15 square units.

  3. Find the upper triangle’s area

    The white triangle at the top left has corners (0, 1), (0, 5) and P. Its square corner, or right angle, is at (0, 5). One side runs up the y-axis from 1 to 5, so it is 4 units. The other runs across from 0 to 3, so it is 3 units.

    A=12×4×3=6A = \frac{1}{2} \times 4 \times 3 = 6

    Its area is 6 square units.

  4. Find the lower triangle’s area

    The white triangle at the bottom right has corners (1, 0), (3, 0) and P. Its square corner is at (3, 0). One side runs along the x-axis from 1 to 3, so it is 2 units. The other goes up from 0 to 5, so it is 5 units.

    A=12×2×5=5A = \frac{1}{2} \times 2 \times 5 = 5

    Its area is 5 square units.

    Watch out: The lines cross the axes at 1, not at O. So these sides are 4 and 2, not 5 and 3.

  5. Subtract the white triangles

    Take both triangles away from the rectangle.

    15−(6+5)=15−11=415 - (6 + 5) = 15 - 11 = 4

    The shaded area is 4 square units.

    So the answer is C, 4.

Check your answer

Cut the shaded region along a line from O to P. One piece has a base of 1 on the x-axis and a height of 5, since P is 5 units up. Its area is 2.5. The other has a base of 1 on the y-axis and a height of 3, since P is 3 units across. Its area is 1.5. Then 2.5 + 1.5 = 4. Using 3 for both heights gives 3, choice B.

More questions like this

The original DOE question
Question 113 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 113. Open page 208 of the handbook (PDF)

Question 80 of 99

2023 Test B Math, question 57

Question

A box contains 14 red marbles, 10 black marbles, and 16 white marbles. How many more white marbles need to be added to the box so that the probability of randomly drawing a white marble is 35\frac{3}{5}?

  • A8
  • B10
  • C14
  • D20
Show the answer

Correct answer: D

Let w represent the number of additional white marbles.

Set the probability of drawing a white marble equal to 35\frac{3}{5} and solve the equation for w:

16+w14+10+16+w=35\frac{16+w}{14+10+16+w}=\frac{3}{5}

16+w40+w=35\frac{16+w}{40+w}=\frac{3}{5}

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

80+5w=120+3w80+5w=120+3w

2w=402w=40

w=20w=20

Question analysis

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

NYS standards
  • 6.SP.8a uniform probability models
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 probability questions usually give the counts and ask for a probability. This runs the other way: the target probability is given, and the number to add changes both the white count and the total.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 8, 34 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    “How many more white marbles need to be added” puts the unknown in two places, (16 + w)/(40 + w), because each white marble added also raises the total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    The unknown number of added marbles has to be carried as a letter through a proportion, 5(16 + w) = 3(40 + w), before it can be found. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Many steps minor Working it out

    Totaling the marbles, setting up the proportion, cross-multiplying and solving 80 + 5w = 120 + 3w make four steps, unless the choices are tested one by one. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the box

    The box has 14 red, 10 black and 16 white marbles.

    14+10+16=4014 + 10 + 16 = 40

    That is 40 marbles. You will add some white marbles. Call that number w. You need w so that the chance of drawing white is 35\frac{3}{5}.

  2. Write the new probability

    Adding w white marbles changes both counts. There will be 16 + w white marbles and 40 + w marbles in all. Use the probability formula.

    16+w40+w=35\frac{16 + w}{40 + w} = \frac{3}{5}

    Watch out: Taking 35\frac{3}{5} of 40 gives 24, and 24 − 16 = 8, choice A. That forgets that the new marbles also make the box bigger.

  3. Cross multiply

    Fractions are hard to solve with. Clear them with cross multiplication.

    5×(16+w)=3×(40+w)5 \times (16 + w) = 3 \times (40 + w)

    Then multiply each outside number by both terms inside the parentheses.

    80+5w=120+3w80 + 5w = 120 + 3w

  4. Solve for w

    Subtract 3w from both sides. Then subtract 80 from both sides.

    2w=402w = 40

    Divide both sides by 2.

    w=20w = 20

    You need to add 20 white marbles.

    So the answer is D, 20.

Check your answer

Add 20 white marbles. Then there are 16 + 20 = 36 white marbles and 40 + 20 = 60 marbles in all. Divide the top and bottom by 12: 3660=35\frac{36}{60} = \frac{3}{5}. It works.

More questions like this

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

Question 81 of 99

2018 Test A Math, question 5

Question

The sum of two consecutive integers is −15. If 1 is added to the smaller integer and 2 is subtracted from the larger integer, what is the product of the two resulting integers?

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

Show the answer

Correct answer: 63

If x is the smaller consecutive integer, then

x+1x+1

is the larger consecutive integer. Use their sum (−15) to find x:

x+(x+1)=−15x+(x+1)=-15

2x+1=−152x+1=-15

2x=−162x=-16

x=−8x=-8

The two consecutive integers are −8 and −7. One is added to the smaller integer:

−8+1=−7-8+1=-7

and 2 is subtracted from the larger integer:

−7−2=−9-7-2=-9

Find the product:

(−7)(−9)=63(-7)(-9)=63

Question analysis

A grid-in problem about two consecutive integers whose sum is −15. Students find the two integers, add 1 to the smaller and subtract 2 from the larger, then multiply the results. Deciding which negative integer is the smaller one, and keeping signs straight through every step, are what make it tricky.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 A typical grade 7 task solves one two-step equation or multiplies signed numbers. This links both: set up a consecutive-integer equation, solve it with negatives, adjust each integer, and multiply, with no choices to check against.
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 9, 30 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, yet the question collects most of the ways students lose points with negatives. The crucial judgment is which of −8 and −7 is the smaller; a student who picks −7 still produces a tidy, wrong product. Gridding the product of the original pair, or giving a negative product, are the other usual losses. Writing each intermediate pair down, rather than holding them in mind, protects against all three.

  1. Error-prone arithmetic moderate Working it out

    Every number in play is negative: −8 and −7, then −7 and −9, and a product of two negatives that must come out positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    "1 is added to the smaller integer and 2 is subtracted from the larger" depends on knowing that −8, not −7, is the smaller, since it lies farther below zero. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Many steps moderate Working it out

    Four linked steps: set up x + (x + 1) = −15, solve for x, adjust each integer, then multiply; an error early in the chain carries through to the product. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Halve the sum to see which two integers sit on either side of it, remember that the smaller negative is farther from zero, then adjust each and multiply.

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.)
  • 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

Consecutive integers
n,n+1n, \; n + 1
Integers are whole numbers and their opposites, like −2, −1, 0, 1 and 2. Consecutive integers come one right after another. If n is the smaller one, the next one is n + 1, because each step up the number line adds 1.
Multiplying two negative numbers
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers, so −a and −b are negative. When you multiply two negative numbers, the answer is positive. The two minus signs cancel out.

  1. Sort out the question

    Two consecutive integers add up to −15. Integers are whole numbers and their negatives. Consecutive means one comes right after the other, like 4 and 5. You need to do three things:

    1. Find the two integers.
    2. Add 1 to the smaller one. Subtract 2 from the larger one.
    3. Find the product of the two new numbers. The product is the answer when you multiply.

