Practice Test

2023 Handbook

2023 Test A Math

Question 1 of 57

2023 Test A Math, question 1

Question

Ms. Li opened a retirement account with a deposit of $2,500. This account earns 4% simple interest annually. How many years will it take her to earn $500 on her $2,500 deposit?

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

Show the answer

Correct answer: 5

Simple interest (y) is calculated by multiplying the initial deposit (p), the interest rate (r), and time (t):

$prt = y$

$(2{,}500)(0.04)t = 500$

$100t = 500$

$t = 5$

Question analysis

Ms. Li deposits $2,500 in a retirement account that pays 4% simple interest each year, and this grid-in asks how many years it takes to earn $500 in interest. Students find one year's interest and work out how many years are needed. It turns the usual simple interest question around, asking for the time rather than the interest.

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 a grade 7 percent application, usually asked as how much interest builds up over a given time. This one gives the interest and asks for the time, a short reverse 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
An answer in an unexpected form minor
Solution: 2 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

Once one year's interest is known, $100, the rest is counting how many years make $500, so the arithmetic is light. The question turns the usual simple interest problem around, giving the interest and asking for the time, and that reversal is where students stall or grid a percent instead of a number of years. Writing 4% as 0.04, not 0.4, keeps the yearly interest from coming out ten times too big.

Something you must already know minor Working it out
The stem says 4% simple interest annually but never says that each year adds the same 4% of the $2,500 deposit, $100, so the yearly amount has to come from memory. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The question gives the interest, $500, and asks for the time, so the familiar formula runs backward: 2,500 × 0.04 × t = 500 has to be solved for t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
As a grid-in it shows no choices, so 20, the interest as a percent of the deposit, can be gridded with nothing on the page to say it is not a number of years. (Enright, Morley & Sheehan, 2002)

Strategy: Work out one year's interest as 4% of $2,500, then find how many of those yearly amounts add up to the $500 of interest.

Traps

Stops one step early: A choice is a number from partway through the solution.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

20 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that the $500 of interest is 20% of the $2,500 deposit and grids 20, stopping before dividing by the 4% earned each year.
0.5 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Writes 4% as 0.4 instead of 0.04, so one year's interest looks like $1,000 and the $500 seems to take only half a year.
125 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Divides the $500 of interest by the 4 in 4%, combining two numbers from the stem without first finding 4% of the deposit.

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 extra money the account pays you. P is the deposit, the money you put in. r is the yearly rate as a decimal. t is the time in years. Simple means each year’s interest is figured on the deposit only, so every year adds the same amount.
Percent to decimal: $p\% = \frac{p}{100}$; p is the percent. Percent means out of 100. So divide by 100 to write it as a decimal. For example, 4% is 4 ÷ 100, which is 0.04.

Sort out the question
Ms. Li puts $2,500 into an account. This first amount is called the deposit. Each year the account pays 4% simple interest. Interest is extra money the bank pays you for keeping your money there. You need the number of years it takes to earn $500 of interest. The answer is a number of years, not dollars and not a percent.
Find one year’s interest
First write 4% as a decimal. 4% means 4 out of 100.
$4\% = \frac{4}{100} = 0.04$
Now use the formula for 1 year, so t = 1.
$I = P \times r \times t$
$I = 2{,}500 \times 0.04 \times 1 = 100$
The account earns $100 each year. Watch out: 4% is 0.04, not 0.4. Using 0.4 makes one year’s interest $1,000, and you would grid 0.5 years.
Find how many years make $500
Simple interest adds the same $100 every year. So the time t must make the interest $500.
$2{,}500 \times 0.04 \times t = 500$
$100 \times t = 500$
$t = 500 \div 100 = 5$
It takes 5 years. Watch out: $500 is 20% of $2,500. But 20 is a percent, not a number of years. Don’t grid 20. So you would grid 5.

Check your answer

Count up one year at a time: $100, $200, $300, $400, $500. That is 5 years. Or put t = 5 back into the formula: $2{,}500 \times 0.04 \times 5 = 500$ . That matches the $500 in the question.

More questions like this

The original DOE question

Question 58 as printed on page 75 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 75, question 58. Open page 75 of the handbook (PDF)

Question 2 of 57

2023 Test A Math, question 2

Question

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

What is the value of the expression shown above?

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

Show the answer

Correct answer: -9

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

Question analysis

This grid-in asks for the value of an arithmetic expression that mixes an absolute value, a negative number raised to the third power, a division and a mixed number. Students apply the order of operations with signed numbers. Cubing a negative base, and doing the division and multiplication before any adding, are where most slips happen.

NYS standards: 7.NS.2a rational number multiplication
6.EE.2c evaluating expressions and order of operations
6.NS.7c absolute value as distance
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students routinely evaluate expressions with negative numbers in the right order. This one combines an absolute value, a cubed negative and a mixed number in a single line, so it takes more steps than usual, each with a sign to watch.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The expression packs four kinds of notation into one line: a subtraction, a division by an absolute value, a negative base raised to the third power and a mixed number. None of the operations is hard alone, but each has a rule that is easy to break under time pressure, and the slips lead to very different totals. Rewriting the power and the mixed number first, then dividing and multiplying before any addition, avoids them.

Error-prone arithmetic moderate Working it out
(−2)³ must stay negative, −8, and the mixed number 1½ must become 3/2 before multiplying, so the last product is −12; a sign or fraction slip here moves the total far from −9. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Six numbers and six operations, including a power and an absolute value, have to be worked in a planned order, and every addition waits until the division and multiplication are done. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
An answer in an unexpected form minor Choosing the answer
The value is negative and the grid-in offers no choices, so the minus sign has to be gridded and nothing on the page confirms that the total falls below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the absolute value and the power first, change the mixed number to an improper fraction, do the division and multiplication, and only then add and subtract from left to right.

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Order of operations: $\text{grouping} \rightarrow \text{powers} \rightarrow \times \text{ and } \div \rightarrow + \text{ and } -$; This is the order for working out an expression. First work inside grouping symbols, such as parentheses and absolute value bars. Next do powers. Then multiply and divide from left to right. Last, add and subtract from left to right. Using one order means everyone gets the same answer.
A power: $a^{3} = a \times a \times a$; a is the base, the number being multiplied. The small 3 is the exponent. It says how many times the base is used as a factor, a number being multiplied. It does not mean a times 3.
Absolute value: $\left|a\right| = \text{the distance from } a \text{ to } 0$; The bars around a number mean absolute value. It is how far the number is from 0 on a number line. A distance is never negative, so the absolute value of −3 is 3.
Mixed number to fraction: $a\frac{b}{c} = \frac{a \times c + b}{c}$; a is the whole number part, and b over c is the fraction part. Each whole is c pieces, so a wholes are a × c pieces. Add the b extra pieces. The bottom number stays c.

Sort out the expression
You need the value of $6 - 9 \div \left|-3\right| + (-2)^{3} \times 1\frac{1}{2}$ . It has many operations, so the order matters. It has absolute value bars, a power of a negative number and a mixed number. The answer may be negative, so keep track of every sign.
Do the absolute value and the power
Grouping symbols and powers come first.
$\left|-3\right| = 3$
$(-2)^{3} = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$
Two negatives multiply to a positive, 4. One more negative makes it negative again. Watch out: (−2)³ is −8, not +8. Using +8 leads to 15. It is also not −2 × 3. That mistake leads to a total of −6.
Change the mixed number to a fraction
A fraction is easier to multiply than a mixed number.
$1\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}$
Now the expression looks like this:
$6 - 9 \div 3 + (-8) \times \frac{3}{2}$
Multiply and divide next
Multiplying and dividing come before adding and subtracting.
$9 \div 3 = 3$
$(-8) \times \frac{3}{2} = \frac{-24}{2} = -12$
A negative times a positive is negative. Watch out: don’t just work from left to right. Doing 6 − 9 first, then dividing by 3, leads to −13.5.
Add and subtract last
Now only adding and subtracting are left. Work from left to right.
$6 - 3 + (-12) = 3 + (-12) = -9$
Adding −12 is the same as taking away 12. So you would grid −9, with the minus sign in its own column.

Check your answer

Look at the size of each part. The part 6 − 3 is a small positive number, 3. The last part, −8 times 1½, is a bigger negative number, −12. A big negative plus a small positive is negative. So an answer below zero, −9, makes sense.

More questions like this

The original DOE question

Question 59 as printed on page 75 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 75, question 59. Open page 75 of the handbook (PDF)

Question 3 of 57

2023 Test A Math, question 3

Question

Solve for x:

$7x + 3 - 2(2x + 1) = 13$

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

Show the answer

Correct answer: 4

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

Question analysis

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

NYS standards: 7.EE.4a two-step equations
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students expand expressions with negative factors and solve equations such as p(x + q) = r. This equation needs both skills plus combining like terms, which is routine work with one negative sign that is easy to lose.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
An answer in an unexpected form minor
Solution: 4 steps
Reading level: Question and choices: about grade 3, 4 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
The −2 in front of (2x + 1) must reach both terms, giving −4x − 2; the second product, −2 × 1, is the one most often written as +2 or left as +1. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in with no choices, a value such as 3 or 8/3 from a distributing slip looks just as finished as the right one; only substituting back exposes it. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

3 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Multiplies only the 2x by −2 and leaves the + 1 untouched, so the equation becomes 3x + 4 = 13 and x = 3.
8/3 Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Gives the minus sign to 4x but not to the 2, writing −4x + 2, so the equation becomes 3x + 5 = 13 and x = 8/3.
12 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Simplifies correctly to 3x + 1 = 13, subtracts 1 to reach 3x = 12, and grids 12 without dividing by 3.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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. A term is a number, a letter, or a number times a letter. Multiply each term inside by a. If a is negative, every product gets its sign too.
Keeping an equation balanced: $\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c$; a and b are the two equal sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal, like a balanced scale. Use it to get the unknown by itself.

Sort out the equation
You need the value of x that makes $7x + 3 - 2(2x + 1) = 13$ true. x is the unknown number. The tricky part is the −2 in front of the parentheses. It multiplies everything inside them.
Multiply out the parentheses
Use the distributive property with a = −2, b = 2x and c = 1.
$-2(2x + 1) = (-2) \times 2x + (-2) \times 1 = -4x - 2$
So the equation becomes this:
$7x + 3 - 4x - 2 = 13$
Watch out: the −2 must multiply the 1 too. Leaving + 1 gives x = 3. Writing + 2 instead of − 2 gives $x = \frac{8}{3}$ .
Combine like terms
Like terms are terms of the same kind. Put the x terms together. Put the plain numbers together.
$7x - 4x = 3x$
$3 - 2 = 1$
$3x + 1 = 13$
Get x by itself
Undo the + 1 first. Subtract 1 from both sides, so the equation stays balanced.
$3x + 1 - 1 = 13 - 1$
$3x = 12$
3x means 3 times x. So divide both sides by 3.
$x = 12 \div 3 = 4$
Watch out: don’t stop at 3x = 12. 12 is three times the answer. So you would grid 4.

Check your answer

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

More questions like this

The original DOE question

Question 60 as printed on page 75 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 75, question 60. Open page 75 of the handbook (PDF)

Question 4 of 57

2023 Test A Math, question 4

Question

$Line l is vertical and line n is horizontal, and a small square marks the right angle where they meet. Line m rises from the lower left to the upper right; it crosses line l below line n and crosses line n to the right of line l. Where m crosses l, the angle between the lower part of l and the lower-left part of m is marked 35°. Where m crosses n, the angle between the right part of n and the upper-right part of m is marked x°.$

In the figure above, line l is perpendicular to line n. What is the value of x?

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

Show the answer

Correct answer: 55

Since vertical angles are congruent, the right triangle in the figure has acute angles with measures of 35° and x°. The interior angles of a triangle sum to 180°, so set up an equation to solve for x:

$90 + 35 + x = 180$

$125 + x = 180$

$x = 55$

Question analysis

Three lines cross in a figure: line l is perpendicular to line n, and a slanted line m meets them at a marked 35° angle and an unknown angle x. This geometry grid-in asks for the value of x. Students must see the right triangle the lines form and carry the marked angles into it using vertical angles.

NYS standards: 8.G.5 angle sums, transversals, angle-angle similarity
7.G.5 angle relationships and equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 angle problems use vertical, supplementary and complementary angles. This one also needs the angle sum of a triangle, which New York places in grade 8, and the triangle is not drawn as a separate shape, so students must spot it.
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
Something you must already know moderate
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The figure hides a small right triangle between the three lines, and the question is easy once students see it. Its angles are the right angle, the vertical partner of the 35° angle and the vertical partner of x, so x is what the right angle and 35° leave of 180°. Students who never spot the triangle tend to guess a direct relationship between the two marked angles, which no angle fact supports.

Reading a figure, table or graph moderate Understanding the question
Neither marked angle is inside the right triangle that lines l, m and n form: the 35° opens below the point where m crosses l, and x opens above line n, so each must be carried in as a vertical angle. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know moderate Working it out
The stem states only that l is perpendicular to n; students must supply that vertical angles are equal and that the three angles of a triangle add to 180°. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
As a grid-in it offers no choices, so 125, the right angle plus the 35° angle, can be gridded with nothing to signal the missing subtraction. (Enright, Morley & Sheehan, 2002)

Strategy: Find the triangle formed by the three lines, use vertical angles to move the 35° and x° angles inside it, and apply the triangle's angle sum with the right angle.

Traps

Stops one step early: A choice is a number from partway through the solution.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

125 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Adds the right angle and the 35° angle, 90 + 35 = 125, and grids that sum without subtracting it from the triangle's 180°.
35 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Assumes line m crosses l and n at equal angles, as it would cross two parallel lines, though l and n are perpendicular.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Vertical angles: $\text{angle } a = \text{angle } b$; When two lines cross, they make an X. The angles straight across from each other in the X are called vertical angles. a and b are one such pair. They are always equal, because each one plus the angle between them makes a straight line.
Angles of a triangle: $A + B + C = 180$; A, B and C are the three angles inside any triangle, in degrees. They always add up to 180°. Use it when you know two of the angles and need the third.

Sort out the figure
Line l is perpendicular to line n. Perpendicular means they meet at a right angle, 90°. The small square in the figure marks it. Line m crosses both lines. The three lines make a triangle with a right angle. You need x, the angle marked x° where m crosses n.
Move the 35° angle into the triangle
The 35° angle is below the point where m crosses l. So it is outside the triangle. Lines l and m make an X there. The triangle’s angle is straight across from the 35° angle. They are vertical angles, so they are equal. The triangle’s angle at that corner is 35°.
Move the x° angle into the triangle
The x° angle is above line n, so it is outside the triangle too. Lines m and n make another X. The triangle’s angle at that crossing is straight across from x°. So the third angle of the triangle is also x°. Watch out: line m does not cross l and n at the same angle. Lines l and n are not parallel, so x is not 35.
Use the triangle’s angle sum
The triangle’s three angles are 90°, 35° and x°.
$A + B + C = 180$
$90 + 35 + x = 180$
$125 + x = 180$
$x = 180 - 125 = 55$
Watch out: 125 is only the two known angles added together. You still subtract it from 180. Gridding 125 is wrong. So you would grid 55.

Check your answer

In a right triangle, the two smaller angles add up to 90°. That is because the right angle uses 90° of the 180°. Check: $35 + 55 = 90$ . Also, in the figure the x° angle looks a bit bigger than the 35° angle, and 55 is.

More questions like this

The original DOE question

Question 61 as printed on page 76 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 76, question 61. Open page 76 of the handbook (PDF)

Question 5 of 57

2023 Test A Math, question 5

Question

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

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

Show the answer

Correct answer: 19.2

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

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

$x = 136$

Next, subtract the 3 discarded numbers from the sum:

$136 - 9 - 11 - 20 = 96$

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 6 of 57

2023 Test A Math, question 6

Question

$3^{4} + 7^{4} =$

A40
B370
C2,482
D10,000

Show the answer

Correct answer: C

$3^{4} + 7^{4} = 81 + 2{,}401 = 2{,}482$

Question analysis

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

NYS standards: 6.EE.1 whole-number exponent expressions
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Evaluating whole-number powers is grade 6 work, and this question asks only for two of them and a sum. The one demand beyond routine practice is computing a fourth power of 7 accurately without a calculator.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 16+, 1 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Error-prone arithmetic moderate Working it out
7⁴ means 7 × 7 × 7 × 7 = 2,401, a four-digit product built by hand, most easily as 49 × 49; 3⁴ = 81 is quick by comparison. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
An exponent counts repeated factors rather than acting as a multiplier, so 3⁴ is 3 × 3 × 3 × 3 = 81, not 3 × 4 = 12; the item assumes that meaning. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Multiplies each base by its exponent, 3 × 4 + 7 × 4 = 12 + 28 = 40, instead of using each base as a factor four times.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Uses each base as a factor three times instead of four, 27 + 343 = 370, so both powers stop one factor short.
D Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Adds the bases before applying the exponent, treating 3⁴ + 7⁴ as (3 + 7)⁴ = 10,000, though powers come before addition.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

Sort out the question
You need the value of $3^{4} + 7^{4}$ . Each number has a small 4 above it. This 4 is an exponent. Work out each power on its own. Then add the two results.
Find 3 to the fourth power
3⁴ means four 3s multiplied together. Pair them up to make it easier.
$3^{4} = 3 \times 3 \times 3 \times 3 = 9 \times 9 = 81$
Watch out: 3⁴ is not 3 × 4 = 12. Doing that to both powers gives 12 + 28 = 40, choice A.
Find 7 to the fourth power
7⁴ means four 7s multiplied together. Pair them up the same way.
$7^{4} = 7 \times 7 \times 7 \times 7 = 49 \times 49 = 2{,}401$
To find 49 × 49, do 49 × 50 = 2,450. Then take away one 49. That gives 2,401. Watch out: count four 7s, not three. Three 7s make 343. Using three factors in both powers gives 370, choice B.
Add the two powers
$81 + 2{,}401 = 2{,}482$
Watch out: don’t add 3 and 7 first. That gives $(3 + 7)^{4} = 10^{4} = 10{,}000$ , choice D. Powers come before adding. So the answer is C, 2,482.

Check your answer

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

More questions like this

The original DOE question

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

Question 7 of 57

2023 Test A Math, question 7

Question

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

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

Show the answer

Correct answer: A

To solve, add the three given fractions:

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

The common denominator is 24:

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

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

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

Question analysis

A rainfall word problem gives the inches of rain on Monday, Tuesday and Wednesday as two mixed numbers and a fraction, and asks for the total over the three days. Students add fractions with unlike denominators. Finding a common denominator for fourths, thirds and eighths, and regrouping the fractional part into wholes, is where errors come in.

NYS standards: 7.NS.3 four operations with rational numbers
7.NS.1d properties for adding and subtracting
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Adding fractions and mixed numbers with unlike denominators is taught before grade 7, and this is a direct word problem. It takes more care than a typical exercise only because three different denominators must be rewritten over a common one.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 35 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The setup is plain addition, so the difficulty is all in the fractions. Students who know to rename over a common denominator rarely miss it, while those who add the fraction parts straight across, or put the numerators over 24 without scaling them, reach choices with a whole part of 3. Since 1 3/4 + 2 2/3 alone is more than 4, any total below 4 can be ruled out at a glance.

Error-prone arithmetic moderate Working it out
The fractional parts have denominators 4, 3 and 8, so they must be renamed as 18, 16 and 21 twenty-fourths, and their sum, 55/24, regrouped as 2 7/24 before the wholes are added. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Renaming, adding the numerators, regrouping the improper fraction and adding the whole numbers are four separate moves, and losing the 2 carried from 55/24 drops the total by two inches. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the three fractional parts over a common denominator of 24, add the whole numbers and the fractions separately, and regroup any extra twenty-fourths as a whole.

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions: $\frac{a}{b} = \frac{a \times k}{b \times k}$; a over b is a fraction, and k is any whole number except 0. Multiplying the top and the bottom by the same number gives an equal fraction. You cut each piece into k smaller pieces. So you have k times as many pieces, each k times smaller.
Adding fractions with the same bottom: $\frac{a}{d} + \frac{b}{d} + \frac{c}{d} = \frac{a + b + c}{d}$; d is the denominator, the bottom number. It tells the size of the pieces. a, b and c are the numerators, the top numbers. They count the pieces. When the pieces are the same size, add the counts. The bottom stays the same.
Fraction to mixed number: $\frac{a}{b} = q\frac{r}{b}$; a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.

Sort out the question
Rain fell on three days: $1\frac{3}{4}$ inches, $2\frac{2}{3}$ inches and $\frac{7}{8}$ inch. You need the total rain, in inches. So add the three amounts. Add the whole numbers and the fractions separately. The whole numbers are 1 and 2, so they add to 3 inches.
Find a common denominator
The fractions have denominators, or bottom numbers, 4, 3 and 8. You can only add pieces of the same size. So find the smallest number that 4, 3 and 8 all divide into. Count by 8s: 8, 16, 24. 24 works, since 24 = 4 × 6 and 24 = 3 × 8. So use 24ths.
Rename each fraction in 24ths
Multiply the top and the bottom by the same number.
$\frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}$
$\frac{2}{3} = \frac{2 \times 8}{3 \times 8} = \frac{16}{24}$
$\frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24}$
Watch out: change the tops too. Keeping 3, 2 and 7 over 24 gives $\frac{12}{24} = \frac{1}{2}$ and a total of $3\frac{1}{2}$ , choice D.
Add the fractions
Now the pieces are the same size, so add the tops.
$\frac{18}{24} + \frac{16}{24} + \frac{21}{24} = \frac{55}{24}$
That is more than 1 whole. 24 goes into 55 twice, with 7 left over.
$\frac{55}{24} = 2\frac{7}{24}$
Watch out: never add the bottoms. Adding straight across gives $\frac{12}{15} = \frac{4}{5}$ and a total of $3\frac{4}{5}$ , choice C.
Put the parts together
Add the 3 whole inches from before.
$3 + 2\frac{7}{24} = 5\frac{7}{24}$
So the answer is A, $5\frac{7}{24}$ inches.

Check your answer

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

More questions like this

The original DOE question

Question 64 as printed on page 77 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 64. Open page 77 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 8 of 57

2023 Test A Math, question 8

Question

Jamel works at a computer store. He is paid an hourly rate plus a 15% commission on all computer products he sells. Last week, Jamel was paid $802.50 for working 30 hours and selling $1,250.00 worth of computer products. What is Jamel’s hourly rate?

A$20.50/hr
B$26.75/hr
C$33.00/hr
D$37.65/hr

Show the answer

Correct answer: A

First, determine the amount of Jamel’s weekly pay that is from commission:

$\$1{,}250.00 \times 15\% = \$187.50$

Next, subtract Jamel’s commission amount from his weekly pay:

$\$802.50 - \$187.50 = \$615.00$

Last, divide the result by 30 hours to find Jamel’s hourly rate:

$\$615.00 \div 30 = \$20.50$

Question analysis

Jamel earns an hourly wage plus a 15% commission on his sales, and one week he is paid $802.50 for 30 hours of work and $1,250 in sales. Students find his hourly rate. The question mixes a percent with a rate, and the commission has to be taken out of the total pay before dividing by the hours.

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 Commission is one of the percent applications named in the grade 7 standards, and the remaining work is a two-step equation with decimals. That matches a typical grade 7 multistep problem, with the order of the steps as the main demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 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

Jamel's pay has two parts, and the question asks for only one of them, so the commission has to be removed before dividing by the hours. Students who divide the whole $802.50 by 30 get a believable hourly figure that still contains the commission. Checking the answer against the story, 30 × $20.50 + $187.50 = $802.50, confirms both parts.

Turning words into math moderate Understanding the question
The phrase "paid an hourly rate plus a 15% commission on all computer products he sells" makes $802.50 the sum of two parts, 30 hours of pay and 15% of $1,250. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps moderate Working it out
Three operations in order: 15% of $1,250 is $187.50, $802.50 − $187.50 = $615.00, and $615.00 ÷ 30 = $20.50; dividing before subtracting gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Every amount carries cents, and 15% of $1,250 and $615 ÷ 30 are worked by hand, so a misplaced decimal point shifts the hourly rate. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Take 15% of the $1,250 in sales to find the commission, remove that from the week's pay, and divide what is left by the 30 hours worked.

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.
Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the whole $802.50 by 30 hours, which gives Jamel's average pay per hour with the commission included, not his hourly rate.
C Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Follows the word plus and adds the $187.50 commission to the $802.50 instead of removing it, so (802.50 + 187.50) ÷ 30 = $33.00.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Swaps the two dollar amounts, taking 15% of the $802.50 pay and subtracting it from the $1,250 of sales: (1,250 − 120.38) ÷ 30 ≈ $37.65.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{p}{100} \times \text{whole}$; p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Pay with commission: $\text{pay} = \text{rate} \times \text{hours} + \text{commission}$; The rate is the money Jamel earns for each hour. Hours is how long he worked. The commission is extra money that is a percent of what he sells. His pay is these two amounts added together.

Sort out the question
Jamel’s pay has two parts. One part is an hourly rate, the money he earns for each hour. The other part is a commission. A commission is extra pay that is a percent of what he sells. He worked 30 hours and sold $1,250.00 of products. In all he got $802.50. You need his hourly rate.
Find the commission
His commission is 15% of what he sold. 15% means 15 out of 100, or 0.15.
$\text{part} = \frac{p}{100} \times \text{whole}$
$\text{part} = 0.15 \times 1{,}250 = 187.50$
His commission was $187.50. Watch out: take 15% of the $1,250 he sold, not of his $802.50 pay. Mixing them up leads to $37.65, choice D.
Take the commission out of his pay
The rest of his pay came from his hours.
$802.50 - 187.50 = 615.00$
He earned $615.00 for 30 hours of work. Watch out: the word plus does not mean add here. Adding the $187.50 to $802.50 counts the commission twice. That leads to $33.00, choice C.
Divide by the hours he worked
Share the $615.00 equally over the 30 hours.
$615 \div 30 = 20.50$
Watch out: don’t divide all $802.50 by 30. That gives $26.75, choice B. That amount still includes the commission. So the answer is A, $20.50 per hour.

Check your answer

Work forward from the answer. 30 hours at $20.50 is $30 \times 20.50 = 615$ dollars. Add the commission: $615 + 187.50 = 802.50$ . That is exactly Jamel’s pay for the week.

