Practice Test

7.NS the number system

Question 1 of 83

2026 Practice Test A Math, question 17

Question

Consider this expression:

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

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 $|(-8) - (-3) + 6| - |6 - 11|$ :

Step 1: Simplify the first absolute value term.

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

$-5 + 6 = 1$

Simplify the second absolute value term.

$6 - 11 = -5$

Therefore the equivalent expression is

$|1| - |-5|$ .

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: 7.NS.1c rational number subtraction
6.NS.7c absolute value as distance
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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
An answer in an unexpected form minor
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.

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)
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)
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

EC A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Takes the absolute value of each number before combining, so the first bars become 8 − 3 + 6 = 11; the same rule still gives 6 − 11 = −5 in the second.
BC Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Believes absolute value bars make everything inside positive, so it drops every minus sign and adds 8 + 3 + 6 = 17, then works the short pair 6 − 11 normally.
CA Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Simplifies both insides correctly but places −5 first and 1 second, reversing the subtraction so the expression equals 5 − 1 = 4 instead of 1 − 5 = −4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number: $a - (-b) = a + b$; 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: $|a| = \text{the distance from } a \text{ to } 0$; 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.

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.
Simplify inside the first bars
Inside the first bars is $(-8) - (-3) + 6$ . Subtracting −3 is the same as adding 3. Work from left to right:
$(-8) - (-3) = -8 + 3 = -5$
$-5 + 6 = 1$
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.
Simplify inside the second bars
Inside the second bars is $6 - 11$ . You take a bigger number from a smaller one, so the result is negative:
$6 - 11 = -5$
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.
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:
$|1| - |-5| = 1 - 5 = -4$
Swapping the tiles gives $|-5| - |1| = 5 - 1 = 4$ , 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 83

2026 Practice Test A Math, question 32

Question

What is the value of x in the equation $6x = 9x + 13$ ?

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 ≥.

x =

Show the answer

Correct answer: x = -13/3

Solution:

Start with $6x = 9x + 13$ .

Subtract $9x$ from both sides to move all terms involving x to one side.

$6x - 9x = 13$

Combine like terms.

$-3x = 13$

Divide both sides by −3 to solve for x.

$x = -\frac{13}{3}$

$x = -4\frac{1}{3}$

Question analysis

This equation question asks for the value of x when 6x equals 9x plus 13, typed into a box after "x =". Students solve a linear equation with the variable on both sides by gathering the x terms and dividing. With no answer choices to check against, keeping the sign of each moved term is the main challenge.

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 The variable appears on both sides, a grade 8 topic, yet collecting it takes one subtraction and one division, much like a grade 7 two-step equation. The typed answer, with no choices to compare, adds some risk.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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

Two moves solve the equation, so the method is not in doubt; the points are lost in the result. Subtracting 6x instead, which gives 0 = 3x + 13, works as well but moves the sign onto the 13, and either route ends in a division that must keep the negative. The answer is not a whole number, which unsettles students who expect one, and a switch to decimals leads to a rounded entry that is not exact.

Error-prone arithmetic moderate Working it out
Collecting the x terms gives 6x − 9x = −3x, a negative coefficient, and 13 ÷ (−3) is a negative fraction that does not come out whole, so both the sign and the fraction must survive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The answer is typed after the printed x =, with no choices to compare, and its decimal form, −4.333…, repeats forever, so only a fraction or mixed number is exact. (Enright, Morley & Sheehan, 2002)

Strategy: Subtract 9x from both sides so that x appears once, divide both sides by the coefficient left on x, and check by substituting the result into both sides of the original equation.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
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

13/3 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Reaches −3x = 13 correctly but drops the negative when dividing 13 by −3, entering a positive 13/3.
-4.33 Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Solves correctly but turns −13/3 into a decimal and rounds it to −4.33, which is close to the exact value but not equal to it.
-3/13 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Gets to −3x = 13 but divides the wrong way round, putting the coefficient −3 over the constant 13 instead of 13 over −3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides: $\text{if } a = b \text{, then } a - c = b - c \text{ and } \frac{a}{c} = \frac{b}{c}$; a and b are the two sides of an equation, and c is any number or term (not zero when you divide). Taking the same thing away from both sides, or dividing both by the same number, keeps the sides equal. That is how you get x by itself.
Dividing by a negative number: $\frac{a}{-b} = -\frac{a}{b}$; a and b are positive numbers, so −b is negative. A positive number divided by a negative number is negative. You can write the minus sign in front of the whole fraction.
Changing a fraction to a mixed number: $\frac{a}{b} = q\frac{r}{b}$; Use this when the top of a fraction is bigger than the bottom. Divide a by b. q is the number of whole times b fits into a. r is the remainder, what is left over. The remainder goes on top of the leftover fraction.

Sort out the question
You need the value of x that makes $6x = 9x + 13$ true. The letter x appears on both sides, so first gather the x terms on one side. There are no choices to check against, so work carefully.
Move the x terms to one side
Subtract 9x from both sides. This clears the x term off the right side:
$6x - 9x = 9x + 13 - 9x$
$-3x = 13$
Since 6 − 9 = −3, the left side is −3 times x.
Divide both sides by −3
x is multiplied by −3, so divide both sides by −3:
$x = \frac{13}{-3} = -\frac{13}{3}$
A positive divided by a negative is negative, so x is negative.
Watch out: dropping the minus sign gives 13/3, which is wrong. Also, the number you divide by goes on the bottom. Writing −3/13 turns the fraction upside down.
Write it as a mixed number
A mixed number is a whole number with a fraction, like 4 1/3. Divide 13 by 3. The 3 fits 4 times, with 1 left over.
$\frac{13}{3} = 4\frac{1}{3}$
So $x = -4\frac{1}{3}$ .
Watch out: as a decimal, −13/3 is −4.333…, and the 3s never end. Rounding to −4.33 is close, but it is not exact. Keep the fraction.
Enter the answer
The box already shows x =, so you enter only the number. On the test, press the minus key first. Then press the fraction key, a small box over a small box. Type 13 in the top box and 3 in the bottom box. On this site, type −13/3 or −4 1/3.
So the finished answer reads x = −13/3. You type only the −13/3, which is the same as −4 1/3.

Check your answer

Put $x = -\frac{13}{3}$ into both sides. The left side is $6 \times \left(-\frac{13}{3}\right) = -26$ . The right side is $9 \times \left(-\frac{13}{3}\right) + 13 = -39 + 13 = -26$ . Both sides equal −26, so the answer is right.

More questions like this

The original DOE question

Question 32 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 A Math, page 103, question 32.

Question 3 of 83

2026 Practice Test A Math, question 50

Question

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

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:

$\frac{7}{5}$

Substitute $b = 3$ , $c = 4$ , and $d = 5$ in $\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}$ : $\frac{\frac{4}{5}}{\frac{4}{3}} + \frac{4}{5}$ .

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

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

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

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
An answer in an unexpected form minor
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.

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)
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)
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

28/15 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Multiplies 4/5 by 4/3 instead of dividing, getting 16/15, then adds 4/5, or 12/15, to reach 28/15.
37/15 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Flips the top fraction instead of the bottom one, so the stacked part becomes 5/4 × 4/3 = 5/3, and 5/3 + 4/5 = 37/15.
1.4 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds the right value but works in decimals, 0.6 + 0.8 = 1.4, and enters that, though the item asks for a fraction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; 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: $\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}$; 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.

Sort out the question
You know $b = 3$ , $c = 4$ and $d = 5$ . You need the value of:
$\frac{\frac{c}{d}}{\frac{c}{b}} + \frac{c}{d}$
The first part is a fraction on top of a fraction. A fraction bar means divide, so this part is $\frac{c}{d} \div \frac{c}{b}$ . The answer must be typed as a fraction.
Put in the numbers
Replace each letter with its value. Look closely: the bottom fraction has b under the c, not d.
$\frac{c}{d} = \frac{4}{5} \qquad \frac{c}{b} = \frac{4}{3}$
So the expression becomes:
$\frac{4}{5} \div \frac{4}{3} + \frac{4}{5}$
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:
$\frac{4}{5} \div \frac{4}{3} = \frac{4}{5} \times \frac{3}{4} = \frac{12}{20} = \frac{3}{5}$
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.
Add the last fraction
Both fractions are now fifths, so add the tops and keep the bottom:
$\frac{3}{5} + \frac{4}{5} = \frac{3 + 4}{5} = \frac{7}{5}$
The value is 7/5.
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. $\frac{4}{5} = 0.8$ , and 4/3 is about 1.33. Then $0.8 \div 1.33$ is about 0.6, which is 3/5. Last, $0.6 + 0.8 = 1.4$ , 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 4 of 83

2026 Practice Test C Math, question 2

Question

If $(8 \div m) + 4 = 20$ , what is the value of m?

A $\frac{1}{3}$
B $\frac{1}{2}$
C2
D3

Show the answer

Correct answer: B

The quantity $(8 \div m)$ means the same as $\frac{8}{m}$ , so the equation can be rewritten:

$\left(\frac{8}{m}\right) + 4 = 20$

Subtract 4 from both sides:

$\left(\frac{8}{m}\right) = 16$

Multiply both sides of the equation by m:

$8 = 16m$

Solve for m:

$\left(\frac{8}{16}\right) = m$

Simplify:

$\left(\frac{8(\div 8)}{16(\div 8)}\right) = \left(\frac{1}{2}\right)$

Question analysis

This equation question asks for the value of m when 8 divided by m, plus 4, equals 20. Students undo the addition, then work out what m must be for 8 divided by it to give the value left. Because m is the divisor rather than the number being divided, the last step needs care.

NYS standards: 7.EE.4a two-step equations
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Undoing an addition and then a division is typical two-step grade 7 equation work. The unknown sitting in a denominator is less common than px + q = r, which makes the last step easier to get backward.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
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

Undoing the + 4 is routine, so nearly all the risk sits in the last step, where m divides 8 rather than being divided by it. Students who read 8 ÷ m = 16 as asking how many 8s make 16 answer 2, and substituting it back, 8 ÷ 2 + 4 = 8, shows at once that it fails. The right value makes 8 ÷ 1/2 = 16, a division that gives a larger number, which feels wrong but is correct.

Error-prone arithmetic moderate Working it out
The unknown is the divisor, so 8 ÷ m = 16 ends in m = 8 ÷ 16, a quotient less than 1; the familiar order, 16 ÷ 8, gives a whole number instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
Students must see that (8 ÷ m) is the fraction 8/m and that freeing m from a denominator means multiplying both sides by it, a move grade 7 equations rarely need. (Anderson & Krathwohl, 2001)

Strategy: Subtract 4 from both sides, multiply both sides by m so the unknown is no longer in a denominator, then divide; check by substituting your value back into the equation.

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).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: weak
Adds 4 to 20 instead of subtracting it, so 8 ÷ m = 24 and m = 8/24, which simplifies to 1/3.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Gets 8 ÷ m = 16 correctly but then divides 16 by 8, putting the divisor and the dividend the wrong way round, so m comes out as 2.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Makes both mistakes, adding 4 to get 8 ÷ m = 24 and then dividing 24 by 8 instead of 8 by 24.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } a \times c = b \times c$; a and b are the two sides of an equation, and c is any number. An equation is like a balanced scale. Subtract the same number from both sides, or multiply both sides by the same number, and it stays balanced. That is how you get the unknown by itself.

Sort out the question
The equation is $(8 \div m) + 4 = 20$ . You need the value of m. Here m is the number you divide 8 by. In words: 8 divided by some number, plus 4, makes 20.
Undo the plus 4
A fraction bar means divide, so $8 \div m$ is the same as $\frac{8}{m}$ . Subtract 4 from both sides to undo the plus 4:
$\frac{8}{m} + 4 - 4 = 20 - 4$
$\frac{8}{m} = 16$
Now 8 divided by m equals 16.
Watch out: adding 4 instead gives 24 on the right. That leads to choice A, 1/3, or to D, 3.
Get m out of the bottom
m is on the bottom of the fraction. Multiply both sides by m to free it:
$8 = 16 \times m$
Now divide both sides by 16:
$m = \frac{8}{16} = \frac{8 \div 8}{16 \div 8} = \frac{1}{2}$
8 goes into both 8 and 16, so 8/16 simplifies to 1/2.
Choose the answer
Watch out: it is tempting to divide 16 by 8 and get 2, choice C. But ask yourself: 8 divided by what number makes 16? To get an answer bigger than 8, you must divide by a number less than 1.
So the answer is B, 1/2.

Check your answer

Put 1/2 back into the equation. How many halves are in 8? There are 16, so $8 \div \frac{1}{2} = 16$ . Then $16 + 4 = 20$ . That matches the equation, so m = 1/2 is right.

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 C Math, page 75, question 2.

Question 5 of 83

2026 Practice Test C Math, question 3

Question

$6x = x - 1{,}680$

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

A336
B240
C−240
D−336

Show the answer

Correct answer: D

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

$6x - x = x - x - 1{,}680$

$5x = -1{,}680$

$\frac{5x}{5} = \frac{-1{,}680}{5}$

$x = -336$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
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.

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)
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

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Reaches 5x = −1,680 correctly but drops the negative sign when dividing by 5, giving a positive 336 of the right size.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Moves x to the left without changing its sign, getting 7x, and also loses the minus on 1,680, so 1,680 ÷ 7 = 240.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Moves x to the left side without changing its sign, writing 6x + x = −1,680, so divides by 7 and gets −240.

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: $\text{if } a = b \text{, then } a - c = b - c \text{ and } \frac{a}{c} = \frac{b}{c}$; 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: $(-a) \div b = -(a \div b)$; 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.

Sort out the question
The equation is $6x = x - 1{,}680$ . 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.
Subtract x from both sides
Take x away from both sides. On the right, x − x is 0, so only −1,680 is left:
$6x - x = x - x - 1{,}680$
$5x = -1{,}680$
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.
Divide both sides by 5
5x means 5 times x. Divide both sides by 5 to undo it:
$\frac{5x}{5} = \frac{-1{,}680}{5}$
Find the size first. Split 1,680 into 1,500 + 180:
$1{,}500 \div 5 + 180 \div 5 = 300 + 36 = 336$
A negative divided by a positive is negative:
$x = -336$
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 $6 \times (-336) = -2{,}016$ . The right side is $-336 - 1{,}680 = -2{,}016$ . 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 6 of 83

2026 Practice Test C Math, question 8

Question

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

Which fractions are equivalent to the decimal 0.75?

Select the four correct answers.

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

A $\frac{3}{4}$
B $\frac{15}{20}$
C $\frac{75}{100}$
D $\frac{2}{5}$
E $\frac{6}{8}$
F $\frac{8}{10}$

Show the answer

Correct answer: A, B, C, and E

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

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

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

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

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

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

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

Question analysis

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

NYS standards: 7.NS.2d fractions to decimals by long division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Matching fractions to a decimal such as 0.75 builds on grade 4 to 6 work, and the denominators here divide easily. The added demand is the format: four of the six must be chosen, and one wrong or missing pick loses the item.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Every choice must be checked moderate
A very close wrong answer minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Every choice must be checked moderate Choosing the answer
Six fractions with six different denominators have to be tested one by one, by dividing or by simplifying, since 15/20 and 8/10 cannot be judged by how they look. (Embretson & Wetzel, 1987)
A very close wrong answer minor Choosing the answer
8/10 is 0.8, only 0.05 above 0.75, and it follows 6/8 by adding 2 to the top and the bottom, so it looks like one more member of the family. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Much to keep track of minor Working it out
Four of the six must be chosen and all four must be right for the item to count, so each verdict has to be kept while the remaining fractions are tested. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

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

The wrong answers, classified

D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No usual working links 2/5 to 0.75, since 2 ÷ 5 = 0.4, so it mostly catches students filling the fourth slot by guessing.
F Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Builds 8/10 from 6/8 by adding 2 to the top and the bottom, as if equal additions kept a fraction's value, or accepts 0.8 as close enough.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Six fractions are given, A to F. You need the four that equal the decimal 0.75. Equivalent means equal in value. Test every choice, because some fractions look alike but are not equal.
Write 0.75 as a fraction
0.75 means 75 hundredths:
$0.75 = \frac{75}{100}$
So choice C works. Now simplify it. 25 goes into both 75 and 100:
$\frac{75 \div 25}{100 \div 25} = \frac{3}{4}$
So choice A, 3/4, works too.
Test choices B and E
Simplify each one to see if it becomes 3/4:
- B: 5 goes into 15 and 20. $\frac{15 \div 5}{20 \div 5} = \frac{3}{4}$ . It works.
- E: 2 goes into 6 and 8. $\frac{6 \div 2}{8 \div 2} = \frac{3}{4}$ . It works.
Test choices D and F
Divide the top by the bottom:
- D: $2 \div 5 = 0.4$ . That is less than 0.75.
- F: $8 \div 10 = 0.8$ . That is more than 0.75.
Watch out: F looks like 6/8 with 2 added to the top and the bottom. Adding the same number to both changes the value. Only multiplying or dividing both keeps it.
Name the four answers
A, B, C and E all equal 0.75. On the test, you check the box beside each of them. On this site, you choose all four, then check your answer.
So the answers are A, B, C and E.

Check your answer

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

More questions like this

The original DOE question

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

Question 7 of 83

2026 Practice Test C Math, question 11

Question

Allison has 5 stamp albums with 576 stamps in each album. If she transfers her stamp collection to 6 albums, each holding 378 stamps, how many stamps will be left over?

A198
B612
C620
D632

Show the answer

Correct answer: B

Determine the total number of stamps in 5 albums holding 576 stamps:

$5 \times 576 = 2{,}880$

Determine the total number of stamps in 6 albums holding 378 stamps:

$6 \times 378 = 2{,}268$

Subtract 2,268 from 2,880 to find the number of stamps that will be left over:

$2{,}880 - 2{,}268 = 612$

Question analysis

Allison moves a stamp collection from 5 albums of 576 stamps into 6 albums that hold 378 each, and students find how many stamps are left over. It tests multiplying and subtracting large whole numbers in a word problem. The question gives no total, so the collection has to be counted before the new albums' capacity is taken away.

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: Below grade 7 Whole-number multiplication and subtraction in a three-step word problem is easier than grade 7 work with fractions, decimals or negatives. Because the products are large, a slip or a rounded estimate easily gives another choice.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The plan is clear from the stem: count the stamps, then take away what the six new albums hold, so the question tests careful arithmetic. Its choices punish the estimating that saves time elsewhere, since rounding either factor shifts the result by only a few tens, onto another listed number. Students who skip the album counts altogether and compare one album with one album find 198 waiting.

A very close wrong answer moderate Choosing the answer
612, 620 and 632 lie within 20 of one another, so rounding 576 and 378 to friendly numbers before multiplying lands on a wrong choice; only exact products separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Two products of three-digit numbers, 5 × 576 = 2,880 and 6 × 378 = 2,268, must be done by hand and then subtracted with regrouping. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply to find the total number of stamps, multiply again to find how many the six new albums hold, subtract, and check the last digit of each product.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Subtracts 378 from 576, comparing one old album with one new album, and ignores that there are 5 old albums and 6 new ones.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Rounds 576 to 580 and 378 to 380 before multiplying, so 2,900 − 2,280 = 620, an estimate rather than the exact number left over.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: weak
Rounds only 576 up to 580, getting 2,900 stamps, then subtracts the exact 2,268, so the count comes out 20 too high.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total in equal groups: $\text{total} = \text{number of groups} \times \text{amount in each group}$; Here the groups are albums. The amount in each group is how many stamps one album holds. When every group is the same size, multiplying is a fast way to add them all up.

Sort out the question
Allison has 5 albums with 576 stamps in each. She moves all her stamps into 6 new albums. Each new album holds 378 stamps. Some stamps will not fit. You need the number left over.
Count all her stamps
Use the equal-groups formula. Split 576 into 500 + 76 to make it easy:
$5 \times 576 = 5 \times 500 + 5 \times 76$
$2{,}500 + 380 = 2{,}880$
Allison has 2,880 stamps.
Count what the new albums hold
Do the same for the 6 new albums. Split 378 into 300 + 78:
$6 \times 378 = 6 \times 300 + 6 \times 78$
$1{,}800 + 468 = 2{,}268$
The new albums hold 2,268 stamps in all.
Subtract to find the leftovers
The stamps left over are the ones she has minus the ones that fit:
$2{,}880 - 2{,}268 = 612$
Watch out: $576 - 378 = 198$ , choice A, compares just one old album with one new one. Rounding to 580 and 380 first gives 620, choice C. Rounding only 576 gives 632, choice D. Use the exact numbers.
So the answer is B, 612.

Check your answer

Add the leftovers to the stamps that fit: $2{,}268 + 612 = 2{,}880$ . That is exactly the number Allison started with. So no stamps were lost or counted twice.

More questions like this

The original DOE question

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

Question 8 of 83

2026 Practice Test C Math, question 14

Question

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

A−18
B−6
C−2
D2

Show the answer

Correct answer: B

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

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

Evaluate the terms involving exponents:

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

Evaluate the terms involving multiplication:

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

Simplify the expression:

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

$\lvert -6 \rvert - \lvert -12 \rvert$

The absolute value of $(-6)$ is 6, and the absolute value of $(-12)$ is 12, so $\lvert -6 \rvert - \lvert -12 \rvert$ is the equivalent of subtracting 12 from 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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
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.

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)
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

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Simplifies both insides correctly but treats the first bars like parentheses, keeping −6, so −6 − 12 = −18.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes (−2)³ as positive 8, as if every power of a negative were positive, so the first bars hold −2 + 4 + 8 = 10, and 10 − 12 = −2.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Takes (−2)² as −4, so the first bars hold −2 − 4 − 8 = −14, and 14 − 12 = 2; dropping the minus from x throughout also gives 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{the distance from } a \text{ to } 0$; 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: $(-2)^{2} = (-2) \times (-2) \quad (-2)^{3} = (-2) \times (-2) \times (-2)$; 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.

Sort out the question
You need the value of $|x + x^{2} + x^{3}| - |x + 2x + 3x|$ when x = −2. The bars | | mean absolute value. Work out each pair of bars on its own. Subtract only at the end.
Find the powers of −2
Put −2 in for x. Keep each negative sign:
$(-2)^{2} = (-2) \times (-2) = 4$
$(-2)^{3} = 4 \times (-2) = -8$
The square is positive, but the cube is negative.
Watch out: taking $(-2)^{3}$ as +8 makes the first bars hold 10. That leads to choice C, −2.
Work out the first bars
Add the three terms inside the first pair of bars:
$|(-2) + 4 + (-8)| = |-6| = 6$
−2 + 4 is 2, and 2 + (−8) is −6. The distance from −6 to 0 is 6.
Work out the second bars
Here 2x means 2 times x, so it is −4. And 3x is −6:
$|(-2) + (-4) + (-6)| = |-12| = 12$
All three terms are negative, so they add to −12. Its absolute value is 12.
Subtract the second value
Now subtract the second value from the first:
$6 - 12 = -6$
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. $x + 2x + 3x = 6x$ , and $6 \times (-2) = -12$ . 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 9 of 83

2026 Practice Test C Math, question 16

Question

$\frac{3+(-3)}{3-(-3)} =$

A−1
B0
C $\frac{1}{6}$
D1

Show the answer

Correct answer: B

Simplify the expression.

$3 + (-3)$ is the equivalent of subtracting 3 from 3, and $3 - (-3)$ is the equivalent of adding 3 to 3:

$\frac{3+(-3)}{3-(-3)} = \frac{3-3}{3+3} = \frac{0}{6}$

The zero property of division states that zero divided by any nonzero number equals zero, so $\frac{0}{6} = 0$ .

Question analysis

This fraction has 3 plus negative 3 on top and 3 minus negative 3 on the bottom, and students find its value. It tests adding and subtracting negative numbers and dividing integers. The top and the bottom look almost alike, so each sign has to be read carefully before anything is divided.

NYS standards: 7.NS.1c rational number subtraction
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding and subtracting a negative integer and dividing are grade 7 basics, and here the numbers are small. It is a short routine calculation, easier than typical grade 7 work, though the two signs invite a quick misreading.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 16+, 2 words.
Corrected from the DOE handbook: A grammatical error in the explanation was corrected.
A rule stated too broadly in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is two one-step sign calculations and a division by 6. The danger is reading the top and the bottom as alike: they differ in a single sign, and treating them the same way gives 1 or −1. Students who rewrite 3 − (−3) as 3 + 3 and 3 + (−3) as 3 − 3 see 0 over 6 at once, and zero divided by a nonzero number is zero.

Error-prone arithmetic minor Working it out
The top adds −3 and the bottom subtracts −3, signs that look nearly the same but give 0 and 6, so each has to be rewritten before anything is divided. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Simplify the top and the bottom separately, rewriting the subtraction of a negative as an addition, and only then divide the top by the bottom.

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

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Adds the sizes on top and keeps the minus, taking 3 + (−3) as −6, so the fraction becomes −6/6 = −1.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No sign rule gives a top of 1, so 1/6 mostly draws students who find the bottom, 6, and are unsure what a top of 0 means.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Treats the top like the bottom, adding 3 and 3 in both places, so the fraction becomes 6/6 = 1.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation says “3 − (−3) is equivalent of adding 3 to 3”, without the article that the same sentence uses in “3 + (−3) is the equivalent of subtracting 3 from 3”. The site now says “is the equivalent of adding 3 to 3”.
Explanation: The DOE’s explanation says “The zero property of division states that zero divided by any number equals zero.” Zero divided by zero is undefined, not zero, so the rule holds only when the divisor is not zero. The site now says “zero divided by any nonzero number equals zero”. Here the divisor is 6, so 0/6 = 0, as before.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding a negative: $a + (-b) = a - b$; a and b are numbers. Adding a negative is the same as subtracting. On a number line, adding −3 moves you 3 steps left, just like subtracting 3.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Subtracting a negative is the same as adding. Taking away a debt of 3 dollars leaves you 3 dollars better off. On a number line, you move right.
Zero divided by a number: $\frac{0}{b} = 0 \text{ when } b \neq 0$; b is any number except 0. Zero split into b equal parts leaves 0 in each part. The rule needs b not to be 0, because dividing by 0 has no answer.

Sort out the question
The top of the fraction is $3 + (-3)$ , and the bottom is $3 - (-3)$ . A fraction bar means divide. Work out the top and the bottom separately first. Then divide the top by the bottom.
Work out the top
Adding −3 is the same as subtracting 3:
$3 + (-3) = 3 - 3 = 0$
A number plus its opposite is always 0.
Watch out: do not add the 3s and keep the minus sign. That makes the top −6 and leads to choice A, −1.
Work out the bottom
Subtracting −3 is the same as adding 3:
$3 - (-3) = 3 + 3 = 6$
Watch out: the top and bottom look alike, but they are not. Making both of them 6 gives 6/6 = 1, choice D.
Divide the top by the bottom
Now the fraction is 0 over 6. Zero divided by any number other than zero is zero:
$\frac{3+(-3)}{3-(-3)} = \frac{0}{6} = 0$
A top of 0 does not make the fraction 1/6, choice C. So the answer is B, 0.

Check your answer

Check by multiplying. A division is right when the answer times the bottom gives the top. Here $0 \times 6 = 0$ , which is the top. So 0/6 really is 0.

More questions like this

The original DOE question

Question 16 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 C Math, page 89, question 16. This is the DOE’s version, before the corrections listed in the analysis.

Question 10 of 83

2026 Practice Test C Math, question 17

Question

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

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

A $\frac{1}{2}$ unit
B1 unit
C $1\frac{1}{2}$ units
D2 units

Show the answer

Correct answer: A

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

$7 - 1 = 6$

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

$6 \div 2 = 3$

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

$1 + 3 = 4$

$7 - 3 = 4$

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

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

$6 - 3 = 3$

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

$3 \div 2 = 1.5$

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

$3 + 1.5 = 4.5$

$6 - 1.5 = 4.5$

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

Determine the distance between the two midpoints by subtracting:

$4.5 - 4 = 0.5$

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

Question analysis

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

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding distances between points on a number line is grade 7 work, here with whole-number positions. Two midpoints must be located before their distance can be found, so the task takes three short steps rather than one.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Reading a figure, table or graph minor Understanding the question
Segment RS lies inside MN, from 3 to 6, and its midpoint, 4 1/2, falls between two ticks of the number line rather than on a labeled point. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Two midpoints have to be found, at 4 and 4 1/2, before their distance is taken, and the half-length of RS met on the way, 1 1/2, is among the choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Sees that RS starts 2 units in from M and ends 1 unit in from N, and takes the difference of those gaps, 1, without halving it.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Halves each segment, 3 units for MN and 1 1/2 for RS, and subtracts the half-lengths instead of comparing the midpoints' positions.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the distance from M to R, 2 units, the gap between the left ends of the two segments, without locating any midpoint.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The number line shows M at 1, R at 3, S at 6 and N at 7. Segment MN goes from 1 to 7. Segment RS goes from 3 to 6. A midpoint is the point exactly halfway along a segment. You need the distance between the two midpoints.
Find the midpoint of MN
Use the midpoint formula with the ends 1 and 7:
$\frac{1 + 7}{2} = \frac{8}{2} = 4$
The midpoint of MN is at 4, which is 3 units from each end.
Find the midpoint of RS
Use the same formula with the ends 3 and 6:
$\frac{3 + 6}{2} = \frac{9}{2} = 4.5$
The midpoint of RS is at 4.5. That is halfway between the marks for 4 and 5, where no point is labeled.
Subtract the two midpoints
Use the distance formula:
$4.5 - 4 = 0.5$
0.5 is one half, so the midpoints are 1/2 unit apart.
Watch out: half of MN is 3, and half of RS is 1 1/2. Subtracting those gives 1 1/2, choice C. But they are half-lengths, not positions. And 2, choice D, is only the distance from M to R.
Choose the answer
The midpoints, 4 and 4.5, are close together because RS sits near the middle of MN. So the answer is A, 1/2 unit.

Check your answer

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

More questions like this

The original DOE question

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

Question 11 of 83

2026 Practice Test C Math, question 18

Question

I. $s - t$
II. $st^{2}$
III. $s^{t}$

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

AII only
BI and II only
CII and III only
DI, II, and III

Show the answer

Correct answer: D

Each of the three expressions must be considered:

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

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

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

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

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

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

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

Question analysis

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

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

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
No values are given for s and t, only their signs, and must means each expression has to be positive for every such pair, not just for one example. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know moderate Working it out
Expression III needs the meaning of a negative exponent, s to the power t equals 1 over s to the power −t, a grade 8 rule the question does not state. (Anderson & Krathwohl, 2001)
Every choice must be checked minor Choosing the answer
Each of the three expressions is judged on its own and the choices combine the verdicts, so one wrong judgment anywhere points to a wrong combination. (Embretson & Wetzel, 1987)

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

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Trusts only the squared expression, believing that subtracting t must make s smaller and that a negative exponent makes a negative number.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Handles the subtraction and the square correctly but believes a negative exponent makes the result negative, though s to the power t is a positive fraction such as 1/8.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reads s − t as taking a number away from s, so thinks it could be negative, missing that subtracting the negative t adds to s.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
An integer is a whole number or its opposite, such as −2, 0 or 5. Here s is a positive integer, like 1, 2 or 3. And t is a negative integer, like −1, −2 or −3. Must means always. An expression counts only if it is positive for every such pair.
Try one pair of numbers
Try s = 2 and t = −3 in each expression:
- I: $2 - (-3) = 2 + 3 = 5$
- II: $2 \times (-3)^{2} = 2 \times 9 = 18$
- III: $2^{-3} = \frac{1}{2^{3}} = \frac{1}{8}$
All three are positive this time. Now see why that always happens.
Explain expression I
Use the rule for subtracting:
$s - t = s + (-t)$
t is negative, so its opposite, −t, is positive. So s − t is a positive plus a positive. That is always positive.
Watch out: subtracting does not always make a number smaller. Subtracting a negative makes it bigger. Missing this leads to choice C.
Explain expression II
$t^{2}$ means t × t. A negative times a negative is positive, so $t^{2}$ is always positive. Then $s \times t^{2}$ is a positive times a positive. That is always positive too.
Explain expression III
Use the rule for a negative exponent:
$s^{t} = \frac{1}{s^{-t}}$
−t is a positive integer, so $s^{-t}$ is a positive integer. So $s^{t}$ is 1 over a positive number. That is a positive fraction.
Watch out: a negative exponent does not make the answer negative. It makes a fraction. Thinking it does leads to choice B, I and II only.
Choose the answer
Each of the three expressions is positive for every positive s and negative t. None of them can ever be zero or negative.
So the answer is D, I, II, and III.

Check your answer

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

More questions like this

The original DOE question

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

Question 12 of 83

2026 Practice Test C Math, question 19

Question

What is the value of $\left(-\frac{3}{4}\right)\left(-\frac{2}{5}\right)$ ?

A $-\frac{7}{20}$
B $-\frac{3}{10}$
C $\frac{3}{10}$
D $\frac{7}{20}$

Show the answer

Correct answer: C

Evaluate the expression by multiplying the terms in the numerator and the terms in the denominator. The product of two negative values is positive:

$\frac{(-3) \times (-2)}{4 \times 5} = \frac{6}{20}$

Simplify the result:

$\frac{6 \div 2}{20 \div 2} = \frac{3}{10}$

Question analysis

This question asks for the product of negative three fourths and negative two fifths. Students multiply the fractions, settle the sign of the result, and simplify. The choices mix positive and negative answers with results from other operations, so both the sign and the operation have to be right.

NYS standards: 7.NS.2a rational number multiplication
7.NS.2c properties for multiplying and dividing
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Multiplying two negative fractions is a standard grade 7 exercise, done here in one product and one simplification. It matches typical classroom work, with the sign rule and lowest terms as the only points of care.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 10, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a single product of two fractions, and most students finish it quickly. The choices are built around two decisions, the sign of the product and the operation itself: two of them come from subtracting the fractions, and one keeps a minus sign that two negative factors cancel. Settling the sign first, positive, rules out half of the choices before any multiplying.

A very close wrong answer minor Choosing the answer
3/10 and −3/10 are identical except for the sign, so the size alone does not separate them; only the rule for multiplying two negatives does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Two negative fractions are multiplied, 3 × 2 over 4 × 5, and the product 6/20 has to be reduced to 3/10 before it matches a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Decide the sign of the product first, then multiply the numerators and the denominators and reduce the fraction to lowest terms.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Subtracts the fractions instead of multiplying them, −3/4 − (−2/5) = −15/20 + 8/20, which gives −7/20.
B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Multiplies the sizes correctly, 3/4 × 2/5 = 6/20 = 3/10, but keeps a minus sign, though two negative factors give a positive product.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Drops both signs and subtracts the sizes, 3/4 − 2/5 = 15/20 − 8/20 = 7/20, instead of multiplying the fractions.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the tops (numerators). b and d are the bottoms (denominators). Multiply the tops together and the bottoms together. You do not need a common denominator to multiply. That is only for adding and subtracting.
Multiplying two negatives: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. So when both factors are negative, the minus signs cancel.

Sort out the question
The two fractions sit side by side in parentheses, with nothing between them. That means multiply. Both fractions are negative. You need their product, the answer to the multiplication.
Decide the sign first
A negative times a negative is positive. So the answer will be positive. That rules out choices A and B.
Watch out: choice B, −3/10, has the right size but the wrong sign. The two minus signs cancel.
Multiply tops and bottoms
Now multiply the sizes. Multiply the tops, then the bottoms:
$\frac{3}{4} \times \frac{2}{5} = \frac{3 \times 2}{4 \times 5} = \frac{6}{20}$
Watch out: do not find a common denominator. That is for adding and subtracting. Subtracting the fractions gives 7/20 or −7/20, choices D and A.
Simplify the fraction
2 goes into both 6 and 20. Divide the top and the bottom by 2:
$\frac{6 \div 2}{20 \div 2} = \frac{3}{10}$
The product is positive 3/10. So the answer is C, 3/10.

Check your answer

Use decimals. 3/4 is 0.75 and 2/5 is 0.4. Then $0.75 \times 0.4 = 0.3$ , which is 3/10. Two negatives give a positive, so the product is positive 3/10.

