Practice Test

6.NS the number system

Question 1 of 112

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 112

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 3 of 112

2026 Practice Test B Math, question 4

Question

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

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

$1\text{ mink} = \frac{1}{4}\text{ bart}$

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

How many minks will Maria receive from the exchange?

Enter your answer in the space.

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

Show the answer

Correct answer: 116

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

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

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

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

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

Question analysis

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

NYS standards: 6.RP.3d converting measurement units
6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Converting units with a given rate is grade 6 ratio reasoning. Invented units remove any real-world sense check, and the two conversions run in opposite directions, one through a fraction, which matches the care grade 7 problems need.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math moderate Understanding the question
Both rates put the mink on the left, so “1 mink = 1/4 bart” has to be turned around into 1 bart = 4 minks, and the invented units give no sense of which is worth more. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Two conversions run in opposite directions, dividing the zogs by 3 and multiplying the barts by 4, and their results are added, so a slip in either step changes the total. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The total is typed into a box with no choices to compare, so a conversion done the wrong way still produces a believable whole number that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

26 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Gets 20 minks from the zogs but divides the barts by 4, 24 ÷ 4 = 6, as if a bart were worth a quarter of a mink, for 20 + 6 = 26.
276 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies the 60 zogs by 3, as if each zog were worth 3 minks, then adds the 96 minks from the barts correctly: 180 + 96 = 276.
84 Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the 60 zogs and 24 barts as they stand, 60 + 24 = 84, without converting either kind of point into minks first.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Maria trades two kinds of points for minks. You know:
- 3 zogs trade for 1 mink.
- 1 mink is worth $\frac{1}{4}$ of a bart.
- Maria has 60 zogs and 24 barts.
You need the total number of minks. Zogs and barts are different units, like dimes and dollars. So you cannot just add them.
Watch out: 60 + 24 = 84 adds zogs to barts. That is not a number of minks.
Find what one zog is worth
It takes 3 zogs to make 1 mink. So one zog is only part of a mink. One zog is worth $\frac{1}{3}$ of a mink.
A zog is smaller than a mink. So Maria will get fewer minks than she has zogs.
Turn the bart rule around
The rule says 1 mink is $\frac{1}{4}$ of a bart. So a mink is a quarter of a bart. It takes 4 quarters to make a whole:
$4 \times \frac{1}{4} = 1$
So 4 minks make 1 bart. One bart is worth 4 minks. A bart is bigger than a mink.
Watch out: the fraction tempts you to divide by 4. Then 24 barts become only 6 minks, and the total is 26.
Change the zogs into minks
Use the rate. Each of the 60 zogs is worth 1/3 of a mink:
$60 \times \frac{1}{3} = 60 \div 3 = 20$
Multiplying by 1/3 is the same as dividing by 3. So 60 zogs trade for 20 minks.
Watch out: multiplying 60 by 3 gives 180. That treats a zog as bigger than a mink, and the total becomes 276.
Change the barts into minks
Each of the 24 barts is worth 4 minks:
$24 \times 4 = 96$
So 24 barts trade for 96 minks.
Add the minks
Both amounts are now in minks, so you can add them:
$20 + 96 = 116$
Maria receives 116 minks. Type only the number in the box, with no units. So you would enter 116.

Check your answer

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

More questions like this

The original DOE question

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

Question 4 of 112

2026 Practice Test C Math, question 1

Question

What is the prime factorization of 756?

A $2^{2} \cdot 3^{2} \cdot 21$
B $2^{2} \cdot 3 \cdot 7 \cdot 9$
C $2^{2} \cdot 7 \cdot 27$
D $2^{2} \cdot 3^{3} \cdot 7$

Show the answer

Correct answer: D

Determine the prime factors of 756 using a factor tree:

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

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

Question analysis

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

NYS standards: 6.NS.4 common factors and multiples
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Factor trees and exponent notation are grade 6 skills, and 756 splits quickly by small divisors. What adds a little is that every choice multiplies out to 756, so checking the product alone settles nothing.
Estimated difficulty: Straightforward: two steps or one careful reading.
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
Solution: 3 steps
Reading level: Question and choices: about grade 11, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

A very close wrong answer moderate Choosing the answer
2² · 3² · 21 shares the right answer's 2² and differs only in 3² · 21 against 3³ · 7, and both equal 756, so neither an estimate nor a product check separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that a prime factorization may hold only primes, and must spot that 21, 9 and 27 are composite even though none of them is even. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Divides out 2 twice and 3 twice, then stops at 21, which is 3 × 7, so one composite factor remains in a product that still equals 756.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Splits off a 3 and a 7 but keeps 9 as if it were prime, perhaps because it is odd, though 9 = 3 × 3.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Splits 756 into 4 × 189 and 189 into 7 × 27, then stops, not seeing that 27 is 3 × 3 × 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
You need the prime factorization of 756. That means writing 756 as a product of prime numbers only. A product is the answer to a multiplication. A prime number has exactly two factors, 1 and itself. Examples are 2, 3, 5 and 7.
All four choices multiply to 756. So multiplying a choice back does not pick the answer. You must check that every factor is prime.
Divide by 2 while you can
A factor tree splits a number into two factors, again and again. Start with the smallest prime, 2. 756 is even, so 2 goes into it:
$756 = 2 \times 378$
378 is even too:
$378 = 2 \times 189$
189 is odd, so 2 does not go into it. You have pulled out two 2s.
Divide by 3 while you can
Add the digits of 189: $1 + 8 + 9 = 18$ . A number can be divided by 3 when its digits add up to a number that 3 goes into. 3 goes into 18, so it goes into 189:
$189 = 3 \times 63$
$63 = 3 \times 21$
$21 = 3 \times 7$
7 is prime, so the tree stops. You have pulled out three 3s and one 7.
Write the factors with exponents
Collect every prime you pulled out:
$756 = 2 \times 2 \times 3 \times 3 \times 3 \times 7$
Use the exponent rule. Two 2s make $2^{2}$ , and three 3s make $3^{3}$ . The choices use a raised dot, ·, for times. So 756 is $2^{2} \cdot 3^{3} \cdot 7$ .
Choose the answer
Watch out: each wrong choice is a factor tree that stopped too soon. A ends in 21, which is 3 × 7. B has 9, which is 3 × 3. C has 27, which is 3 × 3 × 3. Being odd does not make a number prime.
Only D uses primes alone. So the answer is D, $2^{2} \cdot 3^{3} \cdot 7$ .

Check your answer

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

More questions like this

The original DOE question

Question 1 as shown on the DOE’s digital test, page 74 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 74, question 1.

Question 5 of 112

2026 Practice Test C Math, question 7

Question

$-12 > 6x$

For what values of x is the above inequality true?

A $x > -6$
B $x < -6$
C $x < -2$
D $x > -2$

Show the answer

Correct answer: C

Divide to solve for x:

$\frac{(-12)}{6} > \frac{6x}{6}$

$-2 > x$

Rewriting $-2 > x$ with the x term first requires reversing the inequality symbol in order to keep the meaning “x is less than $(-2)$ ”.

$x < -2$

Question analysis

This inequality question asks which values of x make −12 > 6x true. Students solve a one-step inequality by dividing and then state the solution with x written first. A negative number on one side and the variable on the other make the direction of the symbol the main thing to get right.

NYS standards: 7.EE.4b two-step inequalities
6.NS.7a inequality as relative position
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 inequalities usually take two steps, such as 3x + 4 > 19. This one needs a single division, and the care goes into reading the result correctly when the variable ends up on the right of the symbol.
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
Solution: 2 steps
Reading level: Question and choices: about grade 10, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One division finds the boundary, so the question is really about direction. Two things pull students the wrong way: the −12 on the left makes the inequality look like one that must be flipped, and the result −2 > x has x on the right, where it is easy to read backward. Testing x = −3, which gives −12 > −18, shows that values below −2 work.

A very close wrong answer moderate Choosing the answer
x < −2 and x > −2 share the boundary −2 and differ only in direction, so finding the right number settles nothing until the direction is read correctly. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must know that the symbol flips only when both sides are multiplied or divided by a negative number, and that −2 > x says the same as x < −2. (Anderson & Krathwohl, 2001)

Strategy: Divide both sides by 6, asking whether you divided by a negative number, then rewrite the result with x on the left so that the symbol still opens toward the larger side.

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 surface strategy Expected pull: weak
Adds 6 to −12 instead of dividing by 6, getting −6, and also reverses the direction, so both the boundary and the symbol are wrong.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Undoes the 6 by adding it to −12 rather than dividing, as if 6x meant x − 6, getting −6 > x and so x < −6.
D Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Finds −2 > x but flips the symbol because −12 is negative, or reads −2 > x as x > −2, though dividing by a positive 6 keeps the direction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing both sides of an inequality: $\text{if } a > b \text{ and } c > 0 \text{, then } \frac{a}{c} > \frac{b}{c}$; a and b are the two sides of an inequality, and c is the number you divide by. If c is positive, the symbol stays the same. It flips only when you multiply or divide both sides by a negative number. Here c is 6, which is positive.
Reading an inequality the other way: $a > b \text{ means the same as } b < a$; a and b are any two numbers. Saying a is greater than b is the same as saying b is less than a. When you swap the two sides, turn the symbol around. Its point still aims at the smaller number.

Sort out the question
The inequality is $-12 > 6x$ . The symbol > means is greater than. So it says −12 is greater than 6 times x. You need every value of x that makes this true.
Divide both sides by 6
To get x alone, divide both sides by 6:
$\frac{-12}{6} > \frac{6x}{6}$
$-2 > x$
You divided by 6, which is positive. So the symbol does not flip.
Watch out: −12 is negative, but that does not matter. Only dividing by a negative number flips the symbol.
Put x first
$-2 > x$ says −2 is greater than x. That means x is less than −2. Swap the sides and turn the symbol around:
$x < -2$
Watch out: keeping > when you swap the sides gives choice D, $x > -2$ . That is the opposite meaning.
Choose the answer
Choices A and B use −6. That comes from adding 6 to −12 instead of dividing by 6. The right boundary is −2, and x is below it.
So the answer is C, $x < -2$ .

Check your answer

Test a number below −2, such as −3. Then $6 \times (-3) = -18$ , and $-12 > -18$ is true. Now test 0, which is above −2. Then $6 \times 0 = 0$ , and $-12 > 0$ is false. So only numbers below −2 work.

More questions like this

The original DOE question

Question 7 as shown on the DOE’s digital test, page 80 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 80, question 7.

Question 6 of 112

2026 Practice Test C Math, question 13

Question

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

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

What is the result?

A13.5
B135
C135.2
D135.3

Show the answer

Correct answer: D

Begin by multiplying 135.28 by 10:

$135.28 \times 10 = 1{,}352.8$

Add 0.5:

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

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

Divide by 10:

$1{,}353 \div 10 = 135.3$

Question analysis

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

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Multiplying and dividing a decimal by 10 and adding 0.5 are grade 5 and 6 skills. The four steps are spelled out, so the demand is following them exactly and in order, not choosing a method.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Many steps minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 3, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

A very close wrong answer moderate Choosing the answer
135.3 and 135.2 differ only in the tenths digit, and 135 shares their whole-number part, so a rough sense of the result's size rules out nothing. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Many steps minor Working it out
Four listed steps move the decimal point twice, and doing them in another order, or skipping one, still produces a believable number near 135. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Adding 0.5 to 1,352.8 carries into the ones place, since 8 tenths and 5 tenths make 1.3, which turns 1,352 into 1,353 before the decimals are dropped. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Skips step 1, so adds 0.5 to 135.28, drops the decimals to get 135, and then divides by 10 to reach 13.5.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Divides by 10 before dropping the digits after the decimal point, so 1,353.3 becomes 135.33 and then 135, doing steps 3 and 4 in reverse order.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Leaves out step 2, or adds 0.5 without carrying, so 1,352 is kept and 135.2 results: 135.28 cut down instead of rounded up.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
You start with 135.28 and do four steps, in order. Write down the number after each step. Doing them in your head, or in another order, leads to a wrong choice. You need the number at the end.
Multiply by 10
Step 1. Multiplying by 10 moves the decimal point one place to the right:
$135.28 \times 10 = 1{,}352.8$
The digits stay the same. Only the decimal point moves.
Add 0.5
Step 2. Line up the decimal points. 8 tenths plus 5 tenths is 13 tenths. That is 1 whole and 3 tenths, so carry the 1 into the ones place:
$1{,}352.8 + 0.5 = 1{,}353.3$
Watch out: skipping this step, or forgetting to carry, keeps 1,352. That leads to choice C, 135.2.
Drop the digits after the point
Step 3. Cut off the digits after the decimal point. Do not round. So 1,353.3 becomes 1,353.
Watch out: do step 3 before step 4. Dividing first gives 135.33. Dropping its decimals then gives 135, choice B.
Divide by 10
Step 4. Dividing by 10 moves the decimal point one place to the left:
$1{,}353 \div 10 = 135.3$
Skipping step 1 gives 13.5, choice A. So the answer is D, 135.3.

Check your answer

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

More questions like this

The original DOE question

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

Question 7 of 112

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 8 of 112

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 9 of 112

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 10 of 112

2026 Practice Test C Math, question 25

Question

For each $\frac{3}{8}$ foot of curtain width, $\frac{5}{6}$ yard of fabric is needed to make ruffles. How many yards of fabric are needed per foot of curtain width?

A $\frac{5}{16}$ yd
B $\frac{29}{24}$ yd
C $\frac{20}{9}$ yd
D $\frac{7}{2}$ yd

Show the answer

Correct answer: C

It takes $\frac{5}{6}$ yard of fabric to make $\frac{3}{8}$ foot of curtain width.

To find the ratio of 1 foot of curtain width to $\frac{3}{8}$ foot of curtain width $\left(1:\frac{3}{8}\right)$ , divide.

$1 \div \frac{3}{8} = 1 \times \frac{8}{3} = \frac{8}{3}$

So the amount of fabric needed for 1 foot of curtain width is $\frac{8}{3}$ times the amount needed for $\frac{3}{8}$ foot of curtain width.

$\frac{8}{3} \times \frac{5}{6} = \frac{40}{18}$

$\frac{40 \div 2}{18 \div 2} = \frac{20}{9}$

The amount of fabric needed for 1 foot of curtain width is $\frac{20}{9}$ yards.

Question analysis

Making ruffles takes 5/6 yard of fabric for every 3/8 foot of curtain width, and students find how many yards of fabric one foot of curtain width needs. It tests finding a unit rate when both quantities are fractions. The two units differ, and choosing whether to multiply, divide or add the fractions is the crucial decision.

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of two fractions is the grade 7 rate standard itself. Two units, feet and yards, and an unfamiliar sewing setting make it harder to see which fraction divides which.
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
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two fractions and two units, feet and yards, leave students to decide whether to multiply, divide or add, and each wrong choice records one of those decisions. A sense check rules out the most tempting one before any arithmetic: a whole foot is more than 3/8 foot, so it needs more than 5/6 yard, yet multiplying gives only 5/16. Dividing the yards by the feet then picks out the answer among the rest.

Turning words into math moderate Understanding the question
“For each 3/8 foot of curtain width, 5/6 yard of fabric” gives the rate as a pair of fractions, and “per foot” means dividing the yards by the feet. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Dividing by 3/8 means multiplying by 8/3, and the product 40/18 has to be reduced to 20/9, an improper fraction rather than a whole number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Divide the yards of fabric by the feet of curtain width, since a rate per foot puts feet in the denominator, then simplify the 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).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 5/6 by 3/8 instead of dividing, getting 15/48 = 5/16, so a whole foot seems to need less fabric than 3/8 foot.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: moderate
Adds 5/6 and 3/8 over a common denominator, 20/24 + 9/24 = 29/24, combining yards with feet instead of forming a rate.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: weak
Finds that a foot is 8/3 times 3/8 foot, then adds 8/3 and 5/6 instead of multiplying them: 16/6 + 5/6 = 21/6 = 7/2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{yards per foot} = \text{yards} \div \text{feet}$; A unit rate tells how much for exactly 1 of something. Here it is yards of fabric for 1 foot of curtain. Per foot means for each foot, so feet go on the bottom. Divide the yards by the feet.
Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is any number, and b/c is a fraction. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down. It works because b/c fits into 1 exactly c/b times.

Sort out the question
For every 3/8 foot of curtain width, you need 5/6 yard of fabric. You need the yards of fabric for 1 whole foot. That is a unit rate, an amount for exactly 1 foot.
3/8 foot is less than a whole foot. So a whole foot needs more than 5/6 yard.
Set up the division
Per foot means divide the yards by the feet:
$\frac{5}{6} \div \frac{3}{8}$
Watch out: multiplying 5/6 by 3/8 gives 5/16, choice A. That is less fabric for more curtain, which makes no sense. Adding them gives 29/24, choice B, but you can’t add yards and feet.
Multiply by the reciprocal
Flip 3/8 to get 8/3, and multiply:
$\frac{5}{6} \times \frac{8}{3} = \frac{5 \times 8}{6 \times 3} = \frac{40}{18}$
The 8/3 says a whole foot is 8/3 times as long as 3/8 foot. So it needs 8/3 times as much fabric.
Watch out: adding 8/3 and 5/6 gives 7/2, choice D. Multiply them instead.
Simplify the fraction
2 goes into both 40 and 18:
$\frac{40 \div 2}{18 \div 2} = \frac{20}{9}$
So 1 foot of curtain width needs 20/9 yards of fabric, a little more than 2 yards. So the answer is C, 20/9 yd.

Check your answer

Multiply back. 3/8 foot should need 3/8 of the fabric for a whole foot: $\frac{20}{9} \times \frac{3}{8} = \frac{60}{72} = \frac{5}{6}$ yard. That is the amount the question gives. Also, 20/9 is more than 5/6, as it should be.

More questions like this

The original DOE question

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

Question 11 of 112

2026 Practice Test C Math, question 29

Question

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

SMOOTHIE RECIPE

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 0.25

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

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

$1.5 = 6p$

$\frac{1.5}{6} = \frac{6p}{6}$

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

The constant of proportionality is 0.25.

Question analysis

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

NYS standards: 7.RP.2b constant of proportionality
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the constant of proportionality in a table is a core grade 7 skill, and these values are friendly decimals. Typing the answer with no choices, and taking the ratio in the order the question gives, are the added demands.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
An answer in an unexpected form minor
Solution: 1 step
Reading level: Question and choices: about grade 6, 75 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math moderate Understanding the question
The opening sentence speaks of a ratio of fruit to yogurt, but the question asks for the constant of yogurt to fruit, so the division is y ÷ x, against the table’s column order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The constant is typed into a box with no choices, so 4, the ratio taken the other way round, is a believable entry that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

4 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides cups of fruit by cups of yogurt, 6 ÷ 1.5 = 4, following the stem’s “ratio of fruit to yogurt” and turning the requested constant upside down.
0.5 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Takes the steady rise of 0.5 cup of yogurt from row to row as the constant, ignoring that the fruit rises by 2 cups each time.
1.5 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives 1.5, the first yogurt value in the table, as if the constant were an amount in one batch rather than a ratio of two amounts.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The table lists cups of fruit (x) and cups of yogurt (y) for four batches. You need the constant of proportionality of yogurt to fruit.
That is the number of cups of yogurt for each 1 cup of fruit. It is also the number you multiply x by to get y.
Divide yogurt by fruit
Use the first row: 6 cups of fruit and 1.5 cups of yogurt. Yogurt goes on top, because the question says yogurt to fruit:
$k = \frac{y}{x} = \frac{1.5}{6} = 0.25$
So there is 0.25 cup, or one quarter of a cup, of yogurt for each cup of fruit.
Watch out: the first sentence says fruit to yogurt, but the question asks the other way. Dividing $6 \div 1.5 = 4$ turns the answer upside down.
Check two more rows
The constant must be the same in every row. Test two more:
$\frac{2.0}{8} = 0.25 \qquad \frac{3.0}{12} = 0.25$
They match. Watch out: the yogurt goes up by 0.5 cup from row to row. That 0.5 is not the constant. The fruit goes up by 2 cups at the same time, and $0.5 \div 2 = 0.25$ .
Enter the number
Type only the number in the box, with no units. You can type 0.25 or .25. On this site, 1/4 counts too.
So you would enter 0.25.

Check your answer

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

More questions like this

The original DOE question

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

Question 12 of 112

2026 Practice Test C Math, question 30

Question

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

A0.955%
B1.91%
C9.55%
D19.1%

Show the answer

Correct answer: C

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

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

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

Question analysis

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

NYS standards: 6.RP.3c percent as a rate per 100
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Writing a part of a whole as a percent is grade 6 work, done here with one division and one conversion. The awkward numbers, 191 out of 2,000, make placing the decimal point the only real demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Solution: 2 steps
Reading level: Question and choices: about grade 7, 24 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Finds 0.0955 but moves the decimal point only one place when writing the percent, giving 0.955%, ten times too small.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Divides 191 by 100, as if a percent were always out of 100, and never uses the 2,000 seniors, so 191 ÷ 100 = 1.91 becomes 1.91%.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Divides 191 by 1,000 instead of 2,000, dropping the 2 while moving the decimal point, so 0.191 becomes 19.1%, twice the true share.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
There are 2,000 seniors in all. Of them, 191 are in the Honor Society. You need the percent of all 2,000 seniors who are in it.
So 191 is the part, and 2,000 is the whole.
Write the share as a fraction
Put the part over the whole:
$\frac{\text{part}}{\text{whole}} = \frac{191}{2{,}000}$
This fraction says 191 out of every 2,000 seniors. Next, turn it into a decimal by dividing.
Divide to get a decimal
Dividing by 2,000 is the same as dividing by 2 and then by 1,000:
$191 \div 2 = 95.5$
$95.5 \div 1{,}000 = 0.0955$
Dividing by 1,000 moves the decimal point 3 places to the left. So the share is 0.0955.
Change the decimal to a percent
Multiply by 100. That moves the decimal point 2 places to the right:
$0.0955 \times 100 = 9.55$
So the share is 9.55%.
Watch out: moving the point only 1 place gives 0.955%, choice A. Dividing by 1,000 instead of 2,000 gives 19.1%, choice D. Dividing 191 by 100 gives 1.91%, choice B.
Give the answer
The 191 Honor Society members are 9.55% of all 2,000 seniors. That is a little less than one tenth of them.
So the answer is C, 9.55%.

Check your answer

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

More questions like this

The original DOE question

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

Question 13 of 112

2026 Practice Test C Math, question 37

Question

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

PICKING CANDY FROM
A BOWL

Flavor	Probability
Butterscotch	0.57
Lemon	0.095
Maple	0.125
Strawberry	0.21

Which flavor is most likely to be randomly picked?

Abutterscotch
Blemon
Cmaple
Dstrawberry

Show the answer

Correct answer: A

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

Question analysis

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

NYS standards: 6.SP.6 probability from 0 to 1
6.NS.7 ordering and absolute value
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Reading probabilities as likelihoods is grade 6 content, and this needs only a comparison of four decimals. The one demand is comparing decimals with different numbers of digits, a skill from earlier grades.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Error-prone arithmetic minor
Solution: 1 step
Reading level: Question and choices: about grade 9, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads lemon’s 0.095 as 0.95, overlooking the 0 in the tenths place, so it looks almost certain and larger than 0.57.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the decimal with more digits as the larger, or compares 125 with 57 as whole numbers, so 0.125 seems bigger than 0.57.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common misreading makes 0.21 the largest, since it is neither the longest decimal nor the one with the biggest digits, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The table gives the probability of picking each of four flavors. You need the flavor that is most likely to be picked.
The greater the probability, the more likely the pick. So you need the greatest of the four decimals.
Line up the decimals
The decimals have different numbers of digits. Add zeros at the end so each has three decimal places. This does not change their values:
Flavor Probability
Butterscotch 0.570
Lemon 0.095
Maple 0.125
Strawberry 0.210
Compare the tenths digits
Start with the first digit after the decimal point, the tenths place. The tenths digits are 5, 0, 1 and 2. The biggest is 5, so 0.570 is the greatest.
Watch out: 0.125 has more digits than 0.57, but it is smaller. Compare place by place, not by length. That mistake leads to choice C. Also, lemon is 0.095, not 0.95, so choice B is out.
Give the answer
Butterscotch has the greatest probability, 0.57. So it is the flavor most likely to be picked.
So the answer is A, butterscotch.

Flavor	Probability
Butterscotch	0.570
Lemon	0.095
Maple	0.125
Strawberry	0.210

Check your answer

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

More questions like this

The original DOE question

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

Question 14 of 112

2026 Practice Test D Math, question 1

Question

What is the prime factorization of 3,575?

A $5 \cdot 11 \cdot 13$
B $5^{2} \cdot 11 \cdot 13$
C $5 \cdot 715$
D $5^{2} \cdot 143$

Show the answer

Correct answer: B

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

$3{,}575 \div 25 = 143$

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

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

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

The prime factorization of 3,575 is $5^{2} \times 11 \times 13$ .

Question analysis

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

NYS standards: 6.NS.4 common factors and multiples
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Breaking a number into primes is grade 6 number work that relies on divisibility tests. The number here is larger than in most class examples, but the work is still a few divisions, lighter than typical grade 7 problems.
Estimated difficulty: Straightforward: two steps or one careful reading.
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
Solution: 3 steps
Reading level: Question and choices: about grade 11, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

A very close wrong answer moderate Choosing the answer
5² · 143 matches the right factorization in its first factor and multiplies back to 3,575 just as well, so a product check cannot separate them; only splitting 143 into 11 × 13 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Nothing on the page defines a prime factorization or gives a test for 11, and 143 passes the familiar quick tests for 2, 3 and 5, so it looks prime. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Lists each prime that divides 3,575 only once, 5, 11 and 13, as if a factorization named which primes divide it, though their product is only 715.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Divides out one 5 and stops at 5 · 715, a true factorization but not a prime one, though 715 ends in 5 and splits further.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Divides by 25 and stops at 143, which passes the tests for 2, 3 and 5 and looks prime, though 143 = 11 × 13.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
A prime number is a whole number greater than 1 whose only factors are 1 and itself. Examples are 2, 3, 5, 7, 11 and 13. The prime factorization of 3,575 writes it as a product (the answer to a multiplication) of prime numbers only. So you must break 3,575 down until every factor is prime.
Divide by 25 first
3,575 ends in 75. So it is divisible by 25:
$3{,}575 \div 25 = 143$
That means 3,575 = 25 × 143. But 25 is not prime. It is 5 × 5, which you can write as 5².
Test 143 for prime factors
Now see if 143 splits. Try the primes in order. 143 is odd, so 2 does not work. Its digits add to 1 + 4 + 3 = 8, which is not a multiple of 3, so 3 does not work. It does not end in 0 or 5, so 5 does not work. Since 7 × 20 = 140, 143 ÷ 7 leaves a remainder of 3, so 7 does not work. Now try 11:
$143 \div 11 = 13$
So 143 = 11 × 13, and both 11 and 13 are prime.
Watch out: 143 looks prime, but it is not. Stopping at 5² · 143 leads to choice D.
Put all the primes together
Collect every prime you found. The 5 appears twice, then 11 and 13 once each. The dot · means times, just like ×.
$3{,}575 = 5 \times 5 \times 11 \times 13 = 5^{2} \cdot 11 \cdot 13$
Watch out: write the 5 twice, since it divides 3,575 twice. Writing it once gives 5 · 11 · 13, choice A, which equals only 715. Choice C, 5 · 715, stops too early, because 715 ends in 5 and splits further.
So the answer is B, $5^{2} \cdot 11 \cdot 13$ .

Check your answer

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

More questions like this

The original DOE question

Question 15 of 112

2026 Practice Test D Math, question 4

Question

If $x = \frac{1}{4}$ , what is the value of $\frac{2}{2 - x}$ ?

A $\frac{3}{4}$
B $\frac{8}{9}$
C $1\frac{1}{7}$
D $3\frac{1}{2}$

Show the answer

Correct answer: C

The first step is to substitute the value for x into the expression to get $\frac{2}{2 - \frac{1}{4}}$ .

Then, evaluate the denominator and finally divide the numerator by the denominator.

$\frac{2}{\left(\frac{7}{4}\right)} = 2 \div \left(\frac{7}{4}\right) = 2\left(\frac{4}{7}\right) = \frac{8}{7} = 1\frac{1}{7}$

Question analysis

This substitution question sets x equal to one fourth and asks for the value of 2 divided by (2 − x). Students substitute, subtract a fraction from a whole number, and divide by the result. The expression is a stacked fraction, so treating the whole denominator as one quantity and dividing rather than multiplying are where students slip.

NYS standards: 6.EE.2c evaluating expressions and order of operations
6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Substituting a value and dividing by a fraction are grade 6 skills, and here they take three short steps with friendly numbers. The stacked fraction asks for care, but 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 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 10, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Substitution is one step, so the question is about handling a fraction inside a fraction. The bar over 2 − x groups the whole denominator, and the division by 7/4 must flip it; both errors lead to listed choices. A size check helps: 2 − 1/4 is a little less than 2, so the value must be a little more than 1, which rules out every wrong choice.

Error-prone arithmetic moderate Working it out
The denominator 2 − 1/4 has to become 7/4 before anything is divided, and dividing 2 by 7/4 means multiplying by 4/7; multiplying by 7/4 instead is the usual slip. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The work ends at the improper fraction 8/7, but the choices show it only as a mixed number, listed among 3/4, 8/9 and 3 1/2. (Enright, Morley & Sheehan, 2002)

Strategy: Simplify the denominator to a single fraction first, then divide 2 by that fraction by multiplying by its reciprocal; treating the fraction bar as a grouping symbol avoids the common slip.

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).
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

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Lets the fraction bar cover only the first 2, working 2/2 − 1/4 = 1 − 1/4 = 3/4 instead of dividing 2 by the whole denominator.
B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Adds 1/4 to 2 instead of subtracting it, so the denominator becomes 9/4, and 2 ÷ 9/4 = 8/9.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Finds the denominator 7/4 correctly but multiplies 2 by it instead of dividing, 2 × 7/4 = 14/4 = 3 1/2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is any number, and b/c is a fraction. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). It works because dividing asks how many copies of b/c fit into a.
Changing an improper fraction to a mixed number: $\frac{a}{b} = q\frac{r}{b}$; Use it when the top, a, is bigger than the bottom, b. q is how many whole times b fits into a. r is the remainder, what is left over. For 8/7, the 7 fits once with 1 left, so 8/7 = 1 1/7.

Sort out the question
You know $x = \frac{1}{4}$ . You need the value of 2 divided by (2 − x). A fraction bar means divide. It also groups, so you work out the whole bottom, 2 − x, before you divide.
Work out the bottom
Put in 1/4 for x. To subtract, write 2 as fourths, since 2 = 8/4:
$2 - \frac{1}{4} = \frac{8}{4} - \frac{1}{4} = \frac{7}{4}$
The bottom is 7/4.
Watch out: the sign is minus. Adding 1/4 gives 9/4 and then 8/9, choice B.
Divide 2 by 7/4
Now the problem is 2 ÷ 7/4. Use the rule for dividing by a fraction. Flip 7/4 to get 4/7, then multiply:
$2 \div \frac{7}{4} = 2 \times \frac{4}{7} = \frac{8}{7}$
Watch out: multiplying by 7/4 instead of 4/7 gives 14/4 = 3 1/2, choice D.
Write it as a mixed number
The choices use mixed numbers. 7 fits into 8 once, with 1 left over:
$\frac{8}{7} = 1\frac{1}{7}$
Watch out: the fraction bar covers all of 2 − 1/4. Working 2/2 = 1 first and then taking away 1/4 gives 3/4, choice A.
So the answer is C, $1\frac{1}{7}$ .

Check your answer

Estimate. The bottom, 1 3/4, is a bit less than 2. So 2 divided by it should be a bit more than 1, and 1 1/7 fits. Also check by multiplying: 7/4 × 8/7 = 56/28 = 2. That is the top, so the division is right.

More questions like this

The original DOE question

Question 4 as shown on the DOE’s digital test, page 77 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 77, question 4.

Question 16 of 112

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 17 of 112

2026 Practice Test D Math, question 21

Question

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

AMaxie will still owe $2.25.
BMaxie will still owe $1.75.
CMaxie will have $1.75 left.
DMaxie will have $2.25 left.

Show the answer

Correct answer: C

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

$7.75 + 11.00 + 4.50 = 23.25$

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

$25.00 - 23.25 = 1.75$

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

Question analysis

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

NYS standards: 6.NS.3 multi-digit decimal operations
6.NS.5 positive and negative numbers
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Adding and subtracting money amounts is grade 6 decimal work. Here it takes one sum and one difference, and the only further demand is deciding whether the result is money left or money owed, well below typical grade 7 tasks.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 66 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic minor Working it out
The loans add to $23.25, and subtracting that from $25.00 means borrowing across two zeros, a classic place for a digit to go wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Each amount appears twice among the choices, once as money left and once as money owed, so the result must also be read for direction by asking which is larger, the gift or the loans. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Needs two errors at once, a wrong difference such as $2.25 and the reversed outcome, so no single line of working reaches it and it mostly draws guesses.
B Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: moderate
Finds the $1.75 difference correctly but reverses its meaning, as if the $25.00 gift fell short of the $23.25 borrowed instead of covering it.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts $23.25 from $25.00 column by column, taking the smaller digit from the larger instead of borrowing, so 5 − 0 and 2 − 0 give $2.25.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
Maxie borrowed from three people: $7.75, $11.00 and $4.50. Her grandmother gave her $25.00. She uses the $25.00 to pay everyone back. You need to know if money is left over or if she still owes some, and how much.
Add up what she borrowed
Line up the decimal points and add:
$7.75 + 11.00 + 4.50 = 23.25$
She owes $23.25 in all. That is the amount she must pay back.
Compare the gift with the debt
The gift, $25.00, is more than the $23.25 she owes. So she can pay everyone back and still have money left. She will not owe anything.
Watch out: once you find $1.75, keep its meaning straight. Saying she still owes $1.75, choice B, gets the direction backwards.
Subtract to find what is left
Take the debt away from the gift:
$25.00 - 23.25 = 1.75$
You need to regroup (borrow) across the zeros. Think of $25.00 as 24 dollars and 100 cents. Then 100 − 25 = 75 cents, and 24 − 23 = 1 dollar.
Watch out: taking the smaller digit from the larger one in each column gives $2.25, choice D.
So the answer is C: Maxie will have $1.75 left.

Check your answer

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

More questions like this

The original DOE question

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

Question 18 of 112

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 19 of 112

2026 Practice Test D Math, question 24

Question

An athlete runs $\frac{1}{8}$ kilometer in $\frac{3}{4}$ minute. At this rate, how many kilometers would the athlete run in 1 minute?

A $\frac{3}{32}$
B $\frac{1}{6}$
C $1\frac{1}{2}$
D6

Show the answer

Correct answer: B

To determine the number of kilometers in 1 minute, find the unit rate in kilometers per minute: $\frac{\frac{1}{8}\text{ kilometer}}{\frac{3}{4}\text{ minute}}$ .

To evaluate the fraction, use the reciprocal when dividing to get $\frac{1}{6}$ kilometer per 1 minute.

$\frac{1}{8} \div \frac{3}{4} = \frac{1}{8} \times \frac{4}{3} = \frac{4}{24} = \frac{1}{6}$

Question analysis

An athlete runs 1/8 kilometer in 3/4 minute, and students find how many kilometers the athlete would run in 1 minute at the same rate. It tests unit rates with fractions. Both the distance and the time are fractions, so deciding which one to divide by which is the main challenge.

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of fractions is the grade 7 standard itself, and this is a standard example: one fraction divided by another in two short steps, with the order of division the only real decision.
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: 2 steps
Reading level: Question and choices: about grade 9, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is a unit rate with fractions, and its wrong choices are the other ways to combine 1/8 and 3/4. Since 1 minute is only a third longer than 3/4 minute, the distance must be only a third more than 1/8 kilometer, which rules out the product and both large choices at once. Dividing with the units written in, kilometers over minutes, keeps the order right.

Error-prone arithmetic moderate Working it out
The rate is a fraction divided by a fraction, 1/8 ÷ 3/4, and only the 3/4 may be flipped; multiplying straight across or dividing the other way round both give listed choices. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
Kilometers in 1 minute means kilometers per minute, so the 1/8 goes on top even though the time, 3/4, is the larger number. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Divide the distance by the time, kilometers over minutes, by multiplying 1/8 by the reciprocal of 3/4, and check that the units of the result are kilometers per minute.

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 surface strategy Expected pull: moderate
Multiplies 1/8 by 3/4 instead of dividing, getting 3/32, which is less than the 1/8 kilometer already run in under a minute.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: weak
Splits the 3/4 minute into quarters but multiplies 1/8 by 3 instead of dividing, then multiplies by 4 for a full minute, giving 12/8 = 1 1/2.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Divides 3/4 by 1/8, minutes by kilometers, which gives 6 minutes per kilometer, the rate turned upside down.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{rate} = \text{distance} \div \text{time}$; A unit rate tells how much happens in 1 unit of time. Here the distance is in kilometers and the time is in minutes, so the rate is kilometers per minute. You divide because you are sharing the distance equally over the minutes.
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 second fraction, the one you divide by.

Sort out the question
The athlete runs 1/8 kilometer in 3/4 minute, which is less than a full minute. You need how far the athlete runs in 1 whole minute at the same speed. That is the unit rate, in kilometers per minute.
Set up the division
Kilometers per minute means kilometers divided by minutes. So the distance comes first and the time second:
$\text{rate} = \frac{1}{8} \div \frac{3}{4}$
Watch out: dividing the other way, 3/4 ÷ 1/8, gives 6. That is minutes per kilometer, the rate turned upside down, choice D.
Divide the fractions
Flip 3/4 to get 4/3, then multiply. Multiply the tops and multiply the bottoms:
$\frac{1}{8} \times \frac{4}{3} = \frac{4}{24}$
Simplify by dividing the top and bottom by 4:
$\frac{4}{24} = \frac{1}{6}$
Watch out: multiplying 1/8 × 3/4 without flipping gives 3/32, choice A.
Make sense of the answer
In 1 minute, the athlete runs 1/6 kilometer. That is a bit more than 1/8 kilometer, which makes sense. A full minute is a bit longer than 3/4 minute, so the athlete goes a bit farther. Choice C, 1 1/2 kilometers, would be 12 times as far in barely more time.
So the answer is B, $\frac{1}{6}$ kilometer.

