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Practice Test

6.EE.2a writing algebraic expressions

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Question 1 of 6

2026 Practice Test C Math, question 5

Question

Maya read n pages for her English homework. Trevor read three times as many pages as Maya. Together, Maya and Trevor read 12 more pages than Ali read. Write an expression that shows how many pages Ali read, in terms of n.

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

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

Show the answer

Correct answer: 4n – 12

Let n represent the number of pages that Maya read.

Trevor read three times as many pages as Maya: Trevor’s pages =3n= 3n

To find how many pages they read together, add Maya’s and Trevor’s pages: n+3n=4nn + 3n = 4n

This means that together, they read 4n4n pages.

Ali read 12 fewer pages than Maya and Trevor read together. So 12 should be subtracted from the total:

Ali’s pages =4n−12= 4n - 12

Question analysis

Maya reads n pages, Trevor reads three times as many, and together they read 12 more pages than Ali. Students type an expression in n for the number of pages Ali read. It tests writing and simplifying expressions from words, and the comparison with Ali has to be translated with care.

NYS standards
  • 6.EE.6 variables representing unknown numbers
  • 6.EE.2a writing algebraic expressions
  • 6.EE.3 generating equivalent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Writing an expression from words is grade 6 content, but this chains three relationships, and the last compares the pair with Ali rather than describing Ali directly. Typing the answer with no choices to compare adds risk.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 52 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Building the pair's total, n + 3n = 4n, is quick; the question turns on the direction of the last comparison. The pair read more than Ali, so Ali's count is the smaller one, yet the word more pulls toward adding 12. Testing n = 10 settles it: Maya and Trevor read 40 pages together, so Ali must have read 28, not 52.

  1. Turning words into math moderate Understanding the question

    “Together, Maya and Trevor read 12 more pages than Ali read” compares the pair with Ali, so Ali read 12 fewer than 4n, though the word more invites adding 12. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    Every count is written in terms of n, so there is no page total to check an expression against unless the student picks a value for n and tests it. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The expression is typed into a box with no choices to compare, so an entry such as 4n + 12 gets no warning that the comparison was read backward. (Enright, Morley & Sheehan, 2002)

Strategy: Write Trevor's pages and the pair's total in terms of n, then test your expression with a value such as n = 10 to check that the pair really read 12 more than Ali.

Traps

  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding like terms
n+3n=4nn + 3n = 4n
n is the number of pages Maya read. Terms with the same letter are called like terms. Add them by adding the numbers in front: 1 n plus 3 n is 4 n. It works because 1 group of n pages and 3 more groups make 4 groups.
Turning a more-than fact around
if T=A+12, then A=T−12\text{if } T = A + 12 \text{, then } A = T - 12
T is the pages Maya and Trevor read together. A is the pages Ali read. The pair read 12 more than Ali, so Ali read 12 fewer than the pair. Subtracting 12 undoes adding 12.

  1. Sort out the question

    Maya read n pages. n is a variable, a letter that stands for a number you don’t know. Trevor read 3 times as many pages as Maya. Together they read 12 more pages than Ali. You need Ali’s pages, written with n.

  2. Write Trevor’s pages

    Three times as many as n is 3 times n:

    Trevor=3n\text{Trevor} = 3n

    A number written next to a letter means multiply. So 3n means 3 × n pages.

  3. Add Maya’s and Trevor’s pages

    Together means add. Maya read 1 group of n, and Trevor read 3 groups:

    n+3n=4nn + 3n = 4n

    Together they read 4n pages.

    Watch out: do not leave Maya out. Using only Trevor’s 3n gives 3n − 12, which is wrong.

  4. Decide who read more

    The pair read 12 more than Ali. So Ali read 12 fewer than the pair. Take 12 away from their total:

    Ali=4n−12\text{Ali} = 4n - 12

    Watch out: the word more tempts you to add 12. But 4n + 12 would mean Ali read more than the pair. And 12 − 4n subtracts in the wrong order.

  5. Enter the expression

    On the test, the keypad has an n key. Press 4, then n. Then press the minus key beside the plus key, then 1 and 2. You don’t need a times sign. On this site, type 4n-12.

    So you would enter 4n − 12. Any expression equal to it, such as −12 + 4n, is also right.

Check your answer

Pick a number for n, such as 10. Maya read 10 pages and Trevor read 30, so together they read 40. Your expression gives Ali 4×10−12=284 \times 10 - 12 = 28 pages. Is 40 twelve more than 28? Yes, 28+12=4028 + 12 = 40. So the expression works.