More questions like this

The original DOE question

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

Question 13 of 83

2026 Practice Test C Math, question 22

Question

During a hiking trip, a group of students recorded the elevation (in feet) at several points along the trail:

$-120\text{ ft},\ -80\text{ ft},\ 0\text{ ft},\ 45\text{ ft},\ -30\text{ ft},\ 90\text{ ft}$

What is the difference between the highest and lowest elevations recorded on the trail?

Enter your answer in the space.

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

Show the answer

Correct answer: 210

Determine the difference between the highest and lowest elevations:

Determine the highest elevation:

Of −120, −80, 0, 45, −30, and 90, the greatest value is 90 because its location is farthest to the right on a number line.

Highest elevation $= 90\text{ ft}$

Determine the lowest elevation:

Of −120, −80, 0, 45, −30, and 90, the lowest value is −120 because its location is farthest to the left on a number line.

Lowest elevation $= -120\text{ ft}$

Determine the difference between the two elevations:

$90\text{ ft} - (-120)\text{ ft} = 90\text{ ft} + 120\text{ ft} = 210\text{ ft}$

Question analysis

A group of students on a hiking trip records six elevations in feet, some below zero and some above, and finds the difference between the highest and the lowest, typing the result. It tests ordering positive and negative numbers and subtracting integers. Spotting the lowest reading among several negatives and subtracting a negative number correctly are where errors creep in.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7b order statements in context
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the gap between a positive and a negative value by subtracting is core grade 7 work with integers. Picking the extremes from six readings is quick, so the demand is the subtraction of a negative number.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Picking the highest and lowest of six readings is quick, though −120 must be recognized as lower than −80 and −30. The points are lost in the subtraction: the span runs 120 feet up to zero and 90 more above it, and students who treat the readings as plain sizes find only 30. Picturing the two readings on a vertical number line shows the whole span.

Error-prone arithmetic moderate Working it out
The difference is 90 − (−120), and subtracting a negative means adding 120, so the part of the trail below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The difference is typed with no choices, so 30 or −30, the results of mishandling the negative, gets no warning that the two readings lie on opposite sides of zero. (Enright, Morley & Sheehan, 2002)

Strategy: Pick the greatest and the least elevation by their places on a number line, then subtract the least from the greatest, turning the subtraction of a negative into an addition.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

30 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Subtracts the sizes, 120 − 90 = 30, as if both readings were on the same side of zero, leaving out the stretch below zero.
-30 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Combines the two extremes by adding, 90 + (−120) = −30, the net of the readings, instead of subtracting the lowest from the highest.
-210 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Subtracts in the reverse order, lowest minus highest, −120 − 90 = −210, and enters a negative difference.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Difference between highest and lowest: $\text{difference} = \text{highest} - \text{lowest}$; The difference between the highest and lowest values is how far apart they are. Subtract the lowest from the highest, so the answer is never negative.
Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Subtracting a negative is the same as adding a positive. On a number line, it means moving to the right.

Sort out the question
Elevation is height above or below a starting level, here 0 feet. Negative elevations are below 0. The six readings are −120, −80, 0, 45, −30 and 90 feet. You need the difference between the highest and the lowest.
Find the highest and the lowest
Picture the readings on a number line. The highest is farthest right: 90 feet. The lowest is farthest left: −120 feet.
For negative numbers, the bigger the number after the minus sign, the lower it is. So −120 is lower than −80.
Subtract the lowest from the highest
Use the difference formula. Subtracting a negative is the same as adding:
$90 - (-120) = 90 + 120 = 210$
So the two readings are 210 feet apart.
Watch out: $120 - 90 = 30$ treats both readings as if they were above 0. Adding them, $90 + (-120) = -30$ , is wrong too.
Enter the answer
A difference like this is a distance, so it is positive. Subtracting in the other order gives −210, which is wrong. Type only the number, with no units.
So you would enter 210.

Check your answer

Count the distance in two parts. From −120 feet up to 0 is 120 feet. From 0 up to 90 is 90 more feet. Together, $120 + 90 = 210$ feet. That matches.

More questions like this

The original DOE question

Question 22 as shown on the DOE’s digital test, page 95 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 95, question 22.

Question 14 of 83

2026 Practice Test D Math, question 3

Question

Two sets, R and S, are described below. The sum of the elements in set R equals the sum of the elements in set S.

$\mathrm{R} = \{5, x, 3, 8\}$

$\mathrm{S} = \{6, y, 4, 1\}$

What is the value of $x - y$ ?

A−7
B−5
C5
D7

Show the answer

Correct answer: B

First, find the sum for each set by adding the terms in each set separately.

The sum for set R is $5 + x + 3 + 8$ , which is $x + 16$ .

The sum for set S is $6 + y + 4 + 1$ , which is $y + 11$ .

The question states that these two sums are equal to each other, so write the equation and then rearrange using inverse operations.

$x + 16 = y + 11$

$x = y - 5$

$x - y = -5$

If the rearranging to obtain the target value is unclear, use the equation $x + 16 = y + 11$ and find a few possible values for x and y.

For example, if $x = 1$ , then $y = 6$ because $1 + 16 = 6 + 11$ . If $x = 2$ , then $y = 7$ because $2 + 16 = 7 + 11$ . Then find the value of $x - y$ . No matter which values are chosen, $x - y$ will always be $-5$ .

Question analysis

Two sets of four numbers, R and S, each contain one unknown, x and y, and the sums of the two sets are equal. Students find the value of x − y without being able to find x or y alone. It tests writing and rearranging an equation, and deciding which way round the difference goes is the tricky part.

NYS standards: 7.EE.4 equations and inequalities from problems
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students write and solve equations with one unknown. This one has two unknowns that are never found separately; students must rearrange to get their difference, a less familiar goal, though the arithmetic is light.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The sums take seconds; the point is lost on the sign. Because set R's known numbers add to more than set S's, x has to be the smaller unknown, so x − y is negative, but subtracting the two known sums in the order they appear gives a positive 5. Choosing one pair of values that balances the sets, such as x = 1 and y = 6, settles the sign at once.

A very close wrong answer moderate Choosing the answer
The choices pair −5 with 5 and −7 with 7, so the size 5 is easy to reach and only the order of subtraction, x − y rather than y − x, separates the right choice from its opposite. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
Neither x nor y can be found, only the gap between them, so students who look for a value of each letter stall; the question rewards rearranging x + 16 = y + 11. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Add the known numbers in each set, set the two sums equal, then move both letters to one side so it reads x − y; testing one pair of values confirms the sign.

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

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives −7; it needs a slip in a sum, such as 18 for 5 + 3 + 8, so it mostly draws guesses near −5.
C Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Subtracts the known sums in the order the sets appear, 16 − 11 = 5, which is y − x, missing that R's larger known sum makes x the smaller unknown.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Needs two unrelated errors, a slip in a sum and the subtraction reversed, so no single line of working reaches 7 and it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Doing the same to both sides: $\text{if } a = b \text{, then } a - c = b - c$; a and b are two equal amounts, and c is any number or letter. Taking the same thing away from both sides keeps them equal. You use it to move terms around until the equation shows what you want.

Sort out the question
Set R holds 5, x, 3 and 8. Set S holds 6, y, 4 and 1. The elements (the members) of R add up to the same total as the elements of S. You cannot find x or y alone. But you can find $x - y$ , the gap between them.
Add up each set
Add the known numbers in each set, and keep the letter:
$5 + x + 3 + 8 = x + 16$
$6 + y + 4 + 1 = y + 11$
The two sums are equal, so:
$x + 16 = y + 11$
Rearrange to get x − y
You want $x - y$ on one side. Subtract y from both sides:
$x - y + 16 = 11$
Then subtract 16 from both sides:
$x - y = 11 - 16 = -5$
Make sense of the sign
Set R already has 16 from its numbers, but set S has only 11. So y must be 5 more than x to make the sums match. That makes $x - y$ negative.
Watch out: 16 − 11 = 5 is $y - x$ , not $x - y$ . Using it leads to choice C.
So the answer is B, −5.

Check your answer

Try real values. If x = 1, then R adds to 17. So y + 11 = 17, and y = 6. Then $x - y = 1 - 6 = -5$ . Try x = 10: then y = 15, and 10 − 15 = −5 again.

More questions like this

The original DOE question

Question 15 of 83

2026 Practice Test D Math, question 9

Question

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

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

Select the three correct answers.

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

A $-m + k$
B $-m \cdot k$
C $m - k$
D $-k \cdot m$
E $\frac{m}{k}$

Show the answer

Correct answer: A, B, and D

Each of the expressions must be considered:

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

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

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

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

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

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

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

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

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

A negative divided by a positive remains negative.

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

Question analysis

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

NYS standards: 7.NS.2a rational number multiplication
7.NS.1c rational number subtraction
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Sign rules for adding, subtracting, multiplying and dividing integers are core grade 7 content. Here they are applied to letters with known signs instead of numbers, across five expressions, and all three right ones must be chosen with no wrong one.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
Only the signs of m and k are known, so the sign of each expression must be reasoned from the rules or tested with values such as m = −2 and k = 3. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
All five expressions mix the same two letters with minus signs, so none can be dismissed at a glance; each is judged on its own and exactly three must be picked. (Embretson & Wetzel, 1987)
Much to keep track of minor Working it out
The item scores only when all three positive expressions are chosen and both negative ones left out, so five separate verdicts must be kept and combined. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Sees two negatives, m and the minus before k, and applies “two negatives make a positive” to subtraction, though a negative minus a positive moves further below zero.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Left out by students who read the leading minus in −k · m as making the product negative, missing that m is negative too, so the product is positive.
E Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Sees no minus sign in m/k and takes it as positive, forgetting that m itself is negative, so the quotient of a negative and a positive is negative.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
m is a negative integer, a whole number below 0, such as −2. k is a positive integer, such as 3. You must pick the three expressions that are always positive. Check all five, and test m = −2 and k = 3 as you go.
Understand the minus in front
A minus sign in front of a letter means the opposite of that number. Since m is negative, −m is positive. Since k is positive, −k is negative. With the test values, −m = 2 and −k = −3.
Check choices A and B
A: $-m + k$ is a positive plus a positive, so it is always positive. Test: 2 + 3 = 5.
B: $-m \cdot k$ is a positive times a positive, so it is always positive. The dot · means times. Test: 2 × 3 = 6.
Check choices C and E
C: $m - k$ starts below 0 and moves further down, so it is always negative. Test: −2 − 3 = −5.
E: $\frac{m}{k}$ is a negative divided by a positive, so it is always negative. Test: −2 ÷ 3 = −2/3.
Watch out: two negatives make a positive only when you multiply or divide them. Choice C subtracts, so it stays negative. Choice E shows no minus sign, but m itself is negative.
Check choice D
D: $-k \cdot m$ is a negative times a negative, so it is always positive. Test: −3 × (−2) = 6.
Watch out: the minus sign in front makes D look negative. But m is negative too, and the two negatives make a positive. Skipping D leaves you one answer short.
Choose the three answers
Choices C and E are always negative, so leave them out. The three positive ones are A, B and D. On the test, you check the box next to each one.
So the answers are A, B and D.

Check your answer

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

More questions like this

The original DOE question

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

Question 16 of 83

2026 Practice Test D Math, question 14

Question

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

Show the answer

Correct answer: B

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

Pages remaining after the morning: $84 - 21 = 63$

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

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

Question analysis

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

NYS standards: 7.EE.3 multi-step rational number problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Taking fractions of a quantity in stages is typical grade 7 work. The numbers divide evenly and the bold word flags the trap, so the four short steps match routine grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Turning words into math minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Many steps moderate Working it out
Four operations in order, a quarter of 84, a subtraction, a third of the 63 left, and a second subtraction, each feed the next, so a slip early carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The bold “remaining” moves the base of the second fraction from all 84 pages to the 63 left after the morning, and “still need to read” asks for what is left, not what was read. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Takes 1/3 of all 84 pages for the afternoon, 28, instead of 1/3 of the 63 remaining, so 84 − 21 − 28 = 35.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 45; it is not a fraction or a difference of any page count in the problem, so it mostly draws guesses near 42.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Takes the afternoon's third from all 84 pages and subtracts only that, 84 − 28 = 56, overlooking the 21 pages already read in the morning.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
James has 84 pages to read. In the morning he reads 1/4 of all the pages. In the afternoon he reads 1/3 of the remaining pages, the ones still left after the morning. You need the number of pages he still has to read after that.
Find the morning pages
Take 1/4 of 84:
$\frac{1}{4} \times 84 = 84 \div 4 = 21$
He reads 21 pages in the morning. That leaves:
$84 - 21 = 63$
63 pages are left after the morning.
Find the afternoon pages
The afternoon fraction is 1/3 of the 63 pages left, not of all 84:
$\frac{1}{3} \times 63 = 63 \div 3 = 21$
He reads 21 pages in the afternoon.
Watch out: taking 1/3 of all 84 pages gives 28. Then 84 − 21 − 28 = 35, choice A. The bold word remaining tells you to use 63.
Find the pages still to read
Take the afternoon pages away from the 63 pages left after the morning:
$63 - 21 = 42$
He still needs to read 42 pages.
Watch out: subtract both readings, morning and afternoon. Taking only 28 from 84 gives 56, choice D.
So the answer is B, 42.

Check your answer

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

More questions like this

The original DOE question

Question 17 of 83

2026 Practice Test D Math, question 15

Question

$\frac{1}{11} + \frac{1}{22} + \frac{1}{33}$

What is the value of the expression shown above?

A $\frac{1}{22}$
B $\frac{1}{11}$
C $\frac{1}{6}$
D $\frac{3}{22}$

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.

$\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}$

$\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}$

$\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}$

$\frac{6}{66} + \frac{3}{66} + \frac{2}{66} = \frac{11}{66} = \frac{1}{6}$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
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.

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)
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

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the numerators and the denominators straight across, 3 over 11 + 22 + 33 = 66, and simplifies 3/66 to 1/22.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Puts the three 1s over the largest denominator, 33, without renaming the other two fractions, so 3/33 simplifies to 1/11.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Adds the first two fractions correctly, 2/22 + 1/22 = 3/22, and stops before adding the third fraction, 1/33.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; 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: $\frac{a}{d} + \frac{b}{d} + \frac{c}{d} = \frac{a + b + c}{d}$; 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.

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.
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:
$2 \times 3 \times 11 = 66$
Rename each fraction
Multiply the top and bottom of each fraction by the number that makes the bottom 66:
$\frac{1}{11} = \frac{1 \times 6}{11 \times 6} = \frac{6}{66}$
$\frac{1}{22} = \frac{1 \times 3}{22 \times 3} = \frac{3}{66}$
$\frac{1}{33} = \frac{1 \times 2}{33 \times 2} = \frac{2}{66}$
Add the tops
Now every piece is a sixty-sixth. Add the tops and keep the bottom:
$\frac{6}{66} + \frac{3}{66} + \frac{2}{66} = \frac{11}{66}$
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.
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:
$\frac{11}{66} = \frac{11 \div 11}{66 \div 11} = \frac{1}{6}$
Watch out: add all three fractions. Stopping after the first two gives 3/22, choice D.
So the answer is C, $\frac{1}{6}$ .

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 18 of 83

2026 Practice Test D Math, question 18

Question

Which of the following decimals is equivalent to $1\frac{2}{3} + 2\frac{4}{9}$ ?

A $3.\overline{1}$
B $3.\overline{6}$
C $4.\overline{1}$
D $4.\overline{9}$

Show the answer

Correct answer: C

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

$1\frac{2}{3} + 2\frac{4}{9}$

$1\frac{6}{9} + 2\frac{4}{9}$

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

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

$4\frac{1}{9} = 4.111\ldots = 4.\overline{1}$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
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.

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)
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

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Adds the fractions to 10/9 but keeps only the 1/9 left over, forgetting to carry the whole 1 into the 3, so the sum becomes 3 1/9.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Adds the numerators 2 and 4 over 9 without renaming 2/3 as 6/9, so gets 3 6/9, which is 3.666…
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Reaches 4 1/9 but repeats the denominator's 9 instead of the numerator's 1, as if one ninth were 0.999… rather than 0.111…

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; 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: $\frac{1}{9} = 0.111\ldots = 0.\overline{1}$; 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…

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.
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:
$\frac{2}{3} = \frac{2 \times 3}{3 \times 3} = \frac{6}{9}$
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.
Add the mixed numbers
Add the whole numbers, then add the fractions:
$1 + 2 = 3 \qquad \frac{6}{9} + \frac{4}{9} = \frac{10}{9}$
10/9 is more than 1. It is 9/9 plus 1/9, or 1 1/9. Carry the 1 into the whole number:
$3 + 1\frac{1}{9} = 4\frac{1}{9}$
Watch out: forgetting to carry the 1 gives 3 1/9, choice A.
Change it to a decimal
One ninth is 0.111…, with the 1 repeating forever. So:
$4\frac{1}{9} = 4.111\ldots = 4.\overline{1}$
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, $4.\overline{1}$ .

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 19 of 83

2026 Practice Test D Math, question 19

Question

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

A$615.60
B$627.48
C$1,108.08
D$1,846.80

Show the answer

Correct answer: C

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

$456 \times 1.25 = 570$

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

$456 + 570 = 1{,}026$

The sales tax on the selling price is:

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

The customer pays the selling price plus the sales tax:

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

Question analysis

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

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students find markups and sales tax, usually with increases under 100%. An increase of more than 100% is unusual and easy to misread, and two percent changes must be applied in order, which goes beyond routine grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps moderate
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math moderate Understanding the question
“Increases the purchase cost by 125%” adds 125% of $456 to the $456 itself, a multiplier of 2.25; reading it as 125% of the cost leaves the original price out. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps moderate Working it out
Four steps in a row, the increase, the selling price, the tax and the total, each depend on the one before, and two of them are easy to merge or skip. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Taking 8% of $1,026 means multiplying by 0.08, and a decimal point one place off, 0.8, makes the tax ten times too large. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the increase, 125% of $456 = $570, and uses it as the selling price without adding back the $456 cost, then adds 8% tax.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads the 125% increase as $125, a fixed amount, so the selling price is $581, and then adds 8% tax correctly to reach $627.48.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Finds the $1,026 selling price correctly but writes 8% as 0.8, adding $820.80 in tax instead of $82.08.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The manager pays $456.00 for the furniture. The selling price is that cost increased by 125%. The customer then pays 8% sales tax on top. Sales tax is an extra charge, a percent of the price. You need the total the customer pays.
Find the increase
Find 125% of $456. As a decimal, 125% is 1.25, which is 1 and a quarter. A quarter of 456 is 114:
$456 \times 1.25 = 456 + 114 = 570$
The increase is $570.
Find the selling price
Increases by means the increase is added to the cost:
$456 + 570 = 1{,}026$
The selling price is $1,026.
Watch out: $570 is only the increase. Using it as the price gives $615.60, choice A. And 125% is a percent, not $125. Adding $125 leads to $627.48, choice B.
Find the sales tax
As a decimal, 8% is 0.08. Find 8% of the selling price:
$1{,}026 \times 0.08 = 82.08$
The tax is $82.08.
Watch out: 8% is 0.08, not 0.8. Using 0.8 makes the tax ten times too big and gives $1,846.80, choice D.
Add the tax to the price
The customer pays the selling price plus the tax:
$1{,}026 + 82.08 = 1{,}108.08$
So the answer is C, $1,108.08.

Check your answer

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

More questions like this

The original DOE question

Question 20 of 83

2026 Practice Test D Math, question 20

Question

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

A−76
B−4
C4
D148

Show the answer

Correct answer: B

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

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

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

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

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

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

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

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

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

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

$\frac{16}{-4} = -4$

Question analysis

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

NYS standards: 7.NS.2a rational number multiplication
7.NS.2b rational number division
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Order of operations with negative numbers is standard grade 7 practice. This expression stacks a squared negative, two products and a division by a negative into one fraction, so it takes more steps than usual, each with a sign to track.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
A very close wrong answer minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 6 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
Five sign decisions sit in one expression: squaring −4, two products with a negative factor, a sum that climbs from −288 past zero, and a division by −4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The numerator alone takes an exponent, two multiplications and two additions before the single division, and all of it must be finished before dividing. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The right value and the choice 4 differ only in sign, so a student with the right size must still check the sign of a positive numerator over −4. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Squares as if the minus sat outside the parentheses, taking (−4)² as −16, so −9 × (−16) = 144, the numerator becomes 304, and 304 ÷ (−4) = −76.
C Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Works the numerator out correctly as 16 but divides by −4 as if the signs did not matter, giving a positive quotient.
D Negatives and direction (Brown & Burton, 1978)
Reveals a surface strategy Expected pull: weak
Drops every negative sign, adding 144 + 144 + 304 = 592 and dividing by 4, so 148 keeps none of the signs in the problem.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
You need the value of one big fraction. The fraction bar is a grouping symbol, so finish the whole top first. Then divide by −4. Follow the order of operations, and track every minus sign.
Do the exponent first
The only exponent is (−4)². Squaring means multiplying a number by itself:
$(-4)^{2} = (-4) \times (-4) = 16$
Watch out: (−4)² is 16, not −16. Using −16 makes the top 304, which gives −76, choice A.
Do the multiplication
Now multiply. A negative times a positive is negative:
$-9 \times 16 = -144$
$36 \times (-4) = -144$
So the top becomes:
$-144 + (-144) + 304$
Add from left to right
Add the two negatives first:
$-144 + (-144) = -288$
Then add 304. Since 304 is more than 288, the result is positive:
$-288 + 304 = 16$
The whole top is 16.
Divide by −4
Now divide the top by the bottom. A positive divided by a negative is negative:
$\frac{16}{-4} = -4$
Watch out: positive 4, choice C, ignores the negative bottom. Dropping every minus sign gives 592 ÷ 4 = 148, choice D.
So the answer is B, −4.

Check your answer

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

More questions like this

The original DOE question

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

Question 21 of 83

2026 Practice Test D Math, question 22

Question

Seth grows strawberries in his garden.

He picked $2\frac{1}{2}$ cups of strawberries on Monday.
He picked 4 cups of strawberries on Tuesday.
He ate $\frac{3}{4}$ cup of strawberries for breakfast each day on Wednesday, Thursday, and Friday.

How many cups of strawberries did Seth have left?

A $3\frac{1}{2}$
B $4\frac{1}{4}$
C6
D9

Show the answer

Correct answer: B

When Seth picks strawberries, the value represents a positive number, and when he eats the strawberries, the value represents a negative number. Write an expression and then evaluate the expression, where the negative value is multiplied by 3 because he ate the same amount on 3 different days.

$2\frac{1}{2} + 4 + (3)\left(-\frac{3}{4}\right)$

$2\frac{1}{2} + 4 - 2\frac{1}{4}$

$6\frac{1}{2} - 2\frac{1}{4}$

$6\frac{2}{4} - 2\frac{1}{4}$

$4\frac{1}{4}$

If the multiplication is confusing, evaluate the expression below which represents the sum for the 5 days with each day listed separately.

$2\frac{1}{2} + 4 + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right)$

$4\frac{1}{4}$

Question analysis

Seth picked strawberries on Monday and Tuesday, in amounts given as a mixed number and a whole number, then ate the same fraction of a cup at breakfast on three days. Students find how many cups he has left. It tests adding, multiplying and subtracting fractions in a short word problem, and the eaten amount repeats each day.

NYS standards: 7.NS.3 four operations with rational numbers
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 and multiplying a fraction by a whole number are skills from before grade 7. The three steps are short and the numbers friendly, so the demand is below typical grade 7 work with signed rational numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The bulleted list lays out the story day by day, so setting up the problem is easy; the points are lost in the fractions. The number of breakfast days, 3, sits right beside the 3/4 cup, and subtracting 3 cups instead of three quarters of a cup each day is the tempting shortcut. Writing every amount in fourths, 26/4 picked and 9/4 eaten, makes the last step a whole-number subtraction.

Error-prone arithmetic moderate Working it out
Three breakfasts of 3/4 cup make 9/4 = 2 1/4 cups, and taking that from 6 1/2 means renaming the half as 2/4, two fraction steps where a slip is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Picking adds and eating subtracts, and “each day on Wednesday, Thursday, and Friday” means the 3/4 cup counts three times, not once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add the two amounts picked, multiply the breakfast amount by the number of days, and subtract that from the total; writing every amount in fourths keeps the arithmetic simple.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Subtracts 3 cups, the number of breakfast days, instead of 3 × 3/4 = 2 1/4 cups, so 6 1/2 − 3 = 3 1/2.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives exactly 6; it is near the 6 1/2 cups picked and mostly catches students who drop the fractions or guess.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the 2 1/4 cups eaten to the 6 1/2 cups picked instead of subtracting, 8 3/4 cups, and takes the nearest choice, 9.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total of a repeated amount: $\text{total} = \text{number of times} \times \text{amount each time}$; Use it when the same amount happens again and again. Here Seth eats 3/4 cup on each of 3 days, so the number of times is 3 and the amount each time is 3/4 cup.
A whole number times a fraction: $n \times \frac{a}{b} = \frac{n \times a}{b}$; n is a whole number, and a/b is a fraction. Multiply the whole number by the top only, and keep the bottom. It works because n groups of a pieces make n × a pieces, and each piece is still size 1/b.

Sort out the question
Seth picked 2 1/2 cups on Monday and 4 cups on Tuesday. Then he ate 3/4 cup each day on Wednesday, Thursday and Friday. That is 3 days of eating. Picking adds strawberries, and eating takes them away. You need the number of cups he has left.
Add what he picked
Add the two days of picking:
$2\frac{1}{2} + 4 = 6\frac{1}{2}$
He picked 6 1/2 cups in all, before he ate any.
Find how much he ate
He ate 3/4 cup on each of 3 days. Multiply:
$3 \times \frac{3}{4} = \frac{9}{4} = 2\frac{1}{4}$
To change 9/4 to a mixed number, note that 4 fits into 9 twice, with 1 left over. He ate 2 1/4 cups.
Watch out: he ate 3/4 cup three times, not 3 cups. Subtracting 3 gives 3 1/2, choice A.
Take away what he ate
Write the half as fourths, so both fractions have the same bottom: 1/2 = 2/4. Then subtract the wholes and the fourths:
$6\frac{1}{2} - 2\frac{1}{4} = 6\frac{2}{4} - 2\frac{1}{4} = 4\frac{1}{4}$
Watch out: eating takes away. Adding the 2 1/4 cups gives 8 3/4, and the nearest choice is 9, choice D.
So the answer is B, $4\frac{1}{4}$ cups.

Check your answer

Use decimals. 2.5 + 4 = 6.5 cups picked. 3 × 0.75 = 2.25 cups eaten. Then 6.5 − 2.25 = 4.25 cups, which is 4 1/4.

More questions like this

The original DOE question

Question 22 of 83

2026 Practice Test D Math, question 23

Question

Which situation involves quantities that combine to equal zero?

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

Show the answer

Correct answer: A

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

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

Question analysis

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

NYS standards: 7.NS.1a opposite quantities making 0
6.NS.5 positive and negative numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Describing situations where opposite quantities make zero is named in the grade 7 standards, and this is its most direct form: one reading of four everyday situations, easier than typical grade 7 work with signed numbers.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Turning words into math minor
Every choice must be checked minor
Solution: 1 step
Reading level: Question and choices: about grade 5, 60 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math minor Understanding the question
Each situation must be turned into signed numbers, receiving or rising as plus and giving or buying as minus, before its total can be judged. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Every choice must be checked minor Choosing the answer
Every choice repeats one amount, $5, $10, 3 floors or 20 feet, so equal numbers alone do not identify the answer; the direction of each move has to be checked. (Embretson & Wetzel, 1987)

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

Traps

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Treats the two equal $10 amounts as opposites that cancel, though buying a book and buying lunch both take money away, for a total of −$20.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Pairs the two 3s as amounts that cancel, missing that rising and then rising again both go up, ending 6 floors above the ground floor.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Sees that starting 20 feet below ground and climbing 20 feet reaches zero, then overlooks the last 20 feet climbed, which ends 20 feet above ground.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
You need the situation whose amounts add up to zero. Give each amount a sign. Use + for getting money or going up. Use − for spending money or going down. Then add the amounts in each choice. The direction matters, not just the numbers.
Write each choice with signs
Turn each situation into signed numbers and add them up:
Choice Amounts Total
A +5, then −5 0
B −10, then −10 −20
C +3, then +3 +6
D start at −20, then +20, then +20 +20
Look at the tricky choices
In B, buying a book and buying lunch both take money away. Equal amounts cancel only when they go in opposite directions. In C, rising and rising again both go up, so you end 6 floors up.
Watch out: in D, starting at −20 and climbing 20 feet does reach 0, ground level. But then you climb 20 feet more. You end at +20, 20 feet above the ground, so D is wrong.
Choose the zero total
Only choice A has two opposite amounts of the same size. Getting $5 is +5, and giving $5 away is −5:
$5 + (-5) = 0$
So the answer is A: receiving $5 as a gift and then giving $5 to a friend.

Choice	Amounts	Total
A	+5, then −5	0
B	−10, then −10	−20
C	+3, then +3	+6
D	start at −20, then +20, then +20	+20

Check your answer

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

More questions like this

The original DOE question

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

Question 23 of 83

2026 Practice Test D Math, question 33

Question

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

A$14,000
B$15,620
C$15,920
D$25,920

Show the answer

Correct answer: C

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

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

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

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

Question analysis

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

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding sales tax to a price is a standard grade 7 percent task. This one also subtracts a down payment at the end, three routine steps with round numbers, much like a typical classroom problem.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 29 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is the standard one for tax on a purchase, so the question checks order and completeness. Forgetting the tax or forgetting the down payment each produces an amount as reasonable-looking as the answer, and a careless 8 × 24 produces one just $300 below it. Multiplying the price by 1.08 in one step, which gives $25,920, is a second route that catches a slip in the tax.

Many steps minor Working it out
Three operations come in a fixed order, 8% of $24,000, then adding the tax, then subtracting the $10,000, and leaving out the first or the last gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Finding 8% of $24,000 comes down to 8 × 24 = 192 with two zeros added, and a lost carry in that product turns $1,920 into $1,620. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Subtracts the $10,000 down payment from the $24,000 price and never adds the 8% sales tax, so the tax is simply left out.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Drops the carried 3 when multiplying 8 × 24, so the tax comes out $1,620 instead of $1,920 and the balance $300 too low.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Adds the $1,920 tax to the price correctly but stops at $25,920, the full cost, forgetting to subtract the $10,000 Bindu already paid.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
You know:
- The car costs $24,000.
- The sales tax is 8%. Sales tax is extra money added to a price.
- Bindu paid $10,000 at the start. That is the down payment.
You need the amount Bindu still has to pay. Work in order: tax, then full cost, then subtract.
Find the sales tax
Change 8% to the decimal 0.08. Multiply it by the price:
$0.08 \times 24{,}000 = 1{,}920$
A quick way: 1% of 24,000 is 240, so 8% is $8 \times 240 = 1{,}920$ . The tax is $1,920.
Watch out: in 8 × 24 you carry a 3 from 8 × 4 = 32. Dropping it gives 162, so the tax looks like $1,620. That leads to $15,620, choice B.
Find the full cost
Add the tax to the price:
$24{,}000 + 1{,}920 = 25{,}920$
The car costs $25,920 with tax.
Watch out: this is not the answer yet. Stopping here gives $25,920, choice D. Bindu already paid part of it.
Subtract the down payment
Bindu already paid $10,000. Take it away:
$25{,}920 - 10{,}000 = 15{,}920$
She has $15,920 left to pay.
Watch out: leaving out the tax gives $24{,}000 - 10{,}000 = 14{,}000$ , choice A.
So the answer is C, $15,920.

Check your answer

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

More questions like this

The original DOE question

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

Question 24 of 83

2024 Test A Math, question 1

Question

Simplify:

$-3.8 + 2.3 - (-1.1)$

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

Show the answer

Correct answer: -0.4

Subtracting a negative number is the same as adding a positive number. Rewrite the expression to simplify:

$-3.8 + 2.3 - (-1.1)$

$-3.8 + 2.3 + 1.1$

$-3.8 + 3.4$

$-0.4$

Question analysis

This grid-in asks students to simplify a short expression that mixes negative and positive decimals and ends by subtracting a negative number. It tests adding and subtracting rational numbers in order. The double negative, and keeping track of the sign of the final result, are where errors creep in.

NYS standards: 7.NS.1c rational number subtraction
7.NS.1b rational number addition
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding and subtracting negative decimals is core grade 7 content, and this is a routine three-term version. It matches a typical classroom exercise; the only extra demand is gridding a signed decimal correctly.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 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

This is a three-term expression with small decimals, so the method is never in doubt; the points are lost on signs. The double negative at the end is the classic place to slip; the other is the result itself, which sits just below zero and is easy to grid without its minus sign. Grouping the positive terms first, 2.3 + 1.1, and comparing that sum with 3.8 settles both the size and the sign.

Error-prone arithmetic moderate Working it out
The last term, − (−1.1), has to turn into + 1.1, and the total then crosses zero, since 2.3 + 1.1 = 3.4 falls just short of the 3.8 below zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in it shows no choices, so a double negative read as a single one, or a result gridded without its minus sign, gets no warning from the page. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite subtracting −1.1 as adding 1.1, combine the two positive decimals first, then compare their total with 3.8 to decide the sign and size.

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

-2.6 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Treats subtracting −1.1 as subtracting 1.1, so it works out −3.8 + 2.3 − 1.1 = −1.5 − 1.1 and moves further below zero.
0.4 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Rewrites the double negative correctly and finds the gap 3.8 − 3.4 = 0.4, but drops the minus sign that the larger negative term gives the result.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number: $a - (-b) = a + b$; a is any number, and b is a positive number, so −b is negative. Taking away a negative number is the same as adding the positive number. Think of money: if someone takes away a debt of b dollars, you are b dollars better off.
Adding a negative and a positive: $-a + b = -(a - b)$; a and b are positive numbers, and a is the larger one. A negative and a positive partly cancel out. So find the difference of their sizes, a − b. The result keeps the sign of the number farther from zero, which here is the negative one.

Sort out the expression
You need to simplify $-3.8 + 2.3 - (-1.1)$ . Simplify means work it out to one number. Two of the numbers are negative: −3.8 and −1.1. One is positive: 2.3. The tricky part is at the end. It says minus negative 1.1. The answer can be negative, so keep track of every sign.
Change minus a negative to plus
Taking away a negative is the same as adding a positive. Use the rule with b = 1.1.
$a - (-b) = a + b$
$-3.8 + 2.3 - (-1.1) = -3.8 + 2.3 + 1.1$
Now only one negative number is left. Watch out: do not just subtract 1.1. That gives $-3.8 + 2.3 - 1.1 = -2.6$ , which is wrong.
Add the two positive numbers
Group the positive numbers first. This makes the rest of the work short.
$2.3 + 1.1 = 3.4$
Now the problem is $-3.8 + 3.4$ . One number is negative and one is positive.
Combine the negative and the positive
They partly cancel. Use the rule with a = 3.8 and b = 3.4.
$-a + b = -(a - b)$
$-3.8 + 3.4 = -(3.8 - 3.4) = -0.4$
The negative number, 3.8, is farther from zero than 3.4 is. So the result is negative. Watch out: keep the minus sign. Dropping it gives 0.4, which is wrong. So you would grid -0.4, with the minus sign in its own column.

Check your answer

Use a number line. Start at −3.8. Adding 2.3 moves you right, to −1.5. Subtracting −1.1 also moves you right, by 1.1, to −0.4. You end just to the left of zero. So a small negative answer makes sense.

More questions like this

The original DOE question

Question 58 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 2025 Admissions, page 75, question 58. Open page 75 of the handbook (PDF)

Question 25 of 83

2024 Test A Math, question 4

Question

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

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

Show the answer

Correct answer: 5.5

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

$8 \div 2 = 4$

$4 \times 2\frac{1}{8}$

$4 \times 2.125 = 8.5$

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

$8.5 + 4 = 12.5$

Mr. Chan mows his lawn every 2 weeks and cuts off the top $1\frac{3}{4}$ inches. Find the total number of inches that are cut off in 8 weeks.

