SHSAT Sunset

Practice Test

A faulty procedure

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Question 1 of 73

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 2 of 73

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 3 of 73

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 4 of 73

2026 Practice Test A Math, question 50

Question

Let b=3b = 3, c=4c = 4, and d=5d = 5. Determine the value of cdcb+cd\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}.

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

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

Show the answer

Correct answer: 7/5

Solution:

75\frac{7}{5}

Substitute b=3b = 3, c=4c = 4, and d=5d = 5 in cdcb+cd\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}: 4543+45\frac{\frac{4}{5}}{\frac{4}{3}} + \frac{4}{5}.

Then simplify the complex fraction by multiplying the complex fraction by a form of 1, (3434)\left(\frac{\frac{3}{4}}{\frac{3}{4}}\right): 4543+45=4543(3434)+45\frac{\frac{4}{5}}{\frac{4}{3}} + \frac{4}{5} = \frac{\frac{4}{5}}{\frac{4}{3}}\left(\frac{\frac{3}{4}}{\frac{3}{4}}\right) + \frac{4}{5}.

Simplify after multiplying by a form of 1: (45)(34)(43)(34)+45=351+45\frac{\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)}{\left(\frac{4}{3}\right)\left(\frac{3}{4}\right)} + \frac{4}{5} = \frac{\frac{3}{5}}{1} + \frac{4}{5}.

Then simplify again and add the two fractions: 351+45=35+45=75\frac{\frac{3}{5}}{1} + \frac{4}{5} = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}.

Question analysis

This substitution question gives values for b, c and d and asks for the value of an expression made of a fraction divided by a fraction, plus another fraction. Students substitute, simplify the complex fraction and add, then type the result. Dividing fractions in the stacked form is the main hurdle, and the item wants a fraction, not a decimal.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.NS.1 dividing fractions by fractions
  • 7.NS.2c properties for multiplying and dividing
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students evaluate expressions and divide and add fractions. A fraction stacked over another fraction looks unfamiliar and there are no choices to check, but with these values the work is three short steps.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the values are in, the work is one fraction division and one addition, but the division is printed as a fraction over a fraction, in small type. Multiplying the two fractions, or flipping the top one, gives a sum in fifteenths that looks as finished as the right one. Noticing that the 4s cancel, so the stacked part is simply 3/5, turns the last step into adding fifths.

  1. Error-prone arithmetic moderate Working it out

    Dividing 4/5 by 4/3 means multiplying by 3/4, giving 3/5, which is then added to 4/5; multiplying straight across or flipping the top fraction instead both lead to fifteenths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    Nothing on the page says that a fraction over a fraction is a division, or that dividing by 4/3 means multiplying by its reciprocal, and the stacked form is rare in grade 7 work. (Anderson & Krathwohl, 2001)

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

    The value is typed into a fraction box with no choices to check against, and the item asks for a fraction, so the equal decimal 1.4 is not accepted. (Enright, Morley & Sheehan, 2002)

Strategy: Substitute the values, rewrite the stacked fraction as the top fraction times the reciprocal of the bottom one, where the 4s cancel, then add the last fraction and leave the result as a fraction.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Flip only the fraction you divide by. It works because a fraction times its reciprocal is 1, so the reciprocal undoes the fraction, just as dividing does.
Adding fractions with the same bottom
ad+bd=a+bd\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}
d is the denominator, the bottom number, which tells the size of each piece. a and b are the numerators, the top numbers, which count the pieces. When the pieces are the same size, add the tops and keep the bottom.

  1. Sort out the question

    You know b=3b = 3, c=4c = 4 and d=5d = 5. You need the value of:

    cdcb+cd\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}

    The first part is a fraction on top of a fraction. A fraction bar means divide, so this part is cd÷cb\frac{c}{d} \div \frac{c}{b}. The answer must be typed as a fraction.

  2. Put in the numbers

    Replace each letter with its value. Look closely: the bottom fraction has b under the c, not d.

    cd=45cb=43\frac{c}{d} = \frac{4}{5} \qquad \frac{c}{b} = \frac{4}{3}

    So the expression becomes:

    45÷43+45\frac{4}{5} \div \frac{4}{3} + \frac{4}{5}

  3. Divide the two fractions

    Use the rule for dividing fractions. The reciprocal of 4/3 is 3/4. To multiply fractions, multiply the tops and multiply the bottoms:

    45÷43=45×34=1220=35\frac{4}{5} \div \frac{4}{3} = \frac{4}{5} \times \frac{3}{4} = \frac{12}{20} = \frac{3}{5}

    To simplify 12/20, divide its top and bottom by 4.

    Watch out: flip only the bottom fraction. Multiplying without flipping gives 16/15 and a total of 28/15. Flipping the top one gives 5/3 and a total of 37/15.

  4. Add the last fraction

    Both fractions are now fifths, so add the tops and keep the bottom:

    35+45=3+45=75\frac{3}{5} + \frac{4}{5} = \frac{3 + 4}{5} = \frac{7}{5}

    The value is 7/5.

  5. Enter it as a fraction

    The item asks for a fraction. 7/5 equals the decimal 1.4, but 1.4 is not accepted. On the test, the answer box already holds a fraction template, a small box over a small box. Type 7 in the top box. Then click the bottom box and type 5. On this site, type 7/5.

    So you would enter 7/5.

Check your answer

Check with decimals. 45=0.8\frac{4}{5} = 0.8, and 4/3 is about 1.33. Then 0.8÷1.330.8 \div 1.33 is about 0.6, which is 3/5. Last, 0.6+0.8=1.40.6 + 0.8 = 1.4, and 1.4 is the same as 7/5.

More questions like this

The original DOE question
Question 50 as shown on the DOE’s digital test, page 121 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test A Math, page 121, question 50.

Question 5 of 73

2026 Practice Test B Math, question 17

Question

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

Enter your answer in the space.

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

Show the answer

Correct answer: 15

Solve the inequality for x.

x+7<23x + 7 < 23

x<16x < 16

There are 15 positive integers that are less than 16.

Question analysis

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

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

Why it is hard

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

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

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

  2. Much to keep track of minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Get x by itself

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

    x+7−7<23−7x + 7 - 7 < 23 - 7

    x<16x < 16

    So x must be less than 16.

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

  3. Check the boundary number

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

    16+7=2316 + 7 = 23

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

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

  4. Count the positive integers

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

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

    Counting from 1 up to 15 gives 15 numbers.

  5. Enter the count

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

Check your answer

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

More questions like this

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

Question 6 of 73

2026 Practice Test C Math, question 3

Question

6x=x−1,6806x = x - 1{,}680

What is the value of x in the equation shown above?

  • A336
  • B240
  • C−240
  • D−336
Show the answer

Correct answer: D

Simplify 6x=x−1,6806x = x - 1{,}680. First, subtract x from both sides of the equation. Then, divide both sides by 5.

6x−x=x−x−1,6806x - x = x - x - 1{,}680

5x=−1,6805x = -1{,}680

5x5=−1,6805\frac{5x}{5} = \frac{-1{,}680}{5}

x=−336x = -336

Question analysis

This algebra question gives the equation 6x = x − 1,680, with x on both sides, and asks for the value of x. Students collect the x terms on one side and divide by the coefficient that remains. The choices pair two sizes with both signs, so every move across the equals sign has to keep its sign right.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.NS.2b rational number division
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 An unknown on both sides of the equals sign is a grade 8 topic, yet gathering the x terms here takes one subtraction and one division, the same work as a two-step grade 7 equation.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 12 words.
Corrected from the DOE handbook
  • A line of the explanation’s working, printed without its fraction bars and equals sign, 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

Collecting the x terms takes one subtraction, and the only way the method goes wrong is moving x across the equals sign, which turns the 5 into a 7. The sign is the bigger hazard: 6x equals x minus a large number, so x must be negative, yet the division 1,680 ÷ 5 is done on positive numbers. Substituting −336 back gives −2,016 on each side, which confirms both the size and the sign.

  1. Error-prone arithmetic moderate Working it out

    Subtracting x from both sides leaves 5x = −1,680, and 1,680 ÷ 5 = 336 has to be worked out by hand while the negative sign is carried through to the result. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The choices are two sizes, 336 and 240, each offered with both signs, so 336 and −336 are written the same except for the minus; only tracking the sign of 1,680 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Subtract x from both sides so the unknown appears once, divide by the coefficient that is left, and substitute the result into both sides to check that they match.

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

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation page for this question is one of the pages replaced in the appendix, and on it the step between 5x = −1,680 and x = −336 shows only 5x over 5 and −1,680 over 5: the two fraction bars and the equals sign between them are missing, so the line is not an equation. The sentence above it says to divide both sides by 5, and the 2022–23 guide, which first printed this explanation, prints the step as 5x/5 = −1,680/5. The site shows 5x/5 = −1,680/5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced
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 0 when you divide). An equation is like a balanced scale. Subtract the same thing from both sides, or divide both sides by the same number, and it stays balanced.
Dividing a negative by a positive
(−a)÷b=−(a÷b)(-a) \div b = -(a \div b)
a and b are positive numbers, so −a is negative. Divide the sizes as usual. Then give the answer a minus sign, because a negative divided by a positive is negative.

  1. Sort out the question

    The equation is 6x=x−1,6806x = x - 1{,}680. You need the value of x. Here 6x means 6 times x. The unknown x is on both sides. So first gather it on one side.

  2. Subtract x from both sides

    Take x away from both sides. On the right, x − x is 0, so only −1,680 is left:

    6x−x=x−x−1,6806x - x = x - x - 1{,}680

    5x=−1,6805x = -1{,}680

    Six x’s take away one x leaves five x’s.

    Watch out: moving x to the left without changing its sign gives 7x. Dividing by 7 then leads to choice C, −240, or B, 240.

  3. Divide both sides by 5

    5x means 5 times x. Divide both sides by 5 to undo it:

    5x5=−1,6805\frac{5x}{5} = \frac{-1{,}680}{5}

    Find the size first. Split 1,680 into 1,500 + 180:

    1,500÷5+180÷5=300+36=3361{,}500 \div 5 + 180 \div 5 = 300 + 36 = 336

    A negative divided by a positive is negative:

    x=−336x = -336

  4. Choose the answer

    Watch out: the minus sign is easy to drop. Positive 336 is choice A. Carry the sign of −1,680 all the way to the end.

    So the answer is D, −336.

Check your answer

Put −336 into both sides. The left side is 6×(−336)=−2,0166 \times (-336) = -2{,}016. The right side is −336−1,680=−2,016-336 - 1{,}680 = -2{,}016. The two sides match, so x = −336 is right.

More questions like this

The original DOE question
Question 3 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 C Math, page 76, question 3. This is the DOE’s version, before the corrections listed in the analysis.

Question 7 of 73

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 73

2026 Practice Test C Math, question 28

Question

If z+3w4=5w\frac{z+3w}{4} = 5w, what is the value of z in terms of w?

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: z = 17w

The goal is to isolate the variable z by using inverse operations.

First, multiply both sides by 4:

z+3w4=5w\frac{z+3w}{4} = 5w

z+3w=20wz + 3w = 20w

Then subtract 3w3w from both sides:

z=17wz = 17w

Question analysis

This algebra question gives an equation in which z plus 3w, all divided by 4, equals 5w, and students type z in terms of w after a printed "z =". It tests solving a literal equation with inverse operations. The 4 divides the whole sum, so undoing the steps in the right order, with the right signs, is the tricky part.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 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 with numbers, such as (x + 3)/4 = 5. Solving for one letter in terms of another is grade 8 work, and the typed answer gives no choices to check against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 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 equation takes only two inverse operations, but their order is the trap: the 4 divides the whole sum z + 3w, so it has to be cleared first. Students who subtract 3w from 5w straight away, or who move 3w to the right without changing its sign, reach a tidy multiple of w that looks as finished as the right one. Substituting w = 1, which needs z = 17 to make (z + 3)/4 equal 5, checks any entry.

  1. Variables instead of numbers moderate Understanding the question

    With w on both sides and no numbers to substitute, the result has to stay a multiple of w, and a slip such as 23w gives no sign that it is wrong. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The answer is typed after a printed “z =”, with no choices to compare, so the entry must be a single term in w, not an equation or a number. (Enright, Morley & Sheehan, 2002)

Strategy: Treat w like a known number: undo the division by 4 first, multiplying the whole of each side, then undo the added 3w, keeping the terms in w together.

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.
  • 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

Doing the same thing to both sides
if A=B, then 4A=4B and A−C=B−C\text{if } A = B, \text{ then } 4A = 4B \text{ and } A - C = B - C
A and B are the two sides of an equation, so they are equal. C is any amount. An equation is like a balanced scale. If you multiply both sides by the same number, or take the same amount from both, it stays balanced. That is how you get one letter alone.

  1. Sort out the question

    The equation is z+3w4=5w\frac{z + 3w}{4} = 5w. You need z in terms of w. That means getting z alone on one side. The other side will use only w and numbers.

    Treat w like a number you do not know yet. The answer box already shows z =, so you type only what comes after it.

  2. Undo the division first

    The fraction bar means the whole top, z + 3w, is divided by 4. So undo that first. Multiply both sides by 4, because multiplying undoes dividing:

    z+3w4×4=5w×4\frac{z + 3w}{4} \times 4 = 5w \times 4

    z+3w=20wz + 3w = 20w

    Watch out: do not subtract 3w first. The 3w sits inside the fraction, so it is divided by 4 too. Subtracting it from 5w first leads to the wrong answer 8w.

  3. Undo the adding

    Now 3w is added to z. Subtract 3w from both sides:

    z+3w−3w=20w−3wz + 3w - 3w = 20w - 3w

    z=17wz = 17w

    Think of w as a thing, like apples. 20 apples take away 3 apples leaves 17 apples.

    Watch out: moving 3w to the other side without changing its sign gives 20w+3w=23w20w + 3w = 23w. That is wrong.

  4. Enter it after the printed z =

    On the test, the box already shows z =. Press 1, then 7, then the w key. You do not need a times sign between 17 and w. Do not type z = again. On this site, type 17w in the box.

    So you would enter 17w, which makes z = 17w.

Check your answer

Pick a number for w, such as w = 1. Then z = 17. Put both into the first equation. The left side is 17+34=204=5\frac{17 + 3}{4} = \frac{20}{4} = 5. The right side is 5×1=55 \times 1 = 5. The two sides match, so 17w works.

More questions like this

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

Question 9 of 73

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 10 of 73

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 11 of 73

2026 Practice Test C Math, question 40

Question

In basketball, a player can earn 1 point for making a free throw and 2 points for making another basket. Marcus makes a mean of 8 free throws per game, and he scores a total of at least 25 points per game. The inequality below shows this relationship, where b represents the number of 2-point baskets Marcus makes.

2b+8≥252b + 8 \geq 25

What is the number of 2-point baskets Marcus needs to make if he wants to score at least his minimum total points for a game and make his mean number of free throws?

  • Ab≤8b \leq 8
  • Bb≥9b \geq 9
  • Cb≥16b \geq 16
  • Db≤25b \leq 25
Show the answer

Correct answer: B

Solve the inequality 2b+8≥252b + 8 \geq 25.

2b+8−8≥25−82b + 8 - 8 \geq 25 - 8

2b≥172b \geq 17

2b2≥172\frac{2b}{2} \geq \frac{17}{2}

b≥8.5b \geq 8.5

Since the number of baskets made must be a whole number, the fewest 2-point baskets, b, he must make is 9.

Question analysis

In basketball, Marcus makes a mean of 8 one-point free throws a game and wants at least 25 points, shown by the inequality 2b + 8 ≥ 25. Students find how many 2-point baskets he needs. It tests solving a two-step inequality and interpreting the solution in context, since a player cannot make part of a basket.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving a two-step inequality from a word problem is a grade 7 standard. Here the inequality is already written, so the extra demand is interpreting its solution as a count of baskets, typical of grade 7 tasks.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 92 words.
Students, last 30 days
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 inequality is handed over, so the solving is two routine moves; the judgment comes after. Dividing 17 by 2 leaves a half, and since a basket cannot be split, the boundary has to move up to 9: eight baskets give 2(8) + 8 = 24 points, one short of 25. Students who flip the symbol while dividing, or who move the 8 without changing its sign, still find a choice to match.

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

    Solving gives b ≥ 8.5, which no choice shows; it has to be recognized as b ≥ 9, because Marcus cannot make half a basket. (Enright, Morley & Sheehan, 2002)

  2. Complex sentences minor Understanding the question

    The 33-word question restates the given inequality as “at least his minimum total points” and “make his mean number of free throws”, wording harder than the math it describes. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Undo the added 8 and then the 2 to solve for b, then ask which whole numbers of baskets actually reach 25 points, testing the boundary value 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.)
  • 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.
  • 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

Solving an inequality
if A≥B, then A−C≥B−C and A2≥B2\text{if } A \geq B, \text{ then } A - C \geq B - C \text{ and } \frac{A}{2} \geq \frac{B}{2}
A and B are the two sides. The sign ≥ means greater than or equal to. Taking the same amount C from both sides keeps the sign the same. Dividing both sides by a positive number, like 2, also keeps it the same. Only a negative number flips it.

  1. Sort out the question

    A free throw is worth 1 point, and each other basket is worth 2 points. Marcus makes 8 free throws, which give 8 points. The letter b is the number of 2-point baskets, so they give 2b points.

    He wants at least 25 points in all. That is the inequality 2b+8≥252b + 8 \geq 25. An inequality compares two amounts. The sign ≥ means greater than or equal to. You need the values of b that work.

  2. Subtract 8 from both sides

    Undo the added 8 first:

    2b+8−8≥25−82b + 8 - 8 \geq 25 - 8

    2b≥172b \geq 17

    Watch out: adding 8 instead gives 2b≥332b \geq 33. That leads to choice C, which is wrong.

  3. Divide both sides by 2

    2b means 2 times b. Divide both sides by 2. Since 2 is positive, the sign stays the same:

    2b2≥172\frac{2b}{2} \geq \frac{17}{2}

    b≥8.5b \geq 8.5

    Watch out: do not flip the sign here. Flipping it gives b≤8.5b \leq 8.5, which leads to choice A.

  4. Think about whole baskets

    Marcus cannot make half a basket. So test the whole numbers near 8.5:

    2×8+8=242 \times 8 + 8 = 24

    2×9+8=262 \times 9 + 8 = 26

    Eight baskets give 24 points, one short of 25. Nine baskets give 26 points, which is enough. So he needs 9 or more baskets.

  5. Give the answer

    The number of baskets is 9 or more, written b≥9b \geq 9. Choice D, b≤25b \leq 25, just copies the 25 from the points.

    So the answer is B, b≥9b \geq 9.

Check your answer

Try a bigger number, such as b = 12. Then 2×12+8=322 \times 12 + 8 = 32 points, which is at least 25. Any number of baskets above 9 also works, and that is what ≥ says. Below 9, the total stays at 24 points or fewer.

More questions like this

The original DOE question
Question 40 as shown on the DOE’s digital test, page 113 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 113, question 40.

Question 12 of 73

2026 Practice Test C Math, question 50

Question

Terry wants to pour cement around the edge of the circular patio in her backyard. The patio has a radius of 5 feet. What is the distance, in feet, around the edge of the patio? Use 3.14 for π\pi.

  • A15.7
  • B31.4
  • C49.3
  • D78.5
Show the answer

Correct answer: B

To find the distance around a circle if the radius is known, use the formula for circumference, C=2πrC = 2\pi r.

C=2(3.14)(5)=31.4C = 2(3.14)(5) = 31.4

Question analysis

Terry wants to pour cement around the edge of a circular patio with a radius of 5 feet. Students find the distance around the edge, using 3.14 for π. It tests the circumference of a circle, and because the formulas for a circle's area and circumference both use the radius, they are easy to mix up.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Applying a circle formula is a grade 7 standard, and this is a one-formula problem with a given value for π. The only demand beyond routine is recalling the right formula, which the test does not print.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One formula and one multiplication finish the problem once the right formula is chosen. The cement in the story pulls toward area, and the area, 78.5, is listed, as is half the circumference, from multiplying 3.14 by the radius alone. Remembering that the circumference is π times the diameter, here 10 feet, and that the answer is in feet, not square feet, settles it.

  1. Something you must already know moderate Working it out

    The circumference formula is not printed, and the area formula uses the same radius and π, so recalling the wrong one gives a number that looks just as finished. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    Pouring cement suggests covering a surface, but the question asks for “the distance, in feet, around the edge”, which is the circumference, not the area. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Decide which circle measurement the distance around the edge is, recall its formula, and substitute the radius and 3.14, doubling the radius wherever the formula calls for the diameter.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • 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

Circumference of a circle
C=2πrC = 2 \pi r
C is the circumference, the distance around the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, about 3.14. 2r is the diameter, the distance all the way across. The distance around is always about 3.14 times that.
Area of a circle
A=πr2A = \pi r^2
A is the area, the space inside the circle, in square units. r² means r times r. It is here so you can tell the two formulas apart. This question does not use it.

  1. Sort out the question

    The patio is a circle with a radius of 5 feet. The radius is the distance from the center to the edge. Terry wants cement around the edge.

    You need the distance around the edge, in feet. The distance around a circle is called its circumference. Use 3.14 for π.

  2. Choose the right formula

    The distance around is a length, so use the circumference formula, C=2πrC = 2\pi r.

    Watch out: pouring cement may make you think of area. The area, 3.14×5×5=78.53.14 \times 5 \times 5 = 78.5 square feet, is choice D. That is the space inside the patio, not the distance around it.

  3. Put in the numbers

    Put 3.14 in for π and 5 in for r. Multiply 2 by 5 first, because that is easy:

    C=2×3.14×5=10×3.14=31.4C = 2 \times 3.14 \times 5 = 10 \times 3.14 = 31.4

    The distance around the patio is 31.4 feet.

    Watch out: 3.14×5=15.73.14 \times 5 = 15.7, choice A, forgets the 2. It goes only halfway around.

  4. Give the answer

    The distance around the edge of the patio is 31.4 feet.

    So the answer is B, 31.4.

Check your answer

Estimate. The diameter is 10 feet. The distance around a circle is a little more than 3 times its diameter. 3×10=303 \times 10 = 30, so about 31 feet makes sense. Choice D, 78.5, is far too big, and it is in square feet.

More questions like this

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

Question 13 of 73

2026 Practice Test D Math, question 2

Question

What is the value of z in x10+y−x5=z10\frac{x}{10} + \frac{y - x}{5} = \frac{z}{10}?

  • A1
  • B5
  • Cyy
  • D2y−x2y - x
Show the answer

Correct answer: D

First, multiply both sides of the equation by 10 to eliminate the fractions. In other words, multiply each term by 10.

x10+y−x5=z10\frac{x}{10} + \frac{y - x}{5} = \frac{z}{10}

10(x10+y−x5)=10(z10)10\left(\frac{x}{10} + \frac{y - x}{5}\right) = 10\left(\frac{z}{10}\right)

10x10+10(y−x)5=10z10\frac{10x}{10} + \frac{10(y - x)}{5} = \frac{10z}{10}

x+2(y−x)=zx + 2(y - x) = z

Then, distribute and combine like terms to find the value for z.

x+2(y−x)=zx + 2(y - x) = z

x+2y−2x=zx + 2y - 2x = z

2y−x=z2y - x = z

Question analysis

This algebra question gives an equation with three letters and fractions, x over 10 plus (y − x) over 5 equals z over 10, and asks for the value of z. Students clear the denominators, distribute and combine like terms. The two fractions on the left have different denominators, which is where the usual slip happens.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 7.EE.2 rewriting expressions to show relationships
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students add and expand linear expressions with fraction coefficients, usually in one letter. Here three letters appear, one numerator is a difference, and the result is an expression rather than a number, which adds abstraction but no new skill.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 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 equation has three fractions and three letters, and the work is three routine moves: multiply through by 10, distribute, collect the x terms. Most errors come from the middle term, whose denominator 5 differs from the other two; treating it as tenths makes z simply y. Students who test values should avoid x = y, since that pair makes y and 2y − x agree, and x = y = 1 even makes the number 1 fit.

  1. Variables instead of numbers moderate Understanding the question

    The answer is z in terms of x and y, not a number, yet two choices are plain numbers, 1 and 5, which tempt students who expect the equation to give z a single value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic moderate Working it out

    Clearing the denominators turns (y − x)/5 into 2(y − x), and the −2x from distributing must then combine with the lone x to leave −x, two places where a factor or a sign slips. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply every term on both sides by 10 so no fractions remain, distribute across the parentheses, and combine the x terms; substituting small numbers for x and y checks the result.

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.
  • 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

Doing the same to both sides
if a=b, then 10×a=10×b\text{if } a = b \text{, then } 10 \times a = 10 \times b
a and b are the two sides of an equation. Multiplying both sides by the same number keeps them equal, like adding the same weight to both pans of a balance. Here 10 is a good choice, because it clears away every fraction.
The distributive property
a(b−c)=a×b−a×ca(b - c) = a \times b - a \times c
a is the number outside the parentheses, and 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) is a groups of b minus a groups of c.

  1. Sort out the question

    The equation has three letters, x, y and z. You need z written using x and y. The answer will not be one number, because x and y can be anything. The fractions make the work messy, so clear them first.

  2. Multiply every term by 10

    The bottoms are 10, 5 and 10, and each one divides into 10. So multiply every term by 10:

    10×x10+10×y−x5=10×z1010 \times \frac{x}{10} + 10 \times \frac{y - x}{5} = 10 \times \frac{z}{10}

    10 times x/10 is x. Since 10 ÷ 5 = 2, 10 times (y − x)/5 is 2(y − x). And 10 times z/10 is z.

    x+2(y−x)=zx + 2(y - x) = z

  3. Distribute the 2

    Use the distributive property. Multiply the 2 by both y and x:

    2(y−x)=2y−2x2(y - x) = 2y - 2x

    So the equation becomes:

    x+2y−2x=zx + 2y - 2x = z

  4. Combine like terms

    Like terms have the same letter. Here x and −2x are like terms. Since 1 − 2 = −1, they combine to −x:

    x−2x=−xx - 2x = -x

    That leaves:

    2y−x=z2y - x = z

    Watch out: do not add the tops of x/10 and (y − x)/5 straight away. Their bottoms are different. Doing it anyway gives z = y, choice C. Choices A and B are plain numbers, but z changes when x and y do.

    So the answer is D, 2y−x2y - x.