Check your answer

Work in quarter minutes. In 3 quarter minutes, the athlete runs 1/8 km. So in 1 quarter minute, the athlete runs 1/8 ÷ 3 = 1/24 km. A full minute has 4 quarters: 4 × 1/24 = 4/24 = 1/6 km.

More questions like this

The original DOE question

Question 24 as shown on the DOE’s digital test, page 97 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 97, question 24.

Question 20 of 112

2026 Practice Test D Math, question 30

Question

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

A $\frac{8}{15}$
B $\frac{5}{6}$
C $1\frac{7}{8}$
D $3\frac{2}{3}$

Show the answer

Correct answer: A

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

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

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

$\frac{\frac{2}{3}}{\frac{5}{4}} = \frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}$

Question analysis

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

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of fractions is exactly the grade 7 standard. This one divides by a mixed number, and the cook's plan to enlarge the recipe adds wording without adding any numbers to use.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
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: 2 steps
Reading level: Question and choices: about grade 5, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The work is a single fraction division, but it involves a mixed number and an awkward result. Students unsure of the rule multiply the two amounts or divide them the other way round, and each gives a tidy listed fraction. A size check settles it: 1 1/4 cups of flour take only 2/3 cup of sugar, so one cup of flour needs less than 2/3 cup, which leaves a single choice.

Error-prone arithmetic moderate Working it out
The divisor is the mixed number 1 1/4, which must become 5/4 before dividing; then 2/3 × 4/5 gives 8/15, a fraction that does not simplify and looks unfinished. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase “for each cup of flour” puts flour in the denominator, sugar ÷ flour, and the sentence about increasing the recipe adds nothing to the calculation. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies 2/3 by 5/4 instead of dividing, getting 10/12 = 5/6, more sugar for one cup of flour than the recipe uses for 1 1/4 cups.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides the flour by the sugar, 5/4 ÷ 2/3 = 15/8 = 1 7/8, which is cups of flour for each cup of sugar, the rate upside down.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Splits the mixed number apart, dividing 2/3 by only the 1/4 to get 2 2/3, then adding the whole 1 to make 3 2/3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{unit rate} = \frac{\text{cups of sugar}}{\text{cups of flour}}$; A unit rate tells how much of one thing goes with 1 of another. Here it is the cups of sugar for 1 cup of flour. The amount you want goes on top. The amount you want to make 1 goes on the bottom.
Mixed number to fraction: $w\frac{n}{d} = \frac{w \times d + n}{d}$; w is the whole number, and n/d is the fraction part. Multiply the whole number by the bottom, add the top, and keep the bottom. It works because each whole holds d pieces of size 1/d.
Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; 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.

Sort out the question
The recipe uses $\frac{2}{3}$ cup of sugar for every $1\frac{1}{4}$ cups of flour. You need the cups of sugar for each 1 cup of flour. The part about making more of the recipe gives no numbers, so you can set it aside.
Set up the unit rate
You want sugar for each cup of flour. So sugar goes on top and flour goes on the bottom:
$\text{unit rate} = \frac{2}{3} \div 1\frac{1}{4}$
Watch out: dividing the other way, flour by sugar, gives $\frac{15}{8} = 1\frac{7}{8}$ . That is cups of flour for each cup of sugar, choice C.
Change the mixed number
You cannot divide by a mixed number directly. Change it to a fraction first. Each whole has 4 fourths:
$1\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{5}{4}$
Divide the fractions
Multiply by the reciprocal of 5/4, which is 4/5. Multiply the tops and multiply the bottoms:
$\frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}$
The fraction 8/15 cannot be simplified, since 8 and 15 share no factor but 1.
Watch out: multiplying without flipping gives $\frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6}$ , choice B.
So the answer is A, $\frac{8}{15}$ .

Check your answer

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

More questions like this

The original DOE question

Question 21 of 112

2026 Practice Test D Math, question 32

Question

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

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

A $2\frac{1}{4}$
B $4\frac{1}{2}$
C9
D72

Show the answer

Correct answer: B

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

Convert 72 cups to gallons:

$72 \text{ cups} \times \frac{1 \text{ quart}}{4 \text{ cups}} \times \frac{1 \text{ gallon}}{4 \text{ quarts}} =$

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

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

$4\frac{1}{2} \text{ gallons}$

Question analysis

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

NYS standards: 6.RP.3d converting measurement units
6.NS.2 multi-digit division
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting units with given factors is grade 6 work. This adds a first step, the total number of cups, and a two-stage conversion through quarts, but the numbers are friendly and the path is routine for grade 7 students.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 32 words.
Corrected from the DOE handbook: A stray word at the end of the explanation was removed.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Many steps minor Working it out
A multiplication, 36 × 2, comes before two divisions by 4, cups to quarts and quarts to gallons, and stopping after the first step leaves a believable 72. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The word “each” turns 36 people into 72 cups, and the bold word gallons sets the unit of the answer, a change the note supplies in two steps. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Converts 36 cups to gallons, 36 ÷ 16 = 2 1/4, missing the word each, so every person’s second cup of juice is left out.
C Units or scale (King et al., 2004)
Reveals a misconception Expected pull: weak
Uses 8 cups to a gallon, adding the two 4s in the note instead of multiplying them, so 72 cups become 72 ÷ 8 = 9 gallons.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Finds the 72 cups of juice and stops, giving the amount in cups though the bold word gallons asks for it in gallons.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation ends with a line that reads only “lineExtend”, a formatting code left in the document by mistake. The site leaves it out.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
You know:
- 36 people were at the party.
- Each person drank 2 cups of juice.
- 1 gallon = 4 quarts, and 1 quart = 4 cups.
You need the total amount of juice in gallons. The bold word tells you the unit of the answer.
Find the total in cups
Each of the 36 people drank 2 cups:
$36 \times 2 = 72$
They drank 72 cups in all.
Watch out: skipping the word each uses only 36 cups. Then $36 \div 16 = 2\frac{1}{4}$ gallons, choice A.
Find the cups in a gallon
A gallon is 4 quarts, and each quart is 4 cups:
$4 \times 4 = 16$
So 1 gallon holds 16 cups.
Watch out: adding the two 4s gives 8 cups in a gallon. Then $72 \div 8 = 9$ , choice C.
Change cups to gallons
Divide the cups by 16. Simplify the fraction by dividing the top and bottom by 8:
$72 \div 16 = \frac{72}{16} = \frac{9}{2} = 4\frac{1}{2}$
They drank $4\frac{1}{2}$ gallons of juice.
Watch out: 72 is the amount in cups. The question asks for gallons, so 72, choice D, is in the wrong unit.
So the answer is B, $4\frac{1}{2}$ .

Check your answer

Work backward. $4\frac{1}{2}$ gallons is $4\frac{1}{2} \times 4 = 18$ quarts. 18 quarts is $18 \times 4 = 72$ cups. Shared among 36 people, that is $72 \div 36 = 2$ cups each. It matches.

More questions like this

The original DOE question

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

Question 22 of 112

2026 Practice Test D Math, question 37

Question

What 3-digit number between 630 and 650 is divisible by both 5 and 8?

Enter your answer in the space.

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

Show the answer

Correct answer: 640

The final digit of the three-digit number is 0. Each of the digits 0 through 9, inclusive, must be considered.

To narrow the choices, first consider that the number between 630 and 650 must be divisible by both 5 and 8.

A number divisible by 5 must end in 0 or 5. Therefore, the only possibilities are 0 and 5.

If the number ended in 0, the whole number 640 must also be divisible by 8:

$640 \div 8 = 80 \text{ and } 640 \div 5 = 128$

Both result in a remainder of 0.
If the number ended in 5, the whole numbers 635 and 645 must also be divisible by 8:

$635 \div 8 = 79R3 \text{ and } 635 \div 5 = 127$

$645 \div 8 = 80R5 \text{ and } 645 \div 5 = 129$

Dividing 635 by 5 results in a remainder of 0, but dividing 635 by 8 results in a remainder of 3.

Dividing 645 by 5 results in a remainder of 0, but dividing 645 by 8 results in a remainder of 5.

Therefore, 640 is the only whole number between 630 and 650 that is divisible by both 5 and 8.

Question analysis

This number question asks for the 3-digit number between 630 and 650 that is divisible by both 5 and 8, typed into a box. Students use divisibility rules or common multiples to find it. The work is short, but the number has to pass both tests, so checking only one divisor is the easy slip.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Common multiples are grade 6 content. Divisibility by 5 limits the last digit, and only a few numbers lie between 630 and 650, so this is a quick check, shorter than typical grade 7 work.
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: Much to keep track of minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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

Few numbers need checking, so most students succeed; the risk is testing only one divisor. Students who start from multiples of 8 meet 632 first, and those who start from multiples of 5 can slip on 645 ÷ 8. Using the least common multiple, 40, shows at once that the number must be a multiple of 40, which leaves one candidate in the range.

Much to keep track of minor Working it out
Each candidate must pass two tests, divisible by 5 and by 8, and both ends of the range, 630 and 650, pass the first test but fail the second. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
The shortcut that a multiple of 5 ends in 0 or 5 is assumed, and it cuts the search to 635, 640 and 645 before any division by 8. (Anderson & Krathwohl, 2001)

Strategy: Use divisibility by 5 to limit the last digit, then check only those few numbers in the range for divisibility by 8.

Traps

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

The wrong answers, classified

632 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Takes the first multiple of 8 after 630, 8 × 79 = 632, and stops without checking divisibility by 5, which a number ending in 2 fails.
645 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Picks a number ending in 5 so that 5 divides it, then slips on the check for 8, which leaves a remainder of 5.
40 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Finds the least common multiple of 5 and 8, 40, and enters it, one step before finding its multiple between 630 and 650.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisible by 5: $\text{last digit} = 0 \text{ or } 5$; Divisible means it divides evenly, with no remainder. A whole number is divisible by 5 only when its last digit is 0 or 5. That is because counting by 5s always lands on a number ending in 0 or 5.
Least common multiple: $\text{LCM}(5, 8) = 5 \times 8 = 40$; The least common multiple, or LCM, is the smallest number that both 5 and 8 divide evenly. 5 and 8 share no factor except 1, so their LCM is their product. Any number divisible by both is a multiple of 40.

Sort out the question
You need a 3-digit number between 630 and 650. It must be divisible by 5 and also by 8. Between means 630 and 650 themselves do not count.
Use the rule for 5
A number divisible by 5 ends in 0 or 5. Between 630 and 650, only three numbers do:
$635, \; 640, \; 645$
Only these need checking for 8.
Test each number for 8
Divide each one by 8 and look at the remainder, the amount left over:
- 635 ÷ 8 = 79, with 3 left over.
- 640 ÷ 8 = 80, with nothing left over.
- 645 ÷ 8 = 80, with 5 left over.
Only 640 divides evenly by 8.
Watch out: 645 passes the test for 5 but fails for 8. And 632, which is $8 \times 79$ , passes for 8 but ends in 2, so it fails for 5.
Enter the number
640 passes both tests. Type the whole number in the box.
Watch out: 40 is the smallest number divisible by 5 and 8, but it is not between 630 and 650.
So you would enter 640.

Check your answer

Use the LCM. A number divisible by both 5 and 8 must be a multiple of 40. $40 \times 16 = 640$ . The multiples next to it are 600 and 680, which are outside the range. So 640 is the only one.

More questions like this

The original DOE question

Question 23 of 112

2026 Practice Test D Math, question 39

Question

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

HOCKEY GOAL PROBABILITIES

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

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

A0.13
B0.21
C0.34
D0.50

Show the answer

Correct answer: C

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

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

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

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

$0.21 + 0.09 + 0.04 = 0.34.$

Question analysis

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

NYS standards: 6.SP.8 probability models for simple events
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of an event from a given probability model is grade 6 work in New York. Here the model is a table, and the task is choosing the right rows and adding decimals, shorter than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Turning words into math minor Understanding the question
The phrase “three or more goals” has to become the three rows for 3, 4 and 5 goals, starting at the 3-goal row itself. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The table lists six probabilities, two of them both 0.21, so the rows for 3, 4 and 5 goals must be picked out by their goal counts, not by their values. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide which rows the words "three or more" cover, add their probabilities, and check by subtracting the total of the other rows from 1.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Adds only the rows for 4 and 5 goals, 0.09 + 0.04 = 0.13, reading three or more as more than three and leaving out the end row.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Reads only the row for exactly 3 goals, 0.21, overlooking the words “or more” and the two rows below it.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Counts 3 of the 6 rows, for 3, 4 and 5 goals, as half the outcomes, ignoring that the table gives them different probabilities.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding probabilities: $P(A \text{ or } B) = P(A) + P(B)$; A and B are two events that cannot both happen at once, like scoring exactly 3 goals and exactly 4 goals. The chance that one or the other happens is the sum of their chances.
The chance it does not happen: $P(\text{not } A) = 1 - P(A)$; The chances of all possible outcomes add up to 1. So the chance an event does not happen is 1 minus the chance it does. Use it to check an answer from the other side.

Sort out the question
The table gives the chance of each number of goals, from 0 to 5. You need the chance the team scores three or more goals in the next game.
Pick the right rows
Three or more means 3 goals, 4 goals or 5 goals. The row for 3 counts too. Read those three rows:
- 3 goals: 0.21
- 4 goals: 0.09
- 5 goals: 0.04
Watch out: leaving out the 3-goal row gives 0.09 + 0.04 = 0.13, choice A. Using only the 3-goal row gives 0.21, choice B.
Add the chances
A game can end with only one number of goals. So add the three chances:
$0.21 + 0.09 + 0.04 = 0.34$
Watch out: three of the six rows is half the rows, but that does not make the chance 0.50, choice D. The rows have different chances.
So the answer is C, 0.34.

Check your answer

Find the chance of fewer than 3 goals, the other rows: $0.16 + 0.21 + 0.29 = 0.66$ . Then the chance of 3 or more is $1 - 0.66 = 0.34$ . Both ways give 0.34.

More questions like this

The original DOE question

Question 39 as shown on the DOE’s digital test, page 112 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 112, question 39.

Question 24 of 112

2026 Practice Test D Math, question 42

Question

A student reads $\frac{1}{12}$ of a book in $\frac{1}{3}$ hour. What fraction of the book can the student read in 1 hour?

A $\frac{1}{36}$
B $\frac{1}{4}$
C $\frac{5}{12}$
D $\frac{12}{3}$

Show the answer

Correct answer: B

To find the fraction of the book the student can read in 1 hour, find the unit rate per hour. First, set up the ratio.

$\frac{\frac{1}{12} \text{ book}}{\frac{1}{3} \text{ hour}}$

Then use division to find the unit rate.

$\frac{\frac{1}{12}}{\frac{1}{3}} = \frac{1}{12} \div \frac{1}{3} = \frac{1}{12} \times \frac{3}{1} = \frac{3}{12} = \frac{1}{4}$

Another way to solve the problem is to set up the ratio and then multiply both the numerator and denominator by 3, since $\frac{1}{3}$ hour times 3 equals 1 hour.

$\frac{3 \times \frac{1}{12} \text{ book}}{3 \times \frac{1}{3} \text{ hour}} = \frac{\frac{1}{4} \text{ book}}{1 \text{ hour}}$

Question analysis

A student reads 1/12 of a book in 1/3 of an hour, and this rate question asks what fraction of the book the student can read in a whole hour. It tests finding a unit rate when both amounts are fractions. Deciding whether to multiply or divide, and which fraction goes on top, decides the result.

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of fractions is the grade 7 standard itself. This is a short example with unit fractions, where scaling the time up to a whole hour is as quick as dividing.
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
Turning words into math minor
Solution: 2 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

Thinking in thirds of an hour makes this quick: an hour holds three of them, so the student reads three twelfths of the book. Students who reach for a fraction rule instead must choose between multiplying and dividing, and between two orders of division. A sense check rules out two wrong choices at once: a whole hour must cover more than 1/12 of the book, and no more than the whole book.

Error-prone arithmetic minor Working it out
Both amounts are unit fractions, and 1/12 ÷ 1/3 means multiplying by 3, which gives 3/12 that still has to be simplified to match a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase “fraction of the book … in 1 hour” asks for book per hour, so the fraction of the book is divided by the time, not the other way round. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Ask how many thirds of an hour make 1 hour, then scale the fraction of the book by the same factor; or divide the fraction of the book by the time.

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).
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 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 1/12 by 1/3 instead of dividing, getting 1/36, which would mean reading less of the book in an hour than in a third of one.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two fractions, 1/12 + 4/12 = 5/12, combining an amount of the book with an amount of time as if they were alike.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Divides the time by the amount read, 1/3 ÷ 1/12 = 12/3, which is hours per book, the rate turned upside down.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{unit rate} = \frac{\text{part of the book}}{\text{hours}}$; A unit rate tells how much happens in 1 unit, here 1 hour. Put the part of the book on top and the time on the bottom. Dividing finds the part of the book for one whole hour.
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.

Sort out the question
The student reads $\frac{1}{12}$ of a book in $\frac{1}{3}$ of an hour. You need the part of the book read in 1 whole hour. That is a unit rate.
Scale up to one hour
One hour holds three thirds of an hour. So in 1 hour the student reads 3 times as much:
$3 \times \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$
To simplify 3/12, divide the top and bottom by 3. The student reads 1/4 of the book in 1 hour.
Check with the unit rate
Divide the part of the book by the time. Multiply by the reciprocal of 1/3, which is 3/1:
$\frac{1}{12} \div \frac{1}{3} = \frac{1}{12} \times \frac{3}{1} = \frac{3}{12} = \frac{1}{4}$
Both ways give 1/4.
Avoid the wrong operations
Watch out: multiplying gives $\frac{1}{12} \times \frac{1}{3} = \frac{1}{36}$ , choice A. That is less than 1/12. But in a full hour, the student must read more than in a third of an hour. Dividing the time by the book gives 12/3, choice D. That is 4 whole books.
So the answer is B, $\frac{1}{4}$ .

Check your answer

If the student reads 1/4 of the book each hour, the whole book takes 4 hours. In 1/3 of an hour, the student reads $\frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$ of the book. That matches the question.

More questions like this

The original DOE question

Question 42 as shown on the DOE’s digital test, page 115 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 115, question 42.

Question 25 of 112

2026 Practice Test D Math, question 43

Question

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

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

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

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

The difference between the cost of the bars is .

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

Show the answer

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

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

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

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

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

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

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

Question analysis

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

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Comparing unit prices is grade 6 rate work. The menus split it into three short steps, two divisions of money amounts and one subtraction, which is lighter than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Many steps minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 87 words.
Corrected from the DOE handbook: The order of the answers in the DOE’s key was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Many steps minor Working it out
Three menus must all be right to score: two divisions for the costs per bar, then a subtraction that uses both, so a slip in either division spoils the third menu. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Each division is with money, $12.00 ÷ 24 and $6.50 ÷ 10, so the decimal point must be placed carefully, though the menus offer only sensible amounts. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

CBA Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Divides $12.00 by 20 instead of 24, misreading the size of the Art Club’s pack, so $0.60 a bar leaves a gap of only $0.05.
BBB An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Finds both costs per bar correctly, then slips in the subtraction, taking $0.65 − $0.50 as $0.10, the option beside the right one.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
You know:
- The Art Club sells 24 bars for $12.00.
- The Science Club sells 10 bars for $6.50.
There are three menus. The first is the Art Club’s cost per bar. The second is the Science Club’s. The third is the difference, how much more one bar costs than the other.
Find the Art Club’s cost
Share $12.00 among 24 bars:
$12.00 \div 24 = 0.50$
Each bar costs $0.50. That makes sense: 24 half-dollars make $12. Pick $0.50 in the first menu.
Watch out: dividing by 20 instead of 24 gives $0.60, which is also in the menu.
Find the Science Club’s cost
Share $6.50 among 10 bars. Dividing by 10 moves the decimal point one place to the left:
$6.50 \div 10 = 0.65$
Each bar costs $0.65. Pick $0.65 in the second menu.
Find the difference
Subtract the smaller cost from the larger one:
$0.65 - 0.50 = 0.15$
The bars differ by $0.15. Pick $0.15 in the third menu.
Watch out: $0.10 sits right next to $0.15 in the menu. A slip in the subtraction lands there.
Fill in the three menus
On the test and on this site, you choose each answer from a drop-down menu.
So the menus read $0.50, $0.65 and $0.15.

Check your answer

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

More questions like this

The original DOE question

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

Question 26 of 112

2026 Practice Test D Math, question 49

Question

Anise has a piece of cloth measuring 45 inches by 75 inches. She wants to cut it into squares of equal size, without any cloth left over. What is the greatest possible side length, in inches, of one of those squares?

Show the answer

Correct answer: C

Given that each piece will be a square, the length and width will be the same. The greatest possible length can be determined by finding the greatest common factor of the length and width of the original piece of cloth.

Begin by finding the prime factorization of the length (75 in.) and width (45 in.).

$45 = 3 \times 3 \times 5$

$75 = 3 \times 5 \times 5$

The common factors can be seen as $3 \times 5$ .

The greatest possible side length, in inches, will be $3 \times 5 = 15$ .

Question analysis

Anise wants to cut a 45-inch by 75-inch piece of cloth into equal squares with nothing left over. This factors question asks for the greatest possible side length of each square, in inches. Students must see that the side has to divide both lengths evenly, then find the largest number that does.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Greatest common factors of two numbers up to 100 are grade 6 content. The cloth setting asks students to recognize that this is what is needed, but the work is one short factor search, below typical 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: Turning words into math minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 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

Once the side is seen as a common factor, the work is short, since 45 and 75 share both 3 and 5. Students who spot one shared factor and stop reach a listed choice, and those who think only of the largest square that fits use the full 45-inch width. Checking that 45 and 75 both divide by the chosen side, and that no larger number does, closes it.

Turning words into math minor Understanding the question
The words “squares of equal size, without any cloth left over” mean the side must divide both 45 and 75 exactly; nothing names the greatest common factor. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Finding the greatest common factor of 45 and 75 calls for prime factorization or a list of factors, a grade 6 skill the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Write each length as a product of primes, multiply the primes they share, and check that both 45 and 75 divide evenly by the result.

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: weak
Finds a factor that 45 and 75 share, 3, and stops without combining it with the 5 they also share.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Spots the common factor 5, since both lengths end in 5, and stops before multiplying it by the 3 they also share.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Uses the cloth’s full width, 45 inches, as the side, the largest square that fits, but 45 does not divide 75, leaving a 45-by-30 strip.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor: $\text{GCF} = \text{product of the prime factors both numbers share}$; A factor divides a number evenly. The greatest common factor, or GCF, is the biggest number that divides both. Write each number as a product of primes, numbers whose only factors are 1 and themselves. Then multiply the primes they share.

Sort out the question
The cloth is 45 inches by 75 inches. It is cut into equal squares, with no cloth left over. You need the greatest side the squares can have.
No cloth left over means the side must fit evenly into 45 and into 75. So the side is a common factor of both. The greatest one is the GCF.
Break each number into primes
Write each length as a product of prime numbers:
$45 = 3 \times 3 \times 5$
$75 = 3 \times 5 \times 5$
The two lists share one 3 and one 5.
Multiply the shared primes
Multiply the primes the two numbers share:
$\text{GCF} = 3 \times 5 = 15$
The squares can be 15 inches on a side.
Watch out: 3 and 5 are each common factors, choices A and B. But they are not the greatest. Use every prime the numbers share.
Rule out the full width
Watch out: a 45-inch square, choice D, fits the width. But 75 ÷ 45 = 1 with 30 left over. That leaves a strip of cloth 45 inches by 30 inches.
So the answer is C, 15.

Check your answer

Divide each length by 15: $45 \div 15 = 3$ and $75 \div 15 = 5$ . So the cloth cuts into 3 rows of 5 squares, with nothing left. Both divisions come out even.

More questions like this

The original DOE question

Question 49 as shown on the DOE’s digital test, page 122 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 122, question 49.

Question 27 of 112

2024 Test A Math, question 3

Question

A juice mixture contains $\frac{3}{16}$ gallon of apple juice and $\frac{3}{40}$ gallon of cranberry juice. How many gallons of apple juice per gallon of cranberry juice does the mixture contain? (Express your answer as a decimal.)

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

Show the answer

Correct answer: 2.5

Set up a proportion to solve:

$\frac{\frac{3}{16}\text{ apple}}{\frac{3}{40}\text{ cranberry}} = \frac{x\text{ apple}}{1\text{ cranberry}}$

$\frac{3}{16} \times \frac{40}{3} = x$

$\frac{40}{16} = x$

$2.5 = x$

Question analysis

This ratio grid-in describes a juice mixture with a fraction of a gallon of apple juice and a smaller fraction of a gallon of cranberry juice. Students find the unit rate, gallons of apple juice per gallon of cranberry juice, by dividing one fraction by the other and writing it as a decimal. Dividing in the wrong order is the main risk.

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of two fractions is exactly the grade 7 standard, and this is a textbook example of it. The only added demands are dividing in the right order and answering as a decimal.
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
Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is one division of fractions, but the question tests whether the division runs the right way. The word per tells which juice goes on top, and a student who divides the smaller amount by the larger gets a tidy decimal for the reverse rate. A sense check helps: the mixture holds more apple juice than cranberry, so the rate must be greater than 1.

Turning words into math moderate Understanding the question
The phrase "gallons of apple juice per gallon of cranberry juice" fixes the order: apple's 3/16 goes on top and cranberry's 3/40 below, though the two fractions invite either order. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic moderate Working it out
Dividing 3/16 by 3/40 means multiplying by 40/3; the 3s cancel to leave 40/16, which still has to be reduced and written as a decimal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The answer must be gridded as a decimal, and with no choices shown, the upside-down 0.4 gives no sign that it answers the reverse question. (Enright, Morley & Sheehan, 2002)

Strategy: Put the apple juice over the cranberry juice and divide the fractions by multiplying by the reciprocal; the matching 3s cancel, leaving a simple fraction to turn into a decimal.

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

0.4 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides cranberry by apple, 3/40 ÷ 3/16 = 16/40 = 0.4, giving gallons of cranberry per gallon of apple, the rate turned upside down.
0.014 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Multiplies the two amounts instead of dividing them, 3/16 × 3/40 = 9/640 ≈ 0.014, as if comparing the juices called for a product.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $r = a \div b$; r is the rate: how much of one thing there is for each 1 of another. a is the amount of the first thing. b is the amount of the second thing, the one named after the word per. Dividing shares a out evenly over each 1 of b.
Dividing by a fraction: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a/b and c/d are fractions, and b, c and d are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. It works because flipping a fraction gives the number that undoes it.

Sort out the question
The mix has $\frac{3}{16}$ gallon of apple juice and $\frac{3}{40}$ gallon of cranberry juice. You need the “gallons of apple juice per gallon of cranberry juice”. Per means for each. So the question asks: for each 1 gallon of cranberry juice, how many gallons of apple juice are there? The answer must be a decimal.
Set up the division
To find an amount for each 1 gallon, divide. Apple juice goes first, because it comes before per. Cranberry juice comes second.
$r = a \div b$
$r = \frac{3}{16} \div \frac{3}{40}$
Watch out: do not flip the order. Cranberry divided by apple gives 0.4. That is gallons of cranberry juice for each gallon of apple juice, the reverse question.
Multiply by the reciprocal
To divide by $\frac{3}{40}$ , multiply by its reciprocal, $\frac{40}{3}$ .
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
$\frac{3}{16} \times \frac{40}{3} = \frac{3 \times 40}{16 \times 3} = \frac{40}{16}$
The 3 on top and the 3 on the bottom cancel, since $3 \div 3 = 1$ . Watch out: do not just multiply the two amounts. That gives $\frac{9}{640}$ , about 0.014, which is far too small.
Change the fraction to a decimal
Make $\frac{40}{16}$ simpler by dividing the top and bottom by 8. Then divide.
$\frac{40}{16} = \frac{5}{2} = 5 \div 2 = 2.5$
There are 2.5 gallons of apple juice for each gallon of cranberry juice. So you would grid 2.5.

Check your answer

Both fractions have 3 on top. The one with the smaller bottom, $\frac{3}{16}$ , is bigger. So there is more apple juice than cranberry juice, and the answer must be more than 1. That rules out 0.4. Also, $3 \div 40 = 0.075$ and $0.075 \times 2.5 = 0.1875$ , which is $3 \div 16$ .

More questions like this

The original DOE question

Question 60 as printed on page 75 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 75, question 60. Open page 75 of the handbook (PDF)

Question 28 of 112

2024 Test A Math, question 12

Question

If n is a whole number, and 0.01 is between $\frac{1}{n}$ and $\frac{1}{n + 2}$ , what is the value of n?

Show the answer

Correct answer: D

$0.01 = \frac{1}{100}$ , which is between $\frac{1}{101}$ and $\frac{1}{99}$ .

Therefore, $n + 2 = 101$ and $n = 99$ .

Question analysis

A number sense question: 0.01 lies between 1 over n and 1 over n + 2, where n is a whole number, and students find n. It tests comparing a decimal with unit fractions and knowing how a fraction changes as its denominator grows. The unusual form, with n in two denominators, is what slows students down.

NYS standards: 6.NS.7a inequality as relative position
6.EE.5 solutions and substitution
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students compare fractions and decimals routinely, but not with an unknown in the denominators. The comparison is simple once 0.01 is written as a fraction; the unfamiliar setup is the main demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Variables instead of numbers moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 9, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The setup looks unfamiliar, with n in two denominators, but one rewrite makes it routine: 0.01 is 1/100, so the two denominators must sit on either side of 100. The three small choices fail a quick check, since every fraction from 1/1 to 1/4 is far above 0.01, and 0 cannot be a denominator at all. Students who test the choices instead of rewriting the decimal spend a little longer but reach the same place.

Variables instead of numbers moderate Understanding the question
The unknown sits in two denominators, n and n + 2, so students must reason about which unit fractions lie on either side of 0.01 rather than solve an equation. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Two facts are assumed: 0.01 equals 1/100, and a unit fraction shrinks as its denominator grows, so 1/99 is just above 0.01 and 1/101 just below. (Anderson & Krathwohl, 2001)

Strategy: Rewrite 0.01 as a unit fraction, recall that a larger denominator gives a smaller fraction, and test each choice in both 1/n and 1/(n + 2).

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats 1/0 as 0, so 0.01 seems to lie between 0 and 1/2; but division by zero is undefined, so n cannot be 0.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 1 from 0.01 or from 1/n as the value of n, without checking that 0.01 is far below both 1 and 1/3.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the 2 from n + 2, or the two decimal places of 0.01, without testing 1/2 and 1/4, which are both far above 0.01.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions: $\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}$; A unit fraction has 1 on top, like one fifth. a and b are positive whole numbers on the bottom. A bigger bottom cuts the whole into more pieces, so each piece is smaller. So the fraction with the bigger bottom is the smaller number.

Sort out the question
n is a whole number: 0, 1, 2, 3 and so on. The number 0.01 is between $\frac{1}{n}$ and $\frac{1}{n + 2}$ . Between means one fraction is bigger than 0.01 and the other is smaller. You need the value of n.
Write 0.01 as a fraction
0.01 means one hundredth.
$0.01 = \frac{1}{100}$
Now all three numbers are fractions with 1 on top. That makes them easy to compare.
Find fractions on each side
With 1 on top, a bigger bottom makes a smaller fraction.
$\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}$
So $\frac{1}{99}$ is a little more than $\frac{1}{100}$ , and $\frac{1}{101}$ is a little less.
$\frac{1}{101} < \frac{1}{100} < \frac{1}{99}$
The bottoms 99 and 101 are 2 apart, just like n and n + 2.
Match the fractions to n
The smaller bottom is n, and the larger bottom is n + 2.
$n = 99 \text{ and } n + 2 = 101$
Both fit. Watch out: small choices do not work. For n = 1, the fractions are 1 and $\frac{1}{3}$ . Both are far bigger than 0.01. For n = 0, $\frac{1}{0}$ is not a number, because you can’t divide by zero. So the answer is D, 99.

Check your answer

Write the fractions as decimals. $\frac{1}{99} \approx 0.0101$ , a bit more than 0.01. $\frac{1}{101} \approx 0.0099$ , a bit less. The sign ≈ means about. So 0.01 sits right between them.

More questions like this

The original DOE question

Question 69 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 69. Open page 78 of the handbook (PDF)

Question 29 of 112

2024 Test A Math, question 19

Question

8.9, 8.2, 8.5, 9.0, 8.4, 8.6, 8.8

At a skating championship, there are seven judges who each award a score for each skater’s performance. The highest and lowest scores given to each skater are discarded, and the mean of the remaining scores is then calculated and reported as the skater’s final score. What is the final score for the skater who received the scores shown above from the judges?

A8.60
B8.62
C8.64
D8.70

Show the answer

Correct answer: C

The highest and lowest scores are discarded: 8.2 and 9.0.

To find the mean, add the remaining scores and then divide that sum by the total number of remaining scores.

$8.9 + 8.5 + 8.4 + 8.6 + 8.8 = 43.2$

$43.2 \div 5 = 8.64$

Question analysis

A statistics question about a skating championship with seven judges: the highest and lowest of a skater's seven scores are dropped, and the rest are averaged. Students apply that rule to the listed scores, decimals close together. Following the dropping rule exactly and adding the decimals carefully both matter, since the choices are very close.

NYS standards: 6.SP.5c range, center, and overall pattern
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a mean is grade 6 content, and this adds only a rule about which scores to drop. Grade 7 statistics usually compares whole data sets, so this is easier, though the close choices demand exact decimal arithmetic.
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
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 62 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is short, but the choices are spaced so closely that a small slip or a skipped rule lands on a neighbor. Averaging all seven scores, or picking the middle score, gives values within a few hundredths of the answer. The question rewards following the stated rule exactly and adding the five remaining scores with care.

A very close wrong answer moderate Choosing the answer
The choices 8.60, 8.62 and 8.64 lie within 0.04 of one another, so no estimate separates them; only the exact sum of the five kept scores, 43.2, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Five one-decimal scores must be added and divided by 5; subtracting 8 from each first leaves 0.9, 0.5, 0.4, 0.6 and 0.8, which are easier to add. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The scoring rule takes two sentences to state: discard the single highest and the single lowest score, then average the five that remain. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Cross out the single highest and the single lowest score, add the five that remain, and divide that sum by 5.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: moderate
Gives the median, the middle score 8.6, instead of the mean of the five scores left after the highest and lowest are dropped.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Averages all seven scores without discarding any, 60.4 ÷ 7 = 8.628…, and takes 8.62, the nearest choice to that mean.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Drops only the lowest score, 8.2, and averages the other six, (60.4 − 8.2) ÷ 6 = 8.7, keeping the top score of 9.0.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean, or average: $\text{mean} = \frac{\text{sum of the scores}}{\text{number of scores}}$; The mean is the average. The sum is what you get when you add all the scores. Divide it by how many scores you added. It shares the total out evenly, as if every score were the same.

Sort out the rule
Seven judges gave the seven scores shown. The rule has two parts. First, throw out the highest score and the lowest score. Then find the mean of the scores that are left. The mean is the average. That is the final score you need.
Cross out the highest and lowest
The highest score is 9.0. The lowest is 8.2. Cross both out. Five scores are left: 8.9, 8.5, 8.4, 8.6 and 8.8. Watch out: drop both. Dropping only 8.2 gives 8.70, choice D. Keeping all seven gives about 8.63, which is closest to 8.62, choice B.
Add the five scores
$8.9 + 8.5 + 8.4 + 8.6 + 8.8 = 43.2$
A shortcut: each score is 8 plus a bit. The bits are 0.9, 0.5, 0.4, 0.6 and 0.8, and they add to 3.2. Five 8s make 40, and $40 + 3.2 = 43.2$ .
Divide by the number of scores
$\text{mean} = \frac{\text{sum of the scores}}{\text{number of scores}}$
$\text{mean} = \frac{43.2}{5} = 8.64$
Watch out: do not just take the middle score, 8.6. That is the median, choice A. The question asks for the mean. So the answer is C, 8.64.

Check your answer

Multiply back: $5 \times 8.64 = 43.2$ , the sum. Also, the mean should sit between the lowest kept score, 8.4, and the highest kept score, 8.9. 8.64 does.

More questions like this

The original DOE question

Question 76 as printed on page 80 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 76. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 30 of 112

2024 Test A Math, question 24

Question

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

A$1,440
B$14,400
C$144,000
D$14,400,000,000

Show the answer

Correct answer: B

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

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

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

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

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

Question analysis

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

NYS standards: 6.NS.2 multi-digit division
6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Dividing large whole numbers and using place value are grade 5 and 6 skills. Grade 7 work goes further with rational numbers; this asks only for one careful division, with counting zeros as the main risk.
Estimated difficulty: Straightforward: two steps or one careful reading.
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 minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 37 words.
Corrected from the DOE handbook: A misprinted number in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
The debt has 13 digits and the population 9; cancelling seven zeros from each leaves 360,000 ÷ 25, and losing or adding one zero shifts the answer tenfold. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
Three choices share the digits 144 and differ only by powers of ten, so getting the digits right does not settle it; only exact place value does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Cancels one zero too many while dividing and ends with $1,440, a per-person amount ten times too small.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Cancels one zero too few from the debt and ends with $144,000, a per-person amount ten times too large.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Divides the debt by 250 alone, dropping the six zeros that follow 250 in the population without removing any from the debt.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The debt was about $3,600,000,000,000, which is 3.6 trillion dollars. There were about 250,000,000 people, which is 250 million. You need the debt per person. Per person means the amount for each one person.
Set up the division
Share the debt equally by dividing it by the number of people. The debt goes on top, and the people go on the bottom.
$\text{per person} = \text{total} \div \text{number of people}$
$\text{per person} = \frac{3{,}600{,}000{,}000{,}000}{250{,}000{,}000}$
Cancel zeros in pairs
Count the zeros at the end of each number. The debt ends in 11 zeros. The population ends in 7. Cross off 7 zeros from each. Each pair you cross off divides the top and bottom by 10.
$\frac{360{,}000}{25}$
Watch out: cross off the same number from each. One too many from the debt gives $1,440, choice A. One too few gives $144,000, choice C.
Divide what is left
Dividing by 25 is the same as dividing by 100 and then multiplying by 4. That is because 4 groups of 25 make 100.
$360{,}000 \div 100 = 3{,}600$
$3{,}600 \times 4 = 14{,}400$
The debt was about $14,400 per person. Watch out: choice D divides by 250 only and forgets the zeros after it. So the answer is B, $14,400.