$8 \div 2 = 4$

$4 \times 1\frac{3}{4}$

$4 \times 1.75 = 7$

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

$12.5 - 7 = 5.5$

Question analysis

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

NYS standards: 7.EE.3 multi-step rational number problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve multi-step problems with mixed numbers and convert between fractions and decimals. This one strings four such steps together, including counting the 2-week periods, which is typical grade 7 work done carefully.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Error-prone arithmetic moderate
Turning words into math minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 3, 49 words.
Corrected from the DOE handbook: A misspelled name in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Many steps moderate Working it out
Four pieces have to be combined: count the four 2-week periods in 8 weeks, total the growth, total the mowing, and start from the lawn's first 4 inches. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic moderate Working it out
The amounts are mixed numbers with unlike denominators, 2 1/8 and 1 3/4, which must be multiplied by 4 or subtracted and then gridded as a decimal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase "every 2 weeks" governs both the growing and the mowing, so 8 weeks means four rounds of each, not eight. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
As a grid-in it gives no choices, so a height that leaves out the starting 4 inches, or leaves out the mowing, looks as finished as the right one. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

1.5 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the net gain of 3/8 inch per period and multiplies by the 4 periods, 1.5 inches, but stops before adding the lawn's starting height of 4 inches.
12.5 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Adds four periods of growth, 8.5 inches, to the starting 4 inches and stops there, never subtracting the 1 3/4 inches cut at each of the four mowings.
7 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Applies the net gain of 3/8 inch to each of the 8 weeks instead of each 2-week period, 8 × 3/8 = 3, then adds the starting 4.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The lawn starts at 4 inches tall. Every 2 weeks it grows $2\frac{1}{8}$ inches. Every 2 weeks Mr. Chan also cuts off $1\frac{3}{4}$ inches. You need the height after 8 weeks, as a decimal. Three things matter: the start, the growing and the cutting.
Count the 2-week periods
Both the growing and the cutting happen every 2 weeks. So count how many 2-week periods fit in 8 weeks.
$8 \div 2 = 4$
There are 4 periods. The lawn grows 4 times and is cut 4 times. Watch out: do not count 8 rounds, one for each week. That leads to the wrong answer 7.
Change the mixed numbers to decimals
The answer must be a decimal, so change both amounts now.
$2\frac{1}{8} = 2 + 1 \div 8 = 2.125$
$1\frac{3}{4} = 1 + 3 \div 4 = 1.75$
Each period, the lawn grows 2.125 inches and loses 1.75 inches.
Add up the growing
The lawn grows 2.125 inches, 4 times.
$\text{total} = n \times a$
$\text{total} = 4 \times 2.125 = 8.5$
It grows 8.5 inches in all. Add the starting height of 4 inches.
$4 + 8.5 = 12.5$
Without mowing, the lawn would be 12.5 inches tall. Watch out: do not forget the starting 4 inches. Leaving them out leads to 1.5.
Add up the cutting
Mr. Chan cuts off 1.75 inches, 4 times.
$\text{total} = 4 \times 1.75 = 7$
He cuts off 7 inches in all.
Subtract the cutting
$12.5 - 7 = 5.5$
The lawn is 5.5 inches tall after 8 weeks. Watch out: 12.5 is not the answer. It leaves out the mowing. So you would grid 5.5.

Check your answer

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

More questions like this

The original DOE question

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

Question 26 of 83

2024 Test A Math, question 9

Question

A list of consecutive integers begins with m and ends with n. If $n - m = 66$ , how many integers are in the list?

Show the answer

Correct answer: D

Consider the case where $m = 1$ .

$n - m = 66$

$n - 1 = 66$

$n - 1 + 1 = 66 + 1$

$n = 67$

The list of integers from 1 to 67 contains 67 numbers.

Question analysis

A number question about a list of consecutive integers that starts at m and ends at n, where n minus m is 66. Students decide how many integers the list contains. The tricky part is relating the difference between the two ends to the number of integers in the list.

NYS standards: 7.NS.1c rational number subtraction
6.EE.6 variables representing unknown numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Consecutive integers and differences are familiar grade 7 ideas, and the arithmetic is trivial. What keeps this at grade 7 rather than below it is the counting trap, which needs a moment of careful reasoning.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Variables instead of numbers moderate
Solution: 2 steps
Reading level: Question and choices: about grade 7, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is trivial, so the question tests one idea: the number of integers from m to n is one more than their difference, because both ends belong to the list. The stem supplies 66 ready-made, and taking it as the count is the natural first reaction. A five-number list, 1 to 5, whose ends differ by 4, shows the extra one in seconds.

A very close wrong answer moderate Choosing the answer
The choices 66 and 67 differ by one, and 66 is the difference the stem hands over; only counting both ends of the list separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers moderate Understanding the question
The list is given only by its ends, m and n, so the count has to be reasoned out in general, or checked on a small list such as 1 through 5. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Try a small list you can count, such as 1 through 5, and compare its count with the difference between its first and last numbers.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working counts only the two named ends, m and n, in a list said to hold consecutive integers, so 2 mostly draws guesses.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Halves 66, as if consecutive integers went up by 2 like consecutive odd or even numbers, so the list held only every other integer.
C Counting the ends (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes the difference of the ends as the count, forgetting that a list from m to n includes both m and n.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting consecutive integers: $\text{count} = n - m + 1$; m is the first number in the list and n is the last. n − m counts the steps from each number to the next. A list always has one more number than it has steps, so you add 1. This works for any list that goes up by 1.

Sort out the question
Integers are whole numbers and their negatives, like −2, 0 and 5. Consecutive integers come one after another, going up by 1, like 4, 5, 6. The list starts at m and ends at n. The difference $n - m$ is 66. You need how many integers are in the list.
Try a small list
Letters are hard to count, so try a list you can count: 1, 2, 3, 4, 5. Here m = 1 and n = 5.
$n - m = 5 - 1 = 4$
But the list has 5 numbers, not 4. The difference counts the steps: 1 to 2, 2 to 3, 3 to 4 and 4 to 5. Both ends are in the list, so there is one more number than steps.
Use the counting rule
$\text{count} = n - m + 1$
$\text{count} = 66 + 1 = 67$
Watch out: 66 is the difference, not the count. Choosing 66, choice C, leaves out one end of the list. Also, consecutive integers go up by 1, not 2, so halving 66 to get 33 is wrong. So the answer is D, 67.

Check your answer

Try m = 1. Then $n = 1 + 66 = 67$ . The list 1, 2, 3, and so on up to 67, plainly has 67 numbers. Any other start gives the same count, because the list just slides along.

More questions like this

The original DOE question

Question 66 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 66. Open page 78 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 27 of 83

2024 Test A Math, question 11

Question

Jar 1 and Jar 2 each contain $\frac{1}{2}$ cup of water. If $\frac{1}{4}$ of the water in Jar 1 is poured into Jar 2, how much water is now in Jar 2?

A $\frac{1}{8}$ cup
B $\frac{1}{4}$ cup
C $\frac{5}{8}$ cup
D $\frac{3}{4}$ cup

Show the answer

Correct answer: C

Jar 1 starts with $\frac{1}{2}$ cup of water. If $\frac{1}{4}$ of its water is poured into Jar 2, then the amount of water added to Jar 2 is $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$ cup.

Jar 2 starts with $\frac{1}{2}$ cup of water. If $\frac{1}{8}$ cup of water is added to it, then the amount of water now in Jar 2 is $\frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8}$ cup.

Question analysis

A fraction word problem about two jars that each hold half a cup of water; a quarter of the water in the first jar is poured into the second. Students find how much water the second jar now holds. Reading exactly what the quarter is a quarter of is the detail that matters.

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 Multiplying and adding positive fractions is review by grade 7, and the problem takes two short steps. It is easier than typical grade 7 work with negative numbers or longer rational-number problems, apart from careful reading.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The numbers are friendly halves, quarters and eighths, and the work is two steps. The question turns on one phrase: the quarter is a fraction of Jar 1's water, not a measured quarter cup, and reading it the second way gives a choice that looks right. Students who also remember that Jar 2 already held water finish correctly.

Turning words into math moderate Understanding the question
The phrase "1/4 of the water in Jar 1" means a quarter of Jar 1's half cup, an eighth of a cup, not a quarter cup; the word of calls for multiplying. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Adding the 1/8 cup to Jar 2's 1/2 cup needs the common denominator 8, so 1/2 becomes 4/8 before the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the amount poured by taking one quarter of Jar 1's half cup, then add that amount to the half cup already in Jar 2.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the amount poured, 1/4 of 1/2 cup = 1/8 cup, and stops there instead of adding it to the half cup already in Jar 2.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks the fraction 1/4 straight from the question, as if the share poured were the amount of water now in Jar 2.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Reads 1/4 of the water as 1/4 cup and adds it to Jar 2's half cup, 1/2 + 1/4 = 3/4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of an amount: $\frac{a}{b} \text{ of } N = \frac{a}{b} \times N$; N is the whole amount, and a over b is the part of it you want. In math, the word of means multiply. For example, half of 10 is one half times 10, which is 5.
Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the tops, and b and d are the bottoms. Multiply the tops together and the bottoms together. Taking a part of a part makes the pieces smaller, so the bottom gets bigger.
Adding fractions with the same bottom: $\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}$; a and b are the tops, and d is the shared bottom, called the denominator. If the bottoms match, add the tops and keep the bottom. If they do not match, rewrite one fraction first.

Sort out the question
Each jar starts with $\frac{1}{2}$ cup of water. You pour “ $\frac{1}{4}$ of the water in Jar 1” into Jar 2. You need how much water is in Jar 2 after that, in cups.
Find how much water is poured
Jar 1 has $\frac{1}{2}$ cup. You pour $\frac{1}{4}$ of it. Of means multiply.
$\frac{a}{b} \text{ of } N = \frac{a}{b} \times N$
$\frac{1}{4} \times \frac{1}{2} = \frac{1 \times 1}{4 \times 2} = \frac{1}{8}$
You pour $\frac{1}{8}$ cup. Watch out: a quarter of the water is not a quarter cup. It is a quarter of a half cup. Adding $\frac{1}{4}$ cup gives $\frac{3}{4}$ cup, choice D.
Make the bottoms match
Now add the poured water to the $\frac{1}{2}$ cup already in Jar 2. The bottoms are 2 and 8, so rewrite the half with 8 on the bottom. Multiply its top and bottom by 4.
$\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}$
Add the water in Jar 2
$\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}$
$\frac{4}{8} + \frac{1}{8} = \frac{5}{8}$
Jar 2 now holds $\frac{5}{8}$ cup. Watch out: $\frac{1}{8}$ cup is only the water poured in. Stopping there gives choice A. So the answer is C, $\frac{5}{8}$ cup.

Check your answer

Jar 2 started with a half cup and got a little more. So it should hold just over a half cup, and $\frac{5}{8}$ is. Also, the water in both jars still adds up to 1 cup: Jar 1 has $\frac{3}{8}$ cup left, and $\frac{3}{8} + \frac{5}{8} = 1$ .

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 2025 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 28 of 83

2024 Test A Math, question 35

Question

Evaluate:

$\left|(-8) - 12 + (-17) - (-31)\right| - \left|24\right|$

A−30
B−18
C18
D44

Show the answer

Correct answer: B

$\left|(-8) - 12 + (-17) - (-31)\right| - \left|24\right|$

Simplify negatives inside the first absolute value brackets.

$\left|-8 - 12 - 17 + 31\right| - \left|24\right|$

Simplify the first absolute value.

$\left|-6\right| - \left|24\right|$

Calculate the absolute values.

$6 - 24$

Perform the final subtraction.

$6 - 24 = -18$

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: 7.NS.1c rational number subtraction
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 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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Many steps minor
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.

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)
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)
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

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Simplifies the inside to −6 but treats the bars like parentheses, keeping −6, so −6 − 24 = −30.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Assumes an expression made of absolute values cannot be negative, so subtracts 6 from 24 instead of 24 from 6.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Reads − (−31) as − 31, so the inside becomes −8 − 12 − 17 − 31 = −68, and 68 − 24 = 44.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding and subtracting negatives: $a + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b$; 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: $\left|a\right| = \text{the distance from } a \text{ to } 0$; 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.

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.
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.
$a + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b$
$(-8) - 12 + (-17) - (-31) = -8 - 12 - 17 + 31$
Watch out: minus negative 31 is plus 31. Treating it as minus 31 makes the inside −68. That leads to 44, choice D.
Combine the four numbers
Go from left to right, one number at a time.
$-8 - 12 = -20$
$-20 - 17 = -37$
$-37 + 31 = -6$
The inside of the first bars is −6.
Take the absolute values
$\left|a\right| = \text{the distance from } a \text{ to } 0$
$\left|-6\right| = 6 \quad \text{and} \quad \left|24\right| = 24$
Now the expression is $6 - 24$ . Watch out: the bars turn −6 into 6. Keeping −6 gives $-6 - 24 = -30$ , choice A.
Do the last subtraction
$6 - 24 = -18$
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 $24 - 6$ 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 $8 + 12 + 17 = 37$ . Since $31 - 37 = -6$ , 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 29 of 83

2024 Test A Math, question 37

Question

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

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

A $\frac{1}{2}$
B1
C $1\frac{1}{2}$
D3

Show the answer

Correct answer: A

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

$1 + 2 = 3$

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

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

Question analysis

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

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 number line problems usually involve negative numbers and fractions as coordinates. This one has no coordinates, only two whole-number distances, and needs three short steps: a total length, a midpoint and a difference.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 39 words.
Corrected from the DOE handbook: An imprecise statement about the midpoint in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is light, but every intermediate result is waiting among the choices, so the risk is stopping before the last subtraction. Students also have to keep two midpoints apart: the one asked about lies between X and Z, not halfway along the 2-unit stretch from Y. Marking X at 0 and the midpoint at 1.5 on the drawing makes the half-unit gap visible.

Reading a figure, table or graph minor Understanding the question
The number line has no coordinates, only dimension arrows labeled 1 and 2, so students must assign positions themselves, such as X at 0, Y at 1 and Z at 3. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Three short results come in sequence, the length XZ, the midpoint's distance from X, then its distance from Y, and each is also a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Halves the 2-unit segment from Y to Z, the midpoint of the wrong pair of points, which is 1 unit from Y; it also matches the labeled XY.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that the midpoint is 1 1/2 units from X and stops before subtracting the 1 unit between X and Y.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two labeled distances, 1 + 2 = 3, and gives the whole length XZ rather than any distance to the midpoint.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the picture
The points go X, then Y, then Z, from left to right. X to Y is 1 unit, and Y to Z is 2 units. You need the midpoint of X and Z, the point halfway between them. Then you need how far Y is from that midpoint. The line has no numbers, so measure everything from X.
Find the length from X to Z
$XZ = XY + YZ$
$XZ = 1 + 2 = 3$
X to Z is 3 units. Watch out: 3 is the whole length, choice D. It is not a distance to the midpoint.
Place the midpoint
The midpoint of X and Z is half of 3 units from X.
$\text{distance from an end to the midpoint} = \frac{\text{length of the segment}}{2}$
$\frac{3}{2} = 1\frac{1}{2}$
So the midpoint is $1\frac{1}{2}$ units to the right of X. Watch out: this is its distance from X, not from Y. Stopping here gives choice C.
Measure from Y to the midpoint
Y is 1 unit to the right of X. The midpoint is $1\frac{1}{2}$ units to the right of X. Subtract to find the gap between them.
$1\frac{1}{2} - 1 = \frac{1}{2}$
Watch out: use the midpoint of X and Z, not of Y and Z. Halving the 2 units from Y to Z gives 1, choice B.
So the answer is A, $\frac{1}{2}$ unit.

Check your answer

Give each point a number, with X at 0. Then Y is at 1 and Z is at 3. Halfway between 0 and 3 is $1\frac{1}{2}$ . From Y at 1 to $1\frac{1}{2}$ is $\frac{1}{2}$ . This fits the picture too: Y sits a little left of the middle of XZ.

More questions like this

The original DOE question

Question 94 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 86, question 94. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 30 of 83

2024 Test A Math, question 49

Question

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

A $3\frac{5}{6}$
B $4\frac{1}{6}$
C $4\frac{2}{3}$
D $7\frac{1}{6}$

Show the answer

Correct answer: A

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

Carolyn walked $1\frac{2}{3}$ miles back, so she still had $5\frac{1}{2} - 1\frac{2}{3}$ miles left to walk home.

$5\frac{1}{2} - 1\frac{2}{3} = \frac{11}{2} - \frac{5}{3}$

$\frac{33}{6} - \frac{10}{6} = \frac{23}{6} = 3\frac{5}{6}$

So Carolyn had $3\frac{5}{6}$ 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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Turning words into math minor
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.

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)
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)
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

B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Subtracts the whole numbers, 5 − 1 = 4, then the fractions the easy way round, 2/3 − 1/2 = 1/6, giving 4 1/6.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
On the store-to-library stretch, uses the 1 2/3 miles already walked instead of the 5/6 mile left, then adds the 3 miles home.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Adds all three distances, 3 + 2 1/2 + 1 2/3 = 7 1/6, which is how far she has walked so far, not how far is left.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mixed number to fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; 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: $\frac{a}{b} = \frac{a \times n}{b \times n}$; 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: $\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}$; 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.

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 $2\frac{1}{2}$ miles more. Carolyn walks back the same way. After $1\frac{2}{3}$ miles she stops at a friend’s house. You need the miles still left to get home.
Find the distance from home to the store
$3 + 2\frac{1}{2} = 5\frac{1}{2}$
The store is $5\frac{1}{2}$ miles from home. She starts back from the store, so this is the full trip home. Watch out: adding all three distances gives $7\frac{1}{6}$ , choice D. That is how far she has walked, not how far is left.
Change to improper fractions
The miles left are $5\frac{1}{2} - 1\frac{2}{3}$ . An improper fraction has a top bigger than its bottom. Change each mixed number to one.
$a\frac{b}{c} = \frac{a \times c + b}{c}$
$5\frac{1}{2} = \frac{5 \times 2 + 1}{2} = \frac{11}{2}$
$1\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3}$
Use sixths for both
To subtract, the fractions need the same bottom. 2 and 3 both go into 6, so change both to sixths.
$\frac{a}{b} = \frac{a \times n}{b \times n}$
$\frac{11}{2} = \frac{11 \times 3}{2 \times 3} = \frac{33}{6}$
$\frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6}$
Subtract and simplify
$\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}$
$\frac{33}{6} - \frac{10}{6} = \frac{23}{6}$
23 sixths is 3 wholes (18 sixths) and 5 sixths more. So $\frac{23}{6} = 3\frac{5}{6}$ . Watch out: $\frac{1}{2}$ is less than $\frac{2}{3}$ . Subtracting the fraction parts the wrong way round gives $4\frac{1}{6}$ , choice B.
So the answer is A, $3\frac{5}{6}$ miles.

Check your answer

Add back: $3\frac{5}{6} + 1\frac{2}{3} = 3\frac{5}{6} + 1\frac{4}{6} = 4\frac{9}{6} = 5\frac{3}{6} = 5\frac{1}{2}$ . That is the full distance from the store to home. Also estimate: 5.5 minus about 1.7 is about 3.8, and $3\frac{5}{6}$ 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 31 of 83

2024 Test A Math, question 51

Question

$3(0.01) - 3(0.1) =$

A−0.33
B−0.27
C0
D0.33

Show the answer

Correct answer: B

$3(0.01) - 3(0.1) =$

$0.03 - 0.3 = -0.27$

Question analysis

This decimal question asks for the value of 3 times 0.01 minus 3 times 0.1. Students multiply each decimal by 3 and subtract, keeping the order of the subtraction as written. The two decimals look alike, so place value and the sign of the result both need care.

NYS standards: 7.NS.1c rational number subtraction
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students subtract rational numbers and get negative results. This short computation adds place-value care, since 0.01 and 0.1 look alike, but it is a routine two-step grade 7 exercise.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 16+, 1 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two quick multiplications leave 0.03 − 0.3, a subtraction whose result is negative. The usual errors are adding the two products, with or without a minus sign, and treating 0.01 and 0.1 as the same number. Rewriting the subtraction as −(0.3 − 0.03) keeps the place values lined up and makes the sign plain.

Error-prone arithmetic moderate Working it out
The products 0.03 and 0.3 have their digits in different places, and taking the larger from the smaller crosses zero, so both alignment and sign need care. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The choice −0.33 has the right sign and is only 0.06 from the answer, so a sense that the result is a little below zero cannot separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Multiply each decimal by 3, then subtract with the place values lined up; or factor out the 3 and compute 3 times (0.01 − 0.1).

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Sees the result must be negative but adds the sizes, −(0.03 + 0.3) = −0.33, instead of subtracting 0.03 from 0.3.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Treats 0.01 and 0.1 as equal, since each shows a single 1 after the point, so the two products cancel to 0.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: weak
Adds the products, 0.03 + 0.3 = 0.33, instead of subtracting the second from the first.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a bigger number: $a - b = -(b - a)$; a and b are numbers, and b is the bigger one. When you take a bigger number away from a smaller one, the answer is negative. Its size is the gap between them, b minus a. So subtract the other way, then put a minus sign in front.

Sort out the problem
3(0.01) means 3 times 0.01. A number right next to parentheses means multiply. Then subtract 3 times 0.1. Note that 0.01 is one hundredth, and 0.1 is one tenth. They are not equal. Watch out: treating them as equal makes the two parts cancel to 0, choice C.
Multiply each part
$3 \times 0.01 = 0.03$
$3 \times 0.1 = 0.3$
So the problem is $0.03 - 0.3$ . Write 0.3 as 0.30 so both have two decimal places. Now you can see that 0.30 is bigger than 0.03.
Subtract the bigger number
You take a bigger number from a smaller one, so the answer is negative.
$a - b = -(b - a)$
$0.03 - 0.30 = -(0.30 - 0.03) = -0.27$
Watch out: find the gap between the numbers. Don’t add them. Adding gives 0.33, choice D, or −0.33, choice A.
So the answer is B, −0.27.

Check your answer

Pull out the 3 first: $3 \times (0.01 - 0.1) = 3 \times (-0.09) = -0.27$ . Same answer. Also think of a number line. Start at 0.03, just right of 0. Move 0.30 to the left. You land a little left of 0, at −0.27.

More questions like this

The original DOE question

Question 108 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 89, question 108. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 32 of 83

2024 Test A Math, question 52

Question

What is the value of $10\frac{1}{2} + \left(-5\frac{1}{3}\right) - \left(-2\frac{3}{4}\right)$ ?

A $2\frac{5}{12}$
B $7\frac{11}{12}$
C $13\frac{1}{12}$
D $18\frac{7}{12}$

Show the answer

Correct answer: B

Rewrite the mixed numbers so that they all have equivalent denominators.

$10\frac{1}{2} + \left(-5\frac{1}{3}\right) - \left(-2\frac{3}{4}\right)$

$10\frac{6}{12} + \left(-5\frac{4}{12}\right) - \left(-2\frac{9}{12}\right)$

Subtracting a negative is the same as adding.

$10\frac{6}{12} - 5\frac{4}{12} + 2\frac{9}{12}$

Group the whole numbers together and the fractions together.

$10\frac{6}{12} - 5\frac{4}{12} + 2\frac{9}{12}$

$(10 - 5 + 2) + \left(\frac{6}{12} - \frac{4}{12} + \frac{9}{12}\right)$

$7\frac{11}{12}$

Question analysis

This signed-number question asks for the value of 10 1/2 plus negative 5 1/3 minus negative 2 3/4. It tests adding and subtracting negative mixed numbers with unlike denominators. Turning each sign combination into a plain addition or subtraction, and then working in twelfths, is where students most often go wrong.

NYS standards: 7.NS.1c rational number subtraction
7.NS.1d properties for adding and subtracting
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding and subtracting negative mixed numbers is central grade 7 work. This three-term expression is typical in length, but it needs a common denominator of 12 and two different sign rules: adding a negative and subtracting one.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 10, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The expression is short, but it packs both sign rules for negative numbers into one line, and each misapplication has its own choice. Subtracting a negative is the rule most often broken, and the result is a reasonable-looking mixed number. Estimating with decimals, 10.5 − 5.3 + 2.8 is about 8, separates the answer from every wrong choice.

Error-prone arithmetic moderate Working it out
Two different sign rules meet: adding −5 1/3 means subtracting 5 1/3, while subtracting −2 3/4 means adding 2 3/4, and mixing them up gives a listed value. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
After the signs, halves, thirds and quarters become twelfths, and the wholes, 10 − 5 + 2, are combined apart from the fractions, 6/12 − 4/12 + 9/12. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite adding a negative as subtraction and subtracting a negative as addition, convert the fractions to twelfths, then combine the whole numbers and the fractions separately.

Traps

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Handles the added negative but treats subtracting −2 3/4 as subtracting 2 3/4, so 5 1/6 − 2 3/4 = 2 5/12.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Drops the negative signs inside the parentheses and keeps the operations, 10 1/2 + 5 1/3 − 2 3/4 = 13 1/12.
D Negatives and direction (Brown & Burton, 1978)
Reveals a surface strategy Expected pull: weak
Ignores every sign and adds all three mixed numbers, 10 1/2 + 5 1/3 + 2 3/4 = 18 7/12.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding and subtracting negatives: $a + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b$; a is any number, and b is a positive number, so −b is negative. Adding a negative moves you left on the number line, just like subtracting. Subtracting a negative is like taking away a debt, so you end up with more. It is the same as adding.
Equivalent fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; 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.

Sort out the expression
You need $10\frac{1}{2} + \left(-5\frac{1}{3}\right) - \left(-2\frac{3}{4}\right)$ . There are three mixed numbers, and two of them are negative. A mixed number is a whole number plus a fraction. Plan: fix the signs, give the fractions the same bottom, then work with the whole numbers and the fractions one group at a time.
Fix the signs
Adding $-5\frac{1}{3}$ means subtracting $5\frac{1}{3}$ . Subtracting $-2\frac{3}{4}$ means adding $2\frac{3}{4}$ .
$a + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b$
$10\frac{1}{2} - 5\frac{1}{3} + 2\frac{3}{4}$
Watch out: subtracting a negative is adding. Subtracting $2\frac{3}{4}$ instead gives $2\frac{5}{12}$ , choice A. Dropping the minus signs gives C or D.
Use twelfths for the fractions
The bottoms are 2, 3 and 4. All three go into 12, so change each fraction to twelfths.
$\frac{a}{b} = \frac{a \times n}{b \times n}$
$\frac{1}{2} = \frac{6}{12}, \quad \frac{1}{3} = \frac{4}{12}, \quad \frac{3}{4} = \frac{9}{12}$
Now the problem is $10\frac{6}{12} - 5\frac{4}{12} + 2\frac{9}{12}$ .
Combine whole numbers, then fractions
Work with the whole numbers and the fractions one group at a time. Keep each sign.
$10 - 5 + 2 = 7$
$\frac{6}{12} - \frac{4}{12} + \frac{9}{12} = \frac{11}{12}$
The fraction part is less than 1, so just put them together: $7\frac{11}{12}$ .
So the answer is B, $7\frac{11}{12}$ .

Check your answer

Estimate with decimals. $10.5 - 5.33 + 2.75$ is about 7.92. And $7\frac{11}{12}$ is also about 7.92, since $\frac{11}{12}$ is almost 1. They match.

More questions like this

The original DOE question

Question 109 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 89, question 109. Open page 89 of the handbook (PDF)

Question 33 of 83

2024 Test A Math, question 56

Question

$Number line from −3 to 8 with a tick at every integer. Point P is at −3, point Q is at 1 and point R is at 7, each marked with a dot.$

On the number line above, what is the distance, in units, between the midpoint of $\overline{PQ}$ and the midpoint of $\overline{QR}$ ?

Show the answer

Correct answer: C

To find the midpoint of a segment, subtract the value on the left from the value on the right and then divide by 2. Then add the result to the value on the left of the segment.

Find the midpoint of $\overline{PQ}$ :

$1 - (-3) = 4$

$4 \div 2 = 2$

$2 + (-3) = -1$

Find the midpoint of $\overline{QR}$ :

$7 - 1 = 6$

$6 \div 2 = 3$

$3 + 1 = 4$

The distance between −1 and 4 on the number line is 5 units.

Question analysis

A number line marks points P, Q and R at labeled integer positions, with P to the left of zero. This midpoint and distance question asks how far apart the midpoints of segments PQ and QR are. Students locate both midpoints and then measure the distance between them, which takes care with negative positions.

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students measure distances on a number line as the difference of two positions, including across zero. Here two midpoints must be located first, but all three points sit on labeled integers, which keeps it at grade 7 level.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 11, 21 words.
Corrected from the DOE handbook: A reversed instruction in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The points sit on labeled integers, so the midpoints can be found by counting, and the question is mostly about finishing carefully. Dropping the negative in 4 − (−1), stopping at the second midpoint, and counting tick marks instead of spaces each give a listed choice. Marking −1 and 4 on the figure and counting the unit spaces between them confirms the distance.

Error-prone arithmetic minor Working it out
P sits at −3, so the midpoint of PQ is (−3 + 1) ÷ 2 = −1, and the final distance, 4 − (−1), subtracts a negative number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Three results come in sequence, the midpoint of PQ, the midpoint of QR, then the gap between them, and the second of them, 4, is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Read the positions of P, Q and R, average each pair of endpoints to place the two midpoints, then subtract the left midpoint from the right one.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Stops one step early: A choice is a number from partway through the solution.
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds the midpoints −1 and 4 but computes the gap as 4 − 1 = 3, dropping the negative sign instead of finding 4 − (−1).
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the midpoint of QR at 4 and gives its position as the answer, stopping before measuring from the other midpoint at −1.
D Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Counts the six tick marks from −1 to 4 inclusive rather than the five unit spaces between them.

What was corrected from the DOE handbook, and why

Explanation: The handbook’s rule says to “subtract the value on the right from the value on the left and then divide by 2,” but its own working does the opposite: 1 − (−3) = 4 for PQ and 7 − 1 = 6 for QR, the right value minus the left. Following the rule as printed gives (−3 − 1) ÷ 2 = −2 and (1 − 7) ÷ 2 = −3, which would put the midpoints at −5 and −2, outside the segments, and lead to a distance of 3, a wrong choice. The site now says “subtract the value on the left from the value on the right.” The midpoints, −1 and 4, and the distance between them, 5 units, are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint on a number line: $\text{midpoint} = \text{left end} + \frac{\text{right end} - \text{left end}}{2}$; The left end and right end are the numbers at the two ends of a segment. Right end minus left end is the segment’s length. The midpoint is halfway, so it is half the length to the right of the left end.
Distance on a number line: $\text{distance} = \text{right point} - \text{left point}$; On a number line, numbers grow as you move right. The distance between two points is the larger number minus the smaller one. That is the right point minus the left point, so a distance is never negative.

Read the number line
P is at −3, Q is at 1 and R is at 7. A bar over two letters, such as $\overline{PQ}$ , means the segment between them. A segment is the straight piece of line between two points. You need the midpoint of each segment, the point exactly halfway between its ends. Then you need the distance between those two midpoints.
Find the midpoint of PQ
The left end is −3 and the right end is 1.
$\text{midpoint} = \text{left end} + \frac{\text{right end} - \text{left end}}{2}$
$\text{midpoint} = -3 + \frac{1 - (-3)}{2} = -3 + \frac{4}{2} = -3 + 2 = -1$
So the midpoint of PQ is at −1. Note that $1 - (-3)$ is $1 + 3$ , which is 4.
Find the midpoint of QR
The left end is 1 and the right end is 7.
$\text{midpoint} = 1 + \frac{7 - 1}{2} = 1 + \frac{6}{2} = 1 + 3 = 4$
So the midpoint of QR is at 4. Watch out: 4 is where a midpoint is, not the answer. Stopping here gives choice B.
Measure between the midpoints
$\text{distance} = \text{right point} - \text{left point}$
$\text{distance} = 4 - (-1) = 4 + 1 = 5$
Watch out: keep the minus sign on −1. Using $4 - 1 = 3$ gives choice A. Also count the spaces, not the tick marks. From −1 to 4 there are 5 spaces but 6 marks, and 6 is choice D.
So the answer is C, 5 units.

Check your answer

Each midpoint should sit in the middle of its segment. −1 is 2 units from both −3 and 1. 4 is 3 units from both 1 and 7. There is also a shortcut. The gap between the midpoints is half of PR: $\frac{7 - (-3)}{2} = \frac{10}{2} = 5$ .

More questions like this

The original DOE question

Question 113 as printed on page 91 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 91, question 113. Open page 91 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 34 of 83

2024 Test B Math, question 4

Question

At one location, the temperature increased from −7° F to 4° F between 5:00 a.m. and 9:00 a.m. By 11:00 a.m., the temperature was 3.5 times the 9:00 a.m. temperature. What was the total increase in temperature between 5:00 a.m. and 11:00 a.m., in degrees Fahrenheit?

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

Show the answer

Correct answer: 21

The 11:00 a.m. temperature is 3.5 times the 9:00 a.m. temperature. The 9:00 a.m. temperature was 4° F.

$4\times 3.5=14$

The temperature increased from −7° F at 5:00 a.m. to 14° F at 11:00 a.m., which is an increase of 21° F.

Question analysis

A temperature word problem with negative numbers: the temperature rises from −7° F to 4° F by 9:00 a.m., and by 11:00 a.m. it is 3.5 times the 9:00 a.m. reading. Students find the total rise from 5:00 to 11:00 a.m. Subtracting a negative starting temperature is where most slips happen.

NYS standards: 7.NS.3 four operations with rational numbers
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Using signed numbers for temperature change is standard grade 7 work. This adds a decimal multiplier and asks for a change rather than a temperature, two short steps, which keeps it at a typical grade 7 level.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 43 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Finding the 11:00 a.m. temperature is a single multiplication; the points are lost in the subtraction that follows. A rise that starts below zero has two parts, 7 degrees up to zero and 14 more above it, and students who treat −7 as 7 lose the first part. The times in the stem also invite measuring from 9:00 a.m. instead of 5:00 a.m., which covers only the last two hours.

Error-prone arithmetic moderate Working it out
The last step is 14 − (−7): subtracting the negative starting temperature means adding 7, and the part of the rise below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
“3.5 times the 9:00 a.m. temperature” multiplies the reading of 4°, not the 11-degree rise, and “total increase” runs from the 5:00 a.m. reading, not from 9:00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Three times of day carry three temperatures, −7°, 4° and 14°, and the answer compares the first with the last, not either of them with the middle one. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find the 11:00 a.m. temperature from the 9:00 a.m. one, then subtract the 5:00 a.m. temperature from it, remembering that subtracting a negative number adds its size.

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.
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

7 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds 14° at 11:00 a.m. but subtracts 7 instead of −7, so 14 − 7 = 7, leaving out the part of the rise below zero.
14 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the 11:00 a.m. temperature, 4 × 3.5 = 14, and grids it without subtracting the 5:00 a.m. temperature to get the increase.
10 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Measures the rise from 9:00 a.m. instead of 5:00 a.m., 14 − 4 = 10, which is only the last two hours of the total increase.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Change in temperature: $\text{change} = \text{end} - \text{start}$; End is the last temperature, and start is the first one. Subtract start from end to find how much it went up. It works with negative numbers too. A positive change means the temperature rose.
Subtracting a negative number: $a - (-b) = a + b$; a is any number, and −b is a negative number. Taking away a negative is the same as adding its size. On a thermometer, going from −7 up to 0 is 7 degrees, so those 7 degrees get added.