  2. Find the two integers

    Call the smaller integer n. Then the larger one is n + 1. Their sum is −15.

    n+(n+1)=−15n + (n + 1) = -15

    2n+1=−152n + 1 = -15

    Subtract 1 from both sides. Then divide both sides by 2.

    2n=−162n = -16

    n=−8n = -8

    So the integers are −8 and −7.

  3. Decide which one is smaller

    On a number line, −8 is to the left of −7. Numbers to the left are smaller. So −8 is the smaller integer, and −7 is the larger one. Watch out: 8 is bigger than 7, but −8 is smaller than −7. Mixing them up gives −6 and −10, and the wrong answer 60.

  4. Change each integer

    Add 1 to the smaller integer.

    −8+1=−7-8 + 1 = -7

    Subtract 2 from the larger integer.

    −7−2=−9-7 - 2 = -9

    The two new numbers are −7 and −9.

  5. Multiply the new numbers

    Both numbers are negative, so their product is positive.

    (−7)×(−9)=63(-7) \times (-9) = 63

    Watch out: the answer is not −63, because two negatives make a positive. Also, don’t multiply the first pair, −8 and −7. That gives 56 and skips the changes. So you would grid 63.

Check your answer

Check the pair: −8+(−7)=−15-8 + (-7) = -15, and they are one apart. Half of −15 is −7.5, and −8 and −7 sit on either side of it. Then 7×9=637 \times 9 = 63, and the sign is positive.

More questions like this

The original DOE question
Question 62 as printed on page 81 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 62. Open page 81 of the handbook (PDF)

Question 82 of 99

2018 Test A Math, question 17

Question

The sum of the numbers x, y, and z is 50. The ratio of x  to y is 1:4, and the ratio of y to z is 4:5. What is the value of y ?

  • A4
  • B8
  • C10
  • D20
Show the answer

Correct answer: D

Since both ratios have yy in common, solve for xx and zz in terms of yy in both equations. Using x:y=1:4x:y=1:4, solve for xx in terms of yy.

xy=14\frac{x}{y}=\frac{1}{4}

x=y4x=\frac{y}{4}

Using the ratio y:z=4:5y:z=4:5, solve for zz in terms of yy:

yz=45\frac{y}{z}=\frac{4}{5}

z=54yz=\frac{5}{4}y

The question states x+y+z=50x+y+z=50. Substitute from the two equations above and solve for yy.

14y+y+54y=50\frac{1}{4}y+y+\frac{5}{4}y=50

104y=50\frac{10}{4}y=50

10y=20010y=200

y=20y=20

Question analysis

A ratio problem with three unknowns: x, y and z add to 50, x to y is 1 to 4, and y to z is 4 to 5. Students merge the two ratios into one three-part ratio and use the sum to find y. The step that is easy to skip is turning ratio parts into actual values.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 ratio problems usually involve one ratio. Here two ratios share a term and must be merged before the total is used, though the shared term already matches, which keeps it near typical grade 7 work.
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, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards noticing that y has the same term in both ratios, which turns two ratios into one three-part ratio in a single move. Students who miss this often write x and z in terms of y, as the explanation does, which works but is slower. The wrong choices are numbers taken from the ratios themselves, so a student who never scales up to the total of 50 finds an answer waiting.

  1. No obvious method moderate Working it out

    The two ratios have to be merged into one, x : y : z = 1 : 4 : 5, which works only because y is 4 in both; nothing in the wording says to combine them. (Schoenfeld, 1985)

  2. Variables instead of numbers minor Understanding the question

    x, y and z are given only through two ratios and their sum of 50, so there is no actual count to anchor the work until the parts are scaled. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Many steps minor Working it out

    Adding the parts, dividing 50 by their total and scaling y's 4 parts are three steps, and the count of parts after the first step is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Since y is 4 in both ratios, write x : y : z as one combined ratio, find the value of one part from the total of 50, and scale up.

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.
  • 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

Sharing a total in a ratio
one part=total÷number of parts\text{one part} = \text{total} \div \text{number of parts}
In a ratio like 3:2, the amounts are made of equal-sized parts. Add the numbers in the ratio to count the parts. Divide the total by that count to get the size of one part. Then multiply by how many parts each amount has.

  1. Sort out the question

    The three numbers x, y and z add up to 50. The ratio of x to y is 1:4. The ratio of y to z is 4:5. A ratio compares sizes. You need the actual value of y.

  2. Join the two ratios

    y is 4 in both ratios. So the two ratios use the same size parts, and you can join them into one.

    x:y:z=1:4:5x : y : z = 1 : 4 : 5

    This means x is 1 part, y is 4 parts, and z is 5 parts.

  3. Count the parts

    1+4+5=101 + 4 + 5 = 10

    There are 10 parts in all. Watch out: 10 is the number of parts, not the value of y. Picking it is choice C.

  4. Find the size of one part

    Use the sharing formula. The 10 parts share the total of 50.

    one part=50÷10=5\text{one part} = 50 \div 10 = 5

    Each part is worth 5.

  5. Find y

    y is 4 parts.

    y=4×5=20y = 4 \times 5 = 20

    Watch out: the 4 in the ratio counts parts. It is not the value of y. Picking 4 is choice A. So the answer is D, 20.

Check your answer

Find all three: x = 5, y = 20, z = 25. They add up: 5+20+25=505 + 20 + 25 = 50. x to y is 5 to 20, which is 1:4. y to z is 20 to 25, which is 4:5. Everything fits.

More questions like this

The original DOE question
Question 74 as printed on page 84 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 74. Open page 84 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 83 of 99

2018 Test A Math, question 35

Question

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

  • A16
  • B30
  • C33
  • D55
Show the answer

Correct answer: C

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

Deion

520=x60\frac{5}{20}=\frac{x}{60}

20x=30020x=300

x=15x=15

Deion can create 15 slides in 1 hour.

Kyra

310=x60\frac{3}{10}=\frac{x}{60}

10x=18010x=180

x=18x=18

Kyra can create 18 slides in 1 hour.

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

15 + 18 = 33.

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 rate problems often ask for a single unit rate. This one needs two rates over different time periods, each scaled to an hour and then combined, which is typical grade 7 work built on grade 6 content.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 42 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is easy once each person is scaled to an hour on their own; the trouble is the urge to combine first. Adding the slides and the minutes into one rate treats two different rates like fractions added across the top and bottom, and it gives a listed answer. Keeping the two workers separate until the last step avoids every trap.

  1. Much to keep track of moderate Working it out

    Deion's rate is given per 20 minutes and Kyra's per 10, so each needs its own scale factor to an hour, 3 and 6, before the totals can be added. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Many steps minor Working it out

    Two scalings and an addition: 5 × 3 = 15 for Deion, 3 × 6 = 18 for Kyra, then the sum, each a simple step. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The choices 33 and 30 differ by only 3 slides, so a rough sense of the total cannot separate them; only scaling Kyra's own rate to 18 an hour, not 15, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out what you know

    Deion makes 5 slides every 20 minutes. Kyra makes 3 slides every 10 minutes. Each works at a constant rate, which means a steady speed that does not change.