More questions like this

The original DOE question

Question 65 as printed on page 77 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 65. Open page 77 of the handbook (PDF)

Question 9 of 57

2023 Test A Math, question 9

Question

A revolving sign makes 1 complete revolution every 90 minutes. If the sign starts moving at 2:30 p.m., at what time will the sign complete 8 revolutions?

A1:00 a.m.
B2:30 a.m.
C4:00 p.m.
D12:00 midnight

Show the answer

Correct answer: B

First, use a proportion to find out how many minutes it will take the sign to complete 8 revolutions. Let x equal the total number of minutes:

$\frac{x}{8} = \frac{90}{1}$

$x = 720$

Next, convert 720 minutes into hours and minutes:

$\frac{720}{60} = 12\text{ hours}$

Finally, 12 hours after 2:30 p.m. is 2:30 a.m.

Question analysis

A revolving sign turns once every 90 minutes and starts at 2:30 in the afternoon. Students find the clock time when it finishes 8 full revolutions. The question combines a rate with converting minutes to hours, and then adding the elapsed time to the starting time on a 12-hour clock.

NYS standards: 6.RP.3b unit rate problems
6.RP.3d converting measurement units
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 A constant rate, a multiplication and a conversion from minutes to hours are grade 6 skills. The only step beyond routine work is adding the elapsed time to a starting clock time.
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
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 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

Each step is routine, so most points are lost in keeping elapsed time and clock time apart. The 12 hours is a duration, and the clock reading that ends it is 12 hours after 2:30 in the afternoon, not 12:00. A sense check helps: 12 hours after any time shows the same digits on the clock with a.m. and p.m. swapped.

Many steps minor Working it out
Three moves: 8 × 90 = 720 minutes, 720 ÷ 60 = 12 hours, then 12 hours counted on from the 2:30 start, which crosses midnight into the morning. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
Counting hours past midnight means switching from p.m. to a.m. on a 12-hour clock, which the stem leaves to the student; the choices mix both halves of the day. (Anderson & Krathwohl, 2001)

Strategy: Multiply 90 minutes by 8 to get the total time, change it to hours, and then count that many hours forward from 2:30 p.m.

Traps

Stops one step early: A choice is a number from partway through the solution.
Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
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 Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Counts only 7 revolutions, 630 minutes or 10½ hours, as if the first turn were already done at the start, and lands at 1:00 a.m.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds a single revolution's 90 minutes to the 2:30 start, giving 4:00 in the afternoon, without multiplying by the 8 revolutions.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds that 8 revolutions take 12 hours and picks 12:00 midnight, reading the elapsed time as a clock time instead of counting on from 2:30.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total time: $\text{total time} = \text{time for one} \times \text{how many}$; Time for one is how long one revolution takes. How many is the number of revolutions. Each one takes the same time, so multiply to get the total time.
Minutes to hours: $\text{hours} = \text{minutes} \div 60$; There are 60 minutes in 1 hour. So to change minutes to hours, divide by 60. Each group of 60 minutes makes 1 hour.

Sort out the question
The sign turns all the way around once every 90 minutes. One full turn is called a revolution. The sign starts at 2:30 p.m. You need the clock time when it finishes 8 revolutions.
Find the total time in minutes
Each revolution takes 90 minutes, and there are 8 of them.
$\text{total time} = 90 \times 8 = 720$
The 8 revolutions take 720 minutes. Watch out: use all 8 revolutions. Counting only 7 gives 630 minutes and 1:00 a.m., choice A. Adding just one 90 minutes gives 4:00 p.m., choice C.
Change the minutes to hours
Clock times are easier to count in hours. There are 60 minutes in an hour.
$\text{hours} = 720 \div 60 = 12$
The sign turns for 12 hours.
Count forward from 2:30 p.m.
Start at 2:30 p.m. and add 12 hours. When you pass midnight, p.m. changes to a.m. So 12 hours after 2:30 p.m. is 2:30 a.m. the next morning. Watch out: 12 hours is an amount of time, not a clock time. Reading it as 12:00 midnight gives choice D. So the answer is B, 2:30 a.m.

Check your answer

From 2:30 p.m. to 2:30 a.m. is one full trip around the clock, 12 hours. 12 hours is $12 \times 60 = 720$ minutes. And $720 \div 90 = 8$ , the number of revolutions in the question.

More questions like this

The original DOE question

Question 66 as printed on page 77 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 77, question 66. Open page 77 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 10 of 57

2023 Test A Math, question 10

Question

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

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

Show the answer

Correct answer: D

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 11 of 57

2023 Test A Math, question 11

Question

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

What value of n makes the equation above true?

A1
B $1\frac{1}{9}$
C5
D $10\frac{1}{9}$

Show the answer

Correct answer: B

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

Question analysis

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

NYS standards: 6.EE.7 one-step equations
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Solving a proportion is familiar from grades 6 and 7, and this one needs only cross-multiplying and one division. Its one extra demand is simplifying the result and writing it in the form the choices use.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: An answer in an unexpected form moderate
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

An answer in an unexpected form moderate Choosing the answer
Cross-multiplying gives n = 90/81, which has to be reduced to 10/9 and rewritten as the mixed number 1 1/9 before it matches any choice. (Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The choices 1 and 1 1/9 differ by one ninth, and estimating 90 ÷ 81 as about 1 cannot separate them; only keeping the remainder does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

A Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Finds n by dividing 10 by 9 but keeps only the whole-number part of the quotient, 1, dropping the remainder that the exact answer needs.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No cross-multiplication or scaling of 81/10 gives 5, which would make 9/n equal 1.8 instead of 8.1, so it mostly catches guesses.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Reaches n = 10 × 9/81 = 10 × 1/9 but writes the product as the mixed number 10 1/9, as if a whole number beside a fraction meant multiplication.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cross-multiplying: $\frac{a}{b} = \frac{c}{d} \text{, so } a \times d = b \times c$; a over b and c over d are two equal fractions, and b and d are not 0. Multiply each top by the other bottom. The two products are equal. It works because it is the same as multiplying both sides by b and by d.
Simplifying a fraction: $\frac{a \times k}{b \times k} = \frac{a}{b}$; a over b is a fraction, and k is a number that divides both the top and the bottom. Dividing the top and the bottom by the same number gives an equal fraction with smaller numbers.
Fraction to mixed number: $\frac{a}{b} = q\frac{r}{b}$; a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.

Sort out the question
The equation says two fractions are equal: $\frac{81}{10}$ and $\frac{9}{n}$ . n is an unknown bottom number. You need the value of n that makes them equal. Some choices are mixed numbers, so n may not be a whole number.
Cross-multiply the two fractions
Multiply each top by the other fraction’s bottom. This gets n out of the bottom of the fraction.
$\frac{81}{10} = \frac{9}{n}$
$81 \times n = 10 \times 9$
$81n = 90$
Divide both sides by 81
81n means 81 times n. To undo multiplying by 81, divide both sides by 81.
$n = \frac{90}{81}$
Now simplify. 90 and 81 are both in the 9 times table, so divide the top and bottom by 9.
$\frac{90}{81} = \frac{90 \div 9}{81 \div 9} = \frac{10}{9}$
Write it as a mixed number
10 ÷ 9 is 1 with 1 left over. The leftover stays over 9.
$\frac{10}{9} = 1\frac{1}{9}$
Watch out: keep the leftover. Dropping it gives 1, choice A. Also, ten ninths is not 10 and one ninth. That mixed number, $10\frac{1}{9}$ , is choice D, which is far too big. So the answer is B, $1\frac{1}{9}$ .

Check your answer

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

More questions like this

The original DOE question

Question 68 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 68. Open page 78 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 12 of 57

2023 Test A Math, question 12

Question

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

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

Show the answer

Correct answer: A

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

Let $n = 3$ .

Option A:

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

Option B:

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

Option C:

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

Option D:

$n + 2 = 3 + 2 = 5$

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 69 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 69. Open page 78 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 13 of 57

2023 Test A Math, question 13

Question

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

Show the answer

Correct answer: B

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

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

Question analysis

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

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding factor pairs of a two-digit number is review by grade 7. The question takes two quick steps, listing the pairs and adding each, with nothing to compute beyond small products and sums.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the factor pair 5 and 13 and answers with the factor 5, one of the two integers, instead of their sum.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No pair of positive integers with product 65 adds to 24; reaching it needs a wrong pair such as 3 and 21, whose product is 63, so it mostly catches guesses.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No factor pair of 65 adds to 52, which is 65 − 13 or 4 × 13, and no common working combines them that way, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 70 as printed on page 78 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 78, question 70. Open page 78 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 14 of 57

2023 Test A Math, question 14

Question

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

A−1
B−2
C−3
D−5

Show the answer

Correct answer: D

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

Question analysis

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

NYS standards: 6.NS.7a inequality as relative position
6.NS.6c positioning rational numbers and points
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Ordering negative numbers is grade 6 content, and the question adds only the condition that the integer be odd. It is lighter than typical grade 7 work, though the order of negatives still trips up students who compare their sizes.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

Question 71 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 71. Open page 79 of the handbook (PDF)

Question 15 of 57

2023 Test A Math, question 15

Question

A swim instructor used the box plot below to display the distribution of the ages of students who signed up for swim lessons.

$Box plot titled STUDENTS’ AGES above a number line from 5 to 13 with a tick and a label at every whole number and arrows at both ends. The left whisker starts at 7, the box runs from 8 to 10 with the median line at 9, and the right whisker ends at 12.$

Which statement about the distribution of ages is true?

AThe data contain an outlier.
BApproximately half the students are exactly 9 years old.
CApproximately one-fourth of the students are at least 10 years old.
DThe interquartile range is 3 greater than the range of the data.

Show the answer

Correct answer: C

The data points in the box plot are as follows:

minimum: 7
first quartile: 8
median: 9
third quartile: 10
maximum: 12

Since the value of the third quartile is 10, approximately one-fourth of the students are 10–12 years old.

Question analysis

A swim instructor's box plot shows the ages of students who signed up for lessons. Students decide which of four statements about the distribution is true, using the minimum, quartiles, median and maximum. The statements test different box plot ideas, including outliers, the median, the quartiles and the range, so each needs its own check.

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: At grade 7 New York places box plots, the interquartile range and outliers in grade 7. Here four statements test those ideas one at a time; each check is short, so the item matches typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly understand
What makes it hard: Something you must already know moderate
Every choice must be checked moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 66 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Reading the five values off the plot is quick; the work is knowing what a box plot can and cannot say. It shows where quarters of the data begin and end, not how many students share any one age, and a long whisker is not an outlier unless it passes the 1.5 × IQR cutoff, here 13. The true statement uses the third quartile: about a quarter of the students are 10 or older.

Something you must already know moderate Working it out
The statements rely on box plot facts the question does not give: each section holds about a quarter of the data, the middle line is the median, and outliers lie beyond 1.5 IQRs. (Anderson & Krathwohl, 2001)
Every choice must be checked moderate Choosing the answer
Four statements test four separate ideas, an outlier, the share at one age, a quarter of the data and two spreads, so ruling one out says nothing about the rest. (Embretson & Wetzel, 1987)
Reading a figure, table or graph minor Understanding the question
The five values, 7, 8, 9, 10 and 12, are read off a 5-to-13 scale, where the longer right whisker draws the eye more than the box does. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Read the five key values off the plot, then test each statement: compute the range, the interquartile range and the outlier cutoffs, and recall what share of the data each part holds.

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.
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 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Takes the long right whisker, reaching 12, as a sign of an outlier, but 12 is below the cutoff 10 + 1.5 × 2 = 13.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reads the median line at 9 as the age of half the students, though a median only splits the ages into a lower and an upper half.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Swaps the two measures, taking the box's 2 years as the range and the whiskers' 5 years as the interquartile range, so the IQR seems 3 greater.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Range: $\text{range} = \text{maximum} - \text{minimum}$; The maximum is the largest value, the right end of a box plot. The minimum is the smallest, the left end. The range is how far apart they are. It shows how spread out all the data are.
Interquartile range: $\text{IQR} = Q_3 - Q_1$; IQR is short for interquartile range. Q1, the first quartile, is the left edge of the box. Q3, the third quartile, is the right edge. The IQR is the width of the box. It is the spread of the middle half of the data. The four parts of a box plot each hold about one-fourth of the data.
Outlier test: $\text{below } Q_1 - 1.5 \times \text{IQR} \text{ or above } Q_3 + 1.5 \times \text{IQR}$; An outlier is a value far from the rest of the data. A common rule says a value is an outlier if it lies more than 1.5 IQRs below Q1 or above Q3.

Read the five values off the plot
A box plot shows five numbers. The median is the middle value when the ages are in order.
Part Age
Minimum, the left whisker’s end 7
First quartile, Q1, the box’s left edge 8
Median, the line in the box 9
Third quartile, Q3, the box’s right edge 10
Maximum, the right whisker’s end 12
Check choice A, an outlier
$\text{IQR} = 10 - 8 = 2$
$Q_3 + 1.5 \times 2 = 13 \text{ and } Q_1 - 1.5 \times 2 = 5$
An outlier would be above 13 or below 5. The ages only run from 7 to 12. So there is no outlier. Watch out: a long whisker is not an outlier. It just means the top ages are spread out. Choice A is false.
Check choice B, the median
The median, 9, is the middle age. About half the students are 9 or younger. About half are 9 or older. It does not tell how many are exactly 9. Watch out: the median line does not mean half the students are 9. Choice B is false.
Check choice C, the top fourth
The box plot splits the students into four groups of about the same size. Each group holds about one-fourth of them. The last group runs from Q3, 10, to the maximum, 12. At least 10 means 10 or more. So about one-fourth of the students are at least 10 years old. Choice C is true.
Check choice D, the two spreads
$\text{range} = 12 - 7 = 5$
$\text{IQR} = 2$
The IQR is 3 less than the range, not 3 greater. The box always fits inside the whiskers, so the IQR can never be bigger. Watch out: swapping the two measures gives choice D. So the answer is C: about one-fourth of the students are at least 10 years old.

Part	Age
Minimum, the left whisker’s end	7
First quartile, Q1, the box’s left edge	8
Median, the line in the box	9
Third quartile, Q3, the box’s right edge	10
Maximum, the right whisker’s end	12

Check your answer

Picture 20 students. About 5 would be in each fourth of the plot. The top 5 would be ages 10 to 12. That is about one-fourth, just as choice C says. Each other choice fails a fact you can read straight off the plot.

More questions like this

The original DOE question

Question 72 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 72. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 16 of 57

2023 Test A Math, question 16

Question

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

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

Show the answer

Correct answer: C

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

Question analysis

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

NYS standards: 6.NS.6c positioning rational numbers and points
6.NS.7a inequality as relative position
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Changing an improper fraction to a mixed number and placing it between whole numbers is review well before grade 7. This takes one division with a small divisor and nothing more.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 13 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No division of 29 by 7 lands below 3, since 7 × 3 = 21 is far short of 29; only misreading 29/7 as 2.9 gets there, so it mostly catches guesses.
B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Finds the whole-number part, 4, but takes it as the upper end of the pair, as if 29/7 fell just short of 4 instead of just past it.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Nothing puts 29/7 above 5, since 7 × 5 = 35 is more than 29; even rounding 4 1/7 up gives 5 as the upper end, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction to mixed number: $\frac{a}{b} = q\frac{r}{b}$; a over b is a fraction whose top is bigger than its bottom. q is the quotient, how many whole times b fits into a. r is the remainder, what is left over. The leftover stays over the same bottom number.

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

Check your answer

Multiply each end by 7. $4 \times 7 = 28$ and $5 \times 7 = 35$ . The top of the fraction, 29, is between 28 and 35. So $\frac{29}{7}$ is between 4 and 5. As a decimal it is about 4.14.

More questions like this

The original DOE question

Question 73 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 79, question 73. Open page 79 of the handbook (PDF)

Question 17 of 57

2023 Test A Math, question 17

Question

A customer wants to buy a pair of hiking boots.

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

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

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

Show the answer

Correct answer: C

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

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

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

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

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

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

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

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 18 of 57

2023 Test A Math, question 18

Question

Sheila is saving money for her summer vacation. She starts the summer with a balance of $90.00 and plans to save 15% of her earnings each week. She earns the same amount each week. After 12 weeks, Sheila has saved a total of $472.59. How much money does Sheila earn each week?

A$212.55
B$262.55
C$302.55
D$312.55

Show the answer

Correct answer: A

Let x be the amount that Sheila needs to earn each week.

$\$90.00 + 0.15 \times 12x = \$472.59$

First, subtract 90 from both sides:

$0.15 \times 12x = \$382.59$

$1.8x = \$382.59$

Last, divide both sides by 1.8:

$x = \$212.55$

Question analysis

Sheila starts the summer with $90 and saves 15% of what she earns each week, earning the same amount every week, and after 12 weeks she has $472.59. Students find her weekly earnings. They must write an equation that separates the starting balance from the savings and then work with a percent and long decimals.

NYS standards: 7.EE.4a two-step equations
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation of the form px + q = r from a word problem is core grade 7 work. Here the coefficient combines a percent with 12 weeks, and the decimals are long, which adds steps but no new idea.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The story has three layers, a starting balance, a weekly saving rate and a 12-week total, and the unknown sits under the percent. Students lose the point by forgetting the $90 was already there, or by moving it the wrong way. Testing a choice is a dependable alternative: 15% of $212.55 is about $31.88, and 12 of those plus $90 gives $472.59.

Turning words into math moderate Understanding the question
The total of $472.59 includes the $90.00 she started with, and only 15% of each week's pay is saved, so the equation is 90 + 12 × 0.15x = 472.59. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic moderate Working it out
Dividing $382.59 by 1.8 by hand means shifting the decimal point to get 3,825.9 ÷ 18, which comes out to exactly $212.55 only if every digit is placed correctly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
Subtract the $90, then divide by 12 weeks and by 0.15, or by their product 1.8; doing the division before the subtraction gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Subtract the starting $90 from the final total, then divide by 12 weeks and by 0.15, or test each choice by taking 15% of it, multiplying by 12 and adding $90.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
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 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Overlooks the $90 she started with, treating the whole $472.59 as savings from earnings, and divides it by 1.8 to get $262.55.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No setup of the savings gives $302.55, the right weekly pay plus the $90 starting balance, a sum with no meaning here, so it mostly catches guesses.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a slip Expected pull: moderate
Adds the $90 to $472.59 instead of subtracting it, so the savings from earnings look like $562.59, and dividing by 1.8 gives $312.55.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{p}{100} \times \text{whole}$; p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Savings over time: $\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}$; The start is the money she had at the beginning. Weekly savings is what she adds each week. Weeks is how many weeks pass. Each week adds the same amount, so multiply. Then add what she started with.

Sort out the question
Sheila starts with $90.00. Each week she earns the same amount and saves 15% of it. After 12 weeks she has $472.59 in all. That total includes the $90 she started with. You need how much she earns each week, not how much she saves.
Write an equation
Call her weekly earnings x. She saves 15% of x each week. That is $0.15 \times x$ .
$\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}$
$472.59 = 90 + 0.15x \times 12$
Take away the starting $90
Only the money added during the 12 weeks came from her pay. Subtract 90 from both sides.
$0.15x \times 12 = 472.59 - 90 = 382.59$
She saved $382.59 from her pay. Watch out: subtract the $90, don’t add it. Adding leads to $312.55, choice D. Forgetting the $90 leads to $262.55, choice B.
Combine the numbers next to x
$0.15 \times 12 = 1.8$
$1.8x = 382.59$
Each week she saves 0.15 of one week’s pay. Over 12 weeks, that adds up to 1.8 times one week’s pay.
Divide by 1.8
Divide both sides by 1.8. First move the decimal point one place right in both numbers. Then you divide by the whole number 18.
$x = 382.59 \div 1.8 = 3{,}825.9 \div 18 = 212.55$
Sheila earns $212.55 each week. So the answer is A, $212.55.

Check your answer

Work forward. 12 weeks of saving 15% is 1.8 weeks of pay: $1.8 \times 212.55 = 382.59$ . Add the $90 she started with: $382.59 + 90 = 472.59$ . That is the total in the question.

More questions like this

The original DOE question

Question 75 as printed on page 80 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 75. Open page 80 of the handbook (PDF)

Question 19 of 57

2023 Test A Math, question 19

Question

$x:35 = 20:28$

For what value of x is the proportion shown above true?

Show the answer

Correct answer: B

Write the ratios in fraction form and solve for x:

$\frac{x}{35} = \frac{20}{28}$	Cross multiply.
$28x = 700$	Apply the multiplicative inverse property; divide both sides by 28.
$x = \frac{700}{28} = 25$

Question analysis

This question gives the proportion x:35 = 20:28 and asks for the value of x that makes it true. Students use equivalent ratios or cross-multiplication. The trap is reasoning by adding or subtracting the difference between terms rather than multiplying, which gives numbers that look reasonable but break the ratio.

NYS standards: 6.RP.3 ratio and rate reasoning
6.EE.7 one-step equations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a missing term in a proportion is grade 6 ratio reasoning, and here the given ratio simplifies to small numbers. It is easier than typical grade 7 proportional reasoning, which usually involves fractions or a real context.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

This is a single proportion with whole numbers, and the only real trap is reasoning by differences instead of by a common factor. Since 35 is 7 more than 28, adding or subtracting 7 to 20 is tempting, but equal ratios scale; they do not shift. Reducing 20:28 to 5:7 shows at once that the scale factor to 35 is 5.

Error-prone arithmetic minor Working it out
Cross-multiplying gives 28x = 700, and 700 ÷ 28 is not a table fact; simplifying 20:28 to 5:7 first turns it into 5 × 5. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
27 is only 2 more than 25, so a rough sense that x is a little more than 20 fits both; only the exact scale factor, 35/28 = 5/4, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Simplify 20:28 to lowest terms first, then scale it up so its second term becomes 35, or cross-multiply and divide by 28.

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.
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 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Reasons with differences: since 35 is 7 more than 28, adds 7 to 20, keeping the gap between the terms instead of their ratio.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Scales 20 by 28/35 instead of 35/28, so 20 × 28 ÷ 35 = 16, which turns the ratio of the second terms upside down.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Reasons with differences and also goes the wrong way, subtracting 7 from 20, though x must exceed 20 because 35 exceeds 28.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Ratio as a fraction: $a:b = \frac{a}{b}$; a:b is a ratio, read a to b. It compares two amounts by division. You can write it as the fraction a over b. Two equal ratios make a proportion.
Equivalent fractions: $\frac{a}{b} = \frac{a \times k}{b \times k}$; a over b is a fraction, and k is any number except 0. Multiplying or dividing the top and the bottom by the same number gives an equal fraction.

Sort out the question
$x:35 = 20:28$ is a proportion. A proportion says two ratios are equal. A ratio compares two numbers, like 20 to 28. You need the value of x that makes the two ratios equal.
Write the ratios as fractions
Write each ratio as a fraction. The first number of the ratio goes on top. The second goes on the bottom.
$\frac{x}{35} = \frac{20}{28}$
Simplify the known fraction
20 and 28 both divide by 4. Smaller numbers are easier to work with.
$\frac{20}{28} = \frac{20 \div 4}{28 \div 4} = \frac{5}{7}$
So now $\frac{x}{35} = \frac{5}{7}$ .
Scale up so the bottom is 35
7 times 5 is 35. To keep the fraction equal, multiply the top by 5 too.
$\frac{5}{7} = \frac{5 \times 5}{7 \times 5} = \frac{25}{35}$
So x = 25. Watch out: a ratio is about multiplying, not adding. 35 is 7 more than 28, but adding 7 to 20 gives 27, choice A. Also, x must be more than 20, because 35 is more than 28. So 16 and 13 are too small. So the answer is B, 25.

Check your answer

Cross-multiply, as the test’s own explanation does. $28 \times 25 = 700$ and $35 \times 20 = 700$ . The two products are equal, so the ratios 25:35 and 20:28 are equal.

More questions like this

The original DOE question

Question 76 as printed on page 80 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 76. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 20 of 57

2023 Test A Math, question 20

Question

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

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

Show the answer

Correct answer: C

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

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

Question analysis

This arithmetic question asks for the value of a fraction whose numerator is the sum of the squares of 3, −8 and 2, and whose denominator is the square of 3 − 8 + 2. Students apply the order of operations with negative numbers. Squaring a negative number and adding inside parentheses before squaring are the points to watch.

NYS standards: 7.NS.2a rational number multiplication
6.EE.2c evaluating expressions and order of operations
7.NS.1c rational number subtraction
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Squaring negative numbers and simplifying with the order of operations is standard grade 7 practice. Here the fraction bar groups a numerator and a denominator that each involve a negative sign, but the work stays short.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 16+, 3 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The expression is built so that signs decide everything: the top's 3², (−8)² and 2² are all positive, and so is the square of the bottom's −3. Students who keep a negative after squaring, or who cancel the 9 on top against the 9 below, reach the other choices. The result, 77/9, does not reduce, which can make careful students doubt a correct answer.

Error-prone arithmetic moderate Working it out
Two negatives sit inside squares: (−8)² must be +64, and the denominator's 3 − 8 + 2 = −3 must be squared to +9, so every term ends up positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps minor Working it out
The numerator needs three squares and a sum, the denominator a sum and then a square, and the fraction bar groups each side, so neither part can be split or cancelled. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Simplify the top and the bottom separately: square each number in the numerator, add inside the parentheses before squaring in the denominator, then write the fraction in lowest terms.

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Square of a negative: $(-a)^{2} = (-a) \times (-a) = a^{2}$; a is any number. To square means to multiply a number by itself. A negative times a negative is positive. So the square of a negative number is always positive.
The fraction bar groups: $\frac{\text{top}}{\text{bottom}} = (\text{top}) \div (\text{bottom})$; A fraction bar means divide. It also groups: work out the whole top and the whole bottom first, then divide. You can only cancel a number that multiplies the whole top, not one part of a sum.