Sort out what you know
There are three times and three temperatures, in degrees Fahrenheit (°F).
- 5:00 a.m.: −7°F, which is 7 degrees below zero
- 9:00 a.m.: 4°F
- 11:00 a.m.: 3.5 times the 9:00 a.m. temperature
You need the total increase from 5:00 a.m. to 11:00 a.m. That is how much warmer it got from the first time to the last.
Find the 11:00 temperature
“3.5 times the 9:00 a.m. temperature” means 3.5 times 4°F.
$3.5 \times 4 = 14$
At 11:00 a.m. it was 14°F. Watch out: 14 is a temperature, not the increase. Gridding 14 stops too early.
Subtract the start from the end
The increase runs from 5:00 a.m. to 11:00 a.m. So it starts at −7°F and ends at 14°F.
$\text{change} = \text{end} - \text{start}$
$\text{change} = 14 - (-7)$
Watch out: start at 5:00 a.m., not at 9:00 a.m. Starting from 4°F gives only 10.
Subtract the negative number
Taking away a negative number is the same as adding its size.
$a - (-b) = a + b$
$14 - (-7) = 14 + 7 = 21$
The temperature rose 21°F. Watch out: writing 14 − 7 = 7 loses the minus sign. It skips the 7 degrees the air warmed just to reach zero.
So you would grid 21.

Check your answer

Count up on a thermometer. From −7°F up to 0°F is 7 degrees. From 0°F up to 14°F is 14 more. $7 + 14 = 21$ . Or add the two parts of the morning: it rose 11 degrees by 9:00 and 10 more by 11:00, and $11 + 10 = 21$ .

More questions like this

The original DOE question

Question 61 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 199, question 61. Open page 199 of the handbook (PDF)

Question 35 of 83

2024 Test B Math, question 7

Question

The number line shows points E, F, G, and H.

$Number line from −7 to 7 with a tick at every integer and arrows at both ends; the ticks are labeled −7, −5, −3, −1, 0, 1, 3, 5 and 7. Point E is on −7, point F is on the unlabeled tick between −3 and −1, point G is on the unlabeled tick between 1 and 3, and point H is on 7.$

Which point represents the sum of $2.5+(-4.5)$ ?

Apoint E
Bpoint F
Cpoint G
Dpoint H

Show the answer

Correct answer: B

Find the sum of $2.5+(-4.5)$ :

$2.5+(-4.5)=2.5-4.5=-2$

Point F is located at −2.

Question analysis

A number line question with points E, F, G and H marked on a line labeled from −7 to 7. Students add 2.5 and negative 4.5 and find which point shows the sum. It tests adding a positive and a negative decimal, then reading positions on a line whose labels skip some integers.

NYS standards: 7.NS.1b rational number addition
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Adding rational numbers with different signs and showing the sum on a number line is exactly the grade 7 standard. This is one routine sum with simple decimals, a typical classroom exercise.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 1, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a one-line sum followed by a lookup, and most students finish quickly. The four points are placed so that each common sign mistake lands on one of them: −7, 2 and 7 all come from mishandling the negative. Picturing the sum as a move of 4.5 units to the left from 2.5 shows both the size and the direction at once.

Error-prone arithmetic minor Working it out
Adding −4.5 to 2.5 means taking 4.5 away from 2.5, and the larger size of −4.5 makes the sum negative, a sign rule that is easy to apply backwards. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph minor Understanding the question
Only 0 and the odd numbers are labeled, so point F sits at an unlabeled tick between −3 and −1 and must be found by counting. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Rewrite adding −4.5 as subtracting 4.5, work out the sum, then find it on the line by counting tick marks from a labeled number, one tick per integer.

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

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Adds the sizes 2.5 and 4.5 and makes the total negative, treating both numbers as negative, which lands on −7 at point E.
C Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Subtracts 2.5 from 4.5 to get the right size, 2, but keeps it positive, missing that the negative number has the larger size.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Ignores the negative sign on 4.5 and adds the two numbers, 2.5 + 4.5 = 7, landing on point H at the far right.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding a negative number: $a + (-b) = a - b$; a is any number, and −b is a negative number. Adding a negative number is the same as subtracting its size. On a number line, it means moving b steps to the left.

Sort out the question
You need to find $2.5 + (-4.5)$ . Then find that sum (the answer to the addition) on the number line.
The number line has a tick at every whole number from −7 to 7. Only 0 and the odd numbers have labels.
Turn the addition into subtraction
$a + (-b) = a - b$
$2.5 + (-4.5) = 2.5 - 4.5$
You are taking 4.5 away from 2.5. You take away more than you have, so the answer will be negative.
Do the subtraction
The two sizes differ by 2, because $4.5 - 2.5 = 2$ . The larger size, 4.5, belongs to the negative number. So the sum is negative.
$2.5 - 4.5 = -2$
Watch out: the answer is −2, not 2. Point G is at 2.
Find −2 on the line
−2 has no label. It is the tick between −3 and −1. Point F sits on that tick.
Watch out: adding 2.5 + 4.5 = 7 lands on point H. Making both numbers negative gives −7, point E.
So the answer is B, point F.

Check your answer

Start at 2.5 on the number line and move 4.5 steps to the left. The first 2.5 steps take you to 0. You still have 2 steps to go. Two more steps left of 0 lands on −2, which is point F.

More questions like this

The original DOE question

Question 64 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 200, question 64. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 36 of 83

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 $3\frac{1}{3}$ 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?

A $6\frac{2}{3}$ feet
B $7\frac{1}{3}$ feet
C10 feet
D $13\frac{1}{3}$ feet

Show the answer

Correct answer: A

The posts are each 10 feet long, and $3\frac{1}{3}$ 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:

$10-3\frac{1}{3}=$

$\frac{10}{1}-\frac{10}{3}=$

$\frac{30}{3}-\frac{10}{3}=\frac{20}{3}=6\frac{2}{3}\text{ feet}$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
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.

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

B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Subtracts the whole numbers, 10 − 3 = 7, then carries the 1/3 along instead of taking it away too, getting 7 1/3.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives the post's full 10-foot length, the whole post instead of the part above ground, overlooking the 3 1/3 feet sunk in the hole.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the depth of the hole to the length of the post, 10 + 3 1/3, as if digging the hole made the post taller.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts of a whole: $\text{above} = \text{whole} - \text{below}$; 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: $1 = \frac{3}{3}$; 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.

Sort out the question
Each post is 10 feet long. Part of it goes into a hole $3\frac{1}{3}$ feet deep. The rest sticks up above the ground. You need the height of the part above the ground.
Decide to subtract
The part below and the part above make up the whole post. So take the part below away from the whole.
$\text{above} = \text{whole} - \text{below}$
$\text{above} = 10 - 3\frac{1}{3}$
Watch out: do not add. $10 + 3\frac{1}{3} = 13\frac{1}{3}$ , choice D, makes the post longer than it is.
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.
$1 = \frac{3}{3}$
$10 = 9 + \frac{3}{3} = 9\frac{3}{3}$
Now there are thirds to subtract from.
Subtract wholes and thirds
Subtract the whole numbers, then the thirds.
$9\frac{3}{3} - 3\frac{1}{3} = 6\frac{2}{3}$
That is $9 - 3 = 6$ and $\frac{3}{3} - \frac{1}{3} = \frac{2}{3}$ . Watch out: the one third must be taken away too. Doing 10 − 3 = 7 and then keeping the one third gives $7\frac{1}{3}$ , choice B.
So the answer is A, $6\frac{2}{3}$ feet.

Check your answer

Add the two parts back together: $6\frac{2}{3} + 3\frac{1}{3} = 9\frac{3}{3} = 10$ . 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 37 of 83

2024 Test B Math, question 22

Question

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

A $-1.18$
B $-1.\overline{118}$
C $-1.\overline{18}$
D $-1.\overline{81}$

Show the answer

Correct answer: C

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

$-\frac{13}{11}=-1.18181818...=-1.\overline{18}$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know minor
Error-prone arithmetic minor
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.

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)
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)
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

A Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Stops the division after two decimal places and treats −1.18 as exact, missing that the digits 1 and 8 go on repeating.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Counts the whole-number 1 into the block, seeing 1, 18, 18 … as 118 repeating, and bars 118, which means −1.118118… rather than −1.1818….
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Bars 81, the pair in the wrong order, which gives −1.8181…, equal to −1 9/11 rather than the −1 2/11 that −13/11 is.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction as division: $\frac{a}{b} = a \div b$; 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: $0.\overline{18} = 0.181818\ldots$; 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.

Sort out the question
You need the decimal equal to $-\frac{13}{11}$ . The fraction is negative, so the decimal will be negative too. Work with $\frac{13}{11}$ first. Then put the minus sign back.
Divide 13 by 11
$\frac{a}{b} = a \div b$
Use long division for $13 \div 11$ .
- 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.
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.
$\frac{13}{11} = 1.181818\ldots$
Watch out: −1.18, choice A, stops too soon. The digits never end.
Write it with a bar
$0.\overline{18} = 0.181818\ldots$
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.
$-\frac{13}{11} = -1.\overline{18}$
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, $-1.\overline{18}$ .

Check your answer

Multiply back: $11 \times 1.1818 = 12.9998$ . 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 38 of 83

2024 Test B Math, question 24

Question

What is the value of $\frac{\left(1\frac{5}{7}-1\frac{6}{7}\right)}{\left(3\frac{4}{7}-3\frac{6}{7}\right)}$ ?

A $-\frac{1}{2}$
B $-\frac{1}{7}$
C $\frac{1}{7}$
D $\frac{1}{2}$

Show the answer

Correct answer: D

First, subtract to simplify.

$\frac{1\frac{5}{7}-1\frac{6}{7}}{3\frac{4}{7}-3\frac{6}{7}}=\frac{-\frac{1}{7}}{-\frac{2}{7}}=-\frac{1}{7}\div -\frac{2}{7}$

Follow the process for dividing fractions:

$\left(-\frac{1}{7}\right)\times\left(-\frac{7}{2}\right)=\frac{7}{14}=\frac{1}{2}$

Question analysis

A fractions question: find the value of a quotient whose top is 1 5/7 minus 1 6/7 and whose bottom is 3 4/7 minus 3 6/7. Students subtract mixed numbers that share a denominator and then divide one difference by the other. Keeping track of positive and negative signs through both steps is the main hurdle.

NYS standards: 7.NS.2b rational number division
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students subtract and divide rational numbers, including negatives, and use the sign rules for quotients. This item chains three such operations with easy sevenths, a typical grade 7 exercise.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 10, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The mixed numbers look heavy, but the whole parts cancel and only sevenths remain. What decides the question is sign handling: each difference is negative, and one negative divided by another is positive. Students who stop after the top, subtract instead of divide, or keep a minus sign each find their result among the choices.

Error-prone arithmetic moderate Working it out
Both differences are negative, 1 5/7 − 1 6/7 = −1/7 and 3 4/7 − 3 6/7 = −2/7, and dividing them needs both the sign rule and a reciprocal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Two subtractions must be finished before the division, and a sign lost in either one carries into the final quotient. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out the top and the bottom separately, writing each difference as a signed fraction in sevenths, then divide by multiplying the top by the reciprocal of the bottom and apply the sign rule.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Stops one step early: A choice is a number from partway through the solution.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds −1/7 and −2/7 correctly but keeps a negative sign when dividing, forgetting that a negative divided by a negative is positive.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Works out the top, 1 5/7 − 1 6/7 = −1/7, and stops there without dividing it by the bottom difference.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Subtracts the bottom difference from the top one, −1/7 − (−2/7) = 1/7, instead of dividing the two.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Only the second fraction flips.
Dividing two negative numbers: $(-a) \div (-b) = a \div b$; a and b are positive numbers, and b is not zero. A negative number divided by a negative number gives a positive answer. The two minus signs cancel. Multiplying two negatives works the same way.

Sort out the problem
The long fraction bar means divide. So you need the top difference divided by the bottom difference. A difference is the answer to a subtraction. Work out the top and the bottom first, then divide.
Work out the top
Both numbers on top have the same whole number, 1. So the wholes cancel, and you only need $\frac{5}{7} - \frac{6}{7}$ . You take away more than you have, so the answer is negative.
$1\frac{5}{7} - 1\frac{6}{7} = \frac{5}{7} - \frac{6}{7} = -\frac{1}{7}$
Work out the bottom
Both numbers on the bottom have the whole number 3, so those cancel too.
$3\frac{4}{7} - 3\frac{6}{7} = \frac{4}{7} - \frac{6}{7} = -\frac{2}{7}$
Watch out: do not stop after the top. $-\frac{1}{7}$ , choice B, is only part of the problem.
Divide the top by the bottom
Both results are negative. The two minus signs cancel.
$(-a) \div (-b) = a \div b$
$-\frac{1}{7} \div \left(-\frac{2}{7}\right) = \frac{1}{7} \div \frac{2}{7}$
Now flip the second fraction and multiply.
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
$\frac{1}{7} \times \frac{7}{2} = \frac{7}{14} = \frac{1}{2}$
Avoid the traps and answer
Watch out: keeping a minus sign gives $-\frac{1}{2}$ , choice A. Subtracting the bottom from the top, instead of dividing, gives $\frac{1}{7}$ , choice C.
So the answer is D, $\frac{1}{2}$ .

Check your answer

Think in sevenths. The top is −1 seventh and the bottom is −2 sevenths. That is like $(-1) \div (-2) = \frac{1}{2}$ . Or multiply back: $\frac{1}{2} \times \left(-\frac{2}{7}\right) = -\frac{1}{7}$ , which is the top.

More questions like this

The original DOE question

Question 81 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 205, question 81. Open page 205 of the handbook (PDF)

Question 39 of 83

2024 Test B Math, question 33

Question

If $x=-4$ , what is the sum of x and 10% of $|x|$ ?

A−4.4
B−3.6
C0.4
D4.4

Show the answer

Correct answer: B

First, determine 10% of $|x|$ .

$0.10\bullet|x|=$

$0.10\bullet|-4|=$

$0.10\bullet 4=0.4$

Then determine the sum of x and 10% of $|x|$ .

$-4+0.4=-3.6$

The sum is −3.6.

Question analysis

With x equal to a negative number, students find the sum of x and 10 percent of the absolute value of x. It tests absolute value, a simple percent and adding numbers with different signs. The tricky part is keeping track of which term is negative and which is not.

NYS standards: 7.NS.1b rational number addition
6.NS.7c absolute value as distance
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students add positive and negative decimals and find simple percents. This joins those skills with an absolute value in two short steps, so it matches typical practice, with the sign of each term as the main hazard.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 11 words.
Students, last 30 days: 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, but it asks students to treat the same number two ways: x keeps its negative sign while |x| drops it. Each wrong choice gets the sign wrong in one place, by taking 10% of −4, by making both terms positive, or by stopping at the positive 0.4. Writing the two terms separately, −4 and +0.4, before adding prevents each slip.

Error-prone arithmetic moderate Working it out
The sum mixes a negative integer with a positive decimal, −4 + 0.4, and the result has to keep the negative sign while moving 0.4 toward zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
Students must know that |−4| = 4, so the 10% is taken of a positive number even though x itself is negative. (Anderson & Krathwohl, 2001)

Strategy: Evaluate the absolute value first, take 10 percent of that positive number, and then add the result to x itself, keeping x's negative sign.

Traps

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Takes 10% of x instead of |x|, getting −0.4, and adds it to −4 to reach −4.4, overlooking the absolute value bars.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Finds 10% of |x| = 0.4 correctly and stops there, without adding x as the word sum requires.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Uses 4 for x as well as for |x|, so adds 4 + 0.4 = 4.4 and loses the negative sign of x itself.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{distance from } a \text{ to } 0$; The two bars 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 |−4| = 4.
Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; The whole is the amount the percent is taken of. The percent tells how many out of every 100. The part is the piece you get. Use it to find a percent of any number.

Sort out the question
You know that x = −4. You need the sum of two things: x itself, and 10% of |x|. A sum means you add.
The bars around x mean absolute value. So the 10% is taken of |x|, not of x.
Find the absolute value
−4 is 4 steps from 0 on the number line. So its absolute value is 4. An absolute value is never negative.
$|x| = |-4| = 4$
Take 10 percent of it
$\text{part} = \frac{10}{100} \times 4 = 0.4$
So 10% of |x| is 0.4.
Watch out: If you take 10% of x instead, you get −0.4. Then −4 + (−0.4) = −4.4, choice A. The bars make the number positive first.
Add x and the result
Now add x, which is still −4. Its own minus sign stays.
$-4 + 0.4 = -3.6$
Start at −4 on the number line. Move 0.4 to the right, toward 0. You land on −3.6.
Watch out: Stopping at 0.4 gives choice C. Using 4 in place of x gives 4.4, choice D.
So the answer is B, −3.6.

Check your answer

Adding a small positive number should move −4 a little toward 0. The answer −3.6 is just to the right of −4, so that fits. Also, $-3.6 - (-4) = 0.4$ , the amount you added.

More questions like this

The original DOE question

Question 90 as printed on page 209 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 90. Open page 209 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 40 of 83

2024 Test B Math, question 42

Question

Jevon earns $\frac{1}{2}\%$ per year on the money in his savings account. Which expression is not another way to write $\frac{1}{2}\%$ ?

A $\frac{0.005}{100}$
B $\frac{\frac{1}{2}}{100}$
C $\frac{5}{1{,}000}$
D $\frac{1}{200}$

Show the answer

Correct answer: A

Rewrite the percent as a decimal:

$\frac{1}{2}\%=0.5\%=0.005$

So the correct answer must not equal 0.005:

Option A

$\frac{0.005}{100}=0.00005$

Option B

$\frac{\frac{1}{2}}{100}=0.5\div 100=0.005$

Option C

$\frac{5}{1{,}000}=0.005$

Option D

$\frac{1}{200}=0.005$

All answer options except Option A equal 0.005. Therefore, Option A is the correct answer.

Question analysis

Jevon's savings account earns one-half percent a year, and students must find the expression that is not another way to write one-half percent. It tests converting a percent less than 1 into fractions and decimals. Reading the word not carefully, and evaluating each stacked or unfamiliar fraction correctly, are what make it tricky.

NYS standards: 6.RP.3c percent as a rate per 100
7.NS.2d fractions to decimals by long division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Converting between percents, fractions and decimals is grade 6 and 7 work. A percent smaller than 1 is less familiar, and the question asks which form is not equal, so all four must be checked, which keeps it at typical grade 7 demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A question in the negative moderate
Every choice must be checked moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every expression is built from the same idea, so the question tests whether students can convert a percent smaller than 1 without losing track of decimal places. The odd one out, 0.005 over 100, looks like a correct form, because 0.005 is one-half percent and dividing by 100 seems to turn it into a percent. Converting one-half percent to 0.005 once and comparing each expression with it beats reasoning about each form.

A question in the negative moderate Understanding the question
The bold word not asks for the one expression unequal to one-half percent, so the three expressions that check out are exactly the ones to reject. (Haladyna, Downing & Rodriguez, 2002)
Every choice must be checked moderate Choosing the answer
Three of the four forms are equal, so each has to be converted to a decimal and compared with 0.005 before the odd one out can be named. (Embretson & Wetzel, 1987)
Error-prone arithmetic minor Working it out
One-half percent is 0.5 with the point moved two places, 0.005, and dividing that by 100 moves it two more places, to 0.00005. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write one-half percent as a single decimal by dividing one half by 100, then evaluate each expression as a decimal and compare it with that value.

Traps

NOT, LEAST or EXCEPT: The question asks for the exception, the least, or what is not true.
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Misses the bold not and picks the expression that equals one-half percent most plainly, one half over 100, the literal meaning of percent.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Believes a percent must be written over 100, so 5 over 1,000 looks like a different number, though both equal 0.005.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Divides 1 by 200 and misplaces the decimal point, getting 0.05 instead of 0.005, so 1/200 seems not to equal one-half percent.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction: $p\% = \frac{p}{100}$; p is any number. The % sign means out of 100. So p percent is p divided by 100. Dividing by 100 moves the decimal point two places to the left.

Sort out the question
The question asks which expression is not equal to $\frac{1}{2}\%$ . So three choices are equal to it, and one is different. You need the odd one out.
Write the percent as a decimal
One half is 0.5. A percent means out of 100, so divide by 100.
$\frac{1}{2}\% = 0.5\% = \frac{0.5}{100} = 0.005$
So $\frac{1}{2}\%$ is 0.005. It is half of one percent, a very small amount.
Test each choice
Turn each choice into a decimal. Then compare it with 0.005.
Choice Work it out Equal to 0.005?
A $0.005 \div 100 = 0.00005$ No
B $0.5 \div 100 = 0.005$ Yes
C $5 \div 1{,}000 = 0.005$ Yes
D $1 \div 200 = 0.005$ Yes
Pick the one that is different
Choice A divides by 100 twice. The 0.005 was already divided by 100 once. Dividing again makes it 100 times too small.
Watch out: Choice B is the plainest way to write $\frac{1}{2}\%$ : one half over 100. Picking it means missing the word “not.” Also, 1 ÷ 200 is 0.005, not 0.05, so D is equal too.
So the answer is A, $\frac{0.005}{100}$ .

Choice	Work it out	Equal to 0.005?
A	$0.005 \div 100 = 0.00005$	No
B	$0.5 \div 100 = 0.005$	Yes
C	$5 \div 1{,}000 = 0.005$	Yes
D	$1 \div 200 = 0.005$	Yes

Check your answer

Use fractions instead. Half of one hundredth is $\frac{1}{2} \times \frac{1}{100} = \frac{1}{200}$ , which is choice D. And $\frac{5}{1{,}000}$ simplifies to $\frac{1}{200}$ too. Only A is a different amount.

More questions like this

The original DOE question

Question 99 as printed on page 211 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 99. Open page 211 of the handbook (PDF)

Question 41 of 83

2024 Test B Math, question 49

Question

Suppose that w, x, y, and z are integers, none of which are zero. Under which set of conditions would the product $x^{2}yz^{3}|w|$ always be positive?

Ax is positive and w is positive.
Bx is negative and w is negative.
Cy is negative and z is negative.
Dx and y are negative, and z and w are positive.

Show the answer

Correct answer: C

First, analyze the role of each variable in the expression $x^{2}yz^{3}|w|$ .

$x^{2}$ will always be a positive number.

y can be a positive or a negative number.

$z^{3}$ can be a positive or a negative number.

$|w|$ will always be a positive number.

The expressions $x^{2}$ and $|w|$ will always be positive. Therefore, making the values of y and z negative would result in the expression $yz^{3}$ being positive, since a negative multiplied by a negative is positive. Thus the given product would always be positive.

Question analysis

Four nonzero integers appear in a product that includes a square, a cube and an absolute value. Students choose the set of conditions that makes the product positive every time. It tests the sign rules for multiplication together with exponents and absolute value, and the trap is judging the sign by counting negative variables without looking at what is done to each.

NYS standards: 7.NS.2a rational number multiplication
6.EE.1 whole-number exponent expressions
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students learn sign rules for products of signed numbers. Here they must reason about four unknown integers under an even power, an odd power and an absolute value, and decide what is always true, which is more abstract than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 58 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One insight decides the question: x² and |w| are positive whatever x and w are, so only the signs of y and z³ matter. Students who count negative letters instead of negative factors pick conditions that look balanced but leave y as the only negative factor. Students who test a single set of values can be fooled by a choice that leaves y and z unspecified.

Variables instead of numbers moderate Understanding the question
There are four variables and no numbers, and always positive must hold for every allowed value, so one test case can rule a choice out but never confirm it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
Each set of conditions has to be checked against all four factors; the first two fix only x and w and leave y and z free to make the product negative. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
Students must know that a nonzero square and an absolute value are always positive, while an odd power such as z³ keeps the sign of z. (Anderson & Krathwohl, 2001)

Strategy: Work out, factor by factor, which signs each part of the product can take, then test each set of conditions against every factor rather than counting negatives.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Checks only one number (Schoenfeld, 1985)
Reveals partial knowledge Expected pull: moderate
Tests the choice with every variable positive, so the product is positive and it seems to work; y = −1 with z = 1 would make it negative.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Pairs the negatives x and w as if they must cancel, not seeing that x² and |w| are positive anyway, so y and z still decide.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Counts two negative letters, x and y, as an even number of negatives, but x² is positive, so y is the only negative factor.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Signs in multiplication: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.

Sort out the question
w, x, y and z are integers, meaning whole numbers or their negatives. None of them is 0. The product $x^2 y z^3 |w|$ means $x^2 \times y \times z^3 \times |w|$ .
You need the conditions that make the product positive every time, for every number that fits them.
Look at x squared and |w|
$x^2$ means x × x. Two positives or two negatives multiply to a positive. So $x^2$ is always positive.
$|w|$ is the absolute value of w, its distance from 0. It is always positive, since w is not 0.
So x and w never change the sign of the product. Their signs do not matter.
Look at y and z cubed
y can be positive or negative. $z^3$ means z × z × z, and it has the same sign as z. For example, $(-2)^3 = -8$ .
So the sign of the whole product is the sign of $y \times z^3$ . It depends only on y and z.
Test each choice
- A and B say nothing about y and z. If y = −1 and z = 1, the product is negative. So they do not always work.
- C: y is negative, and $z^3$ is negative. A negative times a negative is positive. This always works.
- D: y is negative and z is positive. So the product is always negative.
Watch out: D has two negative letters, x and y. But $x^2$ is positive anyway, so y is the only negative factor.
So the answer is C, y is negative and z is negative.

Check your answer

Try numbers for C. Let x = −2, y = −3, z = −1 and w = −5. Then $x^2 \times y \times z^3 \times |w| = 4 \times (-3) \times (-1) \times 5 = 60$ , which is positive. Even with x and w negative, the product stays positive.

More questions like this

The original DOE question

Question 106 as printed on page 213 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 106. Open page 213 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 42 of 83

2024 Test B Math, question 53

Question

Which graph represents the solution to $|x+1|\geq 2$ ?

A $Number line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1. The line is shaded from −3 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.$
B $Number line from −3 to 3 with a tick at every integer. Closed dots at −1 and 1. The line is shaded from −1 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.$
C $Number line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1, with the segment between them shaded.$
D $Number line from −3 to 3 with a tick at every integer. Closed dots at −3 and 3. The line is shaded from −3 to the left, ending in an arrow, and from 3 to the right, ending in an arrow.$

Show the answer

Correct answer: A

Since $|x+1|\geq 2$ , either $x+1\leq -2$ or $x+1\geq 2$ .

Solving $x+1\leq -2$ for x gives $x\leq -3$ .

Solving $x+1\geq 2$ for x gives $x\geq 1$ .

A filled circle on a number line represents “or equal to.” So the correct number line representation has filled circles at −3 and 1, with the number line filled to the left of −3 for “less than” and to the right of 1 for “greater than.”

Question analysis

Students choose the number line graph that shows every solution of an inequality in which x + 1 sits inside absolute value bars. It tests absolute value as distance and graphing inequalities. The tricky part is turning one absolute value statement into two pieces and placing each boundary point and shaded ray correctly.

NYS standards: 7.NS.1c rational number subtraction
7.EE.4b two-step inequalities
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students solve and graph two-step inequalities and know absolute value as distance. An inequality with an absolute value has to be split into two cases, which is outside the grade 6 to 8 standards and well above typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 7 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Reading |x + 1| as the distance from x to −1 makes this quick: the points 2 units away are −3 and 1, and greater than or equal means outside them. Students who solve only one case and mirror it across 0 land on graphs centered on 0, and students who find both endpoints may still shade between them. The graphs differ in just one dot or one shading choice, so each must be read carefully.

Something you must already know moderate Working it out
Solving |x + 1| ≥ 2 needs two cases, x + 1 ≤ −2 or x + 1 ≥ 2, joined by or; absolute value inequalities go beyond grade 7, and the stem gives no hint. (Anderson & Krathwohl, 2001)
A very close wrong answer moderate Choosing the answer
The right graph and the segment from −3 to 1 share both filled endpoints, so only deciding between outside and between the boundaries separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
Four number lines must be read for filled dots and arrow directions, and two of them differ from the right one only in where a single dot sits. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the absolute value as distance from −1 on the number line, find the points exactly 2 units away, and decide whether the solutions lie between or outside them.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Solves x + 1 ≥ 2 to get x ≥ 1, then mirrors it across 0 to x ≤ −1, as if the solution were centered on 0 rather than on −1.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Finds −3 and 1 but shades between them, the solution of |x + 1| ≤ 2, reversing which side of the boundaries lies at least 2 units from −1.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Solves x + 1 ≤ −2 to get x ≤ −3, then mirrors it across 0 to x ≥ 3, assuming the boundaries are symmetric about 0.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value as distance: $|x - a| = \text{distance from } x \text{ to } a$; x and a are points on the number line. The bars mean absolute value. |x − a| is how far apart the two points are, which is never negative. Since x + 1 = x − (−1), |x + 1| is the distance from x to −1.

Sort out the question
You need the graph of every x that makes $|x + 1| \geq 2$ true. The sign ≥ means greater than or equal to. The bars mean absolute value.
On a number line, a filled dot means that point counts. A thick line with an arrow means all the numbers that way count too.
Read it as a distance
Adding 1 is the same as subtracting −1, so $x + 1 = x - (-1)$ . By the formula, $|x + 1|$ is the distance from x to −1.
So the inequality says that x is at least 2 units away from −1.
Find the two boundary points
Start at −1 and go 2 units each way.
$-1 - 2 = -3$
$-1 + 2 = 1$
At −3 and at 1, the distance is exactly 2. The words or equal to let them count, so they get filled dots.
Decide which side to shade
Being at least 2 away means being farther out. Numbers between −3 and 1 are too close to −1. For example, 0 is only 1 unit away. So shade to the left of −3 and to the right of 1.
In algebra, that is $x + 1 \leq -2$ or $x + 1 \geq 2$ . These give $x \leq -3$ or $x \geq 1$ .
Match the graph
Choice A has filled dots at −3 and 1, with arrows going outward. That matches.
Watch out: Choice C shades between −3 and 1. Those are the points within 2 of −1, the answer to $|x + 1| \leq 2$ . Choices B and D are centered on 0, but the center here is −1.
So the answer is A, the graph shaded outward from −3 and 1.

Check your answer

Test a point from each part. For x = 2, $|2 + 1| = 3$ , and 3 is at least 2. For x = 0, $|0 + 1| = 1$ , which is less than 2. For x = −3, $|-3 + 1| = 2$ , which counts. Graph A agrees with all three.

More questions like this

The original DOE question

Question 110 as printed on page 214 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 110. Open page 214 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 43 of 83

2024 Test B Math, question 57

Question

If x and y are both negative integers with $x<y$ , which expression has the least value?

A $x^{2}y$
B $xy^{2}$
C $-x^{2}y$
D $-xy^{2}$

Show the answer

Correct answer: A

Since x and y are both negative integers, both $x^{2}$ and $y^{2}$ are positive.

Therefore, both $x^{2}y$ and $xy^{2}$ are negative, and both $-x^{2}y$ and $-xy^{2}$ are positive.

Because $x<y$ and both x and y are negative, $|y|<|x|$ and $y^{2}<x^{2}$ . So $|y^{2}|<|x^{2}|$ .

Since $|y|<|x|$ and x and y are integers, $|xy^{2}|<|x^{2}y|$ .

So $x^{2}y<xy^{2}$ .

$-xy^{2}=|xy^{2}|\text{ and }-x^{2}y=|x^{2}y|$

Therefore, $x^{2}y<xy^{2}<-xy^{2}<-x^{2}y$ .

Question analysis

Two negative integers x and y, with x less than y, appear in four expressions built from squares and minus signs. Students find the expression with the least value. It tests multiplying negative numbers and comparing their sizes, and the difficulty is tracking how each minus sign and each square changes both the sign and the size of the result.

NYS standards: 7.NS.2a rational number multiplication
6.NS.7d absolute value versus order
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students multiply negative numbers, but this asks them to compare four variable expressions for any pair of negative integers. Deciding how the squares compare when x is less than y goes beyond typical grade 7 sign practice.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
Every choice must be checked minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two facts settle the question: squares of negatives are positive, and the more negative number has the larger square. Students who track the minus signs carelessly think a leading minus makes an expression negative, and students who assume the larger y has the larger square choose the wrong negative expression. Substituting x = −3 and y = −2 gives −18, −12, 18 and 12, which orders all four at once.

Variables instead of numbers moderate Understanding the question
No values are given, only that x and y are negative integers with x < y, so students must reason about signs or invent a pair such as −3 and −2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
x²y and xy² use the same letters with the exponent moved; both are negative, and only comparing x² with y² separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Every choice must be checked minor Choosing the answer
All four expressions have to be placed in order, because the minus signs in front of the last two turn those expressions positive. (Embretson & Wetzel, 1987)

Strategy: Pick two negative integers with x less than y, such as −3 and −2, evaluate all four expressions, and confirm the order with a second pair.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Knows x²y and xy² are both negative but takes y² as the larger square because y > x, though the more negative x has the bigger square.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reads the leading minus as making −x²y negative, missing that y is negative, so −x²y is in fact positive.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Assumes the leading minus makes −xy² negative, missing that x is negative, so −xy² is in fact positive.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Signs in multiplication: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.
Squaring a number: $x^2 = x \times x$; x is any number. x² (x squared) means x times itself. The square of a number other than 0 is always positive, because two negatives or two positives multiply to a positive.

Sort out the question
x and y are negative integers, such as −3 and −2. Since x < y, x is to the left of y on the number line. So x is more negative, farther from 0.
You need the expression with the least value. That is the most negative one.
Find the sign of each expression
Squares are positive, so $x^2$ and $y^2$ are positive.
- $x^2 y$ is a positive times a negative, so it is negative.
- $x y^2$ is a negative times a positive, so it is negative.
- $-x^2 y$ and $-x y^2$ are the opposites of these, so they are positive.
So the least value must be A or B. Watch out: The minus sign in front of C and D does not make them negative. It turns a negative into a positive.
Compare A and B with numbers
Try x = −3 and y = −2, since −3 < −2.
$x^2 y = (-3)^2 \times (-2) = 9 \times (-2) = -18$
$x y^2 = (-3) \times (-2)^2 = (-3) \times 4 = -12$
Since −18 < −12, A is smaller.
See why it always works
Both expressions contain x × y, which is positive. $x^2 y = (xy) \times x$ and $x y^2 = (xy) \times y$ . Multiplying the same positive number by the more negative x gives the more negative result.
Watch out: y is bigger than x, but $y^2$ is smaller than $x^2$ : 4 is less than 9. Thinking y² is bigger leads to choice B.
So the answer is A, $x^2 y$ .

Check your answer

Try another pair, x = −5 and y = −1. Then A is $25 \times (-1) = -25$ , B is $(-5) \times 1 = -5$ , C is 25 and D is 5. A is still the least.

More questions like this

The original DOE question

Question 114 as printed on page 215 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 114. Open page 215 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 44 of 83

2023 Test A Math, question 2

Question

$6 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}$

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

$6 - 9 \div \left\|-3\right\| + (-2)^{3} \times 1\frac{1}{2}$	Calculate the exponent and convert the mixed number to an improper fraction.
$6 - 9 \div \left\|-3\right\| + (-8) \times \frac{3}{2}$	Next, calculate the absolute value.
$6 - 9 \div 3 + (-8) \times \frac{3}{2}$	Perform the indicated multiplication and division.
$6 - 3 + (-12)$	Perform the addition and subtraction.
$3 + (-12) = -9$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
An answer in an unexpected form minor
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.

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)
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)
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

15 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes (−2)³ as positive 8, as if every power of a negative were positive, so the last product is +12 and 6 − 3 + 12 = 15.
-13.5 Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Works strictly from left to right: 6 − 9 = −3, divided by 3 is −1, plus −8 is −9, and times 1½ gives −13.5.
-6 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Multiplies the base by the exponent, taking (−2)³ as −6, so the product with 3/2 is −9 and the total is 6 − 3 − 9 = −6.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\text{grouping} \rightarrow \text{powers} \rightarrow \times \text{ and } \div \rightarrow + \text{ and } -$; 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: $a^{3} = a \times a \times a$; 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: $\left|a\right| = \text{the distance from } a \text{ to } 0$; 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: $a\frac{b}{c} = \frac{a \times c + b}{c}$; 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.

