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Numbers and operations

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Question 1 of 120

2026 Practice Test A Math, question 17

Question

Consider this expression:

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

Create an equivalent expression.

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

|| − ||

On the test you drag tiles into the boxes; here, choose each box’s tile from its menu. Each tile can be used once. The DOE has not published how answers with several parts are scored; this site counts one right only when every part is right.

Show the answer

Correct answer: blank 1: 1; blank 2: −5

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

Step 1: Simplify the first absolute value term.

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

−5+6=1-5 + 6 = 1

Simplify the second absolute value term.

6−11=−56 - 11 = -5

Therefore the equivalent expression is

|1|−|−5||1| - |-5|.

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. Many steps minor Working it out

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

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

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

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

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

    |□|−|□||\square| - |\square|

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

  2. Simplify inside the first bars

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

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

    −5+6=1-5 + 6 = 1

    The first box gets the tile 1.

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

  3. Simplify inside the second bars

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

    6−11=−56 - 11 = -5

    The second box gets the tile −5.

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

  4. Keep the boxes in order

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

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

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

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

Check your answer

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

More questions like this

The original DOE question
Question 17 as shown on the DOE’s digital test, page 88 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test A Math, page 88, question 17.

Question 2 of 120

2026 Practice Test A Math, question 50

Question

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

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

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

Show the answer

Correct answer: 7/5

Solution:

75\frac{7}{5}

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. Something you must already know minor Working it out

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

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

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

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

  2. Put in the numbers

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

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

    So the expression becomes:

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

  3. Divide the two fractions

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

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

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

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

  4. Add the last fraction

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

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

    The value is 7/5.

  5. Enter it as a fraction

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

    So you would enter 7/5.

Check your answer

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

More questions like this

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

Question 3 of 120

2026 Practice Test C Math, question 1

Question

What is the prime factorization of 756?

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

Correct answer: D

Determine the prime factors of 756 using a factor tree:

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

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

Question analysis

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

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

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

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

  2. Something you must already know minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Divide by 2 while you can

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

    756=2×378756 = 2 \times 378

    378 is even too:

    378=2×189378 = 2 \times 189

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

  3. Divide by 3 while you can

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

    189=3×63189 = 3 \times 63

    63=3×2163 = 3 \times 21

    21=3×721 = 3 \times 7

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

  4. Write the factors with exponents

    Collect every prime you pulled out:

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

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

  5. Choose the answer

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

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

Check your answer

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

More questions like this

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

Question 4 of 120

2026 Practice Test C Math, question 4

Question

A person buys a used car with a down payment of $1,200 and makes monthly payments of $275 for 3 years. What is the total amount the person pays for the car?

  • A$2,025
  • B$9,900
  • C$11,100
  • D$13,500
Show the answer

Correct answer: C

Determine the total cost of the car by adding the down payment to the total monthly payments over 3 years.

There are 12 months in a year, so multiply (12 months/year × 3 years) to get a total of 36 months. Determine the total monthly payments by multiplying the monthly payment ($275) × the total number of months (36):

275×36=9,900275 \times 36 = 9{,}900

Determine the total cost of the car by adding the down payment ($1,200) to the total monthly payments ($9,900):

9,900+1,200=11,1009{,}900 + 1{,}200 = 11{,}100

The total cost of the car is $11,100.

Question analysis

A used car is bought with a down payment of $1,200 followed by monthly payments of $275 for 3 years, and students find the total paid. The work is to count the payments, multiply, and add the down payment. Mixing up years and months, or forgetting the money paid up front, leads to the wrong totals among the choices.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 A total made of one payment up front plus equal monthly payments is a short whole-number calculation, simpler than grade 7 word problems with percents or negatives. The only conversion is from years to months.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 29 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The situation is familiar and the method direct, so this is one of the easier questions on the test. Points are lost in two places: turning 3 years into 36 payments, and remembering the money paid up front after the long multiplication. Each slip, like paying the down payment once a year, leads to a listed total, so checking that the answer holds exactly one $1,200 and 36 payments of $275 is worth the seconds.

  1. Much to keep track of minor Working it out

    The $1,200 down payment is paid once while the $275 repeats every month, so it has to be set aside while 36 payments are totaled and then added at the end. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Error-prone arithmetic minor Working it out

    The product 275 × 36 = 9,900 is a two-digit multiplication done by hand; splitting it as 275 × 4 = 1,100 and then 1,100 × 9 makes it quick and safe. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Turn 3 years into a number of months, multiply the monthly payment by it, and add the down payment once at the end.

Traps

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing years into months
months=12×years\text{months} = 12 \times \text{years}
Each year has 12 months. Multiply the number of years by 12 to get the number of months. You need months here because a payment is made every month.
Total cost with a down payment
total=down payment+monthly payment×months\text{total} = \text{down payment} + \text{monthly payment} \times \text{months}
The down payment is money paid once, at the start. The monthly payment is paid every month. Multiply it by the number of months to get all the monthly payments. Then add the down payment only once, because it is paid only once.

  1. Sort out the question

    The person pays $1,200 once, at the start. This is the down payment. Then the person pays $275 every month for 3 years. You need the total amount paid for the car.

    The payments come once a month, but the time is given in years. So change the years into months first.

  2. Change 3 years into months

    Use the months formula:

    months=12×3=36\text{months} = 12 \times 3 = 36

    So the person makes 36 monthly payments.

    Watch out: multiplying $275 by 3 counts only 3 payments. That leads to choice A, $2,025.

  3. Add up the monthly payments

    Multiply $275 by 36. A quick way is to split 36 into 4 × 9:

    275×4=1,100275 \times 4 = 1{,}100

    1,100×9=9,9001{,}100 \times 9 = 9{,}900

    So the 36 payments add up to $9,900.

    Watch out: stopping here gives choice B, $9,900. The down payment is still missing.

  4. Add the down payment once

    Use the total-cost formula:

    total=1,200+9,900=11,100\text{total} = 1{,}200 + 9{,}900 = 11{,}100

    The person pays $11,100 in all. Adding $1,200 for every year counts it three times. That gives choice D, $13,500.

    So the answer is C, $11,100.

Check your answer

Estimate. $275 is a little less than $300, and 36×300=10,80036 \times 300 = 10{,}800. Add the $1,200: 10,800+1,200=12,00010{,}800 + 1{,}200 = 12{,}000. The real total should be a little less than $12,000. $11,100 fits.

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 C Math, page 77, question 4.

Question 5 of 120

2026 Practice Test C Math, question 8

Question

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

Which fractions are equivalent to the decimal 0.75?

Select the four correct answers.

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

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

Correct answer: A, B, C, and E

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

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

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Every choice must be checked moderate Choosing the answer

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

  2. A very close wrong answer minor Choosing the answer

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

  3. Much to keep track of minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Write 0.75 as a fraction

    0.75 means 75 hundredths:

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

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

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

    So choice A, 3/4, works too.

  3. Test choices B and E

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

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

  4. Test choices D and F

    Divide the top by the bottom:

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

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

  5. Name the four answers

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

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

Check your answer

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

More questions like this

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

Question 6 of 120

2026 Practice Test C Math, question 11

Question

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

  • A198
  • B612
  • C620
  • D632
Show the answer

Correct answer: B

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

5×576=2,8805 \times 576 = 2{,}880

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

6×378=2,2686 \times 378 = 2{,}268

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

2,880−2,268=6122{,}880 - 2{,}268 = 612

Question analysis

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

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Whole-number multiplication and subtraction in a three-step word problem is easier than grade 7 work with fractions, decimals or negatives. Because the products are large, a slip or a rounded estimate easily gives another choice.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 27 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

    612, 620 and 632 lie within 20 of one another, so rounding 576 and 378 to friendly numbers before multiplying lands on a wrong choice; only exact products separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    Two products of three-digit numbers, 5 × 576 = 2,880 and 6 × 378 = 2,268, must be done by hand and then subtracted with regrouping. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Allison has 5 albums with 576 stamps in each. She moves all her stamps into 6 new albums. Each new album holds 378 stamps. Some stamps will not fit. You need the number left over.

  2. Count all her stamps

    Use the equal-groups formula. Split 576 into 500 + 76 to make it easy:

    5×576=5×500+5×765 \times 576 = 5 \times 500 + 5 \times 76

    2,500+380=2,8802{,}500 + 380 = 2{,}880

    Allison has 2,880 stamps.

  3. Count what the new albums hold

    Do the same for the 6 new albums. Split 378 into 300 + 78:

    6×378=6×300+6×786 \times 378 = 6 \times 300 + 6 \times 78

    1,800+468=2,2681{,}800 + 468 = 2{,}268

    The new albums hold 2,268 stamps in all.

  4. Subtract to find the leftovers

    The stamps left over are the ones she has minus the ones that fit:

    2,880−2,268=6122{,}880 - 2{,}268 = 612

    Watch out: 576−378=198576 - 378 = 198, choice A, compares just one old album with one new one. Rounding to 580 and 380 first gives 620, choice C. Rounding only 576 gives 632, choice D. Use the exact numbers.

    So the answer is B, 612.

Check your answer

Add the leftovers to the stamps that fit: 2,268+612=2,8802{,}268 + 612 = 2{,}880. That is exactly the number Allison started with. So no stamps were lost or counted twice.

More questions like this

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

Question 7 of 120

2026 Practice Test C Math, question 13

Question

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

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

What is the result?

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

Correct answer: D

Begin by multiplying 135.28 by 10:

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

Add 0.5:

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

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

Divide by 10:

1,353÷10=135.31{,}353 \div 10 = 135.3

Question analysis

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

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

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

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

  2. Many steps minor Working it out

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

  3. Error-prone arithmetic minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Multiply by 10

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

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

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

  3. Add 0.5

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

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

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

  4. Drop the digits after the point

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

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

  5. Divide by 10

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

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

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

Check your answer

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

More questions like this

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

Question 8 of 120

2026 Practice Test C Math, question 14

Question

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

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

Correct answer: B

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

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

Evaluate the terms involving exponents:

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

Evaluate the terms involving multiplication:

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

Simplify the expression:

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

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

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

6−12=(−6)6 - 12 = (-6)

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. Many steps minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the powers of −2

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

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

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

    The square is positive, but the cube is negative.

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

  3. Work out the first bars

    Add the three terms inside the first pair of bars:

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

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

  4. Work out the second bars

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

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

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

  5. Subtract the second value

    Now subtract the second value from the first:

    6−12=−66 - 12 = -6

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

    So the answer is B, −6.

Check your answer

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

More questions like this

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

Question 9 of 120

2026 Practice Test C Math, question 16

Question

3+(−3)3−(−3)=\frac{3+(-3)}{3-(-3)} =

  • A−1
  • B0
  • C16\frac{1}{6}
  • D1
Show the answer

Correct answer: B

Simplify the expression.

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

3+(−3)3−(−3)=3−33+3=06\frac{3+(-3)}{3-(-3)} = \frac{3-3}{3+3} = \frac{0}{6}

The zero property of division states that zero divided by any nonzero number equals zero, so 06=0\frac{0}{6} = 0.

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding and subtracting a negative integer and dividing are grade 7 basics, and here the numbers are small. It is a short routine calculation, easier than typical grade 7 work, though the two signs invite a quick misreading.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 16+, 2 words.
Corrected from the DOE handbook
  • A grammatical error in the explanation was corrected.
  • A rule stated too broadly in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

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

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The top of the fraction is 3+(−3)3 + (-3), and the bottom is 3−(−3)3 - (-3). A fraction bar means divide. Work out the top and the bottom separately first. Then divide the top by the bottom.

  2. Work out the top

    Adding −3 is the same as subtracting 3:

    3+(−3)=3−3=03 + (-3) = 3 - 3 = 0

    A number plus its opposite is always 0.

    Watch out: do not add the 3s and keep the minus sign. That makes the top −6 and leads to choice A, −1.

  3. Work out the bottom

    Subtracting −3 is the same as adding 3:

    3−(−3)=3+3=63 - (-3) = 3 + 3 = 6

    Watch out: the top and bottom look alike, but they are not. Making both of them 6 gives 6/6 = 1, choice D.

  4. Divide the top by the bottom

    Now the fraction is 0 over 6. Zero divided by any number other than zero is zero:

    3+(−3)3−(−3)=06=0\frac{3+(-3)}{3-(-3)} = \frac{0}{6} = 0

    A top of 0 does not make the fraction 1/6, choice C. So the answer is B, 0.

Check your answer

Check by multiplying. A division is right when the answer times the bottom gives the top. Here 0×6=00 \times 6 = 0, which is the top. So 0/6 really is 0.

More questions like this

The original DOE question
Question 16 as shown on the DOE’s digital test, page 89 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 89, question 16. This is the DOE’s version, before the corrections listed in the analysis.

Question 10 of 120

2026 Practice Test C Math, question 17

Question

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

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

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

Correct answer: A

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

7−1=67 - 1 = 6

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

6÷2=36 \div 2 = 3

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

1+3=41 + 3 = 4

7−3=47 - 3 = 4

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

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

6−3=36 - 3 = 3

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

3÷2=1.53 \div 2 = 1.5

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

3+1.5=4.53 + 1.5 = 4.5

6−1.5=4.56 - 1.5 = 4.5

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

Determine the distance between the two midpoints by subtracting:

4.5−4=0.54.5 - 4 = 0.5

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

Question analysis

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

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

Why it is hard

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

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

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

  2. Many steps minor Working it out

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the midpoint of MN

    Use the midpoint formula with the ends 1 and 7:

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

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

  3. Find the midpoint of RS

    Use the same formula with the ends 3 and 6:

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

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

  4. Subtract the two midpoints

    Use the distance formula:

    4.5−4=0.54.5 - 4 = 0.5

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

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

  5. Choose the answer

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

Check your answer

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

More questions like this

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

Question 11 of 120

2026 Practice Test C Math, question 18

Question

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

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

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

Correct answer: D

Each of the three expressions must be considered:

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

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

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

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

  2. Something you must already know moderate Working it out

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

  3. Every choice must be checked minor Choosing the answer

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

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Try one pair of numbers

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

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

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

  3. Explain expression I

    Use the rule for subtracting:

    s−t=s+(−t)s - t = s + (-t)

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

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

  4. Explain expression II

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

  5. Explain expression III

    Use the rule for a negative exponent:

    st=1s−ts^{t} = \frac{1}{s^{-t}}

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

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

  6. Choose the answer

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

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

Check your answer

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

More questions like this

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

Question 12 of 120

2026 Practice Test C Math, question 19

Question

What is the value of (−34)(−25)\left(-\frac{3}{4}\right)\left(-\frac{2}{5}\right)?

  • A−720-\frac{7}{20}
  • B−310-\frac{3}{10}
  • C310\frac{3}{10}
  • D720\frac{7}{20}
Show the answer

Correct answer: C

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

(−3)×(−2)4×5=620\frac{(-3) \times (-2)}{4 \times 5} = \frac{6}{20}

Simplify the result:

6÷220÷2=310\frac{6 \div 2}{20 \div 2} = \frac{3}{10}

Question analysis

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

NYS standards
  • 7.NS.2a rational number multiplication
  • 7.NS.2c properties for multiplying and dividing
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Multiplying two negative fractions is a standard grade 7 exercise, done here in one product and one simplification. It matches typical classroom work, with the sign rule and lowest terms as the only points of care.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 10, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. A very close wrong answer minor Choosing the answer

    3/10 and −3/10 are identical except for the sign, so the size alone does not separate them; only the rule for multiplying two negatives does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    Two negative fractions are multiplied, 3 × 2 over 4 × 5, and the product 6/20 has to be reduced to 3/10 before it matches a choice. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The two fractions sit side by side in parentheses, with nothing between them. That means multiply. Both fractions are negative. You need their product, the answer to the multiplication.

  2. Decide the sign first

    A negative times a negative is positive. So the answer will be positive. That rules out choices A and B.

    Watch out: choice B, −3/10, has the right size but the wrong sign. The two minus signs cancel.

  3. Multiply tops and bottoms

    Now multiply the sizes. Multiply the tops, then the bottoms:

    34×25=3×24×5=620\frac{3}{4} \times \frac{2}{5} = \frac{3 \times 2}{4 \times 5} = \frac{6}{20}

    Watch out: do not find a common denominator. That is for adding and subtracting. Subtracting the fractions gives 7/20 or −7/20, choices D and A.

  4. Simplify the fraction

    2 goes into both 6 and 20. Divide the top and the bottom by 2:

    6÷220÷2=310\frac{6 \div 2}{20 \div 2} = \frac{3}{10}

    The product is positive 3/10. So the answer is C, 3/10.

Check your answer

Use decimals. 3/4 is 0.75 and 2/5 is 0.4. Then 0.75×0.4=0.30.75 \times 0.4 = 0.3, which is 3/10. Two negatives give a positive, so the product is positive 3/10.

More questions like this

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

Question 13 of 120

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

Determine the difference between the two elevations:

90 ft−(−120) ft=90 ft+120 ft=210 ft90\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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Difference between highest and lowest
difference=highest−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+ba - (-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.

  1. 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.

  2. 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.

  3. Subtract the lowest from the highest

    Use the difference formula. Subtracting a negative is the same as adding:

    90−(−120)=90+120=21090 - (-120) = 90 + 120 = 210

    So the two readings are 210 feet apart.

    Watch out: 120−90=30120 - 90 = 30 treats both readings as if they were above 0. Adding them, 90+(−120)=−3090 + (-120) = -30, is wrong too.

  4. 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=210120 + 90 = 210 feet. That matches.

More questions like this

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

Question 14 of 120

2026 Practice Test C Math, question 25

Question

For each 38\frac{3}{8} foot of curtain width, 56\frac{5}{6} yard of fabric is needed to make ruffles. How many yards of fabric are needed per foot of curtain width?

  • A516\frac{5}{16} yd
  • B2924\frac{29}{24} yd
  • C209\frac{20}{9} yd
  • D72\frac{7}{2} yd
Show the answer

Correct answer: C

It takes 56\frac{5}{6} yard of fabric to make 38\frac{3}{8} foot of curtain width.

To find the ratio of 1 foot of curtain width to 38\frac{3}{8} foot of curtain width (1:38)\left(1:\frac{3}{8}\right), divide.

1÷38=1×83=831 \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 83\frac{8}{3} times the amount needed for 38\frac{3}{8} foot of curtain width.

83×56=4018\frac{8}{3} \times \frac{5}{6} = \frac{40}{18}

40÷218÷2=209\frac{40 \div 2}{18 \div 2} = \frac{20}{9}

The amount of fabric needed for 1 foot of curtain width is 209\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.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
yards per foot=yards÷feet\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÷bc=a×cba \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.

  1. 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.

  2. Set up the division

    Per foot means divide the yards by the feet:

    56÷38\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.

  3. Multiply by the reciprocal

    Flip 3/8 to get 8/3, and multiply:

    56×83=5×86×3=4018\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.

  4. Simplify the fraction

    2 goes into both 40 and 18:

    40÷218÷2=209\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: 209×38=6072=56\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 15 of 120

2026 Practice Test D Math, question 1

Question

What is the prime factorization of 3,575?

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

Correct answer: B

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

3,575÷25=1433{,}575 \div 25 = 143

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

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

  2. Something you must already know minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Divide by 25 first

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

    3,575÷25=1433{,}575 \div 25 = 143

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

  3. Test 143 for prime factors

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

    143÷11=13143 \div 11 = 13

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

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

  4. Put all the primes together

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

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

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

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

Check your answer

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

More questions like this

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

Question 16 of 120

2026 Practice Test D Math, question 4

Question

If x=14x = \frac{1}{4}, what is the value of 22−x\frac{2}{2 - x}?

  • A34\frac{3}{4}
  • B89\frac{8}{9}
  • C1171\frac{1}{7}
  • D3123\frac{1}{2}
Show the answer

Correct answer: C

The first step is to substitute the value for x into the expression to get 22−14\frac{2}{2 - \frac{1}{4}}.

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

2(74)=2÷(74)=2(47)=87=117\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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction
a÷bc=a×cba \div \frac{b}{c} = a \times \frac{c}{b}
a is 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
ab=qrb\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.

  1. Sort out the question

    You know x=14x = \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.

  2. Work out the bottom

    Put in 1/4 for x. To subtract, write 2 as fourths, since 2 = 8/4:

    2−14=84−14=742 - \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.

  3. 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÷74=2×47=872 \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.

  4. Write it as a mixed number

    The choices use mixed numbers. 7 fits into 8 once, with 1 left over:

    87=117\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, 1171\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 17 of 120

2026 Practice Test D Math, question 9

Question

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

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

Select the three correct answers.

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

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

Correct answer: A, B, and D

Each of the expressions must be considered:

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

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

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

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

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

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

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

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

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

A negative divided by a positive remains negative.

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

Question analysis

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

NYS standards
  • 7.NS.2a rational number multiplication
  • 7.NS.1c rational number subtraction
  • 7.NS.2b rational number division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Sign rules for adding, subtracting, multiplying and dividing integers are core grade 7 content. Here they are applied to letters with known signs instead of numbers, across five expressions, and all three right ones must be chosen with no wrong one.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

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

  2. Every choice must be checked moderate Choosing the answer

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

  3. Much to keep track of minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Understand the minus in front

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

  3. Check choices A and B

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

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

  4. Check choices C and E

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

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

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

  5. Check choice D

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

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

  6. Choose the three answers

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

    So the answers are A, B and D.

Check your answer

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

More questions like this

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

Question 18 of 120

2026 Practice Test D Math, question 14

Question

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

  • A35
  • B42
  • C45
  • D56
Show the answer

Correct answer: B

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Many steps moderate Working it out

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

  2. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the morning pages

    Take 1/4 of 84:

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

    He reads 21 pages in the morning. That leaves:

    84−21=6384 - 21 = 63

    63 pages are left after the morning.

  3. Find the afternoon pages

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

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

    He reads 21 pages in the afternoon.

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

  4. Find the pages still to read

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

    63−21=4263 - 21 = 42

    He still needs to read 42 pages.

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

    So the answer is B, 42.

Check your answer

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

More questions like this

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

Question 19 of 120

2026 Practice Test D Math, question 15

Question

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

What is the value of the expression shown above?

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

Correct answer: C

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

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

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

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

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find a common denominator

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

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

  3. Rename each fraction

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

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

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

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

  4. Add the tops

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

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

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

  5. Simplify the fraction

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

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

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

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

Check your answer

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

More questions like this

The original DOE question
Question 15 as shown on the DOE’s digital test, page 88 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 88, question 15.

Question 20 of 120

2026 Practice Test D Math, question 18

Question

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

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

Correct answer: C

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

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

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

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

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

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Rename two thirds as ninths

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

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

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

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

  3. Add the mixed numbers

    Add the whole numbers, then add the fractions:

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

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

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

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

  4. Change it to a decimal

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

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

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

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

Check your answer

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

More questions like this

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

Question 21 of 120

2026 Practice Test D Math, question 20

Question

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

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

Correct answer: B

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

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

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

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

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

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

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

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

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

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

16−4=−4\frac{16}{-4} = -4

Question analysis

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

NYS standards
  • 7.NS.2a rational number multiplication
  • 7.NS.2b rational number division
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Order of operations with negative numbers is standard grade 7 practice. This expression stacks a squared negative, two products and a division by a negative into one fraction, so it takes more steps than usual, each with a sign to track.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 5, 6 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. Many steps minor Working it out

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

  3. A very close wrong answer minor Choosing the answer

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Do the exponent first

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

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

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

  3. Do the multiplication

    Now multiply. A negative times a positive is negative:

    −9×16=−144-9 \times 16 = -144

    36×(−4)=−14436 \times (-4) = -144

    So the top becomes:

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

  4. Add from left to right

    Add the two negatives first:

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

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

    −288+304=16-288 + 304 = 16

    The whole top is 16.

  5. Divide by −4

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

    16−4=−4\frac{16}{-4} = -4

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

    So the answer is B, −4.

Check your answer

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

More questions like this

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

Question 22 of 120

2026 Practice Test D Math, question 21

Question

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

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

Correct answer: C

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

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

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

25.00−23.25=1.7525.00 - 23.25 = 1.75

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

Question analysis

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

NYS standards
  • 6.NS.3 multi-digit decimal operations
  • 6.NS.5 positive and negative numbers
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Adding and subtracting money amounts is grade 6 decimal work. Here it takes one sum and one difference, and the only further demand is deciding whether the result is money left or money owed, well below typical grade 7 tasks.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 66 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

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

  2. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Add up what she borrowed

    Line up the decimal points and add:

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

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

  3. Compare the gift with the debt

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

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

  4. Subtract to find what is left

    Take the debt away from the gift:

    25.00−23.25=1.7525.00 - 23.25 = 1.75

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

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

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

Check your answer

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

More questions like this

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

Question 23 of 120

2026 Practice Test D Math, question 22

Question

Seth grows strawberries in his garden.

  • He picked 2122\frac{1}{2} cups of strawberries on Monday.
  • He picked 4 cups of strawberries on Tuesday.
  • He ate 34\frac{3}{4} cup of strawberries for breakfast each day on Wednesday, Thursday, and Friday.

How many cups of strawberries did Seth have left?

  • A3123\frac{1}{2}
  • B4144\frac{1}{4}
  • C6
  • D9
Show the answer

Correct answer: B

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

212+4+(3)(−34)2\frac{1}{2} + 4 + (3)\left(-\frac{3}{4}\right)

212+4−2142\frac{1}{2} + 4 - 2\frac{1}{4}

612−2146\frac{1}{2} - 2\frac{1}{4}

624−2146\frac{2}{4} - 2\frac{1}{4}

4144\frac{1}{4}

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

212+4+(−34)+(−34)+(−34)2\frac{1}{2} + 4 + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right) + \left(-\frac{3}{4}\right)

4144\frac{1}{4}

Question analysis

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

NYS standards
  • 7.NS.3 four operations with rational numbers
  • 7.EE.3 multi-step rational number problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding mixed numbers and multiplying a fraction by a whole number are skills from before grade 7. The three steps are short and the numbers friendly, so the demand is below typical grade 7 work with signed rational numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    Three breakfasts of 3/4 cup make 9/4 = 2 1/4 cups, and taking that from 6 1/2 means renaming the half as 2/4, two fraction steps where a slip is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    Picking adds and eating subtracts, and “each day on Wednesday, Thursday, and Friday” means the 3/4 cup counts three times, not once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Seth picked 2 1/2 cups on Monday and 4 cups on Tuesday. Then he ate 3/4 cup each day on Wednesday, Thursday and Friday. That is 3 days of eating. Picking adds strawberries, and eating takes them away. You need the number of cups he has left.

  2. Add what he picked

    Add the two days of picking:

    212+4=6122\frac{1}{2} + 4 = 6\frac{1}{2}

    He picked 6 1/2 cups in all, before he ate any.

  3. Find how much he ate

    He ate 3/4 cup on each of 3 days. Multiply:

    3×34=94=2143 \times \frac{3}{4} = \frac{9}{4} = 2\frac{1}{4}

    To change 9/4 to a mixed number, note that 4 fits into 9 twice, with 1 left over. He ate 2 1/4 cups.

    Watch out: he ate 3/4 cup three times, not 3 cups. Subtracting 3 gives 3 1/2, choice A.

  4. Take away what he ate

    Write the half as fourths, so both fractions have the same bottom: 1/2 = 2/4. Then subtract the wholes and the fourths:

    612−214=624−214=4146\frac{1}{2} - 2\frac{1}{4} = 6\frac{2}{4} - 2\frac{1}{4} = 4\frac{1}{4}

    Watch out: eating takes away. Adding the 2 1/4 cups gives 8 3/4, and the nearest choice is 9, choice D.

    So the answer is B, 4144\frac{1}{4} cups.

Check your answer

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

More questions like this

The original DOE question
Question 22 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 D Math, page 95, question 22.

Question 24 of 120

2026 Practice Test D Math, question 23

Question

Which situation involves quantities that combine to equal zero?

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

Correct answer: A

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

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

Question analysis

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

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

Why it is hard

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

  1. Turning words into math minor Understanding the question

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

  2. Every choice must be checked minor Choosing the answer

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Write each choice with signs

    Turn each situation into signed numbers and add them up:

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

  3. Look at the tricky choices

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

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

  4. Choose the zero total

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

    5+(−5)=05 + (-5) = 0

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

Check your answer

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

More questions like this

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

Question 25 of 120

2026 Practice Test D Math, question 24

Question

An athlete runs 18\frac{1}{8} kilometer in 34\frac{3}{4} minute. At this rate, how many kilometers would the athlete run in 1 minute?

  • A332\frac{3}{32}
  • B16\frac{1}{6}
  • C1121\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: 18 kilometer34 minute\frac{\frac{1}{8}\text{ kilometer}}{\frac{3}{4}\text{ minute}}.

To evaluate the fraction, use the reciprocal when dividing to get 16\frac{1}{6} kilometer per 1 minute.

18÷34=18×43=424=16\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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
rate=distance÷time\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
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal (the fraction flipped upside down). Flip only the second fraction, the one you divide by.

  1. 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.

  2. Set up the division

    Kilometers per minute means kilometers divided by minutes. So the distance comes first and the time second:

    rate=18÷34\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.

  3. Divide the fractions

    Flip 3/4 to get 4/3, then multiply. Multiply the tops and multiply the bottoms:

    18×43=424\frac{1}{8} \times \frac{4}{3} = \frac{4}{24}

    Simplify by dividing the top and bottom by 4:

    424=16\frac{4}{24} = \frac{1}{6}

    Watch out: multiplying 1/8 × 3/4 without flipping gives 3/32, choice A.

  4. 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, 16\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 26 of 120

2026 Practice Test D Math, question 30

Question

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

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

Correct answer: A

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Set up the unit rate

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

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

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

  3. Change the mixed number

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

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

  4. Divide the fractions

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

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

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

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

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

Check your answer

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

More questions like this

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

Question 27 of 120

2026 Practice Test D Math, question 36

Question

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

DAILY READING LOG

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 25.6%

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

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

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

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

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

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

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

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

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Finding what percent one number is of another is grade 6 and 7 work. This adds three inclusive page ranges to count, which takes more care than subtracting, then a rounding step, all typed with no choices to check against.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 3, 66 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know moderate Working it out

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

  2. Many steps moderate Working it out

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

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

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

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

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Stops one step early: A choice is a number from partway through the solution.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

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

  2. Count the pages each day

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

    90−45+1=4690 - 45 + 1 = 46

    135−100+1=36135 - 100 + 1 = 36

    190−145+1=46190 - 145 + 1 = 46

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

  3. Add the three days

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

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

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

  4. Change to a percent

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

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

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

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

  5. Enter the answer

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

Check your answer

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

More questions like this

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

Question 28 of 120

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÷8=80 and 640÷5=128640 \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÷8=79R3 and 635÷5=127635 \div 8 = 79R3 \text{ and } 635 \div 5 = 127

    645÷8=80R5 and 645÷5=129645 \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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisible by 5
last digit=0 or 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
LCM(5,8)=5×8=40\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.

  1. 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.

  2. Use the rule for 5

    A number divisible by 5 ends in 0 or 5. Between 630 and 650, only three numbers do:

    635,640,645635, \; 640, \; 645

    Only these need checking for 8.

  3. 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 8×798 \times 79, passes for 8 but ends in 2, so it fails for 5.

  4. 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. 40×16=64040 \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 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 D Math, page 110, question 37.

Question 29 of 120

2026 Practice Test D Math, question 42

Question

A student reads 112\frac{1}{12} of a book in 13\frac{1}{3} hour. What fraction of the book can the student read in 1 hour?

  • A136\frac{1}{36}
  • B14\frac{1}{4}
  • C512\frac{5}{12}
  • D123\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.

112 book13 hour\frac{\frac{1}{12} \text{ book}}{\frac{1}{3} \text{ hour}}

Then use division to find the unit rate.

11213=112÷13=112×31=312=14\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 13\frac{1}{3} hour times 3 equals 1 hour.

3×112 book3×13 hour=14 book1 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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
unit rate=part of the bookhours\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
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a, b, c and d are numbers that are not zero. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Flip only the fraction you divide by.

  1. Sort out the question

    The student reads 112\frac{1}{12} of a book in 13\frac{1}{3} of an hour. You need the part of the book read in 1 whole hour. That is a unit rate.

  2. Scale up to one hour

    One hour holds three thirds of an hour. So in 1 hour the student reads 3 times as much:

    3×112=312=143 \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.

  3. 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:

    112÷13=112×31=312=14\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.

  4. Avoid the wrong operations

    Watch out: multiplying gives 112×13=136\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, 14\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 14×13=112\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 30 of 120

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?

  • A3
  • B5
  • C15
  • D45
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=3×3×545 = 3 \times 3 \times 5

75=3×5×575 = 3 \times 5 \times 5

The common factors can be seen as 3×53 \times 5.

The greatest possible side length, in inches, will be 3×5=153 \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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor
GCF=product of the prime factors both numbers share\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.

  1. 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.

  2. Break each number into primes

    Write each length as a product of prime numbers:

    45=3×3×545 = 3 \times 3 \times 5

    75=3×5×575 = 3 \times 5 \times 5

    The two lists share one 3 and one 5.

  3. Multiply the shared primes

    Multiply the primes the two numbers share:

    GCF=3×5=15\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.

  4. 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÷15=345 \div 15 = 3 and 75÷15=575 \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 31 of 120

2024 Test A Math, question 1

Question

Simplify:

−3.8+2.3−(−1.1)-3.8 + 2.3 - (-1.1)

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

Show the answer

Correct answer: -0.4

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

−3.8+2.3−(−1.1)-3.8 + 2.3 - (-1.1)

−3.8+2.3+1.1-3.8 + 2.3 + 1.1

−3.8+3.4-3.8 + 3.4

−0.4-0.4

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Adding and subtracting negative decimals is core grade 7 content, and this is a routine three-term version. It matches a typical classroom exercise; the only extra demand is gridding a signed decimal correctly.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 16+, 2 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    The last term, − (−1.1), has to turn into + 1.1, and the total then crosses zero, since 2.3 + 1.1 = 3.4 falls just short of the 3.8 below zero. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    As a grid-in it shows no choices, so a double negative read as a single one, or a result gridded without its minus sign, gets no warning from the page. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the expression

    You need to simplify −3.8+2.3−(−1.1)-3.8 + 2.3 - (-1.1). Simplify means work it out to one number. Two of the numbers are negative: −3.8 and −1.1. One is positive: 2.3. The tricky part is at the end. It says minus negative 1.1. The answer can be negative, so keep track of every sign.

  2. Change minus a negative to plus

    Taking away a negative is the same as adding a positive. Use the rule with b = 1.1.

    a−(−b)=a+ba - (-b) = a + b

    −3.8+2.3−(−1.1)=−3.8+2.3+1.1-3.8 + 2.3 - (-1.1) = -3.8 + 2.3 + 1.1

    Now only one negative number is left. Watch out: do not just subtract 1.1. That gives −3.8+2.3−1.1=−2.6-3.8 + 2.3 - 1.1 = -2.6, which is wrong.

  3. Add the two positive numbers

    Group the positive numbers first. This makes the rest of the work short.

    2.3+1.1=3.42.3 + 1.1 = 3.4

    Now the problem is −3.8+3.4-3.8 + 3.4. One number is negative and one is positive.

  4. Combine the negative and the positive

    They partly cancel. Use the rule with a = 3.8 and b = 3.4.

    −a+b=−(a−b)-a + b = -(a - b)

    −3.8+3.4=−(3.8−3.4)=−0.4-3.8 + 3.4 = -(3.8 - 3.4) = -0.4

    The negative number, 3.8, is farther from zero than 3.4 is. So the result is negative. Watch out: keep the minus sign. Dropping it gives 0.4, which is wrong. So you would grid -0.4, with the minus sign in its own column.

Check your answer

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

More questions like this

The original DOE question
Question 58 as printed on page 75 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 75, question 58. Open page 75 of the handbook (PDF)

Question 32 of 120

2024 Test A Math, question 3

Question

A juice mixture contains 316\frac{3}{16} gallon of apple juice and 340\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:

316 apple340 cranberry=x apple1 cranberry\frac{\frac{3}{16}\text{ apple}}{\frac{3}{40}\text{ cranberry}} = \frac{x\text{ apple}}{1\text{ cranberry}}

316×403=x\frac{3}{16} \times \frac{40}{3} = x

4016=x\frac{40}{16} = x

2.5=x2.5 = 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.
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.

  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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
r=a÷br = 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
ab÷cd=ab×dc\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.

  1. Sort out the question

    The mix has 316\frac{3}{16} gallon of apple juice and 340\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.

  2. 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÷br = a \div b

    r=316÷340r = \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.

  3. Multiply by the reciprocal

    To divide by 340\frac{3}{40}, multiply by its reciprocal, 403\frac{40}{3}.

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

    316×403=3×4016×3=4016\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÷3=13 \div 3 = 1. Watch out: do not just multiply the two amounts. That gives 9640\frac{9}{640}, about 0.014, which is far too small.

  4. Change the fraction to a decimal

    Make 4016\frac{40}{16} simpler by dividing the top and bottom by 8. Then divide.

    4016=52=5÷2=2.5\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, 316\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÷40=0.0753 \div 40 = 0.075 and 0.075×2.5=0.18750.075 \times 2.5 = 0.1875, which is 3÷163 \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 33 of 120

2024 Test A Math, question 4

Question

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

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

Show the answer

Correct answer: 5.5

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

8÷2=48 \div 2 = 4

4×2184 \times 2\frac{1}{8}

4×2.125=8.54 \times 2.125 = 8.5

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

8.5+4=12.58.5 + 4 = 12.5

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

8÷2=48 \div 2 = 4

4×1344 \times 1\frac{3}{4}

4×1.75=74 \times 1.75 = 7

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

12.5−7=5.512.5 - 7 = 5.5

Question analysis

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

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

Why it is hard

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

  1. Many steps moderate Working it out

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

  2. Error-prone arithmetic moderate Working it out

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

  3. Turning words into math minor Understanding the question

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

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

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

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

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Count the 2-week periods

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

    8÷2=48 \div 2 = 4

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

  3. Change the mixed numbers to decimals

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

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

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

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

  4. Add up the growing

    The lawn grows 2.125 inches, 4 times.

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

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

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

    4+8.5=12.54 + 8.5 = 12.5

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

  5. Add up the cutting

    Mr. Chan cuts off 1.75 inches, 4 times.

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

    He cuts off 7 inches in all.

