SHSAT Sunset

Practice Test

2019 Handbook

2019 Test B Math

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Question 1 of 57

2019 Test B Math, question 1

Question

LAST YEAR’S TOTAL SALES

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

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

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

Show the answer

Correct answer: 250

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

x1,000=2,2509,000\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000}

x=(2,250)(1,000)9,000=2,2509=250x = \frac{(2,250)(1,000)}{9{,}000} = \frac{2{,}250}{9} = 250

Question analysis

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

NYS standards
  • 6.RP.3 ratio and rate reasoning
  • 6.SP.7 long-run relative frequency
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 proportion problems often take several steps or awkward numbers. This one scales a single share from a table to a new total, with numbers that divide cleanly, which is closer to grade 6 ratio work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 55 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the share is seen as 2,250 out of 9,000, a quarter, the arithmetic is light, which makes this one of the easier grid-ins. Points are lost by entering the share itself as a percent, by reading the wrong row of the table, or by building a monthly average when the question supplies next month’s total. With no choices to compare, checking that the entry is a quarter of 1,000 is the only safeguard.

  1. Turning words into math minor Understanding the question

    The phrase “best estimate (based on last year’s sales)” has to become a proportion: two-door cars keep the same share of next month’s 1,000 vehicles as of last year’s 9,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table lists three vehicle types and a total; only the two-door row, 2,250, and the 9,000 total are needed, so the four-door and truck rows are distractions. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The table shows last year’s sales. In all, 9,000 vehicles were sold, and 2,250 of them were two-door cars. Next month, 1,000 vehicles will be sold. You need about how many of them will be two-door cars.

    “Based on last year’s sales” means two-door cars keep the same share of all sales. A share is a part out of a whole. The four-door and truck rows don’t matter here.

  2. Find last year’s share of two-door cars

    Last year, 2,250 out of 9,000 vehicles were two-door cars. Write that as a fraction and simplify it.

    2,2509,000=14\frac{2{,}250}{9{,}000} = \frac{1}{4}

    This works because 2,250×4=9,0002{,}250 \times 4 = 9{,}000. So 1 out of every 4 vehicles sold was a two-door car.

  3. Set up the proportion and solve

    Let x be the number of two-door cars next month. Next month’s share must match last year’s share.

    x1,000=2,2509,000=14\frac{x}{1{,}000} = \frac{2{,}250}{9{,}000} = \frac{1}{4}

    To get x alone, multiply both sides by 1,000.

    x=14×1,000=250x = \frac{1}{4} \times 1{,}000 = 250

    So about 250 two-door cars will be sold.

  4. Give the answer

    Watch out: one fourth is the same as 25%. Gridding 25 gives the percent, not the number of cars. Also be sure to use the two-door row. The four-door row would give 650. So you would grid 250.

Check your answer

Try it another way. Next month’s 1,000 vehicles are 19\frac{1}{9} of last year’s 9,000. So take 19\frac{1}{9} of the 2,250 two-door cars: 2,250÷9=2502{,}250 \div 9 = 250. Both ways give 250 cars.

More questions like this

The original DOE question
Question 58 as printed on page 191 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 58. Open page 191 of the handbook (PDF)

Question 2 of 57

2019 Test B Math, question 2

Question

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

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

Show the answer

Correct answer: 13.5

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

w9=32\frac{w}{9} = \frac{3}{2}

w=3(9)2=272=13.5w = \frac{3(9)}{2} = \frac{27}{2} = 13.5

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Scale drawings are grade 7 content, and this is a typical one: a single length converted with a stated scale. The only extra demand is using the width rather than the length and entering the result with no choices to check.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

    “2 inches represents 3 yards” must become 1.5 yards for every inch of drawing, so the 9-inch width is multiplied by 3/2, not by 2/3. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The answer is a decimal entered with no choices to compare, so a result from the 15-inch length or from an upside-down scale looks just as finished. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The drawing shows a rectangle, a four-sided shape with square corners. On the drawing, the length is 15 inches and the width is 9 inches. The scale says 2 inches on the drawing stand for 3 yards in real life. You need the real width, in yards.

    Only the width matters. The 15-inch length is extra.

  2. Find what one inch stands for

    If 2 inches stand for 3 yards, then 1 inch stands for half as much.

    3÷2=1.53 \div 2 = 1.5

    So each inch on the drawing is 1.5 yards in real life. That means the number of yards will be bigger than the number of inches.

  3. Set up the proportion and solve

    Let w be the real width, in yards. Put yards on top and inches on the bottom, on both sides.

    w9=32\frac{w}{9} = \frac{3}{2}

    To get w alone, multiply both sides by 9.

    w=3×92=272=13.5w = \frac{3 \times 9}{2} = \frac{27}{2} = 13.5

    The real width is 13.5 yards.

  4. Give the answer

    Watch out: don’t flip the scale. Using 9×239 \times \frac{2}{3} gives 6, which is smaller than the drawing’s number. And don’t use the length: 15 inches would give 22.5 yards. So you would grid 13.5.

Check your answer

Count groups of 2 inches. The 9-inch width holds 4.5 groups of 2 inches. Each group stands for 3 yards, so the real width is 4.5×3=13.54.5 \times 3 = 13.5 yards. That matches.