Sort out the expression
You need the value of $6 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}$ . 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.
Do the absolute value and the power
Grouping symbols and powers come first.
$\left|-3\right| = 3$
$(-2)^{3} = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$
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.
Change the mixed number to a fraction
A fraction is easier to multiply than a mixed number.
$1\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}$
Now the expression looks like this:
$6 - 9 \div 3 + (-8) \times \frac{3}{2}$
Multiply and divide next
Multiplying and dividing come before adding and subtracting.
$9 \div 3 = 3$
$(-8) \times \frac{3}{2} = \frac{-24}{2} = -12$
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.
Add and subtract last
Now only adding and subtracting are left. Work from left to right.
$6 - 3 + (-12) = 3 + (-12) = -9$
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 45 of 83

2023 Test A Math, question 7

Question

In one week, $1\frac{3}{4}$ inches of rain fell on Monday, $2\frac{2}{3}$ inches fell on Tuesday, and $\frac{7}{8}$ inch fell on Wednesday. How many inches of rain fell during those three days?

A $5\frac{7}{24}$
B $5\frac{1}{24}$
C $3\frac{4}{5}$
D $3\frac{1}{2}$

Show the answer

Correct answer: A

To solve, add the three given fractions:

$1\frac{3}{4} + 2\frac{2}{3} + \frac{7}{8}$

The common denominator is 24:

$= 1\frac{18}{24} + 2\frac{16}{24} + \frac{21}{24}$

$= 4\frac{10}{24} + \frac{21}{24}$

$= 5\frac{7}{24}$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
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.

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)
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

B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single renaming mistake gives 49 twenty-fourths; this choice sits exactly 1/4 below the right total, so it mostly catches guesses near it.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the whole numbers to get 3, then adds the fractions straight across, (3 + 2 + 7)/(4 + 3 + 8) = 12/15 = 4/5.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Finds the common denominator 24 but adds the numerators without renaming them, (3 + 2 + 7)/24 = 1/2, and adds that to the wholes, 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions: $\frac{a}{b} = \frac{a \times k}{b \times k}$; 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: $\frac{a}{d} + \frac{b}{d} + \frac{c}{d} = \frac{a + b + c}{d}$; 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: $\frac{a}{b} = q\frac{r}{b}$; 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.

Sort out the question
Rain fell on three days: $1\frac{3}{4}$ inches, $2\frac{2}{3}$ inches and $\frac{7}{8}$ 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.
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.
Rename each fraction in 24ths
Multiply the top and the bottom by the same number.
$\frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}$
$\frac{2}{3} = \frac{2 \times 8}{3 \times 8} = \frac{16}{24}$
$\frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24}$
Watch out: change the tops too. Keeping 3, 2 and 7 over 24 gives $\frac{12}{24} = \frac{1}{2}$ and a total of $3\frac{1}{2}$ , choice D.
Add the fractions
Now the pieces are the same size, so add the tops.
$\frac{18}{24} + \frac{16}{24} + \frac{21}{24} = \frac{55}{24}$
That is more than 1 whole. 24 goes into 55 twice, with 7 left over.
$\frac{55}{24} = 2\frac{7}{24}$
Watch out: never add the bottoms. Adding straight across gives $\frac{12}{15} = \frac{4}{5}$ and a total of $3\frac{4}{5}$ , choice C.
Put the parts together
Add the 3 whole inches from before.
$3 + 2\frac{7}{24} = 5\frac{7}{24}$
So the answer is A, $5\frac{7}{24}$ inches.

Check your answer

Use decimals. $1\frac{3}{4}$ is 1.75, $2\frac{2}{3}$ is about 2.67, and $\frac{7}{8}$ is 0.875. They add to about 5.29. And $\frac{7}{24}$ is about 0.29. So $5\frac{7}{24}$ fits, and $5\frac{1}{24}$ 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 46 of 83

2023 Test A Math, question 10

Question

$A number line from 0 to 10 with a tick and a label at every whole number and arrows at both ends. Point W is marked with a dot at 0, and point X is marked with a dot at 2.$

Points Y and Z are not shown on the number line above. If X is the midpoint of $\overline{WY}$ , and Y is the midpoint of $\overline{WZ}$ , where on the number line would point Z be located?

Show the answer

Correct answer: D

First, find where Y is located on the number line. Point X is the midpoint of $\overline{WY}$ . Since there are 2 units between W and X, there must also be 2 units between X and Y. Thus, Y is located at 4 on the number line.

Point Y is the midpoint of $\overline{WZ}$ . Use the same reasoning as above. Since there are 4 units between W and Y, there must be 4 units between Y and Z. So, Z is located at 8 on the number line.

Question analysis

A number line shows point W at 0 and point X at 2, and two more points, Y and Z, are described but not drawn. Students use two midpoint facts, one after the other, to find where Z lies. Each midpoint fact is about a different segment, so keeping track of which point is the middle one matters.

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Midpoints on a number line rest on distance, which grade 7 students find as a difference. Here every point lies at a whole number of zero or more, so the work is two short steps with no negative numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each midpoint statement doubles a distance from W, so the problem is two short doublings, 2 to 4 and then 4 to 8. The risk is stopping halfway, since Y's position is found first and appears among the choices, or reusing X as an endpoint in the second statement. Rereading which point the question asks about, and checking that Y is halfway between W and Z, catches both.

Many steps minor Working it out
The two midpoint statements must be used in order: the first places Y at 4, and only then can the second, about segment WZ, place Z. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Reading a figure, table or graph minor Understanding the question
Only W at 0 and X at 2 are marked on the 0-to-10 line, and the overbars name WY and WZ as segments, so Y and Z have to be placed from the statements alone. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the first midpoint to place Y as far beyond X as X is from W, then use the second midpoint the same way to place Z beyond Y.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives 2, the position of point X, the only labeled point besides W, without using either midpoint statement.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Uses the first statement to place Y at 4 and answers with Y's position instead of going on to locate Z, the point the question asks about.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Treats Y as the midpoint of XZ instead of WZ, so Z lands 2 units past Y, at 6, continuing the even steps from W to X to Y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint of a segment: $AM = MB$; A and B are the ends of a segment, a straight piece of a line. M is its midpoint, the point exactly halfway. AM is the distance from A to M, and MB is the distance from M to B. A midpoint splits a segment into two equal parts, so these are equal.

Sort out the question
The number line shows W at 0 and X at 2. Points Y and Z are not shown. X is the midpoint of segment WY. Y is the midpoint of segment WZ. A midpoint is the point exactly halfway along a segment. You need the number where Z sits.
Place point Y
X is halfway from W to Y. So X to Y is as long as W to X.
$WX = 2 - 0 = 2$
$XY = WX = 2$
So Y is 2 units past X.
$2 + 2 = 4$
Point Y is at 4. Watch out: this is Y, not Z. Stopping here gives 4, choice B.
Place point Z
Now Y is halfway from W to Z. W to Y is 4 units, so Y to Z is also 4 units.
$YZ = WY = 4 - 0 = 4$
$4 + 4 = 8$
Watch out: the second midpoint is for segment WZ, not XZ. Taking just 2 more units from Y puts Z at 6, choice C. So the answer is D, 8.

Check your answer

Check both midpoints. Halfway from 0 to 4 is $(0 + 4) \div 2 = 2$ , which is X. Halfway from 0 to 8 is $(0 + 8) \div 2 = 4$ , which is Y. Both facts in the question are true.

More questions like this

The original DOE question

Question 67 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 67. Open page 78 of the handbook (PDF)

Question 47 of 83

2023 Test A Math, question 17

Question

A customer wants to buy a pair of hiking boots.

The original price of the boots is $85.75.
The store is offering a 15% discount on all boot purchases.
The customer has a coupon for an additional 25% off the sale price.
The tax rate is 8.5%.

What is the final cost of the boots, including the tax, to the nearest cent?

A$55.82
B$59.08
C$59.32
D$63.17

Show the answer

Correct answer: C

First, determine the sale price after the 15% discount is applied.

Since the customer is saving 15%, this means they are paying 85% $(100 - 15)$ , so multiply the original price by 85% (0.85) to find the discounted price:

$\$85.75 \times 0.85 \approx \$72.89$

Next, find the price after the 25% coupon has been applied:

Since the customer is saving 25% off the sale price, this means they are paying 75% $(100 - 25)$ , so multiply the price found in the first step by 75% (0.75) to find the second discounted price:

$\$72.89 \times 0.75 \approx \$54.67$

Last, find the final cost after an 8.5% tax is added to the second discounted price.

Since the customer is paying 8.5% tax, this means they are paying 108.5% $(100 + 8.5)$ , so multiply the second discounted price by 108.5% (1.085) to find the final cost:

$\$54.67 \times 1.085 \approx \$59.32$

Question analysis

A pair of hiking boots costs $85.75, goes on sale at 15% off, gets an extra 25% off with a coupon, and is then taxed at 8.5%. Students find the final cost to the nearest cent. Each percent applies to the price left after the step before, so the discounts cannot simply be added, and the decimals call for careful multiplication.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 A discount followed by sales tax is standard grade 7 percent work. This item stacks two discounts and a tax, with cents to keep at every stage, so it takes about twice the steps and decimal multiplication of a typical problem.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic major
Turning words into math moderate
Slow to do carefully moderate
Solution: 3 steps
Reading level: Question and choices: about grade 4, 58 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question tests whether students treat successive percents as successive multipliers, and then holds them to three long multiplications. Adding the 15% and 25% into one discount is the tempting shortcut, and a tax read as a dollar amount is the other. The key rounds to the cent at each stage, as a register would; carried exactly, the result is $59.31, so the nearest choice, $59.32, still stands.

Error-prone arithmetic major Working it out
Three products of cents and decimals are worked by hand: 85.75 × 0.85 = 72.8875, then 72.89 × 0.75 = 54.6675, then 54.67 × 1.085 ≈ 59.32, each rounded to the cent. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math moderate Understanding the question
The phrase "an additional 25% off the sale price" means 25% of the already discounted $72.89, not of the $85.75 original, so the two discounts multiply rather than add. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Slow to do carefully moderate Choosing the answer
A reliable method needs three long decimal multiplications without a calculator, well past the minute and a half a question allows, which tempts students to combine the percents. (Lu & Sireci, 2007)

Strategy: Multiply by 0.85 for the sale, by 0.75 for the coupon on the sale price, and by 1.085 for the tax, rounding to the cent at each stage as a register would.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes the 25% coupon off the original $85.75 instead of the sale price, so the discounts add to 40%: 85.75 × 0.60 × 1.085 ≈ $55.82.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives $59.08; it sits 24 cents below the right total and matches no discount or tax mistake, so it mostly catches guesses.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Applies both discounts correctly, reaching $54.67, but adds $8.50 as the tax, reading 8.5% as a dollar amount instead of a percent of the price.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Price after a discount: $\text{new price} = \text{price} \times \left(1 - \frac{p}{100}\right)$; p is the percent off. A discount is money taken off the price. If you save p percent, you pay the rest, 100 − p percent. For 15% off you pay 85%, so multiply by 0.85.
Price with tax: $\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)$; The price is the cost before tax. t is the tax rate, a percent. Tax is extra money added on top. You pay the whole price, 100%, plus t percent more. So for 8.5% tax you multiply by 1.085.

Sort out the question
The boots cost $85.75 at first. The store takes 15% off. Then a coupon takes 25% off the sale price. The sale price is the price after the first discount. Then 8.5% tax is added. You need the final cost to the nearest cent. A cent is $0.01, the second place after the decimal point.
Take off the 15% discount
You save 15%, so you pay 85%. That is 0.85 of the price.
$\text{new price} = \text{price} \times \left(1 - \frac{p}{100}\right)$
$\text{new price} = 85.75 \times \left(1 - \frac{15}{100}\right) = 85.75 \times 0.85 = 72.8875$
A price can’t have part of a cent, so round to two decimal places. The third digit is 7, which is 5 or more, so round up.
$72.8875 \approx 72.89$
The sign ≈ means about. The sale price is $72.89.
Take 25% off the sale price
The coupon takes 25% off $72.89, so you pay 75% of it.
$72.89 \times 0.75 = 54.6675 \approx 54.67$
The price is now $54.67. Watch out: take the coupon off the sale price, not the first price. Adding 15% and 25% to get 40% off $85.75 leads to $55.82, choice A.
Add the 8.5% tax
You pay the price plus 8.5% more, which is 108.5% of it.
$\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)$
$\text{total} = 54.67 \times 1.085 = 59.31695 \approx 59.32$
Watch out: 8.5% is a percent of the price, not $8.50. Adding $8.50 gives $63.17, choice D.
See why rounding each time matters
A store rounds to the cent after each step, as you just did. The test’s answer is worked the same way. Keep every digit and round only at the end, and you get this:
$85.75 \times 0.85 \times 0.75 \times 1.085 = 59.3122\ldots \approx 59.31$
That is $59.31, which is not a choice. The first two roundings each went up by a tiny part of a cent. That was just enough to push the total from $59.31 to $59.32. So here, round at each step, as a store would. So the answer is C, $59.32.

Check your answer

Check the tax a second way. 8.5% of $54.67 is $0.085 \times 54.67 = 4.64695$ , about $4.65. Then $54.67 + 4.65 = 59.32$ . Also estimate: two discounts leave a bit less than two-thirds of $86, near $55. Tax adds under $5, so about $59 or $60 makes sense.

More questions like this

The original DOE question

Question 74 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 74. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 48 of 83

2023 Test A Math, question 20

Question

$\frac{3^{2} + (-8)^{2} + 2^{2}}{(3 - 8 + 2)^{2}} =$

A−60
B $-\frac{17}{3}$
C $\frac{77}{9}$
D68

Show the answer

Correct answer: C

$\frac{3^{2} + (-8)^{2} + 2^{2}}{(3 - 8 + 2)^{2}} =$

$\frac{9 + 64 + 4}{(-3)^{2}} = \frac{77}{9}$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
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.

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)
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

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Squares as if the minus sat outside the parentheses, (−8)² as −64, and also cancels the 9 on top against the 9 below, leaving −64 + 4 = −60.
B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Squares as if the minus sat outside the parentheses, taking (−8)² as −64, so the top is 9 − 64 + 4 = −51 and −51/9 = −17/3.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Cancels the 9 in the numerator, 9 + 64 + 4, against the denominator's 9, leaving 64 + 4 = 68, though only a factor of the whole top can cancel.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Square of a negative: $(-a)^{2} = (-a) \times (-a) = a^{2}$; 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: $\frac{\text{top}}{\text{bottom}} = (\text{top}) \div (\text{bottom})$; 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.

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.
Work out the top
Square each number first. Then add.
$3^{2} = 9$
$(-8)^{2} = (-8) \times (-8) = 64$
$2^{2} = 4$
$9 + 64 + 4 = 77$
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 $-\frac{17}{3}$ , choice B.
Work out the bottom
Work inside the parentheses first. Add and subtract before you square.
$3 - 8 + 2 = -5 + 2 = -3$
$(-3)^{2} = (-3) \times (-3) = 9$
The bottom is 9.
Divide and simplify
$\frac{77}{9}$
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, $\frac{77}{9}$ .

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 $-\frac{17}{3}$ . Also, $\frac{77}{9}$ 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 49 of 83

2023 Test A Math, question 33

Question

If $m = 5$ and $t = -1$ , what is the value of $\frac{6 - 8(2 - t)}{2m + 4(3 - m)}$ ?

A−9
B−1
C1
D9

Show the answer

Correct answer: A

$\frac{6 - 8(2 - t)}{2m + 4(3 - m)} =$

$\frac{6 - 8[2 - (-1)]}{2(5) + 4(3 - 5)} =$

$\frac{6 - 8(3)}{10 + 4(-2)} =$

$\frac{6 - 24}{10 + (-8)} =$

$\frac{-18}{2} = -9$

Question analysis

This algebra question gives m = 5 and t = −1 and asks for the value of a fraction whose numerator and denominator both contain parentheses. Students substitute the values and follow the order of operations. Subtracting a negative number and multiplying by a negative difference make sign mistakes likely, and the four choices differ only in sign and size.

NYS standards: 6.EE.2c evaluating expressions and order of operations
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Substituting values into expressions with negative numbers is everyday grade 7 work. This fraction has a numerator and a denominator to evaluate separately, with a minus sign before a negative and a negative difference, so it is a longer version of a standard exercise.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
A very close wrong answer minor
Solution: 4 steps
Reading level: Question and choices: about grade 8, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a plain evaluation with no story, and every point is lost on signs. Subtracting the negative t, then handling the two negative intermediate results, each invite a dropped minus, and the four choices are built so that each such slip lands on a listed value. Writing −1 in parentheses when substituting, and finishing the top and the bottom before dividing, keeps the signs in view.

Error-prone arithmetic moderate Working it out
The value t = −1 sits inside 2 − t, so the numerator needs 2 − (−1) = 3, and the work then passes through two negatives, 6 − 24 = −18 and 3 − 5 = −2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The numerator and the denominator each take a substitution, a parenthesis, a product and a sum, worked separately before one final division, so a slip in either half changes the result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The choices pair each size with each sign, −9 with 9 and −1 with 1, so getting 18 and 2 right still leaves the sign to decide between two choices. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Substitute with parentheses around −1, simplify the numerator and the denominator separately, innermost parentheses first, and divide only at the end, checking each sign as you go.

Traps

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

The wrong answers, classified

B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Treats 2 − (−1) as 1 instead of 3, so the numerator becomes 6 − 8 = −2 and the fraction −2 over 2 equals −1.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Takes 6 − 24 as 18 and 3 − 5 as 2, subtracting the smaller number from the larger both times, so the fraction is 18 over 10 + 8 = 18, which is 1.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Gets every step right except 6 − 24, which it writes as 18 instead of −18, so the quotient 18 ÷ 2 comes out positive.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number: $a - (-b) = a + b$; a and b are numbers. Taking away a negative number is the same as adding the positive number. Think of a debt of 1 dollar being taken away: you are 1 dollar better off.
Dividing a negative by a positive: $\frac{-a}{b} = -\frac{a}{b}$; a and b are positive numbers. A negative number divided by a positive number gives a negative answer. The answer is positive only when both signs are the same.

Sort out the question
You know m = 5 and t = −1. You need the value of the big fraction. Work out the top, called the numerator, and the bottom, called the denominator, one at a time. Then divide the top by the bottom.
Put in the numbers
$\frac{6 - 8(2 - t)}{2m + 4(3 - m)} = \frac{6 - 8(2 - (-1))}{2(5) + 4(3 - 5)}$
Put each number in parentheses, so the minus signs stay clear. Work inside the parentheses first, then multiply, then add or subtract.
Work out the top
$a - (-b) = a + b$
$2 - (-1) = 2 + 1 = 3$
$6 - 8(3) = 6 - 24 = -18$
The top is −18. It is negative because 24 is more than 6. Watch out: 2 − (−1) is 3, not 1. Using 1 makes the top −2 and the answer −1, choice B.
Work out the bottom
$2(5) + 4(3 - 5) = 10 + 4(-2) = 10 - 8 = 2$
Here 3 − 5 = −2, because 5 is more than 3. Then 4 times −2 is −8. The bottom is 2.
Divide the top by the bottom
$\frac{-a}{b} = -\frac{a}{b}$
$\frac{-18}{2} = -9$
Watch out: if you lost the minus sign in 6 − 24, you would get 9, choice D. So the answer is A, −9.

Check your answer

Check the sign first. The top, 6 − 24, is clearly negative. The bottom, 10 − 8, is positive. Negative over positive is negative, so only −1 or −9 can work. Without the signs, the top is 18 and the bottom is 2, and 18 ÷ 2 = 9. So −9 fits.

More questions like this

The original DOE question

Question 90 as printed on page 84 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 90. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 50 of 83

2023 Test A Math, question 35

Question

$A number line with tick marks every 2 units, labeled −6, −4, −2, 0, 2, 4 and 6. Point J is the dot at −4 and point K is the dot at 6.$

On the number line above, point L (not shown) is located on line segment JK so that $\mathrm{JL} = \frac{2}{3}\mathrm{LK}$ . What is the position of point L?

A−2
B0
C2
D4

Show the answer

Correct answer: B

$\mathrm{JK} = 6 - (-4) = 10$ units. Let $x = \mathrm{JL}$ . Then $\mathrm{LK} = 10 - x$ . Substitute these expressions in the given equation:

$x = \frac{2}{3}(10 - x)$

Solve for x:

$x = \frac{2}{3}(10 - x)$	Apply the multiplicative inverse property; multiply both sides of the equation by 3.
$3x = 2(10 - x)$	Apply the distributive property; multiply the 2 by 10 and x.
$3x = 20 - 2x$	Apply the additive inverse property; add $2x$ to both sides of the equation.
$5x = 20$	Apply the multiplicative inverse property; divide both sides of the equation by 5.
$x = 4$

So, point L is 4 units to the right of point J: $-4 + 4 = 0$ . Point L is located at 0 on the number line.

Question analysis

This number line question puts J at −4 and K at 6 and asks for the position of a point L on segment JK where JL is two thirds of LK. Students find the segment's length, divide it in a 2-to-3 ratio, and turn a distance into a coordinate. Reading the ratio backwards, or giving a distance instead of a position, leads to wrong answers.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Distances on a number line that crosses zero are grade 7 rational-number work, and splitting a length in a given ratio is grade 7 proportional reasoning. Turning a fraction of one part into a split, then into a position, makes a typical multistep task.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Reading a figure, table or graph minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The ratio is stated between two unknown pieces, which is less familiar than a fraction of a whole, and the final step turns a length into a position. Students who stop at JL's length, read 2/3 the wrong way round, or use its 2 as a distance all reach listed points. Sketching five equal shares of 2 units from −4 to 6, two for JL and three for LK, makes the position plain.

Turning words into math moderate Understanding the question
JL = 2/3 LK compares the two pieces with each other, not a piece with the whole JK, so the 10 units must be split in the ratio 2 : 3, giving JL 4 and LK 6. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
J's position, −4, and K's, 6, are read from the number line, whose labeled ticks run every 2 units, and L itself is not drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Three steps follow in order: the length JK = 6 − (−4), the length JL from the ratio, and finally L's position counted from J rather than from 0. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length of JK, split it into five equal shares with two for JL and three for LK, then count JL's length to the right from J.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the 2 in 2/3 as the length of JL and moves 2 units right from J at −4, without relating JL to LK or to JK.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Turns the ratio upside down, making LK two thirds of JL, so JL is 6 and L lands at 2, nearer K than J.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds JL = 4 correctly but gives that length as L's position, stopping before counting the 4 units from J at −4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line: $\text{distance} = \text{right point} - \text{left point}$; Subtract the position on the left from the position on the right. The answer is how many units apart the two points are. A distance is never negative.
Splitting a length in a ratio: $\text{one part} = \frac{\text{whole length}}{a + b}$; A ratio a : b compares two pieces. If a length is split in the ratio a : b, think of it as a + b equal parts. The first piece gets a of them and the second gets b. Divide the whole length by a + b to find one part.

Sort out the question
Read the number line. Point J is at −4 and point K is at 6. Point L sits somewhere between them. JL is the length from J to L, and LK is the length from L to K. The rule says JL is $\frac{2}{3}$ of LK. You need the position of L.
Find the length of JK
$\text{distance} = \text{right point} - \text{left point}$
K is on the right, so subtract J’s position from K’s.
$\mathrm{JK} = 6 - (-4) = 6 + 4 = 10$
Taking away −4 is the same as adding 4. J and K are 10 units apart.
Turn the rule into equal parts
JL is $\frac{2}{3}$ of LK. So for every 2 units in JL, LK has 3 units. That means JK is made of 2 + 3 = 5 equal parts. JL gets 2 parts and LK gets 3.
$\text{one part} = \frac{\text{whole length}}{a + b}$
$\text{one part} = \frac{10}{2 + 3} = 2$
Each part is 2 units long.
Find the two lengths
$\mathrm{JL} = 2 \times 2 = 4$
$\mathrm{LK} = 3 \times 2 = 6$
Watch out: JL is $\frac{2}{3}$ of LK, not the other way round. Flipping the ratio makes JL 6 and puts L at 2, choice C.
Find the position of L
L is 4 units to the right of J. Start at J’s position, −4, and move 4 to the right.
$-4 + 4 = 0$
Watch out: 4 is the length of JL, not where L sits. Answering 4 is choice D. So the answer is B, 0.

Check your answer

Put L at 0 and test the rule. JL = 0 − (−4) = 4, and LK = 6 − 0 = 6. Then $\frac{2}{3} \times 6 = 4$ , which is JL. Also 4 + 6 = 10, the whole length JK.

More questions like this

The original DOE question

Question 92 as printed on page 84 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 92. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 51 of 83

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?

A $2\frac{1}{2}$
B $2\frac{2}{3}$
C $3\frac{1}{3}$
D $5\frac{1}{3}$

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:

$\frac{c}{1} = \frac{8}{1.5}$	Cross multiply.
$1.5c = 8(1)$	Apply the multiplicative inverse property; divide both sides of the equation by 1.5.
$c = 5\frac{1}{3}$

So, the number of laps Carla still had to finish is $8 - 5\frac{1}{3} = 2\frac{2}{3}$ .

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic moderate
A very close wrong answer minor
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.

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)
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)
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

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Works 80 ÷ 15 as 5 remainder 5 and writes 5.5, turning the remainder into a decimal digit, so Carla seems to have run 5 1/2 laps and 2 1/2 remain.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Finds 5 1/3 correctly but subtracts it from 8 by taking 5 from 8 and keeping the 1/3, getting 3 1/3 instead of regrouping a whole.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds how many laps Carla had run when Vicente finished, 8 ÷ 1.5 = 5 1/3, and gives that instead of the laps she still had left.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, rate and time: $d = r \times t$; d is the distance, r is the rate or speed, and t is the time. 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: $a \div \frac{m}{n} = a \times \frac{n}{m}$; 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.

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.
Compare their distances
$d = r \times t$
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.
$8 = 1.5 \times c$
To find c, divide 8 by 1.5.
Divide 8 by 1.5
Write 1.5 as the fraction $\frac{3}{2}$ .
$a \div \frac{m}{n} = a \times \frac{n}{m}$
$c = 8 \div \frac{3}{2} = 8 \times \frac{2}{3} = \frac{16}{3} = 5\frac{1}{3}$
Carla has run $5\frac{1}{3}$ laps. Watch out: that is how far she has run, not what is left. Stopping here gives choice D.
Find the laps she has left
$8 - 5\frac{1}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3} = 2\frac{2}{3}$
Watch out: 8 − $5\frac{1}{3}$ is not $3\frac{1}{3}$ . You must break 1 whole into thirds first. Answering $3\frac{1}{3}$ is choice C.
So the answer is B, $2\frac{2}{3}$ laps.

Check your answer

Multiply back: $1.5 \times 5\frac{1}{3} = \frac{3}{2} \times \frac{16}{3} = 8$ . So when Carla has run $5\frac{1}{3}$ laps, Vicente has run 8. And $5\frac{1}{3} + 2\frac{2}{3} = 8$ , her full race.

More questions like this

The original DOE question

Question 52 of 83

2023 Test B Math, question 5

Question

Solve the equation for x:

$\frac{(21)(14)x}{(49)(48)} = 0.875$

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

Show the answer

Correct answer: 7

$\frac{(21)(14)x}{(49)(48)} = 0.875$

$\left(\frac{21}{49}\right)\left(\frac{14}{48}\right)x = 0.875$

$\left(\frac{3}{7}\right)\left(\frac{7}{24}\right)x = 0.875$

$\left(\frac{1}{8}\right)x = 0.875$

$x = (0.875)(8) = 7$

Question analysis

This equation puts x in a fraction with (21)(14) on top and (49)(48) below, set equal to the decimal 0.875. The grid-in asks for the value of x. Students do best to cancel common factors before multiplying anything out, and it helps to know the fraction that 0.875 equals.

NYS standards: 7.EE.3 multi-step rational number problems
7.NS.2c properties for multiplying and dividing
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Seventh graders solve equations like ax = b, usually with a simple coefficient. Here the coefficient is a product of four numbers that is best cancelled first, and a decimal has to be matched with a fraction.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
No obvious method minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 6 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a one-unknown equation dressed in large numbers, and the work is in reducing them without slipping. Students who cancel early find that the whole coefficient is one eighth and that 0.875 is seven eighths, which makes the last step mental. Those who multiply out face a four-digit product and a decimal division, and with nothing to check against, an early slip goes unnoticed.

Error-prone arithmetic moderate Working it out
Multiplied out, the coefficient is 294 over 2,352 and the right side is the decimal 0.875; both invite slips unless 21/49 and 14/48 are cancelled first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
No obvious method minor Working it out
The quick route pairs 21 with 49 and 14 with 48 before multiplying, so that 3/7 × 7/24 collapses to 1/8; the layout of the fraction does not suggest that pairing. (Schoenfeld, 1985)
An answer in an unexpected form minor Choosing the answer
The result is entered with no choices, so a coefficient’s denominator or a quotient taken the wrong way looks as complete as the right value. (Enright, Morley & Sheehan, 2002)

Strategy: Cancel before multiplying: reduce 21 over 49 and 14 over 48, multiply the simplified fractions, then undo that coefficient, writing 0.875 as a fraction if it helps.

Traps

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

The wrong answers, classified

8 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Simplifies the coefficient correctly to 1/8 but enters the 8 from its denominator, never multiplying it by 0.875.
0.109375 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides 0.875 by 8 instead of multiplying, undoing the coefficient 1/8 the wrong way, so x comes out as 0.109375.
3.5 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Slips in reducing 14/48, taking it as 7/12 instead of 7/24, so the coefficient becomes 1/4 and x = 0.875 × 4 = 3.5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the tops, or numerators. b and d are the bottoms, or denominators. Multiply the tops together and the bottoms together. It also works backward: you can split one fraction into two smaller ones. That lets you simplify before the numbers get big.

Sort out the question
You need the value of x. On top, 21, 14 and x are multiplied. The bottom is 49 times 48. The right side is 0.875.
Multiplying 49 by 48 makes a big number. It is easier to simplify first.
Split the fraction into two
Use the rule for multiplying fractions backward. Pair 21 with 49, and 14 with 48.
$\frac{21 \times 14}{49 \times 48} = \frac{21}{49} \times \frac{14}{48}$
Now the equation says these two fractions times x equal 0.875.
Simplify each fraction
To simplify, divide the top and bottom by the same number. Both 21 and 49 divide by 7. Both 14 and 48 divide by 2.
$\frac{21}{49} = \frac{3}{7} \qquad \frac{14}{48} = \frac{7}{24}$
Watch out: 48 ÷ 2 is 24, not 12. Writing $\frac{7}{12}$ leads to 3.5.
Multiply the simple fractions
$\frac{3}{7} \times \frac{7}{24} = \frac{3 \times 7}{7 \times 24} = \frac{3}{24} = \frac{1}{8}$
The 7 on top and the 7 on the bottom cancel, since 7 ÷ 7 = 1. Then divide 3 and 24 by 3 to get $\frac{1}{8}$ . So the equation becomes:
$\frac{1}{8} \times x = 0.875$
Solve for x
One eighth of x is 0.875. To undo dividing by 8, multiply both sides by 8.
$x = 0.875 \times 8 = 7$
Watch out: 8 is not the answer, and don’t divide 0.875 by 8. So you would grid 7.

Check your answer

0.875 is the same as $\frac{7}{8}$ . One eighth of 7 is $\frac{7}{8}$ , so x = 7 works. You can also multiply it all out: $21 \times 14 \times 7 = 2{,}058$ and $49 \times 48 = 2{,}352$ . Then $2{,}058 \div 2{,}352 = 0.875$ .

More questions like this

The original DOE question

Question 62 as printed on page 192 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 62. Open page 192 of the handbook (PDF)

Question 53 of 83

2023 Test B Math, question 6

Question

What is the value of $3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)$ ?

A $-3\frac{1}{12}$
B $-2\frac{11}{12}$
C $6\frac{1}{12}$
D $9\frac{5}{12}$

Show the answer

Correct answer: C

The first step is to rewrite the problem replacing all subtraction with adding the opposite:

$3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)$

$3\frac{1}{4} + \left(-1\frac{2}{3}\right) + 4\frac{1}{2}$

Then, rewrite all the fractions using a common denominator of 12, which is the least common multiple of the fractions’ denominators, 4, 3, and 2:

$3\frac{3}{12} + \left(-1\frac{8}{12}\right) + 4\frac{6}{12}$

Next, combine the positive numbers and, finally, add the negative number to the sum by subtracting the number with the least absolute value from the number with the greatest absolute value and using the sign of the number with the greatest absolute value:

$7\frac{9}{12} + \left(-1\frac{8}{12}\right)$

$6\frac{1}{12}$

Question analysis

What is 3 1/4 − 1 2/3 − (−4 1/2)? This question on rational-number arithmetic combines mixed numbers with denominators of 4, 3 and 2 and ends by subtracting a negative number. Students must turn that last subtraction into an addition and find a common denominator before combining.

NYS standards: 7.NS.1c rational number subtraction
7.NS.1d properties for adding and subtracting
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Subtracting negative mixed numbers is core grade 7 content. This expression is typical in length, but it needs a common denominator of 12 and careful handling of the rule for subtracting a negative.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 10, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The expression is short, but it mixes a subtraction that stays a subtraction with one that becomes an addition, and that difference is what most wrong answers miss. The fractions add a second load, since twelfths are needed before anything combines. Estimating first, 3 − 1.7 + 4.5 is about 6, shows the result is positive, which rules out both negative values.

Error-prone arithmetic moderate Working it out
Three mixed numbers with denominators 4, 3 and 2 must become twelfths, and the last term, −(−4 1/2), changes sign while the middle subtraction does not. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Rewriting the double negative, renaming each fraction, adding 3 1/4 and 4 1/2, then taking away 1 2/3 are four steps, each able to carry a slip forward. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Change subtracting −4 1/2 into adding 4 1/2, rename every fraction in twelfths, add the two positive numbers, then take away 1 2/3.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single error gives −3 1/12; it needs subtracting 4 1/2 and also miswriting 3 1/4 as 3 1/12, two errors at once, so it mostly draws guesses.
B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Treats subtracting −4 1/2 as subtracting 4 1/2, so 3 1/4 − 1 2/3 − 4 1/2 = −2 11/12.
D Negatives and direction (Brown & Burton, 1978)
Reveals a surface strategy Expected pull: weak
Drops both minus signs and adds the three mixed numbers, 3 1/4 + 1 2/3 + 4 1/2 = 9 5/12, as if every operation were addition.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number: $a - (-b) = a + b$; a and b stand for any numbers. Taking away a negative number is the same as adding its opposite, the same number with the other sign. Think of it this way: taking away a debt leaves you richer.
Equivalent fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; a is the top and b is the bottom of a fraction. Multiplying both by the same number n does not change its value. Use it to give fractions the same bottom number, so you can add or subtract them.

Sort out the question
You need the value of $3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)$ . Each number is a mixed number, a whole number plus a fraction. There are two subtractions. The second one takes away a negative number.
Change the double negative
Use the rule for subtracting a negative. Taking away $-4\frac{1}{2}$ is the same as adding $4\frac{1}{2}$ .
$3\frac{1}{4} - 1\frac{2}{3} + 4\frac{1}{2}$
The first minus sign stays as it is. Watch out: if you subtract $4\frac{1}{2}$ instead, you get $-2\frac{11}{12}$ , choice B.
Give the fractions one denominator
The denominator is the bottom number of a fraction. Here they are 4, 3 and 2. The smallest number all three divide into is 12. So change each fraction to twelfths.
$\frac{1}{4} = \frac{3}{12} \qquad \frac{2}{3} = \frac{8}{12} \qquad \frac{1}{2} = \frac{6}{12}$
The problem is now:
$3\frac{3}{12} - 1\frac{8}{12} + 4\frac{6}{12}$
Add the two positive numbers
You may add and subtract in any order, as long as each number keeps its sign. Add the two positive numbers first.
$3\frac{3}{12} + 4\frac{6}{12} = 7\frac{9}{12}$
That is 3 + 4 = 7 wholes and 3 + 6 = 9 twelfths.
Take away the last number
$7\frac{9}{12} - 1\frac{8}{12} = 6\frac{1}{12}$
That is 7 − 1 = 6 wholes and 9 − 8 = 1 twelfth. So the answer is C, $6\frac{1}{12}$ .

Check your answer

Estimate with decimals: $3.25 - 1.67 + 4.5 \approx 6.08$ . And $6\frac{1}{12}$ is about 6.08 too. The answer must be positive, so A and B are out. Choice D comes from adding all three numbers.