    One hour is 60 minutes. You need how many slides they make together in 60 minutes.

  2. Find Deion’s slides in an hour

    Deion’s rate is 5 slides per 20-minute block. An hour holds 3 of those blocks, since 60 ÷ 20 = 3:

    a=5×3=15a = 5 \times 3 = 15

    Deion makes 15 slides in an hour.

  3. Find Kyra’s slides in an hour

    Kyra’s rate is 3 slides per 10-minute block. An hour holds 6 of those blocks, since 60 ÷ 10 = 6:

    a=3×6=18a = 3 \times 6 = 18

    Kyra makes 18 slides in an hour. Watch out: Kyra has her own rate. Doubling Deion’s 15 to get 30, choice B, treats her as if she worked at his speed.

  4. Add the two amounts

    15+18=3315 + 18 = 33

    Together they make 33 slides. Watch out: do not add the slides and the minutes first. Mixing 5 + 3 slides with 20 + 10 minutes blends two different rates and leads to choice A, 16.

    So the answer is C, 33.

Check your answer

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

More questions like this

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

Question 84 of 99

2018 Test A Math, question 38

Question

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

  • A216 hr2\frac{1}{6}\text{ hr}
  • B314 hr3\frac{1}{4}\text{ hr}
  • C334 hr3\frac{3}{4}\text{ hr}
  • D712 hr7\frac{1}{2}\text{ hr}
Show the answer

Correct answer: B

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

13040=314\frac{130}{40}=3\frac{1}{4}

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Unit rate problems like this are grade 6 content and common in grade 7. It adds a subtraction step and a conversion to hours, and the word 'additional' must be read closely, which keeps it at grade 7 level.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Many steps moderate Working it out

    The solution chains four moves: subtract the 20 pages read, double 20 pages per half hour to 40 per hour, divide 130 by 40, and write the result in hours. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    The choices 3 1/4 and 3 3/4 hours differ by exactly the half hour already spent on the first 20 pages, so only subtracting those pages first separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    A student who works in minutes gets 195 minutes and must convert it to the mixed number 3 1/4 hours to find it among the choices. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out what you know

    Ryan must read 150 pages in all. He read the first 20 pages in 30 minutes. The word additional is in bold. It means extra time, after the 30 minutes he already spent.

    So you need the time for the pages he has not read yet.

  2. Find the pages left

    150−20=130150 - 20 = 130

    He still has 130 pages to read. Watch out: timing all 150 pages gives 3343\frac{3}{4} hours, which is choice C. That counts pages he already read.

  3. Find his rate per hour

    The choices are in hours, so find how many pages he reads in an hour. 30 minutes is half an hour. In a full hour he reads twice as much:

    20×2=4020 \times 2 = 40

    His rate is 40 pages per hour.

  4. Divide the pages by the rate

    Dividing tells you how many hours the 130 pages take at 40 pages an hour:

    t=13040=31040=314t = \frac{130}{40} = 3\frac{10}{40} = 3\frac{1}{4}

    It will take 3143\frac{1}{4} more hours. So the answer is B, 3143\frac{1}{4} hr.

Check your answer

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

More questions like this

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

Question 85 of 99

2018 Test A Math, question 44

Question

Number line with points W, X, Y and Z marked from left to right. W is at −10 and Z is at 8; a tick mark at 0 lies between X and Y. X and Y have no values given.

On the number line above, points WW, XX, YY, and ZZ are integers, and WX:XY:YZ=4:2:3WX:XY:YZ = 4:2:3. What is the value of WY\overline{WY}?

  • A8
  • B11
  • C12
  • D18
Show the answer

Correct answer: C

Convert the ratios into fractions of WZWZ. Use the sum of the ratios for the denominator.

WX:XY:YZ=4:2:3WX:XY:YZ=4:2:3

WX=44+2+3WZ=49WZWX=\frac{4}{4+2+3}WZ=\frac{4}{9}WZ

XY=24+2+3WZ=29WZXY=\frac{2}{4+2+3}WZ=\frac{2}{9}WZ

The part of WZWZ that is WYWY is the sum of those fractions:

WY=49WZ+29WZ=69WZ=23WZWY=\frac{4}{9}WZ+\frac{2}{9}WZ=\frac{6}{9}WZ=\frac{2}{3}WZ

Find the length of WZWZ:

WZ=8−(−10)=18WZ=8-(-10)=18

The value of WYWY is 23(18)=12\frac{2}{3}(18)=12.

Question analysis

This number line question places integer points W, X, Y and Z on a line and gives the ratio of the segments WX, XY and YZ as 4 to 2 to 3. Students find the whole length from W to Z, split it by the ratio, and add the parts that make up WY. Mixing up a part with the whole is the main risk.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students find distances on a number line and use part-to-whole ratios, usually separately. This item joins them: a negative endpoint, a three-term ratio and a choice of which segments to combine.
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 10, 21 words.
Corrected from the DOE handbook
  • Equations in the explanation that mixed up a fraction of a length with the length itself were corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is routine, but the question keeps asking for a different piece of the line: the whole WZ, the part WX, and finally WY, which is two parts added together. Two of the wrong choices are exactly those other pieces. Students who mark the actual positions, X at −2 and Y at 2, can check the ratio before answering.

  1. Combines ideas moderate Working it out

    A three-part ratio, 4:2:3, must be applied to a length found from a negative endpoint, 8 − (−10) = 18, joining ratio and number-line distance. (Embretson, 1998; Daniel & Embretson, 2010)

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

    Only W at −10 and Z at 8 are labeled, and the drawing is not to scale: Y is drawn nearer 5 than 2, so reading WY from the picture fails. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    Find WZ, split it into 9 equal parts of 2, then add the 4 parts of WX and the 2 parts of XY. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length WZ from the endpoints on the number line, divide it by the total number of ratio parts, and multiply by the parts from W to Y.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes WX = 4/9, XY = 2/9 and WY = 2/3, and then says “The value of WY is 2/3(18) = 12”. Those fractions are parts of WZ, not the lengths themselves (WX is 8 and WY is 12), so WY cannot be both 2/3 and 12. The lines now read WX = 4/9 WZ, XY = 2/9 WZ and WY = 2/3 WZ. The method and the answer are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
d=b−ad = b - a
d is the distance between two points. b is the point on the right and a is the point on the left. Subtracting left from right always gives a positive distance. It works even when a is negative.
Size of one part of a ratio
u=Lpu = \frac{L}{p}
u is the size of one part. L is the whole length being shared. p is the total number of parts, found by adding the numbers in the ratio. Every part is the same size, so you share the length equally.

  1. Sort out what you know

    W is at −10 and Z is at 8. X and Y are between them. The ratio 4:2:3 means WX is 4 parts, XY is 2 parts and YZ is 3 parts. All the parts are the same size.

    You need the length WY, from W to Y. The drawing is not exact, so use the numbers, not the picture.

  2. Find the whole length WZ

    d=8−(−10)=8+10=18d = 8 - (-10) = 8 + 10 = 18

    WZ is 18 units long. Subtracting a negative number is the same as adding.