Check your answer

Pick numbers. Let x = 2 and y = 7. The left side is 2/10 + 5/5 = 0.2 + 1 = 1.2. Choice D gives z = 2 × 7 − 2 = 12, and 12/10 = 1.2. Both sides match.

More questions like this

The original DOE question
Question 2 as shown on the DOE’s digital test, page 75 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 75, question 2.

Question 14 of 73

2026 Practice Test D Math, question 5

Question

In a school raffle, 1 out of every 25 students wins a prize. Each prize includes a randomly selected trading card from a set of 100 different cards, labeled C1 through C100, with each card equally likely to be chosen. If one student is selected at random, what is the probability that the student both wins a prize and receives Card C17?

Write an expression that represents the probability that a student wins the prize and receives Card C17.

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: 1/25 · 1/100

To calculate the joint probability, first determine the probability of each independent event.

The probability that a randomly selected student wins the prize is 125\frac{1}{25}.

The probability that a randomly selected student will receive Card C17 is 1100\frac{1}{100}, because Card C17 is 1 card out of 100 labeled cards.

The probability that a student wins the prize and receives Card C17 is the product of those two probabilities, 125⋅1100\frac{1}{25} \cdot \frac{1}{100}.

Question analysis

In a school raffle, 1 out of every 25 students wins a prize, and each prize includes one of 100 equally likely trading cards. Students type an expression for the probability that a randomly chosen student both wins and gets one particular card. It tests the probability of two independent events, and combining the two chances correctly is the main hurdle.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 6.SP.8a uniform probability models
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students find compound probabilities, usually by listing outcomes for coins or number cubes. These two stages have 25 and 100 outcomes, far too many pairs to list, so a rule must be used, and the answer is typed as an expression.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 8, 87 words.
Students, last 30 days
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 chance is stated almost directly, 1 out of 25 and 1 out of 100, so the question tests one rule: both events must happen, and the chances multiply. The word “and” pulls students toward adding, which gives a result larger than either chance, a sign that something is wrong. Others write only the card's chance and forget that the student must first win a prize.

  1. Something you must already know moderate Working it out

    The question never says that the chance of two independent events both happening is the product of their chances, so students must bring that rule to the 1/25 and the 1/100 themselves. (Anderson & Krathwohl, 2001)

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

    The answer is typed as an expression with no choices to compare, so a sum of the two chances, which looks just as finished, gets no warning on screen. (Enright, Morley & Sheehan, 2002)

Strategy: Write each event's chance as a fraction, then apply the rule for two independent events that must both happen, and check that the result is smaller than either chance alone.

Traps

  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • Stops one step early: A choice is a number from partway through the solution.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome
P=outcomes you wantall possible outcomesP = \frac{\text{outcomes you want}}{\text{all possible outcomes}}
P is the probability, the chance that something happens, from 0 to 1. The top counts the outcomes you want. The bottom counts all the outcomes that could happen. Use it when every outcome is equally likely, like one card picked from 100.
Probability that two independent events both happen
P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)
A and B are two events. Independent means one does not change the chance of the other. P(A) is the chance of A, and P(B) is the chance of B. Multiply them to get the chance that both happen. The result is smaller than either chance, since both things must happen.

  1. Sort out the question

    Two things must both happen. First, the student wins a prize. Second, the card in that prize is Card C17. You need an expression for that chance. An expression is a math phrase made of numbers and signs, with no equal sign. You do not have to work it out to one number.

  2. Find the chance of winning

    The question says “1 out of every 25 students wins a prize.” One student is the part that wins, and 25 is the whole group. So:

    P(win)=125P(\text{win}) = \frac{1}{25}

  3. Find the chance of Card C17

    Each prize has one card, picked at random from 100 cards. Every card is equally likely. Card C17 is 1 of those 100 cards:

    P(C17)=1100P(\text{C17}) = \frac{1}{100}

  4. Multiply the two chances

    Winning a prize does not change which card is inside it. So the two events are independent. To find the chance that both happen, multiply:

    P(win and C17)=125⋅1100P(\text{win and C17}) = \frac{1}{25} \cdot \frac{1}{100}

    The dot · means times. Worked out, this is 1/2500.

    Watch out: the word and does not mean add here. 1/25 + 1/100 is bigger than either chance, but needing both things must be less likely. Writing only 1/100 forgets that the student must win a prize first.

  5. Enter the expression

    On the test, press the fraction key, a small box over a small box. Type 1 on top and 25 on the bottom. Move the cursor out of the fraction with the → key. Press the dot key for times. Then use the fraction key again for 1 over 100. On this site, type 1/25 * 1/100 or 1/25 · 1/100. The value 1/2500 also counts.

    So you would enter 1/25 · 1/100.

Check your answer

Picture 2,500 students. 1 in 25 wins, so about 100 students win a prize. Of those 100, about 1 gets Card C17. That is 1 student out of 2,500, which matches 1/25 · 1/100 = 1/2500. It is smaller than 1/25 and 1/100, as it should be.

More questions like this

The original DOE question
Question 5 as shown on the DOE’s digital test, page 78 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 78, question 5.

Question 15 of 73

2026 Practice Test D Math, question 10

Question

2x+11>3x+92x + 11 > 3x + 9

For what values of x is the above inequality true?

  • Ax<2x < 2
  • Bx>2x > 2
  • Cx<20x < 20
  • Dx>20x > 20
Show the answer

Correct answer: A

Use inverse operations to isolate the variable x by first getting the x variable on one side.

2x+11>3x+92x + 11 > 3x + 9

2x−3x+11>3x−3x+92x - 3x + 11 > 3x - 3x + 9

−x+11>9-x + 11 > 9

−x+11−11>9−11-x + 11 - 11 > 9 - 11

−x>−2-x > -2

Now, when dividing by −1-1, the direction of the inequality sign changes.

−x>−2-x > -2

x<2x < 2

Another way to solve is to keep the coefficient of x positive and then recognize that 2>x2 > x is the same as x<2x < 2.

2x+11>3x+92x + 11 > 3x + 9

11>x+911 > x + 9

2>x2 > x

x<2x < 2

Question analysis

This algebra question asks which values of x make the inequality 2x + 11 > 3x + 9 true, choosing from four solution sets. It tests solving a linear inequality with the variable on both sides. Collecting the x terms can leave a negative coefficient, and handling the inequality sign at that point is where most mistakes happen.

NYS standards
  • 7.EE.4b two-step inequalities
  • 8.EE.7b linear equations with rational coefficients
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 Two-step inequalities are grade 7 work, and having x on both sides, a grade 8 idea, takes only one extra subtraction here. The demand matches a careful grade 7 problem, with one sign decision that is easy to get wrong.
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 10, 15 words.
Corrected from the DOE handbook
  • A grammatical error in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The boundary 2 comes out of two routine moves; the direction is where points are lost. Whichever side the x terms are collected on, the last step either divides by a negative or leaves x on the right, and both moments invite a reversed answer. A second risk is moving the 9 without changing its sign, which yields 20. Substituting x = 0, which makes 11 > 9 true, confirms the direction.

  1. Error-prone arithmetic moderate Working it out

    Subtracting 3x from both sides leaves −x > −2, and dividing by −1 must reverse the sign; subtracting 2x instead leaves 2 > x, which still has to be read from the x side. (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, so finding the 2 settles nothing; only the sign rule or a test value such as x = 0 decides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Subtract 2x from both sides so the x term stays positive, undo the + 9, then rewrite the result with x first; testing a value in the original inequality checks the direction.

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

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation says “Another way to solve is to keep the coefficient of x positive and then recognizing that 2 > x is the same as x < 2.” The two verbs joined by “and then” should have the same form, so the site now says “and then recognize that”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides of an inequality
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 or term. Subtracting (or adding) the same amount on both sides keeps the inequality true and its sign the same. That is how you move terms from one side to the other.
Dividing by a negative number
if a>b, then −a<−b\text{if } a > b \text{, then } -a < -b
a and b are any numbers. Multiplying or dividing both sides by a negative number flips the sign, so > becomes <. For example, 3 > 2, but −3 < −2. Use it whenever you divide by a negative number.

  1. Sort out the question

    The sign > means is greater than. You need every value of x that makes 2x + 11 greater than 3x + 9. The answer is a range of values, not one number. Solve it like an equation, but keep an eye on the direction of the sign.

  2. Gather the x terms on one side

    The right side has more x’s, 3x, than the left side, 2x. So subtract 2x from both sides. That keeps the x term positive:

    2x+11−2x>3x+9−2x2x + 11 - 2x > 3x + 9 - 2x

    11>x+911 > x + 9

  3. Get x by itself

    Subtract 9 from both sides:

    11−9>x+9−911 - 9 > x + 9 - 9

    2>x2 > x

    Watch out: to move the 9, subtract it from both sides. Adding it instead gives 20 > x, which leads to choice C or D.

  4. Read it from the x side

    2>x2 > x says 2 is greater than x. That is the same as saying x is less than 2:

    x<2x < 2

    Watch out: when you swap the sides, the sign must point the other way. Writing x>2x > 2 gives choice B. If you subtract 3x instead, you get −x>−2-x > -2. Dividing by −1 then flips the sign, and you get x<2x < 2 again.

    So the answer is A, x<2x < 2.

Check your answer

Test x = 0, which is less than 2. The left side is 11 and the right side is 9, and 11 > 9 is true. Test x = 5, which is more than 2. The left side is 21 and the right side is 24, and 21 > 24 is false. So only values below 2 work.

More questions like this

The original DOE question
Question 10 as shown on the DOE’s digital test, page 83 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 83, question 10. This is the DOE’s version, before the corrections listed in the analysis.

Question 16 of 73

2026 Practice Test D Math, question 15

Question

111+122+133\frac{1}{11} + \frac{1}{22} + \frac{1}{33}

What is the value of the expression shown above?

  • A122\frac{1}{22}
  • B111\frac{1}{11}
  • C16\frac{1}{6}
  • D322\frac{3}{22}
Show the answer

Correct answer: C

To add fractions, find a common denominator and rename the fractions using the common denominator. The least common denominator for 11, 22, and 33 is 66. After adding the fractions, the final fraction can be reduced to lowest terms.

111=1×611×6=666\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}

122=1×322×3=366\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}

133=1×233×2=266\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}

666+366+266=1166=16\frac{6}{66} + \frac{3}{66} + \frac{2}{66} = \frac{11}{66} = \frac{1}{6}

Question analysis

This fractions question asks for the value of 1/11 + 1/22 + 1/33. Students find a common denominator for three fractions, rename each one, add and simplify. The denominators are all multiples of 11, and adding the tops and bottoms straight across, or forgetting to rename a numerator, leads to tempting wrong choices.

NYS standards
  • 7.NS.1d properties for adding and subtracting
  • 6.NS.4 common factors and multiples
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding fractions with unlike denominators is taught before grade 7. Finding a common denominator for three related numbers and simplifying is short, routine work, lighter than typical grade 7 problems with signed rational numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question checks whether students rename fractions before adding, and each wrong choice comes from a shortcut that skips that step or stops early. Seeing that 22 and 33 are multiples of 11 points to 66 as the common denominator. A size check helps too: the sum must be more than 1/11, its largest term, which rules out two of the wrong choices at once.

  1. Error-prone arithmetic moderate Working it out

    Three unlike denominators must be renamed over 66, not over the product of all three, and the numerators 6, 3 and 2 have to be found one by one before adding. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The sum comes out as 11/66, which must be reduced to match a choice; its simplest form has a denominator, 6, unlike any in the expression. (Enright, Morley & Sheehan, 2002)

Strategy: Notice that 22 and 33 are multiples of 11, rename all three fractions over their least common multiple, add the numerators, and simplify the result.

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

Equivalent fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a/b is a fraction, and n is any number except 0. Multiplying the top and the bottom by the same n gives an equal fraction. It works because n/n is 1, and multiplying by 1 changes nothing. Use it to give fractions the same bottom.
Adding fractions with the same denominator
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, which tells the size of each piece. a, b and c are the numerators, the top numbers, which count the pieces. When the pieces are the same size, add the tops and keep the bottom.

  1. Sort out the question

    You need to add three fractions: 1/11, 1/22 and 1/33. Their denominators (bottom numbers) are all different. You can only add fractions when the pieces are the same size. So first give all three the same denominator.

  2. Find a common denominator

    A common denominator is a number that 11, 22 and 33 all divide into. Notice that 22 = 2 × 11 and 33 = 3 × 11. So the smallest number that all three divide into is:

    2×3×11=662 \times 3 \times 11 = 66

  3. Rename each fraction

    Multiply the top and bottom of each fraction by the number that makes the bottom 66:

    111=1×611×6=666\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}

    122=1×322×3=366\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}

    133=1×233×2=266\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}

  4. Add the tops

    Now every piece is a sixty-sixth. Add the tops and keep the bottom:

    666+366+266=1166\frac{6}{66} + \frac{3}{66} + \frac{2}{66} = \frac{11}{66}

    Watch out: never add the bottoms. Adding straight across gives 3/66 = 1/22, choice A. Putting the three 1s over 33 gives 3/33 = 1/11, choice B.

  5. Simplify the fraction

    To simplify, divide the top and bottom by a number that goes into both. Here 11 goes into 11 and into 66:

    1166=11÷1166÷11=16\frac{11}{66} = \frac{11 \div 11}{66 \div 11} = \frac{1}{6}

    Watch out: add all three fractions. Stopping after the first two gives 3/22, choice D.

    So the answer is C, 16\frac{1}{6}.

Check your answer

Estimate with decimals. 1/11 is about 0.091, 1/22 is about 0.045 and 1/33 is about 0.030. They add to about 0.166. And 1/6 is 0.1666…, a match. The other choices are all 0.136 or less.

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 D Math, page 88, question 15.

Question 17 of 73

2026 Practice Test D Math, question 18

Question

Which of the following decimals is equivalent to 123+2491\frac{2}{3} + 2\frac{4}{9}?

  • A3.13.\overline{1}
  • B3.63.\overline{6}
  • C4.14.\overline{1}
  • D4.94.\overline{9}
Show the answer

Correct answer: C

First, add the mixed numbers. Then, convert the fraction 19\frac{1}{9} to a decimal.

123+2491\frac{2}{3} + 2\frac{4}{9}

169+2491\frac{6}{9} + 2\frac{4}{9}

3+1093 + \frac{10}{9}

3+1193 + 1\frac{1}{9}

419=4.111…=4.14\frac{1}{9} = 4.111\ldots = 4.\overline{1}

Question analysis

This question asks which repeating decimal equals the sum of the mixed numbers 1 2/3 and 2 4/9. Students rename the fractions with a common denominator, add, and convert the fraction part to a decimal with a bar over its repeating digit. The choices differ in both the whole number and the repeating digit, so each step has to be right.

NYS standards
  • 7.NS.2d fractions to decimals by long division
  • 7.NS.1d properties for adding and subtracting
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Adding mixed numbers and writing a fraction as a repeating decimal are standard grade 7 skills. This combines them in three short steps; renaming a fraction and reading the bar notation are the only places that need care.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 13, 13 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two fraction moves and one conversion stand between the sum and the choices. The carry is the main hazard: 10/9 has to become 1 1/9, and students who keep only the 1/9 land on a choice with the wrong whole number. Estimating first, about 1.7 + 2.4 = 4.1, rules out both choices that begin with 3 and the one that equals 5.

  1. Error-prone arithmetic moderate Working it out

    Adding 2/3 and 4/9 needs 2/3 renamed as 6/9, and the sum 10/9 is more than 1, so a whole must be carried into the 3 before converting. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The choices are repeating decimals with bars, so a mixed number in ninths has to be recognized as a digit repeating forever; knowing that ninths repeat their numerator makes the match quick. (Enright, Morley & Sheehan, 2002)

Strategy: Rename 2/3 as ninths, add the whole numbers and the fractions separately, regroup any fraction of 9/9 or more as a whole, then divide the leftover numerator by 9.

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.
  • 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

Equivalent fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a/b is a fraction, and n is any number except 0. Multiplying the top and the bottom by the same n gives an equal fraction, because n/n is 1. Use it to give two fractions the same bottom before you add them.
Ninths as repeating decimals
19=0.111…=0.1\frac{1}{9} = 0.111\ldots = 0.\overline{1}
1 divided by 9 is 0.111…, with 1s that go on forever. The bar over a digit means that digit repeats forever. Ninths from 1/9 to 8/9 repeat their top digit, so 4/9 is 0.444… and 2/9 is 0.222…

  1. Sort out the question

    You need to add two mixed numbers, 1 2/3 and 2 4/9. A mixed number is a whole number plus a fraction. Then write the sum as a decimal. The choices are repeating decimals, where the bar over a digit means it repeats forever.

  2. Rename two thirds as ninths

    The fractions 2/3 and 4/9 have different bottoms. Since 3 × 3 = 9, multiply the top and bottom of 2/3 by 3:

    23=2×33×3=69\frac{2}{3} = \frac{2 \times 3}{3 \times 3} = \frac{6}{9}

    So 1 2/3 is the same as 1 6/9.

    Watch out: adding the tops 2 and 4 over 9 without renaming gives 3 6/9, choice B.

  3. Add the mixed numbers

    Add the whole numbers, then add the fractions:

    1+2=369+49=1091 + 2 = 3 \qquad \frac{6}{9} + \frac{4}{9} = \frac{10}{9}

    10/9 is more than 1. It is 9/9 plus 1/9, or 1 1/9. Carry the 1 into the whole number:

    3+119=4193 + 1\frac{1}{9} = 4\frac{1}{9}

    Watch out: forgetting to carry the 1 gives 3 1/9, choice A.

  4. Change it to a decimal

    One ninth is 0.111…, with the 1 repeating forever. So:

    419=4.111…=4.14\frac{1}{9} = 4.111\ldots = 4.\overline{1}

    Watch out: ninths repeat the top digit, not the 9. Choice D, 4.999…, does not match 4 1/9.

    So the answer is C, 4.14.\overline{1}.

Check your answer

Use rough decimals. 1 2/3 is about 1.667, and 2 4/9 is about 2.444. Their sum is about 4.111. That matches 4.111… and rules out both choices that start with 3. And 4.999… is almost 5, which is too big.

More questions like this

The original DOE question
Question 18 as shown on the DOE’s digital test, page 91 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 91, question 18.

Question 18 of 73

2026 Practice Test D Math, question 21

Question

Maxie borrowed $7.75 from her mother, $11.00 from her father, and $4.50 from her brother to purchase a video game. Her grandmother gave her $25.00 as a gift. How much money will Maxie have left or still owe if she uses the money her grandmother gave her to pay back the money she borrowed?

  • AMaxie will still owe $2.25.
  • BMaxie will still owe $1.75.
  • CMaxie will have $1.75 left.
  • DMaxie will have $2.25 left.
Show the answer

Correct answer: C

First, determine the total amount Maxie borrowed, in dollars:

7.75+11.00+4.50=23.257.75 + 11.00 + 4.50 = 23.25

Then use subtraction to determine the difference between the amount her grandmother gave her and the amount she borrowed:

25.00−23.25=1.7525.00 - 23.25 = 1.75

The difference is positive, which means that Maxie will still have $1.75 left from the amount her grandmother gave her.

Question analysis

Maxie borrowed money from her mother, father and brother to buy a video game, then received $25.00 from her grandmother to pay them back. Students decide how much money she will have left over or still owe. It tests adding and subtracting decimals with money, and the choices also test which way the difference goes.

NYS standards
  • 6.NS.3 multi-digit decimal operations
  • 6.NS.5 positive and negative numbers
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Adding and subtracting money amounts is grade 6 decimal work. Here it takes one sum and one difference, and the only further demand is deciding whether the result is money left or money owed, well below typical grade 7 tasks.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 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

Few students miss this one, since it needs only one addition and one subtraction with money. The choices come in pairs, so the right amount must also be matched with the right outcome. Lining up the decimal points, and asking whether $25.00 is more or less than the $23.25 borrowed, settles both the amount and whether Maxie owes or keeps money.

  1. Error-prone arithmetic minor Working it out

    The loans add to $23.25, and subtracting that from $25.00 means borrowing across two zeros, a classic place for a digit to go wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    Each amount appears twice among the choices, once as money left and once as money owed, so the result must also be read for direction by asking which is larger, the gift or the loans. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add the three amounts borrowed, compare the total with the $25.00 gift, and decide from which is larger whether Maxie ends up with money left or still owes some.

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

Money left over
left=money she has−money she owes\text{left} = \text{money she has} - \text{money she owes}
Money she has is the $25.00 gift. Money she owes is the total she borrowed. If the result is positive, she has that much left. If it were negative, she would still owe that much. Comparing the two amounts first tells you which.

  1. Sort out the question

    Maxie borrowed from three people: $7.75, $11.00 and $4.50. Her grandmother gave her $25.00. She uses the $25.00 to pay everyone back. You need to know if money is left over or if she still owes some, and how much.

  2. Add up what she borrowed

    Line up the decimal points and add:

    7.75+11.00+4.50=23.257.75 + 11.00 + 4.50 = 23.25

    She owes $23.25 in all. That is the amount she must pay back.

  3. Compare the gift with the debt

    The gift, $25.00, is more than the $23.25 she owes. So she can pay everyone back and still have money left. She will not owe anything.

    Watch out: once you find $1.75, keep its meaning straight. Saying she still owes $1.75, choice B, gets the direction backwards.

  4. Subtract to find what is left

    Take the debt away from the gift:

    25.00−23.25=1.7525.00 - 23.25 = 1.75

    You need to regroup (borrow) across the zeros. Think of $25.00 as 24 dollars and 100 cents. Then 100 − 25 = 75 cents, and 24 − 23 = 1 dollar.

    Watch out: taking the smaller digit from the larger one in each column gives $2.25, choice D.

    So the answer is C: Maxie will have $1.75 left.

Check your answer

Add back. The $23.25 she pays plus the $1.75 she keeps is 23.25 + 1.75 = 25.00. That is the whole gift, so the subtraction is right.

More questions like this

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

Question 19 of 73

2026 Practice Test D Math, question 30

Question

A recipe uses 23\frac{2}{3} cup of sugar for every 1141\frac{1}{4} cups of flour. A cook wants to increase the recipe. How many cups of sugar are needed for each cup of flour?

  • A815\frac{8}{15}
  • B56\frac{5}{6}
  • C1781\frac{7}{8}
  • D3233\frac{2}{3}
Show the answer

Correct answer: A

To find the cups of sugar for each cup of flour, find the unit rate using this ratio:

23 cup of sugar114 cups of flour\frac{\frac{2}{3} \text{ cup of sugar}}{1\frac{1}{4} \text{ cups of flour}}

Then, convert the denominator to an improper fraction and divide to find the cups of sugar per cup of flour.

2354=23÷54=23×45=815\frac{\frac{2}{3}}{\frac{5}{4}} = \frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}

Question analysis

A recipe uses 2/3 cup of sugar for every 1 1/4 cups of flour, and this unit rate question asks how many cups of sugar go with each cup of flour. Students divide one fraction by a mixed number. Deciding which amount to divide by which, and converting the mixed number first, are where mistakes creep in.

NYS standards
  • 7.RP.1 unit rates with fractions
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Computing a unit rate from a ratio of fractions is exactly the grade 7 standard. This one divides by a mixed number, and the cook's plan to enlarge the recipe adds wording without adding any numbers to use.
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, 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 work is a single fraction division, but it involves a mixed number and an awkward result. Students unsure of the rule multiply the two amounts or divide them the other way round, and each gives a tidy listed fraction. A size check settles it: 1 1/4 cups of flour take only 2/3 cup of sugar, so one cup of flour needs less than 2/3 cup, which leaves a single choice.

  1. Error-prone arithmetic moderate Working it out

    The divisor is the mixed number 1 1/4, which must become 5/4 before dividing; then 2/3 × 4/5 gives 8/15, a fraction that does not simplify and looks unfinished. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The phrase “for each cup of flour” puts flour in the denominator, sugar ÷ flour, and the sentence about increasing the recipe adds nothing to the calculation. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Write the sugar over the flour, change 1 1/4 to an improper fraction, and divide by multiplying by its reciprocal, keeping sugar on top because the question asks for sugar per cup of flour.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
unit rate=cups of sugarcups of flour\text{unit rate} = \frac{\text{cups of sugar}}{\text{cups of flour}}
A unit rate tells how much of one thing goes with 1 of another. Here it is the cups of sugar for 1 cup of flour. The amount you want goes on top. The amount you want to make 1 goes on the bottom.
Mixed number to fraction
wnd=w×d+ndw\frac{n}{d} = \frac{w \times d + n}{d}
w is the whole number, and n/d is the fraction part. Multiply the whole number by the bottom, add the top, and keep the bottom. It works because each whole holds d pieces of size 1/d.
Dividing fractions
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Flip only the fraction you divide by.

  1. Sort out the question

    The recipe uses 23\frac{2}{3} cup of sugar for every 1141\frac{1}{4} cups of flour. You need the cups of sugar for each 1 cup of flour. The part about making more of the recipe gives no numbers, so you can set it aside.

  2. Set up the unit rate

    You want sugar for each cup of flour. So sugar goes on top and flour goes on the bottom:

    unit rate=23÷114\text{unit rate} = \frac{2}{3} \div 1\frac{1}{4}

    Watch out: dividing the other way, flour by sugar, gives 158=178\frac{15}{8} = 1\frac{7}{8}. That is cups of flour for each cup of sugar, choice C.

  3. Change the mixed number

    You cannot divide by a mixed number directly. Change it to a fraction first. Each whole has 4 fourths:

    114=1×4+14=541\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{5}{4}

  4. Divide the fractions

    Multiply by the reciprocal of 5/4, which is 4/5. Multiply the tops and multiply the bottoms:

    23÷54=23×45=815\frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}

    The fraction 8/15 cannot be simplified, since 8 and 15 share no factor but 1.

    Watch out: multiplying without flipping gives 23×54=1012=56\frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6}, choice B.

    So the answer is A, 815\frac{8}{15}.

Check your answer

More than 1 cup of flour takes only 2/3 cup of sugar. So 1 cup of flour needs a bit less than 2/3. Since 23=1015\frac{2}{3} = \frac{10}{15}, the answer 8/15 is a bit less. Also, 815×54=4060=23\frac{8}{15} \times \frac{5}{4} = \frac{40}{60} = \frac{2}{3}, which brings back the recipe.