More questions like this

The original DOE question

Question 63 as printed on page 193 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 63. Open page 193 of the handbook (PDF)

Question 54 of 83

2023 Test B Math, question 21

Question

$-mn$

$(-m)(-n)$

$|mn|$

$|-mn|$

$|(-m)(-n)|$

If m and n are positive integers, how many of the five expressions above are equivalent to $mn$ ?

Show the answer

Correct answer: D

Since m and n are positive, the value of $mn$ is also positive. Of the five options given:

$-mn$ is negative.

$(-m)(-n)$ is positive.

$|mn|$ is positive.

$|-mn|$ is positive.

$|(-m)(-n)|$ is positive.

There are 4 expressions that are positive, and each one is equal to $mn$ .

Question analysis

Five expressions are listed: −mn, (−m)(−n), |mn|, |−mn| and |(−m)(−n)|. If m and n are positive integers, students count how many are equivalent to mn. It tests the sign rules for multiplying negatives together with what absolute value does, and each expression has to be judged on its own.

NYS standards: 7.NS.2a rational number multiplication
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Multiplying signed numbers and absolute value are grade 6 and 7 content. Checking five expressions one by one takes longer than a typical single computation, but each check uses a familiar rule.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Much to keep track of minor
Something you must already know minor
Solution: 5 steps
Reading level: Question and choices: about grade 10, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each expression takes only a moment, but the answer is a count, so one misjudged expression changes it. The minus signs are designed to unsettle: four of the five expressions show at least one, and only one of those actually stays negative. Substituting small positive numbers and comparing each value with mn turns the question into arithmetic.

Variables instead of numbers moderate Understanding the question
The five expressions use only m and n, so each sign rule has to be applied in general unless the student substitutes values such as m = 2 and n = 3. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Five expressions are judged one at a time and a count kept, and two of them combine the product rule for negatives with absolute value bars. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
The rules that two negative factors give a positive product and that absolute value removes a minus sign are assumed, not stated. (Anderson & Krathwohl, 2001)

Strategy: Pick small values such as m = 2 and n = 3, evaluate each of the five expressions, and count the ones equal to 6.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Treats any minus sign as making the value negative, even inside bars or in a product of two negatives, so counts only |mn|.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Accepts |mn| and (−m)(−n) but believes bars cannot cancel a minus sign written inside them, rejecting |−mn| and |(−m)(−n)|.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Thinks a negative times a negative stays negative, so rejects (−m)(−n) along with −mn and counts only the three absolute values.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying two negatives: $(-a) \times (-b) = a \times b$; a and b are positive numbers. A negative number times a negative number gives a positive number. The two minus signs cancel out. One negative and one positive give a negative.
Absolute value: $|a| = \text{the distance from } a \text{ to } 0$; The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so |−6| = 6 and |6| = 6. The bars remove any minus sign.

Sort out the question
m and n are positive integers. That means whole numbers greater than 0. Writing mn means m times n, so mn is positive. You need to count how many of the five expressions always equal mn.
Pick numbers to test
Pick easy values: $m = 2$ and $n = 3$ . Then $mn = 2 \times 3 = 6$ . Any expression that equals mn must come out to 6.
Test the first two expressions
$-mn = -(2 \times 3) = -6$
This is negative, so it does not equal 6.
$(-m)(-n) = (-2)(-3) = 6$
A negative times a negative is positive, so this one equals 6. Watch out: if you think a negative times a negative stays negative, you reject this one and get 3, choice C.
Test the three absolute values
$|mn| = |6| = 6$
$|-mn| = |-6| = 6$
$|(-m)(-n)| = |6| = 6$
All three equal 6, because absolute value bars remove a minus sign. Watch out: the bars work even when the minus sign is inside them. Rejecting the last two gives 2, choice B.
Count the matches
Four expressions equal mn: $(-m)(-n)$ , $|mn|$ , $|-mn|$ and $|(-m)(-n)|$ . Only $-mn$ does not. The sign rules work the same for any positive m and n, not just 2 and 3. So the answer is D, 4.

Check your answer

Try other numbers, such as $m = 1$ and $n = 5$ . Then $mn = 5$ . The five expressions give −5, 5, 5, 5 and 5. Again, four of them match mn.

More questions like this

The original DOE question

Question 78 as printed on page 197 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 78. Open page 197 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 55 of 83

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 $\overline{QR}$ than $\overline{PQ}$ ?

A $\frac{1}{6}$
B $\frac{1}{2}$
C $\frac{2}{3}$
D3

Show the answer

Correct answer: B

First, find the lengths of the given segments:

$\mathrm{QR}=5\frac{1}{6}-3=2\frac{1}{6}$

$\mathrm{PQ}=3-1\frac{1}{3}=1\frac{2}{3}$

Then, find the difference:

$2\frac{1}{6}-1\frac{2}{3}=$

$\frac{13}{6}-\frac{5}{3}=$

$\frac{13}{6}-\frac{10}{6}=$

$\frac{3}{6}=\frac{1}{2}$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Reading a figure, table or graph minor
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.

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)
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

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts only the whole numbers in 3 − 1 1/3 and keeps the 1/3, giving PQ as 2 1/3, then compares it with 2 1/6 and gets 1/6.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2/3; it is only the fraction part of PQ’s length, 1 2/3, so it mostly draws guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 3, the coordinate of point Q, a number read off the line rather than a difference between two lengths.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Length on a number line: $\text{length} = \text{right point} - \text{left point}$; 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: $a\frac{b}{c} = \frac{a \times c + b}{c}$; 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.

Read the number line
Point P is at $1\frac{1}{3}$ . Point Q is at 3. Point R is at $5\frac{1}{6}$ .
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.
Find the length of QR
R is on the right, and Q is on the left. Use the length rule.
$\text{QR} = 5\frac{1}{6} - 3 = 2\frac{1}{6}$
QR is $2\frac{1}{6}$ units long.
Find the length of PQ
Q is on the right, and P is on the left.
$\text{PQ} = 3 - 1\frac{1}{3}$
Take it in two parts. First, 3 − 1 = 2. Then take away the extra $\frac{1}{3}$ . That gives $2 - \frac{1}{3} = 1\frac{2}{3}$ . So PQ is $1\frac{2}{3}$ units long.
Watch out: Don’t just do 3 − 1 and keep the $\frac{1}{3}$ . That gives $2\frac{1}{3}$ , which is too long. It leads to choice A.
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.
$2\frac{1}{6} = \frac{2 \times 6 + 1}{6} = \frac{13}{6}$
$1\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3} = \frac{10}{6}$
Subtract the lengths
Take the shorter length away from the longer one.
$\frac{13}{6} - \frac{10}{6} = \frac{3}{6} = \frac{1}{2}$
QR is $\frac{1}{2}$ unit longer than PQ.
So the answer is B, $\frac{1}{2}$ .

Check your answer

Write every point in sixths. P is $\frac{8}{6}$ , Q is $\frac{18}{6}$ and R is $\frac{31}{6}$ . 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 56 of 83

2023 Test B Math, question 46

Question

$0^{3}+4^{0}+2^{-1}+(-1)^{2}$

What is the value of the expression above, in decimal form?

A0
B0.5
C1.5
D2.5

Show the answer

Correct answer: D

$0^{3}+4^{0}+2^{-1}+(-1)^{2}$

$=0+1+\frac{1}{2}+1$

$=2\frac{1}{2}=2.5$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know major
Error-prone arithmetic minor
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.

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)
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

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Treats 2 to the power −1 as −2 instead of one half, so the four terms 0 + 1 − 2 + 1 add to 0.
B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Squares as if the minus sat outside the parentheses, taking (−1)² as −1, so the terms add to 0 + 1 + 0.5 − 1 = 0.5.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Takes 4 to the power 0 as 0 rather than 1, so the expression becomes 0 + 0 + 0.5 + 1 = 1.5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Zero exponent: $a^0 = 1$; 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: $a^{-n} = \frac{1}{a^n}$; 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.

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.
Find the first two powers
Zero times itself is still zero.
$0^3 = 0 \times 0 \times 0 = 0$
Next, use the zero exponent rule.
$4^0 = 1$
Watch out: $4^0$ is 1, not 0. Taking it as 0 gives choice C.
Find the negative power
Use the negative exponent rule.
$2^{-1} = \frac{1}{2^1} = \frac{1}{2} = 0.5$
Watch out: $2^{-1}$ is not −2. That mistake gives choice A.
Find the last power
The base is −1, inside the parentheses. Multiply it by itself. A negative times a negative is positive.
$(-1)^2 = (-1) \times (-1) = 1$
Watch out: The answer is not −1. That mistake gives choice B.
Add the four values
Now add them. The question wants a decimal, so write one half as 0.5.
$0 + 1 + 0.5 + 1 = 2.5$
So the answer is D, 2.5.

Check your answer

See why $4^0 = 1$ with a pattern. $4^2 = 16$ and $4^1 = 4$ . Each step down divides by 4, so $4^0 = 4 \div 4 = 1$ . In the same way, $2^0 = 1$ and $2^{-1} = 1 \div 2 = \frac{1}{2}$ .

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 83

2023 Test B Math, question 48

Question

The outdoor temperature was 4.2° F at 10 p.m. After $3\frac{1}{2}$ hours, the outdoor temperature was −0.7° F. Which value best represents the average change in temperature per hour?

A−1.4° F
B−1.0° F
C1.0° F
D1.4° F

Show the answer

Correct answer: A

First, recognize that the temperature decreased. This means that the order of subtraction matters and that the result is a negative number:

$-0.7-4.2=-4.9$

The temperature decreased 4.9 degrees. Divide by 3.5 to get the average change per hour:

$-4.9\div 3.5=-1.4$

Question analysis

The temperature was 4.2° F at 10 p.m. and −0.7° F three and a half hours later. Students find the average change in temperature per hour. It tests subtracting and dividing with negative decimals, and the choices pair two sizes of change with both signs, so the sign has to be right as well as the size.

NYS standards: 7.NS.3 four operations with rational numbers
7.NS.1c rational number subtraction
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Signed-number problems about temperature are standard grade 7 work. This one asks for an average rate, so a change that crosses zero must be divided by a mixed-number time, with the sign kept through both steps.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is short, but signs decide every choice: two choices come from treating −0.7 as 0.7, and one comes from getting the size right and the direction wrong. Students who think of the drop in two parts, 4.2 degrees down to zero and 0.7 more, find the 4.9-degree fall; dividing by 3.5 hours and remembering the temperature fell finishes it.

Error-prone arithmetic moderate Working it out
The change crosses zero, −0.7 − 4.2 = −4.9, and then a negative decimal is divided by 3.5, from the 3 1/2 hours, with the sign kept through both steps. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase “average change in temperature per hour” means the change from the 4.2° start to the −0.7° end divided by the time, final minus initial, so a drop is negative. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The answer, −1.4° F, and the choice 1.4° F share every digit and differ only in sign, so a correct division followed by a careless sign cannot be caught by estimating. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Subtract the starting temperature from the ending one, keeping track of the sign, then divide that change by the 3.5 hours.

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

B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Subtracts 0.7 from 4.2 as if both temperatures were above zero, a 3.5-degree drop, so 3.5 ÷ 3.5 gives 1, then marks the change negative.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Treats −0.7 as 0.7, finding a 3.5-degree change, and also loses its direction, reporting a rise of 1 degree an hour for a temperature that fell.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds the 4.9-degree change and divides by 3.5 correctly, or subtracts in the order start minus end, and reports a rise for a temperature that fell.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average change per hour: $\text{average change per hour} = \frac{\text{end} - \text{start}}{\text{number of hours}}$; Start is the first temperature, and end is the last one. End minus start is the total change. It is negative when the temperature drops. Dividing by the number of hours shares that change equally over each hour.

Sort out the question
At 10 p.m. it was 4.2° F. After $3\frac{1}{2}$ hours, it was −0.7° F. That is below zero. The temperature went down, so the change is negative. You need the average change for each hour.
Find the total change
Subtract the start from the end.
$-0.7 - 4.2 = -4.9$
Picture a thermometer. Going from 4.2 down to 0 is a drop of 4.2 degrees. Going on to −0.7 is 0.7 more. In all, it fell 4.9 degrees.
Watch out: 4.2 − 0.7 = 3.5 treats −0.7 as if it were above zero. That leads to choice B or C.
Divide by the hours
Write $3\frac{1}{2}$ hours as 3.5 hours. Use the formula.
$\text{average change per hour} = \frac{-0.7 - 4.2}{3.5} = \frac{-4.9}{3.5} = -1.4$
To divide, think 49 ÷ 35 = 1.4, since 35 × 1.4 = 49. The change is negative, so it is −1.4 degrees per hour.
Keep the sign
The temperature fell, so the answer must be negative.
Watch out: 1.4 has the right size, but it says the temperature went up. That is choice D.
So the answer is A, −1.4° F.

Check your answer

Work forward. Dropping 1.4 degrees each hour for 3.5 hours is $1.4 \times 3.5 = 4.9$ degrees. Then $4.2 - 4.9 = -0.7$ . That is the temperature in the problem.

More questions like this

The original DOE question

Question 105 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 105. Open page 205 of the handbook (PDF)

Question 58 of 83

2023 Test B Math, question 55

Question

$6.\overline{34}-4.\overline{12}$

What is the value of the expression above, written as an improper fraction?

A $\frac{24}{11}$
B $\frac{224}{101}$
C $\frac{111}{50}$
D $\frac{20}{9}$

Show the answer

Correct answer: D

$6.\overline{34}-4.\overline{12}=2.\overline{22}=2.\overline{2}$

$x=2.2222...$	Set x equal to the repeating decimal.
$10x=22.2222...$	Multiply both sides by 10.
$-x=-2.2222...$	Subtract 10x and x.
$9x=20$	Apply the multiplicative inverse property; divide both sides by 9.
$x=\frac{20}{9}$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know moderate
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.

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)
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

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Sets x = 2.222… but adds 10x and x instead of subtracting, then drops the repeating part as if it had cancelled, so 11x = 22 + 2 = 24.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Multiplies by 100 for two-digit blocks, then adds x instead of subtracting it and drops the repeating part, so 101x = 222 + 2 = 224.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Drops the repeating bar and treats the difference as exactly 2.22, which is 222/100 = 111/50, slightly less than the true repeating value.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Repeating decimal to fraction: $10x - x = 9x$; 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.

Sort out the bars
A bar over digits means they repeat forever. So $6.\overline{34}$ means 6.343434… and $4.\overline{12}$ 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.
Subtract the decimals
Line up the digits. Each block of 34 minus 12 gives 22. The whole numbers give 6 − 4 = 2.
$6.343434\ldots - 4.121212\ldots = 2.222222\ldots$
The difference is 2.222…, with the 2 repeating forever. You can write it $2.\overline{2}$ .
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.
$x = 2.222\ldots$
$10x = 22.222\ldots$
Subtract to remove the repeats
Subtract the first line from the second. The repeating 2s after the point match up, so they cancel.
$10x - x = 22.222\ldots - 2.222\ldots$
$9x = 20$
Divide to find x
Divide both sides by 9.
$x = \frac{20}{9}$
Watch out: $\frac{111}{50}$ 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, $\frac{20}{9}$ .

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, $\frac{24}{11}$ , 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 83

2022 Test A Math, question 1

Question

The figure below shows a circle with a hole in the middle. The outer diameter of the circle is $1\frac{1}{4}$ inches. The diameter of the hole is $\frac{1}{4}$ inch.

$A large circle shaded gray, with a smaller white circle, the hole, at its center. A double-headed arrow across the hole is labeled 1/4 in., and a double-headed arrow below the figure, spanning the full width of the large circle, is labeled 1 1/4 in. The shaded region is the part of the large circle outside the hole.$

What is the area of the shaded region, to the nearest hundredth of a square inch?

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

Show the answer

Correct answer: 1.18

To determine the area of a circle, begin by finding the radius of the circle, which is half of the diameter.

The radius of the entire circle is $1\frac{1}{4}\text{ in.} \div 2 = \frac{5}{8}\text{ in.}$ , so the area of the entire circle is $\pi\left(\frac{5}{8}\right)^{2}\text{ in.}^{2} = \frac{25}{64}\pi\text{ in.}^{2}$

The radius of the hole in the circle is $\frac{1}{4}\text{ in.} \div 2 = \frac{1}{8}\text{ in.}$ , so the area of the hole in the circle = $\pi\left(\frac{1}{8}\right)^{2}\text{ in.}^{2} = \frac{1}{64}\pi\text{ in.}^{2}$

To determine the area of the shaded region, subtract the area of the hole in the circle from the area of the entire circle:

$\left(\frac{25}{64}\right)\pi\text{ in.}^{2} - \left(\frac{1}{64}\right)\pi\text{ in.}^{2} =$

$\left(\frac{24}{64}\right)\pi\text{ in.}^{2} =$

$\left(\frac{3}{8}\right)\pi\text{ in.}^{2} =$

$\left(\frac{3}{8}\right)\text{ in.}^{2}(3.14) =$

$1.1775\text{ in.}^{2}$

1.1775 square inches rounded to the nearest hundredth of a square inch = 1.18 square inches

Question analysis

A gray ring is a circle 1 1/4 inches across with a hole 1/4 inch across at its center, and this grid-in asks for the area of the shaded part to the nearest hundredth of a square inch. Students use the area of a circle twice. Turning diameters given as fractions into radii, and squaring them, is where mistakes happen.

NYS standards: 7.G.4 circle area and circumference
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 find circle areas and often subtract a hole from a shape. This ring gives its diameters as a mixed number and a fraction, so halving, squaring fractions and rounding add work, though no step goes beyond grade 7.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Error-prone arithmetic moderate
Combines ideas minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 45 words.
Corrected from the DOE handbook: A missing exponent in the explanation’s area formula was restored.
An arithmetic slip in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is the familiar one for a ring, the big circle's area minus the hole's, but the labels are diameters given as fractions, so each one has to be halved and then squared before π enters. Students lose points by skipping the halving, by stopping at the full disk, or by subtracting the radii before squaring. Keeping both areas as fractions of π, 25/64π − 1/64π = 3/8π, leaves one multiplication and one rounding at the very end.

Something you must already know moderate Working it out
Neither area formula is printed, and both circles are labeled by diameter, so students must recall A = πr² and halve 1 1/4 inches and 1/4 inch to radii of 5/8 and 1/8 first. (Anderson & Krathwohl, 2001)
Error-prone arithmetic moderate Working it out
Every value is a fraction: 5/4 ÷ 2 = 5/8, squares of 25/64 and 1/64, then 3/8 × 3.14 worked by hand, so a slip can enter at each stage. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Combines ideas minor Working it out
The gray ring has no formula of its own; it is the large circle's area minus the hole's, so the area formula is used twice and the results subtracted. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
The grid-in wants the nearest hundredth, so 1.1775 must be rounded to 1.18, and a whole-disk area such as 1.23 gets no warning from a list of choices. (Enright, Morley & Sheehan, 2002)

Strategy: Write 1 1/4 as 5/4 and halve both diameters to get the radii, subtract the hole's area from the area of the full circle, and round to the nearest hundredth only at the very end.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Stops one step early: A choice is a number from partway through the solution.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

4.71 Wrong formula (King et al., 2004)
Reveals a slip Expected pull: strong
Uses the labeled diameters as radii, π(5/4)² − π(1/4)² = 24/16π = 1.5π ≈ 4.71, forgetting to halve them first, which makes the ring four times too big.
1.23 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the area of the whole circle, π(5/8)² = 25/64π ≈ 1.23 square inches, and stops before subtracting the hole's 1/64π.
0.79 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Subtracts the radii first, 5/8 − 1/8 = 1/2, then squares, π(1/2)² ≈ 0.79, as if the difference of two squares were the square of the difference.

What was corrected from the DOE handbook, and why

Explanation: The handbook writes the area of the whole circle as “π(5/8) in.² = 25/64π in.²” and the area of the hole as “π(1/8) in.² = 1/64π in.²” (p. 128, question 58), squaring the unit but not the radius. The area of a circle is πr², and π(5/8) is only 5/8π. The site now reads π(5/8)² in.² = 25/64π in.² and π(1/8)² in.² = 1/64π in.². The areas that follow and the answer, 1.18, are unchanged.
Explanation: The handbook computes (3/8) in.²(3.14) as “1.175 in.²” and then rounds “1.175 square inches” to 1.18 (p. 128, question 58). But 3/8 × 3.14 = 1.1775, so the site now reads 1.1775 in both places. Rounded to the nearest hundredth it is still 1.18 square inches, the answer (with π itself, 3/8 × π = 1.178…, which also rounds to 1.18).

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Radius from diameter: $r = \frac{d}{2}$; d is the diameter, the distance straight across a circle through its center. r is the radius, the distance from the center to the edge. The diameter is two radii laid end to end, so the radius is half of it.
Area of a circle: $A = \pi \times r^{2}$; A is the area, the space inside the circle. π (pi) is a number a little more than 3, about 3.14. r is the radius. r² means r × r. Square the radius first, then multiply by π.

Sort out the question
The gray part is a ring. It is the big circle with the hole cut out. The big circle is $1\frac{1}{4}$ inches across. The hole is $\frac{1}{4}$ inch across. These are diameters. You need the area of the gray ring, in square inches. Round it to the nearest hundredth, two places after the decimal point.
Find both radii
The area formula uses the radius, so halve each diameter first. Write the mixed number as a fraction: $1\frac{1}{4} = \frac{5}{4}$ .
$r = \frac{d}{2}$
$r = \frac{5}{4} \div 2 = \frac{5}{4} \times \frac{1}{2} = \frac{5}{8}$
$r = \frac{1}{4} \div 2 = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$
The big circle’s radius is 5/8 inch. The hole’s radius is 1/8 inch. Watch out: don’t use the diameters as radii. That makes the ring four times too big, and you would grid 4.71.
Find the area of the big circle
$A = \pi \times r^{2}$
$A = \pi \times \left(\frac{5}{8}\right)^{2} = \pi \times \frac{5 \times 5}{8 \times 8} = \frac{25}{64}\pi$
To square a fraction, square the top and square the bottom. The big circle’s area is 25/64 π square inches. Keep π for now. You will use 3.14 at the very end.
Find the area of the hole
$A = \pi \times \left(\frac{1}{8}\right)^{2} = \pi \times \frac{1 \times 1}{8 \times 8} = \frac{1}{64}\pi$
The hole’s area is 1/64 π square inches. Watch out: square each radius on its own. Don’t subtract the radii first. That gives $\frac{5}{8} - \frac{1}{8} = \frac{1}{2}$ , and $\pi \times \left(\frac{1}{2}\right)^{2}$ leads to 0.79.
Subtract the hole
The ring is the big circle minus the hole. Both fractions have 64 on the bottom, so subtract the tops.
$\frac{25}{64}\pi - \frac{1}{64}\pi = \frac{24}{64}\pi = \frac{3}{8}\pi$
To simplify 24/64, divide the top and bottom by 8. Watch out: don’t stop at the big circle. Its area alone is about 1.23 square inches.
Multiply by 3.14 and round
$\frac{3}{8} \times 3.14 = \frac{9.42}{8} = 1.1775$
The ring’s area is 1.1775 square inches. To round to the hundredths place, look at the next digit. It is 7, which is 5 or more, so round up. 1.1775 becomes 1.18. Using the π key instead gives 1.178…, which also rounds to 1.18. So you would grid 1.18.

Check your answer

Estimate. The big circle is about 1.23 square inches. The hole is tiny: $\frac{1}{64} \times 3.14$ is about 0.05 square inch. $1.23 - 0.05 = 1.18$ . That matches. The ring must also be smaller than the whole circle, and it is.

More questions like this

The original DOE question

Question 58 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 79, question 58. Open page 79 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 60 of 83

2022 Test A Math, question 3

Question

The list shows the daily low temperatures in a city during six days in the winter.

7° F, 4° F, 1° F, −2° F, −5° F, 3° F

What is the difference between the lowest and highest temperatures, in degrees Fahrenheit, for the six days?

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

Show the answer

Correct answer: 12

Determine the greatest temperature listed for the six days:

Of 7, 4, 1, (−2), (−5), and 3, the greatest value is 7 because its location is farthest to the right on a number line.

Greatest temperature = 7° F

Determine the lowest temperature listed for the six days:

Of 7, 4, 1, (−2), (−5), and 3, the lowest value is (−5) because its location is farthest to the left on a number line.

Lowest temperature = (−5)° F

Determine the difference between the two temperatures:

$7^{\circ}\text{ F} - (-5)^{\circ}\text{ F} = 7^{\circ}\text{ F} + 5^{\circ}\text{ F} = 12^{\circ}\text{ F}$

Question analysis

A list gives a city's daily low temperatures over six winter days, some above zero and some below. This grid-in asks for the difference between the highest and the lowest of them, in degrees Fahrenheit. Students order positive and negative numbers and subtract integers; picking the right negative reading and handling the double minus sign are the tricky steps.

NYS standards: 7.NS.1c rational number subtraction
6.NS.7b order statements in context
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Subtracting a negative number to find the distance between two temperatures is core grade 7 integer work. Picking the extremes from six one-digit readings is quick, so this is a typical, short grade 7 exercise.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Six readings, two of them below zero, must be sorted before any subtraction, and −5 has to be seen as colder than −2 even though 5 is the bigger number. The subtraction is where most points go: the gap runs 5 degrees up to zero and 7 more above it. A quick sketch of a thermometer from −5 to 7 shows the full span and catches a sign dropped along the way.

Error-prone arithmetic moderate Working it out
The span is 7 − (−5), and subtracting a negative means adding 5, so the five degrees below zero are easy to drop, leaving only 2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
With no choices to compare, nothing flags an entry of 2 or −12; the student must see alone that the gap between the extremes is a positive count of degrees. (Enright, Morley & Sheehan, 2002)

Strategy: Picture the six readings on a number line: the lowest is the one farthest to the left. Subtract the lowest from the highest, remembering that subtracting a negative number means adding.

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.)
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

2 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Subtracts the sizes, 7 − 5 = 2, as if both readings were above zero, leaving out the 5 degrees between −5° F and zero.
9 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Takes −2° F as the lowest reading because 2 is less than 5, then subtracts 7 − (−2) = 9.
-12 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Subtracts in the reverse order, lowest minus highest, −5 − 7 = −12, and enters a negative difference.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Difference between two values: $D = \text{highest} - \text{lowest}$; D is the difference. Subtract the lowest value from the highest, in that order, so D is not negative. It tells how far apart the two values are on a number line.
Subtracting a negative number: $a - (-b) = a + b$; a and b are numbers, and −b is the negative of b. Taking away a negative is the same as adding its positive. On a number line, going from −b up to a passes b steps to zero, then a more.

Sort out the question
The list has six low temperatures, in degrees Fahrenheit (° F). Some are below zero. A minus sign means below zero. You need the highest and the lowest temperatures. Then find the difference between them. The difference tells how many degrees apart they are.
Find the highest and lowest
Picture a number line, or a thermometer. Numbers farther to the right are higher. The highest temperature is 7° F. The lowest is the one farthest to the left. That is −5° F. Watch out: −5 is colder than −2. The number 5 is bigger than 2, but 5 below zero is lower than 2 below. Using −2 as the lowest leads to 9.
Subtract the lowest from the highest
$D = \text{highest} - \text{lowest}$
$D = 7 - (-5) = 7 + 5 = 12$
Subtracting a negative is the same as adding. The temperatures are 12 degrees apart. Watch out: $7 - 5 = 2$ leaves out the 5 degrees below zero. Also, a difference here is not negative, so don’t grid −12. So you would grid 12.

Check your answer

Count on a thermometer. From −5° F up to 0° F is 5 degrees. From 0° F up to 7° F is 7 more degrees. $5 + 7 = 12$ degrees. That matches.

More questions like this

The original DOE question

Question 60 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 79, question 60. Open page 79 of the handbook (PDF)

Question 61 of 83

2022 Test A Math, question 7

Question

$\frac{x^{2}y^{6}}{xy^{2}}$

What is the value of the expression above if $x = -2$ and $y = 2$ ?

A−64
B−32
C16
D32

Show the answer

Correct answer: B

First, substitute −2 for x and 2 for y.

$\frac{(-2)^{2}(2)^{6}}{(-2)(2)^{2}}$

Then, simplify.

$\frac{(-2)(-2)(2)^{2}(2)^{4}}{(-2)(2)^{2}}$

$\frac{(-2)(2)^{4}}{1}$

$\frac{-2 \times 16}{1}$

$-32$

Question analysis

This question asks for the value of a fraction built from powers of x and y when x is −2 and y is 2. Students evaluate an expression with exponents and a negative number, or simplify it first by cancelling common factors. Keeping track of the negative sign through the squaring and the division is the tricky part.

NYS standards: 6.EE.2c evaluating expressions and order of operations
7.NS.2a rational number multiplication
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students evaluate expressions with negative values and whole-number exponents. This one is a single fraction of powers, routine in method, but raising a negative base to powers and then dividing leaves several chances to lose the sign.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 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 routes work: substitute and compute 256 ÷ (−8), or simplify to xy⁴ first and evaluate (−2)(16). Simplifying first is safer, because it leaves a single negative factor to track. Substituting first brings in (−2)², where treating the minus sign as outside the parentheses flips the final sign, and each listed wrong value comes from an exponent or a sign handled wrongly.

Error-prone arithmetic moderate Working it out
Substituting first gives (−2)²(2)⁶ over (−2)(2)², so the work runs through a squared negative, 2⁶ = 64 and a negative divisor, and one factor of −2 decides the sign. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The quotient rule for powers is not stated; students who know that x²/x = x and y⁶/y² = y⁴ can reduce the expression to xy⁴ before any number goes in. (Anderson & Krathwohl, 2001)

Strategy: Cancel the factors of x and y that the top and bottom share before substituting, so there is less to compute, and track the sign of every factor of −2 through to the end.

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

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a slip Expected pull: weak
Drops the square on y in the denominator, dividing 4 × 64 = 256 by (−2)(2) = −4 instead of by (−2)(4) = −8, giving −64.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Cancels x² against x as if x²/x were 1 rather than x, so only y⁶/y² = y⁴ = 16 is left, with no factor of −2.
D Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Squares as if the minus sat outside the parentheses, taking (−2)² as −4, so −4 × 64 = −256 and −256 ÷ (−8) = 32.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A power: $a^{4} = a \times a \times a \times a$; a is the base, the number being multiplied. The small number is the exponent. It says how many times the base is used as a factor, a number being multiplied. So y⁶ means six y’s multiplied together. It does not mean y times 6.
Dividing powers with the same base: $\frac{a^{m}}{a^{n}} = a^{m - n}$; a is the base. m and n are exponents. The top has m factors of a and the bottom has n. Each matching pair cancels, because a ÷ a = 1. So m − n factors are left. A letter with no exponent, like x, has exponent 1.

Sort out the question
You need the value of $\frac{x^{2}y^{6}}{xy^{2}}$ when x = −2 and y = 2. The top means $x^{2} \times y^{6}$ . The bottom means $x \times y^{2}$ . You could put the numbers in right away. But simplifying first is safer. It leaves only one negative number to track.
Simplify the x part
Use the rule for dividing powers. The x on the bottom has exponent 1.
$\frac{x^{2}}{x} = x^{2 - 1} = x^{1} = x$
You can see it by writing it out: $\frac{x \times x}{x}$ leaves one x. Watch out: x² ÷ x is x, not 1. If you cancel every x, you lose the −2 and get 16, choice C.
Simplify the y part
$\frac{y^{6}}{y^{2}} = y^{6 - 2} = y^{4}$
So the whole fraction becomes a simpler expression:
$\frac{x^{2}y^{6}}{xy^{2}} = x \times y^{4}$
Watch out: the bottom has y², not y. Using y leaves y⁵ and leads to −64, choice A.
Put in the numbers
Now substitute x = −2 and y = 2.
$x \times y^{4} = (-2) \times 2^{4}$
$2^{4} = 2 \times 2 \times 2 \times 2 = 16$
$(-2) \times 16 = -32$
A negative times a positive is negative. So the answer is B, −32.

Check your answer

Substitute first, the long way. Top: $(-2)^{2} \times 2^{6} = 4 \times 64 = 256$ . Bottom: $(-2) \times 2^{2} = (-2) \times 4 = -8$ . Then $256 \div (-8) = -32$ . Watch out: (−2)² is +4, since two negatives multiply to a positive. Writing −4 leads to 32, choice D.

More questions like this

The original DOE question

Question 64 as printed on page 81 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 81, question 64. Open page 81 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 62 of 83

2022 Test B Math, question 26

Question

$\frac{\left(\frac{2}{3} - \frac{4}{5} \times \frac{1}{3}\right)}{\left(\frac{5}{3} + \frac{1}{4} \div \frac{3}{4}\right)} =$

A $-\frac{2}{115}$
B $-\frac{1}{45}$
C $\frac{2}{9}$
D $\frac{1}{5}$

Show the answer

Correct answer: D

First, simplify the numerator and the denominator separately.

To simplify the numerator, multiply the last two fractions and then subtract from the first fraction.

$\frac{2}{3} - \left(\frac{4}{5} \times \frac{1}{3}\right)$

$\frac{2}{3} - \frac{4}{15}$

$\frac{10}{15} - \frac{4}{15}$

$\frac{6}{15} = \frac{2}{5}$

To simplify the denominator, divide the last two fractions and then add to the first fraction.

$\frac{5}{3} + \left(\frac{1}{4} \div \frac{3}{4}\right)$

$\frac{5}{3} + \left(\frac{1}{4} \times \frac{4}{3}\right)$

$\frac{5}{3} + \frac{1}{3}$

$\frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2$

Finally, divide the numerator by the denominator.

$\frac{\frac{2}{5}}{2} = \frac{2}{5} \div 2 = \frac{2}{5} \times \frac{1}{2} = \frac{1}{5}$

Question analysis

This fraction arithmetic question gives a complex fraction: a numerator with a subtraction and a multiplication over a denominator with an addition and a division, all with fractions. Students choose its value. It tests order of operations and the four operations with fractions, and working left to right instead of following the order of operations leads to a wrong result.

NYS standards: 7.NS.3 four operations with rational numbers
6.NS.1 dividing fractions by fractions
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students compute with fractions, usually two or three operations at a time. This expression needs five, in an order set by the rules rather than read left to right, so it is longer and more trap-prone than typical work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Many steps moderate
Error-prone arithmetic moderate
Something you must already know minor
Solution: 5 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

Nothing here is beyond grade 7, but the expression stacks five fraction operations, and the top and the bottom each hide an order-of-operations decision. Working left to right in the numerator alone gives a negative value, and doing so in both parts gives an even stranger fraction; both are listed. The sign is a quick check: 4/15 is less than 2/3, so the numerator is positive, and so is the value.

Many steps moderate Working it out
The numerator and the denominator are each simplified on their own, to 2/5 and 2, before the final division, so five operations stand between the expression and its value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic moderate Working it out
Every step is fraction work: 2/3 − 4/15 needs fifteenths, 1/4 ÷ 3/4 needs the divisor flipped, and the last step, 2/5 ÷ 2, is easily turned into 2/5 × 2. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
No parentheses inside the numerator or the denominator show the order, so students must know to multiply before subtracting and to divide before adding. (Anderson & Krathwohl, 2001)

Strategy: Simplify the numerator and the denominator separately, doing the multiplication and the division before the subtraction and the addition, then divide the two results.

Traps

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

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Works left to right in both parts, (2/3 − 4/5) × 1/3 = −2/45 over (5/3 + 1/4) ÷ 3/4 = 23/9, and −2/45 ÷ 23/9 = −2/115.
B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Subtracts before multiplying in the numerator, (2/3 − 4/5) × 1/3 = −2/45, but handles the denominator correctly, so −2/45 ÷ 2 = −1/45.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2/9: it is positive like the answer, but no single order or fraction slip reaches it, so it mostly draws guesses between the two positive choices.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\times \text{ and } \div \text{ before } + \text{ and } -$; When no parentheses show the order, multiply and divide first. Then add and subtract. Everyone follows the same order, so an expression has only one value.
Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the top numbers, and b and d are the bottom numbers. Multiply top times top and bottom times bottom.
Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. The reciprocal of c/d is d/c. It works because c/d times d/c is 1. So multiplying by d/c undoes c/d, just as dividing by c/d does.