  3. Find the size of one part

    Add the ratio numbers to get the number of parts:

    4+2+3=94 + 2 + 3 = 9

    Share the 18 units equally among 9 parts:

    u=189=2u = \frac{18}{9} = 2

    Each part is 2 units long.

  4. Find WY

    WY is WX and XY together, so it has 4 + 2 = 6 parts:

    6×2=126 \times 2 = 12

    Watch out: WX alone is 4 × 2 = 8, which is choice A. And 18 is the whole line, choice D. So the answer is C, 12.

Check your answer

Place the points. W is at −10. X is 8 units to the right, at −2. Y is 4 more, at 2. Z is 6 more, at 8, which matches its label. And the distance from −10 to 2 is 2 − (−10) = 12.

More questions like this

The original DOE question
Question 101 as printed on page 89 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 89, question 101. Open page 89 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 86 of 99

2018 Test A Math, question 54

Question

Number line with points J, K, L and M marked from left to right. Only K has a value: 3/8.

On the number line above, JK=312JK=3\frac{1}{2} , JM=934JM=9\frac{3}{4} , and LM=118LM=1\frac{1}{8} . What is the position of point LL?

  • A5185\frac{1}{8}
  • B5145\frac{1}{4}
  • C5125\frac{1}{2}
  • D6146\frac{1}{4}
Show the answer

Correct answer: C

First, use JK=312JK=3\frac{1}{2} to find the location of JJ:

38−J=312\frac{3}{8}-J=3\frac{1}{2}

J=38−312=−318J=\frac{3}{8}-3\frac{1}{2}=-3\frac{1}{8}

Now, use JM=934JM=9\frac{3}{4} to find the location of MM:

M−(−318)=934M-\left(-3\frac{1}{8}\right)=9\frac{3}{4}

M+318=934M+3\frac{1}{8}=9\frac{3}{4}

M=934−318=658M=9\frac{3}{4}-3\frac{1}{8}=6\frac{5}{8}

Finally, use LM=118LM=1\frac{1}{8} to find the location of LL:

658−L=1186\frac{5}{8}-L=1\frac{1}{8}

L=658−118=548=512L=6\frac{5}{8}-1\frac{1}{8}=5\frac{4}{8}=5\frac{1}{2}

Question analysis

This number line question gives the lengths JK, JM and LM as mixed numbers and asks for the position of point L. Students must chain three distance steps, moving left from K to J, right to M and back to L, with fractions in eighths and a negative position. Losing track of where K sits is the main danger.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students add and subtract rational numbers and find distances on a number line one at a time. This item chains three such steps with mixed numbers and a negative position, more than typical grade 7 work demands.
Estimated difficulty
Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Solution
4 steps
Reading level
Question and choices: about grade 9, 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

Every number is a mixed fraction, one position is negative, and the figure's even spacing hides how long KL really is. The distance route, KL = 9 3/4 − 3 1/2 − 1 1/8, avoids the negative entirely but ends one step short of a position unless K's 3/8 is added back. That last step is the one most students skip, which is why the question sits at the top of the scale.

  1. A very close wrong answer major Choosing the answer

    The choices 5 1/8, 5 1/4 and 5 1/2 lie within 3/8 of one another, so no estimate separates them; only exact eighths and adding K's 3/8 pick out the right one. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic moderate Working it out

    Every step mixes halves, quarters and eighths, and placing J means 3/8 − 3 1/2 = −3 1/8, a subtraction that crosses zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    Only K is labeled, and the points are drawn evenly spaced although KL is over four times LM, so the picture gives no help with sizes. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  4. Many steps minor Working it out

    Three linked moves, K to J, J to M and M back to L, each depend on the last, so an early slip carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Start from K's labeled position, subtract JK to place J, add JM to place M, then subtract LM, keeping every fraction in eighths.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
d=b−ad = b - a
d is the distance between two points. b is the point on the right and a is the point on the left. Subtracting left from right always gives a positive distance. It works even when a is negative.
Equal fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top number (numerator) and b is the bottom number (denominator). Multiplying both by the same number k gives an equal fraction. It is like cutting each piece into k smaller pieces. Use it to give fractions the same bottom before you add or subtract.

  1. Sort out what you know

    Only K has a value: 38\frac{3}{8}. JK means the distance from J to K. You know JK = 3123\frac{1}{2}, JM = 9349\frac{3}{4} and LM = 1181\frac{1}{8}.

    You need the position of L, the number where L sits. The points are drawn evenly spaced, but the distances are not equal. So use the numbers, not the picture.

  2. Write the fractions in eighths

    The fractions use halves, fourths and eighths. Give them all a bottom of 8:

    312=348934=9683\frac{1}{2} = 3\frac{4}{8} \qquad 9\frac{3}{4} = 9\frac{6}{8}

    Watch out: 34\frac{3}{4} is 68\frac{6}{8}, not 48\frac{4}{8}. That slip leads to choice B.

  3. Find the distance from K to M

    JM is JK and KM put together. So take JK away from JM:

    968−348=6289\frac{6}{8} - 3\frac{4}{8} = 6\frac{2}{8}

    KM is 6286\frac{2}{8}, or 6146\frac{1}{4}. Watch out: this is a distance, not where L sits. Choice D stops here.

  4. Find the distance from K to L

    KM is KL and LM put together. So take LM away:

    628−118=5186\frac{2}{8} - 1\frac{1}{8} = 5\frac{1}{8}

    KL is 5185\frac{1}{8}. Watch out: this is still a distance. Choice A stops here.

  5. Find the position of L

    L is 5185\frac{1}{8} to the right of K, and K is at 38\frac{3}{8}. Add:

    38+518=548=512\frac{3}{8} + 5\frac{1}{8} = 5\frac{4}{8} = 5\frac{1}{2}

    So the answer is C, 5125\frac{1}{2}.

Check your answer

Place every point. J is 3483\frac{4}{8} left of K, at −318-3\frac{1}{8}. M is 9689\frac{6}{8} right of J, at 6586\frac{5}{8}. L is 1181\frac{1}{8} left of M, at 5485\frac{4}{8}, which is 5125\frac{1}{2}. Same answer.

More questions like this

The original DOE question
Question 111 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 111. Open page 91 of the handbook (PDF)

Question 87 of 99

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 88 of 99

2018 Test A Math, question 57

Question

A paste used to cover a billboard is made by mixing the following ingredients by weight: 4 parts powder, 3 parts water, 2 parts resin, and 1 part hardener. To cover one billboard requires 30 pounds of this paste. How many total pounds of resin are required to cover 4 billboards?

  • A6
  • B8
  • C24
  • D48
Show the answer

Correct answer: C

The ratio is 4:3:2:14:3:2:1, so the total parts is 10.

Since there are two parts resin, the fraction of resin is

210=15\frac{2}{10}=\frac{1}{5}

So the amount of resin in 30 lb of paste (for 1 billboard) is

15×30=6\frac{1}{5}\times 30=6 lb.

For 4 billboards, that would be 6 × 4 = 24 lb.