More questions like this

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

Question 20 of 73

2026 Practice Test D Math, question 34

Question

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

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

Correct answer: C

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

Question analysis

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

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

Why it is hard

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

  1. Something you must already know moderate Working it out

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

  2. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Last week Emily:

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

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

  2. Write each chance

    Use last week to find each chance:

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

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

  3. Multiply the two chances

    Both rides must happen, so multiply:

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

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

  4. Avoid the tempting wrong choices

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

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

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

Check your answer

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

More questions like this

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

Question 21 of 73

2024 Test A Math, question 7

Question

If 2yx−y2x=□2x\frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x} and x≠0x \neq 0, what expression is represented by □\square?

  • Ayy
  • B2y2y
  • C3y3y
  • D4y4y
Show the answer

Correct answer: C

Use properties of equations to determine the missing expression:

2yx−y2x=□2x\frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}

22∙2yx−y2x=□2x\frac{2}{2} \bullet \frac{2y}{x} - \frac{y}{2x} = \frac{\square}{2x}

4y2x−y2x=□2x\frac{4y}{2x} - \frac{y}{2x} = \frac{\square}{2x}

4y−y2x=□2x\frac{4y - y}{2x} = \frac{\square}{2x}

3y2x=□2x\frac{3y}{2x} = \frac{\square}{2x}

The missing expression is 3y3y.

Question analysis

An algebra question with fractions: two fractions with x and 2x in their denominators are subtracted, and students find the numerator that makes the difference equal a fraction over 2x. It tests rewriting a fraction over a common denominator when the denominators contain a variable. Adjusting only part of a fraction is the common mistake.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 7.EE.2 rewriting expressions to show relationships
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students subtract fractions with unlike denominators and combine like terms, but rarely see a variable in a denominator. The steps are the same two as with numbers, so the demand is typical once the form is familiar.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 13, 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

This is subtraction of fractions with unlike denominators, written with letters. The box over 2x shows the target denominator, which helps, but each wrong choice matches a familiar fraction error: subtracting the tops before matching the bottoms, or rewriting the first fraction and stopping. Students who rewrite 2y/x as 4y/2x and then subtract y from the new numerator finish in seconds.

  1. Variables instead of numbers moderate Understanding the question

    The denominators are x and 2x, so the common denominator is built with a letter, and there is no number to check unless students substitute values such as x = 1 and y = 2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Only the first fraction changes: 2y/x becomes 4y/2x with top and bottom both doubled, while y/2x already has the denominator the box sits over. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Rewrite the first fraction over 2x by multiplying its top and bottom by 2, then subtract the numerators, which now share a denominator.

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

Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top of the fraction, b is the bottom, and k is any number except zero. Multiplying the top and bottom by the same number does not change the value. It is like multiplying by k over k, which is 1.
Subtracting fractions with the same bottom
ad−bd=a−bd\frac{a}{d} - \frac{b}{d} = \frac{a - b}{d}
a and b are the tops, and d is the shared bottom, called the denominator. When two fractions have the same bottom, their pieces are the same size. So subtract the tops and keep the bottom.

  1. Sort out the question

    The box, □\square, stands for a missing expression. An expression is a math phrase made of numbers and letters, like 5y. You need what goes in the box to make the two sides equal. x≠0x \neq 0 means x is not zero, so no fraction divides by zero.

  2. Pick a shared bottom

    You can only subtract fractions when their bottoms match. The bottoms here are x and 2x. The box sits over 2x, so use 2x for both. The second fraction, y2x\frac{y}{2x}, already has it.

  3. Rewrite the first fraction over 2x

    To turn x into 2x, multiply by 2. Multiply the top by 2 as well, so the value stays the same.

    ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}

    2yx=2y×2x×2=4y2x\frac{2y}{x} = \frac{2y \times 2}{x \times 2} = \frac{4y}{2x}

    Watch out: this is only the first fraction. Stopping here and choosing 4y gives choice D.

  4. Subtract the tops

    ad−bd=a−bd\frac{a}{d} - \frac{b}{d} = \frac{a - b}{d}

    4y2x−y2x=4y−y2x=3y2x\frac{4y}{2x} - \frac{y}{2x} = \frac{4y - y}{2x} = \frac{3y}{2x}

    4y take away y is like 4 apples take away 1 apple. That leaves 3y. Watch out: do not subtract the tops, 2y − y, before the bottoms match. That gives y, choice A.

  5. Match the box

    The left side equals 3y2x\frac{3y}{2x}. The right side is □2x\frac{\square}{2x}. The bottoms are the same, so the tops must be equal. The box is 3y. So the answer is C, 3y3y.

Check your answer

Try numbers. Let x = 1 and y = 2. The left side is 41−22=4−1=3\frac{4}{1} - \frac{2}{2} = 4 - 1 = 3. With 3y in the box, the right side is 3×22×1=62=3\frac{3 \times 2}{2 \times 1} = \frac{6}{2} = 3. Both sides are 3, so 3y works.

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 2025 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 22 of 73

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 23 of 73

2024 Test A Math, question 24

Question

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

  • A$1,440
  • B$14,400
  • C$144,000
  • D$14,400,000,000
Show the answer

Correct answer: B

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

3,600,000,000,000250,000,000=\frac{3{,}600{,}000{,}000{,}000}{250{,}000{,}000} =

3,600,000,000,000250,000,000=\frac{3{,}600{,}00\cancel{0}{,}\cancel{0}\cancel{0}\cancel{0}{,}\cancel{0}\cancel{0}\cancel{0}}{25\cancel{0}{,}\cancel{0}\cancel{0}\cancel{0}{,}\cancel{0}\cancel{0}\cancel{0}} =

360,00025=14,400\frac{360{,}000}{25} = 14{,}400

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

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Dividing large whole numbers and using place value are grade 5 and 6 skills. Grade 7 work goes further with rational numbers; this asks only for one careful division, with counting zeros as the main risk.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 37 words.
Corrected from the DOE handbook
  • A misprinted number in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    The debt has 13 digits and the population 9; cancelling seven zeros from each leaves 360,000 ÷ 25, and losing or adding one zero shifts the answer tenfold. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer minor Choosing the answer

    Three choices share the digits 144 and differ only by powers of ten, so getting the digits right does not settle it; only exact place value does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The debt was about $3,600,000,000,000, which is 3.6 trillion dollars. There were about 250,000,000 people, which is 250 million. You need the debt per person. Per person means the amount for each one person.

  2. Set up the division

    Share the debt equally by dividing it by the number of people. The debt goes on top, and the people go on the bottom.

    per person=total÷number of people\text{per person} = \text{total} \div \text{number of people}

    per person=3,600,000,000,000250,000,000\text{per person} = \frac{3{,}600{,}000{,}000{,}000}{250{,}000{,}000}

  3. Cancel zeros in pairs

    Count the zeros at the end of each number. The debt ends in 11 zeros. The population ends in 7. Cross off 7 zeros from each. Each pair you cross off divides the top and bottom by 10.

    360,00025\frac{360{,}000}{25}

    Watch out: cross off the same number from each. One too many from the debt gives $1,440, choice A. One too few gives $144,000, choice C.

  4. Divide what is left

    Dividing by 25 is the same as dividing by 100 and then multiplying by 4. That is because 4 groups of 25 make 100.

    360,000÷100=3,600360{,}000 \div 100 = 3{,}600

    3,600×4=14,4003{,}600 \times 4 = 14{,}400

    The debt was about $14,400 per person. Watch out: choice D divides by 250 only and forgets the zeros after it. So the answer is B, $14,400.

Check your answer

Multiply back. First, 14,400×25=360,00014{,}400 \times 25 = 360{,}000. Then put back the 7 zeros you crossed off: 3,600,000,000,000. That is the debt, so $14,400 is right.

More questions like this

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

Question 24 of 73

2024 Test A Math, question 30

Question

The reciprocal of 14\frac{1}{4} is added to the reciprocal of 3. What is the reciprocal of this sum?

  • A313\frac{3}{13}
  • B34\frac{3}{4}
  • C45\frac{4}{5}
  • D134\frac{13}{4}
Show the answer

Correct answer: A

The reciprocal of 14\frac{1}{4} is added to the reciprocal of 3.

41+13=123+13=133\frac{4}{1} + \frac{1}{3} = \frac{12}{3} + \frac{1}{3} = \frac{13}{3}

The reciprocal of this sum is 313\frac{3}{13}.

Question analysis

This fractions question asks for the reciprocal of a sum: the reciprocal of one fourth is added to the reciprocal of 3, and students must flip the total. It tests knowing what a reciprocal is and adding a whole number to a fraction. Keeping the three flips in the right order, without skipping one, is what makes it tricky.

NYS standards
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students use reciprocals when dividing fractions, usually one at a time. This item chains three reciprocals around one addition, with positive numbers only, so the demand is following the wording in order rather than new content.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 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 step is easy, so the question is about order: flip each number, add, then flip the total. The common failures are flipping at the wrong moment, term by term instead of the whole sum, and adding a whole number to a fraction straight across. Writing the full expression 1 ÷ (4 + 1/3) before doing any arithmetic keeps the three reciprocals in their places.

  1. Turning words into math moderate Understanding the question

    The two sentences hold three reciprocals and a sum, which must become 1 ÷ (4 + 1/3) before any arithmetic; "this sum" means flipping the total, not each part. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Adding 4 and 1/3 means writing 4 as 12/3 to get 13/3, a whole number plus a fraction, where adding tops and bottoms gives 5/4 instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work in the order of the words: flip each number to get its reciprocal, add the two using a common denominator, and then flip that sum.

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reciprocal
reciprocal of ab=ba\text{reciprocal of } \frac{a}{b} = \frac{b}{a}
a is the top of a fraction and b is the bottom. The reciprocal is the fraction flipped upside down. A whole number such as 3 is the fraction 3 over 1, so its reciprocal is 1 over 3. A number times its reciprocal always makes 1.
Adding fractions with the same bottom
ac+bc=a+bc\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}
c is the shared bottom number, called the denominator. a and b are the tops, called numerators. When the pieces are the same size, you just add how many pieces you have. The size of the pieces, c, stays the same.

  1. Sort out the wording

    The question has three reciprocals in it. A reciprocal is a fraction flipped upside down. Do the jobs in the order of the words:

    1. Flip 14\frac{1}{4}.
    2. Flip 3.
    3. Add the two results to get the sum.
    4. Flip that sum.

    The last flip gives the answer.

  2. Flip each number

    reciprocal of ab=ba\text{reciprocal of } \frac{a}{b} = \frac{b}{a}

    Flip 14\frac{1}{4} to get 41\frac{4}{1}, which is 4. The number 3 is the same as 31\frac{3}{1}, so its reciprocal is 13\frac{1}{3}.

  3. Add the two reciprocals

    You need 4+134 + \frac{1}{3}. To add, both numbers need the same bottom. So write 4 in thirds. One whole is 3 thirds, so 4 wholes are 12 thirds.

    ac+bc=a+bc\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}

    123+13=133\frac{12}{3} + \frac{1}{3} = \frac{13}{3}

    The sum is 133\frac{13}{3}. Watch out: don’t add the tops and the bottoms straight across. That gives 54\frac{5}{4}, which flips to 45\frac{4}{5}, choice C.

  4. Flip the sum

    reciprocal of 133=313\text{reciprocal of } \frac{13}{3} = \frac{3}{13}

    Watch out: flip the whole sum, not each part. Flipping 4 and 13\frac{1}{3} one at a time takes you back to 14+3=134\frac{1}{4} + 3 = \frac{13}{4}, choice D.

    So the answer is A, 313\frac{3}{13}.

Check your answer

A number times its reciprocal is 1. Test it: 133×313=3939=1\frac{13}{3} \times \frac{3}{13} = \frac{39}{39} = 1. Also estimate. The sum is a bit more than 4, so its reciprocal should be a bit less than 14\frac{1}{4}. 313\frac{3}{13} is about 0.23, just under 0.25.

More questions like this

The original DOE question
Question 87 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 87. Open page 83 of the handbook (PDF)

Question 25 of 73

2024 Test A Math, question 31

Question

Nura made a square poster with a side length of 13 inches. Latrice made a square poster with a side length of 15 inches. What is the difference, in square inches, between the area of Latrice’s poster and the area of Nura’s poster?

  • A56
  • B8
  • C4
  • D2
Show the answer

Correct answer: A

Nura’s poster is square, so all the side lengths are 13 inches. The area, in square inches, is

A=s2=132=169A = s^{2} = 13^{2} = 169

Latrice’s poster is square, so all the side lengths are 15 inches.

A=152=225A = 15^{2} = 225

Subtract to determine the difference between the areas.

225−169=56225 - 169 = 56

The difference, in square inches, is 56.

Question analysis

This area word problem compares two square posters, one 13 inches on a side and one 15 inches on a side, and asks for the difference between their areas in square inches. Students square each side length and subtract. Working with the difference in the sides, or with perimeters, instead of the areas is the easy mistake.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.G.1 area of polygons
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Area of a square is elementary content, and squaring 13 and 15 is grade 6 exponent work. Typical grade 7 area problems involve circles, composite shapes or scale drawings; this one has two familiar areas and a subtraction.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 41 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is short, but the question invites a shortcut: the sides differ by 2, so it feels as if the areas should differ by 2 squared. Two choices reward working with the side lengths and one rewards perimeters. Squaring each side before subtracting avoids all three, and a sketch of the 13-inch square inside the 15-inch one shows the gap is an L-shaped strip, far more than 4 square inches.

  1. Something you must already know minor Working it out

    The area of a square, side times side, is not given; students must square 13 and 15 themselves rather than work with the side lengths. (Anderson & Krathwohl, 2001)

  2. Error-prone arithmetic minor Working it out

    Squaring 13 and 15 by hand gives 169 and 225, and the subtraction 225 − 169 needs a borrow, where an arithmetic slip creeps in. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find each poster's area by squaring its side length, then subtract the smaller area from the larger; work with the areas, not the side lengths.

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square
A=s2=s×sA = s^{2} = s \times s
A is the area, the flat space inside the shape. s is the length of one side. All four sides of a square are equal, so the area is side times side. The small 2 means squared, or times itself. Area is measured in square units, here square inches.

  1. Write down what you know

    Nura’s poster is a square with sides of 13 inches. Latrice’s poster is a square with sides of 15 inches. You need the difference between their areas, in square inches. The difference is how much bigger one area is than the other. So find each area, then subtract.

  2. Find the area of Nura’s poster

    Use the area formula with the side of 13 inches.

    A=s2A = s^{2}

    A=132=13×13=169A = 13^{2} = 13 \times 13 = 169

    Nura’s poster has an area of 169 square inches.

  3. Find the area of Latrice’s poster

    Use the same formula with the side of 15 inches.

    A=152=15×15=225A = 15^{2} = 15 \times 15 = 225

    Latrice’s poster has an area of 225 square inches.

  4. Subtract the two areas

    Take the smaller area away from the larger one.

    225−169=56225 - 169 = 56

    The difference is 56 square inches.

    Watch out: don’t subtract the sides first and then square. (15−13)2=22=4(15 - 13)^{2} = 2^{2} = 4, choice C, is not the same as 152−13215^{2} - 13^{2}. The side difference, 2 inches, is choice D, a length. Comparing perimeters, the distance around, gives 8, choice B.

    So the answer is A, 56 square inches.

Check your answer

Picture Nura’s poster laid in a corner of Latrice’s. The extra part is an L-shaped strip 2 inches wide. One arm is 15 by 2 and the other is 13 by 2. 30+26=5630 + 26 = 56, so the strip is 56 square inches. That matches.

More questions like this

The original DOE question
Question 88 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 2025 Admissions, page 84, question 88. 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 26 of 73

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 27 of 73

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 28 of 73

2024 Test A Math, question 45

Question

Misha wants to use ribbon to make 2 straps for a backpack. The ribbon costs $5.00 a yard. If each strap requires 1141\frac{1}{4} yards of ribbon, how much will Misha pay for the ribbon (not including tax)?

  • A$4.00
  • B$6.25
  • C$11.25
  • D$12.50
Show the answer

Correct answer: D

Each strap requires 1141\frac{1}{4} yards of ribbon, and ribbon costs $5.00 a yard.

114×5=54×51=254=6.251\frac{1}{4} \times 5 = \frac{5}{4} \times \frac{5}{1} = \frac{25}{4} = 6.25, so the ribbon for each strap costs $6.25.

Misha wants to make 2 straps.

6.25×2=12.506.25 \times 2 = 12.50, so the cost of the ribbon for 2 straps is $12.50.

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Multiplying a mixed number by a whole number and finding a total cost are grade 5 and 6 skills. With two short steps and friendly numbers, this is easier than typical grade 7 problems with percents or negatives.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

    The length per strap is a mixed number, 1 1/4 yards, so the arithmetic is easiest as 2 × 1.25 = 2.5 yards, then 2.5 × $5.00. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Much to keep track of minor Working it out

    The number of straps comes in the first sentence and the length of each in the last, so the 2 has to be carried through to the cost. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

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

The wrong answers, classified

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 $5.00 for one yard. Number of units is how many you buy, here yards of ribbon. Multiply them to get the total before tax.

  1. Sort out the question

    Misha makes 2 straps. Each strap needs 1141\frac{1}{4} yards of ribbon. A yard is a length of 3 feet. Ribbon costs $5.00 a yard. You need the total cost for both straps, with no tax.

  2. Find the total ribbon

    First change the mixed number to a decimal. One quarter is 0.25, so 1141\frac{1}{4} is 1.25. Then double it for 2 straps.

    2×1.25=2.52 \times 1.25 = 2.5

    Misha needs 2.5 yards of ribbon. Watch out: double the whole mixed number, the 1 and the quarter. Doubling only the 1 gives 2142\frac{1}{4} yards and $11.25, choice C.

  3. Find the cost

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

    cost=5×2.5=12.5\text{cost} = 5 \times 2.5 = 12.5

    The ribbon costs $12.50. Watch out: this is for both straps. One strap’s ribbon costs $6.25, choice B. Don’t stop there.

    So the answer is D, $12.50.

Check your answer

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

More questions like this

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

Question 29 of 73

2024 Test A Math, question 49

Question

Carolyn walked 3 miles from her house to the library and then 2122\frac{1}{2} miles farther to the grocery store. Returning home by the same route, she walked 1231\frac{2}{3} miles before stopping at a friend’s house. How many miles did Carolyn have left to walk home?

  • A3563\frac{5}{6}
  • B4164\frac{1}{6}
  • C4234\frac{2}{3}
  • D7167\frac{1}{6}
Show the answer

Correct answer: A

The distance from Carolyn’s house to the grocery store is 3+212=5123 + 2\frac{1}{2} = 5\frac{1}{2} miles.

Carolyn walked 1231\frac{2}{3} miles back, so she still had 512−1235\frac{1}{2} - 1\frac{2}{3} miles left to walk home.

512−123=112−535\frac{1}{2} - 1\frac{2}{3} = \frac{11}{2} - \frac{5}{3}

336−106=236=356\frac{33}{6} - \frac{10}{6} = \frac{23}{6} = 3\frac{5}{6}

So Carolyn had 3563\frac{5}{6} miles left to walk home.

Question analysis

Carolyn walks 3 miles from home to the library and 2 1/2 miles more to the grocery store, then heads back the same way and stops at a friend's house after 1 2/3 miles. This fractions word problem asks how far she still has to walk home. Students add and subtract mixed numbers, and the subtraction needs regrouping across unlike denominators.

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 word problems with mixed numbers. This one has only positive numbers, but it needs a clear picture of the route and a subtraction that requires regrouping across unlike denominators.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The story needs a quick sketch, but the real test is one mixed-number subtraction in which the second fraction is larger than the first. Many students subtract the fractions the easy way round, which gives a choice close to the right one. Converting both numbers to sixths, 33/6 − 10/6, sidesteps the regrouping, and adding the result to 1 2/3 to get back 5 1/2 is a quick check.

  1. Error-prone arithmetic moderate Working it out

    The subtraction 5 1/2 − 1 2/3 needs sixths, 5 3/6 − 1 4/6, and the fraction on top is the smaller one, so a whole must be regrouped. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    Subtracting the fraction parts the wrong way round gives 4 1/6, only 1/3 away from the answer, so a rough estimate cannot tell the two apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Turning words into math minor Understanding the question

    The route has to be pictured: the walk back starts at the grocery store, 5 1/2 miles from home, not at the library. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Sketch the route as a line from home to the library to the store, find the total distance out, then subtract the distance already walked back, using sixths.

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.
  • 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

Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, b is the top of the fraction, and c is the bottom. Each whole holds c pieces, so a wholes hold a times c pieces. Then add the b extra pieces.
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.
Subtracting fractions with the same bottom
ac−bc=a−bc\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}
c is the shared bottom number. a and b are the tops. When the pieces are the same size, you just take away how many pieces you have. The size of the pieces, c, stays the same.

  1. Picture the route

    Draw a line: home, then the library, then the store. Home to the library is 3 miles. The library to the store is 2122\frac{1}{2} miles more. Carolyn walks back the same way. After 1231\frac{2}{3} miles she stops at a friend’s house. You need the miles still left to get home.

  2. Find the distance from home to the store

    3+212=5123 + 2\frac{1}{2} = 5\frac{1}{2}

    The store is 5125\frac{1}{2} miles from home. She starts back from the store, so this is the full trip home. Watch out: adding all three distances gives 7167\frac{1}{6}, choice D. That is how far she has walked, not how far is left.

  3. Change to improper fractions

    The miles left are 512−1235\frac{1}{2} - 1\frac{2}{3}. An improper fraction has a top bigger than its bottom. Change each mixed number to one.

    abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}

    512=5×2+12=1125\frac{1}{2} = \frac{5 \times 2 + 1}{2} = \frac{11}{2}

    123=1×3+23=531\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3}

  4. Use sixths for both

    To subtract, the fractions need the same bottom. 2 and 3 both go into 6, so change both to sixths.

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

    112=11×32×3=336\frac{11}{2} = \frac{11 \times 3}{2 \times 3} = \frac{33}{6}

    53=5×23×2=106\frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6}

  5. Subtract and simplify

    ac−bc=a−bc\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}

    336−106=236\frac{33}{6} - \frac{10}{6} = \frac{23}{6}

    23 sixths is 3 wholes (18 sixths) and 5 sixths more. So 236=356\frac{23}{6} = 3\frac{5}{6}. Watch out: 12\frac{1}{2} is less than 23\frac{2}{3}. Subtracting the fraction parts the wrong way round gives 4164\frac{1}{6}, choice B.

    So the answer is A, 3563\frac{5}{6} miles.

Check your answer

Add back: 356+123=356+146=496=536=5123\frac{5}{6} + 1\frac{2}{3} = 3\frac{5}{6} + 1\frac{4}{6} = 4\frac{9}{6} = 5\frac{3}{6} = 5\frac{1}{2}. That is the full distance from the store to home. Also estimate: 5.5 minus about 1.7 is about 3.8, and 3563\frac{5}{6} is about 3.83.

More questions like this

The original DOE question
Question 106 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 106. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 30 of 73

2024 Test A Math, question 50

Question

A child grows 1141\frac{1}{4} inches in 13\frac{1}{3} of a year. What would be his yearly growth rate in inches per year?

  • A512\frac{5}{12}
  • B3143\frac{1}{4}
  • C3343\frac{3}{4}
  • D4144\frac{1}{4}
Show the answer

Correct answer: C

Set up a proportion to solve.

114 in.13 yr=x in.1 yr\frac{1\frac{1}{4}\text{ in.}}{\frac{1}{3}\text{ yr}} = \frac{x\text{ in.}}{1\text{ yr}}

114=13x1\frac{1}{4} = \frac{1}{3}x

54=13x\frac{5}{4} = \frac{1}{3}x

(31)(54)=x\left(\frac{3}{1}\right)\left(\frac{5}{4}\right) = x

154=x\frac{15}{4} = x

334=x3\frac{3}{4} = x

Question analysis

A child grows 1 1/4 inches in one third of a year, and this unit rate question asks for his growth rate in inches per year. Students divide a mixed number by a fraction, or scale the growth up to a whole year. Choosing between multiplying and dividing, and handling the whole mixed number, are where the wrong rates come from.

NYS standards
  • 7.RP.1 unit rates with fractions
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Computing a unit rate from a ratio of fractions is the grade 7 standard itself, and this is a typical example: one mixed number over one unit fraction, in two short steps with no extra information.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The idea is simple: three thirds make a year, so the yearly growth is three times the growth in a third. Errors come from the mixed number, where students triple the 1 and leave the 1/4 alone, and from multiplying by 1/3 instead of dividing, which shrinks the rate. A sense check helps, since a yearly rate must be larger than the 1 1/4 inches grown in a third of a year.

  1. Error-prone arithmetic minor Working it out

    Dividing 1 1/4 by 1/3 means multiplying by 3, and the mixed number must be multiplied whole, as 5/4 × 3 = 15/4, not just its 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer minor Choosing the answer

    Multiplying only the whole part gives 3 1/4, which differs from the answer only in the fraction's numerator, so a rough size check cannot rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Divide the growth by the time, writing 1 1/4 as 5/4 and dividing by 1/3 as multiplying by 3; or ask how much he grows in three such periods.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rate
rate=amount÷time\text{rate} = \text{amount} \div \text{time}
The amount is how much something changed, here inches grown. The time is how long it took, here part of a year. Dividing tells how much it changes in 1 unit of time, here 1 whole year.
Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, b is the top of the fraction, and c is the bottom. Each whole holds c pieces, so a wholes hold a times c pieces. Then add the b extra pieces.
Dividing by a fraction
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. This works because dividing by a number undoes multiplying by it.

  1. Sort out the question

    The child grows 1141\frac{1}{4} inches in 13\frac{1}{3} of a year. You need his growth rate in inches per year. That is how much he would grow in 1 whole year at the same speed. A year has 3 thirds, so the answer should be bigger than 1141\frac{1}{4}.

  2. Set up the rate

    Divide the growth by the time to get the inches for 1 whole year.

    rate=amount÷time\text{rate} = \text{amount} \div \text{time}

    rate=114÷13\text{rate} = 1\frac{1}{4} \div \frac{1}{3}

    Change the mixed number to a fraction first.

    abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}

    114=1×4+14=541\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{5}{4}

  3. Divide by multiplying

    To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. The reciprocal of 13\frac{1}{3} is 3.

    ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

    54÷13=54×31=154\frac{5}{4} \div \frac{1}{3} = \frac{5}{4} \times \frac{3}{1} = \frac{15}{4}

    Watch out: multiplying by 13\frac{1}{3} instead gives 512\frac{5}{12}, choice A. That is less than he grew in a third of a year.

