Practice Test

2023 Handbook

2023 Test B Math

Question 1 of 57

2023 Test B Math, question 1

Question

LAST YEAR’S TOTAL SALES

Vehicles	Number
Four-door cars	5,850
Two-door cars	2,250
Trucks	900
Total	9,000

The table above shows the number of each type of vehicle sold at a dealership last year. If 1,000 vehicles will be sold next month, what is the best estimate (based on last year’s sales) of the number of two-door cars that will be sold?

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

Show the answer

Correct answer: 250

Let x represent the number of two-door cars that will be sold next month. Use the information in the table to set up a proportion:

$\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000}$

$x = \frac{(2,250)(1,000)}{9{,}000} = \frac{2{,}250}{9} = 250$

Question analysis

A table shows last year's sales at a car dealership: four-door cars, two-door cars and trucks, 9,000 vehicles in all. This grid-in asks for the best estimate of how many two-door cars will be sold next month if 1,000 vehicles are sold. Students set up a proportion from the table; most of its rows are not needed.

NYS standards: 6.RP.3 ratio and rate reasoning
6.SP.7 long-run relative frequency
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 proportion problems often take several steps or awkward numbers. This one scales a single share from a table to a new total, with numbers that divide cleanly, which is closer to grade 6 ratio work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 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 share is seen as 2,250 out of 9,000, a quarter, the arithmetic is light, which makes this one of the easier grid-ins. Points are lost by entering the share itself as a percent, by reading the wrong row of the table, or by building a monthly average when the question supplies next month’s total. With no choices to compare, checking that the entry is a quarter of 1,000 is the only safeguard.

Turning words into math minor Understanding the question
The phrase “best estimate (based on last year’s sales)” has to become a proportion: two-door cars keep the same share of next month’s 1,000 vehicles as of last year’s 9,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The table lists three vehicle types and a total; only the two-door row, 2,250, and the 9,000 total are needed, so the four-door and truck rows are distractions. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find what fraction of last year's 9,000 vehicles were two-door cars, simplify it, then take that fraction of the 1,000 vehicles expected next month.

Traps

Stops one step early: A choice is a number from partway through the solution.
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

25 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds that 2,250 of 9,000 sales is 25 percent and enters that percent, stopping before taking a quarter of next month’s 1,000 vehicles.
650 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads the four-door row instead of the two-door row and scales it correctly: 5,850 of 9,000 is 650 of 1,000.
187.5 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Divides last year’s 2,250 two-door sales by 12 months, a monthly average that ignores the stated 1,000 vehicles for next month.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Proportion: $\frac{\text{part}}{\text{whole}} = \frac{\text{part}}{\text{whole}}$; A proportion says two fractions are equal. Each fraction here is a part over its whole: two-door cars over all vehicles sold. Use it when a share stays the same while the total changes. Then you can find the missing part.

Sort out the question
The table shows last year’s sales. In all, 9,000 vehicles were sold, and 2,250 of them were two-door cars. Next month, 1,000 vehicles will be sold. You need about how many of them will be two-door cars.
“Based on last year’s sales” means two-door cars keep the same share of all sales. A share is a part out of a whole. The four-door and truck rows don’t matter here.
Find last year’s share of two-door cars
Last year, 2,250 out of 9,000 vehicles were two-door cars. Write that as a fraction and simplify it.
$\frac{2{,}250}{9{,}000} = \frac{1}{4}$
This works because $2{,}250 \times 4 = 9{,}000$ . So 1 out of every 4 vehicles sold was a two-door car.
Set up the proportion and solve
Let x be the number of two-door cars next month. Next month’s share must match last year’s share.
$\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000} = \frac{1}{4}$
To get x alone, multiply both sides by 1,000.
$x = \frac{1}{4} \times 1{,}000 = 250$
So about 250 two-door cars will be sold.
Give the answer
Watch out: one fourth is the same as 25%. Gridding 25 gives the percent, not the number of cars. Also be sure to use the two-door row. The four-door row would give 650. So you would grid 250.

Check your answer

Try it another way. Next month’s 1,000 vehicles are $\frac{1}{9}$ of last year’s 9,000. So take $\frac{1}{9}$ of the 2,250 two-door cars: $2{,}250 \div 9 = 250$ . Both ways give 250 cars.

More questions like this

The original DOE question

Question 58 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 58. Open page 191 of the handbook (PDF)

Question 2 of 57

2023 Test B Math, question 2

Question

In a scale drawing of a rectangular garden, the length is 15 inches and the width is 9 inches. In the drawing, 2 inches represents 3 yards. What is the width of the actual garden, in yards?

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

Show the answer

Correct answer: 13.5

Let w represent the width of the actual garden. Set up a proportion based on the information given:

$\frac{w}{9} = \frac{3}{2}$

$w = \frac{3(9)}{2} = \frac{27}{2} = 13.5$

Question analysis

In a scale drawing of a rectangular garden, 2 inches stands for 3 yards, and the drawing measures 15 inches by 9 inches. This grid-in asks for the real width of the garden in yards. Students apply the scale to one side only: the length is given but plays no part, and the scale has to be used the right way round.

NYS standards: 7.G.1 scale drawings
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Scale drawings are grade 7 content, and this is a typical one: a single length converted with a stated scale. The only extra demand is using the width rather than the length and entering the result with no choices to check.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The drawing gives two lengths and only the width matters, so the first decision is which number to scale. The second is direction: the real garden must measure more in yards than the drawing’s 9 in inches, which a student can check before entering anything. Errors come from scaling the length, turning the scale over, or treating 2 to 3 as an increase of one.

Turning words into math minor Understanding the question
“2 inches represents 3 yards” must become 1.5 yards for every inch of drawing, so the 9-inch width is multiplied by 3/2, not by 2/3. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The answer is a decimal entered with no choices to compare, so a result from the 15-inch length or from an upside-down scale looks just as finished. (Enright, Morley & Sheehan, 2002)

Strategy: Find how many yards one inch of the drawing stands for, then multiply the drawing's width, not its length, by that number.

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).
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

22.5 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Applies the scale to the 15-inch length instead of the 9-inch width: 15 × 3 ÷ 2 = 22.5 yards.
6 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Turns the scale upside down, multiplying the 9 inches by 2/3 instead of 3/2, so the real garden comes out smaller than the drawing’s number.
10 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Reads 2 inches to 3 yards as adding 1, keeping a constant difference instead of a constant ratio, so 9 inches becomes 9 + 1 = 10 yards.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scale proportion: $\frac{\text{real size}}{\text{drawing size}} = \frac{\text{real size}}{\text{drawing size}}$; A scale drawing shrinks everything by the same amount. So real size over drawing size makes the same fraction for every part of the drawing. Keep the real sizes on top on both sides, and the drawing sizes on the bottom.

Sort out the question
The drawing shows a rectangle, a four-sided shape with square corners. On the drawing, the length is 15 inches and the width is 9 inches. The scale says 2 inches on the drawing stand for 3 yards in real life. You need the real width, in yards.
Only the width matters. The 15-inch length is extra.
Find what one inch stands for
If 2 inches stand for 3 yards, then 1 inch stands for half as much.
$3 \div 2 = 1.5$
So each inch on the drawing is 1.5 yards in real life. That means the number of yards will be bigger than the number of inches.
Set up the proportion and solve
Let w be the real width, in yards. Put yards on top and inches on the bottom, on both sides.
$\frac{w}{9} = \frac{3}{2}$
To get w alone, multiply both sides by 9.
$w = \frac{3 \times 9}{2} = \frac{27}{2} = 13.5$
The real width is 13.5 yards.
Give the answer
Watch out: don’t flip the scale. Using $9 \times \frac{2}{3}$ gives 6, which is smaller than the drawing’s number. And don’t use the length: 15 inches would give 22.5 yards. So you would grid 13.5.

Check your answer

Count groups of 2 inches. The 9-inch width holds 4.5 groups of 2 inches. Each group stands for 3 yards, so the real width is $4.5 \times 3 = 13.5$ yards. That matches.

More questions like this

The original DOE question

Question 59 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 59. Open page 191 of the handbook (PDF)

Question 3 of 57

2023 Test B Math, question 3

Question

A kindergarten teacher has 72 crayons, 60 pencils, and 84 sheets of paper to distribute to the children in her class. If each child receives an equal number of each item and there are no items remaining, what is the greatest possible number of children in the class?

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

Show the answer

Correct answer: 12

The greatest possible number of children in the class is the greatest common factor of 72, 60, and 84. Write the prime factorization of each number:

$72 = 2^{3} \times 3^{2}$

$60 = 2^{2} \times 3 \times 5$

$84 = 2^{2} \times 3 \times 7$

Then, find the factors that all three have in common, which is $2^{2} \times 3 = 12$ .

Question analysis

A kindergarten teacher shares 72 crayons, 60 pencils and 84 sheets of paper so that every child gets the same number of each item, with nothing left over. This grid-in asks for the greatest possible number of children in the class. Students must recognize a greatest common factor problem and not confuse it with a least common multiple.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Greatest common factors are grade 6 content, normally of two numbers. This one uses three numbers and a word setting that must be translated, but the factoring is short, so the demand stays below typical grade 7 work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 11, 45 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The story has to be recognized as a sharing problem before any arithmetic starts, and the word greatest points to the largest divisor shared by all three counts. Students who spot a small shared factor can stop too soon, while those who think of a common multiple head the wrong way entirely. Checking that 72, 60 and 84 all divide by the entry, and that no larger number does, settles it.

Turning words into math moderate Understanding the question
Nothing names a greatest common factor; “each child receives an equal number of each item and there are no items remaining” means the class size must divide 72, 60 and 84 exactly. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Finding the greatest common factor of three numbers, by prime factorization or by listing factors, is a grade 6 method the question assumes rather than explains. (Anderson & Krathwohl, 2001)

Strategy: Write 72, 60 and 84 as products of primes, multiply the primes all three share, and check that each amount divides evenly by the result.

Traps

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

The wrong answers, classified

6 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Spots that 6 divides 72, 60 and 84 and stops there, a common factor not yet multiplied by the extra 2 that all three share.
2520 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Finds the least common multiple of 72, 60 and 84, 2,520, confusing a number all three counts divide into with one that divides all three.
4 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Takes only the shared 2 × 2 from the prime factorizations and drops the factor 3 that 72, 60 and 84 also share.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor: $\text{GCF} = \text{product of the shared prime factors}$; A factor of a number divides it with nothing left over. The GCF is the greatest number that is a factor of every number in the group. A prime number has only 1 and itself as factors, like 2, 3, 5 and 7. Multiply the primes that all the numbers share, as many times as each number has them.

Sort out the question
The teacher has 72 crayons, 60 pencils and 84 sheets of paper. Every child gets the same number of each item, with none left over.
So the number of children must divide 72, 60 and 84 with no remainder. A number like that is a common factor. You need the greatest one.
Break each number into primes
Write each number as a product of primes. A product is the answer to a multiplication.
$72 = 2 \times 2 \times 2 \times 3 \times 3$
$60 = 2 \times 2 \times 3 \times 5$
$84 = 2 \times 2 \times 3 \times 7$
You can check each one by multiplying it back out.
Pick out the shared primes
Look for the primes that appear in all three lists. Each list has at least two 2s and one 3. Only 72 has a third 2 and a second 3. The 5 and the 7 appear in one list each.
$\text{GCF} = 2 \times 2 \times 3 = 12$
Watch out: keep every shared prime. Using only one 2 gives 6. Leaving out the 3 gives 4.
Give the answer
Watch out: 2,520 is the smallest number that all three numbers divide into. That is the least common multiple, a different idea. The class can’t have 2,520 children when there are only 60 pencils. So you would grid 12.

Check your answer

Share the items among 12 children. Each child gets $72 \div 12 = 6$ crayons, $60 \div 12 = 5$ pencils and $84 \div 12 = 7$ sheets. Nothing is left over. The numbers 6, 5 and 7 share no factor but 1, so no bigger class works.

More questions like this

The original DOE question

Question 60 as printed on page 191 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 60. Open page 191 of the handbook (PDF)

Question 4 of 57

2023 Test B Math, question 4

Question

Mr. Smith opened a retirement account with a deposit of $900. This account earns 5% simple interest annually. How many years will it take for his $900 deposit to earn $360 interest?

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

Show the answer

Correct answer: 8

Calculate simple interest (y) by multiplying the initial deposit (p) by the interest rate (r) by the number of years (t):

$prt = y$

$900(0.05)t = 360$

$45t = 360$

$t = 8$

Question analysis

Mr. Smith deposits $900 in a retirement account that earns 5% simple interest each year. This grid-in asks how many years it takes for the deposit to earn $360 in interest. Students find one year's interest and then how many years add up to the total, and the percent must become the right decimal.

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 Simple interest is grade 7 percent work, and a typical problem asks for the interest earned over a given time. This one runs the formula backward to find the time, one extra step, with friendly numbers.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Turning words into math minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 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 numbers are friendly once one year’s interest is found, so most of the difficulty is setting up the right relationship. Students who divide $360 by 5 treat the rate as a dollar amount, and students who compare $360 with $900 stop at a percent. Multiplying back, eight years of $45, confirms the $360.

Something you must already know minor Working it out
Simple interest is the deposit times the rate times the years; the stem does not give this rule, and it must be known to see that each year adds the same $45. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The question gives the total interest, $360, and asks for the time, so the usual interest calculation has to be run backward, as 45t = 360. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Writing 5% as 0.05, not 0.5 or 0.005, is the one place-value decision, and a slip there changes the yearly interest by a power of ten. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work out the interest for one year, 5 percent of $900, then divide the $360 target by that yearly amount.

Traps

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

The wrong answers, classified

40 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds that $360 is 40 percent of the $900 deposit and enters 40, stopping before dividing by the 5 percent earned each year.
72 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Reads 5% as $5 a year, a fixed amount instead of a share of the $900, and divides 360 by 5.
80 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Writes 5% as 0.005, so a year’s interest seems to be $4.50, and 360 ÷ 4.5 = 80 years.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Simple interest: $I = P \times r \times t$; I is the interest, the money the account earns. P is the deposit you start with. r is the yearly rate, written as a decimal. t is the number of years. With simple interest, every year earns the same amount, because it is always based on the first deposit.

Sort out the question
Mr. Smith puts $900 into an account. It earns 5% simple interest each year. Interest is extra money the bank pays you for keeping your money there. You need the number of years until the interest adds up to $360.
Write 5% as a decimal
Percent means “out of 100.” So 5% is 5 out of 100.
$\frac{5}{100} = 0.05$
Watch out: 5% is not 0.5 or 0.005. Using 0.005 makes a year’s interest $4.50, and that leads to 80 years.
Find the interest for one year
Each year the account earns 5% of $900.
$900 \times 0.05 = 45$
So it earns $45 every year. Watch out: 5% is not $5 a year. Dividing 360 by 5 gives 72, which is wrong.
Use the formula to find the years
Put the numbers into the formula. I is 360, P is 900 and r is 0.05.
$I = P \times r \times t$
$360 = 900 \times 0.05 \times t$
$360 = 45 \times t$
To undo “times 45,” divide both sides by 45.
$t = 360 \div 45 = 8$
Give the answer
It takes 8 years. Watch out: $360 is 40% of $900. Gridding 40 stops too soon, because each year adds only 5%. So you would grid 8.

Check your answer

Add up the interest year by year. $45 a year for 8 years is $45 \times 8 = 360$ dollars. Also, 8 years at 5% a year is 40%, and 40% of $900 is $360.

More questions like this

The original DOE question

Question 61 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 61. Open page 192 of the handbook (PDF)

Question 5 of 57

2023 Test B Math, question 5

Question

Solve the equation for x:

$\frac{(21)(14)x}{(49)(48)} = 0.875$

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

Show the answer

Correct answer: 7

$\frac{(21)(14)x}{(49)(48)} = 0.875$

$\left(\frac{21}{49}\right)\left(\frac{14}{48}\right)x = 0.875$

$\left(\frac{3}{7}\right)\left(\frac{7}{24}\right)x = 0.875$

$\left(\frac{1}{8}\right)x = 0.875$

$x = (0.875)(8) = 7$

Question analysis

This equation puts x in a fraction with (21)(14) on top and (49)(48) below, set equal to the decimal 0.875. The grid-in asks for the value of x. Students do best to cancel common factors before multiplying anything out, and it helps to know the fraction that 0.875 equals.

NYS standards: 7.EE.3 multi-step rational number problems
7.NS.2c properties for multiplying and dividing
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Seventh graders solve equations like ax = b, usually with a simple coefficient. Here the coefficient is a product of four numbers that is best cancelled first, and a decimal has to be matched with a fraction.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
No obvious method minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 6 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a one-unknown equation dressed in large numbers, and the work is in reducing them without slipping. Students who cancel early find that the whole coefficient is one eighth and that 0.875 is seven eighths, which makes the last step mental. Those who multiply out face a four-digit product and a decimal division, and with nothing to check against, an early slip goes unnoticed.

Error-prone arithmetic moderate Working it out
Multiplied out, the coefficient is 294 over 2,352 and the right side is the decimal 0.875; both invite slips unless 21/49 and 14/48 are cancelled first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
No obvious method minor Working it out
The quick route pairs 21 with 49 and 14 with 48 before multiplying, so that 3/7 × 7/24 collapses to 1/8; the layout of the fraction does not suggest that pairing. (Schoenfeld, 1985)
An answer in an unexpected form minor Choosing the answer
The result is entered with no choices, so a coefficient’s denominator or a quotient taken the wrong way looks as complete as the right value. (Enright, Morley & Sheehan, 2002)

Strategy: Cancel before multiplying: reduce 21 over 49 and 14 over 48, multiply the simplified fractions, then undo that coefficient, writing 0.875 as a fraction if it helps.

Traps

Stops one step early: A choice is a number from partway through the solution.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

8 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Simplifies the coefficient correctly to 1/8 but enters the 8 from its denominator, never multiplying it by 0.875.
0.109375 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides 0.875 by 8 instead of multiplying, undoing the coefficient 1/8 the wrong way, so x comes out as 0.109375.
3.5 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Slips in reducing 14/48, taking it as 7/12 instead of 7/24, so the coefficient becomes 1/4 and x = 0.875 × 4 = 3.5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying fractions: $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$; a and c are the tops, or numerators. b and d are the bottoms, or denominators. Multiply the tops together and the bottoms together. It also works backward: you can split one fraction into two smaller ones. That lets you simplify before the numbers get big.

Sort out the question
You need the value of x. On top, 21, 14 and x are multiplied. The bottom is 49 times 48. The right side is 0.875.
Multiplying 49 by 48 makes a big number. It is easier to simplify first.
Split the fraction into two
Use the rule for multiplying fractions backward. Pair 21 with 49, and 14 with 48.
$\frac{21 \times 14}{49 \times 48} = \frac{21}{49} \times \frac{14}{48}$
Now the equation says these two fractions times x equal 0.875.
Simplify each fraction
To simplify, divide the top and bottom by the same number. Both 21 and 49 divide by 7. Both 14 and 48 divide by 2.
$\frac{21}{49} = \frac{3}{7} \qquad \frac{14}{48} = \frac{7}{24}$
Watch out: 48 ÷ 2 is 24, not 12. Writing $\frac{7}{12}$ leads to 3.5.
Multiply the simple fractions
$\frac{3}{7} \times \frac{7}{24} = \frac{3 \times 7}{7 \times 24} = \frac{3}{24} = \frac{1}{8}$
The 7 on top and the 7 on the bottom cancel, since 7 ÷ 7 = 1. Then divide 3 and 24 by 3 to get $\frac{1}{8}$ . So the equation becomes:
$\frac{1}{8} \times x = 0.875$
Solve for x
One eighth of x is 0.875. To undo dividing by 8, multiply both sides by 8.
$x = 0.875 \times 8 = 7$
Watch out: 8 is not the answer, and don’t divide 0.875 by 8. So you would grid 7.

Check your answer

0.875 is the same as $\frac{7}{8}$ . One eighth of 7 is $\frac{7}{8}$ , so x = 7 works. You can also multiply it all out: $21 \times 14 \times 7 = 2{,}058$ and $49 \times 48 = 2{,}352$ . Then $2{,}058 \div 2{,}352 = 0.875$ .

More questions like this

The original DOE question

Question 62 as printed on page 192 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 62. Open page 192 of the handbook (PDF)

Question 6 of 57

2023 Test B Math, question 6

Question

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

A $-3\frac{1}{12}$
B $-2\frac{11}{12}$
C $6\frac{1}{12}$
D $9\frac{5}{12}$

Show the answer

Correct answer: C

The first step is to rewrite the problem replacing all subtraction with adding the opposite:

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

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

Then, rewrite all the fractions using a common denominator of 12, which is the least common multiple of the fractions’ denominators, 4, 3, and 2:

$3\frac{3}{12} + \left(-1\frac{8}{12}\right) + 4\frac{6}{12}$

Next, combine the positive numbers and, finally, add the negative number to the sum by subtracting the number with the least absolute value from the number with the greatest absolute value and using the sign of the number with the greatest absolute value:

$7\frac{9}{12} + \left(-1\frac{8}{12}\right)$

$6\frac{1}{12}$

Question analysis

What is 3 1/4 − 1 2/3 − (−4 1/2)? This question on rational-number arithmetic combines mixed numbers with denominators of 4, 3 and 2 and ends by subtracting a negative number. Students must turn that last subtraction into an addition and find a common denominator before combining.

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

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

Why it is hard

The expression is short, but it mixes a subtraction that stays a subtraction with one that becomes an addition, and that difference is what most wrong answers miss. The fractions add a second load, since twelfths are needed before anything combines. Estimating first, 3 − 1.7 + 4.5 is about 6, shows the result is positive, which rules out both negative values.

Error-prone arithmetic moderate Working it out
Three mixed numbers with denominators 4, 3 and 2 must become twelfths, and the last term, −(−4 1/2), changes sign while the middle subtraction does not. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Rewriting the double negative, renaming each fraction, adding 3 1/4 and 4 1/2, then taking away 1 2/3 are four steps, each able to carry a slip forward. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Change subtracting −4 1/2 into adding 4 1/2, rename every fraction in twelfths, add the two positive numbers, then take away 1 2/3.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No single error gives −3 1/12; it needs subtracting 4 1/2 and also miswriting 3 1/4 as 3 1/12, two errors at once, so it mostly draws guesses.
B Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Treats subtracting −4 1/2 as subtracting 4 1/2, so 3 1/4 − 1 2/3 − 4 1/2 = −2 11/12.
D Negatives and direction (Brown & Burton, 1978)
Reveals a surface strategy Expected pull: weak
Drops both minus signs and adds the three mixed numbers, 3 1/4 + 1 2/3 + 4 1/2 = 9 5/12, as if every operation were addition.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative number: $a - (-b) = a + b$; a and b stand for any numbers. Taking away a negative number is the same as adding its opposite, the same number with the other sign. Think of it this way: taking away a debt leaves you richer.
Equivalent fractions: $\frac{a}{b} = \frac{a \times n}{b \times n}$; a is the top and b is the bottom of a fraction. Multiplying both by the same number n does not change its value. Use it to give fractions the same bottom number, so you can add or subtract them.

Sort out the question
You need the value of $3\frac{1}{4} - 1\frac{2}{3} - \left(-4\frac{1}{2}\right)$ . Each number is a mixed number, a whole number plus a fraction. There are two subtractions. The second one takes away a negative number.
Change the double negative
Use the rule for subtracting a negative. Taking away $-4\frac{1}{2}$ is the same as adding $4\frac{1}{2}$ .
$3\frac{1}{4} - 1\frac{2}{3} + 4\frac{1}{2}$
The first minus sign stays as it is. Watch out: if you subtract $4\frac{1}{2}$ instead, you get $-2\frac{11}{12}$ , choice B.
Give the fractions one denominator
The denominator is the bottom number of a fraction. Here they are 4, 3 and 2. The smallest number all three divide into is 12. So change each fraction to twelfths.
$\frac{1}{4} = \frac{3}{12} \qquad \frac{2}{3} = \frac{8}{12} \qquad \frac{1}{2} = \frac{6}{12}$
The problem is now:
$3\frac{3}{12} - 1\frac{8}{12} + 4\frac{6}{12}$
Add the two positive numbers
You may add and subtract in any order, as long as each number keeps its sign. Add the two positive numbers first.
$3\frac{3}{12} + 4\frac{6}{12} = 7\frac{9}{12}$
That is 3 + 4 = 7 wholes and 3 + 6 = 9 twelfths.
Take away the last number
$7\frac{9}{12} - 1\frac{8}{12} = 6\frac{1}{12}$
That is 7 − 1 = 6 wholes and 9 − 8 = 1 twelfth. So the answer is C, $6\frac{1}{12}$ .

Check your answer

Estimate with decimals: $3.25 - 1.67 + 4.5 \approx 6.08$ . And $6\frac{1}{12}$ is about 6.08 too. The answer must be positive, so A and B are out. Choice D comes from adding all three numbers.

More questions like this

The original DOE question

Question 63 as printed on page 193 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 63. Open page 193 of the handbook (PDF)

Question 7 of 57

2023 Test B Math, question 7

Question

On Fran’s map, 2 centimeters represents 3 kilometers. The points on this map that represent Fran’s home and Bryant State Park are 4.5 centimeters apart. How many kilometers apart are Fran’s home and Bryant State Park, to the nearest kilometer?

Show the answer

Correct answer: C

Let x represent the distance, in kilometers, between Fran’s house and Bryant State Park. Set up a proportion to solve for x:

$\frac{x}{4.5} = \frac{3}{2}$	Cross multiply.
$2x = 3(4.5)$	Apply the multiplicative inverse property; divide both sides of the equation by 2.
$x = \frac{13.5}{2} = 6\frac{3}{4} \approx 7$

Question analysis

On Fran's map, 2 centimeters represent 3 kilometers, and her home and Bryant State Park are 4.5 centimeters apart on the map. Students find the real distance to the nearest kilometer. It is a map-scale proportion problem: the scale must be applied in the right direction, and the result rounded.