Question analysis

This ratio word problem gives a billboard paste recipe of 4 parts powder, 3 parts water, 2 parts resin and 1 part hardener by weight, with 30 pounds of paste per billboard. It asks how many pounds of resin four billboards need. Students must find resin's share of the total parts and scale up to four billboards.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.1 ratio concepts and language
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Part-to-whole ratio problems with several ingredients are typical grade 7 work. This one adds a final scaling step for four billboards, which keeps it a routine multistep ratio problem at grade 7 level.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 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 recipe lists four ingredients but only resin's share matters, so the work is part over whole and then scaling. The question asks for four billboards only in the last sentence, after the per-billboard amount has been set up, which is why stopping early is the main risk. Finding the total paste first, 120 pounds, and taking a fifth of it reaches the answer in one step.

  1. Many steps minor Working it out

    Add the parts to 10, take resin's 2/10 of 30 pounds, then multiply by 4 billboards: three short steps. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the ratio parts to get the total, find resin's fraction of one batch of paste, take that fraction of 30 pounds, and multiply by the number of billboards.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction from a ratio
f=pTf = \frac{p}{T}
f is the fraction of the mix that is one ingredient. p is that ingredient’s number of parts. T is the total number of parts, found by adding all the ratio numbers. Use it to turn a ratio into a fraction of the whole.

  1. Sort out what you know

    The paste is 4 parts powder, 3 parts water, 2 parts resin and 1 part hardener, by weight. A part is just an equal share of the mix.

    One billboard needs 30 pounds of paste. You need the pounds of resin for 4 billboards.

  2. Find resin’s fraction of the paste

    4+3+2+1=104 + 3 + 2 + 1 = 10

    The mix has 10 parts in all, and resin is 2 of them:

    f=210=15f = \frac{2}{10} = \frac{1}{5}

    So resin is one fifth of the paste. Watch out: 2 parts is not 2 pounds. Treating it that way gives 2 × 4 = 8, choice B.

  3. Find the resin for one billboard

    15×30=6\frac{1}{5} \times 30 = 6

    One billboard needs 6 pounds of resin. Watch out: 6 is choice A, but that covers only one billboard.

  4. Multiply by four billboards

    Each billboard needs the same 6 pounds of resin, so multiply by 4:

    6×4=246 \times 4 = 24

    Four billboards need 24 pounds of resin. So the answer is C, 24.

Check your answer

Work it another way. Four billboards need 4 × 30 = 120 pounds of paste. One fifth of 120 is 24. Choice D, 48, is the powder, which has 4 parts out of 10.

More questions like this

The original DOE question
Question 114 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 114. Open page 92 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 99

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 90 of 99

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 91 of 99

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 92 of 99

2018 Test B Math, question 18

Question

PRICES FOR NEWSPAPER ADVERTISING

Page Space Price
14\frac{1}{4} page $200
12\frac{1}{2} page $350
full page $600

The table above shows prices for newspaper advertising. A store purchased 14\frac{1}{4} pages, 12\frac{1}{2} pages, and full pages of page space in equal numbers for a total of $11,500. What is the total amount of page space the store purchased?

  • A1341\frac{3}{4} pages
  • B10 pages
  • C161216\frac{1}{2} pages
  • D171217\frac{1}{2} pages
Show the answer

Correct answer: D

The question says that an equal number (xx) of each type of page space was purchased. To find the number of each type of page space that was purchased, multiply the price per type by xx and set it equal to the total amount spent, then solve for xx:

200x+350x+600x=11,500200x+350x+600x=11,500

1,150x=11,5001,150x=11,500

x=10x=10

Thus, the store purchased 10 units of each type of page space. To find the total amount of page space purchased, multiply each type of page space by 10, and add:

(10×14 page)+(10×12 page)+(10×1 page)\left(10\times\frac{1}{4}\text{ page}\right)+\left(10\times\frac{1}{2}\text{ page}\right)+(10\times 1\text{ page})

=1712 pages=17\frac{1}{2}\text{ pages}

Question analysis

This word problem gives a table of newspaper advertising prices for quarter-page, half-page and full-page ads, and says a store bought equal numbers of each for a set total. Students write and solve an equation for how many ads of each size were bought. The question wants total page space, which takes a second calculation with fractions.

NYS standards
  • 7.EE.4a two-step equations
  • 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 write and solve equations like px = r from word problems. This adds a table, combines three prices into one coefficient and needs a second step with fractions after solving, a longer chain than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 5, 59 words.
Corrected from the DOE handbook
  • An arithmetic slip in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Solving for the number of ads is the obvious goal, and finding it feels like the finish, but the question asks for page space, which needs one more pass through the table. The awkward wording about buying pages in equal numbers slows students before any arithmetic starts. The wrong choices mark each place to stop: one set of ads, the number of each ad, or a slip in adding the fractional pages.

  1. Many steps major Working it out

    The work has two phases: turn the money into a count, 1,150x = 11,500 so x = 10, then turn the count into space by multiplying 10 by 1/4, 1/2 and 1 and adding. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math moderate Understanding the question

    The phrase “purchased 1/4 pages, 1/2 pages, and full pages of page space in equal numbers” has to be read as the same number of ads of each size, so the three prices add to one coefficient. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    Both columns of the table are needed: the price column builds the equation, and the page-space column turns the count of ads into the final total. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Let x be the number of each ad size, add the three prices to write one equation for the total, solve it, then multiply x by the page space in one set of ads.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook adds 200x + 350x + 600x and writes 1,500x = 11,500. The sum is 1,150x, and the handbook’s next line, x = 10, only follows from 1,150x = 11,500. The site shows 1,150x = 11,500.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of equal sets
number of sets=total spentcost of one set\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}}
A set here means one ad of each size. The cost of one set is what one of each costs together. Dividing the total money by it tells how many sets were bought, because every set costs the same.

  1. Sort out the question

    The store bought the same number of each ad size: quarter page, half page and full page. It spent $11,500 in all.

    You need the total page space. That is not the number of ads, and not the money.

  2. Find the cost of one set

    Think of one quarter page, one half page and one full page as a set. The store bought equal numbers of each, so it bought whole sets.

    200+350+600=1,150200 + 350 + 600 = 1{,}150

    One set costs $1,150.

  3. Find how many sets

    number of sets=total spentcost of one set=11,5001,150=10\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}} = \frac{11{,}500}{1{,}150} = 10

    The store bought 10 sets. That means 10 ads of each size.

    Watch out: 10 is the number of each ad, not the page space. Stopping here gives choice B.

  4. Find the page space in one set

    Add the sizes in one set. Write them in quarters so they have the same bottom number.

    14+12+1=14+24+44=74\frac{1}{4} + \frac{1}{2} + 1 = \frac{1}{4} + \frac{2}{4} + \frac{4}{4} = \frac{7}{4}

    One set is 74\frac{7}{4}, or 1341\frac{3}{4}, pages.

    Watch out: 1341\frac{3}{4} pages is only one set. That is choice A.

  5. Multiply by the number of sets

    There are 10 sets, so multiply the space in one set by 10.

    10×74=704=171210 \times \frac{7}{4} = \frac{70}{4} = 17\frac{1}{2}

    So the answer is D, 171217\frac{1}{2} pages.