More questions like this

The original DOE question
Question 5 as shown on the DOE’s digital test, page 78 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 78, question 5.

Question 2 of 6

2026 Practice Test D Math, question 6

Question

The cost of shipping a package is $12, plus $2 for each additional pound over 3 pounds. Which expression represents the total cost, in dollars, of shipping a package that weighs p pounds? Assume that p≥3p \geq 3.

  • A12+2(p+3)12 + 2(p + 3)
  • B12+2(p−3)12 + 2(p - 3)
  • C14p+314p + 3
  • D14p−314p - 3
Show the answer

Correct answer: B

Since the additional cost per pound is not added until the package weighs more than 3 pounds, the total cost for 1, 2, or 3 pounds is $12.

  • 1 pound is $12
  • 2 pounds is $12
  • 3 pounds is $12

Since p is greater than or equal to 3 pounds, the total cost for:

  • 4 pounds is

12+(4−3)(2)=1412 + (4 - 3)(2) = 14

  • 5 pounds is

12+(5−3)(2)=1612 + (5 - 3)(2) = 16

  • p pounds is

12+(p−3)(2)=12+2(p−3)12 + (p - 3)(2) = 12 + 2(p - 3)

You may try to write the expression without generating the pattern by recognizing that subtracting 3 from the number of pounds for the package before multiplying by $2 and then adding $12 results in the total cost.

Question analysis

Shipping a package costs $12, plus $2 for each additional pound over 3 pounds, and students choose the expression for the total cost of a package that weighs p pounds. It tests writing an algebraic expression from a word problem. The per-pound charge starts only after the first 3 pounds, and turning that phrase into algebra is the tricky part.

NYS standards
  • 6.EE.6 variables representing unknown numbers
  • 6.EE.2a writing algebraic expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Writing an expression from a cost description is grade 6 algebra. Because the per-pound charge starts only after the first 3 pounds, it is less routine than a fee plus a simple rate, in line with typical grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 6, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards modeling the fee in two parts: a flat $12 that covers the first 3 pounds, and $2 for each pound beyond them. The two choices that start with 12 look equally reasonable until a weight is tried, and the two that begin with 14p come from merging the fee with the per-pound charge. Testing p = 3, which must cost exactly $12, rules out three choices in one step.

  1. Turning words into math moderate Understanding the question

    “$2 for each additional pound over 3 pounds” must become 2(p − 3), a charge on the pounds beyond the first 3, and the word “over” invites writing p + 3 instead. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    12 + 2(p + 3) and 12 + 2(p − 3) differ only in the sign inside the parentheses, so their shape cannot separate them; only a test weight or a careful reading can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    The weight is the letter p, so there is no cost to work out and compare; choosing a weight such as 5 pounds, which costs $16, turns every choice into a number. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Test a weight such as 5 pounds: work out its cost from the description, then substitute 5 for p in each choice and keep the one that gives the same cost.

Traps

  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost with a base fee and a rate
total=base fee+rate×extra pounds\text{total} = \text{base fee} + \text{rate} \times \text{extra pounds}
The base fee is the set price, $12, which covers the first 3 pounds. The rate is the price for each pound after that, $2. The extra pounds are the pounds over 3, which is p − 3. Use it when a price has a fixed part plus a part per unit.

  1. Sort out the question

    Shipping costs $12 to start. Each pound over 3 pounds adds $2. The package weighs p pounds, and p is at least 3. You need an expression for the total cost in dollars.

  2. Count the extra pounds

    Only the pounds above 3 cost extra. To find them, take 3 away from the weight:

    extra pounds=p−3\text{extra pounds} = p - 3

    For example, a 5-pound package has 5 − 3 = 2 extra pounds.

    Watch out: over 3 pounds means more than 3. It does not tell you to add 3. Writing p + 3 leads to choice A.

  3. Build the expression

    Each extra pound costs $2, so the extra pounds cost 2(p − 3) dollars. Add the $12 base fee:

    total=12+2(p−3)\text{total} = 12 + 2(p - 3)

    Watch out: the $12 is paid once, not for every pound. Choices C and D charge $14 for every pound, which mixes up the fee and the rate.