NYS standards: 7.G.1 scale drawings
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Map scales are standard grade 7 proportional reasoning, and this problem is typical: one distance, one scale, then rounding. Nothing beyond a single proportion is needed.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a one-proportion map question, and the only judgments are the direction of the scale and when to round. The phrase to the nearest kilometer tempts some students to round the scale itself before multiplying, which lands on a listed answer two kilometers too far. Rounding only the final 6.75 keeps the result right.

Turning words into math minor Understanding the question
“2 centimeters represents 3 kilometers” gives 1.5 kilometers per centimeter; set up the other way round, the scale shrinks the 4.5 centimeters instead of enlarging them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
The map distance is a decimal, 4.5 × 3 ÷ 2 = 6.75, and the result must then be rounded to the nearest kilometer, two places where a quick estimate drifts. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work out how many kilometers 1 centimeter on the map stands for, multiply by 4.5, and round only at the very end.

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).
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Turns the scale upside down, multiplying 4.5 by 2/3 instead of 3/2, so the real distance comes out smaller than the map distance.
B Units or scale (King et al., 2004)
Reveals a surface strategy Expected pull: weak
Takes the 4.5 centimeters on the map as kilometers and rounds to 5, never applying the scale that turns each centimeter into 1.5 kilometers.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Rounds the scale of 1.5 kilometers per centimeter up to 2 before multiplying, so 4.5 × 2 = 9 instead of 6.75.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scale proportion: $\frac{\text{real distance}}{\text{map distance}} = \frac{\text{real distance}}{\text{map distance}}$; A map shrinks every distance by the same amount. So real distance over map distance makes the same fraction everywhere on the map. Put kilometers on top on both sides and centimeters on the bottom. Then solve for the missing number.

Sort out the question
On Fran’s map, 2 centimeters stand for 3 kilometers in real life. Her home and the park are 4.5 centimeters apart on the map. You need the real distance in kilometers, rounded to the nearest whole kilometer.
Set up the proportion
Let x be the real distance, in kilometers. Put kilometers on top and centimeters on the bottom, on both sides.
$\frac{x}{4.5} = \frac{3}{2}$
Solve for x
To get x alone, multiply both sides by 4.5.
$x = \frac{3 \times 4.5}{2} = \frac{13.5}{2} = 6.75$
So the real distance is 6.75 kilometers. Watch out: keep the scale exact until the end. Rounding 1.5 kilometers per centimeter up to 2 gives $4.5 \times 2 = 9$ , choice D.
Round to the nearest kilometer
6.75 is between 6 and 7. Its decimal part, 0.75, is more than 0.5, so round up to 7.
Watch out: don’t flip the scale. Multiplying 4.5 by $\frac{2}{3}$ gives 3, choice A. So the answer is C, 7.

Check your answer

Each centimeter stands for $3 \div 2 = 1.5$ kilometers. So 4 centimeters is 6 kilometers, and half a centimeter is 0.75 kilometer. That adds to 6.75, which rounds to 7. The real number should be bigger than 4.5, and it is.

More questions like this

The original DOE question

Question 64 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 64. 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 8 of 57

2023 Test B Math, question 8

Question

Solve for x in terms of y if $2x + 4 = y$ .

A $x = \frac{y - 4}{2}$
B $x = 2(y + 4)$
C $x = \frac{y + 4}{2}$
D $x = 2(4 - y)$

Show the answer

Correct answer: A

$2x + 4 = y$	Apply the additive inverse property; subtract 4 from both sides of the equation.
$2x = y - 4$	Apply the multiplicative inverse property; divide both sides of the equation by 2.
$x = \frac{y - 4}{2}$

Question analysis

Given the equation 2x + 4 = y, students solve for x in terms of y. This literal-equation question offers four choices that mix multiplying and dividing by 2 with adding and subtracting 4. Undoing the two operations on x in the right order, with the right inverse each time, is what it tests.

NYS standards: 8.EE.7b linear equations with rational coefficients
7.EE.4a two-step equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students solve two-step equations such as 2x + 4 = 10. Here the right side is the letter y, so the same two inverse steps leave an expression, a first taste of the grade 8 skill of rearranging.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 13 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The two inverse steps are routine, subtract 4 and then divide by 2, but the letters hide whether they were done correctly. The main hazard is moving the 4 across the equals sign while keeping its plus sign, which produces an expression nearly identical to the right one. Substituting x = 1 and y = 6 tests every choice in seconds.

A very close wrong answer moderate Choosing the answer
(y − 4)/2 and (y + 4)/2 differ only in the sign before 4, and that sign depends on undoing the +4 by subtracting it from both sides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
With y on the right instead of a number, the answer is an expression, so there is no value to check unless the student substitutes a pair such as x = 1 and y = 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Treat y as if it were a number and undo the operations on x in reverse order, using the inverse of each, then check by substituting a value.

Traps

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

The wrong answers, classified

B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Multiplies by 2 instead of dividing and adds 4 instead of subtracting, applying the operations on x again rather than undoing them.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Moves the 4 across the equals sign without changing its sign, getting 2x = y + 4, then divides by 2 correctly.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Needs two unrelated errors at once, the subtraction written backward as 4 − y and a multiplication by 2 in place of division, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } \frac{a}{c} = \frac{b}{c}$; a and b are the two sides of an equation, and c is any number (not 0 when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it balanced. Use this to undo, one at a time, each step that was done to x.

Sort out the question
The equation is $2x + 4 = y$ . “Solve for x in terms of y” means get x alone on one side. The other side will have y in it. Treat y as if it were a number.
Undo the adding of 4
Right now, x is doubled and then 4 is added. Undo those steps in reverse order. First, subtract 4 from both sides.
$2x + 4 - 4 = y - 4$
$2x = y - 4$
Watch out: the 4 must be subtracted from y. Writing $2x = y + 4$ leads to choice C.
Undo the doubling
Next, divide both sides by 2. Divide the whole right side, not only the y.
$\frac{2x}{2} = \frac{y - 4}{2}$
$x = \frac{y - 4}{2}$
Give the answer
Watch out: choices B and D multiply by 2. But x was multiplied by 2, so you undo that by dividing. So the answer is A, $x = \frac{y - 4}{2}$ .

Check your answer

Pick a value, such as x = 1. Then $y = 2 \times 1 + 4 = 6$ . Put y = 6 into choice A: $\frac{6 - 4}{2} = 1$ . That gives back x = 1. Choice C would give $\frac{6 + 4}{2} = 5$ , which is wrong.

More questions like this

The original DOE question

Question 65 as printed on page 193 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 65. Open page 193 of the handbook (PDF)

Question 9 of 57

2023 Test B Math, question 9

Question

Bryan completed a 100-meter race in 11.74 seconds. Luis completed the same race in 11.69 seconds. What was the difference between their times, expressed as a fraction of a second?

A $\frac{1}{20}$
B $\frac{1}{10}$
C $\frac{3}{20}$
D $\frac{1}{5}$

Show the answer

Correct answer: A

First, subtract Luis’s time from Bryan’s time: $11.74 - 11.69 = 0.05$ .

Then, convert to a fraction:

$0.05 = \frac{5}{100} = \frac{1}{20}$ .

Question analysis

Bryan ran a 100-meter race in 11.74 seconds and Luis ran the same race in 11.69 seconds. Students find the difference between their times and express it as a fraction of a second. It tests subtracting decimals and rewriting a decimal as a fraction in lowest terms.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Subtracting decimals and writing a decimal as a fraction are skills from before grade 7. The two times are close, so the subtraction is quick, and the only care needed is with place value and simplifying.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two quick skills settle this question: lining up the decimals to subtract, and naming the result as a fraction in lowest terms. The two times differ only in their last two digits, so the subtraction is short, but it needs a borrow that some students skip. The fraction step is where a decimal’s digit gets mistaken for a denominator.

Error-prone arithmetic minor Working it out
Subtracting 11.69 from 11.74 needs a borrow in the hundredths, 14 − 9, after which the tenths cancel to 0, the step where place value goes wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The difference, 0.05, must be read as five hundredths and reduced as a fraction; the question gives no help with converting a decimal. (Anderson & Krathwohl, 2001)

Strategy: Line up the decimal points, subtract, write the difference as a number of hundredths over 100, and reduce the fraction.

Traps

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

The wrong answers, classified

B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Looks only at the tenths, 11.7 − 11.6 = 0.1, dropping the hundredths, a place-value slip that doubles the true gap.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts the smaller digit from the larger in each place, 9 − 4 = 5 and 7 − 6 = 1, getting 0.15, which is 3/20.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Gets 0.05 but reads its digit 5 as a denominator, writing 1/5, instead of five hundredths, 5/100.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths: $\text{a decimal with two places} = \frac{\text{its digits}}{100}$; The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths, and you can write its digits over 100. For example, 0.37 is 37/100. Then simplify by dividing the top and bottom by the same number.

Sort out the question
Bryan ran the race in 11.74 seconds. Luis ran it in 11.69 seconds. The difference is how much longer one time is than the other. You need it as a fraction of a second.
Subtract the two times
Subtract the smaller time from the larger one. Line up the decimal points.
$11.74 - 11.69 = 0.05$
In the hundredths place, 4 is less than 9. So regroup: trade 1 tenth for 10 hundredths. Then 14 − 9 = 5 hundredths. The tenths are now 6 − 6 = 0.
Watch out: don’t just take the smaller digit from the larger one in each place. That gives 0.15, which is $\frac{3}{20}$ , choice C.
Write the difference as a fraction
The 5 is in the hundredths place, so 0.05 is 5 hundredths.
$0.05 = \frac{5}{100}$
Now simplify. Divide the top and bottom by 5.
$\frac{5 \div 5}{100 \div 5} = \frac{1}{20}$
Give the answer
Watch out: 0.05 is not $\frac{1}{5}$ . One fifth is 0.2, which is four times as big. So the answer is A, $\frac{1}{20}$ .

Check your answer

Turn $\frac{1}{20}$ back into a decimal: $1 \div 20 = 0.05$ . Then add it to Luis’s time: $11.69 + 0.05 = 11.74$ . That is Bryan’s time, so the difference is right.

More questions like this

The original DOE question

Question 66 as printed on page 193 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 66. Open page 193 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 10 of 57

2023 Test B Math, question 10

Question

A total of 24 students in a math class took a 100-point test:

None of the students scored 100 points.
75% of the students scored less than 82 points.
The lowest score in the class was 54 points.

Which box plot best represents the test scores from this math class?

A $Box plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 54, the box runs from 60 to about 72 with the median line at about 63, and the right whisker ends at about 82.5, halfway between 80 and 85.$
B $Box plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 54, the box runs from about 62.5 to about 82.5 with the median line at about 71, and the right whisker ends at about 98.$
C $Box plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 51, the box runs from 60 to about 83 with the median line at about 74, and the right whisker ends at about 99.$
D $Box plot titled TEST SCORES above a number line from 50 to 100 with arrows at both ends and a labeled tick every 5. The left whisker starts at about 54, the box runs from about 63 to about 83.5 with the median line at 70, and the right whisker ends at 100.$

Show the answer

Correct answer: B

Use the information in the bullet points to reason which box plot could represent the data set of test scores.

Since none of the students scored 100, the maximum value on the box plot must be less than 100.

If 75% of the students scored less than 82, then 25% scored 82 points or more. Since the third quartile (Q3) indicates $\frac{3}{4}$ (75%) of the data is to the left and $\frac{1}{4}$ (25%) of the data is to the right, it is reasonable that Q3 would be at or about 82.

Given that the lowest score in the class was 54, the minimum point on the box plot must be 54.

The box plot graph in option B is the only one that meets all three criteria.

Question analysis

Twenty-four students took a 100-point test: nobody scored 100, 75% scored below 82, and the lowest score was 54. Students choose the box plot that fits all three facts. The skill is linking each statement to a part of the plot, the whiskers, the box and the quartiles, and checking every plot closely.

NYS standards: 7.SP.1 box plots and interquartile range
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students usually read values off a box plot. This item runs the other way, turning a percent statement into a quartile and checking the whiskers against two more facts, but the content stays at grade 7.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Every choice must be checked moderate
Reading a figure, table or graph moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 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 question turns three sentences into three features of a box plot, and a plot that matches two of them is easy to accept. The upper-quartile fact is the conceptual one: students who do not know that the box’s right edge marks 75 percent put 82 at the end of the whisker. The other two facts are quick visual checks, but only if every plot is actually checked.

Every choice must be checked moderate Choosing the answer
Each wrong plot fails a different fact, so all four must be tested against all three bullets: the left whisker against 54, the right whisker against 100, and the box’s right edge against 82. (Embretson & Wetzel, 1987)
Reading a figure, table or graph moderate Understanding the question
The axis is labeled every 5 points, and the deciding features fall between labels: one left whisker starts near 51 instead of 54, and box edges sit near 82.5. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Nothing explains that “75% of the students scored less than 82” locates the upper quartile, the right edge of the box, rather than the highest score. (Anderson & Krathwohl, 2001)

Strategy: Turn each fact into a feature to check: where the left whisker starts, where the right whisker ends, and where the box ends. Rule out any plot that breaks one.

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Puts 82 at the end of the right whisker, as if it were the highest score, when 75 percent below 82 marks the box’s right edge.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Accepts the left whisker as starting at the lowest score, 54, when it begins near 51, just past the 50 tick.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Matches the lowest score and the box’s right edge near 82 but overlooks that a whisker reaching 100 means someone scored 100.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts of a box plot: $\text{minimum},\ Q_1,\ \text{median},\ Q_3,\ \text{maximum}$; A box plot shows five numbers. The left whisker starts at the minimum, the lowest value. The box starts at Q1, the first quartile, and ends at Q3, the third quartile. The line inside is the median, the middle value. The right whisker ends at the maximum. About a quarter of the data falls in each of the four sections.

Sort out the question
24 students took a test. You know three facts. Nobody scored 100. 75% scored less than 82. The lowest score was 54. You need the box plot that fits all three facts.
Turn each fact into a feature
Match each fact to a part of the box plot.
- The lowest score is 54, so the left whisker starts at 54.
- Nobody scored 100, so the right whisker ends before 100.
- 75% is three quarters. Three quarters of the scores are below Q3. So the box should end at about 82.
Read the number line
The ticks are 5 points apart: 50, 55, 60 and so on. A score of 54 sits just left of the 55 tick. A score of 82 sits a little less than halfway from 80 to 85. Look closely, because these points fall between labels.
Check plot A
In plot A, the right whisker ends near 82, and the box ends near 72. That would make 82 the highest score. Watch out: 75% below 82 does not make 82 the top score. It tells you where the box ends. So A is out.
Check plot C
In plot C, the left whisker starts near 51, just right of the 50 tick. But the lowest score was 54. So C is out.
Check plot D
Plot D starts at 54, and its box ends near 82. But its right whisker reaches 100. That means someone scored 100, and nobody did. Watch out: D fits two facts, so it is easy to stop early. Check all three. So D is out.
Check plot B and answer
Plot B starts at 54. Its box ends near 82. Its right whisker stops near 98, below 100. It fits all three facts. So the answer is B.

Check your answer

Go over B once more with all three facts. Its lowest point is 54. Its box ends near 82. Its highest point is under 100. Each other plot breaks one fact: A’s box, C’s left whisker and D’s right whisker.

More questions like this

The original DOE question

Question 67 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 67. Open page 194 of the handbook (PDF)

Question 11 of 57

2023 Test B Math, question 11

Question

$n = 0.13 + \frac{207}{100}$

What is the value of n expressed as a fraction?

A $\frac{1}{3}$
B $\frac{11}{5}$
C $\frac{9}{4}$
D $\frac{23}{10}$

Show the answer

Correct answer: B

$n = 0.13 + \frac{207}{100} =$

$\frac{13}{100} + \frac{207}{100} =$

$\frac{220}{100} = \frac{11}{5}$

Question analysis

The number n equals 0.13 plus 207/100, and students must write n as a single fraction. This question mixes a decimal with a fraction in hundredths. Converting both to the same form, adding, and reducing to lowest terms are the steps, and the four fractions offered look quite different at first glance.

NYS standards: 6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Converting decimals to fractions and simplifying are skills taught before grade 7. Both numbers here are in hundredths, so the addition is immediate and only the final reduction takes thought.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: An answer in an unexpected form minor
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 15 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Converting 0.13 to hundredths makes this a single addition, 13 + 207 = 220 hundredths, so the question tests little beyond reducing a fraction. Two of the wrong choices sit close to 2.2 but come from no common working, and the third only echoes the digits of 0.13. A student who checks each choice as a decimal finds only one equal to 2.2.

An answer in an unexpected form minor Choosing the answer
The sum comes out as 220/100 or 2.2, but the choices are in lowest terms, so it must be reduced, or each choice converted, before anything matches. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
One term is a decimal and the other a fraction; writing 0.13 as 13/100 puts both in hundredths, and a misplaced digit there changes the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write 0.13 as 13/100 so both terms are in hundredths, add the numerators, then divide the top and bottom by their greatest common factor.

Traps

Answer in another form: The right answer is written differently from the way you worked it out.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Echoes the digits of 0.13, reading it as the fraction 1/3, and ignores the 207/100 altogether.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 9/4, which is 2.25; adding 13 and 207 hundredths gives 2.2, so this mostly draws guesses near the sum.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Nothing leads to 23/10 except 13 + 207 coming out as 230, a slip with no reason behind it, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths: $\text{a decimal with two places} = \frac{\text{its digits}}{100}$; The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths. Write its digits over 100. For example, 0.47 is 47/100.
Adding fractions with the same bottom: $\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$; a and b are the tops. c is the shared bottom, the denominator. When the bottoms match, the pieces are the same size. So add the tops and keep the bottom.

Sort out the question
n equals 0.13 plus $\frac{207}{100}$ . One part is a decimal and one is a fraction. You need n as a fraction. The choices are in lowest terms, which means they can’t be simplified any more.
Write the decimal as a fraction
The 3 in 0.13 is in the hundredths place. So 0.13 is 13 hundredths.
$0.13 = \frac{13}{100}$
Now both numbers are in hundredths. Watch out: 0.13 is not $\frac{1}{3}$ . That only copies the digits, choice A.
Add the hundredths
Use the rule for adding fractions. The bottoms match, so add the tops and keep the 100 on the bottom.
$\frac{13}{100} + \frac{207}{100} = \frac{13 + 207}{100} = \frac{220}{100}$
Simplify the fraction
$\frac{220}{100}$ is not a choice, so simplify it. Find the greatest number that divides both 220 and 100. That is 20.
$\frac{220 \div 20}{100 \div 20} = \frac{11}{5}$
So the answer is B, $\frac{11}{5}$ .

Check your answer

Change both to decimals. $\frac{11}{5} = 11 \div 5 = 2.2$ . And $0.13 + 2.07 = 2.20$ . They match. Choices C and D are close, but $\frac{9}{4} = 2.25$ and $\frac{23}{10} = 2.3$ .

More questions like this

The original DOE question

Question 68 as printed on page 194 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 68. Open page 194 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 12 of 57

2023 Test B Math, question 12

Question

The perimeter of a 12-sided polygon is 100 centimeters. The length of one side is 25 centimeters. The length of another side is 35 centimeters. The remaining sides are equal in length to one another. What is the length of each of the remaining sides, in centimeters?

Show the answer

Correct answer: A

Let x represent the length of one of the 10 remaining sides:

$25 + 35 + 10x = 100$

$60 + 10x = 100$

$10x = 40$

$x = 4$

Question analysis

A 12-sided polygon has a perimeter of 100 centimeters: one side is 25 cm long, another is 35 cm, and the rest are all equal to one another. Students find the length of each remaining side. This perimeter problem turns into a two-step equation, and the tricky part is counting the sides that are left.

NYS standards: 7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving a two-step equation from a short description is standard grade 7 work. The perimeter setting is simple; the only extra care is counting the sides that remain after the two given ones.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The perimeter and the two named sides make the subtraction obvious, so the real decision comes after it: what to divide the leftover 40 centimeters by. Students who stop at the leftover length, or who lose track of how many sides remain, reach a listed choice. Writing 25 + 35 + 10x = 100 keeps the count of 10 in view.

Turning words into math minor Understanding the question
“The remaining sides are equal in length to one another” hides the count: 12 sides minus the two given ones leaves 10 equal sides to share the rest of the perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Subtracting 25 and 35 from 100, counting the sides left, and dividing are three steps, and the first result, 40, is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Subtract the two known sides from the perimeter, count how many sides are left, and share the remaining length equally among them.

Traps

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

B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 5; sharing the 40 centimeters among 8 sides would, but nothing in the question suggests 8, so it mostly draws guesses.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Shares the leftover 40 centimeters between two sides, as if the figure were four-sided, overlooking that it has 12 sides.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the total of the remaining sides, 100 − 25 − 35 = 40, and stops before sharing it among the 10 equal sides.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a polygon: $P = \text{the sum of all the side lengths}$; P is the perimeter, the total distance around a shape. A polygon is a flat shape with straight sides. A sum is the answer to an addition. So add the lengths of every side to get P.

Sort out the question
The shape is a polygon with 12 sides. Its perimeter is 100 centimeters. One side is 25 centimeters, and another is 35 centimeters. The other sides all have the same length. You need that length.
Count the equal sides
Two of the 12 sides are already known. The rest are the equal sides.
$12 - 2 = 10$
So there are 10 equal sides. Call the length of each one x.
Write an equation for the perimeter
Use the perimeter formula. All 12 sides add up to 100. Ten sides of length x add up to 10x, which means 10 times x.
$P = \text{the sum of all the side lengths}$
$25 + 35 + 10x = 100$
Solve for x
$60 + 10x = 100$
$10x = 100 - 60 = 40$
The 10 equal sides add up to 40 centimeters. Watch out: 40 is all ten sides together, not one side. Stopping here gives choice D. Now share the 40 among the 10 sides.
$x = 40 \div 10 = 4$
Give the answer
Each remaining side is 4 centimeters long. Watch out: sharing the 40 centimeters between only 2 sides gives 20, choice C. That treats the shape as if it had 4 sides, but it has 12. So the answer is A, 4.

Check your answer

Add all 12 sides: $25 + 35 + 10 \times 4 = 25 + 35 + 40 = 100$ . That is the perimeter the question gives, so 4 centimeters is right.

More questions like this

The original DOE question

Question 69 as printed on page 194 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 69. Open page 194 of the handbook (PDF)

Question 13 of 57

2023 Test B Math, question 13

Question

Shiow-Chen has a box that contains 4 green cards, 7 yellow cards, and 6 red cards. She randomly draws one card at a time from the box and does not return the cards to the box after they have been drawn. The first card she draws is yellow. The second card she draws is green. What is the probability that the third card she draws will also be green?

A $\frac{1}{5}$
B $\frac{4}{17}$
C $\frac{1}{4}$
D $\frac{1}{3}$

Show the answer

Correct answer: A

After Shiow-Chen draws a yellow card and a green card, the remaining cards are 3 green, 6 yellow, and 6 red. The probability of drawing a green card from that group is:

$\frac{3}{3 + 6 + 6} = \frac{3}{15} = \frac{1}{5}$

Question analysis

Shiow-Chen's box holds 4 green, 7 yellow and 6 red cards, and she draws cards one at a time without putting them back. Her first card is yellow and her second is green. Students find the probability that the third card is also green, which means updating both the green count and the total first.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Simple probability is grade 6 content. The extra demand is tracking two earlier draws that were not put back, which changes the counts for the third draw, a step typical grade 7 problems often leave out.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 70 words.
Students, last 30 days: 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 probability itself is one fraction, but it has to be built from a box that two draws have already changed. Students who use the original 17 cards, or who compare green with the other colors instead of with all the cards, reach listed choices that look reasonable. Listing the counts after each draw, then putting the green cards over all the cards left, avoids both.

Much to keep track of moderate Working it out
Three color counts and the total must each be updated after two draws, 4 green, 7 yellow and 6 red becoming 3, 6 and 6 out of 15, before the third draw is considered. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math minor Understanding the question
“Does not return the cards” and the two named draws are the conditions that change the box, and the stem states them in passing before asking about the third card. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Take the two drawn cards out of the box: lower the yellow and green counts by one each, recount the total, then put the green cards left over the cards left.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

B Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Uses the full box, 4 green cards out of 17, as if neither of the first two cards had been taken out.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Updates the counts correctly but compares the 3 green cards with the 12 cards of other colors, 3/12 = 1/4, a part-to-part ratio.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the three colors as equally likely, giving green a 1-in-3 chance whatever the number of cards of each color.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of results you want}}{\text{total number of results}}$; P is the probability, how likely something is, from 0 (never) to 1 (always). The top counts the results you want, here green cards. The bottom counts every possible result, here every card in the box. It works when each result is equally likely, as in a random draw.

Sort out the question
The box starts with 4 green, 7 yellow and 6 red cards. Shiow-Chen draws cards one at a time, and she does not put them back. The first card is yellow, and the second is green. You need the probability that the third card is green.
Take out the two cards
The cards are not put back, so the box changes. Lower the yellow count by 1 and the green count by 1.
Color At the start After two draws
Green 4 3
Yellow 7 6
Red 6 6
Total 17 15
Use the probability formula
For the third draw, the results you want are the 3 green cards left. The total is the 15 cards left.
$P = \frac{\text{number of results you want}}{\text{total number of results}}$
$P = \frac{3}{3 + 6 + 6} = \frac{3}{15}$
Watch out: using the full box gives $\frac{4}{17}$ , choice B. But two cards are gone.
Simplify and answer
Divide the top and bottom by 3.
$\frac{3}{15} = \frac{1}{5}$
Watch out: compare the green cards with all 15 cards, not with the 12 cards of other colors. That gives $\frac{1}{4}$ , choice C. And the three colors are not equally likely, so $\frac{1}{3}$ is wrong too. So the answer is A, $\frac{1}{5}$ .