  4. Change back to a mixed number

    15 quarters is 3 wholes (12 quarters) and 3 quarters more.

    154=334\frac{15}{4} = 3\frac{3}{4}

    He grows 3343\frac{3}{4} inches per year. Watch out: triple all of 1141\frac{1}{4}, not just the 1. Tripling only the whole number gives 3143\frac{1}{4}, choice B.

    So the answer is C, 3343\frac{3}{4}.

Check your answer

Think of the year as 3 thirds. He grows 1141\frac{1}{4} inches in each third. Add them: 114+114+1141\frac{1}{4} + 1\frac{1}{4} + 1\frac{1}{4}. The wholes make 3, and the quarters make 34\frac{3}{4}. That is 3343\frac{3}{4} inches, the same answer.

More questions like this

The original DOE question
Question 107 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 107. Open page 89 of the handbook (PDF)

Question 31 of 73

2024 Test B Math, question 1

Question

35(2x+5)−2x\frac{3}{5}(2x+5)-2x

After the expression above is simplified, what is the coefficient of x, expressed as a decimal?

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

Show the answer

Correct answer: -0.8

First, distribute 35\frac{3}{5} over (2x+5)(2x+5) to get the expression 65x+3−2x\frac{6}{5}x+3-2x.

Express −2x-2x as −105x-\frac{10}{5}x and then add like terms:

65x+3+(−105x)=−45x+3\frac{6}{5}x+3+\left(-\frac{10}{5}x\right)=-\frac{4}{5}x+3.

The coefficient of x is −45-\frac{4}{5}.

When expressed as a decimal,

−45=−0.8-\frac{4}{5}=-0.8.

Question analysis

An algebra grid-in on simplifying an expression: three fifths times the quantity 2x + 5, minus 2x. Students distribute a fraction, combine the x terms and report the coefficient of x as a decimal. Multiplying by a fraction, the subtraction that follows, and gridding the decimal correctly all leave room for slips.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.2b parts of an expression
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Expanding and simplifying linear expressions with fraction coefficients is a core grade 7 skill, and this is a typical short example. Writing the coefficient as a decimal in the grid adds care but no new idea.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 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 distribution itself is routine; points are lost after it. Students who stop at 6/5x forget the −2x outside the parentheses, and those who combine the x terms correctly can still drop the sign of a negative result. Rewriting 2x as 10/5x keeps the subtraction in fifths, and substituting x = 1 into both forms confirms the coefficient before gridding.

  1. Error-prone arithmetic moderate Working it out

    Multiplying 2x by 3/5 gives 6/5x, and taking away 2x means rewriting it as 10/5x; the result is a negative fraction that must then become a decimal with its sign. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    There are no choices to check against, and the coefficient must be gridded as a decimal with its minus sign, so −4/5 has to become −0.8. (Enright, Morley & Sheehan, 2002)

Strategy: Multiply both terms inside the parentheses by 3/5, rewrite 2x in fifths so the x terms share a denominator, combine them, and convert that fraction to a decimal.

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.
  • 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. A term is a number, a letter, or a number times a letter. To multiply out the parentheses, multiply each term inside by a. It works because a groups of b + c is a groups of b plus a groups of c.
Multiplying a fraction by a whole number
ab×c=a×cb\frac{a}{b} \times c = \frac{a \times c}{b}
a over b is the fraction, and c is the whole number. Multiply only the top number, the numerator, by c. The bottom number, the denominator, stays the same. The size of each piece does not change. You just have more pieces.

  1. Sort out the question

    You start with this expression. An expression is a math phrase with numbers, letters and signs, but no equals sign.

    35(2x+5)−2x\frac{3}{5}(2x + 5) - 2x

    To simplify it means to write it in its shortest form. The coefficient of x is the number multiplied by x. In 7x, the coefficient is 7. You need the coefficient of x, written as a decimal.

  2. Multiply out the parentheses

    The 3/5 multiplies everything inside the parentheses. Use the distributive property.

    a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c

    35(2x+5)=35×2x+35×5\frac{3}{5}(2x + 5) = \frac{3}{5} \times 2x + \frac{3}{5} \times 5

    Multiply only the top of 3/5 by 2: 35×2=65\frac{3}{5} \times 2 = \frac{6}{5}. Also, 35×5=3\frac{3}{5} \times 5 = 3. Now the whole expression is:

    65x+3−2x\frac{6}{5}x + 3 - 2x

    Watch out: do not double the bottom number too. Writing 6/10 leads to −1.4.

  3. Write 2x in fifths

    There are now two x terms: 65x\frac{6}{5}x and −2x-2x. To combine them, they need the same denominator (bottom number). One whole is 5/5, so 2 wholes are 10/5.

    2x=105x2x = \frac{10}{5}x

    Watch out: the −2x is outside the parentheses, but it still counts. Stopping at 6/5 and gridding 1.2 is a common mistake.

  4. Combine the x terms

    Six fifths take away ten fifths leaves negative four fifths.

    65x−105x=−45x\frac{6}{5}x - \frac{10}{5}x = -\frac{4}{5}x

    The 3 has no x, so it stays as it is. The simplified expression is −45x+3-\frac{4}{5}x + 3. So the coefficient of x is −45-\frac{4}{5}. Watch out: 6 − 10 is negative. Dropping the minus sign gives 0.8.

  5. Change it to a decimal

    A fraction means the top divided by the bottom: 4÷5=0.84 \div 5 = 0.8. Keep the minus sign.

    −45=−0.8-\frac{4}{5} = -0.8

    So you would grid -0.8, with the minus sign in its own column.

Check your answer

Pick an easy value, x = 1, and put it into both forms. The first form: 35×7−2=4.2−2=2.2\frac{3}{5} \times 7 - 2 = 4.2 - 2 = 2.2. The simplified form: −0.8+3=2.2-0.8 + 3 = 2.2. Both give 2.2, so the simplifying worked.

More questions like this

The original DOE question
Question 58 as printed on page 198 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 198, question 58. Open page 198 of the handbook (PDF)

Question 32 of 73

2024 Test B Math, question 10

Question

423÷212=4\frac{2}{3}\div 2\frac{1}{2}=

  • A1528\frac{15}{28}
  • B113151\frac{13}{15}
  • C2132\frac{1}{3}
  • D3133\frac{1}{3}
Show the answer

Correct answer: B

423÷212=143÷52=4\frac{2}{3}\div 2\frac{1}{2}=\frac{14}{3}\div\frac{5}{2}=

143×25=2815=11315\frac{14}{3}\times\frac{2}{5}=\frac{28}{15}=1\frac{13}{15}

Question analysis

A fractions question asks for 4 2/3 divided by 2 1/2, with mixed numbers on both sides of the division sign. Students convert to improper fractions, multiply by the reciprocal and simplify. Mixed numbers invite shortcuts, so every step has to follow the rule for dividing fractions.

NYS standards
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Dividing fractions is a grade 6 standard, and this is one division with small mixed numbers. Grade 7 work usually adds negative numbers or a word problem around it, so this sits below grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 16+, 5 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Dividing mixed numbers has a fixed procedure, and every wrong choice is a shortcut that skips part of it. Estimating first helps: 4 2/3 is a little less than twice 2 1/2, so the quotient is a little under 2, which already rules out 15/28 and 3 1/3. The remaining near miss, 2 1/3, comes from dividing by the whole number 2 alone.

  1. Error-prone arithmetic moderate Working it out

    Both mixed numbers must become improper fractions, 14/3 and 5/2, before the second is flipped, and 14/3 × 2/5 = 28/15 leaves an improper fraction to deal with. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The choices are written as mixed numbers and one proper fraction, so 28/15 has to be converted to a mixed number before it can be matched. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite both mixed numbers as improper fractions, multiply the first by the reciprocal of the second, then simplify and turn the improper fraction back into a mixed number.

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

Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number, and b over c is the fraction part. Multiply the whole number by c to turn the wholes into pieces. Then add the b extra pieces. The bottom number stays c.
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.

  1. Sort out the problem

    You need to divide 4234\frac{2}{3} by 2122\frac{1}{2}. Both are mixed numbers, a whole number and a fraction together. Mixed numbers are hard to divide as they are. So first change each one into an improper fraction. That is a fraction whose top is bigger than its bottom.

  2. Change the mixed numbers

    abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}

    423=4×3+23=1434\frac{2}{3} = \frac{4 \times 3 + 2}{3} = \frac{14}{3}

    212=2×2+12=522\frac{1}{2} = \frac{2 \times 2 + 1}{2} = \frac{5}{2}

    Watch out: do not divide the whole parts and the fraction parts on their own. Doing 4÷2=24 \div 2 = 2 and 23÷12=43\frac{2}{3} \div \frac{1}{2} = \frac{4}{3}, then adding, gives 3133\frac{1}{3}, choice D.

  3. Flip the second fraction and multiply

    ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

    143÷52=143×25=2815\frac{14}{3} \div \frac{5}{2} = \frac{14}{3} \times \frac{2}{5} = \frac{28}{15}

    Multiply the tops, 14×2=2814 \times 2 = 28, and the bottoms, 3×5=153 \times 5 = 15. Watch out: flip only the second fraction. Flipping the first one gives 1528\frac{15}{28}, choice A.

  4. Change back to a mixed number

    The choices are mixed numbers, so change 28/15 back. 15 goes into 28 one time, with 13 left over.

    2815=11315\frac{28}{15} = 1\frac{13}{15}

    Watch out: dividing by just 2 and dropping the one half gives 2132\frac{1}{3}, choice C.

    So the answer is B, 113151\frac{13}{15}.

Check your answer

Estimate. 4234\frac{2}{3} is about 4.7, and 2122\frac{1}{2} is 2.5. Two groups of 2.5 make 5, so the answer is a bit less than 2. 113151\frac{13}{15} is about 1.9, which fits. Or multiply back: 2815×52=14030=143\frac{28}{15} \times \frac{5}{2} = \frac{140}{30} = \frac{14}{3}.

More questions like this

The original DOE question
Question 67 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 201, question 67. Open page 201 of the handbook (PDF)

Question 33 of 73

2024 Test B Math, question 12

Question

Lena is building a fence. She will need to dig holes to help support the posts that hold up the fence. The holes need to have a depth of 3133\frac{1}{3} feet below the ground. Each post is 10 feet long. What is the height of the part of the post that is above the ground?

  • A6236\frac{2}{3} feet
  • B7137\frac{1}{3} feet
  • C10 feet
  • D131313\frac{1}{3} feet
Show the answer

Correct answer: A

The posts are each 10 feet long, and 3133\frac{1}{3} feet of each post is below the ground, so the height of the part of the posts that is above the ground is the difference between the two amounts:

10−313=10-3\frac{1}{3}=

101−103=\frac{10}{1}-\frac{10}{3}=

303−103=203=623 feet\frac{30}{3}-\frac{10}{3}=\frac{20}{3}=6\frac{2}{3}\text{ feet}

Question analysis

A fence-building word problem: each post is 10 feet long, and 3 1/3 feet of it goes into a hole in the ground. Students find how much of the post stands above ground by subtracting a mixed number from a whole number. Regrouping the whole number into thirds is the step that trips students up.

NYS standards
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Subtracting a mixed number from a whole number is fraction work from before grade 7, set here in a one-step story. It asks much less than grade 7 problems, which add negative numbers or several steps.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 2, 61 words.
Students, last 30 days
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 situation is plain, a post partly buried in a hole, and the subtraction 10 − 3 1/3 is the whole task. The only real trap is the regrouping, since there are no thirds to take the 1/3 from until one foot is split into three. Checking that the two parts of the post add back to 10 feet catches the slip.

  1. Error-prone arithmetic minor Working it out

    Taking 3 1/3 from the whole number 10 requires regrouping 10 as 9 3/3, the step where students subtract the whole numbers and simply carry the 1/3 along. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the buried part from the whole post, rewriting 10 as 9 and 3/3, or as 30/3, so that the thirds can be taken away.

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.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts of a whole
above=whole−below\text{above} = \text{whole} - \text{below}
The whole is the full length of the post. Below is the part in the hole, and above is the part you can see. The two parts add up to the whole, so subtract to find one part.
One whole in thirds
1=331 = \frac{3}{3}
One whole can be cut into 3 equal thirds. So 1 is the same as 3 over 3. Use this when you need to take thirds away from a whole number.

  1. Sort out the question

    Each post is 10 feet long. Part of it goes into a hole 3133\frac{1}{3} feet deep. The rest sticks up above the ground. You need the height of the part above the ground.

  2. Decide to subtract

    The part below and the part above make up the whole post. So take the part below away from the whole.

    above=whole−below\text{above} = \text{whole} - \text{below}

    above=10−313\text{above} = 10 - 3\frac{1}{3}

    Watch out: do not add. 10+313=131310 + 3\frac{1}{3} = 13\frac{1}{3}, choice D, makes the post longer than it is.

  3. Turn one whole into thirds

    The 10 has no thirds to take 1/3 from. So take 1 whole from the 10 and cut it into 3 thirds.

    1=331 = \frac{3}{3}

    10=9+33=93310 = 9 + \frac{3}{3} = 9\frac{3}{3}

    Now there are thirds to subtract from.

  4. Subtract wholes and thirds

    Subtract the whole numbers, then the thirds.

    933−313=6239\frac{3}{3} - 3\frac{1}{3} = 6\frac{2}{3}

    That is 9−3=69 - 3 = 6 and 33−13=23\frac{3}{3} - \frac{1}{3} = \frac{2}{3}. Watch out: the one third must be taken away too. Doing 10 − 3 = 7 and then keeping the one third gives 7137\frac{1}{3}, choice B.

    So the answer is A, 6236\frac{2}{3} feet.

Check your answer

Add the two parts back together: 623+313=933=106\frac{2}{3} + 3\frac{1}{3} = 9\frac{3}{3} = 10. That is the whole post. Also, the answer must be less than 10 feet, since part of the post is under the ground.

More questions like this

The original DOE question
Question 69 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 69. Open page 202 of the handbook (PDF)

Question 34 of 73

2024 Test B Math, question 22

Question

What decimal is equivalent to −1311-\frac{13}{11}?

  • A−1.18-1.18
  • B−1.118-1.\overline{118}
  • C−1.18-1.\overline{18}
  • D−1.81-1.\overline{81}
Show the answer

Correct answer: C

Calculate the decimal equivalent of −1311-\frac{13}{11}:

−1311=−1.18181818...=−1.18-\frac{13}{11}=-1.18181818...=-1.\overline{18}

Question analysis

A fractions and decimals question asks which decimal equals negative 13/11. The choices differ in which digits repeat and how the repetition is marked. Students divide 13 by 11, spot the block of digits that repeats, and read the bar notation carefully.

NYS standards
  • 7.NS.2d fractions to decimals by long division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Converting a fraction to a repeating decimal by long division is a named grade 7 skill. This is one short division, typical of grade 7 work, with care needed only in reading the bar notation.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 15, 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

Dividing 13 by 11 is quick; the question is really about reading and writing repeating decimals. Three choices differ only in where the bar goes, so students who know the digits run 1.1818… can still pick the wrong bar. Writing out each barred choice for a few digits and comparing it with the quotient is a fast, sure check.

  1. A very close wrong answer moderate Choosing the answer

    The bars over 18 and over 81 cover the same two digits, so only checking which digit comes right after the decimal point, 1 before 8, tells them apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Reading bar notation is assumed: the bar marks exactly the block of digits that repeats, and the whole-number 1 never belongs under it. (Anderson & Krathwohl, 2001)

  3. Error-prone arithmetic minor Working it out

    The long division 13 ÷ 11 has to run until the remainder 2 comes back, which is where the repeating pair is confirmed. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Do the long division of 13 by 11 until a remainder comes back, note which digits repeat from that point, and match that block to the bar in each choice.

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.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • 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

Fraction as division
ab=a÷b\frac{a}{b} = a \div b
a is the top number, the numerator. b is the bottom number, the denominator. A fraction bar means divide. To turn a fraction into a decimal, divide the top by the bottom.
Bar for repeating digits
0.18=0.181818…0.\overline{18} = 0.181818\ldots
The bar sits over the digits that repeat forever, in the order they repeat. The dots mean the pattern goes on and on. Digits that are not under the bar show up only once.

  1. Sort out the question

    You need the decimal equal to −1311-\frac{13}{11}. The fraction is negative, so the decimal will be negative too. Work with 1311\frac{13}{11} first. Then put the minus sign back.

  2. Divide 13 by 11

    ab=a÷b\frac{a}{b} = a \div b

    Use long division for 13÷1113 \div 11.

    • 11 goes into 13 once, with 2 left over. Write 1 and a decimal point.
    • Bring down a 0 to make 20. 11 goes into 20 once, with 9 left over. Write 1.
    • Bring down a 0 to make 90. 11 goes into 90 eight times, with 2 left over. Write 8.

    So far you have 1.18.

  3. Spot the repeating pattern

    The remainder (the amount left over) is 2 again. That is where you were after the first step. So the same digits come back: 1, 8, 1, 8, and so on forever.

    1311=1.181818…\frac{13}{11} = 1.181818\ldots

    Watch out: −1.18, choice A, stops too soon. The digits never end.

  4. Write it with a bar

    0.18=0.181818…0.\overline{18} = 0.181818\ldots

    A bar goes over only the digits that repeat. Here 18 repeats, starting right after the decimal point. The whole number 1 does not repeat, so it is not under the bar. Put the minus sign back.

    −1311=−1.18-\frac{13}{11} = -1.\overline{18}

    Watch out: a bar over 81, choice D, means −1.8181…. A bar over 118, choice B, means −1.118118….

    So the answer is C, −1.18-1.\overline{18}.

Check your answer

Multiply back: 11×1.1818=12.999811 \times 1.1818 = 12.9998. That is almost exactly 13, and more 18s bring it even closer. Also, 13/11 is a bit more than 1, so a decimal between 1 and 2 makes sense.

More questions like this

The original DOE question
Question 79 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 79. Open page 205 of the handbook (PDF)

Question 35 of 73

2024 Test B Math, question 29

Question

Jar Q contains 12 balls, and Jar R contains 8 balls. The probability of drawing a yellow ball at random from Jar Q is 13\frac{1}{3}. The probability of drawing a yellow ball at random from Jar R is 34\frac{3}{4}. All the balls are poured into an empty jar, Jar S. What is the probability of drawing a yellow ball at random from Jar S?

  • A14\frac{1}{4}
  • B13\frac{1}{3}
  • C12\frac{1}{2}
  • D47\frac{4}{7}
Show the answer

Correct answer: C

Jar Q contains 12 balls, and the probability of drawing a yellow ball at random is 13\frac{1}{3}.

13×44=412\frac{1}{3}\times\frac{4}{4}=\frac{4}{12}

There are 4 yellow balls in Jar Q.

Jar R contains 8 balls, and the probability of drawing a yellow ball at random is 34\frac{3}{4}.

34×22=68\frac{3}{4}\times\frac{2}{2}=\frac{6}{8}

There are 6 yellow balls in Jar R.

Jar S contains all the balls from Jar Q and Jar R.

12+8=2012+8=20

There are 20 balls in Jar S.

4+6=104+6=10

There are 10 yellow balls in Jar S.

1020÷1010=12\frac{10}{20}\div\frac{10}{10}=\frac{1}{2}

The probability of drawing a yellow ball at random from Jar S is 12\frac{1}{2}.

Question analysis

Two jars hold different numbers of balls and give different chances of drawing a yellow ball, and then everything is poured into a third jar. Students find the probability of drawing yellow from the combined jar. The tricky part is that the jars are different sizes, so their probabilities cannot be merged without first counting the yellow balls.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 probability tasks usually give the counts and ask for a fraction. Here each jar's probability must be turned back into a count of yellow balls before the jars are pooled, a reverse step that typical classwork rarely asks for.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 6, 66 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light; the hurdle is seeing that chances from jars of different sizes cannot be combined as fractions. Every wrong choice works on 1/3 and 3/4 directly, by multiplying them, by keeping one, or by adding tops and bottoms, while the right path runs through counts of balls. Asking how many yellow balls each jar holds before anything is poured avoids all three.

  1. Turning words into math moderate Understanding the question

    Each jar's probability, 1/3 of 12 balls and 3/4 of 8 balls, has to be turned back into a count of yellow balls, 4 and 6, before the two jars can be combined. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    The straight-across 4/7, about 0.57, sits just above the right 1/2, and both lie between the jars' chances of 1/3 and 3/4, so a rough estimate does not rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Turn each jar's probability into a number of yellow balls by multiplying it by that jar's total, then divide all the yellow balls by all the balls.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P(yellow)=yellow ballsall ballsP(\text{yellow}) = \frac{\text{yellow balls}}{\text{all balls}}
P(yellow) is the probability, or chance, of drawing a yellow ball. It is a number from 0 to 1. The top counts the yellow balls. The bottom counts all the balls. It works because every ball is equally likely to be drawn.

  1. Sort out what you know

    Jar Q holds 12 balls. The probability, or chance, of drawing yellow from it is 13\frac{1}{3}.

    Jar R holds 8 balls. The chance of drawing yellow from it is 34\frac{3}{4}.

    All the balls are poured into Jar S. You need the chance of drawing yellow from Jar S.

  2. Count the yellow balls in Jar Q

    The jars are different sizes, so you can’t mix their chances directly. Count the yellow balls instead. The chance tells what fraction of the balls are yellow. So multiply it by the number of balls.

    yellow balls=13×12=4\text{yellow balls} = \frac{1}{3} \times 12 = 4

    Jar Q has 4 yellow balls.

  3. Count the yellow balls in Jar R

    Do the same for Jar R. Three fourths of its 8 balls are yellow.

    yellow balls=34×8=6\text{yellow balls} = \frac{3}{4} \times 8 = 6

    Jar R has 6 yellow balls.

  4. Count the balls in Jar S

    Jar S gets every ball from both jars. Add all the balls. Then add the yellow balls.

    12+8=2012 + 8 = 20

    4+6=104 + 6 = 10

    Jar S has 20 balls, and 10 of them are yellow.

  5. Find the chance for Jar S

    Use the probability formula with the Jar S counts.

    P(yellow)=1020=12P(\text{yellow}) = \frac{10}{20} = \frac{1}{2}

    Watch out: Adding the tops and the bottoms, 1 + 3 over 3 + 4, gives 47\frac{4}{7}, choice D. Those numbers are parts of fractions, not balls. Multiplying 13×34\frac{1}{3} \times \frac{3}{4} gives 14\frac{1}{4}, choice A. That is the chance of two yellow draws, one from each jar.

    So the answer is C, 12\frac{1}{2}.

Check your answer

Half of the 20 balls is 10, and 4 + 6 = 10 yellow balls. That matches. Also, the mixed jar’s chance should land between the two jars’ chances, 13\frac{1}{3} and 34\frac{3}{4}. One half does.

More questions like this

The original DOE question
Question 86 as printed on page 207 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 207, question 86. Open page 207 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 36 of 73

2024 Test B Math, question 31

Question

If y=4xy=4x, what is the value of 3y+2(3y+5)−x3y+2(3y+5)-x in terms of x?

  • A8x+108x+10
  • B32x+1032x+10
  • C35x+1035x+10
  • D36x+1036x+10
Show the answer

Correct answer: C

First, substitute 4x4x for each y in the expression:

3y+2(3y+5)−x=3y+2(3y+5)-x=

3(4x)+2[3(4x)+5]−x3(4x)+2[3(4x)+5]-x

Then simplify the expression:

3(4x)+2[3(4x)+5]−x3(4x)+2[3(4x)+5]-x

12x+2(12x+5)−x12x+2(12x+5)-x

12x+24x+10−x12x+24x+10-x

12x+24x−x+1012x+24x-x+10

35x+1035x+10

The simplified expression is 35x+1035x+10.

Question analysis

Given that y equals 4x, students rewrite the expression 3y + 2(3y + 5) − x using only x. It tests substitution, the distributive property and combining like terms. Errors creep in when a term is dropped or when x terms and y terms are merged before y has been replaced.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Substituting an expression for a variable and simplifying with the distributive property is standard grade 7 work. Putting the substitution inside parentheses calls for care, but the steps are routine for students who combine like terms fluently.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Substitution and distribution are routine here, so points are lost to bookkeeping rather than method. The expression has three x pieces, 12x, 24x and a lone − x, and the last is easy to overlook after the larger terms. The other two wrong choices come from collecting 3y, 6y and x as if they were like terms, a reason to replace y with 4x before simplifying anything.

  1. A very close wrong answer moderate Choosing the answer

    Every choice ends in + 10, and 36x + 10 differs from 35x + 10 by a single x, so only remembering the last − x in the expression separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    The substitution sits inside nested grouping, 2[3(4x) + 5], so the 2 must reach both the 12x and the 5 before the x terms are collected. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Replace every y with 4x in parentheses before doing anything else, then distribute the 2 and collect the x terms and the constant separately.

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
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. Terms are the parts joined by plus or minus signs. Multiply a by each term inside. It works because 2 groups of (12x + 5) hold 2 of the 12x and 2 of the 5.

  1. Sort out the question

    y=4xy = 4x means y is always 4 times x. You need to rewrite the expression with no y in it. That is what “in terms of x” means. The answer will use only x and numbers.

  2. Replace each y with 4x

    Put 4x in parentheses everywhere you see y. The parentheses keep the multiplying clear.

    3y+2(3y+5)−x=3(4x)+2(3(4x)+5)−x3y + 2(3y + 5) - x = 3(4x) + 2(3(4x) + 5) - x

    3(4x) means 3 times 4x, which is 12x. It shows up twice.

    =12x+2(12x+5)−x= 12x + 2(12x + 5) - x

  3. Multiply out the parentheses

    Use the distributive property. The 2 must multiply both the 12x and the 5.

    2(12x+5)=2×12x+2×5=24x+102(12x + 5) = 2 \times 12x + 2 \times 5 = 24x + 10

    The whole expression is now:

    12x+24x+10−x12x + 24x + 10 - x

  4. Combine the like terms

    Like terms have the same letter part. Here the x terms are 12x, 24x and −x. The −x means take away one x.

    12x+24x−x=35x12x + 24x - x = 35x

    The 10 has no x, so it stays as it is. The result is 35x+1035x + 10.

    Watch out: Don’t forget the −x at the end. Stopping at 12x + 24x gives 36x + 10, choice D. Also, swap in 4x before you combine. Adding 3y, 6y and −x as if they were all y’s mixes unlike terms. It leads to choice B.

    So the answer is C, 35x+1035x + 10.