Check your answer

Add the sizes another way. 10 quarter pages make 2.5 pages, not 1.5. 10 half pages make 5 pages. 10 full pages make 10 pages. Then 2.5+5+10=17.52.5 + 5 + 10 = 17.5, the same answer.

More questions like this

The original DOE question
Question 75 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 75. Open page 156 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 93 of 99

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 94 of 99

2018 Test B Math, question 31

Question

Simplify this expression:

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

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

Correct answer: B

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

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. Many steps minor Working it out

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

  3. A very close wrong answer minor Choosing the answer

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Multiply the 4 into the first group

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

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

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

  3. Take away the second group

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

    −(5−x)=−5+x-(5-x) = -5 + x

    Taking away a minus xx gives plus xx.

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

  4. Combine the like terms

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

    28−5=2328 - 5 = 23

    −12x+x=−11x-12x + x = -11x

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

  5. Give the answer

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

Check your answer

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

More questions like this

The original DOE question
Question 88 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 88. Open page 158 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 95 of 99

2018 Test B Math, question 33

Question

A large container is partially filled with nn liters of water. Ito adds 10 liters of water to the container, making it 60% full. If Ignacio adds 6 more liters of water, the container will be 75% full. What is the value of nn ?

  • A14
  • B15
  • C26
  • D30
Show the answer

Correct answer: A

Let xx be the total number of liters the container can hold.

n+10x=60% and n+16x=75%\frac{n+10}{x}=60\%\text{ and }\frac{n+16}{x}=75\%

First, solve each equation for xx:

Equation 1:

n+10x=60100\frac{n+10}{x}=\frac{60}{100}

n+10x=35\frac{n+10}{x}=\frac{3}{5}

3x=5(n+10)3x=5(n+10)

x=5n+503x=\frac{5n+50}{3}

Equation 2:

n+16x=75100\frac{n+16}{x}=\frac{75}{100}

n+16x=34\frac{n+16}{x}=\frac{3}{4}

3x=4(n+16)3x=4(n+16)

x=4n+643x=\frac{4n+64}{3}

Now, set the two equations equal to each other and solve for nn.

5n+503=4n+643\frac{5n+50}{3}=\frac{4n+64}{3}

5n+50=4n+645n+50=4n+64

n+50=64n+50=64

n=14 litersn=14\text{ liters}

Question analysis

Water is poured into a large container in two stages: 10 liters make it 60% full, and 6 more liters bring it to 75% full. Students must work out how much water was in it to begin with. This percent word problem hides two unknowns, the starting amount and the capacity, and several in-between numbers look like answers.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here the whole is unknown and must be found from the change between two percents before a second step reaches the answer.
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, 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 container's size is the hidden key, and nothing in the stem names it. Students who write an equation for each fill level, or who notice that 6 liters equal 15 percent, get there; others grab a number from the story or stop at the 75% level. Because every wrong choice is a number that appears in the problem or arises while solving it, a careless finish is rewarded with a match.

  1. Turning words into math major Understanding the question

    Two percent-full statements about a container of unknown size have to become equations, n + 10 = 60% of the capacity and n + 16 = 75% of it, and the capacity is never given. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method moderate Working it out

    The quick route is to see that Ignacio's 6 liters raise the level by 15 percentage points, which fixes the capacity at 40 liters; the wording does not point to it. (Schoenfeld, 1985)

  3. Many steps moderate Working it out

    After the capacity, the solution still needs the 60% amount, 24 liters, and a subtraction of Ito's 10 liters, so two unknowns are chased through four steps. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  4. A very close wrong answer minor Choosing the answer

    15 sits one unit from the right 14, so a rough estimate of the starting amount cannot choose between them; only solving exactly, 24 − 10, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Compare the two fill levels: the extra 6 liters account for the change in percent, which gives the capacity. Then use the 60% level to work back to the starting amount.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the full amount, here how much the container holds. The part is p percent of it. Percent means out of 100, so you divide p by 100 and then multiply.

  1. Sort out what you know

    The container starts with nn liters of water. Ito adds 10 liters, and it is 60% full. Then 6 more liters make it 75% full. You do not know how much the container holds when full. Call that its capacity. You need nn, the starting amount.

  2. See what the 6 liters did

    The extra 6 liters raised the level from 60% to 75%.

    75−60=1575 - 60 = 15

    So 6 liters is 15% of the capacity.

    Watch out: 15 is a percent, not a number of liters. Picking 15 gives choice B.

  3. Find the full capacity

    If 15% is 6 liters, then 1% is 6÷15=0.46 \div 15 = 0.4 liters. A full container is 100%.

    100×0.4=40100 \times 0.4 = 40

    The container holds 40 liters when full.

  4. Find the water at 60 percent

    Use the percent formula with 60 for p and 40 liters for the whole.

    part=60100×40=24\text{part} = \frac{60}{100} \times 40 = 24

    After Ito adds his water, there are 24 liters.

  5. Undo Ito’s 10 liters

    Ito added 10 liters to the starting amount. To go back, subtract them.

    n=24−10=14n = 24 - 10 = 14

    The container started with 14 liters.

    Watch out: 75% of 40 is 30 liters. That is the amount after both people add water, not at the start. It is choice D.

  6. Give the answer

    The question asks for nn, the amount before anyone added water. So the answer is A, 14.

Check your answer

Run the story forward. Start with 14 liters. Add 10 to get 24 liters, and 24 is 60% of 40. Add 6 more to get 30 liters, and 30 is 75% of 40. Both facts match.

More questions like this

The original DOE question
Question 90 as printed on page 159 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 90. Open page 159 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 96 of 99

2018 Test B Math, question 40

Question

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

  • A1
  • B5
  • C7
  • D9
Show the answer

Correct answer: B

Calculate each mean speed:

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

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

Then calculate the difference of both mean speeds:

S−R=18−13=5 kphS-R=18-13=5\text{ kph}

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Unit rates and speed are grade 6 topics, and this needs only two divisions and a subtraction with whole-number results. It is easier than typical grade 7 rate problems, which often involve fractions.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 29 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Many steps minor Working it out

    Two divisions, 65 ÷ 5 and 72 ÷ 4, have to be done before the subtraction, so no single operation on the given numbers answers the question. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    "How much less was Rajiv's mean speed" asks for a difference of two rates, while the listed distances and times invite subtracting them directly. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out what you know

    Rajiv went 65 kilometers in 5 hours. Shaina went 72 kilometers in 4 hours. Mean speed means average speed. It is measured in kilometers per hour (kph), the distance in one hour. You need how much slower Rajiv was than Shaina.

  2. Find Rajiv’s speed

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

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

    Rajiv’s mean speed is 13 kph.

  3. Find Shaina’s speed

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

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

    Shaina’s mean speed is 18 kph.

  4. Subtract the two speeds

    Find how much less 13 is than 18.

    18−13=518 - 13 = 5

    Rajiv was 5 kph slower.

    Watch out: do not subtract the distances or the times. 72−65=772 - 65 = 7 gives choice C, and 5−4=15 - 4 = 1 gives choice A. Neither one is a speed.