Color	At the start	After two draws
Green	4	3
Yellow	7	6
Red	6	6
Total	17	15

Check your answer

17 cards minus the 2 drawn leaves 15. One fifth of 15 is 3, the number of green cards left. Also, the chance went down from $\frac{4}{17}$ . That makes sense, because a green card was taken out.

More questions like this

The original DOE question

Question 70 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 70. Open page 195 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 14 of 57

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?

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:

$n = 3 + 2x$

$93 = 3 + 2x$

$90 = 2x$

$45 = x$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
No obvious method minor
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.

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)
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

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Reads “the number that comes just before it” as the whole number one less than 93, ignoring the sequence’s rule.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Undoes the adding of 3, 93 − 3 = 90, and stops before undoing the doubling.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds 3 instead of subtracting it before halving, (93 + 3) ÷ 2 = 48, undoing the rule in the wrong direction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule: $n = 2x + 3$; 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.

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.
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.
$n = 2x + 3$
$93 = 2x + 3$
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.
$93 - 3 = 2x$
$90 = 2x$
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.
Undo the doubling
Now divide both sides by 2.
$x = 90 \div 2 = 45$
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: $2 \times 45 + 3 = 93$ . 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 15 of 57

2023 Test B Math, question 15

Question

On a map, the distance between Deshawn’s house and Xin’s house is 5.4 inches. If 2 inches on the map represents 3 miles, what is the distance between these two houses, in miles?

A2.6
B3.6
C7.5
D8.1

Show the answer

Correct answer: D

Let d represent the distance between the two houses, in miles. Write a proportion to solve for d:

$\frac{d}{5.4} = \frac{3}{2}$

$d = \frac{3(5.4)}{2} = \frac{16.2}{2} = 8.1$

Question analysis

A map shows Deshawn's and Xin's houses 5.4 inches apart, with 2 inches on the map standing for 3 miles. Students find the actual distance between the houses in miles. This scale question uses a decimal map length, so the multiplication must be exact and the inches-to-miles ratio kept the right way up.

NYS standards: 7.G.1 scale drawings
7.RP.2b constant of proportionality
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Converting a map distance with a given scale is routine grade 7 work. This one needs a single proportion and a decimal multiplication, with no rounding or second step, so it matches a typical classroom exercise.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 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

Unlike most map questions, this one asks for an exact distance, so the decimal 5.4 has to be carried all the way through. The two tempting wrong answers come from opposite places: an upside-down scale shrinks the distance, and a rounded map length gives a result close to the true one. Checking that the answer is one and a half times 5.4 catches both.

Error-prone arithmetic minor Working it out
The map distance is the decimal 5.4, and 5.4 × 3 = 16.2 then halved is a decimal multiplication where rounding 5.4 to make it easier changes the answer. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
7.5 and 8.1 miles are close enough that a rough estimate, 5 × 1.5, lands on the wrong one; only the exact product separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Set up miles over inches equal to 3 over 2, then multiply 5.4 by 3 and divide by 2, keeping the decimal exact throughout.

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2.6, which sits just below half of 5.4 and matches no step with the scale, so it mostly draws guesses.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Inverts the scale, multiplying 5.4 by 2/3 instead of 3/2, as if each mile took up 1.5 inches on the map.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Rounds 5.4 inches down to 5 before scaling, so 5 × 3 ÷ 2 = 7.5 miles instead of the exact distance.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scale proportion: $\frac{\text{real distance}}{\text{map distance}} = \frac{\text{real distance}}{\text{map distance}}$; A map shrinks every distance by the same amount. So real distance over map distance makes the same fraction everywhere on the map. Put miles on top on both sides and inches on the bottom. Then solve for the missing distance.

Sort out the question
On the map, Deshawn’s house and Xin’s house are 5.4 inches apart. The scale says 2 inches on the map stand for 3 miles. You need the real distance between the houses, in miles.
Set up the proportion
Let d be the real distance, in miles. Put miles on top and inches on the bottom, on both sides.
$\frac{d}{5.4} = \frac{3}{2}$
Solve for d
To get d alone, multiply both sides by 5.4.
$d = \frac{3 \times 5.4}{2} = \frac{16.2}{2} = 8.1$
The houses are 8.1 miles apart. Watch out: keep 5.4 exact. Rounding it to 5 gives 7.5 miles, choice C.
Give the answer
Watch out: don’t flip the scale. Multiplying 5.4 by $\frac{2}{3}$ gives 3.6, choice B. Each inch stands for more than a mile, so the miles must be more than 5.4. So the answer is D, 8.1.

Check your answer

Each inch stands for $3 \div 2 = 1.5$ miles. So 5 inches is 7.5 miles, and 0.4 inch is $0.4 \times 1.5 = 0.6$ mile. Together, $7.5 + 0.6 = 8.1$ miles.

More questions like this

The original DOE question

Question 72 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 72. Open page 195 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 16 of 57

2023 Test B Math, question 16

Question

$A wheel: a circle with a small solid hub at its center and 15 straight spokes running from the hub to the rim, equally spaced around it.$

This wheel has 15 spokes, equally spaced. What is the measure of the angle between adjacent spokes?

A30°
B24°
C15°
D12°

Show the answer

Correct answer: B

The total number of degrees in a circle is 360. The measure of the angle between adjacent spokes is $\frac{360}{15} = 24^{\circ}$ .

Question analysis

A wheel has 15 spokes, equally spaced around its hub, as the figure shows. Students find the measure of the angle between two neighboring spokes. The question tests knowing how many degrees make a full turn around a point and sharing them equally among the spokes.

NYS standards: 7.G.5 angle relationships and equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Finding an angle by dividing a full turn into equal parts is early angle measurement. It takes one division, and the only fact needed is the number of degrees in a circle, so it sits below grade 7.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
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

One fact and one division answer this question, so it is among the easiest on the test. The wrong choices catch students who use a straight angle’s 180° for the full turn, who take the number of spokes as the angle, or who guess a familiar angle. Multiplying the answer by 15 to get back to 360° is a quick check.

Something you must already know minor Working it out
The question never says that a full turn is 360°; knowing it, and that the 15 spokes split it into 15 equal angles, is the whole method. (Anderson & Krathwohl, 2001)

Strategy: The spokes split the full turn around the hub into 15 equal angles, so divide the number of degrees in a circle by 15.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working with 15 spokes gives 30°, which is 360 ÷ 12, the angle between hours on a clock, so it mostly draws guesses at a familiar angle.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes the number of spokes, 15, as the angle in degrees, matching a number in the question without any calculation.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Divides 180 by 15, using a straight angle’s 180° as the full turn around the hub instead of 360°.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal angles around a point: $\text{each angle} = 360 \div \text{number of equal angles}$; One full turn around a point is 360 degrees. If lines from the point split that turn into equal angles, each angle gets the same share. So divide 360 by how many angles there are.

Sort out the question
The wheel has 15 spokes. They all start at the hub, the center of the wheel, and they are equally spaced. Adjacent spokes are spokes right next to each other. You need the angle between two adjacent spokes.
Count the angles around the hub
Go once around the hub. Between each spoke and the next is one angle. So 15 spokes make 15 angles, all the same size. Together they make one full turn, which is 360 degrees.
Watch out: a full turn is 360 degrees, not 180. Using 180 gives 12 degrees, choice D.
Share the full turn
Use the formula for equal angles.
$\text{each angle} = 360 \div \text{number of equal angles}$
$\text{each angle} = 360 \div 15 = 24$
Each angle is 24 degrees. Watch out: 15 is the number of spokes, not the size of the angle, so choice C is wrong. So the answer is B, 24°.

Check your answer

Multiply back: 15 angles of 24 degrees make $15 \times 24 = 360$ degrees, a full turn. Also, a clock’s 12 hour marks are 30 degrees apart. This wheel has more spokes than that, so each angle must be smaller than 30 degrees.

More questions like this

The original DOE question

Question 73 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 73. Open page 195 of the handbook (PDF)

Question 17 of 57

2023 Test B Math, question 17

Question

$m \bullet t = 72$

For the equation above, what is the sum of the values of m when $t = 1$ , when $t = 2$ , and when $t = 3$ ?

A12
B72
C132
D135

Show the answer

Correct answer: C

First, find the value of m for each value of t:

For $t = 1$ :

$1m = 72$

$m = 72$

For $t = 2$ :

$2m = 72$

$m = 36$

For $t = 3$ :

$3m = 72$

$m = 24$

Then, add those three values together:

$72 + 36 + 24 = 132$

Question analysis

The equation m • t = 72 links two variables, and the raised dot means multiplication. Students find m when t is 1, 2 and 3, and then add those three values of m. Each value is a one-step division, and the trap is combining the t values before dividing.

NYS standards: 6.EE.7 one-step equations
6.EE.2c evaluating expressions and order of operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Solving equations like 3m = 72 is grade 6 work. This item repeats that step three times and adds the results, more bookkeeping than difficulty, so it is below typical grade 7 demand.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 21 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each value of m comes from one division, so the work is bookkeeping rather than algebra. The trap is structural: adding 1, 2 and 3 first and dividing once feels efficient but answers a different question, since m does not change in step with t. Writing the three values of m in a row before adding them keeps the steps separate.

Turning words into math minor Understanding the question
“The sum of the values of m when t = 1, when t = 2, and when t = 3” asks for three separate solutions added together, not one solution at a combined t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Three divisions, 72 ÷ 1, 72 ÷ 2 and 72 ÷ 3, and an addition make four small steps, each easy but each a chance to drop a value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Divide 72 by each value of t in turn to get three separate values of m, then add the three quotients.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Adds the three t values to get 6 and divides 72 by 6, as if the sum of the m values were m at the sum of the t values.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Gives back 72, the constant in the equation and the value of m only when t = 1, without the values for t = 2 and t = 3.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 135, which is 3 more than the right sum; it would need a slip such as 72 ÷ 3 = 27, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division undoes multiplication: $\text{if } a \times b = c \text{, then } a = c \div b$; a, b and c are numbers, and b is not 0. If you know a product and one of the numbers that made it, divide to find the other one. For example, 4 × 5 = 20, so 4 = 20 ÷ 5.

Sort out the question
The equation is $m \bullet t = 72$ . The dot means times. You need m three times: when t is 1, when t is 2 and when t is 3. Then add the three values of m.
Get m by itself
Since m times t is 72, m is 72 divided by t.
$m = 72 \div t$
Now put in each value of t, one at a time.
Find m for each value of t
Each time, divide 72 by the value of t. A table keeps the three answers in order.
Value of t Value of m
1 $m = 72 \div 1 = 72$
2 $m = 72 \div 2 = 36$
3 $m = 72 \div 3 = 24$
Add the three values
$72 + 36 + 24 = 132$
Watch out: don’t add the t values first. $72 \div (1 + 2 + 3) = 12$ is choice A, and it is not the same thing. Also, 72 alone is only the first value, choice B. So the answer is C, 132.

Value of t	Value of m
1	$m = 72 \div 1 = 72$
2	$m = 72 \div 2 = 36$
3	$m = 72 \div 3 = 24$

Check your answer

Check each pair by multiplying: $72 \times 1 = 72$ , $36 \times 2 = 72$ and $24 \times 3 = 72$ . All three pairs make 72. Then add again: 72 + 36 = 108, and 108 + 24 = 132.

More questions like this

The original DOE question

Question 74 as printed on page 196 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 74. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 18 of 57

2023 Test B Math, question 18

Question

$A rectangle whose bottom side is labeled 16 ft and whose left side is labeled 8 ft. It is shaded gray except for a white triangle inside it: the triangle’s vertex is at the middle of the left side, and its base lies along the right side and is labeled 4 ft.$

The diagram above shows a rectangle with an inscribed triangle. The triangle’s vertex and base touch the sides of the rectangle as shown. What is the total area of the shaded portion?

A32 sq ft
B64 sq ft
C96 sq ft
D128 sq ft

Show the answer

Correct answer: C

First, find the area of the rectangle, in square feet:

$16 \times 8 = 128$

Then, find the area of the triangle:

$\frac{1}{2} \times 4 \times 16 = 32$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle:

$128 - 32 = 96\text{ sq ft}$

Question analysis

A 16-foot by 8-foot rectangle has a white triangle inside it, with one vertex at the middle of the left side and a 4-foot base along the right side. Students find the total area of the shaded part. They subtract the triangle's area, whose height is not labeled and must be read from the figure.

NYS standards: 6.G.1 area of polygons
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Composite areas built from rectangles and triangles appear in grades 6 and 7. This one is typical, but the triangle lies on its side, so students must see which length is its height rather than read it from a label.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The shaded region is a rectangle with a triangle cut out, so the plan is standard, but the triangle is turned so that its height runs across the figure. Students who see that the height is the 16-foot side have two short areas to find and one subtraction. The listed wrong answers stop at either area or leave out the one half.

Reading a figure, table or graph moderate Understanding the question
The triangle lies on its side: its 4-foot base runs along the right edge, so its height is the rectangle’s full 16-foot length, which the figure labels only as the bottom side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
The rectangle’s area, the triangle’s area and their difference are three separate results, and the first two, 128 and 32, are themselves listed choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the rectangle's area, find the triangle's area from its base on the right side and its height across the rectangle, then subtract.

Traps

Stops one step early: A choice is a number from partway through the solution.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the triangle’s area, ½ × 4 × 16 = 32, and gives the white region instead of the shaded part around it.
B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Leaves out the ½ in the triangle’s area, 4 × 16 = 64, and subtracts that from 128, which leaves 64.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the rectangle’s area, 16 × 8 = 128, and stops without subtracting the white triangle.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle: $A = l \times w$; A is the area, the space inside a flat shape, in square units. l is the length and w is the width. It works because the rectangle is filled by w rows of unit squares, with l squares in each row.
Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, the side you measure from. h is the height, the distance from the base to the opposite corner, at a right angle to the base. You take half, because a triangle is half of a rectangle with the same base and height.

Sort out the figure
The figure is a rectangle 16 feet long and 8 feet tall. Inside it is a white triangle. Its tip touches the middle of the left side. Its base lies along the right side and is 4 feet long. The shaded part is the gray area around the triangle. You need its area.
Plan: find the rectangle’s area, then take away the triangle’s area.
Find the area of the rectangle
$A = l \times w$
$A = 16 \times 8 = 128$
The rectangle’s area is 128 square feet. Watch out: this includes the white triangle. Stopping here gives choice D.
Find the triangle’s height
The triangle lies on its side. Its base is the 4-foot piece on the right side, which runs up and down. The height must meet the base at a right angle, a square corner. So the height runs straight across, from the base to the tip on the left side. That is the full length of the rectangle, 16 feet.
Find the area of the triangle
$A = \frac{1}{2} \times b \times h$
$A = \frac{1}{2} \times 4 \times 16 = 32$
The triangle’s area is 32 square feet. Watch out: don’t forget the $\frac{1}{2}$ . Using $4 \times 16 = 64$ leaves 64 for the shaded part, choice B.
Subtract to find the shaded area
$128 - 32 = 96$
The shaded part is 96 square feet. Watch out: 32 is the white triangle, not the shaded part, so choice A is wrong. So the answer is C, 96 sq ft.

Check your answer

The triangle’s base, 4 feet, is half the rectangle’s height of 8 feet. Its height is the full 16 feet. So the triangle is half of half the rectangle, or one fourth of it. One fourth of 128 is 32, and $128 - 32 = 96$ .

More questions like this

The original DOE question

Question 75 as printed on page 196 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 75. Open page 196 of the handbook (PDF)

Question 19 of 57

2023 Test B Math, question 19

Question

Orange juice costs $2 per gallon, and grapefruit juice costs $3 per gallon. A recent shipment of equal amounts of the two types of juice costs a total of $250. How many gallons of orange juice are in the shipment?

A50
B62.5
C100
D125

Show the answer

Correct answer: A

Let x represent the number of gallons of orange juice. Since the shipment contains equal amounts of the two types of juice, x also represents the number of gallons of grapefruit juice. Write an equation to solve for x:

$2x + 3x = 250$

$5x = 250$

$x = 50$

Question analysis

Orange juice costs $2 a gallon and grapefruit juice costs $3 a gallon, and a shipment with equal amounts of each costs $250. Students find how many gallons of orange juice are in the shipment. The word problem leads to a simple equation; the key phrase is equal amounts of juice, not equal amounts of money.

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 Writing and solving an equation from a short word problem is standard grade 7 work. This one has a single unknown, since both juices come in the same number of gallons, and only two short steps.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Solution: 3 steps
Reading level: Question and choices: about grade 6, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The equation is short once the key phrase is read as equal gallons, since one gallon of each juice costs $5 together and $250 buys 50 such pairs. Read as equal spending, the same phrase leads to a listed answer that is not a whole number of gallons. The other trap is giving the gallons of both juices together when only orange juice is asked for.

Turning words into math moderate Understanding the question
“Equal amounts of the two types of juice” means the same number of gallons of each, x, so the total cost is 2x + 3x, not two equal halves of $250. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Price one gallon of each juice together, then see how many such pairs $250 buys; that is the number of gallons of each kind.

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

B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Reads “equal amounts” as equal spending, $125 on each juice, and divides $125 by $2 a gallon to get 62.5.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the right count for each juice but gives the whole shipment, 50 + 50 = 100 gallons, instead of the orange juice alone.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides $250 by 2, the orange juice price or the number of juices, combining numbers from the question without equal gallons of each.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost: $\text{cost} = \text{price per gallon} \times \text{number of gallons}$; The price per gallon is what one gallon costs. Multiply it by the number of gallons to get what you pay in all. For example, 3 gallons at $2 a gallon cost $6.

Sort out the question
Orange juice costs $2 a gallon. Grapefruit juice costs $3 a gallon. The shipment has equal amounts of each, so it has the same number of gallons of each juice. It costs $250 in all. You need the number of gallons of orange juice.
Write the cost of each juice
Let x be the number of gallons of each juice. Use the cost formula for each one.
$\text{cost} = \text{price per gallon} \times \text{number of gallons}$
$\text{orange juice cost} = 2 \times x = 2x$
$\text{grapefruit juice cost} = 3 \times x = 3x$
Add the costs and solve
The two costs add up to $250. Two x’s plus three x’s make five x’s.
$2x + 3x = 250$
$5x = 250$
$x = 250 \div 5 = 50$
There are 50 gallons of each juice.
Give the answer
Watch out: equal amounts means equal gallons, not equal dollars. Spending $125 on each juice gives 62.5 gallons, choice B. Also, 100 gallons is both juices together, choice C. So the answer is A, 50.

Check your answer

Find each cost. 50 gallons of orange juice cost $50 \times 2 = 100$ dollars. 50 gallons of grapefruit juice cost $50 \times 3 = 150$ dollars. Together, $100 + 150 = 250$ dollars, the total in the question.

More questions like this

The original DOE question

Question 76 as printed on page 196 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 196, question 76. Open page 196 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 20 of 57

2023 Test B Math, question 20

Question

Five consecutive numbers have a sum of 110. What is the least of these consecutive numbers?

Show the answer

Correct answer: C

Let n represent the least of the five numbers. Then the next four consecutive numbers are $(n + 1)$ , $(n + 2)$ , $(n + 3)$ , $(n + 4)$ . Write an equation to solve for n:

$n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 110$

$5n + 10 = 110$

$5n = 100$

$n = 20$

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Variables instead of numbers minor
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.

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)
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

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 11, which is 110 ÷ 10; five consecutive numbers starting at 11 add to only 65, so it mostly draws guesses.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Steps by 2 as if the numbers were consecutive even numbers, n through n + 8, so 5n + 20 = 110 gives 18.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides 110 by 5 to get 22, the middle number, and gives it without counting down to the least number.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive numbers: $n,\ n + 1,\ n + 2,\ n + 3,\ n + 4$; 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.

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.
Name the five numbers
Let n be the least number. Then the five numbers are:
$n,\ n + 1,\ n + 2,\ n + 3,\ n + 4$
Each one is 1 more than the one before it.
Write the sum as an equation
Add all five and set the sum equal to 110.
$n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 110$
There are five n’s, which make 5n. The extra numbers add up to 1 + 2 + 3 + 4 = 10.
$5n + 10 = 110$
Solve for n
Subtract 10 from both sides. Then divide both sides by 5.
$5n = 100$
$n = 100 \div 5 = 20$
So the least number is 20.
Give the answer
Watch out: $110 \div 5 = 22$ 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: $20 + 21 + 22 + 23 + 24 = 110$ . 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 21 of 57

2023 Test B Math, question 21

Question

$-mn$

$(-m)(-n)$

$|mn|$

$|-mn|$

$|(-m)(-n)|$

If m and n are positive integers, how many of the five expressions above are equivalent to $mn$ ?

Show the answer

Correct answer: D

Since m and n are positive, the value of $mn$ is also positive. Of the five options given:

$-mn$ is negative.

$(-m)(-n)$ is positive.

$|mn|$ is positive.

$|-mn|$ is positive.

$|(-m)(-n)|$ is positive.

There are 4 expressions that are positive, and each one is equal to $mn$ .

Question analysis

Five expressions are listed: −mn, (−m)(−n), |mn|, |−mn| and |(−m)(−n)|. If m and n are positive integers, students count how many are equivalent to mn. It tests the sign rules for multiplying negatives together with what absolute value does, and each expression has to be judged on its own.

NYS standards: 7.NS.2a rational number multiplication
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Multiplying signed numbers and absolute value are grade 6 and 7 content. Checking five expressions one by one takes longer than a typical single computation, but each check uses a familiar rule.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Much to keep track of minor
Something you must already know minor
Solution: 5 steps
Reading level: Question and choices: about grade 10, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each expression takes only a moment, but the answer is a count, so one misjudged expression changes it. The minus signs are designed to unsettle: four of the five expressions show at least one, and only one of those actually stays negative. Substituting small positive numbers and comparing each value with mn turns the question into arithmetic.

Variables instead of numbers moderate Understanding the question
The five expressions use only m and n, so each sign rule has to be applied in general unless the student substitutes values such as m = 2 and n = 3. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Much to keep track of minor Working it out
Five expressions are judged one at a time and a count kept, and two of them combine the product rule for negatives with absolute value bars. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
The rules that two negative factors give a positive product and that absolute value removes a minus sign are assumed, not stated. (Anderson & Krathwohl, 2001)

Strategy: Pick small values such as m = 2 and n = 3, evaluate each of the five expressions, and count the ones equal to 6.

Traps

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Treats any minus sign as making the value negative, even inside bars or in a product of two negatives, so counts only |mn|.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Accepts |mn| and (−m)(−n) but believes bars cannot cancel a minus sign written inside them, rejecting |−mn| and |(−m)(−n)|.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Thinks a negative times a negative stays negative, so rejects (−m)(−n) along with −mn and counts only the three absolute values.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying two negatives: $(-a) \times (-b) = a \times b$; a and b are positive numbers. A negative number times a negative number gives a positive number. The two minus signs cancel out. One negative and one positive give a negative.
Absolute value: $|a| = \text{the distance from } a \text{ to } 0$; The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so |−6| = 6 and |6| = 6. The bars remove any minus sign.

Sort out the question
m and n are positive integers. That means whole numbers greater than 0. Writing mn means m times n, so mn is positive. You need to count how many of the five expressions always equal mn.
Pick numbers to test
Pick easy values: $m = 2$ and $n = 3$ . Then $mn = 2 \times 3 = 6$ . Any expression that equals mn must come out to 6.
Test the first two expressions
$-mn = -(2 \times 3) = -6$
This is negative, so it does not equal 6.
$(-m)(-n) = (-2)(-3) = 6$
A negative times a negative is positive, so this one equals 6. Watch out: if you think a negative times a negative stays negative, you reject this one and get 3, choice C.
Test the three absolute values
$|mn| = |6| = 6$
$|-mn| = |-6| = 6$
$|(-m)(-n)| = |6| = 6$
All three equal 6, because absolute value bars remove a minus sign. Watch out: the bars work even when the minus sign is inside them. Rejecting the last two gives 2, choice B.
Count the matches
Four expressions equal mn: $(-m)(-n)$ , $|mn|$ , $|-mn|$ and $|(-m)(-n)|$ . Only $-mn$ does not. The sign rules work the same for any positive m and n, not just 2 and 3. So the answer is D, 4.

Check your answer

Try other numbers, such as $m = 1$ and $n = 5$ . Then $mn = 5$ . The five expressions give −5, 5, 5, 5 and 5. Again, four of them match mn.

More questions like this

The original DOE question

Question 78 as printed on page 197 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 78. Open page 197 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 22 of 57

2023 Test B Math, question 22

Question

A vendor is selling 10 packages of homemade fudge at a bake sale. The vendor weighed and labeled each package. The package weights are given in ounces:

4.6	6.0	3.8	3.5	4.5
5.4	3.9	6.5	4.8	6.2

The vendor created a box plot of the package weights. Which box plot best represents the data?

A $Box plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at about 4.1, the box runs from 5 to 6 with the median line at 5.5, and the right whisker ends at about 6.3.$
B $Box plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from 4 to 6 with the median line at 4.5, and the right whisker ends at 6.5.$
C $Box plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from about 3.9, just left of 4, to 6, with the median line halfway between 4.5 and 5, and the right whisker ends at 6.5.$
D $Box plot titled PACKAGE WEIGHTS above a number line labeled Ounces, from 3 to 7 with arrows at both ends and a labeled tick every 0.5. The left whisker starts at 3.5, the box runs from 4 to 6 with the median line at 5, and the right whisker ends at 6.5.$

Show the answer

Correct answer: C

A set of 10 data points is given. The first step for determining the quartiles and minimum and maximum values of the box plot is to order the data from smallest to greatest value. For this data set the order is:

3.5	3.8	3.9	4.5	4.6
4.8	5.4	6.0	6.2	6.5

The minimum value is the smallest decimal, 3.5, and the maximum value is the greatest decimal, 6.5.