  6. Subtract the cutting

    12.5−7=5.512.5 - 7 = 5.5

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

Check your answer

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

More questions like this

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

Question 34 of 120

2024 Test A Math, question 9

Question

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

  • A2
  • B33
  • C66
  • D67
Show the answer

Correct answer: D

Consider the case where m=1m = 1.

n−m=66n - m = 66

n−1=66n - 1 = 66

n−1+1=66+1n - 1 + 1 = 66 + 1

n=67n = 67

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

Question analysis

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

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

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

    The choices 66 and 67 differ by one, and 66 is the difference the stem hands over; only counting both ends of the list separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Variables instead of numbers moderate Understanding the question

    The list is given only by its ends, m and n, so the count has to be reasoned out in general, or checked on a small list such as 1 through 5. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Integers are whole numbers and their negatives, like −2, 0 and 5. Consecutive integers come one after another, going up by 1, like 4, 5, 6. The list starts at m and ends at n. The difference n−mn - m is 66. You need how many integers are in the list.

  2. Try a small list

    Letters are hard to count, so try a list you can count: 1, 2, 3, 4, 5. Here m = 1 and n = 5.

    n−m=5−1=4n - m = 5 - 1 = 4

    But the list has 5 numbers, not 4. The difference counts the steps: 1 to 2, 2 to 3, 3 to 4 and 4 to 5. Both ends are in the list, so there is one more number than steps.

  3. Use the counting rule

    count=n−m+1\text{count} = n - m + 1

    count=66+1=67\text{count} = 66 + 1 = 67

    Watch out: 66 is the difference, not the count. Choosing 66, choice C, leaves out one end of the list. Also, consecutive integers go up by 1, not 2, so halving 66 to get 33 is wrong. So the answer is D, 67.

Check your answer

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

More questions like this

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

Question 35 of 120

2024 Test A Math, question 11

Question

Jar 1 and Jar 2 each contain 12\frac{1}{2} cup of water. If 14\frac{1}{4} of the water in Jar 1 is poured into Jar 2, how much water is now in Jar 2?

  • A18\frac{1}{8} cup
  • B14\frac{1}{4} cup
  • C58\frac{5}{8} cup
  • D34\frac{3}{4} cup
Show the answer

Correct answer: C

Jar 1 starts with 12\frac{1}{2} cup of water. If 14\frac{1}{4} of its water is poured into Jar 2, then the amount of water added to Jar 2 is 12×14=18\frac{1}{2} \times \frac{1}{4} = \frac{1}{8} cup.

Jar 2 starts with 12\frac{1}{2} cup of water. If 18\frac{1}{8} cup of water is added to it, then the amount of water now in Jar 2 is 12+18=48+18=58\frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8} cup.

Question analysis

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

NYS standards
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Multiplying and adding positive fractions is review by grade 7, and the problem takes two short steps. It is easier than typical grade 7 work with negative numbers or longer rational-number problems, apart from careful reading.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    The phrase "1/4 of the water in Jar 1" means a quarter of Jar 1's half cup, an eighth of a cup, not a quarter cup; the word of calls for multiplying. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Adding the 1/8 cup to Jar 2's 1/2 cup needs the common denominator 8, so 1/2 becomes 4/8 before the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Each jar starts with 12\frac{1}{2} cup of water. You pour “14\frac{1}{4} of the water in Jar 1” into Jar 2. You need how much water is in Jar 2 after that, in cups.

  2. Find how much water is poured

    Jar 1 has 12\frac{1}{2} cup. You pour 14\frac{1}{4} of it. Of means multiply.

    ab of N=ab×N\frac{a}{b} \text{ of } N = \frac{a}{b} \times N

    14×12=1×14×2=18\frac{1}{4} \times \frac{1}{2} = \frac{1 \times 1}{4 \times 2} = \frac{1}{8}

    You pour 18\frac{1}{8} cup. Watch out: a quarter of the water is not a quarter cup. It is a quarter of a half cup. Adding 14\frac{1}{4} cup gives 34\frac{3}{4} cup, choice D.

  3. Make the bottoms match

    Now add the poured water to the 12\frac{1}{2} cup already in Jar 2. The bottoms are 2 and 8, so rewrite the half with 8 on the bottom. Multiply its top and bottom by 4.

    12=1×42×4=48\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}

  4. Add the water in Jar 2

    ad+bd=a+bd\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}

    48+18=58\frac{4}{8} + \frac{1}{8} = \frac{5}{8}

    Jar 2 now holds 58\frac{5}{8} cup. Watch out: 18\frac{1}{8} cup is only the water poured in. Stopping there gives choice A. So the answer is C, 58\frac{5}{8} cup.

Check your answer

Jar 2 started with a half cup and got a little more. So it should hold just over a half cup, and 58\frac{5}{8} is. Also, the water in both jars still adds up to 1 cup: Jar 1 has 38\frac{3}{8} cup left, and 38+58=1\frac{3}{8} + \frac{5}{8} = 1.

More questions like this

The original DOE question
Question 68 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 68. Open page 78 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 36 of 120

2024 Test A Math, question 12

Question

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

  • A0
  • B1
  • C2
  • D99
Show the answer

Correct answer: D

0.01=11000.01 = \frac{1}{100}, which is between 1101\frac{1}{101} and 199\frac{1}{99}.

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

Question analysis

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

NYS standards
  • 6.NS.7a inequality as relative position
  • 6.EE.5 solutions and substitution
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students compare fractions and decimals routinely, but not with an unknown in the denominators. The comparison is simple once 0.01 is written as a fraction; the unfamiliar setup is the main demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 9, 18 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    The unknown sits in two denominators, n and n + 2, so students must reason about which unit fractions lie on either side of 0.01 rather than solve an equation. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know minor Working it out

    Two facts are assumed: 0.01 equals 1/100, and a unit fraction shrinks as its denominator grows, so 1/99 is just above 0.01 and 1/101 just below. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    n is a whole number: 0, 1, 2, 3 and so on. The number 0.01 is between 1n\frac{1}{n} and 1n+2\frac{1}{n + 2}. Between means one fraction is bigger than 0.01 and the other is smaller. You need the value of n.

  2. Write 0.01 as a fraction

    0.01 means one hundredth.

    0.01=11000.01 = \frac{1}{100}

    Now all three numbers are fractions with 1 on top. That makes them easy to compare.

  3. Find fractions on each side

    With 1 on top, a bigger bottom makes a smaller fraction.

    if a<b, then 1a>1b\text{if } a < b \text{, then } \frac{1}{a} > \frac{1}{b}

    So 199\frac{1}{99} is a little more than 1100\frac{1}{100}, and 1101\frac{1}{101} is a little less.

    1101<1100<199\frac{1}{101} < \frac{1}{100} < \frac{1}{99}

    The bottoms 99 and 101 are 2 apart, just like n and n + 2.

  4. Match the fractions to n

    The smaller bottom is n, and the larger bottom is n + 2.

    n=99 and n+2=101n = 99 \text{ and } n + 2 = 101

    Both fit. Watch out: small choices do not work. For n = 1, the fractions are 1 and 13\frac{1}{3}. Both are far bigger than 0.01. For n = 0, 10\frac{1}{0} is not a number, because you can’t divide by zero. So the answer is D, 99.

Check your answer

Write the fractions as decimals. 199≈0.0101\frac{1}{99} \approx 0.0101, a bit more than 0.01. 1101≈0.0099\frac{1}{101} \approx 0.0099, a bit less. The sign ≈ means about. So 0.01 sits right between them.

More questions like this

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

Question 37 of 120

2024 Test A Math, question 18

Question

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

  • A96
  • B111
  • C112
  • D127
Show the answer

Correct answer: C

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

2x+1=y2x + 1 = y

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

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

126+1=y126 + 1 = y

127=y127 = y

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

2x+1=632x + 1 = 63

Subtract one from both sides of the equation.

2x+(1−1)=63−12x + (1 - 1) = 63 - 1

2x=622x = 62

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

x=31x = 31

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

2x+1=312x + 1 = 31

2x+(1−1)=31−12x + (1 - 1) = 31 - 1

2x=302x = 30

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

x=15x = 15

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

127−15=112127 - 15 = 112

The difference is 112.

Question analysis

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

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve two-step equations but rarely have to run a rule backward through a sequence. This mixes forward and backward steps with position counting, so it asks for more planning than a typical grade 7 task.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 7, 32 words.
Corrected from the DOE handbook
  • A misplaced parenthesis in the explanation’s working was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. No obvious method moderate Working it out

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

  2. Many steps moderate Working it out

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

  3. Much to keep track of minor Working it out

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

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

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the 10th number

    Use the rule with x = 63.

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

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

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

  3. Plan how to go back

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

    x=(y−1)÷2x = (y - 1) \div 2

  4. Find the 8th number

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

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

  5. Find the 7th number

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

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

    The 7th number is 15.

  6. Subtract the 7th from the 10th

    127−15=112127 - 15 = 112

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

Check your answer

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

More questions like this

The original DOE question
Question 75 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 75. Open page 80 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 38 of 120

2024 Test A Math, question 23

Question

Let x be an odd number. In terms of x, what is the sum of the two even numbers closest to x?

  • Axx
  • B2x2x
  • C2x−22x - 2
  • D2x−42x - 4
Show the answer

Correct answer: B

If x is odd, then the two even numbers closest to x are (x−1)(x - 1) and (x+1)(x + 1). Find the sum of these two expressions:

(x−1)+(x+1)=2x(x - 1) + (x + 1) = 2x

Question analysis

An algebra question about odd and even numbers: x is odd, and students write the sum of the two even numbers closest to x in terms of x. It tests representing neighboring integers with expressions and adding them. Deciding exactly which even numbers are closest to an odd number is where errors start.

NYS standards
  • 6.EE.2a writing algebraic expressions
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students write and combine linear expressions; here the expressions describe neighboring integers, a familiar idea from consecutive-integer problems. Two short steps and a quick number check put it at typical grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 3, 26 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The algebra is a single addition, so the work is in naming the right two numbers. An odd x sits between two even numbers, one step away on each side, and the minus one and plus one cancel. Wrong choices come from picking both neighbors below x, from treating x as even, or from giving the middle of the pair; a test value catches each of them.

  1. Variables instead of numbers moderate Understanding the question

    Here x stands for any odd number, so the two even neighbors must be written as x − 1 and x + 1; trying x = 7, with neighbors 6 and 8, makes it concrete. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The phrase "the two even numbers closest to x" means one just below and one just above x, not two in a row on the same side. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Substitute a real odd number such as 7 for x, find its two closest even numbers and their sum, then see which expression gives that sum.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Neighbors of a number
x−1 and x+1x - 1 \text{ and } x + 1
x is any whole number. The number just below it is x − 1, and the number just above it is x + 1. Whole numbers take turns being odd and even. So if x is odd, both of its neighbors are even.
Combining like terms
x+x=2xx + x = 2x
Like terms have the same letter, such as x and 2x. To add them, add the numbers in front. x by itself means 1x, so x + x is 2x, which is two x’s.

  1. Sort out the question

    x is an odd number. You need the sum of the two even numbers closest to x. Sum means the result of adding. In terms of x means write the answer using the letter x.

  2. Try a real number first

    Pick an odd number, like 7. The even numbers closest to 7 are 6 and 8, one on each side.

    6+8=146 + 8 = 14

    Notice that 14 is 2×72 \times 7.

  3. Write the neighbors with x

    Odd and even numbers take turns. So both numbers right next to an odd x are even.

    x−1 and x+1x - 1 \text{ and } x + 1

    For x = 7, these are 6 and 8. Watch out: take one even number below x and one above. Using two even numbers below x, x−1x - 1 and x−3x - 3, gives 2x−42x - 4, choice D.

  4. Add the two neighbors

    (x−1)+(x+1)=x+x−1+1=2x(x - 1) + (x + 1) = x + x - 1 + 1 = 2x

    The −1 and the +1 cancel, and x+x=2xx + x = 2x. Watch out: x is the number in the middle, not the sum, so choice A is wrong. So the answer is B, 2x2x.

Check your answer

Try another odd number, 11. Its even neighbors are 10 and 12, and 10+12=2210 + 12 = 22. The answer 2x gives 2×11=222 \times 11 = 22. They match.

More questions like this

The original DOE question
Question 80 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 80. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 39 of 120

2024 Test A Math, question 24

Question

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

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

Correct answer: B

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. A very close wrong answer minor Choosing the answer

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

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Set up the division

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

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

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

  3. Cancel zeros in pairs

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

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

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

  4. Divide what is left

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

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

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

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

Check your answer

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

More questions like this

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

Question 40 of 120

2024 Test A Math, question 25

Question

Points M, N, P and Q, each marked with a dot, lie in that order from left to right on a line. A dimension arrow from M to N is labeled x cm, and a longer dimension arrow from M to P is labeled y cm. No lengths are marked for NP, PQ or MQ.

In the figure above, N is the midpoint of MQ\overline{MQ}. Which segment has length (2x−y)(2x - y) centimeters?

  • APQ\overline{PQ}
  • BNP\overline{NP}
  • CMQ\overline{MQ}
  • DMP\overline{MP}
Show the answer

Correct answer: A

These facts are given:

  • N is the midpoint of MQ\overline{MQ}.
  • MN\overline{MN} has a length of x centimeters.
  • MP\overline{MP} has a length of y centimeters.

Use the first two facts to determine that MQ\overline{MQ} has a length of 2x2x centimeters. So (2x−y)(2x - y) can be represented by MQ−MPMQ - MP, which is PQ\overline{PQ}.

Question analysis

A geometry and algebra question with a labeled figure: points M, N, P and Q lie on a line, MN is x cm, MP is y cm, and N is the midpoint of MQ. Students find which segment has length 2x − y. They must turn the midpoint fact into an expression and read a difference of lengths on the figure.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.EE.6 variables representing unknown numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students evaluate and rewrite expressions but rarely read one as a length on a figure. Combining the midpoint idea from geometry with a difference of expressions, and no numbers, makes this more abstract than typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 9, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question reverses the usual task: instead of finding a length, students match an expression to a segment. The midpoint fact turns into MQ = 2x, and taking away the y of MP leaves the piece from P to Q. The wrong choices reward stopping at 2x, reading only the labeled arrows, or matching the letter y; picking x = 5 and y = 7 lets students check with numbers.

  1. Variables instead of numbers moderate Understanding the question

    Every length is a letter: MN is x, MP is y, and the target is the expression 2x − y, so the matching segment cannot be found by measuring. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The figure marks only MN and MP with arrows; the length of MQ comes from the stem's midpoint fact, not from the drawing, and has to be written as 2x. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the midpoint to write the length of the whole segment in terms of x, or pick sample values for x and y, then see which segment's length matches 2x − y.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint
MQ=2×MNMQ = 2 \times MN
N is the midpoint of segment MQ, the point exactly in the middle. MN is the half from M to N, and MQ is the whole segment. The whole is two equal halves, so it is 2 times MN.
Adding segments
MP+PQ=MQMP + PQ = MQ
When points lie in order on a line, the short segments add up to the long one. MP runs from M to P, and PQ runs from P to Q. Put end to end, they make MQ.

  1. Sort out the figure

    Points M, N, P and Q lie in that order on a line. Segment MN is x cm long, and segment MP is y cm long. N is the midpoint of MQ\overline{MQ}, the point exactly halfway from M to Q. You need the segment that is 2x−y2x - y cm long.

  2. Find the length of MQ

    MN is half of MQ, so MQ is twice as long.

    MQ=2×MN=2xMQ = 2 \times MN = 2x

    MQ is 2x cm long. Watch out: 2x is only part of the expression. Stopping here gives MQ, choice C.

  3. Take MP away from MQ

    Subtracting y from 2x means taking MP away from MQ. MP covers the part from M to P. What is left of MQ is the piece from P to Q.

    MP+PQ=MQMP + PQ = MQ

    PQ=MQ−MP=2x−yPQ = MQ - MP = 2x - y

    So PQ is 2x−y2x - y cm long. Watch out: do not pick MP just because y appears. MP is y, choice D. NP is y−xy - x, choice B. So the answer is A, PQ\overline{PQ}.

Check your answer

Try numbers. Say x = 5, so MN = 5 and MQ = 10. Say y = 7, so MP = 7. Then PQ is 10−7=310 - 7 = 3. And 2x−y=10−7=32x - y = 10 - 7 = 3. They match. NP would be 7−5=27 - 5 = 2, which does not.

More questions like this

The original DOE question
Question 82 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 82. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 41 of 120

2024 Test A Math, question 30

Question

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

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

Correct answer: A

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

  2. Error-prone arithmetic minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the wording

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

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

    The last flip gives the answer.

  2. Flip each number

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

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

  3. Add the two reciprocals

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

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

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

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

  4. Flip the sum

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

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

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

Check your answer

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

More questions like this

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

Question 42 of 120

2024 Test A Math, question 32

Question

INGREDIENTS FOR 4 SERVINGS OF OATMEAL

Ingredient Cups
Oats 23\frac{2}{3}
Water 3143\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?

  • A16\frac{1}{6}
  • B839\frac{8}{39}
  • C1316\frac{13}{16}
  • D138\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:

23314\frac{\frac{2}{3}}{3\frac{1}{4}}

The denominator can be rewritten as 134\frac{13}{4}.

The quotient 23134\frac{\frac{2}{3}}{\frac{13}{4}} can be rewritten as:

23×413=839.\frac{2}{3} \times \frac{4}{13} = \frac{8}{39}.

Therefore, the unit rate for cups of oats per cup of water is 839\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.
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Unit rate
unit rate=first amountsecond amount\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
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, b is the top of the fraction, and c is the bottom. Each whole holds c pieces, so a wholes hold a times c pieces. Then add the b extra pieces.
Dividing by a fraction
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. This works because dividing by a number undoes multiplying by it.

  1. Sort out the question

    The table shows 23\frac{2}{3} cup of oats and 3143\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 16\frac{1}{6} cup for each serving, choice A.

  2. Set up the rate

    Oats comes before per, so oats goes on top. Water goes on the bottom.

    unit rate=first amountsecond amount\text{unit rate} = \frac{\text{first amount}}{\text{second amount}}

    unit rate=23÷314\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.

  3. Change the mixed number

    A mixed number is a whole number plus a fraction. Change 3143\frac{1}{4} to a fraction so you can divide.

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

    314=3×4+14=1343\frac{1}{4} = \frac{3 \times 4 + 1}{4} = \frac{13}{4}

    So 3 and one quarter cups is 13 quarters of a cup.

  4. Divide by multiplying

    To divide by 134\frac{13}{4}, multiply by its reciprocal, 413\frac{4}{13}.

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

    23÷134=23×413=2×43×13=839\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 839\frac{8}{39} cup of oats per cup of water.

    So the answer is B, 839\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 839\frac{8}{39} is about 0.2. Also multiply back: 839×134=104156=23\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 43 of 120

2024 Test A Math, question 33

Question

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

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

Correct answer: D

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

Rewrite with equivalent fractions that have like denominators.

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

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

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

Multiply both sides of the equation by 10.

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

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

Question analysis

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

NYS standards
  • 7.EE.4a two-step equations
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve equations such as (2/3)x = 4/5 with fractional coefficients. Here both sides must first be simplified over unlike denominators, so four fraction operations come before x is found: on level, but long.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 8, 12 words.
Corrected from the DOE handbook
  • A minus sign printed in place of an equals sign in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. Many steps moderate Working it out

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

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the equation

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

  2. Simplify the left side

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

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

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

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

  3. Simplify the right side

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

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

    Now the equation is much simpler:

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

  4. Get x by itself

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

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

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

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

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

  5. Simplify the fraction

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

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

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

Check your answer

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

More questions like this

The original DOE question
Question 90 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 85, question 90. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 44 of 120

2024 Test A Math, question 35

Question

Evaluate:

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

  • A−30
  • B−18
  • C18
  • D44
Show the answer

Correct answer: B

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

Simplify negatives inside the first absolute value brackets.

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

Simplify the first absolute value.

|−6|−|24|\left|-6\right| - \left|24\right|

Calculate the absolute values.

6−246 - 24

Perform the final subtraction.

6−24=−186 - 24 = -18

Question analysis

This integer question asks students to evaluate an expression with two absolute values: the first holds four positive and negative numbers, and the second is the absolute value of 24. It tests subtracting negative numbers and understanding absolute value. Subtracting a negative number and handling each pair of bars separately are where most slips happen.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students add and subtract integers and know absolute value as distance from zero. This expression puts four signed terms inside one pair of bars and subtracts a second absolute value, a typical but sign-heavy grade 7 exercise.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 16+, 2 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is routine, but the question stacks sign decisions, and one slip anywhere produces a clean wrong number. The last step is the subtle one: each pair of bars gives a positive number, but 6 minus 24 is still negative. Students who believe bars guarantee a positive result flip that subtraction, and the choices offer the flipped number beside the right one.

  1. Error-prone arithmetic moderate Working it out

    The first bars hold four signed terms, and two must be rewritten, + (−17) as − 17 and − (−31) as + 31, before they combine to −6. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The answer, −18, and the choice 18 differ only in sign, so estimating cannot separate them; only doing the last step as 6 − 24 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Many steps minor Working it out

    Each pair of bars is simplified inside and made positive before the outer subtraction, so three separate jobs must be done in the right order. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite each subtraction of a negative as addition, combine the four numbers inside the first bars, take both absolute values, and only then do the final subtraction.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding and subtracting negatives
a+(−b)=a−banda−(−b)=a+ba + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b
a is any number, and b is a positive number, so −b is negative. Adding a negative moves you left on the number line, just like subtracting. Subtracting a negative is like taking away a debt, so you end up with more. It is the same as adding.
Absolute value
|a|=the distance from a to 0\left|a\right| = \text{the distance from } a \text{ to } 0
The bars around a number mean absolute value. a is any number. Its absolute value is how far a is from 0 on the number line. A distance is never negative, so the absolute value of −6 is 6.

  1. Sort out the expression

    The expression has two absolute values. The bars mean the distance from 0, so what comes out of them is never negative. Work in this order. First simplify inside the first bars. Then take both absolute values. Last, do the subtraction between them.

  2. Rewrite the signs inside

    Two signs inside the first bars are tricky. Adding −17 is the same as subtracting 17. Subtracting −31 is the same as adding 31.

    a+(−b)=a−banda−(−b)=a+ba + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b

    (−8)−12+(−17)−(−31)=−8−12−17+31(-8) - 12 + (-17) - (-31) = -8 - 12 - 17 + 31

    Watch out: minus negative 31 is plus 31. Treating it as minus 31 makes the inside −68. That leads to 44, choice D.

  3. Combine the four numbers

    Go from left to right, one number at a time.

    −8−12=−20-8 - 12 = -20

    −20−17=−37-20 - 17 = -37

    −37+31=−6-37 + 31 = -6

    The inside of the first bars is −6.

  4. Take the absolute values

    |a|=the distance from a to 0\left|a\right| = \text{the distance from } a \text{ to } 0

    |−6|=6and|24|=24\left|-6\right| = 6 \quad \text{and} \quad \left|24\right| = 24

    Now the expression is 6−246 - 24. Watch out: the bars turn −6 into 6. Keeping −6 gives −6−24=−30-6 - 24 = -30, choice A.

  5. Do the last subtraction

    6−24=−186 - 24 = -18

    You take 24 away from 6. Taking a bigger number from a smaller one gives a negative answer. Watch out: absolute values are never negative, but this answer can be. Doing 24−624 - 6 instead gives 18, choice C.

    So the answer is B, −18.

Check your answer

Check the inside a new way. The positive part is 31. The negative parts add to 8+12+17=378 + 12 + 17 = 37. Since 31−37=−631 - 37 = -6, the inside is right. Then, on a number line, start at 6 and move 24 to the left. Six steps reach 0, and 18 more land on −18.

More questions like this

The original DOE question
Question 92 as printed on page 85 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 85, question 92. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 45 of 120

2024 Test A Math, question 37

Question

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

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

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

Correct answer: A

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

1+2=31 + 2 = 3

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

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

Question analysis

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

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

Why it is hard

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

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

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

  2. Many steps minor Working it out

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

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

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the picture

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

  2. Find the length from X to Z

    XZ=XY+YZXZ = XY + YZ

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

    X to Z is 3 units. Watch out: 3 is the whole length, choice D. It is not a distance to the midpoint.

  3. Place the midpoint

    The midpoint of X and Z is half of 3 units from X.

    distance from an end to the midpoint=length of the segment2\text{distance from an end to the midpoint} = \frac{\text{length of the segment}}{2}

    32=112\frac{3}{2} = 1\frac{1}{2}

    So the midpoint is 1121\frac{1}{2} units to the right of X. Watch out: this is its distance from X, not from Y. Stopping here gives choice C.

  4. Measure from Y to the midpoint

    Y is 1 unit to the right of X. The midpoint is 1121\frac{1}{2} units to the right of X. Subtract to find the gap between them.

    112−1=121\frac{1}{2} - 1 = \frac{1}{2}

    Watch out: use the midpoint of X and Z, not of Y and Z. Halving the 2 units from Y to Z gives 1, choice B.

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

Check your answer

Give each point a number, with X at 0. Then Y is at 1 and Z is at 3. Halfway between 0 and 3 is 1121\frac{1}{2}. From Y at 1 to 1121\frac{1}{2} is 12\frac{1}{2}. This fits the picture too: Y sits a little left of the middle of XZ.

More questions like this

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

Question 46 of 120

2024 Test A Math, question 43

Question

3.6÷0.018=3.6 \div 0.018 =

  • A0.005
  • B0.648
  • C20
  • D200
Show the answer

Correct answer: D

3.60.018×(1,0001,000)=3,60018=200\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.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 16+, 1 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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.

  1. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scaling a division
a÷b=(a×n)÷(b×n)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.

  1. Sort out the problem

    You need 3.6÷0.0183.6 \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.

  2. 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÷b=(a×n)÷(b×n)a \div b = (a \times n) \div (b \times n)

    3.6÷0.018=(3.6×1,000)÷(0.018×1,000)=3,600÷183.6 \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÷18=20360 \div 18 = 20, choice C.

  3. Divide the whole numbers

    3,600÷18=2003{,}600 \div 18 = 200

    This works because 18×2=3618 \times 2 = 36, so 18×200=3,60018 \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: 200×0.018=3.6200 \times 0.018 = 3.6. Also estimate. 0.018 is close to 0.02, and 3.6÷0.02=1803.6 \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 47 of 120

2024 Test A Math, question 45

Question

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

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

Correct answer: D

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

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

Misha wants to make 2 straps.

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

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

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

  2. Much to keep track of minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost
cost=price per unit×number of units\text{cost} = \text{price per unit} \times \text{number of units}
Price per unit is the cost of one unit, here $5.00 for one yard. Number of units is how many you buy, here yards of ribbon. Multiply them to get the total before tax.

  1. Sort out the question

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

  2. Find the total ribbon

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

    2×1.25=2.52 \times 1.25 = 2.5

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

  3. Find the cost

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

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

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

    So the answer is D, $12.50.

Check your answer

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

More questions like this

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

Question 48 of 120

2024 Test A Math, question 49

Question

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

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

Correct answer: A

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

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

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

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

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

Question analysis

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

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve multi-step word problems with mixed numbers. This one has only positive numbers, but it needs a clear picture of the route and a subtraction that requires regrouping across unlike denominators.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. A very close wrong answer moderate Choosing the answer

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

  3. Turning words into math minor Understanding the question

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

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

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, b is the top of the fraction, and c is the bottom. Each whole holds c pieces, so a wholes hold a times c pieces. Then add the b extra pieces.
Equivalent fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a is the top of a fraction and b is the bottom. n is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, cut into smaller pieces. Use it to give two fractions the same bottom before you add or subtract.
Subtracting fractions with the same bottom
ac−bc=a−bc\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}
c is the shared bottom number. a and b are the tops. When the pieces are the same size, you just take away how many pieces you have. The size of the pieces, c, stays the same.

  1. Picture the route

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

  2. Find the distance from home to the store

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

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

  3. Change to improper fractions

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

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

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

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

  4. Use sixths for both

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

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

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

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

  5. Subtract and simplify

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

    336−106=236\frac{33}{6} - \frac{10}{6} = \frac{23}{6}

    23 sixths is 3 wholes (18 sixths) and 5 sixths more. So 236=356\frac{23}{6} = 3\frac{5}{6}. Watch out: 12\frac{1}{2} is less than 23\frac{2}{3}. Subtracting the fraction parts the wrong way round gives 4164\frac{1}{6}, choice B.

    So the answer is A, 3563\frac{5}{6} miles.

Check your answer

Add back: 356+123=356+146=496=536=5123\frac{5}{6} + 1\frac{2}{3} = 3\frac{5}{6} + 1\frac{4}{6} = 4\frac{9}{6} = 5\frac{3}{6} = 5\frac{1}{2}. That is the full distance from the store to home. Also estimate: 5.5 minus about 1.7 is about 3.8, and 3563\frac{5}{6} is about 3.83.

More questions like this

The original DOE question
Question 106 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 89, question 106. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 49 of 120

2024 Test A Math, question 50

Question

A child grows 1141\frac{1}{4} inches in 13\frac{1}{3} of a year. What would be his yearly growth rate in inches per year?

  • A512\frac{5}{12}
  • B3143\frac{1}{4}
  • C3343\frac{3}{4}
  • D4144\frac{1}{4}
Show the answer

Correct answer: C

Set up a proportion to solve.

114 in.13 yr=x in.1 yr\frac{1\frac{1}{4}\text{ in.}}{\frac{1}{3}\text{ yr}} = \frac{x\text{ in.}}{1\text{ yr}}

114=13x1\frac{1}{4} = \frac{1}{3}x

54=13x\frac{5}{4} = \frac{1}{3}x

(31)(54)=x\left(\frac{3}{1}\right)\left(\frac{5}{4}\right) = x

154=x\frac{15}{4} = x

334=x3\frac{3}{4} = x

Question analysis

A child grows 1 1/4 inches in one third of a year, and this unit rate question asks for his growth rate in inches per year. Students divide a mixed number by a fraction, or scale the growth up to a whole year. Choosing between multiplying and dividing, and handling the whole mixed number, are where the wrong rates come from.

NYS standards
  • 7.RP.1 unit rates with fractions
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Computing a unit rate from a ratio of fractions is the grade 7 standard itself, and this is a typical example: one mixed number over one unit fraction, in two short steps with no extra information.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The idea is simple: three thirds make a year, so the yearly growth is three times the growth in a third. Errors come from the mixed number, where students triple the 1 and leave the 1/4 alone, and from multiplying by 1/3 instead of dividing, which shrinks the rate. A sense check helps, since a yearly rate must be larger than the 1 1/4 inches grown in a third of a year.

  1. Error-prone arithmetic minor Working it out

    Dividing 1 1/4 by 1/3 means multiplying by 3, and the mixed number must be multiplied whole, as 5/4 × 3 = 15/4, not just its 1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer minor Choosing the answer

    Multiplying only the whole part gives 3 1/4, which differs from the answer only in the fraction's numerator, so a rough size check cannot rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Divide the growth by the time, writing 1 1/4 as 5/4 and dividing by 1/3 as multiplying by 3; or ask how much he grows in three such periods.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Rate
rate=amount÷time\text{rate} = \text{amount} \div \text{time}
The amount is how much something changed, here inches grown. The time is how long it took, here part of a year. Dividing tells how much it changes in 1 unit of time, here 1 whole year.
Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, b is the top of the fraction, and c is the bottom. Each whole holds c pieces, so a wholes hold a times c pieces. Then add the b extra pieces.
Dividing by a fraction
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. This works because dividing by a number undoes multiplying by it.

  1. Sort out the question

    The child grows 1141\frac{1}{4} inches in 13\frac{1}{3} of a year. You need his growth rate in inches per year. That is how much he would grow in 1 whole year at the same speed. A year has 3 thirds, so the answer should be bigger than 1141\frac{1}{4}.

  2. Set up the rate

    Divide the growth by the time to get the inches for 1 whole year.

    rate=amount÷time\text{rate} = \text{amount} \div \text{time}

    rate=114÷13\text{rate} = 1\frac{1}{4} \div \frac{1}{3}

    Change the mixed number to a fraction first.

    abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}

    114=1×4+14=541\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{5}{4}

  3. Divide by multiplying

    To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. The reciprocal of 13\frac{1}{3} is 3.

    ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

    54÷13=54×31=154\frac{5}{4} \div \frac{1}{3} = \frac{5}{4} \times \frac{3}{1} = \frac{15}{4}

    Watch out: multiplying by 13\frac{1}{3} instead gives 512\frac{5}{12}, choice A. That is less than he grew in a third of a year.

  4. Change back to a mixed number

    15 quarters is 3 wholes (12 quarters) and 3 quarters more.

    154=334\frac{15}{4} = 3\frac{3}{4}

    He grows 3343\frac{3}{4} inches per year. Watch out: triple all of 1141\frac{1}{4}, not just the 1. Tripling only the whole number gives 3143\frac{1}{4}, choice B.

    So the answer is C, 3343\frac{3}{4}.

Check your answer

Think of the year as 3 thirds. He grows 1141\frac{1}{4} inches in each third. Add them: 114+114+1141\frac{1}{4} + 1\frac{1}{4} + 1\frac{1}{4}. The wholes make 3, and the quarters make 34\frac{3}{4}. That is 3343\frac{3}{4} inches, the same answer.

More questions like this

The original DOE question
Question 107 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 89, question 107. Open page 89 of the handbook (PDF)

Question 50 of 120

2024 Test A Math, question 51

Question

3(0.01)−3(0.1)=3(0.01) - 3(0.1) =

  • A−0.33
  • B−0.27
  • C0
  • D0.33
Show the answer

Correct answer: B

3(0.01)−3(0.1)=3(0.01) - 3(0.1) =

0.03−0.3=−0.270.03 - 0.3 = -0.27

Question analysis

This decimal question asks for the value of 3 times 0.01 minus 3 times 0.1. Students multiply each decimal by 3 and subtract, keeping the order of the subtraction as written. The two decimals look alike, so place value and the sign of the result both need care.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students subtract rational numbers and get negative results. This short computation adds place-value care, since 0.01 and 0.1 look alike, but it is a routine two-step grade 7 exercise.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 16+, 1 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two quick multiplications leave 0.03 − 0.3, a subtraction whose result is negative. The usual errors are adding the two products, with or without a minus sign, and treating 0.01 and 0.1 as the same number. Rewriting the subtraction as −(0.3 − 0.03) keeps the place values lined up and makes the sign plain.

  1. Error-prone arithmetic moderate Working it out

    The products 0.03 and 0.3 have their digits in different places, and taking the larger from the smaller crosses zero, so both alignment and sign need care. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer minor Choosing the answer

    The choice −0.33 has the right sign and is only 0.06 from the answer, so a sense that the result is a little below zero cannot separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Multiply each decimal by 3, then subtract with the place values lined up; or factor out the 3 and compute 3 times (0.01 − 0.1).

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a bigger number
a−b=−(b−a)a - b = -(b - a)
a and b are numbers, and b is the bigger one. When you take a bigger number away from a smaller one, the answer is negative. Its size is the gap between them, b minus a. So subtract the other way, then put a minus sign in front.

  1. Sort out the problem

    3(0.01) means 3 times 0.01. A number right next to parentheses means multiply. Then subtract 3 times 0.1. Note that 0.01 is one hundredth, and 0.1 is one tenth. They are not equal. Watch out: treating them as equal makes the two parts cancel to 0, choice C.

  2. Multiply each part

    3×0.01=0.033 \times 0.01 = 0.03

    3×0.1=0.33 \times 0.1 = 0.3

    So the problem is 0.03−0.30.03 - 0.3. Write 0.3 as 0.30 so both have two decimal places. Now you can see that 0.30 is bigger than 0.03.

  3. Subtract the bigger number

    You take a bigger number from a smaller one, so the answer is negative.

    a−b=−(b−a)a - b = -(b - a)

    0.03−0.30=−(0.30−0.03)=−0.270.03 - 0.30 = -(0.30 - 0.03) = -0.27

    Watch out: find the gap between the numbers. Don’t add them. Adding gives 0.33, choice D, or −0.33, choice A.

    So the answer is B, −0.27.

Check your answer

Pull out the 3 first: 3×(0.01−0.1)=3×(−0.09)=−0.273 \times (0.01 - 0.1) = 3 \times (-0.09) = -0.27. Same answer. Also think of a number line. Start at 0.03, just right of 0. Move 0.30 to the left. You land a little left of 0, at −0.27.

More questions like this

The original DOE question
Question 108 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 89, question 108. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 51 of 120

2024 Test A Math, question 52

Question

What is the value of 1012+(−513)−(−234)10\frac{1}{2} + \left(-5\frac{1}{3}\right) - \left(-2\frac{3}{4}\right)?

  • A25122\frac{5}{12}
  • B711127\frac{11}{12}
  • C1311213\frac{1}{12}
  • D1871218\frac{7}{12}
Show the answer

Correct answer: B

Rewrite the mixed numbers so that they all have equivalent denominators.

1012+(−513)−(−234)10\frac{1}{2} + \left(-5\frac{1}{3}\right) - \left(-2\frac{3}{4}\right)

10612+(−5412)−(−2912)10\frac{6}{12} + \left(-5\frac{4}{12}\right) - \left(-2\frac{9}{12}\right)

Subtracting a negative is the same as adding.

10612−5412+291210\frac{6}{12} - 5\frac{4}{12} + 2\frac{9}{12}

Group the whole numbers together and the fractions together.

10612−5412+291210\frac{6}{12} - 5\frac{4}{12} + 2\frac{9}{12}

(10−5+2)+(612−412+912)(10 - 5 + 2) + \left(\frac{6}{12} - \frac{4}{12} + \frac{9}{12}\right)

711127\frac{11}{12}

Question analysis

This signed-number question asks for the value of 10 1/2 plus negative 5 1/3 minus negative 2 3/4. It tests adding and subtracting negative mixed numbers with unlike denominators. Turning each sign combination into a plain addition or subtraction, and then working in twelfths, is where students most often go wrong.

NYS standards
  • 7.NS.1c rational number subtraction
  • 7.NS.1d properties for adding and subtracting
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Adding and subtracting negative mixed numbers is central grade 7 work. This three-term expression is typical in length, but it needs a common denominator of 12 and two different sign rules: adding a negative and subtracting one.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 10, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The expression is short, but it packs both sign rules for negative numbers into one line, and each misapplication has its own choice. Subtracting a negative is the rule most often broken, and the result is a reasonable-looking mixed number. Estimating with decimals, 10.5 − 5.3 + 2.8 is about 8, separates the answer from every wrong choice.

  1. Error-prone arithmetic moderate Working it out

    Two different sign rules meet: adding −5 1/3 means subtracting 5 1/3, while subtracting −2 3/4 means adding 2 3/4, and mixing them up gives a listed value. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    After the signs, halves, thirds and quarters become twelfths, and the wholes, 10 − 5 + 2, are combined apart from the fractions, 6/12 − 4/12 + 9/12. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite adding a negative as subtraction and subtracting a negative as addition, convert the fractions to twelfths, then combine the whole numbers and the fractions separately.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding and subtracting negatives
a+(−b)=a−banda−(−b)=a+ba + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b
a is any number, and b is a positive number, so −b is negative. Adding a negative moves you left on the number line, just like subtracting. Subtracting a negative is like taking away a debt, so you end up with more. It is the same as adding.
Equivalent fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a is the top of a fraction and b is the bottom. n is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, cut into smaller pieces. Use it to give two fractions the same bottom before you add or subtract.