  4. Test each choice with 5 pounds

    A 5-pound package costs $12 plus two $2 charges, which is $16. Put p = 5 into each choice:

    ChoiceValue when p = 5
    A12 + 2(8) = 28
    B12 + 2(2) = 16
    C14(5) + 3 = 73
    D14(5) − 3 = 67

    Only B gives $16. So the answer is B, 12+2(p−3)12 + 2(p - 3).

Check your answer

Try p = 3, the smallest weight allowed. There are no extra pounds, so it should cost just $12. Choice B gives 12 + 2(0) = 12. Try p = 4: 12 + 2(1) = 14, which is $12 plus one $2 charge.

More questions like this

The original DOE question
Question 6 as shown on the DOE’s digital test, page 79 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 79, question 6.

Question 3 of 6

2024 Test A Math, question 23

Question

Let x be an odd number. In terms of x, what is the sum of the two even numbers closest to x?

  • Axx
  • B2x2x
  • C2x−22x - 2
  • D2x−42x - 4
Show the answer

Correct answer: B

If x is odd, then the two even numbers closest to x are (x−1)(x - 1) and (x+1)(x + 1). Find the sum of these two expressions:

(x−1)+(x+1)=2x(x - 1) + (x + 1) = 2x

Question analysis

An algebra question about odd and even numbers: x is odd, and students write the sum of the two even numbers closest to x in terms of x. It tests representing neighboring integers with expressions and adding them. Deciding exactly which even numbers are closest to an odd number is where errors start.

NYS standards
  • 6.EE.2a writing algebraic expressions
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students write and combine linear expressions; here the expressions describe neighboring integers, a familiar idea from consecutive-integer problems. Two short steps and a quick number check put it at typical grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 3, 26 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The algebra is a single addition, so the work is in naming the right two numbers. An odd x sits between two even numbers, one step away on each side, and the minus one and plus one cancel. Wrong choices come from picking both neighbors below x, from treating x as even, or from giving the middle of the pair; a test value catches each of them.

  1. Variables instead of numbers moderate Understanding the question

    Here x stands for any odd number, so the two even neighbors must be written as x − 1 and x + 1; trying x = 7, with neighbors 6 and 8, makes it concrete. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The phrase "the two even numbers closest to x" means one just below and one just above x, not two in a row on the same side. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Substitute a real odd number such as 7 for x, find its two closest even numbers and their sum, then see which expression gives that sum.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Neighbors of a number
x−1 and x+1x - 1 \text{ and } x + 1
x is any whole number. The number just below it is x − 1, and the number just above it is x + 1. Whole numbers take turns being odd and even. So if x is odd, both of its neighbors are even.
Combining like terms
x+x=2xx + x = 2x
Like terms have the same letter, such as x and 2x. To add them, add the numbers in front. x by itself means 1x, so x + x is 2x, which is two x’s.

  1. Sort out the question

    x is an odd number. You need the sum of the two even numbers closest to x. Sum means the result of adding. In terms of x means write the answer using the letter x.

  2. Try a real number first

    Pick an odd number, like 7. The even numbers closest to 7 are 6 and 8, one on each side.

    6+8=146 + 8 = 14

    Notice that 14 is 2×72 \times 7.

  3. Write the neighbors with x

    Odd and even numbers take turns. So both numbers right next to an odd x are even.

    x−1 and x+1x - 1 \text{ and } x + 1

    For x = 7, these are 6 and 8. Watch out: take one even number below x and one above. Using two even numbers below x, x−1x - 1 and x−3x - 3, gives 2x−42x - 4, choice D.

  4. Add the two neighbors

    (x−1)+(x+1)=x+x−1+1=2x(x - 1) + (x + 1) = x + x - 1 + 1 = 2x

    The −1 and the +1 cancel, and x+x=2xx + x = 2x. Watch out: x is the number in the middle, not the sum, so choice A is wrong. So the answer is B, 2x2x.

Check your answer

Try another odd number, 11. Its even neighbors are 10 and 12, and 10+12=2210 + 12 = 22. The answer 2x gives 2×11=222 \times 11 = 22. They match.

More questions like this

The original DOE question
Question 80 as printed on page 82 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 82, question 80. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 4 of 6

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?

  • Ak−6=2(j+6)k - 6 = 2(j + 6)
  • Bk−6=2j+6k - 6 = 2j + 6
  • C2(k−6)=j2(k - 6) = j
  • D2(k−6)=j+62(k - 6) = j + 6
Show the answer

Correct answer: A

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

k−6=2(j+6)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.
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Twice as many
A=2×BA = 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+aca(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.

  1. 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.