More questions like this

The original DOE question
Question 59 as printed on page 191 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 191, question 59. Open page 191 of the handbook (PDF)

Question 3 of 57

2019 Test B Math, question 3

Question

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

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

Show the answer

Correct answer: 12

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

72=23×3272 = 2^{3} \times 3^{2}

60=22×3×560 = 2^{2} \times 3 \times 5

84=22×3×784 = 2^{2} \times 3 \times 7

Then, find the factors that all three have in common, which is 22×3=122^{2} \times 3 = 12.

Question analysis

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

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Greatest common factors are grade 6 content, normally of two numbers. This one uses three numbers and a word setting that must be translated, but the factoring is short, so the demand stays below typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 45 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    Nothing names a greatest common factor; “each child receives an equal number of each item and there are no items remaining” means the class size must divide 72, 60 and 84 exactly. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Finding the greatest common factor of three numbers, by prime factorization or by listing factors, is a grade 6 method the question assumes rather than explains. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The teacher has 72 crayons, 60 pencils and 84 sheets of paper. Every child gets the same number of each item, with none left over.

    So the number of children must divide 72, 60 and 84 with no remainder. A number like that is a common factor. You need the greatest one.

  2. Break each number into primes

    Write each number as a product of primes. A product is the answer to a multiplication.

    72=2×2×2×3×372 = 2 \times 2 \times 2 \times 3 \times 3

    60=2×2×3×560 = 2 \times 2 \times 3 \times 5

    84=2×2×3×784 = 2 \times 2 \times 3 \times 7

    You can check each one by multiplying it back out.

  3. Pick out the shared primes

    Look for the primes that appear in all three lists. Each list has at least two 2s and one 3. Only 72 has a third 2 and a second 3. The 5 and the 7 appear in one list each.

    GCF=2×2×3=12\text{GCF} = 2 \times 2 \times 3 = 12

    Watch out: keep every shared prime. Using only one 2 gives 6. Leaving out the 3 gives 4.

  4. Give the answer

    Watch out: 2,520 is the smallest number that all three numbers divide into. That is the least common multiple, a different idea. The class can’t have 2,520 children when there are only 60 pencils. So you would grid 12.

Check your answer

Share the items among 12 children. Each child gets 72÷12=672 \div 12 = 6 crayons, 60÷12=560 \div 12 = 5 pencils and 84÷12=784 \div 12 = 7 sheets. Nothing is left over. The numbers 6, 5 and 7 share no factor but 1, so no bigger class works.

More questions like this

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

Question 4 of 57

2019 Test B Math, question 4

Question

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

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

Show the answer

Correct answer: 8

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

prt=yprt = y

900(0.05)t=360900(0.05)t = 360

45t=36045t = 360

t=8t = 8

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Simple interest is grade 7 percent work, and a typical problem asks for the interest earned over a given time. This one runs the formula backward to find the time, one extra step, with friendly numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 7, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The numbers are friendly once one year’s interest is found, so most of the difficulty is setting up the right relationship. Students who divide $360 by 5 treat the rate as a dollar amount, and students who compare $360 with $900 stop at a percent. Multiplying back, eight years of $45, confirms the $360.

  1. Something you must already know minor Working it out

    Simple interest is the deposit times the rate times the years; the stem does not give this rule, and it must be known to see that each year adds the same $45. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The question gives the total interest, $360, and asks for the time, so the usual interest calculation has to be run backward, as 45t = 360. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Writing 5% as 0.05, not 0.5 or 0.005, is the one place-value decision, and a slip there changes the yearly interest by a power of ten. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Mr. Smith puts $900 into an account. It earns 5% simple interest each year. Interest is extra money the bank pays you for keeping your money there. You need the number of years until the interest adds up to $360.

  2. Write 5% as a decimal

    Percent means “out of 100.” So 5% is 5 out of 100.

    5100=0.05\frac{5}{100} = 0.05

    Watch out: 5% is not 0.5 or 0.005. Using 0.005 makes a year’s interest $4.50, and that leads to 80 years.

  3. Find the interest for one year

    Each year the account earns 5% of $900.

    900×0.05=45900 \times 0.05 = 45

    So it earns $45 every year. Watch out: 5% is not $5 a year. Dividing 360 by 5 gives 72, which is wrong.

  4. Use the formula to find the years

    Put the numbers into the formula. I is 360, P is 900 and r is 0.05.

    I=P×r×tI = P \times r \times t

    360=900×0.05×t360 = 900 \times 0.05 \times t

    360=45×t360 = 45 \times t

    To undo “times 45,” divide both sides by 45.

    t=360÷45=8t = 360 \div 45 = 8

  5. Give the answer

    It takes 8 years. Watch out: $360 is 40% of $900. Gridding 40 stops too soon, because each year adds only 5%. So you would grid 8.

Check your answer

Add up the interest year by year. $45 a year for 8 years is 45×8=36045 \times 8 = 360 dollars. Also, 8 years at 5% a year is 40%, and 40% of $900 is $360.

More questions like this

The original DOE question
Question 61 as printed on page 192 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 192, question 61. Open page 192 of the handbook (PDF)

Question 5 of 57

2019 Test B Math, question 5

Question

Solve the equation for x:

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

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

Show the answer

Correct answer: 7

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

(2149)(1448)x=0.875\left(\frac{21}{49}\right)\left(\frac{14}{48}\right)x = 0.875

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

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

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

Question analysis

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

NYS standards
  • 7.EE.3 multi-step rational number problems
  • 7.NS.2c properties for multiplying and dividing
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Seventh graders solve equations like ax = b, usually with a simple coefficient. Here the coefficient is a product of four numbers that is best cancelled first, and a decimal has to be matched with a fraction.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 6 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    Multiplied out, the coefficient is 294 over 2,352 and the right side is the decimal 0.875; both invite slips unless 21/49 and 14/48 are cancelled first. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. No obvious method minor Working it out

    The quick route pairs 21 with 49 and 14 with 48 before multiplying, so that 3/7 × 7/24 collapses to 1/8; the layout of the fraction does not suggest that pairing. (Schoenfeld, 1985)

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

    The result is entered with no choices, so a coefficient’s denominator or a quotient taken the wrong way looks as complete as the right value. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    You need the value of x. On top, 21, 14 and x are multiplied. The bottom is 49 times 48. The right side is 0.875.