  1. Sort out the expression

    You need 1012+(−513)−(−234)10\frac{1}{2} + \left(-5\frac{1}{3}\right) - \left(-2\frac{3}{4}\right). There are three mixed numbers, and two of them are negative. A mixed number is a whole number plus a fraction. Plan: fix the signs, give the fractions the same bottom, then work with the whole numbers and the fractions one group at a time.

  2. Fix the signs

    Adding −513-5\frac{1}{3} means subtracting 5135\frac{1}{3}. Subtracting −234-2\frac{3}{4} means adding 2342\frac{3}{4}.

    a+(−b)=a−banda−(−b)=a+ba + (-b) = a - b \quad \text{and} \quad a - (-b) = a + b

    1012−513+23410\frac{1}{2} - 5\frac{1}{3} + 2\frac{3}{4}

    Watch out: subtracting a negative is adding. Subtracting 2342\frac{3}{4} instead gives 25122\frac{5}{12}, choice A. Dropping the minus signs gives C or D.

  3. Use twelfths for the fractions

    The bottoms are 2, 3 and 4. All three go into 12, so change each fraction to twelfths.

    ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}

    12=612,13=412,34=912\frac{1}{2} = \frac{6}{12}, \quad \frac{1}{3} = \frac{4}{12}, \quad \frac{3}{4} = \frac{9}{12}

    Now the problem is 10612−5412+291210\frac{6}{12} - 5\frac{4}{12} + 2\frac{9}{12}.

  4. Combine whole numbers, then fractions

    Work with the whole numbers and the fractions one group at a time. Keep each sign.

    10−5+2=710 - 5 + 2 = 7

    612−412+912=1112\frac{6}{12} - \frac{4}{12} + \frac{9}{12} = \frac{11}{12}

    The fraction part is less than 1, so just put them together: 711127\frac{11}{12}.

    So the answer is B, 711127\frac{11}{12}.

Check your answer

Estimate with decimals. 10.5−5.33+2.7510.5 - 5.33 + 2.75 is about 7.92. And 711127\frac{11}{12} is also about 7.92, since 1112\frac{11}{12} is almost 1. They match.

More questions like this

The original DOE question
Question 109 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 89, question 109. Open page 89 of the handbook (PDF)

Question 52 of 120

2024 Test A Math, question 56

Question

Number line from −3 to 8 with a tick at every integer. Point P is at −3, point Q is at 1 and point R is at 7, each marked with a dot.

On the number line above, what is the distance, in units, between the midpoint of PQ\overline{PQ} and the midpoint of QR\overline{QR}?

  • A3
  • B4
  • C5
  • D6
Show the answer

Correct answer: C

To find the midpoint of a segment, subtract the value on the left from the value on the right and then divide by 2. Then add the result to the value on the left of the segment.

Find the midpoint of PQ\overline{PQ}:

1−(−3)=41 - (-3) = 4

4÷2=24 \div 2 = 2

2+(−3)=−12 + (-3) = -1

Find the midpoint of QR\overline{QR}:

7−1=67 - 1 = 6

6÷2=36 \div 2 = 3

3+1=43 + 1 = 4

The distance between −1 and 4 on the number line is 5 units.

Question analysis

A number line marks points P, Q and R at labeled integer positions, with P to the left of zero. This midpoint and distance question asks how far apart the midpoints of segments PQ and QR are. Students locate both midpoints and then measure the distance between them, which takes care with negative positions.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students measure distances on a number line as the difference of two positions, including across zero. Here two midpoints must be located first, but all three points sit on labeled integers, which keeps it at grade 7 level.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 11, 21 words.
Corrected from the DOE handbook
  • A reversed instruction in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The points sit on labeled integers, so the midpoints can be found by counting, and the question is mostly about finishing carefully. Dropping the negative in 4 − (−1), stopping at the second midpoint, and counting tick marks instead of spaces each give a listed choice. Marking −1 and 4 on the figure and counting the unit spaces between them confirms the distance.

  1. Error-prone arithmetic minor Working it out

    P sits at −3, so the midpoint of PQ is (−3 + 1) ÷ 2 = −1, and the final distance, 4 − (−1), subtracts a negative number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Three results come in sequence, the midpoint of PQ, the midpoint of QR, then the gap between them, and the second of them, 4, is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Read the positions of P, Q and R, average each pair of endpoints to place the two midpoints, then subtract the left midpoint from the right one.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Stops one step early: A choice is a number from partway through the solution.
  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook’s rule says to “subtract the value on the right from the value on the left and then divide by 2,” but its own working does the opposite: 1 − (−3) = 4 for PQ and 7 − 1 = 6 for QR, the right value minus the left. Following the rule as printed gives (−3 − 1) ÷ 2 = −2 and (1 − 7) ÷ 2 = −3, which would put the midpoints at −5 and −2, outside the segments, and lead to a distance of 3, a wrong choice. The site now says “subtract the value on the left from the value on the right.” The midpoints, −1 and 4, and the distance between them, 5 units, are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint on a number line
midpoint=left end+right end−left end2\text{midpoint} = \text{left end} + \frac{\text{right end} - \text{left end}}{2}
The left end and right end are the numbers at the two ends of a segment. Right end minus left end is the segment’s length. The midpoint is halfway, so it is half the length to the right of the left end.
Distance on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
On a number line, numbers grow as you move right. The distance between two points is the larger number minus the smaller one. That is the right point minus the left point, so a distance is never negative.

  1. Read the number line

    P is at −3, Q is at 1 and R is at 7. A bar over two letters, such as PQ\overline{PQ}, means the segment between them. A segment is the straight piece of line between two points. You need the midpoint of each segment, the point exactly halfway between its ends. Then you need the distance between those two midpoints.

  2. Find the midpoint of PQ

    The left end is −3 and the right end is 1.

    midpoint=left end+right end−left end2\text{midpoint} = \text{left end} + \frac{\text{right end} - \text{left end}}{2}

    midpoint=−3+1−(−3)2=−3+42=−3+2=−1\text{midpoint} = -3 + \frac{1 - (-3)}{2} = -3 + \frac{4}{2} = -3 + 2 = -1

    So the midpoint of PQ is at −1. Note that 1−(−3)1 - (-3) is 1+31 + 3, which is 4.

  3. Find the midpoint of QR

    The left end is 1 and the right end is 7.

    midpoint=1+7−12=1+62=1+3=4\text{midpoint} = 1 + \frac{7 - 1}{2} = 1 + \frac{6}{2} = 1 + 3 = 4

    So the midpoint of QR is at 4. Watch out: 4 is where a midpoint is, not the answer. Stopping here gives choice B.

  4. Measure between the midpoints

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    distance=4−(−1)=4+1=5\text{distance} = 4 - (-1) = 4 + 1 = 5

    Watch out: keep the minus sign on −1. Using 4−1=34 - 1 = 3 gives choice A. Also count the spaces, not the tick marks. From −1 to 4 there are 5 spaces but 6 marks, and 6 is choice D.

    So the answer is C, 5 units.

Check your answer

Each midpoint should sit in the middle of its segment. −1 is 2 units from both −3 and 1. 4 is 3 units from both 1 and 7. There is also a shortcut. The gap between the midpoints is half of PR: 7−(−3)2=102=5\frac{7 - (-3)}{2} = \frac{10}{2} = 5.

More questions like this

The original DOE question
Question 113 as printed on page 91 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 91, question 113. Open page 91 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 53 of 120

2024 Test B Math, question 4

Question

At one location, the temperature increased from −7° F to 4° F between 5:00 a.m. and 9:00 a.m. By 11:00 a.m., the temperature was 3.5 times the 9:00 a.m. temperature. What was the total increase in temperature between 5:00 a.m. and 11:00 a.m., in degrees Fahrenheit?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 21

The 11:00 a.m. temperature is 3.5 times the 9:00 a.m. temperature. The 9:00 a.m. temperature was 4° F.

4×3.5=144\times 3.5=14

The temperature increased from −7° F at 5:00 a.m. to 14° F at 11:00 a.m., which is an increase of 21° F.

Question analysis

A temperature word problem with negative numbers: the temperature rises from −7° F to 4° F by 9:00 a.m., and by 11:00 a.m. it is 3.5 times the 9:00 a.m. reading. Students find the total rise from 5:00 to 11:00 a.m. Subtracting a negative starting temperature is where most slips happen.

NYS standards
  • 7.NS.3 four operations with rational numbers
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Using signed numbers for temperature change is standard grade 7 work. This adds a decimal multiplier and asks for a change rather than a temperature, two short steps, which keeps it at a typical grade 7 level.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 5, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Finding the 11:00 a.m. temperature is a single multiplication; the points are lost in the subtraction that follows. A rise that starts below zero has two parts, 7 degrees up to zero and 14 more above it, and students who treat −7 as 7 lose the first part. The times in the stem also invite measuring from 9:00 a.m. instead of 5:00 a.m., which covers only the last two hours.

  1. Error-prone arithmetic moderate Working it out

    The last step is 14 − (−7): subtracting the negative starting temperature means adding 7, and the part of the rise below zero is easy to leave out. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    “3.5 times the 9:00 a.m. temperature” multiplies the reading of 4°, not the 11-degree rise, and “total increase” runs from the 5:00 a.m. reading, not from 9:00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Much to keep track of minor Working it out

    Three times of day carry three temperatures, −7°, 4° and 14°, and the answer compares the first with the last, not either of them with the middle one. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find the 11:00 a.m. temperature from the 9:00 a.m. one, then subtract the 5:00 a.m. temperature from it, remembering that subtracting a negative number adds its size.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Stops one step early: A choice is a number from partway through the solution.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Change in temperature
change=end−start\text{change} = \text{end} - \text{start}
End is the last temperature, and start is the first one. Subtract start from end to find how much it went up. It works with negative numbers too. A positive change means the temperature rose.
Subtracting a negative number
a−(−b)=a+ba - (-b) = a + b
a is any number, and −b is a negative number. Taking away a negative is the same as adding its size. On a thermometer, going from −7 up to 0 is 7 degrees, so those 7 degrees get added.

  1. Sort out what you know

    There are three times and three temperatures, in degrees Fahrenheit (°F).

    • 5:00 a.m.: −7°F, which is 7 degrees below zero
    • 9:00 a.m.: 4°F
    • 11:00 a.m.: 3.5 times the 9:00 a.m. temperature

    You need the total increase from 5:00 a.m. to 11:00 a.m. That is how much warmer it got from the first time to the last.

  2. Find the 11:00 temperature

    “3.5 times the 9:00 a.m. temperature” means 3.5 times 4°F.

    3.5×4=143.5 \times 4 = 14

    At 11:00 a.m. it was 14°F. Watch out: 14 is a temperature, not the increase. Gridding 14 stops too early.

  3. Subtract the start from the end

    The increase runs from 5:00 a.m. to 11:00 a.m. So it starts at −7°F and ends at 14°F.

    change=end−start\text{change} = \text{end} - \text{start}

    change=14−(−7)\text{change} = 14 - (-7)

    Watch out: start at 5:00 a.m., not at 9:00 a.m. Starting from 4°F gives only 10.

  4. Subtract the negative number

    Taking away a negative number is the same as adding its size.

    a−(−b)=a+ba - (-b) = a + b

    14−(−7)=14+7=2114 - (-7) = 14 + 7 = 21

    The temperature rose 21°F. Watch out: writing 14 − 7 = 7 loses the minus sign. It skips the 7 degrees the air warmed just to reach zero.

    So you would grid 21.

Check your answer

Count up on a thermometer. From −7°F up to 0°F is 7 degrees. From 0°F up to 14°F is 14 more. 7+14=217 + 14 = 21. Or add the two parts of the morning: it rose 11 degrees by 9:00 and 10 more by 11:00, and 11+10=2111 + 10 = 21.

More questions like this

The original DOE question
Question 61 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 199, question 61. Open page 199 of the handbook (PDF)

Question 54 of 120

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)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)=2.5−4.5=−22.5+(-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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding a negative number
a+(−b)=a−ba + (-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.

  1. Sort out the question

    You need to find 2.5+(−4.5)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.

  2. Turn the addition into subtraction

    a+(−b)=a−ba + (-b) = a - b

    2.5+(−4.5)=2.5−4.52.5 + (-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.

  3. Do the subtraction

    The two sizes differ by 2, because 4.5−2.5=24.5 - 2.5 = 2. The larger size, 4.5, belongs to the negative number. So the sum is negative.

    2.5−4.5=−22.5 - 4.5 = -2

    Watch out: the answer is −2, not 2. Point G is at 2.

  4. 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 55 of 120

2024 Test B Math, question 10

Question

423÷212=4\frac{2}{3}\div 2\frac{1}{2}=

  • A1528\frac{15}{28}
  • B113151\frac{13}{15}
  • C2132\frac{1}{3}
  • D3133\frac{1}{3}
Show the answer

Correct answer: B

423÷212=143÷52=4\frac{2}{3}\div 2\frac{1}{2}=\frac{14}{3}\div\frac{5}{2}=

143×25=2815=11315\frac{14}{3}\times\frac{2}{5}=\frac{28}{15}=1\frac{13}{15}

Question analysis

A fractions question asks for 4 2/3 divided by 2 1/2, with mixed numbers on both sides of the division sign. Students convert to improper fractions, multiply by the reciprocal and simplify. Mixed numbers invite shortcuts, so every step has to follow the rule for dividing fractions.

NYS standards
  • 6.NS.1 dividing fractions by fractions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Dividing fractions is a grade 6 standard, and this is one division with small mixed numbers. Grade 7 work usually adds negative numbers or a word problem around it, so this sits below grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 16+, 5 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Dividing mixed numbers has a fixed procedure, and every wrong choice is a shortcut that skips part of it. Estimating first helps: 4 2/3 is a little less than twice 2 1/2, so the quotient is a little under 2, which already rules out 15/28 and 3 1/3. The remaining near miss, 2 1/3, comes from dividing by the whole number 2 alone.

  1. Error-prone arithmetic moderate Working it out

    Both mixed numbers must become improper fractions, 14/3 and 5/2, before the second is flipped, and 14/3 × 2/5 = 28/15 leaves an improper fraction to deal with. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. An answer in an unexpected form minor Choosing the answer

    The choices are written as mixed numbers and one proper fraction, so 28/15 has to be converted to a mixed number before it can be matched. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite both mixed numbers as improper fractions, multiply the first by the reciprocal of the second, then simplify and turn the improper fraction back into a mixed number.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number, and b over c is the fraction part. Multiply the whole number by c to turn the wholes into pieces. Then add the b extra pieces. The bottom number stays c.
Dividing fractions
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Only the second fraction flips.

  1. Sort out the problem

    You need to divide 4234\frac{2}{3} by 2122\frac{1}{2}. Both are mixed numbers, a whole number and a fraction together. Mixed numbers are hard to divide as they are. So first change each one into an improper fraction. That is a fraction whose top is bigger than its bottom.

  2. Change the mixed numbers

    abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}

    423=4×3+23=1434\frac{2}{3} = \frac{4 \times 3 + 2}{3} = \frac{14}{3}

    212=2×2+12=522\frac{1}{2} = \frac{2 \times 2 + 1}{2} = \frac{5}{2}

    Watch out: do not divide the whole parts and the fraction parts on their own. Doing 4÷2=24 \div 2 = 2 and 23÷12=43\frac{2}{3} \div \frac{1}{2} = \frac{4}{3}, then adding, gives 3133\frac{1}{3}, choice D.

  3. Flip the second fraction and multiply

    ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

    143÷52=143×25=2815\frac{14}{3} \div \frac{5}{2} = \frac{14}{3} \times \frac{2}{5} = \frac{28}{15}

    Multiply the tops, 14×2=2814 \times 2 = 28, and the bottoms, 3×5=153 \times 5 = 15. Watch out: flip only the second fraction. Flipping the first one gives 1528\frac{15}{28}, choice A.

  4. Change back to a mixed number

    The choices are mixed numbers, so change 28/15 back. 15 goes into 28 one time, with 13 left over.

    2815=11315\frac{28}{15} = 1\frac{13}{15}

    Watch out: dividing by just 2 and dropping the one half gives 2132\frac{1}{3}, choice C.

    So the answer is B, 113151\frac{13}{15}.

Check your answer

Estimate. 4234\frac{2}{3} is about 4.7, and 2122\frac{1}{2} is 2.5. Two groups of 2.5 make 5, so the answer is a bit less than 2. 113151\frac{13}{15} is about 1.9, which fits. Or multiply back: 2815×52=14030=143\frac{28}{15} \times \frac{5}{2} = \frac{140}{30} = \frac{14}{3}.

More questions like this

The original DOE question
Question 67 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 201, question 67. Open page 201 of the handbook (PDF)

Question 56 of 120

2024 Test B Math, question 12

Question

Lena is building a fence. She will need to dig holes to help support the posts that hold up the fence. The holes need to have a depth of 3133\frac{1}{3} feet below the ground. Each post is 10 feet long. What is the height of the part of the post that is above the ground?

  • A6236\frac{2}{3} feet
  • B7137\frac{1}{3} feet
  • C10 feet
  • D131313\frac{1}{3} feet
Show the answer

Correct answer: A

The posts are each 10 feet long, and 3133\frac{1}{3} feet of each post is below the ground, so the height of the part of the posts that is above the ground is the difference between the two amounts:

10−313=10-3\frac{1}{3}=

101−103=\frac{10}{1}-\frac{10}{3}=

303−103=203=623 feet\frac{30}{3}-\frac{10}{3}=\frac{20}{3}=6\frac{2}{3}\text{ feet}

Question analysis

A fence-building word problem: each post is 10 feet long, and 3 1/3 feet of it goes into a hole in the ground. Students find how much of the post stands above ground by subtracting a mixed number from a whole number. Regrouping the whole number into thirds is the step that trips students up.

NYS standards
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Subtracting a mixed number from a whole number is fraction work from before grade 7, set here in a one-step story. It asks much less than grade 7 problems, which add negative numbers or several steps.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 2, 61 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The situation is plain, a post partly buried in a hole, and the subtraction 10 − 3 1/3 is the whole task. The only real trap is the regrouping, since there are no thirds to take the 1/3 from until one foot is split into three. Checking that the two parts of the post add back to 10 feet catches the slip.

  1. Error-prone arithmetic minor Working it out

    Taking 3 1/3 from the whole number 10 requires regrouping 10 as 9 3/3, the step where students subtract the whole numbers and simply carry the 1/3 along. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the buried part from the whole post, rewriting 10 as 9 and 3/3, or as 30/3, so that the thirds can be taken away.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts of a whole
above=whole−below\text{above} = \text{whole} - \text{below}
The whole is the full length of the post. Below is the part in the hole, and above is the part you can see. The two parts add up to the whole, so subtract to find one part.
One whole in thirds
1=331 = \frac{3}{3}
One whole can be cut into 3 equal thirds. So 1 is the same as 3 over 3. Use this when you need to take thirds away from a whole number.

  1. Sort out the question

    Each post is 10 feet long. Part of it goes into a hole 3133\frac{1}{3} feet deep. The rest sticks up above the ground. You need the height of the part above the ground.

  2. Decide to subtract

    The part below and the part above make up the whole post. So take the part below away from the whole.

    above=whole−below\text{above} = \text{whole} - \text{below}

    above=10−313\text{above} = 10 - 3\frac{1}{3}

    Watch out: do not add. 10+313=131310 + 3\frac{1}{3} = 13\frac{1}{3}, choice D, makes the post longer than it is.

  3. Turn one whole into thirds

    The 10 has no thirds to take 1/3 from. So take 1 whole from the 10 and cut it into 3 thirds.

    1=331 = \frac{3}{3}

    10=9+33=93310 = 9 + \frac{3}{3} = 9\frac{3}{3}

    Now there are thirds to subtract from.

  4. Subtract wholes and thirds

    Subtract the whole numbers, then the thirds.

    933−313=6239\frac{3}{3} - 3\frac{1}{3} = 6\frac{2}{3}

    That is 9−3=69 - 3 = 6 and 33−13=23\frac{3}{3} - \frac{1}{3} = \frac{2}{3}. Watch out: the one third must be taken away too. Doing 10 − 3 = 7 and then keeping the one third gives 7137\frac{1}{3}, choice B.

    So the answer is A, 6236\frac{2}{3} feet.

Check your answer

Add the two parts back together: 623+313=933=106\frac{2}{3} + 3\frac{1}{3} = 9\frac{3}{3} = 10. That is the whole post. Also, the answer must be less than 10 feet, since part of the post is under the ground.

More questions like this

The original DOE question
Question 69 as printed on page 202 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 202, question 69. Open page 202 of the handbook (PDF)

Question 57 of 120

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?

  • A5
  • B6
  • C15
  • D16
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=1,8502{,}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÷120≈15.4171{,}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
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.
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance left
left=goal−done\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
laps=distance÷one lap\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.

  1. 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.

  2. Find the distance left

    left=goal−done\text{left} = \text{goal} - \text{done}

    left=2,500−650=1,850\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.

  3. Divide by the length of a lap

    laps=distance÷one lap\text{laps} = \text{distance} \div \text{one lap}

    1,850÷120≈15.41{,}850 \div 120 \approx 15.4

    The sign ≈ means about. Check with whole laps: 15×120=1,80015 \times 120 = 1{,}800. That leaves 1,850−1,800=501{,}850 - 1{,}800 = 50 meters still to go. So 15 laps are not quite enough.

  4. 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=2,450650 + 1{,}800 = 2{,}450 meters, which is short of 2,500. With 16 more laps: 650+16×120=650+1,920=2,570650 + 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 58 of 120

2024 Test B Math, question 21

Question

−149, −49, −17, −1-\frac{1}{49},\ -49,\ -\frac{1}{7},\ -1

Which number shown above is the greatest?

  • A−149-\frac{1}{49}
  • B−49-49
  • C−17-\frac{1}{7}
  • D−1-1
Show the answer

Correct answer: A

All the numbers are negative, so they are all less than zero. Consider the absolute value of each number and where it falls on a number line. The negative number closest to 0 will be the greatest.

The negative number with the greatest absolute value is −49, so it has the least value.

The negative number with the next greatest absolute value is −1, so it would fall next in order from least to greatest.

−49, −1-49,\ -1

Compare the absolute values of −149-\frac{1}{49} and −17-\frac{1}{7}.

149<17\frac{1}{49}<\frac{1}{7}

Therefore, the numbers listed from least to greatest are:

−49, −1, −17, −149-49,\ -1,\ -\frac{1}{7},\ -\frac{1}{49}

So the number −149-\frac{1}{49} is the greatest.

Question analysis

Four negative numbers are listed: negative 1/49, negative 49, negative 1/7 and negative 1. Students decide which of them is the greatest. Comparing negative fractions with negative whole numbers tests whether students order numbers by their position on the number line or by their size without the sign.

NYS standards
  • 6.NS.7a inequality as relative position
  • 6.NS.7d absolute value versus order
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Ordering positive and negative rational numbers on a number line is grade 6 content in New York. This is a one-step comparison of four numbers, lighter than typical grade 7 work with signed numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 10, 12 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

With no arithmetic to do, the question tests how firmly students picture negative numbers on a line. Comparing sizes without the sign picks the number farthest from 0, and comparing the two fractions by habit picks the wrong one of the pair. Placing all four as points left of 0, with −1 between −49 and the fractions, makes the order plain.

  1. A very close wrong answer moderate Choosing the answer

    −1/49 and −1/7 differ only in the denominator, and for negative numbers the larger denominator gives the greater value, the reverse of what their sizes suggest. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Students must bring the number-line fact that a negative number's value falls as its distance from 0 grows, so −49 is the least of the four, not the greatest. (Anderson & Krathwohl, 2001)

Strategy: Picture each number's place on the number line: among negative numbers, the one closest to 0 is the greatest, so compare their distances from 0.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing negative numbers
if a<b, then −a>−b\text{if } a \lt b \text{, then } -a \gt -b
a and b are positive numbers. Putting a minus sign on both flips their order. The one with the smaller size is closer to 0, so it is greater. For example, 2 is less than 5, but −2 is greater than −5.

  1. Sort out the question

    All four numbers are negative, which means less than 0. You need the greatest one, the one with the highest value. On a number line, greater numbers are farther to the right.

  2. Think about distance from 0

    For negative numbers, the one closest to 0 is the greatest. The size of a number without its sign is its absolute value. It is the number’s distance from 0. So look for the smallest size.

    if a<b, then −a>−b\text{if } a \lt b \text{, then } -a \gt -b

  3. Compare the sizes

    The sizes are 1/49, 49, 1/7 and 1. Both 1/49 and 1/7 have 1 on top. Cutting 1 into 49 pieces gives much smaller pieces than cutting it into 7. So 1/49 is less than 1/7.

    149<17<1<49\frac{1}{49} \lt \frac{1}{7} \lt 1 \lt 49

    So 1/49 is the smallest size. That puts −149-\frac{1}{49} closest to 0.

  4. Put the numbers in order

    Flip the order of the sizes to order the negative numbers, from least to greatest.

    −49<−1<−17<−149-49 \lt -1 \lt -\frac{1}{7} \lt -\frac{1}{49}

    Watch out: −49 has the biggest digits, but it is the least, so choice B is wrong. Watch out: 1/7 is more than 1/49, but −1/7 is less than −1/49. So choice C is wrong.

    So the answer is A, −149-\frac{1}{49}.

Check your answer

Change the fractions to decimals: −149≈−0.02-\frac{1}{49} \approx -0.02 and −17≈−0.14-\frac{1}{7} \approx -0.14. Think of money you owe. Owing 2 cents is better than owing 14 cents, $1 or $49. So −1/49 is the greatest.

More questions like this

The original DOE question
Question 78 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 205, question 78. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 59 of 120

2024 Test B Math, question 22

Question

What decimal is equivalent to −1311-\frac{13}{11}?

  • A−1.18-1.18
  • B−1.118-1.\overline{118}
  • C−1.18-1.\overline{18}
  • D−1.81-1.\overline{81}
Show the answer

Correct answer: C

Calculate the decimal equivalent of −1311-\frac{13}{11}:

−1311=−1.18181818...=−1.18-\frac{13}{11}=-1.18181818...=-1.\overline{18}

Question analysis

A fractions and decimals question asks which decimal equals negative 13/11. The choices differ in which digits repeat and how the repetition is marked. Students divide 13 by 11, spot the block of digits that repeats, and read the bar notation carefully.

NYS standards
  • 7.NS.2d fractions to decimals by long division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Converting a fraction to a repeating decimal by long division is a named grade 7 skill. This is one short division, typical of grade 7 work, with care needed only in reading the bar notation.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 15, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Dividing 13 by 11 is quick; the question is really about reading and writing repeating decimals. Three choices differ only in where the bar goes, so students who know the digits run 1.1818… can still pick the wrong bar. Writing out each barred choice for a few digits and comparing it with the quotient is a fast, sure check.

  1. A very close wrong answer moderate Choosing the answer

    The bars over 18 and over 81 cover the same two digits, so only checking which digit comes right after the decimal point, 1 before 8, tells them apart. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know minor Working it out

    Reading bar notation is assumed: the bar marks exactly the block of digits that repeats, and the whole-number 1 never belongs under it. (Anderson & Krathwohl, 2001)

  3. Error-prone arithmetic minor Working it out

    The long division 13 ÷ 11 has to run until the remainder 2 comes back, which is where the repeating pair is confirmed. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Do the long division of 13 by 11 until a remainder comes back, note which digits repeat from that point, and match that block to the bar in each choice.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction as division
ab=a÷b\frac{a}{b} = a \div b
a is the top number, the numerator. b is the bottom number, the denominator. A fraction bar means divide. To turn a fraction into a decimal, divide the top by the bottom.
Bar for repeating digits
0.18=0.181818…0.\overline{18} = 0.181818\ldots
The bar sits over the digits that repeat forever, in the order they repeat. The dots mean the pattern goes on and on. Digits that are not under the bar show up only once.

  1. Sort out the question

    You need the decimal equal to −1311-\frac{13}{11}. The fraction is negative, so the decimal will be negative too. Work with 1311\frac{13}{11} first. Then put the minus sign back.

  2. Divide 13 by 11

    ab=a÷b\frac{a}{b} = a \div b

    Use long division for 13÷1113 \div 11.

    • 11 goes into 13 once, with 2 left over. Write 1 and a decimal point.
    • Bring down a 0 to make 20. 11 goes into 20 once, with 9 left over. Write 1.
    • Bring down a 0 to make 90. 11 goes into 90 eight times, with 2 left over. Write 8.

    So far you have 1.18.

  3. Spot the repeating pattern

    The remainder (the amount left over) is 2 again. That is where you were after the first step. So the same digits come back: 1, 8, 1, 8, and so on forever.

    1311=1.181818…\frac{13}{11} = 1.181818\ldots

    Watch out: −1.18, choice A, stops too soon. The digits never end.

  4. Write it with a bar

    0.18=0.181818…0.\overline{18} = 0.181818\ldots

    A bar goes over only the digits that repeat. Here 18 repeats, starting right after the decimal point. The whole number 1 does not repeat, so it is not under the bar. Put the minus sign back.

    −1311=−1.18-\frac{13}{11} = -1.\overline{18}

    Watch out: a bar over 81, choice D, means −1.8181…. A bar over 118, choice B, means −1.118118….

    So the answer is C, −1.18-1.\overline{18}.

Check your answer

Multiply back: 11×1.1818=12.999811 \times 1.1818 = 12.9998. That is almost exactly 13, and more 18s bring it even closer. Also, 13/11 is a bit more than 1, so a decimal between 1 and 2 makes sense.

More questions like this

The original DOE question
Question 79 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 205, question 79. Open page 205 of the handbook (PDF)

Question 60 of 120

2024 Test B Math, question 23

Question

Each loaf of bread a baker makes requires 15\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 15\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:

5×40=2005\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.
What makes it hard
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.

  1. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of equal groups
groups=total÷size of one group\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÷1n=a×na \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.

  1. Sort out what you know

    Each loaf uses 15\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.

  2. 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.

    groups=total÷size of one group\text{groups} = \text{total} \div \text{size of one group}

    loaves=40÷15\text{loaves} = 40 \div \frac{1}{5}

    Watch out: do not multiply by 1/5. 40×15=840 \times \frac{1}{5} = 8, choice A, is a fifth of the salt, not a number of loaves.

  3. 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÷1n=a×na \div \frac{1}{n} = a \times n

    40÷15=40×5=20040 \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 15\frac{1}{5} ounce each use 200×15=40200 \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 61 of 120

2024 Test B Math, question 24

Question

What is the value of (157−167)(347−367)\frac{\left(1\frac{5}{7}-1\frac{6}{7}\right)}{\left(3\frac{4}{7}-3\frac{6}{7}\right)}?

  • A−12-\frac{1}{2}
  • B−17-\frac{1}{7}
  • C17\frac{1}{7}
  • D12\frac{1}{2}
Show the answer

Correct answer: D

First, subtract to simplify.

157−167347−367=−17−27=−17÷−27\frac{1\frac{5}{7}-1\frac{6}{7}}{3\frac{4}{7}-3\frac{6}{7}}=\frac{-\frac{1}{7}}{-\frac{2}{7}}=-\frac{1}{7}\div -\frac{2}{7}

Follow the process for dividing fractions:

(−17)×(−72)=714=12\left(-\frac{1}{7}\right)\times\left(-\frac{7}{2}\right)=\frac{7}{14}=\frac{1}{2}

Question analysis

A fractions question: find the value of a quotient whose top is 1 5/7 minus 1 6/7 and whose bottom is 3 4/7 minus 3 6/7. Students subtract mixed numbers that share a denominator and then divide one difference by the other. Keeping track of positive and negative signs through both steps is the main hurdle.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students subtract and divide rational numbers, including negatives, and use the sign rules for quotients. This item chains three such operations with easy sevenths, a typical grade 7 exercise.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 10, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The mixed numbers look heavy, but the whole parts cancel and only sevenths remain. What decides the question is sign handling: each difference is negative, and one negative divided by another is positive. Students who stop after the top, subtract instead of divide, or keep a minus sign each find their result among the choices.

  1. Error-prone arithmetic moderate Working it out

    Both differences are negative, 1 5/7 − 1 6/7 = −1/7 and 3 4/7 − 3 6/7 = −2/7, and dividing them needs both the sign rule and a reciprocal. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Two subtractions must be finished before the division, and a sign lost in either one carries into the final quotient. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out the top and the bottom separately, writing each difference as a signed fraction in sevenths, then divide by multiplying the top by the reciprocal of the bottom and apply the sign rule.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing fractions
ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}
a over b is the number you divide. c over d is the number you divide by. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. Only the second fraction flips.
Dividing two negative numbers
(−a)÷(−b)=a÷b(-a) \div (-b) = a \div b
a and b are positive numbers, and b is not zero. A negative number divided by a negative number gives a positive answer. The two minus signs cancel. Multiplying two negatives works the same way.

  1. Sort out the problem

    The long fraction bar means divide. So you need the top difference divided by the bottom difference. A difference is the answer to a subtraction. Work out the top and the bottom first, then divide.

  2. Work out the top

    Both numbers on top have the same whole number, 1. So the wholes cancel, and you only need 57−67\frac{5}{7} - \frac{6}{7}. You take away more than you have, so the answer is negative.

    157−167=57−67=−171\frac{5}{7} - 1\frac{6}{7} = \frac{5}{7} - \frac{6}{7} = -\frac{1}{7}

  3. Work out the bottom

    Both numbers on the bottom have the whole number 3, so those cancel too.

    347−367=47−67=−273\frac{4}{7} - 3\frac{6}{7} = \frac{4}{7} - \frac{6}{7} = -\frac{2}{7}

    Watch out: do not stop after the top. −17-\frac{1}{7}, choice B, is only part of the problem.

  4. Divide the top by the bottom

    Both results are negative. The two minus signs cancel.

    (−a)÷(−b)=a÷b(-a) \div (-b) = a \div b

    −17÷(−27)=17÷27-\frac{1}{7} \div \left(-\frac{2}{7}\right) = \frac{1}{7} \div \frac{2}{7}

    Now flip the second fraction and multiply.

    ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

    17×72=714=12\frac{1}{7} \times \frac{7}{2} = \frac{7}{14} = \frac{1}{2}

  5. Avoid the traps and answer

    Watch out: keeping a minus sign gives −12-\frac{1}{2}, choice A. Subtracting the bottom from the top, instead of dividing, gives 17\frac{1}{7}, choice C.

    So the answer is D, 12\frac{1}{2}.

Check your answer

Think in sevenths. The top is −1 seventh and the bottom is −2 sevenths. That is like (−1)÷(−2)=12(-1) \div (-2) = \frac{1}{2}. Or multiply back: 12×(−27)=−17\frac{1}{2} \times \left(-\frac{2}{7}\right) = -\frac{1}{7}, which is the top.

More questions like this

The original DOE question
Question 81 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 205, question 81. Open page 205 of the handbook (PDF)

Question 62 of 120

2024 Test B Math, question 33

Question

If x=−4x=-4, what is the sum of x and 10% of |x||x|?

  • A−4.4
  • B−3.6
  • C0.4
  • D4.4
Show the answer

Correct answer: B

First, determine 10% of |x||x|.

0.10∙|x|=0.10\bullet|x|=

0.10∙|−4|=0.10\bullet|-4|=

0.10∙4=0.40.10\bullet 4=0.4

Then determine the sum of x and 10% of |x||x|.

−4+0.4=−3.6-4+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
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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value
|a|=distance from a to 0|a| = \text{distance from } a \text{ to } 0
The two bars 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
part=percent100×whole\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.

  1. 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.

  2. 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|x| = |-4| = 4

  3. Take 10 percent of it

    part=10100×4=0.4\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.

  4. Add x and the result

    Now add x, which is still −4. Its own minus sign stays.

    −4+0.4=−3.6-4 + 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)=0.4-3.6 - (-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 63 of 120

2024 Test B Math, question 34

Question

The volume of a cube is 512 cubic centimeters. What is the length, in centimeters, of one edge of this cube?

  • A8
  • B422342\frac{2}{3}
  • C851385\frac{1}{3}
  • D128
Show the answer

Correct answer: A

The volume of a cube is V=s3V=s^{3}, where s is the length of one edge.

So 512=s3512=s^{3}, and s=5123s=\sqrt[3]{512}.

If you are not sure how to find the cube root, you can estimate.

We know 10×10×10=103=1,00010\times 10\times 10=10^{3}=1{,}000, so 1,0003=10\sqrt[3]{1{,}000}=10.

Since 512 is less than 1,000, the answer must be less than 10. The only given option less than 10 is 8. Double-check to make sure this is correct: 83=8×8×8=5128^{3}=8\times 8\times 8=512.

Question analysis

A cube has a volume of 512 cubic centimeters, and students work backward to find the length of one edge. It tests the volume formula for a cube and the idea of a cube root. The tricky part is going from a volume to a length, the reverse of the usual volume problem.

NYS standards
  • 8.EE.2 square and cube roots
  • 6.G.5 perfect squares and perfect cubes
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 volume work goes from edge lengths to a volume. Going backward needs a cube root, a grade 8 idea, and 512 is beyond the perfect cubes students are expected to know, though checking a guess by multiplying keeps it short.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-fact question: a cube's volume is its edge used three times as a factor. Students who remember that need only the number whose cube is 512, and the choices help, because only one is under 10 while 10 × 10 × 10 = 1,000. The wrong choices divide 512 by numbers tied to other measures of a cube or a square, so they catch students who do not recall the formula.

  1. Something you must already know moderate Working it out

    Neither V = s³ nor the cube root of 512 is given, so students must recall the formula and know, or find by trial, that 8 × 8 × 8 = 512. (Anderson & Krathwohl, 2001)

Strategy: Write the volume as edge × edge × edge, then find the number that, multiplied by itself three times, gives 512, and confirm it by multiplying.

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a cube
V=s3V = s^3
V is the volume, the space inside, in cubic units. s is the length of one edge. s³ means s × s × s. A cube’s length, width and height are all s, so length × width × height becomes s × s × s.

  1. Sort out the question

    The cube’s volume is 512 cubic centimeters. Volume is the space inside a solid. You need the length of one edge.

    A cube has 12 edges, and they are all the same length. So one edge tells you all of them.

  2. Write the volume formula

    V=s3V = s^3

    512=s×s×s512 = s \times s \times s

    You need a number that gives 512 when you multiply it by itself three times. That number is called the cube root of 512.