Sort out the question
The big fraction bar means divide: the top part divided by the bottom part. Work out the top and the bottom on their own. Then divide.
Inside each part, no parentheses show the order. So follow the order of operations.
Multiply first in the top
The top is $\frac{2}{3} - \frac{4}{5} \times \frac{1}{3}$ . Multiply before you subtract.
$\frac{4}{5} \times \frac{1}{3} = \frac{4 \times 1}{5 \times 3} = \frac{4}{15}$
Watch out: subtracting first, 2/3 − 4/5, makes the top negative. That leads to −1/45, choice B. Working left to right in the bottom as well gives −2/115, choice A.
Subtract to finish the top
To subtract, the bottoms must match. Change 2/3 to fifteenths by multiplying its top and bottom by 5.
$\frac{2}{3} - \frac{4}{15} = \frac{10}{15} - \frac{4}{15} = \frac{6}{15} = \frac{2}{5}$
To simplify 6/15, divide its top and bottom by 3. The top part is 2/5.
Divide first in the bottom
The bottom is $\frac{5}{3} + \frac{1}{4} \div \frac{3}{4}$ . Divide before you add. Flip 3/4 and multiply.
$\frac{1}{4} \div \frac{3}{4} = \frac{1}{4} \times \frac{4}{3} = \frac{4}{12} = \frac{1}{3}$
To simplify 4/12, divide its top and bottom by 4.
Add to finish the bottom
The bottoms already match, so add the tops and keep the bottom.
$\frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2$
Six thirds make 2 wholes. The bottom part is 2.
Divide the top by the bottom
Now divide 2/5 by 2. Flip 2, which is 2/1, to get 1/2.
$\frac{2}{5} \div 2 = \frac{2}{5} \times \frac{1}{2} = \frac{2}{10} = \frac{1}{5}$
Dividing by 2 is the same as taking half, and half of 2/5 is 1/5. So the answer is D, $\frac{1}{5}$ .

Check your answer

Look at the signs. 4/15 is less than 2/3, so the top is positive. The bottom is positive too. So the answer must be positive, which rules out A and B. Then multiply back: $\frac{1}{5} \times 2 = \frac{2}{5}$ , the top. It matches.

More questions like this

The original DOE question

Question 83 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 2023 Admissions, page 204, question 83. Open page 204 of the handbook (PDF)

Question 63 of 83

2022 Test B Math, question 38

Question

WINTER TEMPERATURES IN CHICAGO

Day	1	2	3	4	5	6	7
Temperature (°F)	17	7	2	4	−3	−5	1

The table above shows temperatures for a one-week period during the winter in Chicago. What is the range of temperatures shown in the table?

Show the answer

Correct answer: C

Range is the difference between the largest and the smallest values in a given set of data. In this case, the highest temperature is 17 and the lowest temperature is −5. Therefore, the range is $17-(-5)=17+5=22$ .

Question analysis

A table gives Chicago's temperatures in degrees Fahrenheit for seven winter days, some above zero and some below. Students find the range of the temperatures. It tests the meaning of range and subtracting a negative number, and treating a below-zero reading as if it were positive gives a wrong spread.

NYS standards: 7.NS.1c rational number subtraction
6.SP.5c range, center, and overall pattern
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Range is a grade 6 idea, and subtracting a negative number is grade 7. Here they meet in a single subtraction from a short table, typical grade 7 practice with integers rather than anything longer.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Reading a figure, table or graph minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Finding a range is routine, but the lowest reading is below zero, so the subtraction crosses zero: from −5 up to 0 is 5 degrees, and from 0 up to 17 is 17 more. Students who treat −5 as 5 find a smaller gap that is listed. Picturing a thermometer, or a number line, shows the full spread at a glance.

Error-prone arithmetic moderate Working it out
The lowest reading is −5°F, so the range is 17 − (−5), and subtracting a negative number means adding 5, the step where a gap of 12 appears instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph minor Understanding the question
The seven readings run across one table row in day order, 17, 7, 2, 4, −3, −5, 1, so the extremes have to be picked out, including the lower of the two below zero. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The word range is not defined in the question; students must know it means the highest value minus the lowest, not the highest value alone. (Anderson & Krathwohl, 2001)

Strategy: Pick out the highest and the lowest temperature in the table, then subtract the lowest from the highest, writing the subtraction of a negative as an addition.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Subtracts 5 from 17 as if −5 were 5, so the range comes out 12, missing that a reading below zero lies further from 17, not closer.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: weak
Gives the highest temperature, 17°F, as the range, taking the top of the readings for the spread between the highest and the lowest.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 27: it is 5 more than the true spread and matches no sign or subtraction mistake with these readings, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range: $\text{range} = \text{highest} - \text{lowest}$; The range tells how spread out the data are. Find the highest value and the lowest value, then subtract. It is the distance between the two ends of the data on a number line.
Subtracting a negative: $a - (-b) = a + b$; a is any number, and −b is a negative number. Taking away a negative is the same as adding the positive number b. On a number line, the gap from −5 up to 17 is 5 to reach 0, then 17 more.

Sort out the question
The table gives seven temperatures, in degrees Fahrenheit (°F): 17, 7, 2, 4, −3, −5 and 1. A minus sign means below zero. You need the range, the highest temperature minus the lowest.
Find the highest and lowest
The highest is 17°F, on day 1. For the lowest, look at the numbers below zero: −3 and −5. −5 is colder, since it is further below zero. So the lowest is −5°F, on day 6.
Subtract the lowest from the highest
$\text{range} = 17 - (-5) = 17 + 5 = 22$
Subtracting a negative is the same as adding. The range is 22 degrees Fahrenheit.
Watch out: −5 is not 5. Doing 17 − 5 = 12 gives choice A. And 17 alone is just the highest temperature, choice B. So the answer is C, 22.

Check your answer

Picture a thermometer. From −5°F up to 0°F is 5 degrees. From 0°F up to 17°F is 17 more. In all, $5 + 17 = 22$ degrees. That matches.

More questions like this

The original DOE question

Question 95 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 2023 Admissions, page 207, question 95. Open page 207 of the handbook (PDF)

Question 64 of 83

2018 Test A Math, question 3

Question

For what value of $w$ is $4w=2w-8$ ?

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

Show the answer

Correct answer: -4

$4w=2w-8$

$2w=-8$

$w=-4$

Question analysis

A one-line algebra grid-in: find the value of w that makes 4w equal to 2w minus 8. Students solve a linear equation with the variable on both sides by collecting the w terms and dividing. Errors come from handling the subtraction sign or stopping before w stands alone.

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 Equations with the variable on both sides are formally a grade 8 topic, but this one takes two short steps with whole numbers. It asks about as much as a typical grade 7 two-step equation.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 7 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One of the easiest questions on the test: collect the w terms on one side and divide. Errors come from carelessness rather than method, such as dropping the negative sign or gridding −8 after the first step because it already looks like a finished answer. Substituting the result back into both sides of the equation takes seconds and catches both.

Error-prone arithmetic minor Working it out
The only constant is −8, so the answer is negative; dividing −8 by the coefficient 2 is the step where the sign is most often lost. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The grid-in needs the negative sign filled in its own column, and there are no choices to show that the answer lies below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Subtract 2w from both sides so the variable appears only once, divide by the remaining coefficient, and check by substituting into both sides.

Traps

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

The wrong answers, classified

4 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Gets 2w = −8 correctly but drops the sign when dividing by 2, gridding 4 instead of a negative value.
-8 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Subtracts 2w from both sides to reach 2w = −8 and grids −8 without dividing by the coefficient 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c$; a and b are the two sides of an equation, and c is any number or term. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. The same is true if you add the same thing, or divide both sides by the same number that is not zero.

Sort out the equation
An equation says two things are equal. You need the value of w that makes $4w$ equal to $2w - 8$ . Here 4w means 4 times w, and 2w means 2 times w. The answer may be a negative number, so keep track of the signs.
Get w on one side
w is on both sides. Subtract 2w from both sides. Then w is only on the left, and the equation stays balanced.
$4w - 2w = 2w - 8 - 2w$
$2w = -8$
Now 2 times w is −8.
Divide both sides by 2
To undo times 2, divide both sides by 2. A negative number divided by a positive number is negative.
$w = -8 \div 2 = -4$
Watch out: keep the minus sign. Dropping it gives 4. Also, don’t stop at $2w = -8$ and grid −8. So you would grid -4, with the minus sign in its own column.

Check your answer

Put −4 back in for w. Left side: $4 \times (-4) = -16$ . Right side: $2 \times (-4) - 8 = -8 - 8 = -16$ . Both sides are −16, so −4 works.

More questions like this

The original DOE question

Question 60 as printed on page 81 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 60. Open page 81 of the handbook (PDF)

Question 65 of 83

2018 Test A Math, question 5

Question

The sum of two consecutive integers is −15. If 1 is added to the smaller integer and 2 is subtracted from the larger integer, what is the product of the two resulting integers?

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

Show the answer

Correct answer: 63

If x is the smaller consecutive integer, then

$x+1$

is the larger consecutive integer. Use their sum (−15) to find x:

$x+(x+1)=-15$

$2x+1=-15$

$2x=-16$

$x=-8$

The two consecutive integers are −8 and −7. One is added to the smaller integer:

$-8+1=-7$

and 2 is subtracted from the larger integer:

$-7-2=-9$

Find the product:

$(-7)(-9)=63$

Question analysis

A grid-in problem about two consecutive integers whose sum is −15. Students find the two integers, add 1 to the smaller and subtract 2 from the larger, then multiply the results. Deciding which negative integer is the smaller one, and keeping signs straight through every step, are what make it tricky.

NYS standards: 7.EE.4a two-step equations
7.NS.2a rational number multiplication
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 A typical grade 7 task solves one two-step equation or multiplies signed numbers. This links both: set up a consecutive-integer equation, solve it with negatives, adjust each integer, and multiply, with no choices to check against.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math moderate
Many steps moderate
Solution: 4 steps
Reading level: Question and choices: about grade 9, 30 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every step is routine, yet the question collects most of the ways students lose points with negatives. The crucial judgment is which of −8 and −7 is the smaller; a student who picks −7 still produces a tidy, wrong product. Gridding the product of the original pair, or giving a negative product, are the other usual losses. Writing each intermediate pair down, rather than holding them in mind, protects against all three.

Error-prone arithmetic moderate Working it out
Every number in play is negative: −8 and −7, then −7 and −9, and a product of two negatives that must come out positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math moderate Understanding the question
"1 is added to the smaller integer and 2 is subtracted from the larger" depends on knowing that −8, not −7, is the smaller, since it lies farther below zero. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps moderate Working it out
Four linked steps: set up x + (x + 1) = −15, solve for x, adjust each integer, then multiply; an error early in the chain carries through to the product. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Halve the sum to see which two integers sit on either side of it, remember that the smaller negative is farther from zero, then adjust each and multiply.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

60 Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Takes −7 as the smaller integer because 7 is less than 8, so adjusts to −7 + 1 = −6 and −8 − 2 = −10, and multiplies to 60.
56 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds −8 and −7 and multiplies them to get 56, skipping the sentence that adds 1 to one integer and subtracts 2 from the other.
-63 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Finds the adjusted integers −7 and −9 correctly but gives their product a negative sign, gridding −63.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive integers: $n, \; n + 1$; Integers are whole numbers and their opposites, like −2, −1, 0, 1 and 2. Consecutive integers come one right after another. If n is the smaller one, the next one is n + 1, because each step up the number line adds 1.
Multiplying two negative numbers: $(-a) \times (-b) = a \times b$; a and b are positive numbers, so −a and −b are negative. When you multiply two negative numbers, the answer is positive. The two minus signs cancel out.

Sort out the question
Two consecutive integers add up to −15. Integers are whole numbers and their negatives. Consecutive means one comes right after the other, like 4 and 5. You need to do three things:
1. Find the two integers.
2. Add 1 to the smaller one. Subtract 2 from the larger one.
3. Find the product of the two new numbers. The product is the answer when you multiply.
Find the two integers
Call the smaller integer n. Then the larger one is n + 1. Their sum is −15.
$n + (n + 1) = -15$
$2n + 1 = -15$
Subtract 1 from both sides. Then divide both sides by 2.
$2n = -16$
$n = -8$
So the integers are −8 and −7.
Decide which one is smaller
On a number line, −8 is to the left of −7. Numbers to the left are smaller. So −8 is the smaller integer, and −7 is the larger one. Watch out: 8 is bigger than 7, but −8 is smaller than −7. Mixing them up gives −6 and −10, and the wrong answer 60.
Change each integer
Add 1 to the smaller integer.
$-8 + 1 = -7$
Subtract 2 from the larger integer.
$-7 - 2 = -9$
The two new numbers are −7 and −9.
Multiply the new numbers
Both numbers are negative, so their product is positive.
$(-7) \times (-9) = 63$
Watch out: the answer is not −63, because two negatives make a positive. Also, don’t multiply the first pair, −8 and −7. That gives 56 and skips the changes. So you would grid 63.

Check your answer

Check the pair: $-8 + (-7) = -15$ , and they are one apart. Half of −15 is −7.5, and −8 and −7 sit on either side of it. Then $7 \times 9 = 63$ , and the sign is positive.

More questions like this

The original DOE question

Question 62 as printed on page 81 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 62. Open page 81 of the handbook (PDF)

Question 66 of 83

2018 Test A Math, question 7

Question

$7+\left ( 3n+6 \right )-\left ( 4n+8 \right )=$

A $5-n$
B $5+n$
C $21-n$
D $21+n$

Show the answer

Correct answer: A

$7+3n+6-4n-8=(7+6-8)+(3n-4n)=5-n$

Question analysis

An expression question: simplify a sum and difference of two linear expressions in n. Students remove the parentheses, subtract every term inside the second group, and combine like terms. The minus sign in front of the second parentheses is the trap, since it has to change the sign of both terms inside.

NYS standards: 7.EE.1 linear expressions with rational coefficients
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Simplifying linear expressions with a subtracted group is standard grade 7 work. This is a typical textbook exercise in length, with the usual sign trap and no extra steps or real-world context.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Solution: 2 steps
Reading level: Question and choices: about grade 16+, 5 words.
Corrected from the DOE handbook: The handbook’s answer key for this question was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

There are only five terms, but two signs decide the answer: the minus in front of the second parentheses and the sign left after 3n − 4n. Each wrong choice comes from one or both of those sign decisions, so all of them look plausible, and checking needs care rather than time. Even the DOE handbook keyed the version with the wrong sign on n.

Error-prone arithmetic moderate Working it out
The minus sign before (4n + 8) has to reach both terms, making −4n and −8, and 3n − 4n leaves a negative n term. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
5 − n and 5 + n are written alike and share the constant 5; only the sign of the n term, from 3n − 4n, separates them, and the DOE handbook's own key picked the wrong one. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Rewrite the subtraction as adding the opposite of each term in the second parentheses, then group the constants and the n terms separately.

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.)

The wrong answers, classified

B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Gets the constants right, 7 + 6 − 8 = 5, but writes 3n − 4n as +n, subtracting the smaller coefficient from the larger and keeping a plus sign.
C Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Applies the minus sign to 4n but not to 8, so the constants become 7 + 6 + 8 = 21 while the n term comes out right.
D Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Makes both mistakes: adds the 8 instead of subtracting it, for 21, and also writes 3n − 4n as +n.

What was corrected from the DOE handbook, and why

Answer key: The handbook (its question 64) keys this question as F, which is 5 + n (choice B here), in its answer key and in the heading of its explanation. Its own working gives 7 + 3n + 6 − 4n − 8 = 5 − n, which is the handbook’s choice E and choice A here, and that is correct. The key here is A.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a group in parentheses: $-(a + b) = -a - b$; a and b are the terms inside the parentheses. A term is a number, a letter, or a number times a letter. The minus sign in front means you subtract the whole group. So you subtract every term inside, not just the first one.
Combining like terms: $3n - 4n = (3 - 4)n$; n stands for an unknown number, and 3n means 3 times n. Like terms have the same letter part, like 3n and 4n. To combine them, add or subtract the numbers in front and keep the letter. Plain numbers combine only with plain numbers.

Sort out the question
You need to simplify the expression, which means write it in a shorter form. An expression is a math phrase made of numbers, letters and signs. Here n stands for an unknown number. Find the choice that matches the short form.
Remove the parentheses
The first group has a plus sign in front. Adding a group changes nothing, so $+(3n + 6)$ is just $+3n + 6$ . The second group has a minus sign in front. You must subtract both terms inside it.
$-(4n + 8) = -4n - 8$
$7 + 3n + 6 - 4n - 8$
Watch out: the minus reaches the 8 too. Adding the 8 instead gives 21 for the plain numbers. That leads to choice C or D.
Combine the plain numbers
Group the numbers without a letter. Work them out from left to right.
$7 + 6 - 8 = 5$
The plain numbers make 5.
Combine the n terms
You have 3 n’s, and you take away 4 n’s. The numbers in front give 3 − 4 = −1. So you are left with −1n, which is written −n.
$3n - 4n = (3 - 4)n = -1n = -n$
Watch out: it is easy to write +n here. That gives 5 + n, choice B.
Put the two parts together
Join the number part and the n part. Adding a negative is the same as subtracting.
$5 + (-n) = 5 - n$
So the answer is A, 5 − n.

Check your answer

Pick any number for n and test it. Try n = 1. The original: $7 + (3 + 6) - (4 + 8) = 7 + 9 - 12 = 4$ . Choice A: $5 - 1 = 4$ . They match. Choice B gives $5 + 1 = 6$ , which does not.

More questions like this

The original DOE question

Question 64 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 64. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 67 of 83

2018 Test A Math, question 20

Question

Mrs. Cranston bought five bottles of water for $0.90 each and 8 pounds of meat. She paid a total of $26.90 for these items, not including tax. What was the price per pound of the meat?

A$2.80
B$3.25
C$14.40
D$22.40

Show the answer

Correct answer: A

Let $x$ be the price per pound for the meat. Set up an equation to show what Mrs. Cranston spent:

$5(0.90)+8x=26.90$

$4.50+8x=26.90$

$8x=22.40$

$x=2.80$

The price per pound for the meat is $2.80.

Question analysis

This money word problem has a shopper buying five bottles of water at a set price plus 8 pounds of meat, for a known total. Students write and solve a two-step equation to find the price per pound of the meat. The decimals call for care, and the question wants a price per pound rather than a total.

NYS standards: 7.EE.4a two-step equations
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Word problems that lead to an equation like px + q = r are the core of grade 7 algebra. This is a standard example with money decimals and no extra twist.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a two-step money problem: take away the water, then divide by the pounds. The listed wrong answers come from doing one of those steps wrongly or not at all, especially stopping at the total cost of the meat, which looks like a finished dollar amount. Checking that 8 pounds at the chosen price plus $4.50 makes $26.90 confirms the answer.

Turning words into math minor Understanding the question
"five bottles of water for $0.90 each" has to become 5 × 0.90 before the total is split, and the unknown is a price per pound, not the meat's total cost. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Every amount is a decimal of dollars, so 26.90 − 4.50 and 22.40 ÷ 8 must be lined up and worked carefully without a calculator. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the cost of all five bottles from the total, then divide what is left by the 8 pounds of meat.

Traps

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

The wrong answers, classified

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Misses the word each and subtracts one bottle's $0.90, then divides: (26.90 − 0.90) ÷ 8 = 3.25.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Finds the $22.40 spent on meat and subtracts the 8 pounds from it instead of dividing, getting 14.40.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Subtracts the five bottles from the total, 26.90 − 4.50 = 22.40, and stops at the meat's total cost without dividing by 8 pounds.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost: $C = p \times q$; C is the total cost. p is the price of one item, or of one pound. q is the quantity, how many items or pounds you buy. Each one costs p, so q of them cost p times q.

Sort out the question
Mrs. Cranston bought five bottles of water “for $0.90 each”. She also bought 8 pounds of meat. Everything cost $26.90 in all. You need the price per pound of the meat, the cost of just 1 pound.
Find the cost of the water
Each bottle costs $0.90, and there are 5 bottles. Use the cost formula.
$C = p \times q$
$C = 0.90 \times 5 = 4.50$
The water cost $4.50. Watch out: each means every bottle costs $0.90. Taking away just one $0.90 leads to $3.25, choice B.
Find the cost of the meat
Take the water away from the total. What is left paid for the meat.
$26.90 - 4.50 = 22.40$
All 8 pounds of meat cost $22.40. Watch out: this is the cost of all the meat, not of one pound. Stopping here gives choice D.
Find the price per pound
Use the cost formula again. Now C is 22.40 and q is 8 pounds. Divide to find p.
$22.40 = p \times 8$
$p = 22.40 \div 8 = 2.80$
Each pound costs $2.80. So the answer is A, $2.80.

Check your answer

Put it all back together. Meat: $8 \times 2.80 = 22.40$ . Water: 4.50. Total: $22.40 + 4.50 = 26.90$ . That is exactly what she paid.

More questions like this

The original DOE question

Question 77 as printed on page 85 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 77. Open page 85 of the handbook (PDF)

Question 68 of 83

2018 Test A Math, question 28

Question

$2\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}$

What is the value of the expression shown above?

A $14\frac{7}{20}$
B $14\frac{2}{5}$
C $15\frac{7}{20}$
D $15\frac{2}{5}$

Show the answer

Correct answer: D

$2\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}$

Convert all the fractions to a common denominator (10):

$2\frac{2}{10}+3\frac{3}{10}+4\frac{4}{10}+5\frac{5}{10}$

$=(2+3+4+5)+\left(\frac{2+3+4+5}{10}\right)$

$=14+1\frac{4}{10}=15\frac{2}{5}$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
An answer in an unexpected form minor
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.

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)
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)
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

Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Adds the numerators straight across, 1 + 3 + 2 + 1 = 7, and puts them over 20, a common denominator (10 × 2), without rescaling them, so nothing carries.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Adds the fractions to 14/10 but keeps only the 4/10 = 2/5 left over, forgetting to carry the whole 1 into 14.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Pairs a correct carry into 15 with the 7/20 from adding numerators, two steps no single line of working combines; it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; 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.

Sort out the question
You need to add four mixed numbers. A mixed number is a whole number plus a fraction, like $2\frac{1}{5}$ . The fractions have different denominators, or bottom numbers: 5, 10, 5 and 2. To add fractions, they need the same denominator.
Add the whole numbers
Add the whole numbers first. You will add the fractions on their own after that.
$2 + 3 + 4 + 5 = 14$
The whole numbers make 14.
Rewrite each fraction in tenths
Use 10 as the denominator, since 5 and 2 both go into 10.
$\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10}$
$\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}$
$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$
$\frac{3}{10}$ 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.
Add the fractions
$\frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} = \frac{14}{10}$
14 tenths is more than 1 whole. Ten tenths make 1, so 14 tenths is 1 whole and 4 tenths.
$\frac{14}{10} = 1\frac{4}{10} = 1\frac{2}{5}$
To simplify 4/10, divide the top and bottom by 2.
Put it all together
Add the 14 wholes to the sum of the fractions.
$14 + 1\frac{2}{5} = 15\frac{2}{5}$
Watch out: carry the extra whole into the 14. Missing it gives $14\frac{2}{5}$ , choice B. So the answer is D, $15\frac{2}{5}$ .

Check your answer

Change each mixed number to a decimal: $2.2 + 3.3 + 4.4 + 5.5 = 15.4$ . The 0.4 is 4 tenths, which is 2/5. So the sum is $15\frac{2}{5}$ .

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 69 of 83

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 = $\frac{1}{8}$ . 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 $LN=\frac{1}{8}$ , point $N$ is located at

$4\frac{5}{16}+\frac{1}{8}=4\frac{7}{16}$ .

So $M$ must be between point $L$ , $4\frac{5}{16}$ , and point $N$ , $4\frac{7}{16}$ . Point $L$ can also be written as 4.3125, and point $N$ 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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
A very close wrong answer moderate
Reading a figure, table or graph minor
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.

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)
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)
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

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Measures the one-eighth gap to the left of L, giving 4.1875 to 4.3125, an interval that contains 4.26 but contradicts the figure.
B Rounds too early (King et al., 2004)
Reveals a slip Expected pull: strong
Rounds L's 4.3125 to 4.31 and treats 4.31 as inside the interval, missing that it lies slightly to the left of L.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Writes 4 5/16 as 4.516 by placing the fraction's digits after the decimal point, so N becomes about 4.641 and 4.58 appears to fit.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions: $\frac{a}{b} = \frac{a \times k}{b \times k}$; 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: $\frac{a}{b} = a \div b$; 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.

Sort out what you know
Point L is at $4\frac{5}{16}$ . Point N is to the right of L. The distance from L to N, called LN, is $\frac{1}{8}$ .
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.
Find where N is
N is $\frac{1}{8}$ to the right of L, so add. First give $\frac{1}{8}$ a bottom of 16, to match L:
$\frac{1}{8} = \frac{1 \times 2}{8 \times 2} = \frac{2}{16}$
$4\frac{5}{16} + \frac{2}{16} = 4\frac{7}{16}$
Watch out: N is to the right of L in the figure. Going left instead gives a gap that holds 4.26, choice A.
Change both points to decimals
Divide the top of each fraction by the bottom:
$\frac{5}{16} = 5 \div 16 = 0.3125$
$\frac{7}{16} = 7 \div 16 = 0.4375$
So L is at 4.3125 and N is at 4.4375.
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 70 of 83

2018 Test A Math, question 39

Question

Suppose $M=\frac{w}{x}$ , $N=\frac{y}{z}$ , and $w,x,y$ , and $z$ do not equal 0. What is $\frac{M}{N}$ in terms of $w,x,y$ , and $z$ ?

A $\frac{wx}{yz}$
B $\frac{wy}{xz}$
C $\frac{wz}{xy}$
D $\frac{xy}{wz}$

Show the answer

Correct answer: C

It is easier to rewrite

$\frac{M}{N}$

$M\div N$

since they are both fractions.

$M\div N=\frac{w}{x}\div \frac{y}{z}=\frac{w}{x}\times \frac{z}{y}=\frac{wz}{xy}$

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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
A very close wrong answer moderate
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.

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)
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

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Stacks M's letters over N's letters, writing wx over yz, without carrying out any fraction division.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies M by N straight across, top times top and bottom times bottom, instead of dividing.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Flips the first fraction instead of the second, computing x/w × y/z, which gives N divided by M.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; 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: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; 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.

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 $\frac{M}{N}$ means M divided by N. So you must divide one fraction by another.
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:
$\frac{M}{N} = M \div N = \frac{w}{x} \div \frac{y}{z}$
Flip the second fraction
Use the rule for dividing fractions. Keep the first fraction. Flip only the second one, and change divide to multiply:
$\frac{w}{x} \times \frac{z}{y}$
Watch out: flipping the first fraction instead gives choice D, which is N ÷ M. Multiplying without flipping at all gives choice B.
Multiply across
Multiply the tops and the bottoms:
$\frac{w \times z}{x \times y} = \frac{wz}{xy}$
Here wz means w times z. So the answer is C, $\frac{wz}{xy}$ .

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 71 of 83

2018 Test A Math, question 44

Question

$Number line with points W, X, Y and Z marked from left to right. W is at −10 and Z is at 8; a tick mark at 0 lies between X and Y. X and Y have no values given.$

On the number line above, points $W$ , $X$ , $Y$ , and $Z$ are integers, and $WX:XY:YZ = 4:2:3$ . What is the value of $\overline{WY}$ ?

Show the answer

Correct answer: C

Convert the ratios into fractions of $WZ$ . Use the sum of the ratios for the denominator.

$WX:XY:YZ=4:2:3$

$WX=\frac{4}{4+2+3}WZ=\frac{4}{9}WZ$

$XY=\frac{2}{4+2+3}WZ=\frac{2}{9}WZ$

The part of $WZ$ that is $WY$ is the sum of those fractions:

$WY=\frac{4}{9}WZ+\frac{2}{9}WZ=\frac{6}{9}WZ=\frac{2}{3}WZ$

Find the length of $WZ$ :

$WZ=8-(-10)=18$

The value of $WY$ is $\frac{2}{3}(18)=12$ .

Question analysis

This number line question places integer points W, X, Y and Z on a line and gives the ratio of the segments WX, XY and YZ as 4 to 2 to 3. Students find the whole length from W to Z, split it by the ratio, and add the parts that make up WY. Mixing up a part with the whole is the main risk.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students find distances on a number line and use part-to-whole ratios, usually separately. This item joins them: a negative endpoint, a three-term ratio and a choice of which segments to combine.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Reading a figure, table or graph moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 10, 21 words.
Corrected from the DOE handbook: Equations in the explanation that mixed up a fraction of a length with the length itself were corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is routine, but the question keeps asking for a different piece of the line: the whole WZ, the part WX, and finally WY, which is two parts added together. Two of the wrong choices are exactly those other pieces. Students who mark the actual positions, X at −2 and Y at 2, can check the ratio before answering.

Combines ideas moderate Working it out
A three-part ratio, 4:2:3, must be applied to a length found from a negative endpoint, 8 − (−10) = 18, joining ratio and number-line distance. (Embretson, 1998; Daniel & Embretson, 2010)
Reading a figure, table or graph moderate Understanding the question
Only W at −10 and Z at 8 are labeled, and the drawing is not to scale: Y is drawn nearer 5 than 2, so reading WY from the picture fails. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Find WZ, split it into 9 equal parts of 2, then add the 4 parts of WX and the 2 parts of XY. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length WZ from the endpoints on the number line, divide it by the total number of ratio parts, and multiply by the parts from W to Y.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds WX, 4/9 of 18 = 8, and stops without adding XY's 4 to reach Y; 8 also matches Z's label.
B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Places Y at 2 but counts the integers strictly between −10 and 2, which are −9 through 1, getting 11 instead of the distance 12.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Gives the whole length WZ, 8 − (−10) = 18, instead of the part from W to Y.

What was corrected from the DOE handbook, and why

Explanation: The handbook writes WX = 4/9, XY = 2/9 and WY = 2/3, and then says “The value of WY is 2/3(18) = 12”. Those fractions are parts of WZ, not the lengths themselves (WX is 8 and WY is 12), so WY cannot be both 2/3 and 12. The lines now read WX = 4/9 WZ, XY = 2/9 WZ and WY = 2/3 WZ. The method and the answer are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line: $d = b - a$; d is the distance between two points. b is the point on the right and a is the point on the left. Subtracting left from right always gives a positive distance. It works even when a is negative.
Size of one part of a ratio: $u = \frac{L}{p}$; u is the size of one part. L is the whole length being shared. p is the total number of parts, found by adding the numbers in the ratio. Every part is the same size, so you share the length equally.

Sort out what you know
W is at −10 and Z is at 8. X and Y are between them. The ratio 4:2:3 means WX is 4 parts, XY is 2 parts and YZ is 3 parts. All the parts are the same size.
You need the length WY, from W to Y. The drawing is not exact, so use the numbers, not the picture.
Find the whole length WZ
$d = 8 - (-10) = 8 + 10 = 18$
WZ is 18 units long. Subtracting a negative number is the same as adding.
Find the size of one part
Add the ratio numbers to get the number of parts:
$4 + 2 + 3 = 9$
Share the 18 units equally among 9 parts:
$u = \frac{18}{9} = 2$
Each part is 2 units long.
Find WY
WY is WX and XY together, so it has 4 + 2 = 6 parts:
$6 \times 2 = 12$
Watch out: WX alone is 4 × 2 = 8, which is choice A. And 18 is the whole line, choice D. So the answer is C, 12.

Check your answer

Place the points. W is at −10. X is 8 units to the right, at −2. Y is 4 more, at 2. Z is 6 more, at 8, which matches its label. And the distance from −10 to 2 is 2 − (−10) = 12.

More questions like this

The original DOE question

Question 101 as printed on page 89 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 89, question 101. Open page 89 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 72 of 83

2018 Test A Math, question 48

Question

$\frac{10}{13}=0.\overline{769230}$

In the infinitely repeating decimal above, 7 is the first digit in the repeating pattern. What is the 391st digit?

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
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.

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

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Sees that 65 blocks fill 390 places and lands on the 390th digit, the 0 that ends a block, one place short.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Uses the quotient 65 instead of the remainder, takes its last digit 5, and reads the fifth digit of 769230, which is 3.
C Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Treats the remainder 1 as one step past the block's first digit, moving from 7 to the second digit, 6.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division with a remainder: $n = d \times q + r$; 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.

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.
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.
Divide 391 by 6
Find how many full blocks fit into 391 digits, and what is left:
$391 = 6 \times 65 + 1$
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.
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 73 of 83

2018 Test A Math, question 54

Question

$Number line with points J, K, L and M marked from left to right. Only K has a value: 3/8.$

On the number line above, $JK=3\frac{1}{2}$ , $JM=9\frac{3}{4}$ , and $LM=1\frac{1}{8}$ . What is the position of point $L$ ?

A $5\frac{1}{8}$
B $5\frac{1}{4}$
C $5\frac{1}{2}$
D $6\frac{1}{4}$

Show the answer

Correct answer: C

First, use $JK=3\frac{1}{2}$ to find the location of $J$ :

$\frac{3}{8}-J=3\frac{1}{2}$

$J=\frac{3}{8}-3\frac{1}{2}=-3\frac{1}{8}$

Now, use $JM=9\frac{3}{4}$ to find the location of $M$ :

$M-\left(-3\frac{1}{8}\right)=9\frac{3}{4}$

$M+3\frac{1}{8}=9\frac{3}{4}$

$M=9\frac{3}{4}-3\frac{1}{8}=6\frac{5}{8}$

Finally, use $LM=1\frac{1}{8}$ to find the location of $L$ :

$6\frac{5}{8}-L=1\frac{1}{8}$

$L=6\frac{5}{8}-1\frac{1}{8}=5\frac{4}{8}=5\frac{1}{2}$

Question analysis

This number line question gives the lengths JK, JM and LM as mixed numbers and asks for the position of point L. Students must chain three distance steps, moving left from K to J, right to M and back to L, with fractions in eighths and a negative position. Losing track of where K sits is the main danger.

NYS standards: 7.NS.1c rational number subtraction
7.NS.1b rational number addition
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students add and subtract rational numbers and find distances on a number line one at a time. This item chains three such steps with mixed numbers and a negative position, more than typical grade 7 work demands.
Estimated difficulty: Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer major
Error-prone arithmetic moderate
Reading a figure, table or graph moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 9, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every number is a mixed fraction, one position is negative, and the figure's even spacing hides how long KL really is. The distance route, KL = 9 3/4 − 3 1/2 − 1 1/8, avoids the negative entirely but ends one step short of a position unless K's 3/8 is added back. That last step is the one most students skip, which is why the question sits at the top of the scale.

A very close wrong answer major Choosing the answer
The choices 5 1/8, 5 1/4 and 5 1/2 lie within 3/8 of one another, so no estimate separates them; only exact eighths and adding K's 3/8 pick out the right one. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic moderate Working it out
Every step mixes halves, quarters and eighths, and placing J means 3/8 − 3 1/2 = −3 1/8, a subtraction that crosses zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph moderate Understanding the question
Only K is labeled, and the points are drawn evenly spaced although KL is over four times LM, so the picture gives no help with sizes. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Three linked moves, K to J, J to M and M back to L, each depend on the last, so an early slip carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Start from K's labeled position, subtract JK to place J, add JM to place M, then subtract LM, keeping every fraction in eighths.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds KL = 9 3/4 − 3 1/2 − 1 1/8 = 5 1/8 and stops one step short, never adding K's position of 3/8.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Rewrites 9 3/4 as 9 4/8 instead of 9 6/8, so M lands at 6 3/8 and L at 6 3/8 − 1 1/8 = 5 1/4.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds KM = 9 3/4 − 3 1/2 = 6 1/4, the distance from K to M, and gives it without moving back 1 1/8 to L.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line: $d = b - a$; d is the distance between two points. b is the point on the right and a is the point on the left. Subtracting left from right always gives a positive distance. It works even when a is negative.
Equal fractions: $\frac{a}{b} = \frac{a \times k}{b \times k}$; 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.