    Multiplying 49 by 48 makes a big number. It is easier to simplify first.

  2. Split the fraction into two

    Use the rule for multiplying fractions backward. Pair 21 with 49, and 14 with 48.

    21×1449×48=2149×1448\frac{21 \times 14}{49 \times 48} = \frac{21}{49} \times \frac{14}{48}

    Now the equation says these two fractions times x equal 0.875.

  3. Simplify each fraction

    To simplify, divide the top and bottom by the same number. Both 21 and 49 divide by 7. Both 14 and 48 divide by 2.

    2149=371448=724\frac{21}{49} = \frac{3}{7} \qquad \frac{14}{48} = \frac{7}{24}

    Watch out: 48 ÷ 2 is 24, not 12. Writing 712\frac{7}{12} leads to 3.5.

  4. Multiply the simple fractions

    37×724=3×77×24=324=18\frac{3}{7} \times \frac{7}{24} = \frac{3 \times 7}{7 \times 24} = \frac{3}{24} = \frac{1}{8}

    The 7 on top and the 7 on the bottom cancel, since 7 ÷ 7 = 1. Then divide 3 and 24 by 3 to get 18\frac{1}{8}. So the equation becomes:

    18×x=0.875\frac{1}{8} \times x = 0.875

  5. Solve for x

    One eighth of x is 0.875. To undo dividing by 8, multiply both sides by 8.

    x=0.875×8=7x = 0.875 \times 8 = 7

    Watch out: 8 is not the answer, and don’t divide 0.875 by 8. So you would grid 7.

Check your answer

0.875 is the same as 78\frac{7}{8}. One eighth of 7 is 78\frac{7}{8}, so x = 7 works. You can also multiply it all out: 21×14×7=2,05821 \times 14 \times 7 = 2{,}058 and 49×48=2,35249 \times 48 = 2{,}352. Then 2,058÷2,352=0.8752{,}058 \div 2{,}352 = 0.875.

More questions like this

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

Question 6 of 57

2019 Test B Math, question 6

Question

Two glasses can hold the same amount of liquid. Glass A is 12\frac{1}{2} filled, and Glass B is 13\frac{1}{3} filled. If the liquid in Glass B is poured into Glass A, what fraction of Glass A will then be filled?

  • A56\frac{5}{6}
  • B45\frac{4}{5}
  • C34\frac{3}{4}
  • D15\frac{1}{5}
Show the answer

Correct answer: A

Add the fractions to find the answer:

12+13=36+26=56\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}

Question analysis

Two glasses of the same size hold liquid: one is half full and the other a third full. When the second is poured into the first, students find what fraction of the first glass is filled. The fraction word problem tests adding fractions with different denominators, and the care lies in finding a common denominator before adding.

NYS standards
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Adding two unlike fractions in a word problem is grade 7 rational-number work at its simplest, with halves and thirds. Typical grade 7 problems use mixed numbers, negative values or several steps.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 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

The question is a single sum of unlike fractions dressed as pouring, and halves and thirds are among the friendliest denominators. A quick sense check carries most of the weight: the result must exceed the 1/2 already in Glass A and stay below a full glass. Only one wrong choice comes from a real fraction error, adding the denominators, and it fails that check; the other two match no working.

  1. Error-prone arithmetic minor Working it out

    Halves and thirds cannot be added as written; both must become sixths, 3/6 and 2/6, and combining numerators or denominators straight across gives less than the half already in Glass A. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    The opening sentence, that both glasses hold the same amount, is what makes 1/2 and 1/3 fractions of the same whole, so pouring one into the other is a sum. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Because the two glasses are the same size, both fractions refer to the same whole, so the amount in Glass A is just their sum; check that it stays under one full glass.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top number, the numerator. b is the bottom number, the denominator. k is any whole number except 0. Multiplying the top and bottom by the same number gives an equal fraction. The pieces get smaller, but you get more of them.
Adding fractions with the same denominator
an+bn=a+bn\frac{a}{n} + \frac{b}{n} = \frac{a + b}{n}
n is the denominator. It tells how many equal parts make one whole. a and b are the numerators, how many of those parts you have. Parts of the same size can be added, so add the tops and keep the bottom.

  1. Sort out the question

    The two glasses can hold the same amount, so they are the same size. Glass A is 12\frac{1}{2} full. Glass B is 13\frac{1}{3} full. Glass B is poured into Glass A. You need how full Glass A is then. Both fractions are parts of the same whole, one glass. That means you add them.

  2. Give both fractions the same denominator

    The denominator is the bottom number. You can’t add halves and thirds as they are, because the pieces are different sizes. Find a number that 2 and 3 both go into. The smallest is 6, so change both fractions to sixths.

    ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}

    12=1×32×3=36\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}

    13=1×23×2=26\frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6}

  3. Add the sixths

    Now the pieces are the same size. Add the tops and keep the bottom.

    an+bn=a+bn\frac{a}{n} + \frac{b}{n} = \frac{a + b}{n}

    36+26=3+26=56\frac{3}{6} + \frac{2}{6} = \frac{3 + 2}{6} = \frac{5}{6}

    Glass A is now five sixths full.

    Watch out: don’t add the bottoms. Adding 2 and 3 to get one fifth is choice D. One fifth is less than the half Glass A already had, so it can’t be right.

    So the answer is A, 56\frac{5}{6}.

Check your answer

Estimate. Glass A starts half full. Adding more makes it more than half full. A third is less than a half, so the total is less than two halves, one full glass. Five sixths is more than one half and less than 1, so it fits.

More questions like this

The original DOE question
Question 63 as printed on page 178 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 178, question 63. Open page 178 of the handbook (PDF)

Question 7 of 57

2019 Test B Math, question 7

Question

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

  • A3
  • B5
  • C7
  • D9
Show the answer

Correct answer: C

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

x4.5=32\frac{x}{4.5} = \frac{3}{2} Cross multiply.
2x=3(4.5)2x = 3(4.5) Apply the multiplicative inverse property; divide both sides of the equation by 2.
x=13.52=634≈7x = \frac{13.5}{2} = 6\frac{3}{4} \approx 7

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Map scales are standard grade 7 proportional reasoning, and this problem is typical: one distance, one scale, then rounding. Nothing beyond a single proportion is needed.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

    “2 centimeters represents 3 kilometers” gives 1.5 kilometers per centimeter; set up the other way round, the scale shrinks the 4.5 centimeters instead of enlarging them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    The map distance is a decimal, 4.5 × 3 ÷ 2 = 6.75, and the result must then be rounded to the nearest kilometer, two places where a quick estimate drifts. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    On Fran’s map, 2 centimeters stand for 3 kilometers in real life. Her home and the park are 4.5 centimeters apart on the map. You need the real distance in kilometers, rounded to the nearest whole kilometer.

  2. Set up the proportion

    Let x be the real distance, in kilometers. Put kilometers on top and centimeters on the bottom, on both sides.

    x4.5=32\frac{x}{4.5} = \frac{3}{2}

  3. Solve for x

    To get x alone, multiply both sides by 4.5.

    x=3×4.52=13.52=6.75x = \frac{3 \times 4.5}{2} = \frac{13.5}{2} = 6.75

    So the real distance is 6.75 kilometers. Watch out: keep the scale exact until the end. Rounding 1.5 kilometers per centimeter up to 2 gives 4.5×2=94.5 \times 2 = 9, choice D.

  4. Round to the nearest kilometer

    6.75 is between 6 and 7. Its decimal part, 0.75, is more than 0.5, so round up to 7.

    Watch out: don’t flip the scale. Multiplying 4.5 by 23\frac{2}{3} gives 3, choice A. So the answer is C, 7.

Check your answer

Each centimeter stands for 3÷2=1.53 \div 2 = 1.5 kilometers. So 4 centimeters is 6 kilometers, and half a centimeter is 0.75 kilometer. That adds to 6.75, which rounds to 7. The real number should be bigger than 4.5, and it is.

More questions like this

The original DOE question
Question 64 as printed on page 193 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 193, question 64. Open page 193 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 8 of 57

2019 Test B Math, question 8

Question

Solve for x in terms of y if 2x+4=y2x + 4 = y.

  • Ax=y−42x = \frac{y - 4}{2}
  • Bx=2(y+4)x = 2(y + 4)
  • Cx=y+42x = \frac{y + 4}{2}
  • Dx=2(4−y)x = 2(4 - y)
Show the answer

Correct answer: A

2x+4=y2x + 4 = y Apply the additive inverse property; subtract 4 from both sides of the equation.
2x=y−42x = y - 4 Apply the multiplicative inverse property; divide both sides of the equation by 2.
x=y−42x = \frac{y - 4}{2}

Question analysis

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

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.EE.4a two-step equations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve two-step equations such as 2x + 4 = 10. Here the right side is the letter y, so the same two inverse steps leave an expression, a first taste of the grade 8 skill of rearranging.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 13 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

    (y − 4)/2 and (y + 4)/2 differ only in the sign before 4, and that sign depends on undoing the +4 by subtracting it from both sides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Variables instead of numbers minor Understanding the question

    With y on the right instead of a number, the answer is an expression, so there is no value to check unless the student substitutes a pair such as x = 1 and y = 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The equation is 2x+4=y2x + 4 = y. “Solve for x in terms of y” means get x alone on one side. The other side will have y in it. Treat y as if it were a number.

  2. Undo the adding of 4

    Right now, x is doubled and then 4 is added. Undo those steps in reverse order. First, subtract 4 from both sides.

    2x+4−4=y−42x + 4 - 4 = y - 4

    2x=y−42x = y - 4

    Watch out: the 4 must be subtracted from y. Writing 2x=y+42x = y + 4 leads to choice C.

  3. Undo the doubling

    Next, divide both sides by 2. Divide the whole right side, not only the y.

    2x2=y−42\frac{2x}{2} = \frac{y - 4}{2}

    x=y−42x = \frac{y - 4}{2}

  4. Give the answer

    Watch out: choices B and D multiply by 2. But x was multiplied by 2, so you undo that by dividing. So the answer is A, x=y−42x = \frac{y - 4}{2}.