  3. Try numbers to find the edge

    Start with an easy one: 10×10×10=1,00010 \times 10 \times 10 = 1{,}000. That is too big, so the edge is less than 10. Try 8.

    8×8×8=64×8=5128 \times 8 \times 8 = 64 \times 8 = 512

    It works. Each edge is 8 centimeters.

  4. Rule out the other choices

    Watch out: The volume is not shared out among the faces or the edges. Dividing 512 by the 6 faces gives 851385\frac{1}{3}, choice C. Dividing by the 12 edges gives 422342\frac{2}{3}, choice B. Dividing by 4 gives 128, choice D.

    So the answer is A, 8.

Check your answer

Test the biggest choice. 128×128×128128 \times 128 \times 128 is over 2 million, far more than 512. Only an edge less than 10 can work, and 8 is the only choice less than 10. And 83=5128^3 = 512.

More questions like this

The original DOE question
Question 91 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 91. Open page 209 of the handbook (PDF)

Question 64 of 120

2024 Test B Math, question 36

Question

The integers (n−1)(n-1), n, and (n+1)(n+1) are factors of 168. What is the greatest possible value of n?

  • A3
  • B7
  • C14
  • D84
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-1), n, and (n+1)(n+1) where n is as large as possible. This group is 6, 7, and 8, where n=7n=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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pair
a×b=168a \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.

  1. 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.

  2. 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 factorPartner
    1168
    284
    356
    442
    628
    724
    821
    1214

  3. 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.

  4. 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÷6=28168 \div 6 = 28, 168÷7=24168 \div 7 = 24 and 168÷8=21168 \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 65 of 120

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÷7=12, with a remainder of 286\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
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.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Weeks and leftover days
days=7×weeks+leftover\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.

  1. 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.

  2. 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÷7=12 remainder 286 \div 7 = 12 \text{ remainder } 2

    86=7×12+286 = 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.

  3. 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 66 of 120

2024 Test B Math, question 42

Question

Jevon earns 12%\frac{1}{2}\% per year on the money in his savings account. Which expression is not another way to write 12%\frac{1}{2}\%?

  • A0.005100\frac{0.005}{100}
  • B12100\frac{\frac{1}{2}}{100}
  • C51,000\frac{5}{1{,}000}
  • D1200\frac{1}{200}
Show the answer

Correct answer: A

Rewrite the percent as a decimal:

12%=0.5%=0.005\frac{1}{2}\%=0.5\%=0.005

So the correct answer must not equal 0.005:

Option A

0.005100=0.00005\frac{0.005}{100}=0.00005

Option B

12100=0.5÷100=0.005\frac{\frac{1}{2}}{100}=0.5\div 100=0.005

Option C

51,000=0.005\frac{5}{1{,}000}=0.005

Option D

1200=0.005\frac{1}{200}=0.005

All answer options except Option A equal 0.005. Therefore, Option A is the correct answer.

Question analysis

Jevon's savings account earns one-half percent a year, and students must find the expression that is not another way to write one-half percent. It tests converting a percent less than 1 into fractions and decimals. Reading the word not carefully, and evaluating each stacked or unfamiliar fraction correctly, are what make it tricky.

NYS standards
  • 6.RP.3c percent as a rate per 100
  • 7.NS.2d fractions to decimals by long division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Converting between percents, fractions and decimals is grade 6 and 7 work. A percent smaller than 1 is less familiar, and the question asks which form is not equal, so all four must be checked, which keeps it at typical grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 7, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every expression is built from the same idea, so the question tests whether students can convert a percent smaller than 1 without losing track of decimal places. The odd one out, 0.005 over 100, looks like a correct form, because 0.005 is one-half percent and dividing by 100 seems to turn it into a percent. Converting one-half percent to 0.005 once and comparing each expression with it beats reasoning about each form.

  1. A question in the negative moderate Understanding the question

    The bold word not asks for the one expression unequal to one-half percent, so the three expressions that check out are exactly the ones to reject. (Haladyna, Downing & Rodriguez, 2002)

  2. Every choice must be checked moderate Choosing the answer

    Three of the four forms are equal, so each has to be converted to a decimal and compared with 0.005 before the odd one out can be named. (Embretson & Wetzel, 1987)

  3. Error-prone arithmetic minor Working it out

    One-half percent is 0.5 with the point moved two places, 0.005, and dividing that by 100 moves it two more places, to 0.00005. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write one-half percent as a single decimal by dividing one half by 100, then evaluate each expression as a decimal and compare it with that value.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction
p%=p100p\% = \frac{p}{100}
p is any number. The % sign means out of 100. So p percent is p divided by 100. Dividing by 100 moves the decimal point two places to the left.

  1. Sort out the question

    The question asks which expression is not equal to 12%\frac{1}{2}\%. So three choices are equal to it, and one is different. You need the odd one out.

  2. Write the percent as a decimal

    One half is 0.5. A percent means out of 100, so divide by 100.

    12%=0.5%=0.5100=0.005\frac{1}{2}\% = 0.5\% = \frac{0.5}{100} = 0.005

    So 12%\frac{1}{2}\% is 0.005. It is half of one percent, a very small amount.

  3. Test each choice

    Turn each choice into a decimal. Then compare it with 0.005.

    ChoiceWork it outEqual to 0.005?
    A0.005÷100=0.000050.005 \div 100 = 0.00005No
    B0.5÷100=0.0050.5 \div 100 = 0.005Yes
    C5÷1,000=0.0055 \div 1{,}000 = 0.005Yes
    D1÷200=0.0051 \div 200 = 0.005Yes

  4. Pick the one that is different

    Choice A divides by 100 twice. The 0.005 was already divided by 100 once. Dividing again makes it 100 times too small.

    Watch out: Choice B is the plainest way to write 12%\frac{1}{2}\%: one half over 100. Picking it means missing the word “not.” Also, 1 ÷ 200 is 0.005, not 0.05, so D is equal too.

    So the answer is A, 0.005100\frac{0.005}{100}.

Check your answer

Use fractions instead. Half of one hundredth is 12×1100=1200\frac{1}{2} \times \frac{1}{100} = \frac{1}{200}, which is choice D. And 51,000\frac{5}{1{,}000} simplifies to 1200\frac{1}{200} too. Only A is a different amount.

More questions like this

The original DOE question
Question 99 as printed on page 211 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 99. Open page 211 of the handbook (PDF)

Question 67 of 120

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÷7≈68.57480\div 7\approx 68.57

Option D:

420 is not evenly divisible by 8.

420÷8=52.5420\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.
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.

  1. 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)

  2. 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

The wrong answers, classified

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÷b=a whole numbera \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.

  1. 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.

  2. 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.

    60×7=42060 \times 7 = 420

  3. Bring in the 8

    8 is 2×2×22 \times 2 \times 2, so the number needs three 2s as factors. But 420 has only two: 420=2×2×105420 = 2 \times 2 \times 105, and 105 is odd. Multiply by one more 2.

    420×2=840420 \times 2 = 840

    Now 840÷8=105840 \div 8 = 105 and 840÷7=120840 \div 7 = 120. Both are whole numbers.

  4. Test the choices from smallest up

    Choice D: 420÷8=52.5420 \div 8 = 52.5. That is not a whole number, so 420 fails.

    Choice C: 480÷7≈68.57480 \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=60×14840 = 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 68 of 120

2024 Test B Math, question 49

Question

Suppose that w, x, y, and z are integers, none of which are zero. Under which set of conditions would the product x2yz3|w|x^{2}yz^{3}|w| always be positive?

  • Ax is positive and w is positive.
  • Bx is negative and w is negative.
  • Cy is negative and z is negative.
  • Dx and y are negative, and z and w are positive.
Show the answer

Correct answer: C

First, analyze the role of each variable in the expression x2yz3|w|x^{2}yz^{3}|w|.

x2x^{2} will always be a positive number.

y can be a positive or a negative number.

z3z^{3} can be a positive or a negative number.

|w||w| will always be a positive number.

The expressions x2x^{2} and |w||w| will always be positive. Therefore, making the values of y and z negative would result in the expression yz3yz^{3} being positive, since a negative multiplied by a negative is positive. Thus the given product would always be positive.

Question analysis

Four nonzero integers appear in a product that includes a square, a cube and an absolute value. Students choose the set of conditions that makes the product positive every time. It tests the sign rules for multiplication together with exponents and absolute value, and the trap is judging the sign by counting negative variables without looking at what is done to each.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.EE.1 whole-number exponent expressions
  • 6.NS.7c absolute value as distance
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students learn sign rules for products of signed numbers. Here they must reason about four unknown integers under an even power, an odd power and an absolute value, and decide what is always true, which is more abstract than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 6, 58 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One insight decides the question: x² and |w| are positive whatever x and w are, so only the signs of y and z³ matter. Students who count negative letters instead of negative factors pick conditions that look balanced but leave y as the only negative factor. Students who test a single set of values can be fooled by a choice that leaves y and z unspecified.

  1. Variables instead of numbers moderate Understanding the question

    There are four variables and no numbers, and always positive must hold for every allowed value, so one test case can rule a choice out but never confirm it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    Each set of conditions has to be checked against all four factors; the first two fix only x and w and leave y and z free to make the product negative. (Embretson & Wetzel, 1987)

  3. Something you must already know minor Working it out

    Students must know that a nonzero square and an absolute value are always positive, while an odd power such as z³ keeps the sign of z. (Anderson & Krathwohl, 2001)

Strategy: Work out, factor by factor, which signs each part of the product can take, then test each set of conditions against every factor rather than counting negatives.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Signs in multiplication
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.

  1. Sort out the question

    w, x, y and z are integers, meaning whole numbers or their negatives. None of them is 0. The product x2yz3|w|x^2 y z^3 |w| means x2×y×z3×|w|x^2 \times y \times z^3 \times |w|.

    You need the conditions that make the product positive every time, for every number that fits them.

  2. Look at x squared and |w|

    x2x^2 means x × x. Two positives or two negatives multiply to a positive. So x2x^2 is always positive.

    |w||w| is the absolute value of w, its distance from 0. It is always positive, since w is not 0.

    So x and w never change the sign of the product. Their signs do not matter.

  3. Look at y and z cubed

    y can be positive or negative. z3z^3 means z × z × z, and it has the same sign as z. For example, (−2)3=−8(-2)^3 = -8.

    So the sign of the whole product is the sign of y×z3y \times z^3. It depends only on y and z.

  4. Test each choice

    • A and B say nothing about y and z. If y = −1 and z = 1, the product is negative. So they do not always work.
    • C: y is negative, and z3z^3 is negative. A negative times a negative is positive. This always works.
    • D: y is negative and z is positive. So the product is always negative.

    Watch out: D has two negative letters, x and y. But x2x^2 is positive anyway, so y is the only negative factor.

    So the answer is C, y is negative and z is negative.

Check your answer

Try numbers for C. Let x = −2, y = −3, z = −1 and w = −5. Then x2×y×z3×|w|=4×(−3)×(−1)×5=60x^2 \times y \times z^3 \times |w| = 4 \times (-3) \times (-1) \times 5 = 60, which is positive. Even with x and w negative, the product stays positive.

More questions like this

The original DOE question
Question 106 as printed on page 213 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 106. Open page 213 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 69 of 120

2024 Test B Math, question 51

Question

Rita used 125 small solid cubes to build a large solid cube. The measure of each edge of the small cubes equals 1 inch. What is the measure of each edge of the large cube, in inches?

  • A5
  • B25
  • C41.33
  • D125
Show the answer

Correct answer: A

The volume of a cube with a side length of s is

V=s3V=s^{3}

Since 125 small cubes were used to build the large cube, 125=s3125=s^{3}.

Each side of the large cube is made up of 5 small cubes.

125=53125=5^{3}

Since 5 small cubes make up each edge of the large cube, and the edge of each small cube measures 1 inch, the measure of each edge of the large cube is 1×5=5 inches1\times 5=5\text{ inches}.

Question analysis

Rita stacks 125 one-inch cubes into a single large solid cube. Students find the length of each edge of the large cube in inches. It tests volume and perfect cubes with a building-block model, and the care needed is in telling apart the cubes along an edge, across a face and in the whole solid.

NYS standards
  • 6.G.5 perfect squares and perfect cubes
  • 6.G.2 volume with fractional edge lengths
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Building a large cube from unit cubes is the model grade 6 uses to introduce perfect cubes. It is a single step with small numbers, easier than typical grade 7 geometry, which works with composite solids and formulas.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 3, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is one of the easiest questions on the test, since 125 is a familiar cube and the small cubes make the large one easy to picture as 5 rows of 5 in 5 layers. The wrong choices name a layer instead of an edge, a division by 3 for three dimensions, or the total count itself, and multiplying 5 × 5 × 5 as a check rules all of them out.

  1. Something you must already know minor Working it out

    Students must see that 125 one-inch cubes make a volume of 125 cubic inches, and know or find that 5 × 5 × 5 = 125. (Anderson & Krathwohl, 2001)

Strategy: Think of the large cube as equal rows, columns and layers of small cubes, and find the number that, used three times as a factor, gives 125.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a cube
V=s3V = s^3
V is the volume, the space inside, in cubic units. s is the length of one edge. s³ means s × s × s. A cube’s length, width and height are all s, so length × width × height becomes s × s × s.

  1. Sort out the question

    Rita stacks 125 small cubes into one big solid cube. Each small cube has edges of 1 inch. You need the length of one edge of the big cube, in inches.

  2. Find the big cube’s volume

    Each small cube is 1×1×1=11 \times 1 \times 1 = 1 cubic inch. The big cube is solid, with no gaps. So its volume is 125 cubic inches.

  3. Find the edge length

    V=s3V = s^3

    125=s×s×s125 = s \times s \times s

    Find the number that makes 125 when you use it three times. Try 5.

    5×5×5=25×5=1255 \times 5 \times 5 = 25 \times 5 = 125

    So 5 small cubes sit along each edge. Each one is 1 inch long, so the edge is 5 inches.

  4. Rule out the other choices

    Watch out: 25, choice B, is the number of small cubes in one layer, 5 rows of 5. 125, choice D, is the number of cubes, or the volume. Dividing 125 by 3 does not give an edge either. That is about 41.67, near choice C.

    So the answer is A, 5.

Check your answer

Picture building it. A row of 5 cubes is 5 inches long. Five rows make a layer of 25 cubes. Five layers make 25×5=12525 \times 5 = 125 cubes. That matches.

More questions like this

The original DOE question
Question 108 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 108. 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 70 of 120

2024 Test B Math, question 53

Question

Which graph represents the solution to |x+1|≥2|x+1|\geq 2?

  • ANumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1. The line is shaded from −3 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.
  • BNumber line from −3 to 3 with a tick at every integer. Closed dots at −1 and 1. The line is shaded from −1 to the left, ending in an arrow, and from 1 to the right, ending in an arrow.
  • CNumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 1, with the segment between them shaded.
  • DNumber line from −3 to 3 with a tick at every integer. Closed dots at −3 and 3. The line is shaded from −3 to the left, ending in an arrow, and from 3 to the right, ending in an arrow.
Show the answer

Correct answer: A

Since |x+1|≥2|x+1|\geq 2, either x+1≤−2x+1\leq -2 or x+1≥2x+1\geq 2.

Solving x+1≤−2x+1\leq -2 for x gives x≤−3x\leq -3.

Solving x+1≥2x+1\geq 2 for x gives x≥1x\geq 1.

A filled circle on a number line represents “or equal to.” So the correct number line representation has filled circles at −3 and 1, with the number line filled to the left of −3 for “less than” and to the right of 1 for “greater than.”

Question analysis

Students choose the number line graph that shows every solution of an inequality in which x + 1 sits inside absolute value bars. It tests absolute value as distance and graphing inequalities. The tricky part is turning one absolute value statement into two pieces and placing each boundary point and shaded ray correctly.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph two-step inequalities and know absolute value as distance. An inequality with an absolute value has to be split into two cases, which is outside the grade 6 to 8 standards and well above typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 9, 7 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Reading |x + 1| as the distance from x to −1 makes this quick: the points 2 units away are −3 and 1, and greater than or equal means outside them. Students who solve only one case and mirror it across 0 land on graphs centered on 0, and students who find both endpoints may still shade between them. The graphs differ in just one dot or one shading choice, so each must be read carefully.

  1. Something you must already know moderate Working it out

    Solving |x + 1| ≥ 2 needs two cases, x + 1 ≤ −2 or x + 1 ≥ 2, joined by or; absolute value inequalities go beyond grade 7, and the stem gives no hint. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The right graph and the segment from −3 to 1 share both filled endpoints, so only deciding between outside and between the boundaries separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Reading a figure, table or graph minor Understanding the question

    Four number lines must be read for filled dots and arrow directions, and two of them differ from the right one only in where a single dot sits. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the absolute value as distance from −1 on the number line, find the points exactly 2 units away, and decide whether the solutions lie between or outside them.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Absolute value as distance
|x−a|=distance from x to a|x - a| = \text{distance from } x \text{ to } a
x and a are points on the number line. The bars mean absolute value. |x − a| is how far apart the two points are, which is never negative. Since x + 1 = x − (−1), |x + 1| is the distance from x to −1.

  1. Sort out the question

    You need the graph of every x that makes |x+1|≥2|x + 1| \geq 2 true. The sign ≥ means greater than or equal to. The bars mean absolute value.

    On a number line, a filled dot means that point counts. A thick line with an arrow means all the numbers that way count too.

  2. Read it as a distance

    Adding 1 is the same as subtracting −1, so x+1=x−(−1)x + 1 = x - (-1). By the formula, |x+1||x + 1| is the distance from x to −1.

    So the inequality says that x is at least 2 units away from −1.

  3. Find the two boundary points

    Start at −1 and go 2 units each way.

    −1−2=−3-1 - 2 = -3

    −1+2=1-1 + 2 = 1

    At −3 and at 1, the distance is exactly 2. The words or equal to let them count, so they get filled dots.

  4. Decide which side to shade

    Being at least 2 away means being farther out. Numbers between −3 and 1 are too close to −1. For example, 0 is only 1 unit away. So shade to the left of −3 and to the right of 1.

    In algebra, that is x+1≤−2x + 1 \leq -2 or x+1≥2x + 1 \geq 2. These give x≤−3x \leq -3 or x≥1x \geq 1.

  5. Match the graph

    Choice A has filled dots at −3 and 1, with arrows going outward. That matches.

    Watch out: Choice C shades between −3 and 1. Those are the points within 2 of −1, the answer to |x+1|≤2|x + 1| \leq 2. Choices B and D are centered on 0, but the center here is −1.

    So the answer is A, the graph shaded outward from −3 and 1.

Check your answer

Test a point from each part. For x = 2, |2+1|=3|2 + 1| = 3, and 3 is at least 2. For x = 0, |0+1|=1|0 + 1| = 1, which is less than 2. For x = −3, |−3+1|=2|-3 + 1| = 2, which counts. Graph A agrees with all three.

More questions like this

The original DOE question
Question 110 as printed on page 214 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 110. Open page 214 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 71 of 120

2024 Test B Math, question 57

Question

If x and y are both negative integers with x<yx<y, which expression has the least value?

  • Ax2yx^{2}y
  • Bxy2xy^{2}
  • C−x2y-x^{2}y
  • D−xy2-xy^{2}
Show the answer

Correct answer: A

Since x and y are both negative integers, both x2x^{2} and y2y^{2} are positive.

Therefore, both x2yx^{2}y and xy2xy^{2} are negative, and both −x2y-x^{2}y and −xy2-xy^{2} are positive.

Because x<yx<y and both x and y are negative, |y|<|x||y|<|x| and y2<x2y^{2}<x^{2}. So |y2|<|x2||y^{2}|<|x^{2}|.

Since |y|<|x||y|<|x| and x and y are integers, |xy2|<|x2y||xy^{2}|<|x^{2}y|.

So x2y<xy2x^{2}y<xy^{2}.

−xy2=|xy2| and −x2y=|x2y|-xy^{2}=|xy^{2}|\text{ and }-x^{2}y=|x^{2}y|

Therefore, x2y<xy2<−xy2<−x2yx^{2}y<xy^{2}<-xy^{2}<-x^{2}y.

Question analysis

Two negative integers x and y, with x less than y, appear in four expressions built from squares and minus signs. Students find the expression with the least value. It tests multiplying negative numbers and comparing their sizes, and the difficulty is tracking how each minus sign and each square changes both the sign and the size of the result.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.NS.7d absolute value versus order
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students multiply negative numbers, but this asks them to compare four variable expressions for any pair of negative integers. Deciding how the squares compare when x is less than y goes beyond typical grade 7 sign practice.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 8, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two facts settle the question: squares of negatives are positive, and the more negative number has the larger square. Students who track the minus signs carelessly think a leading minus makes an expression negative, and students who assume the larger y has the larger square choose the wrong negative expression. Substituting x = −3 and y = −2 gives −18, −12, 18 and 12, which orders all four at once.

  1. Variables instead of numbers moderate Understanding the question

    No values are given, only that x and y are negative integers with x < y, so students must reason about signs or invent a pair such as −3 and −2. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    x²y and xy² use the same letters with the exponent moved; both are negative, and only comparing x² with y² separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Every choice must be checked minor Choosing the answer

    All four expressions have to be placed in order, because the minus signs in front of the last two turn those expressions positive. (Embretson & Wetzel, 1987)

Strategy: Pick two negative integers with x less than y, such as −3 and −2, evaluate all four expressions, and confirm the order with a second pair.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Signs in multiplication
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers, so −a and −b are negative. A negative times a negative is positive. A negative times a positive is negative. So a product is positive when it has an even number of negative factors.
Squaring a number
x2=x×xx^2 = x \times x
x is any number. x² (x squared) means x times itself. The square of a number other than 0 is always positive, because two negatives or two positives multiply to a positive.

  1. Sort out the question

    x and y are negative integers, such as −3 and −2. Since x < y, x is to the left of y on the number line. So x is more negative, farther from 0.

    You need the expression with the least value. That is the most negative one.

  2. Find the sign of each expression

    Squares are positive, so x2x^2 and y2y^2 are positive.

    • x2yx^2 y is a positive times a negative, so it is negative.
    • xy2x y^2 is a negative times a positive, so it is negative.
    • −x2y-x^2 y and −xy2-x y^2 are the opposites of these, so they are positive.

    So the least value must be A or B. Watch out: The minus sign in front of C and D does not make them negative. It turns a negative into a positive.

  3. Compare A and B with numbers

    Try x = −3 and y = −2, since −3 < −2.

    x2y=(−3)2×(−2)=9×(−2)=−18x^2 y = (-3)^2 \times (-2) = 9 \times (-2) = -18

    xy2=(−3)×(−2)2=(−3)×4=−12x y^2 = (-3) \times (-2)^2 = (-3) \times 4 = -12

    Since −18 < −12, A is smaller.

  4. See why it always works

    Both expressions contain x × y, which is positive. x2y=(xy)×xx^2 y = (xy) \times x and xy2=(xy)×yx y^2 = (xy) \times y. Multiplying the same positive number by the more negative x gives the more negative result.

    Watch out: y is bigger than x, but y2y^2 is smaller than x2x^2: 4 is less than 9. Thinking y² is bigger leads to choice B.

    So the answer is A, x2yx^2 y.

Check your answer

Try another pair, x = −5 and y = −1. Then A is 25×(−1)=−2525 \times (-1) = -25, B is (−5)×1=−5(-5) \times 1 = -5, C is 25 and D is 5. A is still the least.

More questions like this

The original DOE question
Question 114 as printed on page 215 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 114. Open page 215 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 72 of 120

2023 Test A Math, question 2

Question

6−9÷|−3|+(−2)3×1126 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}

What is the value of the expression shown above?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: -9

6−9÷|−3|+(−2)3×1126 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2} Calculate the exponent and convert the mixed number to an improper fraction.
6−9÷|−3|+(−8)×326 - 9 \div \left|-3\right| + (-8) \times \frac{3}{2} Next, calculate the absolute value.
6−9÷3+(−8)×326 - 9 \div 3 + (-8) \times \frac{3}{2} Perform the indicated multiplication and division.
6−3+(−12)6 - 3 + (-12) Perform the addition and subtraction.
3+(−12)=−93 + (-12) = -9

Question analysis

This grid-in asks for the value of an arithmetic expression that mixes an absolute value, a negative number raised to the third power, a division and a mixed number. Students apply the order of operations with signed numbers. Cubing a negative base, and doing the division and multiplication before any adding, are where most slips happen.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.EE.2c evaluating expressions and order of operations
  • 6.NS.7c absolute value as distance
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students routinely evaluate expressions with negative numbers in the right order. This one combines an absolute value, a cubed negative and a mixed number in a single line, so it takes more steps than usual, each with a sign to watch.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 5, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The expression packs four kinds of notation into one line: a subtraction, a division by an absolute value, a negative base raised to the third power and a mixed number. None of the operations is hard alone, but each has a rule that is easy to break under time pressure, and the slips lead to very different totals. Rewriting the power and the mixed number first, then dividing and multiplying before any addition, avoids them.

  1. Error-prone arithmetic moderate Working it out

    (−2)³ must stay negative, −8, and the mixed number 1½ must become 3/2 before multiplying, so the last product is −12; a sign or fraction slip here moves the total far from −9. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Six numbers and six operations, including a power and an absolute value, have to be worked in a planned order, and every addition waits until the division and multiplication are done. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. An answer in an unexpected form minor Choosing the answer

    The value is negative and the grid-in offers no choices, so the minus sign has to be gridded and nothing on the page confirms that the total falls below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the absolute value and the power first, change the mixed number to an improper fraction, do the division and multiplication, and only then add and subtract from left to right.

Traps

  • Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations
grouping→powers→× and ÷→+ and −\text{grouping} \rightarrow \text{powers} \rightarrow \times \text{ and } \div \rightarrow + \text{ and } -
This is the order for working out an expression. First work inside grouping symbols, such as parentheses and absolute value bars. Next do powers. Then multiply and divide from left to right. Last, add and subtract from left to right. Using one order means everyone gets the same answer.
A power
a3=a×a×aa^{3} = a \times a \times a
a is the base, the number being multiplied. The small 3 is the exponent. It says how many times the base is used as a factor, a number being multiplied. It does not mean a times 3.
Absolute value
|a|=the distance from a to 0\left|a\right| = \text{the distance from } a \text{ to } 0
The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so the absolute value of −3 is 3.
Mixed number to fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part, and b over c is the fraction part. Each whole is c pieces, so a wholes are a × c pieces. Add the b extra pieces. The bottom number stays c.

  1. Sort out the expression

    You need the value of 6−9÷|−3|+(−2)3×1126 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}. It has many operations, so the order matters. It has absolute value bars, a power of a negative number and a mixed number. The answer may be negative, so keep track of every sign.

  2. Do the absolute value and the power

    Grouping symbols and powers come first.

    |−3|=3\left|-3\right| = 3

    (−2)3=(−2)×(−2)×(−2)=4×(−2)=−8(-2)^{3} = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8

    Two negatives multiply to a positive, 4. One more negative makes it negative again. Watch out: (−2)³ is −8, not +8. Using +8 leads to 15. It is also not −2 × 3. That mistake leads to a total of −6.

  3. Change the mixed number to a fraction

    A fraction is easier to multiply than a mixed number.

    112=1×2+12=321\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}

    Now the expression looks like this:

    6−9÷3+(−8)×326 - 9 \div 3 + (-8) \times \frac{3}{2}

  4. Multiply and divide next

    Multiplying and dividing come before adding and subtracting.

    9÷3=39 \div 3 = 3

    (−8)×32=−242=−12(-8) \times \frac{3}{2} = \frac{-24}{2} = -12

    A negative times a positive is negative. Watch out: don’t just work from left to right. Doing 6 − 9 first, then dividing by 3, leads to −13.5.

  5. Add and subtract last

    Now only adding and subtracting are left. Work from left to right.

    6−3+(−12)=3+(−12)=−96 - 3 + (-12) = 3 + (-12) = -9

    Adding −12 is the same as taking away 12. So you would grid −9, with the minus sign in its own column.

Check your answer

Look at the size of each part. The part 6 − 3 is a small positive number, 3. The last part, −8 times 1½, is a bigger negative number, −12. A big negative plus a small positive is negative. So an answer below zero, −9, makes sense.

More questions like this

The original DOE question
Question 59 as printed on page 75 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 75, question 59. Open page 75 of the handbook (PDF)

Question 73 of 120

2023 Test A Math, question 6

Question

34+74=3^{4} + 7^{4} =

  • A40
  • B370
  • C2,482
  • D10,000
Show the answer

Correct answer: C

34+74=81+2,401=2,4823^{4} + 7^{4} = 81 + 2{,}401 = 2{,}482

Question analysis

This exponents question asks for the sum of 3 to the fourth power and 7 to the fourth power. Students evaluate each power separately and then add. The traps are treating an exponent as a multiplier and adding the bases before raising them to a power, and the larger power takes careful multiplication.

NYS standards
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Evaluating whole-number powers is grade 6 work, and this question asks only for two of them and a sum. The one demand beyond routine practice is computing a fourth power of 7 accurately without a calculator.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 16+, 1 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The sum is short, but it tests exactly what an exponent means and when to apply it. The wrong choices come from three misreadings of the notation: treating the power as a multiplier, using one factor too few, and adding the bases before raising them. A size check settles it: 7⁴ alone is 49 × 49, a little under 2,500, so 40 and 370 are far too small and 10,000 far too large.

  1. Error-prone arithmetic moderate Working it out

    7⁴ means 7 × 7 × 7 × 7 = 2,401, a four-digit product built by hand, most easily as 49 × 49; 3⁴ = 81 is quick by comparison. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    An exponent counts repeated factors rather than acting as a multiplier, so 3⁴ is 3 × 3 × 3 × 3 = 81, not 3 × 4 = 12; the item assumes that meaning. (Anderson & Krathwohl, 2001)

Strategy: Compute each power on its own, squaring and then squaring again, and add the two results only at the end.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A power
a4=a×a×a×aa^{4} = a \times a \times a \times a
a is the base, the number being multiplied. The small 4 is the exponent. It tells how many times the base is used as a factor, a number being multiplied. So the exponent counts factors. It does not mean multiply by 4.
Powers come before adding
a4+b4≠(a+b)4a^{4} + b^{4} \neq (a + b)^{4}
a and b are any two positive numbers. The sign ≠ means is not equal to. Powers come before adding. So raise each number to its power first, then add. Adding the bases first gives a much bigger number.

  1. Sort out the question

    You need the value of 34+743^{4} + 7^{4}. Each number has a small 4 above it. This 4 is an exponent. Work out each power on its own. Then add the two results.

  2. Find 3 to the fourth power

    3⁴ means four 3s multiplied together. Pair them up to make it easier.

    34=3×3×3×3=9×9=813^{4} = 3 \times 3 \times 3 \times 3 = 9 \times 9 = 81

    Watch out: 3⁴ is not 3 × 4 = 12. Doing that to both powers gives 12 + 28 = 40, choice A.

  3. Find 7 to the fourth power

    7⁴ means four 7s multiplied together. Pair them up the same way.

    74=7×7×7×7=49×49=2,4017^{4} = 7 \times 7 \times 7 \times 7 = 49 \times 49 = 2{,}401

    To find 49 × 49, do 49 × 50 = 2,450. Then take away one 49. That gives 2,401. Watch out: count four 7s, not three. Three 7s make 343. Using three factors in both powers gives 370, choice B.

  4. Add the two powers

    81+2,401=2,48281 + 2{,}401 = 2{,}482

    Watch out: don’t add 3 and 7 first. That gives (3+7)4=104=10,000(3 + 7)^{4} = 10^{4} = 10{,}000, choice D. Powers come before adding. So the answer is C, 2,482.

Check your answer

Estimate. 7⁴ is 49 × 49, a little less than 50 × 50 = 2,500. Adding 81 keeps it near 2,500. Only 2,482 is close. 40 and 370 are far too small, and 10,000 is far too big.

More questions like this

The original DOE question
Question 63 as printed on page 77 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 63. Open page 77 of the handbook (PDF)

Question 74 of 120

2023 Test A Math, question 7

Question

In one week, 1341\frac{3}{4} inches of rain fell on Monday, 2232\frac{2}{3} inches fell on Tuesday, and 78\frac{7}{8} inch fell on Wednesday. How many inches of rain fell during those three days?

  • A57245\frac{7}{24}
  • B51245\frac{1}{24}
  • C3453\frac{4}{5}
  • D3123\frac{1}{2}
Show the answer

Correct answer: A

To solve, add the three given fractions:

134+223+781\frac{3}{4} + 2\frac{2}{3} + \frac{7}{8}

The common denominator is 24:

=11824+21624+2124= 1\frac{18}{24} + 2\frac{16}{24} + \frac{21}{24}

=41024+2124= 4\frac{10}{24} + \frac{21}{24}

=5724= 5\frac{7}{24}

Question analysis

A rainfall word problem gives the inches of rain on Monday, Tuesday and Wednesday as two mixed numbers and a fraction, and asks for the total over the three days. Students add fractions with unlike denominators. Finding a common denominator for fourths, thirds and eighths, and regrouping the fractional part into wholes, is where errors come in.

NYS standards
  • 7.NS.3 four operations with rational numbers
  • 7.NS.1d properties for adding and subtracting
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding fractions and mixed numbers with unlike denominators is taught before grade 7, and this is a direct word problem. It takes more care than a typical exercise only because three different denominators must be rewritten over a common one.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The setup is plain addition, so the difficulty is all in the fractions. Students who know to rename over a common denominator rarely miss it, while those who add the fraction parts straight across, or put the numerators over 24 without scaling them, reach choices with a whole part of 3. Since 1 3/4 + 2 2/3 alone is more than 4, any total below 4 can be ruled out at a glance.

  1. Error-prone arithmetic moderate Working it out

    The fractional parts have denominators 4, 3 and 8, so they must be renamed as 18, 16 and 21 twenty-fourths, and their sum, 55/24, regrouped as 2 7/24 before the wholes are added. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Renaming, adding the numerators, regrouping the improper fraction and adding the whole numbers are four separate moves, and losing the 2 carried from 55/24 drops the total by two inches. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the three fractional parts over a common denominator of 24, add the whole numbers and the fractions separately, and regroup any extra twenty-fourths as a whole.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a over b is a fraction, and k is any whole number except 0. Multiplying the top and the bottom by the same number gives an equal fraction. You cut each piece into k smaller pieces. So you have k times as many pieces, each k times smaller.
Adding fractions with the same bottom
ad+bd+cd=a+b+cd\frac{a}{d} + \frac{b}{d} + \frac{c}{d} = \frac{a + b + c}{d}
d is the denominator, the bottom number. It tells the size of the pieces. a, b and c are the numerators, the top numbers. They count the pieces. When the pieces are the same size, add the counts. The bottom stays the same.
Fraction to mixed number
ab=qrb\frac{a}{b} = q\frac{r}{b}
a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.

  1. Sort out the question

    Rain fell on three days: 1341\frac{3}{4} inches, 2232\frac{2}{3} inches and 78\frac{7}{8} inch. You need the total rain, in inches. So add the three amounts. Add the whole numbers and the fractions separately. The whole numbers are 1 and 2, so they add to 3 inches.

  2. Find a common denominator

    The fractions have denominators, or bottom numbers, 4, 3 and 8. You can only add pieces of the same size. So find the smallest number that 4, 3 and 8 all divide into. Count by 8s: 8, 16, 24. 24 works, since 24 = 4 × 6 and 24 = 3 × 8. So use 24ths.

  3. Rename each fraction in 24ths

    Multiply the top and the bottom by the same number.

    34=3×64×6=1824\frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}

    23=2×83×8=1624\frac{2}{3} = \frac{2 \times 8}{3 \times 8} = \frac{16}{24}

    78=7×38×3=2124\frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24}

    Watch out: change the tops too. Keeping 3, 2 and 7 over 24 gives 1224=12\frac{12}{24} = \frac{1}{2} and a total of 3123\frac{1}{2}, choice D.

  4. Add the fractions

    Now the pieces are the same size, so add the tops.

    1824+1624+2124=5524\frac{18}{24} + \frac{16}{24} + \frac{21}{24} = \frac{55}{24}

    That is more than 1 whole. 24 goes into 55 twice, with 7 left over.

    5524=2724\frac{55}{24} = 2\frac{7}{24}

    Watch out: never add the bottoms. Adding straight across gives 1215=45\frac{12}{15} = \frac{4}{5} and a total of 3453\frac{4}{5}, choice C.

  5. Put the parts together

    Add the 3 whole inches from before.

    3+2724=57243 + 2\frac{7}{24} = 5\frac{7}{24}

    So the answer is A, 57245\frac{7}{24} inches.

Check your answer

Use decimals. 1341\frac{3}{4} is 1.75, 2232\frac{2}{3} is about 2.67, and 78\frac{7}{8} is 0.875. They add to about 5.29. And 724\frac{7}{24} is about 0.29. So 57245\frac{7}{24} fits, and 51245\frac{1}{24} is too small.

More questions like this

The original DOE question
Question 64 as printed on page 77 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 64. Open page 77 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 75 of 120

2023 Test A Math, question 10

Question

A number line from 0 to 10 with a tick and a label at every whole number and arrows at both ends. Point W is marked with a dot at 0, and point X is marked with a dot at 2.

Points Y and Z are not shown on the number line above. If X is the midpoint of WY\overline{WY}, and Y is the midpoint of WZ\overline{WZ}, where on the number line would point Z be located?

  • A2
  • B4
  • C6
  • D8
Show the answer

Correct answer: D

First, find where Y is located on the number line. Point X is the midpoint of WY\overline{WY}. Since there are 2 units between W and X, there must also be 2 units between X and Y. Thus, Y is located at 4 on the number line.

Point Y is the midpoint of WZ\overline{WZ}. Use the same reasoning as above. Since there are 4 units between W and Y, there must be 4 units between Y and Z. So, Z is located at 8 on the number line.

Question analysis

A number line shows point W at 0 and point X at 2, and two more points, Y and Z, are described but not drawn. Students use two midpoint facts, one after the other, to find where Z lies. Each midpoint fact is about a different segment, so keeping track of which point is the middle one matters.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Midpoints on a number line rest on distance, which grade 7 students find as a difference. Here every point lies at a whole number of zero or more, so the work is two short steps with no negative numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 36 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each midpoint statement doubles a distance from W, so the problem is two short doublings, 2 to 4 and then 4 to 8. The risk is stopping halfway, since Y's position is found first and appears among the choices, or reusing X as an endpoint in the second statement. Rereading which point the question asks about, and checking that Y is halfway between W and Z, catches both.