Check your answer

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

More questions like this

The original DOE question

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

Question 31 of 112

2024 Test A Math, question 30

Question

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

A $\frac{3}{13}$
B $\frac{3}{4}$
C $\frac{4}{5}$
D $\frac{13}{4}$

Show the answer

Correct answer: A

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

$\frac{4}{1} + \frac{1}{3} = \frac{12}{3} + \frac{1}{3} = \frac{13}{3}$

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

Question analysis

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

NYS standards: 6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students use reciprocals when dividing fractions, usually one at a time. This item chains three reciprocals around one addition, with positive numbers only, so the demand is following the wording in order rather than new content.
Estimated difficulty: Straightforward: two steps or one careful reading.
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: 3 steps
Reading level: Question and choices: about grade 8, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is easy, so the question is about order: flip each number, add, then flip the total. The common failures are flipping at the wrong moment, term by term instead of the whole sum, and adding a whole number to a fraction straight across. Writing the full expression 1 ÷ (4 + 1/3) before doing any arithmetic keeps the three reciprocals in their places.

Turning words into math moderate Understanding the question
The two sentences hold three reciprocals and a sum, which must become 1 ÷ (4 + 1/3) before any arithmetic; "this sum" means flipping the total, not each part. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Adding 4 and 1/3 means writing 4 as 12/3 to get 13/3, a whole number plus a fraction, where adding tops and bottoms gives 5/4 instead. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: weak
Multiplies the two reciprocals instead of adding them, 4 × 1/3 = 4/3, then flips that product to 3/4.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Adds 4/1 and 1/3 straight across, numerators and denominators, getting 5/4, then flips it to 4/5.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Flips each term of 4 + 1/3 instead of the whole sum, as if the reciprocal of a sum were the sum of reciprocals: 1/4 + 3 = 13/4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the wording
The question has three reciprocals in it. A reciprocal is a fraction flipped upside down. Do the jobs in the order of the words:
1. Flip $\frac{1}{4}$ .
2. Flip 3.
3. Add the two results to get the sum.
4. Flip that sum.
The last flip gives the answer.
Flip each number
$\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$
Flip $\frac{1}{4}$ to get $\frac{4}{1}$ , which is 4. The number 3 is the same as $\frac{3}{1}$ , so its reciprocal is $\frac{1}{3}$ .
Add the two reciprocals
You need $4 + \frac{1}{3}$ . To add, both numbers need the same bottom. So write 4 in thirds. One whole is 3 thirds, so 4 wholes are 12 thirds.
$\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$
$\frac{12}{3} + \frac{1}{3} = \frac{13}{3}$
The sum is $\frac{13}{3}$ . Watch out: don’t add the tops and the bottoms straight across. That gives $\frac{5}{4}$ , which flips to $\frac{4}{5}$ , choice C.
Flip the sum
$\text{reciprocal of } \frac{13}{3} = \frac{3}{13}$
Watch out: flip the whole sum, not each part. Flipping 4 and $\frac{1}{3}$ one at a time takes you back to $\frac{1}{4} + 3 = \frac{13}{4}$ , choice D.
So the answer is A, $\frac{3}{13}$ .

Check your answer

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

More questions like this

The original DOE question

Question 87 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 87. Open page 83 of the handbook (PDF)

Question 32 of 112

2024 Test A Math, question 32

Question

INGREDIENTS FOR 4 SERVINGS OF OATMEAL

Ingredient	Cups
Oats	$\frac{2}{3}$
Water	$3\frac{1}{4}$

The table shows the cups of ingredients used to make 4 servings of oatmeal. What is the unit rate for cups of oats per cup of water?

A $\frac{1}{6}$
B $\frac{8}{39}$
C $\frac{13}{16}$
D $\frac{13}{8}$

Show the answer

Correct answer: B

The unit rate for cups of oats per cup of water can be found by dividing the number of cups of oats by the number of cups of water:

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

The denominator can be rewritten as $\frac{13}{4}$ .

The quotient $\frac{\frac{2}{3}}{\frac{13}{4}}$ can be rewritten as:

$\frac{2}{3} \times \frac{4}{13} = \frac{8}{39}.$

Therefore, the unit rate for cups of oats per cup of water is $\frac{8}{39}$ .

Question analysis

This unit rate question uses a recipe table for 4 servings of oatmeal, with 2/3 cup of oats and 3 1/4 cups of water, and asks for cups of oats per cup of water. Students divide one fraction by another, including a mixed number. Deciding which quantity goes on top, and whether the number of servings matters, is the tricky part.

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Unit rates from fractions, such as 1/2 mile in 1/4 hour, are exactly grade 7 work. This one adds a mixed number, a fraction divided by a fraction, and a table titled for 4 servings, so students must decide which numbers the rate uses.
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 moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question rewards reading the rate's wording exactly and then carrying out a fraction division without slips. The 4 servings in the title is the hook: dividing by 4 gives a per-serving amount that sounds like a unit rate and is listed. Students who write oats over water as one fraction, convert the mixed number and multiply by the reciprocal finish in a few lines.

Error-prone arithmetic moderate Working it out
The rate is a fraction over a mixed number, 2/3 ÷ 3 1/4, so 3 1/4 must become 13/4 and the division a multiplication by 4/13. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math moderate Understanding the question
In "cups of oats per cup of water" the quantity before "per" goes on top; reading the phrase loosely puts water over oats or brings in servings. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The table's title, "for 4 servings", supplies the only other number on the page, and it plays no part in a rate of oats to water. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Write the rate as cups of oats over cups of water, change the mixed number to an improper fraction, and divide by multiplying by its reciprocal.

Traps

Information you don't need: The question gives numbers or facts that play no part in the answer.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the 2/3 cup of oats by the 4 servings, 2/3 ÷ 4 = 1/6, giving oats per serving instead of oats per cup of water.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the 3 1/4 cups of water by the 4 servings, 13/4 ÷ 4 = 13/16, leaving the oats out altogether.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Turns the rate upside down, water over oats, and also drops the 3 of 2/3, dividing 13/4 by 2 to get 13/8.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{unit rate} = \frac{\text{first amount}}{\text{second amount}}$; A unit rate tells how much of the first amount goes with 1 of the second amount. The word per means for each. The amount named before per goes on top, and the amount after it goes on the bottom.
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.
Dividing by a fraction: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. This works because dividing by a number undoes multiplying by it.

Sort out the question
The table shows $\frac{2}{3}$ cup of oats and $3\frac{1}{4}$ cups of water. You need the unit rate for “cups of oats per cup of water.” That is how many cups of oats go with 1 cup of water.
Watch out: the 4 servings play no part here. Dividing the oats by 4 servings gives $\frac{1}{6}$ cup for each serving, choice A.
Set up the rate
Oats comes before per, so oats goes on top. Water goes on the bottom.
$\text{unit rate} = \frac{\text{first amount}}{\text{second amount}}$
$\text{unit rate} = \frac{2}{3} \div 3\frac{1}{4}$
Watch out: water over oats flips the rate. That gives a number bigger than 1, like choice D.
Change the mixed number
A mixed number is a whole number plus a fraction. Change $3\frac{1}{4}$ to a fraction so you can divide.
$a\frac{b}{c} = \frac{a \times c + b}{c}$
$3\frac{1}{4} = \frac{3 \times 4 + 1}{4} = \frac{13}{4}$
So 3 and one quarter cups is 13 quarters of a cup.
Divide by multiplying
To divide by $\frac{13}{4}$ , multiply by its reciprocal, $\frac{4}{13}$ .
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
$\frac{2}{3} \div \frac{13}{4} = \frac{2}{3} \times \frac{4}{13} = \frac{2 \times 4}{3 \times 13} = \frac{8}{39}$
No whole number bigger than 1 divides both 8 and 39, so the fraction can’t be made simpler. The rate is $\frac{8}{39}$ cup of oats per cup of water.
So the answer is B, $\frac{8}{39}$ .

Check your answer

There is much more water than oats. So the oats for 1 cup of water should be small, and $\frac{8}{39}$ is about 0.2. Also multiply back: $\frac{8}{39} \times \frac{13}{4} = \frac{104}{156} = \frac{2}{3}$ . That is the cups of oats in the table.

More questions like this

The original DOE question

Question 89 as printed on page 84 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 84, question 89. Open page 84 of the handbook (PDF)

Question 33 of 112

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 34 of 112

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 35 of 112

2024 Test A Math, question 38

Question

By what percent did the price of a cup of coffee increase if its price was increased from $1.25 to $1.35?

A7%
B8%
C10%
D12%

Show the answer

Correct answer: B

Subtract to determine the amount of the increase.

$1.35 - 1.25 = 0.10$

Divide the increase by the original price.

$0.10 \div 1.25 = 0.08$

Multiply the quotient by 100 to express the decimal answer as a percent.

$0.08 \times 100 = 8$

Question analysis

This percent increase question gives the price of a cup of coffee before and after a raise, from $1.25 to $1.35, and asks by what percent the price went up. Students find the change in price and compare it with a price. Deciding which price to compare the change with is the tricky part.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Percent increase is grade 7 content, and this is a standard case with two prices. What makes it harder than routine is deciding which price the change is measured against and dividing small decimals accurately.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
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 moderate
Solution: 3 steps
Reading level: Question and choices: about grade 6, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Most students see that the price rose by 10 cents; the question is what to compare that with. Percent change is measured from the original amount, and dividing by the new price gives a number close enough to the answer that rounding hides the mistake. Reading the 10 cents directly as 10 percent is the other quick trap, since 10 cents is only 10 percent of a dollar.

A very close wrong answer moderate Choosing the answer
Dividing the 10-cent rise by either price gives a single-digit percent, 8% from $1.25 and about 7.4% from $1.35, so only the choice of base separates the two listed answers. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic moderate Working it out
Without a calculator, 0.10 ÷ 1.25 is a division by a decimal; rewriting it in cents as 10/125, which simplifies to 8/100, is the clean way through. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract to get the increase, then divide it by the price before the raise and multiply by 100; working in cents keeps the decimals out of the way.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Divides the 10-cent increase by the new price, $1.35, measuring from the wrong amount; 0.10 ÷ 1.35 is about 7.4%, so 7%.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads the 10-cent increase as 10 percent, as if every percent change were a percent of one dollar rather than of the $1.25 price.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Divides the wrong way round, $1.25 ÷ $0.10 = 12.5, and takes the nearest choice, 12%.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$; New is the price after the change, and original is the price before it. The top is the amount of the increase. Dividing by the original compares the increase with where you started. Multiplying by 100 turns that decimal into a percent.

Sort out the question
The coffee cost $1.25 before. That is the original price. Now it costs $1.35, the new price. You need the percent increase. That compares how much the price went up with the original price.
Find the amount of the increase
$1.35 - 1.25 = 0.10$
The price went up $0.10, or 10 cents. Watch out: 10 cents is not 10%. It would be 10% only if the price started at $1.00. That mistake gives choice C.
Divide by the original price
$\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$
$\frac{1.35 - 1.25}{1.25} \times 100 = \frac{0.10}{1.25} \times 100$
Dividing by a decimal is easier in cents: 10 cents out of 125 cents. Divide the top and bottom by 5. Then multiply both by 4 to get hundredths.
$\frac{10}{125} = \frac{2}{25} = \frac{8}{100} = 0.08$
Change to a percent
$0.08 \times 100 = 8$
The price went up 8%. Watch out: divide by the original price, $1.25, not the new one. Dividing by $1.35 gives about 7.4%, which rounds to choice A.
So the answer is B, 8%.

Check your answer

Work forward from the answer. 8% of $1.25 is $0.08 \times 1.25 = 0.10$ , or 10 cents. Then $1.25 + 0.10 = 1.35$ . That is the new price, so 8% is right.

More questions like this

The original DOE question

Question 95 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 86, question 95. Open page 86 of the handbook (PDF)

Question 36 of 112

2024 Test A Math, question 40

Question

The price of a sandwich was raised from $6.25 to $6.75. What was the percent increase in the price?

A5%
B8%
C7%
D50%

Show the answer

Correct answer: B

Use the percent increase expression to find the answer:

$\frac{\text{new price} - \text{original price}}{\text{original price}}$

$\frac{6.75 - 6.25}{6.25} = \frac{0.50}{6.25} = 0.08 = 8\%$

Question analysis

A sandwich's price went up from $6.25 to $6.75, and this percent change question asks for the percent increase in the price. Students divide the 50-cent increase by one of the prices and write the result as a percent. Picking which price is the base, and doing the decimal division accurately, decide whether the answer comes out right.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a percent increase is a core grade 7 skill, and this is the plain version with no context beyond two prices. Its demand lies in measuring the change against the correct price and dividing by $6.25 without a calculator.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
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 moderate
Solution: 3 steps
Reading level: Question and choices: about grade 2, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The 50-cent rise is easy to find; the percent depends on what it is compared with. Dividing by the new price, $6.75, gives a value that rounds to a listed choice one point below the answer, and reading 50 cents as 50 percent gives another. The choices are printed out of order, 8% before 7%, so students should read each one rather than scan the list.

A very close wrong answer moderate Choosing the answer
Measured against $6.75 the 50-cent rise is about 7.4%, and against $6.25 it is 8%; both are listed side by side, so only the right base separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic moderate Working it out
The division 0.50 ÷ 6.25 is easiest as 50/625, which reduces to 2/25 and then to 8/100; a three-digit decimal divisor invites a slip. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Put the 50-cent change over the original price, 0.50 over 6.25, then scale the fraction to hundredths or divide to read off the percent.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 5%: dividing the 50-cent rise by either price does not come near it, so it mostly draws guesses.
C Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Divides the 50-cent increase by the new price, $6.75, instead of the original, which gives about 7.4%, so 7%.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads the 50-cent increase as 50 percent, treating cents as a percent of one dollar rather than of the $6.25 price.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$; New is the price after the change, and original is the price before it. The top is the amount of the increase. Dividing by the original compares the increase with where you started. Multiplying by 100 turns that decimal into a percent.

Sort out the question
The sandwich cost $6.25 before. That is the original price. Now it costs $6.75, the new price. You need the percent increase. That compares the increase with the original price. Note that the choices are not in order: C, 7%, comes after B, 8%.
Find the amount of the increase
$6.75 - 6.25 = 0.50$
The price went up $0.50, or 50 cents. Watch out: 50 cents is not 50%. It would be 50% only if the price started at $1.00. That mistake gives choice D.
Divide by the original price
$\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$
$\frac{6.75 - 6.25}{6.25} \times 100 = \frac{0.50}{6.25} \times 100$
In cents, this is 50 out of 625. Divide the top and bottom by 25. Then multiply both by 4 to get hundredths.
$\frac{50}{625} = \frac{2}{25} = \frac{8}{100} = 0.08$
Change to a percent
$0.08 \times 100 = 8$
The price went up 8%. Watch out: divide by the original price, $6.25, not the new one. Dividing by $6.75 gives about 7.4%, which is choice C.
So the answer is B, 8%.

Check your answer

Work forward from the answer. 8% of $6.25 is $0.08 \times 6.25 = 0.50$ , or 50 cents. Then $6.25 + 0.50 = 6.75$ . That is the new price, so 8% is right.

More questions like this

The original DOE question

Question 97 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 86, question 97. Open page 86 of the handbook (PDF)

Question 37 of 112

2024 Test A Math, question 42

Question

Lian bought enough oranges to fill 4 bags. Each bag contains 8 oranges. The total cost was $11.52. At that rate, how much would Lian pay for 42 oranges?

A$17.28
B$15.12
C$15.02
D$12.52

Show the answer

Correct answer: B

First, find the amount, in dollars, that Lian paid for each orange. There are 4 bags with 8 oranges each.

$4 \times 8 = 32$

Divide the total cost by the total number of oranges to get the cost per orange:

$11.52 \div 32 = 0.36$

Multiply to find the cost of 42 oranges:

$0.36 \times 42 = 15.12$

Question analysis

This rate word problem says Lian paid $11.52 for 4 bags of 8 oranges each and asks how much 42 oranges would cost at the same rate. Students find the price of one orange and multiply. Since 42 oranges is not a whole number of bags, working by the bag instead of by the orange invites mistakes.

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a unit price and scaling it up is grade 6 rate work. The decimals are multi-digit, but the plan is routine, with less to manage than grade 7 problems built on percents or 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: Error-prone arithmetic moderate
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 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 plan is familiar, total oranges, price per orange, then the new cost, so the work lies in the decimals. Because 42 oranges is five bags and two oranges, students who price by the bag either round up to six bags or must handle the leftover. Estimating, since 42 is about a third more than 32, gives roughly $15 and rules out the far choices but not the near one.

Error-prone arithmetic moderate Working it out
Two decimal operations are done by hand: 11.52 ÷ 32 = 0.36 for one orange, then 0.36 × 42 = 15.12, where a dropped carry is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The choices $15.12 and $15.02 differ by 10 cents, so a rough estimate of about $15 cannot choose between them; only the exact product can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Multiply to find how many oranges were bought, divide the total cost by that number to get the price of one orange, then multiply by 42.

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Prices whole bags, rounding 42 oranges up to 6 bags of 8, and pays 6 × $2.88 = $17.28, missing that "at that rate" means per orange.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Uses the right method but slips one digit in multiplying 0.36 by 42, a tens-place error that lands 10 cents short of the total.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives $12.52, exactly one dollar more than $11.52; no price per orange or per bag leads there, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit price: $\text{unit price} = \frac{\text{total cost}}{\text{number of items}}$; The unit price is what one item costs. Divide the total cost by how many items you got. It works because the total is shared equally among the items.
Cost: $\text{cost} = \text{unit price} \times \text{number of items}$; The unit price is the cost of one item, here one orange. Multiply it by how many items you buy to get the total cost. This works when every item costs the same.

Sort out the question
Lian bought 4 bags with 8 oranges in each bag. She paid $11.52 in all. You need the cost of 42 oranges. “At that rate” means each orange costs the same as before. So find the price of one orange first.
Count the oranges
Each of the 4 bags holds 8 oranges, so multiply to find how many she bought.
$4 \times 8 = 32$
Lian bought 32 oranges for $11.52.
Find the price of one orange
$\text{unit price} = \frac{\text{total cost}}{\text{number of items}}$
$\text{unit price} = \frac{11.52}{32} = 0.36$
This is easier in cents: $1{,}152 \div 32 = 36$ . So one orange costs $0.36, or 36 cents.
Find the cost of 42 oranges
$\text{cost} = \text{unit price} \times \text{number of items}$
$\text{cost} = 0.36 \times 42$
Break 42 into 40 and 2 to make it easier.
$0.36 \times 40 = 14.40$
$0.36 \times 2 = 0.72$
$14.40 + 0.72 = 15.12$
The 42 oranges cost $15.12.
Avoid the traps and answer
Watch out: $15.02, choice C, is only 10 cents away. Multiply and add with care. Watch out: Lian pays for each orange, not for full bags. Buying 6 whole bags would cost $17.28, choice A.
So the answer is B, $15.12.

Check your answer

42 oranges is 10 more than 32. Those 10 oranges cost $3.60, since $10 \times 0.36 = 3.60$ . Then $11.52 + 3.60 = 15.12$ . Same answer, so $15.12 is right.

More questions like this

The original DOE question

Question 99 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 87, question 99. Open page 87 of the handbook (PDF)

Question 38 of 112

2024 Test A Math, question 43

Question

$3.6 \div 0.018 =$

A0.005
B0.648
C20
D200

Show the answer

Correct answer: D

$\frac{3.6}{0.018} \times \left(\frac{1{,}000}{1{,}000}\right) = \frac{3{,}600}{18} = 200$

Question analysis

This decimal division question asks for 3.6 divided by 0.018, with no context. Students rewrite the division so the divisor is a whole number and then divide. Because the divisor has three decimal places, moving the decimal point the right number of places in both numbers is where students go wrong.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Dividing multi-digit decimals is a grade 6 fluency standard, and this is a single division with no context. It is easier than typical grade 7 work, though shifting three decimal places invites a place-value slip.
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: 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

This is a single division, and the only real hazard is the decimal point. Shifting it three places in the divisor but only two in the dividend gives a listed answer ten times too small. An estimate settles it: 0.018 is about 0.02, and 3.6 ÷ 0.02 = 180, so the answer must be in the hundreds.

Error-prone arithmetic moderate Working it out
The divisor 0.018 has three decimal places, so both numbers must be multiplied by 1,000, turning 3.6 into 3,600, not 360, before dividing by 18. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply both numbers by 1,000 so the divisor becomes a whole number, then divide; estimate first by asking roughly how many 0.02s fit into 3.6.

Traps

An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
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: weak
Divides the wrong way round, 0.018 ÷ 3.6 = 0.005, putting the smaller number on top as if division always shrinks.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Multiplies instead of dividing, 36 × 18 = 648, and misplaces the decimal point too, since 3.6 × 0.018 is really 0.0648.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Moves the decimal point two places instead of three, dividing 360 by 18 to get 20, a result ten times too small.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scaling a division: $a \div b = (a \times n) \div (b \times n)$; a is the number being divided, and b is the number you divide by. n is any number except 0. Multiplying both numbers by the same n keeps the answer the same, just like an equal fraction. Use it to turn a decimal you divide by into a whole number.

Sort out the problem
You need $3.6 \div 0.018$ . The number you divide by, 0.018, has three decimal places. That means three digits after the decimal point. Dividing by a decimal is hard, so first turn 0.018 into a whole number.
Multiply both numbers by 1,000
Multiplying by 1,000 moves the decimal point 3 places to the right. Do it to both numbers, so the answer does not change.
$a \div b = (a \times n) \div (b \times n)$
$3.6 \div 0.018 = (3.6 \times 1{,}000) \div (0.018 \times 1{,}000) = 3{,}600 \div 18$
Watch out: 3.6 has only one decimal place, so fill in zeros: 3.6 becomes 3,600. Moving its point only two places gives $360 \div 18 = 20$ , choice C.
Divide the whole numbers
$3{,}600 \div 18 = 200$
This works because $18 \times 2 = 36$ , so $18 \times 200 = 3{,}600$ . Watch out: divide 3.6 by 0.018, not the other way around. Dividing 0.018 by 3.6 gives 0.005, choice A.
So the answer is D, 200.

Check your answer

Multiply back: $200 \times 0.018 = 3.6$ . Also estimate. 0.018 is close to 0.02, and $3.6 \div 0.02 = 180$ . The answer should be near that, and 200 is. Dividing by a number less than 1 gives a bigger answer, so 200 makes sense.

More questions like this

The original DOE question

Question 100 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 87, question 100. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 39 of 112

2024 Test A Math, question 45

Question

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

A$4.00
B$6.25
C$11.25
D$12.50

Show the answer

Correct answer: D

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

$1\frac{1}{4} \times 5 = \frac{5}{4} \times \frac{5}{1} = \frac{25}{4} = 6.25$ , so the ribbon for each strap costs $6.25.

Misha wants to make 2 straps.

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

Question analysis

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

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Multiplying a mixed number by a whole number and finding a total cost are grade 5 and 6 skills. With two short steps and friendly numbers, this is easier than typical grade 7 problems with percents or negatives.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic minor Working it out
The length per strap is a mixed number, 1 1/4 yards, so the arithmetic is easiest as 2 × 1.25 = 2.5 yards, then 2.5 × $5.00. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Much to keep track of minor Working it out
The number of straps comes in the first sentence and the length of each in the last, so the 2 has to be carried through to the cost. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

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

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides the price by the length, $5.00 ÷ 1.25 = 4, instead of multiplying yards by dollars per yard.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Prices the ribbon for one strap, 1 1/4 × $5.00 = $6.25, and forgets that Misha is making 2 straps.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Doubles only the whole-number part of 1 1/4, getting 2 1/4 yards instead of 2 1/2, and pays 2.25 × $5.00 = $11.25.

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

Sort out the question
Misha makes 2 straps. Each strap needs $1\frac{1}{4}$ yards of ribbon. A yard is a length of 3 feet. Ribbon costs $5.00 a yard. You need the total cost for both straps, with no tax.
Find the total ribbon
First change the mixed number to a decimal. One quarter is 0.25, so $1\frac{1}{4}$ is 1.25. Then double it for 2 straps.
$2 \times 1.25 = 2.5$
Misha needs 2.5 yards of ribbon. Watch out: double the whole mixed number, the 1 and the quarter. Doubling only the 1 gives $2\frac{1}{4}$ yards and $11.25, choice C.
Find the cost
$\text{cost} = \text{price per unit} \times \text{number of units}$
$\text{cost} = 5 \times 2.5 = 12.5$
The ribbon costs $12.50. Watch out: this is for both straps. One strap’s ribbon costs $6.25, choice B. Don’t stop there.
So the answer is D, $12.50.

Check your answer

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

More questions like this

The original DOE question

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

Question 40 of 112

2024 Test A Math, question 50

Question

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

A $\frac{5}{12}$
B $3\frac{1}{4}$
C $3\frac{3}{4}$
D $4\frac{1}{4}$

Show the answer

Correct answer: C

Set up a proportion to solve.

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

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

$\frac{5}{4} = \frac{1}{3}x$

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

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

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

Question analysis

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

NYS standards: 7.RP.1 unit rates with fractions
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Computing a unit rate from a ratio of fractions is the grade 7 standard itself, and this is a typical example: one mixed number over one unit fraction, in two short steps with no extra information.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 25 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic minor Working it out
Dividing 1 1/4 by 1/3 means multiplying by 3, and the mixed number must be multiplied whole, as 5/4 × 3 = 15/4, not just its 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
Multiplying only the whole part gives 3 1/4, which differs from the answer only in the fraction's numerator, so a rough size check cannot rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies 5/4 by 1/3 instead of dividing by it, getting 5/12, a yearly rate smaller than the growth in a third of a year.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Triples only the whole-number part of 1 1/4, writing 3 × 1 1/4 as 3 1/4 and leaving the 1/4 unchanged.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds 3 to 1 1/4 instead of multiplying it by 3, getting 4 1/4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rate: $\text{rate} = \text{amount} \div \text{time}$; The amount is how much something changed, here inches grown. The time is how long it took, here part of a year. Dividing tells how much it changes in 1 unit of time, here 1 whole year.
Mixed number to fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; 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.
Dividing by a fraction: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. This works because dividing by a number undoes multiplying by it.

Sort out the question
The child grows $1\frac{1}{4}$ inches in $\frac{1}{3}$ of a year. You need his growth rate in inches per year. That is how much he would grow in 1 whole year at the same speed. A year has 3 thirds, so the answer should be bigger than $1\frac{1}{4}$ .
Set up the rate
Divide the growth by the time to get the inches for 1 whole year.
$\text{rate} = \text{amount} \div \text{time}$
$\text{rate} = 1\frac{1}{4} \div \frac{1}{3}$
Change the mixed number to a fraction first.
$a\frac{b}{c} = \frac{a \times c + b}{c}$
$1\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{5}{4}$
Divide by multiplying
To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. The reciprocal of $\frac{1}{3}$ is 3.
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
$\frac{5}{4} \div \frac{1}{3} = \frac{5}{4} \times \frac{3}{1} = \frac{15}{4}$
Watch out: multiplying by $\frac{1}{3}$ instead gives $\frac{5}{12}$ , choice A. That is less than he grew in a third of a year.
Change back to a mixed number
15 quarters is 3 wholes (12 quarters) and 3 quarters more.
$\frac{15}{4} = 3\frac{3}{4}$
He grows $3\frac{3}{4}$ inches per year. Watch out: triple all of $1\frac{1}{4}$ , not just the 1. Tripling only the whole number gives $3\frac{1}{4}$ , choice B.
So the answer is C, $3\frac{3}{4}$ .

Check your answer

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

More questions like this

The original DOE question

Question 107 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 89, question 107. Open page 89 of the handbook (PDF)

Question 41 of 112

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 42 of 112

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 43 of 112

2024 Test B Math, question 3

Question

$Scale drawing of a garden shaped like a rectangle with its lower right corner cut out; all six sides are horizontal or vertical. The top side is labeled 13 cm and the left side 6 cm. From the top right corner, the right side, labeled 3 cm, goes down to an unlabeled horizontal side running left, then an unlabeled vertical side runs down to the bottom side, which is labeled 8 cm.$

The figure above shows a scale drawing of a garden, where 1 centimeter represents 2.5 meters. What is the perimeter of the actual garden in meters?

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

Show the answer

Correct answer: 95

First, find the perimeter of the figure.

Two sides do not have labels. The top of the figure measures 13 cm and the bottom shows a length measuring 8 cm, so the horizontal side without a label measures 5 cm $(13-8)$ .

The left side of the figure measures 6 cm and the right side shows a length measuring 3 cm, so the vertical side without a label measures 3 cm $(6-3)$ .

So the total perimeter of the scale drawing is

$13+3+5+3+8+6=38\text{ cm}$

Use the scale to find the perimeter of the actual garden:

$\frac{1\text{ cm}}{2.5\text{ m}}=\frac{38\text{ cm}}{p\text{ m}}$

$p=2.5\times 38=95$

Question analysis

A grid-in on a scale drawing of an L-shaped garden, where each centimeter stands for 2.5 meters and two sides carry no labels. Students find the perimeter of the drawing, then convert it to the real garden's perimeter in meters. The unlabeled sides are easy to leave out.

NYS standards: 7.G.1 scale drawings
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings are grade 7 content, and a typical task converts one or two given lengths. Here two missing sides must be worked out and the perimeter assembled before the scale applies, more steps but nothing 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: Reading a figure, table or graph moderate
Combines ideas minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 9, 24 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The figure hides the work: it labels four of the six sides, and the missing two are short pieces of the notch that are easy to overlook. Once all six are found, the scale step is a single multiplication, but it must go the right way, since the real garden is larger than the drawing. Tracing the outline and counting six sides guards against the most common loss.

Reading a figure, table or graph moderate Understanding the question
Two sides of the notch carry no labels, so students must find them from the opposite sides: 13 − 8 = 5 cm across and 6 − 3 = 3 cm down. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas minor Working it out
The perimeter of an L-shaped drawing has to be joined to a scale of 2.5 meters per centimeter, two separate skills in one answer. (Embretson, 1998; Daniel & Embretson, 2010)
An answer in an unexpected form minor Choosing the answer
With no choices, nothing flags a perimeter that is missing two sides or still in centimeters; the student must check the six-sided total alone. (Enright, Morley & Sheehan, 2002)

Strategy: Use the opposite sides to work out the two unlabeled lengths, add all six sides of the drawing, then multiply the total in centimeters by 2.5 meters per centimeter.

Traps

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

75 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: strong
Adds only the four labeled sides, 13 + 3 + 8 + 6 = 30 cm, and scales that to 75 m, missing the two unlabeled sides of the notch.
38 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the drawing's perimeter, 38 cm, correctly and grids it without applying the scale of 2.5 meters per centimeter.
15.2 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides the 38 cm perimeter by 2.5 instead of multiplying, shrinking the garden though each centimeter of the drawing stands for 2.5 meters.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter: $P = \text{sum of all the side lengths}$; P is the perimeter, the distance all the way around a shape. Add the length of every side, once each. It works for any shape with straight sides, even one with a corner cut out.
Using a scale: $\text{actual length} = \text{drawing length} \times s$; s is the scale, the number of meters that 1 centimeter on the drawing stands for. Here s is 2.5. Each centimeter on the drawing is 2.5 meters in real life, so you multiply.

Sort out what you know
The figure is a scale drawing, a small copy of the real garden. Each 1 cm on the drawing stands for 2.5 m in the real garden. You need the perimeter of the real garden in meters. The perimeter is the distance all the way around.
The shape has 6 sides. Four are labeled: 13 cm, 3 cm, 8 cm and 6 cm. The two sides around the cut-out corner have no labels.
Find the missing side across
Look at the top side, 13 cm. Below it, the bottom side (8 cm) and the short unlabeled side across the notch together reach just as far. So the missing side is the difference.
$13 - 8 = 5$
The missing side across is 5 cm.
Find the missing side going down
Now look at the left side, 6 cm. On the right, the 3 cm side and the unlabeled side going down together make the same height.
$6 - 3 = 3$
The missing side going down is 3 cm.
Add all six sides
$P = \text{sum of all the side lengths}$
$P = 13 + 3 + 5 + 3 + 8 + 6 = 38$
The perimeter of the drawing is 38 cm. Watch out: adding only the four labeled sides gives 30 cm. That leads to the wrong answer 75.
Change to real meters
Each centimeter stands for 2.5 meters, so multiply by 2.5.
$\text{actual length} = \text{drawing length} \times s$
$\text{actual length} = 38 \times 2.5 = 76 + 19 = 95$
Here $38 \times 2 = 76$ and $38 \times 0.5 = 19$ . The real perimeter is 95 meters. Watch out: 38 is still in centimeters, so do not grid it. Dividing by 2.5 gives 15.2, which makes the garden smaller than its drawing.
So you would grid 95.

Check your answer

Slide the two notch sides out to the corner. The distance around stays the same, so it equals a 13 cm by 6 cm rectangle: $2 \times (13 + 6) = 38$ . Also, 95 meters is more than 38, which fits, because each centimeter stands for more than 1 meter.

More questions like this

The original DOE question

Question 60 as printed on page 198 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 198, question 60. Open page 198 of the handbook (PDF)

Question 44 of 112

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 45 of 112

2024 Test B Math, question 10

Question

$4\frac{2}{3}\div 2\frac{1}{2}=$

A $\frac{15}{28}$
B $1\frac{13}{15}$
C $2\frac{1}{3}$
D $3\frac{1}{3}$

Show the answer

Correct answer: B

$4\frac{2}{3}\div 2\frac{1}{2}=\frac{14}{3}\div\frac{5}{2}=$

$\frac{14}{3}\times\frac{2}{5}=\frac{28}{15}=1\frac{13}{15}$

Question analysis

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

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

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
Both mixed numbers must become improper fractions, 14/3 and 5/2, before the second is flipped, and 14/3 × 2/5 = 28/15 leaves an improper fraction to deal with. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The choices are written as mixed numbers and one proper fraction, so 28/15 has to be converted to a mixed number before it can be matched. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Flips the first fraction instead of the second, computing 3/14 × 5/2 = 15/28, which is 2 1/2 divided by 4 2/3.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Divides 4 2/3 by the whole number 2 and drops the 1/2 of the divisor, so 14/3 ÷ 2 = 7/3, or 2 1/3.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Divides the whole parts and the fraction parts separately, 4 ÷ 2 = 2 and 2/3 ÷ 1/2 = 4/3, then adds them to get 3 1/3.

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, and b over c is the fraction part. Multiply the whole number by c to turn the wholes into pieces. Then add the b extra pieces. The bottom number stays c.
Dividing fractions: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$; 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.

Sort out the problem
You need to divide $4\frac{2}{3}$ by $2\frac{1}{2}$ . Both are mixed numbers, a whole number and a fraction together. Mixed numbers are hard to divide as they are. So first change each one into an improper fraction. That is a fraction whose top is bigger than its bottom.
Change the mixed numbers
$a\frac{b}{c} = \frac{a \times c + b}{c}$
$4\frac{2}{3} = \frac{4 \times 3 + 2}{3} = \frac{14}{3}$
$2\frac{1}{2} = \frac{2 \times 2 + 1}{2} = \frac{5}{2}$
Watch out: do not divide the whole parts and the fraction parts on their own. Doing $4 \div 2 = 2$ and $\frac{2}{3} \div \frac{1}{2} = \frac{4}{3}$ , then adding, gives $3\frac{1}{3}$ , choice D.
Flip the second fraction and multiply
$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$
$\frac{14}{3} \div \frac{5}{2} = \frac{14}{3} \times \frac{2}{5} = \frac{28}{15}$
Multiply the tops, $14 \times 2 = 28$ , and the bottoms, $3 \times 5 = 15$ . Watch out: flip only the second fraction. Flipping the first one gives $\frac{15}{28}$ , choice A.
Change back to a mixed number
The choices are mixed numbers, so change 28/15 back. 15 goes into 28 one time, with 13 left over.
$\frac{28}{15} = 1\frac{13}{15}$
Watch out: dividing by just 2 and dropping the one half gives $2\frac{1}{3}$ , choice C.
So the answer is B, $1\frac{13}{15}$ .

Check your answer

Estimate. $4\frac{2}{3}$ is about 4.7, and $2\frac{1}{2}$ is 2.5. Two groups of 2.5 make 5, so the answer is a bit less than 2. $1\frac{13}{15}$ is about 1.9, which fits. Or multiply back: $\frac{28}{15} \times \frac{5}{2} = \frac{140}{30} = \frac{14}{3}$ .

More questions like this

The original DOE question

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

Question 46 of 112

2024 Test B Math, question 15

Question

Claire is running on a circular track. She has already run 650 meters. Each lap around the track is a distance of 120 meters. Her goal is to run at least 2,500 meters in all. How many more laps does she need to complete in order to reach her goal?

Show the answer

Correct answer: D

Claire’s goal is to run at least 2,500 meters, and she has already run 650 meters.

$2{,}500-650=1{,}850$

Claire has at least 1,850 meters left to run. Divide by the distance around the track to find the number of laps she needs to complete:

$1{,}850\div 120\approx 15.417$

Since she wants to run at least 2,500 meters, round 15.417 up to 16. Claire needs to run 16 laps to meet her goal.