Sort out the expression
The top, the numerator, is three squares added together. The bottom, the denominator, is a sum in parentheses that is then squared. To square means to multiply a number by itself. Work out the top and the bottom on their own. Then divide.
Work out the top
Square each number first. Then add.
$3^{2} = 9$
$(-8)^{2} = (-8) \times (-8) = 64$
$2^{2} = 4$
$9 + 64 + 4 = 77$
Watch out: (−8)² is +64, not −64. The minus sign is inside the parentheses, so it is squared too. Using −64 makes the top −51 and gives $-\frac{17}{3}$ , choice B.
Work out the bottom
Work inside the parentheses first. Add and subtract before you square.
$3 - 8 + 2 = -5 + 2 = -3$
$(-3)^{2} = (-3) \times (-3) = 9$
The bottom is 9.
Divide and simplify
$\frac{77}{9}$
77 is 7 × 11 and 9 is 3 × 3. They share no factor, so the fraction is in lowest terms. Watch out: don’t cancel the 9 on top with the 9 below. That 9 is only one part of a sum. Canceling it leaves 64 + 4 = 68, choice D. So the answer is C, $\frac{77}{9}$ .

Check your answer

Every square is positive or zero. So the top and the bottom are both positive, and the answer must be positive. That rules out −60 and $-\frac{17}{3}$ . Also, $\frac{77}{9}$ is about 8.6, which is far from 68.

More questions like this

The original DOE question

Question 77 as printed on page 80 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 77. Open page 80 of the handbook (PDF)

Question 21 of 57

2023 Test A Math, question 21

Question

A soccer coach purchased 15 pairs of cleats for team members and spent a total of $805.95, including tax. If the tax rate is 8%, what is the price of each pair of cleats before tax?

A$49.43
B$49.75
C$53.73
D$58.03

Show the answer

Correct answer: B

Since the total cost of $805.95 includes an 8% tax, this means that the cost of 15 pairs of cleats before tax was multiplied by 1.08 to find the total cost of the cleats including the tax. So, first, use division to find the cost of the 15 pairs of cleats before tax:

$\$805.95 \div 1.08 = \$746.25$

Then, divide that quotient by 15 to find the price of 1 pair of cleats:

$\$746.25 \div 15 = \$49.75$

Question analysis

A soccer coach pays $805.95, tax included, for 15 pairs of cleats, and the tax rate is 8%. Students find the price of one pair before tax. The total already includes the tax, so students must work backward from a percent increase, and a common shortcut lands very close to the right price.

NYS standards: 7.RP.3 multistep ratio and percent problems
6.RP.3b unit rate problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Typical grade 7 tax problems add tax to a price. This one works backward from a total that already includes the tax, a reverse percent step many students find unfamiliar, and then shares the cost among 15 pairs.
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
A very close wrong answer moderate
Error-prone arithmetic moderate
Solution: 2 steps
Reading level: Question and choices: about grade 5, 33 words.
Corrected from the DOE handbook: A repeated word in the question was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Removing a tax is harder than adding one, because the 8% was taken of an amount the problem never states. Taking 8% off the $805.95 total feels natural and lands within 32 cents of the answer, which is why this question is hard. A quick check exposes it: 15 pairs at $49.75 cost $746.25, and adding 8% of that gives exactly $805.95.

Turning words into math moderate Understanding the question
The phrase "spent a total of $805.95, including tax" makes $805.95 equal to 108% of the pre-tax cost, so the tax is removed by dividing by 1.08, not by taking 8% off. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
$49.43 and $49.75 are 32 cents apart, and taking 8% off the total lands on the lower one; no estimate separates them, only dividing by 1.08. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic moderate Working it out
805.95 ÷ 1.08 = 746.25 and 746.25 ÷ 15 = 49.75 are long decimal divisions by hand; dividing by 15 first, to 53.73, and then by 1.08 is no easier. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Treat the total as 108% of the pre-tax cost: divide $805.95 by 1.08, then divide by 15 pairs, and check by adding 8% back.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
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 Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals a misconception Expected pull: strong
Takes 8% of the $805.95 total and subtracts it, 805.95 × 0.92 ÷ 15 ≈ $49.43, though the tax was 8% of the smaller pre-tax cost.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Divides the $805.95 total by 15 pairs without removing the tax, which gives the price of a pair with tax rather than before it.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Divides the total by 15 and then multiplies by 1.08 instead of dividing, adding the tax a second time: 53.73 × 1.08 ≈ $58.03.

What was corrected from the DOE handbook, and why

Stem: The handbook’s question reads “what is the the price of each pair of cleats before tax?”, with “the” printed twice (p. 80, question 78). The site now reads “what is the price of each pair of cleats before tax?”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Price with tax: $\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)$; The price is the cost before tax. t is the tax rate, a percent. Tax is extra money added on top. You pay the whole price, 100%, plus t percent more. So for 8% tax you multiply by 1.08.
Undoing a multiplication: $\text{if } a \times b = c \text{, then } a = c \div b$; a is the unknown, b is the number it was multiplied by, and c is the result. Dividing by b undoes multiplying by b. Use it to work backward to the amount before tax.

Sort out the question
The coach bought 15 pairs of cleats, which are sports shoes. He paid $805.95 in all, and that includes 8% tax. Tax is extra money added to the price. You need the price of one pair before tax.
Write the total with tax
The tax is 8% of the cost before tax. So the coach paid 100% + 8% = 108% of that cost. 108% is 1.08. Call the pre-tax cost of all 15 pairs c.
$\text{total} = \text{price} \times \left(1 + \frac{t}{100}\right)$
$c \times 1.08 = 805.95$
Divide to remove the tax
To undo multiplying by 1.08, divide by 1.08. Move both decimal points two places right first.
$c = 805.95 \div 1.08 = 80{,}595 \div 108 = 746.25$
All 15 pairs cost $746.25 before tax. Watch out: don’t take 8% of $805.95 and subtract it. The tax was 8% of the smaller pre-tax cost. That way leads to $49.43, choice A.
Find the price of one pair
All 15 pairs cost the same, so share the cost equally.
$746.25 \div 15 = 49.75$
Watch out: dividing $805.95 by 15 without removing the tax gives $53.73, choice C. That price still includes the tax. So the answer is B, $49.75.

Check your answer

Work forward. 15 pairs at $49.75 cost $15 \times 49.75 = 746.25$ dollars. The tax is $0.08 \times 746.25 = 59.70$ dollars. Then $746.25 + 59.70 = 805.95$ , the total in the question.

More questions like this

The original DOE question

Question 78 as printed on page 80 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 80, question 78. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 22 of 57

2023 Test A Math, question 22

Question

The numbers m, n, p, and q are different, and each is equal to one of the numbers 1, 2, 3, 6, or 12. If $2m = 6q = \frac{1}{2}n = p$ , what is the value of p?

Show the answer

Correct answer: C

Substitute each answer option for p to see which value will work:

Option A:

If $p = 2$ , then $\frac{1}{2}n = 2$ , so $n = 4$ . This cannot be the answer, because n cannot equal 4.

Option B:

If $p = 3$ , then $6q = 3$ , so $q = 0.5$ . This cannot be the answer, because q cannot equal 0.5.

Option C:

If $p = 6$ , then $2m = 6$ , so $m = 3$ ; $6q = 6$ , so $q = 1$ ; and $\frac{1}{2}n = 6$ , so $n = 12$ . This is the correct answer.

Option D:

If $p = 12$ , then $\frac{1}{2}n = 12$ , so $n = 24$ . This cannot be the answer, because n cannot equal 24.

Question analysis

Four letters, m, n, p and q, stand for different numbers from the list 1, 2, 3, 6 and 12, and the expressions 2m, 6q, half of n, and p are all equal. Students find the value of p. The puzzle-like setup has no single equation to solve, so testing values against every condition is the reliable route.

NYS standards: 6.EE.5 solutions and substitution
6.EE.7 one-step equations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 The equations involve only doubling, halving and multiplying by 6, all grade 6 skills. What goes beyond grade 7 practice is the chain of four equal expressions, with every letter limited to a short list of different numbers.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Solution: 3 steps
Reading level: Question and choices: about grade 5, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The chain looks forbidding, but it gives each letter as a simple multiple of p: m is half of p, q a sixth of it, and n twice it. Seen that way, q = p/6 forces p to be 6 or 12, and n = 2p rules out 12, so only one value fits. Students who check one or two of the letters, but not all three, can accept 2 or 12.

Variables instead of numbers moderate Understanding the question
Four letters are tied by one chain of equalities, 2m = 6q = ½n = p, and the only numbers are the five allowed values, so each letter has to be written in terms of p. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
Each choice for p must be tested against three letters, m = p/2, q = p/6 and n = 2p, and all of them must land on different numbers from 1, 2, 3, 6 and 12. (Embretson & Wetzel, 1987)

Strategy: Test each choice as p: find m, q and n from it, and keep the value for which all three are different numbers from the list 1, 2, 3, 6 and 12.

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Checks only m, finding that p = 2 gives m = 1 on the list, and misses that q would be 1/3 and n would be 4.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds the solution, m = 3, q = 1, n = 12 and p = 6, then answers with m's value, 3, instead of the value of p.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Tests p = 12 against 2m and 6q, getting m = 6 and q = 2, both on the list, and stops before seeing that ½n = 12 needs n = 24.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing a multiplication: $\text{if } a \times m = p \text{, then } m = p \div a$; a is a number that multiplies m, and p is the result. Dividing by a undoes multiplying by a. Use it to write each letter using p.

Sort out the question
m, n, p and q are four different numbers. Each one is 1, 2, 3, 6 or 12. The chain $2m = 6q = \frac{1}{2}n = p$ says that four amounts are all equal to p. You need the value of p.
Write each letter using p
Each part of the chain equals p. Undo each multiplication.
$2m = p \text{, so } m = p \div 2$
$6q = p \text{, so } q = p \div 6$
$\frac{1}{2}n = p \text{, so } n = 2 \times p$
Half of n is p, so n is twice p.
Test each choice for p
p m q n All on the list?
2 1 $\frac{1}{3}$ 4 No
3 1.5 0.5 6 No
6 3 1 12 Yes
12 6 2 24 No
Only p = 6 puts every letter on the list.
Check that the letters are different
With p = 6, you get m = 3, q = 1 and n = 12. All four numbers are on the list, and no two are the same. Watch out: p = 12 makes m = 6 and q = 2, which look fine. But n = 24 is not on the list, so choice D fails. Also, the question asks for p, not m. Picking 3 is choice B. So the answer is C, 6.

p	m	q	n	All on the list?
2	1	$\frac{1}{3}$	4	No
3	1.5	0.5	6	No
6	3	1	12	Yes
12	6	2	24	No

Check your answer

Put the values into the chain: $2 \times 3 = 6$ , $6 \times 1 = 6$ , $\frac{1}{2} \times 12 = 6$ , and p = 6. All four parts equal 6. The numbers 3, 1, 12 and 6 are all different and all on the list.

More questions like this

The original DOE question

Question 79 as printed on page 81 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 79. Open page 81 of the handbook (PDF)

Question 23 of 57

2023 Test A Math, question 23

Question

$\{0.1, 0.01, 0.001, 0.0001, 0.00001\}$

If a person chooses a number at random from the set above, what is the probability that the number is less than 0.005?

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

Show the answer

Correct answer: C

Three of the five numbers listed (0.001, 0.0001, and 0.00001) are less than 0.005. So, the probability of a person selecting a number less than 0.005 is $\frac{3}{5}$ .

Question analysis

A set holds five decimals, from one tenth down to one hundred-thousandth, and one number is chosen at random. Students find the probability that the number is less than 0.005. Comparing decimals with many zeros is the real task here, along with putting the count over the right total.

NYS standards: 6.SP.8a uniform probability models
6.NS.7a inequality as relative position
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Probability with equally likely outcomes is grade 6 content in New York, and this needs one comparison per number and one fraction. The only real demand is comparing decimals with many zeros correctly.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each number in the set is a tenth of the one before, so only the three smallest fall below 0.005, which lies between 0.01 and 0.001. Students lose the point either by counting the numbers above 0.005 instead of below it or by writing the count over the wrong total. Lining the decimals up with the same number of places makes every comparison easy.

Error-prone arithmetic moderate Working it out
0.005 has to be compared with decimals from 0.1 down to 0.00001; written to five places, 0.00500 against 0.01000 and 0.00100, the order becomes a comparison of whole numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The probability is the count of numbers below 0.005 over all 5 numbers in the set; the stem leaves that part-to-whole rule to the student. (Anderson & Krathwohl, 2001)

Strategy: Write each number with five decimal places, compare it with 0.00500, count those that are smaller, and divide by the five numbers in the set.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Gives 1/5, the chance of picking any one particular number from the set, without counting how many of the five are below 0.005.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Counts the 2 numbers greater than 0.005, 0.1 and 0.01, instead of those less than it, finding the chance of the opposite event.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Writes a part-to-part ratio, the 2 numbers above 0.005 over the 3 below it, instead of dividing by all 5 numbers in the set.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{wanted results}}{\text{possible results}}$; P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.
Adding zeros to a decimal: $0.1 = 0.10 = 0.100$; Zeros added to the right end of a decimal don’t change its value. One tenth is the same as ten hundredths. Giving every number the same number of decimal places lets you compare them like whole numbers.

Sort out the question
The set has five numbers. A person picks one at random, so each number has the same chance. You need the probability, or chance, that the number picked is less than 0.005.
Give every number five decimal places
Decimals are easier to compare with the same number of places. Add zeros to the end of each.
Number With five places
0.1 0.10000
0.01 0.01000
0.001 0.00100
0.0001 0.00010
0.00001 0.00001
0.005 0.00500
Count the numbers below 0.005
Compare each number with 0.00500 like whole numbers: 10000, 1000, 100, 10 and 1 against 500. Two numbers are bigger: 0.1 and 0.01. Three are smaller: 0.001, 0.0001 and 0.00001. Watch out: 0.01 has fewer digits than 0.005, but it is bigger. Counting the 2 bigger numbers gives $\frac{2}{5}$ , choice B.
Write the probability
$P = \frac{\text{wanted results}}{\text{possible results}} = \frac{3}{5}$
Watch out: divide by all 5 numbers in the set. Comparing the 2 bigger numbers with the 3 smaller ones gives $\frac{2}{3}$ , choice D. So the answer is C, $\frac{3}{5}$ .

Number	With five places
0.1	0.10000
0.01	0.01000
0.001	0.00100
0.0001	0.00010
0.00001	0.00001
0.005	0.00500

Check your answer

The numbers below 0.005 and the numbers above it make up the whole set: 3 + 2 = 5. So the two chances, $\frac{3}{5}$ and $\frac{2}{5}$ , add up to 1, as they should.

More questions like this

The original DOE question

Question 80 as printed on page 81 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 80. Open page 81 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 24 of 57

2023 Test A Math, question 24

Question

A student uses Spinner R and Spinner T to generate a list of two-digit numbers.

$Two spinners, each with an arrow pointing out from its center. Spinner R is a circle divided into 4 equal sections, labeled 1 and 2 across the top and 3 and 4 across the bottom. Spinner T is a circle divided into 6 equal sections, labeled, clockwise from the top left, 1, 1, 2, 3, 4 and 4.$

Spinner R determines the digit in the tens place, and Spinner T determines the digit in the ones place. What is the probability that the two-digit number determined by spinning each spinner one time is a prime number?

A $\frac{3}{8}$
B $\frac{5}{12}$
C $\frac{3}{5}$
D $\frac{5}{8}$

Show the answer

Correct answer: A

Spinner R has 4 equal sections and Spinner T has 6 equal sections, so the number of possible two-digit numbers that can be created is found by multiplying 4 and 6, which is 24. List the possible outcomes:

11	11	12	13	14	14
21	21	22	23	24	24
31	31	32	33	34	34
41	41	42	43	44	44

There are 9 prime numbers out of 24 possible outcomes: 11, 11, 13, 23, 31, 31, 41, 41, and 43.

So the probability that the two-digit number created is a prime number is $\frac{9}{24}$ , which is equivalent to $\frac{3}{8}$ .

Question analysis

Spinner R, with four equal sections, gives the tens digit and Spinner T, with six equal sections and some labels repeated, gives the ones digit of a two-digit number. Students find the probability that the number is prime. Building the full table of outcomes, counting repeated sections separately, and checking whether each number is prime all take care.

NYS standards: 7.SP.8b sample spaces for compound events
7.SP.8a compound event as fraction of outcomes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Listing a sample space for two spinners is grade 7 work, usually with distinct labels and a simple event. Here one spinner repeats two numbers, the space has 24 outcomes, and each two-digit result must be tested for being prime.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Much to keep track of moderate
Something you must already know minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 57 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question combines a compound sample space with a test for primes, and its figure adds a twist: two labels repeat on Spinner T. Counting the 24 section pairs rather than the 16 different numbers is the careful route, though here both happen to give the same fraction. The points are lost in the prime test, where 33 is easy to count as prime, and in writing the count as a share of all 24 outcomes.

Reading a figure, table or graph moderate Understanding the question
Spinner T has six equal sections but only four labels, with 1 and 4 each appearing twice, so the 24 equally likely outcomes include repeats such as 11 and 11. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Much to keep track of moderate Working it out
Twenty-four two-digit numbers have to be built and each tested for primality while the repeated sections are kept track of; a 4-by-6 table is the reliable way to hold them. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
Which two-digit numbers are prime is not given; students must test 21, 33, 43 and the rest themselves, and 21 = 3 × 7 and 33 = 3 × 11 look prime at a glance. (Anderson & Krathwohl, 2001)

Strategy: Make a 4-by-6 table with Spinner R's digits as tens and Spinner T's six sections as ones, mark each prime, and count every section, including repeated labels.

Traps

An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Counts 33 as prime, missing that 33 = 3 × 11, so 10 of the 24 outcomes look prime and 10/24 reduces to 5/12.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Finds the 9 prime outcomes but compares them with the 15 that are not prime, a part-to-part ratio, 9/15 = 3/5, instead of dividing by all 24.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Counts the 15 outcomes that are not prime and gives 15/24 = 5/8, the probability of the opposite event.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting outcomes: $\text{outcomes} = a \times b$; a is the number of equal sections on the first spinner, and b is the number on the second. Each section of the first can go with each section of the second. So the number of pairs is a times b.
Probability: $P = \frac{\text{wanted results}}{\text{possible results}}$; P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.

Sort out the question
Spinner R gives the tens digit. Spinner T gives the ones digit. Together they make a two-digit number, like 23. You need the chance that the number is prime. A prime number has exactly two factors, 1 and itself. For example, 13 is prime, but 21 = 3 × 7 is not.
Count all the outcomes
Spinner R has 4 equal sections. Spinner T has 6 equal sections. Look closely: T has two 1s and two 4s. Each section counts as its own outcome.
$\text{outcomes} = 4 \times 6 = 24$
There are 24 outcomes, all equally likely.
Make a table of the numbers
Each row is a tens digit from R. Each column is a section of T. Prime numbers are in bold.
R 1 1 2 3 4 4
1 11 11 12 13 14 14
2 21 21 22 23 24 24
3 31 31 32 33 34 34
4 41 41 42 43 44 44
Count the prime numbers
Numbers ending in 2 or 4 are even, so 2 divides them. They are not prime. Check the rest. 11, 13, 23, 31, 41 and 43 are prime. But 21 = 3 × 7 and 33 = 3 × 11 are not. Count every section, so 11, 31 and 41 each count twice. That makes 3 + 1 + 2 + 3 = 9 primes. Watch out: counting 33 as prime gives $\frac{10}{24} = \frac{5}{12}$ , choice B.
Find the probability
$P = \frac{9}{24} = \frac{9 \div 3}{24 \div 3} = \frac{3}{8}$
Watch out: divide by all 24 outcomes. Comparing the 9 primes with the 15 others gives $\frac{9}{15} = \frac{3}{5}$ , choice C. So the answer is A, $\frac{3}{8}$ .

R	1	1	2	3	4	4
1	11	11	12	13	14	14
2	21	21	22	23	24	24
3	31	31	32	33	34	34
4	41	41	42	43	44	44

Check your answer

There are $24 - 9 = 15$ numbers that are not prime. Their chance is $\frac{15}{24} = \frac{5}{8}$ . And $\frac{3}{8} + \frac{5}{8} = 1$ , as it should be. Notice that $\frac{5}{8}$ , choice D, is the chance of not prime.

More questions like this

The original DOE question

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

Question 25 of 57

2023 Test A Math, question 25

Question

$Line h is horizontal, with arrows at both ends. Line j rises from the lower left to the upper right, and line k rises steeply from the bottom, crossing line h to the right of line j; lines j and k meet at point Q above line h. Where j crosses h, the angle above h and to the left of j is marked 135°. Where k crosses h, the angle below h and to the right of k is marked 100°. At Q, the angle between the parts of lines k and j above Q is marked x°.$

Lines j and k intersect at point Q, and h is a straight line. What is the value of x?

Show the answer

Correct answer: C

First, find the measures of the three angles inside the triangle. The lower left angle is supplementary to the one marked 135°, so the measure of that angle is $180 - 135 = 45$ . The angle on top is a vertical angle with the one marked x°, so the measure of that angle is also x°. The angle on the lower right is a vertical angle with the one marked 100°, so the measure of that angle is also 100°. Since the measures of the three interior angles of a triangle sum to 180°, set up an equation and solve for x:

$45 + x + 100 = 180$

$145 + x = 180$

$x = 35$

Question analysis

Lines j and k cross at point Q above a straight line h, forming a triangle, with a 135° angle marked where j meets h and a 100° angle where k meets h. Students find the angle x at Q. Each marked angle lies outside the triangle, so it has to be turned into an inside angle before the triangle's angle sum can be used.

NYS standards: 8.G.5 angle sums, transversals, angle-angle similarity
7.G.5 angle relationships and equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students use supplementary and vertical angles to find unknown angles. This figure needs both, applied at three different vertices, plus the angle sum of a triangle, which New York teaches in grade 8.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph major
Combines ideas moderate
Something you must already know minor
Solution: 4 steps
Reading level: Question and choices: about grade 2, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every angle the solution needs has to be carried into the triangle from outside, and the two marked angles need different treatment. The 135° is supplementary to the triangle's angle, but the 100° is vertical to it, so subtracting both from 180° out of habit builds the wrong triangle. Checking which side of each line a marked angle sits on, not just how near the triangle it is, separates the two cases.

Reading a figure, table or graph major Understanding the question
None of the three marked angles is inside the triangle formed by h, j and k: the 135° is outside on the left, the 100° is below line h, and x is above Q. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Combines ideas moderate Working it out
Two different angle facts are needed: the 135° gives a supplementary interior angle of 45°, while the 100° and x carry into the triangle unchanged as vertical angles. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know minor Working it out
Supplementary angles, vertical angles and the triangle's 180° sum are all assumed, and the question names none of them. (Anderson & Krathwohl, 2001)

Strategy: Mark the three angles of the triangle formed by lines h, j and k: one is supplementary to 135°, one is vertical to 100°, and the one at Q is vertical to x°.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: strong
Takes the triangle's angle at line k as 180 − 100 = 80, the supplement of the marked angle instead of its vertical angle, so x = 180 − 45 − 80 = 55.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the triangle's angle where line j meets h, 180 − 135 = 45, and gives it instead of going on to the angle at Q.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No angle fact in the figure gives 30, and the angle at Q is drawn close to its true size, so this choice mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line: $a + b = 180$; a and b are two angles side by side on a straight line. Together they make a half turn, a straight angle. So they add up to 180°. Angles like this are called supplementary.
Vertical angles: $\text{angle } a = \text{angle } b$; When two lines cross, they make an X. The angles straight across from each other in the X are called vertical angles. a and b are one such pair. They are always equal, because each one plus the angle between them makes a straight line.
Angles of a triangle: $A + B + C = 180$; A, B and C are the three angles inside any triangle, in degrees. They always add up to 180°. Use it when you know two of the angles and need the third.

Sort out the figure
Lines h, j and k make a triangle. Its corners are where j crosses h, where k crosses h, and point Q. The figure marks two angles outside the triangle, 135° and 100°. You need x, the angle marked above Q.
Find the angle where j crosses h
The 135° angle is above h and to the left of j. The triangle’s angle is right next to it, on the other side of j. Together they make the straight line h. Call the triangle’s angle a.
$a + b = 180$
$135 + a = 180$
$a = 180 - 135 = 45$
So that corner of the triangle is 45°. Watch out: this is only one corner, not x. Stopping here gives 45, choice B.
Find the angle where k crosses h
The 100° angle is below h and to the right of k. The triangle’s angle is above h and to the left of k. Lines h and k make an X there, and these two angles are straight across from each other. They are vertical angles, so the triangle’s angle is 100°. Watch out: it is not 180 − 100 = 80. That angle is beside the 100°, not across from it. Using 80 gives 55, choice A.
Find the angle at Q
The x° angle is above Q, between the top parts of lines j and k. The triangle’s angle at Q is between the parts of j and k below Q. These two angles are straight across from each other. So they are vertical angles, and the triangle’s angle at Q is also x°.
Use the triangle’s angle sum
$A + B + C = 180$
$45 + 100 + x = 180$
$145 + x = 180$
$x = 180 - 145 = 35$
The three angles of the triangle must add up to 180°. Only x is unknown. So the answer is C, 35.

Check your answer

Add the triangle’s three angles: $45 + 100 + 35 = 180$ . Also, the angle at Q looks narrow in the figure, much less than a right angle of 90°. 35° fits that.

More questions like this

The original DOE question

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

Question 26 of 57

2023 Test A Math, question 26

Question

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

A $\frac{1}{x + 1} > \frac{1}{x + 2}$
B $\frac{1}{x + 1} < \frac{1}{x + 2}$
C $\frac{1}{x + 1} - \frac{1}{x + 2} > 1$
D $\frac{1}{x + 1} - \frac{1}{x + 2} > \frac{1}{x}$

Show the answer

Correct answer: A

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

Option A:

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

Option B:

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

Option C:

$\frac{1}{3} - \frac{1}{4} > 1$ , or $\frac{1}{12} > 1$ . This is false.

Option D:

$\frac{1}{3} - \frac{1}{4} > \frac{1}{2}$ , or $\frac{1}{12} > \frac{1}{2}$ . This is false.

Option A is the only true inequality.