  1. Many steps minor Working it out

    The two midpoint statements must be used in order: the first places Y at 4, and only then can the second, about segment WZ, place Z. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Reading a figure, table or graph minor Understanding the question

    Only W at 0 and X at 2 are marked on the 0-to-10 line, and the overbars name WY and WZ as segments, so Y and Z have to be placed from the statements alone. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the first midpoint to place Y as far beyond X as X is from W, then use the second midpoint the same way to place Z beyond Y.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint of a segment
AM=MBAM = MB
A and B are the ends of a segment, a straight piece of a line. M is its midpoint, the point exactly halfway. AM is the distance from A to M, and MB is the distance from M to B. A midpoint splits a segment into two equal parts, so these are equal.

  1. Sort out the question

    The number line shows W at 0 and X at 2. Points Y and Z are not shown. X is the midpoint of segment WY. Y is the midpoint of segment WZ. A midpoint is the point exactly halfway along a segment. You need the number where Z sits.

  2. Place point Y

    X is halfway from W to Y. So X to Y is as long as W to X.

    WX=2−0=2WX = 2 - 0 = 2

    XY=WX=2XY = WX = 2

    So Y is 2 units past X.

    2+2=42 + 2 = 4

    Point Y is at 4. Watch out: this is Y, not Z. Stopping here gives 4, choice B.

  3. Place point Z

    Now Y is halfway from W to Z. W to Y is 4 units, so Y to Z is also 4 units.

    YZ=WY=4−0=4YZ = WY = 4 - 0 = 4

    4+4=84 + 4 = 8

    Watch out: the second midpoint is for segment WZ, not XZ. Taking just 2 more units from Y puts Z at 6, choice C. So the answer is D, 8.

Check your answer

Check both midpoints. Halfway from 0 to 4 is (0+4)÷2=2(0 + 4) \div 2 = 2, which is X. Halfway from 0 to 8 is (0+8)÷2=4(0 + 8) \div 2 = 4, which is Y. Both facts in the question are true.

More questions like this

The original DOE question
Question 67 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 67. Open page 78 of the handbook (PDF)

Question 76 of 120

2023 Test A Math, question 12

Question

If n is an integer and 3n+33n + 3 is an even number, which expression must also represent an even number?

  • A5n+15n + 1
  • B4n+54n + 5
  • C2n+32n + 3
  • Dn+2n + 2
Show the answer

Correct answer: A

The expression 3n+33n + 3 is equal to 3(n+1)3(n + 1), and it is given that this is an even number. The expression (n+1)(n + 1) must also represent an even number, because 3 is an odd number: if (n+1)(n + 1) were odd, 3(n+1)3(n + 1) would be the product of two odd numbers, which is an odd number. If (n+1)(n + 1) is an even number, then n must be an odd number. Choose an odd number to substitute for n in the answer options to find which expression represents an even number:

Let n=3n = 3.

Option A:

5n+1=5(3)+1=165n + 1 = 5(3) + 1 = 16

Option B:

4n+5=4(3)+5=174n + 5 = 4(3) + 5 = 17

Option C:

2n+3=2(3)+3=92n + 3 = 2(3) + 3 = 9

Option D:

n+2=3+2=5n + 2 = 3 + 2 = 5

Since 16 is the only even number, the correct answer is option A.

Question analysis

This number properties question says that 3n + 3 is even for an integer n and asks which of four expressions must also be even. Students reason about odd and even numbers written with a variable. The hidden step is deciding what the given fact says about n itself before testing the expressions.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.4 common factors and multiples
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students rewrite and evaluate expressions, but rarely reason about whether an expression is odd or even for every value of a variable. Here that reasoning comes in two layers: first about n, then about each choice.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 23 words.
Corrected from the DOE handbook
  • The explanation gave the wrong reason why n + 1 must be even; the reason was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question hides its key step: 3n + 3 = 3(n + 1), and since 3 is odd, n + 1 must be even, so n is odd. A student who skips that step and tests a convenient even value never checks the condition and lands on a wrong expression. Once n is known to be odd, any odd test value, such as 1, picks out 5n + 1 at once.

  1. Variables instead of numbers moderate Understanding the question

    No value of n is given; the only fact is that 3n + 3 is even, and it has to be turned into a fact about n itself, that n is odd, before any choice is judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know minor Working it out

    The parity rules are assumed, not stated: an odd number times an odd number is odd, and an even number plus an odd number is odd. (Anderson & Krathwohl, 2001)

  3. Every choice must be checked minor Choosing the answer

    Each expression has to be tested with an odd n or by parity rules; three of the four come out odd whenever n is odd, so none can be dismissed on sight. (Embretson & Wetzel, 1987)

Strategy: First work out what 3n + 3 being even says about n, by factoring out 3 or testing small integers, then check each expression with a value of n that fits.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation said that n + 1 must be even “because 3 is an odd number and the product of an odd number and an even number is an even number.” That fact does not show that n + 1 is even; the reason is that if n + 1 were odd, 3(n + 1) would be the product of two odd numbers, which is odd, and it is given as even. The site now gives that reason. The answer, A, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Pulling out a common factor
a×b+a×c=a(b+c)a \times b + a \times c = a(b + c)
a is a factor, a number being multiplied, that both terms share. b and c are what is left. This is the distributive property run backward. It turns a sum into a product, which is easier to reason about.
Multiplying odd numbers
odd×odd=odd\text{odd} \times \text{odd} = \text{odd}
An even number splits into pairs, like 2, 4 or 6. An odd number has one left over, like 1, 3 or 5. An odd number of odd groups always leaves one extra. So the product is odd.
Adding odd and even numbers
odd+odd=even,even+odd=odd\text{odd} + \text{odd} = \text{even}, \quad \text{even} + \text{odd} = \text{odd}
Two odd numbers each have one left over. Together the two extras make a pair, so the sum is even. An even number plus an odd number still has one left over, so the sum is odd.

  1. Sort out the question

    n is an integer, a whole number that can be positive, negative or 0. You are told that 3n+33n + 3 is even. You need the expression that must be even too. Must means it is even every time. First find out what the fact tells you about n.

  2. Pull out the 3

    Both terms in 3n+33n + 3 have a factor of 3. Pull it out.

    3n+3=3(n+1)3n + 3 = 3(n + 1)

    So 3 times (n+1)(n + 1) is an even number.

  3. Decide whether n is odd or even

    3 is odd. If n+1n + 1 were odd, 3(n+1)3(n + 1) would be odd times odd. That is always odd. But we are told it is even. So n+1n + 1 must be even. One less than an even number is odd. So n is odd. Watch out: don’t test n = 2. Then 3n+3=93n + 3 = 9, which is odd. With n = 2, only n+2n + 2 comes out even, choice D.

  4. Test each choice with an odd n

    Pick an odd number, n = 3. It fits, since 3×3+3=123 \times 3 + 3 = 12 is even.

    • A: 5n+1=5×3+1=165n + 1 = 5 \times 3 + 1 = 16, even
    • B: 4n+5=4×3+5=174n + 5 = 4 \times 3 + 5 = 17, odd
    • C: 2n+3=2×3+3=92n + 3 = 2 \times 3 + 3 = 9, odd
    • D: n+2=3+2=5n + 2 = 3 + 2 = 5, odd

    Only A is even.

  5. Make sure A always works

    Must means every odd n, not just 3. If n is odd, 5n is odd times odd, so it is odd. Odd plus 1 is even. So 5n+15n + 1 is always even. Watch out: 4n is even, but adding the odd 5 makes 4n+54n + 5 odd, choice B. In the same way, 2n+32n + 3 is always odd, choice C. So the answer is A, 5n+15n + 1.

Check your answer

Try another odd number, n = 1. Then 3n+3=63n + 3 = 6, which is even, so n = 1 fits. And 5n+1=65n + 1 = 6 is even too. The other three choices give 9, 5 and 3, all odd. Choice A wins again.

More questions like this

The original DOE question
Question 69 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 69. Open page 78 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 77 of 120

2023 Test A Math, question 13

Question

The product of two positive integers is 65. Which number could be the sum of the two integers?

  • A5
  • B18
  • C24
  • D52
Show the answer

Correct answer: B

First, find the pairs of positive integers that multiply to 65: 1×651 \times 65 and 5×135 \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.
What makes it hard
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.

  1. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pair
a×b=65a \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
last digit 0 or 5→divides 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.

  1. 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.

  2. List the factor pairs of 65

    Try each small number. See if it divides 65 with nothing left over.

    • 1 works: 1×65=651 \times 65 = 65.
    • 2 and 4 don’t work, because 65 is odd. 3 doesn’t work either: 3×21=633 \times 21 = 63, with 2 left over.
    • 5 works, since 65 ends in 5: 5×13=655 \times 13 = 65.
    • 6 and 8 are even, so they don’t work. 7 doesn’t either: 7×9=637 \times 9 = 63.

    Since 8×8=648 \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.

  3. Add each pair

    1+65=661 + 65 = 66

    5+13=185 + 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=185 + 13 = 18 and 5×13=655 \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 78 of 120

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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing negative numbers
if a>b>0, then −a<−b\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.

  1. 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.

  2. 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<−3-5 < -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.

  3. 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 79 of 120

2023 Test A Math, question 16

Question

Between which two consecutive integers is the fraction 297\frac{29}{7}?

  • A2 and 3
  • B3 and 4
  • C4 and 5
  • D5 and 6
Show the answer

Correct answer: C

First, convert 297\frac{29}{7} to a mixed number: 297=417\frac{29}{7} = 4\frac{1}{7}. The two consecutive integers on either side of 4174\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.
What makes it hard
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.

  1. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction to mixed number
ab=qrb\frac{a}{b} = q\frac{r}{b}
a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.

  1. Sort out the question

    Consecutive integers are whole numbers that come one right after the other, like 4 and 5. You need the two that 297\frac{29}{7} falls between. So find the whole number just below it and the one just above it.

  2. Divide 29 by 7

    A fraction bar means divide. How many 7s fit in 29?

    7×4=287 \times 4 = 28

    7×5=357 \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=129 - 28 = 1. So 29 ÷ 7 is 4 with 1 left over.

  3. Write it as a mixed number

    297=417\frac{29}{7} = 4\frac{1}{7}

    4174\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. 4×7=284 \times 7 = 28 and 5×7=355 \times 7 = 35. The top of the fraction, 29, is between 28 and 35. So 297\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 80 of 120

2023 Test A Math, question 20

Question

32+(−8)2+22(3−8+2)2=\frac{3^{2} + (-8)^{2} + 2^{2}}{(3 - 8 + 2)^{2}} =

  • A−60
  • B−173-\frac{17}{3}
  • C779\frac{77}{9}
  • D68
Show the answer

Correct answer: C

32+(−8)2+22(3−8+2)2=\frac{3^{2} + (-8)^{2} + 2^{2}}{(3 - 8 + 2)^{2}} =

9+64+4(−3)2=779\frac{9 + 64 + 4}{(-3)^{2}} = \frac{77}{9}

Question analysis

This arithmetic question asks for the value of a fraction whose numerator is the sum of the squares of 3, −8 and 2, and whose denominator is the square of 3 − 8 + 2. Students apply the order of operations with negative numbers. Squaring a negative number and adding inside parentheses before squaring are the points to watch.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.EE.2c evaluating expressions and order of operations
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Squaring negative numbers and simplifying with the order of operations is standard grade 7 practice. Here the fraction bar groups a numerator and a denominator that each involve a negative sign, but the work stays short.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 16+, 3 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The expression is built so that signs decide everything: the top's 3², (−8)² and 2² are all positive, and so is the square of the bottom's −3. Students who keep a negative after squaring, or who cancel the 9 on top against the 9 below, reach the other choices. The result, 77/9, does not reduce, which can make careful students doubt a correct answer.

  1. Error-prone arithmetic moderate Working it out

    Two negatives sit inside squares: (−8)² must be +64, and the denominator's 3 − 8 + 2 = −3 must be squared to +9, so every term ends up positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    The numerator needs three squares and a sum, the denominator a sum and then a square, and the fraction bar groups each side, so neither part can be split or cancelled. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Simplify the top and the bottom separately: square each number in the numerator, add inside the parentheses before squaring in the denominator, then write the fraction in lowest terms.

Traps

  • Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Square of a negative
(−a)2=(−a)×(−a)=a2(-a)^{2} = (-a) \times (-a) = a^{2}
a is any number. To square means to multiply a number by itself. A negative times a negative is positive. So the square of a negative number is always positive.
The fraction bar groups
topbottom=(top)÷(bottom)\frac{\text{top}}{\text{bottom}} = (\text{top}) \div (\text{bottom})
A fraction bar means divide. It also groups: work out the whole top and the whole bottom first, then divide. You can only cancel a number that multiplies the whole top, not one part of a sum.

  1. Sort out the expression

    The top, the numerator, is three squares added together. The bottom, the denominator, is a sum in parentheses that is then squared. To square means to multiply a number by itself. Work out the top and the bottom on their own. Then divide.

  2. Work out the top

    Square each number first. Then add.

    32=93^{2} = 9

    (−8)2=(−8)×(−8)=64(-8)^{2} = (-8) \times (-8) = 64

    22=42^{2} = 4

    9+64+4=779 + 64 + 4 = 77

    Watch out: (−8)² is +64, not −64. The minus sign is inside the parentheses, so it is squared too. Using −64 makes the top −51 and gives −173-\frac{17}{3}, choice B.

  3. Work out the bottom

    Work inside the parentheses first. Add and subtract before you square.

    3−8+2=−5+2=−33 - 8 + 2 = -5 + 2 = -3

    (−3)2=(−3)×(−3)=9(-3)^{2} = (-3) \times (-3) = 9

    The bottom is 9.

  4. Divide and simplify

    779\frac{77}{9}

    77 is 7 × 11 and 9 is 3 × 3. They share no factor, so the fraction is in lowest terms. Watch out: don’t cancel the 9 on top with the 9 below. That 9 is only one part of a sum. Canceling it leaves 64 + 4 = 68, choice D. So the answer is C, 779\frac{77}{9}.

Check your answer

Every square is positive or zero. So the top and the bottom are both positive, and the answer must be positive. That rules out −60 and −173-\frac{17}{3}. Also, 779\frac{77}{9} is about 8.6, which is far from 68.

More questions like this

The original DOE question
Question 77 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 77. Open page 80 of the handbook (PDF)

Question 81 of 120

2023 Test A Math, question 23

Question

{0.1,0.01,0.001,0.0001,0.00001}\{0.1, 0.01, 0.001, 0.0001, 0.00001\}

If a person chooses a number at random from the set above, what is the probability that the number is less than 0.005?

  • A15\frac{1}{5}
  • B25\frac{2}{5}
  • C35\frac{3}{5}
  • D23\frac{2}{3}
Show the answer

Correct answer: C

Three of the five numbers listed (0.001, 0.0001, and 0.00001) are less than 0.005. So, the probability of a person selecting a number less than 0.005 is 35\frac{3}{5}.

Question analysis

A set holds five decimals, from one tenth down to one hundred-thousandth, and one number is chosen at random. Students find the probability that the number is less than 0.005. Comparing decimals with many zeros is the real task here, along with putting the count over the right total.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.NS.7a inequality as relative position
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Probability with equally likely outcomes is grade 6 content in New York, and this needs one comparison per number and one fraction. The only real demand is comparing decimals with many zeros correctly.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 27 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each number in the set is a tenth of the one before, so only the three smallest fall below 0.005, which lies between 0.01 and 0.001. Students lose the point either by counting the numbers above 0.005 instead of below it or by writing the count over the wrong total. Lining the decimals up with the same number of places makes every comparison easy.

  1. Error-prone arithmetic moderate Working it out

    0.005 has to be compared with decimals from 0.1 down to 0.00001; written to five places, 0.00500 against 0.01000 and 0.00100, the order becomes a comparison of whole numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    The probability is the count of numbers below 0.005 over all 5 numbers in the set; the stem leaves that part-to-whole rule to the student. (Anderson & Krathwohl, 2001)

Strategy: Write each number with five decimal places, compare it with 0.00500, count those that are smaller, and divide by the five numbers in the set.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=wanted resultspossible resultsP = \frac{\text{wanted results}}{\text{possible results}}
P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.
Adding zeros to a decimal
0.1=0.10=0.1000.1 = 0.10 = 0.100
Zeros added to the right end of a decimal don’t change its value. One tenth is the same as ten hundredths. Giving every number the same number of decimal places lets you compare them like whole numbers.

  1. Sort out the question

    The set has five numbers. A person picks one at random, so each number has the same chance. You need the probability, or chance, that the number picked is less than 0.005.

  2. Give every number five decimal places

    Decimals are easier to compare with the same number of places. Add zeros to the end of each.

    NumberWith five places
    0.10.10000
    0.010.01000
    0.0010.00100
    0.00010.00010
    0.000010.00001
    0.0050.00500

  3. Count the numbers below 0.005

    Compare each number with 0.00500 like whole numbers: 10000, 1000, 100, 10 and 1 against 500. Two numbers are bigger: 0.1 and 0.01. Three are smaller: 0.001, 0.0001 and 0.00001. Watch out: 0.01 has fewer digits than 0.005, but it is bigger. Counting the 2 bigger numbers gives 25\frac{2}{5}, choice B.

  4. Write the probability

    P=wanted resultspossible results=35P = \frac{\text{wanted results}}{\text{possible results}} = \frac{3}{5}

    Watch out: divide by all 5 numbers in the set. Comparing the 2 bigger numbers with the 3 smaller ones gives 23\frac{2}{3}, choice D. So the answer is C, 35\frac{3}{5}.

Check your answer

The numbers below 0.005 and the numbers above it make up the whole set: 3 + 2 = 5. So the two chances, 35\frac{3}{5} and 25\frac{2}{5}, add up to 1, as they should.

More questions like this

The original DOE question
Question 80 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 80. Open page 81 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 82 of 120

2023 Test A Math, question 24

Question

A student uses Spinner R and Spinner T to generate a list of two-digit numbers.

Two spinners, each with an arrow pointing out from its center. Spinner R is a circle divided into 4 equal sections, labeled 1 and 2 across the top and 3 and 4 across the bottom. Spinner T is a circle divided into 6 equal sections, labeled, clockwise from the top left, 1, 1, 2, 3, 4 and 4.

Spinner R determines the digit in the tens place, and Spinner T determines the digit in the ones place. What is the probability that the two-digit number determined by spinning each spinner one time is a prime number?

  • A38\frac{3}{8}
  • B512\frac{5}{12}
  • C35\frac{3}{5}
  • D58\frac{5}{8}
Show the answer

Correct answer: A

Spinner R has 4 equal sections and Spinner T has 6 equal sections, so the number of possible two-digit numbers that can be created is found by multiplying 4 and 6, which is 24. List the possible outcomes:

11 11 12 13 14 14
21 21 22 23 24 24
31 31 32 33 34 34
41 41 42 43 44 44

There are 9 prime numbers out of 24 possible outcomes: 11, 11, 13, 23, 31, 31, 41, 41, and 43.

So the probability that the two-digit number created is a prime number is 924\frac{9}{24}, which is equivalent to 38\frac{3}{8}.

Question analysis

Spinner R, with four equal sections, gives the tens digit and Spinner T, with six equal sections and some labels repeated, gives the ones digit of a two-digit number. Students find the probability that the number is prime. Building the full table of outcomes, counting repeated sections separately, and checking whether each number is prime all take care.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.SP.8a compound event as fraction of outcomes
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Listing a sample space for two spinners is grade 7 work, usually with distinct labels and a simple event. Here one spinner repeats two numbers, the space has 24 outcomes, and each two-digit result must be tested for being prime.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 7, 57 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question combines a compound sample space with a test for primes, and its figure adds a twist: two labels repeat on Spinner T. Counting the 24 section pairs rather than the 16 different numbers is the careful route, though here both happen to give the same fraction. The points are lost in the prime test, where 33 is easy to count as prime, and in writing the count as a share of all 24 outcomes.

  1. Reading a figure, table or graph moderate Understanding the question

    Spinner T has six equal sections but only four labels, with 1 and 4 each appearing twice, so the 24 equally likely outcomes include repeats such as 11 and 11. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Much to keep track of moderate Working it out

    Twenty-four two-digit numbers have to be built and each tested for primality while the repeated sections are kept track of; a 4-by-6 table is the reliable way to hold them. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Something you must already know minor Working it out

    Which two-digit numbers are prime is not given; students must test 21, 33, 43 and the rest themselves, and 21 = 3 × 7 and 33 = 3 × 11 look prime at a glance. (Anderson & Krathwohl, 2001)

Strategy: Make a 4-by-6 table with Spinner R's digits as tens and Spinner T's six sections as ones, mark each prime, and count every section, including repeated labels.

Traps

  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting outcomes
outcomes=a×b\text{outcomes} = a \times b
a is the number of equal sections on the first spinner, and b is the number on the second. Each section of the first can go with each section of the second. So the number of pairs is a times b.
Probability
P=wanted resultspossible resultsP = \frac{\text{wanted results}}{\text{possible results}}
P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.

  1. Sort out the question

    Spinner R gives the tens digit. Spinner T gives the ones digit. Together they make a two-digit number, like 23. You need the chance that the number is prime. A prime number has exactly two factors, 1 and itself. For example, 13 is prime, but 21 = 3 × 7 is not.

  2. Count all the outcomes

    Spinner R has 4 equal sections. Spinner T has 6 equal sections. Look closely: T has two 1s and two 4s. Each section counts as its own outcome.

    outcomes=4×6=24\text{outcomes} = 4 \times 6 = 24

    There are 24 outcomes, all equally likely.

  3. Make a table of the numbers

    Each row is a tens digit from R. Each column is a section of T. Prime numbers are in bold.

    R112344
    1111112131414
    2212122232424
    3313132333434
    4414142434444

  4. Count the prime numbers

    Numbers ending in 2 or 4 are even, so 2 divides them. They are not prime. Check the rest. 11, 13, 23, 31, 41 and 43 are prime. But 21 = 3 × 7 and 33 = 3 × 11 are not. Count every section, so 11, 31 and 41 each count twice. That makes 3 + 1 + 2 + 3 = 9 primes. Watch out: counting 33 as prime gives 1024=512\frac{10}{24} = \frac{5}{12}, choice B.

  5. Find the probability

    P=924=9÷324÷3=38P = \frac{9}{24} = \frac{9 \div 3}{24 \div 3} = \frac{3}{8}

    Watch out: divide by all 24 outcomes. Comparing the 9 primes with the 15 others gives 915=35\frac{9}{15} = \frac{3}{5}, choice C. So the answer is A, 38\frac{3}{8}.

Check your answer

There are 24−9=1524 - 9 = 15 numbers that are not prime. Their chance is 1524=58\frac{15}{24} = \frac{5}{8}. And 38+58=1\frac{3}{8} + \frac{5}{8} = 1, as it should be. Notice that 58\frac{5}{8}, choice D, is the chance of not prime.

More questions like this

The original DOE question
Question 81 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 81. Open page 81 of the handbook (PDF)

Question 83 of 120

2023 Test A Math, question 26

Question

Which statement must be true if x is a whole number greater than or equal to 1?

  • A1x+1>1x+2\frac{1}{x + 1} > \frac{1}{x + 2}
  • B1x+1<1x+2\frac{1}{x + 1} < \frac{1}{x + 2}
  • C1x+1−1x+2>1\frac{1}{x + 1} - \frac{1}{x + 2} > 1
  • D1x+1−1x+2>1x\frac{1}{x + 1} - \frac{1}{x + 2} > \frac{1}{x}
Show the answer

Correct answer: A

Let x=2x = 2, and substitute that value for x in each inequality to determine which one is correct:

Option A:

13>14\frac{1}{3} > \frac{1}{4}. This is true.

Option B:

13<14\frac{1}{3} < \frac{1}{4}. This is false.

Option C:

13−14>1\frac{1}{3} - \frac{1}{4} > 1, or 112>1\frac{1}{12} > 1. This is false.

Option D:

13−14>12\frac{1}{3} - \frac{1}{4} > \frac{1}{2}, or 112>12\frac{1}{12} > \frac{1}{2}. This is false.

Option A is the only true inequality.

Question analysis

This question asks which of four inequalities must be true when x is a whole number of at least 1. The inequalities compare fractions with denominators x + 1 and x + 2, or their difference. Students reason about how the size of a fraction changes as its denominator grows, and must be sure a statement holds for every allowed value.

NYS standards
  • 6.NS.7a inequality as relative position
  • 6.EE.5 solutions and substitution
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Comparing fractions is familiar, but these fractions have variable denominators and the question asks what must be true for every whole number from 1 up. That abstraction makes it as demanding as typical grade 7 work, though the arithmetic is light.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 6, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Choosing a small value, such as x = 1 or 2, and testing every statement settles this quickly, because three of the four fail for any value tried. The trap is the belief that a bigger denominator makes a bigger fraction, which turns the true comparison around. Students unsure of fraction subtraction lose time on the two difference statements, though both differences are tiny, never more than 1/6.

  1. Variables instead of numbers moderate Understanding the question

    All four statements are about the unknown x, and the bold must means each has to hold for every whole number from 1 up, not just for a single test value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    The statements are four separate inequalities, two comparisons and two differences, so each has to be tested; one counterexample is enough to rule a statement out. (Embretson & Wetzel, 1987)

  3. Error-prone arithmetic minor Working it out

    Testing the differences means subtracting unlike fractions, such as 1/3 − 1/4 = 1/12 when x = 2, and comparing the result with 1 and with 1/x. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute a small whole number such as 1 or 2 for x in all four statements, rule out any that fail, and confirm the survivor by thinking about larger denominators.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions
if 0<a<b, then 1a>1b\text{if } 0 < a < b \text{, then } \frac{1}{a} > \frac{1}{b}
a and b are positive numbers, and b is bigger. One over a means one whole cut into a equal pieces. Cutting the same whole into more pieces makes each piece smaller. So the fraction with the bigger bottom is smaller.
Subtracting fractions
1a−1b=b−aa×b\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}
a and b are the bottoms of the two fractions. Rewrite both fractions over the common bottom a × b. Then subtract the tops. The bottom stays a × b.

  1. Sort out the question

    x is a whole number that is 1 or more, so x can be 1, 2, 3 and so on. You need the statement that must be true. That means it is true for every such x. One x that makes a statement false rules it out.

  2. Test x = 1 in each statement

    With x = 1, x + 1 is 2 and x + 2 is 3.

    • A: 12>13\frac{1}{2} > \frac{1}{3}. True.
    • B: 12<13\frac{1}{2} < \frac{1}{3}. False.
    • C: 12−13=36−26=16\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}, which is not more than 1. False.
    • D: 16\frac{1}{6} is not more than 11\frac{1}{1}. False.

    Only A is true when x = 1. So B, C and D are out.

  3. Make sure A is always true

    Must means A has to work for every x, not just 1. In 1x+1\frac{1}{x + 1} and 1x+2\frac{1}{x + 2}, both tops are 1. The bottom x + 2 is always 1 more than x + 1. Cutting a whole into more pieces makes each piece smaller. So A is always true. Watch out: a bigger bottom makes a smaller fraction, not a bigger one. Thinking the other way gives choice B.

  4. See why C and D always fail

    Use the subtracting rule with a = x + 1 and b = x + 2.

    1a−1b=b−aa×b\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}

    1x+1−1x+2=(x+2)−(x+1)(x+1)(x+2)=1(x+1)(x+2)\frac{1}{x + 1} - \frac{1}{x + 2} = \frac{(x + 2) - (x + 1)}{(x + 1)(x + 2)} = \frac{1}{(x + 1)(x + 2)}

    The bottom is at least 2 × 3 = 6, so the difference is at most 16\frac{1}{6}. That is less than 1. Its bottom is also bigger than x, so it is less than 1x\frac{1}{x}. Watch out: keep the bottom when you subtract. Keeping only the top, 1, leads to choice D. So the answer is A.

Check your answer

Try a bigger number, x = 9. A tenth of a pizza is more than an eleventh, so 110>111\frac{1}{10} > \frac{1}{11}. The difference, 1110\frac{1}{110}, is tiny. It is less than 1 and less than 19\frac{1}{9}. A still holds, and C and D still fail.

More questions like this

The original DOE question
Question 83 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 83. Open page 82 of the handbook (PDF)

Question 84 of 120

2023 Test A Math, question 29

Question

Integer x is evenly divisible by 3. Which expression below is also evenly divisible by 3?

  • A2x+12x + 1
  • B3x−53x - 5
  • C4x−14x - 1
  • D4x+64x + 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=3x = 3.

Option A:

2x+1=2(3)+1=72x + 1 = 2(3) + 1 = 7

Option B:

3x−5=3(3)−5=43x - 5 = 3(3) - 5 = 4

Option C:

4x−1=4(3)−1=114x - 1 = 4(3) - 1 = 11

Option D:

4x+6=4(3)+6=184x + 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.
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.

  1. 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)

  2. 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

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)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.

  1. 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.

  2. 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.

    ChoiceWorkDivisible by 3?
    A2(3)+1=72(3) + 1 = 7No
    B3(3)−5=43(3) - 5 = 4No
    C4(3)−1=114(3) - 1 = 11No
    D4(3)+6=184(3) + 6 = 18Yes

    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.

  3. 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)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.

  4. 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+64x + 6.

Check your answer

Try two more multiples of 3. With x = 6, choice D gives 4(6)+6=304(6) + 6 = 30, and 30 ÷ 3 = 10. With x = 9, it gives 4(9)+6=424(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 85 of 120

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?

  • AR
  • BS
  • CT
  • DU
Show the answer

Correct answer: C

There are 6 spaces, so first divide 100 by 6: 100÷6=16100 \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
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.
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division with a remainder
a=b×q+ra = 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.

  1. 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.

  2. 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.

  3. Divide 100 by 6

    a=b×q+ra = b \times q + r

    100=6×16+4100 = 6 \times 16 + 4

    The arrow makes 16 full trips. That takes 6×16=966 \times 16 = 96 minutes, and it is back on P. There are 4 minutes left over.

  4. 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 86 of 120

2023 Test A Math, question 35

Question

A number line with tick marks every 2 units, labeled −6, −4, −2, 0, 2, 4 and 6. Point J is the dot at −4 and point K is the dot at 6.

On the number line above, point L (not shown) is located on line segment JK so that JL=23LK\mathrm{JL} = \frac{2}{3}\mathrm{LK}. What is the position of point L?

  • A−2
  • B0
  • C2
  • D4
Show the answer

Correct answer: B

JK=6−(−4)=10\mathrm{JK} = 6 - (-4) = 10 units. Let x=JLx = \mathrm{JL}. Then LK=10−x\mathrm{LK} = 10 - x. Substitute these expressions in the given equation:

x=23(10−x)x = \frac{2}{3}(10 - x)

Solve for x:

x=23(10−x)x = \frac{2}{3}(10 - x) Apply the multiplicative inverse property; multiply both sides of the equation by 3.
3x=2(10−x)3x = 2(10 - x) Apply the distributive property; multiply the 2 by 10 and x.
3x=20−2x3x = 20 - 2x Apply the additive inverse property; add 2x2x to both sides of the equation.
5x=205x = 20 Apply the multiplicative inverse property; divide both sides of the equation by 5.
x=4x = 4

So, point L is 4 units to the right of point J: −4+4=0-4 + 4 = 0. Point L is located at 0 on the number line.

Question analysis

This number line question puts J at −4 and K at 6 and asks for the position of a point L on segment JK where JL is two thirds of LK. Students find the segment's length, divide it in a 2-to-3 ratio, and turn a distance into a coordinate. Reading the ratio backwards, or giving a distance instead of a position, leads to wrong answers.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.1c rational number subtraction
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Distances on a number line that crosses zero are grade 7 rational-number work, and splitting a length in a given ratio is grade 7 proportional reasoning. Turning a fraction of one part into a split, then into a position, makes a typical multistep task.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The ratio is stated between two unknown pieces, which is less familiar than a fraction of a whole, and the final step turns a length into a position. Students who stop at JL's length, read 2/3 the wrong way round, or use its 2 as a distance all reach listed points. Sketching five equal shares of 2 units from −4 to 6, two for JL and three for LK, makes the position plain.

  1. Turning words into math moderate Understanding the question

    JL = 2/3 LK compares the two pieces with each other, not a piece with the whole JK, so the 10 units must be split in the ratio 2 : 3, giving JL 4 and LK 6. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Reading a figure, table or graph minor Understanding the question

    J's position, −4, and K's, 6, are read from the number line, whose labeled ticks run every 2 units, and L itself is not drawn. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    Three steps follow in order: the length JK = 6 − (−4), the length JL from the ratio, and finally L's position counted from J rather than from 0. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the length of JK, split it into five equal shares with two for JL and three for LK, then count JL's length to the right from J.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
Subtract the position on the left from the position on the right. The answer is how many units apart the two points are. A distance is never negative.
Splitting a length in a ratio
one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}
A ratio a : b compares two pieces. If a length is split in the ratio a : b, think of it as a + b equal parts. The first piece gets a of them and the second gets b. Divide the whole length by a + b to find one part.

  1. Sort out the question

    Read the number line. Point J is at −4 and point K is at 6. Point L sits somewhere between them. JL is the length from J to L, and LK is the length from L to K. The rule says JL is 23\frac{2}{3} of LK. You need the position of L.

  2. Find the length of JK

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    K is on the right, so subtract J’s position from K’s.

    JK=6−(−4)=6+4=10\mathrm{JK} = 6 - (-4) = 6 + 4 = 10

    Taking away −4 is the same as adding 4. J and K are 10 units apart.

  3. Turn the rule into equal parts

    JL is 23\frac{2}{3} of LK. So for every 2 units in JL, LK has 3 units. That means JK is made of 2 + 3 = 5 equal parts. JL gets 2 parts and LK gets 3.

    one part=whole lengtha+b\text{one part} = \frac{\text{whole length}}{a + b}

    one part=102+3=2\text{one part} = \frac{10}{2 + 3} = 2

    Each part is 2 units long.

  4. Find the two lengths

    JL=2×2=4\mathrm{JL} = 2 \times 2 = 4

    LK=3×2=6\mathrm{LK} = 3 \times 2 = 6

    Watch out: JL is 23\frac{2}{3} of LK, not the other way round. Flipping the ratio makes JL 6 and puts L at 2, choice C.

  5. Find the position of L

    L is 4 units to the right of J. Start at J’s position, −4, and move 4 to the right.

    −4+4=0-4 + 4 = 0

    Watch out: 4 is the length of JL, not where L sits. Answering 4 is choice D. So the answer is B, 0.

Check your answer

Put L at 0 and test the rule. JL = 0 − (−4) = 4, and LK = 6 − 0 = 6. Then 23×6=4\frac{2}{3} \times 6 = 4, which is JL. Also 4 + 6 = 10, the whole length JK.

More questions like this

The original DOE question
Question 92 as printed on page 84 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 92. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 87 of 120

2023 Test A Math, question 41

Question

Yesterday Sarah read 15% of her entire book. Today she read another 17% of the entire book. In lowest terms, what fraction of the book is left for her to read?

  • A725\frac{7}{25}
  • B310\frac{3}{10}
  • C1725\frac{17}{25}
  • D710\frac{7}{10}
Show the answer

Correct answer: C

So far, Sarah has read 15%+17%=32%15\% + 17\% = 32\% of the entire book. That means she has 100%−32%=68%100\% - 32\% = 68\% left to read.

68%=68100=172568\% = \frac{68}{100} = \frac{17}{25}

Question analysis

This percent word problem says Sarah read 15% of her book yesterday and 17% today, and asks what fraction of the book she still has to read, in lowest terms. Students combine percents of the same whole, find the part remaining, and convert a percent to a simplified fraction. Rounding, or giving the part already read, leads to wrong choices.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 percent work often takes a percent of a percent or finds a percent change. Here both percents are of the same whole book, so they combine directly; the rest is a subtraction and simplifying a fraction, lighter than typical.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 33 words.
Corrected from the DOE handbook
  • Missing percent signs in the explanation were 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 percent work is a sum and a subtraction from 100, so the question mostly tests the last step, writing 68% as a fraction in lowest terms. Students who round to 30% read and 70% left to get a friendlier fraction land on a listed choice, as do those who give the part read. Dividing 68 and 100 by their common factor 4 keeps the answer exact.

  1. Turning words into math minor Understanding the question

    Both percents are “of the entire book,” so they simply add to 32%, and “left for her to read” asks for the rest, 100% − 32%, not the part already read. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. An answer in an unexpected form minor Choosing the answer

    The result has to be written as a fraction in lowest terms, so 68/100 must be divided by 4 to reach a form that appears among the choices. (Enright, Morley & Sheehan, 2002)

Strategy: Add the two percents, since both are of the whole book, subtract the total from 100%, then write the result over 100 and simplify it fully.

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

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes “So far, Sarah has read 15 + 17 = 32 of the entire book. That means she has 100 − 32 = 68 left to read.” (p. 143, question 98), with no percent signs, so 32 and 68 read as amounts of the book (pages, say) that the question never gives. The question gives 15% and 17%, and the handbook’s next line begins “68% =”, so the site now reads 15% + 17% = 32% and 100% − 32% = 68%. The answer, 17/25, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction
p%=p100p\% = \frac{p}{100}
p is the percent. Percent means “out of 100,” so p percent is p hundredths. For example, 50% is 50 out of 100, which is one half.
Simplifying a fraction
ab=a÷nb÷n\frac{a}{b} = \frac{a \div n}{b \div n}
a is the top number, b is the bottom, and n is a number that divides both evenly. Dividing the top and bottom by the same number keeps the value the same. A fraction is in lowest terms when no number bigger than 1 divides both.

  1. Sort out the question

    Sarah read 15% of her book yesterday and 17% today. Both are percents of the whole book. You need the part still left to read, as a fraction in lowest terms. That means a fraction that can’t be simplified any more.

  2. Find how much she has read

    15%+17%=32%15\% + 17\% = 32\%

    Both percents are of the same whole book, so you can add them. She has read 32% of it.

  3. Find how much is left

    The whole book is 100%.

    100%−32%=68%100\% - 32\% = 68\%

    Watch out: 32% is the part she has read, not the part left. Rounding 32% to 30% gives 310\frac{3}{10}, choice B.

  4. Write 68% as a fraction

    p%=p100p\% = \frac{p}{100}

    68%=6810068\% = \frac{68}{100}

    This fraction is not in lowest terms yet, because 68 and 100 are both even.

  5. Write it in lowest terms

    The number 4 divides both 68 and 100 evenly.

    ab=a÷nb÷n\frac{a}{b} = \frac{a \div n}{b \div n}

    68100=68÷4100÷4=1725\frac{68}{100} = \frac{68 \div 4}{100 \div 4} = \frac{17}{25}

    Watch out: don’t round 68% to 70%. That gives 710\frac{7}{10}, choice D. So the answer is C, 1725\frac{17}{25}.