Question analysis

A word problem about running laps: Claire has already run 650 meters on a 120-meter track and wants to run at least 2,500 meters in all. Students find how many more laps she must complete. Subtracting and dividing are straightforward; deciding what to do with the leftover part of a lap is the real test.

NYS standards: 7.EE.4b two-step inequalities
6.NS.2 multi-digit division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve word problems that lead to inequalities like px + q ≥ r and interpret the solution in context. This is a typical three-step example, where the solution must be turned into a whole number of laps.
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
A very close wrong answer moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic leads to about 15.4 laps, and the question is entirely about what to do with the fraction of a lap. Rounding to the nearest whole number is a habit that fails here, because the goal must be met or passed. Multiplying back, 15 × 120 = 1,800 and 16 × 120 = 1,920, shows which count actually covers the 1,850 meters still to run.

Turning words into math moderate Understanding the question
“At least 2,500 meters in all” means a part lap must still be run in full, so 1,850 ÷ 120 ≈ 15.4 has to be rounded up, against the usual rounding rule. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
15 and 16 laps are one apart and both sit next to the quotient 15.4, so estimating cannot decide between them; only the condition that the goal be reached can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Dividing 1,850 by 120 by hand does not come out even, and long division by a three-digit number invites slips in placing the quotient. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract what she has run from 2,500, divide by 120, and check your whole number by multiplying it back: those laps must cover all of the remaining distance.

Traps

Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the 650 meters already run by 120 and rounds down to 5, counting laps finished so far instead of laps still needed.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Divides the 650 meters already run by 120, about 5.4, and rounds up as the goal requires, but for the distance done rather than the distance left.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Finds 1,850 ÷ 120 ≈ 15.4 and rounds down to 15 by the usual rule, but 15 more laps leave her 50 meters short of the goal.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance left: $\text{left} = \text{goal} - \text{done}$; Goal is the distance she wants to reach in all. Done is the distance she has already run. Subtract to find how far she still has to go.
Number of laps: $\text{laps} = \text{distance} \div \text{one lap}$; Distance is how far she still has to run. One lap is the length of the track, 120 meters. Dividing tells how many laps fit into that distance. If it does not come out even, part of a lap is left over.

Sort out what you know
Claire has run 650 meters. One lap is 120 meters. Her goal is “at least 2,500 meters in all”. At least means 2,500 or more, so she must not fall short. You need how many more laps she must run.
Find the distance left
$\text{left} = \text{goal} - \text{done}$
$\text{left} = 2{,}500 - 650 = 1{,}850$
She still needs 1,850 meters. Watch out: divide the distance left, not the 650 meters she already ran. Dividing 650 by 120 leads to 5 or 6, choices A and B.
Divide by the length of a lap
$\text{laps} = \text{distance} \div \text{one lap}$
$1{,}850 \div 120 \approx 15.4$
The sign ≈ means about. Check with whole laps: $15 \times 120 = 1{,}800$ . That leaves $1{,}850 - 1{,}800 = 50$ meters still to go. So 15 laps are not quite enough.
Round up to reach the goal
She needs part of one more lap. She has to complete whole laps, so she runs a 16th lap. Here you round up, even though 15.4 would usually round down. Watch out: 15 laps leave her 50 meters short, so choice C is wrong.
So the answer is D, 16.

Check your answer

Add up her total. With 15 more laps: $650 + 1{,}800 = 2{,}450$ meters, which is short of 2,500. With 16 more laps: $650 + 16 \times 120 = 650 + 1{,}920 = 2{,}570$ meters. That reaches the goal.

More questions like this

The original DOE question

Question 72 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 72. Open page 202 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 47 of 112

2024 Test B Math, question 21

Question

$-\frac{1}{49},\ -49,\ -\frac{1}{7},\ -1$

Which number shown above is the greatest?

A $-\frac{1}{49}$
B $-49$
C $-\frac{1}{7}$
D $-1$

Show the answer

Correct answer: A

All the numbers are negative, so they are all less than zero. Consider the absolute value of each number and where it falls on a number line. The negative number closest to 0 will be the greatest.

The negative number with the greatest absolute value is −49, so it has the least value.

The negative number with the next greatest absolute value is −1, so it would fall next in order from least to greatest.

$-49,\ -1$

Compare the absolute values of $-\frac{1}{49}$ and $-\frac{1}{7}$ .

$\frac{1}{49}<\frac{1}{7}$

Therefore, the numbers listed from least to greatest are:

$-49,\ -1,\ -\frac{1}{7},\ -\frac{1}{49}$

So the number $-\frac{1}{49}$ is the greatest.

Question analysis

Four negative numbers are listed: negative 1/49, negative 49, negative 1/7 and negative 1. Students decide which of them is the greatest. Comparing negative fractions with negative whole numbers tests whether students order numbers by their position on the number line or by their size without the sign.

NYS standards: 6.NS.7a inequality as relative position
6.NS.7d absolute value versus order
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Ordering positive and negative rational numbers on a number line is grade 6 content in New York. This is a one-step comparison of four numbers, lighter than typical grade 7 work with signed numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
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
Solution: 2 steps
Reading level: Question and choices: about grade 10, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

With no arithmetic to do, the question tests how firmly students picture negative numbers on a line. Comparing sizes without the sign picks the number farthest from 0, and comparing the two fractions by habit picks the wrong one of the pair. Placing all four as points left of 0, with −1 between −49 and the fractions, makes the order plain.

A very close wrong answer moderate Choosing the answer
−1/49 and −1/7 differ only in the denominator, and for negative numbers the larger denominator gives the greater value, the reverse of what their sizes suggest. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know minor Working it out
Students must bring the number-line fact that a negative number's value falls as its distance from 0 grows, so −49 is the least of the four, not the greatest. (Anderson & Krathwohl, 2001)

Strategy: Picture each number's place on the number line: among negative numbers, the one closest to 0 is the greatest, so compare their distances from 0.

Traps

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

The wrong answers, classified

B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Picks the number with the largest digits, 49, judging its size without the minus sign that puts it farthest below 0.
C Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Knows 1/7 is greater than 1/49 and keeps that order for the negatives, missing that the order reverses on the left of 0.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Places −1/7 and −1/49 below −1, as if dividing a negative number made it more negative, so −1 looks like the greatest.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing negative numbers: $\text{if } a \lt b \text{, then } -a \gt -b$; a and b are positive numbers. Putting a minus sign on both flips their order. The one with the smaller size is closer to 0, so it is greater. For example, 2 is less than 5, but −2 is greater than −5.

Sort out the question
All four numbers are negative, which means less than 0. You need the greatest one, the one with the highest value. On a number line, greater numbers are farther to the right.
Think about distance from 0
For negative numbers, the one closest to 0 is the greatest. The size of a number without its sign is its absolute value. It is the number’s distance from 0. So look for the smallest size.
$\text{if } a \lt b \text{, then } -a \gt -b$
Compare the sizes
The sizes are 1/49, 49, 1/7 and 1. Both 1/49 and 1/7 have 1 on top. Cutting 1 into 49 pieces gives much smaller pieces than cutting it into 7. So 1/49 is less than 1/7.
$\frac{1}{49} \lt \frac{1}{7} \lt 1 \lt 49$
So 1/49 is the smallest size. That puts $-\frac{1}{49}$ closest to 0.
Put the numbers in order
Flip the order of the sizes to order the negative numbers, from least to greatest.
$-49 \lt -1 \lt -\frac{1}{7} \lt -\frac{1}{49}$
Watch out: −49 has the biggest digits, but it is the least, so choice B is wrong. Watch out: 1/7 is more than 1/49, but −1/7 is less than −1/49. So choice C is wrong.
So the answer is A, $-\frac{1}{49}$ .

Check your answer

Change the fractions to decimals: $-\frac{1}{49} \approx -0.02$ and $-\frac{1}{7} \approx -0.14$ . Think of money you owe. Owing 2 cents is better than owing 14 cents, $1 or $49. So −1/49 is the greatest.

More questions like this

The original DOE question

Question 78 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 205, question 78. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 48 of 112

2024 Test B Math, question 23

Question

Each loaf of bread a baker makes requires $\frac{1}{5}$ ounce of salt. How many loaves can be made if 40 ounces of salt are used?

A8
B20
C45
D200

Show the answer

Correct answer: D

Each loaf of bread requires $\frac{1}{5}$ ounce of salt. Therefore, the baker can make 5 loaves of bread with 1 ounce of salt. Multiply to find the number of loaves that can be made with 40 ounces of salt:

$5\times 40=200$

Question analysis

A baking word problem: each loaf of bread needs 1/5 ounce of salt, and the baker uses 40 ounces. Students find how many loaves that much salt makes, working with a whole number and a unit fraction. Deciding whether to multiply or divide is the whole question.

NYS standards: 6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Dividing a whole number by a fraction in a word problem is a grade 6 standard. This takes one step with a unit fraction, simpler than typical grade 7 rational number problems.
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: Turning words into math minor
Solution: 1 step
Reading level: Question and choices: about grade 3, 24 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The only decision is whether to multiply or divide, and thinking about one ounce first settles it: one ounce holds five fifths, so it makes five loaves. Students who multiply 40 by 1/5 get a small number, but a tiny amount of salt per loaf should give many loaves, not few. That reasonableness check leaves a single choice.

Turning words into math minor Understanding the question
“How many loaves” from 40 ounces at 1/5 ounce each asks how many fifths fit in 40, a division that many students turn into multiplying by 1/5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work out how many loaves one ounce of salt makes, then scale up to 40 ounces; dividing 40 by 1/5 gives the same count.

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).
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 40 by 1/5 instead of dividing, finding a fifth of the salt rather than how many fifths of an ounce it holds.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Knows to multiply 40 by 5 but drops a zero, writing the product as 20 instead of 200.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the 5 from 1/5 to the 40 ounces, combining the numbers in the problem without a model of the situation.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of equal groups: $\text{groups} = \text{total} \div \text{size of one group}$; Total is the whole amount you have. Size of one group is how much each one uses. Dividing tells how many groups fit into the total. Here one group is the salt for one loaf.
Dividing by a unit fraction: $a \div \frac{1}{n} = a \times n$; a is any number. 1 over n is a unit fraction, a fraction with 1 on top. Dividing by it asks how many nths fit into a. Each whole holds n of them, so you multiply by n.

Sort out what you know
Each loaf uses $\frac{1}{5}$ ounce of salt. The baker uses 40 ounces of salt. You need how many loaves that makes. That means how many fifths of an ounce fit into 40 ounces.
Decide to divide
You are splitting the salt into equal groups of 1/5 ounce. Each group makes one loaf. So divide the total salt by the salt for one loaf.
$\text{groups} = \text{total} \div \text{size of one group}$
$\text{loaves} = 40 \div \frac{1}{5}$
Watch out: do not multiply by 1/5. $40 \times \frac{1}{5} = 8$ , choice A, is a fifth of the salt, not a number of loaves.
Divide by the fraction
Each ounce holds 5 fifths, so 1 ounce makes 5 loaves. Dividing by 1/5 is the same as multiplying by 5.
$a \div \frac{1}{n} = a \times n$
$40 \div \frac{1}{5} = 40 \times 5 = 200$
The salt makes 200 loaves. Watch out: keep both zeros. 20, choice B, drops one.
So the answer is D, 200.

Check your answer

Multiply back. 200 loaves at $\frac{1}{5}$ ounce each use $200 \times \frac{1}{5} = 40$ ounces. That is all the salt. Also, each loaf uses less than 1 ounce, so there should be more loaves than ounces. 200 is more than 40.

More questions like this

The original DOE question

Question 80 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 80. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 49 of 112

2024 Test B Math, question 26

Question

If 1 sind = 0.75 plunk, how many sinds are equivalent to 8 plunks, rounded to the nearest hundredth?

A1.33
B6.00
C7.25
D10.67

Show the answer

Correct answer: D

Set up a proportion:

$\frac{1\text{ sind}}{0.75\text{ plunk}}=\frac{x\text{ sinds}}{8\text{ plunks}}$

$1(8)=0.75x$

$8=\frac{75}{100}x$

$8=\frac{3}{4}x$

$8\left(\frac{4}{3}\right)=x$

$\frac{32}{3}=x$

$10\frac{2}{3}=x$

$10.67\approx x$

Question analysis

A unit conversion question with made-up units: 1 sind equals 0.75 plunk, and students find how many sinds equal 8 plunks, rounded to the nearest hundredth. With unfamiliar unit names there is no everyday sense to lean on, so deciding whether to multiply or divide by 0.75 is the hard part.

NYS standards: 6.RP.3d converting measurement units
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Converting units with ratio reasoning is grade 6 content, and made-up unit names are a familiar SHSAT twist. Working with a decimal rate and rounding to hundredths put it at about the level of typical 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
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Made-up units remove the everyday sense that usually tells students which way a conversion goes, so the relationship must be reasoned out. The most tempting error multiplies by 0.75 and goes the wrong way; another finds the rate for one plunk and stops. A size check settles it: each sind is less than a plunk, so the number of sinds must exceed 8.

Turning words into math moderate Understanding the question
“1 sind = 0.75 plunk” says a sind is smaller than a plunk, so 8 plunks hold more than 8 sinds; seeing that decides between dividing and multiplying by 0.75. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
8 ÷ 0.75 is easiest as 8 × 4/3 = 32/3, and the repeating decimal 10.666… must then be rounded to the nearest hundredth. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Set up a proportion with sinds over plunks on both sides, put the 8 plunks in the matching place, solve for the sinds, and round to the nearest hundredth.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Stops one step early: A choice is a number from partway through the solution.
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 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds how many sinds make one plunk, 1 ÷ 0.75 ≈ 1.33, and stops before scaling up to 8 plunks.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 8 by 0.75 instead of dividing, converting in the wrong direction and finding the plunks in 8 sinds.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts 0.75 from 8, combining the two numbers in the question without setting up a conversion.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a, b, c and d are numbers or letters, and b and d are not zero. When two fractions are equal, multiply each top by the other bottom. The two products are equal. It works because it is the same as multiplying both sides by b and by d.
Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is any number, and b over c is the fraction you divide by. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down. It asks how many of that fraction fit into a.

Sort out what you know
Sinds and plunks are made-up units, like inches and feet. One sind equals 0.75 plunk, so a sind is smaller than a plunk. You need how many sinds equal 8 plunks. Round to the nearest hundredth, two places after the decimal point.
Decide if the answer is more than 8
A sind is only 3/4 of a plunk. So it takes more than 8 sinds to make 8 plunks. The answer must be more than 8. Watch out: $8 \times 0.75 = 6$ , choice B, goes the wrong way. That is the number of plunks in 8 sinds.
Set up a proportion
A proportion says that two ratios are equal. Put sinds on top and plunks on the bottom on both sides. Let x be the number of sinds.
$\frac{1}{0.75} = \frac{x}{8}$
$\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$
$1 \times 8 = 0.75 \times x$
So $0.75x = 8$ .
Divide by 0.75
0.75 is the same as $\frac{3}{4}$ . Divide both sides by it. To divide by a fraction, multiply by its reciprocal.
$a \div \frac{b}{c} = a \times \frac{c}{b}$
$x = 8 \div \frac{3}{4} = 8 \times \frac{4}{3} = \frac{32}{3} = 10.666\ldots$
Watch out: 1.33, choice A, is the number of sinds in just 1 plunk. Keep going to 8 plunks.
Round to the nearest hundredth
Look at the third digit after the point, the thousandths digit. It is 6, which is 5 or more. So round the hundredths digit up, from 6 to 7.
$10.666\ldots \approx 10.67$
So the answer is D, 10.67.

Check your answer

Multiply back. Each sind is 0.75 plunk, so 10.67 sinds are $10.67 \times 0.75 \approx 8.00$ plunks. That matches the 8 plunks. The answer is also more than 8, as it must be.

More questions like this

The original DOE question

Question 83 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 206, question 83. Open page 206 of the handbook (PDF)

Question 50 of 112

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 51 of 112

2024 Test B Math, question 36

Question

The integers $(n-1)$ , n, and $(n+1)$ are factors of 168. What is the greatest possible value of n?

Show the answer

Correct answer: B

First, determine all possible factors of 168. One method of doing this is to use a table to list pairs of factors.

Factors of 168
1	168
2	84
3	56
4	42
6	28
7	24
8	21
12	14

Then determine the group of three sequential factors $(n-1)$ , n, and $(n+1)$ where n is as large as possible. This group is 6, 7, and 8, where $n=7$ .

Question analysis

Three consecutive integers, written as n − 1, n and n + 1, must all be factors of 168, and students find the greatest possible value of n. It tests factors and consecutive integers. The catch is that a number can be a factor while its neighbors are not, so every member of the run needs checking.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Listing factors is grade 6 work, but here students must read n − 1, n and n + 1 as three consecutive integers and search the factor list for the largest such run. Pairing algebraic notation with factors matches 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: Variables instead of numbers minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once the factors of 168 are listed in order, 1, 2, 3, 4, 6, 7, 8, 12 and up, the runs of three in a row stand out, and the last of them is 6, 7, 8. Students lose points by stopping at the first run they see or by checking only n itself. Testing each choice with its two neighbors, starting from the largest, settles it in under a minute.

Variables instead of numbers minor Understanding the question
The stem names the numbers only as (n − 1), n and (n + 1), which has to be read as three whole numbers in a row that all divide 168. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Each candidate must pass three divisibility tests at once, for itself and for both neighbors, while the search continues for the greatest such n. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: List the factor pairs of 168 in order, then look along the list for three whole numbers in a row; or test each choice together with its two neighbors.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
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 the run 2, 3, 4 among the small factors and stops searching, missing the later run 6, 7, 8 that 168 also contains.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Checks only that 14 divides 168, since 12 × 14 = 168, without testing its neighbors; neither 13 nor 15 is a factor.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Takes the largest factor below 168 itself, 84, for the greatest n, ignoring that 83 and 85 would also have to divide 168.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pair: $a \times b = 168$; a and b are whole numbers. If a times b equals 168, then a and b are both factors of 168. A factor divides a number with no remainder. Listing every pair, from small a upward, finds every factor.

Sort out the question
A factor of 168 divides 168 with no remainder. The numbers n − 1, n and n + 1 are three whole numbers in a row, such as 6, 7 and 8. The middle one is n.
All three must be factors of 168. You need the greatest n that works.
List the factor pairs of 168
Try each whole number from 1 up. When it divides 168, write it with its partner. Stop when the pairs meet, after 12 and 14.
Small factor Partner
1 168
2 84
3 56
4 42
6 28
7 24
8 21
12 14
Look for three in a row
Here are all the factors in order:
1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168
The runs of three in a row are 1, 2, 3, then 2, 3, 4, then 6, 7, 8. No other three factors sit side by side.
Pick the greatest middle number
The middle numbers of the runs are 2, 3 and 7. The greatest is 7, so n = 7.
Watch out: Stopping at the run 2, 3, 4 gives 3, choice A. Keep looking. Also, 14 and 84 are factors, but their neighbors are not. Neither 13 nor 15 divides 168, so 14, choice C, fails. For 84, choice D, neither 83 nor 85 is a factor.
So the answer is B, 7.

Check your answer

Divide 168 by each number in the run: $168 \div 6 = 28$ , $168 \div 7 = 24$ and $168 \div 8 = 21$ . All three are whole numbers, so 6, 7 and 8 are all factors.

More questions like this

The original DOE question

Question 93 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 93. Open page 209 of the handbook (PDF)

Question 52 of 112

2024 Test B Math, question 37

Question

Raoul gave two musical performances. His second performance occurred 86 days after his first performance. If Raoul’s second performance was on a Tuesday, on what day of the week was his first performance?

ASunday
BMonday
CWednesday
DThursday

Show the answer

Correct answer: A

The total number of days (86) is divided by the number of days in a week (7).

$86\div 7=12\text{, with a remainder of }2$

The 2 days are counted back from Tuesday, so the first performance was on a Sunday.

Question analysis

Raoul's second musical performance came 86 days after his first and fell on a Tuesday. Students work out the day of the week of the first performance. It tests division with remainders in a repeating weekly cycle, and the traps are counting in the wrong direction or miscounting the leftover days.

NYS standards: 6.NS.2 multi-digit division
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Dividing with a remainder is grade 6 arithmetic, but using the remainder to step through a seven-day cycle, and counting backward, is less routine than grade 7 classwork. It needs care with direction rather than new 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: No obvious method moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Counting 86 days one at a time is hopeless in the time allowed, so the question rewards seeing that whole weeks return to the same weekday. Two decisions remain after that, and each has a listed wrong answer: which way to count, and whether Tuesday itself is one of the 2 days. Saying the days aloud, Monday is 1 back and Sunday is 2 back, avoids both.

No obvious method moderate Working it out
Nothing in the stem says to divide by 7; the key is that 84 of the 86 days are exactly 12 weeks, which leaves only 2 days to count. (Schoenfeld, 1985)
Turning words into math minor Understanding the question
"His second performance occurred 86 days after his first" puts the unknown day before the Tuesday, so the leftover days are counted backward. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Remove whole weeks from the 86 days, since each full week lands on the same weekday, then count the leftover days backward from Tuesday.

Traps

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

The wrong answers, classified

B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts Tuesday itself as the first of the 2 leftover days, so moves back only one day, to Monday.
C Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Counts forward from Tuesday instead of back and also counts Tuesday as one of the 2 days, landing on Wednesday.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Counts the 2 leftover days forward from Tuesday, finding the day 86 days after the second performance instead of before it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Weeks and leftover days: $\text{days} = 7 \times \text{weeks} + \text{leftover}$; days is the total number of days. weeks is the number of full weeks inside it. leftover is the days that remain, from 0 to 6. The day of the week repeats every 7 days, so only the leftover days change it.

Sort out the question
The second show was on a Tuesday. It came 86 days after the first show. So the first show was 86 days before that Tuesday.
You need the day of the week of the first show. So you count back 86 days from Tuesday.
Take out the full weeks
Counting back 86 days one at a time is slow. Here is a shortcut. Every 7 days, the same weekday comes back. So divide 86 by 7.
$86 \div 7 = 12 \text{ remainder } 2$
$86 = 7 \times 12 + 2$
That is 12 full weeks and 2 extra days. Going back 12 weeks from a Tuesday lands on a Tuesday again.
Count back the 2 extra days
Start on Tuesday and go back one day at a time. One day back is Monday. Two days back is Sunday.
Watch out: Do not count Tuesday itself as one of the 2 days. That lands on Monday, choice B. And count backward, not forward. Counting forward 2 days gives Thursday, choice D.
So the answer is A, Sunday.

Check your answer

Count forward to be sure. Start on a Sunday. After 84 days, which is 12 weeks, it is Sunday again. Two more days are Monday and Tuesday. That is day 86, a Tuesday. It matches.

More questions like this

The original DOE question

Question 94 as printed on page 210 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 94. Open page 210 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 53 of 112

2024 Test B Math, question 38

Question

Michael received a check for $213 for working 20 hours at his part-time job. The amount on the check was his total pay after $32 in taxes was deducted. What was Michael’s hourly pay rate before taxes were deducted?

A$9.05
B$10.65
C$12.25
D$21.35

Show the answer

Correct answer: C

Add the taxes to the amount of the check to find Michael’s total pay before taxes were deducted:

$213+32=245$

Divide the total pay by the number of hours to find the hourly pay rate:

$245\div 20=12.25$

Question analysis

Michael's paycheck shows what he earned for 20 hours after taxes were taken out, and the tax amount is given. Students find his hourly pay rate before taxes. It tests unit rates in a money word problem, where students must decide what to do with the tax that was deducted.

NYS standards: 6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Unit rate problems are grade 6 content, and this one needs one addition and one division with a decimal result. It is easier than typical grade 7 work, apart from deciding which amount is the pay before taxes.
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 5, 36 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two operations in the right order, adding the tax back and dividing by the hours, are all this takes. The danger is the word deducted, which pulls students toward subtracting the $32 a second time, and the phrase before taxes, which rules out the take-home rate from dividing the check alone. Checking that $12.25 × 20 − $32 gives back $213 confirms the answer.

Turning words into math minor Understanding the question
"The amount on the check was his total pay after $32 in taxes was deducted" makes $213 what remained, so the $32 is added back, not taken away. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Rebuild Michael's pay before taxes from the check and the tax that was taken out, then divide that total by the 20 hours worked.

Traps

Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
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 Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Follows the word deducted and subtracts the $32 from the $213 check again, (213 − 32) ÷ 20 = 9.05, instead of adding it back.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides the $213 check by 20 hours, which gives Michael's hourly pay after taxes rather than before.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard working gives $21.35, which equals 32 − 10.65; dividing $213 by 10 hours instead of 20, $21.30, comes closest, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Hourly pay rate: $\text{rate} = \frac{\text{total pay}}{\text{hours}}$; rate is the pay for one hour. total pay is the money earned for all the hours. hours is the time worked. Dividing shares the total pay equally among the hours. Use it when the pay for each hour is the same.

Sort out the question
Michael’s check was for $213. That is what was left after $32 in taxes was deducted, which means taken away. He worked 20 hours.
You need his hourly pay before taxes. So first find his full pay before the $32 was taken out.
Add the taxes back
The $32 was taken away from his full pay to leave $213. To undo taking away, add it back.
$213 + 32 = 245$
His pay before taxes was $245. Watch out: The word “deducted” tempts you to subtract again. That gives $(213 - 32) \div 20 = 9.05$ , choice A.
Divide by the hours
$\text{rate} = \frac{245}{20} = 12.25$
He earned $12.25 for each hour before taxes. Watch out: Dividing the $213 check by 20 gives $10.65, choice B. That is his pay for each hour after taxes.
So the answer is C, $12.25.

Check your answer

Work forward. $12.25 \times 20 = 245$ , so 20 hours pay $245. Take away $32 in taxes: $245 - 32 = 213$ . That is the amount on the check.

More questions like this

The original DOE question

Question 95 as printed on page 210 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 95. Open page 210 of the handbook (PDF)

Question 54 of 112

2024 Test B Math, question 46

Question

The least positive integer evenly divisible by the first 6 positive integers is 60. What is the least positive integer evenly divisible by the first 8 positive integers?

A1,680
B840
C480
D420

Show the answer

Correct answer: B

Each option is divisible by 1, 2, 3, 4, 5, and 6.

Option A:

1,680 is evenly divisible by the first 8 positive integers.

Option B:

840 is evenly divisible by the first 8 positive integers and is less than 1,680.

Option C:

480 is not evenly divisible by 7.

$480\div 7\approx 68.57$

Option D:

420 is not evenly divisible by 8.

$420\div 8=52.5$

Option B (840) is the correct answer.

Question analysis

Knowing that 60 is the smallest number divisible by each of 1 through 6, students find the smallest positive integer divisible by each of 1 through 8. It tests least common multiples and prime factors. The challenge is adding only the factors that 7 and 8 really need, so the result works for all eight numbers and is still the least.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Least common multiples are grade 6 content, usually for two small numbers. Extending the given result from six numbers to eight means deciding what 7 and 8 add, more reasoning than typical classwork but no new 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: Every choice must be checked moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 13, 25 words.
Corrected from the DOE handbook: A rounded decimal in the explanation is now marked as approximate (≈ instead of =).
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The given fact, 60 for the first six integers, is a head start, but only for students who see what 7 and 8 add. Multiplying 60 by 7 is needed; multiplying by all of 8 is not, since 60 already supplies two of its three factors of 2. Students who multiply by the new numbers whole get answers that are too large or that miss one divisor, so checking the smaller choices first is the efficient route.

Every choice must be checked moderate Choosing the answer
Both 1,680 and 840 are divisible by 7 and by 8, so testing the choices is safe only from the smallest up; 420 fails for 8 and 480 for 7. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
Building on 60 needs prime factors: 7 adds a new prime, while 8 = 2 × 2 × 2 needs only one more 2, since 60 already holds two. (Anderson & Krathwohl, 2001)

Strategy: Start from 60 and ask which extra prime factors 7 and 8 require, or test the choices from the smallest up for divisibility by both 7 and 8.

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
Multiplies 60 by 7 and by 4 to cover 8, getting 1,680, which all of 1 to 8 divide, but overlooks that the question wants the least.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Multiplies 60 by 8 to bring in the 8 but never checks divisibility by 7, which 480 fails.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Multiplies 60 by 7 to bring in the 7 and stops, but 420 has only two factors of 2, so 8 does not divide it.

What was corrected from the DOE handbook, and why

Explanation: To show that 480 is not evenly divisible by 7, the handbook writes 480 ÷ 7 = 68.57. The quotient is really 68.571428…, a decimal that never ends, so 68.57 is only a rounded value. The site writes 480 ÷ 7 ≈ 68.57, as the handbook itself does for other rounded results in this test. The point of the step is unchanged: the quotient is not a whole number, so 480 is not evenly divisible by 7 and is not divisible by all of the first 8 positive integers.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Evenly divisible: $a \div b = \text{a whole number}$; a is the number you test, and b is the number you divide by. If the answer is a whole number, with no remainder, a is evenly divisible by b. Then b is a factor of a.

Sort out the question
The first 8 positive integers are 1, 2, 3, 4, 5, 6, 7 and 8. “Evenly divisible” means dividing leaves no remainder.
You need the least, or smallest, number that all 8 of them divide. You are told that 60 is the smallest for 1 to 6. So build on 60.
Bring in the 7
60 is not divisible by 7. And 7 is prime, which means its only factors are 1 and 7. So the new number needs a factor of 7. Multiply 60 by 7.
$60 \times 7 = 420$
Bring in the 8
8 is $2 \times 2 \times 2$ , so the number needs three 2s as factors. But 420 has only two: $420 = 2 \times 2 \times 105$ , and 105 is odd. Multiply by one more 2.
$420 \times 2 = 840$
Now $840 \div 8 = 105$ and $840 \div 7 = 120$ . Both are whole numbers.
Test the choices from smallest up
Choice D: $420 \div 8 = 52.5$ . That is not a whole number, so 420 fails.
Choice C: $480 \div 7 \approx 68.57$ . The ≈ sign means about, since the decimal never ends. It is not a whole number, so 480 fails.
Choice B: 840 works for 7 and 8. It also works for 1 to 6, since $840 = 60 \times 14$ .
Watch out: 1,680, choice A, works too. But it is twice 840, so it is not the least.
So the answer is B, 840.

Check your answer

Divide 840 by each number from 1 to 8. You get 840, 420, 280, 210, 168, 140, 120 and 105. Every answer is a whole number, so 840 works for all eight.

More questions like this

The original DOE question

Question 103 as printed on page 212 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 103. Open page 212 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 55 of 112

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 56 of 112

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 57 of 112

2024 Test B Math, question 54

Question

A rectangular kitchen floor is 12 feet long and 7.5 feet wide. It will be covered completely with new square tiles measuring 0.75 foot on each side. How many tiles will cover the floor?

A68
B78
C120
D160

Show the answer

Correct answer: D

First, determine the area of the kitchen floor in square feet by multiplying the length and width:

$12\times 7.5=90\text{ square feet}$

Next, determine the area of each square tile:

$0.75\times 0.75=\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}$

The number of tiles needed is found by dividing the area of the floor by the area of the tiles:

$90\div\frac{9}{16}=90\times\frac{16}{9}=160$

Question analysis

A rectangular kitchen floor measuring 12 feet by 7.5 feet will be covered with square tiles 0.75 foot on a side. Students find how many tiles are needed. It tests area and division with decimals, and working with a decimal side length while keeping lengths and areas apart is what makes it tricky.

NYS standards: 6.G.1 area of polygons
6.NS.3 multi-digit decimal operations
6.NS.1 dividing fractions by fractions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Area of a rectangle is grade 6 content, but here it combines with dividing by a decimal tile area, or by the tile's side along each wall. Three decimal steps and a tempting shortcut make it typical grade 7 demand.
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
Solution: 3 steps
Reading level: Question and choices: about grade 6, 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 floor's 90 square feet comes quickly; the trap is what to divide it by. A tile is described by its side, and dividing by 0.75 instead of 0.5625 gives a listed answer that looks reasonable. Counting tiles along each wall, 16 along the length and 10 along the width, avoids both the decimal division and the side-versus-area mistake.

Error-prone arithmetic moderate Working it out
The tile's area, 0.75 × 0.75 = 0.5625 square foot, must divide 90, a decimal division that is easier as 90 × 16/9 or as 16 tiles by 10 tiles. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The tiles are "square tiles measuring 0.75 foot on each side", a side length, which has to be squared before it can be compared with the floor's area in square feet. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find how many tiles fit along the length and how many along the width by dividing each by 0.75, then multiply the two counts.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
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: weak
Multiplies the 90-square-foot area by the tile's 0.75-foot side instead of dividing, getting 67.5 and rounding to 68.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard working gives 78, which is twice the floor's 39-foot perimeter, 4 × (12 + 7.5), so it mostly draws guesses.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Divides the 90 square feet by the tile's side, 0.75 foot, instead of its area, 0.5625 square foot, getting 120 tiles.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle: $A = l \times w$; A is the area, the flat space inside. l is the length and w is the width. Multiplying counts how many unit squares fit, row by row.
Area of a square: $A = s^2$; A is the area. s is the length of one side. s² means s × s. A square is a rectangle whose length and width are both s.

Sort out the question
The floor is a rectangle 12 feet long and 7.5 feet wide. Each tile is a square, 0.75 foot on each side. You need the number of tiles that cover the whole floor.
Find the floor’s area
Area tells how much flat space the tiles must cover. Use the rectangle formula.
$A = 12 \times 7.5 = 90$
The floor is 90 square feet.
Find one tile’s area
The tile’s side is a length. You need its area to compare it with the floor’s area. Use the square formula.
$A = 0.75^2 = 0.75 \times 0.75 = 0.5625$
As a fraction, $0.75 = \frac{3}{4}$ , so the area is $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$ square foot.
Divide the floor by one tile
The number of tiles is the floor’s area divided by one tile’s area. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down.
$90 \div \frac{9}{16} = 90 \times \frac{16}{9} = 10 \times 16 = 160$
Watch out: Dividing 90 by 0.75, the side, gives 120, choice C. You must divide by the tile’s area. Multiplying 90 by 0.75 gives about 68, choice A.
So the answer is D, 160.

Check your answer

Count the tiles in rows. Along the 12-foot length, $12 \div 0.75 = 16$ tiles fit. Along the 7.5-foot width, $7.5 \div 0.75 = 10$ tiles fit. That makes $16 \times 10 = 160$ tiles.

More questions like this

The original DOE question

Question 111 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 111. Open page 214 of the handbook (PDF)

Question 58 of 112

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 59 of 112

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 60 of 112

2023 Test A Math, question 5

Question

The mean value of 8 numbers is 17. Three of these numbers (9, 11, and 20) are discarded. What is the mean of the 5 remaining numbers?

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

Show the answer

Correct answer: 19.2

First, find the sum of the original 8 numbers. Let x equal the sum of those numbers and use the mean formula to solve for x:

$\frac{x}{8} = 17$

$x = 136$

Next, subtract the 3 discarded numbers from the sum:

$136 - 9 - 11 - 20 = 96$

Finally, calculate the mean of the remaining 5 numbers by dividing the new sum, 96, by 5:

$\frac{96}{5} = 19.2$

Question analysis

Eight numbers have a mean of 17, and three of them, 9, 11 and 20, are removed. This statistics grid-in asks for the mean of the five numbers that are left. Students have to rebuild the total from the mean before removing anything, and then divide by the new count rather than the old one.

NYS standards: 6.SP.5c range, center, and overall pattern
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Finding a mean is grade 6 content, but working back from a mean to a total, removing values and averaging again is a typical grade 7 multistep task. As a grid-in, it offers no choices to check a wrong total against.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Much to keep track of minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

A mean cannot be adjusted directly when numbers are removed, so the question rewards switching from the mean to the total and back. The removed numbers, 9, 11 and 20, average less than 17, so the new mean must rise above 17; that quick check rules out an answer such as 12. Points are lost by dividing by the wrong count or by stopping at the new total.

Many steps moderate Working it out
Three operations come in a fixed order: multiply the mean by 8 to get 136, subtract 9, 11 and 20 to leave 96, then divide by the numbers that remain. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
The count changes partway through, from 8 numbers to 5, and the divisor at the end must be the new count, not the 8 from the first sentence. (Sweller, 1988; Carpenter, Just & Shell, 1990)
An answer in an unexpected form minor Choosing the answer
The answer, 19.2, is a decimal that must be gridded with its point, and with no choices a total of 96 or a mean over 8 numbers looks complete. (Enright, Morley & Sheehan, 2002)

Strategy: Turn the mean into a total by multiplying by 8, subtract the three discarded numbers, and divide what is left by the count of numbers that remain.

Traps

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

The wrong answers, classified

12 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds the remaining total, 136 − 40 = 96, correctly but divides it by the original 8 numbers instead of the 5 that are left.
96 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Subtracts the three discarded numbers from the total of 136 and grids the new sum, 96, without dividing by the 5 numbers left.
27.2 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Turns the mean into the total, 136, and divides it by 5 without first subtracting the three discarded numbers, 9, 11 and 20.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mean, or average: $\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}$; The mean is the average. The sum is what you get when you add all the numbers. Divide it by how many numbers there are. The mean is the value each number would have if the total were shared out equally.
Sum from a mean: $\text{sum} = \text{mean} \times \text{how many numbers}$; This is the mean formula turned around. If you know the mean and how many numbers there are, multiply them to get the sum. It works even when you don’t know the numbers themselves.

Sort out the question
There are 8 numbers. Their mean, or average, is 17. You don’t know what the 8 numbers are. Three of them, 9, 11 and 20, are discarded, or thrown away. That leaves 5 numbers. You need the mean of those 5 numbers.
Find the sum of all 8 numbers
You can’t add the 8 numbers, because you don’t know them. But the mean tells you their sum.
$\text{sum} = \text{mean} \times \text{how many numbers}$
$\text{sum} = 17 \times 8 = 136$
The 8 numbers add up to 136.
Take away the three discarded numbers
Subtract the three numbers that were thrown away.
$136 - 9 - 11 - 20 = 136 - 40 = 96$
The 5 numbers left add up to 96. Watch out: 96 is a sum, not a mean, so gridding 96 is wrong. Also, dividing 136 by 5 without subtracting first gives 27.2, which is wrong.
Divide by the 5 numbers left
Now use the mean formula for the numbers that are left.
$\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}$
$\text{mean} = \frac{96}{5} = 19.2$
Watch out: divide by 5, not by 8. Only 5 numbers are left. Dividing by 8 gives 12. So you would grid 19.2, with the decimal point in its own column.