  2. 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.

  3. Write “twice as many”

    After the gift, Ken has more games. So Jeff’s new amount is the one that gets doubled.

    A=2×BA = 2 \times B

    k−6=2(j+6)k - 6 = 2(j + 6)

    Watch out: putting the 2 on Ken’s side, as in 2(k−6)=j+62(k - 6) = j + 6, would mean Jeff has twice as many. That is choice D.

  4. Keep the parentheses

    a(b+c)=ab+aca(b + c) = ab + ac

    2(j+6)=2j+122(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)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)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 5 of 6

2022 Test A Math, question 13

Question

Jay sold n tickets. Pilar sold three times as many tickets as Jay did. Together, Jay and Pilar sold ten more tickets than Amie sold. Which expression shows how many tickets Amie sold?

  • A10−4n10 - 4n
  • B3n2+103n^{2} + 10
  • C4n+104n + 10
  • D4n−104n - 10
Show the answer

Correct answer: D

Since Jay sold n tickets and Pilar sold 3 times the number of tickets Jay sold, Pilar sold 3n3n tickets.

Then, since together Jay and Pilar sold 10 more tickets than Amie, subtract 10 from (n+3n)(n + 3n) to get the expression that shows how many tickets Amie sold.

(n+3n)−10(n + 3n) - 10

Then combine like terms.

4n−104n - 10

Question analysis

Jay sells n tickets, Pilar sells three times as many, and together they sell ten more tickets than Amie. Students choose the expression in n for the number of tickets Amie sold. It tests writing and combining algebraic expressions from words, and the last sentence needs a careful reading, because it describes Amie only through the other two.

NYS standards
  • 6.EE.6 variables representing unknown numbers
  • 6.EE.2a writing algebraic expressions
  • 6.EE.3 generating equivalent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Writing an expression from words is grade 6 content, but this one links three people in a chain, and the last sentence compares the pair's total with Amie instead of describing Amie directly, which grade 7 students often misread.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Building the pair's total, n + 3n = 4n, is routine; the question is decided by which way the last comparison runs. Because the pair sold more than Amie, her expression must come out smaller than 4n, yet the most tempting choice adds the 10. Trying n = 5 settles it: the pair sold 20 tickets, so Amie sold 10, and only one expression gives 10.

  1. Turning words into math moderate Understanding the question

    “Together, Jay and Pilar sold ten more tickets than Amie sold” compares the pair's 4n with Amie, so her count is 10 less, though the word more suggests adding. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    Every count is written in terms of n, so no ticket total can be checked unless the student substitutes a value such as n = 5 into each expression. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Write Pilar's tickets and the pair's total in terms of n, then test each choice with a small value such as n = 5 against the sentence comparing the pair with Amie.

Traps

  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Combining like terms
a×n+b×n=(a+b)×na \times n + b \times n = (a + b) \times n
n is the unknown number. a and b are the numbers that multiply it. Like terms are terms with the same letter. Add the numbers in front: one n plus three n’s is four n’s, just as 1 apple plus 3 apples is 4 apples.
Ten more than
if A=B+10, then B=A−10\text{if } A = B + 10 \text{, then } B = A - 10
A and B are two amounts. If A is 10 more than B, A is the bigger one. To get the smaller amount, B, take 10 away from A. The words more than tell you which amount is bigger. They don’t always mean add.

  1. Sort out the question

    Jay sold n tickets. The letter n stands for a number you don’t know. Pilar sold three times as many as Jay. Jay and Pilar together sold ten more than Amie. You need an expression for Amie’s tickets. An expression is a math phrase with numbers and letters, like 2n + 1.

  2. Write Pilar’s and the pair’s tickets

    Three times as many as n is 3n. So Pilar sold 3n tickets. Together, Jay and Pilar sold n + 3n. Combine like terms:

    n+3n=1×n+3×n=4nn + 3n = 1 \times n + 3 \times n = 4n

    So the pair sold 4n tickets.

  3. Turn the last sentence into math

    The pair sold ten more than Amie. So the pair has the bigger number. Amie’s tickets plus 10 make the pair’s 4n.

    Amie+10=4n\text{Amie} + 10 = 4n

    To find Amie’s tickets, take 10 away from 4n.

    Amie=4n−10\text{Amie} = 4n - 10

    Watch out: the word more tempts you to add 10. That gives 4n+104n + 10, choice C. But Amie sold fewer tickets than the pair, not more.