Check your answer

Turn the answer back into a percent: 1725=68100\frac{17}{25} = \frac{68}{100}, which is 68%. Then 68% + 32% = 100%, the whole book. Also, 17 and 25 share no factor except 1, so the fraction is in lowest terms.

More questions like this

The original DOE question
Question 98 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 98. 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 88 of 120

2023 Test A Math, question 42

Question

{1,2,3,4,5,…,198,199,200}\{1,2,3,4,5,\ldots,198,199,200\}

How many members of the set shown above are multiples of 6 but not multiples of 9?

  • A11
  • B13
  • C20
  • D22
Show the answer

Correct answer: D

First, list the first several multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54 . . .

Every third multiple (i.e., 18, 36, 54, . . .) is also a multiple of 9. Thus, 23\frac{2}{3} of the multiples of 6 are not multiples of 9.

Next, calculate how many multiples of 6 are in the given number range. The last multiple of 6 would be 198. So, there are a total of 1986=33\frac{198}{6} = 33 multiples of 6. Since 23\frac{2}{3} of those are not multiples of 9, the answer is 23(33)=22\frac{2}{3}(33) = 22.

Question analysis

This number properties question shows the set of whole numbers from 1 to 200 and asks how many of them are multiples of 6 but not multiples of 9. Students count multiples in a range and remove the overlap, the numbers divisible by both. Finding the right common multiple, and removing only the overlap, is where it gets tricky.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Multiples and common multiples are grade 6 content, but counting the members of a set that meet one condition and fail another is rare in grade 7 work. The overlap between the two kinds of multiples must be found and removed exactly.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 6, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Counting multiples of 6 is routine; the harder decision is how to remove the ones that are also multiples of 9. Those are the multiples of 18, the least common multiple, not of 54 or 15, and students who count them correctly must still subtract them rather than report them. Listing 6, 12, 18, 24, 30, 36 shows the pattern, two kept for every one removed, which also gives the answer as two thirds of 33.

  1. No obvious method moderate Working it out

    The stem never mentions a common multiple: students must see that the multiples of 6 that are also multiples of 9 are the multiples of 18, every third multiple of 6. (Schoenfeld, 1985)

  2. Much to keep track of minor Working it out

    Three counts are in play at once, 33 multiples of 6 up to 198, 11 multiples of 18 up to 198, and their difference, and two of them are listed choices. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Turning words into math minor Understanding the question

    The bold not in “multiples of 6 but not multiples of 9” asks for what is left after a removal, a set difference rather than a count of the numbers removed. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Count the multiples of 6 up to 200, find the smallest number that is a multiple of both 6 and 9, count its multiples in the set, and subtract.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting multiples
last multiple=n×k\text{last multiple} = n \times k
n is the number whose multiples you count, such as 6. The first multiple is n × 1, the second is n × 2, and so on. So if the largest multiple in the set is n × k, there are k multiples in all.
Least common multiple
LCM(6,9)=18\text{LCM}(6, 9) = 18
The least common multiple, or LCM, is the smallest number that two numbers both divide evenly. A number is a multiple of both 6 and 9 exactly when it is a multiple of 18.

  1. Sort out the question

    The set holds the whole numbers from 1 to 200. A multiple of 6 is a number that 6 divides evenly, like 6, 12 or 18. You need the multiples of 6 that are not multiples of 9. So count all the multiples of 6. Then take away those that are also multiples of 9.

  2. Count the multiples of 6

    200 is not a multiple of 6, but 198 is. So 198 is the last multiple of 6 in the set.

    last multiple=n×k\text{last multiple} = n \times k

    198=6×33198 = 6 \times 33

    There are 33 multiples of 6.

  3. Find the numbers to take away

    List the multiples of 6: 6, 12, 18, 24, 30, 36. The first one 9 also divides is 18, then 36. So every third multiple of 6 is also a multiple of 9.

    LCM(6,9)=18\text{LCM}(6, 9) = 18

    Watch out: the shared multiples are multiples of 18, not of 6 + 9 = 15. Using 15 leads to 13 or 20, choices B and C.

  4. Count the multiples of 18

    The last multiple of 18 in the set is also 198.

    198=18×11198 = 18 \times 11

    There are 11 of them. Watch out: these are the numbers you take away, not the answer. Stopping here gives 11, choice A.

  5. Take away the multiples of 18

    33−11=2233 - 11 = 22

    There are 22 multiples of 6 left over. So the answer is D, 22.

Check your answer

Every third multiple of 6 is a multiple of 18, so two out of every three multiples of 6 stay. Two thirds of 33 is 23×33=22\frac{2}{3} \times 33 = 22. That matches. Test one: 30 is a multiple of 6, and 30 ÷ 9 is not a whole number, so 30 counts.

More questions like this

The original DOE question
Question 99 as printed on page 86 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 99. Open page 86 of the handbook (PDF)

Question 89 of 120

2023 Test A Math, question 55

Question

What is the least of four consecutive integers whose sum is 58?

  • A1
  • B2
  • C12
  • D13
Show the answer

Correct answer: D

Let n represent the least of the four integers. The other three integers are (n+1)(n + 1), (n+2)(n + 2), and (n+3)(n + 3). Write an equation to solve for n:

n+(n+1)+(n+2)+(n+3)=58n + (n + 1) + (n + 2) + (n + 3) = 58

4n+6=584n + 6 = 58

4n=524n = 52

n=13n = 13

Question analysis

This number properties question asks for the least of four consecutive integers that add up to 58. Students write the integers in terms of one unknown and solve a two-step equation, or reason from the average of the four. Starting the list at n + 1 instead of n, which shifts every integer by one, is the main trap.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.6 variables representing unknown numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an equation of the form px + q = r is core grade 7 work. Expressing four consecutive integers in terms of the least one is a familiar setup, so this is a routine three-step grade 7 problem.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 6, 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

One equation, 4n + 6 = 58, solves the question, and dividing 58 by 4 to find the middle of the four, 14.5, is quicker still. The only real risk is the count: writing the integers as n + 1 through n + 4 and reporting n gives a least integer one too small. The two smallest choices match no reasonable working, so they draw only guesses. Adding the four integers back up to 58 is a ten-second check.

  1. Turning words into math minor Understanding the question

    The phrase “four consecutive integers” has to become n, n + 1, n + 2 and n + 3, with n the least, so the sum is 4n + 6, not 4n + 10. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    The choices 12 and 13 differ by one, the size of a fence-post error, so only checking that 13 + 14 + 15 + 16 = 58 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Call the least integer n, write the next three as n + 1, n + 2 and n + 3, add them and solve; or divide 58 by 4 to find the middle of the four.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive integers
n,n+1,n+2,n+3n,\; n + 1,\; n + 2,\; n + 3
Integers are whole numbers and their negatives, like −2, 0 and 7. Consecutive integers come one right after another, like 4, 5, 6, 7. If n is the least one, each next one is 1 more than the one before.

  1. Sort out the question

    You need four consecutive integers, numbers in a row like 7, 8, 9, 10. Together they add up to 58. You need the least, or smallest, of the four.

  2. Name the four numbers

    Let n be the least number. Then the other three are 1, 2 and 3 more than n.

    n,n+1,n+2,n+3n,\; n + 1,\; n + 2,\; n + 3

  3. Write and simplify the equation

    Add the four numbers and set the sum equal to 58.

    n+(n+1)+(n+2)+(n+3)=58n + (n + 1) + (n + 2) + (n + 3) = 58

    4n+6=584n + 6 = 58

    To combine, count the n’s: there are four of them. The plain numbers make 1 + 2 + 3 = 6.

  4. Solve for n

    Subtract 6 from both sides. Then divide both sides by 4.

    4n=58−6=524n = 58 - 6 = 52

    n=52÷4=13n = 52 \div 4 = 13

    Watch out: if you call the numbers n + 1 to n + 4, you get n = 12. Then 12 is not one of the four numbers; the least one is 13. Answering 12 is choice C.

    So the answer is D, 13.

Check your answer

Add them up: 13+14+15+16=5813 + 14 + 15 + 16 = 58. It works. Another way: 58÷4=14.558 \div 4 = 14.5 is their average. It falls halfway between the two middle numbers, 14 and 15. So the four numbers are 13, 14, 15 and 16.

More questions like this

The original DOE question
Question 112 as printed on page 90 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 112. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 90 of 120

2023 Test A Math, question 56

Question

Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. The line is shaded from the unlabeled tick halfway between 0 and 1 to the unlabeled tick halfway between 1 and 2, with closed dots at both ends.

If all possible values of x are indicated by the shaded part of the number line above, which number line best shows all possible values of 1x\frac{1}{x}?

  • ANumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 1 and 2 to the tick halfway between 2 and 3, with closed dots at both ends.
  • BNumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from 1 to 2, with closed dots at both ends.
  • CNumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 0 and 1 to 2, with closed dots at both ends.
  • DNumber line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from a closed dot between the tick halfway between 0 and 1 and the tick at 1, about two-thirds of the way from 0 to 1, to a closed dot at 2.
Show the answer

Correct answer: D

The shaded region goes from 12\frac{1}{2} to 32\frac{3}{2}. So, the possible values of 1x\frac{1}{x} would be between the reciprocal of 12\frac{1}{2} and the reciprocal of 32\frac{3}{2}, which are 2 and 23\frac{2}{3}, respectively. Find the number line in the answer options that has a shaded region that ranges from 23\frac{2}{3} to 2, which is option D.

Question analysis

This number line question shades all possible values of x from one half to three halves and asks which number line shows all possible values of 1/x. Students find reciprocals of fractions and work out how a whole range of values changes. Keeping track of what happens to each end of the range, and placing fractions accurately on the line, makes it tricky.

NYS standards
  • 6.NS.1 dividing fractions by fractions
  • 6.NS.7a inequality as relative position
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Reciprocals of single fractions are grade 6 work. Applying them to every value in a shaded range at once, and working out where the new range begins and ends, is reasoning about a whole set of numbers that grade 7 classes rarely practice.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 12, 27 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Few students have flipped an interval before, so the question rewards testing its two ends and noticing that the order reverses. The harder part comes at the end: 2/3 falls between the tick marks, and a graph ending at the nearby 1/2 tick looks right at a glance. Students who believe a reciprocal is always at least 1, or who shift the interval instead of flipping it, reach the other graphs.

  1. Reading a figure, table or graph moderate Understanding the question

    The shaded segment's ends sit on unlabeled half-unit ticks, 1/2 and 3/2, and each choice must be read the same way, with 2/3 falling between ticks rather than on one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. A very close wrong answer moderate Choosing the answer

    The graph from 1/2 to 2 matches the right graph's right end and differs at the left end by only a sixth of a unit, 1/2 against 2/3. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    The question asks about 1/x for every x in a range, so a whole interval must be transformed, and its ends swap: the smallest x gives the largest 1/x. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Take the reciprocal of each end of the shaded part, ask whether a bigger x gives a bigger or a smaller 1/x, and look for the graph whose endpoints are exactly those values.

Traps

  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reciprocal
reciprocal of ab=ba\text{reciprocal of } \frac{a}{b} = \frac{b}{a}
a and b are the top and bottom of a fraction. The reciprocal flips the fraction upside down. The number 1 over x is the reciprocal of x. A number times its reciprocal always makes 1.

  1. Sort out the question

    x can be any number in the shaded part of the first number line. You need the number line that shows every possible value of 1x\frac{1}{x}, the reciprocal of x. The reciprocal is the number flipped upside down, like 12\frac{1}{2} for 2.

  2. Read the shaded part

    The ticks come every half unit. The shading starts at the tick halfway between 0 and 1, which is 12\frac{1}{2}. It ends halfway between 1 and 2, which is 32\frac{3}{2}, or 1121\frac{1}{2}. Both dots are filled in, so both ends count.

  3. Find the reciprocals of the ends

    Flip each end of the shaded part.

    reciprocal of ab=ba\text{reciprocal of } \frac{a}{b} = \frac{b}{a}

    reciprocal of 12=21=2\text{reciprocal of } \frac{1}{2} = \frac{2}{1} = 2

    reciprocal of 32=23\text{reciprocal of } \frac{3}{2} = \frac{2}{3}

    So the ends of the new shading are 2 and 23\frac{2}{3}.

  4. Notice that the ends swap

    When x gets bigger, 1x\frac{1}{x} gets smaller. So the smallest x, 12\frac{1}{2}, gives the biggest value, 2. The biggest x, 32\frac{3}{2}, gives the smallest, 23\frac{2}{3}. So 1x\frac{1}{x} runs from 23\frac{2}{3} to 2, ends included.

  5. Find the matching number line

    23\frac{2}{3} lies between the half tick and 1, a third of the way from 12\frac{1}{2} to 1. Line D starts there and ends at 2. Watch out: line C starts at the half tick, 12\frac{1}{2}, not at 23\frac{2}{3}. Line B starts at 1.

    So the answer is D, the line shaded from 23\frac{2}{3} to 2.

Check your answer

Test a value in the middle. x = 1 is shaded, and 11=1\frac{1}{1} = 1. On line D, 1 is shaded too. Test an end: 32×23=1\frac{3}{2} \times \frac{2}{3} = 1, so 23\frac{2}{3} really is the reciprocal of 32\frac{3}{2}.

More questions like this

The original DOE question
Question 113 as printed on page 91 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 91, question 113. Open page 91 of the handbook (PDF)

Question 91 of 120

2023 Test A Math, question 57

Question

Chanelle selects a marble from a container and then returns it to the container. When she does this 3 times, the probability of choosing a red marble all 3 times is 1216\frac{1}{216}. Based on this information, what is the probability of Chanelle choosing a red marble the first time she selects a marble?

  • A172\frac{1}{72}
  • B136\frac{1}{36}
  • C18\frac{1}{8}
  • D16\frac{1}{6}
Show the answer

Correct answer: D

Let 1x\frac{1}{x} be the probability of Chanelle selecting a red marble the first time. The probability of selecting a red marble three times is 1x×1x×1x\frac{1}{x} \times \frac{1}{x} \times \frac{1}{x}, which is given as 1216\frac{1}{216}. Since 216=6×6×6216 = 6 \times 6 \times 6, the value of x is 6, and the probability of Chanelle selecting a red marble the first time is 16\frac{1}{6}.

Question analysis

This probability question says Chanelle draws a marble and puts it back, three times, and the chance of red all three times is 1/216. It asks for the chance of red on her first draw. Students reverse the multiplication rule for repeated independent draws, which means finding a cube root, and dividing by 3 is the tempting shortcut.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 8.EE.2 square and cube roots
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Multiplying probabilities for repeated draws with replacement is grade 7 compound probability. Running it backward, from three draws to one, needs a cube root, a grade 8 idea, though the numbers are friendly and the work is short once set up.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 55 words.
Students, last 30 days
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 three draws are seen as one chance cubed, the question is a cube root, and 216 is a perfect cube many students know. The risk lies in undoing the cube the wrong way: dividing by 3 because there are three draws, or stopping at 36 after one division by 6. A quick check, 1/6 × 1/6 × 1/6 = 1/216, confirms the answer.

  1. Something you must already know moderate Working it out

    Students must know that three draws with the marble returned each time multiply, p × p × p, and must recognize 216 as 6 × 6 × 6; the stem states neither. (Anderson & Krathwohl, 2001)

  2. No obvious method minor Working it out

    The usual direction runs from one draw to three; here the three-draw chance is given and the bold first asks for one draw, so students work backward and undo a cube. (Schoenfeld, 1985)

Strategy: Call the chance of red on one draw p, so three draws with replacement give p × p × p; then find the number whose cube equals the given probability.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of repeated picks
P(red 3 times)=p×p×pP(\text{red 3 times}) = p \times p \times p
p is the chance of red on one pick. Chanelle puts the marble back each time, so every pick has the same chance p, and one pick doesn’t change the next. The chance of red on all three picks is the three chances multiplied.

  1. Sort out the question

    Chanelle picks a marble and puts it back, 3 times. The chance of red all 3 times is 1216\frac{1}{216}. You need the chance of red on just the first pick. Call that chance p.

  2. Write the three picks as a product

    She puts each marble back, so the container is the same every time. Each pick has the same chance, p.

    P(red 3 times)=p×p×pP(\text{red 3 times}) = p \times p \times p

    p×p×p=1216p \times p \times p = \frac{1}{216}

  3. Find the number that fits

    You need a fraction that, multiplied by itself three times, makes 1216\frac{1}{216}. Its top is 1, since 1 × 1 × 1 = 1. For the bottom, find a number that times itself three times makes 216.

    6×6×6=36×6=2166 \times 6 \times 6 = 36 \times 6 = 216

    So the bottom is 6, and p is 16\frac{1}{6}.

  4. Avoid the tempting shortcuts

    Watch out: dividing 216 by 3 gives 72, choice A. Three picks means multiplying three times, not dividing by 3. Also, 136\frac{1}{36} is the chance of red twice in a row, choice B.

    So the answer is D, 16\frac{1}{6}.

Check your answer

Multiply it out: 16×16×16=1216\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216}. That matches the question. Choice A fails fast: 172×172\frac{1}{72} \times \frac{1}{72} is already far smaller than 1216\frac{1}{216}.

More questions like this

The original DOE question
Question 114 as printed on page 91 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 91, question 114. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 92 of 120

2023 Test B Math, question 3

Question

A kindergarten teacher has 72 crayons, 60 pencils, and 84 sheets of paper to distribute to the children in her class. If each child receives an equal number of each item and there are no items remaining, what is the greatest possible number of children in the class?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 12

The greatest possible number of children in the class is the greatest common factor of 72, 60, and 84. Write the prime factorization of each number:

72=23×3272 = 2^{3} \times 3^{2}

60=22×3×560 = 2^{2} \times 3 \times 5

84=22×3×784 = 2^{2} \times 3 \times 7

Then, find the factors that all three have in common, which is 22×3=122^{2} \times 3 = 12.

Question analysis

A kindergarten teacher shares 72 crayons, 60 pencils and 84 sheets of paper so that every child gets the same number of each item, with nothing left over. This grid-in asks for the greatest possible number of children in the class. Students must recognize a greatest common factor problem and not confuse it with a least common multiple.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Greatest common factors are grade 6 content, normally of two numbers. This one uses three numbers and a word setting that must be translated, but the factoring is short, so the demand stays below typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 45 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The story has to be recognized as a sharing problem before any arithmetic starts, and the word greatest points to the largest divisor shared by all three counts. Students who spot a small shared factor can stop too soon, while those who think of a common multiple head the wrong way entirely. Checking that 72, 60 and 84 all divide by the entry, and that no larger number does, settles it.

  1. Turning words into math moderate Understanding the question

    Nothing names a greatest common factor; “each child receives an equal number of each item and there are no items remaining” means the class size must divide 72, 60 and 84 exactly. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Finding the greatest common factor of three numbers, by prime factorization or by listing factors, is a grade 6 method the question assumes rather than explains. (Anderson & Krathwohl, 2001)

Strategy: Write 72, 60 and 84 as products of primes, multiply the primes all three share, and check that each amount divides evenly by the result.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor
GCF=product of the shared prime factors\text{GCF} = \text{product of the shared prime factors}
A factor of a number divides it with nothing left over. The GCF is the greatest number that is a factor of every number in the group. A prime number has only 1 and itself as factors, like 2, 3, 5 and 7. Multiply the primes that all the numbers share, as many times as each number has them.

  1. Sort out the question

    The teacher has 72 crayons, 60 pencils and 84 sheets of paper. Every child gets the same number of each item, with none left over.

    So the number of children must divide 72, 60 and 84 with no remainder. A number like that is a common factor. You need the greatest one.

  2. Break each number into primes

    Write each number as a product of primes. A product is the answer to a multiplication.

    72=2×2×2×3×372 = 2 \times 2 \times 2 \times 3 \times 3

    60=2×2×3×560 = 2 \times 2 \times 3 \times 5

    84=2×2×3×784 = 2 \times 2 \times 3 \times 7

    You can check each one by multiplying it back out.

  3. Pick out the shared primes

    Look for the primes that appear in all three lists. Each list has at least two 2s and one 3. Only 72 has a third 2 and a second 3. The 5 and the 7 appear in one list each.

    GCF=2×2×3=12\text{GCF} = 2 \times 2 \times 3 = 12

    Watch out: keep every shared prime. Using only one 2 gives 6. Leaving out the 3 gives 4.

  4. Give the answer

    Watch out: 2,520 is the smallest number that all three numbers divide into. That is the least common multiple, a different idea. The class can’t have 2,520 children when there are only 60 pencils. So you would grid 12.

Check your answer

Share the items among 12 children. Each child gets 72÷12=672 \div 12 = 6 crayons, 60÷12=560 \div 12 = 5 pencils and 84÷12=784 \div 12 = 7 sheets. Nothing is left over. The numbers 6, 5 and 7 share no factor but 1, so no bigger class works.

More questions like this

The original DOE question
Question 60 as printed on page 191 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 60. Open page 191 of the handbook (PDF)

Question 93 of 120

2023 Test B Math, question 5

Question

Solve the equation for x:

(21)(14)x(49)(48)=0.875\frac{(21)(14)x}{(49)(48)} = 0.875

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 7

(21)(14)x(49)(48)=0.875\frac{(21)(14)x}{(49)(48)} = 0.875

(2149)(1448)x=0.875\left(\frac{21}{49}\right)\left(\frac{14}{48}\right)x = 0.875

(37)(724)x=0.875\left(\frac{3}{7}\right)\left(\frac{7}{24}\right)x = 0.875

(18)x=0.875\left(\frac{1}{8}\right)x = 0.875

x=(0.875)(8)=7x = (0.875)(8) = 7

Question analysis

This equation puts x in a fraction with (21)(14) on top and (49)(48) below, set equal to the decimal 0.875. The grid-in asks for the value of x. Students do best to cancel common factors before multiplying anything out, and it helps to know the fraction that 0.875 equals.

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.2c properties for multiplying and dividing
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Seventh graders solve equations like ax = b, usually with a simple coefficient. Here the coefficient is a product of four numbers that is best cancelled first, and a decimal has to be matched with a fraction.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 6 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-unknown equation dressed in large numbers, and the work is in reducing them without slipping. Students who cancel early find that the whole coefficient is one eighth and that 0.875 is seven eighths, which makes the last step mental. Those who multiply out face a four-digit product and a decimal division, and with nothing to check against, an early slip goes unnoticed.

  1. Error-prone arithmetic moderate Working it out

    Multiplied out, the coefficient is 294 over 2,352 and the right side is the decimal 0.875; both invite slips unless 21/49 and 14/48 are cancelled first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. No obvious method minor Working it out

    The quick route pairs 21 with 49 and 14 with 48 before multiplying, so that 3/7 × 7/24 collapses to 1/8; the layout of the fraction does not suggest that pairing. (Schoenfeld, 1985)

  3. An answer in an unexpected form minor Choosing the answer

    The result is entered with no choices, so a coefficient’s denominator or a quotient taken the wrong way looks as complete as the right value. (Enright, Morley & Sheehan, 2002)

Strategy: Cancel before multiplying: reduce 21 over 49 and 14 over 48, multiply the simplified fractions, then undo that coefficient, writing 0.875 as a fraction if it helps.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying fractions
ab×cd=a×cb×d\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}
a and c are the tops, or numerators. b and d are the bottoms, or denominators. Multiply the tops together and the bottoms together. It also works backward: you can split one fraction into two smaller ones. That lets you simplify before the numbers get big.

  1. Sort out the question

    You need the value of x. On top, 21, 14 and x are multiplied. The bottom is 49 times 48. The right side is 0.875.

    Multiplying 49 by 48 makes a big number. It is easier to simplify first.

  2. Split the fraction into two

    Use the rule for multiplying fractions backward. Pair 21 with 49, and 14 with 48.

    21×1449×48=2149×1448\frac{21 \times 14}{49 \times 48} = \frac{21}{49} \times \frac{14}{48}

    Now the equation says these two fractions times x equal 0.875.

  3. Simplify each fraction

    To simplify, divide the top and bottom by the same number. Both 21 and 49 divide by 7. Both 14 and 48 divide by 2.

    2149=371448=724\frac{21}{49} = \frac{3}{7} \qquad \frac{14}{48} = \frac{7}{24}

    Watch out: 48 ÷ 2 is 24, not 12. Writing 712\frac{7}{12} leads to 3.5.

  4. Multiply the simple fractions

    37×724=3×77×24=324=18\frac{3}{7} \times \frac{7}{24} = \frac{3 \times 7}{7 \times 24} = \frac{3}{24} = \frac{1}{8}

    The 7 on top and the 7 on the bottom cancel, since 7 ÷ 7 = 1. Then divide 3 and 24 by 3 to get 18\frac{1}{8}. So the equation becomes:

    18×x=0.875\frac{1}{8} \times x = 0.875

  5. Solve for x

    One eighth of x is 0.875. To undo dividing by 8, multiply both sides by 8.

    x=0.875×8=7x = 0.875 \times 8 = 7

    Watch out: 8 is not the answer, and don’t divide 0.875 by 8. So you would grid 7.

Check your answer

0.875 is the same as 78\frac{7}{8}. One eighth of 7 is 78\frac{7}{8}, so x = 7 works. You can also multiply it all out: 21×14×7=2,05821 \times 14 \times 7 = 2{,}058 and 49×48=2,35249 \times 48 = 2{,}352. Then 2,058÷2,352=0.8752{,}058 \div 2{,}352 = 0.875.

More questions like this

The original DOE question
Question 62 as printed on page 192 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 62. Open page 192 of the handbook (PDF)

Question 94 of 120

2023 Test B Math, question 6

Question

What is the value of 314−123−(−412)3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)?

  • A−3112-3\frac{1}{12}
  • B−21112-2\frac{11}{12}
  • C61126\frac{1}{12}
  • D95129\frac{5}{12}
Show the answer

Correct answer: C

The first step is to rewrite the problem replacing all subtraction with adding the opposite:

314−123−(−412)3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)

314+(−123)+4123\frac{1}{4} + \left(-1\frac{2}{3}\right) + 4\frac{1}{2}

Then, rewrite all the fractions using a common denominator of 12, which is the least common multiple of the fractions’ denominators, 4, 3, and 2:

3312+(−1812)+46123\frac{3}{12} + \left(-1\frac{8}{12}\right) + 4\frac{6}{12}

Next, combine the positive numbers and, finally, add the negative number to the sum by subtracting the number with the least absolute value from the number with the greatest absolute value and using the sign of the number with the greatest absolute value:

7912+(−1812)7\frac{9}{12} + \left(-1\frac{8}{12}\right)

61126\frac{1}{12}

Question analysis

What is 3 1/4 − 1 2/3 − (−4 1/2)? This question on rational-number arithmetic combines mixed numbers with denominators of 4, 3 and 2 and ends by subtracting a negative number. Students must turn that last subtraction into an addition and find a common denominator before combining.

NYS standards
  • 7.NS.1c rational number subtraction
  • 7.NS.1d properties for adding and subtracting
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Subtracting negative mixed numbers is core grade 7 content. This expression is typical in length, but it needs a common denominator of 12 and careful handling of the rule for subtracting a negative.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 10, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The expression is short, but it mixes a subtraction that stays a subtraction with one that becomes an addition, and that difference is what most wrong answers miss. The fractions add a second load, since twelfths are needed before anything combines. Estimating first, 3 − 1.7 + 4.5 is about 6, shows the result is positive, which rules out both negative values.

  1. Error-prone arithmetic moderate Working it out

    Three mixed numbers with denominators 4, 3 and 2 must become twelfths, and the last term, −(−4 1/2), changes sign while the middle subtraction does not. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Rewriting the double negative, renaming each fraction, adding 3 1/4 and 4 1/2, then taking away 1 2/3 are four steps, each able to carry a slip forward. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Change subtracting −4 1/2 into adding 4 1/2, rename every fraction in twelfths, add the two positive numbers, then take away 1 2/3.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number
a−(−b)=a+ba - (-b) = a + b
a and b stand for any numbers. Taking away a negative number is the same as adding its opposite, the same number with the other sign. Think of it this way: taking away a debt leaves you richer.
Equivalent fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a is the top and b is the bottom of a fraction. Multiplying both by the same number n does not change its value. Use it to give fractions the same bottom number, so you can add or subtract them.

  1. Sort out the question

    You need the value of 314−123−(−412)3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right). Each number is a mixed number, a whole number plus a fraction. There are two subtractions. The second one takes away a negative number.

  2. Change the double negative

    Use the rule for subtracting a negative. Taking away −412-4\frac{1}{2} is the same as adding 4124\frac{1}{2}.

    314−123+4123\frac{1}{4} - 1\frac{2}{3} + 4\frac{1}{2}

    The first minus sign stays as it is. Watch out: if you subtract 4124\frac{1}{2} instead, you get −21112-2\frac{11}{12}, choice B.

  3. Give the fractions one denominator

    The denominator is the bottom number of a fraction. Here they are 4, 3 and 2. The smallest number all three divide into is 12. So change each fraction to twelfths.

    14=31223=81212=612\frac{1}{4} = \frac{3}{12} \qquad \frac{2}{3} = \frac{8}{12} \qquad \frac{1}{2} = \frac{6}{12}

    The problem is now:

    3312−1812+46123\frac{3}{12} - 1\frac{8}{12} + 4\frac{6}{12}

  4. Add the two positive numbers

    You may add and subtract in any order, as long as each number keeps its sign. Add the two positive numbers first.

    3312+4612=79123\frac{3}{12} + 4\frac{6}{12} = 7\frac{9}{12}

    That is 3 + 4 = 7 wholes and 3 + 6 = 9 twelfths.

  5. Take away the last number

    7912−1812=61127\frac{9}{12} - 1\frac{8}{12} = 6\frac{1}{12}

    That is 7 − 1 = 6 wholes and 9 − 8 = 1 twelfth. So the answer is C, 61126\frac{1}{12}.

Check your answer

Estimate with decimals: 3.25−1.67+4.5≈6.083.25 - 1.67 + 4.5 \approx 6.08. And 61126\frac{1}{12} is about 6.08 too. The answer must be positive, so A and B are out. Choice D comes from adding all three numbers.

More questions like this

The original DOE question
Question 63 as printed on page 193 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 63. Open page 193 of the handbook (PDF)

Question 95 of 120

2023 Test B Math, question 9

Question

Bryan completed a 100-meter race in 11.74 seconds. Luis completed the same race in 11.69 seconds. What was the difference between their times, expressed as a fraction of a second?

  • A120\frac{1}{20}
  • B110\frac{1}{10}
  • C320\frac{3}{20}
  • D15\frac{1}{5}
Show the answer

Correct answer: A

First, subtract Luis’s time from Bryan’s time: 11.74−11.69=0.0511.74 - 11.69 = 0.05.

Then, convert to a fraction:

0.05=5100=1200.05 = \frac{5}{100} = \frac{1}{20}.

Question analysis

Bryan ran a 100-meter race in 11.74 seconds and Luis ran the same race in 11.69 seconds. Students find the difference between their times and express it as a fraction of a second. It tests subtracting decimals and rewriting a decimal as a fraction in lowest terms.

NYS standards
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Subtracting decimals and writing a decimal as a fraction are skills from before grade 7. The two times are close, so the subtraction is quick, and the only care needed is with place value and simplifying.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 5, 32 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two quick skills settle this question: lining up the decimals to subtract, and naming the result as a fraction in lowest terms. The two times differ only in their last two digits, so the subtraction is short, but it needs a borrow that some students skip. The fraction step is where a decimal’s digit gets mistaken for a denominator.

  1. Error-prone arithmetic minor Working it out

    Subtracting 11.69 from 11.74 needs a borrow in the hundredths, 14 − 9, after which the tenths cancel to 0, the step where place value goes wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    The difference, 0.05, must be read as five hundredths and reduced as a fraction; the question gives no help with converting a decimal. (Anderson & Krathwohl, 2001)

Strategy: Line up the decimal points, subtract, write the difference as a number of hundredths over 100, and reduce the fraction.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths
a decimal with two places=its digits100\text{a decimal with two places} = \frac{\text{its digits}}{100}
The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths, and you can write its digits over 100. For example, 0.37 is 37/100. Then simplify by dividing the top and bottom by the same number.

  1. Sort out the question

    Bryan ran the race in 11.74 seconds. Luis ran it in 11.69 seconds. The difference is how much longer one time is than the other. You need it as a fraction of a second.

  2. Subtract the two times

    Subtract the smaller time from the larger one. Line up the decimal points.

    11.74−11.69=0.0511.74 - 11.69 = 0.05

    In the hundredths place, 4 is less than 9. So regroup: trade 1 tenth for 10 hundredths. Then 14 − 9 = 5 hundredths. The tenths are now 6 − 6 = 0.

    Watch out: don’t just take the smaller digit from the larger one in each place. That gives 0.15, which is 320\frac{3}{20}, choice C.

  3. Write the difference as a fraction

    The 5 is in the hundredths place, so 0.05 is 5 hundredths.

    0.05=51000.05 = \frac{5}{100}

    Now simplify. Divide the top and bottom by 5.

    5÷5100÷5=120\frac{5 \div 5}{100 \div 5} = \frac{1}{20}

  4. Give the answer

    Watch out: 0.05 is not 15\frac{1}{5}. One fifth is 0.2, which is four times as big. So the answer is A, 120\frac{1}{20}.

Check your answer

Turn 120\frac{1}{20} back into a decimal: 1÷20=0.051 \div 20 = 0.05. Then add it to Luis’s time: 11.69+0.05=11.7411.69 + 0.05 = 11.74. That is Bryan’s time, so the difference is right.

More questions like this

The original DOE question
Question 66 as printed on page 193 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 66. Open page 193 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 96 of 120

2023 Test B Math, question 11

Question

n=0.13+207100n = 0.13 + \frac{207}{100}

What is the value of n expressed as a fraction?

  • A13\frac{1}{3}
  • B115\frac{11}{5}
  • C94\frac{9}{4}
  • D2310\frac{23}{10}
Show the answer

Correct answer: B

n=0.13+207100=n = 0.13 + \frac{207}{100} =

13100+207100=\frac{13}{100} + \frac{207}{100} =

220100=115\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.
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.

  1. 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)

  2. 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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths
a decimal with two places=its digits100\text{a decimal with two places} = \frac{\text{its digits}}{100}
The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths. Write its digits over 100. For example, 0.47 is 47/100.
Adding fractions with the same bottom
ac+bc=a+bc\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}
a and b are the tops. c is the shared bottom, the denominator. When the bottoms match, the pieces are the same size. So add the tops and keep the bottom.

  1. Sort out the question

    n equals 0.13 plus 207100\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.

  2. Write the decimal as a fraction

    The 3 in 0.13 is in the hundredths place. So 0.13 is 13 hundredths.

    0.13=131000.13 = \frac{13}{100}

    Now both numbers are in hundredths. Watch out: 0.13 is not 13\frac{1}{3}. That only copies the digits, choice A.

  3. Add the hundredths

    Use the rule for adding fractions. The bottoms match, so add the tops and keep the 100 on the bottom.

    13100+207100=13+207100=220100\frac{13}{100} + \frac{207}{100} = \frac{13 + 207}{100} = \frac{220}{100}

  4. Simplify the fraction

    220100\frac{220}{100} is not a choice, so simplify it. Find the greatest number that divides both 220 and 100. That is 20.

    220÷20100÷20=115\frac{220 \div 20}{100 \div 20} = \frac{11}{5}

    So the answer is B, 115\frac{11}{5}.

Check your answer

Change both to decimals. 115=11÷5=2.2\frac{11}{5} = 11 \div 5 = 2.2. And 0.13+2.07=2.200.13 + 2.07 = 2.20. They match. Choices C and D are close, but 94=2.25\frac{9}{4} = 2.25 and 2310=2.3\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 97 of 120

2023 Test B Math, question 14

Question

Each number in a sequence is 3 more than twice the number that comes just before it. If 93 is a number in the sequence, what number comes just before it?

  • A92
  • B90
  • C48
  • D45
Show the answer

Correct answer: D

Let n represent a number in the sequence, and let x represent the number that comes just before n:

n=3+2xn = 3 + 2x

93=3+2x93 = 3 + 2x

90=2x90 = 2x

45=x45 = x

Question analysis

In a sequence, each number is 3 more than twice the number just before it, and 93 is one of the numbers. Students find the number that comes just before 93. That means running the rule backward, undoing the adding and the doubling in reverse order.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students routinely solve two-step equations. The demand here is translating the wordy sequence rule into one, or running the rule backward in the right order, rather than any harder content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 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 arithmetic is two short steps, so the question rests on reading a wordy rule correctly and running it in reverse. Each wrong choice marks a different place to go wrong: ignoring the rule, stopping after the subtraction, or undoing the adding with more adding. Checking the answer forward, twice the number plus 3, confirms it reaches 93.

  1. Turning words into math moderate Understanding the question

    “Each number … is 3 more than twice the number that comes just before it” must become 2x + 3 = 93, with x the earlier number, not the later one. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method minor Working it out

    The rule is stated forward, from a number to the next, but the question runs backward, so each operation has to be undone in reverse order: subtract 3, then halve. (Schoenfeld, 1985)

Strategy: Call the earlier number x, write the rule as an expression in x set equal to 93, and solve it by undoing each operation in reverse order.

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.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule
n=2x+3n = 2x + 3
x is any number in the sequence, a list that follows a rule. n is the number right after it. Twice means times 2, so twice x is 2x. Then 3 more means add 3. The same rule links every number to the one after it.

  1. Sort out the question

    A sequence is a list of numbers that follows a rule. Here, each number is “3 more than twice the number that comes just before it.” 93 is in the list. You need the number right before 93. Call it x.

  2. Write the rule as an equation

    Twice x is 2x. Three more than that is 2x + 3. The rule says this equals the next number, which is 93.

    n=2x+3n = 2x + 3

    93=2x+393 = 2x + 3

  3. Undo the adding of 3

    The rule doubles x first, then adds 3. To go backward, undo the steps in reverse order. First, subtract 3 from both sides.

    93−3=2x93 - 3 = 2x

    90=2x90 = 2x

    Watch out: 90 is twice the number, not the number itself. Stopping here gives choice B. And don’t add 3, which leads to 48, choice C.

  4. Undo the doubling

    Now divide both sides by 2.

    x=90÷2=45x = 90 \div 2 = 45

    Watch out: “just before” means just before in this list. It does not mean one less than 93, so 92 is wrong. So the answer is D, 45.

Check your answer

Run the rule forward from 45. Twice 45 is 90, and 3 more is 93: 2×45+3=932 \times 45 + 3 = 93. So 45 really does come just before 93.

More questions like this

The original DOE question
Question 71 as printed on page 195 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 71. Open page 195 of the handbook (PDF)

Question 98 of 120

2023 Test B Math, question 20

Question

Five consecutive numbers have a sum of 110. What is the least of these consecutive numbers?