Check your answer

Pick a value, such as x = 1. Then y=2×1+4=6y = 2 \times 1 + 4 = 6. Put y = 6 into choice A: 6−42=1\frac{6 - 4}{2} = 1. That gives back x = 1. Choice C would give 6+42=5\frac{6 + 4}{2} = 5, which is wrong.

More questions like this

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

Question 9 of 57

2019 Test B Math, question 9

Question

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

  • A120\frac{1}{20}
  • B110\frac{1}{10}
  • C320\frac{3}{20}
  • D15\frac{1}{5}
Show the answer

Correct answer: A

First, subtract Luis’s time from Bryan’s time: 11.74−11.69=0.0511.74 - 11.69 = 0.05.

Then, convert to a fraction:

0.05=5100=1200.05 = \frac{5}{100} = \frac{1}{20}.

Question analysis

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

NYS standards
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Subtracting decimals and writing a decimal as a fraction are skills from before grade 7. The two times are close, so the subtraction is quick, and the only care needed is with place value and simplifying.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 5, 32 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

    Subtracting 11.69 from 11.74 needs a borrow in the hundredths, 14 − 9, after which the tenths cancel to 0, the step where place value goes wrong. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    The difference, 0.05, must be read as five hundredths and reduced as a fraction; the question gives no help with converting a decimal. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths
a decimal with two places=its digits100\text{a decimal with two places} = \frac{\text{its digits}}{100}
The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths, and you can write its digits over 100. For example, 0.37 is 37/100. Then simplify by dividing the top and bottom by the same number.

  1. Sort out the question

    Bryan ran the race in 11.74 seconds. Luis ran it in 11.69 seconds. The difference is how much longer one time is than the other. You need it as a fraction of a second.

  2. Subtract the two times

    Subtract the smaller time from the larger one. Line up the decimal points.

    11.74−11.69=0.0511.74 - 11.69 = 0.05

    In the hundredths place, 4 is less than 9. So regroup: trade 1 tenth for 10 hundredths. Then 14 − 9 = 5 hundredths. The tenths are now 6 − 6 = 0.

    Watch out: don’t just take the smaller digit from the larger one in each place. That gives 0.15, which is 320\frac{3}{20}, choice C.

  3. Write the difference as a fraction

    The 5 is in the hundredths place, so 0.05 is 5 hundredths.

    0.05=51000.05 = \frac{5}{100}

    Now simplify. Divide the top and bottom by 5.

    5÷5100÷5=120\frac{5 \div 5}{100 \div 5} = \frac{1}{20}

  4. Give the answer

    Watch out: 0.05 is not 15\frac{1}{5}. One fifth is 0.2, which is four times as big. So the answer is A, 120\frac{1}{20}.

Check your answer

Turn 120\frac{1}{20} back into a decimal: 1÷20=0.051 \div 20 = 0.05. Then add it to Luis’s time: 11.69+0.05=11.7411.69 + 0.05 = 11.74. That is Bryan’s time, so the difference is right.

More questions like this

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

Question 10 of 57

2019 Test B Math, question 10

Question

A recipe uses 14\frac{1}{4} teaspoon of salt and 2122\frac{1}{2} teaspoons of sugar. What is the ratio of salt to sugar in this recipe?

  • A1:201:20
  • B1:101:10
  • C1:51:5
  • D1:41:4
Show the answer

Correct answer: B

14:212\frac{1}{4} : 2\frac{1}{2}

14:52\frac{1}{4} : \frac{5}{2}

4(14):4(52)4\left(\frac{1}{4}\right) : 4\left(\frac{5}{2}\right)

1:101 : 10

Question analysis

A recipe calls for 1/4 teaspoon of salt and 2 1/2 teaspoons of sugar, and students write the ratio of salt to sugar in whole numbers. It tests ratios whose terms are a fraction and a mixed number. Both terms have to be rescaled together, and the mixed number is where slips creep in.

NYS standards
  • 7.RP.1 unit rates with fractions
  • 6.RP.1 ratio concepts and language
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Ratios of fractions are grade 7 content, and this one is routine: convert a single mixed number, then scale both terms up to whole numbers. It asks for no further use of the ratio.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 27 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

A ratio of a fraction to a mixed number is unfamiliar enough that students reach for shortcuts, and each shortcut has a choice waiting. Pairing numerators after converting, or reading the salt's 1/4 as a ratio by itself, never compares the two amounts properly. Scaling both terms by the same factor, 4, or asking how many quarter teaspoons fit into 2 1/2 teaspoons, keeps the comparison intact.

  1. Error-prone arithmetic moderate Working it out

    The sugar is a mixed number, so 2 1/2 must become 5/2, and both terms must then be multiplied by 4, or 5/2 divided by 1/4, without dropping a denominator. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The recipe gives a fraction and a mixed number, but every choice is a whole-number ratio, so 1/4 : 2 1/2 has to be recognized in its scaled form. (Enright, Morley & Sheehan, 2002)

Strategy: Change 2 1/2 to 5/2, then divide the salt amount by the sugar amount; the quotient, written as a fraction with numerator 1, gives the ratio directly.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
  • An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Changing a mixed number to a fraction
212=2×2+122\frac{1}{2} = \frac{2 \times 2 + 1}{2}
A mixed number is a whole number plus a fraction. Multiply the whole number by the bottom number to count the halves in the wholes. Then add the top number, and keep the same bottom. Two wholes are 4 halves, so 2 and a half is 5 halves.
Equivalent ratios
a:b=(a×k):(b×k)a : b = (a \times k) : (b \times k)
a and b are the two amounts in a ratio, in order. k is any number except 0. Multiplying both amounts by the same number keeps the comparison the same. Use it to turn the fractions in a ratio into whole numbers.