Check your answer

Test a number. Let x = 1, so y = 4. The original expression gives 3(4)+2(3×4+5)−1=12+34−1=453(4) + 2(3 \times 4 + 5) - 1 = 12 + 34 - 1 = 45. Your answer gives 35(1)+10=4535(1) + 10 = 45. They match. Choice D would give 46.

More questions like this

The original DOE question
Question 88 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 208, question 88. 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 37 of 73

2024 Test B Math, question 32

Question

Two circular plates have the dimensions shown.

Two circles side by side. The larger circle has a dashed diameter labeled 18 in., and the smaller circle has a dashed diameter labeled 12 in.

What is the difference between the areas of the two plates, in square inches?

  • A6π6\pi
  • B9π9\pi
  • C45π45\pi
  • D180π180\pi
Show the answer

Correct answer: C

First, determine the area of each circle using the formula A=πr2A=\pi r^{2}:

A circle with a diameter of 18 inches has a radius of 9 inches, so the area is π(9)2=81π\pi(9)^{2}=81\pi.

A circle with a diameter of 12 inches has a radius of 6 inches, so the area is π(6)2=36π\pi(6)^{2}=36\pi.

The difference between the two areas is 81π−36π=45π81\pi-36\pi=45\pi.

Question analysis

Two circular plates are drawn with their diameters marked in inches, and students find the difference between their areas in square inches, in terms of pi. It tests the area of a circle. The diameters, not the radii, are marked, so a step comes before the area formula, which the test does not provide.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding a circle's area is a grade 7 standard, and this asks for two areas and their difference. It matches typical classwork: the formula must be recalled and each diameter halved first, with no further twist.
Estimated difficulty
Straightforward: two steps or one careful reading.
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

Two area calculations and a subtraction, kept in terms of π, make this routine for students who know the area formula. Points are lost by using the labeled diameters as radii, by subtracting circumferences, or by subtracting the radii before squaring, and each of these has a listed answer. Writing both areas, 81π and 36π, before subtracting avoids all three.

  1. Something you must already know moderate Working it out

    The area formula A = πr² is not given, and the figure labels each plate by its diameter, so 18 and 12 must be halved to radii of 9 and 6 first. (Anderson & Krathwohl, 2001)

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

    The dashed lines run all the way across each circle, which marks them as diameters; reading the 18 in. and 12 in. as radii leads to a listed choice. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Halve each diameter to get the radius, find each plate's area as pi times the radius squared, and subtract the smaller area from the larger.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • 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.
  • 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

Area of a circle
A=πr2A = \pi r^2
A is the area, the flat space inside the circle. r is the radius, the distance from the center to the edge. It is half the diameter. π (pi) is a fixed number, about 3.14. r² means r × r. Use it for the space inside any circle.

  1. Sort out the question

    Each dashed line goes all the way across its circle, through the center. A line like that is a diameter. The big plate has a diameter of 18 inches. The small plate has a diameter of 12 inches.

    You need the difference between the two areas. That means the big area minus the small area, in square inches.

  2. Find each radius

    The area formula uses the radius, not the diameter. The radius is half the diameter.

    18÷2=918 \div 2 = 9

    12÷2=612 \div 2 = 6

    The radii are 9 inches and 6 inches. Watch out: If you use 18 and 12 as the radii, you get 180π180\pi, choice D.

  3. Find the big plate’s area

    A=πr2=π×92=π×81=81πA = \pi r^2 = \pi \times 9^2 = \pi \times 81 = 81\pi

    The big plate’s area is 81π81\pi square inches. Keep π as a symbol, since all the choices use π.

  4. Find the small plate’s area

    A=π×62=π×36=36πA = \pi \times 6^2 = \pi \times 36 = 36\pi

    The small plate’s area is 36π36\pi square inches. Squaring means multiplying a number by itself, so 62=6×6=366^2 = 6 \times 6 = 36.

  5. Subtract the two areas

    81π−36π=45π81\pi - 36\pi = 45\pi

    Subtract the numbers in front of π, just as 81 apples minus 36 apples is 45 apples. The difference is 45π45\pi square inches.

    Watch out: Don’t subtract the radii first. Squaring 9 − 6 = 3 gives 9π9\pi, choice B. Subtracting the distances around the plates, 18π−12π18\pi - 12\pi, gives 6π6\pi, choice A. That uses the wrong formula.

    So the answer is C, 45π45\pi.

Check your answer

Estimate with π about 3. The big plate is about 3×81=2433 \times 81 = 243 square inches. The small one is about 3×36=1083 \times 36 = 108. The difference is about 135, and 45×3=13545 \times 3 = 135. It matches.

More questions like this

The original DOE question
Question 89 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 208, question 89. Open page 208 of the handbook (PDF)

Question 38 of 73

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 39 of 73

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 40 of 73

2023 Test A Math, question 3

Question

Solve for x:

7x+3−2(2x+1)=137x + 3 - 2(2x + 1) = 13

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

Show the answer

Correct answer: 4

7x+3−2(2x+1)=137x + 3 - 2(2x + 1) = 13 Apply the distributive property; multiply the −2 by 2x2x and 1.
7x+3−4x−2=137x + 3 - 4x - 2 = 13 Combine like terms.
3x+1=133x + 1 = 13 Apply the additive inverse property; subtract 1 from both sides of the equation.
3x=123x = 12 Apply the multiplicative inverse property; divide both sides of the equation by 3.
x=4x = 4

Question analysis

This grid-in gives the equation 7x + 3 − 2(2x + 1) = 13 and asks students to solve for x. It tests the distributive property with a negative factor, combining like terms and solving a two-step equation. The minus sign in front of the parentheses has to reach both terms inside, a step many students rush.

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 expand expressions with negative factors and solve equations such as p(x + q) = r. This equation needs both skills plus combining like terms, which is routine work with one negative sign that is easy to lose.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
4 steps
Reading level
Question and choices: about grade 3, 4 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a standard multi-step linear equation, and the work is short once the parentheses are cleared. Everything rests on distributing −2 correctly; the later steps, combining 7x with −4x and 3 with −2, then undoing the + 1 and the × 3, are routine. Substituting 4 back into the left side, 28 + 3 − 18 = 13, is a fast check that catches a sign slip.

  1. Error-prone arithmetic moderate Working it out

    The −2 in front of (2x + 1) must reach both terms, giving −4x − 2; the second product, −2 × 1, is the one most often written as +2 or left as +1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    As a grid-in with no choices, a value such as 3 or 8/3 from a distributing slip looks just as finished as the right one; only substituting back exposes it. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute −2 to both terms inside the parentheses, combine the x terms and the constants, then undo the addition and the multiplication in turn.

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. A term is a number, a letter, or a number times a letter. Multiply each term inside by a. If a is negative, every product gets its sign too.
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 equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

  1. Sort out the equation

    You need the value of x that makes 7x+3−2(2x+1)=137x + 3 - 2(2x + 1) = 13 true. x is the unknown number. The tricky part is the −2 in front of the parentheses. It multiplies everything inside them.

  2. Multiply out the parentheses

    Use the distributive property with a = −2, b = 2x and c = 1.

    −2(2x+1)=(−2)×2x+(−2)×1=−4x−2-2(2x + 1) = (-2) \times 2x + (-2) \times 1 = -4x - 2

    So the equation becomes this:

    7x+3−4x−2=137x + 3 - 4x - 2 = 13

    Watch out: the −2 must multiply the 1 too. Leaving + 1 gives x = 3. Writing + 2 instead of − 2 gives x=83x = \frac{8}{3}.

  3. Combine like terms

    Like terms are terms of the same kind. Put the x terms together. Put the plain numbers together.

    7x−4x=3x7x - 4x = 3x

    3−2=13 - 2 = 1

    3x+1=133x + 1 = 13

  4. Get x by itself

    Undo the + 1 first. Subtract 1 from both sides, so the equation stays balanced.

    3x+1−1=13−13x + 1 - 1 = 13 - 1

    3x=123x = 12

    3x means 3 times x. So divide both sides by 3.

    x=12÷3=4x = 12 \div 3 = 4

    Watch out: don’t stop at 3x = 12. 12 is three times the answer. So you would grid 4.

Check your answer

Put x = 4 back into the first equation: 7×4+3−2(2×4+1)=28+3−2×9=31−18=137 \times 4 + 3 - 2(2 \times 4 + 1) = 28 + 3 - 2 \times 9 = 31 - 18 = 13. That matches the 13 on the right side, so x = 4 is right.

More questions like this

The original DOE question
Question 60 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 60. Open page 75 of the handbook (PDF)

Question 41 of 73

2023 Test A Math, question 6

Question

34+74=3^{4} + 7^{4} =

  • A40
  • B370
  • C2,482
  • D10,000
Show the answer

Correct answer: C

34+74=81+2,401=2,4823^{4} + 7^{4} = 81 + 2{,}401 = 2{,}482

Question analysis

This exponents question asks for the sum of 3 to the fourth power and 7 to the fourth power. Students evaluate each power separately and then add. The traps are treating an exponent as a multiplier and adding the bases before raising them to a power, and the larger power takes careful multiplication.

NYS standards
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Evaluating whole-number powers is grade 6 work, and this question asks only for two of them and a sum. The one demand beyond routine practice is computing a fourth power of 7 accurately without a calculator.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 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

The sum is short, but it tests exactly what an exponent means and when to apply it. The wrong choices come from three misreadings of the notation: treating the power as a multiplier, using one factor too few, and adding the bases before raising them. A size check settles it: 7⁴ alone is 49 × 49, a little under 2,500, so 40 and 370 are far too small and 10,000 far too large.

  1. Error-prone arithmetic moderate Working it out

    7⁴ means 7 × 7 × 7 × 7 = 2,401, a four-digit product built by hand, most easily as 49 × 49; 3⁴ = 81 is quick by comparison. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    An exponent counts repeated factors rather than acting as a multiplier, so 3⁴ is 3 × 3 × 3 × 3 = 81, not 3 × 4 = 12; the item assumes that meaning. (Anderson & Krathwohl, 2001)

Strategy: Compute each power on its own, squaring and then squaring again, and add the two results only at the end.

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

A power
a4=a×a×a×aa^{4} = a \times a \times a \times a
a is the base, the number being multiplied. The small 4 is the exponent. It tells how many times the base is used as a factor, a number being multiplied. So the exponent counts factors. It does not mean multiply by 4.
Powers come before adding
a4+b4≠(a+b)4a^{4} + b^{4} \neq (a + b)^{4}
a and b are any two positive numbers. The sign ≠ means is not equal to. Powers come before adding. So raise each number to its power first, then add. Adding the bases first gives a much bigger number.

  1. Sort out the question

    You need the value of 34+743^{4} + 7^{4}. Each number has a small 4 above it. This 4 is an exponent. Work out each power on its own. Then add the two results.

  2. Find 3 to the fourth power

    3⁴ means four 3s multiplied together. Pair them up to make it easier.

    34=3×3×3×3=9×9=813^{4} = 3 \times 3 \times 3 \times 3 = 9 \times 9 = 81

    Watch out: 3⁴ is not 3 × 4 = 12. Doing that to both powers gives 12 + 28 = 40, choice A.

  3. Find 7 to the fourth power

    7⁴ means four 7s multiplied together. Pair them up the same way.

    74=7×7×7×7=49×49=2,4017^{4} = 7 \times 7 \times 7 \times 7 = 49 \times 49 = 2{,}401

    To find 49 × 49, do 49 × 50 = 2,450. Then take away one 49. That gives 2,401. Watch out: count four 7s, not three. Three 7s make 343. Using three factors in both powers gives 370, choice B.

  4. Add the two powers

    81+2,401=2,48281 + 2{,}401 = 2{,}482

    Watch out: don’t add 3 and 7 first. That gives (3+7)4=104=10,000(3 + 7)^{4} = 10^{4} = 10{,}000, choice D. Powers come before adding. So the answer is C, 2,482.

Check your answer

Estimate. 7⁴ is 49 × 49, a little less than 50 × 50 = 2,500. Adding 81 keeps it near 2,500. Only 2,482 is close. 40 and 370 are far too small, and 10,000 is far too big.

More questions like this

The original DOE question
Question 63 as printed on page 77 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 63. Open page 77 of the handbook (PDF)

Question 42 of 73

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 43 of 73

2023 Test A Math, question 11

Question

8110=9n\frac{81}{10} = \frac{9}{n}

What value of n makes the equation above true?

  • A1
  • B1191\frac{1}{9}
  • C5
  • D101910\frac{1}{9}
Show the answer

Correct answer: B

8110=9n\frac{81}{10} = \frac{9}{n} Cross multiply.
81n=9(10)81n = 9(10) Apply the multiplicative inverse property; divide both sides of the equation by 81.
n=9081=1981=119n = \frac{90}{81} = 1\frac{9}{81} = 1\frac{1}{9} Then simplify the fraction.

Question analysis

This question gives the proportion 81/10 = 9/n and asks for the value of n that makes it true. Students solve a proportion by cross-multiplying and dividing, then simplify the result. Reducing the resulting fraction and matching its form to the choices is where students slip.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Solving a proportion is familiar from grades 6 and 7, and this one needs only cross-multiplying and one division. Its one extra demand is simplifying the result and writing it in the form the choices use.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 7, 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

The proportion solves in one cross-multiplication, but the result is an improper fraction that must be simplified and rewritten to match. Students who notice that 81 is 9 × 9 can shortcut it: the numerator shrinks by a factor of 9 from left to right, so the denominator must shrink too, giving 10 ÷ 9. The mixed-number form is where answers drift, by dropping the leftover ninth or by writing 10 × 1/9 as 10 1/9.

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

    Cross-multiplying gives n = 90/81, which has to be reduced to 10/9 and rewritten as the mixed number 1 1/9 before it matches any choice. (Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer minor Choosing the answer

    The choices 1 and 1 1/9 differ by one ninth, and estimating 90 ÷ 81 as about 1 cannot separate them; only keeping the remainder does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Cross-multiply to get 81n = 90, divide both sides by 81, and simplify the fraction before comparing it with the choices.

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.
  • 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

Cross-multiplying
ab=cd, so a×d=b×c\frac{a}{b} = \frac{c}{d} \text{, so } a \times d = b \times c
a over b and c over d are two equal fractions, and b and d are not 0. Multiply each top by the other bottom. The two products are equal. It works because it is the same as multiplying both sides by b and by d.
Simplifying a fraction
a×kb×k=ab\frac{a \times k}{b \times k} = \frac{a}{b}
a over b is a fraction, and k is a number that divides both the top and the bottom. Dividing the top and the bottom by the same number gives an equal fraction with smaller numbers.
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

    The equation says two fractions are equal: 8110\frac{81}{10} and 9n\frac{9}{n}. n is an unknown bottom number. You need the value of n that makes them equal. Some choices are mixed numbers, so n may not be a whole number.

  2. Cross-multiply the two fractions

    Multiply each top by the other fraction’s bottom. This gets n out of the bottom of the fraction.

    8110=9n\frac{81}{10} = \frac{9}{n}

    81×n=10×981 \times n = 10 \times 9

    81n=9081n = 90

  3. Divide both sides by 81

    81n means 81 times n. To undo multiplying by 81, divide both sides by 81.

    n=9081n = \frac{90}{81}

    Now simplify. 90 and 81 are both in the 9 times table, so divide the top and bottom by 9.

    9081=90÷981÷9=109\frac{90}{81} = \frac{90 \div 9}{81 \div 9} = \frac{10}{9}

  4. Write it as a mixed number

    10 ÷ 9 is 1 with 1 left over. The leftover stays over 9.

    109=119\frac{10}{9} = 1\frac{1}{9}

    Watch out: keep the leftover. Dropping it gives 1, choice A. Also, ten ninths is not 10 and one ninth. That mixed number, 101910\frac{1}{9}, is choice D, which is far too big. So the answer is B, 1191\frac{1}{9}.

Check your answer

Estimate. 8110\frac{81}{10} is about 8. For 9n\frac{9}{n} to be about 8, n must be a little more than 1. Check exactly: 9÷109=9×910=81109 \div \frac{10}{9} = 9 \times \frac{9}{10} = \frac{81}{10}. It matches.

More questions like this

The original DOE question
Question 68 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 68. Open page 78 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 44 of 73

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 45 of 73

2023 Test A Math, question 26

Question

Which statement must be true if x is a whole number greater than or equal to 1?

  • A1x+1>1x+2\frac{1}{x + 1} > \frac{1}{x + 2}
  • B1x+1<1x+2\frac{1}{x + 1} < \frac{1}{x + 2}
  • C1x+1−1x+2>1\frac{1}{x + 1} - \frac{1}{x + 2} > 1
  • D1x+1−1x+2>1x\frac{1}{x + 1} - \frac{1}{x + 2} > \frac{1}{x}
Show the answer

Correct answer: A

Let x=2x = 2, and substitute that value for x in each inequality to determine which one is correct:

Option A:

13>14\frac{1}{3} > \frac{1}{4}. This is true.

Option B:

13<14\frac{1}{3} < \frac{1}{4}. This is false.

Option C:

13−14>1\frac{1}{3} - \frac{1}{4} > 1, or 112>1\frac{1}{12} > 1. This is false.

Option D:

13−14>12\frac{1}{3} - \frac{1}{4} > \frac{1}{2}, or 112>12\frac{1}{12} > \frac{1}{2}. This is false.

Option A is the only true inequality.

Question analysis

This question asks which of four inequalities must be true when x is a whole number of at least 1. The inequalities compare fractions with denominators x + 1 and x + 2, or their difference. Students reason about how the size of a fraction changes as its denominator grows, and must be sure a statement holds for every allowed value.

NYS standards
  • 6.NS.7a inequality as relative position
  • 6.EE.5 solutions and substitution
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Comparing fractions is familiar, but these fractions have variable denominators and the question asks what must be true for every whole number from 1 up. That abstraction makes it as demanding as typical grade 7 work, though the arithmetic is light.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 6, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Choosing a small value, such as x = 1 or 2, and testing every statement settles this quickly, because three of the four fail for any value tried. The trap is the belief that a bigger denominator makes a bigger fraction, which turns the true comparison around. Students unsure of fraction subtraction lose time on the two difference statements, though both differences are tiny, never more than 1/6.

  1. Variables instead of numbers moderate Understanding the question

    All four statements are about the unknown x, and the bold must means each has to hold for every whole number from 1 up, not just for a single test value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    The statements are four separate inequalities, two comparisons and two differences, so each has to be tested; one counterexample is enough to rule a statement out. (Embretson & Wetzel, 1987)

  3. Error-prone arithmetic minor Working it out

    Testing the differences means subtracting unlike fractions, such as 1/3 − 1/4 = 1/12 when x = 2, and comparing the result with 1 and with 1/x. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute a small whole number such as 1 or 2 for x in all four statements, rule out any that fail, and confirm the survivor by thinking about larger denominators.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions
if 0<a<b, then 1a>1b\text{if } 0 < a < b \text{, then } \frac{1}{a} > \frac{1}{b}
a and b are positive numbers, and b is bigger. One over a means one whole cut into a equal pieces. Cutting the same whole into more pieces makes each piece smaller. So the fraction with the bigger bottom is smaller.
Subtracting fractions
1a−1b=b−aa×b\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}
a and b are the bottoms of the two fractions. Rewrite both fractions over the common bottom a × b. Then subtract the tops. The bottom stays a × b.

  1. Sort out the question

    x is a whole number that is 1 or more, so x can be 1, 2, 3 and so on. You need the statement that must be true. That means it is true for every such x. One x that makes a statement false rules it out.

  2. Test x = 1 in each statement

    With x = 1, x + 1 is 2 and x + 2 is 3.

    • A: 12>13\frac{1}{2} > \frac{1}{3}. True.
    • B: 12<13\frac{1}{2} < \frac{1}{3}. False.
    • C: 12−13=36−26=16\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}, which is not more than 1. False.
    • D: 16\frac{1}{6} is not more than 11\frac{1}{1}. False.

    Only A is true when x = 1. So B, C and D are out.

  3. Make sure A is always true

    Must means A has to work for every x, not just 1. In 1x+1\frac{1}{x + 1} and 1x+2\frac{1}{x + 2}, both tops are 1. The bottom x + 2 is always 1 more than x + 1. Cutting a whole into more pieces makes each piece smaller. So A is always true. Watch out: a bigger bottom makes a smaller fraction, not a bigger one. Thinking the other way gives choice B.

  4. See why C and D always fail

    Use the subtracting rule with a = x + 1 and b = x + 2.

    1a−1b=b−aa×b\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}

    1x+1−1x+2=(x+2)−(x+1)(x+1)(x+2)=1(x+1)(x+2)\frac{1}{x + 1} - \frac{1}{x + 2} = \frac{(x + 2) - (x + 1)}{(x + 1)(x + 2)} = \frac{1}{(x + 1)(x + 2)}

    The bottom is at least 2 × 3 = 6, so the difference is at most 16\frac{1}{6}. That is less than 1. Its bottom is also bigger than x, so it is less than 1x\frac{1}{x}. Watch out: keep the bottom when you subtract. Keeping only the top, 1, leads to choice D. So the answer is A.

Check your answer

Try a bigger number, x = 9. A tenth of a pizza is more than an eleventh, so 110>111\frac{1}{10} > \frac{1}{11}. The difference, 1110\frac{1}{110}, is tiny. It is less than 1 and less than 19\frac{1}{9}. A still holds, and C and D still fail.

More questions like this

The original DOE question
Question 83 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 83. Open page 82 of the handbook (PDF)

Question 46 of 73

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 47 of 73

2023 Test A Math, question 52

Question

Vicente and Carla each ran 8 laps around a track. They started at the same time and place. If Vicente ran 1.5 times as fast as Carla, how many laps did Carla have left to finish when Vicente finished his 8th lap?

  • A2122\frac{1}{2}
  • B2232\frac{2}{3}
  • C3133\frac{1}{3}
  • D5135\frac{1}{3}
Show the answer

Correct answer: B

Vicente ran 1.5 times as fast as Carla, so that means for every lap Carla finished, Vicente finished 1.5 laps. Set up a proportion to calculate c, the number of laps Carla had completed when Vicente finished 8 laps:

c1=81.5\frac{c}{1} = \frac{8}{1.5} Cross multiply.
1.5c=8(1)1.5c = 8(1) Apply the multiplicative inverse property; divide both sides of the equation by 1.5.
c=513c = 5\frac{1}{3}

So, the number of laps Carla still had to finish is 8−513=2238 - 5\frac{1}{3} = 2\frac{2}{3}.

Question analysis

This ratio word problem says Vicente and Carla each ran 8 laps, starting at the same time and place, and Vicente ran 1.5 times as fast. It asks how many laps Carla still had to run when Vicente finished. Students reason with a speed ratio and mixed numbers, and giving the laps run rather than the laps remaining is an easy slip.

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 rate problems usually give a speed and a time. Here only the ratio of two speeds is given, so students must reason that distances run in the same time share that ratio, then work with a fractional number of laps.
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, 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

No time is given, so students must reason that in equal time distance scales with speed, which turns the race into one division and one subtraction. Both steps hide a trap: the division by 1.5 leaves a remainder that is easy to misread, and subtracting a mixed number from 8 needs regrouping. The result of the first step, Carla's laps run, is itself a choice, so the question also rewards rereading what is asked.

  1. Turning words into math moderate Understanding the question

    The phrase “1.5 times as fast” has to become distances in the same time: while Vicente runs 8 laps, Carla runs 8 ÷ 1.5, not 8 − 1.5 or 8 × 1.5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic moderate Working it out

    Dividing 8 by 1.5 gives 16/3, or 5 1/3, a division by a decimal with a remainder, and 8 − 5 1/3 then needs a whole regrouped into thirds. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. A very close wrong answer minor Choosing the answer

    The choice 2 1/2 is only a sixth of a lap from 2 2/3, so a rough estimate of how far Carla trails cannot separate them; only the exact quotient does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: In the same amount of time the faster runner covers 1.5 times the distance, so divide Vicente's laps by 1.5 to get Carla's, then subtract that from 8.

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 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

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. When two runners run for the same time, the faster one goes farther by the same factor as the speed. So 1.5 times as fast means 1.5 times as far.
Dividing by a fraction
a÷mn=a×nma \div \frac{m}{n} = a \times \frac{n}{m}
a is any number and m over n is a fraction. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down.

  1. Sort out the question

    Vicente and Carla each run 8 laps, starting together. Vicente runs 1.5 times as fast as Carla. You need how many laps Carla still has to run when Vicente finishes. First find how far Carla has run at that moment.

  2. Compare their distances

    d=r×td = r \times t

    They have run for the same time. So Vicente’s distance is 1.5 times Carla’s. Let c be the laps Carla has run.

    8=1.5×c8 = 1.5 \times c

    To find c, divide 8 by 1.5.

  3. Divide 8 by 1.5

    Write 1.5 as the fraction 32\frac{3}{2}.

    a÷mn=a×nma \div \frac{m}{n} = a \times \frac{n}{m}

    c=8÷32=8×23=163=513c = 8 \div \frac{3}{2} = 8 \times \frac{2}{3} = \frac{16}{3} = 5\frac{1}{3}

    Carla has run 5135\frac{1}{3} laps. Watch out: that is how far she has run, not what is left. Stopping here gives choice D.

  4. Find the laps she has left

    8−513=243−163=83=2238 - 5\frac{1}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3} = 2\frac{2}{3}

    Watch out: 8 − 5135\frac{1}{3} is not 3133\frac{1}{3}. You must break 1 whole into thirds first. Answering 3133\frac{1}{3} is choice C.

    So the answer is B, 2232\frac{2}{3} laps.

Check your answer

Multiply back: 1.5×513=32×163=81.5 \times 5\frac{1}{3} = \frac{3}{2} \times \frac{16}{3} = 8. So when Carla has run 5135\frac{1}{3} laps, Vicente has run 8. And 513+223=85\frac{1}{3} + 2\frac{2}{3} = 8, her full race.

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 2024 Admissions, page 89, question 109. Open page 89 of the handbook (PDF)

Question 48 of 73

2023 Test B Math, question 8

Question

Solve for x in terms of y if 2x+4=y2x + 4 = y.