  • A11
  • B18
  • C20
  • D22
Show the answer

Correct answer: C

Let n represent the least of the five numbers. Then the next four consecutive numbers are (n+1)(n + 1), (n+2)(n + 2), (n+3)(n + 3), (n+4)(n + 4). Write an equation to solve for n:

n+(n+1)+(n+2)+(n+3)+(n+4)=110n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 110

5n+10=1105n + 10 = 110

5n=1005n = 100

n=20n = 20

Question analysis

Five consecutive numbers add up to 110, and students find the least of them. This number-properties question can be solved with an equation in one unknown, or by noting that the middle of an odd number of consecutive numbers equals their average. The bold word least matters, because the middle number is easier to find.

NYS standards
  • 7.EE.4a two-step equations
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 A sum of consecutive numbers leads to a two-step equation, standard grade 7 work. The average shortcut makes it quicker, and the only extra demand is answering for the least number rather than the middle one.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Dividing 110 by 5 is the natural first move, and it gives the middle number, which is a listed choice. The bold word least signals that one more step is needed, counting down two from the middle. The equation route, 5n + 10 = 110, goes straight to the least number but takes longer to set up.

  1. Turning words into math minor Understanding the question

    “Consecutive numbers” means each is 1 more than the last, and the bold “least” asks for the first of the five, not the middle one that dividing by 5 gives. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    None of the five numbers is given, so they must be written as n through n + 4, or pictured as two on each side of a middle value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Divide 110 by 5 to find the middle of the five numbers, then count down to the least of them.

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive numbers
n, n+1, n+2, n+3, n+4n,\ n + 1,\ n + 2,\ n + 3,\ n + 4
Consecutive numbers come one right after another when you count, like 7, 8, 9. Each is 1 more than the one before. If n is the least number, the next ones are n + 1, n + 2, and so on.

  1. Sort out the question

    Five consecutive numbers add up to 110. Consecutive means each number is 1 more than the one before. You need the least, or smallest, of the five.

  2. Name the five numbers

    Let n be the least number. Then the five numbers are:

    n, n+1, n+2, n+3, n+4n,\ n + 1,\ n + 2,\ n + 3,\ n + 4

    Each one is 1 more than the one before it.

  3. Write the sum as an equation

    Add all five and set the sum equal to 110.

    n+(n+1)+(n+2)+(n+3)+(n+4)=110n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 110

    There are five n’s, which make 5n. The extra numbers add up to 1 + 2 + 3 + 4 = 10.

    5n+10=1105n + 10 = 110

  4. Solve for n

    Subtract 10 from both sides. Then divide both sides by 5.

    5n=1005n = 100

    n=100÷5=20n = 100 \div 5 = 20

    So the least number is 20.

  5. Give the answer

    Watch out: 110÷5=22110 \div 5 = 22 is the middle number, not the least. Picking 22 is choice D. The question asks for the least of the five. So the answer is C, 20.

Check your answer

Write out the five numbers and add them: 20+21+22+23+24=11020 + 21 + 22 + 23 + 24 = 110. The sum matches. Also, the middle number, 22, is 110 ÷ 5, and the least is 2 less than the middle.

More questions like this

The original DOE question
Question 77 as printed on page 197 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 77. Open page 197 of the handbook (PDF)

Question 99 of 120

2023 Test B Math, question 21

Question

−mn-mn

(−m)(−n)(-m)(-n)

|mn||mn|

|−mn||-mn|

|(−m)(−n)||(-m)(-n)|

If m and n are positive integers, how many of the five expressions above are equivalent to mnmn?

  • A1
  • B2
  • C3
  • D4
Show the answer

Correct answer: D

Since m and n are positive, the value of mnmn is also positive. Of the five options given:

−mn-mn is negative.

(−m)(−n)(-m)(-n) is positive.

|mn||mn| is positive.

|−mn||-mn| is positive.

|(−m)(−n)||(-m)(-n)| is positive.

There are 4 expressions that are positive, and each one is equal to mnmn.

Question analysis

Five expressions are listed: −mn, (−m)(−n), |mn|, |−mn| and |(−m)(−n)|. If m and n are positive integers, students count how many are equivalent to mn. It tests the sign rules for multiplying negatives together with what absolute value does, and each expression has to be judged on its own.

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.NS.7c absolute value as distance
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Multiplying signed numbers and absolute value are grade 6 and 7 content. Checking five expressions one by one takes longer than a typical single computation, but each check uses a familiar rule.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
5 steps
Reading level
Question and choices: about grade 10, 23 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each expression takes only a moment, but the answer is a count, so one misjudged expression changes it. The minus signs are designed to unsettle: four of the five expressions show at least one, and only one of those actually stays negative. Substituting small positive numbers and comparing each value with mn turns the question into arithmetic.

  1. Variables instead of numbers moderate Understanding the question

    The five expressions use only m and n, so each sign rule has to be applied in general unless the student substitutes values such as m = 2 and n = 3. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    Five expressions are judged one at a time and a count kept, and two of them combine the product rule for negatives with absolute value bars. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Something you must already know minor Working it out

    The rules that two negative factors give a positive product and that absolute value removes a minus sign are assumed, not stated. (Anderson & Krathwohl, 2001)

Strategy: Pick small values such as m = 2 and n = 3, evaluate each of the five expressions, and count the ones equal to 6.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying two negatives
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers. A negative number times a negative number gives a positive number. The two minus signs cancel out. One negative and one positive give a negative.
Absolute value
|a|=the distance from a to 0|a| = \text{the distance from } a \text{ to } 0
The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so |−6| = 6 and |6| = 6. The bars remove any minus sign.

  1. Sort out the question

    m and n are positive integers. That means whole numbers greater than 0. Writing mn means m times n, so mn is positive. You need to count how many of the five expressions always equal mn.

  2. Pick numbers to test

    Pick easy values: m=2m = 2 and n=3n = 3. Then mn=2×3=6mn = 2 \times 3 = 6. Any expression that equals mn must come out to 6.

  3. Test the first two expressions

    −mn=−(2×3)=−6-mn = -(2 \times 3) = -6

    This is negative, so it does not equal 6.

    (−m)(−n)=(−2)(−3)=6(-m)(-n) = (-2)(-3) = 6

    A negative times a negative is positive, so this one equals 6. Watch out: if you think a negative times a negative stays negative, you reject this one and get 3, choice C.

  4. Test the three absolute values

    |mn|=|6|=6|mn| = |6| = 6

    |−mn|=|−6|=6|-mn| = |-6| = 6

    |(−m)(−n)|=|6|=6|(-m)(-n)| = |6| = 6

    All three equal 6, because absolute value bars remove a minus sign. Watch out: the bars work even when the minus sign is inside them. Rejecting the last two gives 2, choice B.

  5. Count the matches

    Four expressions equal mn: (−m)(−n)(-m)(-n), |mn||mn|, |−mn||-mn| and |(−m)(−n)||(-m)(-n)|. Only −mn-mn does not. The sign rules work the same for any positive m and n, not just 2 and 3. So the answer is D, 4.

Check your answer

Try other numbers, such as m=1m = 1 and n=5n = 5. Then mn=5mn = 5. The five expressions give −5, 5, 5, 5 and 5. Again, four of them match mn.

More questions like this

The original DOE question
Question 78 as printed on page 197 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 78. Open page 197 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 100 of 120

2023 Test B Math, question 27

Question

N={6,7,…,21,22}\mathrm{N} = \{6, 7, \ldots, 21, 22\}

Set L (not shown) is a list of 25 consecutive integers. The median of the integers in set N above is equal to the least integer in set L. What is the median of the integers in set L?

  • A25
  • B26
  • C27
  • D28
Show the answer

Correct answer: B

First, find the median of set N. There are 17 integers in set N, so the 9th integer is the median. The 9th integer is 14.

The least integer in set L is equal to the median of set N, so the least integer in set L is 14.

There are 25 integers in set L, so the median is the 13th integer. That integer is 26.

Question analysis

Set N lists the consecutive integers from 6 to 22, and set L is a list of 25 consecutive integers whose least member equals the median of N. Students find the median of set L. It chains two median questions together, and counting positions in a run of consecutive integers invites off-by-one slips.

NYS standards
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Medians of short lists are grade 6 content. This item hides both lists, one behind set notation with an ellipsis and one not shown at all, and links them, so it needs more planning and careful counting than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 5, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains two medians, and both depend on counting members correctly in runs of consecutive integers. The easy slip is between position and distance: the 13th member of L is 12 steps above its first, not 13. With four consecutive choices, every counting error has a home, so writing out the first and last members of L, 14 and 38, and averaging them is the safest check.

  1. A very close wrong answer moderate Choosing the answer

    The choices 25, 26, 27 and 28 are consecutive, so a count that is off by one position lands on a neighboring choice with nothing to flag it. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. No obvious method moderate Working it out

    Set L is never listed; its median has to be found from its least member, 14, and its size, 25, by working out that the middle is the 13th member. (Schoenfeld, 1985)

  3. Reading a figure, table or graph minor Understanding the question

    Set N is written with an ellipsis, {6, 7, …, 21, 22}, so its 17 members must be counted, and 22 − 6 = 16 undercounts by one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find N's median by averaging its least and greatest members. L's median is its middle member, so work out that member's position and count up from L's least member.

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).
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting the integers in a list
count=last−first+1\text{count} = \text{last} - \text{first} + 1
This counts the whole numbers in a list that goes up by 1, like 6, 7, 8. Subtracting gives the number of steps from the first number to the last. You add 1 because the first number counts too.
Position of the median
position=count+12\text{position} = \frac{\text{count} + 1}{2}
The count is how many numbers are in a list that is in order. When the count is odd, the median is at this position. It has the same number of values before it and after it.

  1. Sort out the question

    Set N is 6, 7, and so on up to 22. The dots stand for every integer in between. Integers are whole numbers. Set L has 25 consecutive integers, one right after another. Its least number equals the median of N. You need the median of L.

  2. Count the numbers in set N

    count=last−first+1\text{count} = \text{last} - \text{first} + 1

    22−6+1=1722 - 6 + 1 = 17

    Set N has 17 integers. Watch out: 22−6=1622 - 6 = 16 misses one, because the first number counts too.

  3. Find the median of set N

    position=count+12\text{position} = \frac{\text{count} + 1}{2}

    position=17+12=9\text{position} = \frac{17 + 1}{2} = 9

    The median is the 9th number. 6 is the 1st, so the 9th is 8 steps up: 6 + 8 = 14. The median of N is 14.

  4. Find the middle place in set L

    So set L starts at 14 and has 25 integers. Find the position of its median.

    position=25+12=13\text{position} = \frac{25 + 1}{2} = 13

    The median of L is its 13th number.

  5. Count up to the 13th number

    14 is the 1st number. The 13th is 12 steps after it.

    14+12=2614 + 12 = 26

    Watch out: the 13th number is 12 steps up, not 13. Adding 13 gives 27, choice C. And 25 is how many numbers L has, not its median. So the answer is B, 26.

Check your answer

Set L runs from 14 to 14+24=3814 + 24 = 38. For consecutive integers, the median is halfway between the ends: (14+38)÷2=26(14 + 38) \div 2 = 26. That matches. The same trick gives N’s median: (6+22)÷2=14(6 + 22) \div 2 = 14.

More questions like this

The original DOE question
Question 84 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 84. Open page 199 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 101 of 120

2023 Test B Math, question 29

Question

Which number could be the sum of 4 consecutive integers?

  • A−12
  • B−10
  • C−8
  • D−4
Show the answer

Correct answer: B

The sum of 4 consecutive integers can be expressed as:

n+(n+1)+(n+2)+(n+3)n+(n+1)+(n+2)+(n+3)

=4n+6=4n+6

Set that expression equal to each of the options and solve for n:

Option A:

4n+6=−124n+6=-12

4n=−184n=-18

n=−184n=-\frac{18}{4} is not an integer.

Option B:

4n+6=−104n+6=-10

4n=−164n=-16

n=−4n=-4 is an integer.

Option C:

4n+6=−84n+6=-8

4n=−144n=-14

n=−144n=-\frac{14}{4} is not an integer.

Option D:

4n+6=−44n+6=-4

4n=−104n=-10

n=−104n=-\frac{10}{4} is not an integer.

The only option for which n is an integer is −10.

Question analysis

A number properties question about consecutive integers: which of four negative numbers could be the sum of 4 consecutive integers? Students write the sum with a variable, or add runs of negative integers, and test each choice. Keeping signs straight with negative numbers, and remembering that the integers must be whole numbers, is the tricky part.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 consecutive-integer problems usually give the sum and ask for the integers. This one reverses the task: students test which negative total is possible at all, using the structure of the sum rather than one equation.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 9 words.
Corrected from the DOE handbook
  • A stray word in the explanation’s last sentence was removed.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Three of the four choices are multiples of 4, and that is the whole trap: a student who pictures the sum as four times some integer finds three choices that seem to work. The sum of four consecutive integers is 4n + 6, so it is always even but never a multiple of 4. Trying the most natural run, −4, −3, −2 and −1, lands on −10 at once and is the fastest check.

  1. Every choice must be checked moderate Choosing the answer

    The stem asks which number “could be” the sum, so no single equation gives the answer; each of −12, −10, −8 and −4 has to be tested against the form a four-integer sum must take. (Embretson & Wetzel, 1987)

  2. No obvious method minor Working it out

    Nothing points to the shortcut that n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 is always 2 more than a multiple of 4, or to simply adding a run such as −1, −2, −3 and −4. (Schoenfeld, 1985)

  3. Error-prone arithmetic minor Working it out

    Every choice is negative, so solving 4n + 6 = −12 or adding a run of integers means subtracting 6 from a negative number and dividing a negative by 4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the integers as n, n + 1, n + 2 and n + 3, add them, and set the sum equal to each choice to see which one gives a whole-number n.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook ends with “The only option for which n is an integer answer is −10.” The word “answer” does not belong in that sentence; the site reads “The only option for which n is an integer is −10.” The working above it is unchanged: only 4n + 6 = −10 gives an integer, n = −4, and −4 + (−3) + (−2) + (−1) = −10.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of 4 consecutive integers
n+(n+1)+(n+2)+(n+3)=4n+6n + (n + 1) + (n + 2) + (n + 3) = 4n + 6
n is the smallest of the four integers. Each next integer is 1 more than the one before, so they are n, n + 1, n + 2 and n + 3. Adding them gives four n’s, plus 1 + 2 + 3, which is 6. Use it to add any four integers in a row.

  1. Sort out the question

    Integers are whole numbers and their negatives, like −2, 0 and 5. They have no fractions or decimals. Consecutive integers come one right after another, like 7, 8, 9, 10 or −3, −2, −1, 0.

    You need the choice that can be the sum of four consecutive integers. The sum is what you get when you add them.

  2. Write the sum with a letter

    You don’t know the integers yet. So call the smallest one n. Each next one is 1 more: n + 1, n + 2 and n + 3. Use the formula. Collect the four n’s, then add the numbers.

    n+n+n+n+1+2+3=4n+6n + n + n + n + 1 + 2 + 3 = 4n + 6

    Any four consecutive integers add up to 4n + 6.

  3. Test each choice

    Set 4n + 6 equal to each choice and solve for n. First subtract 6 from both sides. Then divide both sides by 4.

    Choice4n + 64nn
    A−12−18−4.5
    B−10−16−4
    C−8−14−3.5
    D−4−10−2.5

    n must be an integer. Only choice B gives one: n = −4.

  4. Name the four integers

    Start at n = −4 and count up by 1. The integers are −4, −3, −2 and −1.

    Watch out: Choices A, C and D are multiples of 4. A multiple of 4 is 4 times an integer, like 4 × (−3) = −12. It is tempting to think four integers must add to one of those. But the sum is 4n + 6, and the extra 6 gets in the way. Choice D works only with n = −2.5, and that is not an integer.

    So the answer is B, −10.

Check your answer

Add the integers: −4+(−3)+(−2)+(−1)=−10-4 + (-3) + (-2) + (-1) = -10. That is choice B. Also, 4n + 6 is always 2 more than a multiple of 4. And −10 is 2 more than −12, which is 4 × (−3).

More questions like this

The original DOE question
Question 86 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 86. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 102 of 120

2023 Test B Math, question 32

Question

A number line with arrows at both ends and three points marked on it. From left to right, point P is at 1 1/3, point Q is at 3, and point R is at 5 1/6.

On the number line above, how many units longer is QR\overline{QR} than PQ\overline{PQ}?

  • A16\frac{1}{6}
  • B12\frac{1}{2}
  • C23\frac{2}{3}
  • D3
Show the answer

Correct answer: B

First, find the lengths of the given segments:

QR=516−3=216\mathrm{QR}=5\frac{1}{6}-3=2\frac{1}{6}

PQ=3−113=123\mathrm{PQ}=3-1\frac{1}{3}=1\frac{2}{3}

Then, find the difference:

216−123=2\frac{1}{6}-1\frac{2}{3}=

136−53=\frac{13}{6}-\frac{5}{3}=

136−106=\frac{13}{6}-\frac{10}{6}=

36=12\frac{3}{6}=\frac{1}{2}

Question analysis

A number line shows three points labeled with mixed numbers and a whole number, and students find how many units longer one segment is than another. It tests subtracting mixed numbers with unlike denominators to find distances. Regrouping when the fraction being subtracted is the larger one is where most errors happen.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding distances between points on a number line is grade 7 work. Here the points are mixed numbers with different denominators, so finding two distances and their difference makes it a typical but careful grade 7 task.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is three subtractions, so most errors are in the fractions rather than the plan. Taking 1 1/3 from a whole number such as 3 tempts students to subtract only the whole parts and keep the 1/3, which reverses which segment looks longer. Rewriting every coordinate in sixths, 8/6, 18/6 and 31/6, turns the work into whole-number subtraction of numerators.

  1. Error-prone arithmetic moderate Working it out

    Both lengths subtract mixed numbers with different denominators, 5 1/6 − 3 and 3 − 1 1/3, and the second needs regrouping before the two results are compared in sixths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Reading a figure, table or graph minor Understanding the question

    The three coordinates come from the labels under P, Q and R, and the segments QR and PQ must each be measured from Q at 3, the point they share. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find each length by subtracting the left endpoint from the right one, writing the mixed numbers in sixths, then subtract the shorter length from the longer.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Length on a number line
length=right point−left point\text{length} = \text{right point} - \text{left point}
The length of a segment is the distance between its two end points. Subtract the number on the left from the number on the right. Subtracting tells how many units you move to get from one point to the other.
Mixed number to improper fraction
abc=a×c+bca\frac{b}{c} = \frac{a \times c + b}{c}
a is the whole number part. b over c is the fraction part, with b on top and c on the bottom. Each whole holds c parts, so a wholes hold a × c parts. Then add the b extra parts. An improper fraction has a top bigger than its bottom.

  1. Read the number line

    Point P is at 1131\frac{1}{3}. Point Q is at 3. Point R is at 5165\frac{1}{6}.

    The bar over QR means segment QR, the piece of the line from Q to R. You need how much longer QR is than PQ. So find both lengths, then subtract.

  2. Find the length of QR

    R is on the right, and Q is on the left. Use the length rule.

    QR=516−3=216\text{QR} = 5\frac{1}{6} - 3 = 2\frac{1}{6}

    QR is 2162\frac{1}{6} units long.

  3. Find the length of PQ

    Q is on the right, and P is on the left.

    PQ=3−113\text{PQ} = 3 - 1\frac{1}{3}

    Take it in two parts. First, 3 − 1 = 2. Then take away the extra 13\frac{1}{3}. That gives 2−13=1232 - \frac{1}{3} = 1\frac{2}{3}. So PQ is 1231\frac{2}{3} units long.

    Watch out: Don’t just do 3 − 1 and keep the 13\frac{1}{3}. That gives 2132\frac{1}{3}, which is too long. It leads to choice A.

  4. Write both lengths in sixths

    The fractions have different bottoms, 6 and 3. Change both lengths to sixths so you can subtract. Use the mixed number rule.

    216=2×6+16=1362\frac{1}{6} = \frac{2 \times 6 + 1}{6} = \frac{13}{6}

    123=1×3+23=53=1061\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3} = \frac{10}{6}

  5. Subtract the lengths

    Take the shorter length away from the longer one.

    136−106=36=12\frac{13}{6} - \frac{10}{6} = \frac{3}{6} = \frac{1}{2}

    QR is 12\frac{1}{2} unit longer than PQ.

    So the answer is B, 12\frac{1}{2}.

Check your answer

Write every point in sixths. P is 86\frac{8}{6}, Q is 186\frac{18}{6} and R is 316\frac{31}{6}. QR is 31 − 18 = 13 sixths. PQ is 18 − 8 = 10 sixths. The difference is 3 sixths, which is one half.

More questions like this

The original DOE question
Question 89 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 89. Open page 200 of the handbook (PDF)

Question 103 of 120

2023 Test B Math, question 33

Question

A line segment with point W at its left end and point Y at its right end. Point X lies on the segment between W and Y, closer to Y.

The distance from W to X is 70% of the distance from W to Y. If the distance from X to Y is 15 miles, what is the distance, in miles, from W to X?

  • A10.5
  • B35
  • C45
  • D50
Show the answer

Correct answer: B

Since the distance from W to X is 70% of the distance from W to Y, the distance from X to Y is 30% of the distance from W to Y. Let d represent the distance from W to Y. Then 15 miles is 30% of d.

0.30d=150.30d=15

d=150.30=50d=\frac{15}{0.30}=50

So, the distance from W to X, in miles, is 50−15=3550-15=35.

Question analysis

A percent question with a line segment: point X lies between W and Y, the distance from W to X is 70% of the whole, and the distance from X to Y is 15 miles. Students find the distance from W to X. Matching the 15 miles to the right percent before finding the whole is the step students most often skip.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 percent problems often give a part and its percent and ask for the whole. This one gives the distance for the part the percent does not describe and asks for the other part, so the whole has to be found on the way.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The percent and the given distance describe different pieces of the road, which is the whole difficulty. Students who apply the 70% straight to the 15 miles get a small number fast, and students who solve correctly for the whole of WY must still remember to take XY away. Checking that 35 is 70% of 50 confirms the answer and catches both slips.

  1. Turning words into math moderate Understanding the question

    The 70% describes WX, but the 15 miles given is XY, so the sentence has to be recast as “XY is 30% of WY” before the 15 miles can be used. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Finding the complement, 30%, then the whole, 15 ÷ 0.30 = 50, then the part, 50 − 15, gives three results, and the middle one is itself a choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out what percent of WY the 15-mile stretch from X to Y must be, use it to find the whole of WY, then find the part from W to X.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}
The whole is the full amount. The part is a piece of it. The percent tells how many hundredths of the whole the part is. For example, 30% means 30 out of every 100, or 0.30. Use it whenever a percent describes a piece of a whole.

  1. Sort out the distances

    Point X sits between W and Y. So the trip from W to Y has two pieces: W to X, then X to Y.

    W to X is 70% of the whole distance from W to Y. X to Y is 15 miles. You need the distance from W to X, in miles.

  2. Find the percent for X to Y

    The whole distance from W to Y is 100%. W to X uses 70% of it. The rest is X to Y.

    100−70=30100 - 70 = 30

    So the 15 miles from X to Y is 30% of the whole distance.

  3. Find the whole distance

    Let d be the distance from W to Y. Use the percent formula. The part is 15, and the percent is 30.

    15=30100×d=0.30×d15 = \frac{30}{100} \times d = 0.30 \times d

    Divide both sides by 0.30 to get d by itself.

    d=15÷0.30=50d = 15 \div 0.30 = 50

    The distance from W to Y is 50 miles.

  4. Find the distance from W to X

    Take the 15 miles from X to Y away from the whole.

    50−15=3550 - 15 = 35

    W to X is 35 miles.

    Watch out: 50 miles is the whole distance from W to Y, choice D. You still have to subtract. Also, the 70% is of the whole distance, not of the 15 miles. Taking 70% of 15 gives 10.5, choice A.

    So the answer is B, 35 miles.

Check your answer

Is 35 really 70% of 50? 0.70×50=350.70 \times 50 = 35. Yes. And 35 + 15 = 50, the whole distance. The figure also shows X closer to Y. So W to X should be longer than 15 miles, and it is.

More questions like this

The original DOE question
Question 90 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 90. Open page 201 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 104 of 120

2023 Test B Math, question 42

Question

1 gallon = 16 cups

1 cup = 8 fluid ounces

If 1 gallon = 2x2^{x} fluid ounces, what is the value of x?

  • A4
  • B7
  • C8
  • D64
Show the answer

Correct answer: B

If 1 gallon is equal to 16 cups, and 1 cup is equal to 8 fluid ounces, then

1 gallon=1\text{ gallon}=

(16)(8 fluid ounces)=(16)(8\text{ fluid ounces})=

128 fluid ounces128\text{ fluid ounces}

128=27128=2^{7}

1 gallon=27 fluid ounces1\text{ gallon}=2^{7}\text{ fluid ounces}

So, x=7x=7.

Question analysis

This question gives two facts, 16 cups in a gallon and 8 fluid ounces in a cup, and says a gallon equals 2 to the power x fluid ounces. Students find the exponent x. It combines a unit conversion with writing a number as a power of 2, and the exponent is easy to confuse with a multiplier.

NYS standards
  • 6.RP.3d converting measurement units
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Converting units with given facts is grade 6 work, and this conversion is a single multiplication. The twist is writing the result as a power of 2, which calls for fluency with powers that grade 7 work rarely asks for.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The conversion is one multiplication, so the question really tests whether students can name the exponent for 128. Stopping at the cups, where 16 = 2⁴, or treating 2 to the power x as 2 times x gives a listed choice. Writing 16 as 2⁴ and 8 as 2³ and adding the exponents is the quickest route, and doubling from 1 seven times is a sure check.

  1. Combines ideas minor Working it out

    A unit conversion, 16 cups × 8 fluid ounces = 128 fluid ounces, has to be joined to exponents by recognizing 128 as a power of 2. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Something you must already know minor Working it out

    Students must know or build the powers of 2 up to 128, and see that the exponent counts factors of 2, not the number of terms in a list starting at 1. (Anderson & Krathwohl, 2001)

Strategy: Convert 1 gallon to fluid ounces by multiplying the two facts, then keep halving the result, or keep doubling from 2, counting how many factors of 2 it takes.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Meaning of an exponent
2x=2×2×⋯×22^x = 2 \times 2 \times \cdots \times 2
The 2 is the base, the number being multiplied. x is the exponent, the small raised number. It tells how many 2s are multiplied together. For example, 2 to the third power means 2 × 2 × 2, which is 8.
Multiplying powers of the same base
2a×2b=2a+b2^a \times 2^b = 2^{a + b}
a and b are exponents. The first power has a twos multiplied together. The second has b more twos. Multiplying puts all the twos together, so you add the exponents.

  1. Sort out the question

    A gallon is 16 cups. A cup is 8 fluid ounces. A gallon is also 2x2^x fluid ounces. You need x, the exponent. It tells how many 2s are multiplied together.

  2. Change a gallon to fluid ounces

    Each of the 16 cups holds 8 fluid ounces. So multiply.

    16×8=12816 \times 8 = 128

    One gallon is 128 fluid ounces.

  3. Write 128 as a power of 2

    Now you need 2x=1282^x = 128. Keep multiplying by 2, and count the 2s.

    Number of 2s1234567
    Product248163264128

    It takes seven 2s to make 128.

    27=1282^7 = 128

    So x = 7.

  4. Give the answer

    Watch out: A gallon is 16 cups, and 16=2416 = 2^4. Stopping there gives 4, choice A. And 2x2^x does not mean 2 times x. Solving 2 × x = 128 gives 64, choice D.

    So the answer is B, 7.

Check your answer

Use the exponent rule. 16 is 242^4 and 8 is 232^3. So 16×8=24×23=24+3=2716 \times 8 = 2^4 \times 2^3 = 2^{4 + 3} = 2^7. The exponent is 7 again.

More questions like this

The original DOE question
Question 99 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 99. Open page 203 of the handbook (PDF)

Question 105 of 120

2023 Test B Math, question 43

Question

A certain square puzzle has an area of 64 square inches, and a cubic block has a volume of 64 cubic inches. What is the ratio of the length of a side of the puzzle to the length of a side of the block?

  • A1:31:3
  • B1:21:2
  • C2:32:3
  • D2:12:1
Show the answer

Correct answer: D

The length of one side of the puzzle is 64=8\sqrt{64}=8 inches. The length of one side of the block is 643=4\sqrt[3]{64}=4 inches.

The ratio of the side of the puzzle to the side of the block is 8:48:4, which reduces to 2:12:1.

Question analysis

A square puzzle has an area of 64 square inches, and a cube-shaped block has a volume of 64 cubic inches. Students find the ratio of a side of the puzzle to a side of the block. Both figures are described by the number 64, but turning an area and a volume back into lengths takes different steps, and the order of the ratio matters.

NYS standards
  • 8.EE.2 square and cube roots
  • 6.G.5 perfect squares and perfect cubes
  • 6.RP.1 ratio concepts and language
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 area and volume problems go from side lengths to an area or a volume. Going backward needs a square root and a cube root, and cube roots are grade 8 content, although 64 is both a perfect square and a perfect cube.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 46 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Because 64 is both a perfect square and a perfect cube, the computation is easy once students know to undo squaring and cubing. The risk is at the end: with sides of 8 and 4 inches in hand, writing the block first turns the ratio upside down. Students who never find the side lengths are left with ideas such as two dimensions against three.

  1. Something you must already know moderate Working it out

    Neither formula is given: the puzzle’s side is the number whose square is 64 and the block’s edge the number whose cube is 64, and cube roots are grade 8 work. (Anderson & Krathwohl, 2001)

  2. Combines ideas minor Working it out

    A square and a cube share the number 64 in different units, so two different inverse operations feed one ratio, which must list the puzzle’s side first. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Find the side of the square from its area and the edge of the cube from its volume by testing small whole numbers, then write the puzzle's side first and simplify.

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square
A=s×sA = s \times s
A is the area, in square units. s is the length of one side. All sides of a square are equal, so the area is the side times itself. To find the side from the area, ask which number times itself gives the area.
Volume of a cube
V=s×s×sV = s \times s \times s
V is the volume, the space inside, in cubic units. s is the length of one edge. A cube’s length, width and height are all equal, so the volume is the edge used three times. To go backward, ask which number used three times gives the volume.

  1. Sort out the shapes

    The puzzle is a square with an area of 64 square inches. The block is a cube with a volume of 64 cubic inches. A cube is a box with every edge the same length. You need the ratio of the puzzle’s side to the block’s side, with the puzzle first.

  2. Find the puzzle’s side

    Use the area formula backward. Which number times itself makes 64?

    8×8=648 \times 8 = 64

    The puzzle’s side is 8 inches. This number is the square root of 64.

  3. Find the block’s side

    Use the volume formula backward. Which number used three times makes 64? Try small numbers. 3 × 3 × 3 = 27 is too small.

    4×4×4=644 \times 4 \times 4 = 64

    The block’s side is 4 inches. This number is the cube root of 64.

  4. Write and simplify the ratio

    Put the puzzle first, as the question asks. Then divide both parts by 4.

    8:4=(8÷4):(4÷4)=2:18 : 4 = (8 \div 4) : (4 \div 4) = 2 : 1

    Watch out: Writing the block’s side first gives 1 : 2, choice B. The order in a ratio matters.

    So the answer is D, 2 : 1.

Check your answer

Check both sides. 8 × 8 = 64 square inches, and 4 × 4 × 4 = 64 cubic inches. The puzzle’s side is longer, so its number in the ratio must be bigger. Only choice D puts the bigger number first.

More questions like this

The original DOE question
Question 100 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 100. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 106 of 120

2023 Test B Math, question 45

Question

The Barnes family and the Ramirez family each have a pizza of equal size. The first pizza is cut into 18 equal slices, and the second pizza is cut into 15 equal slices. If the Barnes family eats 11 slices from the first pizza, what is the greatest number of whole slices from the second pizza that the Ramirez family can eat without eating a greater percentage of a pizza than the Barnes family ate?

  • A6
  • B7
  • C8
  • D9
Show the answer

Correct answer: D

The fraction of the pizza that the Barnes family eats is 1118\frac{11}{18}. Let x represent the number of slices of pizza that the Ramirez family eats. Then the fraction of the pizza that the Ramirez family eats is x15\frac{x}{15}. Write an inequality and solve for x:

x15≤1118\frac{x}{15}\leq\frac{11}{18}

x≤11×1518x\leq\frac{11\times 15}{18}

x≤16518x\leq\frac{165}{18}

x≤916x\leq 9\frac{1}{6}

So, the greatest number of whole slices of pizza the Ramirez family can eat is 9.

Question analysis

Two pizzas of the same size are cut into 18 and 15 equal slices, and one family eats 11 slices of the first. Students find the most whole slices of the second pizza the other family can eat without eating a larger share. It tests comparing fractions with different denominators and rounding to a whole number in the right direction.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 inequality problems usually ask for the greatest whole number that fits a budget. This one wraps that idea in a long stem about shares of two pizzas sliced differently, with several conditions to keep straight at once.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 11, 72 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The math is one inequality between fractions, but the long final sentence hides it, so students who skim fall back on comparing slice counts. Reasoning that the second pizza has 3 fewer slices, so the Ramirez family should eat 3 fewer, is the strongest pull. Comparing 11/18 with x/15, or turning 11/18 into about 61% and taking 61% of 15, keeps the comparison about shares of a pizza.

  1. Complex sentences moderate Understanding the question

    The question is one 42-word sentence ending in a nested condition, “without eating a greater percentage of a pizza than the Barnes family ate”, a reading hurdle rather than more math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

  2. Turning words into math moderate Understanding the question

    The limit of no “greater percentage of a pizza” must become the inequality x/15 ≤ 11/18, comparing fractions of two pizzas cut into different numbers of slices, not slice counts. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Solving gives x ≤ 11 × 15 ÷ 18 = 165/18, a division by 18 that leaves 9 1/6, which must then be cut back to whole slices. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the Barnes family's share as a fraction of their pizza, then find the largest whole number of fifteenths that does not exceed it, by cross-multiplying or by converting to decimals.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a pizza
fraction eaten=slices eatenslices in the pizza\text{fraction eaten} = \frac{\text{slices eaten}}{\text{slices in the pizza}}
The top counts the slices eaten. The bottom counts all the equal slices the pizza was cut into. The fraction tells how much of the whole pizza was eaten. Use fractions to compare pizzas cut into different numbers of slices.

  1. Sort out the question

    The two pizzas are the same size. The Barnes pizza has 18 slices, and they eat 11. The Ramirez pizza has 15 slices, so its slices are bigger. You need the most whole slices the Ramirez family can eat without eating a bigger share of their pizza.

  2. Find the Barnes family’s share

    Their share is the slices they ate out of all the slices in their pizza. Use the fraction formula.

    fraction eaten=1118\text{fraction eaten} = \frac{11}{18}

    The Barnes family ate 1118\frac{11}{18} of their pizza.

  3. Write an inequality

    Let x be the number of Ramirez slices. Their share is x15\frac{x}{15}. It must be no more than the Barnes share. The sign ≤ means less than or equal to.

    x15≤1118\frac{x}{15} \leq \frac{11}{18}

  4. Solve for x

    Multiply both sides by 15 to get x by itself.

    x≤11×1518=16518x \leq \frac{11 \times 15}{18} = \frac{165}{18}

    Now divide. 18 goes into 165 nine times, since 18 × 9 = 162. There are 3 left over.

    x≤9318=916x \leq 9\frac{3}{18} = 9\frac{1}{6}

  5. Pick the whole number

    x must be a whole number no bigger than 9169\frac{1}{6}. So the most is 9 slices. 10 slices would be too many.

    Watch out: The Ramirez pizza has 3 fewer slices, but that does not mean they eat 3 fewer. Their slices are bigger. Eating 11 − 3 = 8 slices is choice C, which is not the most they can eat.

    So the answer is D, 9.

Check your answer

Use decimals. The Barnes share is 11÷18≈0.61111 \div 18 \approx 0.611. Nine Ramirez slices are 9÷15=0.69 \div 15 = 0.6, just under it. Ten slices are 10÷15≈0.66710 \div 15 \approx 0.667, too much. So 9 is the most.

More questions like this

The original DOE question
Question 102 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 102. Open page 204 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 107 of 120

2023 Test B Math, question 46

Question

03+40+2−1+(−1)20^{3}+4^{0}+2^{-1}+(-1)^{2}

What is the value of the expression above, in decimal form?

  • A0
  • B0.5
  • C1.5
  • D2.5
Show the answer

Correct answer: D

03+40+2−1+(−1)20^{3}+4^{0}+2^{-1}+(-1)^{2}

=0+1+12+1=0+1+\frac{1}{2}+1

=212=2.5=2\frac{1}{2}=2.5

Question analysis

An exponents question: students find the decimal value of a sum of four powers, zero cubed, four to the zero power, two to the negative first power and negative one squared. Each of the four terms follows a different rule, and a slip on any one of them changes the total.

NYS standards
  • 8.EE.1 integer exponent properties
  • 7.NS.2a rational number multiplication
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 work uses whole-number exponents and signed numbers. Zero and negative exponents are grade 8 content, and this expression tests both alongside a negative base, each with its own common misconception.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 6, 12 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every term tests a different exponent rule, and each wrong choice comes from breaking exactly one of them. The zero exponent is the most common casualty, since 4⁰ looks as if it should be 0 like 0³; a negative exponent read as a negative number is close behind. Writing each term’s value above it before adding, 0, 1, 1/2 and 1, keeps the four rules separate.

  1. Something you must already know major Working it out

    Three rules the test does not give decide the sum: any nonzero number to the 0 power is 1, 2 to the −1 power is 1/2, and (−1)² is positive 1. (Anderson & Krathwohl, 2001)

  2. Error-prone arithmetic minor Working it out

    The four terms carry a zero base, a zero exponent, a negative exponent and a negative base, so a sign or a stray 0 can slip in at each one. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Evaluate each power on its own, using the rules for a zero exponent and a negative exponent and the sign of a squared negative number, then add the four values.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Zero exponent
a0=1a^0 = 1
a is any number except 0. An exponent of 0 always gives 1. It works because each step down in the exponent divides by a. For example, 4 to the first power is 4, and 4 ÷ 4 = 1.
Negative exponent
a−n=1ana^{-n} = \frac{1}{a^n}
a is the base, and n is a positive whole number. A negative exponent does not make the number negative. It means 1 divided by the base to the positive power. For example, 2 to the −1 power is one half.

  1. Sort out the question

    The expression adds four powers. A power is a base with an exponent, a small raised number. The exponent tells how many times to use the base as a factor, a number you multiply. Find each power, then add.

  2. Find the first two powers

    Zero times itself is still zero.