  1. Sort out the question

    The recipe uses 14\frac{1}{4} teaspoon of salt and 2122\frac{1}{2} teaspoons of sugar. A ratio compares two amounts. “Salt to sugar” means salt first, then sugar. Every choice is a ratio of whole numbers. So you need to turn the two amounts into whole numbers that compare the same way.

  2. Change the mixed number to a fraction

    A mixed number is hard to work with. Write the sugar as a fraction.

    212=2×2+122\frac{1}{2} = \frac{2 \times 2 + 1}{2}

    212=4+12=522\frac{1}{2} = \frac{4 + 1}{2} = \frac{5}{2}

    So the ratio of salt to sugar is 14:52\frac{1}{4} : \frac{5}{2}.

  3. Multiply both parts by 4

    Multiply both parts by the same number to clear the fractions. Use 4, because both bottom numbers, 4 and 2, go into 4.

    a:b=(a×k):(b×k)a : b = (a \times k) : (b \times k)

    14:52=(14×4):(52×4)=1:10\frac{1}{4} : \frac{5}{2} = \left(\frac{1}{4} \times 4\right) : \left(\frac{5}{2} \times 4\right) = 1 : 10

    For the sugar, 5×4=205 \times 4 = 20, and then 20÷2=1020 \div 2 = 10.

    Watch out: forgetting to divide by 2 gives 1 : 20, choice A. Pairing the top numbers, 1 : 5, choice C, compares quarters with halves. And 1 : 4, choice D, is only the salt amount. It leaves out the sugar.

    So the answer is B, 1 : 10.

Check your answer

Think in quarter teaspoons. The salt is 1 quarter teaspoon. One teaspoon is 4 quarters, so the sugar is 2.5×4=102.5 \times 4 = 10 quarter teaspoons. That is 1 quarter of salt for every 10 quarters of sugar, or 1 : 10.

More questions like this

The original DOE question
Question 67 as printed on page 178 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 178, question 67. Open page 178 of the handbook (PDF)

Question 11 of 57

2019 Test B Math, question 11

Question

n=0.13+207100n = 0.13 + \frac{207}{100}

What is the value of n expressed as a fraction?

  • A13\frac{1}{3}
  • B115\frac{11}{5}
  • C94\frac{9}{4}
  • D2310\frac{23}{10}
Show the answer

Correct answer: B

n=0.13+207100=n = 0.13 + \frac{207}{100} =

13100+207100=\frac{13}{100} + \frac{207}{100} =

220100=115\frac{220}{100} = \frac{11}{5}

Question analysis

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

NYS standards
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Converting decimals to fractions and simplifying are skills taught before grade 7. Both numbers here are in hundredths, so the addition is immediate and only the final reduction takes thought.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 7, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

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

    The sum comes out as 220/100 or 2.2, but the choices are in lowest terms, so it must be reduced, or each choice converted, before anything matches. (Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    One term is a decimal and the other a fraction; writing 0.13 as 13/100 puts both in hundredths, and a misplaced digit there changes the sum. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Decimals as hundredths
a decimal with two places=its digits100\text{a decimal with two places} = \frac{\text{its digits}}{100}
The second place after the decimal point is the hundredths place. So a decimal with two places counts hundredths. Write its digits over 100. For example, 0.47 is 47/100.
Adding fractions with the same bottom
ac+bc=a+bc\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}
a and b are the tops. c is the shared bottom, the denominator. When the bottoms match, the pieces are the same size. So add the tops and keep the bottom.

  1. Sort out the question

    n equals 0.13 plus 207100\frac{207}{100}. One part is a decimal and one is a fraction. You need n as a fraction. The choices are in lowest terms, which means they can’t be simplified any more.

  2. Write the decimal as a fraction

    The 3 in 0.13 is in the hundredths place. So 0.13 is 13 hundredths.

    0.13=131000.13 = \frac{13}{100}

    Now both numbers are in hundredths. Watch out: 0.13 is not 13\frac{1}{3}. That only copies the digits, choice A.

  3. Add the hundredths

    Use the rule for adding fractions. The bottoms match, so add the tops and keep the 100 on the bottom.

    13100+207100=13+207100=220100\frac{13}{100} + \frac{207}{100} = \frac{13 + 207}{100} = \frac{220}{100}

  4. Simplify the fraction

    220100\frac{220}{100} is not a choice, so simplify it. Find the greatest number that divides both 220 and 100. That is 20.

    220÷20100÷20=115\frac{220 \div 20}{100 \div 20} = \frac{11}{5}

    So the answer is B, 115\frac{11}{5}.

Check your answer

Change both to decimals. 115=11÷5=2.2\frac{11}{5} = 11 \div 5 = 2.2. And 0.13+2.07=2.200.13 + 2.07 = 2.20. They match. Choices C and D are close, but 94=2.25\frac{9}{4} = 2.25 and 2310=2.3\frac{23}{10} = 2.3.