  • Ax=y−42x = \frac{y - 4}{2}
  • Bx=2(y+4)x = 2(y + 4)
  • Cx=y+42x = \frac{y + 4}{2}
  • Dx=2(4−y)x = 2(4 - y)
Show the answer

Correct answer: A

2x+4=y2x + 4 = y Apply the additive inverse property; subtract 4 from both sides of the equation.
2x=y−42x = y - 4 Apply the multiplicative inverse property; divide both sides of the equation by 2.
x=y−42x = \frac{y - 4}{2}

Question analysis

Given the equation 2x + 4 = y, students solve for x in terms of y. This literal-equation question offers four choices that mix multiplying and dividing by 2 with adding and subtracting 4. Undoing the two operations on x in the right order, with the right inverse each time, is what it tests.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.EE.4a two-step equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve two-step equations such as 2x + 4 = 10. Here the right side is the letter y, so the same two inverse steps leave an expression, a first taste of the grade 8 skill of rearranging.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 13 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The two inverse steps are routine, subtract 4 and then divide by 2, but the letters hide whether they were done correctly. The main hazard is moving the 4 across the equals sign while keeping its plus sign, which produces an expression nearly identical to the right one. Substituting x = 1 and y = 6 tests every choice in seconds.

  1. A very close wrong answer moderate Choosing the answer

    (y − 4)/2 and (y + 4)/2 differ only in the sign before 4, and that sign depends on undoing the +4 by subtracting it from both sides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Variables instead of numbers minor Understanding the question

    With y on the right instead of a number, the answer is an expression, so there is no value to check unless the student substitutes a pair such as x = 1 and y = 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Treat y as if it were a number and undo the operations on x in reverse order, using the inverse of each, then check by substituting a value.

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.
  • 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

Keeping an equation balanced
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 0 when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it balanced. Use this to undo, one at a time, each step that was done to x.

  1. Sort out the question

    The equation is 2x+4=y2x + 4 = y. “Solve for x in terms of y” means get x alone on one side. The other side will have y in it. Treat y as if it were a number.

  2. Undo the adding of 4

    Right now, x is doubled and then 4 is added. Undo those steps in reverse order. First, subtract 4 from both sides.

    2x+4−4=y−42x + 4 - 4 = y - 4

    2x=y−42x = y - 4

    Watch out: the 4 must be subtracted from y. Writing 2x=y+42x = y + 4 leads to choice C.

  3. Undo the doubling

    Next, divide both sides by 2. Divide the whole right side, not only the y.

    2x2=y−42\frac{2x}{2} = \frac{y - 4}{2}

    x=y−42x = \frac{y - 4}{2}

  4. Give the answer

    Watch out: choices B and D multiply by 2. But x was multiplied by 2, so you undo that by dividing. So the answer is A, x=y−42x = \frac{y - 4}{2}.

Check your answer

Pick a value, such as x = 1. Then y=2×1+4=6y = 2 \times 1 + 4 = 6. Put y = 6 into choice A: 6−42=1\frac{6 - 4}{2} = 1. That gives back x = 1. Choice C would give 6+42=5\frac{6 + 4}{2} = 5, which is wrong.

More questions like this

The original DOE question
Question 65 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 65. Open page 193 of the handbook (PDF)

Question 49 of 73

2023 Test B Math, question 9

Question

Bryan completed a 100-meter race in 11.74 seconds. Luis completed the same race in 11.69 seconds. What was the difference between their times, expressed as a fraction of a second?

  • A120\frac{1}{20}
  • B110\frac{1}{10}
  • C320\frac{3}{20}
  • D15\frac{1}{5}
Show the answer

Correct answer: A

First, subtract Luis’s time from Bryan’s time: 11.74−11.69=0.0511.74 - 11.69 = 0.05.

Then, convert to a fraction:

0.05=5100=1200.05 = \frac{5}{100} = \frac{1}{20}.

Question analysis

Bryan ran a 100-meter race in 11.74 seconds and Luis ran the same race in 11.69 seconds. Students find the difference between their times and express it as a fraction of a second. It tests subtracting decimals and rewriting a decimal as a fraction in lowest terms.

NYS standards
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Subtracting decimals and writing a decimal as a fraction are skills from before grade 7. The two times are close, so the subtraction is quick, and the only care needed is with place value and simplifying.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 5, 32 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two quick skills settle this question: lining up the decimals to subtract, and naming the result as a fraction in lowest terms. The two times differ only in their last two digits, so the subtraction is short, but it needs a borrow that some students skip. The fraction step is where a decimal’s digit gets mistaken for a denominator.

  1. Error-prone arithmetic minor Working it out

    Subtracting 11.69 from 11.74 needs a borrow in the hundredths, 14 − 9, after which the tenths cancel to 0, the step where place value goes wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    The difference, 0.05, must be read as five hundredths and reduced as a fraction; the question gives no help with converting a decimal. (Anderson & Krathwohl, 2001)

Strategy: Line up the decimal points, subtract, write the difference as a number of hundredths over 100, and reduce the fraction.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths
a decimal with two places=its digits100\text{a decimal with two places} = \frac{\text{its digits}}{100}
The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths, and you can write its digits over 100. For example, 0.37 is 37/100. Then simplify by dividing the top and bottom by the same number.

  1. Sort out the question

    Bryan ran the race in 11.74 seconds. Luis ran it in 11.69 seconds. The difference is how much longer one time is than the other. You need it as a fraction of a second.

  2. Subtract the two times

    Subtract the smaller time from the larger one. Line up the decimal points.

    11.74−11.69=0.0511.74 - 11.69 = 0.05

    In the hundredths place, 4 is less than 9. So regroup: trade 1 tenth for 10 hundredths. Then 14 − 9 = 5 hundredths. The tenths are now 6 − 6 = 0.

    Watch out: don’t just take the smaller digit from the larger one in each place. That gives 0.15, which is 320\frac{3}{20}, choice C.

  3. Write the difference as a fraction

    The 5 is in the hundredths place, so 0.05 is 5 hundredths.

    0.05=51000.05 = \frac{5}{100}

    Now simplify. Divide the top and bottom by 5.

    5÷5100÷5=120\frac{5 \div 5}{100 \div 5} = \frac{1}{20}

  4. Give the answer

    Watch out: 0.05 is not 15\frac{1}{5}. One fifth is 0.2, which is four times as big. So the answer is A, 120\frac{1}{20}.

Check your answer

Turn 120\frac{1}{20} back into a decimal: 1÷20=0.051 \div 20 = 0.05. Then add it to Luis’s time: 11.69+0.05=11.7411.69 + 0.05 = 11.74. That is Bryan’s time, so the difference is right.

More questions like this

The original DOE question
Question 66 as printed on page 193 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 66. Open page 193 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 50 of 73

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 51 of 73

2023 Test B Math, question 24

Question

A prism is made up of 9 congruent cubes. Each of the cubes has side lengths measuring 4 units. What is the volume, in cubic units, of the prism?

  • A108
  • B144
  • C480
  • D576
Show the answer

Correct answer: D

First, determine the volume of each congruent cube. All the side lengths of the cubes are equal, so the length, width, and height of each cube is 4 units. To determine the volume of each cube, multiply length×width×height\text{length} \times \text{width} \times \text{height}:

4×4×4=644 \times 4 \times 4 = 64

So, the volume of each cube is 64 cubic units.

Left: a prism built from 9 cubes, three rows of three, one cube deep, with one cube’s height, width and depth each labeled 4. Right: a single cube whose height, width and depth are each labeled 4.

The prism is comprised of 9 of these cubes, so multiply the volume of each cube (64) by the total number of cubes (9) to determine the volume, in cubic units, of the prism:

64×9=57664 \times 9 = 576

Question analysis

A prism is built from 9 identical cubes, and each cube has edges 4 units long. Students find the volume of the prism in cubic units. It tests the volume of a cube and how volumes add when solids are put together, and it rewards keeping area, surface area and volume apart.

NYS standards
  • 6.G.2 volume with fractional edge lengths
  • 6.G.5 perfect squares and perfect cubes
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Volume of cubes and rectangular prisms is grade 5 and 6 content. This item needs one cube's volume and a multiplication by 9, two short steps, so it is below typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 3, 27 words.
Corrected from the DOE handbook
  • A letter x printed in place of a multiplication sign in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question needs one formula and one multiplication, and the stem does not even say how the cubes are arranged, because the volume does not depend on it. The wrong choices come from mishandling the cube’s volume: the exponent multiplied instead of applied, the area of one face, or a surface area. Writing the volume out as edge times edge times edge, in cubic units, rules all three out.

  1. Something you must already know minor Working it out

    A cube’s volume, edge × edge × edge, is not given; with it, each 4-unit cube holds 64 cubic units, and the arrangement of the 9 cubes does not matter. (Anderson & Krathwohl, 2001)

Strategy: Find the volume of one cube by multiplying its three edges, then multiply by the number of cubes; how the cubes are arranged does not matter.

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

What was corrected from the DOE handbook, and why

  • Explanation: The handbook gives the volume of a cube as “length x width x height”, with the letter x where it means “times”. In a math explanation the letter x reads as a variable, so the site writes length × width × height, the sign the handbook itself uses on the next line (4 × 4 × 4 = 64). Nothing else changes: each cube is 64 cubic units, and the 9 cubes make 64 × 9 = 576.

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 a solid, in cubic units. s is the length of one edge. All the edges of a cube are equal, so its length, width and height are all s. Multiplying them counts how many unit cubes fit inside.

  1. Sort out the question

    A prism, a solid with flat faces, is made of 9 congruent cubes. Congruent means exactly the same size and shape. Each cube has sides of 4 units. You need the volume of the whole prism, in cubic units.

    The question doesn’t say how the cubes are stacked. You don’t need to know. Volume is the space inside, and 9 cubes fill the same space however they are arranged.

  2. Find the volume of one cube

    V=s×s×sV = s \times s \times s

    V=4×4×4=64V = 4 \times 4 \times 4 = 64

    Each cube has a volume of 64 cubic units. Watch out: 434^3 means 4×4×44 \times 4 \times 4, not 4×3=124 \times 3 = 12. That slip leads to 108, choice A. Also, 4×4=164 \times 4 = 16 is only the area of one face. It leads to 144, choice B.

  3. Multiply by the number of cubes

    The prism is 9 of these cubes, so its volume is 9 times as big.

    64×9=57664 \times 9 = 576

    The prism’s volume is 576 cubic units.

  4. Give the answer

    Watch out: volume counts the space inside, not the outside faces. Suppose the cubes formed one flat 3-by-3 layer. Adding up the areas of its outside faces gives 480, choice C. That is surface area, not volume. So the answer is D, 576.

Check your answer

Estimate: 64×10=64064 \times 10 = 640, and one cube fewer is 640−64=576640 - 64 = 576. You can also test one arrangement. A flat 3-by-3 layer is 12 by 12 by 4 units, and 12×12×4=57612 \times 12 \times 4 = 576. Same answer.

More questions like this

The original DOE question
Question 81 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 81. Open page 198 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 52 of 73

2023 Test B Math, question 29

Question

Which number could be the sum of 4 consecutive integers?

  • A−12
  • B−10
  • C−8
  • D−4
Show the answer

Correct answer: B

The sum of 4 consecutive integers can be expressed as:

n+(n+1)+(n+2)+(n+3)n+(n+1)+(n+2)+(n+3)

=4n+6=4n+6

Set that expression equal to each of the options and solve for n:

Option A:

4n+6=−124n+6=-12

4n=−184n=-18

n=−184n=-\frac{18}{4} is not an integer.

Option B:

4n+6=−104n+6=-10

4n=−164n=-16

n=−4n=-4 is an integer.

Option C:

4n+6=−84n+6=-8

4n=−144n=-14

n=−144n=-\frac{14}{4} is not an integer.

Option D:

4n+6=−44n+6=-4

4n=−104n=-10

n=−104n=-\frac{10}{4} is not an integer.

The only option for which n is an integer is −10.

Question analysis

A number properties question about consecutive integers: which of four negative numbers could be the sum of 4 consecutive integers? Students write the sum with a variable, or add runs of negative integers, and test each choice. Keeping signs straight with negative numbers, and remembering that the integers must be whole numbers, is the tricky part.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 consecutive-integer problems usually give the sum and ask for the integers. This one reverses the task: students test which negative total is possible at all, using the structure of the sum rather than one equation.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 9 words.
Corrected from the DOE handbook
  • A stray word in the explanation’s last sentence was removed.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Three of the four choices are multiples of 4, and that is the whole trap: a student who pictures the sum as four times some integer finds three choices that seem to work. The sum of four consecutive integers is 4n + 6, so it is always even but never a multiple of 4. Trying the most natural run, −4, −3, −2 and −1, lands on −10 at once and is the fastest check.

  1. Every choice must be checked moderate Choosing the answer

    The stem asks which number “could be” the sum, so no single equation gives the answer; each of −12, −10, −8 and −4 has to be tested against the form a four-integer sum must take. (Embretson & Wetzel, 1987)

  2. No obvious method minor Working it out

    Nothing points to the shortcut that n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 is always 2 more than a multiple of 4, or to simply adding a run such as −1, −2, −3 and −4. (Schoenfeld, 1985)

  3. Error-prone arithmetic minor Working it out

    Every choice is negative, so solving 4n + 6 = −12 or adding a run of integers means subtracting 6 from a negative number and dividing a negative by 4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the integers as n, n + 1, n + 2 and n + 3, add them, and set the sum equal to each choice to see which one gives a whole-number n.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook ends with “The only option for which n is an integer answer is −10.” The word “answer” does not belong in that sentence; the site reads “The only option for which n is an integer is −10.” The working above it is unchanged: only 4n + 6 = −10 gives an integer, n = −4, and −4 + (−3) + (−2) + (−1) = −10.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of 4 consecutive integers
n+(n+1)+(n+2)+(n+3)=4n+6n + (n + 1) + (n + 2) + (n + 3) = 4n + 6
n is the smallest of the four integers. Each next integer is 1 more than the one before, so they are n, n + 1, n + 2 and n + 3. Adding them gives four n’s, plus 1 + 2 + 3, which is 6. Use it to add any four integers in a row.

  1. Sort out the question

    Integers are whole numbers and their negatives, like −2, 0 and 5. They have no fractions or decimals. Consecutive integers come one right after another, like 7, 8, 9, 10 or −3, −2, −1, 0.

    You need the choice that can be the sum of four consecutive integers. The sum is what you get when you add them.

  2. Write the sum with a letter

    You don’t know the integers yet. So call the smallest one n. Each next one is 1 more: n + 1, n + 2 and n + 3. Use the formula. Collect the four n’s, then add the numbers.

    n+n+n+n+1+2+3=4n+6n + n + n + n + 1 + 2 + 3 = 4n + 6

    Any four consecutive integers add up to 4n + 6.

  3. Test each choice

    Set 4n + 6 equal to each choice and solve for n. First subtract 6 from both sides. Then divide both sides by 4.

    Choice4n + 64nn
    A−12−18−4.5
    B−10−16−4
    C−8−14−3.5
    D−4−10−2.5

    n must be an integer. Only choice B gives one: n = −4.

  4. Name the four integers

    Start at n = −4 and count up by 1. The integers are −4, −3, −2 and −1.

    Watch out: Choices A, C and D are multiples of 4. A multiple of 4 is 4 times an integer, like 4 × (−3) = −12. It is tempting to think four integers must add to one of those. But the sum is 4n + 6, and the extra 6 gets in the way. Choice D works only with n = −2.5, and that is not an integer.

    So the answer is B, −10.

Check your answer

Add the integers: −4+(−3)+(−2)+(−1)=−10-4 + (-3) + (-2) + (-1) = -10. That is choice B. Also, 4n + 6 is always 2 more than a multiple of 4. And −10 is 2 more than −12, which is 4 × (−3).

More questions like this

The original DOE question
Question 86 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 86. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 53 of 73

2023 Test B Math, question 32

Question

A number line with arrows at both ends and three points marked on it. From left to right, point P is at 1 1/3, point Q is at 3, and point R is at 5 1/6.

On the number line above, how many units longer is QR\overline{QR} than PQ\overline{PQ}?

  • A16\frac{1}{6}
  • B12\frac{1}{2}
  • C23\frac{2}{3}
  • D3
Show the answer

Correct answer: B

First, find the lengths of the given segments:

QR=516−3=216\mathrm{QR}=5\frac{1}{6}-3=2\frac{1}{6}

PQ=3−113=123\mathrm{PQ}=3-1\frac{1}{3}=1\frac{2}{3}

Then, find the difference:

216−123=2\frac{1}{6}-1\frac{2}{3}=

136−53=\frac{13}{6}-\frac{5}{3}=

136−106=\frac{13}{6}-\frac{10}{6}=

36=12\frac{3}{6}=\frac{1}{2}

Question analysis

A number line shows three points labeled with mixed numbers and a whole number, and students find how many units longer one segment is than another. It tests subtracting mixed numbers with unlike denominators to find distances. Regrouping when the fraction being subtracted is the larger one is where most errors happen.

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 the points are mixed numbers with different denominators, so finding two distances and their difference makes it a typical but careful grade 7 task.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is three subtractions, so most errors are in the fractions rather than the plan. Taking 1 1/3 from a whole number such as 3 tempts students to subtract only the whole parts and keep the 1/3, which reverses which segment looks longer. Rewriting every coordinate in sixths, 8/6, 18/6 and 31/6, turns the work into whole-number subtraction of numerators.

  1. Error-prone arithmetic moderate Working it out

    Both lengths subtract mixed numbers with different denominators, 5 1/6 − 3 and 3 − 1 1/3, and the second needs regrouping before the two results are compared in sixths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The three coordinates come from the labels under P, Q and R, and the segments QR and PQ must each be measured from Q at 3, the point they share. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find each length by subtracting the left endpoint from the right one, writing the mixed numbers in sixths, then subtract the shorter length from the longer.

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.
  • 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

Length on a number line
length=right point−left point\text{length} = \text{right point} - \text{left point}
The length of a segment is the distance between its two end points. Subtract the number on the left from the number on the right. Subtracting tells how many units you move to get from one point to the other.
Mixed number to improper fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part. b over c is the fraction part, with b on top and c on the bottom. Each whole holds c parts, so a wholes hold a × c parts. Then add the b extra parts. An improper fraction has a top bigger than its bottom.

  1. Read the number line

    Point P is at 1131\frac{1}{3}. Point Q is at 3. Point R is at 5165\frac{1}{6}.

    The bar over QR means segment QR, the piece of the line from Q to R. You need how much longer QR is than PQ. So find both lengths, then subtract.

  2. Find the length of QR

    R is on the right, and Q is on the left. Use the length rule.

    QR=516−3=216\text{QR} = 5\frac{1}{6} - 3 = 2\frac{1}{6}

    QR is 2162\frac{1}{6} units long.

  3. Find the length of PQ

    Q is on the right, and P is on the left.

    PQ=3−113\text{PQ} = 3 - 1\frac{1}{3}

    Take it in two parts. First, 3 − 1 = 2. Then take away the extra 13\frac{1}{3}. That gives 2−13=1232 - \frac{1}{3} = 1\frac{2}{3}. So PQ is 1231\frac{2}{3} units long.

    Watch out: Don’t just do 3 − 1 and keep the 13\frac{1}{3}. That gives 2132\frac{1}{3}, which is too long. It leads to choice A.

  4. Write both lengths in sixths

    The fractions have different bottoms, 6 and 3. Change both lengths to sixths so you can subtract. Use the mixed number rule.

    216=2×6+16=1362\frac{1}{6} = \frac{2 \times 6 + 1}{6} = \frac{13}{6}

    123=1×3+23=53=1061\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3} = \frac{10}{6}

  5. Subtract the lengths

    Take the shorter length away from the longer one.

    136−106=36=12\frac{13}{6} - \frac{10}{6} = \frac{3}{6} = \frac{1}{2}

    QR is 12\frac{1}{2} unit longer than PQ.

    So the answer is B, 12\frac{1}{2}.

Check your answer

Write every point in sixths. P is 86\frac{8}{6}, Q is 186\frac{18}{6} and R is 316\frac{31}{6}. QR is 31 − 18 = 13 sixths. PQ is 18 − 8 = 10 sixths. The difference is 3 sixths, which is one half.

More questions like this

The original DOE question
Question 89 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 89. Open page 200 of the handbook (PDF)

Question 54 of 73

2023 Test B Math, question 38

Question

How is 1x−12x\frac{1}{x}-\frac{1}{2x} written as a single fraction? (Assume x≠0x\neq 0.)

  • A12x\frac{1}{2x}
  • B13x\frac{1}{3x}
  • C−1x-\frac{1}{x}
  • D−12x-\frac{1}{2x}
Show the answer

Correct answer: A

1x−12x=22x−12x=12x\frac{1}{x}-\frac{1}{2x}=\frac{2}{2x}-\frac{1}{2x}=\frac{1}{2x}

Question analysis

This algebra question asks students to write 1/x minus 1/(2x) as a single fraction, assuming x is not zero. It tests finding a common denominator when the denominators contain a variable. Getting both fractions over the same denominator before subtracting, and keeping track of the sign, are what matter.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 7.EE.2 rewriting expressions to show relationships
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students subtract fractions with unlike denominators and work with linear expressions, but seldom see a variable in a denominator. The steps match those for numerical fractions, so it is a short task once the common denominator is found.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 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

This is subtraction of unlike fractions with a letter in the bottom, and every wrong choice is a familiar fraction error: combining the denominators straight across by adding or subtracting, or subtracting the numerators the wrong way round. Substituting x = 1 makes it 1 − 1/2 = 1/2, which rules out the negative choices and 1/3 at once.

  1. Variables instead of numbers moderate Understanding the question

    The denominators are x and 2x, so the common denominator is built with a letter, and no number checks the result unless students substitute a value such as x = 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Only the first fraction changes: 1/x becomes 2/2x, and then the subtraction 2 − 1 must be done in the order written, first numerator minus second. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Rewrite 1/x with the denominator 2x by multiplying its top and bottom by 2, then subtract the numerators and keep the common denominator.

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.
  • 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

Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top, or numerator. b is the bottom, or denominator. k is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, because k over k is 1.
Subtracting fractions with the same bottom
ac−bc=a−bc\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}
a and b are the tops. c is the shared bottom. When the bottoms match, the pieces are the same size. So subtract the tops and keep the bottom.

  1. Sort out the question

    You need to write 1x−12x\frac{1}{x} - \frac{1}{2x} as one fraction. The problem says x is not 0. That matters because you can’t divide by 0.

  2. Find a common bottom

    To subtract fractions, their bottoms must match. 2x is 2 times x. So multiply the top and bottom of 1x\frac{1}{x} by 2.

    1x=1×2x×2=22x\frac{1}{x} = \frac{1 \times 2}{x \times 2} = \frac{2}{2x}

    Now both fractions have the bottom 2x.

  3. Subtract the tops

    Subtract the tops in order. Keep the bottom.

    22x−12x=2−12x=12x\frac{2}{2x} - \frac{1}{2x} = \frac{2 - 1}{2x} = \frac{1}{2x}

    Watch out: Never add or subtract the bottoms. Adding them gives 13x\frac{1}{3x}, choice B. Subtracting them gives −1x-\frac{1}{x}, choice C. And keep the order: 2 − 1, not 1 − 2. The wrong order gives choice D.

    So the answer is A, 12x\frac{1}{2x}.

Check your answer

Try x = 1. The problem becomes 1−12=121 - \frac{1}{2} = \frac{1}{2}. Choice A gives 12×1=12\frac{1}{2 \times 1} = \frac{1}{2}. It matches. Choices B, C and D give 13\frac{1}{3}, −1 and −12-\frac{1}{2}.

More questions like this

The original DOE question
Question 95 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 95. Open page 202 of the handbook (PDF)

Question 55 of 73

2023 Test B Math, question 42

Question

1 gallon = 16 cups

1 cup = 8 fluid ounces

If 1 gallon = 2x2^{x} fluid ounces, what is the value of x?

  • A4
  • B7
  • C8
  • D64
Show the answer

Correct answer: B

If 1 gallon is equal to 16 cups, and 1 cup is equal to 8 fluid ounces, then

1 gallon=1\text{ gallon}=

(16)(8 fluid ounces)=(16)(8\text{ fluid ounces})=

128 fluid ounces128\text{ fluid ounces}

128=27128=2^{7}

1 gallon=27 fluid ounces1\text{ gallon}=2^{7}\text{ fluid ounces}

So, x=7x=7.

Question analysis

This question gives two facts, 16 cups in a gallon and 8 fluid ounces in a cup, and says a gallon equals 2 to the power x fluid ounces. Students find the exponent x. It combines a unit conversion with writing a number as a power of 2, and the exponent is easy to confuse with a multiplier.

NYS standards
  • 6.RP.3d converting measurement units
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Converting units with given facts is grade 6 work, and this conversion is a single multiplication. The twist is writing the result as a power of 2, which calls for fluency with powers that grade 7 work rarely asks for.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The conversion is one multiplication, so the question really tests whether students can name the exponent for 128. Stopping at the cups, where 16 = 2⁴, or treating 2 to the power x as 2 times x gives a listed choice. Writing 16 as 2⁴ and 8 as 2³ and adding the exponents is the quickest route, and doubling from 1 seven times is a sure check.

  1. Combines ideas minor Working it out

    A unit conversion, 16 cups × 8 fluid ounces = 128 fluid ounces, has to be joined to exponents by recognizing 128 as a power of 2. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Something you must already know minor Working it out

    Students must know or build the powers of 2 up to 128, and see that the exponent counts factors of 2, not the number of terms in a list starting at 1. (Anderson & Krathwohl, 2001)

Strategy: Convert 1 gallon to fluid ounces by multiplying the two facts, then keep halving the result, or keep doubling from 2, counting how many factors of 2 it takes.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Meaning of an exponent
2x=2×2×⋯×22^x = 2 \times 2 \times \cdots \times 2
The 2 is the base, the number being multiplied. x is the exponent, the small raised number. It tells how many 2s are multiplied together. For example, 2 to the third power means 2 × 2 × 2, which is 8.
Multiplying powers of the same base
2a×2b=2a+b2^a \times 2^b = 2^{a + b}
a and b are exponents. The first power has a twos multiplied together. The second has b more twos. Multiplying puts all the twos together, so you add the exponents.

  1. Sort out the question

    A gallon is 16 cups. A cup is 8 fluid ounces. A gallon is also 2x2^x fluid ounces. You need x, the exponent. It tells how many 2s are multiplied together.

  2. Change a gallon to fluid ounces

    Each of the 16 cups holds 8 fluid ounces. So multiply.

    16×8=12816 \times 8 = 128

    One gallon is 128 fluid ounces.

  3. Write 128 as a power of 2

    Now you need 2x=1282^x = 128. Keep multiplying by 2, and count the 2s.