    So the answer is B, 5.

Check your answer

Work backward. At 13 kph for 5 hours, Rajiv goes 13×5=6513 \times 5 = 65 kilometers. At 18 kph for 4 hours, Shaina goes 18×4=7218 \times 4 = 72 kilometers. Both match the question, so the speeds are right.

More questions like this

The original DOE question
Question 97 as printed on page 160 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 160, question 97. Open page 160 of the handbook (PDF)

Question 97 of 99

2018 Test B Math, question 42

Question

There are 1,000 cubic centimeters in 1 liter, and 1,000 cubic millimeters in 1 milliliter. How many cubic millimeters are there in 1,000 cubic centimeters?

  • A1,000
  • B10,000
  • C100,000
  • D1,000,000
Show the answer

Correct answer: D

If 1 L = 1,000 cu cm, then 1 L = 1,000 mL. Set up a proportion, letting xx = the amount of cubic millimeters in 1,000 cubic centimeters.

1,000 cu mm1 mL=x cu mm1,000 mL\frac{1{,}000\text{ cu mm}}{1\text{ mL}}=\frac{x\text{ cu mm}}{1{,}000\text{ mL}}

Solve for xx: 1,000,000 cubic millimeters are in 1,000 cubic centimeters.

Question analysis

The question gives two metric facts, cubic centimeters per liter and cubic millimeters per milliliter, and asks how many cubic millimeters are in 1,000 cubic centimeters. It tests chaining unit conversions with volume units. The hard part is that the two facts do not connect without a link the question leaves out.

NYS standards
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 6 and 7 students convert units with one given factor. This needs three linked conversions, one using a fact the question does not state, and involves cubic units, which most grade 7 students find unfamiliar.
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 13, 20 words.
Corrected from the DOE handbook
  • The units in the explanation’s proportion were corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question gives two conversion facts that do not connect until the student supplies the link between liters and milliliters. Students who instead reach for 1 centimeter = 10 millimeters often scale a volume like a length or an area and land short by a factor of 10 or 100. Writing the unit at every step, so the units cancel, keeps both factors of 1,000 in view.

  1. Something you must already know moderate Working it out

    The chain needs 1 liter = 1,000 milliliters, which the stem never states, yet it is the only link between its two conversion facts. (Anderson & Krathwohl, 2001)

  2. Many steps moderate Working it out

    Three conversions in a row, cubic centimeters to liters, liters to milliliters and milliliters to cubic millimeters, multiply out to 1,000 × 1,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Build a chain from cubic centimeters to liters, liters to milliliters, and milliliters to cubic millimeters, writing the units at every step so they cancel.

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook’s proportion is 1,000 cu mm / 1 L = x cu mm / 1,000 cu mm. But 1,000 cubic millimeters make 1 milliliter, not 1 liter, and the second denominator should be the 1,000 mL that equal 1,000 cubic centimeters. The site uses 1,000 cu mm / 1 mL = x cu mm / 1,000 mL, which gives x = 1,000,000 cubic millimeters.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a cube
V=s×s×sV = s \times s \times s
V is the volume, the space inside the cube. s is the length of one side. A cube has the same length, width and height, so you multiply the side by itself three times.

  1. Sort out what you know

    You are told two facts. 1 liter (L) is 1,000 cubic centimeters (cu cm). 1 milliliter (mL) is 1,000 cubic millimeters (cu mm). You need the number of cu mm in 1,000 cu cm.

    One fact is missing. 1 liter is 1,000 milliliters. The word milli means one thousandth.

  2. Change cubic centimeters to liters

    The first fact says 1,000 cu cm is exactly 1 liter. So the amount in the question is 1 liter of space.

  3. Change liters to milliliters

    One liter is 1,000 milliliters. So 1,000 cu cm is the same as 1,000 mL.

    Watch out: you are not done yet. The question asks for cubic millimeters. Stopping at 1,000 gives choice A.

  4. Change milliliters to cubic millimeters

    Each milliliter is 1,000 cu mm. There are 1,000 milliliters, so multiply.

    1,000×1,000=1,000,0001{,}000 \times 1{,}000 = 1{,}000{,}000

    So 1,000 cu cm is 1,000,000 cu mm. The answer is D, 1,000,000.

Check your answer

Use the cube formula. A cube 1 cm on each side is 10 mm on each side. Its volume is V=10×10×10=1,000V = 10 \times 10 \times 10 = 1{,}000 cu mm. So 1,000 cu cm is 1,000×1,000=1,000,0001{,}000 \times 1{,}000 = 1{,}000{,}000 cu mm. Using only one 10 gives choice B. Using two 10s gives choice C.

More questions like this

The original DOE question
Question 99 as printed on page 160 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 160, question 99. Open page 160 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 98 of 99

2018 Test B Math, question 46

Question

In a sample of 50 cars at a local dealership, there are 12 red cars and 10 cars with backup cameras. Of the 12 red cars, 4 have backup cameras. If a car is selected at random from the given sample, what is the probability that both of the following are true: the car is not red and does not have a backup camera?

  • A35\frac{3}{5}
  • B1625\frac{16}{25}
  • C1925\frac{19}{25}
  • D45\frac{4}{5}
Show the answer

Correct answer: B

Create a chart using the given information and use subtraction to figure out how many cars are not red and do not have a back-up camera:

Red Not Red TOTAL
Back-up Camera 4 6
(10-4)		 10
No back-up Camera 32
(38-6)
TOTAL 12 38
(50-12)		 50

The probability of selecting a car that meets both conditions from the total of 50 cars at the dealership is 3250=1625\frac{32}{50}=\frac{16}{25} .

Question analysis

In a sample of 50 cars, 12 are red, 10 have backup cameras, and 4 of the red cars have cameras. Students find the probability that a randomly chosen car is neither red nor equipped with a camera. It tests compound probability with overlapping groups, and the overlap between red cars and camera cars is what makes it tricky.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 compound probability usually uses a ready-made table or tree. Here students must build the two-way table from words, handle an overlap, and use a "not this and not that" condition, several steps beyond typical work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 6, 63 words.
Corrected from the DOE handbook
  • A grammar slip in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is an overlap problem dressed as probability: the hard part is counting the cars that avoid both features, not the division by 50. A two-way table makes it routine, but students working in their heads often keep only one of the two conditions. The question rewards placing the 4 red cars with cameras in the table before doing any subtraction.

  1. Much to keep track of major Working it out

    Three overlapping counts, 12 red, 10 with cameras and 4 that are both, must be sorted into four groups before the wanted one, neither red nor with a camera, can be found. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Many steps moderate Working it out

    Find the non-red cars with cameras, 10 − 4 = 6, and the non-red cars, 50 − 12 = 38, subtract to get 32, then simplify 32/50. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer moderate Choosing the answer

    19/25, or 38/50, has the same denominator as the right 16/25 and sits just above it; only removing the 6 non-red cars that have cameras from the 38 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  4. Complex sentences minor Understanding the question

    The last sentence stacks "both of the following are true", a colon and two negatives, "not red" and "does not have"; untangling it is reading work, not extra math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Draw a two-way table with red and not red across, camera and no camera down, fill in the given counts, and complete the missing cells by subtraction.