To determine the median in a set with an even number of values, locate the two middle values (5th and 6th in this case): 4.6 and 4.8. Determine the mean, or average, of the two middle numbers by adding them together and dividing the sum by 2:

$\frac{4.6 + 4.8}{2} = \frac{9.4}{2} = 4.7$

To determine the value of the first quartile (Q1), locate the middle value of the 5 terms to the left of the median:

3.5

3.8

3.9

4.5

4.6

To determine the value of the third quartile (Q3), locate the middle value of the 5 terms to the right of the median:

4.8

5.4

6.0

6.2

6.5

Box plot C identifies the correct minimum, maximum, median, and quartiles of the data set.

Question analysis

A vendor weighs 10 packages of homemade fudge, from 3.5 to 6.5 ounces, and makes a box plot of the weights. Students choose the box plot that matches the data. They must sort the decimals, find the median of an even number of values and both quartiles, then compare plots that differ only slightly.

NYS standards: 7.SP.1 box plots and interquartile range
6.SP.5c range, center, and overall pattern
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Building a box plot from data is grade 7 content. This version needs all five summary values from ten unsorted decimals, including a median between two values, then a close comparison of plots, longer and finer than typical.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
A very close wrong answer moderate
Reading a figure, table or graph moderate
Slow to do carefully minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 43 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Most of the work happens before the plots are even read: the weights must be sorted, and the median of ten values is the average of two of them. After that, three plots look almost identical, so a careless median, taken from the unsorted list or placed halfway between the extremes, settles on a wrong one. The careful student checks the median line and the box’s left edge against values that fall between the tick marks.

Many steps moderate Working it out
Ten unsorted decimals must be ordered, then five values found: the median as the mean of the fifth and sixth, 4.6 and 4.8, and each quartile as the middle of its half. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
Three plots share whiskers at 3.5 and 6.5 and a box ending at 6; they differ only in the median line, at 4.5, near 4.75 or at 5, and in where the box starts. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph moderate Understanding the question
The axis is ticked every half ounce, and the correct plot’s median, 4.7, and lower quartile, 3.9, both fall between ticks, so each line’s position must be judged by eye. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Slow to do carefully minor Choosing the answer
Sorting ten decimals, computing a five-number summary and comparing four plots takes well over a minute and a half when done carefully. (Lu & Sireci, 2007)

Strategy: Sort the ten weights, then find the least and greatest, the median as the mean of the fifth and sixth values, and each quartile as the middle of its half.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working from these weights gives whiskers near 4.1 and 6.3 or a median of 5.5, since the data run from 3.5 to 6.5, so it mostly draws guesses.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Skips sorting and takes 4.5, the fifth weight as printed, for the median, so the plot with its median line on 4.5 seems to fit.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Finds the median with the midrange formula, (3.5 + 6.5) ÷ 2 = 5, halfway between the lightest and heaviest packages instead of the middle of the sorted weights.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Median of an even number of values: $\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}$; First put the values in order. With an even number of values, two sit in the middle. The median is their mean, or average: add them and divide by 2. The median splits the data in half.
Quartiles: $Q_1 = \text{middle of the lower half} \qquad Q_3 = \text{middle of the upper half}$; Q1 is the first quartile and Q3 is the third quartile. Split the ordered data in half at the median. Q1 is the middle of the lower half, and Q3 is the middle of the upper half. A box plot’s box runs from Q1 to Q3.

Sort out the question
The vendor weighed 10 packages of fudge, in ounces. You need the box plot that matches the weights. A box plot is built from five numbers. They are the minimum (least), the first quartile Q1, the median, the third quartile Q3 and the maximum (greatest). So find those five numbers first.
Put the weights in order
List the 10 weights from least to greatest.
3.5 3.8 3.9 4.5 4.6
4.8 5.4 6.0 6.2 6.5
Watch out: you must sort first. The fifth weight as printed is 4.5. Using it as the median points to choice B.
Find the least and greatest
The minimum is 3.5 ounces and the maximum is 6.5 ounces. So the whiskers must run from 3.5 to 6.5. Plot A’s whiskers run from about 4.1 to about 6.3, so A is out.
Find the median
With 10 weights, the two middle ones are the 5th and 6th: 4.6 and 4.8.
$\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}$
$\text{median} = \frac{4.6 + 4.8}{2} = \frac{9.4}{2} = 4.7$
The median is 4.7 ounces. Watch out: the median is not halfway between the least and greatest. That gives $(3.5 + 6.5) \div 2 = 5$ and points to choice D.
Find the two quartiles
The lower half is the 5 weights below the median: 3.5, 3.8, 3.9, 4.5 and 4.6. Its middle value is $Q_1 = 3.9$ .
The upper half is the 5 weights above the median: 4.8, 5.4, 6.0, 6.2 and 6.5. Its middle value is $Q_3 = 6.0$ .
Match the summary to a plot
The five numbers are 3.5, 3.9, 4.7, 6.0 and 6.5. Plots B, C and D all run from 3.5 to 6.5, and their boxes end at 6. They differ in the median line and where the box starts.
- B: the box starts at 4, and the median line is at 4.5.
- D: the box starts at 4, and the median line is at 5.
- C: the box starts just left of 4, and the median line is between 4.5 and 5.
C is the closest match to a Q1 of 3.9 and a median of 4.7. So the answer is C.

Check your answer

A median splits the data in half. Five weights, 3.5 to 4.6, are below 4.7. Five, 4.8 to 6.5, are above it. A median line at 5, as in D, would have six weights below it. A line at 4.5, as in B, would have only three below it.

More questions like this

The original DOE question

Question 79 as printed on page 198 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 198, question 79. Open page 198 of the handbook (PDF)

Question 23 of 57

2023 Test B Math, question 23

Question

FRUIT SOLD
AT STORE XYZ

Fruit	Number Sold
Apples	8
Bananas	6
Cherries	12
Oranges	6
Peaches	?
Total	50

Peaches made up what percentage of the total number of pieces of fruit sold at Store XYZ, according to the table above?

A18%
B28%
C36%
D64%

Show the answer

Correct answer: C

First, find the number of peaches sold:

$50 - (8 + 6 + 12 + 6) =$

$50 - 32 = 18$

Then, find the percentage:

$\frac{18}{50} = 0.36 = 36\%$

Question analysis

A table lists the fruit sold at Store XYZ, apples, bananas, cherries and oranges, with the number of peaches shown as a question mark and a total of 50. Students find what percentage of the fruit sold were peaches. They first work out the missing count, then turn it into a percent of the total.

NYS standards: 6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a missing value and writing a part as a percent of a whole are grade 6 skills. With a total of 50, the percent step is quick, so the task is lighter than typical grade 7 percent problems.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The total of 50 makes the percent step easy, since each piece of fruit is 2 percent, but the question first hides the count it needs. Students who find the 18 peaches and stop, or who give the share of the four listed fruits, reach listed choices. Checking that the peaches’ percent and the other fruits’ percent add to 100 confirms the answer.

Reading a figure, table or graph minor Understanding the question
The table gives four counts, a question mark for peaches and a total of 50, so the missing count must be found by subtraction before any percent. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Adding 8, 6, 12 and 6, subtracting from 50, and turning 18 out of 50 into a percent are three steps, and the count of 18 is itself a listed percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the four known counts, subtract from 50 to get the peaches, then double that count, since each piece out of 50 is 2 percent.

Traps

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

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the 18 peaches but gives that count as the percent, forgetting to compare it with the total of 50 pieces.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 28%, which would need 14 peaches; it takes a slip such as 36 for 8 + 6 + 12 + 6, so it mostly draws guesses.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the share of the other four fruits, 32 of 50 or 64 percent, instead of the share of the peaches.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a total: $\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$; The part is the amount you are asking about, here the peaches. The whole is the total, here all 50 pieces of fruit. Dividing gives the fraction of the whole. Multiplying by 100 turns it into a percent, which means out of 100.

Sort out the question
The table shows how many of each fruit Store XYZ sold. The number of peaches is missing, shown by a question mark. The total is 50 pieces of fruit. You need the percent of the total that were peaches.
Add the fruits you know
$8 + 6 + 12 + 6 = 32$
Apples, bananas, cherries and oranges add up to 32 pieces of fruit. That is every fruit except the peaches.
Find the number of peaches
The peaches are whatever is left of the 50.
$50 - 32 = 18$
So 18 peaches were sold. Watch out: 18 is a count, not a percent. Choosing 18% is choice A.
Turn the count into a percent
$\text{percent} = \frac{\text{part}}{\text{whole}} \times 100$
$\frac{18}{50} \times 100 = 0.36 \times 100 = 36$
Peaches were 36% of the fruit sold. Watch out: 32 out of 50 is 64%, choice D. That is the other four fruits, not the peaches. So the answer is C, 36%.

Check your answer

Out of 50 pieces, each piece is 2% of the total, because $100 \div 50 = 2$ . So 18 peaches are $18 \times 2 = 36$ percent. Also, 36% for peaches plus 64% for the rest makes 100%.

More questions like this

The original DOE question

Question 80 as printed on page 198 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 198, question 80. Open page 198 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 24 of 57

2023 Test B Math, question 24

Question

A prism is made up of 9 congruent cubes. Each of the cubes has side lengths measuring 4 units. What is the volume, in cubic units, of the prism?

A108
B144
C480
D576

Show the answer

Correct answer: D

First, determine the volume of each congruent cube. All the side lengths of the cubes are equal, so the length, width, and height of each cube is 4 units. To determine the volume of each cube, multiply $\text{length} \times \text{width} \times \text{height}$ :

$4 \times 4 \times 4 = 64$

So, the volume of each cube is 64 cubic units.

$Left: a prism built from 9 cubes, three rows of three, one cube deep, with one cube’s height, width and depth each labeled 4. Right: a single cube whose height, width and depth are each labeled 4.$

The prism is comprised of 9 of these cubes, so multiply the volume of each cube (64) by the total number of cubes (9) to determine the volume, in cubic units, of the prism:

$64 \times 9 = 576$

Question analysis

A prism is built from 9 identical cubes, and each cube has edges 4 units long. Students find the volume of the prism in cubic units. It tests the volume of a cube and how volumes add when solids are put together, and it rewards keeping area, surface area and volume apart.

NYS standards: 6.G.2 volume with fractional edge lengths
6.G.5 perfect squares and perfect cubes
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Volume of cubes and rectangular prisms is grade 5 and 6 content. This item needs one cube's volume and a multiplication by 9, two short steps, so it is below typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 27 words.
Corrected from the DOE handbook: A letter x printed in place of a multiplication sign in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question needs one formula and one multiplication, and the stem does not even say how the cubes are arranged, because the volume does not depend on it. The wrong choices come from mishandling the cube’s volume: the exponent multiplied instead of applied, the area of one face, or a surface area. Writing the volume out as edge times edge times edge, in cubic units, rules all three out.

Something you must already know minor Working it out
A cube’s volume, edge × edge × edge, is not given; with it, each 4-unit cube holds 64 cubic units, and the arrangement of the 9 cubes does not matter. (Anderson & Krathwohl, 2001)

Strategy: Find the volume of one cube by multiplying its three edges, then multiply by the number of cubes; how the cubes are arranged does not matter.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Takes 4³ as 4 × 3 = 12, multiplying the edge by the exponent, so each cube seems to hold 12 cubic units and 9 × 12 = 108.
B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Uses the area of one face, 4 × 4 = 16 square units, as each cube’s volume, so 9 × 16 = 144.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Pictures the 9 cubes as a 3-by-3 slab, 12 by 12 by 4 units, and finds its surface area, 480 square units, instead of its volume.

What was corrected from the DOE handbook, and why

Explanation: The handbook gives the volume of a cube as “length x width x height”, with the letter x where it means “times”. In a math explanation the letter x reads as a variable, so the site writes length × width × height, the sign the handbook itself uses on the next line (4 × 4 × 4 = 64). Nothing else changes: each cube is 64 cubic units, and the 9 cubes make 64 × 9 = 576.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a cube: $V = s \times s \times s$; V is the volume, the space inside a solid, in cubic units. s is the length of one edge. All the edges of a cube are equal, so its length, width and height are all s. Multiplying them counts how many unit cubes fit inside.

Sort out the question
A prism, a solid with flat faces, is made of 9 congruent cubes. Congruent means exactly the same size and shape. Each cube has sides of 4 units. You need the volume of the whole prism, in cubic units.
The question doesn’t say how the cubes are stacked. You don’t need to know. Volume is the space inside, and 9 cubes fill the same space however they are arranged.
Find the volume of one cube
$V = s \times s \times s$
$V = 4 \times 4 \times 4 = 64$
Each cube has a volume of 64 cubic units. Watch out: $4^3$ means $4 \times 4 \times 4$ , not $4 \times 3 = 12$ . That slip leads to 108, choice A. Also, $4 \times 4 = 16$ is only the area of one face. It leads to 144, choice B.
Multiply by the number of cubes
The prism is 9 of these cubes, so its volume is 9 times as big.
$64 \times 9 = 576$
The prism’s volume is 576 cubic units.
Give the answer
Watch out: volume counts the space inside, not the outside faces. Suppose the cubes formed one flat 3-by-3 layer. Adding up the areas of its outside faces gives 480, choice C. That is surface area, not volume. So the answer is D, 576.

Check your answer

Estimate: $64 \times 10 = 640$ , and one cube fewer is $640 - 64 = 576$ . You can also test one arrangement. A flat 3-by-3 layer is 12 by 12 by 4 units, and $12 \times 12 \times 4 = 576$ . Same answer.

More questions like this

The original DOE question

Question 81 as printed on page 198 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 198, question 81. Open page 198 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 25 of 57

2023 Test B Math, question 25

Question

The town of Waterville covers an area of 50 square miles. In 2005, its population was 84,000. In 2006, its population had decreased so that the average population per square mile was equal to 1,500. What was the total decrease in population from 2005 to 2006?

A7,500
B8,400
C9,000
D9,200

Show the answer

Correct answer: C

Let p represent the population of Waterville in 2006. The population per square mile is:

$\frac{p}{50} = 1{,}500$

Solve for p:

$p = 1{,}500(50) = 75{,}000$

The population decrease from 2005 to 2006 is $84{,}000 - 75{,}000 = 9{,}000$ .

Question analysis

The town of Waterville covers 50 square miles and had 84,000 people in 2005. By 2006 the population had dropped so that there were 1,500 people per square mile on average. Students find how much the population decreased, turning the population density into a total population before subtracting.

NYS standards: 6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Using a rate such as people per square mile is grade 6 content. Here it is one step inside a two-step problem with large round numbers, which matches typical grade 7 multi-step rate work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question gives a total for 2005 but only a rate for 2006, so the two years cannot be compared until the rate becomes a population. After that, one subtraction finishes it. Students who stop at the 2006 population, or who reach for a number shaped like 84,000, land on listed choices, while the near miss above the answer comes from no sensible working.

Turning words into math moderate Understanding the question
“The average population per square mile was equal to 1,500” is a density, which has to become a total, 1,500 × 50 = 75,000, before it can be compared with 84,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
The 2006 population is only a midpoint; subtracting it from 84,000 is a second step, and the question asks for that decrease, not either population. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply 1,500 people per square mile by 50 square miles to get the 2006 population, then subtract it from the 2005 population.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the 2006 population, 1,500 × 50 = 75,000, and stops; 7,500, one zero short, is the choice whose digits match it.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes a tenth of the 2005 population, 84,000 ÷ 10, a number built from the question rather than from the density.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 9,200; it lies just above the right decrease and needs an arithmetic slip with no reason behind it, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Population per square mile: $\text{people per square mile} = \frac{\text{population}}{\text{area}}$; The population is the number of people who live in a place. The area is its size, here in square miles. Dividing gives the average number of people for each square mile. To get back the population, multiply people per square mile by the area.

Sort out the question
Waterville covers 50 square miles. In 2005, 84,000 people lived there. In 2006, there were 1,500 people per square mile, on average. You need how many fewer people lived there in 2006 than in 2005.
Find the 2006 population
Let p be the 2006 population. Use the formula.
$\text{people per square mile} = \frac{\text{population}}{\text{area}}$
$1{,}500 = \frac{p}{50}$
To undo dividing by 50, multiply both sides by 50.
$p = 1{,}500 \times 50 = 75{,}000$
So 75,000 people lived there in 2006.
Find the decrease
A decrease is how much a number went down. Subtract the 2006 population from the 2005 population.
$84{,}000 - 75{,}000 = 9{,}000$
The population went down by 9,000 people.
Give the answer
Watch out: 75,000 is the 2006 population, not the decrease. Choice A, 7,500, looks like it but has one zero fewer. The question asks how much the population dropped. So the answer is C, 9,000.

Check your answer

Work with people per square mile instead. In 2005, there were $84{,}000 \div 50 = 1{,}680$ people per square mile. In 2006, there were 1,500. That is 180 fewer per square mile, and $180 \times 50 = 9{,}000$ fewer people.

More questions like this

The original DOE question

Question 82 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 82. Open page 199 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 26 of 57

2023 Test B Math, question 26

Question

The box plot below represents a set of data.

$Box plot above a number line with arrows at both ends, a tick every 3 units from 0 to 30, and labels at 0, 6, 12, 18, 24 and 30. The left whisker starts at 3, the box runs from 6 to the unlabeled tick for 27 with the median line at 15, and the right whisker ends at 30.$

What is the interquartile range of the set of data?

Show the answer

Correct answer: C

First, determine the values of the upper and lower quartiles of the set of data:

$The same box plot, with an arrow labeled lower quartile pointing down to the left edge of the box, at 6, and an arrow labeled upper quartile pointing down to the right edge of the box, at the unlabeled tick for 27.$

The upper quartile value is indicated by the vertical line segment at the far right edge of the box plot above the hash mark representing 27.
The lower quartile value is indicated by the vertical line segment at the far left edge of the box plot above the hash mark representing 6.

To determine the interquartile range, subtract the lower quartile value (6) from the upper quartile value (27):

$27 - 6 = 21$

Question analysis

A box plot on a number line marked every 3 units, with labels every 6, represents a set of data. Students find the interquartile range. They have to read the two ends of the box, one of which falls on an unlabeled tick, and remember exactly which part of the plot the interquartile range measures.

NYS standards: 7.SP.1 box plots and interquartile range
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the interquartile range from a box plot is a grade 7 skill, and this is a typical example. Read two values and subtract; the only extra care is a box edge that falls on a tick with no label.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 19 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once the two edges of the box are read, one subtraction finishes the question, so the work is reading the scale and knowing which part of the plot is meant. The upper quartile falls on an unlabeled tick, which costs care. The wrong answers measure the whole spread or only half of the box, all common mix-ups of what the interquartile range covers.

Reading a figure, table or graph moderate Understanding the question
The axis is labeled only every 6 units but ticked every 3, and the box’s right edge sits on the unlabeled tick for 27, so it has to be counted from 24. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Nothing defines the interquartile range; it is the width of the box, the upper quartile minus the lower, not the span of the whiskers. (Anderson & Krathwohl, 2001)

Strategy: Find the values at the left and right edges of the box, counting tick marks in steps of 3 where there is no label, and subtract the smaller from the larger.

Traps

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

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Gives the median minus the lower quartile, 15 − 6 = 9, as the interquartile range, measuring only the left part of the box.
B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Gives the upper quartile minus the median, 27 − 15 = 12, as the interquartile range, measuring only the right part of the box.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Subtracts the ends of the whiskers, 30 − 3 = 27, giving the range of the data instead of the interquartile range.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range: $\text{IQR} = Q_3 - Q_1$; IQR stands for interquartile range. Q3, the third quartile, is the right edge of a box plot’s box. Q1, the first quartile, is the left edge. Subtracting gives the width of the box, where the middle half of the data lies.

Sort out the question
You see a box plot and need the interquartile range, or IQR. That is the width of the box: the upper quartile minus the lower quartile. The whiskers don’t matter here.
Read the number line
The labels are 0, 6, 12, 18, 24 and 30. But there is a tick mark halfway between each pair of labels. So the ticks go up by 3: 0, 3, 6, 9 and so on. The tick between 24 and 30 stands for 27.
Find the edges of the box
The left edge of the box is at 6. That is the lower quartile, Q1. The right edge is at the unlabeled tick after 24, which is 27. That is the upper quartile, Q3.
Subtract to find the IQR
$\text{IQR} = Q_3 - Q_1$
$\text{IQR} = 27 - 6 = 21$
Watch out: $30 - 3 = 27$ uses the ends of the whiskers. That is the range, choice D. Also, the median line at 15 splits the box. $15 - 6 = 9$ and $27 - 15 = 12$ are each only part of the box. So the answer is C, 21.

Check your answer

Count the tick spaces across the box, from 6 to 27. There are 7 spaces, and each is worth 3. So the box is $7 \times 3 = 21$ units wide. That matches.

More questions like this

The original DOE question

Question 83 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 83. Open page 199 of the handbook (PDF)

Question 27 of 57

2023 Test B Math, question 27

Question

$\mathrm{N} = \{6, 7, \ldots, 21, 22\}$

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?

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer moderate
No obvious method moderate
Reading a figure, table or graph minor
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.

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)
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)
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

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives 25, the number of integers in set L stated in the question, instead of finding which integer sits in the middle.
C Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts 13 places up from 14 instead of 12, mixing up the median’s position, 13th, with its distance from the first integer.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 28, which is twice N’s median of 14 and matches no count along set L, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting the integers in a list: $\text{count} = \text{last} - \text{first} + 1$; 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: $\text{position} = \frac{\text{count} + 1}{2}$; 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.

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.
Count the numbers in set N
$\text{count} = \text{last} - \text{first} + 1$
$22 - 6 + 1 = 17$
Set N has 17 integers. Watch out: $22 - 6 = 16$ misses one, because the first number counts too.
Find the median of set N
$\text{position} = \frac{\text{count} + 1}{2}$
$\text{position} = \frac{17 + 1}{2} = 9$
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.
Find the middle place in set L
So set L starts at 14 and has 25 integers. Find the position of its median.
$\text{position} = \frac{25 + 1}{2} = 13$
The median of L is its 13th number.
Count up to the 13th number
14 is the 1st number. The 13th is 12 steps after it.
$14 + 12 = 26$
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 $14 + 24 = 38$ . For consecutive integers, the median is halfway between the ends: $(14 + 38) \div 2 = 26$ . That matches. The same trick gives N’s median: $(6 + 22) \div 2 = 14$ .

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 28 of 57

2023 Test B Math, question 28

Question

The sum of the numbers x, y, and z is 40. The ratio of x to y is 1:2, and the ratio of y to z is 2:5. What is the value of y?

Show the answer

Correct answer: D

The ratio of x to y is 1:2, so $\frac{x}{y} = \frac{1}{2}$ or $x = \frac{1}{2}y$ . Similarly, the ratio of y to z is 2:5, so $\frac{y}{z} = \frac{2}{5}$ . Solve for z in terms of y:

$5y = 2z$

$\frac{5}{2}y = z$

Use the given sum of x, y, and z to find the value of y:

$x + y + z = 40$

$\frac{1}{2}y + y + \frac{5}{2}y = 40$

$4y = 40$

$y = 10$

Question analysis

This ratio question gives two ratios that share y: x to y is 1:2 and y to z is 2:5, and x, y and z total 40. Students find the value of y. The numbers in a ratio count parts, not amounts, so the work is to combine the ratios and find what one part is worth.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Most grade 7 ratio tasks split a total using one ratio. Here two ratios linked through y must first become a single three-part ratio; since y's term already matches, that step is short and the problem stays at grade 7.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Variables instead of numbers minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 2, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question rewards seeing that the two ratios share y’s term, which makes one three-part ratio. From there, one part is worth 5 and y is two parts. The wrong choices are numbers that live inside the ratios, and adding all four ratio numbers, which counts y twice, also leads to the part count, so that choice draws students along two paths.

No obvious method moderate Working it out
Nothing says to merge the ratios, but y is 2 in both, so x : y : z becomes 1 : 2 : 5 at once; the explanation’s route through fractions of y is slower. (Schoenfeld, 1985)
Variables instead of numbers minor Understanding the question
x, y and z appear only in two ratios and a sum of 40, so the ratio numbers 1, 2 and 5 are easy to mistake for the values themselves. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Adding the 8 parts, dividing 40 by 8 to get 5 per part, and taking y’s 2 parts are three steps, and the part count is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Write x : y : z as one ratio using y's matching term, add the parts, divide 40 by that sum to get one part, and multiply by y's parts.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Takes y’s term in the ratios, 2, as y’s actual value, never using the sum of 40.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two 2s that stand for y in the two ratios, 2 + 2 = 4, treating ratio terms as amounts.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Combines the ratios into 1 : 2 : 5 and gives the total number of parts, 1 + 2 + 5 = 8, instead of y’s share of the 40.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Splitting a total by a ratio: $\text{one part} = \text{total} \div \text{number of parts}$; Each number in a ratio counts parts that are all the same size. Add the numbers in the ratio to find how many parts make up the total. Dividing the total by that number gives the size of one part. Each amount is its number of parts times that size.