Check your answer

The three numbers thrown away add to 40, so their mean is about 13. That is less than 17. Removing small numbers should make the mean go up, and 19.2 is more than 17. Also, $19.2 \times 5 = 96$ , the sum of the 5 numbers left.

More questions like this

The original DOE question

Question 62 as printed on page 76 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 76, question 62. Open page 76 of the handbook (PDF)

Question 61 of 112

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 62 of 112

2023 Test A Math, question 12

Question

If n is an integer and $3n + 3$ is an even number, which expression must also represent an even number?

A $5n + 1$
B $4n + 5$
C $2n + 3$
D $n + 2$

Show the answer

Correct answer: A

The expression $3n + 3$ is equal to $3(n + 1)$ , and it is given that this is an even number. The expression $(n + 1)$ must also represent an even number, because 3 is an odd number: if $(n + 1)$ were odd, $3(n + 1)$ would be the product of two odd numbers, which is an odd number. If $(n + 1)$ is an even number, then n must be an odd number. Choose an odd number to substitute for n in the answer options to find which expression represents an even number:

Let $n = 3$ .

Option A:

$5n + 1 = 5(3) + 1 = 16$

Option B:

$4n + 5 = 4(3) + 5 = 17$

Option C:

$2n + 3 = 2(3) + 3 = 9$

Option D:

$n + 2 = 3 + 2 = 5$

Since 16 is the only even number, the correct answer is option A.

Question analysis

This number properties question says that 3n + 3 is even for an integer n and asks which of four expressions must also be even. Students reason about odd and even numbers written with a variable. The hidden step is deciding what the given fact says about n itself before testing the expressions.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.4 common factors and multiples
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students rewrite and evaluate expressions, but rarely reason about whether an expression is odd or even for every value of a variable. Here that reasoning comes in two layers: first about n, then about each choice.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
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 minor
Every choice must be checked minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 23 words.
Corrected from the DOE handbook: The explanation gave the wrong reason why n + 1 must be even; the reason was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question hides its key step: 3n + 3 = 3(n + 1), and since 3 is odd, n + 1 must be even, so n is odd. A student who skips that step and tests a convenient even value never checks the condition and lands on a wrong expression. Once n is known to be odd, any odd test value, such as 1, picks out 5n + 1 at once.

Variables instead of numbers moderate Understanding the question
No value of n is given; the only fact is that 3n + 3 is even, and it has to be turned into a fact about n itself, that n is odd, before any choice is judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
The parity rules are assumed, not stated: an odd number times an odd number is odd, and an even number plus an odd number is odd. (Anderson & Krathwohl, 2001)
Every choice must be checked minor Choosing the answer
Each expression has to be tested with an odd n or by parity rules; three of the four come out odd whenever n is odd, so none can be dismissed on sight. (Embretson & Wetzel, 1987)

Strategy: First work out what 3n + 3 being even says about n, by factoring out 3 or testing small integers, then check each expression with a value of n that fits.

Traps

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Sees the even term 4n and treats the whole expression as even, overlooking that adding the odd 5 makes 4n + 5 odd for every integer n.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Assumes an expression that starts with 2n must be even, though 2n + 3 is an even number plus 3 and so is always odd.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Tests n = 2 without checking the condition, though 3(2) + 3 = 9 is odd; with n = 2, only n + 2 = 4 comes out even.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation said that n + 1 must be even “because 3 is an odd number and the product of an odd number and an even number is an even number.” That fact does not show that n + 1 is even; the reason is that if n + 1 were odd, 3(n + 1) would be the product of two odd numbers, which is odd, and it is given as even. The site now gives that reason. The answer, A, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Pulling out a common factor: $a \times b + a \times c = a(b + c)$; a is a factor, a number being multiplied, that both terms share. b and c are what is left. This is the distributive property run backward. It turns a sum into a product, which is easier to reason about.
Multiplying odd numbers: $\text{odd} \times \text{odd} = \text{odd}$; An even number splits into pairs, like 2, 4 or 6. An odd number has one left over, like 1, 3 or 5. An odd number of odd groups always leaves one extra. So the product is odd.
Adding odd and even numbers: $\text{odd} + \text{odd} = \text{even}, \quad \text{even} + \text{odd} = \text{odd}$; Two odd numbers each have one left over. Together the two extras make a pair, so the sum is even. An even number plus an odd number still has one left over, so the sum is odd.

Sort out the question
n is an integer, a whole number that can be positive, negative or 0. You are told that $3n + 3$ is even. You need the expression that must be even too. Must means it is even every time. First find out what the fact tells you about n.
Pull out the 3
Both terms in $3n + 3$ have a factor of 3. Pull it out.
$3n + 3 = 3(n + 1)$
So 3 times $(n + 1)$ is an even number.
Decide whether n is odd or even
3 is odd. If $n + 1$ were odd, $3(n + 1)$ would be odd times odd. That is always odd. But we are told it is even. So $n + 1$ must be even. One less than an even number is odd. So n is odd. Watch out: don’t test n = 2. Then $3n + 3 = 9$ , which is odd. With n = 2, only $n + 2$ comes out even, choice D.
Test each choice with an odd n
Pick an odd number, n = 3. It fits, since $3 \times 3 + 3 = 12$ is even.
- A: $5n + 1 = 5 \times 3 + 1 = 16$ , even
- B: $4n + 5 = 4 \times 3 + 5 = 17$ , odd
- C: $2n + 3 = 2 \times 3 + 3 = 9$ , odd
- D: $n + 2 = 3 + 2 = 5$ , odd
Only A is even.
Make sure A always works
Must means every odd n, not just 3. If n is odd, 5n is odd times odd, so it is odd. Odd plus 1 is even. So $5n + 1$ is always even. Watch out: 4n is even, but adding the odd 5 makes $4n + 5$ odd, choice B. In the same way, $2n + 3$ is always odd, choice C. So the answer is A, $5n + 1$ .

Check your answer

Try another odd number, n = 1. Then $3n + 3 = 6$ , which is even, so n = 1 fits. And $5n + 1 = 6$ is even too. The other three choices give 9, 5 and 3, all odd. Choice A wins again.

More questions like this

The original DOE question

Question 63 of 112

2023 Test A Math, question 13

Question

The product of two positive integers is 65. Which number could be the sum of the two integers?

Show the answer

Correct answer: B

First, find the pairs of positive integers that multiply to 65: $1 \times 65$ and $5 \times 13$ .

The sums of the pairs are 66 and 18. 18 is the option that is provided in the answer choices.

Question analysis

Two positive integers multiply to 65, and this question asks which number in the choices could be their sum. Students list the factor pairs of 65 and add each pair. The words could be signal that more than one pair exists, and only one of the sums appears among the choices.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding factor pairs of a two-digit number is review by grade 7. The question takes two quick steps, listing the pairs and adding each, with nothing to compute beyond small products and sums.
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: Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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

With only two factor pairs, 1 × 65 and 5 × 13, the work is a short list and two additions. The word could matters: the sum depends on which pair is chosen, and only one of the two sums appears among the choices. The main slip is answering with a factor rather than with the sum the question asks for.

Something you must already know minor Working it out
The pairs come from factoring 65 as 5 × 13, a fact the stem leaves to the student; the pair 1 and 65 also counts, but its sum, 66, is not offered. (Anderson & Krathwohl, 2001)

Strategy: List every pair of positive integers whose product is 65, starting from 1, and add the two numbers in each pair.

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: moderate
Finds the factor pair 5 and 13 and answers with the factor 5, one of the two integers, instead of their sum.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No pair of positive integers with product 65 adds to 24; reaching it needs a wrong pair such as 3 and 21, whose product is 63, so it mostly catches guesses.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No factor pair of 65 adds to 52, which is 65 − 13 or 4 × 13, and no common working combines them that way, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pair: $a \times b = 65$; a and b are positive whole numbers that multiply to make 65. Such a pair is called a factor pair of 65. To find them all, try 1, 2, 3 and so on. Stop when the factors start to repeat.
Dividing by 5: $\text{last digit } 0 \text{ or } 5 \rightarrow \text{divides by } 5$; A whole number splits into groups of 5, with none left over, when its last digit is 0 or 5. So 65 divides by 5, but 64 does not.

Sort out the question
Two positive integers multiply to make 65. Positive integers are the counting numbers 1, 2, 3 and so on. The product is the answer to a multiplication. You need a number that could be their sum, the answer when you add them.
List the factor pairs of 65
Try each small number. See if it divides 65 with nothing left over.
- 1 works: $1 \times 65 = 65$ .
- 2 and 4 don’t work, because 65 is odd. 3 doesn’t work either: $3 \times 21 = 63$ , with 2 left over.
- 5 works, since 65 ends in 5: $5 \times 13 = 65$ .
- 6 and 8 are even, so they don’t work. 7 doesn’t either: $7 \times 9 = 63$ .
Since $8 \times 8 = 64$ , any factor bigger than 8 pairs with one smaller than 8. You already tried those. So the pairs are 1 and 65, and 5 and 13.
Add each pair
$1 + 65 = 66$
$5 + 13 = 18$
The possible sums are 66 and 18. Only 18 is a choice. Watch out: the question asks for the sum, not one of the numbers. Picking the factor 5 is choice A. So the answer is B, 18.

Check your answer

Work back from 18. Look for two numbers that add to 18 and multiply to 65. $5 + 13 = 18$ and $5 \times 13 = 65$ . Both match. For 24 or 52, no pair of whole numbers that multiplies to 65 adds up to them.

More questions like this

The original DOE question

Question 70 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 70. 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 64 of 112

2023 Test A Math, question 14

Question

If n is an odd integer that is less than −3.25, what is the greatest possible value of n?

A−1
B−2
C−3
D−5

Show the answer

Correct answer: D

The two greatest integers less than −3.25 are −4 and −5. Since −4 is even, the answer is −5.

Question analysis

This question asks for the greatest odd integer that is less than −3.25. Students compare negative numbers on a number line and apply two conditions at once, odd and less than. With negatives, the greater number is the one closer to zero, which makes the two conditions easy to misapply.

NYS standards: 6.NS.7a inequality as relative position
6.NS.6c positioning rational numbers and points
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Ordering negative numbers is grade 6 content, and the question adds only the condition that the integer be odd. It is lighter than typical grade 7 work, though the order of negatives still trips up students who compare their sizes.
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 minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is short, but it asks students to reason on the negative side of 0, where greater means closer to 0 and the size of the digits misleads. The integers just below −3.25 are −4 and then −5, and the first of those is even. Sketching a number line, or reading less than as to the left of, keeps the direction straight.

Error-prone arithmetic minor Working it out
On the negative side, less means farther from 0, so the integers below −3.25 run −4, −5, −6; −3, though its digit is closer to 3.25, lies to the right. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Much to keep track of minor Working it out
Three conditions apply at once, odd, less than −3.25 and the greatest such value, and checking only two of them leads to a listed choice. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Picture −3.25 on a number line, list the integers to its left starting with the closest, and take the first one that is odd.

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

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Picks the greatest odd number offered, −1, without checking the condition that n be less than −3.25; −1 lies to its right on the number line.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
The choice −2 is even, so it breaks the condition that n be odd, and it is also greater than −3.25; it mostly catches guesses.
C Negatives and direction (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Looks for the odd integer just past −3.25 but steps toward 0, taking −3, though −3 is greater than −3.25, not less.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing negative numbers: $\text{if } a > b > 0 \text{, then } -a < -b$; a and b are positive numbers, and a is bigger. On a number line, numbers get smaller as you move left. A negative number farther from 0 is farther left. So it is smaller, even though its digits look bigger.

Sort out the question
n has to pass three tests. It is an integer, a whole number with no decimal part. It is odd, so it is not a multiple of 2. And it is less than −3.25. Of all the numbers that pass, you need the greatest one. That is the one farthest to the right on a number line.
Picture −3.25 on a number line
−3.25 sits between −4 and −3. Less than −3.25 means to the left of it.
$-5 < -4 < -3.25 < -3$
So −4 is less than −3.25, but −3 is greater. Watch out: −3 is not less than −3.25. It is to the right, closer to 0. For negative numbers, the one closer to 0 is greater. Picking −3 is choice C.
Find the odd integers to the left
The integers less than −3.25, starting with the closest, are −4, −5, −6 and so on. The greatest one is −4. But −4 is even, since 4 = 2 × 2. So skip it. The next one, −5, is odd. Watch out: −1 is odd, but it is greater than −3.25, so choice A breaks the rule. So the answer is D, −5.

Check your answer

Check all three tests for −5. It is an integer. It is odd. It is to the left of −3.25, so it is less. The only integer between −5 and −3.25 is −4, and it is even. So no greater number works.

More questions like this

The original DOE question

Question 71 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 71. Open page 79 of the handbook (PDF)

Question 65 of 112

2023 Test A Math, question 16

Question

Between which two consecutive integers is the fraction $\frac{29}{7}$ ?

A2 and 3
B3 and 4
C4 and 5
D5 and 6

Show the answer

Correct answer: C

First, convert $\frac{29}{7}$ to a mixed number: $\frac{29}{7} = 4\frac{1}{7}$ . The two consecutive integers on either side of $4\frac{1}{7}$ are 4 and 5.

Question analysis

This question asks between which two consecutive whole numbers the fraction 29/7 lies. Students divide to turn the improper fraction into a mixed number, or compare it with nearby multiples of 7. It is a quick check of number sense with fractions, where a remainder must be read correctly.

NYS standards: 6.NS.6c positioning rational numbers and points
6.NS.7a inequality as relative position
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Changing an improper fraction to a mixed number and placing it between whole numbers is review well before grade 7. This takes one division with a small divisor and nothing more.
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
Solution: 2 steps
Reading level: Question and choices: about grade 5, 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

Dividing 29 by 7 gives 4 with a remainder of 1, so the fraction is 4 1/7 and lies just past 4. The one decision is which side of 4 the remainder puts it on, and a positive remainder always moves the value up. Writing the mixed number before looking at the choices makes the pair obvious.

A very close wrong answer minor Choosing the answer
The pairs 3 and 4 and 4 and 5 both contain 4, so knowing that 29/7 is about 4 is not enough; only the remainder, 29 − 28 = 1, puts it above 4. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Find the largest multiple of 7 that is not more than 29, write the fraction as a mixed number, and read off the whole numbers on each side.

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 division of 29 by 7 lands below 3, since 7 × 3 = 21 is far short of 29; only misreading 29/7 as 2.9 gets there, so it mostly catches guesses.
B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds the whole-number part, 4, but takes it as the upper end of the pair, as if 29/7 fell just short of 4 instead of just past it.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Nothing puts 29/7 above 5, since 7 × 5 = 35 is more than 29; even rounding 4 1/7 up gives 5 as the upper end, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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
Consecutive integers are whole numbers that come one right after the other, like 4 and 5. You need the two that $\frac{29}{7}$ falls between. So find the whole number just below it and the one just above it.
Divide 29 by 7
A fraction bar means divide. How many 7s fit in 29?
$7 \times 4 = 28$
$7 \times 5 = 35$
28 is the biggest multiple of 7 that is not more than 29. A multiple of 7 is 7 times a whole number. Then $29 - 28 = 1$ . So 29 ÷ 7 is 4 with 1 left over.
Write it as a mixed number
$\frac{29}{7} = 4\frac{1}{7}$
$4\frac{1}{7}$ is a little more than 4 and less than 5. Watch out: it is just past 4, not just short of it. So 4 is the lower number of the pair. Picking 3 and 4 is choice B. So the answer is C, 4 and 5.

Check your answer

Multiply each end by 7. $4 \times 7 = 28$ and $5 \times 7 = 35$ . The top of the fraction, 29, is between 28 and 35. So $\frac{29}{7}$ is between 4 and 5. As a decimal it is about 4.14.

More questions like this

The original DOE question

Question 73 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 73. Open page 79 of the handbook (PDF)

Question 66 of 112

2023 Test A Math, question 23

Question

$\{0.1, 0.01, 0.001, 0.0001, 0.00001\}$

If a person chooses a number at random from the set above, what is the probability that the number is less than 0.005?

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

Show the answer

Correct answer: C

Three of the five numbers listed (0.001, 0.0001, and 0.00001) are less than 0.005. So, the probability of a person selecting a number less than 0.005 is $\frac{3}{5}$ .

Question analysis

A set holds five decimals, from one tenth down to one hundred-thousandth, and one number is chosen at random. Students find the probability that the number is less than 0.005. Comparing decimals with many zeros is the real task here, along with putting the count over the right total.

NYS standards: 6.SP.8a uniform probability models
6.NS.7a inequality as relative position
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Probability with equally likely outcomes is grade 6 content in New York, and this needs one comparison per number and one fraction. The only real demand is comparing decimals with many zeros correctly.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · 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 6, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each number in the set is a tenth of the one before, so only the three smallest fall below 0.005, which lies between 0.01 and 0.001. Students lose the point either by counting the numbers above 0.005 instead of below it or by writing the count over the wrong total. Lining the decimals up with the same number of places makes every comparison easy.

Error-prone arithmetic moderate Working it out
0.005 has to be compared with decimals from 0.1 down to 0.00001; written to five places, 0.00500 against 0.01000 and 0.00100, the order becomes a comparison of whole numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The probability is the count of numbers below 0.005 over all 5 numbers in the set; the stem leaves that part-to-whole rule to the student. (Anderson & Krathwohl, 2001)

Strategy: Write each number with five decimal places, compare it with 0.00500, count those that are smaller, and divide by the five numbers in the set.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Gives 1/5, the chance of picking any one particular number from the set, without counting how many of the five are below 0.005.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Counts the 2 numbers greater than 0.005, 0.1 and 0.01, instead of those less than it, finding the chance of the opposite event.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Writes a part-to-part ratio, the 2 numbers above 0.005 over the 3 below it, instead of dividing by all 5 numbers in the set.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{wanted results}}{\text{possible results}}$; P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.
Adding zeros to a decimal: $0.1 = 0.10 = 0.100$; Zeros added to the right end of a decimal don’t change its value. One tenth is the same as ten hundredths. Giving every number the same number of decimal places lets you compare them like whole numbers.

Sort out the question
The set has five numbers. A person picks one at random, so each number has the same chance. You need the probability, or chance, that the number picked is less than 0.005.
Give every number five decimal places
Decimals are easier to compare with the same number of places. Add zeros to the end of each.
Number With five places
0.1 0.10000
0.01 0.01000
0.001 0.00100
0.0001 0.00010
0.00001 0.00001
0.005 0.00500
Count the numbers below 0.005
Compare each number with 0.00500 like whole numbers: 10000, 1000, 100, 10 and 1 against 500. Two numbers are bigger: 0.1 and 0.01. Three are smaller: 0.001, 0.0001 and 0.00001. Watch out: 0.01 has fewer digits than 0.005, but it is bigger. Counting the 2 bigger numbers gives $\frac{2}{5}$ , choice B.
Write the probability
$P = \frac{\text{wanted results}}{\text{possible results}} = \frac{3}{5}$
Watch out: divide by all 5 numbers in the set. Comparing the 2 bigger numbers with the 3 smaller ones gives $\frac{2}{3}$ , choice D. So the answer is C, $\frac{3}{5}$ .

Number	With five places
0.1	0.10000
0.01	0.01000
0.001	0.00100
0.0001	0.00010
0.00001	0.00001
0.005	0.00500

Check your answer

The numbers below 0.005 and the numbers above it make up the whole set: 3 + 2 = 5. So the two chances, $\frac{3}{5}$ and $\frac{2}{5}$ , add up to 1, as they should.

More questions like this

The original DOE question

Question 80 as printed on page 81 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 80. Open page 81 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 67 of 112

2023 Test A Math, question 26

Question

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

A $\frac{1}{x + 1} > \frac{1}{x + 2}$
B $\frac{1}{x + 1} < \frac{1}{x + 2}$
C $\frac{1}{x + 1} - \frac{1}{x + 2} > 1$
D $\frac{1}{x + 1} - \frac{1}{x + 2} > \frac{1}{x}$

Show the answer

Correct answer: A

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

Option A:

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

Option B:

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

Option C:

$\frac{1}{3} - \frac{1}{4} > 1$ , or $\frac{1}{12} > 1$ . This is false.

Option D:

$\frac{1}{3} - \frac{1}{4} > \frac{1}{2}$ , or $\frac{1}{12} > \frac{1}{2}$ . This is false.

Option A is the only true inequality.

Question analysis

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

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

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
All four statements are about the unknown x, and the bold must means each has to hold for every whole number from 1 up, not just for a single test value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
The statements are four separate inequalities, two comparisons and two differences, so each has to be tested; one counterexample is enough to rule a statement out. (Embretson & Wetzel, 1987)
Error-prone arithmetic minor Working it out
Testing the differences means subtracting unlike fractions, such as 1/3 − 1/4 = 1/12 when x = 2, and comparing the result with 1 and with 1/x. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute a small whole number such as 1 or 2 for x in all four statements, rule out any that fail, and confirm the survivor by thinking about larger denominators.

Traps

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Believes a bigger denominator makes a bigger fraction, so 1/(x + 2) seems larger than 1/(x + 1), the comparison turned around.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Both fractions are at most 1/2, so their difference can never exceed 1, and no common working makes it do so; this choice mostly draws guesses.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Subtracts over a common denominator but keeps only the numerator, (x + 2) − (x + 1) = 1, so the difference looks like 1, larger than 1/x once x is 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions: $\text{if } 0 < a < b \text{, then } \frac{1}{a} > \frac{1}{b}$; a and b are positive numbers, and b is bigger. One over a means one whole cut into a equal pieces. Cutting the same whole into more pieces makes each piece smaller. So the fraction with the bigger bottom is smaller.
Subtracting fractions: $\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}$; a and b are the bottoms of the two fractions. Rewrite both fractions over the common bottom a × b. Then subtract the tops. The bottom stays a × b.

Sort out the question
x is a whole number that is 1 or more, so x can be 1, 2, 3 and so on. You need the statement that must be true. That means it is true for every such x. One x that makes a statement false rules it out.
Test x = 1 in each statement
With x = 1, x + 1 is 2 and x + 2 is 3.
- A: $\frac{1}{2} > \frac{1}{3}$ . True.
- B: $\frac{1}{2} < \frac{1}{3}$ . False.
- C: $\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$ , which is not more than 1. False.
- D: $\frac{1}{6}$ is not more than $\frac{1}{1}$ . False.
Only A is true when x = 1. So B, C and D are out.
Make sure A is always true
Must means A has to work for every x, not just 1. In $\frac{1}{x + 1}$ and $\frac{1}{x + 2}$ , both tops are 1. The bottom x + 2 is always 1 more than x + 1. Cutting a whole into more pieces makes each piece smaller. So A is always true. Watch out: a bigger bottom makes a smaller fraction, not a bigger one. Thinking the other way gives choice B.
See why C and D always fail
Use the subtracting rule with a = x + 1 and b = x + 2.
$\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}$
$\frac{1}{x + 1} - \frac{1}{x + 2} = \frac{(x + 2) - (x + 1)}{(x + 1)(x + 2)} = \frac{1}{(x + 1)(x + 2)}$
The bottom is at least 2 × 3 = 6, so the difference is at most $\frac{1}{6}$ . That is less than 1. Its bottom is also bigger than x, so it is less than $\frac{1}{x}$ . Watch out: keep the bottom when you subtract. Keeping only the top, 1, leads to choice D. So the answer is A.

Check your answer

Try a bigger number, x = 9. A tenth of a pizza is more than an eleventh, so $\frac{1}{10} > \frac{1}{11}$ . The difference, $\frac{1}{110}$ , is tiny. It is less than 1 and less than $\frac{1}{9}$ . A still holds, and C and D still fail.

More questions like this

The original DOE question

Question 83 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 83. Open page 82 of the handbook (PDF)

Question 68 of 112

2023 Test A Math, question 29

Question

Integer x is evenly divisible by 3. Which expression below is also evenly divisible by 3?

A $2x + 1$
B $3x - 5$
C $4x - 1$
D $4x + 6$

Show the answer

Correct answer: D

Since integer x is evenly divisible by 3, substitute 3 for x in the answer options to determine which expression is also divisible by 3:

Let $x = 3$ .

Option A:

$2x + 1 = 2(3) + 1 = 7$

Option B:

$3x - 5 = 3(3) - 5 = 4$

Option C:

$4x - 1 = 4(3) - 1 = 11$

Option D:

$4x + 6 = 4(3) + 6 = 18$

Option D is the only value that is divisible by 3.

Question analysis

This number properties question says an integer x is divisible by 3 and asks which of four linear expressions in x is also divisible by 3, whatever multiple of 3 the integer is. Students can test a multiple of 3 in each expression or think about remainders. Testing a value that is not a multiple of 3 can make a wrong expression look right.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.4 common factors and multiples
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students rewrite and evaluate linear expressions, but they seldom ask whether one is divisible by 3 for every allowed value. One well-chosen substitution settles it, so the demand is about typical for grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
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 minor
Solution: 2 steps
Reading level: Question and choices: about grade 13, 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 algebra is light; the question turns on choosing a test value that obeys the condition. The habit of trying x = 1 backfires here, because 1 is not a multiple of 3 and it makes two wrong expressions come out divisible. Students who substitute 3, or who see that 4x and 6 are both multiples of 3 whenever x is, settle it quickly, and a second multiple such as 6 confirms it.

Variables instead of numbers moderate Understanding the question
Nothing pins x down except that it is a multiple of 3, so each expression must be judged for every such x, and any test value has to be a multiple of 3 itself, such as 3 or 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked minor Choosing the answer
The four expressions differ only in their coefficients and constants, from 2x + 1 to 4x + 6, so none can be dismissed at a glance; each needs a test or a look at its terms. (Embretson & Wetzel, 1987)

Strategy: Substitute a multiple of 3, such as 3 or 6, for x in every expression and keep the one whose value divides evenly by 3; a second multiple confirms it.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Tests x = 1, which is not a multiple of 3, finds that 2(1) + 1 = 3 divides evenly, and stops at the first choice; with x = 3 it gives 7.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Sees the 3 in 3x and takes the whole expression to be a multiple of 3, overlooking that subtracting 5 spoils it: with x = 3 it gives 4.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Tries a value that is not a multiple of 3, such as x = 4, where 4x − 1 = 15 divides evenly, though x = 3 gives 11.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding multiples of 3: $3a + 3b = 3(a + b)$; a and b are whole numbers, so 3a and 3b are multiples of 3. A multiple of 3 is a number that 3 divides evenly, with nothing left over. Adding two multiples of 3 always gives another multiple of 3, because you can pull the 3 out of both parts.

Sort out the question
“Evenly divisible by 3” means 3 goes into the number with nothing left over. Such a number is called a multiple of 3, like 3, 6, 9 or 12.
You know that x is a multiple of 3. You need the expression that is a multiple of 3 every time, whatever multiple of 3 x is.
Test x = 3 in every choice
Pick a number for x that follows the rule. The easiest one is 3. Put it into each expression.
Choice Work Divisible by 3?
A $2(3) + 1 = 7$ No
B $3(3) - 5 = 4$ No
C $4(3) - 1 = 11$ No
D $4(3) + 6 = 18$ Yes
Watch out: your test number must be a multiple of 3. With x = 1, choice A gives 3 and looks right. That leads to the wrong choice, A.
See why D always works
One test could be lucky, so see why D works for every x. Since x is a multiple of 3, 4x is one too. For example, 4 × 3 = 12 and 4 × 6 = 24. The number 6 is also a multiple of 3, since 6 = 3 × 2.
$3a + 3b = 3(a + b)$
So 4x + 6 is a multiple of 3 plus a multiple of 3. That is always a multiple of 3.
Rule out the other choices
Watch out: choice B starts with 3x, but then it subtracts 5. The 5 is not a multiple of 3, so it spoils the result, and 3(3) − 5 = 4. Choices A and C also add or take away a number that is not a multiple of 3.
So the answer is D, $4x + 6$ .

Choice	Work	Divisible by 3?
A	$2(3) + 1 = 7$	No
B	$3(3) - 5 = 4$	No
C	$4(3) - 1 = 11$	No
D	$4(3) + 6 = 18$	Yes

Check your answer

Try two more multiples of 3. With x = 6, choice D gives $4(6) + 6 = 30$ , and 30 ÷ 3 = 10. With x = 9, it gives $4(9) + 6 = 42$ , and 42 ÷ 3 = 14. Both divide evenly, so D keeps working.

More questions like this

The original DOE question

Question 86 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 86. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 69 of 112

2023 Test A Math, question 31

Question

$A circle divided into 6 equal spaces, labeled clockwise from the top right P, Q, R, S, T and U. An arrow from the center of the circle points into Space P.$

The arrow starts on Space P and moves clockwise around the circle. It moves through one space each minute. What space will the arrow point to in 100 minutes?

Show the answer

Correct answer: C

There are 6 spaces, so first divide 100 by 6: $100 \div 6 = 16$ , remainder 4. So, the arrow goes around 16 full times and then 4 more spaces. Four spaces from Space P is Space T.

Question analysis

This patterns question shows a circle split into six lettered spaces, with an arrow that starts on P and moves one space clockwise each minute, and asks where it points after 100 minutes. Students use division with a remainder to skip the full trips around. Counting the leftover moves without including the starting space, and in the right direction, is the tricky part.

NYS standards: 6.NS.2 multi-digit division
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Division with a remainder is grade 6 arithmetic, but using that remainder to step around a six-part circle is a pattern task that grade 7 classes seldom practice. It needs careful counting rather than new content.
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: No obvious method minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 1, 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 division is short; the points are lost in turning the remainder into a space. The arrow starts on P, so the 4 leftover moves land on Q, R, S and T, and counting P itself as a move stops one space short. Going around the wrong way, or one move too far, gives the other two choices. Checking a small case first, such as 7 minutes landing on Q, fixes the rule before it is used.

No obvious method minor Working it out
The stem never mentions division; the step that unlocks it is seeing that every 6 minutes brings the arrow back to P, so only the remainder of 100 ÷ 6 matters. (Schoenfeld, 1985)
Reading a figure, table or graph minor Understanding the question
The six spaces and their clockwise order, P, Q, R, S, T, U, appear only in the circle diagram, so the cycle length and the direction of counting both come from the figure. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Divide 100 by 6: every full trip around the circle brings the arrow back to P, so only the remainder matters; count that many moves clockwise, starting with the space after P.

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).
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: moderate
Counts the 4 leftover moves counterclockwise from P, through U, T and S to R, instead of clockwise as the stem says the arrow moves.
B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts Space P itself as the first of the 4 leftover moves, P, Q, R, S, so the arrow stops one space short of where it really points.
D Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Makes 5 moves past P instead of the 4 left after 16 full turns, overshooting by one space, the fence-post slip in the other direction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division with a remainder: $a = b \times q + r$; a is the number you divide, here the minutes. b is the number you divide by, here the spaces in one trip around. q is how many whole groups of b fit into a. r is the remainder, the amount left over. The remainder is always less than b.

Sort out the question
The circle has 6 spaces. Going clockwise, the way a clock’s hands turn, they are P, Q, R, S, T and U. The arrow starts on P and moves one space each minute. You need the space it points to after 100 minutes.
Notice that the pattern repeats
After 6 minutes, the arrow has gone all the way around and points to P again. After 12 minutes it is back on P again, and so on. So full trips of 6 minutes don’t change where it points. Only the leftover minutes matter.
Divide 100 by 6
$a = b \times q + r$
$100 = 6 \times 16 + 4$
The arrow makes 16 full trips. That takes $6 \times 16 = 96$ minutes, and it is back on P. There are 4 minutes left over.
Count the last 4 moves
From P, move one space each minute: Q (minute 97), R (minute 98), S (minute 99), T (minute 100).
Watch out: don’t count P itself as the first move. That stops on S, choice B. Also count clockwise. Counting the other way lands on R, choice A.
So the answer is C, T.

Check your answer

Use a smaller number with the same remainder. 100 − 96 = 4, and 96 minutes is a whole number of trips. So after 100 minutes the arrow points where it does after 4 minutes. Four moves from P are Q, R, S, T. That is T again.

More questions like this

The original DOE question

Question 88 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 88. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 70 of 112

2023 Test A Math, question 42

Question

$\{1,2,3,4,5,\ldots,198,199,200\}$

How many members of the set shown above are multiples of 6 but not multiples of 9?

Show the answer

Correct answer: D

First, list the first several multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54 . . .

Every third multiple (i.e., 18, 36, 54, . . .) is also a multiple of 9. Thus, $\frac{2}{3}$ of the multiples of 6 are not multiples of 9.

Next, calculate how many multiples of 6 are in the given number range. The last multiple of 6 would be 198. So, there are a total of $\frac{198}{6} = 33$ multiples of 6. Since $\frac{2}{3}$ of those are not multiples of 9, the answer is $\frac{2}{3}(33) = 22$ .

Question analysis

This number properties question shows the set of whole numbers from 1 to 200 and asks how many of them are multiples of 6 but not multiples of 9. Students count multiples in a range and remove the overlap, the numbers divisible by both. Finding the right common multiple, and removing only the overlap, is where it gets tricky.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Multiples and common multiples are grade 6 content, but counting the members of a set that meet one condition and fail another is rare in grade 7 work. The overlap between the two kinds of multiples must be found and removed exactly.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Much to keep track of minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Counting multiples of 6 is routine; the harder decision is how to remove the ones that are also multiples of 9. Those are the multiples of 18, the least common multiple, not of 54 or 15, and students who count them correctly must still subtract them rather than report them. Listing 6, 12, 18, 24, 30, 36 shows the pattern, two kept for every one removed, which also gives the answer as two thirds of 33.

No obvious method moderate Working it out
The stem never mentions a common multiple: students must see that the multiples of 6 that are also multiples of 9 are the multiples of 18, every third multiple of 6. (Schoenfeld, 1985)
Much to keep track of minor Working it out
Three counts are in play at once, 33 multiples of 6 up to 198, 11 multiples of 18 up to 198, and their difference, and two of them are listed choices. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math minor Understanding the question
The bold not in “multiples of 6 but not multiples of 9” asks for what is left after a removal, a set difference rather than a count of the numbers removed. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Count the multiples of 6 up to 200, find the smallest number that is a multiple of both 6 and 9, count its multiples in the set, and subtract.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Counts the 11 multiples of 18, the numbers to be removed, and gives that count instead of the multiples of 6 that remain.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Takes 6 + 9 = 15 as the step between numbers divisible by both, and gives its 13 multiples up to 195, the numbers it would remove.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Uses 6 + 9 = 15 instead of 18 as the common multiple and removes its 13 multiples from the 33 multiples of 6, leaving 20.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting multiples: $\text{last multiple} = n \times k$; n is the number whose multiples you count, such as 6. The first multiple is n × 1, the second is n × 2, and so on. So if the largest multiple in the set is n × k, there are k multiples in all.
Least common multiple: $\text{LCM}(6, 9) = 18$; The least common multiple, or LCM, is the smallest number that two numbers both divide evenly. A number is a multiple of both 6 and 9 exactly when it is a multiple of 18.

Sort out the question
The set holds the whole numbers from 1 to 200. A multiple of 6 is a number that 6 divides evenly, like 6, 12 or 18. You need the multiples of 6 that are not multiples of 9. So count all the multiples of 6. Then take away those that are also multiples of 9.
Count the multiples of 6
200 is not a multiple of 6, but 198 is. So 198 is the last multiple of 6 in the set.
$\text{last multiple} = n \times k$
$198 = 6 \times 33$
There are 33 multiples of 6.
Find the numbers to take away
List the multiples of 6: 6, 12, 18, 24, 30, 36. The first one 9 also divides is 18, then 36. So every third multiple of 6 is also a multiple of 9.
$\text{LCM}(6, 9) = 18$
Watch out: the shared multiples are multiples of 18, not of 6 + 9 = 15. Using 15 leads to 13 or 20, choices B and C.
Count the multiples of 18
The last multiple of 18 in the set is also 198.
$198 = 18 \times 11$
There are 11 of them. Watch out: these are the numbers you take away, not the answer. Stopping here gives 11, choice A.
Take away the multiples of 18
$33 - 11 = 22$
There are 22 multiples of 6 left over. So the answer is D, 22.

Check your answer

Every third multiple of 6 is a multiple of 18, so two out of every three multiples of 6 stay. Two thirds of 33 is $\frac{2}{3} \times 33 = 22$ . That matches. Test one: 30 is a multiple of 6, and 30 ÷ 9 is not a whole number, so 30 counts.

More questions like this

The original DOE question

Question 99 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 99. Open page 86 of the handbook (PDF)

Question 71 of 112

2023 Test A Math, question 54

Question

A train travels 2,200 miles from Phoenix to New York City. The train covers the first 240 miles in 5 hours. If the train continues to travel at this rate, how many more hours will it take to reach New York City? Round your answer to the nearest whole hour.

Show the answer

Correct answer: D

First, find the speed of the train in miles per hour: $240 \div 5 = 48$ .

The number of miles left to travel is $2{,}200 - 240 = 1{,}960$ .

To find the number of hours left, use the equation $rt = d$ :

$48t = 1{,}960$	Apply the multiplicative inverse property; divide both sides of the equation by 48.
$t = \frac{1{,}960}{48} = 40\frac{5}{6}$

That number rounded to the nearest whole hour is 41 hours.

Question analysis

This rate word problem describes a 2,200-mile train trip from Phoenix to New York City in which the first 240 miles take 5 hours, and asks how many more hours the rest will take, rounded to the nearest hour. Students find a unit rate and the remaining distance. The choices are close together, so the time for the whole trip is a tempting trap.