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says “a car that meet both conditions”; the site says “a car that meets both conditions.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out what you know

    There are 50 cars. Of these, 12 are red and 10 have backup cameras. Of the red cars, 4 have cameras. You pick one car at random. You need the chance that “both of the following are true”: the car is not red, and it has no camera.

  2. Set up a two-way table

    A two-way table sorts each car by two things at once: its color and its camera. Put in the numbers you know.

    RedNot redTotal
    Camera4?10
    No camera?
    Total12?50

  3. Fill in the missing numbers

    Not red with a camera: 10−4=610 - 4 = 6 cars.

    All cars that are not red: 50−12=3850 - 12 = 38 cars.

    Not red with no camera: 38−6=3238 - 6 = 32 cars.

    Watch out: the 38 cars that are not red include 6 with cameras. Using 38 gives 3850=1925\frac{38}{50} = \frac{19}{25}, choice C. Using all 40 cars with no camera gives 4050=45\frac{40}{50} = \frac{4}{5}, choice D.

  4. Write the probability

    Use the probability formula. There are 32 cars you want out of 50 cars in all.

    P=3250P = \frac{32}{50}

    Divide the top and bottom by 2.

    3250=1625\frac{32}{50} = \frac{16}{25}

    So the answer is B, 1625\frac{16}{25}.

Check your answer

Finish the table. Red cars with no camera: 12−4=812 - 4 = 8. Now add the four inside boxes: 4+6+8+32=504 + 6 + 8 + 32 = 50. That is every car, so the numbers fit. Also, the cars with no camera are 8+32=408 + 32 = 40, and 50−10=4050 - 10 = 40 too.

More questions like this

The original DOE question
Question 103 as printed on page 162 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 103. Open page 162 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 99 of 99

2018 Test B Math, question 50

Question

Which number line below shows the solution set for 2x−2≤y≤4x+102x-2\leq y\leq 4x+10 when y=1y=1?

  • AA number line from −3 to 3 marked in quarter units, with a solid segment from −2 1/4 to −1 1/2; both ends are closed dots.
  • BA number line from −3 to 3 marked in quarter units, with a solid segment from −1 1/2 to 2 1/4; both ends are closed dots.
  • CA number line from −3 to 3 marked in quarter units, with a solid segment from −2 1/4 to 1 1/2; both ends are closed dots.
  • DA number line from −3 to 3 marked in quarter units, with a closed dot at 1 1/2 and a solid ray from it running left past −2 1/2 to an arrowhead.
Show the answer

Correct answer: C

Separate the compound inequality into two pieces:

2x−2≤y2x-2\leq y and y≤4x+10y\leq 4x+10

Substitute y=1y=1 into each inequality and solve for xx:

2x−2≤12x-2\leq1

2x≤32x\leq 3

x≤32x\leq\frac{3}{2}

1≤4x+101\leq 4x+10

−9≤4x-9\leq4x

−94≤x-\frac{9}{4}\leq x

The solution is the number line that shows that xx is greater than or equal to −214-2\frac{1}{4} and less than or equal to 1121\frac{1}{2} .

Question analysis

Students substitute y = 1 into the compound inequality 2x − 2 ≤ y ≤ 4x + 10 and pick the number line that shows the values of x that work. It tests solving two-step inequalities and graphing the solution. The four graphs differ in their endpoints and in whether the shading is closed off on both sides, so each endpoint has to be found exactly.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph one two-step inequality. This compound inequality in two variables requires substituting, solving two inequalities and combining them into one interval, a step toward algebra 1.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 9, 12 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Substituting y = 1 is easy; the work is solving two inequalities and matching a fractional interval to a graph. Students who solve only the left inequality find a matching ray, and sign slips on either endpoint produce intervals that shift or flip across zero. The graphs differ by quarter-unit ticks, so reading them carefully matters as much as the algebra.

  1. Many steps moderate Working it out

    The compound inequality must be split into 2x − 2 ≤ 1 and 1 ≤ 4x + 10, each solved for x, and the two results joined into one interval. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic moderate Working it out

    Both endpoints are fractions, 3/2 and −9/4, and the second comes from dividing −9 by 4, where a lost sign moves the endpoint across zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The graphs are marked in quarter units, so −2 1/4 and 1 1/2 have to be found by counting small ticks between the labeled integers. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  4. A very close wrong answer minor Choosing the answer

    The ray ending at 1 1/2 looks almost like the right segment, sharing its right end; only the left end, an arrow instead of a closed dot at −2 1/4, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Split the compound inequality into two separate inequalities, solve each for x after putting 1 in for y, then combine the two results into a single interval and match it to a graph.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Solving an inequality
if a≤b, then a+c≤b+c\text{if } a \leq b, \text{ then } a + c \leq b + c
a, b and c are numbers. The sign ≤ means is less than or equal to. Adding or subtracting the same number c on both sides keeps it true. Dividing both sides by the same positive number also keeps it true.

  1. Sort out the question

    The sign ≤\leq means is less than or equal to. An inequality compares two amounts that may not be equal. The statement 2x−2≤y≤4x+102x-2 \leq y \leq 4x+10 is two inequalities joined together. You put 1 in place of yy. Then you need every xx that makes both parts true.

  2. Split it into two parts

    With y=1y = 1, the statement becomes two inequalities.

    2x−2≤12x - 2 \leq 1

    1≤4x+101 \leq 4x + 10

    The answer must make both of them true at the same time.

  3. Solve the first part

    Add 2 to both sides. Then divide both sides by 2.

    2x≤32x \leq 3

    x≤32x \leq \frac{3}{2}

    So xx is at most 1121\frac{1}{2}. This is the right end of the answer.

    Watch out: the right end is a positive number. Choice A ends its bar at −112-1\frac{1}{2} instead.

  4. Solve the second part

    Subtract 10 from both sides. Then divide both sides by 4.

    −9≤4x-9 \leq 4x

    −94≤x-\frac{9}{4} \leq x

    So xx is at least −214-2\frac{1}{4}. This is the left end.

    Watch out: keep the minus sign. Choice B uses the opposite signs and ends at 2142\frac{1}{4}.

  5. Put the two parts together

    The answer is every number from −214-2\frac{1}{4} to 1121\frac{1}{2}, with both ends included. On the graph, that is a solid bar with a dot at each end. The small ticks are quarters. So count 1 tick left of −2 and 2 ticks right of 1.

    Watch out: choice D uses only the first part. Its arrow goes left forever.

  6. Give the answer

    Choice C shows a bar from −214-2\frac{1}{4} to 1121\frac{1}{2}, with dots at both ends. So the answer is C.

Check your answer

Test a number on the bar, like x=0x = 0. Then 2(0)−2=−22(0)-2 = -2 and 4(0)+10=104(0)+10 = 10, and 1 is between them. Now test x=−3x = -3, which is off the bar. Then 4(−3)+10=−24(-3)+10 = -2, and 1 is not less than −2. So the bar must stop on the left, which rules out D.

More questions like this

The original DOE question
Question 107 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 107. Open page 163 of the handbook (PDF)