Question analysis

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

NYS standards: 6.NS.7a inequality as relative position
6.EE.5 solutions and substitution
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Comparing fractions is familiar, but these fractions have variable denominators and the question asks what must be true for every whole number from 1 up. That abstraction makes it as demanding as typical grade 7 work, though the arithmetic is light.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 20 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

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

Variables instead of numbers moderate Understanding the question
All four statements are about the unknown x, and the bold must means each has to hold for every whole number from 1 up, not just for a single test value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked moderate Choosing the answer
The statements are four separate inequalities, two comparisons and two differences, so each has to be tested; one counterexample is enough to rule a statement out. (Embretson & Wetzel, 1987)
Error-prone arithmetic minor Working it out
Testing the differences means subtracting unlike fractions, such as 1/3 − 1/4 = 1/12 when x = 2, and comparing the result with 1 and with 1/x. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute a small whole number such as 1 or 2 for x in all four statements, rule out any that fail, and confirm the survivor by thinking about larger denominators.

Traps

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Believes a bigger denominator makes a bigger fraction, so 1/(x + 2) seems larger than 1/(x + 1), the comparison turned around.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Both fractions are at most 1/2, so their difference can never exceed 1, and no common working makes it do so; this choice mostly draws guesses.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Subtracts over a common denominator but keeps only the numerator, (x + 2) − (x + 1) = 1, so the difference looks like 1, larger than 1/x once x is 2.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Comparing unit fractions: $\text{if } 0 < a < b \text{, then } \frac{1}{a} > \frac{1}{b}$; a and b are positive numbers, and b is bigger. One over a means one whole cut into a equal pieces. Cutting the same whole into more pieces makes each piece smaller. So the fraction with the bigger bottom is smaller.
Subtracting fractions: $\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}$; a and b are the bottoms of the two fractions. Rewrite both fractions over the common bottom a × b. Then subtract the tops. The bottom stays a × b.

Sort out the question
x is a whole number that is 1 or more, so x can be 1, 2, 3 and so on. You need the statement that must be true. That means it is true for every such x. One x that makes a statement false rules it out.
Test x = 1 in each statement
With x = 1, x + 1 is 2 and x + 2 is 3.
- A: $\frac{1}{2} > \frac{1}{3}$ . True.
- B: $\frac{1}{2} < \frac{1}{3}$ . False.
- C: $\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$ , which is not more than 1. False.
- D: $\frac{1}{6}$ is not more than $\frac{1}{1}$ . False.
Only A is true when x = 1. So B, C and D are out.
Make sure A is always true
Must means A has to work for every x, not just 1. In $\frac{1}{x + 1}$ and $\frac{1}{x + 2}$ , both tops are 1. The bottom x + 2 is always 1 more than x + 1. Cutting a whole into more pieces makes each piece smaller. So A is always true. Watch out: a bigger bottom makes a smaller fraction, not a bigger one. Thinking the other way gives choice B.
See why C and D always fail
Use the subtracting rule with a = x + 1 and b = x + 2.
$\frac{1}{a} - \frac{1}{b} = \frac{b - a}{a \times b}$
$\frac{1}{x + 1} - \frac{1}{x + 2} = \frac{(x + 2) - (x + 1)}{(x + 1)(x + 2)} = \frac{1}{(x + 1)(x + 2)}$
The bottom is at least 2 × 3 = 6, so the difference is at most $\frac{1}{6}$ . That is less than 1. Its bottom is also bigger than x, so it is less than $\frac{1}{x}$ . Watch out: keep the bottom when you subtract. Keeping only the top, 1, leads to choice D. So the answer is A.

Check your answer

Try a bigger number, x = 9. A tenth of a pizza is more than an eleventh, so $\frac{1}{10} > \frac{1}{11}$ . The difference, $\frac{1}{110}$ , is tiny. It is less than 1 and less than $\frac{1}{9}$ . A still holds, and C and D still fail.

More questions like this

The original DOE question

Question 83 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 83. Open page 82 of the handbook (PDF)

Question 27 of 57

2023 Test A Math, question 27

Question

A basket contains red balls, green balls, and white balls. There are 12 red balls in the basket. The probability of randomly choosing a red ball is 1 in 3. If the probability of randomly choosing a green ball is 1 in 4, how many green balls are in the basket?

Show the answer

Correct answer: C

First, use the information about the red balls to calculate the total number of balls in the basket:

$\frac{1}{3} = \frac{12}{x}$	Cross multiply.
$x = 36$

There are a total of 36 balls in the basket. The probability of selecting a green ball is 1 in 4. Use that information to calculate how many green balls are in the basket:

$\frac{g}{36} = \frac{1}{4}$	Cross multiply.
$4g = 36$	Apply the multiplicative inverse property; divide both sides of the equation by 4.
$g = 9$

There are 9 green balls in the basket.

Question analysis

A basket holds red, green and white balls, including 12 red ones, and the chances of drawing red and green are 1 in 3 and 1 in 4. Students find how many green balls are in the basket. Getting the total from the red count is the step students skip when they apply a probability to the wrong number.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Simple probability is grade 6 content, but here it runs backward: a count and a probability give the total, which then gives another count. Two linked steps like this are typical of 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 minor
Many steps minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 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

The basket's total is never stated, and that missing number is the whole problem: it has to be recovered from the red balls before the green probability means anything. Applying the green chance straight to the 12 red balls is the quick wrong move. Checking that 12 red and 9 green fit in 36 balls with 15 white left over confirms the setup.

Turning words into math minor Understanding the question
The sentence "The probability of randomly choosing a red ball is 1 in 3" has to become 12 = one third of the total, which makes the basket hold 36 balls. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Two linked steps: the red balls give the total, 12 × 3 = 36, and only then does the green probability give a count, 36 ÷ 4 = 9. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use the red balls to find the total, since 12 is one third of it, then take one fourth of that total for the green balls.

Traps

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

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Takes one fourth of the 12 red balls, 12 ÷ 4 = 3, applying the green probability to the only count the stem gives instead of to the total.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No use of the two probabilities gives 8, which is one third of the 24 balls that are not red, so it mostly catches guesses.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Scales the 12 red balls by 4/3 instead of 3/4, taking the 1-in-4 chance to mean more balls than the 1-in-3 chance: 12 × 4/3 = 16.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{balls of one color}}{\text{all the balls}}$; P is the probability, the chance of picking that color. Each ball is equally likely to be picked. So the chance is the fraction of all the balls that have that color. A chance of 1 in 3 means one third of the balls.

Sort out the question
The basket has red, green and white balls. There are 12 red balls. The chance of picking red is 1 in 3. The chance of picking green is 1 in 4. You need the number of green balls. You don’t know the total number of balls yet, so find it first.
Find the total number of balls
1 in 3 means one third of all the balls are red. So 12 is one third of the total. Call the total number of balls b.
$\frac{12}{b} = \frac{1}{3}$
$b = 12 \times 3 = 36$
There are 36 balls in all.
Find the number of green balls
1 in 4 means one fourth of all 36 balls are green. Call the number of green balls g.
$\frac{g}{36} = \frac{1}{4}$
$g = 36 \div 4 = 9$
Watch out: take one fourth of all 36 balls, not of the 12 red ones. $12 \div 4 = 3$ gives choice A. Multiplying 12 by $\frac{4}{3}$ gives 16, choice D. So the answer is C, 9.

Check your answer

A 1 in 4 chance is smaller than a 1 in 3 chance. So there should be fewer green balls than red ones, and 9 is less than 12. Check: $\frac{9}{36} = \frac{1}{4}$ . The rest, $36 - 12 - 9 = 15$ , are white.

More questions like this

The original DOE question

Question 84 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 84. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 28 of 57

2023 Test A Math, question 28

Question

It took Lars 2 hours to ride his bicycle 48 kilometers. What was his average speed in miles per hour? (Use the approximation 1 mile = 1.6 kilometers.)

A1.5
B15.0
C30.0
D38.4

Show the answer

Correct answer: B

First, calculate Lars’s average speed in kilometers per hour:

$\frac{48}{2} = 24$

Then, convert kilometers to miles:

$\frac{m}{24} = \frac{1}{1.6}$	Cross multiply.
$m(1.6) = 24(1)$	Apply the multiplicative inverse property; divide both sides of the equation by 1.6.
$m = \frac{24}{1.6} = 15$

Question analysis

Lars rides 48 kilometers in 2 hours, and this rate question asks for his average speed in miles per hour, using 1 mile = 1.6 kilometers. Students find a unit rate and convert it to other units. Deciding whether to multiply or divide by 1.6, and remembering the 2 hours, are where students go wrong.

NYS standards: 6.RP.3d converting measurement units
6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a speed and converting units with a given rate are grade 6 ratio skills. The question takes two short steps with friendly numbers, lighter than typical grade 7 rate problems with fractional quantities.
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
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 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 tests the direction of a conversion more than the arithmetic. A mile is longer than a kilometer, so a distance has fewer miles than kilometers, and multiplying by 1.6 runs the conversion backward. The other trap is converting the 48 kilometers to 30 miles and forgetting that the ride took 2 hours, which gives a distance instead of a speed.

Combines ideas minor Working it out
A rate and a unit conversion are combined: 48 ÷ 2 = 24 kilometers per hour, then 24 ÷ 1.6 = 15 miles per hour, and the two steps can come in either order. (Embretson, 1998; Daniel & Embretson, 2010)
Error-prone arithmetic minor Working it out
Dividing by 1.6 means dividing 240 by 16; a decimal point in the wrong place makes the speed ten times too small, and that value is offered. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the speed in kilometers per hour first, then divide by 1.6, since a mile is longer than a kilometer and the number of miles must be smaller.

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.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Divides 24 by 16 instead of 1.6, a decimal point in the wrong place that makes the speed ten times too small, 1.5.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Converts the 48 kilometers to 30 miles correctly and stops, forgetting to divide by the 2 hours, so it gives a distance rather than a speed.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Multiplies 24 kilometers per hour by 1.6 instead of dividing, converting in the wrong direction and getting more miles than kilometers.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Average speed: $\text{speed} = \frac{\text{distance}}{\text{time}}$; Distance is how far you go. Time is how long it takes. Speed is how far you go in each unit of time, such as each hour. Average speed shares the whole trip out evenly over the time.
Kilometers to miles: $\text{miles} = \text{kilometers} \div 1.6$; A kilometer is a unit of length used in most countries. 1 mile is about 1.6 kilometers, so a mile is longer. To find how many miles, find how many groups of 1.6 kilometers fit. So divide by 1.6.

Sort out the question
Lars rode 48 kilometers in 2 hours. You need his average speed in miles per hour. That means miles, not kilometers. Use the fact that 1 mile is about 1.6 kilometers.
Find the speed in kilometers per hour
Divide the distance by the time.
$\text{speed} = \frac{48}{2} = 24$
Lars went 24 kilometers each hour. This speed is still in kilometers, so you are not done yet.
Change kilometers to miles
Divide by 1.6. To make it easier, multiply both numbers by 10 first. Then you divide by the whole number 16.
$24 \div 1.6 = 240 \div 16 = 15$
Watch out: divide by 1.6, don’t multiply. A mile is longer, so there are fewer miles than kilometers. $24 \times 1.6 = 38.4$ is choice D. Dividing 24 by 16 gives 1.5, choice A, with the decimal point in the wrong place.
Name the answer
Lars rode 15 miles each hour. Watch out: changing 48 kilometers to 30 miles is only part of the work. 30 is a distance, choice C. You still have to divide by the 2 hours. So the answer is B, 15.0 miles per hour.

Check your answer

Work backward. At 15 miles per hour for 2 hours, Lars rides 30 miles. And $30 \times 1.6 = 48$ kilometers, the distance in the question. Also, 15 is smaller than 24, as it should be, since miles are longer than kilometers.

More questions like this

The original DOE question

Question 85 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 85. Open page 82 of the handbook (PDF)

Question 29 of 57

2023 Test A Math, question 29

Question

Integer x is evenly divisible by 3. Which expression below is also evenly divisible by 3?

A $2x + 1$
B $3x - 5$
C $4x - 1$
D $4x + 6$

Show the answer

Correct answer: D

Since integer x is evenly divisible by 3, substitute 3 for x in the answer options to determine which expression is also divisible by 3:

Let $x = 3$ .

Option A:

$2x + 1 = 2(3) + 1 = 7$

Option B:

$3x - 5 = 3(3) - 5 = 4$

Option C:

$4x - 1 = 4(3) - 1 = 11$

Option D:

$4x + 6 = 4(3) + 6 = 18$

Option D is the only value that is divisible by 3.

Question analysis

This number properties question says an integer x is divisible by 3 and asks which of four linear expressions in x is also divisible by 3, whatever multiple of 3 the integer is. Students can test a multiple of 3 in each expression or think about remainders. Testing a value that is not a multiple of 3 can make a wrong expression look right.

NYS standards: 7.EE.2 rewriting expressions to show relationships
6.NS.4 common factors and multiples
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students rewrite and evaluate linear expressions, but they seldom ask whether one is divisible by 3 for every allowed value. One well-chosen substitution settles it, so the demand is about typical for grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Every choice must be checked minor
Solution: 2 steps
Reading level: Question and choices: about grade 13, 18 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The algebra is light; the question turns on choosing a test value that obeys the condition. The habit of trying x = 1 backfires here, because 1 is not a multiple of 3 and it makes two wrong expressions come out divisible. Students who substitute 3, or who see that 4x and 6 are both multiples of 3 whenever x is, settle it quickly, and a second multiple such as 6 confirms it.

Variables instead of numbers moderate Understanding the question
Nothing pins x down except that it is a multiple of 3, so each expression must be judged for every such x, and any test value has to be a multiple of 3 itself, such as 3 or 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Every choice must be checked minor Choosing the answer
The four expressions differ only in their coefficients and constants, from 2x + 1 to 4x + 6, so none can be dismissed at a glance; each needs a test or a look at its terms. (Embretson & Wetzel, 1987)

Strategy: Substitute a multiple of 3, such as 3 or 6, for x in every expression and keep the one whose value divides evenly by 3; a second multiple confirms it.

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Tests x = 1, which is not a multiple of 3, finds that 2(1) + 1 = 3 divides evenly, and stops at the first choice; with x = 3 it gives 7.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Sees the 3 in 3x and takes the whole expression to be a multiple of 3, overlooking that subtracting 5 spoils it: with x = 3 it gives 4.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Tries a value that is not a multiple of 3, such as x = 4, where 4x − 1 = 15 divides evenly, though x = 3 gives 11.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding multiples of 3: $3a + 3b = 3(a + b)$; a and b are whole numbers, so 3a and 3b are multiples of 3. A multiple of 3 is a number that 3 divides evenly, with nothing left over. Adding two multiples of 3 always gives another multiple of 3, because you can pull the 3 out of both parts.

Sort out the question
“Evenly divisible by 3” means 3 goes into the number with nothing left over. Such a number is called a multiple of 3, like 3, 6, 9 or 12.
You know that x is a multiple of 3. You need the expression that is a multiple of 3 every time, whatever multiple of 3 x is.
Test x = 3 in every choice
Pick a number for x that follows the rule. The easiest one is 3. Put it into each expression.
Choice Work Divisible by 3?
A $2(3) + 1 = 7$ No
B $3(3) - 5 = 4$ No
C $4(3) - 1 = 11$ No
D $4(3) + 6 = 18$ Yes
Watch out: your test number must be a multiple of 3. With x = 1, choice A gives 3 and looks right. That leads to the wrong choice, A.
See why D always works
One test could be lucky, so see why D works for every x. Since x is a multiple of 3, 4x is one too. For example, 4 × 3 = 12 and 4 × 6 = 24. The number 6 is also a multiple of 3, since 6 = 3 × 2.
$3a + 3b = 3(a + b)$
So 4x + 6 is a multiple of 3 plus a multiple of 3. That is always a multiple of 3.
Rule out the other choices
Watch out: choice B starts with 3x, but then it subtracts 5. The 5 is not a multiple of 3, so it spoils the result, and 3(3) − 5 = 4. Choices A and C also add or take away a number that is not a multiple of 3.
So the answer is D, $4x + 6$ .

Choice	Work	Divisible by 3?
A	$2(3) + 1 = 7$	No
B	$3(3) - 5 = 4$	No
C	$4(3) - 1 = 11$	No
D	$4(3) + 6 = 18$	Yes

Check your answer

Try two more multiples of 3. With x = 6, choice D gives $4(6) + 6 = 30$ , and 30 ÷ 3 = 10. With x = 9, it gives $4(9) + 6 = 42$ , and 42 ÷ 3 = 14. Both divide evenly, so D keeps working.

More questions like this

The original DOE question

Question 86 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 86. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 30 of 57

2023 Test A Math, question 30

Question

Lamel has a jar containing 6 red chips, 10 blue chips, and 4 yellow chips. If he removes one chip at random, what is the probability that it will not be red?

A $\frac{4}{5}$
B $\frac{7}{10}$
C $\frac{3}{10}$
D $\frac{1}{5}$

Show the answer

Correct answer: B

There are currently 20 chips in the jar ( $6 + 10 + 4$ ). Of those chips, 14 are not red. So the probability of choosing a chip that is not red is $\frac{14}{20} = \frac{7}{10}$ .

Question analysis

In this probability word problem, Lamel's jar holds red, blue and yellow chips, and students find the chance that one chip drawn at random is not red. It tests counting the favorable outcomes against the total and writing the probability as a fraction. Missing the bold word not, or counting the wrong colors, leads to tempting wrong fractions.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Probability of a single event with equally likely outcomes is grade 6 content in New York. Students add three counts and form one fraction, and the only extra care is the bold word not, so it sits below grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
An answer in an unexpected form minor
Solution: 2 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

This is a one-event probability with small whole numbers, so the bold not carries the question. A student who skims past it computes the chance of red, a tidy fraction that is waiting among the choices. The other wrong fractions come from removing the wrong color or keeping only one of the two colors that are not red. Adding blue and yellow first, then dividing by all 20 chips, avoids each of them.

Turning words into math minor Understanding the question
The bold not turns the question around: the chips that count are the 10 blue and 4 yellow ones, 14 in all, or equivalently 1 minus the red share of 6/20. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The count of 14 out of 20 matches no choice until it is reduced to 7/10, so a correct fraction left unsimplified can look missing from the list. (Enright, Morley & Sheehan, 2002)

Strategy: Add all the chips for the total, then count the chips that are blue or yellow, or subtract the chance of red from 1, and simplify the fraction.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Leaves out the 4 yellow chips instead of the 6 red ones, finding the chance of not yellow, 16/20 = 4/5.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Overlooks the bold not and gives the chance that the chip is red, 6 out of 20, which reduces to 3/10.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Counts only the 4 yellow chips as not red and forgets the 10 blue ones, so it finds 4/20 = 1/5, the chance of yellow alone.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all outcomes}}$; P is the probability, the chance that something happens. The top counts the outcomes you want. The bottom counts every outcome that could happen. It works when each outcome is equally likely, like pulling one chip without looking.

Sort out the question
Lamel’s jar holds 6 red chips, 10 blue chips and 4 yellow chips. He takes one chip without looking. You need the probability that it is not red. That means the chip is blue or yellow.
Count all the chips
$6 + 10 + 4 = 20$
There are 20 chips in all. That is the bottom of the fraction, because any of the 20 could come out.
Count the chips that are not red
The blue and yellow chips are the ones that are not red.
$10 + 4 = 14$
So 14 chips are not red. Watch out: the question asks for not red. Counting the 6 red chips gives $\frac{6}{20} = \frac{3}{10}$ , choice C. Counting only the 4 yellow chips gives $\frac{1}{5}$ , choice D.
Find and simplify the probability
$P = \frac{\text{outcomes you want}}{\text{all outcomes}}$
$P = \frac{14}{20} = \frac{14 \div 2}{20 \div 2} = \frac{7}{10}$
Dividing the top and bottom by 2 puts the fraction in lowest terms. So the answer is B, $\frac{7}{10}$ .

Check your answer

A chip is either red or not red, so the two chances must add up to 1. The chance of red is $\frac{6}{20} = \frac{3}{10}$ . Then $\frac{3}{10} + \frac{7}{10} = 1$ . It works.

More questions like this

The original DOE question

Question 87 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 87. Open page 83 of the handbook (PDF)

Question 31 of 57

2023 Test A Math, question 31

Question

$A circle divided into 6 equal spaces, labeled clockwise from the top right P, Q, R, S, T and U. An arrow from the center of the circle points into Space P.$

The arrow starts on Space P and moves clockwise around the circle. It moves through one space each minute. What space will the arrow point to in 100 minutes?

Show the answer

Correct answer: C

There are 6 spaces, so first divide 100 by 6: $100 \div 6 = 16$ , remainder 4. So, the arrow goes around 16 full times and then 4 more spaces. Four spaces from Space P is Space T.

Question analysis

This patterns question shows a circle split into six lettered spaces, with an arrow that starts on P and moves one space clockwise each minute, and asks where it points after 100 minutes. Students use division with a remainder to skip the full trips around. Counting the leftover moves without including the starting space, and in the right direction, is the tricky part.

NYS standards: 6.NS.2 multi-digit division
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Division with a remainder is grade 6 arithmetic, but using that remainder to step around a six-part circle is a pattern task that grade 7 classes seldom practice. It needs careful counting rather than new content.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 1, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The division is short; the points are lost in turning the remainder into a space. The arrow starts on P, so the 4 leftover moves land on Q, R, S and T, and counting P itself as a move stops one space short. Going around the wrong way, or one move too far, gives the other two choices. Checking a small case first, such as 7 minutes landing on Q, fixes the rule before it is used.

No obvious method minor Working it out
The stem never mentions division; the step that unlocks it is seeing that every 6 minutes brings the arrow back to P, so only the remainder of 100 ÷ 6 matters. (Schoenfeld, 1985)
Reading a figure, table or graph minor Understanding the question
The six spaces and their clockwise order, P, Q, R, S, T, U, appear only in the circle diagram, so the cycle length and the direction of counting both come from the figure. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Divide 100 by 6: every full trip around the circle brings the arrow back to P, so only the remainder matters; count that many moves clockwise, starting with the space after P.

Traps

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

The wrong answers, classified

A Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Counts the 4 leftover moves counterclockwise from P, through U, T and S to R, instead of clockwise as the stem says the arrow moves.
B Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts Space P itself as the first of the 4 leftover moves, P, Q, R, S, so the arrow stops one space short of where it really points.
D Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Makes 5 moves past P instead of the 4 left after 16 full turns, overshooting by one space, the fence-post slip in the other direction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Division with a remainder: $a = b \times q + r$; a is the number you divide, here the minutes. b is the number you divide by, here the spaces in one trip around. q is how many whole groups of b fit into a. r is the remainder, the amount left over. The remainder is always less than b.

Sort out the question
The circle has 6 spaces. Going clockwise, the way a clock’s hands turn, they are P, Q, R, S, T and U. The arrow starts on P and moves one space each minute. You need the space it points to after 100 minutes.
Notice that the pattern repeats
After 6 minutes, the arrow has gone all the way around and points to P again. After 12 minutes it is back on P again, and so on. So full trips of 6 minutes don’t change where it points. Only the leftover minutes matter.
Divide 100 by 6
$a = b \times q + r$
$100 = 6 \times 16 + 4$
The arrow makes 16 full trips. That takes $6 \times 16 = 96$ minutes, and it is back on P. There are 4 minutes left over.
Count the last 4 moves
From P, move one space each minute: Q (minute 97), R (minute 98), S (minute 99), T (minute 100).
Watch out: don’t count P itself as the first move. That stops on S, choice B. Also count clockwise. Counting the other way lands on R, choice A.
So the answer is C, T.

Check your answer

Use a smaller number with the same remainder. 100 − 96 = 4, and 96 minutes is a whole number of trips. So after 100 minutes the arrow points where it does after 4 minutes. Four moves from P are Q, R, S, T. That is T again.

More questions like this

The original DOE question

Question 88 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 88. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 32 of 57

2023 Test A Math, question 32

Question

Mei-Ling is one of 6 members of a committee. If 2 members of that committee are selected to go to a conference, how many of the possible pairs of members would include Mei-Ling?

Show the answer

Correct answer: A

Call the 6 members A, B, C, D, E, and M (for Mei-Ling). The possible combinations are AB, AC, AD, AE, AM, BC, BD, BE, BM, CD, CE, CM, DE, DM, and EM. Of those pairs, 5 include Mei-Ling.

Question analysis

This counting question describes a six-member committee sending two members to a conference and asks how many of the possible pairs include Mei-Ling. Students count combinations in which order does not matter. The traps are counting a pair twice, pairing Mei-Ling with herself, or multiplying the numbers in the question instead of listing the pairs.

NYS standards: 7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students count outcomes with organized lists and tree diagrams. Here the list of pairs is short and a single condition, that one named person belongs to the pair, filters it, which matches routine grade 7 counting 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
Much to keep track of minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 31 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The count is small enough to list, so the question rewards a clear picture of what a pair is: two different people, in no particular order. The tempting errors count Mei-Ling as her own partner, count each pair twice, or multiply the two numbers in the stem. Writing the five pairs out, M with each of A through E, takes seconds and rules out all three.

Turning words into math minor Understanding the question
Pairs that include Mei-Ling are simply Mei-Ling with each of the other 5 members, a count of her possible partners, though the phrase “possible pairs” invites counting all 15 pairs. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Much to keep track of minor Working it out
Listing pairs by letter, as the explanation does with A to E and M, means tracking which pairs are already written, so that AM and MA are not both counted. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Label the members with letters and list every pair that contains Mei-Ling, remembering that a pair is two different people and that order does not matter.

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).
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
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 Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Counts all 6 committee members as possible partners, forgetting that Mei-Ling cannot be paired with herself, one more than the real count.
C Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: moderate
Counts ordered pairs, Mei-Ling with a partner and that partner with Mei-Ling, as different, so each of the 5 pairs is counted twice.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies the 6 members by the 2 people chosen, 6 × 2 = 12, a product of the stem's numbers that counts no actual pairs.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of pairs in a group: $\text{pairs} = \frac{n \times (n - 1)}{2}$; n is the number of people in the group. Each person can pair up with the n − 1 others. Multiplying counts every pair twice, once from each partner’s side. So you divide by 2.