NYS standards: 6.RP.3b unit rate problems
6.NS.2 multi-digit division
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Unit rates are grade 6 content, but this chains four steps: a speed, the distance left, a long division that does not come out even, and rounding. The word more must be read closely, which puts it at typical grade 7 demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Turning words into math minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is routine, but the question has a long chain and asks for only part of the trip. Dividing the whole 2,200 miles by the speed is the natural shortcut and gives a listed choice, and the rounding instruction adds a second place to go wrong. Checking that 41 hours at 48 miles per hour covers about 1,970 miles, close to the 1,960 left, confirms the answer.

Many steps moderate Working it out
Four steps follow in order, the speed 240 ÷ 5 = 48, the miles left 2,200 − 240 = 1,960, the time 1,960 ÷ 48, and the rounding, and an early slip carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The bold more in “how many more hours” limits the question to the 1,960 miles still ahead, though 2,200, the whole trip, is the distance the stem names first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Dividing 1,960 by 48 leaves 40 with a remainder of 40, so the time, 40 5/6 hours, must be rounded up to the nearest hour rather than cut short. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the train's speed from the first 240 miles, subtract those miles from the total to get the distance left, divide by the speed, and round only at the end.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the whole 2,200 miles by 48 miles per hour, about 45.8, or 46 hours, the time for the entire trip rather than the hours still to go.
B Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: weak
Divides the whole 2,200 miles by 48 and then cuts 45.8 down to 45 instead of rounding to the nearest hour, two errors at once.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Subtracts 240 from 2,200 without reducing the hundreds after borrowing, getting 2,060 instead of 1,960, and 2,060 ÷ 48 ≈ 42.9 rounds to 43.

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. If you know two of them, you can find the third. To find the speed, divide the distance by the time. To find the time, divide the distance by the speed.

Sort out the question
The whole trip is 2,200 miles. The train goes the first 240 miles in 5 hours, and it keeps that speed. You need how many more hours it takes, rounded to the nearest whole hour. “More” means the time after the first 240 miles.
Find the train’s speed
$d = r \times t$
$240 = r \times 5$
$r = 240 \div 5 = 48$
Divide the 240 miles by the 5 hours. The train goes 48 miles per hour.
Find the miles still to go
$2{,}200 - 240 = 1{,}960$
There are 1,960 miles left. Watch out: use the miles left, not the whole trip. Dividing 2,200 by 48 gives about 45.8 hours, which leads to 46 or 45, choices A and B.
Find the hours still to go
$1{,}960 = 48 \times t$
$t = 1{,}960 \div 48 = 40\frac{40}{48} = 40\frac{5}{6}$
Here 48 × 40 = 1,920, and 40 miles are left over. That leftover is $\frac{40}{48} = \frac{5}{6}$ of an hour.
Round to the nearest hour
$\frac{5}{6}$ of an hour is more than half an hour, so round up to 41. Watch out: don’t just drop the fraction.
So the answer is D, 41 hours.

Check your answer

At 48 miles per hour, 41 hours covers $48 \times 41 = 1{,}968$ miles. That is just a bit more than 1,960, as it should be after rounding up. In 40 hours, the train would cover only 1,920 miles, which is not enough.

More questions like this

The original DOE question

Question 111 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 111. Open page 90 of the handbook (PDF)

Question 72 of 112

2023 Test A Math, question 56

Question

$Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. The line is shaded from the unlabeled tick halfway between 0 and 1 to the unlabeled tick halfway between 1 and 2, with closed dots at both ends.$

If all possible values of x are indicated by the shaded part of the number line above, which number line best shows all possible values of $\frac{1}{x}$ ?

A $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 1 and 2 to the tick halfway between 2 and 3, with closed dots at both ends.$
B $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from 1 to 2, with closed dots at both ends.$
C $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 0 and 1 to 2, with closed dots at both ends.$
D $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from a closed dot between the tick halfway between 0 and 1 and the tick at 1, about two-thirds of the way from 0 to 1, to a closed dot at 2.$

Show the answer

Correct answer: D

The shaded region goes from $\frac{1}{2}$ to $\frac{3}{2}$ . So, the possible values of $\frac{1}{x}$ would be between the reciprocal of $\frac{1}{2}$ and the reciprocal of $\frac{3}{2}$ , which are 2 and $\frac{2}{3}$ , respectively. Find the number line in the answer options that has a shaded region that ranges from $\frac{2}{3}$ to 2, which is option D.

Question analysis

This number line question shades all possible values of x from one half to three halves and asks which number line shows all possible values of 1/x. Students find reciprocals of fractions and work out how a whole range of values changes. Keeping track of what happens to each end of the range, and placing fractions accurately on the line, makes it tricky.

NYS standards: 6.NS.1 dividing fractions by fractions
6.NS.7a inequality as relative position
6.NS.6c positioning rational numbers and points
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Reciprocals of single fractions are grade 6 work. Applying them to every value in a shaded range at once, and working out where the new range begins and ends, is reasoning about a whole set of numbers that grade 7 classes rarely practice.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 12, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Few students have flipped an interval before, so the question rewards testing its two ends and noticing that the order reverses. The harder part comes at the end: 2/3 falls between the tick marks, and a graph ending at the nearby 1/2 tick looks right at a glance. Students who believe a reciprocal is always at least 1, or who shift the interval instead of flipping it, reach the other graphs.

Reading a figure, table or graph moderate Understanding the question
The shaded segment's ends sit on unlabeled half-unit ticks, 1/2 and 3/2, and each choice must be read the same way, with 2/3 falling between ticks rather than on one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer moderate Choosing the answer
The graph from 1/2 to 2 matches the right graph's right end and differs at the left end by only a sixth of a unit, 1/2 against 2/3. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
The question asks about 1/x for every x in a range, so a whole interval must be transformed, and its ends swap: the smallest x gives the largest 1/x. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Take the reciprocal of each end of the shaded part, ask whether a bigger x gives a bigger or a smaller 1/x, and look for the graph whose endpoints are exactly those values.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds 1 to each endpoint instead of taking reciprocals, sliding the shaded part from 1/2–3/2 one unit right, to 3/2–5/2.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Gets 2 from 1/2 but assumes a reciprocal cannot be less than 1, as flipping a proper fraction suggests, so it starts the shading at 1 instead of 2/3.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: strong
Finds the reciprocals 2/3 and 2 but places 2/3 at the half-unit tick, 1/2, though 2/3 lies a third of the way from that tick toward 1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reciprocal: $\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$; a and b are the top and bottom of a fraction. The reciprocal flips the fraction upside down. The number 1 over x is the reciprocal of x. A number times its reciprocal always makes 1.

Sort out the question
x can be any number in the shaded part of the first number line. You need the number line that shows every possible value of $\frac{1}{x}$ , the reciprocal of x. The reciprocal is the number flipped upside down, like $\frac{1}{2}$ for 2.
Read the shaded part
The ticks come every half unit. The shading starts at the tick halfway between 0 and 1, which is $\frac{1}{2}$ . It ends halfway between 1 and 2, which is $\frac{3}{2}$ , or $1\frac{1}{2}$ . Both dots are filled in, so both ends count.
Find the reciprocals of the ends
Flip each end of the shaded part.
$\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$
$\text{reciprocal of } \frac{1}{2} = \frac{2}{1} = 2$
$\text{reciprocal of } \frac{3}{2} = \frac{2}{3}$
So the ends of the new shading are 2 and $\frac{2}{3}$ .
Notice that the ends swap
When x gets bigger, $\frac{1}{x}$ gets smaller. So the smallest x, $\frac{1}{2}$ , gives the biggest value, 2. The biggest x, $\frac{3}{2}$ , gives the smallest, $\frac{2}{3}$ . So $\frac{1}{x}$ runs from $\frac{2}{3}$ to 2, ends included.
Find the matching number line
$\frac{2}{3}$ lies between the half tick and 1, a third of the way from $\frac{1}{2}$ to 1. Line D starts there and ends at 2. Watch out: line C starts at the half tick, $\frac{1}{2}$ , not at $\frac{2}{3}$ . Line B starts at 1.
So the answer is D, the line shaded from $\frac{2}{3}$ to 2.

Check your answer

Test a value in the middle. x = 1 is shaded, and $\frac{1}{1} = 1$ . On line D, 1 is shaded too. Test an end: $\frac{3}{2} \times \frac{2}{3} = 1$ , so $\frac{2}{3}$ really is the reciprocal of $\frac{3}{2}$ .

More questions like this

The original DOE question

Question 73 of 112

2023 Test B Math, question 3

Question

A kindergarten teacher has 72 crayons, 60 pencils, and 84 sheets of paper to distribute to the children in her class. If each child receives an equal number of each item and there are no items remaining, what is the greatest possible number of children in the class?

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

Show the answer

Correct answer: 12

The greatest possible number of children in the class is the greatest common factor of 72, 60, and 84. Write the prime factorization of each number:

$72 = 2^{3} \times 3^{2}$

$60 = 2^{2} \times 3 \times 5$

$84 = 2^{2} \times 3 \times 7$

Then, find the factors that all three have in common, which is $2^{2} \times 3 = 12$ .

Question analysis

A kindergarten teacher shares 72 crayons, 60 pencils and 84 sheets of paper so that every child gets the same number of each item, with nothing left over. This grid-in asks for the greatest possible number of children in the class. Students must recognize a greatest common factor problem and not confuse it with a least common multiple.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Greatest common factors are grade 6 content, normally of two numbers. This one uses three numbers and a word setting that must be translated, but the factoring is short, so the demand stays below typical grade 7 work.
Estimated difficulty: Straightforward: two steps or one careful reading.
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: 2 steps
Reading level: Question and choices: about grade 11, 45 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The story has to be recognized as a sharing problem before any arithmetic starts, and the word greatest points to the largest divisor shared by all three counts. Students who spot a small shared factor can stop too soon, while those who think of a common multiple head the wrong way entirely. Checking that 72, 60 and 84 all divide by the entry, and that no larger number does, settles it.

Turning words into math moderate Understanding the question
Nothing names a greatest common factor; “each child receives an equal number of each item and there are no items remaining” means the class size must divide 72, 60 and 84 exactly. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Finding the greatest common factor of three numbers, by prime factorization or by listing factors, is a grade 6 method the question assumes rather than explains. (Anderson & Krathwohl, 2001)

Strategy: Write 72, 60 and 84 as products of primes, multiply the primes all three share, and check that each amount divides evenly by the result.

Traps

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

The wrong answers, classified

6 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Spots that 6 divides 72, 60 and 84 and stops there, a common factor not yet multiplied by the extra 2 that all three share.
2520 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Finds the least common multiple of 72, 60 and 84, 2,520, confusing a number all three counts divide into with one that divides all three.
4 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Takes only the shared 2 × 2 from the prime factorizations and drops the factor 3 that 72, 60 and 84 also share.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor: $\text{GCF} = \text{product of the shared prime factors}$; A factor of a number divides it with nothing left over. The GCF is the greatest number that is a factor of every number in the group. A prime number has only 1 and itself as factors, like 2, 3, 5 and 7. Multiply the primes that all the numbers share, as many times as each number has them.

Sort out the question
The teacher has 72 crayons, 60 pencils and 84 sheets of paper. Every child gets the same number of each item, with none left over.
So the number of children must divide 72, 60 and 84 with no remainder. A number like that is a common factor. You need the greatest one.
Break each number into primes
Write each number as a product of primes. A product is the answer to a multiplication.
$72 = 2 \times 2 \times 2 \times 3 \times 3$
$60 = 2 \times 2 \times 3 \times 5$
$84 = 2 \times 2 \times 3 \times 7$
You can check each one by multiplying it back out.
Pick out the shared primes
Look for the primes that appear in all three lists. Each list has at least two 2s and one 3. Only 72 has a third 2 and a second 3. The 5 and the 7 appear in one list each.
$\text{GCF} = 2 \times 2 \times 3 = 12$
Watch out: keep every shared prime. Using only one 2 gives 6. Leaving out the 3 gives 4.
Give the answer
Watch out: 2,520 is the smallest number that all three numbers divide into. That is the least common multiple, a different idea. The class can’t have 2,520 children when there are only 60 pencils. So you would grid 12.

Check your answer

Share the items among 12 children. Each child gets $72 \div 12 = 6$ crayons, $60 \div 12 = 5$ pencils and $84 \div 12 = 7$ sheets. Nothing is left over. The numbers 6, 5 and 7 share no factor but 1, so no bigger class works.

More questions like this

The original DOE question

Question 60 as printed on page 191 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 60. Open page 191 of the handbook (PDF)

Question 74 of 112

2023 Test B Math, question 9

Question

Bryan completed a 100-meter race in 11.74 seconds. Luis completed the same race in 11.69 seconds. What was the difference between their times, expressed as a fraction of a second?

A $\frac{1}{20}$
B $\frac{1}{10}$
C $\frac{3}{20}$
D $\frac{1}{5}$

Show the answer

Correct answer: A

First, subtract Luis’s time from Bryan’s time: $11.74 - 11.69 = 0.05$ .

Then, convert to a fraction:

$0.05 = \frac{5}{100} = \frac{1}{20}$ .

Question analysis

Bryan ran a 100-meter race in 11.74 seconds and Luis ran the same race in 11.69 seconds. Students find the difference between their times and express it as a fraction of a second. It tests subtracting decimals and rewriting a decimal as a fraction in lowest terms.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Subtracting decimals and writing a decimal as a fraction are skills from before grade 7. The two times are close, so the subtraction is quick, and the only care needed is with place value and simplifying.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two quick skills settle this question: lining up the decimals to subtract, and naming the result as a fraction in lowest terms. The two times differ only in their last two digits, so the subtraction is short, but it needs a borrow that some students skip. The fraction step is where a decimal’s digit gets mistaken for a denominator.

Error-prone arithmetic minor Working it out
Subtracting 11.69 from 11.74 needs a borrow in the hundredths, 14 − 9, after which the tenths cancel to 0, the step where place value goes wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The difference, 0.05, must be read as five hundredths and reduced as a fraction; the question gives no help with converting a decimal. (Anderson & Krathwohl, 2001)

Strategy: Line up the decimal points, subtract, write the difference as a number of hundredths over 100, and reduce the fraction.

Traps

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

The wrong answers, classified

B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Looks only at the tenths, 11.7 − 11.6 = 0.1, dropping the hundredths, a place-value slip that doubles the true gap.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts the smaller digit from the larger in each place, 9 − 4 = 5 and 7 − 6 = 1, getting 0.15, which is 3/20.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Gets 0.05 but reads its digit 5 as a denominator, writing 1/5, instead of five hundredths, 5/100.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths: $\text{a decimal with two places} = \frac{\text{its digits}}{100}$; The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths, and you can write its digits over 100. For example, 0.37 is 37/100. Then simplify by dividing the top and bottom by the same number.

Sort out the question
Bryan ran the race in 11.74 seconds. Luis ran it in 11.69 seconds. The difference is how much longer one time is than the other. You need it as a fraction of a second.
Subtract the two times
Subtract the smaller time from the larger one. Line up the decimal points.
$11.74 - 11.69 = 0.05$
In the hundredths place, 4 is less than 9. So regroup: trade 1 tenth for 10 hundredths. Then 14 − 9 = 5 hundredths. The tenths are now 6 − 6 = 0.
Watch out: don’t just take the smaller digit from the larger one in each place. That gives 0.15, which is $\frac{3}{20}$ , choice C.
Write the difference as a fraction
The 5 is in the hundredths place, so 0.05 is 5 hundredths.
$0.05 = \frac{5}{100}$
Now simplify. Divide the top and bottom by 5.
$\frac{5 \div 5}{100 \div 5} = \frac{1}{20}$
Give the answer
Watch out: 0.05 is not $\frac{1}{5}$ . One fifth is 0.2, which is four times as big. So the answer is A, $\frac{1}{20}$ .

Check your answer

Turn $\frac{1}{20}$ back into a decimal: $1 \div 20 = 0.05$ . Then add it to Luis’s time: $11.69 + 0.05 = 11.74$ . That is Bryan’s time, so the difference is right.

More questions like this

The original DOE question

Question 66 as printed on page 193 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 66. Open page 193 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 75 of 112

2023 Test B Math, question 11

Question

$n = 0.13 + \frac{207}{100}$

What is the value of n expressed as a fraction?

A $\frac{1}{3}$
B $\frac{11}{5}$
C $\frac{9}{4}$
D $\frac{23}{10}$

Show the answer

Correct answer: B

$n = 0.13 + \frac{207}{100} =$

$\frac{13}{100} + \frac{207}{100} =$

$\frac{220}{100} = \frac{11}{5}$

Question analysis

The number n equals 0.13 plus 207/100, and students must write n as a single fraction. This question mixes a decimal with a fraction in hundredths. Converting both to the same form, adding, and reducing to lowest terms are the steps, and the four fractions offered look quite different at first glance.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting decimals to fractions and simplifying are skills taught before grade 7. Both numbers here are in hundredths, so the addition is immediate and only the final reduction takes thought.
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: An answer in an unexpected form minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Converting 0.13 to hundredths makes this a single addition, 13 + 207 = 220 hundredths, so the question tests little beyond reducing a fraction. Two of the wrong choices sit close to 2.2 but come from no common working, and the third only echoes the digits of 0.13. A student who checks each choice as a decimal finds only one equal to 2.2.

An answer in an unexpected form minor Choosing the answer
The sum comes out as 220/100 or 2.2, but the choices are in lowest terms, so it must be reduced, or each choice converted, before anything matches. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
One term is a decimal and the other a fraction; writing 0.13 as 13/100 puts both in hundredths, and a misplaced digit there changes the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write 0.13 as 13/100 so both terms are in hundredths, add the numerators, then divide the top and bottom by their greatest common factor.

Traps

Answer in another form: The right answer is written differently from the way you worked it out.
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: weak
Echoes the digits of 0.13, reading it as the fraction 1/3, and ignores the 207/100 altogether.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 9/4, which is 2.25; adding 13 and 207 hundredths gives 2.2, so this mostly draws guesses near the sum.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Nothing leads to 23/10 except 13 + 207 coming out as 230, a slip with no reason behind it, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths: $\text{a decimal with two places} = \frac{\text{its digits}}{100}$; The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths. Write its digits over 100. For example, 0.47 is 47/100.
Adding fractions with the same bottom: $\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$; a and b are the tops. c is the shared bottom, the denominator. When the bottoms match, the pieces are the same size. So add the tops and keep the bottom.

Sort out the question
n equals 0.13 plus $\frac{207}{100}$ . One part is a decimal and one is a fraction. You need n as a fraction. The choices are in lowest terms, which means they can’t be simplified any more.
Write the decimal as a fraction
The 3 in 0.13 is in the hundredths place. So 0.13 is 13 hundredths.
$0.13 = \frac{13}{100}$
Now both numbers are in hundredths. Watch out: 0.13 is not $\frac{1}{3}$ . That only copies the digits, choice A.
Add the hundredths
Use the rule for adding fractions. The bottoms match, so add the tops and keep the 100 on the bottom.
$\frac{13}{100} + \frac{207}{100} = \frac{13 + 207}{100} = \frac{220}{100}$
Simplify the fraction
$\frac{220}{100}$ is not a choice, so simplify it. Find the greatest number that divides both 220 and 100. That is 20.
$\frac{220 \div 20}{100 \div 20} = \frac{11}{5}$
So the answer is B, $\frac{11}{5}$ .

Check your answer

Change both to decimals. $\frac{11}{5} = 11 \div 5 = 2.2$ . And $0.13 + 2.07 = 2.20$ . They match. Choices C and D are close, but $\frac{9}{4} = 2.25$ and $\frac{23}{10} = 2.3$ .

More questions like this

The original DOE question

Question 68 as printed on page 194 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 68. Open page 194 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 76 of 112

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 77 of 112

2023 Test B Math, question 31

Question

If $xy=30$ , what is the value of $10\div x$ in terms of y?

A $\frac{1}{10}y$
B $\frac{1}{3}y$
C $3y$
D $10y$

Show the answer

Correct answer: B

Solve $xy=30$ , the given equation, for x:

$xy=30$

$x=\frac{30}{y}$

Then:

$10\div x=10\div\frac{30}{y}=10\times\frac{y}{30}=\frac{1}{3}y$

Question analysis

An algebra question: given that x times y equals 30, students express 10 divided by x using only y. It tests solving for one variable and dividing by an algebraic fraction. Working with letters instead of numbers, and dividing in the right order, are what trip students up.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.1 dividing fractions by fractions
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students rewrite expressions that have numbers and one variable. This one links two variables through their product and asks for a quotient, with no numbers given to check an answer against.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The algebra is three short moves, but each is a place to turn a fraction upside down: solving xy = 30 for x, dividing 10 by that fraction, and reducing 10/30. Students who reach for the bigger-over-smaller division they know from whole numbers land on 3y. Choosing two numbers with a product of 30, such as 3 and 10, and testing every choice settles the question in seconds.

Variables instead of numbers moderate Understanding the question
x and y are tied only by their product, 30, and the answer must be an expression in y, so there is no number to check unless students pick a pair such as x = 3 and y = 10. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Substituting x = 30/y turns 10 ÷ x into 10 divided by a fraction, which must be flipped to 10 × y/30 and then reduced. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The answer, (1/3)y, and the choice 3y use the same 3, one dividing y and one multiplying it, so a student unsure whether 10 ÷ 30 or 30 ÷ 10 applies sees both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Solve xy = 30 for x, substitute that fraction into 10 ÷ x, and divide by multiplying by its reciprocal; or choose values of x and y with a product of 30 and test each choice.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Writes the 10 from 10 ÷ x under y as one tenth of y, never using the product xy = 30 to link the two letters.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Replaces x with 30/y correctly but divides 30 by 10 instead of 10 by 30, taking the larger number over the smaller, and gets 3y.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats dividing by x as multiplying by y, as if x and y were reciprocals, though their product is 30, not 1, which gives 10y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is the number you divide. b over c is the fraction you divide by: b is its top and c is its bottom. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. For example, 6 ÷ 1/2 = 6 × 2 = 12, because 12 halves fit in 6.

Sort out the question
$xy$ means x times y. So x times y is 30. You need $10 \div x$ “in terms of y.” That means your answer can use y and numbers, but no x.
Solve for x
To swap x out, first get x by itself. Divide both sides of $xy = 30$ by y.
$x = \frac{30}{y}$
This says x is 30 divided by y.
Put the fraction in place of x
Replace x in $10 \div x$ with this fraction.
$10 \div x = 10 \div \frac{30}{y}$
Now use the rule for dividing by a fraction. Flip the fraction and multiply.
$10 \div \frac{30}{y} = 10 \times \frac{y}{30} = \frac{10y}{30}$
Simplify the fraction
Both 10 and 30 can be divided by 10.
$\frac{10y}{30} = \frac{y}{3} = \frac{1}{3}y$
Watch out: 10 ÷ 30 is not 3. Dividing the bigger number by the smaller one gives 3y, choice C. Also, dividing by x is not the same as multiplying by y. That only works when xy is 1, not 30. It leads to choice D.
So the answer is B, $\frac{1}{3}y$ .

Check your answer

Pick numbers that multiply to 30, like x = 3 and y = 10. Then $10 \div x = 10 \div 3 = \frac{10}{3}$ . Choice B gives $\frac{1}{3} \times 10 = \frac{10}{3}$ . They match. Choice C would give 30, and choice D would give 100.

More questions like this

The original DOE question

Question 88 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 88. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 78 of 112

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 79 of 112

2023 Test B Math, question 39

Question

In a certain competition, 1 person finished in first place, 5 people tied for second place, 10 people tied for third place, and the other 4,984 people did not place. What percentage of the people in the competition placed first, second, or third altogether?

A0.0032%
B0.032%
C0.32%
D3.2%

Show the answer

Correct answer: C

The total number of people who participated in the competition was $1+5+10+4{,}984=5{,}000$ . The total number who placed first, second, or third was $1+5+10=16$ .

$\frac{16}{5{,}000}=\frac{32}{10{,}000}=0.0032=0.32\%$

Question analysis

A percent question about a competition with 1 first-place finisher, 5 tied for second, 10 tied for third and 4,984 who did not place. Students find what percentage of everyone placed. All four choices use the same digits, so the challenge is placing the decimal point when a decimal becomes a percent.

NYS standards: 6.RP.3c percent as a rate per 100
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Writing part of a whole as a percent is grade 6 content. This one first totals four groups, then turns a fraction of a large total into a percent, where a slip of one place value is easy to make.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
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 10, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is short, so the question tests decimal places. All four choices share the digits 32 and differ only in where the point falls, so a slip of one place, or forgetting to multiply by 100, still finds a choice waiting. A rough check helps: 16 is a bit more than 15, and 15 of 5,000 is 0.3%, so the answer must sit just above 0.3%.

Error-prone arithmetic moderate Working it out
16 ÷ 5,000 = 0.0032 has two zeros after the decimal point, and writing it as a percent moves the point two more places; each move can land one place off. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The total of 5,000 is not given; it has to be built from 1 + 5 + 10 + 4,984 before the 16 people who placed can be compared with it. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the people who placed and the people who did not to get the total, write the placers as a fraction of it, then change that fraction to a decimal and multiply by 100.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Finds 16 ÷ 5,000 = 0.0032 and adds a percent sign without multiplying by 100, as if a decimal were already a percent.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Finds 0.0032 but moves the decimal point only one place when writing the percent, giving 0.032%, ten times too small.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Divides 16 by 500 instead of 5,000, dropping a zero, or moves the point three places, giving 3.2%, ten times too large.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the groups
Some people placed: 1 in first, 5 in second and 10 in third. The other 4,984 people did not place. You need the percent of all the people who placed.
Count the people who placed
These are the people who finished first, second or third. Add them.
$1 + 5 + 10 = 16$
So 16 people placed.
Count everyone
The whole is everyone in the competition: the people who placed plus the people who did not.
$16 + 4{,}984 = 5{,}000$
So 5,000 people took part. Notice that 4,984 is not the whole. It leaves out the 16 who placed.
Turn the share into a percent
Use the percent formula.
$\text{percent} = \frac{16}{5{,}000} \times 100$
First divide: $16 \div 5{,}000 = 0.0032$ . Then multiply by 100. That moves the decimal point two places to the right.
$0.0032 \times 100 = 0.32$
So 0.32% of the people placed.
Avoid the decimal traps
Watch out: 0.0032 is not a percent yet. Putting a % sign on it gives choice A, which is 100 times too small. Moving the point only one place gives choice B. Moving it three places gives choice D.
So the answer is C, 0.32%.

Check your answer

Estimate. 1% of 5,000 is 50. The 16 people are about a third of 50. So the answer should be about a third of 1%, or about 0.3%. Choice C, 0.32%, fits.

More questions like this

The original DOE question

Question 96 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 96. Open page 202 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 80 of 112

2023 Test B Math, question 51

Question

A number, n, is divided by 25. If the quotient is 10 with a remainder of 5, what is the quotient expressed as a decimal?

A10.20
B10.25
C10.50
D10.75

Show the answer

Correct answer: A

A quotient of 10 with a remainder of 5 means the quotient is

$10+\frac{5}{25}=10\frac{1}{5}=10.20$

Question analysis

A number is divided by 25, giving a quotient of 10 with a remainder of 5, and students write the quotient as a decimal. It tests what a remainder means and how it becomes part of a decimal quotient. Connecting the remainder to the decimal part of the answer is the key idea.

NYS standards: 6.NS.2 multi-digit division
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Division with remainders and decimal quotients are grade 6 skills. This takes two short steps, with no word problem to translate, though students must know what a remainder means once a quotient is written as a decimal.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know moderate
Solution: 2 steps
Reading level: Question and choices: about grade 5, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question checks one idea: a remainder is a fraction of the divisor, not a string of digits to place after the point. Students who write the remainder 5 as tenths, or set the divisor 25 after the point, land on listed choices. Rebuilding n = 25 × 10 + 5 = 255 and dividing by 25 confirms the decimal.

Something you must already know moderate Working it out
A remainder of 5 means 5 out of the 25 needed for another whole, so it becomes 5/25 = 0.2, a link between remainders and decimals the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Rebuild n from the quotient and the remainder, then divide it by 25 to a decimal; or write the remainder as a fraction of the divisor.

Traps

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

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Writes the divisor, 25, after the decimal point, setting the stem’s numbers 10 and 25 side by side instead of turning the remainder into a fraction.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Writes the remainder 5 as the tenths digit, 10.5, believing a remainder becomes the digits after the decimal point.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 10.75; it treats the remainder as three quarters of the divisor, which nothing in the stem suggests, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Quotient with a remainder: $\text{quotient} = q + \frac{r}{d}$; d is the divisor, the number you divide by. q is the whole-number part of the quotient, the answer to the division. r is the remainder, what is left over. The leftover still has to be divided by d, so it becomes the fraction r over d.

Sort out the words
n is divided by 25. So 25 is the divisor, the number you divide by. The quotient is the answer to a division. Here it is 10 with a remainder of 5. The remainder is the amount left over. You need the exact quotient as a decimal.
Turn the remainder into a fraction
The 5 left over still has to be divided by 25. So it becomes the fraction $\frac{5}{25}$ . Use the formula.
$\text{quotient} = 10 + \frac{5}{25}$
Change the fraction to a decimal
Simplify first. Divide the top and bottom by 5. Then write the fraction in hundredths.
$\frac{5}{25} = \frac{1}{5} = \frac{20}{100} = 0.20$
So the quotient is 10 + 0.20 = 10.20.
Give the answer
Watch out: A remainder is not the digits after the decimal point. Writing the 5 after the point gives 10.50, choice C. Writing the 25 after it gives 10.25, choice B.
So the answer is A, 10.20.

Check your answer

Rebuild n. Ten 25s plus the 5 left over is $25 \times 10 + 5 = 255$ . Now divide: $255 \div 25 = 10.2$ . That matches 10.20.

More questions like this

The original DOE question

Question 108 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 108. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 81 of 112

2023 Test B Math, question 54

Question

The speed 10 meters per second is equivalent to how many feet per hour? (Use the approximation 1 foot = 0.3 meter.)

A10,800
B12,000
C36,000
D120,000

Show the answer

Correct answer: D

$\frac{10\text{ m}}{1\text{ sec}}\times\frac{3{,}600\text{ sec}}{1\text{ h}}\times\frac{1\text{ ft}}{0.3\text{ m}}=$

$\frac{36{,}000\text{ ft}}{0.3\text{ h}}=120{,}000\,\frac{\text{ft}}{\text{hr}}$

Question analysis

This unit conversion question gives a speed of 10 meters per second and asks for the same speed in feet per hour, using 1 foot = 0.3 meter. Students convert both the distance unit and the time unit. Setting up each conversion factor so that the unwanted units cancel is what makes it tricky.

NYS standards: 6.RP.3d converting measurement units
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students usually convert one unit with a given factor. This converts both parts of a rate, seconds to hours and meters to feet, with a decimal conversion factor and large numbers to keep track of.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Error-prone arithmetic moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every wrong choice is a conversion handled badly: multiplying by 0.3 when the fact 1 foot = 0.3 meter calls for dividing, dividing by 3 instead of 0.3, or stopping in meters per hour. Writing the units as fractions, so that seconds and meters cancel, shows which way each factor goes. A size check also helps: a foot is shorter than a meter, so the number of feet must be larger than the 36,000 meters.

Many steps moderate Working it out
Both parts of the rate change: seconds become hours by multiplying by 3,600, and meters become feet by dividing by 0.3, two conversions that each have a wrong direction. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic moderate Working it out
Dividing 36,000 by 0.3 means moving the decimal point in the divisor, and a slip of one place gives 12,000 instead of 120,000. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The stem gives 1 foot = 0.3 meter but not the 3,600 seconds in an hour, which students must build from 60 × 60. (Anderson & Krathwohl, 2001)

Strategy: Write the speed as a fraction, multiply by 3,600 seconds per hour, then by 1 foot over 0.3 meter, checking that seconds and meters cancel.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 36,000 meters per hour by 0.3 instead of dividing, turning the fact 1 foot = 0.3 meter upside down and getting 10,800.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Divides 36,000 by 3 instead of 0.3, putting the decimal point in the wrong place and getting an answer ten times too small.
C Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Converts seconds to hours, 10 × 3,600 = 36,000, and stops, leaving the speed in meters per hour rather than feet per hour.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Seconds in an hour: $\text{seconds in 1 hour} = 60 \times 60 = 3{,}600$; An hour has 60 minutes, and each minute has 60 seconds. So 1 hour holds 60 groups of 60 seconds. To change a speed per second into a speed per hour, multiply by 3,600, because an hour is 3,600 seconds long.
Meters to feet: $\text{feet} = \text{meters} \div 0.3$; The problem says 1 foot is about 0.3 meter. So each foot uses up 0.3 meter of the distance. To count the feet, divide the meters by 0.3. A foot is shorter than a meter, so you get more feet than meters.

Sort out the question
The speed is 10 meters per second. That means it moves 10 meters every second. You need the speed in feet per hour. Two units change: seconds become hours, and meters become feet. Change one at a time. Use 1 foot = 0.3 meter.
Find the seconds in an hour
An hour is much longer than a second. So in one hour, the object moves much farther. First find how many seconds are in an hour.
$60 \times 60 = 3{,}600$
One hour is 3,600 seconds.
Find the meters per hour
Each second it moves 10 meters. In 3,600 seconds it moves 3,600 times as far.
$10 \times 3{,}600 = 36{,}000$
The speed is 36,000 meters per hour.
Watch out: Stopping here gives 36,000, choice C. That number is still in meters, not feet.
Change meters to feet
Each foot is 0.3 meter. To find how many feet fit in 36,000 meters, divide by 0.3. Multiplying both numbers by 10 makes the division easier.
$36{,}000 \div 0.3 = 360{,}000 \div 3 = 120{,}000$
So 36,000 meters is 120,000 feet. The speed is 120,000 feet per hour.
Give the answer
Watch out: Multiplying by 0.3 instead of dividing gives 10,800, choice A. Dividing by 3 instead of 0.3 gives 12,000, choice B.
So the answer is D, 120,000 feet per hour.

Check your answer

A foot is shorter than a meter. So any distance is more feet than meters. The answer must be bigger than 36,000, and only choice D is. Also, 120,000 feet at 0.3 meter each is 120,000 × 0.3 = 36,000 meters. That matches.

More questions like this

The original DOE question

Question 111 as printed on page 207 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 207, question 111. Open page 207 of the handbook (PDF)

Question 82 of 112

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 83 of 112

2022 Test A Math, question 24

Question

$12.96 \div 0.08 =$

A1.62
B16.2
C162
D1,620

Show the answer

Correct answer: C

Since the greatest number of decimal places is 2 (hundredths), multiply both terms by 100, and then divide:

$12.96(100) \div 0.08(100) =$

$1{,}296 \div 8 =$

$162$

Question analysis

This question asks students to divide 12.96 by 0.08, a decimal division with no story around it. It tests dividing multi-digit decimals with the standard algorithm. The choices differ only in where the decimal point falls, so moving the point the same number of places in both numbers is the whole challenge.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Dividing multi-digit decimals is grade 6 fluency that grade 7 work builds on rather than teaches. The division comes out exactly, so the real demand is placing the decimal point correctly, a single routine skill.
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: 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

Every choice uses the digits 1, 6 and 2, so the question tests only where the decimal point goes. Dividing by a number less than one gives a result larger than 12.96, which feels wrong to students who expect division to shrink a number. Estimating first, 12.96 ÷ 0.08 is about 13 × 12.5, puts the result in the low hundreds and rules out the other three at once.

Error-prone arithmetic moderate Working it out
The divisor 0.08 has two decimal places, so both numbers must be multiplied by 100, turning the problem into 1,296 ÷ 8, and the quotient comes out larger than 12.96. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Multiply the dividend and the divisor by the same power of 10 so that the divisor becomes a whole number, then carry out the long division.

Traps

An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Divides 12.96 by 8, leaving out the decimal point in the divisor 0.08, so the quotient, 1.62, is 100 times too small.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Moves the decimal point one place instead of two, in effect dividing 12.96 by 0.8, so the quotient, 16.2, is ten times too small.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Shifts the decimal point three places in 12.96 but only two in 0.08, dividing 12,960 by 8, so the quotient is ten times too big.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a decimal: $a \div b = (a \times 100) \div (b \times 100)$; a is the number being divided. b is the divisor, the number you divide by. Multiplying both by the same number, like 100, does not change the answer. Use 100 when the divisor has two decimal places, so it becomes a whole number.

Sort out the question
You need $12.96 \div 0.08$ . The divisor, the number you divide by, is 0.08. It has two decimal places. Dividing by a decimal is hard, so make the divisor a whole number first. All four choices use the digits 1, 6 and 2. So the real question is where the decimal point goes.
Make the divisor a whole number
Multiply both numbers by 100. This moves each decimal point two places to the right.
$12.96 \div 0.08 = (12.96 \times 100) \div (0.08 \times 100)$
$= 1{,}296 \div 8$
The new problem has the same answer. Watch out: move both points the same number of places. Moving the point three places in 12.96 but two in 0.08 gives 1,620, choice D.
Divide by 8
Split 1,296 into parts that 8 divides easily.
$1{,}200 \div 8 = 150 \qquad 96 \div 8 = 12$
$1{,}296 \div 8 = 150 + 12 = 162$
Watch out: ignoring the point in 0.08 means dividing 12.96 by 8. That gives 1.62, choice A. Moving the point only one place gives 16.2, choice B. So the answer is C, 162.

Check your answer

Multiply back: $162 \times 0.08 = 12.96$ . That matches. Also, 0.08 is less than 1. Dividing by a number less than 1 makes the answer bigger. And $1{,}620 \times 0.08 = 129.6$ , which is far too big.

More questions like this

The original DOE question

Question 81 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 2023 Admissions, page 85, question 81. Open page 85 of the handbook (PDF)

Question 84 of 112

2022 Test A Math, question 36

Question

When the three-digit whole number 27 ? is divided by 3 or by 4, the remainder is zero. What is the missing digit?