    Number of 2s1234567
    Product248163264128

    It takes seven 2s to make 128.

    27=1282^7 = 128

    So x = 7.

  4. Give the answer

    Watch out: A gallon is 16 cups, and 16=2416 = 2^4. Stopping there gives 4, choice A. And 2x2^x does not mean 2 times x. Solving 2 × x = 128 gives 64, choice D.

    So the answer is B, 7.

Check your answer

Use the exponent rule. 16 is 242^4 and 8 is 232^3. So 16×8=24×23=24+3=2716 \times 8 = 2^4 \times 2^3 = 2^{4 + 3} = 2^7. The exponent is 7 again.

More questions like this

The original DOE question
Question 99 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 99. Open page 203 of the handbook (PDF)

Question 56 of 73

2023 Test B Math, question 46

Question

03+40+2−1+(−1)20^{3}+4^{0}+2^{-1}+(-1)^{2}

What is the value of the expression above, in decimal form?

  • A0
  • B0.5
  • C1.5
  • D2.5
Show the answer

Correct answer: D

03+40+2−1+(−1)20^{3}+4^{0}+2^{-1}+(-1)^{2}

=0+1+12+1=0+1+\frac{1}{2}+1

=212=2.5=2\frac{1}{2}=2.5

Question analysis

An exponents question: students find the decimal value of a sum of four powers, zero cubed, four to the zero power, two to the negative first power and negative one squared. Each of the four terms follows a different rule, and a slip on any one of them changes the total.

NYS standards
  • 8.EE.1 integer exponent properties
  • 7.NS.2a rational number multiplication
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 work uses whole-number exponents and signed numbers. Zero and negative exponents are grade 8 content, and this expression tests both alongside a negative base, each with its own common misconception.
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, 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

Every term tests a different exponent rule, and each wrong choice comes from breaking exactly one of them. The zero exponent is the most common casualty, since 4⁰ looks as if it should be 0 like 0³; a negative exponent read as a negative number is close behind. Writing each term’s value above it before adding, 0, 1, 1/2 and 1, keeps the four rules separate.

  1. Something you must already know major Working it out

    Three rules the test does not give decide the sum: any nonzero number to the 0 power is 1, 2 to the −1 power is 1/2, and (−1)² is positive 1. (Anderson & Krathwohl, 2001)

  2. Error-prone arithmetic minor Working it out

    The four terms carry a zero base, a zero exponent, a negative exponent and a negative base, so a sign or a stray 0 can slip in at each one. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Evaluate each power on its own, using the rules for a zero exponent and a negative exponent and the sign of a squared negative number, then add the four values.

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

Zero exponent
a0=1a^0 = 1
a is any number except 0. An exponent of 0 always gives 1. It works because each step down in the exponent divides by a. For example, 4 to the first power is 4, and 4 ÷ 4 = 1.
Negative exponent
a−n=1ana^{-n} = \frac{1}{a^n}
a is the base, and n is a positive whole number. A negative exponent does not make the number negative. It means 1 divided by the base to the positive power. For example, 2 to the −1 power is one half.

  1. Sort out the question

    The expression adds four powers. A power is a base with an exponent, a small raised number. The exponent tells how many times to use the base as a factor, a number you multiply. Find each power, then add.

  2. Find the first two powers

    Zero times itself is still zero.

    03=0×0×0=00^3 = 0 \times 0 \times 0 = 0

    Next, use the zero exponent rule.

    40=14^0 = 1

    Watch out: 404^0 is 1, not 0. Taking it as 0 gives choice C.

  3. Find the negative power

    Use the negative exponent rule.

    2−1=121=12=0.52^{-1} = \frac{1}{2^1} = \frac{1}{2} = 0.5

    Watch out: 2−12^{-1} is not −2. That mistake gives choice A.

  4. Find the last power

    The base is −1, inside the parentheses. Multiply it by itself. A negative times a negative is positive.

    (−1)2=(−1)×(−1)=1(-1)^2 = (-1) \times (-1) = 1

    Watch out: The answer is not −1. That mistake gives choice B.

  5. Add the four values

    Now add them. The question wants a decimal, so write one half as 0.5.

    0+1+0.5+1=2.50 + 1 + 0.5 + 1 = 2.5

    So the answer is D, 2.5.

Check your answer

See why 40=14^0 = 1 with a pattern. 42=164^2 = 16 and 41=44^1 = 4. Each step down divides by 4, so 40=4÷4=14^0 = 4 \div 4 = 1. In the same way, 20=12^0 = 1 and 2−1=1÷2=122^{-1} = 1 \div 2 = \frac{1}{2}.

More questions like this

The original DOE question
Question 103 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 103. Open page 204 of the handbook (PDF)

Question 57 of 73

2023 Test B Math, question 53

Question

Micha is playing a game with five cards numbered 1 through 5. He will place the cards in a bag and draw one card at random three times, replacing the card each time. To win a prize, he must draw the number 5 all three times. What is the probability he will draw the number 5 all three times?

  • A1125\frac{1}{125}
  • B3125\frac{3}{125}
  • C115\frac{1}{15}
  • D35\frac{3}{5}
Show the answer

Correct answer: A

Each time Micha draws a card, the probability of drawing the number 5 is 15\frac{1}{5}. To find the probability of drawing the number 5 three times, multiply:

15×15×15=1125\frac{1}{5}\times\frac{1}{5}\times\frac{1}{5}=\frac{1}{125}

Question analysis

Micha draws a card three times from five cards numbered 1 to 5, putting the card back each time, and wins only if he draws a 5 every time. Students find the probability that he wins. It tests the probability of independent events, where the chances for the separate draws have to be combined correctly.

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
At grade 7 Finding the probability of a compound event is a grade 7 standard, and repeated draws with replacement are a standard case. Three equal factors make this a short, typical grade 7 task.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 3, 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

With replacement each draw is the same 1 in 5, so the work is one product, 1/5 × 1/5 × 1/5. The wrong choices mark three ways to mishandle repetition: adding the chances, multiplying 5 by 3 instead of cubing it, and multiplying by 3 for the three draws. A quick sense check helps, since needing three 5s in a row must be far less likely than needing one.

  1. Something you must already know minor Working it out

    The rule that the chance of independent events all happening is the product of their chances is assumed, as is that replacing the card keeps each draw at 1/5. (Anderson & Krathwohl, 2001)

Strategy: Find the probability of drawing a 5 on one draw, then, because the card goes back each time, multiply that probability by itself once for each draw.

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.
  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=results you wantall possible resultsP = \frac{\text{results you want}}{\text{all possible results}}
P is the probability, the chance something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row
P(A and B and C)=P(A)×P(B)×P(C)P(\text{A and B and C}) = P(\text{A}) \times P(\text{B}) \times P(\text{C})
A, B and C are events, things that might happen, like drawing a 5. When they do not change each other’s chances, the chance that all of them happen is the product of their chances. Putting each card back keeps the draws from changing each other.

  1. Sort out the question

    The bag has 5 cards, numbered 1 to 5. Micha draws a card three times. He puts it back each time, so every draw starts with all 5 cards. He wins only if he draws a 5 all three times. You need the probability, or chance, of that.

  2. Find the chance for one draw

    One card out of 5 is a 5. Use the probability formula.

    P(5)=15P(5) = \frac{1}{5}

    Each draw has a 1 in 5 chance of being a 5.

  3. Multiply for three draws

    He needs a 5 on the first draw, and the second, and the third. The card goes back each time, so the draws don’t change each other. Use the rule for events in a row.

    P(5 and 5 and 5)=15×15×15=15×5×5=1125P(\text{5 and 5 and 5}) = \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = \frac{1}{5 \times 5 \times 5} = \frac{1}{125}

  4. Give the answer

    Watch out: Adding the chances gives 35\frac{3}{5}, choice D. That would make winning likely, which makes no sense. Multiplying 5 × 3 instead of 5 × 5 × 5 gives 115\frac{1}{15}, choice C.

    So the answer is A, 1125\frac{1}{125}.

Check your answer

Count all the results. Each draw has 5 possible cards, so three draws have 5 × 5 × 5 = 125 possible orders. Only one of them is 5, 5, 5. That gives 1 out of 125. Needing three 5s should also be much less likely than needing one.

More questions like this

The original DOE question
Question 110 as printed on page 206 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 206, question 110. Open page 206 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 73

2023 Test B Math, question 55

Question

6.34−4.126.\overline{34}-4.\overline{12}

What is the value of the expression above, written as an improper fraction?

  • A2411\frac{24}{11}
  • B224101\frac{224}{101}
  • C11150\frac{111}{50}
  • D209\frac{20}{9}
Show the answer

Correct answer: D

6.34−4.12=2.22=2.26.\overline{34}-4.\overline{12}=2.\overline{22}=2.\overline{2}

x=2.2222...x=2.2222... Set x equal to the repeating decimal.
10x=22.2222...10x=22.2222... Multiply both sides by 10.
−x=−2.2222...-x=-2.2222... Subtract 10x and x.
9x=209x=20 Apply the multiplicative inverse property; divide both sides by 9.
x=209x=\frac{20}{9}

Question analysis

Two repeating decimals are subtracted, and students write the difference as an improper fraction. It tests subtracting repeating decimals and converting a repeating decimal into a fraction, a grade 8 skill. The repeating part has to be handled exactly, so a careful method matters more than speed.

NYS standards
  • 8.NS.1 decimal expansions and irrational numbers
  • 7.NS.2d fractions to decimals by long division
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students change fractions into repeating decimals by long division. Going the other way, from a repeating decimal to a fraction, is grade 8 work, though the subtraction that comes first here is simple.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 18 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The subtraction is easy, since the repeating digits line up and give 2.222…; the conversion is the test. Every choice sits within a few hundredths of the answer, so rounding or estimating cannot separate them, and the exact 2.22 of 111/50 is the strongest pull. Students who know that 0.222… is 2/9 can write 2 2/9 = 20/9 directly.

  1. A very close wrong answer moderate Choosing the answer

    All four choices lie between 2.18 and 2.23: 111/50 is exactly 2.22 and 224/101 is about 2.218, so only 20/9 equals the repeating 2.222… exactly. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know moderate Working it out

    Turning 2.222… into a fraction, by setting x equal to it and subtracting x from 10x, is grade 8 work that the question does not explain. (Anderson & Krathwohl, 2001)

Strategy: Subtract the decimals digit by digit to see which digits repeat, then turn that repeating decimal into a fraction, or divide each choice's numerator by its denominator and compare.

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.
  • 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

Repeating decimal to fraction
10x−x=9x10x - x = 9x
x is the repeating decimal. Multiplying by 10 moves the decimal point one place to the right. The endless repeating digits still line up after the point. So when you subtract x from 10x, they cancel, and only whole numbers are left.

  1. Sort out the bars

    A bar over digits means they repeat forever. So 6.346.\overline{34} means 6.343434… and 4.124.\overline{12} means 4.121212… with no end. You need the difference as an improper fraction. That is a fraction whose top is bigger than its bottom.

  2. Subtract the decimals

    Line up the digits. Each block of 34 minus 12 gives 22. The whole numbers give 6 − 4 = 2.

    6.343434…−4.121212…=2.222222…6.343434\ldots - 4.121212\ldots = 2.222222\ldots

    The difference is 2.222…, with the 2 repeating forever. You can write it 2.22.\overline{2}.

  3. Name the decimal x

    Let x be the repeating decimal. Then multiply both sides by 10. This moves the decimal point one place to the right.

    x=2.222…x = 2.222\ldots

    10x=22.222…10x = 22.222\ldots

  4. Subtract to remove the repeats

    Subtract the first line from the second. The repeating 2s after the point match up, so they cancel.

    10x−x=22.222…−2.222…10x - x = 22.222\ldots - 2.222\ldots

    9x=209x = 20

  5. Divide to find x

    Divide both sides by 9.

    x=209x = \frac{20}{9}

    Watch out: 11150\frac{111}{50} is exactly 2.22, which stops. The real difference never stops. That trap is choice C. Every choice is close to 2.2, so estimating won’t help.

    So the answer is D, 209\frac{20}{9}.

Check your answer

Divide 20 by 9. 9 goes into 20 two times, with 2 left over. Each next step leaves 2 again, so you get 2.222… forever. That matches the difference. Choice A, 2411\frac{24}{11}, is 2.1818… instead.

More questions like this

The original DOE question
Question 112 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 112. Open page 207 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 59 of 73

2018 Test A Math, question 10

Question

In a scale diagram, 0.1250.125 inch represents 125125 feet. How many inches represent 11 foot?

  • A0.0010.001
  • B0.010.01
  • C0.10.1
  • D0.120.12
Show the answer

Correct answer: A

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

x1=0.125125\frac{x}{1}=\frac{0.125}{125}

x=0.001x=0.001 in.

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Scale drawings and unit rates are core grade 7 topics, and this is a single division. The demand matches typical grade 7 work, with decimal place value as the only real hurdle.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 11, 19 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    The one step is 0.125 ÷ 125, a decimal divided by a large whole number, where the place of the decimal point decides between 0.001, 0.01 and 0.1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    0.001 and 0.01 look almost the same, differing by one zero after the decimal point; only exact placement of the point in 0.125 ÷ 125 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the scale

    A scale diagram is a drawing where small lengths stand for real lengths. Here 0.125 inch on the drawing stands for 125 real feet. You need how many inches stand for just 1 foot. That is a unit rate, an amount for 1 unit.

  2. Set up the division

    Divide the inches by the feet. That splits the 0.125 inch into 125 equal shares, one for each foot.

    r=abr = \frac{a}{b}

    r=0.125125r = \frac{0.125}{125}

  3. Divide carefully

    Write 0.125 as a fraction. It is 125 thousandths.

    0.125=12510000.125 = \frac{125}{1000}

    Now divide by 125. Split 125 thousandths into 125 shares, and each share is 1 thousandth.

    1251000÷125=11000=0.001\frac{125}{1000} \div 125 = \frac{1}{1000} = 0.001

    So 1 foot is shown by 0.001 inch.

  4. Match the decimal places

    0.001 has the 1 in the third place after the decimal point. That place is thousandths. Watch out: 0.01 and 0.1 put the decimal point in the wrong place. They are choices B and C, and each is too big. So the answer is A, 0.001.

Check your answer

Multiply back. 0.001×125=0.1250.001 \times 125 = 0.125 inch, which matches the scale. Choice B would give 0.01×125=1.250.01 \times 125 = 1.25 inches. That is ten times too long.

More questions like this

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

Question 60 of 73

2018 Test A Math, question 16

Question

A box of colored pencils contains exactly 6 red pencils. The probability of choosing a red pencil from the box is 27\frac{2}{7} . How many of the pencils in the box are not red?

  • A5
  • B15
  • C21
  • D30
Show the answer

Correct answer: B

Let xx be the total number of colored pencils in the box. Set up a proportion to find xx:

27=6x\frac{2}{7}=\frac{6}{x}

2x=422x=42

x=21x=21

If there are 6 red pencils, then the number of pencils that are not red is

21 − 6 = 15.

Question analysis

A probability word problem: a box holds exactly 6 red pencils, and the chance of drawing a red one is 2/7. Students work backward from the probability to the number of pencils in the box, then answer the question actually asked, which is about the pencils that are not red.

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 in New York. Reversing it to find a total is a step beyond plugging counts into a fraction, which puts this at about the level of typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 32 words.
Corrected from the DOE handbook
  • A spelling mistake in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Students have to treat a probability as a ratio and scale it up: 6 red pencils is three times the 2 in 2/7. The common loss is choosing the total number of pencils, which appears naturally halfway through the work and is a listed choice. Reading the 7 as a pencil count, or scaling by 6 instead of 3, leads to the other wrong answers.

  1. Turning words into math moderate Understanding the question

    "The probability of choosing a red pencil from the box is 2/7" has to become a part-to-whole ratio: 2 of every 7 pencils are red, so the 6 red pencils are 2 parts. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    The question assumes students know that a probability is favorable outcomes over all outcomes, here used in reverse to find the total from the red count. (Anderson & Krathwohl, 2001)

Strategy: Since 2 parts out of 7 are red, find how many pencils one part stands for, then count the parts that are not red.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says “the number of pencils that are not read is 21 − 6 = 15”. “read” now reads “red”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=favorabletotalP = \frac{\text{favorable}}{\text{total}}
P is the probability, the chance that something happens. Favorable is the number of ways to get what you want, here a red pencil. Total is the number of all possible picks, here every pencil in the box. It works when every pick is equally likely.

  1. Sort out the question

    A probability is a number that tells how likely something is. The box has exactly 6 red pencils. The chance of picking a red one is 27\frac{2}{7}. You need the number of pencils that are not red.

  2. Read the probability as parts

    27\frac{2}{7} means 2 out of every 7 pencils are red. Picture the box split into 7 equal parts. Red pencils fill 2 of the parts. The other 5 parts are not red. Watch out: the box does not hold just 7 pencils. It has 6 red ones alone. Taking 7−2=57 - 2 = 5 is choice A.

  3. Find the size of one part

    The 6 red pencils fill 2 parts. Divide to find one part.

    6÷2=36 \div 2 = 3

    Each part is 3 pencils. So the whole box has 7×3=217 \times 3 = 21 pencils.

  4. Check with the probability formula

    Put the counts into the formula to make sure.

    P=favorabletotal=621=27P = \frac{\text{favorable}}{\text{total}} = \frac{6}{21} = \frac{2}{7}

    Divide the top and bottom of 6/21 by 3 to get 2/7. It matches.

  5. Count the pencils that are not red

    21−6=1521 - 6 = 15

    That is the same as 5 parts of 3 pencils each. Watch out: 21 is all the pencils, choice C. The question asks only for the ones that are not red. So the answer is B, 15.

Check your answer

Red and not red should make the whole box: 6+15=216 + 15 = 21. Red is less likely than not red, since 2/7 is less than half. 6 red against 15 not red fits that.

More questions like this

The original DOE question
Question 73 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 73. Open page 84 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 61 of 73

2018 Test A Math, question 28

Question

215+3310+425+5122\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}

What is the value of the expression shown above?

  • A1472014\frac{7}{20}
  • B142514\frac{2}{5}
  • C1572015\frac{7}{20}
  • D152515\frac{2}{5}
Show the answer

Correct answer: D

215+3310+425+5122\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}

Convert all the fractions to a common denominator (10):

2210+3310+4410+55102\frac{2}{10}+3\frac{3}{10}+4\frac{4}{10}+5\frac{5}{10}

=(2+3+4+5)+(2+3+4+510)=(2+3+4+5)+\left(\frac{2+3+4+5}{10}\right)

=14+1410=1525=14+1\frac{4}{10}=15\frac{2}{5}

Question analysis

A fractions question: add four mixed numbers whose fractions have denominators of 5, 10, 5 and 2. Students rewrite the fractions over a common denominator, add the whole numbers and the fractions, regroup any extra whole, and simplify. Regrouping and lowest terms both matter, since the choices differ in exactly those places.

NYS standards
  • 7.NS.1d properties for adding and subtracting
  • 7.EE.3 multi-step rational number problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding mixed numbers with unlike denominators is grade 5 content. With four terms and a regrouping step it takes care, but it asks less than typical grade 7 work with negative rational numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are friendly: every fraction converts to tenths, and the whole parts add to 14. The risk is in the bookkeeping, either forgetting to carry the extra whole from the fractions or adding numerators over a made-up denominator. Converting to decimals, 2.2 + 3.3 + 4.4 + 5.5, is a fast check that avoids both.

  1. Error-prone arithmetic moderate Working it out

    Four mixed numbers with denominators 5, 10, 5 and 2 must be rewritten as tenths, 2 + 3 + 4 + 5 tenths, and the 14/10 regrouped into a whole and a fraction. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    15 2/5 and 14 2/5 are written alike and share the fraction 2/5; only the whole carried from 14/10 separates them, so a missed carry is invisible at a glance. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    The fractions total 1 4/10, but every choice is in lowest terms, so the 4/10 has to be reduced to 2/5 before it matches. (Enright, Morley & Sheehan, 2002)

Strategy: Convert each mixed number to a decimal, since these fractions all convert exactly, add them, and turn the decimal part back into a fraction in lowest terms.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a is the top number, called the numerator. b is the bottom number, called the denominator. n is any whole number except zero. Multiplying the top and bottom by the same n gives an equal fraction. Use it to give fractions the same denominator so you can add them.

  1. Sort out the question

    You need to add four mixed numbers. A mixed number is a whole number plus a fraction, like 2152\frac{1}{5}. The fractions have different denominators, or bottom numbers: 5, 10, 5 and 2. To add fractions, they need the same denominator.

  2. Add the whole numbers

    Add the whole numbers first. You will add the fractions on their own after that.

    2+3+4+5=142 + 3 + 4 + 5 = 14

    The whole numbers make 14.

  3. Rewrite each fraction in tenths

    Use 10 as the denominator, since 5 and 2 both go into 10.

    15=1×25×2=210\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10}

    25=2×25×2=410\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}

    12=1×52×5=510\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}

    310\frac{3}{10} is already in tenths. Watch out: when you change a bottom number, change the top too. Adding the old tops, 1 + 3 + 2 + 1 = 7, and putting them over 20 gives 7/20. That is in choices A and C.

  4. Add the fractions

    210+310+410+510=1410\frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} = \frac{14}{10}

    14 tenths is more than 1 whole. Ten tenths make 1, so 14 tenths is 1 whole and 4 tenths.

    1410=1410=125\frac{14}{10} = 1\frac{4}{10} = 1\frac{2}{5}

    To simplify 4/10, divide the top and bottom by 2.

  5. Put it all together

    Add the 14 wholes to the sum of the fractions.

    14+125=152514 + 1\frac{2}{5} = 15\frac{2}{5}

    Watch out: carry the extra whole into the 14. Missing it gives 142514\frac{2}{5}, choice B. So the answer is D, 152515\frac{2}{5}.

Check your answer

Change each mixed number to a decimal: 2.2+3.3+4.4+5.5=15.42.2 + 3.3 + 4.4 + 5.5 = 15.4. The 0.4 is 4 tenths, which is 2/5. So the sum is 152515\frac{2}{5}.

More questions like this

The original DOE question
Question 85 as printed on page 86 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 85. Open page 86 of the handbook (PDF)

Question 62 of 73

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 63 of 73

2018 Test A Math, question 36

Question

Number line with two points: L, labeled 4 and 5/16, and N to the right of L, with no value given.

On the number line above, LN = 18\frac{1}{8} . Point M (not shown) is located between point L and point N. Which value below is a possible value for M?

  • A4.26
  • B4.31
  • C4.35
  • D4.58
Show the answer

Correct answer: C

Since LN=18LN=\frac{1}{8}, point NN is located at

4516+18=47164\frac{5}{16}+\frac{1}{8}=4\frac{7}{16}.

So MM must be between point LL, 45164\frac{5}{16}, and point NN, 47164\frac{7}{16}. Point LL can also be written as 4.3125, and point NN can be written as 4.4375. The only option given that lies between those two points is 4.35.

Question analysis

This number line question says the distance from point L to point N is one eighth and asks which decimal could be the position of a point M between them. Students take L's position, a mixed number in sixteenths, find N, and compare with decimal choices. Converting sixteenths to decimals precisely is the tricky part.

NYS standards
  • 7.NS.2d fractions to decimals by long division
  • 7.NS.1b rational number addition
  • 6.NS.7a inequality as relative position
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students convert fractions to decimals and compare them. This item needs an exact conversion of sixteenths and a comparison to the thousandths, where rounding too early can mislead, but it stays within grade 7 content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 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 interval from L to N is only an eighth wide, and three of the four choices crowd near it, so precision decides the question. Students comfortable working in sixteenths can avoid decimals until the end, since 4.35 lies between 4 5/16 and 4 7/16. Those who convert loosely, or who misread a mixed number as a decimal, are pulled to a neighboring choice.

  1. Error-prone arithmetic moderate Working it out

    Converting 4 5/16 to 4.3125 and N to 4.4375 needs four decimal places, while the choices stop at hundredths, so early rounding blurs the left boundary. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The choices 4.31 and 4.35 both lie within 0.04 of L, so estimating cannot separate them; only the exact 4.3125 shows 4.31 just outside the interval and 4.35 inside. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

    L's value appears only on the number line, and the picture shows N to the right of L, which fixes which way the one-eighth gap runs. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add one eighth, or two sixteenths, to L's position to locate N, convert both endpoints to four-place decimals, and keep only a choice strictly between them.

Traps

  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • Answer in another form: The right answer is written differently from the way you worked it out.
  • 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

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.
Fraction to decimal
ab=a÷b\frac{a}{b} = a \div b
A fraction bar means divide. a is the top number and b is the bottom number. Dividing the top by the bottom turns the fraction into a decimal. Then you can compare it with decimal choices.

  1. Sort out what you know

    Point L is at 45164\frac{5}{16}. Point N is to the right of L. The distance from L to N, called LN, is 18\frac{1}{8}.

    Point M is somewhere between L and N. You need a choice that fits between them. The choices are decimals, so you will change the fractions into decimals.

  2. Find where N is

    N is 18\frac{1}{8} to the right of L, so add. First give 18\frac{1}{8} a bottom of 16, to match L:

    18=1×28×2=216\frac{1}{8} = \frac{1 \times 2}{8 \times 2} = \frac{2}{16}

    4516+216=47164\frac{5}{16} + \frac{2}{16} = 4\frac{7}{16}

    Watch out: N is to the right of L in the figure. Going left instead gives a gap that holds 4.26, choice A.

  3. Change both points to decimals

    Divide the top of each fraction by the bottom:

    516=5÷16=0.3125\frac{5}{16} = 5 \div 16 = 0.3125

    716=7÷16=0.4375\frac{7}{16} = 7 \div 16 = 0.4375

    So L is at 4.3125 and N is at 4.4375.

  4. Test each choice

    M must be bigger than 4.3125 and smaller than 4.4375.

    • 4.26 is smaller than 4.3125. It is left of L.
    • 4.31 is 4.3100, just smaller than 4.3125. It is a hair left of L.
    • 4.35 is between them. It fits.
    • 4.58 is bigger than 4.4375. It is right of N.

    Watch out: if you round L to 4.31, choice B seems to fit. Keep all four decimal places. So the answer is C, 4.35.

Check your answer

Subtract to be sure. 4.35 − 4.3125 = 0.0375, which is more than 0, so 4.35 is right of L. 4.4375 − 4.35 = 0.0875, which is also more than 0, so 4.35 is left of N. It sits between them.