Sort out the question
The committee has 6 members, and Mei-Ling is one of them. Two members go to the conference together. That makes a pair. You don’t need every possible pair. You only need the pairs that include Mei-Ling.
Find Mei-Ling’s possible partners
A pair with Mei-Ling is Mei-Ling plus one other member. Call Mei-Ling M, and call the other members A, B, C, D and E. Her pairs are M with A, M with B, M with C, M with D and M with E. That is 5 pairs.
Watch out: she can’t be paired with herself. Counting all 6 members as partners gives 6, choice B.
Don’t count a pair twice
Mei-Ling with A is the same pair as A with Mei-Ling. The same two people go either way. Watch out: counting both orders doubles the count to 10, choice C. Multiplying 6 × 2 = 12 gives choice D, but that does not count any pairs.
Check against all the pairs
Count every pair in the committee.
$\text{pairs} = \frac{n \times (n - 1)}{2}$
$\text{pairs} = \frac{6 \times 5}{2} = 15$
Each pair has 2 people, so the 15 pairs have 30 name spots. All 6 members get the same share: $30 \div 6 = 5$ . So the answer is A, 5.

Check your answer

List all 15 pairs: AB, AC, AD, AE, AM, BC, BD, BE, BM, CD, CE, CM, DE, DM, EM. Now circle the ones with M in them: AM, BM, CM, DM, EM. There are exactly 5.

More questions like this

The original DOE question

Question 89 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 89. Open page 83 of the handbook (PDF)

Question 33 of 57

2023 Test A Math, question 33

Question

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

A−9
B−1
C1
D9

Show the answer

Correct answer: A

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

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

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

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 34 of 57

2023 Test A Math, question 34

Question

$Parallelogram PQRS, with P at the bottom left, Q at the top left, R at the top right and S at the bottom right. The angle at P is marked 56°, the angle at Q is marked x°, and the angle at S is marked y°. The angle at R is not marked.$

In the parallelogram above, what is the value of $x + y$ ?

A112
B124
C148
D248

Show the answer

Correct answer: D

In a parallelogram, opposite angles are congruent, so the measure of the unmarked angle is 56°. The sum of the interior angles of a parallelogram is equal to 360°. Use that information to solve the problem:

$x + y + 56 + 56 = 360$

$x + y + 112 = 360$

$x + y = 248$

Question analysis

This geometry question shows parallelogram PQRS with a 56° angle at P and unknown angles x and y at Q and S, and asks for the sum x + y. Students need angle facts about parallelograms: opposite angles match, and the four angles total 360°. Stopping after finding only one angle, or adding the wrong pair, are the common mistakes.

NYS standards: 7.G.5 angle relationships and equations
8.G.5 angle sums, transversals, angle-angle similarity
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: At grade 7 The parallelogram facts needed, equal opposite angles and a 360° angle sum, rest on grade 8 reasoning about parallel lines and triangles. Once they are known, the arithmetic is short, about the size of a typical grade 7 angle problem.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 10, 10 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Two short angle facts finish the question, but the figure marks three of the four corners, and the unmarked one at R is easy to forget. Students who find one obtuse angle stop with half the total, and those who add the two 56° angles give the part already known. Checking that the four angles, 56, 124, 56 and 124, add to 360 confirms the sum.

Something you must already know moderate Working it out
The figure gives one angle and no rules: students must bring both that opposite angles of a parallelogram are equal and that a quadrilateral's angles total 360°, or that neighboring angles add to 180°. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The labels x and y sit at Q and S, opposite corners, while the corner R across from the 56° angle at P carries no label at all. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Mark the angle at R equal to the angle at P, since opposite angles of a parallelogram are equal, then take both away from the total of a quadrilateral's four angles.

Traps

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

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Adds the two 56° angles at P and R, 56 + 56 = 112, giving the part of the 360° already known instead of the two angles that remain.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds one unmarked angle, 180 − 56 = 124, and stops before adding the equal angle at the opposite corner.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No angle fact in the parallelogram gives 148, which is exactly 100 less than the right total, so it mostly catches guesses or a stray subtraction slip.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Opposite angles of a parallelogram: $\angle P = \angle R$; A parallelogram is a four-sided shape whose opposite sides are parallel. P and R are opposite corners: they don’t share a side. Opposite angles of a parallelogram are always equal.
Angles of a four-sided shape: $\angle P + \angle Q + \angle R + \angle S = 360$; P, Q, R and S are the four corners. Their angles always add up to 360 degrees. A line from one corner to the opposite corner cuts the shape into two triangles, and each triangle’s angles add up to 180 degrees.
Neighboring angles of a parallelogram: $\angle P + \angle Q = 180$; P and Q are corners next to each other, joined by one side. In a parallelogram, their angles add up to 180 degrees. The same is true of any two neighboring corners.

Sort out the figure
PQRS is a parallelogram. The angle at corner P is 56°. The angle at Q is x°, and the angle at S is y°. The angle at R has no label. You need x + y, the angles at Q and S added together.
Find the angle at R
R is across from P, so they are opposite corners.
$\angle P = \angle R$
$\angle R = 56$
So the angle at R is 56°, the same as at P.
Use the 360° total
$\angle P + \angle Q + \angle R + \angle S = 360$
$56 + x + 56 + y = 360$
$x + y + 112 = 360$
$x + y = 360 - 112 = 248$
Watch out: 112 is the two angles you already know, not x + y. Answering 112 is choice A.
Find x and y one at a time
Q is next to P, so use the neighbor rule.
$\angle P + \angle Q = 180$
$56 + x = 180, \quad x = 124$
S is also next to P, so y = 124 too. Then $124 + 124 = 248$ , the same total. Watch out: stopping at one angle, 124, gives choice B. The question asks for both added together.
So the answer is D, 248.

Check your answer

Add all four angles: $56 + 124 + 56 + 124 = 360$ . That is the right total for a four-sided shape. Also, the angles at Q and S look wider than a right angle in the figure, and 124° is wider than 90°.

More questions like this

The original DOE question

Question 91 as printed on page 84 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 91. Open page 84 of the handbook (PDF)

Question 35 of 57

2023 Test A Math, question 35

Question

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

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

A−2
B0
C2
D4

Show the answer

Correct answer: B

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

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

Solve for x:

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

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

Question analysis

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

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

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

Why it is hard

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

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

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

Traps

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

The wrong answers, classified

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

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

Check your answer

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

More questions like this

The original DOE question

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

Question 36 of 57

2023 Test A Math, question 36

Question

Josef and Mai divided some stamps between themselves. Josef got 60% of the stamps. If Josef received 500 more stamps than Mai, how many stamps did Josef receive?

A300
B800
C1,500
D2,500

Show the answer

Correct answer: C

Let j be the number of stamps Josef has. Then Mai has $j - 500$ stamps. The total number of stamps is $j + (j - 500) = 2j - 500$ . Since Josef has 60% of the stamps, set up an equation to solve for j:

$\frac{j}{2j - 500} = \frac{60}{100}$	Cross multiply.
$100j = 60(2j - 500)$	Apply the distributive property; multiply the 60 by $2j$ and 500.
$100j = 120j - 30{,}000$	Apply the additive inverse property; subtract $120j$ from both sides of the equation.
$-20j = -30{,}000$	Apply the multiplicative inverse property; divide both sides of the equation by −20.
$j = 1{,}500$

Question analysis

This percent word problem says Josef and Mai split some stamps, Josef got 60% of them, and he received 500 more than Mai; it asks how many stamps Josef received. Students must connect a difference in counts to a difference in percents. Stopping at the total, or taking a percent of 500, gives tempting wrong answers.

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: Above grade 7 Grade 7 percent problems usually give the whole and ask for a part. Here neither share nor the total is known; students must link the 500-stamp gap to the gap between two percents before finding the total and then Josef's part.
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
No obvious method moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 26 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is easy once the 500 is seen as the gap between two percents of one total; seeing that is the hard part. The tempting move is to apply the 60% to the only number given, the 500, which treats a difference as a whole. Students who do find the total must still take Josef's share of it, since the total waits among the choices. Checking that 1,500 and 1,000 differ by 500 confirms the answer.

Turning words into math moderate Understanding the question
The stem describes one split in two ways, Josef's 60% of the total and his 500-stamp lead over Mai, and the link, that the lead is 60% − 40% = 20% of the total, is never stated. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
No obvious method moderate Working it out
No equation is suggested: the quick route treats the 500 as 20 percentage points and scales up to 100%, while the handbook's route, j over 2j − 500, needs cross-multiplying and a negative coefficient. (Schoenfeld, 1985)
Many steps minor Working it out
Even the short route takes three moves, the 20% gap, the 2,500 total and Josef's 60% of it, and the total is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: If Josef has 60%, Mai has the other 40%, so the 500-stamp difference is a known percent of the total; use it to find the total, then take Josef's share.

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.
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 a misconception Expected pull: moderate
Takes 60% of 500, applying the percent to the difference between the two shares as if the 500 stamps were the whole collection.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Takes 60% of 500 to get 300 and then adds the 500 extra stamps, combining the stem's numbers without a model of the two shares.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Sees that the 500-stamp gap is 20% of the total, finds the total of 2,500, and stops before taking Josef's 60% of it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number: $\text{part} = \frac{p}{100} \times \text{whole}$; p is the percent. The whole is the total amount, and the part is the piece you want. Percent means “out of 100,” so 60% is 60 out of every 100. Use it to find a part, or work backward to find the whole.

Sort out the question
Josef and Mai split some stamps. Josef got 60% of them. He also got 500 more stamps than Mai. You need how many stamps Josef got. You don’t know the total yet, so find it first.
Find Mai’s percent
All the stamps make 100%. Josef has 60%, so Mai has the rest.
$100\% - 60\% = 40\%$
Mai has 40% of the stamps.
Find what the 500 stamps are
Josef’s extra 500 stamps are the gap between his share and Mai’s share.
$60\% - 40\% = 20\%$
So 500 stamps are 20% of all the stamps.
Find the total number of stamps
$\text{part} = \frac{p}{100} \times \text{whole}$
$500 = \frac{20}{100} \times \text{whole}$
$\text{whole} = 500 \div \frac{20}{100} = 500 \times 5 = 2{,}500$
There are 2,500 stamps in all. Watch out: that is everyone’s stamps, choice D. Don’t stop here.
Find Josef’s share
$\text{part} = \frac{60}{100} \times 2{,}500 = 1{,}500$
Watch out: taking 60% of 500 gives 300, choice A. The 500 is the gap between them, not the total. So the answer is C, 1,500 stamps.

Check your answer

Mai gets 40% of 2,500, which is 1,000 stamps. Josef’s 1,500 minus Mai’s 1,000 is 500, just as the question says. And 1,500 + 1,000 = 2,500, the whole amount.

More questions like this

The original DOE question

Question 93 as printed on page 84 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 84, question 93. Open page 84 of the handbook (PDF)

Question 37 of 57

2023 Test A Math, question 37

Question

$A circle with a shaded sector between two radii, one pointing straight up and one pointing up and to the right. The central angle of the shaded sector is marked 45°.$

The shaded sector of the circle shown above has an area of $18\pi$ square feet. What is the circumference of the circle?

A $144\pi$ ft
B $24\pi$ ft
C $18\pi$ ft
D $9\pi$ ft

Show the answer

Correct answer: B

The area of the entire circle is $\pi r^{2}$ .

The fraction of the circle that is shaded is $\frac{45}{360} = \frac{1}{8}$ . So, the area of the sector is $\frac{1}{8}\pi r^{2}$ . Use that information to find r:

$\frac{1}{8}\pi r^{2} = 18\pi$

$r^{2} = 144$

$r = 12$

Use r to calculate the circumference of the circle:

$c = 2\pi r = 2\pi(12) = 24\pi$

The circumference is $24\pi$ feet.

Question analysis

This circle question shows a shaded 45° sector with an area of 18π square feet and asks for the circumference of the whole circle. Students work backward: from the sector to the full area, from the area to the radius, and from the radius to the circumference. Mixing up area and circumference, or stopping partway, gives wrong choices.

NYS standards: 7.G.4 circle area and circumference
6.G.5 perfect squares and perfect cubes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 circle problems usually give the radius and ask for area or circumference. This one runs backward from the area of a 45° sector, needing the sector's share of the circle, a square root and a switch to circumference: four linked steps.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Something you must already know moderate
Reading a figure, table or graph minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 30 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question moves from a part of the circle's area to the whole circle's length, so area and circumference both appear and are easy to swap. Many students scale the sector up to 144π and stop, giving an area when a length was asked. Solving r² = 144 also needs a square root, after which 2πr is routine. Asking at the end whether the result is a length, measured in feet rather than square feet, catches the main trap.

Combines ideas moderate Working it out
Three ideas chain together: the 45° sector is 1/8 of the circle, so the whole area is 8 × 18π = 144π, then the radius comes from the area, and the circumference from the radius. (Embretson, 1998; Daniel & Embretson, 2010)
Something you must already know moderate Working it out
Neither πr² nor 2πr is given, nor that a full turn is 360°, so every link of the chain rests on formulas students must already know. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The stem says only “shaded sector”; its 45° angle is marked at the center in the figure, the one place the fraction of the circle comes from. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Work out what fraction of the circle a 45° sector is, scale its area up to the whole circle, solve for the radius, and only then use the circumference formula.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
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: strong
Scales the sector up to the whole circle, 8 × 18π = 144π, and gives that area, though the question asks for the circumference.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Repeats the sector's area from the stem, 18π, as if the shaded part's area were the circle's circumference.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No step with the 45° angle or the circle formulas gives 9π, which is half the sector's area, so it mostly attracts guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a circle in a sector: $\text{fraction of the circle} = \frac{\text{angle}}{360}$; A sector is a slice of a circle, like a slice of pizza. Its angle is the angle at the center, in degrees. A full turn around the center is 360 degrees. So the slice is that fraction of the whole circle.
Area of a circle: $A = \pi r^{2}$; A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π, called pi, is a number a little more than 3. r² means r times r.
Circumference of a circle: $C = 2\pi r$; C is the circumference, the distance around the circle. r is the radius and π is pi. Two radii make the diameter, the distance across, so this is the same as π times the diameter.

Sort out the question
The shaded slice has a 45° angle at the center. Its area is $18\pi$ square feet. You need the circumference, the distance around the whole circle. For that you need the radius, so work back to it.
Find what fraction is shaded
$\text{fraction of the circle} = \frac{\text{angle}}{360}$
$\frac{45}{360} = \frac{1}{8}$
The shaded slice is one eighth of the circle. Eight slices like it would fill the circle.
Find the area of the whole circle
$8 \times 18\pi = 144\pi$
The whole circle has an area of $144\pi$ square feet. Watch out: this is an area, not the circumference. Stopping here gives choice A.
Find the radius
$A = \pi r^{2}$
$\pi r^{2} = 144\pi$
$r^{2} = 144$
$r = 12$
Divide both sides by π. Then ask what number times itself makes 144. It is 12, so the radius is 12 feet.
Find the circumference
$C = 2\pi r$
$C = 2 \times \pi \times 12 = 24\pi$
Leave π as it is, because the choices are written with π. The circumference is $24\pi$ feet. So the answer is B, $24\pi$ ft.

Check your answer

Work forward. A radius of 12 feet gives an area of $\pi \times 12 \times 12 = 144\pi$ square feet. One eighth of that is $144\pi \div 8 = 18\pi$ square feet, the shaded area in the question.

More questions like this

The original DOE question

Question 94 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 94. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 38 of 57

2023 Test A Math, question 38

Question

Which graph represents the solution to $-3x - 7 > -4$ ?

A $Number line from −4 to 4 with a tick at every integer. An open circle at −1, with the line shaded from −1 to the left, ending in an arrow.$
B $Number line from −4 to 4 with a tick at every integer. An open circle at 1, with the line shaded from 1 to the left, ending in an arrow.$
C $Number line from −4 to 4 with a tick at every integer. An open circle at −1, with the line shaded from −1 to the right, ending in an arrow.$
D $Number line from −4 to 4 with a tick at every integer. An open circle at 1, with the line shaded from 1 to the right, ending in an arrow.$

Show the answer

Correct answer: A

$-3x - 7 > -4$	Apply the additive inverse property; add 7 to both sides of the inequality.
$-3x > 3$	Apply the multiplicative inverse property; divide both sides of the inequality by −3. Note: when dividing an inequality by a negative number, the inequality sign is reversed.
$x < -1$

The solution is that x is less than −1, so choose the graph that starts at −1 and continues to the left.

Question analysis

This inequality question asks which number line graph shows the solution of −3x − 7 > −4. Students solve a two-step inequality whose x term has a negative coefficient, then match the boundary point and the direction of the shading. Handling the inequality symbol at the division step is where most mistakes happen, and each slip has a matching graph.

NYS standards: 7.EE.4b two-step inequalities
6.EE.8 writing simple inequalities
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Solving and graphing a two-step inequality is a core grade 7 standard. This one has a negative coefficient on x, the classic grade 7 inequality trap, and four similar graphs test whether the sign and the direction were both handled.
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: Something you must already know moderate
A very close wrong answer moderate
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 7 words.
Corrected from the DOE handbook: A wrong term 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 solving takes two moves, but the second carries the rule the question tests: dividing an inequality by a negative number flips its direction. The four graphs cross two features, boundary at −1 or 1 and shading left or right, so forgetting the flip, dropping a sign, or both each land on a graph. Testing a shaded value such as x = −2 in the original inequality, where 6 − 7 = −1 is greater than −4, confirms the choice.

Something you must already know moderate Working it out
The step that decides the question, reversing the symbol when dividing both sides by −3, is a rule students must bring themselves; −3x > 3 becomes x < −1, not x > −1. (Anderson & Krathwohl, 2001)
A very close wrong answer moderate Choosing the answer
The graph shaded to the right from an open circle at −1 matches the right graph's boundary exactly and differs only in direction, the result of skipping the reversal. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Reading a figure, table or graph minor Understanding the question
Each graph must be read for two things, where its open circle sits, at −1 or 1, and which way the arrow runs, so a boundary alone does not identify the answer. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add 7 to both sides, divide by −3 and reverse the inequality symbol, then match both the boundary point and the shading direction; test one shaded value in the original.

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 slip Expected pull: moderate
Reverses the symbol correctly but drops the negative in 3 ÷ (−3), so the boundary lands at 1 with the shading to its left.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: strong
Divides both sides by −3 without reversing the symbol, solving the inequality as if it were an equation, so x > −1 and the shading runs right.
D Negatives and direction (Brown & Burton, 1978)
Reveals a slip Expected pull: weak
Treats −3x as 3x, so it solves 3x > 3 to get x > 1 with no reversal, which puts both the boundary and the direction wrong.

What was corrected from the DOE handbook, and why

Explanation: The handbook’s steps say “add 7 to both sides of the equation” and “divide both sides of the equation by −3” (p. 142, question 95). But −3x − 7 > −4 is an inequality, not an equation, and the step’s own note is the point of the question: dividing an inequality by a negative number reverses the sign. The site now says “both sides of the inequality” in both steps, as the DOE’s later guides word this step (“Subtract 4 from both sides of the inequality” in the 2024–2025 guide). The working and the answer are unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding to both sides of an inequality: $\text{if } a > b\text{, then } a + c > b + c$; a and b are the two sides, and c is any number. Adding the same amount to both sides keeps the comparison true, like adding the same weight to both pans of a scale.
Dividing an inequality by a negative number: $\text{if } a > b \text{ and } c < 0\text{, then } \frac{a}{c} < \frac{b}{c}$; a and b are the two sides, and c is a negative number. When you divide both sides by a negative number, the greater-than sign flips to less-than. That is because negatives reverse order: 2 is more than 1, but −2 is less than −1.

Sort out the question
An inequality compares two amounts. The sign > means is greater than. You need the graph of every x that makes $-3x - 7 > -4$ true. The graphs differ in two ways: where the open circle sits, at −1 or 1, and which way the shading goes.
Add 7 to both sides
$\text{if } a > b\text{, then } a + c > b + c$
$-3x - 7 + 7 > -4 + 7$
$-3x > 3$
Adding 7 undoes the −7, so the x term is alone on the left.
Divide by −3 and flip the sign
$\text{if } a > b \text{ and } c < 0\text{, then } \frac{a}{c} < \frac{b}{c}$
$\frac{-3x}{-3} < \frac{3}{-3}$
$x < -1$
So x is any number less than −1. Watch out: if you forget to flip the sign, you get x > −1, and that graph is choice C. Also, 3 ÷ (−3) is −1, not 1. Dropping that minus sign leads to B or D.
Match the solution to a graph
Less than −1 means to the left of −1 on the number line. The circle at −1 is open, or empty, because −1 itself does not work. The sign means greater than, not greater than or equal to.
Graph A has an open circle at −1 and is shaded to the left. So the answer is A, the graph shaded left from an open circle at −1.

Check your answer

Test a number on each side of −1. For x = −2: $-3(-2) - 7 = 6 - 7 = -1$ , and −1 is greater than −4, so −2 works. For x = 0: $-3(0) - 7 = -7$ , and −7 is less than −4, so 0 does not work. The shading must go left.

More questions like this

The original DOE question

Question 95 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 95. Open page 85 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 39 of 57

2023 Test A Math, question 39

Question

A scientist mixed three chemicals, R, S, and T, in a glass container. The amount of R is 3 times the amount of S, and the amount of T is $\frac{1}{6}$ the amount of S. What is the ratio of the amount of R to the amount of T?

A1:18
B2:1
C3:1
D18:1

Show the answer

Correct answer: D

First, set up the equations based on the information in the question.

$\mathrm{R} = 3\mathrm{S}$

$\mathrm{T} = \frac{1}{6}\mathrm{S}$

So, the ratio of R:T is

$3\mathrm{S}:\frac{1}{6}\mathrm{S}$

Divide both sides by S to eliminate the variable. Then, multiply both sides by 6 to eliminate the fraction. The resulting ratio is 18:1.

Question analysis

This ratio question describes three chemicals mixed in a container: R is 3 times the amount of S, and T is one sixth of S. Students must combine the two relationships through S to find the ratio of R to T. Writing the ratio in the wrong order, or using only one of the relationships, produces the tempting wrong choices.

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 often link two quantities through a third. Here R and T are each described relative to S, one as a multiple and one as a fraction, and combining them into one ratio is typical grade 7 work with an order trap.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question chains two comparisons through S, and the 1/6 makes the chain awkward to do with letters. Choosing S = 6 turns it into whole numbers at once. The listed wrong ratios come from writing the right numbers in the wrong order, combining 3 and 6 by division, or stopping at the first comparison. Rereading which amount the question names first settles the order.

Turning words into math moderate Understanding the question
Both sentences compare a chemical with S, R as 3 times S and T as 1/6 of S, so S must serve as the link before R can be compared with T. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
The ratios 1:18 and 18:1 use the same two numbers in opposite order, so only checking that R, the larger amount, comes first separates the right ratio from its reverse. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
No amounts are given, only relationships, so students either reason with S as a letter or choose a number for it, such as 6, which makes R 18 and T 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Pick a convenient amount for S, such as 6, so that both R and T come out as whole numbers, then write R to T in that order 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).
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals partial knowledge Expected pull: strong
Works out the amounts correctly but writes the ratio upside down, T to R, putting the smaller amount first.
B Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Divides the 6 by the 3 instead of multiplying them, combining the two comparisons with S the wrong way to get 2:1.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Uses only the first sentence, R is 3 times S, and gives the ratio of R to S instead of R to T.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equal ratios: $a : b = (a \times n) : (b \times n)$; a and b are two amounts being compared, and n is any number except 0. Multiplying or dividing both parts of a ratio by the same number gives an equal ratio. For example, 2 : 1 is the same as 4 : 2.

Sort out the question
There are three chemicals, R, S and T. R is 3 times S. T is $\frac{1}{6}$ of S. You need the ratio of R to T, written R : T. Both facts compare a chemical with S, so S links R and T.
Pick an easy amount for S
No amounts are given, so you may choose one. Pick 6 units for S, because 6 divides evenly by 6.
$\mathrm{R} = 3 \times 6 = 18$
$\mathrm{T} = \frac{1}{6} \times 6 = 1$
So R is 18 units and T is 1 unit.
Write the ratio of R to T
$\mathrm{R} : \mathrm{T} = 18 : 1$
R comes first because the question asks for R to T. Watch out: writing it the other way round gives 1:18, choice A. Using only the first fact gives R : S = 3 : 1, choice C.
Check with letters instead
Without picking a number, R = 3S and T is $\frac{1}{6}\mathrm{S}$ . Divide both parts by S, then multiply both parts by 6.
$a : b = (a \times n) : (b \times n)$
$3\mathrm{S} : \frac{1}{6}\mathrm{S} = 3 : \frac{1}{6} = 18 : 1$
So the answer is D, 18:1.

Check your answer

R should be much bigger than T. R is 3 times S, while T is only a small piece of S. Try another amount, S = 12. Then R = 36 and T = 2, and 36 : 2 is also 18 : 1.

More questions like this

The original DOE question

Question 96 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 96. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 40 of 57

2023 Test A Math, question 40

Question

Ken has k video games, and Jeff has j video games. If Ken gives 6 video games to Jeff, Ken will have twice as many video games as Jeff. Which equation shows the relationship between k and j?

A $k - 6 = 2(j + 6)$
B $k - 6 = 2j + 6$
C $2(k - 6) = j$
D $2(k - 6) = j + 6$

Show the answer

Correct answer: A

If Ken gives 6 video games to Jeff, the number Ken has is $k - 6$ , and the number Jeff has is $j + 6$ . After the exchange, Ken has twice as many video games as Jeff. The equation is:

$k - 6 = 2(j + 6)$

Question analysis

This algebra word problem says Ken has k video games and Jeff has j, and if Ken gives Jeff 6 games, Ken will have twice as many as Jeff. Students choose the equation that models the situation. Getting the comparison the right way round, and grouping each new total correctly, is where most errors happen.

NYS standards: 7.EE.4 equations and inequalities from problems
6.EE.2a writing algebraic expressions
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students write equations from word problems. This one tracks two people's counts after a transfer and states a times-as-many comparison, so both counts change and the factor of 2 must go on the right side: typical but error-prone.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The question is all setup: no solving is needed, just an equation that says what the story says. The trade moves games out of one count and into the other, and the comparison has to be written with the 2 on the smaller side. The wrong equations reverse the comparison, forget Jeff's gain, or drop the parentheses, and all four look alike. Plugging in sample counts is the surest check.