Sort out what you know
Only K has a value: $\frac{3}{8}$ . JK means the distance from J to K. You know JK = $3\frac{1}{2}$ , JM = $9\frac{3}{4}$ and LM = $1\frac{1}{8}$ .
You need the position of L, the number where L sits. The points are drawn evenly spaced, but the distances are not equal. So use the numbers, not the picture.
Write the fractions in eighths
The fractions use halves, fourths and eighths. Give them all a bottom of 8:
$3\frac{1}{2} = 3\frac{4}{8} \qquad 9\frac{3}{4} = 9\frac{6}{8}$
Watch out: $\frac{3}{4}$ is $\frac{6}{8}$ , not $\frac{4}{8}$ . That slip leads to choice B.
Find the distance from K to M
JM is JK and KM put together. So take JK away from JM:
$9\frac{6}{8} - 3\frac{4}{8} = 6\frac{2}{8}$
KM is $6\frac{2}{8}$ , or $6\frac{1}{4}$ . Watch out: this is a distance, not where L sits. Choice D stops here.
Find the distance from K to L
KM is KL and LM put together. So take LM away:
$6\frac{2}{8} - 1\frac{1}{8} = 5\frac{1}{8}$
KL is $5\frac{1}{8}$ . Watch out: this is still a distance. Choice A stops here.
Find the position of L
L is $5\frac{1}{8}$ to the right of K, and K is at $\frac{3}{8}$ . Add:
$\frac{3}{8} + 5\frac{1}{8} = 5\frac{4}{8} = 5\frac{1}{2}$
So the answer is C, $5\frac{1}{2}$ .

Check your answer

Place every point. J is $3\frac{4}{8}$ left of K, at $-3\frac{1}{8}$ . M is $9\frac{6}{8}$ right of J, at $6\frac{5}{8}$ . L is $1\frac{1}{8}$ left of M, at $5\frac{4}{8}$ , which is $5\frac{1}{2}$ . Same answer.

More questions like this

The original DOE question

Question 111 as printed on page 91 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 91, question 111. Open page 91 of the handbook (PDF)

Question 74 of 83

2018 Test B Math, question 3

Question

$\left | \left ( -6 \right )-\left ( -5 \right )+4.2 \right |-\left | 3-9.6 \right |$

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

Show the answer

Correct answer: -3.4

$=3.2-6.6=-3.4$

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: 7.NS.1c rational number subtraction
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 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.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps moderate
An answer in an unexpected form minor
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.

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)
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)
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

3.4 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Believes an expression built from absolute values cannot be negative, so reports 6.6 − 3.2 instead of 3.2 − 6.6.
9.8 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Treats the bars like parentheses, leaving 3 − 9.6 as −6.6, then subtracts that negative: 3.2 + 6.6.
0.2 Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Reads (−6) − (−5) as −6 − 5 = −11, so the first bars give |−6.8| = 6.8, and 6.8 − 6.6 = 0.2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value: $|a| = \text{distance from } a \text{ to } 0$; 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: $a - (-b) = a + b$; 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.

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.
Work inside the first bars
Start with $(-6) - (-5) + 4.2$ . Subtracting −5 is the same as adding 5.
$(-6) - (-5) + 4.2 = -6 + 5 + 4.2 = -1 + 4.2 = 3.2$
3.2 is already positive, so $|3.2| = 3.2$ .
Watch out: If you read the start as −6 − 5, you get −11. That path ends at the wrong answer 0.2.
Work inside the second bars
Next, $3 - 9.6 = -6.6$ . This result is negative. Its distance from 0 is 6.6, so the absolute value is positive.
$|-6.6| = 6.6$
Subtract the two results
Now take the second result away from the first.
$3.2 - 6.6 = -3.4$
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 75 of 83

2018 Test B Math, question 6

Question

If $x=9$ and $y=-7$ , what is the value of $x(x-2y)$ ?

A18
B45
C144
D207

Show the answer

Correct answer: D

$x(x-2y)=9[9-2(-7)]=9(9+14)=9(23)=207$

Question analysis

This algebra question asks students to evaluate the expression x(x − 2y) when x is 9 and y is negative 7. It practices substitution and the order of operations with integers. The risk is the sign, since subtracting twice a negative number turns into addition, and dropping that sign or the factor of 2 is easy.

NYS standards: 6.EE.2c evaluating expressions and order of operations
7.NS.2a rational number multiplication
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Substituting values, including a negative one, into an expression is everyday grade 7 work with integers. This has three short steps and one sign decision, matching a typical classroom exercise.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 8, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Substitution itself is routine, so the question tests one thing: subtracting a negative inside the parentheses. A student who turns 9 − 2(−7) into 9 − 14 gets −45 and may then choose 45, since it is the only choice that looks close. Working the parentheses first and writing −(−14) as +14 before multiplying by 9 avoids every wrong choice.

Error-prone arithmetic moderate Working it out
The middle step is 9 − 2(−7): the product 2(−7) is −14, and subtracting it means adding 14, so two negatives must be handled in one short move. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work inside the parentheses first: find 2y, subtract it from x by rewriting minus a negative as plus, and multiply that result by x last.

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

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Drops the factor 2 and adds y instead of subtracting it, working out 9(9 + (−7)) = 9 × 2 = 18.
B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Writes 9 − 2(−7) as 9 − 14 = −5, gets 9 × (−5) = −45, and picks 45 as the nearest choice.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Handles the negative correctly but drops the coefficient 2, computing 9 − (−7) = 16 and then 9 × 16 = 144.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative: $a - (-b) = a + b$; a and b are numbers. Taking away a negative number is the same as adding the positive number. For example, 9 − (−14) = 9 + 14. Think of taking away a debt: you end up richer.
Multiplying by a negative: $\text{positive} \times \text{negative} = \text{negative}$; When one number is positive and the other is negative, their product is negative. The product is the answer to a multiplication. So 2 × (−7) = −14.

Write down what you know
You know x = 9 and y = −7. You need the value of $x(x - 2y)$ .
A number written right next to parentheses means multiply. So this is x times the amount inside the parentheses.
Put in the numbers
Replace each letter with its value. Put −7 in parentheses so you do not lose its minus sign.
$x(x - 2y) = 9 \times (9 - 2 \times (-7))$
Work inside the parentheses
Always work inside parentheses first. Multiply before you subtract. A positive times a negative is negative, so $2 \times (-7) = -14$ .
$9 - (-14) = 9 + 14 = 23$
Taking away a negative is the same as adding.
Watch out: If you write 9 − 14 = −5, you end with −45. Then choice B, 45, looks close, but it is wrong.
Multiply by x last
Now multiply 9 by the 23 you found inside the parentheses.
$9 \times 23 = 207$
Watch out: Do not drop the 2. Using 9 − (−7) = 16 gives 144, choice C.
So the answer is D, 207.

Check your answer

Check the inside another way. $9 - 2 \times (-7)$ means start at 9 and move 14 to the right on the number line. You land on 23. Then $9 \times 23 = 180 + 27 = 207$ .

More questions like this

The original DOE question

Question 63 as printed on page 154 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 63. Open page 154 of the handbook (PDF)

Question 76 of 83

2018 Test B Math, question 11

Question

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

A−13°
B−10°
C−7°
D5°

Show the answer

Correct answer: B

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

3 × 6 = 18 degrees

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

8 − 18 = −10 degrees

Question analysis

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

NYS standards: 7.NS.3 four operations with rational numbers
7.NS.1b rational number addition
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students use positive and negative numbers for real changes such as temperature. This is a standard three-step example of that skill, with a counting slip as its only real trap.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is light; the question turns on counting the hours from 1:00 to 7:00 correctly. Counting the clock times instead of the gaps between them gives seven hours, starting the drop an hour late gives five, and both errors are built into the choices. Subtracting the times, 7 − 1, is safer than counting on fingers.

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

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

Traps

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

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts the clock times 1, 2, 3, 4, 5, 6, 7 as seven hours, so subtracts 21 degrees from 8.
C Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Starts the drop at 2:00 instead of 1:00, counting five hours and subtracting 15 degrees from 8.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts a single 3-degree drop from 8, combining the two numbers in the question without counting the hours.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Write down what you know
At 1:00 p.m. it was 8 degrees above zero. Each hour after that, it dropped 3 degrees.
You need the temperature at 7:00 p.m.
Count the hours
Subtract the clock times: 7 − 1 = 6. So 6 hours pass.
Watch out: Do not count the clock numbers 1, 2, 3, 4, 5, 6, 7. That is 7 numbers but only 6 hours. Using 7 hours gives −13°, choice A. Starting the count at 2:00 gives 5 hours and −7°, choice C.
Find the total drop
The temperature drops the same amount each hour, so multiply.
$\text{change} = \text{rate} \times \text{time} = 3 \times 6 = 18$
The temperature fell 18 degrees in all.
Subtract from the start
Start at 8 degrees and take away the 18-degree drop.
$8 - 18 = -10$
The drop is bigger than 8, so the temperature goes below zero. A temperature below zero is a negative number. It is 10 degrees below zero.
So the answer is B, −10°.

Check your answer

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

More questions like this

The original DOE question

Question 68 as printed on page 155 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 155, question 68. Open page 155 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 77 of 83

2018 Test B Math, question 18

Question

PRICES FOR NEWSPAPER ADVERTISING

Page Space	Price
$\frac{1}{4}$ page	$200
$\frac{1}{2}$ page	$350
full page	$600

The table above shows prices for newspaper advertising. A store purchased $\frac{1}{4}$ pages, $\frac{1}{2}$ pages, and full pages of page space in equal numbers for a total of $11,500. What is the total amount of page space the store purchased?

A $1\frac{3}{4}$ pages
B10 pages
C $16\frac{1}{2}$ pages
D $17\frac{1}{2}$ pages

Show the answer

Correct answer: D

The question says that an equal number ( $x$ ) of each type of page space was purchased. To find the number of each type of page space that was purchased, multiply the price per type by $x$ and set it equal to the total amount spent, then solve for $x$ :

$200x+350x+600x=11,500$

$1,150x=11,500$

$x=10$

Thus, the store purchased 10 units of each type of page space. To find the total amount of page space purchased, multiply each type of page space by 10, and add:

$\left(10\times\frac{1}{4}\text{ page}\right)+\left(10\times\frac{1}{2}\text{ page}\right)+(10\times 1\text{ page})$

$=17\frac{1}{2}\text{ pages}$

Question analysis

This word problem gives a table of newspaper advertising prices for quarter-page, half-page and full-page ads, and says a store bought equal numbers of each for a set total. Students write and solve an equation for how many ads of each size were bought. The question wants total page space, which takes a second calculation with fractions.

NYS standards: 7.EE.4a two-step equations
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students write and solve equations like px = r from word problems. This adds a table, combines three prices into one coefficient and needs a second step with fractions after solving, a longer chain than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps major
Turning words into math moderate
Reading a figure, table or graph minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 59 words.
Corrected from the DOE handbook: An arithmetic slip in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Solving for the number of ads is the obvious goal, and finding it feels like the finish, but the question asks for page space, which needs one more pass through the table. The awkward wording about buying pages in equal numbers slows students before any arithmetic starts. The wrong choices mark each place to stop: one set of ads, the number of each ad, or a slip in adding the fractional pages.

Many steps major Working it out
The work has two phases: turn the money into a count, 1,150x = 11,500 so x = 10, then turn the count into space by multiplying 10 by 1/4, 1/2 and 1 and adding. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math moderate Understanding the question
The phrase “purchased 1/4 pages, 1/2 pages, and full pages of page space in equal numbers” has to be read as the same number of ads of each size, so the three prices add to one coefficient. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
Both columns of the table are needed: the price column builds the equation, and the page-space column turns the count of ads into the final total. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Let x be the number of each ad size, add the three prices to write one equation for the total, solve it, then multiply x by the page space in one set of ads.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Adds 1/4 + 1/2 + 1, the page space in one set of the three ads, without using the $11,500 to find how many sets were bought.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Solves 1,150x = 11,500 for the number of ads of each size and stops there instead of converting the count to page space.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Takes ten quarter pages as 1 1/2 pages instead of 2 1/2, so adds 1 1/2 + 5 + 10.

What was corrected from the DOE handbook, and why

Explanation: The handbook adds 200x + 350x + 600x and writes 1,500x = 11,500. The sum is 1,150x, and the handbook’s next line, x = 10, only follows from 1,150x = 11,500. The site shows 1,150x = 11,500.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of equal sets: $\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}}$; A set here means one ad of each size. The cost of one set is what one of each costs together. Dividing the total money by it tells how many sets were bought, because every set costs the same.

Sort out the question
The store bought the same number of each ad size: quarter page, half page and full page. It spent $11,500 in all.
You need the total page space. That is not the number of ads, and not the money.
Find the cost of one set
Think of one quarter page, one half page and one full page as a set. The store bought equal numbers of each, so it bought whole sets.
$200 + 350 + 600 = 1{,}150$
One set costs $1,150.
Find how many sets
$\text{number of sets} = \frac{\text{total spent}}{\text{cost of one set}} = \frac{11{,}500}{1{,}150} = 10$
The store bought 10 sets. That means 10 ads of each size.
Watch out: 10 is the number of each ad, not the page space. Stopping here gives choice B.
Find the page space in one set
Add the sizes in one set. Write them in quarters so they have the same bottom number.
$\frac{1}{4} + \frac{1}{2} + 1 = \frac{1}{4} + \frac{2}{4} + \frac{4}{4} = \frac{7}{4}$
One set is $\frac{7}{4}$ , or $1\frac{3}{4}$ , pages.
Watch out: $1\frac{3}{4}$ pages is only one set. That is choice A.
Multiply by the number of sets
There are 10 sets, so multiply the space in one set by 10.
$10 \times \frac{7}{4} = \frac{70}{4} = 17\frac{1}{2}$
So the answer is D, $17\frac{1}{2}$ pages.

Check your answer

Add the sizes another way. 10 quarter pages make 2.5 pages, not 1.5. 10 half pages make 5 pages. 10 full pages make 10 pages. Then $2.5 + 5 + 10 = 17.5$ , the same answer.

More questions like this

The original DOE question

Question 75 as printed on page 156 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 75. Open page 156 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 78 of 83

2018 Test B Math, question 21

Question

Points $X$ , $Y$ , and $Z$ are on a straight line, and $Y$ is between $X$ and $Z$ . Length $\overline{YZ}=\frac{3}{5}\overline{XY}$ , and length $\overline{XY}=20$ centimeters. What is the length of $\overline{XZ}$ ?

A12 cm
B24 cm
C30 cm
D32 cm

Show the answer

Correct answer: D

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

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

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

Question analysis

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

NYS standards: 7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a fraction of a length and adding segments takes only two short steps with friendly numbers. Grade 7 rational-number problems are usually longer or involve negatives, so this sits below grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds YZ, three fifths of 20, and gives it as the answer without adding XY to get the whole segment.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Takes one fifth of 20 but forgets to multiply by 3, so adds 4 cm instead of 12 cm to XY.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard route gives 30; a student would have to take YZ as half of XY rather than three fifths, adding 10 cm.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Picture the line
Points X, Y and Z are on one straight line, with Y between X and Z. So the order is X, then Y, then Z.
A segment is the straight piece between two points. The bar over two letters means the length of that segment.
XY is 20 cm. You need XZ, the whole length.
Find the length of YZ
YZ is three fifths of XY. To find three fifths of 20, first find one fifth. Then take 3 of those.
$\frac{3}{5} \text{ of } 20 = 20 \div 5 \times 3 = 4 \times 3 = 12$
So YZ is 12 cm.
Watch out: 12 cm is only one piece of the line. Stopping here gives choice A.
Add the two pieces
Y is between X and Z, so the two pieces join to make the whole.
$\overline{XZ} = \overline{XY} + \overline{YZ} = 20 + 12 = 32$
XZ is 32 cm.
So the answer is D, 32 cm.

Check your answer

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

More questions like this

The original DOE question

Question 78 as printed on page 157 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 78. Open page 157 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 79 of 83

2018 Test B Math, question 22

Question

Bryana bought $1\frac{3}{4}$ yards of cloth at $8.00 per yard. If there was an 8% sales tax, what was the total cost of the cloth?

A$12.96
B$14.08
C$15.12
D$16.08

Show the answer

Correct answer: C

Calculate the cost of the cloth before tax:

$1\frac{3}{4}\times 8=\frac{7}{4}\times 8=\$14$

Now find the tax for $14 worth of cloth:

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

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

$14 + $1.12 = $15.12

Question analysis

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

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Sales tax on a purchase is a standard grade 7 percent problem. This one adds a mixed-number multiplication first, but the numbers are friendly and the steps are routine for grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is standard for a sales-tax question, so the work lies in the number handling. Converting 1 3/4 yards correctly and taking 8% of the cost, rather than adding 8 cents, are the two places students slip. The choices punish each slip but not stopping early, since neither $14 nor $1.12 appears, which helps students who check their result against the list.

Error-prone arithmetic moderate Working it out
Two number forms meet: the mixed number 1 3/4 times $8 must become $14, and 8% of $14 must become $1.12, a decimal with two places. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The steps, price before tax, then tax, then total, must be done in order, and stopping after any of them gives a number that is not among the choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Multiplies $8 by 1.5 instead of 1.75, pricing the cloth at $12, then adds the 8% tax correctly to reach $12.96.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Prices the cloth correctly at $14 but adds 8 cents, treating 8% as $0.08 instead of 8 hundredths of the price.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Needs two unrelated errors, rounding 1 3/4 yards up to 2 for $16 and then adding 8 cents instead of 8%, so few students land here.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Write down what you know
Bryana bought $1\frac{3}{4}$ yards of cloth. Each yard costs $8.00. There is an 8% sales tax. Sales tax is an extra amount added to the price.
You need the total cost, with the tax.
Find the cost before tax
First change the mixed number to a fraction. A mixed number is a whole number plus a fraction. $1\frac{3}{4}$ is the same as $\frac{7}{4}$ , because 1 whole is 4 quarters.
$\text{cost} = \text{price per unit} \times \text{number of units}$
$\text{cost} = 8 \times \frac{7}{4} = \frac{56}{4} = 14$
The cloth costs $14 before tax.
Watch out: $1\frac{3}{4}$ is 1.75, not 1.5. Using 1.5 gives $12.96, choice A.
Find the tax
$\text{tax} = \frac{\text{tax rate}}{100} \times \text{price}$
$\text{tax} = \frac{8}{100} \times 14 = 0.08 \times 14 = 1.12$
The tax is $1.12.
Watch out: 8% of $14 is not 8 cents. Adding only $0.08 gives $14.08, choice B.
Add the tax to the price
The total is the price before tax plus the tax.
$14 + 1.12 = 15.12$
The total is $15.12.
So the answer is C, $15.12.

Check your answer

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

More questions like this

The original DOE question

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

Question 80 of 83

2018 Test B Math, question 23

Question

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

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

A $-7\frac{1}{6}$
B $-4\frac{1}{2}$
C $4\frac{1}{2}$
D $7\frac{1}{6}$

Show the answer

Correct answer: B

To find $M$ , subtract $N-M$ and set it equal to the length:

$1\frac{1}{3}-M=5\frac{5}{6}$

$-M=5\frac{5}{6}-1\frac{1}{3}$

$-M=5\frac{5}{6}-1\frac{2}{6}$

$-M=4\frac{3}{6}$

$M=-4\frac{1}{2}$

Question analysis

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

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students find distances on a number line as the absolute value of a difference. Doing it with mixed numbers that have unlike denominators, and crossing zero, is a typical and slightly demanding grade 7 task.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Error-prone arithmetic moderate
A very close wrong answer moderate
Solution: 3 steps
Reading level: Question and choices: about grade 8, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The distance is given and one endpoint is known, so the question is really about direction and sign. Because the drawing shows no zero, students must reason that moving 5 5/6 to the left from 1 1/3 carries M past zero. The choices pair each size with each sign, so getting the size right is only half the work; the sign decides between two choices.

Reading a figure, table or graph moderate Understanding the question
The number line labels only N, at 1 1/3, and shows no zero, so students must see from the picture that M lies left of N and work out whether it passes zero. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic moderate Working it out
The subtraction 1 1/3 − 5 5/6 mixes unlike denominators and crosses zero: in sixths it is 8/6 − 35/6 = −27/6, which then reduces to a mixed number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
4 1/2 and −4 1/2 are written the same except for the sign; only noticing that M lies left of N, so below 1 1/3, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Adds the length to N's coordinate, 1 1/3 + 5 5/6 = 7 1/6, and makes it negative because M is on the left.
C Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Computes 5 5/6 − 1 1/3, the right size, but keeps it positive, which would put M to the right of N.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two numbers in the question, 1 1/3 + 5 5/6, placing M to the right of N against the figure.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line: $\text{distance} = \text{right point} - \text{left point}$; On a number line, numbers grow as you move right. The distance between two points is the larger number minus the smaller one. That is the right point minus the left point, so a distance is never negative.
Mixed number to fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; 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.

Read the number line
Point N is at $1\frac{1}{3}$ . Point M is to the left of N. MN means the distance from M to N, and it is $5\frac{5}{6}$ .
You need the number at point M.
Decide which way to move
On a number line, numbers get smaller as you move left. M is left of N, so N is the right point.
$\text{distance} = \text{right point} - \text{left point}$
$5\frac{5}{6} = 1\frac{1}{3} - M$
So M is what you get when you take the distance away from N.
$M = 1\frac{1}{3} - 5\frac{5}{6}$
The distance is bigger than $1\frac{1}{3}$ , so you will pass 0. M will be negative.
Change both numbers to sixths
To subtract, the fractions need the same bottom number. Use sixths, since 3 goes into 6. Change each mixed number to an improper fraction, a fraction whose top is bigger than its bottom.
$1\frac{1}{3} = \frac{1 \times 3 + 1}{3} = \frac{4}{3} = \frac{8}{6}$
$5\frac{5}{6} = \frac{5 \times 6 + 5}{6} = \frac{35}{6}$
Subtract and simplify
$M = \frac{8}{6} - \frac{35}{6} = -\frac{27}{6}$
27 sixths is 4 wholes (24 sixths) and 3 sixths more. And $\frac{3}{6} = \frac{1}{2}$ . So M is $-4\frac{1}{2}$ .
Watch out: $4\frac{1}{2}$ without the minus sign is choice C. That point would be to the right of N, not the left.
So the answer is B, $-4\frac{1}{2}$ .

Check your answer

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

More questions like this

The original DOE question

Question 80 as printed on page 157 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 157, question 80. Open page 157 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 81 of 83

2018 Test B Math, question 26

Question

A cylindrical oil drum can hold 4,320 liters when it is completely full. Currently, the drum is $\frac{1}{3}$ full of oil. How many kiloliters of oil need to be added in order to fill the drum completely?

A1.44
B2.88
C4.32
D14.10

Show the answer

Correct answer: B

Find the number of liters that need to be added. Since $\frac{1}{3}$ of the oil drum is full, $\frac{2}{3}$ of the drum remains empty:

$\frac{2}{3}\times 4{,}320=2{,}880\text{ liters}$

Use the conversion 1 kL = 1,000 L to find the number of kL:

$\frac{2{,}880}{1{,}000}=2.88\text{ kL}$

Question analysis

This fractions and unit conversion problem describes a cylindrical oil drum that holds 4,320 liters when full and is now one third full, and asks how many kiloliters must be added to fill it. Students find the empty fraction, multiply and convert liters to kiloliters. Answering for the right part of the drum, in the right unit, is the challenge.

NYS standards: 7.NS.3 four operations with rational numbers
6.RP.3d converting measurement units
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Taking a fraction of a quantity and converting metric units are both familiar from grades 5 and 6. Together they make a three-step problem at about the level of routine grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is short, but the question describes the drum one way and asks about it another. Students who multiply by the fraction they are given find the oil already in the drum, and that result, converted to kiloliters, is a choice. Since no choice is in liters, students who forget to convert are warned before they answer.

Turning words into math moderate Understanding the question
The stem gives how full the drum is, 1/3, but asks how much must be added, so students must switch to the empty 2/3 before multiplying by 4,320. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
The bold word “kiloliters” calls for the conversion 1 kiloliter = 1,000 liters, which the question does not give. (Anderson & Krathwohl, 2001)

Strategy: Work out what fraction of the drum is still empty, take that fraction of 4,320 liters, and divide by 1,000 to change liters to kiloliters.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Multiplies 4,320 by the given 1/3, finding the oil already in the drum instead of the amount still needed.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Converts the full capacity of 4,320 liters to kiloliters without taking away the third of the drum that is already full.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No conversion of 1,440, 2,880 or 4,320 liters gives 14.10; it is over three times the full drum's capacity, so only guessers choose it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of an amount: $\frac{a}{b} \text{ of } n = \frac{a}{b} \times n$; n is the whole amount, here the full drum. The fraction a over b tells what share of it you want. Multiply to find that share. For example, 2/3 of 9 is 6.
Liters to kiloliters: $\text{kiloliters} = \text{liters} \div 1{,}000$; Kilo means 1,000, so 1 kiloliter holds 1,000 liters. To change liters into kiloliters, divide by 1,000. That moves the decimal point 3 places to the left.

Sort out the question
The full drum holds 4,320 liters. It is now $\frac{1}{3}$ full. You need how much oil to add to fill it.
The answer must be in kiloliters, not liters.
Find the empty fraction
The drum is $\frac{1}{3}$ full. The rest is empty, and that is the part you must fill.
$1 - \frac{1}{3} = \frac{2}{3}$
Watch out: $\frac{1}{3}$ of 4,320 is 1,440 liters. That is the oil already in the drum. It leads to choice A.
Find the liters to add
Take two thirds of the full drum.
$\frac{2}{3} \text{ of } 4{,}320 = \frac{2}{3} \times 4{,}320 = 2{,}880$
One third of 4,320 is 1,440, and two thirds is double that. You need to add 2,880 liters.
Change liters to kiloliters
$\text{kiloliters} = 2{,}880 \div 1{,}000 = 2.88$
You need 2.88 kiloliters.
Watch out: 4.32 is the whole drum in kiloliters. Choice C forgets that part of the drum is already full.
So the answer is B, 2.88.

Check your answer

Add the oil already in the drum to the oil you add: $1.44 + 2.88 = 4.32$ kiloliters. That is 4,320 liters, the full drum. It checks out.

More questions like this

The original DOE question

Question 83 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 83. Open page 158 of the handbook (PDF)

Question 82 of 83

2018 Test B Math, question 41

Question

Points $P$ , $Q$ , $R$ , and $S$ represent −3, −1, 0, and 2, respectively, on a number line. How many units is the midpoint of $\overline{PQ}$ from the midpoint of $\overline{RS}$ ?

Show the answer

Correct answer: C

$Number line from −3 to 3 with a tick at each integer. P is at −3, Q is at −1, R is at 0 and S is at 2.$

Find the midpoint of $PQ$ and $RS$ :

Half the length of $PQ$ is $\frac{-1-(-3)}{2}=\frac{2}{2}=1$ unit.

The midpoint of $PQ$ is located 1 unit from each endpoint, so the midpoint is at −2.

Half the length of $RS$ is $\frac{2-0}{2}=\frac{2}{2}=1$ unit.

The midpoint of RS is located 1 unit from each endpoint, so the midpoint is at 1. The distance between the two midpoints is 1 −(−2) = 3 units.

Question analysis

Four points on a number line are given by their coordinates, some of them negative, and students find the distance between the midpoints of two segments. It tests locating midpoints and measuring distance with negative numbers. The care needed is in placing each midpoint correctly and in measuring a distance that crosses zero.

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Distance on a number line with negative numbers is grade 7 content. Finding two midpoints first adds a step, but the numbers are small integers, so the demand matches typical grade 7 practice.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 12, 26 words.
Corrected from the DOE handbook: Two mislabeled steps in the explanation were corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is simple, but the negative coordinates make the last subtraction easy to get wrong, and there is no picture to check against. Students who sketch the line and mark −2 and 1 can count the three units directly. The wrong choices come from confusing a segment's length with the gap between midpoints, or from counting tick marks instead of spaces.

Error-prone arithmetic moderate Working it out
The midpoint of PQ is at −2, so the final distance is 1 − (−2); dropping the negative sign at that step turns 3 into 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Much to keep track of minor Working it out
Four labeled points and two midpoints have to be held at once, and the question draws no number line, so sketching one is the safe move. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Average the coordinates of each segment's endpoints to place its midpoint, then find the distance between the two midpoints as the absolute value of their difference.

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Computes the distance between the midpoints −2 and 1 as 2 − 1 = 1, dropping the negative sign instead of finding 1 − (−2).
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Gives the length of PQ or of RS, 2 units each, rather than the distance between the two segments' midpoints.
D Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Counts the tick marks from −2 to 1, which are four, instead of the three unit spaces between them.

What was corrected from the DOE handbook, and why

Explanation: The handbook labels (−1 − (−3))/2 = 1 unit as the “Midpoint of PQ” and (2 − 0)/2 = 1 unit as the “Midpoint of RS.” Those are half the lengths of the segments, not their midpoints: as the next sentences say, the midpoint of PQ is at −2 and the midpoint of RS is at 1. The site labels them “Half the length of PQ” and “Half the length of RS.” The distance between the midpoints is still 3 units.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint on a number line: $\text{midpoint} = \frac{a + b}{2}$; a and b are the numbers at the two ends of a segment. The midpoint is the point exactly halfway between them. Adding the ends and dividing by 2 finds their average, which is the middle.
Distance on a number line: $\text{distance} = \text{right number} - \text{left number}$; The right number is the point farther to the right on the number line. The left number is the other point. Subtracting tells how many units apart they are. Taking the left from the right keeps the answer positive.

Sort out what you know
P is at −3, Q is at −1, R is at 0 and S is at 2. A segment is the piece of a line between two points. So $\overline{PQ}$ runs from −3 to −1. You need the distance between the midpoint of PQ and the midpoint of RS. A quick sketch of a number line helps.
Find the midpoint of PQ
Use the midpoint formula with −3 and −1.
$\frac{-3 + (-1)}{2} = \frac{-4}{2} = -2$
The midpoint of PQ is at −2.
Find the midpoint of RS
Use the midpoint formula with 0 and 2.
$\frac{0 + 2}{2} = \frac{2}{2} = 1$
The midpoint of RS is at 1.
Find the distance between them
The midpoint at 1 is to the right of the one at −2. Use the distance formula.
$\text{distance} = 1 - (-2) = 1 + 2 = 3$
Subtracting a negative number is the same as adding.
Watch out: dropping the minus sign gives $2 - 1 = 1$ , choice A. Counting the tick marks −2, −1, 0 and 1 gives 4, choice D. Count the spaces between marks instead.
Give the answer
The midpoints are at −2 and 1, which are 3 units apart. So the answer is C, 3.

Check your answer

Count the jumps on a number line. Going from −2 to −1 is one jump. Going from −1 to 0 is two, and going from 0 to 1 is three. So the midpoints are 3 units apart.

More questions like this

The original DOE question

Question 98 as printed on page 160 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 160, question 98. Open page 160 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 83 of 83

2018 Test B Math, question 44

Question

$A number line with eight evenly spaced hash marks. Point M is on the second mark, labeled −1/4, and point T is on the seventh mark, labeled 5/8. Point R, with no label, is on the fifth mark, three marks to the right of M.$

The hash marks on the number line above are evenly spaced. What is the coordinate of point $R$ ?

A $\frac{7}{40}$
B $\frac{9}{40}$
C $\frac{11}{40}$
D $\frac{21}{40}$

Show the answer

Correct answer: C

There are 5 sections between M and T. To find the length of one of these sections, find the distance between M and T and divide by 5:

$\left(\frac{5}{8}-\left(-\frac{1}{4}\right)\right)\div\frac{5}{1}=$

$\left(\frac{5}{8}+\frac{2}{8}\right)\div\frac{5}{1}=$

$\frac{7}{8}\div\frac{5}{1}=\frac{7}{8}\times\frac{1}{5}=\frac{7}{40}$

R is 3 sections away from M, so add:

$-\frac{1}{4}+3(\frac{7}{40})=-\frac{10}{40}+\frac{21}{40}=\frac{11}{40}$

R is located at $\frac{11}{40}$ .

Question analysis

Students find the fraction at point R on a number line with evenly spaced hash marks, where M is at −1/4 and T is at 5/8. It tests fraction arithmetic with negative numbers. Working out the size of one gap and counting gaps from M to R takes several careful steps.

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students place fractions and negatives on number lines, but here they must subtract a negative fraction, divide by five, and then add back. Four fraction steps with unlike denominators make it harder than typical classwork.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic major
Reading a figure, table or graph moderate
Solution: 4 steps
Reading level: Question and choices: about grade 6, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Fractions, a negative starting point and a count of spaces all have to come out right together. Students lose points by counting tick marks instead of spaces, by subtracting −1/4 as if it were positive, or by measuring R from 0 instead of from M. Converting both labels to fortieths first, −10/40 and 25/40, makes each gap a whole number of fortieths.

Error-prone arithmetic major Working it out
The gap length needs 5/8 − (−1/4) = 7/8, then 7/8 ÷ 5 = 7/40, then −10/40 + 21/40: negatives and three different denominators meet in one problem. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph moderate Understanding the question
The hash marks have to be counted as spaces: five from M to T, with R three spaces past M, while the unlabeled marks outside M and T play no part. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the distance from M to T by subtracting their coordinates, divide it by the number of equal gaps, then step from M toward T by the right number of gaps.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the length of one gap, 7/8 ÷ 5 = 7/40, and gives it as R's coordinate without stepping from M.
B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Subtracts as 5/8 − 1/4 = 3/8 instead of adding, gets gaps of 3/40, and counts three gaps from 0.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Multiplies the 7/40 gap by 3 to get the distance from M to R, then forgets to add it to M's −1/4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line: $\text{distance} = \text{right number} - \text{left number}$; The right number is the point farther to the right on the number line. The left number is the other point. Subtracting tells how many units apart they are. Taking the left from the right keeps the answer positive.
Dividing a fraction by a whole number: $\frac{a}{b} \div c = \frac{a}{b \times c}$; a over b is the fraction, and c is the whole number you divide by. Splitting a fraction into c equal parts makes each part c times smaller. So you multiply the bottom number, b, by c.

Read the number line
M is at $-\frac{1}{4}$ and T is at $\frac{5}{8}$ . The hash marks are evenly spaced, so every space is the same length. Count the spaces. From M to T there are 5 spaces. From M to R there are 3 spaces. You need the number at point R.
Find the distance from M to T
Use the distance formula. First write $-\frac{1}{4}$ in eighths, as $-\frac{2}{8}$ .
$\frac{5}{8} - \left(-\frac{2}{8}\right) = \frac{5}{8} + \frac{2}{8} = \frac{7}{8}$
Taking away a negative is the same as adding. M and T are $\frac{7}{8}$ apart.
Find the length of one space
The 5 spaces are equal, so divide by 5.
$\frac{7}{8} \div 5 = \frac{7}{8 \times 5} = \frac{7}{40}$
Each space is $\frac{7}{40}$ long.
Watch out: this is the length of one space, not where R is. Stopping here gives choice A.
Step from M to R
R is 3 spaces to the right of M.
$3 \times \frac{7}{40} = \frac{21}{40}$
Now add that to M. Write $-\frac{1}{4}$ as $-\frac{10}{40}$ .
$-\frac{10}{40} + \frac{21}{40} = \frac{11}{40}$
Watch out: $\frac{21}{40}$ is how far R is from M, not where R is. Choosing it gives choice D.
Give the answer
Point R is at $\frac{11}{40}$ , a little to the right of 0. So the answer is C, $\frac{11}{40}$ .

Check your answer

Go backward from T instead. T is 2 spaces past R, and $\frac{5}{8} = \frac{25}{40}$ . So R is at $\frac{25}{40} - \frac{14}{40} = \frac{11}{40}$ . Both ways give the same point.

More questions like this

The original DOE question

Question 101 as printed on page 161 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 161, question 101. Open page 161 of the handbook (PDF)