More questions like this

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

Question 12 of 57

2019 Test B Math, question 12

Question

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

  • A4
  • B5
  • C20
  • D40
Show the answer

Correct answer: A

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

25+35+10x=10025 + 35 + 10x = 100

60+10x=10060 + 10x = 100

10x=4010x = 40

x=4x = 4

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving a two-step equation from a short description is standard grade 7 work. The perimeter setting is simple; the only extra care is counting the sides that remain after the two given ones.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 44 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

    “The remaining sides are equal in length to one another” hides the count: 12 sides minus the two given ones leaves 10 equal sides to share the rest of the perimeter. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Subtracting 25 and 35 from 100, counting the sides left, and dividing are three steps, and the first result, 40, is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The shape is a polygon with 12 sides. Its perimeter is 100 centimeters. One side is 25 centimeters, and another is 35 centimeters. The other sides all have the same length. You need that length.

  2. Count the equal sides

    Two of the 12 sides are already known. The rest are the equal sides.

    12−2=1012 - 2 = 10

    So there are 10 equal sides. Call the length of each one x.

  3. Write an equation for the perimeter

    Use the perimeter formula. All 12 sides add up to 100. Ten sides of length x add up to 10x, which means 10 times x.

    P=the sum of all the side lengthsP = \text{the sum of all the side lengths}

    25+35+10x=10025 + 35 + 10x = 100

  4. Solve for x

    60+10x=10060 + 10x = 100

    10x=100−60=4010x = 100 - 60 = 40

    The 10 equal sides add up to 40 centimeters. Watch out: 40 is all ten sides together, not one side. Stopping here gives choice D. Now share the 40 among the 10 sides.

    x=40÷10=4x = 40 \div 10 = 4

  5. Give the answer

    Each remaining side is 4 centimeters long. Watch out: sharing the 40 centimeters between only 2 sides gives 20, choice C. That treats the shape as if it had 4 sides, but it has 12. So the answer is A, 4.

Check your answer

Add all 12 sides: 25+35+10×4=25+35+40=10025 + 35 + 10 \times 4 = 25 + 35 + 40 = 100. That is the perimeter the question gives, so 4 centimeters is right.

More questions like this

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

Question 13 of 57

2019 Test B Math, question 13

Question

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

  • A15\frac{1}{5}
  • B417\frac{4}{17}
  • C14\frac{1}{4}
  • D13\frac{1}{3}
Show the answer

Correct answer: A

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

33+6+6=315=15\frac{3}{3 + 6 + 6} = \frac{3}{15} = \frac{1}{5}

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Simple probability is grade 6 content. The extra demand is tracking two earlier draws that were not put back, which changes the counts for the third draw, a step typical grade 7 problems often leave out.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 70 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The probability itself is one fraction, but it has to be built from a box that two draws have already changed. Students who use the original 17 cards, or who compare green with the other colors instead of with all the cards, reach listed choices that look reasonable. Listing the counts after each draw, then putting the green cards over all the cards left, avoids both.

  1. Much to keep track of moderate Working it out

    Three color counts and the total must each be updated after two draws, 4 green, 7 yellow and 6 red becoming 3, 6 and 6 out of 15, before the third draw is considered. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    “Does not return the cards” and the two named draws are the conditions that change the box, and the stem states them in passing before asking about the third card. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The box starts with 4 green, 7 yellow and 6 red cards. Shiow-Chen draws cards one at a time, and she does not put them back. The first card is yellow, and the second is green. You need the probability that the third card is green.

  2. Take out the two cards

    The cards are not put back, so the box changes. Lower the yellow count by 1 and the green count by 1.

    ColorAt the startAfter two draws
    Green43
    Yellow76
    Red66
    Total1715

  3. Use the probability formula

    For the third draw, the results you want are the 3 green cards left. The total is the 15 cards left.

    P=number of results you wanttotal number of resultsP = \frac{\text{number of results you want}}{\text{total number of results}}

    P=33+6+6=315P = \frac{3}{3 + 6 + 6} = \frac{3}{15}

    Watch out: using the full box gives 417\frac{4}{17}, choice B. But two cards are gone.

  4. Simplify and answer

    Divide the top and bottom by 3.

    315=15\frac{3}{15} = \frac{1}{5}

    Watch out: compare the green cards with all 15 cards, not with the 12 cards of other colors. That gives 14\frac{1}{4}, choice C. And the three colors are not equally likely, so 13\frac{1}{3} is wrong too. So the answer is A, 15\frac{1}{5}.

Check your answer

17 cards minus the 2 drawn leaves 15. One fifth of 15 is 3, the number of green cards left. Also, the chance went down from 417\frac{4}{17}. That makes sense, because a green card was taken out.

More questions like this

The original DOE question
Question 70 as printed on page 195 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 70. Open page 195 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 14 of 57

2019 Test B Math, question 14

Question

Each number in a sequence is 3 more than twice the number that comes just before it. If 93 is a number in the sequence, what number comes just before it?

  • A92
  • B90
  • C48
  • D45
Show the answer

Correct answer: D

Let n represent a number in the sequence, and let x represent the number that comes just before n:

n=3+2xn = 3 + 2x

93=3+2x93 = 3 + 2x

90=2x90 = 2x

45=x45 = x

Question analysis

In a sequence, each number is 3 more than twice the number just before it, and 93 is one of the numbers. Students find the number that comes just before 93. That means running the rule backward, undoing the adding and the doubling in reverse order.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students routinely solve two-step equations. The demand here is translating the wordy sequence rule into one, or running the rule backward in the right order, rather than any harder content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 29 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is two short steps, so the question rests on reading a wordy rule correctly and running it in reverse. Each wrong choice marks a different place to go wrong: ignoring the rule, stopping after the subtraction, or undoing the adding with more adding. Checking the answer forward, twice the number plus 3, confirms it reaches 93.