More questions like this

The original DOE question
Question 93 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 93. Open page 88 of the handbook (PDF)

Question 64 of 73

2018 Test A Math, question 39

Question

Suppose M=wxM=\frac{w}{x} , N=yzN=\frac{y}{z} , and w,x,yw,x,y, and zz do not equal 0. What is MN\frac{M}{N} in terms of w,x,yw,x,y, and zz ?

  • Awxyz\frac{wx}{yz}
  • Bwyxz\frac{wy}{xz}
  • Cwzxy\frac{wz}{xy}
  • Dxywz\frac{xy}{wz}
Show the answer

Correct answer: C

It is easier to rewrite

MN\frac{M}{N}

as

M÷NM\div N

since  they are both fractions.

M÷N=wx÷yz=wx×zy=wzxyM\div N=\frac{w}{x}\div \frac{y}{z}=\frac{w}{x}\times \frac{z}{y}=\frac{wz}{xy}

Question analysis

This algebra question defines M and N as fractions built from the letters w, x, y and z, and asks for M divided by N in terms of those letters. Students must divide one fraction by another using only variables. Without numbers to check against, it is easy to multiply instead of divide or flip the wrong fraction.

NYS standards
  • 6.NS.1 dividing fractions by fractions
  • 7.NS.2c properties for multiplying and dividing
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students divide numeric fractions such as 2/3 ÷ 4/5. This item applies the same rule with four letters instead of numbers, which adds abstraction but no new steps.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 10, 23 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is fraction division in disguise, and students who recognize M/N as M ÷ N need one rule. The difficulty is that every choice is built from the same four letters, so a wrong rule still produces something that looks like a reasonable answer. Substituting small numbers, such as w = 1, x = 2, y = 3 and z = 4, checks the result in seconds.

  1. Variables instead of numbers moderate Understanding the question

    All four letters stand for unknown nonzero numbers, so dividing w/x by y/z must be done symbolically, with no value to check the result against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    All four choices arrange w, x, y and z as one fraction, and xy/wz is wz/xy turned over, so only carrying out the division in the right order tells them apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Rewrite M/N as M ÷ N, then multiply by the reciprocal of the second fraction; substituting small numbers for the letters checks the result.

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.
  • 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, b, c and d are numbers that are not 0. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). It works because dividing by a number is the same as multiplying by 1 over that number.
Multiplying fractions
ab×cd=a×cb×d\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}
a and c are the tops and b and d are the bottoms. Multiply the tops together and the bottoms together. Use it after you flip the second fraction.

  1. Sort out what you know

    M is the fraction w over x. N is the fraction y over z. The letters stand for numbers that are not 0.

    The fraction bar in MN\frac{M}{N} means M divided by N. So you must divide one fraction by another.

  2. Write it as a division

    Rewrite the big fraction as a division. Then put in the fractions for M and N. This lets you use the rule for dividing fractions:

    MN=M÷N=wx÷yz\frac{M}{N} = M \div N = \frac{w}{x} \div \frac{y}{z}

  3. Flip the second fraction

    Use the rule for dividing fractions. Keep the first fraction. Flip only the second one, and change divide to multiply:

    wx×zy\frac{w}{x} \times \frac{z}{y}

    Watch out: flipping the first fraction instead gives choice D, which is N ÷ M. Multiplying without flipping at all gives choice B.

  4. Multiply across

    Multiply the tops and the bottoms:

    w×zx×y=wzxy\frac{w \times z}{x \times y} = \frac{wz}{xy}

    Here wz means w times z. So the answer is C, wzxy\frac{wz}{xy}.

Check your answer

Try small numbers. Let w = 1, x = 2, y = 3 and z = 4. Then M = 1/2 and N = 3/4. So M ÷ N = 1/2 × 4/3 = 4/6. Choice C gives (1 × 4) ÷ (2 × 3) = 4/6 too. They match.

More questions like this

The original DOE question
Question 96 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 96. Open page 88 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 65 of 73

2018 Test A Math, question 48

Question

1013=0.769230\frac{10}{13}=0.\overline{769230}

In the infinitely repeating decimal above, 7 is the first digit in the repeating pattern. What is the 391st digit?

  • A0
  • B3
  • C6
  • D7
Show the answer

Correct answer: D

There are 6 digits in the repeating decimal (769230), so 7 would be the first, seventh, thirteenth digit and so on.

To find the 391st digit, divide 391 by 6.

391 ÷ 6 = 65 R1

Since the remainder is 1, that means the 391st digit is the same as the 1st digit, which is 7.

Question analysis

This repeating decimal question shows the six-digit repeating pattern of 10/13 and asks for the 391st digit after the decimal point. Students must use the length of the repeating block and division with a remainder to find the position within the cycle. Matching the remainder to the right digit is where students slip.

NYS standards
  • 7.NS.2d fractions to decimals by long division
  • 6.NS.2 multi-digit division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Seventh graders learn that fractions have repeating decimals. Using the length of the repeating block and a remainder to find a digit hundreds of places away is a pattern-reasoning step beyond typical seventh-grade classwork.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 8, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the cycle length is used, the arithmetic is a single division, 391 ÷ 6 = 65 remainder 1. The difficulty is turning the remainder into a position: remainder 1 is the first digit of a new block, not the last digit of the previous one or the second of the new one. Checking the rule on small positions, the 1st and 7th digits are both 7, removes the doubt.

  1. No obvious method moderate Working it out

    Nothing in the stem mentions division; students must see that the six-digit block 769230 repeats, so position 391 depends only on the remainder of 391 divided by 6. (Schoenfeld, 1985)

Strategy: Divide 391 by the length of the repeating block and look only at the remainder; a remainder of r points to the r-th digit of the block.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division with a remainder
n=d×q+rn = d \times q + r
n is the number you divide. d is what you divide by. q is the quotient, the number of whole groups. r is the remainder, what is left over. Here it tells how many full blocks of digits fit, and where you land after them.

  1. Sort out what you know

    The bar over 769230 means those six digits repeat forever: 769230769230 and so on. The 7 is digit number 1. You need digit number 391.

  2. Spot the pattern

    The block 769230 has 6 digits, so the pattern starts over every 6 digits. Digit 1 is 7, and so are digits 7, 13, 19 and so on. Each full block ends with a 0, at digits 6, 12, 18 and so on.

  3. Divide 391 by 6

    Find how many full blocks fit into 391 digits, and what is left:

    391=6×65+1391 = 6 \times 65 + 1

    So 65 full blocks use up 6 × 65 = 390 digits. The remainder is 1. That means digit 391 is the 1st digit of the next block.

  4. Read the digit

    The 1st digit of the block is 7. Watch out: digit 390 is the last digit of a full block, which is 0. That is choice A. Digit 391 is one step past it, so it starts a new block.

    So the answer is D, 7.

Check your answer

Test the idea on a small number. Digit 7: 7 = 6 × 1 + 1, so the remainder is 1 and the digit should be 7. Write it out: 7692307. The 7th digit is indeed 7. The rule works.

More questions like this

The original DOE question
Question 105 as printed on page 90 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 105. Open page 90 of the handbook (PDF)

Question 66 of 73

2018 Test A Math, question 50

Question

100(2+0.1)2−100=100\left ( 2+0.1 \right )^{2}-100=

  • A101
  • B200
  • C301
  • D341
Show the answer

Correct answer: D

100(2+0.1)2−100=100(2+0.1)^2-100=

100(2.1)2−100=100(4.41)−100=441−100=341100(2.1)^2-100=100(4.41)-100=441-100=341

Question analysis

This order of operations question asks students to evaluate 100(2 + 0.1)² − 100, an expression with a decimal, a square and a subtraction. Students must add inside the parentheses first, square the result, multiply by 100, and then subtract. Squaring the two parts of the sum separately is the trap.

NYS standards
  • 6.EE.1 whole-number exponent expressions
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Evaluating an expression with an exponent and decimals in the correct order is grade 6 content. This is a short routine calculation, so it asks less than typical grade 7 work with negative numbers and multi-step equations.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 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

This is a direct order-of-operations calculation with no word problem around it, so the only real risk is the square of a sum. Students who apply the exponent to each part, or only to the 0.1, find a listed answer waiting. Adding inside the parentheses before squaring, and then squaring 2.1 exactly, is the whole task.

  1. A very close wrong answer moderate Choosing the answer

    The choices 301 and 341 differ by 40, and squaring 2.1 roughly as 4 lands nearer 301; only the exact square, 4.41, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    Squaring 2.1 by hand to get 4.41 puts two decimal places in the product, an easy place to misplace the point. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add inside the parentheses to get a single decimal, square it carefully, then multiply by 100 and subtract 100, following the order of operations.

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

Order of operations
parentheses→exponents→×,÷→+,−\text{parentheses} \rightarrow \text{exponents} \rightarrow \times, \div \rightarrow +, -
This is the agreed order for doing math steps. Work inside parentheses first, then exponents, then multiply and divide, then add and subtract. Everyone follows it, so a problem has only one answer.
Squaring a number
a2=a×aa^2 = a \times a
a is any number. The small 2, called an exponent, means multiply the number by itself. So 2.1 squared means 2.1 times 2.1.

  1. Sort out the problem

    You need to work out 100(2+0.1)2−100100(2 + 0.1)^2 - 100. The 100 right next to the parentheses means 100 times. Follow the order of operations, one step at a time.

  2. Add inside the parentheses

    Parentheses come first in the order of operations. So add the numbers inside them:

    2+0.1=2.12 + 0.1 = 2.1

    Now the problem is:

    100×2.12−100100 \times 2.1^2 - 100

  3. Square 2.1

    2.12=2.1×2.1=4.412.1^2 = 2.1 \times 2.1 = 4.41

    Work out 21 × 21 = 441, then put in two decimal places, one for each 2.1.

    Watch out: do not square the 2 and the 0.1 one at a time. That gives 4 + 0.01 = 4.01, which is wrong. It leads to 301, choice C.

  4. Multiply, then subtract

    The exponent is done, so multiply next. Multiplying by 100 moves the decimal point two places to the right. Subtracting comes last:

    100×4.41=441100 \times 4.41 = 441

    441−100=341441 - 100 = 341

    So the answer is D, 341.

Check your answer

Estimate. 2.1 is a bit more than 2, so its square is a bit more than 4. 100 times that is a bit more than 400. Take away 100 and you get a bit more than 300. 341 fits. Choices A and B are too small.

More questions like this

The original DOE question
Question 107 as printed on page 90 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 107. Open page 90 of the handbook (PDF)

Question 67 of 73

2018 Test A Math, question 55

Question

If 4x−3y=124x-3y=12 , what is xx in terms of yy?

  • Ax=34y+12x=\frac{3}{4}y+12
  • Bx=−34y+12x=-\frac{3}{4}y+12
  • Cx=34y+3x=\frac{3}{4}y+3
  • Dx=−34y+3x=-\frac{3}{4}y+3
Show the answer

Correct answer: C

4x−3y=124x-3y=12

4x=12+3y4x=12+3y

x=34y+124x=\frac{3}{4}y+\frac{12}{4}

x=34y+3x=\frac{3}{4}y+3

Question analysis

This algebra question gives the linear equation 4x − 3y = 12 and asks for x in terms of y. Students must isolate the x term and then divide to solve for x. With two variables and a negative coefficient, small sign and division slips each lead to a different wrong choice.

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 one-variable equations like 4x − 3 = 12. Rearranging an equation with two variables to express one in terms of the other is grade 8 algebra, with sign and division slips to avoid.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 10, 13 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The solution is two moves, add 3y to both sides and divide everything by 4, yet each move has a classic slip attached. The four choices are the four combinations of getting the sign right or wrong and dividing the 12 or not, so every partial mistake has a home. Checking one point that satisfies the original equation catches any of them.

  1. Variables instead of numbers moderate Understanding the question

    With two variables there is no number to check against unless students invent one; the point x = 6, y = 4 satisfies the equation and separates the choices. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    The choices x = 3/4 y + 3 and x = −3/4 y + 3 are identical except for the sign of the y term, which only correctly moving −3y across the equals sign settles. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Error-prone arithmetic minor Working it out

    Dividing 12 + 3y by 4 must reach both terms, and the negative coefficient −3 has to change sign when it moves, two slips in a short line. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Move the y term to the right side, watching its sign, so only the x term remains on the left, then divide the entire right side, term by term, by 4.

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

Keeping an equation balanced
a=b means a+c=b+ca = b \text{ means } a + c = b + c
a and b are the two sides of an equation, and c is any number. An equation is like a balance. If you do the same thing to both sides, such as adding 3y or dividing by 4, they stay equal.
Dividing a sum
a+bc=ac+bc\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}
a and b are the two parts of the top, and c is the number you divide by. To divide a sum, divide each part. It works because sharing a + b into c groups shares a and b each into c groups.

  1. Sort out what you need

    The question asks for x in terms of y. That means get x alone on one side. The other side may have y and numbers, but no x. Start with 4x−3y=124x - 3y = 12.

  2. Move the y term

    The 3y is being subtracted on the left. Add 3y to both sides to clear it:

    4x−3y+3y=12+3y4x - 3y + 3y = 12 + 3y

    4x=12+3y4x = 12 + 3y

    Watch out: the term was minus 3y. Adding 3y to both sides makes it plus 3y on the right. Keeping the minus sign leads to choice D.

  3. Divide both sides by 4

    x is being multiplied by 4. Divide both sides by 4 to undo it:

    x=12+3y4x = \frac{12 + 3y}{4}

  4. Split the fraction

    Divide each part of the top by 4:

    x=124+3y4=3+34yx = \frac{12}{4} + \frac{3y}{4} = 3 + \frac{3}{4}y

    Watch out: the 12 must be divided by 4 too. Leaving it as 12 gives choice A. The choices put the y term first. So the answer is C, x=34y+3x = \frac{3}{4}y + 3.

Check your answer

Pick y = 4. Choice C gives x = 3 + 3 = 6. Put x = 6 and y = 4 into the start: 4 × 6 − 3 × 4 = 24 − 12 = 12. It works. Choice D gives x = 0, and 4 × 0 − 12 = −12, not 12.

More questions like this

The original DOE question
Question 112 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 112. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 68 of 73

2018 Test B Math, question 2

Question

147−x12=12\frac{147-x}{12}=12

What is the value of xx in the equation shown above?

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

Show the answer

Correct answer: 3

147−x12=12\frac{147-x}{12}=12

147−x=144147-x=144

x=3x=3

Question analysis

This grid-in asks students to solve a two-step linear equation in which an expression containing x is divided by 12 and set equal to 12. The algebra is short, but x is being subtracted from 147, so undoing the operations in the right order and keeping the minus sign straight is where students slip.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 This is exactly the two-step equation practice of grade 7: undo a division, then undo a subtraction. The only twist is that the unknown is being subtracted, so the last step needs care with its sign.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 12 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a two-move equation that most students finish quickly: multiply both sides by 12, then find what must be taken from 147 to leave 144. Points are lost only at the last move, where the minus sign in front of x invites 144 − 147 or 147 + 144. Substituting the result back into the fraction catches either slip in a few seconds.

  1. Error-prone arithmetic minor Working it out

    After multiplying by 12, the unknown sits behind a minus sign in 147 − x = 144, so students must ask what is taken from 147 rather than subtract in the order the numbers appear. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply both sides by 12 to clear the fraction, then ask what number must be taken away from 147 to leave that product.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing division
ab=c means a=b×c\frac{a}{b} = c \text{ means } a = b \times c
a is the top of the fraction, b is the bottom, and c is the result. Division and multiplication undo each other. So if a divided by b gives c, then b times c gives back a. This is the same as multiplying both sides by b.

  1. Read the equation in words

    The equation says: start with 147, take away a mystery number x, then divide by 12. The result is 12.

    You need to find x. To get x alone, undo the steps in reverse order. Dividing by 12 was done last, so undo it first.

  2. Undo the division by 12

    Multiply both sides by 12. Whatever you do to one side, you must do to the other. That keeps the equation balanced.

    147−x12=12\frac{147 - x}{12} = 12

    147−x=12×12=144147 - x = 12 \times 12 = 144

    Watch out: Do not divide 12 by 12. That gives 147 − x = 1 and the wrong answer 146.

  3. Find what was taken away

    Now ask: 147 minus what number leaves 144? The gap between the two numbers is x.

    x=147−144=3x = 147 - 144 = 3

    Watch out: Do not work out 144 − 147. That gives −3. Since 147 is bigger than 144, x must be positive.

    So you would grid 3.

Check your answer

Put 3 back in for x. First, 147−3=144147 - 3 = 144. Then 144÷12=12144 \div 12 = 12. That matches the right side of the equation, so x = 3 is correct.

More questions like this

The original DOE question
Question 59 as printed on page 153 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 59. Open page 153 of the handbook (PDF)

Question 69 of 73

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 70 of 73

2018 Test B Math, question 4

Question

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

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

Show the answer

Correct answer: 300

Let xx be the total number of pages in the workbook.

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

20100=60x\frac{20}{100}=\frac{60}{x}

20x=6,00020x=6{,}000

x=6,00020=300 pagesx=\frac{6{,}000}{20}=300\text{ pages}

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the whole from a part and a percent is a grade 6 skill. It takes a single division here, so it sits below grade 7, although students who reflexively take 20 percent of 60 will go wrong.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 3, 26 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

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

Traps

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, here all the pages in the book. The part is a piece of it, here the pages done. Percent means how many out of every 100. Use this when you know two of the three numbers and need the third.

  1. Sort out the part and whole

    Tyler has done 60 pages. That is 20% of the whole book. So 60 is the part. The total number of pages is the whole.

    The whole is what you need to find. Call it x.

  2. Write the percent equation

    Put the numbers into the percent formula.

    part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}

    60=20100×x60 = \frac{20}{100} \times x

    The fraction 20100\frac{20}{100} simplifies to 15\frac{1}{5}. So 60 pages is one fifth of the book.

  3. Solve for the whole book

    If one fifth of the book is 60 pages, the whole book is 5 times as big.

    x=60×5=300x = 60 \times 5 = 300

    The book has 300 pages.

    Watch out: Finding 20% of 60 gives 12. That treats 60 as the whole, but 60 is the part. Also, 300 − 60 = 240 is the pages still left to do, not the total.

    So you would grid 300.

Check your answer

Find 20% of your answer: 20100×300=60\frac{20}{100} \times 300 = 60. That matches the 60 pages Tyler finished. Also, 300 is bigger than 60, as a whole should be.

More questions like this

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

Question 71 of 73

2018 Test B Math, question 19

Question

How many positive odd numbers satisfy the inequality 3x+8≤923x+8\leq 92 ?

  • A13
  • B14
  • C17
  • D28
Show the answer

Correct answer: B

Solve the inequality for x.

3x + 8 ≤ 92

3x ≤ 84

x ≤ 28

The positive odd numbers less than or equal to 28 are 1, 3, 5, …, 25, and 27. There are 14 of them.

Question analysis

This question asks how many positive odd numbers satisfy the two-step inequality 3x + 8 ≤ 92. Students solve the inequality and then count the odd whole numbers in the solution. The counting after the algebra is where errors creep in, especially at the end of the list.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving a two-step inequality is grade 7 work. Asking for a count of the odd solutions adds a number-property step and an endpoint check, but the total demand stays close to a typical grade 7 problem.
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 16, 9 words.
Corrected from the DOE handbook
  • The explanation’s description of the solutions 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 the inequality is routine, and that is the trap: x ≤ 28 looks like an answer and 28 is among the choices. The real task is counting odd numbers from 1 to 27, where the usual shortcut, (27 − 1) ÷ 2, misses the first term. Pairing each odd number with the even number after it, up to 28, gives the count directly and avoids the fencepost slip.

  1. A very close wrong answer moderate Choosing the answer

    13 and 14 differ by one, and a rough count of the odd numbers up to 27 cannot separate them; only counting both ends, 1 and 27, gives the exact 14. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Combines ideas minor Working it out

    Solving 3x + 8 ≤ 92 is only half the work; counting the positive odd whole numbers up to 28 is a separate number-sense task. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Solve the inequality as you would an equation, keeping the direction of the sign, then count odd numbers from 1 up to the boundary by pairing or halving.

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: The handbook solves the inequality to x ≤ 28 and then lists “the positive odd numbers less than 28.” The solutions include 28 itself, so the site says “less than or equal to 28.” The count does not change, because 28 is even: the odd numbers are still 1, 3, 5, …, 27, which is 14 numbers.

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 less than or equal to. You may subtract the same number from both sides. You may also divide both sides by the same positive number. The sign stays the same either way.
Counting odd numbers
odd numbers from 1 to n=n2\text{odd numbers from 1 to } n = \frac{n}{2}
n is an even number. Split the numbers from 1 to n into pairs: 1 and 2, 3 and 4, and so on. There are n ÷ 2 pairs, and each pair holds exactly one odd number.

  1. Sort out the question

    You need to count the positive odd numbers that make 3x+8≤923x + 8 \leq 92 true.

    The sign ≤ means is less than or equal to. Positive means greater than 0. Odd numbers are 1, 3, 5, 7, and so on.

  2. Solve the inequality

    Solve it like an equation. Subtract 8 from both sides. Then divide both sides by 3. Dividing by a positive number keeps the sign the same.

    3x+8≤923x + 8 \leq 92

    3x≤843x \leq 84

    x≤28x \leq 28

    Watch out: 28 is the largest x allowed, not the count. That trap is choice D.

  3. List the odd numbers

    The positive odd numbers up to 28 are 1, 3, 5, and so on, up to 25 and 27. The number 28 is even, so it does not count.

  4. Count them by pairing

    Split 1 to 28 into pairs: 1 and 2, 3 and 4, and so on, up to 27 and 28. Each pair has one odd number.

    odd numbers from 1 to 28=282=14\text{odd numbers from 1 to 28} = \frac{28}{2} = 14

    Watch out: Working out (27−1)÷2=13(27 - 1) \div 2 = 13 counts the jumps of 2 from 1 to 27, not the numbers. You must add 1 for the first number. That mistake gives choice A.

    So the answer is B, 14.

Check your answer

Test the edges. For x = 27: 3×27+8=893 \times 27 + 8 = 89, which is at most 92, so it works. For x = 29: 3×29+8=953 \times 29 + 8 = 95, which is too big. So 27 is the last odd number that works.

More questions like this

The original DOE question
Question 76 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 76. 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). This is the DOE’s version, before the corrections listed in the analysis.

Question 72 of 73

2018 Test B Math, question 29

Question

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

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

Correct answer: C

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

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

Therefore, 2w−22w-2 must be even.

Question analysis

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

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

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

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

  2. Every choice must be checked moderate Choosing the answer

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

  3. Something you must already know minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out what you know

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

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

  2. Decide whether w is odd or even

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

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

  3. Show that 2w − 2 is even

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

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

  4. Check the other three choices

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

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

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

  5. Give the answer

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

Check your answer

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

More questions like this

The original DOE question
Question 86 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 86. Open page 158 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 73 of 73

2018 Test B Math, question 54

Question

Raoul is at least 3 years older than Vahn. Which of the following inequalities gives the relationship between Raoul’s age (rr ) and Vahn’s age ( vv )?

  • Ar−v≥3r-v\geq 3
  • Br−v≤3r-v\leq 3
  • C3−v≤r3-v\leq r
  • D3−r≤v3-r\leq v
Show the answer

Correct answer: A

Raoul is at least 3 years older than Vahn, which can be written as

r ≥ v + 3

Rewrite this inequality to match the answer options:

r − v ≥ 3

Question analysis

Raoul is at least 3 years older than Vahn, and students choose the inequality that links their ages, r and v. It tests translating a phrase like "at least" into an inequality. The correct form may not match the one students write first, so they need to rearrange carefully and keep the direction of the inequality.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.8 writing simple inequalities
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing inequalities from words is grade 6 and 7 work. The extra demand is that "at least" must become the right symbol and the choices are rearranged forms, so students must manipulate inequalities rather than just recognize one.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 8, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two skills are tested at once: translating "at least … older than" into an inequality, and recognizing that inequality after a rearrangement. The two choices that begin with 3 − v and 3 − r look like legitimate moves but say only that the two ages add up to at least 3. Testing one pair of ages that fits the sentence and one that does not, such as 20 and 12 against 13 and 12, sorts the choices without algebra.

  1. Turning words into math major Understanding the question

    "Raoul is at least 3 years older than Vahn" has to become r ≥ v + 3; "at least" calls for ≥, though the word "least" seems to point the other way. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    No choice is written as r ≥ v + 3, so the natural translation has to be rearranged to r − v ≥ 3 before it can be matched. (Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer moderate Choosing the answer

    r − v ≤ 3 is the right r − v ≥ 3 with the symbol reversed; only deciding that "at least" means greater than or equal to separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the sentence directly as Raoul's age compared with Vahn's age plus 3, then rearrange each choice into that form to see which one matches.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greater than or equal to
a≥ba \geq b
a and b are two amounts. The sign ≥ means a is greater than or equal to b, so a is b or more. The words at least mean the same thing. The sign ≤ means less than or equal to, or at most.

  1. Sort out the question

    The letter rr is Raoul’s age, and vv is Vahn’s age. Raoul is “at least 3 years older than Vahn”. That means the gap between their ages is 3 years or more. An inequality is a math sentence that uses a sign like ≥\geq instead of =. You need the one that says this.

  2. Write the age gap

    Raoul is older, so the gap is his age minus Vahn’s age. The gap is r−vr - v. For example, ages 15 and 10 have a gap of 5.

  3. Turn at least into a sign

    At least 3 means 3 or more. So use ≥\geq, which means greater than or equal to.

    r−v≥3r - v \geq 3

    This is choice A.

    Watch out: the word least does not mean less. Choice B, r−v≤3r - v \leq 3, says the gap is 3 or less. Then Raoul could even be younger.

  4. Check choices C and D

    Add vv to both sides of choice C. You get 3≤r+v3 \leq r + v. That is about the sum of their ages, not the gap. Choice D works the same way. Neither one says Raoul is older.

  5. Give the answer

    Only one choice says the gap is 3 years or more. So the answer is A, r−v≥3r-v \geq 3.

Check your answer

Test ages that do not fit: Raoul 11 and Vahn 10. He is only 1 year older. Choice A gives 11−10=111 - 10 = 1, and 1 is not 3 or more. So A rightly says no. Choice B says yes, so B is wrong.

More questions like this

The original DOE question
Question 111 as printed on page 164 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 111. Open page 164 of the handbook (PDF)