Turning words into math moderate Understanding the question
After the gift both counts change, Ken's to k − 6 and Jeff's to j + 6, and “twice as many video games as Jeff” must double Jeff's whole new count, putting the 2 on his side. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer moderate Choosing the answer
All four equations share the pieces k − 6, j + 6 and a 2; the right one differs from 2(k − 6) = j + 6 only in where the 2 sits, and from k − 6 = 2j + 6 by parentheses. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
There are no numbers to check against, only k and j, so testing an equation means inventing counts that fit the story, such as Jeff with 4 games and Ken with 26. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Write each boy's number of games after the exchange as an expression, then translate "Ken will have twice as many as Jeff" into an equation and test it with sample numbers.

Traps

Equation written backwards: Writes a comparison the wrong way round, word for word: "six times as many students as professors" as 6S = P. The bigger quantity gets the bigger number.
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Leaves out the parentheses, doubling only Jeff's old count j and then adding 6, instead of doubling his whole new count j + 6.
C Equation written backwards (Clement, Lochhead & Monk, 1981)
Reveals a misconception Expected pull: weak
Puts the 2 on Ken's side, writing the comparison backwards, and also forgets that Jeff gains the 6 games, two errors in one equation.
D Equation written backwards (Clement, Lochhead & Monk, 1981)
Reveals a misconception Expected pull: strong
Writes the comparison backwards word for word, doubling Ken's new count, which would mean Jeff ends with twice as many games as Ken.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Twice as many: $A = 2 \times B$; A and B are two amounts. “A has twice as many as B” means A equals 2 times B. The 2 goes with the smaller amount, because that is the one you double to match the bigger one.
The distributive property: $a(b + c) = ab + ac$; a, b and c are numbers. Multiplying a by a sum means multiplying a by each part, then adding. For example, 2(4 + 6) = 8 + 12 = 20.

Sort out the question
Ken has k video games, and Jeff has j. Then Ken gives 6 of his games to Jeff. After that, Ken has twice as many games as Jeff. You need the equation that links k and j.
Write each new amount
Ken gives away 6 games, so he has k − 6 left. Jeff gets those 6 games, so he has j + 6. Both amounts change, so use both new amounts.
Write “twice as many”
After the gift, Ken has more games. So Jeff’s new amount is the one that gets doubled.
$A = 2 \times B$
$k - 6 = 2(j + 6)$
Watch out: putting the 2 on Ken’s side, as in $2(k - 6) = j + 6$ , would mean Jeff has twice as many. That is choice D.
Keep the parentheses
$a(b + c) = ab + ac$
$2(j + 6) = 2j + 12$
Twice Jeff’s new amount is 2j + 12, not 2j + 6. Watch out: dropping the parentheses gives choice B. So the answer is A, $k - 6 = 2(j + 6)$ .

Check your answer

Make up numbers that fit the story. Say Jeff has 4 games and Ken has 26. After the gift, Ken has 20 and Jeff has 10, and 20 is twice 10. Choice A gives $26 - 6 = 2(4 + 6)$ , or 20 = 20. Choice B gives 20 = 14, which is false.

More questions like this

The original DOE question

Question 97 as printed on page 85 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 85, question 97. Open page 85 of the handbook (PDF)

Question 41 of 57

2023 Test A Math, question 41

Question

Yesterday Sarah read 15% of her entire book. Today she read another 17% of the entire book. In lowest terms, what fraction of the book is left for her to read?

A $\frac{7}{25}$
B $\frac{3}{10}$
C $\frac{17}{25}$
D $\frac{7}{10}$

Show the answer

Correct answer: C

So far, Sarah has read $15\% + 17\% = 32\%$ of the entire book. That means she has $100\% - 32\% = 68\%$ left to read.

$68\% = \frac{68}{100} = \frac{17}{25}$

Question analysis

This percent word problem says Sarah read 15% of her book yesterday and 17% today, and asks what fraction of the book she still has to read, in lowest terms. Students combine percents of the same whole, find the part remaining, and convert a percent to a simplified fraction. Rounding, or giving the part already read, leads to wrong choices.

NYS standards: 7.EE.3 multi-step rational number problems
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Below grade 7 Grade 7 percent work often takes a percent of a percent or finds a percent change. Here both percents are of the same whole book, so they combine directly; the rest is a subtraction and simplifying a fraction, lighter than typical.
Estimated difficulty: Straightforward: two steps or one careful reading.
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: 3 steps
Reading level: Question and choices: about grade 5, 33 words.
Corrected from the DOE handbook: Missing percent signs in the explanation were restored.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The percent work is a sum and a subtraction from 100, so the question mostly tests the last step, writing 68% as a fraction in lowest terms. Students who round to 30% read and 70% left to get a friendlier fraction land on a listed choice, as do those who give the part read. Dividing 68 and 100 by their common factor 4 keeps the answer exact.

Turning words into math minor Understanding the question
Both percents are “of the entire book,” so they simply add to 32%, and “left for her to read” asks for the rest, 100% − 32%, not the part already read. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The result has to be written as a fraction in lowest terms, so 68/100 must be divided by 4 to reach a form that appears among the choices. (Enright, Morley & Sheehan, 2002)

Strategy: Add the two percents, since both are of the whole book, subtract the total from 100%, then write the result over 100 and simplify it fully.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No sum or difference of 15% and 17% gives 28%, so 7/25 mostly draws students who match the right fraction's 25 without checking the numerator.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Gives the part already read rather than the part left, and rounds that 32% to 30% to write it as 3/10.
D Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Finds that 68% is left but rounds it to 70% to make an easier fraction, giving 7/10 instead of simplifying 68/100 exactly.

What was corrected from the DOE handbook, and why

Explanation: The handbook writes “So far, Sarah has read 15 + 17 = 32 of the entire book. That means she has 100 − 32 = 68 left to read.” (p. 143, question 98), with no percent signs, so 32 and 68 read as amounts of the book (pages, say) that the question never gives. The question gives 15% and 17%, and the handbook’s next line begins “68% =”, so the site now reads 15% + 17% = 32% and 100% − 32% = 68%. The answer, 17/25, is unchanged.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction: $p\% = \frac{p}{100}$; p is the percent. Percent means “out of 100,” so p percent is p hundredths. For example, 50% is 50 out of 100, which is one half.
Simplifying a fraction: $\frac{a}{b} = \frac{a \div n}{b \div n}$; a is the top number, b is the bottom, and n is a number that divides both evenly. Dividing the top and bottom by the same number keeps the value the same. A fraction is in lowest terms when no number bigger than 1 divides both.

Sort out the question
Sarah read 15% of her book yesterday and 17% today. Both are percents of the whole book. You need the part still left to read, as a fraction in lowest terms. That means a fraction that can’t be simplified any more.
Find how much she has read
$15\% + 17\% = 32\%$
Both percents are of the same whole book, so you can add them. She has read 32% of it.
Find how much is left
The whole book is 100%.
$100\% - 32\% = 68\%$
Watch out: 32% is the part she has read, not the part left. Rounding 32% to 30% gives $\frac{3}{10}$ , choice B.
Write 68% as a fraction
$p\% = \frac{p}{100}$
$68\% = \frac{68}{100}$
This fraction is not in lowest terms yet, because 68 and 100 are both even.
Write it in lowest terms
The number 4 divides both 68 and 100 evenly.
$\frac{a}{b} = \frac{a \div n}{b \div n}$
$\frac{68}{100} = \frac{68 \div 4}{100 \div 4} = \frac{17}{25}$
Watch out: don’t round 68% to 70%. That gives $\frac{7}{10}$ , choice D. So the answer is C, $\frac{17}{25}$ .

Check your answer

Turn the answer back into a percent: $\frac{17}{25} = \frac{68}{100}$ , which is 68%. Then 68% + 32% = 100%, the whole book. Also, 17 and 25 share no factor except 1, so the fraction is in lowest terms.

More questions like this

The original DOE question

Question 98 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 98. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 42 of 57

2023 Test A Math, question 42

Question

$\{1,2,3,4,5,\ldots,198,199,200\}$

How many members of the set shown above are multiples of 6 but not multiples of 9?

Show the answer

Correct answer: D

First, list the first several multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54 . . .

Every third multiple (i.e., 18, 36, 54, . . .) is also a multiple of 9. Thus, $\frac{2}{3}$ of the multiples of 6 are not multiples of 9.

Next, calculate how many multiples of 6 are in the given number range. The last multiple of 6 would be 198. So, there are a total of $\frac{198}{6} = 33$ multiples of 6. Since $\frac{2}{3}$ of those are not multiples of 9, the answer is $\frac{2}{3}(33) = 22$ .

Question analysis

This number properties question shows the set of whole numbers from 1 to 200 and asks how many of them are multiples of 6 but not multiples of 9. Students count multiples in a range and remove the overlap, the numbers divisible by both. Finding the right common multiple, and removing only the overlap, is where it gets tricky.

NYS standards: 6.NS.4 common factors and multiples
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Multiples and common multiples are grade 6 content, but counting the members of a set that meet one condition and fail another is rare in grade 7 work. The overlap between the two kinds of multiples must be found and removed exactly.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: No obvious method moderate
Much to keep track of minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 16 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Counting multiples of 6 is routine; the harder decision is how to remove the ones that are also multiples of 9. Those are the multiples of 18, the least common multiple, not of 54 or 15, and students who count them correctly must still subtract them rather than report them. Listing 6, 12, 18, 24, 30, 36 shows the pattern, two kept for every one removed, which also gives the answer as two thirds of 33.

No obvious method moderate Working it out
The stem never mentions a common multiple: students must see that the multiples of 6 that are also multiples of 9 are the multiples of 18, every third multiple of 6. (Schoenfeld, 1985)
Much to keep track of minor Working it out
Three counts are in play at once, 33 multiples of 6 up to 198, 11 multiples of 18 up to 198, and their difference, and two of them are listed choices. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math minor Understanding the question
The bold not in “multiples of 6 but not multiples of 9” asks for what is left after a removal, a set difference rather than a count of the numbers removed. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Count the multiples of 6 up to 200, find the smallest number that is a multiple of both 6 and 9, count its multiples in the set, and subtract.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Counts the 11 multiples of 18, the numbers to be removed, and gives that count instead of the multiples of 6 that remain.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Takes 6 + 9 = 15 as the step between numbers divisible by both, and gives its 13 multiples up to 195, the numbers it would remove.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Uses 6 + 9 = 15 instead of 18 as the common multiple and removes its 13 multiples from the 33 multiples of 6, leaving 20.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting multiples: $\text{last multiple} = n \times k$; n is the number whose multiples you count, such as 6. The first multiple is n × 1, the second is n × 2, and so on. So if the largest multiple in the set is n × k, there are k multiples in all.
Least common multiple: $\text{LCM}(6, 9) = 18$; The least common multiple, or LCM, is the smallest number that two numbers both divide evenly. A number is a multiple of both 6 and 9 exactly when it is a multiple of 18.

Sort out the question
The set holds the whole numbers from 1 to 200. A multiple of 6 is a number that 6 divides evenly, like 6, 12 or 18. You need the multiples of 6 that are not multiples of 9. So count all the multiples of 6. Then take away those that are also multiples of 9.
Count the multiples of 6
200 is not a multiple of 6, but 198 is. So 198 is the last multiple of 6 in the set.
$\text{last multiple} = n \times k$
$198 = 6 \times 33$
There are 33 multiples of 6.
Find the numbers to take away
List the multiples of 6: 6, 12, 18, 24, 30, 36. The first one 9 also divides is 18, then 36. So every third multiple of 6 is also a multiple of 9.
$\text{LCM}(6, 9) = 18$
Watch out: the shared multiples are multiples of 18, not of 6 + 9 = 15. Using 15 leads to 13 or 20, choices B and C.
Count the multiples of 18
The last multiple of 18 in the set is also 198.
$198 = 18 \times 11$
There are 11 of them. Watch out: these are the numbers you take away, not the answer. Stopping here gives 11, choice A.
Take away the multiples of 18
$33 - 11 = 22$
There are 22 multiples of 6 left over. So the answer is D, 22.

Check your answer

Every third multiple of 6 is a multiple of 18, so two out of every three multiples of 6 stay. Two thirds of 33 is $\frac{2}{3} \times 33 = 22$ . That matches. Test one: 30 is a multiple of 6, and 30 ÷ 9 is not a whole number, so 30 counts.

More questions like this

The original DOE question

Question 99 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 99. Open page 86 of the handbook (PDF)

Question 43 of 57

2023 Test A Math, question 43

Question

Kim jogs 8 kilometers in 1 hour 40 minutes. At that rate, how many meters does she jog per minute?

A0.08
B80
C800
D8,000

Show the answer

Correct answer: B

First, calculate Kim’s jogging time in minutes:

1 hour 40 minutes = 100 minutes

Then, convert kilometers to meters:

8 kilometers = 8,000 meters

Now simplify the ratio to find the answer:

$\frac{8{,}000}{100} = 80$

Question analysis

This rate question says Kim jogs 8 kilometers in 1 hour 40 minutes and asks how many meters she jogs per minute. Students convert a mixed time to minutes and kilometers to meters before finding a unit rate. Skipping a conversion, or stopping before dividing, gives answers that are off by factors of ten or more.

NYS standards: 6.RP.3d converting measurement units
6.RP.3b unit rate problems
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Unit rates and metric conversions are grade 6 topics. This needs two conversions, hours and minutes to minutes and kilometers to meters, then one division with whole numbers, simpler than grade 7 rate problems built on fractions.
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
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The rate itself is one division; what makes the question work is that the distance and the time both arrive in the wrong units. Converting only the time gives a speed in kilometers per minute, and converting only the distance and stopping gives the whole trip in meters. Writing the units beside every number, meters over minutes, shows when the answer is ready.

Something you must already know minor Working it out
The conversions are not given: 1 kilometer is 1,000 meters and 1 hour is 60 minutes, so 8 kilometers becomes 8,000 meters and 1 hour 40 minutes becomes 100 minutes. (Anderson & Krathwohl, 2001)
Many steps minor Working it out
Two conversions, one for distance and one for time, come before the single division, and skipping the distance conversion leaves an answer a thousand times too small. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Rewrite the time entirely in minutes and the distance entirely in meters, then divide distance by time; check that the answer's units are meters per minute.

Traps

Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
Stops one step early: A choice is a number from partway through the solution.
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 Units or scale (King et al., 2004)
Reveals partial knowledge Expected pull: strong
Divides the 8 kilometers by the 100 minutes but never changes kilometers to meters, giving a speed of 0.08 kilometer per minute.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Converts both units correctly but cancels only one zero when dividing 8,000 by 100, getting a rate ten times too large.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: weak
Converts 8 kilometers to 8,000 meters and stops, giving the whole distance before dividing by the 100 minutes it took.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Kilometers to meters: $\text{meters} = \text{kilometers} \times 1{,}000$; Each kilometer is 1,000 meters, so multiply the number of kilometers by 1,000 to get meters. The word part kilo means a thousand.
Rate: $\text{rate} = \frac{\text{distance}}{\text{time}}$; The rate, or speed, is how far you go in one unit of time. Distance is how far, and time is how long. Dividing shares the distance equally among the minutes, so you get the distance for one minute.

Sort out the question
Kim jogs 8 kilometers in 1 hour 40 minutes. You need her speed in meters per minute: how many meters she jogs in one minute. The question gives kilometers and hours, so change both units first.
Change the time to minutes
One hour is 60 minutes. Add the extra 40 minutes.
$60 + 40 = 100$
Kim jogs for 100 minutes.
Change the distance to meters
$\text{meters} = \text{kilometers} \times 1{,}000$
$8 \times 1{,}000 = 8{,}000$
Kim jogs 8,000 meters. Watch out: 8,000 is the whole distance, not meters per minute. Stopping here gives choice D.
Divide the distance by the time
$\text{rate} = \frac{\text{distance}}{\text{time}}$
$\text{rate} = \frac{8{,}000}{100} = 80$
She jogs 80 meters each minute. Watch out: if you skip the change to meters, you get 8 ÷ 100 = 0.08 kilometer per minute, choice A. So the answer is B, 80 meters per minute.

Check your answer

Work backward. At 80 meters a minute, 100 minutes gives $80 \times 100 = 8{,}000$ meters. That is 8 kilometers, just as the question says.

More questions like this

The original DOE question

Question 100 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 100. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 44 of 57

2023 Test A Math, question 44

Question

For what value of x is the equation $\frac{x}{5} - 4 = 3(4 - 2x) - 1$ true?

A $\frac{75}{11}$
B $\frac{75}{31}$
C $\frac{15}{7}$
D $\frac{65}{31}$

Show the answer

Correct answer: B

$\frac{x}{5} - 4 = 3(4 - 2x) - 1$	Apply the distributive property; multiply the 3 by 4 and $-2x$ .
$\frac{x}{5} - 4 = 12 - 6x - 1$	Combine like terms.
$\frac{x}{5} - 4 = 11 - 6x$	Apply the additive inverse property; add $6x$ and 4 to both sides of the equation.
$\frac{x}{5} + 6x = 15$	Get a common denominator.
$\frac{x + 30x}{5} = 15$	Apply the multiplicative inverse property; multiply both sides of the equation by 5.
$31x = 75$	Apply the multiplicative inverse property; divide both sides of the equation by 31.
$x = \frac{75}{31}$

Question analysis

This equation question asks for the value of x that makes x/5 − 4 = 3(4 − 2x) − 1 true. Students use the distributive property, collect variable terms on one side, and clear a fraction before solving. Each wrong choice matches a common slip with the parentheses or with the fraction, so careful bookkeeping matters more than speed.

NYS standards: 8.EE.7b linear equations with rational coefficients
7.EE.1 linear expressions with rational coefficients
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 equations have the form px + q = r. This one has x on both sides, a fractional coefficient and a product to expand, which are grade 8 equation skills, and every wrong choice matches a plausible slip along the way.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Error-prone arithmetic moderate
Many steps moderate
A very close wrong answer minor
Solution: 5 steps
Reading level: Question and choices: about grade 8, 14 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Every step is a routine move in solving a linear equation, but there are five of them and the fraction raises the cost of each. The listed wrong answers come from distributing the 3 to only one term, clearing the fraction from one term only, and stretching the parentheses over the −1. Substituting the result back is slow with a denominator of 31, so careful step-by-step work matters more than checking.

Error-prone arithmetic moderate Working it out
The term x/5 sits beside whole-number terms and a negative −6x, so clearing the fraction means multiplying every term by 5, and the answer is the unfriendly 75/31. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Many steps moderate Working it out
Five moves in a row, distributing the 3, combining 12 and −1, collecting the x terms, clearing the fifth and dividing by 31, each give a slip a place to enter. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer minor Choosing the answer
The fraction 75/11 shares its numerator with 75/31 and differs by one digit of the denominator, so only the coefficient of x, 6 1/5 rather than 2 1/5, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Distribute the 3 to both terms in the parentheses, combine constants, collect the x terms on one side, then multiply every term by 5 to clear the fraction; check by substituting.

Traps

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

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Distributes the 3 only to the 4, writing 12 − 2x, so the equation becomes x/5 + 2x = 15 and then 11x = 75.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Multiplies only x/5 by 5 to clear the fraction, leaving 6x and 15 unchanged, so it solves x + 6x = 15 and gets 7x = 15.
D Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Applies the 3 to the −1 as well, as if it sat inside the parentheses, so the right side becomes 9 − 6x and the constant 13 instead of 15.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The distributive property: $a(b - c) = ab - ac$; a, b and c are numbers or terms. Multiplying a by a difference means multiplying a by each part. The number outside must reach every term inside the parentheses.
Doing the same to both sides: $\text{if } a = b\text{, then } a + c = b + c \text{ and } a \times c = b \times c$; a and b are the two sides of an equation, and c is any number. An equation is like a balanced scale. Adding the same amount to both sides, or multiplying both sides by the same number, keeps it balanced.

Sort out the question
You need the value of x that makes both sides of the equation equal. The plan: clear the parentheses, gather the x terms on one side, and get rid of the fraction. Then divide.
Clear the parentheses
$a(b - c) = ab - ac$
$3(4 - 2x) = 12 - 6x$
$\frac{x}{5} - 4 = 12 - 6x - 1 = 11 - 6x$
Watch out: the 3 multiplies both 4 and 2x. Writing 12 − 2x leads to $\frac{75}{11}$ , choice A. The −1 is outside the parentheses, so the 3 does not touch it.
Gather the x terms
Add 6x to both sides to move the x terms to the left. Add 4 to both sides to move the plain numbers to the right.
$\text{if } a = b\text{, then } a + c = b + c \text{ and } a \times c = b \times c$
$\frac{x}{5} + 6x = 11 + 4 = 15$
Clear the fraction
Multiply both sides by 5. Every term gets multiplied, not only the fraction.
$5 \times \frac{x}{5} + 5 \times 6x = 5 \times 15$
$x + 30x = 75$
$31x = 75$
Watch out: multiplying only the fraction by 5 gives x + 6x = 15. That leads to $\frac{15}{7}$ , choice C.
Divide by 31
$x = \frac{75}{31}$
75 and 31 share no factor except 1, so the fraction is already simplified. So the answer is B, $\frac{75}{31}$ .

Check your answer

Put $x = \frac{75}{31}$ back in. Then $\frac{x}{5} = \frac{15}{31}$ and $2x = \frac{150}{31}$ . Write 4 as $\frac{124}{31}$ . The left side is $\frac{15}{31} - \frac{124}{31} = -\frac{109}{31}$ . On the right, $4 - \frac{150}{31} = -\frac{26}{31}$ , and $3 \times \left(-\frac{26}{31}\right) - 1 = -\frac{78}{31} - \frac{31}{31} = -\frac{109}{31}$ . Both sides match.

More questions like this

The original DOE question

Question 101 as printed on page 86 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 86, question 101. Open page 86 of the handbook (PDF)

Question 45 of 57

2023 Test A Math, question 45

Question

$A rectangle. Point R is on the top side, near its right end. Line segment RS runs down and to the right from R, crosses the right side of the rectangle, and ends at point S outside it. At R, the angle between the top side, to the left of R, and segment RS is marked 130°. Where RS crosses the right side, the angle outside the rectangle between the right side, above the crossing, and the part of RS going on to S is marked y.$

The diagram above shows line segment RS intersecting a rectangle. What is the measure of angle y?

A140°
B130°
C50°
D40°

Show the answer

Correct answer: A

First, find the measures of the interior angles of the triangle formed by the line segment and the rectangle. The top angle on the right is a vertex of the rectangle, so the measure of that angle is 90°. The top angle on the left is supplementary to the given angle of 130°. So, the measure of that angle is $180 - 130 = 50$ . The measure of the lower angle is 40°, since $180 - 90 - 50 = 40$ . Angle y is supplementary to the lower angle of 40°, so the measure of angle y is $180 - 40 = 140$ .

Question analysis

This geometry question shows line segment RS cutting across a corner of a rectangle, with a 130° angle marked where RS meets the top side, and asks for angle y where it crosses the right side. Students combine supplementary angles, the right angle of a rectangle and the angle sum of a triangle. Stopping at an angle inside the triangle is the main trap.

NYS standards: 8.G.5 angle sums, transversals, angle-angle similarity
7.G.5 angle relationships and equations
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 angle problems use supplementary, vertical and adjacent angles. This one adds the angle sum of a triangle, which New York places in grade 8, and chains three facts around a small triangle that students must first spot in the figure.
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
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 17 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The angle facts are basic, but the route runs through a triangle the figure does not point out, and the answer lies outside it. Students who stop inside the triangle give its 40° angle or its 50° angle, and those who assume y matches the marked 130° skip the work entirely. Checking that y and the triangle's angle beside it form a straight line, 140 + 40 = 180, confirms the result.

Reading a figure, table or graph moderate Understanding the question
The small triangle cut off at the rectangle's top right corner is never outlined or labeled, its 90° corner is unmarked, and y sits outside it, between RS and the rectangle's right side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Many steps minor Working it out
Three steps follow in order, a straight angle at R, the triangle's angle sum, and a straight angle at the right side, and the first two results are both listed choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Something you must already know minor Working it out
The rectangle's corner must be taken as 90° although no right-angle mark is drawn, and the 180° totals for a straight line and a triangle are assumed. (Anderson & Krathwohl, 2001)

Strategy: Look at the small triangle cut off at the rectangle's corner: get its angle at R from the straight top side, use the 90° corner and the triangle's angle sum, then use the straight right side.

Traps

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

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Takes y to equal the marked 130° angle, as if RS crossed two parallel sides, though the rectangle's top and right sides meet at a right angle.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds the triangle's angle at R, 180 − 130 = 50, and stops after the first of the three steps.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the triangle's third angle, 180 − 90 − 50 = 40, but y lies outside the triangle, beside that angle on the straight right side.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Angles on a straight line: $a + b = 180$; a and b are two angles side by side that together make a straight line. A straight line makes an angle of 180 degrees. So if you know one of them, subtract it from 180 to get the other.
Angles in a triangle: $a + b + c = 180$; a, b and c are the three angles inside any triangle. They always add up to 180 degrees. If you know two of them, subtract both from 180 to find the third.
Outside angle of a triangle: $\text{outside angle} = a + b$; An outside angle sits next to a triangle’s angle, along a straight line. a and b are the two inside angles at the other corners. The outside angle equals their sum.

Sort out the figure
Segment RS starts at R on the top side of the rectangle. It goes down to the right, crosses the right side, and ends at S. Call the rectangle’s top right corner P. Call the point where RS crosses the right side Q. Then R, P and Q make a small triangle. Angle y is at Q, outside the rectangle.
Find the triangle’s angle at R
The 130° angle and the triangle’s angle at R sit side by side along the top side, a straight line.
$a + b = 180$
$130 + \angle \mathrm{R} = 180$
$\angle \mathrm{R} = 180 - 130 = 50$
Watch out: 50° is only the first step. Stopping here gives choice C.
Find the triangle’s angle at Q
P is a corner of the rectangle, so its angle is 90°.
$a + b + c = 180$
$50 + 90 + \angle \mathrm{Q} = 180$
$\angle \mathrm{Q} = 180 - 140 = 40$
Watch out: y is outside the triangle, so this 40° angle, choice D, is not y.
Find angle y
Angle y and the triangle’s 40° angle at Q sit side by side along RS, a straight line.
$a + b = 180$
$y + 40 = 180$
$y = 140$
So the answer is A, 140°.