Sort out the question
x, y and z add up to 40. A ratio compares two amounts. The ratio of x to y is 1:2, so y is twice x. The ratio of y to z is 2:5. You need the actual value of y, not its number in a ratio.
Combine the two ratios
y is 2 in both ratios. So both ratios count the same size of part, and they join into one.
$x : y : z = 1 : 2 : 5$
So x is 1 part, y is 2 parts and z is 5 parts.
Find the total number of parts
$1 + 2 + 5 = 8$
The 40 is split into 8 equal parts. Watch out: 8 counts parts. It is not y, so choice C is wrong.
Find the size of one part
$\text{one part} = \text{total} \div \text{number of parts}$
$\text{one part} = 40 \div 8 = 5$
The 40 is shared equally among the 8 parts. So each part is worth 5.
Find y
y is 2 parts, so multiply the size of one part by 2.
$y = 2 \times 5 = 10$
Watch out: the 2 in the ratio counts parts, so y is not 2, choice A. Adding the two 2s to get 4 is wrong too, choice B. So the answer is D, 10.

Check your answer

Find all three: x = 5, y = 10 and z = 25. They add to $5 + 10 + 25 = 40$ . x to y is 5 to 10, which is 1:2. y to z is 10 to 25, which is 2:5.

More questions like this

The original DOE question

Question 85 as printed on page 199 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 85. Open page 199 of the handbook (PDF)

Question 29 of 57

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:

$n+(n+1)+(n+2)+(n+3)$

$=4n+6$

Set that expression equal to each of the options and solve for n:

Option A:

$4n+6=-12$

$4n=-18$

$n=-\frac{18}{4}$ is not an integer.

Option B:

$4n+6=-10$

$4n=-16$

$n=-4$ is an integer.

Option C:

$4n+6=-8$

$4n=-14$

$n=-\frac{14}{4}$ is not an integer.

Option D:

$4n+6=-4$

$4n=-10$

$n=-\frac{10}{4}$ 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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Every choice must be checked moderate
No obvious method minor
Error-prone arithmetic minor
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.

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)
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)
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

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Assumes four consecutive integers add to a multiple of 4, as if they had a whole-number middle, and takes −12 = 4 × (−3), the first such choice.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Writes n + (n + 1) + (n + 2) + (n + 3) as 4n, dropping the 1 + 2 + 3, so −8 = 4 × (−2) seems to give a whole-number n.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Solves 4n + 6 = −4 to get n = −2.5 and accepts it, since −2.5, −1.5, −0.5 and 0.5 add to −4, though they are not integers.

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: $n + (n + 1) + (n + 2) + (n + 3) = 4n + 6$; 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.

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.
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.
$n + n + n + n + 1 + 2 + 3 = 4n + 6$
Any four consecutive integers add up to 4n + 6.
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.
Choice 4n + 6 4n n
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.
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.

Choice	4n + 6	4n	n
A	−12	−18	−4.5
B	−10	−16	−4
C	−8	−14	−3.5
D	−4	−10	−2.5

Check your answer

Add the integers: $-4 + (-3) + (-2) + (-1) = -10$ . 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 30 of 57

2023 Test B Math, question 30

Question

PRICE OF A USED VIDEO GAME AT VARIOUS STORES

Price of Used Video Game	Number of Stores
$10.99	5
$11.99	2
$12.99	1
$13.99	4

The table above shows the price of a certain used video game at 12 different stores. What is the median price of this video game?

A$10.99
B$11.49
C$11.99
D$12.32

Show the answer

Correct answer: C

There are 5 stores where the video game is priced at $10.99 and a total of 5 stores where it is priced at either $12.99 or $13.99, so the median price must be $11.99.

Question analysis

A statistics question with a table showing the price of a used video game at 12 stores, grouped by price, and asking for the median price. Students put the prices in order using the number of stores at each price and find the middle. The table gives counts rather than a list, so the position of the middle value has to be worked out from the running totals.

NYS standards: 6.SP.5c range, center, and overall pattern
6.SP.5a number of observations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Finding a median is grade 6 work that grade 7 students meet often. The twist is the frequency table: the 12 prices are not listed one by one, so students count positions instead of reading down a list.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 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 arithmetic is trivial once the positions are found: counting down 5 stores at $10.99 and then 2 at $11.99 puts both the 6th and 7th prices at $11.99. The wrong choices reward three different misreadings of a frequency table: the most common price, a middle pair one place too early, and the mean, which takes a long computation to reach. A student who writes the running totals 5, 7, 8 and 12 beside the table avoids all three.

Reading a figure, table or graph moderate Understanding the question
The 12 prices are never listed; the table gives four prices with store counts of 5, 2, 1 and 4, so the 6th and 7th prices must be found by a running count. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
With 12 values the median is the mean of the 6th and 7th in order, a rule the question assumes; here both fall in the $11.99 row, so no averaging is needed. (Anderson & Krathwohl, 2001)

Strategy: Keep a running total of the store counts down the table; with 12 prices in order, the median is the average of the 6th and 7th, so find the rows where they fall.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: moderate
Gives the mode, $10.99, the price that 5 stores charge, more than any other, instead of the middle price of the 12.
B Counting the ends (Brown & Burton, 1978)
Reveals partial knowledge Expected pull: moderate
Knows to average two middle prices but takes the 5th and 6th instead of the 6th and 7th, averaging $10.99 and $11.99 to get $11.49.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: weak
Computes the mean instead, $147.88 for all 12 prices divided by 12, about $12.32, a different measure of center from the median asked for.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Median of an even number of values: $\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}$; The median is the middle value when all the values are listed from least to greatest. With an even number of values, no single value sits in the middle. So take the two middle values and find their average: add them and divide by 2.

Sort out the table
Each row gives a price and how many stores charge it. For example, 5 stores charge $10.99. So $10.99 shows up 5 times in the list of prices. Add the stores to check the count.
$5 + 2 + 1 + 4 = 12$
There are 12 prices. You need the median. That is the middle price when the prices are in order from least to greatest.
Find the middle places
12 is an even number, so no single price sits in the middle. The two middle prices are the 6th and the 7th. Five prices come before the 6th one. Five prices come after the 7th one.
Count down the table
The table already lists the prices from least to greatest. Keep a running total of the stores. It shows which places in the list each price fills.
Price Stores Places in the list
$10.99 5 1st to 5th
$11.99 2 6th and 7th
$12.99 1 8th
$13.99 4 9th to 12th
The 6th and 7th prices are both $11.99.
Find the median
Use the formula with the 6th and 7th prices.
$\text{median} = \frac{11.99 + 11.99}{2} = \frac{23.98}{2} = 11.99$
The median price is $11.99.
Watch out: $10.99 is the mode, the price that shows up most often. That is choice A, not the median. Averaging the 5th and 6th prices gives $11.49, choice B. Adding all 12 prices and dividing by 12 gives the mean, about $12.32. That is choice D.
So the answer is C, $11.99.

Price	Stores	Places in the list
$10.99	5	1st to 5th
$11.99	2	6th and 7th
$12.99	1	8th
$13.99	4	9th to 12th

Check your answer

Count the prices on each side of $11.99. Five stores charge less: the five at $10.99. Five stores charge more: one at $12.99 and four at $13.99. The two sides match, so $11.99 is in the middle.

More questions like this

The original DOE question

Question 87 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 87. Open page 200 of the handbook (PDF)

Question 31 of 57

2023 Test B Math, question 31

Question

If $xy=30$ , what is the value of $10\div x$ in terms of y?

A $\frac{1}{10}y$
B $\frac{1}{3}y$
C $3y$
D $10y$

Show the answer

Correct answer: B

Solve $xy=30$ , the given equation, for x:

$xy=30$

$x=\frac{30}{y}$

Then:

$10\div x=10\div\frac{30}{y}=10\times\frac{y}{30}=\frac{1}{3}y$

Question analysis

An algebra question: given that x times y equals 30, students express 10 divided by x using only y. It tests solving for one variable and dividing by an algebraic fraction. Working with letters instead of numbers, and dividing in the right order, are what trip students up.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.1 dividing fractions by fractions
6.EE.7 one-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students rewrite expressions that have numbers and one variable. This one links two variables through their product and asks for a quotient, with no numbers given to check an answer against.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The algebra is three short moves, but each is a place to turn a fraction upside down: solving xy = 30 for x, dividing 10 by that fraction, and reducing 10/30. Students who reach for the bigger-over-smaller division they know from whole numbers land on 3y. Choosing two numbers with a product of 30, such as 3 and 10, and testing every choice settles the question in seconds.

Variables instead of numbers moderate Understanding the question
x and y are tied only by their product, 30, and the answer must be an expression in y, so there is no number to check unless students pick a pair such as x = 3 and y = 10. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Substituting x = 30/y turns 10 ÷ x into 10 divided by a fraction, which must be flipped to 10 × y/30 and then reduced. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The answer, (1/3)y, and the choice 3y use the same 3, one dividing y and one multiplying it, so a student unsure whether 10 ÷ 30 or 30 ÷ 10 applies sees both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Solve xy = 30 for x, substitute that fraction into 10 ÷ x, and divide by multiplying by its reciprocal; or choose values of x and y with a product of 30 and test each choice.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Writes the 10 from 10 ÷ x under y as one tenth of y, never using the product xy = 30 to link the two letters.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Replaces x with 30/y correctly but divides 30 by 10 instead of 10 by 30, taking the larger number over the smaller, and gets 3y.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats dividing by x as multiplying by y, as if x and y were reciprocals, though their product is 30, not 1, which gives 10y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Dividing by a fraction: $a \div \frac{b}{c} = a \times \frac{c}{b}$; a is the number you divide. b over c is the fraction you divide by: b is its top and c is its bottom. To divide by a fraction, multiply by its reciprocal, the fraction flipped upside down. For example, 6 ÷ 1/2 = 6 × 2 = 12, because 12 halves fit in 6.

Sort out the question
$xy$ means x times y. So x times y is 30. You need $10 \div x$ “in terms of y.” That means your answer can use y and numbers, but no x.
Solve for x
To swap x out, first get x by itself. Divide both sides of $xy = 30$ by y.
$x = \frac{30}{y}$
This says x is 30 divided by y.
Put the fraction in place of x
Replace x in $10 \div x$ with this fraction.
$10 \div x = 10 \div \frac{30}{y}$
Now use the rule for dividing by a fraction. Flip the fraction and multiply.
$10 \div \frac{30}{y} = 10 \times \frac{y}{30} = \frac{10y}{30}$
Simplify the fraction
Both 10 and 30 can be divided by 10.
$\frac{10y}{30} = \frac{y}{3} = \frac{1}{3}y$
Watch out: 10 ÷ 30 is not 3. Dividing the bigger number by the smaller one gives 3y, choice C. Also, dividing by x is not the same as multiplying by y. That only works when xy is 1, not 30. It leads to choice D.
So the answer is B, $\frac{1}{3}y$ .

Check your answer

Pick numbers that multiply to 30, like x = 3 and y = 10. Then $10 \div x = 10 \div 3 = \frac{10}{3}$ . Choice B gives $\frac{1}{3} \times 10 = \frac{10}{3}$ . They match. Choice C would give 30, and choice D would give 100.

More questions like this

The original DOE question

Question 88 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 88. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 32 of 57

2023 Test B Math, question 32

Question

$A number line with arrows at both ends and three points marked on it. From left to right, point P is at 1 1/3, point Q is at 3, and point R is at 5 1/6.$

On the number line above, how many units longer is $\overline{QR}$ than $\overline{PQ}$ ?

A $\frac{1}{6}$
B $\frac{1}{2}$
C $\frac{2}{3}$
D3

Show the answer

Correct answer: B

First, find the lengths of the given segments:

$\mathrm{QR}=5\frac{1}{6}-3=2\frac{1}{6}$

$\mathrm{PQ}=3-1\frac{1}{3}=1\frac{2}{3}$

Then, find the difference:

$2\frac{1}{6}-1\frac{2}{3}=$

$\frac{13}{6}-\frac{5}{3}=$

$\frac{13}{6}-\frac{10}{6}=$

$\frac{3}{6}=\frac{1}{2}$

Question analysis

A number line shows three points labeled with mixed numbers and a whole number, and students find how many units longer one segment is than another. It tests subtracting mixed numbers with unlike denominators to find distances. Regrouping when the fraction being subtracted is the larger one is where most errors happen.

NYS standards: 7.NS.1c rational number subtraction
6.NS.6c positioning rational numbers and points
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding distances between points on a number line is grade 7 work. Here the points are mixed numbers with different denominators, so finding two distances and their difference makes it a typical but careful grade 7 task.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is three subtractions, so most errors are in the fractions rather than the plan. Taking 1 1/3 from a whole number such as 3 tempts students to subtract only the whole parts and keep the 1/3, which reverses which segment looks longer. Rewriting every coordinate in sixths, 8/6, 18/6 and 31/6, turns the work into whole-number subtraction of numerators.

Error-prone arithmetic moderate Working it out
Both lengths subtract mixed numbers with different denominators, 5 1/6 − 3 and 3 − 1 1/3, and the second needs regrouping before the two results are compared in sixths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Reading a figure, table or graph minor Understanding the question
The three coordinates come from the labels under P, Q and R, and the segments QR and PQ must each be measured from Q at 3, the point they share. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find each length by subtracting the left endpoint from the right one, writing the mixed numbers in sixths, then subtract the shorter length from the longer.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts only the whole numbers in 3 − 1 1/3 and keeps the 1/3, giving PQ as 2 1/3, then compares it with 2 1/6 and gets 1/6.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2/3; it is only the fraction part of PQ’s length, 1 2/3, so it mostly draws guesses.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 3, the coordinate of point Q, a number read off the line rather than a difference between two lengths.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Length on a number line: $\text{length} = \text{right point} - \text{left point}$; The length of a segment is the distance between its two end points. Subtract the number on the left from the number on the right. Subtracting tells how many units you move to get from one point to the other.
Mixed number to improper fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; a is the whole number part. b over c is the fraction part, with b on top and c on the bottom. Each whole holds c parts, so a wholes hold a × c parts. Then add the b extra parts. An improper fraction has a top bigger than its bottom.

Read the number line
Point P is at $1\frac{1}{3}$ . Point Q is at 3. Point R is at $5\frac{1}{6}$ .
The bar over QR means segment QR, the piece of the line from Q to R. You need how much longer QR is than PQ. So find both lengths, then subtract.
Find the length of QR
R is on the right, and Q is on the left. Use the length rule.
$\text{QR} = 5\frac{1}{6} - 3 = 2\frac{1}{6}$
QR is $2\frac{1}{6}$ units long.
Find the length of PQ
Q is on the right, and P is on the left.
$\text{PQ} = 3 - 1\frac{1}{3}$
Take it in two parts. First, 3 − 1 = 2. Then take away the extra $\frac{1}{3}$ . That gives $2 - \frac{1}{3} = 1\frac{2}{3}$ . So PQ is $1\frac{2}{3}$ units long.
Watch out: Don’t just do 3 − 1 and keep the $\frac{1}{3}$ . That gives $2\frac{1}{3}$ , which is too long. It leads to choice A.
Write both lengths in sixths
The fractions have different bottoms, 6 and 3. Change both lengths to sixths so you can subtract. Use the mixed number rule.
$2\frac{1}{6} = \frac{2 \times 6 + 1}{6} = \frac{13}{6}$
$1\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3} = \frac{10}{6}$
Subtract the lengths
Take the shorter length away from the longer one.
$\frac{13}{6} - \frac{10}{6} = \frac{3}{6} = \frac{1}{2}$
QR is $\frac{1}{2}$ unit longer than PQ.
So the answer is B, $\frac{1}{2}$ .

Check your answer

Write every point in sixths. P is $\frac{8}{6}$ , Q is $\frac{18}{6}$ and R is $\frac{31}{6}$ . QR is 31 − 18 = 13 sixths. PQ is 18 − 8 = 10 sixths. The difference is 3 sixths, which is one half.

More questions like this

The original DOE question

Question 89 as printed on page 200 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 89. Open page 200 of the handbook (PDF)

Question 33 of 57

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.

$0.30d=15$

$d=\frac{15}{0.30}=50$

So, the distance from W to X, in miles, is $50-15=35$ .

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps minor
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.

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)
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

Stops one step early: A choice is a number from partway through the solution.
Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: moderate
Takes 70% of the 15 miles, 0.7 × 15 = 10.5, applying the percent to XY instead of to the whole distance from W to Y.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 45; it would make XY a quarter of WY instead of 30%, so it mostly draws guesses.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the whole distance from W to Y, 15 ÷ 0.30 = 50 miles, and stops before subtracting the 15 miles from X to Y.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{\text{percent}}{100} \times \text{whole}$; The whole is the 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.

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.
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.
$100 - 70 = 30$
So the 15 miles from X to Y is 30% of the whole distance.
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.
$15 = \frac{30}{100} \times d = 0.30 \times d$
Divide both sides by 0.30 to get d by itself.
$d = 15 \div 0.30 = 50$
The distance from W to Y is 50 miles.
Find the distance from W to X
Take the 15 miles from X to Y away from the whole.
$50 - 15 = 35$
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? $0.70 \times 50 = 35$ . 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 34 of 57

2023 Test B Math, question 34

Question

There are 3 different highways from City W to City X, 4 different highways from City X to City Y, and 3 different highways from City Y to City Z. How many different routes are there for a trip from City W to City X to City Y to City Z?

Show the answer

Correct answer: D

Multiply the numbers of highways between each pair of cities:

$3\times 4\times 3=36$

Question analysis

A counting question about highways: there are 3 highways from City W to City X, 4 from X to Y and 3 from Y to Z. Students count the different routes from W to Z through X and Y. It tests the counting principle, and deciding whether to add or multiply the numbers of highways is the key.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 students build tree diagrams and lists for compound events. This item needs only the counting principle across three stages, with no restriction on the choices, which makes it shorter than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 11, 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

Once a route is seen as one highway on each of three stretches, the count is a single product, 3 × 4 × 3. The wrong choices come from adding the counts, from stopping after two stretches, and from adding two-stretch products. A small tree for the first two stretches shows why the numbers multiply: each of the 3 highways to City X branches into 4.

Something you must already know minor Working it out
The counting principle, multiplying the 3, 4 and 3 highway choices because each route uses one highway on every stretch, is assumed; the stem never says to multiply. (Anderson & Krathwohl, 2001)

Strategy: A route uses one highway on each of the three stretches, so count the choices for each stretch and multiply; a small tree for two stretches shows why.

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

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds the highways, 3 + 4 + 3 = 10, counting the roads available rather than the ways to choose one road on each stretch.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Multiplies the first two stretches, 3 × 4 = 12 routes from City W to City Y, and stops before the 3 highways on to City Z.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Counts routes two stretches at a time, 3 × 4 = 12 for W to Y and 4 × 3 = 12 for X to Z, and adds them instead of multiplying all three.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle: $\text{total ways} = a \times b \times c$; a, b and c are the numbers of choices for the first, second and third parts of a task. Every choice for one part can go with every choice for the others. So you multiply. Use it when you make one choice at each stage and want to count all the ways.

Sort out the trip
A route goes from City W to X, then to Y, then to Z. It uses exactly one highway on each of three stretches:
- W to X: 3 highways
- X to Y: 4 highways
- Y to Z: 3 highways
You need to count all the different routes.
Count routes from W to Y
Start small. Pick any of the 3 highways to City X. From there, you can take any of the 4 highways to City Y. So each first highway leads to 4 routes.
$3 \times 4 = 12$
There are 12 routes from City W to City Y.
Add the last stretch
Each of those 12 routes can end with any of the 3 highways to City Z. So multiply again. This is the counting principle with all three stretches.
$3 \times 4 \times 3 = 12 \times 3 = 36$
There are 36 routes in all.
Avoid the common traps
Watch out: Adding 3 + 4 + 3 = 10 counts highways, not routes. That gives choice A. Stopping at 12 forgets the last stretch, choice B. Adding 12 + 12 = 24 mixes two partial counts, choice C. A route is built by choosing one highway after another, so you multiply.
So the answer is D, 36.

Check your answer

Picture one highway from W to X. After it, there are 4 × 3 = 12 ways to finish the trip. Each of the 3 highways from W to X gives 12 ways. That makes 12 + 12 + 12 = 36 routes.

More questions like this

The original DOE question

Question 91 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 91. Open page 201 of the handbook (PDF)

Question 35 of 57

2023 Test B Math, question 35

Question

In Center City, the ratio of people to dogs is $4:1$ , and the ratio of dogs to cats is $3:2$ . What is the ratio of people to cats?

A $2:1$
B $7:3$
C $6:1$
D $12:1$

Show the answer

Correct answer: C

Take each ratio and solve for dogs:

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

$4d=p$

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

And

$\frac{d}{c}=\frac{3}{2}$

$d=\frac{3}{2}c$

Set the two expressions for d equal to each other and solve for $\frac{p}{c}$ :

$\frac{1}{4}p=\frac{3}{2}c$

$p=6c$

$\frac{p}{c}=\frac{6}{1}$

So, the ratio of people to cats is $6:1$ .

Question analysis

A ratio question about Center City, where the ratio of people to dogs and the ratio of dogs to cats are both given. Students link the two ratios through the dogs to find the ratio of people to cats. It tests combining ratios, a step that goes wrong when the given numbers are paired up as written.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.1 ratio concepts and language
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 ratio problems usually work with a single ratio. Here two ratios share the dogs but give them different numbers of parts, so one ratio must be rescaled before people and cats can be compared.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Much to keep track of minor
Solution: 3 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

The question rewards seeing that the dogs link the two ratios but are counted differently in each. Reading the people’s 4 and the cats’ 2 straight off the page gives a tidy 2 : 1, and it is the trap most students fall into. Rewriting people to dogs as 12 : 3 makes one chain, 12 : 3 : 2, from which people to cats is 12 : 2.

No obvious method moderate Working it out
Dogs are 1 part in the first ratio and 3 parts in the second, so the ratios cannot be joined until 4 : 1 is rescaled to 12 : 3; nothing in the stem says to do this. (Schoenfeld, 1985)
Much to keep track of minor Working it out
Three groups, people, dogs and cats, run through two ratios, and students must keep track of which number belongs to which group while scaling. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Rewrite both ratios with the same number for dogs, so that people to dogs to cats becomes one three-part ratio, then read off people to cats and simplify.

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Pairs the people’s 4 from the first ratio with the cats’ 2 from the second, 4 : 2 = 2 : 1, without first matching the dogs’ parts.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the two ratios term by term, (4 + 3) : (1 + 2) = 7 : 3, treating ratio terms as amounts that can be combined.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Multiplies people per dog by the dogs’ 3, 4 × 3 = 12, and stops before dividing by the cats’ 2, giving 12 : 1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent ratios: $a : b = (a \times k) : (b \times k)$; a and b are the two amounts in a ratio, in order. k is any number you choose, except 0. Multiplying both amounts by the same number keeps the comparison the same. For example, 4 : 1 and 12 : 3 both mean 4 people for every dog.

Sort out the question
A ratio compares two amounts. People to dogs is 4 : 1. That means 4 people for every 1 dog. Dogs to cats is 3 : 2. That means 3 dogs for every 2 cats.
You need people to cats. Dogs show up in both ratios, so they are the link.
Make the dog numbers match
The dogs are 1 in the first ratio and 3 in the second. To join the ratios, the dog numbers must be the same. So multiply both parts of the first ratio by 3.
$4 : 1 = (4 \times 3) : (1 \times 3) = 12 : 3$
Now there are 12 people for every 3 dogs.
Join the two ratios
Both ratios now have 3 dogs. So you can write all three groups in one ratio.
$\text{people} : \text{dogs} : \text{cats} = 12 : 3 : 2$
For every 12 people, there are 3 dogs and 2 cats.
Read off and simplify
People to cats is 12 : 2. Divide both parts by 2.
$12 : 2 = (12 \div 2) : (2 \div 2) = 6 : 1$
Watch out: Taking the 4 people and the 2 cats straight from the problem gives 4 : 2 = 2 : 1, choice A. That skips matching the dogs. Stopping at 12 : 1, without using the 2 cats, gives choice D.
So the answer is C, 6 : 1.

Check your answer

Build a small town. Start with 2 cats. The second ratio gives 3 dogs. The first ratio gives 4 people for each dog, so 4 × 3 = 12 people. That is 12 people for 2 cats, or 6 people for each cat.

More questions like this

The original DOE question

Question 92 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 92. 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 36 of 57

2023 Test B Math, question 36

Question

The side length of square S is shorter than the side length of square T by 1 centimeter. The area of square S is 9 square centimeters. What is the difference, in square centimeters, between the area of square T and the area of square S?

Show the answer

Correct answer: B

Let x represent the side length of square T, in centimeters. Then the side length of square S is $(x-1)$ .

The area of square S is:

$(x-1)(x-1)=9$

$(x-1)^{2}=9$

$x-1=3$

$x=4$

So the side length of square T is 4 cm, and the area of square T is 16 sq cm.

The difference between the areas, in square centimeters, is $16-9=7$ .

Question analysis

Two squares differ in side length by 1 centimeter, and the smaller one has an area of 9 square centimeters. Students find how much greater the larger square's area is. It tests the link between a square's side and its area, and the extra area is easy to miscount when picturing the border that the longer sides add.

NYS standards: 6.G.5 perfect squares and perfect cubes
6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 The area of a square and recognizing 9 as a perfect square are grade 6 ideas. This takes three short steps with small whole numbers, lighter than grade 7 work on composite figures, though the first step runs backward from area to side.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every number is small, so the work is keeping track of which quantity the question wants: a difference of areas, not a side or a single area. The added area forms an L-shaped strip around square S, and students who picture it as two strips 4 centimeters long count the corner square twice. Finding both areas, 9 and 16, and subtracting avoids reasoning about the strip at all.

Many steps minor Working it out
The work runs from area to side and back: the square root of 9 gives S’s side, adding 1 gives T’s, squaring gives 16, and subtracting gives the difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The phrase “shorter than the side length of square T by 1 centimeter” describes S from T’s point of view, so T’s side is S’s side plus 1, not minus 1. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the side of square S from its area, add 1 centimeter to get the side of square T, find that square's area, and subtract the smaller area from the larger.