  4. Rule out the other choices

    Choice A, 10−4n10 - 4n, subtracts in the wrong order. It is negative once n is 3 or more. Choice B, 3n2+103n^{2} + 10, multiplies n by 3n instead of adding them. So the answer is D, 4n−104n - 10.

Check your answer

Try n = 5. Jay sold 5 and Pilar sold 15, so together they sold 20. Amie sold 10 fewer, which is 10. Choice D gives 4×5−10=104 \times 5 - 10 = 10. That matches. Choice C gives 30, more than the pair sold.

More questions like this

The original DOE question
Question 70 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 2023 Admissions, page 82, question 70. 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 6 of 6

2018 Test B Math, question 48

Question

Coordinate axes with origin O. A shaded triangle stands on the x-axis: its base runs from the point marked m to the point marked n, both to the right of O, and its top vertex touches a dashed horizontal line drawn at 4 on the y-axis.

What is the area of the shaded triangle shown above?

  • Am+nm+n
  • Bn−mn-m
  • C2(n−m)2\left ( n-m \right )
  • D4(n−m)4\left ( n-m \right )
Show the answer

Correct answer: C

The height of the triangle is 4 units. The length of the base is n−mn-m. So the area is

A=12(n−m)(4)=2(n−m)A=\frac{1}{2}(n-m)(4)=2(n-m)

Question analysis

Students find the area of a shaded triangle drawn on coordinate axes, with its base between the points m and n on the x-axis and its top vertex at a height of 4. It tests the triangle area formula with an algebraic side length. Writing the base as an expression in m and n is the step that needs care.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Triangle area is grade 6 content, but a base written with two variables is less familiar. With only two steps, the demand matches typical grade 7 work that mixes geometry with algebraic expressions.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 7, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is trivial, but students must treat a length as a difference of coordinates and keep a formula's one half when the answer is a letter expression. Those who forget the half, or who stop at the base length, find their result among the choices. Trying m = 1 and n = 4 gives a triangle of area 6 that checks the expressions quickly.

  1. Variables instead of numbers moderate Understanding the question

    The base runs from m to n on the x-axis, so its length is the expression n − m, and the area stays in letters with no number to check. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Reading a figure, table or graph moderate Understanding the question

    The height, 4, is read from the dashed line through the apex rather than labeled on the triangle, and the base endpoints appear only as tick labels. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. A very close wrong answer moderate Choosing the answer

    4(n − m) is the right 2(n − m) doubled, written the same way except for the coefficient; only remembering the one half in a triangle's area separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the base as the difference between its endpoint values, read the height from the figure, and apply one-half times base times height.

Traps

  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.
  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height, the straight-up distance from the base to the top corner. You take half because a triangle is half of a rectangle with the same base and height.
Length along the x-axis
length=right end−left end\text{length} = \text{right end} - \text{left end}
The right end and the left end are the numbers where a segment starts and stops on the x-axis. Subtracting the smaller from the larger tells how long the segment is.

  1. Read the figure

    The shaded triangle sits on the x-axis, the flat number line. Its base runs from mm to nn. Its top corner touches the dashed line at height 4. You need the area, written with the letters mm and nn.

  2. Find the length of the base

    The base starts at mm and ends at nn. Use the length rule: right end minus left end.

    base=n−m\text{base} = n - m

    Watch out: adding the ends, m+nm+n, does not give a length. That is choice A.

  3. Find the height

    The height is the straight-up distance from the base to the top corner. The top corner is on the dashed line, which is at 4 on the y-axis. So the height is 4.

  4. Find the area

    Use the triangle area formula.

    A=12×(n−m)×4=2(n−m)A = \frac{1}{2} \times (n - m) \times 4 = 2(n - m)

    Half of 4 is 2, so the area is 2(n−m)2(n-m).

    Watch out: forgetting the one half gives 4(n−m)4(n-m), choice D. Stopping at the base gives n−mn-m, choice B.

  5. Give the answer

    The area is 2 times the length of the base. So the answer is C, 2(n−m)2(n-m).

Check your answer

Try numbers. Say m=1m = 1 and n=4n = 4. The base is 3 and the height is 4, so the area is 12×3×4=6\frac{1}{2} \times 3 \times 4 = 6. Choice C gives 2(4−1)=62(4-1) = 6. It matches. Choice D would give 12.

More questions like this

The original DOE question
Question 105 as printed on page 162 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 105. Open page 162 of the handbook (PDF)