Show the answer

Correct answer: D

Each of the options 0, 2, 3, and 6 must be considered:

Option A: In the case of 0, the three-digit whole number is 270. Divide 270 by both 3 and 4:

$270 \div 3 = 90 \text{ (or } 90R0\text{)}$

$270 \div 4 = 67\frac{1}{2} \text{ (or } 67R2\text{)}$

Dividing 270 by 3 results in a remainder of 0, but dividing 270 by 4 results in a remainder of 2, so 0 cannot be the missing digit.

Option B: In the case of 2, the three-digit whole number is 272. Divide 272 by both 3 and 4:

$272 \div 3 = 90\frac{2}{3} \text{ (or } 90R2\text{)}$

$272 \div 4 = 68 \text{ (or } 68R0\text{)}$

Dividing 272 by 4 results in a remainder of 0, but dividing 272 by 3 results in a remainder of 2, so 2 cannot be the missing digit.

Option C: In the case of 3, the three-digit whole number is 273. Divide 273 by both 3 and 4:

$273 \div 3 = 91 \text{ (or } 91R0\text{)}$

$273 \div 4 = 68\frac{1}{4} \text{ (or } 68R1\text{)}$

Dividing 273 by 3 results in a remainder of 0, but dividing 273 by 4 results in a remainder of 1, so 3 cannot be the missing digit.

Option D: In the case of 6, the three-digit whole number is 276. Divide 276 by both 3 and 4:

$276 \div 3 = 92 \text{ (or } 92R0\text{)}$

$276 \div 4 = 69 \text{ (or } 69R0\text{)}$

Dividing 276 by 3 results in a remainder of 0, and dividing 276 by 4 results in a remainder of 0, so 6 is the missing digit.

Question analysis

A three-digit number begins with 27 and is missing its last digit, and it divides evenly by both 3 and 4. Students find the missing digit. It tests factors, multiples and divisibility rules; a digit that works for one divisor may fail the other, so both tests must be applied.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Divisibility by 3 and by 4 is grade 6 number sense. Testing four possible digits against two conditions is quick, much like a grade 6 exercise, as long as both divisors are actually checked.
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
Every choice must be checked minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every candidate passes at least one of the two tests, which makes the question a matter of finishing the checks rather than finding a method. Students who test only by 3, or only by 4, find a fitting number and stop. Seeing that a number divisible by both 3 and 4 is divisible by 12, and that 276 = 12 × 23, settles it in one step.

Turning words into math minor Understanding the question
The words “divided by 3 or by 4, the remainder is zero” mean that both divisions come out even, though or can be read as either one being enough. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Every choice must be checked minor Choosing the answer
Each candidate, 270, 272, 273 and 276, has to be tested against two divisors, and three of them pass one test, so stopping after one check is costly. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
The divisibility tests are not given: a number is divisible by 3 when its digit sum is, and by 4 when the number formed by its last two digits is. (Anderson & Krathwohl, 2001)

Strategy: Use the divisibility tests: a number is divisible by 4 when its last two digits are, and by 3 when its digit sum is. Keep only a digit that passes both.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Checks only division by 3, which 270 passes with a digit sum of 9, and never tests 4, though 270 ÷ 4 leaves a remainder of 2.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Checks only division by 4, which 272 passes because 72 is a multiple of 4, but its digit sum is 11, so 272 ÷ 3 leaves 2.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the missing digit to the divisor 3 in the question; 273 is a multiple of 3 but leaves a remainder of 1 when divided by 4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Test for dividing by 3: $\text{digit sum divides by 3} \rightarrow \text{number divides by 3}$; The digit sum is what you get when you add up a number’s digits. Divides by 3 means it divides evenly, with a remainder of 0. If the digit sum divides evenly by 3, so does the whole number. Use it to test a big number quickly.
Test for dividing by 4: $\text{last two digits divide by 4} \rightarrow \text{number divides by 4}$; Look only at the number made by the last two digits. If it divides evenly by 4, so does the whole number. This works because every 100 divides evenly by 4, so only the last two digits matter.

Sort out the question
The number is 27 followed by a missing digit. Remainder zero means the division comes out even, with nothing left over. The remainder is zero when the number is “divided by 3 or by 4.” That means both divisions come out even. So the number must pass both tests. The choices give four digits to try: 0, 2, 3 and 6.
Test each number for 3
Add the digits of each number.
Number Digit sum Divides by 3?
270 $2 + 7 + 0 = 9$ yes
272 $2 + 7 + 2 = 11$ no
273 $2 + 7 + 3 = 12$ yes
276 $2 + 7 + 6 = 15$ yes
So 270, 273 and 276 pass. 272 fails.
Test each number for 4
Look at the last two digits: 70, 72, 73 and 76.
$70 \div 4 = 17 \text{ remainder } 2 \qquad 72 \div 4 = 18$
$73 \div 4 = 18 \text{ remainder } 1 \qquad 76 \div 4 = 19$
So 272 and 276 pass. 270 and 273 fail.
Keep the digit that passes both
Only 276 passes both tests. Check by dividing:
$276 \div 3 = 92 \qquad 276 \div 4 = 69$
Watch out: 0 and 3 pass only the test for 3, so A and C fail. 2 passes only the test for 4, so B fails. And 3 is not right just because the question says 3. So the answer is D, 6.

Number	Digit sum	Divides by 3?
270	$2 + 7 + 0 = 9$	yes
272	$2 + 7 + 2 = 11$	no
273	$2 + 7 + 3 = 12$	yes
276	$2 + 7 + 6 = 15$	yes

Check your answer

A number that divides evenly by both 3 and 4 also divides evenly by 12, since $3 \times 4 = 12$ . And $12 \times 23 = 276$ . So 276 is a multiple of 12. None of 270, 272 or 273 is.

More questions like this

The original DOE question

Question 93 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 2023 Admissions, page 89, question 93. Open page 89 of the handbook (PDF)

Question 85 of 112

2022 Test A Math, question 50

Question

Nadia paid $1.00 to mail a letter. The cost to mail a letter is $0.55 for the first ounce of weight plus $0.15 for each additional ounce. What was the weight, in ounces, of the letter Nadia mailed?

A1.4
B1.5
C3
D4

Show the answer

Correct answer: D

Let x represent the number of ounces Nadia’s letter weighed. The first ounce costs $0.55, so $(x - 1)$ represents the number of ounces that cost $0.15 each.

Given that the total cost is $1.00, the equation that describes the situation can be written and solved as shown.

$0.55 + 0.15(x - 1) = 1.00$

$0.55 + 0.15x - 0.15 = 1.00$

$0.15x = 1.00 - 0.55 + 0.15$

$0.15x = 0.60$

$\frac{0.15x}{0.15} = \frac{0.60}{0.15}$

$x = 4$

So the letter Nadia mailed weighed 4 ounces.

Question analysis

Nadia pays $1.00 to mail a letter, and postage costs $0.55 for the first ounce plus $0.15 for each additional ounce. Students find the weight of the letter in ounces. It tests writing and solving a two-step equation from a pricing rule in which the two prices apply to different ounces.

NYS standards: 7.EE.4a two-step equations
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 solve word problems that lead to equations such as px + q = r. This one takes that form once each price is matched to the right ounces, a step that makes it a little more demanding than routine practice.
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
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Pricing the letter ounce by ounce, $0.55, $0.70, $0.85, $1.00, reaches the total at the fourth ounce without any algebra. The equation route works too, but the first ounce is easy to lose: dividing the $0.45 left by $0.15 counts only the ounces after the first. The two decimal choices come from merging the two prices or from treating dollars as ounces.

Turning words into math moderate Understanding the question
“$0.55 for the first ounce of weight plus $0.15 for each additional ounce” makes the cost 0.55 + 0.15(x − 1), so the $0.15 covers every ounce except the first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
The money is in decimals, $1.00 − $0.55 = $0.45 and then $0.45 ÷ $0.15 = 3, and dividing by a decimal invites a misplaced point. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the cost of a letter weighing x ounces, $0.55 plus $0.15 for each ounce after the first, set it equal to $1.00 and solve, then check by pricing the letter ounce by ounce.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Merges the two prices into $0.70 for every ounce and divides $1.00 by $0.70, about 1.43, which rounds to 1.4.
B Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Counts the first ounce, then treats the $0.45 left over as 0.45 ounce, getting 1.45 ounces, which rounds to 1.5.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the 3 additional ounces that the $0.45 left over pays for, 0.45 ÷ 0.15 = 3, and stops before adding the first ounce.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost to mail a letter: $C = 0.55 + 0.15 \times (w - 1)$; C is the cost in dollars. w is the weight in ounces. The first ounce costs $0.55. Each ounce after the first costs $0.15, and there are w − 1 of them. Subtracting 1 leaves out the first ounce, which is already paid for.
Keeping an equation balanced: $\text{if } a = b \text{, then } a + c = b + c \text{, } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

Sort out the question
Nadia paid $1.00 to mail a letter. The first ounce costs $0.55. Each additional ounce, meaning each ounce after the first, costs $0.15. You need the weight of the letter in ounces.
Write the equation
Let w be the weight in ounces. The first ounce is 1 of the w ounces. So the extra ounces number w − 1. The cost is $1.00.
$0.55 + 0.15 \times (w - 1) = 1.00$
Take off the first ounce’s cost
Subtract 0.55 from both sides. This keeps the equation balanced.
$0.15 \times (w - 1) = 1.00 - 0.55 = 0.45$
So $0.45 pays for the extra ounces.
Find the extra ounces
Divide both sides by 0.15. Multiply both numbers by 100 to make them whole.
$w - 1 = 0.45 \div 0.15 = 45 \div 15 = 3$
There are 3 extra ounces after the first one.
Add back the first ounce
$w = 3 + 1 = 4$
The letter weighed 4 ounces. Watch out: 3 counts only the ounces after the first. It is choice C. Also, don’t join $0.55 and $0.15 into one price. $1.00 \div 0.70$ is about 1.4, choice A. So the answer is D, 4 ounces.

Check your answer

Price the letter one ounce at a time. 1 ounce: $0.55. 2 ounces: $0.70. 3 ounces: $0.85. 4 ounces: $1.00. The cost reaches $1.00 at 4 ounces. That matches.

More questions like this

The original DOE question

Question 107 as printed on page 95 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 95, question 107. Open page 95 of the handbook (PDF)

Question 86 of 112

2022 Test A Math, question 51

Question

COST OF ORANGES

Number of Oranges	Cost
4	$1.80
6	$2.70
9	$4.05

The table shows the cost of oranges. What is the unit rate for the cost per orange?

A$0.45
B$0.90
C$1.35
D$1.80

Show the answer

Correct answer: A

The unit rate (cost per orange) can be determined by dividing the cost of x oranges by x.

$\frac{1.80}{4} = \frac{2.70}{6} = \frac{4.05}{9} = 0.45$

The unit rate is $0.45.

Question analysis

A table shows what 4, 6 and 9 oranges cost, in dollars and cents. Students find the unit rate, the cost of a single orange. It tests identifying a unit rate in a proportional table, and the tricky part is keeping straight which column holds the cost and which holds the number of oranges.

NYS standards: 7.RP.2b constant of proportionality
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a unit rate in a table belongs to grade 7 proportional relationships, but here it takes one division with money in a clearly proportional table, so the task is easier than typical grade 7 work.
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: Reading a figure, table or graph minor
Something you must already know minor
Solution: 1 step
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

Any row gives the rate in one division, and the three rows agree, so the question is easy once unit rate is read as the cost of one orange. The traps are differences between rows, which look like a change per orange but cover two or three oranges, and the first cost in the table, which is for four. Dividing a second row confirms the result.

Reading a figure, table or graph minor Understanding the question
The table lists 4, 6 and 9 oranges but never a single orange, so the cost of one has to be computed from a row, such as $1.80 ÷ 4. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The term unit rate is not defined; students must know it means the cost of one orange, a row's cost divided by its number of oranges. (Anderson & Krathwohl, 2001)

Strategy: Pick any row and divide its cost by its number of oranges; doing the same with a second row confirms that the rate stays the same.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Subtracts the first two costs, $2.70 − $1.80 = $0.90, the price of 2 more oranges, without dividing by the 2 oranges it covers.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Subtracts the last two costs, $4.05 − $2.70 = $1.35, the price of 3 more oranges, rather than the price of one.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes $1.80, the first cost in the table, which is what 4 oranges cost together, not one.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{unit rate} = \frac{\text{cost}}{\text{number of oranges}}$; The unit rate is the cost of one item, here one orange. The cost is what a group of oranges costs together. Divide by the number of oranges in that group. Any row of the table gives the same rate, because each orange costs the same.

Sort out the question
The table gives the cost of 4, 6 and 9 oranges. It never shows the cost of one orange. A unit rate is the amount for one unit. Here that means the cost of one orange. You need to find it.
Divide one row
Use the first row: 4 oranges cost $1.80.
$\text{unit rate} = \frac{\text{cost}}{\text{number of oranges}}$
$\text{unit rate} = \frac{1.80}{4} = 0.45$
One orange costs $0.45. Watch out: $1.80 is the cost of 4 oranges, not one. Don’t pick choice D.
Check with the other rows
$\frac{2.70}{6} = 0.45 \qquad \frac{4.05}{9} = 0.45$
Every row gives $0.45, so the rate stays the same. Watch out: $2.70 - 1.80 = 0.90$ is the cost of 2 more oranges, choice B. $4.05 - 2.70 = 1.35$ is the cost of 3 more, choice C. Neither is for one orange. So the answer is A, $0.45.

Check your answer

Multiply back: $0.45 \times 4 = 1.80$ and $0.45 \times 9 = 4.05$ . Both match the table. Even a difference works once you divide: $0.90 \div 2 = 0.45$ .

More questions like this

The original DOE question

Question 108 as printed on page 95 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 95, question 108. Open page 95 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 87 of 112

2022 Test B Math, question 3

Question

Kelly had 7.3 grams of sugar. She used 2.7 grams in her coffee. How many grams of sugar does she have left?

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

Show the answer

Correct answer: 4.6

Subtract the number of grams of sugar Kelly used in her coffee from the original amount she had.

$7.3 - 2.7 = 4.6\text{ grams left}$

Question analysis

Kelly has 7.3 grams of sugar and uses 2.7 grams of it in her coffee. This short grid-in word problem asks how many grams she has left. It tests subtracting decimals, and because the tenths digit being taken away is the larger one, the subtraction needs regrouping.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Subtracting decimals is grade 6 fluency work, and this needs one subtraction in a short, familiar setting. Grade 7 problems usually bring in negative numbers or several steps; the only care here is regrouping in the tenths.
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
An answer in an unexpected form minor
Solution: 1 step
Reading level: Question and choices: about grade 2, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One subtraction in a two-sentence story, so the only real risk is the regrouping in the tenths column. Students who skip the borrow, or who borrow and forget to reduce the ones, still produce a tidy decimal, and nothing on the page flags it. Adding the result back to the 2.7 grams used should return the 7.3 grams Kelly started with.

Error-prone arithmetic minor Working it out
The 7 tenths being taken away is more than the 3 tenths in 7.3, so one of the 7 wholes must be regrouped as ten tenths before 13 − 7 = 6. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
The difference is gridded with no choices to compare, so a regrouping slip that gives 5.4 or 5.6 looks just as finished as the right amount. (Enright, Morley & Sheehan, 2002)

Strategy: Line up the decimal points and subtract the grams used from the grams Kelly started with, regrouping one whole as ten tenths when the tenths column needs it.

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.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
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

5.4 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts the smaller digit from the larger in each column, 7 − 3 = 4 tenths and 7 − 2 = 5 ones, instead of regrouping a whole as ten tenths.
5.6 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Regroups to get 13 − 7 = 6 tenths but forgets to reduce the ones after borrowing, so the ones digit stays 7 − 2 = 5.
10 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds the two amounts, 7.3 + 2.7 = 10, instead of taking the sugar used in the coffee away from the 7.3 grams Kelly had.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Amount left: $\text{left} = \text{start} - \text{used}$; Start is the amount you had at first. Used is the amount taken away. Left is what remains. Taking part of an amount away is subtraction, so you subtract the amount used from the start.

Sort out the question
Kelly started with 7.3 grams of sugar. She used 2.7 grams in her coffee. You need the grams she has left.
Left means what remains after some is taken away. So subtract the amount used from the amount she started with.
$\text{left} = 7.3 - 2.7$
Regroup to subtract the tenths
Line up the decimal points, so tenths sit over tenths. A tenth is one of ten equal parts of a whole. Start with the tenths. You can’t take 7 tenths from 3 tenths. So break 1 of the 7 wholes into 10 tenths. Now you have 6 wholes and 13 tenths.
$13 - 7 = 6$
That leaves 6 tenths.
Watch out: taking the smaller digit from the bigger one, 7 − 3, skips the regrouping. It gives 5.4, which is wrong.
Subtract the whole grams
After regrouping, the wholes column has 6, not 7.
$6 - 2 = 4$
That leaves 4 wholes. Put the decimal point back: 4 wholes and 6 tenths is 4.6 grams.
Watch out: forgetting to change the 7 to a 6 gives 5.6. Adding instead of subtracting gives 10. So you would grid 4.6.

Check your answer

Add back the sugar she used: $4.6 + 2.7 = 7.3$ . That is the amount Kelly started with, so 4.6 grams is right. An estimate agrees too: about 7 − 3 = 4 grams.

More questions like this

The original DOE question

Question 60 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 2023 Admissions, page 197, question 60. Open page 197 of the handbook (PDF)

Question 88 of 112

2022 Test B Math, question 21

Question

The decimal 0.4 can be written as the fraction $\frac{x}{25}$ . What is the value of x?

A0.016
B0.16
C10
D16

Show the answer

Correct answer: C

First, convert the decimal 0.4, which is read as four-tenths, to the fraction $\frac{4}{10}$ , and then set up a proportion to solve for x using cross products.

$\frac{4}{10} = \frac{x}{25}$

$10x = 100$

$x = 10$

Or, reduce the fraction to $\frac{2}{5}$ before setting up the proportion and then use proportional reasoning.

$\frac{4}{10} = \frac{2}{5}$

$\frac{2 \times 5}{5 \times 5} = \frac{10}{25}$

Therefore, $x = 10$ .

Question analysis

This question asks students to write the decimal 0.4 as an equivalent fraction whose denominator is 25 and to choose the missing numerator. It tests converting between decimals and fractions. Tenths do not turn into twenty-fifths in one obvious step, so the numerator has to be scaled with care, in the same way as the denominator.

NYS standards: 6.NS.3 multi-digit decimal operations
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Rewriting a decimal as an equivalent fraction is upper elementary and grade 6 work. It is one conversion with no context, and the friendly denominator keeps it well below the demand of typical grade 7 problems.
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 5, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

A single conversion: 0.4 is 4/10, or 2/5, and 2/5 renamed over 25 is 10/25. The only risk is the direction of the step, since x/25 = 0.4 means multiplying 0.4 by 25, and dividing instead leads to the small decimals among the choices. Any choice can be checked by dividing it by 25, and only one gives back 0.4.

Error-prone arithmetic minor Working it out
Four tenths has to be renamed in twenty-fifths, and 10 does not divide 25, so the numerator is reached through 2/5 = 10/25 or by multiplying 0.4 by 25. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the decimal as a fraction in tenths, simplify it, then multiply the numerator and denominator by the same number until the denominator is 25.

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 misconception Expected pull: moderate
Divides 0.4 by 25 instead of multiplying, undoing the division by 25 with another division, which gives 0.016.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Divides by 25 instead of multiplying and also drops the decimal point of 0.4, so 4 ÷ 25 = 0.16, ten times 0.016.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single mistake gives 16; it is 4 ÷ 25 = 0.16 with the decimal point moved two places, a chain with no reason behind it, so it mostly draws 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 of the fraction, and b is the bottom number. n is any number except 0. Multiplying the top and the bottom by the same number gives an equal fraction. It is like cutting every piece into n smaller pieces.

Sort out the question
You need a fraction with 25 on the bottom that equals 0.4. Its top is x. So you need the x that makes $\frac{x}{25}$ the same amount as 0.4.
Write the decimal as a fraction
0.4 is read four tenths, so it is 4 over 10. Simplify it by dividing the top and the bottom by 2.
$0.4 = \frac{4}{10} = \frac{2}{5}$
Rename the fraction over 25
To turn the bottom 5 into 25, multiply it by 5. Do the same to the top, so the fraction keeps its value.
$\frac{2}{5} = \frac{2 \times 5}{5 \times 5} = \frac{10}{25}$
So x = 10.
Give the answer
Another way: x/25 = 0.4 means x divided by 25 is 0.4. Undo the division by multiplying: $x = 0.4 \times 25 = 10$ .
Watch out: dividing 0.4 by 25 instead gives 0.016, choice A. Dropping the decimal point as well gives 0.16, choice B. So the answer is C, 10.

Check your answer

Divide the answer by 25 and see if you get 0.4. $10 \div 25 = 0.4$ , because 25 × 0.4 = 10. It works. Choice D fails: 16 ÷ 25 = 0.64, not 0.4.

More questions like this

The original DOE question

Question 78 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 2023 Admissions, page 202, question 78. Open page 202 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 89 of 112

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 90 of 112

2022 Test B Math, question 49

Question

A machine part is supposed to be 0.38 millimeters thick. The manufacturer allows a variation of −2.5% to +2.5%. Which interval shows the range of acceptable thicknesses, in millimeters, for the part?

A0.285–0.475
B0.355–0.405
C0.3705–0.3895
D0.3724–0.3876

Show the answer

Correct answer: C

A variation of −2.5% to +2.5% means that the thickness of the machine part can be from 2.5% less than the standard of 0.38 mm to 2.5% greater than 0.38 mm.

The least allowed measure is

$0.38-0.025(0.38)=0.3705.$

The greatest allowed measure is

$0.38+0.025(0.38)=0.3895.$

Therefore, the allowable range of thicknesses is 0.3705 mm to 0.3895 mm.

Question analysis

A machine part should be 0.38 millimeters thick, and the manufacturer accepts a variation from −2.5% to +2.5%. Students choose the interval of acceptable thicknesses. It tests finding a percent of a decimal and applying it as a decrease and an increase; the percent must be turned into millimeters first, and the four-decimal-place choices need exact work.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3c percent as a rate per 100
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Taking a percent of a number and adding or subtracting it is grade 7 percent work. The decimal percent and results to four decimal places make the arithmetic fussier than usual, so careful work matters more than the idea.
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: 3 steps
Reading level: Question and choices: about grade 9, 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 question rewards turning the percent into millimeters before anything else. Students who take ±2.5% as ±0.025 mm find a tidy interval, 0.355 to 0.405, and the other listed intervals come from a misplaced decimal point or from 2% in place of 2.5%. Because the right interval and its near twin differ only from the third decimal place, the product 0.38 × 0.025 has to be exact; multiplying 0.38 by 0.975 and by 1.025 is a quick double-check.

Error-prone arithmetic moderate Working it out
Two and a half percent of 0.38 is 0.025 × 0.38 = 0.0095, a product of two decimals with four decimal places, and the limits 0.3705 and 0.3895 need the same care. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
“A variation of −2.5% to +2.5%” has to become 2.5% of 0.38 millimeters, a length, before it is subtracted for the least thickness and added for the greatest. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The intervals 0.3705–0.3895 and 0.3724–0.3876 both sit about 0.01 mm either side of 0.38, so an estimate cannot separate them and only the exact 0.0095 decides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write 2.5% as a decimal, multiply it by 0.38 to find the allowed change in millimeters, then subtract it for the least thickness and add it for the greatest.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Writes 2.5% as 0.25 instead of 0.025, so the change is 0.38 × 0.25 = 0.095 mm and the interval, 0.285 to 0.475, is ten times too wide.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reads ±2.5% as ±0.025 mm, a fixed amount, adding it to and subtracting it from 0.38 instead of first taking 2.5% of 0.38.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Uses 2% instead of 2.5%, dropping the half percent, so the change is 0.38 × 0.02 = 0.0076 and the interval, 0.3724 to 0.3876, is too narrow.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; 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. Dividing the percent by 100 turns it into a decimal, so you can multiply.

Sort out the question
The part should be 0.38 millimeters thick. It may be off by up to 2.5%, thinner or thicker. You need the thinnest and thickest sizes allowed, in millimeters.
The 2.5% is a percent of 0.38 mm. So first find how many millimeters 2.5% is.
Write 2.5% as a decimal
Percent means out of 100. So divide by 100, which moves the decimal point two places left.
$2.5\% = \frac{2.5}{100} = 0.025$
Watch out: 2.5% is 0.025, not 0.25. Using 0.25 makes the interval ten times too wide, choice A.
Find 2.5% of 0.38
$\text{part} = 0.025 \times 0.38 = 0.0095$
First, 25 × 38 = 950. The two numbers have 3 + 2 = 5 decimal places in all. So the answer has 5 decimal places: 0.00950, or 0.0095. The size may change by 0.0095 millimeters.
Watch out: adding and subtracting 0.025 mm itself gives 0.355 to 0.405, choice B.
Find the two limits
Subtract the change for the thinnest size. Add it for the thickest.
$0.38 - 0.0095 = 0.3705$
$0.38 + 0.0095 = 0.3895$
The part may be from 0.3705 to 0.3895 millimeters thick.
Watch out: using 2% instead of 2.5% gives 0.3724 to 0.3876, choice D. So the answer is C, 0.3705–0.3895.

Check your answer

A part 2.5% thinner is 97.5% of 0.38: $0.975 \times 0.38 = 0.3705$ . A part 2.5% thicker is 102.5% of it: $1.025 \times 0.38 = 0.3895$ . Both match. Also, 0.38 sits exactly in the middle of the interval.

More questions like this

The original DOE question

Question 106 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 2023 Admissions, page 211, question 106. Open page 211 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 91 of 112

2018 Test A Math, question 10

Question

In a scale diagram, $0.125$ inch represents $125$ feet. How many inches represent $1$ foot?

A $0.001$
B $0.01$
C $0.1$
D $0.12$

Show the answer

Correct answer: A

Let $x$ be the number of inches representing 1 foot. Set up a proportion and solve for $x$ :

$\frac{x}{1}=\frac{0.125}{125}$

$x=0.001$ in.

Question analysis

This scale drawing question uses decimals: on a diagram, 0.125 inch stands for 125 feet. Students find how many inches represent just 1 foot, which is the unit rate of the scale. Dividing a decimal by a large whole number makes the decimal point easy to misplace.

NYS standards: 7.G.1 scale drawings
6.RP.3b unit rate problems
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings and unit rates are core grade 7 topics, and this is a single division. The demand matches typical grade 7 work, with decimal place value as the only real hurdle.
Estimated difficulty: Straightforward: two steps or one careful reading.
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
Solution: 1 step
Reading level: Question and choices: about grade 11, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

A single division settles this question, and the choices are built to punish a misplaced decimal point, since three of them are each a power of ten apart. Students who see that 0.125 is one-thousandth of 125 can answer at once; students who divide by hand often lose a zero. Multiplying the chosen answer back by 125 is a quick check.

Error-prone arithmetic moderate Working it out
The one step is 0.125 ÷ 125, a decimal divided by a large whole number, where the place of the decimal point decides between 0.001, 0.01 and 0.1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer moderate Choosing the answer
0.001 and 0.01 look almost the same, differing by one zero after the decimal point; only exact placement of the point in 0.125 ÷ 125 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Divide the inches by the feet to get inches per foot, then check by multiplying your result by 125 to see whether you get back 0.125.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Divides 0.125 by 125 but loses one zero in placing the decimal point, giving 0.01, ten times too large.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Sees that 125 ÷ 125 = 1 and puts the 1 in the tenths place, where 0.125 begins, instead of the thousandths place.
D Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Treats the 0.125 inch as if it already stood for 1 foot, never dividing by the 125 feet it represents, and picks the choice closest to it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $r = \frac{a}{b}$; r is the unit rate, the amount that goes with just 1 unit. a is the first amount, here 0.125 inch. b is the matching second amount, here 125 feet. Dividing shares the inches out equally, one share for each foot.

Sort out the scale
A scale diagram is a drawing where small lengths stand for real lengths. Here 0.125 inch on the drawing stands for 125 real feet. You need how many inches stand for just 1 foot. That is a unit rate, an amount for 1 unit.
Set up the division
Divide the inches by the feet. That splits the 0.125 inch into 125 equal shares, one for each foot.
$r = \frac{a}{b}$
$r = \frac{0.125}{125}$
Divide carefully
Write 0.125 as a fraction. It is 125 thousandths.
$0.125 = \frac{125}{1000}$
Now divide by 125. Split 125 thousandths into 125 shares, and each share is 1 thousandth.
$\frac{125}{1000} \div 125 = \frac{1}{1000} = 0.001$
So 1 foot is shown by 0.001 inch.
Match the decimal places
0.001 has the 1 in the third place after the decimal point. That place is thousandths. Watch out: 0.01 and 0.1 put the decimal point in the wrong place. They are choices B and C, and each is too big. So the answer is A, 0.001.

Check your answer

Multiply back. $0.001 \times 125 = 0.125$ inch, which matches the scale. Choice B would give $0.01 \times 125 = 1.25$ inches. That is ten times too long.

More questions like this

The original DOE question

Question 67 as printed on page 82 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 82, question 67. Open page 82 of the handbook (PDF)

Question 92 of 112

2018 Test A Math, question 15

Question

1 dollar = 7 lorgs

1 dollar = 0.5 dalt

Kevin has 140 lorgs and 16 dalts. If he exchanges the lorgs and dalts for dollars according to the rates above, how many dollars will he receive?

A$28
B$52
C$182
D$282

Show the answer

Correct answer: B

Use proportions to make the conversions:

Lorgs to dollars:

$\frac{140}{x}=\frac{7}{1}$

$7x=140$

$x=\$20$

Dalts to dollars:

$\frac{16}{x}=\frac{0.5}{1}$

$0.5x=16$

$x=\$32$

Total dollars = 20 + 32 = $52

Question analysis

A unit conversion word problem with made-up currencies, lorgs and dalts, each with an exchange rate to dollars. Students convert two amounts into dollars and add them. One rate is a decimal less than 1, which makes it easy to multiply where dividing is needed.

NYS standards: 6.RP.3d converting measurement units
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Converting units with ratio reasoning is grade 6 content. Two conversions and a sum make it typical grade 7 work, with the decimal rate as the step most likely to trip students up.
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
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 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 two conversions look alike but run in different directions in students' minds: 140 lorgs at 7 per dollar is plainly a division, while 0.5 dalt per dollar looks like a price. Most wrong answers come from the dalts. The safest approach is to ask how many dollars one coin is worth, a seventh of a dollar for a lorg and 2 dollars for a dalt, before multiplying.

Turning words into math moderate Understanding the question
"1 dollar = 0.5 dalt" means a dalt is worth 2 dollars, but read word by word it suggests a dalt is half a dollar, inviting multiplication by 0.5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Dividing 16 by 0.5 means dividing by a decimal less than 1, which makes the result larger, a step many students do the other way by habit. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Read each rate as coins per dollar, divide each coin amount by its own rate, and only then add the two dollar amounts.

Traps

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

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Converts the lorgs correctly, 140 ÷ 7 = 20, but multiplies the dalts by the rate, 16 × 0.5 = 8, and adds to get $28.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies each coin amount by the other coin's rate, 140 × 0.5 + 16 × 7 = 70 + 112 = 182, combining the numbers without a model.
D Units or scale (King et al., 2004)
Reveals a slip Expected pull: weak
Divides each amount by the other coin's rate, 140 ÷ 0.5 = 280 and 16 ÷ 7 ≈ 2.3, and rounds the total to $282.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing coins into dollars: $d = \frac{c}{r}$; d is the number of dollars. c is how many coins you have. r is the rate: how many of those coins make 1 dollar. Each dollar uses up r coins, so dividing tells how many dollars the coins fill.

Sort out the rates
The rates tell how many coins make 1 dollar. 1 dollar = 7 lorgs. 1 dollar = 0.5 dalt, which is half a dalt. Kevin has 140 lorgs and 16 dalts. You need how many dollars he gets for all of them.
Change the lorgs to dollars
It takes 7 lorgs to make 1 dollar. So divide the lorgs by 7.
$d = \frac{c}{r}$
$d = \frac{140}{7} = 20$
The lorgs are worth 20 dollars.
Think about what a dalt is worth
Half a dalt makes 1 dollar. So a whole dalt makes 2 dollars. A dalt is worth more than a dollar, not less. Watch out: don’t read the rate as a dalt being worth half a dollar. That leads to $16 \times 0.5 = 8$ and a total of $28, choice A.
Change the dalts to dollars
Use the same formula, with a rate of 0.5 dalt for each dollar.
$d = \frac{16}{0.5} = 32$
Dividing by one half is the same as multiplying by 2, so this is $16 \times 2 = 32$ . The dalts are worth 32 dollars.
Add the two amounts
Add the dollars from the lorgs and the dollars from the dalts.
$20 + 32 = 52$
Kevin gets 52 dollars. So the answer is B, $52.

Check your answer

Work backward. 20 dollars trade for $20 \times 7 = 140$ lorgs. 32 dollars trade for $32 \times 0.5 = 16$ dalts. Both match what Kevin has, so $52 is right.

More questions like this

The original DOE question

Question 72 as printed on page 84 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 72. 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 93 of 112

2018 Test A Math, question 25

Question

Which of the following numbers has factors that include the smallest factor (other than 1) of 91?

Show the answer

Correct answer: B

To find the smallest factor of 91, list the factors:

1, 7, 13, and 91

The smallest factor (other than 1) is 7.

Of the options listed (30, 35, 39, and 44), only 35 is a multiple of 7.

Question analysis

A number properties question about factors: students find the smallest factor of 91 other than 1, then pick the number that shares it. Because 91 looks like a prime number, many students skip the factoring, and the choices include numbers built from factors that 91 might seem to have.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Factors and divisibility are grade 4 to 6 skills, and this takes two quick checks. It is easier than typical grade 7 work, though recognizing that 91 is not prime catches many students.
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: Complex sentences moderate
Something you must already know moderate
Solution: 2 steps
Reading level: Question and choices: about grade 8, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question tests a grade 6 skill inside a hard-to-parse sentence. Once the stem is unpacked, the work is finding the smallest prime factor of 91 and checking the choices for it. Students lose points in two ways: 91 looks prime or divisible by 3, so they guess at its factors, or they find 13 before 7 and accept the choice that shares it.

Complex sentences moderate Understanding the question
The stem nests three ideas, "has factors that include the smallest factor (other than 1) of 91", so it has to be unpacked before any math; this is reading difficulty, not number difficulty. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)
Something you must already know moderate Working it out
Students must find that 91 = 7 × 13, and 91 is a number many students wrongly take to be prime or a multiple of 3. (Anderson & Krathwohl, 2001)

Strategy: Test 91 against the small primes in order, starting with 2, until one divides it evenly, then check which choice is a multiple of that prime.

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 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Wrongly takes 91 as a multiple of 3, as 90 is, and picks 30, which 3 divides; in fact 91 ÷ 3 is not a whole number.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Factors 91 as 7 × 13 but matches 39 = 3 × 13, which shares the larger factor, overlooking the word smallest.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Neither of its prime factors, 2 and 11, divides 91, an odd number, so no common working leads to 44; it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

What a factor is: $b = a \times c$; a, b and c are whole numbers. If b = a × c, then a and c are factors of b. That means a divides b evenly, with no remainder. Every whole number has 1 and itself as factors.

Unpack the long sentence
A factor of a number divides it evenly, with nothing left over. For example, 3 is a factor of 12, because $12 \div 3 = 4$ . The question has two steps. First, find “the smallest factor (other than 1) of 91”. Then find the choice that also has that factor.
Test small numbers on 91
Try dividing 91 by 2, 3, 5 and so on, until one works.
- 2: no. 91 is odd. That also rules out 4 and 6, which are even.
- 3: no. A number is a multiple of 3 only if its digits add to a multiple of 3. Here 9 + 1 = 10, which is not.
- 5: no. 91 does not end in 0 or 5.
- 7: yes!
$91 = 7 \times 13$
So 7 is the smallest factor of 91 other than 1.
Avoid the 3 trap
Watch out: 91 is close to 90, and 3 goes into 90. But $91 \div 3$ is 30 with 1 left over. So 3 is not a factor of 91. That trap leads to 30, choice A.
Check each choice for 7
Now look for a multiple of 7 among the choices.
- 30: no. $7 \times 4 = 28$ and $7 \times 5 = 35$ .
- 35: yes. $35 = 7 \times 5$ .
- 39: no. $39 = 3 \times 13$ .
- 44: no. $7 \times 6 = 42$ .
Watch out: 39 shares 13 with 91, but 13 is the larger factor, not the smallest. That is choice C. So the answer is B, 35.

Check your answer

Multiply to be sure: $7 \times 13 = 91$ and $7 \times 5 = 35$ . So 7 goes into both numbers evenly. No number from 2 to 6 goes into 91, so 7 really is the smallest.

More questions like this

The original DOE question

Question 82 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 82. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 94 of 112

2018 Test A Math, question 31

Question

How many positive even factors of 48 are greater than 24 and less than 48?

Show the answer

Correct answer: A

List the factors of 48:

1 and 48, 2 and 24, 3 and 16, 4 and 12, 6 and 8

There are no factors greater than 24 and less than 48.

Question analysis

This number properties question asks how many positive even factors of 48 are greater than 24 and less than 48. Students list the factors of 48 and test which ones fall strictly inside that range. Reading both conditions precisely, and listing every factor without skipping any, is what matters.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Listing the factors of a two-digit number is grade 6 work, and here it takes one short list. The only demand beyond routine is reading both conditions exactly, which grade 7 students already do regularly.
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
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The factor list is short, so the question is decided by the conditions. The pairs 2 and 24 and 1 and 48 put the only candidates exactly on the boundaries, and the strict words "greater than" and "less than" exclude them. Students also tend to distrust an answer of zero, which nudges them toward counting an endpoint.