Traps

Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds the side of square S, 3 centimeters, from its area of 9 and gives that length instead of a difference between two areas.
C Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: moderate
Pictures the added area as two 4-by-1 strips along two sides of square S, 8 in all, counting the 1-by-1 corner where they overlap twice.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the stem’s two numbers, 9 + 1 = 10, as if a side 1 centimeter longer made the area 1 square centimeter larger, and gives that area.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s \times s$; A is the area, the space inside the square, in square units. s is the length of one side. All four sides of a square are equal, so the area is the side times itself. To go backward from an area, find the number that times itself gives the area.

Sort out the two squares
Square S has an area of 9 square centimeters. Its side is 1 centimeter shorter than the side of square T. You need the area of square T minus the area of square S, in square centimeters.
Find the side of square S
Use the area formula backward. Which number times itself makes 9?
$3 \times 3 = 9$
So each side of square S is 3 centimeters. This number is called the square root of 9.
Find the side of square T
Square T’s side is 1 centimeter longer than square S’s side.
$3 + 1 = 4$
Each side of square T is 4 centimeters.
Find the area of square T
Use the area formula with a side of 4 centimeters.
$A = 4 \times 4 = 16$
The area of square T is 16 square centimeters.
Subtract the areas
Take the smaller area away from the larger one.
$16 - 9 = 7$
The difference is 7 square centimeters.
Watch out: Picturing the new area as two strips, each 4 by 1, gives 8, choice C. But those strips share a 1-by-1 corner, so it gets counted twice. Also, 3 is a side length, not a difference of areas. That is choice A.
So the answer is B, 7 square centimeters.

Check your answer

Draw square T and mark square S in one corner. The extra space is an L shape. It is a 3-by-1 strip, another 3-by-1 strip and a 1-by-1 corner. That is 3 + 3 + 1 = 7 square centimeters. It matches.

More questions like this

The original DOE question

Question 93 as printed on page 201 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 93. Open page 201 of the handbook (PDF)

Question 37 of 57

2023 Test B Math, question 37

Question

Which expression is equivalent to $\frac{1}{2}x+7-\left(x-\frac{9}{2}\right)$ ?

A $\frac{1}{2}(x-5)$
B $\frac{1}{2}(x+5)$
C $-\frac{1}{2}(x+23)$
D $-\frac{1}{2}(x-23)$

Show the answer

Correct answer: D

First, distribute the negative sign, which is the same as multiplying both terms in the parentheses by −1.

Next, combine like terms:

$\frac{1}{2}x+7-\left(x-\frac{9}{2}\right)$

$\frac{1}{2}x+7-x+\frac{9}{2}$

$-\frac{1}{2}x+\frac{23}{2}$

Then, factor the common factor of $-\frac{1}{2}$ from both terms, paying careful attention to the signs:

$-\frac{1}{2}x+\frac{23}{2}=-\frac{1}{2}(x-23)$

Question analysis

An algebra question asking which factored expression equals one half x plus 7 minus the quantity x minus nine halves. Students distribute the subtraction, combine like terms with fractions, and match a factored form. Signs are the danger at every step, and the choices differ only in the signs and numbers inside the parentheses.

NYS standards: 7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students expand and simplify linear expressions with fractions. Here a subtracted group must be distributed and the result factored by a negative fraction, two places where a sign is easily lost, a longer chain than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form moderate
A very close wrong answer moderate
Solution: 3 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

Two separate sign decisions decide this question: distributing the subtraction over x − 9/2, and factoring a negative half out of +23/2. Students who handle the first correctly often stumble on the second and choose the look-alike with +23. Expanding each choice, rather than factoring by hand, is quicker and safer, since it turns the comparison into matching −1/2x + 23/2 term by term.

Error-prone arithmetic moderate Working it out
The minus before (x − 9/2) must change two signs, and the result mixes halves with whole numbers: 1/2x − x and 7 + 9/2 both need care. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form moderate Choosing the answer
The simplified result, −1/2x + 23/2, appears nowhere; every choice is factored by ±1/2, so students must factor out −1/2 themselves or expand each choice to compare. (Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
The answer, −1/2(x − 23), and −1/2(x + 23) differ only in the sign inside the parentheses, and the simplified form’s +23/2 makes the plus sign look right. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Change the sign of both terms inside the parentheses, combine the x terms and the constants, then expand each choice to see which one matches rather than factoring yourself.

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.)
Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

A Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Gives the minus to x but not to −9/2, so the constants make 7 − 9/2 = 5/2, then drops the leading negative when factoring: 1/2(x − 5).
B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Combines 1/2x − x as +1/2x, subtracting the smaller coefficient from the larger, and also leaves −9/2 unchanged, so the result is 1/2x + 5/2 = 1/2(x + 5).
C Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Simplifies correctly to −1/2x + 23/2 but keeps +23 when factoring out −1/2, though dividing +23/2 by −1/2 gives −23.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a group: $-(b - c) = -b + c$; b and c are the two terms inside the parentheses. Terms are the parts joined by plus or minus signs. A minus sign in front of parentheses takes away the whole group. So it changes the sign of every term inside, not only the first one.
Distributive property: $a(b + c) = a \times b + a \times c$; a is the number outside the parentheses. b and c are the terms inside. Multiply a by each term inside. It works because a groups of (b + c) hold a of each. Use it to multiply out the choices and compare them.

Sort out the question
Equivalent means equal for every value of x. The choices are written in a different form from the problem. So simplify the problem first. Then multiply out the choices and compare.
Remove the parentheses
The minus sign in front of the parentheses takes away both terms inside. So x becomes −x, and $-\frac{9}{2}$ becomes $+\frac{9}{2}$ .
$\frac{1}{2}x + 7 - \left(x - \frac{9}{2}\right) = \frac{1}{2}x + 7 - x + \frac{9}{2}$
Watch out: If you change only the sign of x, the numbers become $7 - \frac{9}{2} = \frac{5}{2}$ . That mistake leads to choice A or B.
Combine the x terms
Like terms have the same letter part. The x terms are $\frac{1}{2}x$ and −x. Write x as $\frac{2}{2}x$ so the fractions match.
$\frac{1}{2}x - \frac{2}{2}x = -\frac{1}{2}x$
The result is negative, because you take away more x than you have.
Combine the numbers
Now combine the plain numbers, 7 and $\frac{9}{2}$ . Write 7 as fourteen halves so the bottoms match. Then add.
$7 + \frac{9}{2} = \frac{14}{2} + \frac{9}{2} = \frac{23}{2}$
The expression simplifies to $-\frac{1}{2}x + \frac{23}{2}$ .
Multiply out choices C and D
Only C and D start with $-\frac{1}{2}$ , so test those. Use the distributive property. A negative times a negative is positive.
$-\frac{1}{2}(x + 23) = -\frac{1}{2}x - \frac{23}{2}$
$-\frac{1}{2}(x - 23) = -\frac{1}{2}x + \frac{23}{2}$
The first line is choice C, and the second is choice D. Only D matches.
Watch out: Choice C looks almost the same. But it gives $-\frac{23}{2}$ , not $+\frac{23}{2}$ .
So the answer is D, $-\frac{1}{2}(x - 23)$ .

Check your answer

Try x = 1 in the problem: $\frac{1}{2} + 7 - \left(1 - \frac{9}{2}\right) = 7\frac{1}{2} + 3\frac{1}{2} = 11$ . Now put x = 1 into choice D: $-\frac{1}{2}(1 - 23) = -\frac{1}{2} \times (-22) = 11$ . Both give 11, so they match.

More questions like this

The original DOE question

Question 94 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 94. 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 38 of 57

2023 Test B Math, question 38

Question

How is $\frac{1}{x}-\frac{1}{2x}$ written as a single fraction? (Assume $x\neq 0$ .)

A $\frac{1}{2x}$
B $\frac{1}{3x}$
C $-\frac{1}{x}$
D $-\frac{1}{2x}$

Show the answer

Correct answer: A

$\frac{1}{x}-\frac{1}{2x}=\frac{2}{2x}-\frac{1}{2x}=\frac{1}{2x}$

Question analysis

This algebra question asks students to write 1/x minus 1/(2x) as a single fraction, assuming x is not zero. It tests finding a common denominator when the denominators contain a variable. Getting both fractions over the same denominator before subtracting, and keeping track of the sign, are what matter.

NYS standards: 7.EE.1 linear expressions with rational coefficients
7.EE.2 rewriting expressions to show relationships
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students subtract fractions with unlike denominators and work with linear expressions, but seldom see a variable in a denominator. The steps match those for numerical fractions, so it is a short task once the common denominator is found.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Variables instead of numbers moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 11, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is subtraction of unlike fractions with a letter in the bottom, and every wrong choice is a familiar fraction error: combining the denominators straight across by adding or subtracting, or subtracting the numerators the wrong way round. Substituting x = 1 makes it 1 − 1/2 = 1/2, which rules out the negative choices and 1/3 at once.

Variables instead of numbers moderate Understanding the question
The denominators are x and 2x, so the common denominator is built with a letter, and no number checks the result unless students substitute a value such as x = 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Only the first fraction changes: 1/x becomes 2/2x, and then the subtraction 2 − 1 must be done in the order written, first numerator minus second. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Rewrite 1/x with the denominator 2x by multiplying its top and bottom by 2, then subtract the numerators and keep the common denominator.

Traps

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

The wrong answers, classified

B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Adds the denominators, x + 2x = 3x, and keeps a numerator of 1, as if two unit fractions combined by adding their bottoms.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Subtracts the denominators straight across, x − 2x = −x, and keeps 1 on top, so the result 1/(−x) is written −1/x.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Rewrites 1/x as 2/2x correctly but subtracts the numerators in reverse order, 1 − 2 = −1, giving a negative result.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions: $\frac{a}{b} = \frac{a \times k}{b \times k}$; a is the top, or numerator. b is the bottom, or denominator. k is any number except 0. Multiplying the top and bottom by the same number gives an equal fraction, because k over k is 1.
Subtracting fractions with the same bottom: $\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}$; a and b are the tops. c is the shared bottom. When the bottoms match, the pieces are the same size. So subtract the tops and keep the bottom.

Sort out the question
You need to write $\frac{1}{x} - \frac{1}{2x}$ as one fraction. The problem says x is not 0. That matters because you can’t divide by 0.
Find a common bottom
To subtract fractions, their bottoms must match. 2x is 2 times x. So multiply the top and bottom of $\frac{1}{x}$ by 2.
$\frac{1}{x} = \frac{1 \times 2}{x \times 2} = \frac{2}{2x}$
Now both fractions have the bottom 2x.
Subtract the tops
Subtract the tops in order. Keep the bottom.
$\frac{2}{2x} - \frac{1}{2x} = \frac{2 - 1}{2x} = \frac{1}{2x}$
Watch out: Never add or subtract the bottoms. Adding them gives $\frac{1}{3x}$ , choice B. Subtracting them gives $-\frac{1}{x}$ , choice C. And keep the order: 2 − 1, not 1 − 2. The wrong order gives choice D.
So the answer is A, $\frac{1}{2x}$ .

Check your answer

Try x = 1. The problem becomes $1 - \frac{1}{2} = \frac{1}{2}$ . Choice A gives $\frac{1}{2 \times 1} = \frac{1}{2}$ . It matches. Choices B, C and D give $\frac{1}{3}$ , −1 and $-\frac{1}{2}$ .

More questions like this

The original DOE question

Question 95 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 95. Open page 202 of the handbook (PDF)

Question 39 of 57

2023 Test B Math, question 39

Question

In a certain competition, 1 person finished in first place, 5 people tied for second place, 10 people tied for third place, and the other 4,984 people did not place. What percentage of the people in the competition placed first, second, or third altogether?

A0.0032%
B0.032%
C0.32%
D3.2%

Show the answer

Correct answer: C

The total number of people who participated in the competition was $1+5+10+4{,}984=5{,}000$ . The total number who placed first, second, or third was $1+5+10=16$ .

$\frac{16}{5{,}000}=\frac{32}{10{,}000}=0.0032=0.32\%$

Question analysis

A percent question about a competition with 1 first-place finisher, 5 tied for second, 10 tied for third and 4,984 who did not place. Students find what percentage of everyone placed. All four choices use the same digits, so the challenge is placing the decimal point when a decimal becomes a percent.

NYS standards: 6.RP.3c percent as a rate per 100
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Writing part of a whole as a percent is grade 6 content. This one first totals four groups, then turns a fraction of a large total into a percent, where a slip of one place value is easy to make.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 10, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The method is short, so the question tests decimal places. All four choices share the digits 32 and differ only in where the point falls, so a slip of one place, or forgetting to multiply by 100, still finds a choice waiting. A rough check helps: 16 is a bit more than 15, and 15 of 5,000 is 0.3%, so the answer must sit just above 0.3%.

Error-prone arithmetic moderate Working it out
16 ÷ 5,000 = 0.0032 has two zeros after the decimal point, and writing it as a percent moves the point two more places; each move can land one place off. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The total of 5,000 is not given; it has to be built from 1 + 5 + 10 + 4,984 before the 16 people who placed can be compared with it. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the people who placed and the people who did not to get the total, write the placers as a fraction of it, then change that fraction to a decimal and multiply by 100.

Traps

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

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Finds 16 ÷ 5,000 = 0.0032 and adds a percent sign without multiplying by 100, as if a decimal were already a percent.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Finds 0.0032 but moves the decimal point only one place when writing the percent, giving 0.032%, ten times too small.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Divides 16 by 500 instead of 5,000, dropping a zero, or moves the point three places, giving 3.2%, ten times too large.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the groups
Some people placed: 1 in first, 5 in second and 10 in third. The other 4,984 people did not place. You need the percent of all the people who placed.
Count the people who placed
These are the people who finished first, second or third. Add them.
$1 + 5 + 10 = 16$
So 16 people placed.
Count everyone
The whole is everyone in the competition: the people who placed plus the people who did not.
$16 + 4{,}984 = 5{,}000$
So 5,000 people took part. Notice that 4,984 is not the whole. It leaves out the 16 who placed.
Turn the share into a percent
Use the percent formula.
$\text{percent} = \frac{16}{5{,}000} \times 100$
First divide: $16 \div 5{,}000 = 0.0032$ . Then multiply by 100. That moves the decimal point two places to the right.
$0.0032 \times 100 = 0.32$
So 0.32% of the people placed.
Avoid the decimal traps
Watch out: 0.0032 is not a percent yet. Putting a % sign on it gives choice A, which is 100 times too small. Moving the point only one place gives choice B. Moving it three places gives choice D.
So the answer is C, 0.32%.

Check your answer

Estimate. 1% of 5,000 is 50. The 16 people are about a third of 50. So the answer should be about a third of 1%, or about 0.3%. Choice C, 0.32%, fits.

More questions like this

The original DOE question

Question 96 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 96. Open page 202 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 40 of 57

2023 Test B Math, question 40

Question

Angle A and angle B are complementary. If the measure of angle B is three times as great as the measure of angle A, what is the measure of angle B?

A45°
B67.5°
C75°
D135°

Show the answer

Correct answer: B

Let A represent the measure of angle A, and let B represent the measure of angle B, in degrees. Since the angles are complementary, $A+B=90$ . The question states that $B=3A$ , or $A=\frac{B}{3}$ . Substitute that expression for A in the first equation and solve for B:

$A+B=90$

$\frac{B}{3}+B=90$

$\frac{4B}{3}=90$

$4B=270$

$B=67.5$

Question analysis

A geometry question about two complementary angles, where angle B is three times as large as angle A. Students find the measure of angle B. They must recall what complementary means, since the question does not say, and set up an equation that splits the total in the right proportion.

NYS standards: 7.G.5 angle relationships and equations
7.EE.4a two-step equations
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation from complementary angles is a grade 7 standard. This is the familiar version without a figure: one equation from the definition, then a division and a multiplication, typical of grade 7 classwork.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 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

Once complementary is read as a 90° total, the problem is one equation with a decimal answer, 4A = 90. The 22.5° for angle A is not listed, so stopping at the smaller angle is not rewarded here, but using 180° is: it gives 45° and 135°, and both appear among the choices. Checking that the chosen angle and one third of it add to 90° settles it.

Something you must already know moderate Working it out
The stem never says what complementary means; the 90° total has to be recalled, and a student who confuses it with the supplementary 180° doubles both angles. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The clause “the measure of angle B is three times as great as the measure of angle A” has to become B = 3A, so the 90° splits into four equal parts, not three. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Recall the total that complementary angles share, think of angle A as one part and angle B as three parts, find one part, then multiply for angle B.

Traps

Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Splits the 90° evenly, 90 ÷ 2 = 45, using the complementary total but ignoring that angle B is three times angle A.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 75°, which would need angle A to be 25° and the pair to add to 100°; it mostly draws guesses.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses 180°, the supplementary total, so 4A = 180 gives angle A = 45° and angle B = 135°, a pair that is not complementary.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Complementary angles: $A + B = 90$; A and B are the measures of the two angles, in degrees. Complementary angles are two angles that add up to 90 degrees, the size of a square corner. Use this whenever a problem says two angles are complementary.

Sort out the question
Complementary angles are two angles that add up to 90 degrees. Angle B is three times as big as angle A. You need the measure of angle B, in degrees.
Write the facts as equations
Let A and B stand for the two measures. The angles are complementary, so use the formula.
$A + B = 90$
Angle B is three times angle A, so:
$B = 3A$
Put 3A in place of B
Replace B in the sum with 3A. Then there is only one letter to find.
$A + 3A = 90$
$4A = 90$
Think of it as 4 equal parts. Angle A is 1 part, and angle B is 3 parts.
Find angle A
Divide both sides by 4.
$A = 90 \div 4 = 22.5$
Angle A measures 22.5 degrees. That is not the answer yet, because the question asks for angle B.
Find angle B
Angle B is three times angle A.
$B = 3 \times 22.5 = 67.5$
Angle B measures 67.5 degrees.
Watch out: Angles that add to 180 degrees are called supplementary. Using 180 by mistake gives 45 and 135, choice D. Splitting 90 in half gives 45, choice A. That ignores the fact that B is three times A.
So the answer is B, 67.5°.

Check your answer

Add the two angles: 22.5 + 67.5 = 90. So they are complementary. Now divide: 67.5 ÷ 22.5 = 3. So angle B is three times angle A. Both facts check out.

More questions like this

The original DOE question

Question 97 as printed on page 202 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 202, question 97. Open page 202 of the handbook (PDF)

Question 41 of 57

2023 Test B Math, question 41

Question

The height of a tree in 2013 is 300% of the height it was in 2003. By what percentage did its height increase over this ten-year period?

A $33\frac{1}{3}\%$
B100%
C200%
D300%

Show the answer

Correct answer: C

Let h represent the height of the tree in 2003. Then the height of the tree in 2013 is $3h$ . Calculate the percent increase:

$\frac{3h-h}{h}=\frac{2h}{h}=2=200\%$

Question analysis

A percent question about a tree whose height in 2013 is 300% of its height in 2003. Students find the percent increase over the ten years. No heights are given, so the key is telling the difference between a percent of an amount and a percent change.

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 Grade 7 percent-increase problems give two amounts to compare. This gives only a percent of the earlier height and no heights at all, so students must work out what that percent means for the change with nothing concrete to calculate with.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 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 stem hands students a percent and asks for a different percent, and the given 300% looks like a finished answer. The real step is separating “300% of” from “increased by”: a height that triples has grown by twice its original size. Picking a 2003 height of 100 units makes it concrete, since 300 − 100 = 200 units of growth on a base of 100.

Turning words into math moderate Understanding the question
“300% of the height it was in 2003” describes the new height, three times the old, not the increase; the question then asks for the change, which is a different percent. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers moderate Understanding the question
No heights are given, only a percent, so students must invent a height such as 100 units or work with h and 3h to measure the change. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
A percent increase divides the change by the starting amount, (3h − h) ÷ h, a rule the question assumes without stating. (Anderson & Krathwohl, 2001)

Strategy: Call the 2003 height 100 units, find the 2013 height from the percent given, then divide the change by the 2003 height and write the result as a percent.

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

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides the 2003 height by the 2013 height, h ÷ 3h = 1/3, turning the comparison upside down and reporting 33 1/3%.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 100%, which would mean the height only doubled; it echoes the 100% that the 2003 height stands for, so it mostly draws guesses.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a misconception Expected pull: strong
Reads “300% of” as an increase of 300%, giving the 2013 height as a percent of the 2003 height instead of the percent increase.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase: $\text{percent increase} = \frac{\text{new} - \text{original}}{\text{original}} \times 100$; Original is the starting amount. New is the amount after the change. New minus original is how much it grew. Dividing by the original compares the growth with the start. Multiplying by 100 turns that into a percent.

Sort out the wording
The 2013 height is 300% of the 2003 height. That means it is 3 times the 2003 height. You need the percent increase. That is how much the tree grew, compared with its 2003 height.
Pick an easy height
The question gives no real height, so choose one. Say the tree was 100 feet tall in 2003. Then the 2013 height is 300% of 100 feet.
$\frac{300}{100} \times 100 = 300$
In 2013, the tree was 300 feet tall.
Find how much it grew
The growth is how much taller the tree got. Subtract the old height from the new one.
$300 - 100 = 200$
The tree grew 200 feet in ten years.
Find the percent increase
Use the formula. The original height is the 2003 height.
$\text{percent increase} = \frac{300 - 100}{100} \times 100 = 2 \times 100 = 200$
The height went up by 200%.
Watch out: 300% compares the new height with the old one. It is not the increase. That trap is choice D. Dividing the old height by the new one gives $33\frac{1}{3}\%$ , choice A.
So the answer is C, 200%.

Check your answer

Think about doubling. A tree that doubles adds 100% of its height. A tree that triples adds 200% of its height. So 100 feet plus 200% of 100 feet is 100 + 200 = 300 feet. That is 300% of 100 feet.

More questions like this

The original DOE question

Question 98 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 98. 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 42 of 57

2023 Test B Math, question 42

Question

1 gallon = 16 cups

1 cup = 8 fluid ounces

If 1 gallon = $2^{x}$ fluid ounces, what is the value of x?

Show the answer

Correct answer: B

If 1 gallon is equal to 16 cups, and 1 cup is equal to 8 fluid ounces, then

$1\text{ gallon}=$

$(16)(8\text{ fluid ounces})=$

$128\text{ fluid ounces}$

$128=2^{7}$

$1\text{ gallon}=2^{7}\text{ fluid ounces}$

So, $x=7$ .

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas minor
Something you must already know minor
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.

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)
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

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Writes the 16 cups in a gallon as 2⁴ and stops before multiplying by the 8 fluid ounces in each cup, which adds 3 to the exponent.
C Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Lists the powers of 2 starting at 1, so counts 1, 2, 4, …, 128 as eight numbers and takes 8 as the exponent, though 1 is 2⁰.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Treats 2 to the power x as 2 times x, so 2x = 128 gives 64, confusing an exponent with a multiplier.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Meaning of an exponent: $2^x = 2 \times 2 \times \cdots \times 2$; 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: $2^a \times 2^b = 2^{a + b}$; 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.

Sort out the question
A gallon is 16 cups. A cup is 8 fluid ounces. A gallon is also $2^x$ fluid ounces. You need x, the exponent. It tells how many 2s are multiplied together.
Change a gallon to fluid ounces
Each of the 16 cups holds 8 fluid ounces. So multiply.
$16 \times 8 = 128$
One gallon is 128 fluid ounces.
Write 128 as a power of 2
Now you need $2^x = 128$ . Keep multiplying by 2, and count the 2s.
Number of 2s 1 2 3 4 5 6 7
Product 2 4 8 16 32 64 128
It takes seven 2s to make 128.
$2^7 = 128$
So x = 7.
Give the answer
Watch out: A gallon is 16 cups, and $16 = 2^4$ . Stopping there gives 4, choice A. And $2^x$ 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 $2^4$ and 8 is $2^3$ . So $16 \times 8 = 2^4 \times 2^3 = 2^{4 + 3} = 2^7$ . 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 43 of 57

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?

A $1:3$
B $1:2$
C $2:3$
D $2:1$

Show the answer

Correct answer: D

The length of one side of the puzzle is $\sqrt{64}=8$ inches. The length of one side of the block is $\sqrt[3]{64}=4$ inches.

The ratio of the side of the puzzle to the side of the block is $8:4$ , which reduces to $2:1$ .

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Combines ideas minor
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.

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)
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

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 1 : 3; it makes the puzzle’s side the shorter one, the reverse of 8 and 4 inches, so it mostly draws guesses.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: moderate
Finds both lengths correctly, 8 and 4 inches, but writes the block’s side first, turning the ratio upside down to 1 : 2.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the square to 2 and the cube to 3, the powers behind square inches and cubic inches, and writes 2 : 3 without finding any length.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square: $A = s \times s$; 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: $V = s \times s \times s$; 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.

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.
Find the puzzle’s side
Use the area formula backward. Which number times itself makes 64?
$8 \times 8 = 64$
The puzzle’s side is 8 inches. This number is the square root of 64.
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.
$4 \times 4 \times 4 = 64$
The block’s side is 4 inches. This number is the cube root of 64.
Write and simplify the ratio
Put the puzzle first, as the question asks. Then divide both parts by 4.
$8 : 4 = (8 \div 4) : (4 \div 4) = 2 : 1$
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 44 of 57

2023 Test B Math, question 44

Question

In a certain class, Rosa will take a total of 4 tests. She has already taken 2 of the tests and earned scores of 81 and 83. What is the least possible score Rosa can earn on the third test and still be able to finish the class with an average score of 85 on all 4 tests? (Assume that test scores can range from 0 to 100.)