Check your answer

Use the outside angle rule. Angle y is outside the triangle at Q. The two inside angles at the other corners are 50° and 90°, so $y = 50 + 90 = 140$ . Also, y looks wider than a right angle in the figure, so it must be more than 90°.

More questions like this

The original DOE question

Question 102 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 87, question 102. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 46 of 57

2023 Test A Math, question 46

Question

ESSAY LENGTH

Number of Words	Number of Essays
<100	6
100–250	4
251–500	11
>500	9

All 150 students in Grade 8 at a school are assigned to write an essay on the same topic. A teacher records the number of words in a random sample of the essays, as shown in the table above. Based on this sample, how many students in the entire grade would be expected to write essays with at least 100 words?

A20
B30
C100
D120

Show the answer

Correct answer: D

The number of students in the sample who wrote an essay with at least 100 words is 24 out of the 30 students. Write it as a fraction: $\frac{24}{30} = \frac{4}{5}$ . To predict the number of students in the entire grade who would write an essay with at least 100 words, find $\frac{4}{5}$ of 150, which is $\frac{4}{5}(150) = 120$ .

Question analysis

This statistics question gives a table of word counts for a random sample of Grade 8 essays and asks how many of all 150 students would be expected to write at least 100 words. Students find the sample size, the share meeting the condition, and scale it to the whole grade. Matching the words at least to the table's ranges is the tricky part.

NYS standards: 6.SP.1b samples and populations
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Using a random sample to predict a count for a whole population is typical grade 7 statistics and ratio work. Students total the sample from a table, read at least 100 against the table's ranges, and scale by a whole-number factor.
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
Turning words into math minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 65 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The proportion at the end is routine; the question is decided at the table. Students must turn “at least 100 words” into the right rows and also find the sample size, which is printed nowhere. Each listed wrong answer scales up a different group of rows, so a careless reading still produces a believable number. Noting that only 6 of 30 essays fall short, one fifth, gives a quick check.

Reading a figure, table or graph moderate Understanding the question
The table never prints the sample size; it has to be built by adding all four rows, 6 + 4 + 11 + 9 = 30, and the qualifying essays sit in three rows, not one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Turning words into math minor Understanding the question
The bold “at least 100 words” must be matched to the rows 100–250, 251–500 and >500, keeping the row that begins at 100 and leaving out only the <100 row. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Four steps follow in order: the sample total, the qualifying count of 24, the fraction 24/30 = 4/5, and that fraction of all 150 students. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the table's counts to find the sample size, add the rows with 100 or more words, and scale that fraction of the sample up to all 150 students.

Traps

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

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Counts only the 100–250 row as at least 100 words, then scales its 4 essays up to the grade: 4/30 of 150 = 20.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Scales up the 6 essays with fewer than 100 words, 6/30 of 150 = 30, the very students the question leaves out.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Leaves out the 100–250 row, as if at least 100 meant more than 250, so it scales 11 + 9 = 20 essays up to 100.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from a sample: $\text{expected number} = \frac{\text{count in sample}}{\text{sample size}} \times \text{whole group}$; A sample is a smaller group picked at random from the whole group. The fraction of the sample with some feature is your best guess for the fraction of the whole group. So multiply that fraction by the size of the whole group.

Sort out the question
The table shows a random sample of the essays from a grade of 150 students. You need to predict how many of the 150 students wrote at least 100 words. At least 100 means 100 or more.
Find the sample size
The table doesn’t give the total, so add all four rows.
$6 + 4 + 11 + 9 = 30$
The sample has 30 essays.
Count the essays with enough words
Three rows count: 100–250, 251–500 and >500. Only the <100 row is left out.
$4 + 11 + 9 = 24$
Watch out: the 100–250 row counts, because 100 is at least 100. Leaving it out gives 100, choice C. Using the 6 short essays instead gives 30, choice B.
Scale up to the whole grade
$\text{expected number} = \frac{\text{count in sample}}{\text{sample size}} \times \text{whole group}$
$\frac{24}{30} \times 150 = \frac{4}{5} \times 150 = 120$
About 4 out of every 5 students write at least 100 words. So the answer is D, 120 students.

Check your answer

The whole grade, 150, is 5 times the sample of 30. So multiply the sample count by 5: $24 \times 5 = 120$ . The short essays scale to $6 \times 5 = 30$ , and 120 + 30 = 150, the whole grade.

More questions like this

The original DOE question

Question 103 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 87, question 103. Open page 87 of the handbook (PDF)

Question 47 of 57

2023 Test A Math, question 47

Question

Which percentage is closest in value to 0.0099?

A0%
B0.1%
C1%
D100%

Show the answer

Correct answer: C

Since 0.0099 is equal to 0.99%, the answer option closest to that value is 1%.

Question analysis

This percent question asks which of four percentages is closest in value to the decimal 0.0099. Students convert a small decimal to a percent by multiplying by 100, then compare it with choices that differ by factors of ten. Moving the decimal point the wrong number of places, or reading the digits 99 as a percent, leads to wrong answers.

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 Converting a decimal to a percent is grade 6 work. This decimal has two zeros before its first digit, so the point must move carefully, but it is a single conversion followed by picking the nearest choice, 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: Error-prone arithmetic minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 7 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One conversion answers the question, and the choices are spaced by powers of ten, so the work is all in the decimal places. A decimal this small looks like nothing, and students who forget to multiply by 100 see it as nearly 0%. Others move the point the wrong number of places or read the digits 99 as a percent. Seeing that 0.01 is exactly 1% puts 0.0099 just below it.

Error-prone arithmetic minor Working it out
The decimal 0.0099 has two zeros before its digits, so moving the point two places gives 0.99, and miscounting the places lands on 0.1% or 99%. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
Students must know that a percent is a number of hundredths, so 0.0099 is multiplied by 100 to become 0.99%, a rule the question takes for granted. (Anderson & Krathwohl, 2001)

Strategy: Multiply the decimal by 100 to write it as a percent, moving the point two places to the right, then choose the percent nearest to that value.

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.
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: strong
Reads 0.0099 as if it were already a percent, 0.0099%, which is nearest 0%, never multiplying by 100 to change the decimal into a percent.
B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Moves the decimal point one place instead of two, turning 0.0099 into about 0.1, and writes that as 0.1%, ten times too small.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Sees the digits 99, thinks of 99 percent, and picks the nearest choice, ignoring the zeros that make 0.0099 less than one hundredth.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimal to percent: $\text{percent} = \text{decimal} \times 100$; A percent is a number of hundredths. To change a decimal into a percent, multiply by 100. That moves the decimal point two places to the right. For example, 0.25 × 100 = 25, so 0.25 is 25%.

Sort out the question
The number 0.0099 is a decimal, but the choices are percents. To compare them, change 0.0099 into a percent first. Then find the choice nearest to it.
Change the decimal to a percent
$\text{percent} = \text{decimal} \times 100$
$0.0099 \times 100 = 0.99$
So 0.0099 is 0.99%. The decimal point moved two places to the right. Watch out: 0.0099 is not already a percent. Reading it as 0.0099% points to 0%, choice A.
Find the nearest choice
Find how far 0.99% is from each choice.
Choice Distance from 0.99%
A, 0% 0.99
B, 0.1% 0.89
C, 1% 0.01
D, 100% 99.01
Watch out: moving the point only one place gives about 0.1, choice B. So the answer is C, 1%.

Choice	Distance from 0.99%
A, 0%	0.99
B, 0.1%	0.89
C, 1%	0.01
D, 100%	99.01

Check your answer

Work the other way. 1% is 1 out of 100, or 0.01. And 0.0099 is just a tiny bit less than 0.0100. So 0.0099 is very close to 1%, just a little less.

More questions like this

The original DOE question

Question 104 as printed on page 87 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 87, question 104. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 48 of 57

2023 Test A Math, question 48

Question

The manager of a shipping company recorded the weights, in pounds, of the last 9 packages the company shipped. The manager displayed the distribution of the weights in a box plot with the five-number summary shown below:

minimum: 29
first quartile: 31
median: 42
third quartile: 73
maximum: 98

Which statement about the distribution of weights is supported by the box plot the manager created?

AThe minimum, 29, is an outlier.
BThe maximum, 98, is an outlier.
CThe interquartile range of the data is 11.
DThe interquartile range and the median are equal.

Show the answer

Correct answer: D

First, identify the first quartile and the third quartile:

first quartile: 31
third quartile: 73

Then, find the difference between the first quartile and the third quartile:

$73 - 31 = 42$

The median is given: 42.

So, the interquartile range and the median are equal.

Question analysis

This statistics question gives the five-number summary of nine package weights shown in a box plot and asks which statement about the distribution is supported. Students compute the interquartile range and decide whether the minimum or the maximum is an outlier. Judging outliers by eye, or subtracting the wrong pair of values for the IQR, makes wrong statements look true.

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 and testing for outliers are grade 7 standards in New York. There is no plot to read here, only five numbers, but every statement has to be checked by calculation, which makes it a careful grade 7 task.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Every choice must be checked moderate
Solution: 3 steps
Reading level: Question and choices: about grade 7, 84 words.
Students, last 30 days: 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 given outright, so the work is knowing what each statement claims and testing it. The outlier statements sound right because 98 looks far from 73 and 29 is the smallest value, but the 1.5 × IQR rule clears both. The true statement needs only one subtraction, 73 − 31, and a look at the median, which happens to have the same value.

Something you must already know moderate Working it out
Neither the interquartile range, the third quartile minus the first, nor the outlier rule, more than 1.5 × IQR beyond a quartile, is defined on the page. (Anderson & Krathwohl, 2001)
Every choice must be checked moderate Choosing the answer
The four statements test different ideas, two possible outliers, a value for the IQR and a comparison of the IQR with the median, so ruling out one says nothing about the next. (Embretson & Wetzel, 1987)

Strategy: Compute the interquartile range from the two quartiles, use it to set the outlier cutoffs 1.5 IQR beyond each quartile, and test every statement against those numbers.

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.
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 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Calls 29 an outlier just because it is the minimum, though it sits only 2 below the first quartile and far inside the cutoff 31 − 63 = −32.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Judges 98 an outlier because it lies 25 above the third quartile, without testing it against the cutoff 73 + 1.5 × 42 = 136.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: moderate
Subtracts the first quartile from the median, 42 − 31 = 11, instead of from the third quartile, measuring only the lower half of the box.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range: $\text{IQR} = Q_3 - Q_1$; Q1, the first quartile, is the middle of the lower half of the data. Q3, the third quartile, is the middle of the upper half. The interquartile range, or IQR, is the distance between them. It is the width of the box in a box plot. It shows how spread out the middle half of the data is.
Outlier rule: $\text{outlier: below } Q_1 - 1.5 \times \text{IQR} \text{ or above } Q_3 + 1.5 \times \text{IQR}$; An outlier is a value far away from the rest of the data. Q1 and Q3 are the quartiles, and IQR is the interquartile range. A value is an outlier if it lies more than 1.5 times the IQR below Q1 or above Q3.

Sort out the question
The manager’s box plot is drawn from the five numbers in the list, so you can work from the list. The weights, in pounds, are: minimum 29, first quartile 31, median 42, third quartile 73, maximum 98. You need the one statement that is true. So check each choice.
Find the interquartile range
$\text{IQR} = Q_3 - Q_1$
$\text{IQR} = 73 - 31 = 42$
The interquartile range is 42 pounds. That makes choice C, which says 11, false.
Compare the IQR with the median
The median is 42 pounds. The interquartile range is also 42 pounds. They are equal, so statement D is true.
Watch out: 42 − 31 = 11 subtracts Q1 from the median, not from Q3. That mistake makes choice C look right.
Test the ends for outliers
$\text{outlier: below } Q_1 - 1.5 \times \text{IQR} \text{ or above } Q_3 + 1.5 \times \text{IQR}$
$1.5 \times 42 = 63$
$31 - 63 = -32 \qquad 73 + 63 = 136$
An outlier would be below −32 or above 136. The minimum, 29, is not below −32. The maximum, 98, is not above 136. So A and B are false. Watch out: 98 looks far away, but not far enough to be an outlier.
Choose the answer
Only statement D is supported by the five numbers. So the answer is D, the interquartile range and the median are equal.

Check your answer

Check the subtraction by adding back: $31 + 42 = 73$ , which is Q3. The median, 42, sits inside the box, between 31 and 73, as it must. The two measures mean different things, but here they happen to be the same number.

More questions like this

The original DOE question

Question 105 as printed on page 88 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 88, question 105. Open page 88 of the handbook (PDF)

Question 49 of 57

2023 Test A Math, question 49

Question

$A shaded rectangle inside a larger, unshaded outer rectangle. The bottom side of the outer rectangle is labeled 4 ft. The left side of the shaded rectangle is labeled 2 ft. The shaded rectangle is wider than it is tall.$

In the figure above, the shaded rectangle is similar to the outer rectangle. The length of the outer rectangle is 4 feet, and the perimeter of the outer rectangle is 14 feet. If the width of the shaded rectangle is 2 feet, what is the area of the shaded rectangle?

A $5\frac{1}{3}$ sq ft
B6 sq ft
C $9\frac{1}{3}$ sq ft
D12 sq ft

Show the answer

Correct answer: A

For the outer rectangle, the length is 4 feet and the perimeter is 14 feet. Write and solve an equation to find the width:

$2(4) + 2w = 14$	Apply the additive inverse property; subtract 8 from both sides of the equation.
$2w = 6$	Apply the multiplicative inverse property; divide both sides of the equation by 2.
$w = 3$

Use that information in a proportion to find the length of the shaded rectangle:

$\frac{x}{4} = \frac{2}{3}$	Cross multiply.
$3x = 4(2)$	Apply the multiplicative inverse property; divide both sides of the equation by 3.
$x = \frac{8}{3}$

Now find the area of the shaded rectangle:

$2 \times \frac{8}{3} = \frac{16}{3} = 5\frac{1}{3}$

Question analysis

This geometry question shows a shaded rectangle similar to an outer rectangle that is 4 feet long with a perimeter of 14 feet; the shaded rectangle is 2 feet wide. Students find the shaded area. They must find a missing side, match corresponding sides, and scale by a factor rather than a fixed amount, which makes it a multistep similar-figures problem.

NYS standards: 8.G.4 similarity and dilations
7.G.1 scale drawings
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Similar figures are grade 8 geometry, though grade 7 scale drawings use the same proportions. Students must first find a missing side from the perimeter, match corresponding sides, solve a proportion with a fractional result, and then find an area.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 57 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is familiar, but the three together, perimeter to width, width to scale factor, and sides to area, demand a plan. The strongest pull is scaling by subtraction, shrinking each side by 1 foot because the width drops from 3 to 2, which gives a neat whole-number area. Others stop at the outer rectangle or add the sides instead of multiplying. Checking that 8/3 to 2 matches 4 to 3 confirms the shape is similar.

Combines ideas moderate Working it out
Perimeter, similarity and area are chained: 14 feet gives the outer width of 3 feet, the 2-foot side sets a scale factor of 2/3, and only then can the area be found. (Embretson, 1998; Daniel & Embretson, 2010)
Reading a figure, table or graph minor Understanding the question
The figure marks the shaded rectangle's 2-foot side running up and down and the outer 4-foot length along the bottom, so the 2 feet must be paired with the outer 3-foot side, which is not labeled. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
The scale factor 2/3 makes the shaded length 8/3 feet, and the area 2 × 8/3 = 16/3 has to be written as the mixed number 5 1/3. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Use the perimeter to find the outer rectangle's width, match the shaded width of 2 feet to it, scale the 4-foot length by the same factor, then multiply length by width.

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.
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.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Shrinks each side by the same amount instead of the same factor: the width drops by 1 foot, from 3 to 2, so the length drops from 4 to 3, giving 6.
C Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Finds the shaded sides, 2 and 8/3 feet, correctly but computes the perimeter, 2 × (2 + 8/3) = 28/3 = 9 1/3, instead of the area.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Finds the outer width of 3 feet and gives the outer rectangle's area, 4 × 3 = 12, instead of the smaller shaded one.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a rectangle: $P = 2l + 2w$; P is the perimeter, the distance all the way around. l is the length and w is the width. A rectangle has two lengths and two widths, so you add each one twice.
Scale factor of similar shapes: $\text{scale factor} = \frac{\text{small side}}{\text{matching large side}}$; Similar shapes have the same shape but different sizes. Each side of the small shape equals its matching side on the large shape times one number, the scale factor. So one pair of matching sides tells you the scale factor.
Area of a rectangle: $A = l \times w$; A is the area, the space inside. l is the length and w is the width. You multiply because the rectangle is made of rows of unit squares: w rows with l squares in each.

Sort out the question
The outer rectangle is 4 feet long, and its perimeter is 14 feet. The shaded rectangle is similar to it, so it has the same shape. The shaded rectangle’s width, its left side, is 2 feet. You need the shaded area, so you also need the shaded length.
Find the outer rectangle’s width
$P = 2l + 2w$
$14 = 2 \times 4 + 2w$
$14 = 8 + 2w$
$2w = 6, \quad w = 3$
The outer rectangle is 3 feet wide. That is its up-and-down side.
Find the scale factor
The shaded width, 2 feet, matches the outer width, 3 feet.
$\text{scale factor} = \frac{\text{small side}}{\text{matching large side}}$
$\text{scale factor} = \frac{2}{3}$
Every shaded side is $\frac{2}{3}$ of the matching outer side. Watch out: similar shapes shrink by the same factor, not by the same number of feet. Taking 1 foot off each side gives 2 by 3, an area of 6, choice B.
Find the shaded length
Multiply the outer length, 4 feet, by the scale factor.
$l = \frac{2}{3} \times 4 = \frac{8}{3} = 2\frac{2}{3}$
The shaded rectangle is $2\frac{2}{3}$ feet long.
Find the shaded area
$A = l \times w$
$A = \frac{8}{3} \times 2 = \frac{16}{3} = 5\frac{1}{3}$
Watch out: 4 × 3 = 12 is the outer area, choice D. So the answer is A, $5\frac{1}{3}$ sq ft.

Check your answer

Check the shape. For the outer rectangle, length ÷ width is $4 \div 3 = \frac{4}{3}$ . For the shaded one, it is $\frac{8}{3} \div 2 = \frac{4}{3}$ . They match, so the shapes are similar. Also, $5\frac{1}{3}$ square feet is less than the outer 12, as it should be.

More questions like this

The original DOE question

Question 106 as printed on page 88 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 88, question 106. Open page 88 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 50 of 57

2023 Test A Math, question 50

Question

A rectangular concrete driveway is 30 feet long, 8 feet wide, and 6 inches thick. What is the volume of the concrete?

A44 cu ft
B48 cu ft
C120 cu ft
D240 cu ft

Show the answer

Correct answer: C

First convert 6 inches to 0.5 foot. Then, calculate the volume, in cubic feet:

$30 \times 8 \times 0.5 = 120$

Question analysis

This volume word problem describes a concrete driveway 30 feet long, 8 feet wide and 6 inches thick, and asks for the volume of the concrete in cubic feet. Students convert inches to feet and multiply three dimensions. Mixing units, or stopping at the area of the driveway's surface, is the main way to go wrong.

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 Volume of a rectangular prism with a fractional edge is grade 6 content. The only extra step is changing inches to feet before multiplying, so it is lighter than grade 7 volume work with composite or triangular prisms.
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
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 2, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Volume of a box is familiar, so the question is decided by the mix of units: two dimensions in feet and one in inches. Multiplying the three numbers as printed gives 1,440, which is not a choice, and that should send students back to convert. The listed wrong answers instead come from leaving out the thickness or combining the numbers without the volume rule.

Something you must already know minor Working it out
The stem gives no formula and no conversion, so students must bring both the volume rule, length times width times height, and the 12 inches in a foot. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
The thickness becomes the fraction 1/2 foot, so the product is 30 × 8 × 1/2, or, if the inches are kept, 1,440 divided by 12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Change 6 inches to a fraction of a foot first so that all three measurements share a unit, then multiply length, width and thickness.

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 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Adds the three numbers in the stem, 30 + 8 + 6 = 44, instead of multiplying the dimensions.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Multiplies only the width and the thickness, 8 × 6 = 48, combining two of the stem's numbers and leaving out the 30-foot length.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the area of the driveway's surface, 30 × 8 = 240, and stops before multiplying by the half-foot thickness.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Inches to feet: $\text{feet} = \text{inches} \div 12$; There are 12 inches in 1 foot. So to change inches into feet, divide the number of inches by 12.
Volume of a rectangular box: $V = l \times w \times h$; V is the volume, the space inside a box shape. l is the length, w is the width and h is the height, here the thickness. All three must use the same unit. The answer is in cubic units, such as cubic feet.

Sort out the question
The driveway is a flat box of concrete. It is 30 feet long, 8 feet wide and 6 inches thick. You need its volume, the space the concrete fills, in cubic feet. The thickness is in inches, so change it to feet first.
Change the thickness to feet
$\text{feet} = \text{inches} \div 12$
$6 \div 12 = \frac{1}{2}$
So 6 inches is half a foot. Now all three measures are in feet.
Multiply the three measures
$V = l \times w \times h$
$V = 30 \times 8 \times \frac{1}{2} = 240 \times \frac{1}{2} = 120$
The volume is 120 cubic feet. Watch out: 30 × 8 = 240 is only the area of the top surface, in square feet. Stopping there gives choice D. Adding 30 + 8 + 6 = 44 gives choice A.
So the answer is C, 120 cu ft.

Check your answer

Think of it another way. A slab 30 feet by 8 feet and a full foot thick would hold 240 cubic feet. This one is only half a foot thick, so it holds half as much, 120 cubic feet. Using 6 instead of one half would give 1,440, far too big.

More questions like this

The original DOE question

Question 107 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 89, question 107. Open page 89 of the handbook (PDF)

Question 51 of 57

2023 Test A Math, question 51

Question

At a carnival, visitors can win a prize if an employee cannot correctly guess their age. The employee records the first 15 ages that he guesses one Saturday. The box plot shown below displays data about the ages he recorded.

$A box plot above a number line from 7 to 18 with a tick at every integer. The left whisker starts at 7, the box runs from 8 to 17 with the median line at 10, and the right whisker ends at 18.$

What is the interquartile range of the box plot?

Show the answer

Correct answer: B

Quartile 1 is the median of the lower half of the data, 8.

Quartile 3 is the median of the upper half of the data, 17.

$\text{IQR} = \text{Q3} - \text{Q1} = 17 - 8 = 9$

The interquartile range is 9.

Question analysis

This statistics question shows a box plot of 15 ages that a carnival employee guessed and asks for its interquartile range. Students read the quartiles from the edges of the box and subtract. Confusing the interquartile range with the full range, or using the median line in place of a quartile, gives the wrong choices.

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 Reading a box plot and finding its interquartile range is a grade 7 standard in New York. The values sit on labeled whole-number ticks and one subtraction is needed, a routine grade 7 task once the box's edges are known as the quartiles.
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
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

The arithmetic is one subtraction, so the question tests whether students know which parts of a box plot to use. The whiskers' ends give the range, the line inside the box gives the median, and both are easy to grab in place of the quartiles. The story about the carnival and the 15 ages plays no part in the answer.

Something you must already know minor Working it out
The plot shows five values with no labels, so students must know that the box's ends mark the quartiles and that the interquartile range is the box's width, not the whiskers'. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The values 7, 8, 10, 17 and 18 are read off a scale ticked at every whole number, with the box's ends at 8 and 17 and the median line at 10. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the first and third quartiles on the plot by recalling which parts of a box plot mark them, then subtract the smaller quartile 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.
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.
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

A Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: weak
Subtracts the median from the third quartile, 17 − 10 = 7, measuring only the part of the box right of the middle line.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Gives the median, 10, read off the line inside the box, taking a single value on the plot for the spread between the quartiles.
D Wrong formula (King et al., 2004)
Reveals a misconception Expected pull: strong
Subtracts the minimum from the maximum, 18 − 7 = 11, which is the range shown by the whiskers, not the width of the box.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Interquartile range: $\text{IQR} = Q_3 - Q_1$; Q1, the first quartile, is the middle of the lower half of the data. Q3, the third quartile, is the middle of the upper half. In a box plot, Q1 is the left end of the box and Q3 is the right end. So the IQR is the width of the box.

Sort out the question
A box plot shows five numbers. The whiskers, the lines sticking out, end at the smallest and largest values. The box covers the middle half of the data. The line inside the box is the median, the middle value. You need the interquartile range of the ages.
Read the five numbers
Read each one off the number line below the plot.
- Smallest age, the left whisker: 7
- Q1, the left end of the box: 8
- Median, the line inside the box: 10
- Q3, the right end of the box: 17
- Largest age, the right whisker: 18
Find the interquartile range
$\text{IQR} = Q_3 - Q_1$
$\text{IQR} = 17 - 8 = 9$
The interquartile range is 9 years. Watch out: 18 − 7 = 11 is the range, the whole spread from whisker to whisker. That is choice D. The median, 10, is choice C.
So the answer is B, 9.

Check your answer

Count the spaces across the box. From 8 to 17 there are 9 spaces of 1 year each, so the box is 9 years wide. The answer is also less than the whole range, 11, as it must be.

More questions like this

The original DOE question

Question 108 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 89, question 108. Open page 89 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 52 of 57

2023 Test A Math, question 52

Question

Vicente and Carla each ran 8 laps around a track. They started at the same time and place. If Vicente ran 1.5 times as fast as Carla, how many laps did Carla have left to finish when Vicente finished his 8th lap?

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

Show the answer

Correct answer: B

Vicente ran 1.5 times as fast as Carla, so that means for every lap Carla finished, Vicente finished 1.5 laps. Set up a proportion to calculate c, the number of laps Carla had completed when Vicente finished 8 laps:

$\frac{c}{1} = \frac{8}{1.5}$	Cross multiply.
$1.5c = 8(1)$	Apply the multiplicative inverse property; divide both sides of the equation by 1.5.
$c = 5\frac{1}{3}$

So, the number of laps Carla still had to finish is $8 - 5\frac{1}{3} = 2\frac{2}{3}$ .