    03=0×0×0=00^3 = 0 \times 0 \times 0 = 0

    Next, use the zero exponent rule.

    40=14^0 = 1

    Watch out: 404^0 is 1, not 0. Taking it as 0 gives choice C.

  3. Find the negative power

    Use the negative exponent rule.

    2−1=121=12=0.52^{-1} = \frac{1}{2^1} = \frac{1}{2} = 0.5

    Watch out: 2−12^{-1} is not −2. That mistake gives choice A.

  4. Find the last power

    The base is −1, inside the parentheses. Multiply it by itself. A negative times a negative is positive.

    (−1)2=(−1)×(−1)=1(-1)^2 = (-1) \times (-1) = 1

    Watch out: The answer is not −1. That mistake gives choice B.

  5. Add the four values

    Now add them. The question wants a decimal, so write one half as 0.5.

    0+1+0.5+1=2.50 + 1 + 0.5 + 1 = 2.5

    So the answer is D, 2.5.

Check your answer

See why 40=14^0 = 1 with a pattern. 42=164^2 = 16 and 41=44^1 = 4. Each step down divides by 4, so 40=4÷4=14^0 = 4 \div 4 = 1. In the same way, 20=12^0 = 1 and 2−1=1÷2=122^{-1} = 1 \div 2 = \frac{1}{2}.

More questions like this

The original DOE question
Question 103 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 103. Open page 204 of the handbook (PDF)

Question 108 of 120

2023 Test B Math, question 48

Question

The outdoor temperature was 4.2° F at 10 p.m. After 3123\frac{1}{2} hours, the outdoor temperature was −0.7° F. Which value best represents the average change in temperature per hour?

  • A−1.4° F
  • B−1.0° F
  • C1.0° F
  • D1.4° F
Show the answer

Correct answer: A

First, recognize that the temperature decreased. This means that the order of subtraction matters and that the result is a negative number:

−0.7−4.2=−4.9-0.7-4.2=-4.9

The temperature decreased 4.9 degrees. Divide by 3.5 to get the average change per hour:

−4.9÷3.5=−1.4-4.9\div 3.5=-1.4

Question analysis

The temperature was 4.2° F at 10 p.m. and −0.7° F three and a half hours later. Students find the average change in temperature per hour. It tests subtracting and dividing with negative decimals, and the choices pair two sizes of change with both signs, so the sign has to be right as well as the size.

NYS standards
  • 7.NS.3 four operations with rational numbers
  • 7.NS.1c rational number subtraction
  • 7.NS.2b rational number division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Signed-number problems about temperature are standard grade 7 work. This one asks for an average rate, so a change that crosses zero must be divided by a mixed-number time, with the sign kept through both steps.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 5, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is short, but signs decide every choice: two choices come from treating −0.7 as 0.7, and one comes from getting the size right and the direction wrong. Students who think of the drop in two parts, 4.2 degrees down to zero and 0.7 more, find the 4.9-degree fall; dividing by 3.5 hours and remembering the temperature fell finishes it.

  1. Error-prone arithmetic moderate Working it out

    The change crosses zero, −0.7 − 4.2 = −4.9, and then a negative decimal is divided by 3.5, from the 3 1/2 hours, with the sign kept through both steps. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The phrase “average change in temperature per hour” means the change from the 4.2° start to the −0.7° end divided by the time, final minus initial, so a drop is negative. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. A very close wrong answer minor Choosing the answer

    The answer, −1.4° F, and the choice 1.4° F share every digit and differ only in sign, so a correct division followed by a careless sign cannot be caught by estimating. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Subtract the starting temperature from the ending one, keeping track of the sign, then divide that change by the 3.5 hours.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average change per hour
average change per hour=end−startnumber of hours\text{average change per hour} = \frac{\text{end} - \text{start}}{\text{number of hours}}
Start is the first temperature, and end is the last one. End minus start is the total change. It is negative when the temperature drops. Dividing by the number of hours shares that change equally over each hour.

  1. Sort out the question

    At 10 p.m. it was 4.2° F. After 3123\frac{1}{2} hours, it was −0.7° F. That is below zero. The temperature went down, so the change is negative. You need the average change for each hour.

  2. Find the total change

    Subtract the start from the end.

    −0.7−4.2=−4.9-0.7 - 4.2 = -4.9

    Picture a thermometer. Going from 4.2 down to 0 is a drop of 4.2 degrees. Going on to −0.7 is 0.7 more. In all, it fell 4.9 degrees.

    Watch out: 4.2 − 0.7 = 3.5 treats −0.7 as if it were above zero. That leads to choice B or C.

  3. Divide by the hours

    Write 3123\frac{1}{2} hours as 3.5 hours. Use the formula.

    average change per hour=−0.7−4.23.5=−4.93.5=−1.4\text{average change per hour} = \frac{-0.7 - 4.2}{3.5} = \frac{-4.9}{3.5} = -1.4

    To divide, think 49 ÷ 35 = 1.4, since 35 × 1.4 = 49. The change is negative, so it is −1.4 degrees per hour.

  4. Keep the sign

    The temperature fell, so the answer must be negative.

    Watch out: 1.4 has the right size, but it says the temperature went up. That is choice D.

    So the answer is A, −1.4° F.

Check your answer

Work forward. Dropping 1.4 degrees each hour for 3.5 hours is 1.4×3.5=4.91.4 \times 3.5 = 4.9 degrees. Then 4.2−4.9=−0.74.2 - 4.9 = -0.7. That is the temperature in the problem.

More questions like this

The original DOE question
Question 105 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 105. Open page 205 of the handbook (PDF)

Question 109 of 120

2023 Test B Math, question 50

Question

{3,4,5,6,7}\{3, 4, 5, 6, 7\}

A student made a list of all possible products of 2 different numbers in the set above. What fraction of the products are odd?

  • A320\frac{3}{20}
  • B310\frac{3}{10}
  • C12\frac{1}{2}
  • D35\frac{3}{5}
Show the answer

Correct answer: B

First, list the products:

3×4=123\times 4=12

3×5=153\times 5=15

3×6=183\times 6=18

3×7=213\times 7=21

4×5=204\times 5=20

4×6=244\times 6=24

4×7=284\times 7=28

5×6=305\times 6=30

5×7=355\times 7=35

6×7=426\times 7=42

Of the 10 products, 3 are odd. The fraction is 310\frac{3}{10}.

Question analysis

A student lists every product of two different numbers from the set 3, 4, 5, 6, 7. Students find what fraction of those products are odd. It tests listing outcomes systematically without counting a pair twice, and reasoning about odd and even numbers.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.SP.8a compound event as fraction of outcomes
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Organized lists of outcomes are grade 7 work. Here students list the products of pairs from five numbers, without repeating a pair or pairing a number with itself, then count the ones that fit.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The count of odd products is easy, since only 3, 5 and 7 can make one; the total is where students go wrong. Counting ordered pairs, 5 × 4 = 20, while listing odd products once mixes two kinds of counting. Writing the 10 pairs in order, 3 with each larger number, then 4, and so on, keeps the total honest.

  1. Much to keep track of moderate Working it out

    Listing the products of 2 different numbers from five means tracking 10 pairs, with no pair repeated in reverse order and no number paired with itself. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Something you must already know minor Working it out

    The shortcut, that a product is odd only when both factors are odd, cuts the work to counting pairs from 3, 5 and 7, but it must be known. (Anderson & Krathwohl, 2001)

Strategy: List each pair of different numbers once, then find each product, or use the fact that a product is odd only when both factors are odd, and count.

Traps

  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a list
fraction=odd productsall products\text{fraction} = \frac{\text{odd products}}{\text{all products}}
The top counts the products that are odd. The bottom counts all the products in the list. A product is the answer when you multiply. The fraction tells what share of the list is odd.
Odd times odd
odd×odd=odd\text{odd} \times \text{odd} = \text{odd}
An odd number can’t be split into pairs, like 3 or 7. An even number can, like 4. A product is odd only when both numbers are odd. If either number is even, the product is even, because it splits into pairs.

  1. Sort out the question

    The set has 3, 4, 5, 6 and 7. The student multiplies 2 different numbers, every possible way. A product is the answer to a multiplication. Each pair counts once, since 3 × 4 and 4 × 3 give the same product. You need the fraction of the products that are odd.

  2. List every pair once

    Pair 3 with each bigger number. Then pair 4 with each bigger number, and so on. This way no pair is counted twice.

    3 × 4 = 123 × 5 = 153 × 6 = 183 × 7 = 21
    4 × 5 = 204 × 6 = 244 × 7 = 28
    5 × 6 = 305 × 7 = 35
    6 × 7 = 42

    There are 10 products.

  3. Count the odd products

    The odd products are 15, 21 and 35. That makes 3. They come from the only pairs where both numbers are odd: 3 and 5, 3 and 7, and 5 and 7.

  4. Write the fraction

    Use the fraction formula.

    fraction=310\text{fraction} = \frac{3}{10}

    Watch out: Counting 4 × 3 as well as 3 × 4 makes 20 pairs. Then 320\frac{3}{20}, choice A, mixes two ways of counting. And 35\frac{3}{5}, choice D, is the share of the numbers that are odd, not the products.

    So the answer is B, 310\frac{3}{10}.

Check your answer

Count without listing. Each of the 5 numbers can pair with 4 others, which makes 20. But that counts every pair twice, so there are 20 ÷ 2 = 10 pairs. Only the odd numbers 3, 5 and 7 make odd products, and they form 3 pairs.

More questions like this

The original DOE question
Question 107 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 107. Open page 205 of the handbook (PDF)

Question 110 of 120

2023 Test B Math, question 51

Question

A number, n, is divided by 25. If the quotient is 10 with a remainder of 5, what is the quotient expressed as a decimal?

  • A10.20
  • B10.25
  • C10.50
  • D10.75
Show the answer

Correct answer: A

A quotient of 10 with a remainder of 5 means the quotient is

10+525=1015=10.2010+\frac{5}{25}=10\frac{1}{5}=10.20

Question analysis

A number is divided by 25, giving a quotient of 10 with a remainder of 5, and students write the quotient as a decimal. It tests what a remainder means and how it becomes part of a decimal quotient. Connecting the remainder to the decimal part of the answer is the key idea.

NYS standards
  • 6.NS.2 multi-digit division
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Division with remainders and decimal quotients are grade 6 skills. This takes two short steps, with no word problem to translate, though students must know what a remainder means once a quotient is written as a decimal.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question checks one idea: a remainder is a fraction of the divisor, not a string of digits to place after the point. Students who write the remainder 5 as tenths, or set the divisor 25 after the point, land on listed choices. Rebuilding n = 25 × 10 + 5 = 255 and dividing by 25 confirms the decimal.

  1. Something you must already know moderate Working it out

    A remainder of 5 means 5 out of the 25 needed for another whole, so it becomes 5/25 = 0.2, a link between remainders and decimals the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Rebuild n from the quotient and the remainder, then divide it by 25 to a decimal; or write the remainder as a fraction of the divisor.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Quotient with a remainder
quotient=q+rd\text{quotient} = q + \frac{r}{d}
d is the divisor, the number you divide by. q is the whole-number part of the quotient, the answer to the division. r is the remainder, what is left over. The leftover still has to be divided by d, so it becomes the fraction r over d.

  1. Sort out the words

    n is divided by 25. So 25 is the divisor, the number you divide by. The quotient is the answer to a division. Here it is 10 with a remainder of 5. The remainder is the amount left over. You need the exact quotient as a decimal.

  2. Turn the remainder into a fraction

    The 5 left over still has to be divided by 25. So it becomes the fraction 525\frac{5}{25}. Use the formula.

    quotient=10+525\text{quotient} = 10 + \frac{5}{25}

  3. Change the fraction to a decimal

    Simplify first. Divide the top and bottom by 5. Then write the fraction in hundredths.

    525=15=20100=0.20\frac{5}{25} = \frac{1}{5} = \frac{20}{100} = 0.20

    So the quotient is 10 + 0.20 = 10.20.

  4. Give the answer

    Watch out: A remainder is not the digits after the decimal point. Writing the 5 after the point gives 10.50, choice C. Writing the 25 after it gives 10.25, choice B.

    So the answer is A, 10.20.

Check your answer

Rebuild n. Ten 25s plus the 5 left over is 25×10+5=25525 \times 10 + 5 = 255. Now divide: 255÷25=10.2255 \div 25 = 10.2. That matches 10.20.

More questions like this

The original DOE question
Question 108 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 108. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 111 of 120

2023 Test B Math, question 55

Question

6.34−4.126.\overline{34}-4.\overline{12}

What is the value of the expression above, written as an improper fraction?

  • A2411\frac{24}{11}
  • B224101\frac{224}{101}
  • C11150\frac{111}{50}
  • D209\frac{20}{9}
Show the answer

Correct answer: D

6.34−4.12=2.22=2.26.\overline{34}-4.\overline{12}=2.\overline{22}=2.\overline{2}

x=2.2222...x=2.2222... Set x equal to the repeating decimal.
10x=22.2222...10x=22.2222... Multiply both sides by 10.
−x=−2.2222...-x=-2.2222... Subtract 10x and x.
9x=209x=20 Apply the multiplicative inverse property; divide both sides by 9.
x=209x=\frac{20}{9}

Question analysis

Two repeating decimals are subtracted, and students write the difference as an improper fraction. It tests subtracting repeating decimals and converting a repeating decimal into a fraction, a grade 8 skill. The repeating part has to be handled exactly, so a careful method matters more than speed.

NYS standards
  • 8.NS.1 decimal expansions and irrational numbers
  • 7.NS.2d fractions to decimals by long division
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students change fractions into repeating decimals by long division. Going the other way, from a repeating decimal to a fraction, is grade 8 work, though the subtraction that comes first here is simple.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 18 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The subtraction is easy, since the repeating digits line up and give 2.222…; the conversion is the test. Every choice sits within a few hundredths of the answer, so rounding or estimating cannot separate them, and the exact 2.22 of 111/50 is the strongest pull. Students who know that 0.222… is 2/9 can write 2 2/9 = 20/9 directly.

  1. A very close wrong answer moderate Choosing the answer

    All four choices lie between 2.18 and 2.23: 111/50 is exactly 2.22 and 224/101 is about 2.218, so only 20/9 equals the repeating 2.222… exactly. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know moderate Working it out

    Turning 2.222… into a fraction, by setting x equal to it and subtracting x from 10x, is grade 8 work that the question does not explain. (Anderson & Krathwohl, 2001)

Strategy: Subtract the decimals digit by digit to see which digits repeat, then turn that repeating decimal into a fraction, or divide each choice's numerator by its denominator and compare.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Repeating decimal to fraction
10x−x=9x10x - x = 9x
x is the repeating decimal. Multiplying by 10 moves the decimal point one place to the right. The endless repeating digits still line up after the point. So when you subtract x from 10x, they cancel, and only whole numbers are left.

  1. Sort out the bars

    A bar over digits means they repeat forever. So 6.346.\overline{34} means 6.343434… and 4.124.\overline{12} means 4.121212… with no end. You need the difference as an improper fraction. That is a fraction whose top is bigger than its bottom.

  2. Subtract the decimals

    Line up the digits. Each block of 34 minus 12 gives 22. The whole numbers give 6 − 4 = 2.

    6.343434…−4.121212…=2.222222…6.343434\ldots - 4.121212\ldots = 2.222222\ldots

    The difference is 2.222…, with the 2 repeating forever. You can write it 2.22.\overline{2}.

  3. Name the decimal x

    Let x be the repeating decimal. Then multiply both sides by 10. This moves the decimal point one place to the right.

    x=2.222…x = 2.222\ldots

    10x=22.222…10x = 22.222\ldots

  4. Subtract to remove the repeats

    Subtract the first line from the second. The repeating 2s after the point match up, so they cancel.

    10x−x=22.222…−2.222…10x - x = 22.222\ldots - 2.222\ldots

    9x=209x = 20

  5. Divide to find x

    Divide both sides by 9.

    x=209x = \frac{20}{9}

    Watch out: 11150\frac{111}{50} is exactly 2.22, which stops. The real difference never stops. That trap is choice C. Every choice is close to 2.2, so estimating won’t help.

    So the answer is D, 209\frac{20}{9}.

Check your answer

Divide 20 by 9. 9 goes into 20 two times, with 2 left over. Each next step leaves 2 again, so you get 2.222… forever. That matches the difference. Choice A, 2411\frac{24}{11}, is 2.1818… instead.

More questions like this

The original DOE question
Question 112 as printed on page 207 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 207, question 112. Open page 207 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 112 of 120

2018 Test A Math, question 5

Question

The sum of two consecutive integers is −15. If 1 is added to the smaller integer and 2 is subtracted from the larger integer, what is the product of the two resulting integers?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 63

If x is the smaller consecutive integer, then

x+1x+1

is the larger consecutive integer. Use their sum (−15) to find x:

x+(x+1)=−15x+(x+1)=-15

2x+1=−152x+1=-15

2x=−162x=-16

x=−8x=-8

The two consecutive integers are −8 and −7. One is added to the smaller integer:

−8+1=−7-8+1=-7

and 2 is subtracted from the larger integer:

−7−2=−9-7-2=-9

Find the product:

(−7)(−9)=63(-7)(-9)=63

Question analysis

A grid-in problem about two consecutive integers whose sum is −15. Students find the two integers, add 1 to the smaller and subtract 2 from the larger, then multiply the results. Deciding which negative integer is the smaller one, and keeping signs straight through every step, are what make it tricky.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 A typical grade 7 task solves one two-step equation or multiplies signed numbers. This links both: set up a consecutive-integer equation, solve it with negatives, adjust each integer, and multiply, with no choices to check against.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 9, 30 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is routine, yet the question collects most of the ways students lose points with negatives. The crucial judgment is which of −8 and −7 is the smaller; a student who picks −7 still produces a tidy, wrong product. Gridding the product of the original pair, or giving a negative product, are the other usual losses. Writing each intermediate pair down, rather than holding them in mind, protects against all three.

  1. Error-prone arithmetic moderate Working it out

    Every number in play is negative: −8 and −7, then −7 and −9, and a product of two negatives that must come out positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    "1 is added to the smaller integer and 2 is subtracted from the larger" depends on knowing that −8, not −7, is the smaller, since it lies farther below zero. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Many steps moderate Working it out

    Four linked steps: set up x + (x + 1) = −15, solve for x, adjust each integer, then multiply; an error early in the chain carries through to the product. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Halve the sum to see which two integers sit on either side of it, remember that the smaller negative is farther from zero, then adjust each and multiply.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive integers
n,n+1n, \; n + 1
Integers are whole numbers and their opposites, like −2, −1, 0, 1 and 2. Consecutive integers come one right after another. If n is the smaller one, the next one is n + 1, because each step up the number line adds 1.
Multiplying two negative numbers
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers, so −a and −b are negative. When you multiply two negative numbers, the answer is positive. The two minus signs cancel out.

  1. Sort out the question

    Two consecutive integers add up to −15. Integers are whole numbers and their negatives. Consecutive means one comes right after the other, like 4 and 5. You need to do three things:

    1. Find the two integers.
    2. Add 1 to the smaller one. Subtract 2 from the larger one.
    3. Find the product of the two new numbers. The product is the answer when you multiply.

  2. Find the two integers

    Call the smaller integer n. Then the larger one is n + 1. Their sum is −15.

    n+(n+1)=−15n + (n + 1) = -15

    2n+1=−152n + 1 = -15

    Subtract 1 from both sides. Then divide both sides by 2.

    2n=−162n = -16

    n=−8n = -8

    So the integers are −8 and −7.

  3. Decide which one is smaller

    On a number line, −8 is to the left of −7. Numbers to the left are smaller. So −8 is the smaller integer, and −7 is the larger one. Watch out: 8 is bigger than 7, but −8 is smaller than −7. Mixing them up gives −6 and −10, and the wrong answer 60.

  4. Change each integer

    Add 1 to the smaller integer.

    −8+1=−7-8 + 1 = -7

    Subtract 2 from the larger integer.

    −7−2=−9-7 - 2 = -9

    The two new numbers are −7 and −9.

  5. Multiply the new numbers

    Both numbers are negative, so their product is positive.

    (−7)×(−9)=63(-7) \times (-9) = 63

    Watch out: the answer is not −63, because two negatives make a positive. Also, don’t multiply the first pair, −8 and −7. That gives 56 and skips the changes. So you would grid 63.

Check your answer

Check the pair: −8+(−7)=−15-8 + (-7) = -15, and they are one apart. Half of −15 is −7.5, and −8 and −7 sit on either side of it. Then 7×9=637 \times 9 = 63, and the sign is positive.

More questions like this

The original DOE question
Question 62 as printed on page 81 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 62. Open page 81 of the handbook (PDF)

Question 113 of 120

2018 Test A Math, question 25

Question

Which of the following numbers has factors that include the smallest factor (other than 1) of 91?

  • A30
  • B35
  • C39
  • D44
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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

What a factor is
b=a×cb = 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.

  1. 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÷3=412 \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.

  2. 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=7×1391 = 7 \times 13

    So 7 is the smallest factor of 91 other than 1.

  3. Avoid the 3 trap

    Watch out: 91 is close to 90, and 3 goes into 90. But 91÷391 \div 3 is 30 with 1 left over. So 3 is not a factor of 91. That trap leads to 30, choice A.

  4. Check each choice for 7

    Now look for a multiple of 7 among the choices.

    • 30: no. 7×4=287 \times 4 = 28 and 7×5=357 \times 5 = 35.
    • 35: yes. 35=7×535 = 7 \times 5.
    • 39: no. 39=3×1339 = 3 \times 13.
    • 44: no. 7×6=427 \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: 7×13=917 \times 13 = 91 and 7×5=357 \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 114 of 120

2018 Test A Math, question 28

Question

215+3310+425+5122\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}

What is the value of the expression shown above?

  • A1472014\frac{7}{20}
  • B142514\frac{2}{5}
  • C1572015\frac{7}{20}
  • D152515\frac{2}{5}
Show the answer

Correct answer: D

215+3310+425+5122\frac{1}{5}+3\frac{3}{10}+4\frac{2}{5}+5\frac{1}{2}

Convert all the fractions to a common denominator (10):

2210+3310+4410+55102\frac{2}{10}+3\frac{3}{10}+4\frac{4}{10}+5\frac{5}{10}

=(2+3+4+5)+(2+3+4+510)=(2+3+4+5)+\left(\frac{2+3+4+5}{10}\right)

=14+1410=1525=14+1\frac{4}{10}=15\frac{2}{5}

Question analysis

A fractions question: add four mixed numbers whose fractions have denominators of 5, 10, 5 and 2. Students rewrite the fractions over a common denominator, add the whole numbers and the fractions, regroup any extra whole, and simplify. Regrouping and lowest terms both matter, since the choices differ in exactly those places.

NYS standards
  • 7.NS.1d properties for adding and subtracting
  • 7.EE.3 multi-step rational number problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding mixed numbers with unlike denominators is grade 5 content. With four terms and a regrouping step it takes care, but it asks less than typical grade 7 work with negative rational numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are friendly: every fraction converts to tenths, and the whole parts add to 14. The risk is in the bookkeeping, either forgetting to carry the extra whole from the fractions or adding numerators over a made-up denominator. Converting to decimals, 2.2 + 3.3 + 4.4 + 5.5, is a fast check that avoids both.

  1. Error-prone arithmetic moderate Working it out

    Four mixed numbers with denominators 5, 10, 5 and 2 must be rewritten as tenths, 2 + 3 + 4 + 5 tenths, and the 14/10 regrouped into a whole and a fraction. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    15 2/5 and 14 2/5 are written alike and share the fraction 2/5; only the whole carried from 14/10 separates them, so a missed carry is invisible at a glance. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. An answer in an unexpected form minor Choosing the answer

    The fractions total 1 4/10, but every choice is in lowest terms, so the 4/10 has to be reduced to 2/5 before it matches. (Enright, Morley & Sheehan, 2002)

Strategy: Convert each mixed number to a decimal, since these fractions all convert exactly, add them, and turn the decimal part back into a fraction in lowest terms.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions
ab=a×nb×n\frac{a}{b} = \frac{a \times n}{b \times n}
a is the top number, called the numerator. b is the bottom number, called the denominator. n is any whole number except zero. Multiplying the top and bottom by the same n gives an equal fraction. Use it to give fractions the same denominator so you can add them.

  1. Sort out the question

    You need to add four mixed numbers. A mixed number is a whole number plus a fraction, like 2152\frac{1}{5}. The fractions have different denominators, or bottom numbers: 5, 10, 5 and 2. To add fractions, they need the same denominator.

  2. Add the whole numbers

    Add the whole numbers first. You will add the fractions on their own after that.

    2+3+4+5=142 + 3 + 4 + 5 = 14

    The whole numbers make 14.

  3. Rewrite each fraction in tenths

    Use 10 as the denominator, since 5 and 2 both go into 10.

    15=1×25×2=210\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10}

    25=2×25×2=410\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}

    12=1×52×5=510\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}

    310\frac{3}{10} is already in tenths. Watch out: when you change a bottom number, change the top too. Adding the old tops, 1 + 3 + 2 + 1 = 7, and putting them over 20 gives 7/20. That is in choices A and C.

  4. Add the fractions

    210+310+410+510=1410\frac{2}{10} + \frac{3}{10} + \frac{4}{10} + \frac{5}{10} = \frac{14}{10}

    14 tenths is more than 1 whole. Ten tenths make 1, so 14 tenths is 1 whole and 4 tenths.

    1410=1410=125\frac{14}{10} = 1\frac{4}{10} = 1\frac{2}{5}

    To simplify 4/10, divide the top and bottom by 2.

  5. Put it all together

    Add the 14 wholes to the sum of the fractions.

    14+125=152514 + 1\frac{2}{5} = 15\frac{2}{5}

    Watch out: carry the extra whole into the 14. Missing it gives 142514\frac{2}{5}, choice B. So the answer is D, 152515\frac{2}{5}.

Check your answer

Change each mixed number to a decimal: 2.2+3.3+4.4+5.5=15.42.2 + 3.3 + 4.4 + 5.5 = 15.4. The 0.4 is 4 tenths, which is 2/5. So the sum is 152515\frac{2}{5}.

More questions like this

The original DOE question
Question 85 as printed on page 86 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 85. Open page 86 of the handbook (PDF)

Question 115 of 120

2018 Test A Math, question 31

Question

How many positive even factors of 48 are greater than 24 and less than 48?

  • A0
  • B1
  • C2
  • D12
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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pairs
a×b=na \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.

  1. 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.

  2. List the factors of 48 in pairs

    Start with 1 and work up. Each pair multiplies to 48:

    1×482×243×164×126×81 \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.

  3. 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.

  4. 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 116 of 120

2018 Test A Math, question 32

Question

The least of 5 consecutive integers is ll, and the greatest is gg. What is the value of l+g2\frac{l+g}{2} in terms of ll?

  • A2l2l
  • B3l3l
  • Cl+2l+2
  • Dl+5l+5
Show the answer

Correct answer: C

The first integer is ll, so the second is l+1l+1, the third is l+2l+2, then l+3l+3, and finally l+4l+4.

Since gg is the fifth and greatest of the integers,

g=l+4g=l+4

Substitute l+4l+4 for gg and simplify:

l+g2=l+l+42=2l+42=l+2\frac{l+g}{2}=\frac{l+l+4}{2}=\frac{2l+4}{2}=l+2

Question analysis

This algebra question about five consecutive integers calls the least one l and the greatest g, then asks for (l + g)/2 written in terms of l. Students must express g using l and simplify a fraction that contains a variable. Counting how far the fifth integer is from the first is where many students slip.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 7.EE.2 rewriting expressions to show relationships
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students combine like terms in expressions such as 3(x + 4) − x. This item first asks them to build the expression from a description of consecutive integers, then simplify it, with only letters to work with.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 26 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The algebra is short once g is written as l + 4, so most errors come from the fence-post count or from testing numbers carelessly. The wrong choices in front of the answer are exactly the ones a single test value cannot rule out, which punishes students who stop at the first match. Using two test values, or simplifying (2l + 4)/2 directly, avoids both traps.

  1. Variables instead of numbers moderate Understanding the question

    Every expression is in l, so the simplification of (l + l + 4)/2 has to be done with letters and cannot be checked against a single number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The description "the least of 5 consecutive integers is l, and the greatest is g" must become g = l + 4, four steps up, not five. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Every choice must be checked minor Choosing the answer

    A single test value fits two choices: l = 1 makes 3l and l + 2 both 3, and l = 2 makes 2l and l + 2 both 4. (Embretson & Wetzel, 1987)

Strategy: Write the integers as l, l + 1, and so on up to the fifth, substitute for g and simplify; then test two different starting values, since one value can match a wrong choice by coincidence.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest of 5 consecutive integers
g=l+4g = l + 4
l is the least (smallest) integer and g is the greatest. Consecutive integers go up by 1 each time. Five integers have only 4 jumps from the first to the last, so you add 4, not 5.
Dividing a sum
a+bc=ac+bc\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}
a and b are the two parts of the top, and c is the number you divide by. To divide a sum, divide each part. You must divide every part of the top, not just one.

  1. Sort out the words

    Integers are whole numbers, like −2, 0 and 7. Consecutive integers come one right after another, like 8, 9, 10. There are 5 of them here.

    The smallest is called l and the biggest is called g. You must write l+g2\frac{l + g}{2} using only the letter l.

  2. Write all five integers

    Each integer is 1 more than the one before:

    l, l+1, l+2, l+3, l+4l,\ l+1,\ l+2,\ l+3,\ l+4

    The fifth one is the greatest, so g = l + 4. Watch out: five integers have only four jumps between them. Writing g = l + 5 leads to choice D.

  3. Put l + 4 in place of g

    Swap g for l + 4 in the fraction:

    l+g2=l+l+42=2l+42\frac{l + g}{2} = \frac{l + l + 4}{2} = \frac{2l + 4}{2}

    Here l + l is 2l, which means 2 times l.

  4. Split the fraction

    Divide each part of the top by 2:

    2l2+42=l+2\frac{2l}{2} + \frac{4}{2} = l + 2

    So the answer is C, l+2l + 2.

Check your answer

Test two starting numbers. If l = 1, the integers are 1 to 5, and (1 + 5) ÷ 2 = 3, which is 1 + 2. If l = 10, they are 10 to 14, and (10 + 14) ÷ 2 = 12, which is 10 + 2. Use two tests, because l = 1 also makes 3l equal 3, choice B.

More questions like this

The original DOE question
Question 89 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 89. Open page 87 of the handbook (PDF)

Question 117 of 120

2018 Test A Math, question 33

Question

Johan leased a car for three years. He paid a one-time fee of $1,000, and an additional $300 per month for the full three years. At the end of the three years, what is the total amount Johan paid for leasing this car?

  • A$1,900
  • B$4,600
  • C$10,800
  • D$11,800
Show the answer

Correct answer: D

Three years is 36 months (12 × 3).

Set up an expression to find the total amount Johan paid:

1,000+300(36)=$11,8001{,}000+300(36)=\$11{,}800

Question analysis

This word problem about leasing a car combines a one-time fee with a monthly payment over three years and asks for the total amount paid. Students must convert years to months, multiply by the monthly payment and add the fee. Leaving out one piece of the cost, or mixing years with months, is the usual slip.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 This is a short total-cost calculation with whole numbers, easier than typical grade 7 word problems, which bring in percents, negative numbers or fractions. The only care needed is turning years into months.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
3 steps
Reading level
Question and choices: about grade 4, 41 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is the easiest kind of multi-step word problem: whole numbers, a familiar money situation and a direct method. The only places to slip are changing three years into 36 months and remembering the one-time fee after the long multiplication. Each slip has its own choice, so a quick check that both costs are in the total is worth the seconds.

  1. Much to keep track of minor Working it out

    The $1,000 fee is paid once while the $300 repeats monthly, so it must be held aside while 36 × 300 is computed. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. A very close wrong answer minor Choosing the answer

    The choices $11,800 and $10,800 differ by only $1,000 out of about $11,000, so a rough total cannot separate them; only including the one-time fee does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Count the months in the lease first, multiply the monthly payment by that number, and then add the one-time fee.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a fee
T=f+m×nT = f + m \times n
T is the total paid. f is the one-time fee, paid only once. m is the monthly payment. n is the number of months. You multiply first because the monthly payment repeats every month. Then you add the fee once.

  1. Sort out what you know

    Johan paid $1,000 one time. He also paid $300 every month for 3 years. You need the total he paid.

    The payments are counted by the month, but the time is given in years. So first change the years to months.

  2. Count the months

    Each year has 12 months:

    3×12=363 \times 12 = 36

    So Johan made 36 monthly payments. Watch out: multiplying $300 by 3 years instead of 36 months gives $900. Adding the fee then gives $1,900, which is choice A.

  3. Add up the monthly payments

    Multiply the monthly payment by the number of months:

    300×36=10,800300 \times 36 = 10{,}800

    The monthly payments add up to $10,800.

  4. Add the one-time fee

    Use the formula. The fee goes in once:

    T=1,000+300×36=1,000+10,800=11,800T = 1{,}000 + 300 \times 36 = 1{,}000 + 10{,}800 = 11{,}800

    Watch out: $10,800 is choice C. It leaves out the $1,000 fee. So the answer is D, $11,800.

Check your answer

Round to check. $300 a month for 36 months is a bit more than $10,000. Add the $1,000 fee and you get a bit more than $11,000. $11,800 fits. Choices A and B are far too small for three years of payments.

More questions like this

The original DOE question
Question 90 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 90. 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 118 of 120

2018 Test A Math, question 36

Question

Number line with two points: L, labeled 4 and 5/16, and N to the right of L, with no value given.

On the number line above, LN = 18\frac{1}{8} . Point M (not shown) is located between point L and point N. Which value below is a possible value for M?

  • A4.26
  • B4.31
  • C4.35
  • D4.58
Show the answer

Correct answer: C

Since LN=18LN=\frac{1}{8}, point NN is located at

4516+18=47164\frac{5}{16}+\frac{1}{8}=4\frac{7}{16}.

So MM must be between point LL, 45164\frac{5}{16}, and point NN, 47164\frac{7}{16}. Point LL can also be written as 4.3125, and point NN can be written as 4.4375. The only option given that lies between those two points is 4.35.

Question analysis

This number line question says the distance from point L to point N is one eighth and asks which decimal could be the position of a point M between them. Students take L's position, a mixed number in sixteenths, find N, and compare with decimal choices. Converting sixteenths to decimals precisely is the tricky part.

NYS standards
  • 7.NS.2d fractions to decimals by long division
  • 7.NS.1b rational number addition
  • 6.NS.7a inequality as relative position
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students convert fractions to decimals and compare them. This item needs an exact conversion of sixteenths and a comparison to the thousandths, where rounding too early can mislead, but it stays within grade 7 content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The interval from L to N is only an eighth wide, and three of the four choices crowd near it, so precision decides the question. Students comfortable working in sixteenths can avoid decimals until the end, since 4.35 lies between 4 5/16 and 4 7/16. Those who convert loosely, or who misread a mixed number as a decimal, are pulled to a neighboring choice.

  1. Error-prone arithmetic moderate Working it out

    Converting 4 5/16 to 4.3125 and N to 4.4375 needs four decimal places, while the choices stop at hundredths, so early rounding blurs the left boundary. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The choices 4.31 and 4.35 both lie within 0.04 of L, so estimating cannot separate them; only the exact 4.3125 shows 4.31 just outside the interval and 4.35 inside. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Reading a figure, table or graph minor Understanding the question

    L's value appears only on the number line, and the picture shows N to the right of L, which fixes which way the one-eighth gap runs. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add one eighth, or two sixteenths, to L's position to locate N, convert both endpoints to four-place decimals, and keep only a choice strictly between them.

Traps

  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • Answer in another form: The right answer is written differently from the way you worked it out.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top number (numerator) and b is the bottom number (denominator). Multiplying both by the same number k gives an equal fraction. It is like cutting each piece into k smaller pieces. Use it to give fractions the same bottom before you add or subtract.
Fraction to decimal
ab=a÷b\frac{a}{b} = a \div b
A fraction bar means divide. a is the top number and b is the bottom number. Dividing the top by the bottom turns the fraction into a decimal. Then you can compare it with decimal choices.

  1. Sort out what you know

    Point L is at 45164\frac{5}{16}. Point N is to the right of L. The distance from L to N, called LN, is 18\frac{1}{8}.

    Point M is somewhere between L and N. You need a choice that fits between them. The choices are decimals, so you will change the fractions into decimals.

  2. Find where N is

    N is 18\frac{1}{8} to the right of L, so add. First give 18\frac{1}{8} a bottom of 16, to match L:

    18=1×28×2=216\frac{1}{8} = \frac{1 \times 2}{8 \times 2} = \frac{2}{16}

    4516+216=47164\frac{5}{16} + \frac{2}{16} = 4\frac{7}{16}

    Watch out: N is to the right of L in the figure. Going left instead gives a gap that holds 4.26, choice A.

  3. Change both points to decimals

    Divide the top of each fraction by the bottom:

    516=5÷16=0.3125\frac{5}{16} = 5 \div 16 = 0.3125

    716=7÷16=0.4375\frac{7}{16} = 7 \div 16 = 0.4375

    So L is at 4.3125 and N is at 4.4375.

  4. Test each choice

    M must be bigger than 4.3125 and smaller than 4.4375.

    • 4.26 is smaller than 4.3125. It is left of L.
    • 4.31 is 4.3100, just smaller than 4.3125. It is a hair left of L.
    • 4.35 is between them. It fits.
    • 4.58 is bigger than 4.4375. It is right of N.

    Watch out: if you round L to 4.31, choice B seems to fit. Keep all four decimal places. So the answer is C, 4.35.

Check your answer

Subtract to be sure. 4.35 − 4.3125 = 0.0375, which is more than 0, so 4.35 is right of L. 4.4375 − 4.35 = 0.0875, which is also more than 0, so 4.35 is left of N. It sits between them.

More questions like this

The original DOE question
Question 93 as printed on page 88 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 88, question 93. Open page 88 of the handbook (PDF)

Question 119 of 120

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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of sticks
n=dsn = \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.

  1. 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.

  2. 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.

  3. 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÷6=1230÷6=572 \div 6 = 12 \qquad 30 \div 6 = 5

    So 6 is the greatest common factor.

  4. 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 120 of 120

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?

  • A5
  • B6
  • C13
  • D15
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.
What makes it hard
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.

  1. 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)

  2. 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

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+1n = 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.

  1. 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.

  2. Count the numbers in the set

    n=30−12+1=19n = 30 - 12 + 1 = 19

    There are 19 integers. Watch out: 30 − 12 = 18 forgets one end. That slip leads to choice A, 5.

  3. 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.

  4. 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.

  5. 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)