Show the answer

Correct answer: A

Let x represent Rosa’s score on the third test, and let y represent her score on the fourth test.

$\frac{81+83+x+y}{4}=85$

$164+x+y=340$

$x+y=176$

If Rosa scores 100 on the fourth test, then the least score she can earn on the third test is 76.

Question analysis

Rosa has scores of 81 and 83 on the first two of four tests and wants a mean of 85 on all four. Students find the lowest score she can earn on the third test and still reach that average, with scores capped at 100. Two tests are still unknown, so the problem needs a plan, not a single equation.

NYS standards: 6.SP.5c range, center, and overall pattern
6.EE.8 writing simple inequalities
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 mean problems usually find one missing score. Here two tests remain, and the least possible score depends on how high the other unknown score is allowed to go, a planning step beyond typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 60 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Most students can find that Rosa needs 176 more points, but the question is about the least third score, which only comes from pushing the fourth score to its limit. Splitting the 176 evenly gives a natural-looking 88 that ignores the word least. The parenthesis about the 0 to 100 range is the key information, so it pays to ask what each stated limit is for.

No obvious method moderate Working it out
The least third score depends on how high the fourth can go; only the parenthesis “test scores can range from 0 to 100” tells students to set the fourth test at 100. (Schoenfeld, 1985)
Many steps minor Working it out
The 340 points needed, the 164 already earned, the 176 still needed and the 100 allowed on the last test make four steps, each a chance to lose track. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Multiply the target mean by 4 for the total points needed, subtract the two scores so far, then decide how the remaining points can be split so the third score is as low as possible.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes the mean of the two scores Rosa already has, (81 + 83) ÷ 2 = 82, a number that answers nothing the question asks.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds the 176 points still needed and splits them evenly, 88 on each test, missing that the word least lets the fourth test carry more.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 90; leaving out the fourth test gives (81 + 83 + x) ÷ 3 = 85 and x = 91, the nearest route, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average: $\text{average} = \frac{\text{sum of the scores}}{\text{number of scores}}$; The sum is all the scores added together. The number of scores is how many there are. The average, or mean, shares the total equally among them. Turned around, the sum equals the average times the number of scores.

Sort out the question
Rosa takes 4 tests. She scored 81 and 83 on the first two. She wants an average of 85 on all 4. You need the lowest score she can get on test 3 and still reach that average. Each score is from 0 to 100.
Find the total points needed
Turn the average formula around. The sum is the average times the number of tests.
$\text{sum of the scores} = 85 \times 4 = 340$
Rosa needs 340 points in all.
Find the points still needed
Subtract the points she already has.
$81 + 83 = 164$
$340 - 164 = 176$
Tests 3 and 4 must add up to 176 points.
Make test 4 as high as possible
To make test 3 as low as it can be, test 4 must carry as many points as it can. The highest score is 100.
$176 - 100 = 76$
So the lowest score on test 3 is 76.
Watch out: Splitting 176 evenly gives 88 on each test, choice C. That works, but it is not the least score. And 82 is just the average of her first two tests, choice B.
So the answer is A, 76.

Check your answer

Add all four scores: 81 + 83 + 76 + 100 = 340. Then 340 ÷ 4 = 85, the average she wants. A score of 75 on test 3 would need 101 on test 4, which is not possible.

More questions like this

The original DOE question

Question 101 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 101. Open page 203 of the handbook (PDF)

Question 45 of 57

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?

Show the answer

Correct answer: D

The fraction of the pizza that the Barnes family eats is $\frac{11}{18}$ . 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 $\frac{x}{15}$ . Write an inequality and solve for x:

$\frac{x}{15}\leq\frac{11}{18}$

$x\leq\frac{11\times 15}{18}$

$x\leq\frac{165}{18}$

$x\leq 9\frac{1}{6}$

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: 7.EE.4b two-step inequalities
6.RP.3 ratio and rate reasoning
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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Complex sentences moderate
Turning words into math moderate
Error-prone arithmetic minor
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.

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)
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)
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

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds that 9 slices is the limit but gives the slices left on the second pizza, 15 − 9 = 6, answering a different question.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Subtracts the stem’s numbers, 18 − 11 = 7, the slices left on the Barnes family’s pizza, without comparing shares of the two pizzas.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reasons that the second pizza has 3 fewer slices, so the Ramirez family should eat 3 fewer, 11 − 3 = 8, comparing by subtraction instead of by fractions.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a 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.

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.
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.
$\text{fraction eaten} = \frac{11}{18}$
The Barnes family ate $\frac{11}{18}$ of their pizza.
Write an inequality
Let x be the number of Ramirez slices. Their share is $\frac{x}{15}$ . It must be no more than the Barnes share. The sign ≤ means less than or equal to.
$\frac{x}{15} \leq \frac{11}{18}$
Solve for x
Multiply both sides by 15 to get x by itself.
$x \leq \frac{11 \times 15}{18} = \frac{165}{18}$
Now divide. 18 goes into 165 nine times, since 18 × 9 = 162. There are 3 left over.
$x \leq 9\frac{3}{18} = 9\frac{1}{6}$
Pick the whole number
x must be a whole number no bigger than $9\frac{1}{6}$ . 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 $11 \div 18 \approx 0.611$ . Nine Ramirez slices are $9 \div 15 = 0.6$ , just under it. Ten slices are $10 \div 15 \approx 0.667$ , 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 46 of 57

2023 Test B Math, question 46

Question

$0^{3}+4^{0}+2^{-1}+(-1)^{2}$

What is the value of the expression above, in decimal form?

A0
B0.5
C1.5
D2.5

Show the answer

Correct answer: D

$0^{3}+4^{0}+2^{-1}+(-1)^{2}$

$=0+1+\frac{1}{2}+1$

$=2\frac{1}{2}=2.5$

Question analysis

An exponents question: students find the decimal value of a sum of four powers, zero cubed, four to the zero power, two to the negative first power and negative one squared. Each of the four terms follows a different rule, and a slip on any one of them changes the total.

NYS standards: 8.EE.1 integer exponent properties
7.NS.2a rational number multiplication
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 work uses whole-number exponents and signed numbers. Zero and negative exponents are grade 8 content, and this expression tests both alongside a negative base, each with its own common misconception.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know major
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every term tests a different exponent rule, and each wrong choice comes from breaking exactly one of them. The zero exponent is the most common casualty, since 4⁰ looks as if it should be 0 like 0³; a negative exponent read as a negative number is close behind. Writing each term’s value above it before adding, 0, 1, 1/2 and 1, keeps the four rules separate.

Something you must already know major Working it out
Three rules the test does not give decide the sum: any nonzero number to the 0 power is 1, 2 to the −1 power is 1/2, and (−1)² is positive 1. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
The four terms carry a zero base, a zero exponent, a negative exponent and a negative base, so a sign or a stray 0 can slip in at each one. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Evaluate each power on its own, using the rules for a zero exponent and a negative exponent and the sign of a squared negative number, then add the four values.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Treats 2 to the power −1 as −2 instead of one half, so the four terms 0 + 1 − 2 + 1 add to 0.
B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Squares as if the minus sat outside the parentheses, taking (−1)² as −1, so the terms add to 0 + 1 + 0.5 − 1 = 0.5.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Takes 4 to the power 0 as 0 rather than 1, so the expression becomes 0 + 0 + 0.5 + 1 = 1.5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Zero exponent: $a^0 = 1$; a is any number except 0. An exponent of 0 always gives 1. It works because each step down in the exponent divides by a. For example, 4 to the first power is 4, and 4 ÷ 4 = 1.
Negative exponent: $a^{-n} = \frac{1}{a^n}$; a is the base, and n is a positive whole number. A negative exponent does not make the number negative. It means 1 divided by the base to the positive power. For example, 2 to the −1 power is one half.

Sort out the question
The expression adds four powers. A power is a base with an exponent, a small raised number. The exponent tells how many times to use the base as a factor, a number you multiply. Find each power, then add.
Find the first two powers
Zero times itself is still zero.
$0^3 = 0 \times 0 \times 0 = 0$
Next, use the zero exponent rule.
$4^0 = 1$
Watch out: $4^0$ is 1, not 0. Taking it as 0 gives choice C.
Find the negative power
Use the negative exponent rule.
$2^{-1} = \frac{1}{2^1} = \frac{1}{2} = 0.5$
Watch out: $2^{-1}$ is not −2. That mistake gives choice A.
Find the last power
The base is −1, inside the parentheses. Multiply it by itself. A negative times a negative is positive.
$(-1)^2 = (-1) \times (-1) = 1$
Watch out: The answer is not −1. That mistake gives choice B.
Add the four values
Now add them. The question wants a decimal, so write one half as 0.5.
$0 + 1 + 0.5 + 1 = 2.5$
So the answer is D, 2.5.

Check your answer

See why $4^0 = 1$ with a pattern. $4^2 = 16$ and $4^1 = 4$ . Each step down divides by 4, so $4^0 = 4 \div 4 = 1$ . In the same way, $2^0 = 1$ and $2^{-1} = 1 \div 2 = \frac{1}{2}$ .

More questions like this

The original DOE question

Question 103 as printed on page 204 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 103. Open page 204 of the handbook (PDF)

Question 47 of 57

2023 Test B Math, question 47

Question

Four slips of paper numbered 1, 2, 3, and 4 were placed in a bag:

One slip was selected from the bag, and the number recorded.
Without replacement, a second slip was chosen.
This process continued until no more slips remained in the bag.
One part of a tree diagram representing the sample space is shown.

$A partial tree diagram. The top number, 1, branches down to 2, 3, and 4. Below 2 are 3 and 4, below 3 are 2 and 4, and below 4 are 2 and 3. The 4 under the 2 has one branch leading down to an empty box.$

Which number should be placed in the box to complete this part of the tree diagram?

Show the answer

Correct answer: C

The “without replacement” statement means that the bag contained four slips for the first draw, three slips for the second draw, two slips for the third draw, and one slip for the last draw. Following the branch from the top and navigating down to the box, the 1 was chosen first, the 2 was chosen second, and the 4 was chosen third. The only remaining number is 3. The element would be 1243.

Question analysis

Four slips numbered 1 to 4 are drawn from a bag one at a time without replacement, and part of a tree diagram of the outcomes is shown. Students decide which number belongs in the empty box at the end of one branch. It tests reading a tree diagram and understanding what drawing without replacement means.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Tree diagrams for compound events are a grade 7 topic, but here the tree is already drawn and only one box is empty. Following a single branch to that box is shorter than typical grade 7 work with sample spaces.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 68 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Only one branch matters, and once it is read as the slips drawn so far, 1, 2 and 4, the last slip left in the bag is the only one missing. Errors come from not counting the top 1 as a draw, from forgetting that a drawn slip stays out, or from copying a nearby number in the diagram instead of following the path.

Reading a figure, table or graph minor Understanding the question
The box hangs below the branch 1, then 2, then 4, and students must read that path from the top as the order of the first three draws. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Follow the branch from the top down to the box, list the slips already drawn on that path, and name the one still left in the bag.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals partial knowledge Expected pull: moderate
Treats the 1 at the top as the start of the diagram rather than the first slip drawn, so 1 seems still to be in the bag.
B Without replacement (Fischbein & Schnarch, 1997)
Reveals a misconception Expected pull: weak
Lets slip 2, already drawn second on this branch, come up again, missing that a slip drawn without replacement stays out of the bag.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Copies 4, the number directly above the box, reading the diagram as a pattern of numbers rather than as the order of the draws.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Sort out the question
The bag holds four slips: 1, 2, 3 and 4. Slips are drawn one at a time. “Without replacement” means a drawn slip does not go back in the bag. So each number can be drawn only once.
Read the tree diagram
A tree diagram shows the orders the draws can happen in. Each row going down is one draw. The 1 at the top is the first slip drawn. The numbers below it are the choices for the second draw, and so on.
Follow the path to the box
Start at the top and go down the lines to the box.
- First draw: 1
- Second draw: 2
- Third draw: 4
- Fourth draw: the box
Slips 1, 2 and 4 are already out of the bag.
Find the slip that is left
The bag started with 1, 2, 3 and 4. Only slip 3 has not been drawn. So it must be the fourth draw.
Watch out: The 1 at the top counts as a draw, so 1 is not left. That trap is choice A. Slip 2 was already drawn, so choice B can’t be right. And 4 is just the number above the box, choice D.
So the answer is C, 3.

Check your answer

Read the whole path: 1, 2, 4, 3. Each number from 1 to 4 shows up exactly once. That is just what should happen when no slip goes back in the bag.

More questions like this

The original DOE question

Question 104 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 104. 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 48 of 57

2023 Test B Math, question 48

Question

The outdoor temperature was 4.2° F at 10 p.m. After $3\frac{1}{2}$ hours, the outdoor temperature was −0.7° F. Which value best represents the average change in temperature per hour?

A−1.4° F
B−1.0° F
C1.0° F
D1.4° F

Show the answer

Correct answer: A

First, recognize that the temperature decreased. This means that the order of subtraction matters and that the result is a negative number:

$-0.7-4.2=-4.9$

The temperature decreased 4.9 degrees. Divide by 3.5 to get the average change per hour:

$-4.9\div 3.5=-1.4$

Question analysis

The temperature was 4.2° F at 10 p.m. and −0.7° F three and a half hours later. Students find the average change in temperature per hour. It tests subtracting and dividing with negative decimals, and the choices pair two sizes of change with both signs, so the sign has to be right as well as the size.

NYS standards: 7.NS.3 four operations with rational numbers
7.NS.1c rational number subtraction
7.NS.2b rational number division
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Signed-number problems about temperature are standard grade 7 work. This one asks for an average rate, so a change that crosses zero must be divided by a mixed-number time, with the sign kept through both steps.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Turning words into math minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is short, but signs decide every choice: two choices come from treating −0.7 as 0.7, and one comes from getting the size right and the direction wrong. Students who think of the drop in two parts, 4.2 degrees down to zero and 0.7 more, find the 4.9-degree fall; dividing by 3.5 hours and remembering the temperature fell finishes it.

Error-prone arithmetic moderate Working it out
The change crosses zero, −0.7 − 4.2 = −4.9, and then a negative decimal is divided by 3.5, from the 3 1/2 hours, with the sign kept through both steps. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Turning words into math minor Understanding the question
The phrase “average change in temperature per hour” means the change from the 4.2° start to the −0.7° end divided by the time, final minus initial, so a drop is negative. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The answer, −1.4° F, and the choice 1.4° F share every digit and differ only in sign, so a correct division followed by a careless sign cannot be caught by estimating. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Subtract the starting temperature from the ending one, keeping track of the sign, then divide that change by the 3.5 hours.

Traps

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

The wrong answers, classified

B Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Subtracts 0.7 from 4.2 as if both temperatures were above zero, a 3.5-degree drop, so 3.5 ÷ 3.5 gives 1, then marks the change negative.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Treats −0.7 as 0.7, finding a 3.5-degree change, and also loses its direction, reporting a rise of 1 degree an hour for a temperature that fell.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds the 4.9-degree change and divides by 3.5 correctly, or subtracts in the order start minus end, and reports a rise for a temperature that fell.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average change per hour: $\text{average change per hour} = \frac{\text{end} - \text{start}}{\text{number of hours}}$; Start is the first temperature, and end is the last one. End minus start is the total change. It is negative when the temperature drops. Dividing by the number of hours shares that change equally over each hour.

Sort out the question
At 10 p.m. it was 4.2° F. After $3\frac{1}{2}$ hours, it was −0.7° F. That is below zero. The temperature went down, so the change is negative. You need the average change for each hour.
Find the total change
Subtract the start from the end.
$-0.7 - 4.2 = -4.9$
Picture a thermometer. Going from 4.2 down to 0 is a drop of 4.2 degrees. Going on to −0.7 is 0.7 more. In all, it fell 4.9 degrees.
Watch out: 4.2 − 0.7 = 3.5 treats −0.7 as if it were above zero. That leads to choice B or C.
Divide by the hours
Write $3\frac{1}{2}$ hours as 3.5 hours. Use the formula.
$\text{average change per hour} = \frac{-0.7 - 4.2}{3.5} = \frac{-4.9}{3.5} = -1.4$
To divide, think 49 ÷ 35 = 1.4, since 35 × 1.4 = 49. The change is negative, so it is −1.4 degrees per hour.
Keep the sign
The temperature fell, so the answer must be negative.
Watch out: 1.4 has the right size, but it says the temperature went up. That is choice D.
So the answer is A, −1.4° F.

Check your answer

Work forward. Dropping 1.4 degrees each hour for 3.5 hours is $1.4 \times 3.5 = 4.9$ degrees. Then $4.2 - 4.9 = -0.7$ . That is the temperature in the problem.

More questions like this

The original DOE question

Question 105 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 105. Open page 205 of the handbook (PDF)

Question 49 of 57

2023 Test B Math, question 49

Question

A rectangular swimming pool of uniform depth is 20 yards long and 14 yards wide. If the water is 6 feet deep, what is the volume of water in the pool, in cubic yards?

A40
B286
C560
D1,680

Show the answer

Correct answer: C

First, change 6 feet to yards:

$6\text{ feet}\times\frac{1\text{ yard}}{3\text{ feet}}=2\text{ yards}$

Then, multiply the dimensions of the pool to find the volume, in cubic yards:

$20\times 14\times 2=560$

Question analysis

A rectangular swimming pool is 20 yards long and 14 yards wide, and the water is 6 feet deep. Students find the volume of water in cubic yards. It tests the volume of a rectangular prism, and the depth is in a different unit from the other measurements, so it has to be converted first.

NYS standards: 6.G.2 volume with fractional edge lengths
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 The volume of a rectangular prism is grade 5 and 6 content, and the conversion here gives a whole number. Grade 7 volume work is usually longer, with composite or triangular prisms; here the only extra step is one unit conversion.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The only real trap is the mixed units: 20 and 14 are yards, 6 is feet, and the answer is wanted in cubic yards. Multiplying the three numbers as printed gives a listed choice three times too large. Underlining the unit of each measurement before multiplying, and converting the odd one out, takes seconds.

Combines ideas minor Working it out
The length and width are in yards but the depth is in feet, so a conversion, 6 feet = 2 yards, must come before the volume formula is used. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
Neither 3 feet = 1 yard nor volume = length × width × depth is given; both are assumed in a stem that asks for cubic yards. (Anderson & Krathwohl, 2001)

Strategy: Change the depth from feet to yards first, so that all three measurements are in yards, then multiply the length, width and depth.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the three measurements, 20 + 14 + 6 = 40, with no conversion and no volume formula, combining the stem’s numbers.
B Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Finds the area of the water’s surface, 20 × 14 = 280, then adds the depth of 6 instead of multiplying by it.
D Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Multiplies 20 × 14 × 6 without changing 6 feet to 2 yards, mixing yards with feet and getting a volume three times too large.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a rectangular box: $V = l \times w \times h$; V is the volume, the space inside, in cubic units. l is the length, w is the width, and h is the height or depth. All three must be in the same unit. Multiplying them counts the unit cubes that fill the box.
Feet to yards: $\text{yards} = \text{feet} \div 3$; There are 3 feet in 1 yard. So to change feet to yards, divide the number of feet by 3. A yard is longer than a foot, so you get a smaller number.

Sort out the measurements
The pool is shaped like a box. Uniform depth means the water is equally deep everywhere. The pool is 20 yards long and 14 yards wide. The water is 6 feet deep. You need the volume, the space the water fills, in cubic yards.
Change the depth to yards
Two sides are in yards, but the depth is in feet. All three must use the same unit before you multiply. The answer is in cubic yards, so change feet to yards.
$6 \div 3 = 2$
The water is 2 yards deep.
Find the volume
Use the volume formula with all three measurements in yards.
$V = 20 \times 14 \times 2 = 560$
The pool holds 560 cubic yards of water.
Give the answer
Watch out: Multiplying 20 × 14 × 6 mixes yards with feet. It gives 1,680, choice D, which is three times too big.
So the answer is C, 560 cubic yards.

Check your answer

Check the units. Yards times yards times yards gives cubic yards, just what the question asks for. Also, 1,680 ÷ 3 = 560. So using 6 feet unchanged makes the answer exactly three times too big.

More questions like this

The original DOE question

Question 106 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 106. 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 50 of 57

2023 Test B Math, question 50

Question

$\{3, 4, 5, 6, 7\}$

A student made a list of all possible products of 2 different numbers in the set above. What fraction of the products are odd?

A $\frac{3}{20}$
B $\frac{3}{10}$
C $\frac{1}{2}$
D $\frac{3}{5}$

Show the answer

Correct answer: B

First, list the products:

$3\times 4=12$

$3\times 5=15$

$3\times 6=18$

$3\times 7=21$

$4\times 5=20$

$4\times 6=24$

$4\times 7=28$

$5\times 6=30$

$5\times 7=35$

$6\times 7=42$

Of the 10 products, 3 are odd. The fraction is $\frac{3}{10}$ .

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.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of moderate
Something you must already know minor
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.

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)
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

A Overlapping groups (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Counts ordered pairs for the total, 5 × 4 = 20, so each pair appears twice, but counts the odd products once, 3, giving 3/20.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Assumes about half of all products are odd, as odd and even numbers come equally often, without listing the products or checking any.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives the share of the numbers in the set that are odd, 3 of 5, instead of the share of the products, never forming any products.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a list: $\text{fraction} = \frac{\text{odd products}}{\text{all products}}$; 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: $\text{odd} \times \text{odd} = \text{odd}$; 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.

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.
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 = 12 3 × 5 = 15 3 × 6 = 18 3 × 7 = 21
4 × 5 = 20 4 × 6 = 24 4 × 7 = 28
5 × 6 = 30 5 × 7 = 35
6 × 7 = 42
There are 10 products.
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.
Write the fraction
Use the fraction formula.
$\text{fraction} = \frac{3}{10}$
Watch out: Counting 4 × 3 as well as 3 × 4 makes 20 pairs. Then $\frac{3}{20}$ , choice A, mixes two ways of counting. And $\frac{3}{5}$ , choice D, is the share of the numbers that are odd, not the products.
So the answer is B, $\frac{3}{10}$ .

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 51 of 57

2023 Test B Math, question 51

Question

A number, n, is divided by 25. If the quotient is 10 with a remainder of 5, what is the quotient expressed as a decimal?

A10.20
B10.25
C10.50
D10.75

Show the answer

Correct answer: A

A quotient of 10 with a remainder of 5 means the quotient is

$10+\frac{5}{25}=10\frac{1}{5}=10.20$

Question analysis

A number is divided by 25, giving a quotient of 10 with a remainder of 5, and students write the quotient as a decimal. It tests what a remainder means and how it becomes part of a decimal quotient. Connecting the remainder to the decimal part of the answer is the key idea.

NYS standards: 6.NS.2 multi-digit division
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Division with remainders and decimal quotients are grade 6 skills. This takes two short steps, with no word problem to translate, though students must know what a remainder means once a quotient is written as a decimal.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know moderate
Solution: 2 steps
Reading level: Question and choices: about grade 5, 22 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question checks one idea: a remainder is a fraction of the divisor, not a string of digits to place after the point. Students who write the remainder 5 as tenths, or set the divisor 25 after the point, land on listed choices. Rebuilding n = 25 × 10 + 5 = 255 and dividing by 25 confirms the decimal.

Something you must already know moderate Working it out
A remainder of 5 means 5 out of the 25 needed for another whole, so it becomes 5/25 = 0.2, a link between remainders and decimals the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Rebuild n from the quotient and the remainder, then divide it by 25 to a decimal; or write the remainder as a fraction of the divisor.

Traps

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

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Writes the divisor, 25, after the decimal point, setting the stem’s numbers 10 and 25 side by side instead of turning the remainder into a fraction.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Writes the remainder 5 as the tenths digit, 10.5, believing a remainder becomes the digits after the decimal point.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 10.75; it treats the remainder as three quarters of the divisor, which nothing in the stem suggests, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Quotient with a remainder: $\text{quotient} = q + \frac{r}{d}$; d is the divisor, the number you divide by. q is the whole-number part of the quotient, the answer to the division. r is the remainder, what is left over. The leftover still has to be divided by d, so it becomes the fraction r over d.

Sort out the words
n is divided by 25. So 25 is the divisor, the number you divide by. The quotient is the answer to a division. Here it is 10 with a remainder of 5. The remainder is the amount left over. You need the exact quotient as a decimal.
Turn the remainder into a fraction
The 5 left over still has to be divided by 25. So it becomes the fraction $\frac{5}{25}$ . Use the formula.
$\text{quotient} = 10 + \frac{5}{25}$
Change the fraction to a decimal
Simplify first. Divide the top and bottom by 5. Then write the fraction in hundredths.
$\frac{5}{25} = \frac{1}{5} = \frac{20}{100} = 0.20$
So the quotient is 10 + 0.20 = 10.20.
Give the answer
Watch out: A remainder is not the digits after the decimal point. Writing the 5 after the point gives 10.50, choice C. Writing the 25 after it gives 10.25, choice B.
So the answer is A, 10.20.

Check your answer

Rebuild n. Ten 25s plus the 5 left over is $25 \times 10 + 5 = 255$ . Now divide: $255 \div 25 = 10.2$ . That matches 10.20.

More questions like this

The original DOE question

Question 108 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 108. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 52 of 57

2023 Test B Math, question 52

Question

$An equilateral triangle drawn with dashed sides. On each side a semicircle bulges outward, with that side as its diameter. The three curved arcs together form the outline of the garden.$

The garden design shown above is created in the shape of three semicircles whose diameters lie on the sides of an equilateral triangle. If each diameter is 2 meters long, what is the perimeter of the garden, in meters?