Question analysis

This ratio word problem says Vicente and Carla each ran 8 laps, starting at the same time and place, and Vicente ran 1.5 times as fast. It asks how many laps Carla still had to run when Vicente finished. Students reason with a speed ratio and mixed numbers, and giving the laps run rather than the laps remaining is an easy slip.

NYS standards: 7.RP.3 multistep ratio and percent problems
7.NS.3 four operations with rational numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 rate problems usually give a speed and a time. Here only the ratio of two speeds is given, so students must reason that distances run in the same time share that ratio, then work with a fractional number of laps.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Error-prone arithmetic moderate
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 44 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

No time is given, so students must reason that in equal time distance scales with speed, which turns the race into one division and one subtraction. Both steps hide a trap: the division by 1.5 leaves a remainder that is easy to misread, and subtracting a mixed number from 8 needs regrouping. The result of the first step, Carla's laps run, is itself a choice, so the question also rewards rereading what is asked.

Turning words into math moderate Understanding the question
The phrase “1.5 times as fast” has to become distances in the same time: while Vicente runs 8 laps, Carla runs 8 ÷ 1.5, not 8 − 1.5 or 8 × 1.5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic moderate Working it out
Dividing 8 by 1.5 gives 16/3, or 5 1/3, a division by a decimal with a remainder, and 8 − 5 1/3 then needs a whole regrouped into thirds. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
A very close wrong answer minor Choosing the answer
The choice 2 1/2 is only a sixth of a lap from 2 2/3, so a rough estimate of how far Carla trails cannot separate them; only the exact quotient does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: In the same amount of time the faster runner covers 1.5 times the distance, so divide Vicente's laps by 1.5 to get Carla's, then subtract that from 8.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Works 80 ÷ 15 as 5 remainder 5 and writes 5.5, turning the remainder into a decimal digit, so Carla seems to have run 5 1/2 laps and 2 1/2 remain.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Finds 5 1/3 correctly but subtracts it from 8 by taking 5 from 8 and keeping the 1/3, getting 3 1/3 instead of regrouping a whole.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Finds how many laps Carla had run when Vicente finished, 8 ÷ 1.5 = 5 1/3, and gives that instead of the laps she still had left.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, rate and time: $d = r \times t$; d is the distance, r is the rate or speed, and t is the time. When two runners run for the same time, the faster one goes farther by the same factor as the speed. So 1.5 times as fast means 1.5 times as far.
Dividing by a fraction: $a \div \frac{m}{n} = a \times \frac{n}{m}$; a is any number and m over n is a fraction. Dividing by a fraction is the same as multiplying by its reciprocal, the fraction flipped upside down.

Sort out the question
Vicente and Carla each run 8 laps, starting together. Vicente runs 1.5 times as fast as Carla. You need how many laps Carla still has to run when Vicente finishes. First find how far Carla has run at that moment.
Compare their distances
$d = r \times t$
They have run for the same time. So Vicente’s distance is 1.5 times Carla’s. Let c be the laps Carla has run.
$8 = 1.5 \times c$
To find c, divide 8 by 1.5.
Divide 8 by 1.5
Write 1.5 as the fraction $\frac{3}{2}$ .
$a \div \frac{m}{n} = a \times \frac{n}{m}$
$c = 8 \div \frac{3}{2} = 8 \times \frac{2}{3} = \frac{16}{3} = 5\frac{1}{3}$
Carla has run $5\frac{1}{3}$ laps. Watch out: that is how far she has run, not what is left. Stopping here gives choice D.
Find the laps she has left
$8 - 5\frac{1}{3} = \frac{24}{3} - \frac{16}{3} = \frac{8}{3} = 2\frac{2}{3}$
Watch out: 8 − $5\frac{1}{3}$ is not $3\frac{1}{3}$ . You must break 1 whole into thirds first. Answering $3\frac{1}{3}$ is choice C.
So the answer is B, $2\frac{2}{3}$ laps.

Check your answer

Multiply back: $1.5 \times 5\frac{1}{3} = \frac{3}{2} \times \frac{16}{3} = 8$ . So when Carla has run $5\frac{1}{3}$ laps, Vicente has run 8. And $5\frac{1}{3} + 2\frac{2}{3} = 8$ , her full race.

More questions like this

The original DOE question

Question 109 as printed on page 89 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 89, question 109. Open page 89 of the handbook (PDF)

Question 53 of 57

2023 Test A Math, question 53

Question

At West College, students are randomly assigned to one of 20 dormitories and one of 6 dining rooms. Kharleen likes 6 of the dormitories and 2 of the dining rooms. What is the probability that she is assigned to both a dormitory and a dining room that she likes?

A10%
B12%
C19%
D38%

Show the answer

Correct answer: A

Kharleen likes $\frac{6}{20} = \frac{3}{10}$ of the dormitories. She likes $\frac{2}{6} = \frac{1}{3}$ of the dining rooms. Multiply those fractions together to calculate the probability of Kharleen being assigned to both a dormitory and a dining room that she likes: $\frac{3}{10} \times \frac{1}{3} = \frac{1}{10} = 10\%$ .

Question analysis

This probability word problem says students at West College are randomly assigned one of 20 dormitories and one of 6 dining rooms, and Kharleen likes 6 dormitories and 2 dining rooms. It asks for the chance that she gets both a dormitory and a dining room she likes. Students combine two independent events and write the result as a percent.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the chance that two independent events both happen is grade 7 compound probability. Each event's chance comes straight from the counts given, and joining them and writing the result as a percent is a typical grade 7 task.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 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 on one rule, multiplying the chances of two independent events. Students who know it reduce the fractions first and finish in seconds; students who don't tend to combine the counts, and 6 × 2 = 12 reads naturally as 12%. The other two choices match no common working, so the real contest is between the product of the counts and the product of the chances.

Something you must already know moderate Working it out
The rule that the chance of two independent events both happening is the product of their chances, 3/10 × 1/3, is assumed; nothing in the stem says to multiply. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The phrase “both a dormitory and a dining room that she likes” describes two separate random assignments that must both go her way, so the two chances are combined, not compared. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The choice 12% sits only 2 points above 10%, so a rough sense that the answer is small cannot separate them; only computing 3/10 × 1/3 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the chance of a liked dormitory and the chance of a liked dining room as fractions, multiply them because both must happen, and convert the product to a percent.

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.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Multiplies the numbers of liked options, 6 × 2 = 12, and reads that product as 12%, without dividing by the 20 × 6 possible assignments.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No combination of 6/20 and 2/6 gives 19%, a value between the right answer and either single chance, so it mostly draws estimates and guesses.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 38%, which is larger than either single chance, 30% or about 33%, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all outcomes}}$; P is the probability, the chance that something happens. The top counts the outcomes you want. The bottom counts every outcome that could happen. Each outcome must be equally likely.
Probability of two separate events: $P(\text{A and B}) = P(\text{A}) \times P(\text{B})$; A and B are two events that don’t affect each other, like two separate random picks. P(A) is the chance of A, and P(B) is the chance of B. The chance that both happen is the two chances multiplied.

Sort out the question
Kharleen gets one of 20 dorms, or dormitories, where students live. She also gets one of 6 dining rooms. Each is picked at random. She likes 6 dorms and 2 dining rooms. You need the chance that she gets a dorm she likes and a dining room she likes.
Find the chance of a liked dorm
$P = \frac{\text{outcomes you want}}{\text{all outcomes}}$
$P(\text{dorm}) = \frac{6}{20} = \frac{3}{10}$
She likes 6 of the 20 dorms. Divide the top and bottom by 2 to simplify.
Find the chance of a liked dining room
Use the same formula for the dining rooms.
$P(\text{dining room}) = \frac{2}{6} = \frac{1}{3}$
She likes 2 of the 6 dining rooms. Divide the top and bottom by 2.
Multiply the two chances
The two picks are separate. One doesn’t change the other.
$P(\text{A and B}) = P(\text{A}) \times P(\text{B})$
$\frac{3}{10} \times \frac{1}{3} = \frac{3}{30} = \frac{1}{10}$
Both must happen, so the chance is smaller than either one alone.
Change to a percent
$\frac{1}{10} = \frac{10}{100} = 10\%$
Watch out: multiplying 6 × 2 = 12 and calling it 12% skips dividing by all the possible assignments. That is choice B. So the answer is A, 10%.

Check your answer

Count every possible assignment. There are $20 \times 6 = 120$ dorm and dining room pairs, all equally likely. She likes $6 \times 2 = 12$ of them. Then $\frac{12}{120} = \frac{1}{10}$ , which is 10%, the same answer.

More questions like this

The original DOE question

Question 110 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 110. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 54 of 57

2023 Test A Math, question 54

Question

A train travels 2,200 miles from Phoenix to New York City. The train covers the first 240 miles in 5 hours. If the train continues to travel at this rate, how many more hours will it take to reach New York City? Round your answer to the nearest whole hour.

Show the answer

Correct answer: D

First, find the speed of the train in miles per hour: $240 \div 5 = 48$ .

The number of miles left to travel is $2{,}200 - 240 = 1{,}960$ .

To find the number of hours left, use the equation $rt = d$ :

$48t = 1{,}960$	Apply the multiplicative inverse property; divide both sides of the equation by 48.
$t = \frac{1{,}960}{48} = 40\frac{5}{6}$

That number rounded to the nearest whole hour is 41 hours.

Question analysis

This rate word problem describes a 2,200-mile train trip from Phoenix to New York City in which the first 240 miles take 5 hours, and asks how many more hours the rest will take, rounded to the nearest hour. Students find a unit rate and the remaining distance. The choices are close together, so the time for the whole trip is a tempting trap.

NYS standards: 6.RP.3b unit rate problems
6.NS.2 multi-digit division
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Unit rates are grade 6 content, but this chains four steps: a speed, the distance left, a long division that does not come out even, and rounding. The word more must be read closely, which puts it at typical grade 7 demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
Turning words into math minor
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 4, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Each step is routine, but the question has a long chain and asks for only part of the trip. Dividing the whole 2,200 miles by the speed is the natural shortcut and gives a listed choice, and the rounding instruction adds a second place to go wrong. Checking that 41 hours at 48 miles per hour covers about 1,970 miles, close to the 1,960 left, confirms the answer.

Many steps moderate Working it out
Four steps follow in order, the speed 240 ÷ 5 = 48, the miles left 2,200 − 240 = 1,960, the time 1,960 ÷ 48, and the rounding, and an early slip carries through. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
Turning words into math minor Understanding the question
The bold more in “how many more hours” limits the question to the 1,960 miles still ahead, though 2,200, the whole trip, is the distance the stem names first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Error-prone arithmetic minor Working it out
Dividing 1,960 by 48 leaves 40 with a remainder of 40, so the time, 40 5/6 hours, must be rounded up to the nearest hour rather than cut short. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find the train's speed from the first 240 miles, subtract those miles from the total to get the distance left, divide by the speed, and round only at the end.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Divides the whole 2,200 miles by 48 miles per hour, about 45.8, or 46 hours, the time for the entire trip rather than the hours still to go.
B Rounds too early (King et al., 2004)
Reveals partial knowledge Expected pull: weak
Divides the whole 2,200 miles by 48 and then cuts 45.8 down to 45 instead of rounding to the nearest hour, two errors at once.
C An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Subtracts 240 from 2,200 without reducing the hundreds after borrowing, getting 2,060 instead of 1,960, and 2,060 ÷ 48 ≈ 42.9 rounds to 43.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance, rate and time: $d = r \times t$; d is the distance, r is the rate or speed, and t is the time. If you know two of them, you can find the third. To find the speed, divide the distance by the time. To find the time, divide the distance by the speed.

Sort out the question
The whole trip is 2,200 miles. The train goes the first 240 miles in 5 hours, and it keeps that speed. You need how many more hours it takes, rounded to the nearest whole hour. “More” means the time after the first 240 miles.
Find the train’s speed
$d = r \times t$
$240 = r \times 5$
$r = 240 \div 5 = 48$
Divide the 240 miles by the 5 hours. The train goes 48 miles per hour.
Find the miles still to go
$2{,}200 - 240 = 1{,}960$
There are 1,960 miles left. Watch out: use the miles left, not the whole trip. Dividing 2,200 by 48 gives about 45.8 hours, which leads to 46 or 45, choices A and B.
Find the hours still to go
$1{,}960 = 48 \times t$
$t = 1{,}960 \div 48 = 40\frac{40}{48} = 40\frac{5}{6}$
Here 48 × 40 = 1,920, and 40 miles are left over. That leftover is $\frac{40}{48} = \frac{5}{6}$ of an hour.
Round to the nearest hour
$\frac{5}{6}$ of an hour is more than half an hour, so round up to 41. Watch out: don’t just drop the fraction.
So the answer is D, 41 hours.

Check your answer

At 48 miles per hour, 41 hours covers $48 \times 41 = 1{,}968$ miles. That is just a bit more than 1,960, as it should be after rounding up. In 40 hours, the train would cover only 1,920 miles, which is not enough.

More questions like this

The original DOE question

Question 111 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 111. Open page 90 of the handbook (PDF)

Question 55 of 57

2023 Test A Math, question 55

Question

What is the least of four consecutive integers whose sum is 58?

Show the answer

Correct answer: D

Let n represent the least of the four integers. The other three integers are $(n + 1)$ , $(n + 2)$ , and $(n + 3)$ . Write an equation to solve for n:

$n + (n + 1) + (n + 2) + (n + 3) = 58$

$4n + 6 = 58$

$4n = 52$

$n = 13$

Question analysis

This number properties question asks for the least of four consecutive integers that add up to 58. Students write the integers in terms of one unknown and solve a two-step equation, or reason from the average of the four. Starting the list at n + 1 instead of n, which shifts every integer by one, is the main trap.

NYS standards: 7.EE.4a two-step equations
6.EE.6 variables representing unknown numbers
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Writing and solving an equation of the form px + q = r is core grade 7 work. Expressing four consecutive integers in terms of the least one is a familiar setup, so this is a routine three-step grade 7 problem.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 11 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

One equation, 4n + 6 = 58, solves the question, and dividing 58 by 4 to find the middle of the four, 14.5, is quicker still. The only real risk is the count: writing the integers as n + 1 through n + 4 and reporting n gives a least integer one too small. The two smallest choices match no reasonable working, so they draw only guesses. Adding the four integers back up to 58 is a ten-second check.

Turning words into math minor Understanding the question
The phrase “four consecutive integers” has to become n, n + 1, n + 2 and n + 3, with n the least, so the sum is 4n + 6, not 4n + 10. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The choices 12 and 13 differ by one, the size of a fence-post error, so only checking that 13 + 14 + 15 + 16 = 58 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Call the least integer n, write the next three as n + 1, n + 2 and n + 3, add them and solve; or divide 58 by 4 to find the middle of the four.

Traps

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

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No equation for four consecutive integers gives 1; it may echo the step of 1 between them, but it mostly catches guesses.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working with a sum of 58 leads to 2, which may echo the + 2 in n + 2, so it mostly draws guesses.
C Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: strong
Writes the four integers as n + 1 through n + 4, solves 4n + 10 = 58 to get n = 12, and reports n instead of the least integer, n + 1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive integers: $n,\; n + 1,\; n + 2,\; n + 3$; Integers are whole numbers and their negatives, like −2, 0 and 7. Consecutive integers come one right after another, like 4, 5, 6, 7. If n is the least one, each next one is 1 more than the one before.

Sort out the question
You need four consecutive integers, numbers in a row like 7, 8, 9, 10. Together they add up to 58. You need the least, or smallest, of the four.
Name the four numbers
Let n be the least number. Then the other three are 1, 2 and 3 more than n.
$n,\; n + 1,\; n + 2,\; n + 3$
Write and simplify the equation
Add the four numbers and set the sum equal to 58.
$n + (n + 1) + (n + 2) + (n + 3) = 58$
$4n + 6 = 58$
To combine, count the n’s: there are four of them. The plain numbers make 1 + 2 + 3 = 6.
Solve for n
Subtract 6 from both sides. Then divide both sides by 4.
$4n = 58 - 6 = 52$
$n = 52 \div 4 = 13$
Watch out: if you call the numbers n + 1 to n + 4, you get n = 12. Then 12 is not one of the four numbers; the least one is 13. Answering 12 is choice C.
So the answer is D, 13.

Check your answer

Add them up: $13 + 14 + 15 + 16 = 58$ . It works. Another way: $58 \div 4 = 14.5$ is their average. It falls halfway between the two middle numbers, 14 and 15. So the four numbers are 13, 14, 15 and 16.

More questions like this

The original DOE question

Question 112 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 112. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 56 of 57

2023 Test A Math, question 56

Question

$Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. The line is shaded from the unlabeled tick halfway between 0 and 1 to the unlabeled tick halfway between 1 and 2, with closed dots at both ends.$

If all possible values of x are indicated by the shaded part of the number line above, which number line best shows all possible values of $\frac{1}{x}$ ?

A $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 1 and 2 to the tick halfway between 2 and 3, with closed dots at both ends.$
B $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from 1 to 2, with closed dots at both ends.$
C $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from the tick halfway between 0 and 1 to 2, with closed dots at both ends.$
D $Number line from 0 to 3 with a tick every half unit, labeled at 0, 1, 2 and 3. Shaded from a closed dot between the tick halfway between 0 and 1 and the tick at 1, about two-thirds of the way from 0 to 1, to a closed dot at 2.$

Show the answer

Correct answer: D

The shaded region goes from $\frac{1}{2}$ to $\frac{3}{2}$ . So, the possible values of $\frac{1}{x}$ would be between the reciprocal of $\frac{1}{2}$ and the reciprocal of $\frac{3}{2}$ , which are 2 and $\frac{2}{3}$ , respectively. Find the number line in the answer options that has a shaded region that ranges from $\frac{2}{3}$ to 2, which is option D.

Question analysis

This number line question shades all possible values of x from one half to three halves and asks which number line shows all possible values of 1/x. Students find reciprocals of fractions and work out how a whole range of values changes. Keeping track of what happens to each end of the range, and placing fractions accurately on the line, makes it tricky.

NYS standards: 6.NS.1 dividing fractions by fractions
6.NS.7a inequality as relative position
6.NS.6c positioning rational numbers and points
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Reciprocals of single fractions are grade 6 work. Applying them to every value in a shaded range at once, and working out where the new range begins and ends, is reasoning about a whole set of numbers that grade 7 classes rarely practice.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
A very close wrong answer moderate
Variables instead of numbers minor
Solution: 3 steps
Reading level: Question and choices: about grade 12, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Few students have flipped an interval before, so the question rewards testing its two ends and noticing that the order reverses. The harder part comes at the end: 2/3 falls between the tick marks, and a graph ending at the nearby 1/2 tick looks right at a glance. Students who believe a reciprocal is always at least 1, or who shift the interval instead of flipping it, reach the other graphs.

Reading a figure, table or graph moderate Understanding the question
The shaded segment's ends sit on unlabeled half-unit ticks, 1/2 and 3/2, and each choice must be read the same way, with 2/3 falling between ticks rather than on one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
A very close wrong answer moderate Choosing the answer
The graph from 1/2 to 2 matches the right graph's right end and differs at the left end by only a sixth of a unit, 1/2 against 2/3. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Variables instead of numbers minor Understanding the question
The question asks about 1/x for every x in a range, so a whole interval must be transformed, and its ends swap: the smallest x gives the largest 1/x. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Take the reciprocal of each end of the shaded part, ask whether a bigger x gives a bigger or a smaller 1/x, and look for the graph whose endpoints are exactly those values.

Traps

Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Adds 1 to each endpoint instead of taking reciprocals, sliding the shaded part from 1/2–3/2 one unit right, to 3/2–5/2.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Gets 2 from 1/2 but assumes a reciprocal cannot be less than 1, as flipping a proper fraction suggests, so it starts the shading at 1 instead of 2/3.
C Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: strong
Finds the reciprocals 2/3 and 2 but places 2/3 at the half-unit tick, 1/2, though 2/3 lies a third of the way from that tick toward 1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Reciprocal: $\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$; a and b are the top and bottom of a fraction. The reciprocal flips the fraction upside down. The number 1 over x is the reciprocal of x. A number times its reciprocal always makes 1.

Sort out the question
x can be any number in the shaded part of the first number line. You need the number line that shows every possible value of $\frac{1}{x}$ , the reciprocal of x. The reciprocal is the number flipped upside down, like $\frac{1}{2}$ for 2.
Read the shaded part
The ticks come every half unit. The shading starts at the tick halfway between 0 and 1, which is $\frac{1}{2}$ . It ends halfway between 1 and 2, which is $\frac{3}{2}$ , or $1\frac{1}{2}$ . Both dots are filled in, so both ends count.
Find the reciprocals of the ends
Flip each end of the shaded part.
$\text{reciprocal of } \frac{a}{b} = \frac{b}{a}$
$\text{reciprocal of } \frac{1}{2} = \frac{2}{1} = 2$
$\text{reciprocal of } \frac{3}{2} = \frac{2}{3}$
So the ends of the new shading are 2 and $\frac{2}{3}$ .
Notice that the ends swap
When x gets bigger, $\frac{1}{x}$ gets smaller. So the smallest x, $\frac{1}{2}$ , gives the biggest value, 2. The biggest x, $\frac{3}{2}$ , gives the smallest, $\frac{2}{3}$ . So $\frac{1}{x}$ runs from $\frac{2}{3}$ to 2, ends included.
Find the matching number line
$\frac{2}{3}$ lies between the half tick and 1, a third of the way from $\frac{1}{2}$ to 1. Line D starts there and ends at 2. Watch out: line C starts at the half tick, $\frac{1}{2}$ , not at $\frac{2}{3}$ . Line B starts at 1.
So the answer is D, the line shaded from $\frac{2}{3}$ to 2.

Check your answer

Test a value in the middle. x = 1 is shaded, and $\frac{1}{1} = 1$ . On line D, 1 is shaded too. Test an end: $\frac{3}{2} \times \frac{2}{3} = 1$ , so $\frac{2}{3}$ really is the reciprocal of $\frac{3}{2}$ .

More questions like this

The original DOE question

Question 113 as printed on page 91 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 91, question 113. Open page 91 of the handbook (PDF)

Question 57 of 57

2023 Test A Math, question 57

Question

Chanelle selects a marble from a container and then returns it to the container. When she does this 3 times, the probability of choosing a red marble all 3 times is $\frac{1}{216}$ . Based on this information, what is the probability of Chanelle choosing a red marble the first time she selects a marble?

A $\frac{1}{72}$
B $\frac{1}{36}$
C $\frac{1}{8}$
D $\frac{1}{6}$

Show the answer

Correct answer: D

Let $\frac{1}{x}$ be the probability of Chanelle selecting a red marble the first time. The probability of selecting a red marble three times is $\frac{1}{x} \times \frac{1}{x} \times \frac{1}{x}$ , which is given as $\frac{1}{216}$ . Since $216 = 6 \times 6 \times 6$ , the value of x is 6, and the probability of Chanelle selecting a red marble the first time is $\frac{1}{6}$ .

Question analysis

This probability question says Chanelle draws a marble and puts it back, three times, and the chance of red all three times is 1/216. It asks for the chance of red on her first draw. Students reverse the multiplication rule for repeated independent draws, which means finding a cube root, and dividing by 3 is the tempting shortcut.

NYS standards: 7.SP.8a compound event as fraction of outcomes
8.EE.2 square and cube roots
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Multiplying probabilities for repeated draws with replacement is grade 7 compound probability. Running it backward, from three draws to one, needs a cube root, a grade 8 idea, though the numbers are friendly and the work is short once set up.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
No obvious method minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 55 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

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

Why it is hard

Once the three draws are seen as one chance cubed, the question is a cube root, and 216 is a perfect cube many students know. The risk lies in undoing the cube the wrong way: dividing by 3 because there are three draws, or stopping at 36 after one division by 6. A quick check, 1/6 × 1/6 × 1/6 = 1/216, confirms the answer.

Something you must already know moderate Working it out
Students must know that three draws with the marble returned each time multiply, p × p × p, and must recognize 216 as 6 × 6 × 6; the stem states neither. (Anderson & Krathwohl, 2001)
No obvious method minor Working it out
The usual direction runs from one draw to three; here the three-draw chance is given and the bold first asks for one draw, so students work backward and undo a cube. (Schoenfeld, 1985)

Strategy: Call the chance of red on one draw p, so three draws with replacement give p × p × p; then find the number whose cube equals the given probability.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Divides 216 by 3 because there are three draws, instead of finding the number that multiplied by itself three times makes 216.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Divides 216 by 6 to get 36 and stops one division short, keeping 1/36, which is the chance of two reds in a row, not one.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working with 1/216 gives 1/8, the chance of three heads in three coin tosses, so it mostly attracts guesses at a familiar fraction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of repeated picks: $P(\text{red 3 times}) = p \times p \times p$; p is the chance of red on one pick. Chanelle puts the marble back each time, so every pick has the same chance p, and one pick doesn’t change the next. The chance of red on all three picks is the three chances multiplied.

Sort out the question
Chanelle picks a marble and puts it back, 3 times. The chance of red all 3 times is $\frac{1}{216}$ . You need the chance of red on just the first pick. Call that chance p.
Write the three picks as a product
She puts each marble back, so the container is the same every time. Each pick has the same chance, p.
$P(\text{red 3 times}) = p \times p \times p$
$p \times p \times p = \frac{1}{216}$
Find the number that fits
You need a fraction that, multiplied by itself three times, makes $\frac{1}{216}$ . Its top is 1, since 1 × 1 × 1 = 1. For the bottom, find a number that times itself three times makes 216.
$6 \times 6 \times 6 = 36 \times 6 = 216$
So the bottom is 6, and p is $\frac{1}{6}$ .
Avoid the tempting shortcuts
Watch out: dividing 216 by 3 gives 72, choice A. Three picks means multiplying three times, not dividing by 3. Also, $\frac{1}{36}$ is the chance of red twice in a row, choice B.
So the answer is D, $\frac{1}{6}$ .

Check your answer

Multiply it out: $\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216}$ . That matches the question. Choice A fails fast: $\frac{1}{72} \times \frac{1}{72}$ is already far smaller than $\frac{1}{216}$ .

More questions like this

The original DOE question

Question 114 as printed on page 91 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 91, question 114. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).