A very close wrong answer moderate Choosing the answer
The choices 0 and 1 differ by a single factor, so the count cannot be sized up; only checking whether 48 itself, or 24, falls strictly inside the range separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Much to keep track of minor Working it out
Each factor must pass three tests at once, even, greater than 24 and less than 48, and the factor list itself must be complete. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: List the factors of 48 in pairs, starting from 1 × 48, then check each one against both conditions: it must be even and lie strictly between 24 and 48.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts one boundary factor, usually 48 itself, as inside the range, reading "less than 48" as "48 or less".
C Counting the ends (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Counts both 24 and 48, treating "greater than 24 and less than 48" as if the range included both of its ends.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Drops the condition that the numbers be factors of 48 and counts the even numbers from 26 up to and including 48, which makes 12.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pairs: $a \times b = n$; n is the number you start with, here 48. If two whole numbers a and b multiply to make n, both are factors of n. A factor divides n with nothing left over. Listing factors in pairs makes sure you skip none.

Sort out what the question asks
A factor of 48 is a whole number that divides 48 with nothing left over. You need factors that pass three tests. They must be even. They must be greater than 24. They must be less than 48.
“Less than 48” means 48 itself does not count. “Greater than 24” means 24 does not count either.
List the factors of 48 in pairs
Start with 1 and work up. Each pair multiplies to 48:
$1 \times 48 \quad 2 \times 24 \quad 3 \times 16 \quad 4 \times 12 \quad 6 \times 8$
The next number, 7, does not divide 48. After 6 and 8 the pairs repeat. So the factors are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Look between 24 and 48
In the list, the factor right after 24 is 48. Nothing falls between them.
Here is why. A factor bigger than 24 must pair with a number smaller than 2. The only such number is 1, and it pairs with 48. So no factor of 48 is between 24 and 48, even or odd.
Give the answer
Watch out: 24 and 48 are both factors, but they sit on the edges. Counting 48 gives 1, which is choice B. Counting both gives 2, which is choice C.
So the answer is A, 0.

Check your answer

Try the even numbers 26, 28, and so on up to 46. Dividing 48 by any of them gives more than 1 but less than 2. That is never a whole number, so none of them is a factor. Choice D, 12, counts even numbers without checking that they are factors.

More questions like this

The original DOE question

Question 88 as printed on page 87 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 88. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 95 of 112

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 96 of 112

2018 Test A Math, question 37

Question

An unmarked straight stick will be laid end over end to measure a distance of exactly 72 feet. The same stick will be used in the same way to measure a distance of exactly 30 feet. What is the length of the longest possible stick that can be used for both measurements?

A3 ft
B4 ft
C6 ft
D8 ft

Show the answer

Correct answer: C

The length of the stick must be the greatest common factor of 72 and 30.

The factors of 30 are

1, 2, 3, 5, 6, 10, 15, and 30.

Of those, only 1, 2, 3, and 6 are also factors of 72. The greatest of these is 6.

Question analysis

This word problem asks for the longest stick that can be laid end over end to measure exactly 72 feet and exactly 30 feet. Students must recognize that the stick's length has to divide both distances evenly, so the task is finding a greatest common factor. The story wording hides the factor idea.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a greatest common factor is a sixth-grade skill, and the arithmetic is light. The question is a little less routine than a bare factor problem only because students must see that the story calls for 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: No obvious method moderate
Hard words minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 54 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once the story is recognized as a greatest common factor problem, the arithmetic is light and the choices can simply be tested. The trap is checking only one of the two distances, because the longest choice works for 72 feet but not for 30. A student who lists the factors of 30 first, the smaller number, sees the answer quickly.

No obvious method moderate Working it out
Nothing in the stick story says "common factor"; students must see that laying the stick a whole number of times along 72 and 30 feet means it divides both. (Schoenfeld, 1985)
Hard words minor Understanding the question
The phrases "unmarked straight stick" and "laid end over end" are unusual wording for a math item; they describe whole-stick lengths, a reading hurdle rather than more math. (Perfetti, 2007; Freedle & Kostin, 1993)

Strategy: Translate 'measures both distances exactly' into 'divides both numbers evenly', list the factors of the smaller number, 30, and take the largest that also divides 72.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds a length that divides both 72 and 30, 3 feet, but not the longest one, missing that 6 also divides both.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Checks only that 4 divides 72, giving 18 sticks, and misses that 30 feet would take 7.5 sticks.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Takes the longest choice that divides 72 evenly, nine 8-foot lengths, without checking that 30 feet is not a whole number of 8-foot sticks.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of sticks: $n = \frac{d}{s}$; d is the distance to measure. s is the length of the stick. n is how many times you lay the stick down. For an exact measure, n must be a whole number. So s must divide d evenly.

Sort out the story
You lay a stick down again and again, end to end, like taking steps. The stick has no marks, so you can only count whole stick lengths.
It must fit exactly into 72 feet and exactly into 30 feet. You need the longest stick that does both.
Turn it into factors
Fitting exactly means no piece is left over. So the stick length must divide 72 evenly and divide 30 evenly.
A number that divides both is a common factor. You want the greatest common factor: the biggest number that divides both.
List the factors of 30
Start with the smaller number, 30, since its list is shorter. Its factors are 1, 2, 3, 5, 6, 10, 15 and 30.
Now test them on 72, starting from the biggest. 30, 15, 10 and 5 leave a remainder. But 6 works:
$72 \div 6 = 12 \qquad 30 \div 6 = 5$
So 6 is the greatest common factor.
Check the other choices
Watch out: 8 feet fits 72 feet, 9 times. But 30 ÷ 8 = 3.75, not a whole number, so choice D fails. Choice B fails too, since 30 ÷ 4 = 7.5. A 3-foot stick works, but it is not the longest, so choice A is wrong.
So the answer is C, 6 ft.

Check your answer

Use the stick. A 6-foot stick goes down 12 times for 72 feet, since 12 × 6 = 72. It goes down 5 times for 30 feet, since 5 × 6 = 30. Both counts are whole numbers, so 6 feet works.

More questions like this

The original DOE question

Question 94 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 94. 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 97 of 112

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 98 of 112

2018 Test A Math, question 40

Question

In the set of consecutive integers from 12 to 30, inclusive, there are four integers that are multiples of both 2 and 3. How many integers in this set are multiples of neither 2 nor 3?

Show the answer

Correct answer: B

The question asks for integers from 12 to 30 that are not divisible by 2 or 3. The set of consecutive integers is

{12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}

Since all even numbers are divisible by 2, eliminate all even numbers, leaving the odd numbers in the set:

{13, 15, 17, 19, 21, 23, 25, 27, 29}

Eliminate those integers that are multiples of 3 (15, 21, and 27). The remaining integers are:

{13, 17, 19, 23, 25, 29}

Therefore, there are 6 numbers in the set that are multiples of neither 2 nor 3.

Question analysis

This number properties question takes the integers from 12 to 30, inclusive, and asks how many are multiples of neither 2 nor 3. Students must count the whole set, remove multiples of 2 and of 3 without removing any number twice, and count what remains. Inclusive endpoints and the overlap make it easy to miscount.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Multiples are sixth-grade content, but grade 7 problems rarely ask students to count a set under two overlapping conditions. The inclusive range and the numbers that are multiples of both make this longer and trickier than usual.
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: Much to keep track of moderate
A very close wrong answer moderate
Solution: 3 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

The question gives the overlap, four multiples of both 2 and 3, which invites a subtraction method with three places to go wrong: the count of the set, the overlap, and the final complement. Each of those errors has its own choice. Listing the 19 numbers and crossing out evens and then odd multiples of 3 is slower to describe but faster and safer to do.

Much to keep track of moderate Working it out
Each of the 19 integers must be checked against two conditions, and the four multiples of 6 meet both, so they must be removed once, not twice. (Sweller, 1988; Carpenter, Just & Shell, 1990)
A very close wrong answer moderate Choosing the answer
The choices 5 and 6 differ by one, the result of counting 30 − 12 = 18 integers instead of 19, so only counting the inclusive set exactly separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Cross out every even number first, then cross out the odd multiples of 3 that remain, and count the survivors; listing all nineteen numbers is quick enough.

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.
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: moderate
Counts 30 − 12 = 18 integers instead of 19, then subtracts the 13 multiples of 2 or 3 and gets 5.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Counts the integers that are multiples of 2 or 3, the 13 to be crossed out, and gives that count instead of the 6 left over.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Subtracts only the four multiples of both 2 and 3 from the 19 integers, reading "neither 2 nor 3" as "not both 2 and 3".

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting whole numbers in a list: $n = b - a + 1$; a is the first number and b is the last. n is how many whole numbers there are, counting both ends. You add 1 because subtracting counts the gaps, and there is one more number than gaps.

Sort out the question
The set is every integer (whole number) from 12 to 30. “Inclusive” means 12 and 30 are both in.
A multiple of 2 is an even number. A multiple of 3 is a number in the 3 times table. You need the numbers that are neither: not even, and not in the 3 times table.
Count the numbers in the set
$n = 30 - 12 + 1 = 19$
There are 19 integers. Watch out: 30 − 12 = 18 forgets one end. That slip leads to choice A, 5.
Cross out the even numbers
Every even number is a multiple of 2, so remove them all. The odd numbers are left:
13, 15, 17, 19, 21, 23, 25, 27, 29
That is 9 numbers.
Cross out the multiples of 3
Check each odd number for 3. A quick test: add its digits. If the sum is in the 3 times table, so is the number. That removes 15, 21 and 27. These are left:
13, 17, 19, 23, 25, 29
Watch out: only 12, 18, 24 and 30 are multiples of both. Removing just those gives 19 − 4 = 15, choice D. “Neither” means you remove every multiple of 2 and every multiple of 3.
Count what is left
Six numbers are left. Watch out: 13 is how many you crossed out, choice C. The question asks for the ones left. So the answer is B, 6.

Check your answer

Count another way. Of the 19 numbers, 10 are even. Of the 9 odd ones, 3 are multiples of 3. So 10 + 3 = 13 are crossed out, and 19 − 13 = 6 are left. Same answer.

More questions like this

The original DOE question

Question 97 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 97. Open page 88 of the handbook (PDF)

Question 99 of 112

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 100 of 112

2018 Test A Math, question 49

Question

A car travels at 4,400 feet per minute. The radius of each tire on the car is 1 foot. How many revolutions does one of these tires make in 1 minute? (Use the approximation $\frac{22}{7}$ for $\pi$ .)

A700
B1,925
C13,828
D15,400

Show the answer

Correct answer: A

One revolution is equal to the circumference of the tire:

$C=2\pi r=2(1)(\frac{22}{7})=\frac{44}{7}$ ft

The car travels at 4,400 ft per minute. To calculate the number of revolutions, divide the speed by the circumference:

$4,400\div \frac{44}{7}=4,400\times \frac{7}{44}=700$

700 revolutions.

Question analysis

This circle word problem says a car travels 4,400 feet per minute on tires with a radius of 1 foot and asks how many revolutions each tire makes in a minute, using 22/7 for pi. Students must see that one revolution covers the tire's circumference, then divide. Dividing by a fraction is the step to check.

NYS standards: 7.G.4 circle area and circumference
6.NS.1 dividing fractions by fractions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Circumference is grade 7 content. This item adds the idea that each revolution rolls the tire one circumference forward and requires dividing by a fraction, which is typical of a solid grade 7 application problem.
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 moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question tests whether students connect distance traveled to turns of a wheel, a link the stem leaves unstated. Every wrong choice is larger than the answer, and two exceed 4,400, which is impossible when each turn covers more than 6 feet. A sense check on size rules those out before any fraction work, leaving only the division by 44/7 to do carefully.

Turning words into math moderate Understanding the question
The stem never says that one revolution rolls the car forward one circumference; students must supply that link to turn 4,400 feet into a count of turns. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know moderate Working it out
The circumference formula, C = 2πr, is not given; with r = 1 and 22/7 for π it gives 44/7 feet per turn. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
Dividing 4,400 by 44/7 means multiplying by 7/44, which is clean only if students see that 4,400 ÷ 44 = 100 before multiplying by 7. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the distance covered in one turn with the circumference formula, 2πr, then divide the distance traveled in a minute by that circumference.

Traps

A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard path gives 1,925, which equals 4,400 × 7 ÷ 16, so it mostly draws students guessing a mid-sized count.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: moderate
Multiplies 4,400 by π, 22/7, instead of dividing by the circumference, and truncates 13,828.57 to 13,828.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Divides 4,400 by 2/7 instead of 44/7, dropping the 22 of π while flipping 2 × 22/7, which gives 4,400 × 7 ÷ 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = 2 \times \pi \times r$; C is the circumference, the distance around a circle. r is the radius, from the center to the edge. π (pi) is about 22/7. The distance around is a bit more than 3 times the distance across, which is 2 × r.
Number of turns: $n = \frac{d}{C}$; n is the number of turns. d is the distance the car travels. C is the circumference. Each full turn rolls the car forward one circumference, so you divide the distance by it.

Sort out what you know
The car goes 4,400 feet in 1 minute. Each tire has a radius of 1 foot. A revolution is one full turn of the tire. You need how many turns a tire makes in 1 minute.
Find how far one turn goes
When a tire turns once, the car rolls forward the distance around the tire. That distance is the circumference. Use $\frac{22}{7}$ for π:
$C = 2 \times \frac{22}{7} \times 1 = \frac{44}{7}$
One turn moves the car $\frac{44}{7}$ feet, about 6.3 feet.
Divide the distance by one turn
$n = 4{,}400 \div \frac{44}{7}$
To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down):
$4{,}400 \times \frac{7}{44} = 100 \times 7 = 700$
Notice that 4,400 ÷ 44 = 100. That makes it easy.
Give the answer
The tire makes 700 turns. Watch out: multiplying 4,400 by π instead of dividing by the circumference gives about 13,828, choice C. So the answer is A, 700.

Check your answer

Estimate. One turn is about 6 feet. 4,400 ÷ 6 is about 730, close to 700. There should be fewer turns than feet, since each turn covers several feet. Choices C and D are bigger than 4,400, so they cannot be right.

More questions like this

The original DOE question

Question 106 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 106. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 101 of 112

2018 Test A Math, question 50

Question

$100\left ( 2+0.1 \right )^{2}-100=$

A101
B200
C301
D341

Show the answer

Correct answer: D

$100(2+0.1)^2-100=$

$100(2.1)^2-100=100(4.41)-100=441-100=341$

Question analysis

This order of operations question asks students to evaluate 100(2 + 0.1)² − 100, an expression with a decimal, a square and a subtraction. Students must add inside the parentheses first, square the result, multiply by 100, and then subtract. Squaring the two parts of the sum separately is the trap.

NYS standards: 6.EE.1 whole-number exponent expressions
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Evaluating an expression with an exponent and decimals in the correct order is grade 6 content. This is a short routine calculation, so it asks less than typical grade 7 work with negative numbers and multi-step equations.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · 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 16+, 1 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a direct order-of-operations calculation with no word problem around it, so the only real risk is the square of a sum. Students who apply the exponent to each part, or only to the 0.1, find a listed answer waiting. Adding inside the parentheses before squaring, and then squaring 2.1 exactly, is the whole task.

A very close wrong answer moderate Choosing the answer
The choices 301 and 341 differ by 40, and squaring 2.1 roughly as 4 lands nearer 301; only the exact square, 4.41, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
Squaring 2.1 by hand to get 4.41 puts two decimal places in the product, an easy place to misplace the point. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add inside the parentheses to get a single decimal, square it carefully, then multiply by 100 and subtract 100, following the order of operations.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Squares only the 0.1, as if the parentheses were not there, so 100 × 2.01 − 100 gives 101.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single slip gives 200; it requires (2 + 0.1)² to come out as 3, so it mostly catches guesses.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Squares each term of the sum separately, 4 + 0.01 = 4.01, instead of squaring 2.1, so 100 × 4.01 − 100 gives 301.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\text{parentheses} \rightarrow \text{exponents} \rightarrow \times, \div \rightarrow +, -$; This is the agreed order for doing math steps. Work inside parentheses first, then exponents, then multiply and divide, then add and subtract. Everyone follows it, so a problem has only one answer.
Squaring a number: $a^2 = a \times a$; a is any number. The small 2, called an exponent, means multiply the number by itself. So 2.1 squared means 2.1 times 2.1.

Sort out the problem
You need to work out $100(2 + 0.1)^2 - 100$ . The 100 right next to the parentheses means 100 times. Follow the order of operations, one step at a time.
Add inside the parentheses
Parentheses come first in the order of operations. So add the numbers inside them:
$2 + 0.1 = 2.1$
Now the problem is:
$100 \times 2.1^2 - 100$
Square 2.1
$2.1^2 = 2.1 \times 2.1 = 4.41$
Work out 21 × 21 = 441, then put in two decimal places, one for each 2.1.
Watch out: do not square the 2 and the 0.1 one at a time. That gives 4 + 0.01 = 4.01, which is wrong. It leads to 301, choice C.
Multiply, then subtract
The exponent is done, so multiply next. Multiplying by 100 moves the decimal point two places to the right. Subtracting comes last:
$100 \times 4.41 = 441$
$441 - 100 = 341$
So the answer is D, 341.

Check your answer

Estimate. 2.1 is a bit more than 2, so its square is a bit more than 4. 100 times that is a bit more than 400. Take away 100 and you get a bit more than 300. 341 fits. Choices A and B are too small.

More questions like this

The original DOE question

Question 107 as printed on page 90 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 90, question 107. Open page 90 of the handbook (PDF)

Question 102 of 112

2018 Test A Math, question 53

Question

Using the approximation 2.54 centimeters = 1 inch, how many centimeters are in 4 feet 7 inches?

A21.65
B119.38
C121.92
D139.70

Show the answer

Correct answer: D

Convert 4 ft 7 in. to inches. Since 12 in. = 1 ft :

4(12) + 7 = 55 inches

Multiply that by the conversion

2.54 cm = 1 in.

55 × 2.54 = 139.70 cm

Question analysis

This unit conversion question uses 2.54 centimeters per inch and asks how many centimeters are in 4 feet 7 inches. Students must first turn feet and inches into inches alone, then multiply by the conversion factor. Mixing feet and inches into a single number is the easy slip.

NYS standards: 6.RP.3d converting measurement units
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting measurement units with a given factor is grade 6 content. This item takes two short steps and one decimal multiplication, so it asks a little less than typical grade 7 work, though the mixed units 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: Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 13, 12 words.
Corrected from the DOE handbook: A missing decimal point in the explanation was restored.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The conversion factor is given and the multiplication is mechanical, so the question rests on the mixed measurement. Feet and inches are not a decimal system, and treating 4 feet 7 inches as 47, or dropping the 7, both lead to listed answers. A rough estimate, 55 inches × 2.5 ≈ 137, points straight to the right choice.

Something you must already know minor Working it out
The stem gives centimeters per inch but not inches per foot; students must supply 12 to turn 4 feet into 48 inches. (Anderson & Krathwohl, 2001)

Strategy: Change 4 feet to inches using 12 inches per foot, add the extra 7 inches, then multiply the total number of inches by 2.54.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
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 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Divides 55 inches by 2.54 instead of multiplying, converting in the wrong direction and getting fewer centimeters than inches.
B Units or scale (King et al., 2004)
Reveals a misconception Expected pull: strong
Reads 4 feet 7 inches as 47 inches, as if feet were tens of inches, then multiplies 47 × 2.54 = 119.38.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Converts the 4 feet to 48 inches and multiplies by 2.54, getting 121.92, but leaves out the extra 7 inches.

What was corrected from the DOE handbook, and why

Explanation: The handbook writes the conversion as “254 cm = 1 in.”, dropping the decimal point. The question gives 2.54 centimeters = 1 inch and the handbook’s next line multiplies by 2.54, so it now reads 2.54 cm = 1 in.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Feet to inches: $i = 12 \times f$; i is the number of inches and f is the number of feet. Each foot has 12 inches, so you multiply the feet by 12.
Inches to centimeters: $c = 2.54 \times i$; c is the number of centimeters and i is the number of inches. Each inch is about 2.54 centimeters, so you multiply. A centimeter is smaller than an inch, so you get a bigger number.

Sort out what you know
The length is 4 feet 7 inches. That means 4 feet plus 7 more inches. You know 1 inch is about 2.54 centimeters.
So change everything to inches first, then to centimeters. The question does not say that 1 foot is 12 inches. You have to know it.
Change the length to inches
$i = 12 \times 4 = 48$
4 feet is 48 inches. Now add the extra 7 inches:
$48 + 7 = 55$
The length is 55 inches. Watch out: 4 feet 7 inches is not 47 inches. Using 47 gives choice B. Leaving out the 7 inches gives 121.92, choice C.
Change inches to centimeters
$c = 2.54 \times 55 = 139.70$
To multiply in your head, split 55 into 50 and 5: 127 + 12.7 = 139.7.
Watch out: multiply, do not divide. Dividing 55 by 2.54 gives 21.65, choice A. So the answer is D, 139.70.

Check your answer

A centimeter is smaller than an inch. So the number of centimeters must be bigger than 55. About 2.5 × 55 is 137.5, close to 139.70. Choice A, 21.65, is smaller than 55, so it cannot be right.

More questions like this

The original DOE question

Question 110 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 110. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 103 of 112

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 104 of 112

2018 Test B Math, question 10

Question

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

Show the answer

Correct answer: C

Set up an inequality to compare the costs:

0.15x < 10.50
x < 70

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

Question analysis

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

NYS standards: 7.EE.4b two-step inequalities
6.NS.3 multi-digit decimal operations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an inequality from a real-world situation is a grade 7 standard. This one is short, but it asks students to interpret a strict inequality as a count of sheets, the kind of interpretation grade 7 tests expect.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
Turning words into math minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 10, 45 words.
Corrected from the DOE handbook: An inequality sign in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

A very close wrong answer moderate Choosing the answer
70 and 69 sheets differ by one, and 70 is exactly where the two costs meet at $10.50; only the strict “less expensive” separates them, ruling out 70. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math minor Understanding the question
The words “less expensive than buying the package” must become a strict inequality, 0.15x < 10.50, and “largest value” must then pick the biggest whole number below the boundary. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Dividing 10.50 by 0.15 means shifting the decimal point in both numbers, a step where a misplaced digit gives a wrong boundary. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Checks a round number, 60 sheets for $9.00, sees it is cheaper than the package, and stops without seeking the largest such number.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Divides only the $10 by $0.15, about 66.7, ignoring the 50 cents, and settles on the nearest smaller choice.
D Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: strong
Solves 0.15x = 10.50 and keeps the boundary 70, but 70 sheets cost exactly $10.50, the same as the package, not less.

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
The package costs $10.50. Single sheets cost $0.15 each. Macie buys x sheets.
You need the largest x that makes the sheets cost less than the package. Costing the same is not enough.
Write the inequality
An inequality is a math sentence that compares two amounts. The sign < means is less than.
The cost of x sheets is 0.15 times x. It must be less than 10.50.
$\text{cost} = \text{price of one} \times \text{number bought}$
$0.15 \times x < 10.50$
Find where the costs are equal
First find how many sheets cost exactly $10.50. Divide the package price by the price of one sheet. Multiply both numbers by 100 to clear the decimals.
$10.50 \div 0.15 = 1050 \div 15 = 70$
So 70 sheets cost exactly $10.50.
Pick the largest number below 70
x must be less than 70. You can only buy whole sheets. The largest whole number below 70 is 69.
Watch out: 70 sheets cost the same as the package, not less. That is why choice D is wrong. Choices A and B are cheaper too, but they are not the largest.
So the answer is C, 69.

Check your answer

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

More questions like this

The original DOE question

Question 67 as printed on page 154 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 67. Open page 154 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 105 of 112

2018 Test B Math, question 14

Question

What is the least common multiple of 24, 6, and 18?

A36
B48
C72
D144

Show the answer

Correct answer: C

Since 18 and 24 are both multiples of 6, find the least common multiple of only 18 and 24.

Multiples of 18:

18, 36, 54, 72…

Multiples of 24:

24, 48, 72…

The least common multiple of 6, 18, and 24 is 72.

Question analysis

This number properties question asks for the least common multiple of three numbers: 24, 6 and 18. Students can list multiples or use prime factors. Each candidate has to be checked against all three numbers, and it must be the smallest number that works, not just any shared multiple.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Least common multiples are grade 6 content, where New York's standard uses two numbers up to 12. This uses three larger numbers, but it is still a short factor task below typical grade 7 work.
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 minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 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

Finding a least common multiple is routine, and listing multiples of 24 reaches the answer by the third one. The wrong choices are built to catch incomplete checking: each passes the test for some of the numbers but fails for one, or passes for all and is simply not the smallest. Testing every candidate against all three numbers is the whole skill.

Every choice must be checked minor Choosing the answer
Each wrong choice passes part of the test: 36 is a multiple of 6 and 18, 48 of 6 and 24, and 144 of all three, so every candidate needs checking against all three numbers. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
Students must know what “least common multiple” means and see that 6 can be dropped, since it already divides both 24 and 18. (Anderson & Krathwohl, 2001)

Strategy: Note that 6 divides both 24 and 18, so only those two matter; list multiples of 24 and stop at the first one that 18 also divides.

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 the least common multiple of 6 and 18 and stops, never checking that 36 is not a multiple of 24.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Lists multiples of 24, stops at 48 and misjudges it as a multiple of 18, though 48 ÷ 18 is not a whole number.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Multiplies 24 by 6 to get 144, which all three numbers divide, overlooking that the question asks for the least such multiple.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiples of a number: $n \times 1,\ n \times 2,\ n \times 3,\ \ldots$; n is any whole number. Its multiples are what you get when you multiply n by 1, 2, 3 and so on. The least common multiple of some numbers is the smallest number that is a multiple of every one of them.

Understand the question
A multiple of a number is that number times 1, 2, 3, and so on. A common multiple is a multiple of every number in the list.
You need the least, or smallest, common multiple of 24, 6 and 18.
Drop the 6
6 goes evenly into 24, since $6 \times 4 = 24$ . It also goes into 18, since $6 \times 3 = 18$ .
So any multiple of 24 is already a multiple of 6. You only need to work with 24 and 18.
List the multiples of 24
Start with the bigger number, since its list grows faster. Check each multiple to see if 18 divides it evenly.
- $24 \times 1 = 24$ , and 24 ÷ 18 is not a whole number.
- $24 \times 2 = 48$ , and 48 ÷ 18 is not a whole number.
- $24 \times 3 = 72$ , and $72 \div 18 = 4$ . It works.
Give the answer
72 is the first multiple of 24 that 18 also divides. It is a multiple of 6 too, since $72 \div 6 = 12$ .
Watch out: 36 works for 6 and 18, but not for 24, so choice A is wrong. 144 works for all three, but it is not the smallest.
So the answer is C, 72.

Check your answer

Test 72 with all three numbers: $72 \div 24 = 3$ , $72 \div 6 = 12$ and $72 \div 18 = 4$ . All are whole numbers. The smaller choices, 36 and 48, each fail for one of the numbers.

More questions like this

The original DOE question

Question 71 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 71. Open page 155 of the handbook (PDF)

Question 106 of 112

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 107 of 112

2018 Test B Math, question 29

Question

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

A $w+1$
B $2w-1$
C $2w-2$
D $2w+1$

Show the answer

Correct answer: C

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

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

Therefore, $2w-2$ must be even.

Question analysis

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

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

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
The stem gives no number at all: w − 1 is only known to be odd, so the parity of w itself has to be worked out before any choice can be judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
All four choices are built from w or 2w plus or minus 1 or 2, so each one has to be tested for evenness; three of them come out odd for every allowed w. (Embretson & Wetzel, 1987)
Something you must already know minor Working it out
The parity rules are assumed, not stated: one more than an odd number is even, any multiple of 2 is even, and an even number plus or minus 1 is odd. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Treats w itself as the odd number, so w + 1 looks even; in fact w − 1 being odd makes w even and w + 1 odd.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Doubles the odd number w − 1 to make it even but distributes the 2 to w only, writing 2(w − 1) as 2w − 1, which is odd.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Assumes any expression that starts with 2w is even, overlooking that 2w + 1 is one more than an even number and so always odd.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out what you know
You know that $w-1$ is an odd integer. An integer is a whole number, like 3, 0 or −5. An even number can be split into two equal groups, like 4 or 10. An odd number is 1 more than an even number, like 3 or 11.
You need the choice that must be even. Must means it is even for every possible value of $w$ , not just one.
Decide whether w is odd or even
The number $w$ is 1 more than $w-1$ . One more than an odd number is always even. For example, if $w-1$ is 3, then $w$ is 4. So $w$ is even.
Watch out: do not treat $w$ itself as the odd number. That mistake makes $w+1$ look even and leads to choice A.
Show that 2w − 2 is even
Two times any whole number is even, so $2w$ is even. Taking 2 away from an even number moves you back two spots. You land on another even number. So $2w-2$ is even.
Here is another way to see it. $2w-2$ is the same as $2 \times (w-1)$ . That is 2 times a whole number, so it must be even.
Check the other three choices
Each one adds or takes away 1 from an even number. That always gives an odd number.
- A: $w+1$ is even plus 1, so it is odd.
- B: $2w-1$ is even minus 1, so it is odd.
- D: $2w+1$ is even plus 1, so it is odd.
Watch out: starting with $2w$ does not make a number even. The +1 in choice D makes it odd.
Give the answer
Only one choice is even for every possible $w$ . So the answer is C, $2w-2$ .

Check your answer

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

More questions like this

The original DOE question

Question 86 as printed on page 158 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 86. Open page 158 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 108 of 112

2018 Test B Math, question 30

Question

Three students stand at the starting line of a running track and begin running laps at the same time. Ann completes 1 lap every 2 minutes, Jack completes 1 lap every 3 minutes, and Lee completes 1 lap every 4 minutes. How many laps does Ann complete before all three runners are once again at the starting line at the same time?

Show the answer

Correct answer: B

Find the least common multiple of 2, 3, and 4 — which is 12. So, it takes 12 minutes before all three are back at the starting line. Ann completes 1 lap every 2 minutes, so in 12 minutes she has completed 6 laps.

Question analysis

Three runners start together and finish laps every 2, 3 and 4 minutes, and the question asks how many laps Ann, the fastest, completes before all three are at the starting line together again. It tests least common multiples in a word problem. The catch is keeping track of which quantity the question wants once the common time is found.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Sixth and seventh graders find least common multiples of two numbers in isolation; this applies the idea to three runners in a story and adds a conversion from minutes to laps.
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: 2 steps
Reading level: Question and choices: about grade 8, 56 words.
Students, last 30 days: 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 quantities are in play, minutes and laps, and the question asks for the second while the natural first result is the first. Students who find the 12-minute meeting time often stop there, and students who do count laps can count Jack's or Lee's by mistake. The question rewards tracking whose laps are asked for and converting the time into laps at the very end.

Turning words into math moderate Understanding the question
"All three runners are once again at the starting line at the same time" has to be recognized as the first common multiple of 2, 3 and 4 minutes; the wording never says multiple. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Finding a least common multiple is assumed; multiplying the three lap times instead gives 24 minutes, which is a common time but not the first one. (Anderson & Krathwohl, 2001)

Strategy: Find the first time that is a multiple of all three lap times, then divide that time by Ann's lap time to count her laps.

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 a slip Expected pull: moderate
Divides the 12-minute meeting time by Jack's 3-minute lap, giving his 4 laps instead of Ann's.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the least common multiple, 12 minutes, and gives it as the answer without dividing by Ann's 2-minute lap to count her laps.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No method using the lap times 2, 3 and 4 gives 20; it would mean 40 minutes for Ann, not the first time all three meet.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of laps: $\text{laps} = \text{total time} \div \text{time for one lap}$; Total time is how long the runner runs. Time for one lap is how long one trip around the track takes. Dividing tells you how many of those trips fit into the total time.

Sort out what you know
Ann runs 1 lap every 2 minutes. Jack runs 1 lap every 3 minutes. Lee runs 1 lap every 4 minutes. They all start together.
A runner is back at the start line at the end of every lap. First find when all three are “once again at the starting line at the same time”. Then count Ann’s laps up to that moment.
List when each runner is back
Write the minutes when each runner reaches the line.
- Ann: 2, 4, 6, 8, 10, 12 …
- Jack: 3, 6, 9, 12 …
- Lee: 4, 8, 12 …
The first number in all three lists is 12. This is the least common multiple of 2, 3 and 4. It is the smallest number that all three divide into with nothing left over. So they meet after 12 minutes.
Count Ann’s laps
Use the laps formula. Ann runs for 12 minutes, and each lap takes her 2 minutes.
$\text{laps} = 12 \div 2 = 6$
Ann runs 6 laps.
Watch out: 12 is the number of minutes, not laps. Choosing it gives C. Dividing by Jack’s 3 minutes gives his 4 laps, which is A.
Give the answer
The question asks only about Ann, and she runs 6 laps in 12 minutes. So the answer is B, 6.

Check your answer

Check that everyone is back at 12 minutes. Ann: $6 \times 2 = 12$ . Jack: $4 \times 3 = 12$ . Lee: $3 \times 4 = 12$ . No earlier time works. At 6 minutes, Lee is only halfway through his second lap.

More questions like this

The original DOE question

Question 87 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 87. Open page 158 of the handbook (PDF)

Question 109 of 112

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 110 of 112

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)

Question 111 of 112

2018 Test B Math, question 47

Question

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

Show the answer

Correct answer: A

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

Simplify the fraction to find the answer:

$\frac{6}{100}=\frac{3}{50}$

$x=3$

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Writes 0.06 as 6/100 and keeps 6 as the numerator even though the denominator has changed from 100 to 50.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Doubles the 6 instead of halving it, scaling the numerator up while the denominator goes down from 100 to 50.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Reads 0.06 as 0.6 and multiplies by 50, a decimal-place slip that makes the numerator ten times too large.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
You need to write 0.06 as a fraction with 50 on the bottom. The top number of that fraction is $x$ . You need to find $x$ .
Write the decimal as a fraction
0.06 has two digits after the decimal point. That means it counts hundredths. So 0.06 is six hundredths.
$0.06 = \frac{6}{100}$
Change the bottom to 50
To turn 100 into 50, divide by 2. The rule for equivalent fractions says to do the same to the top.
$\frac{6}{100} = \frac{6 \div 2}{100 \div 2} = \frac{3}{50}$
So $x = 3$ .
Watch out: the 6 must change too. Keeping the 6 gives choice B. Doubling it to 12 gives choice C.
Give the answer
The fraction $\frac{3}{50}$ is the same amount as 0.06, and its top number is 3. So the answer is A, 3.

Check your answer

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

More questions like this

The original DOE question

Question 104 as printed on page 162 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 104. Open page 162 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 112 of 112

2018 Test B Math, question 55

Question

1 sind = 5.6 ricks

1 sind = 12.88 dalts

Using the conversion above, how many dalts are equal to 1 rick?

A0.43
B2.30
C7.28
D18.48

Show the answer

Correct answer: B

Since 5.6 ricks and 12.88 dalts are both equal to 1 sind, then 5.6 ricks = 12.88 dalts.

To calculate the number of dalts (d) in 1 rick, set up a proportion:

$\frac{5.6}{12.88}=\frac{1}{d}$

$5.6d=12.88$

$d=2.3$

Question analysis

Two made-up conversion facts set 1 sind equal to 5.6 ricks and to 12.88 dalts, and students find how many dalts equal 1 rick. It tests unit conversion with invented units and decimal division. The risk is dividing in the wrong order, since both quotients look like reasonable answers.

NYS standards: 6.RP.3d converting measurement units
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Converting units with a given rate is grade 6 ratio reasoning. Invented units remove any real-world sense check and the division involves decimals, which matches the care typical grade 7 problems require.
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
Solution: 2 steps
Reading level: Question and choices: about grade 3, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Because both amounts equal one sind, the problem reduces to 5.6 ricks = 12.88 dalts, but the made-up units remove any feel for which way to divide. The inverted ratio is the strongest pull, while the sum and the difference of the given numbers catch students who combine them without a plan. Checking that 5.6 × 2.3 returns 12.88 confirms the direction.

Error-prone arithmetic moderate Working it out
Dividing 12.88 by 5.6 means dividing by a decimal, best done as 128.8 ÷ 56 = 2.3, where a misplaced decimal point is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Since both amounts equal 1 sind, set the ricks equal to the dalts, then divide both sides by whichever number leaves exactly one rick on its side.

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).
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 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides 5.6 by 12.88, turning the rate upside down: 0.43 is the number of ricks in one dalt, not dalts in one rick.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Subtracts the two given numbers, 12.88 − 5.6 = 7.28, as if the gap between them were the conversion rate.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two given numbers, 12.88 + 5.6 = 18.48, combining the conversion figures without a model of what they mean.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate: $\text{amount for 1} = \text{total amount} \div \text{number of units}$; The total amount is what you have for several units. The number of units is how many of them there are. Dividing shares the total equally, so you get the amount for just one unit.

Sort out what you know
Sinds, ricks and dalts are made-up units. 1 sind is 5.6 ricks. 1 sind is also 12.88 dalts. You need the number of dalts in 1 rick.
Link ricks and dalts
Both 5.6 ricks and 12.88 dalts equal 1 sind. Two things equal to the same thing are equal to each other. So 5.6 ricks are the same amount as 12.88 dalts.
Set up the division
You want 1 rick, not 5.6 ricks. So share the 12.88 dalts among the 5.6 ricks. Use the unit rate formula.
$\text{amount for 1} = 12.88 \div 5.6$
Watch out: do not divide the other way. $5.6 \div 12.88$ is about 0.43. That is ricks in one dalt, choice A. Adding or subtracting the numbers gives D or C.
Do the division
Move the decimal point one place right in both numbers. This makes the number you divide by a whole number.
$12.88 \div 5.6 = 128.8 \div 56 = 2.3$
So 1 rick is 2.3 dalts.
Give the answer
One rick equals 2.3 dalts, written 2.30 in the choices. So the answer is B, 2.30.

Check your answer

Multiply back: $5.6 \times 2.3 = 12.88$ . That matches the dalts in 1 sind. Also, a sind holds more dalts than ricks, so a rick is bigger than a dalt. One rick must be more than 1 dalt. That rules out 0.43.

More questions like this

The original DOE question

Question 112 as printed on page 164 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 112. Open page 164 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Small factor	Partner
1	168
2	84
3	56
4	42
6	28
7	24
8	21
12	14