A6
B12
C $3\pi$
D $6\pi$

Show the answer

Correct answer: C

The perimeter (or circumference) of one of the three semicircles is $\frac{1}{2}(\text{diameter})\pi=\frac{1}{2}(2\pi)=\pi$ . Since there are 3 semicircles that make up the perimeter, the perimeter is $3\pi$ .

Question analysis

A garden is outlined by three semicircles whose diameters, each 2 meters long, are the sides of an equilateral triangle. Students find the perimeter of the garden in meters. It tests the circumference of a circle applied to half circles, and the figure mixes straight dashed sides with curved ones, so deciding what counts as the outline matters.

NYS standards: 7.G.4 circle area and circumference
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Applying the circumference formula is a grade 7 standard. Here it is used for a semicircle and repeated three times, a short extension of routine classwork, with a dashed triangle inside the figure to look past.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph minor
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The perimeter is just three half-circles, and each is half of π × 2. The figure’s dashed triangle invites measuring straight sides instead, and remembering the whole circle’s circumference without halving it gives a choice twice too large. Tracing the garden’s outer edge with a finger before computing shows that only the curves count.

Reading a figure, table or graph minor Understanding the question
The dashed triangle lies inside the garden, so its three 2-meter sides are not part of the perimeter; only the three curved arcs bound the garden. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
The circumference formula is not given, and each arc is half of a circle with a 2-meter diameter, so students must take half of π × 2. (Anderson & Krathwohl, 2001)

Strategy: Find the circumference of a full circle with a 2-meter diameter, take half of it for one semicircle, and multiply by the number of semicircles in the outline.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Adds the three dashed sides of the triangle, 3 × 2 = 6, measuring the triangle inside the garden instead of its curved outline.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 12, which is twice the triangle’s perimeter and matches no arc length, so it mostly draws guesses.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Uses a whole circumference, π × 2, for each of the three arcs, forgetting that each one is only half a circle, which doubles the perimeter.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle: $C = \pi \times d$; C is the circumference, the distance around a circle. d is the diameter, the straight line across the circle through its center. π is a number a little more than 3.14. The distance around any circle is π times as long as the distance across it.
Arc of a semicircle: $\text{arc} = \frac{1}{2} \times \pi \times d$; A semicircle is half of a circle. Its arc is the curved part. d is the diameter, the straight side. The arc is half of the full circle’s circumference, so take half of π times d.

Sort out the figure
The garden’s edge is made of three semicircles, or half circles. Each one sits on a side of the dashed triangle. Each side is a diameter, 2 meters long. The perimeter is the distance around the outside. Here that is only the three arcs, the curved parts. The dashed triangle is inside, so it does not count.
Find a full circle’s circumference
Picture the whole circle that one semicircle comes from. Its diameter is 2 meters. Use the circumference formula.
$C = \pi \times 2 = 2\pi$
The full circle would measure $2\pi$ meters around.
Find the length of one arc
Each semicircle’s arc is half of that circle. Use the arc formula.
$\text{arc} = \frac{1}{2} \times \pi \times 2 = \pi$
One arc is $\pi$ meters long.
Watch out: Using the full circle’s $2\pi$ for each arc doubles the answer. That gives $6\pi$ , choice D.
Add the three arcs
The garden’s edge is three equal arcs.
$\pi + \pi + \pi = 3\pi$
The perimeter is $3\pi$ meters.
Watch out: The dashed sides add to 2 + 2 + 2 = 6 meters, choice A. That is the triangle, not the curved edge.
So the answer is C, $3\pi$ meters.

Check your answer

Estimate. $3\pi$ is about 3 × 3.14 = 9.42 meters. A curved path is longer than the straight line across it. So the edge should be longer than the triangle’s 6 meters, and 9.42 is.

More questions like this

The original DOE question

Question 109 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 206, question 109. Open page 206 of the handbook (PDF)

Question 53 of 57

2023 Test B Math, question 53

Question

Micha is playing a game with five cards numbered 1 through 5. He will place the cards in a bag and draw one card at random three times, replacing the card each time. To win a prize, he must draw the number 5 all three times. What is the probability he will draw the number 5 all three times?

A $\frac{1}{125}$
B $\frac{3}{125}$
C $\frac{1}{15}$
D $\frac{3}{5}$

Show the answer

Correct answer: A

Each time Micha draws a card, the probability of drawing the number 5 is $\frac{1}{5}$ . To find the probability of drawing the number 5 three times, multiply:

$\frac{1}{5}\times\frac{1}{5}\times\frac{1}{5}=\frac{1}{125}$

Question analysis

Micha draws a card three times from five cards numbered 1 to 5, putting the card back each time, and wins only if he draws a 5 every time. Students find the probability that he wins. It tests the probability of independent events, where the chances for the separate draws have to be combined correctly.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the probability of a compound event is a grade 7 standard, and repeated draws with replacement are a standard case. Three equal factors make this a short, typical grade 7 task.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 59 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

With replacement each draw is the same 1 in 5, so the work is one product, 1/5 × 1/5 × 1/5. The wrong choices mark three ways to mishandle repetition: adding the chances, multiplying 5 by 3 instead of cubing it, and multiplying by 3 for the three draws. A quick sense check helps, since needing three 5s in a row must be far less likely than needing one.

Something you must already know minor Working it out
The rule that the chance of independent events all happening is the product of their chances is assumed, as is that replacing the card keeps each draw at 1/5. (Anderson & Krathwohl, 2001)

Strategy: Find the probability of drawing a 5 on one draw, then, because the card goes back each time, multiply that probability by itself once for each draw.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

B Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: weak
Finds 1/125 for three 5s, then multiplies by 3 for the three draws, counting the single outcome 5, 5, 5 as if it could happen in three orders.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Multiplies 5 by 3 instead of raising it to the third power, so the denominator becomes 15 rather than 5 × 5 × 5 = 125.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Adds the three probabilities, 1/5 + 1/5 + 1/5 = 3/5, as if drawing a 5 on any one draw were enough to win.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{results you want}}{\text{all possible results}}$; P is the probability, the chance something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row: $P(\text{A and B and C}) = P(\text{A}) \times P(\text{B}) \times P(\text{C})$; A, B and C are events, things that might happen, like drawing a 5. When they do not change each other’s chances, the chance that all of them happen is the product of their chances. Putting each card back keeps the draws from changing each other.

Sort out the question
The bag has 5 cards, numbered 1 to 5. Micha draws a card three times. He puts it back each time, so every draw starts with all 5 cards. He wins only if he draws a 5 all three times. You need the probability, or chance, of that.
Find the chance for one draw
One card out of 5 is a 5. Use the probability formula.
$P(5) = \frac{1}{5}$
Each draw has a 1 in 5 chance of being a 5.
Multiply for three draws
He needs a 5 on the first draw, and the second, and the third. The card goes back each time, so the draws don’t change each other. Use the rule for events in a row.
$P(\text{5 and 5 and 5}) = \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = \frac{1}{5 \times 5 \times 5} = \frac{1}{125}$
Give the answer
Watch out: Adding the chances gives $\frac{3}{5}$ , choice D. That would make winning likely, which makes no sense. Multiplying 5 × 3 instead of 5 × 5 × 5 gives $\frac{1}{15}$ , choice C.
So the answer is A, $\frac{1}{125}$ .

Check your answer

Count all the results. Each draw has 5 possible cards, so three draws have 5 × 5 × 5 = 125 possible orders. Only one of them is 5, 5, 5. That gives 1 out of 125. Needing three 5s should also be much less likely than needing one.

More questions like this

The original DOE question

Question 110 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 206, question 110. Open page 206 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 54 of 57

2023 Test B Math, question 54

Question

The speed 10 meters per second is equivalent to how many feet per hour? (Use the approximation 1 foot = 0.3 meter.)

A10,800
B12,000
C36,000
D120,000

Show the answer

Correct answer: D

$\frac{10\text{ m}}{1\text{ sec}}\times\frac{3{,}600\text{ sec}}{1\text{ h}}\times\frac{1\text{ ft}}{0.3\text{ m}}=$

$\frac{36{,}000\text{ ft}}{0.3\text{ h}}=120{,}000\,\frac{\text{ft}}{\text{hr}}$

Question analysis

This unit conversion question gives a speed of 10 meters per second and asks for the same speed in feet per hour, using 1 foot = 0.3 meter. Students convert both the distance unit and the time unit. Setting up each conversion factor so that the unwanted units cancel is what makes it tricky.

NYS standards: 6.RP.3d converting measurement units
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students usually convert one unit with a given factor. This converts both parts of a rate, seconds to hours and meters to feet, with a decimal conversion factor and large numbers to keep track of.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Error-prone arithmetic moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every wrong choice is a conversion handled badly: multiplying by 0.3 when the fact 1 foot = 0.3 meter calls for dividing, dividing by 3 instead of 0.3, or stopping in meters per hour. Writing the units as fractions, so that seconds and meters cancel, shows which way each factor goes. A size check also helps: a foot is shorter than a meter, so the number of feet must be larger than the 36,000 meters.

Many steps moderate Working it out
Both parts of the rate change: seconds become hours by multiplying by 3,600, and meters become feet by dividing by 0.3, two conversions that each have a wrong direction. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic moderate Working it out
Dividing 36,000 by 0.3 means moving the decimal point in the divisor, and a slip of one place gives 12,000 instead of 120,000. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The stem gives 1 foot = 0.3 meter but not the 3,600 seconds in an hour, which students must build from 60 × 60. (Anderson & Krathwohl, 2001)

Strategy: Write the speed as a fraction, multiply by 3,600 seconds per hour, then by 1 foot over 0.3 meter, checking that seconds and meters cancel.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 36,000 meters per hour by 0.3 instead of dividing, turning the fact 1 foot = 0.3 meter upside down and getting 10,800.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Divides 36,000 by 3 instead of 0.3, putting the decimal point in the wrong place and getting an answer ten times too small.
C Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Converts seconds to hours, 10 × 3,600 = 36,000, and stops, leaving the speed in meters per hour rather than feet per hour.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Seconds in an hour: $\text{seconds in 1 hour} = 60 \times 60 = 3{,}600$; An hour has 60 minutes, and each minute has 60 seconds. So 1 hour holds 60 groups of 60 seconds. To change a speed per second into a speed per hour, multiply by 3,600, because an hour is 3,600 seconds long.
Meters to feet: $\text{feet} = \text{meters} \div 0.3$; The problem says 1 foot is about 0.3 meter. So each foot uses up 0.3 meter of the distance. To count the feet, divide the meters by 0.3. A foot is shorter than a meter, so you get more feet than meters.

Sort out the question
The speed is 10 meters per second. That means it moves 10 meters every second. You need the speed in feet per hour. Two units change: seconds become hours, and meters become feet. Change one at a time. Use 1 foot = 0.3 meter.
Find the seconds in an hour
An hour is much longer than a second. So in one hour, the object moves much farther. First find how many seconds are in an hour.
$60 \times 60 = 3{,}600$
One hour is 3,600 seconds.
Find the meters per hour
Each second it moves 10 meters. In 3,600 seconds it moves 3,600 times as far.
$10 \times 3{,}600 = 36{,}000$
The speed is 36,000 meters per hour.
Watch out: Stopping here gives 36,000, choice C. That number is still in meters, not feet.
Change meters to feet
Each foot is 0.3 meter. To find how many feet fit in 36,000 meters, divide by 0.3. Multiplying both numbers by 10 makes the division easier.
$36{,}000 \div 0.3 = 360{,}000 \div 3 = 120{,}000$
So 36,000 meters is 120,000 feet. The speed is 120,000 feet per hour.
Give the answer
Watch out: Multiplying by 0.3 instead of dividing gives 10,800, choice A. Dividing by 3 instead of 0.3 gives 12,000, choice B.
So the answer is D, 120,000 feet per hour.

Check your answer

A foot is shorter than a meter. So any distance is more feet than meters. The answer must be bigger than 36,000, and only choice D is. Also, 120,000 feet at 0.3 meter each is 120,000 × 0.3 = 36,000 meters. That matches.

More questions like this

The original DOE question

Question 111 as printed on page 207 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 207, question 111. Open page 207 of the handbook (PDF)

Question 55 of 57

2023 Test B Math, question 55

Question

$6.\overline{34}-4.\overline{12}$

What is the value of the expression above, written as an improper fraction?

A $\frac{24}{11}$
B $\frac{224}{101}$
C $\frac{111}{50}$
D $\frac{20}{9}$

Show the answer

Correct answer: D

$6.\overline{34}-4.\overline{12}=2.\overline{22}=2.\overline{2}$

$x=2.2222...$	Set x equal to the repeating decimal.
$10x=22.2222...$	Multiply both sides by 10.
$-x=-2.2222...$	Subtract 10x and x.
$9x=20$	Apply the multiplicative inverse property; divide both sides by 9.
$x=\frac{20}{9}$

Question analysis

Two repeating decimals are subtracted, and students write the difference as an improper fraction. It tests subtracting repeating decimals and converting a repeating decimal into a fraction, a grade 8 skill. The repeating part has to be handled exactly, so a careful method matters more than speed.

NYS standards: 8.NS.1 decimal expansions and irrational numbers
7.NS.2d fractions to decimals by long division
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students change fractions into repeating decimals by long division. Going the other way, from a repeating decimal to a fraction, is grade 8 work, though the subtraction that comes first here is simple.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer moderate
Something you must already know moderate
Solution: 3 steps
Reading level: Question and choices: about grade 9, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The subtraction is easy, since the repeating digits line up and give 2.222…; the conversion is the test. Every choice sits within a few hundredths of the answer, so rounding or estimating cannot separate them, and the exact 2.22 of 111/50 is the strongest pull. Students who know that 0.222… is 2/9 can write 2 2/9 = 20/9 directly.

A very close wrong answer moderate Choosing the answer
All four choices lie between 2.18 and 2.23: 111/50 is exactly 2.22 and 224/101 is about 2.218, so only 20/9 equals the repeating 2.222… exactly. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Something you must already know moderate Working it out
Turning 2.222… into a fraction, by setting x equal to it and subtracting x from 10x, is grade 8 work that the question does not explain. (Anderson & Krathwohl, 2001)

Strategy: Subtract the decimals digit by digit to see which digits repeat, then turn that repeating decimal into a fraction, or divide each choice's numerator by its denominator and compare.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Sets x = 2.222… but adds 10x and x instead of subtracting, then drops the repeating part as if it had cancelled, so 11x = 22 + 2 = 24.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Multiplies by 100 for two-digit blocks, then adds x instead of subtracting it and drops the repeating part, so 101x = 222 + 2 = 224.
C Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Drops the repeating bar and treats the difference as exactly 2.22, which is 222/100 = 111/50, slightly less than the true repeating value.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Repeating decimal to fraction: $10x - x = 9x$; x is the repeating decimal. Multiplying by 10 moves the decimal point one place to the right. The endless repeating digits still line up after the point. So when you subtract x from 10x, they cancel, and only whole numbers are left.

Sort out the bars
A bar over digits means they repeat forever. So $6.\overline{34}$ means 6.343434… and $4.\overline{12}$ means 4.121212… with no end. You need the difference as an improper fraction. That is a fraction whose top is bigger than its bottom.
Subtract the decimals
Line up the digits. Each block of 34 minus 12 gives 22. The whole numbers give 6 − 4 = 2.
$6.343434\ldots - 4.121212\ldots = 2.222222\ldots$
The difference is 2.222…, with the 2 repeating forever. You can write it $2.\overline{2}$ .
Name the decimal x
Let x be the repeating decimal. Then multiply both sides by 10. This moves the decimal point one place to the right.
$x = 2.222\ldots$
$10x = 22.222\ldots$
Subtract to remove the repeats
Subtract the first line from the second. The repeating 2s after the point match up, so they cancel.
$10x - x = 22.222\ldots - 2.222\ldots$
$9x = 20$
Divide to find x
Divide both sides by 9.
$x = \frac{20}{9}$
Watch out: $\frac{111}{50}$ is exactly 2.22, which stops. The real difference never stops. That trap is choice C. Every choice is close to 2.2, so estimating won’t help.
So the answer is D, $\frac{20}{9}$ .

Check your answer

Divide 20 by 9. 9 goes into 20 two times, with 2 left over. Each next step leaves 2 again, so you get 2.222… forever. That matches the difference. Choice A, $\frac{24}{11}$ , is 2.1818… instead.

More questions like this

The original DOE question

Question 112 as printed on page 207 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 207, question 112. Open page 207 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 56 of 57

2023 Test B Math, question 56

Question

$A coordinate grid with the origin O, the x-axis marked 1, 2, and 3, and the y-axis marked 1 through 5. Line t passes through the point (0, 1) and line s passes through the point (1, 0); they intersect at P (3, 5), and dashed lines run from P across to 5 on the y-axis and down to 3 on the x-axis. The shaded region is the four-sided figure with vertices O, (0, 1), P (3, 5), and (1, 0).$

Lines s and t intersect at point P (3, 5). What is the area, in square units, of the shaded region in the graph above?

Show the answer

Correct answer: C

First, find the area of the rectangle in the graph:

$3\times 5=15$

Then, find the areas of the two right triangles that are not shaded:

Upper triangle:

$\frac{1}{2}\times 4\times 3=6$

Lower triangle:

$\frac{1}{2}\times 2\times 5=5$

The area of the shaded region, in square units, is calculated by subtracting the total area of the two right triangles from the area of the rectangle:

$15-(6+5)=15-11=4$

Question analysis

Two lines meet at the point (3, 5) on a coordinate grid, and the shaded region between them and the axes is a four-sided shape. Students find its area in square units. The shape is not a standard figure, so no single area formula applies to it directly.

NYS standards: 6.G.1 area of polygons
6.G.3 polygons in the coordinate plane
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Areas of triangles and rectangles are grade 6 content, and grade 7 students often find composite areas. This region is an irregular four-sided shape with two slanted sides, so students need a plan, and the lengths come from coordinates rather than labels.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Reading a figure, table or graph moderate
Many steps minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question rewards a plan more than computation: the region has no name, so students must see it inside the dashed rectangle from O to P, with two right triangles cut away. Reading each triangle’s legs from the axes is the second hurdle, since the 1-unit intercepts shorten one leg of each, to 4 instead of 5 and to 2 instead of 3. Only one wrong choice, from using one coordinate as both heights, has a clear working behind it, so most errors here are guesses after a failed plan.

No obvious method moderate Working it out
The shaded shape has four sides, two of them slanted, and no formula of its own; it must be boxed in the 3-by-5 rectangle with two triangles removed, or cut along OP. (Schoenfeld, 1985)
Reading a figure, table or graph moderate Understanding the question
No lengths are labeled; every one comes from the coordinates of P (3, 5), the y-intercept 1 of line s and the x-intercept 1 of line t. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
The rectangle, two triangle areas and a subtraction make four computations, 15, 6 and 5, then 15 − 11, each built from different coordinates. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Draw the rectangle from O to P, find the areas of the two unshaded right triangles inside it from the coordinates, and subtract both from the rectangle's area.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 2; no rectangle or triangle built from the coordinates leaves that area, so it mostly draws students who judge the narrow region by eye.
B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Cuts the region along OP into two triangles with bases of 1 but uses 3, the x-coordinate of P, as the height of both: 1.5 + 1.5 = 3.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 7; half of the 3-by-5 rectangle, 7.5, comes closest, so it mostly draws students who guess the region fills about half.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle: $A = l \times w$; A is the area, in square units. l is the length and w is the width. Multiplying them counts the unit squares that fill the rectangle.
Area of a triangle: $A = \frac{1}{2} \times b \times h$; A is the area. b is the base, one side. h is the height: the distance from the base straight up to the far corner. A right triangle has a square corner, and the two sides that meet there are the base and height. A triangle is half of a rectangle with the same base and height.

Sort out the shaded region
The shaded region has four corners: O at (0, 0), (0, 1) on the y-axis, P at (3, 5), and (1, 0) on the x-axis. Line t meets the y-axis at (0, 1), and line s meets the x-axis at (1, 0). There is no easy formula for this shape. But the dashed lines make a rectangle around it. So find the rectangle’s area and take away the white parts.
Find the rectangle’s area
The rectangle has corners O, (3, 0), P and (0, 5). It is 3 units wide and 5 units tall.
$A = 3 \times 5 = 15$
The rectangle’s area is 15 square units.
Find the upper triangle’s area
The white triangle at the top left has corners (0, 1), (0, 5) and P. Its square corner, or right angle, is at (0, 5). One side runs up the y-axis from 1 to 5, so it is 4 units. The other runs across from 0 to 3, so it is 3 units.
$A = \frac{1}{2} \times 4 \times 3 = 6$
Its area is 6 square units.
Find the lower triangle’s area
The white triangle at the bottom right has corners (1, 0), (3, 0) and P. Its square corner is at (3, 0). One side runs along the x-axis from 1 to 3, so it is 2 units. The other goes up from 0 to 5, so it is 5 units.
$A = \frac{1}{2} \times 2 \times 5 = 5$
Its area is 5 square units.
Watch out: The lines cross the axes at 1, not at O. So these sides are 4 and 2, not 5 and 3.
Subtract the white triangles
Take both triangles away from the rectangle.
$15 - (6 + 5) = 15 - 11 = 4$
The shaded area is 4 square units.
So the answer is C, 4.

Check your answer

Cut the shaded region along a line from O to P. One piece has a base of 1 on the x-axis and a height of 5, since P is 5 units up. Its area is 2.5. The other has a base of 1 on the y-axis and a height of 3, since P is 3 units across. Its area is 1.5. Then 2.5 + 1.5 = 4. Using 3 for both heights gives 3, choice B.

More questions like this

The original DOE question

Question 113 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 113. Open page 208 of the handbook (PDF)

Question 57 of 57

2023 Test B Math, question 57

Question

A box contains 14 red marbles, 10 black marbles, and 16 white marbles. How many more white marbles need to be added to the box so that the probability of randomly drawing a white marble is $\frac{3}{5}$ ?

Show the answer

Correct answer: D

Let w represent the number of additional white marbles.

Set the probability of drawing a white marble equal to $\frac{3}{5}$ and solve the equation for w:

$\frac{16+w}{14+10+16+w}=\frac{3}{5}$

$\frac{16+w}{40+w}=\frac{3}{5}$

$5(16+w)=3(40+w)$

$80+5w=120+3w$

$2w=40$

$w=20$

Question analysis

A box holds 14 red, 10 black and 16 white marbles, and students find how many white marbles must be added so that the chance of drawing a white one is 3/5. Working backward from a target probability to a number of marbles is the unusual step.

NYS standards: 6.SP.8a uniform probability models
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 probability questions usually give the counts and ask for a probability. This runs the other way: the target probability is given, and the number to add changes both the white count and the total.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The tempting shortcut is to take 3/5 of the 40 marbles already in the box, which forgets that the new white marbles join the total. Students who set up the probability with the unknown in both the top and the bottom solve it in a few lines, and testing choices works too: adding 20 gives 36 white of 60, exactly 3/5. The other two choices are counts copied from the stem.

Turning words into math moderate Understanding the question
“How many more white marbles need to be added” puts the unknown in two places, (16 + w)/(40 + w), because each white marble added also raises the total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
The unknown number of added marbles has to be carried as a letter through a proportion, 5(16 + w) = 3(40 + w), before it can be found. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Totaling the marbles, setting up the proportion, cross-multiplying and solving 80 + 5w = 120 + 3w make four steps, unless the choices are tested one by one. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Let w be the number of white marbles added, write the new probability with w in both the top and the bottom, set it equal to 3/5 and solve; or test each choice.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals partial knowledge Expected pull: strong
Takes 3/5 of the 40 marbles now in the box, 24, and subtracts the 16 white ones, forgetting that added marbles raise the total.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 10, the number of black marbles in the stem, without setting up the new probability or testing it.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 14, the number of red marbles in the stem, without checking that adding 14 white marbles gives 30 of 54, not 3/5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P(\text{white}) = \frac{\text{white marbles}}{\text{all marbles}}$; P(white) is the probability, or chance, of drawing a white marble. The top counts the white marbles. The bottom counts every marble in the box. It works because each marble is equally likely to be drawn.
Cross multiplication: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a and c are the tops. b and d are the bottoms. When two fractions are equal, each top times the other bottom gives the same product. It works because you multiply both sides by both bottoms. Use it to clear fractions from an equation.

Sort out the box
The box has 14 red, 10 black and 16 white marbles.
$14 + 10 + 16 = 40$
That is 40 marbles. You will add some white marbles. Call that number w. You need w so that the chance of drawing white is $\frac{3}{5}$ .
Write the new probability
Adding w white marbles changes both counts. There will be 16 + w white marbles and 40 + w marbles in all. Use the probability formula.
$\frac{16 + w}{40 + w} = \frac{3}{5}$
Watch out: Taking $\frac{3}{5}$ of 40 gives 24, and 24 − 16 = 8, choice A. That forgets that the new marbles also make the box bigger.
Cross multiply
Fractions are hard to solve with. Clear them with cross multiplication.
$5 \times (16 + w) = 3 \times (40 + w)$
Then multiply each outside number by both terms inside the parentheses.
$80 + 5w = 120 + 3w$
Solve for w
Subtract 3w from both sides. Then subtract 80 from both sides.
$2w = 40$
Divide both sides by 2.
$w = 20$
You need to add 20 white marbles.
So the answer is D, 20.

Check your answer

Add 20 white marbles. Then there are 16 + 20 = 36 white marbles and 40 + 20 = 60 marbles in all. Divide the top and bottom by 12: $\frac{36}{60} = \frac{3}{5}$ . It works.

More questions like this

The original DOE question

Question 114 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 114. Open page 208 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

3 × 4 = 12	3 × 5 = 15	3 × 6 = 18	3 × 7 = 21
4 × 5 = 20	4 × 6 = 24	4 × 7 = 28
5 × 6 = 30	5 × 7 = 35
6 × 7 = 42