  1. Turning words into math moderate Understanding the question

    “Each number … is 3 more than twice the number that comes just before it” must become 2x + 3 = 93, with x the earlier number, not the later one. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. No obvious method minor Working it out

    The rule is stated forward, from a number to the next, but the question runs backward, so each operation has to be undone in reverse order: subtract 3, then halve. (Schoenfeld, 1985)

Strategy: Call the earlier number x, write the rule as an expression in x set equal to 93, and solve it by undoing each operation in reverse order.

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Stops one step early: A choice is a number from partway through the solution.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule
n=2x+3n = 2x + 3
x is any number in the sequence, a list that follows a rule. n is the number right after it. Twice means times 2, so twice x is 2x. Then 3 more means add 3. The same rule links every number to the one after it.

  1. Sort out the question

    A sequence is a list of numbers that follows a rule. Here, each number is “3 more than twice the number that comes just before it.” 93 is in the list. You need the number right before 93. Call it x.

  2. Write the rule as an equation

    Twice x is 2x. Three more than that is 2x + 3. The rule says this equals the next number, which is 93.

    n=2x+3n = 2x + 3

    93=2x+393 = 2x + 3

  3. Undo the adding of 3

    The rule doubles x first, then adds 3. To go backward, undo the steps in reverse order. First, subtract 3 from both sides.

    93−3=2x93 - 3 = 2x

    90=2x90 = 2x

    Watch out: 90 is twice the number, not the number itself. Stopping here gives choice B. And don’t add 3, which leads to 48, choice C.

  4. Undo the doubling

    Now divide both sides by 2.

    x=90÷2=45x = 90 \div 2 = 45

    Watch out: “just before” means just before in this list. It does not mean one less than 93, so 92 is wrong. So the answer is D, 45.

Check your answer

Run the rule forward from 45. Twice 45 is 90, and 3 more is 93: 2×45+3=932 \times 45 + 3 = 93. So 45 really does come just before 93.

More questions like this

The original DOE question
Question 71 as printed on page 195 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 71. Open page 195 of the handbook (PDF)

Question 15 of 57

2019 Test B Math, question 15

Question

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

  • A2.6
  • B3.6
  • C7.5
  • D8.1
Show the answer

Correct answer: D

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

d5.4=32\frac{d}{5.4} = \frac{3}{2}

d=3(5.4)2=16.22=8.1d = \frac{3(5.4)}{2} = \frac{16.2}{2} = 8.1

Question analysis

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

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Converting a map distance with a given scale is routine grade 7 work. This one needs a single proportion and a decimal multiplication, with no rounding or second step, so it matches a typical classroom exercise.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 30 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

    The map distance is the decimal 5.4, and 5.4 × 3 = 16.2 then halved is a decimal multiplication where rounding 5.4 to make it easier changes the answer. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer minor Choosing the answer

    7.5 and 8.1 miles are close enough that a rough estimate, 5 × 1.5, lands on the wrong one; only the exact product separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    On the map, Deshawn’s house and Xin’s house are 5.4 inches apart. The scale says 2 inches on the map stand for 3 miles. You need the real distance between the houses, in miles.

  2. Set up the proportion

    Let d be the real distance, in miles. Put miles on top and inches on the bottom, on both sides.

    d5.4=32\frac{d}{5.4} = \frac{3}{2}

  3. Solve for d

    To get d alone, multiply both sides by 5.4.

    d=3×5.42=16.22=8.1d = \frac{3 \times 5.4}{2} = \frac{16.2}{2} = 8.1

    The houses are 8.1 miles apart. Watch out: keep 5.4 exact. Rounding it to 5 gives 7.5 miles, choice C.

  4. Give the answer

    Watch out: don’t flip the scale. Multiplying 5.4 by 23\frac{2}{3} gives 3.6, choice B. Each inch stands for more than a mile, so the miles must be more than 5.4. So the answer is D, 8.1.

Check your answer

Each inch stands for 3÷2=1.53 \div 2 = 1.5 miles. So 5 inches is 7.5 miles, and 0.4 inch is 0.4×1.5=0.60.4 \times 1.5 = 0.6 mile. Together, 7.5+0.6=8.17.5 + 0.6 = 8.1 miles.

More questions like this

The original DOE question
Question 72 as printed on page 195 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 72. Open page 195 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 16 of 57

2019 Test B Math, question 16

Question

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

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

  • A30°
  • B24°
  • C15°
  • D12°
Show the answer

Correct answer: B

The total number of degrees in a circle is 360. The measure of the angle between adjacent spokes is 36015=24∘\frac{360}{15} = 24^{\circ}.

Question analysis

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

NYS standards
  • 7.G.5 angle relationships and equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Finding an angle by dividing a full turn into equal parts is early angle measurement. It takes one division, and the only fact needed is the number of degrees in a circle, so it sits below grade 7.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 5, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The wheel has 15 spokes. They all start at the hub, the center of the wheel, and they are equally spaced. Adjacent spokes are spokes right next to each other. You need the angle between two adjacent spokes.

  2. Count the angles around the hub

    Go once around the hub. Between each spoke and the next is one angle. So 15 spokes make 15 angles, all the same size. Together they make one full turn, which is 360 degrees.

    Watch out: a full turn is 360 degrees, not 180. Using 180 gives 12 degrees, choice D.

  3. Share the full turn

    Use the formula for equal angles.

    each angle=360÷number of equal angles\text{each angle} = 360 \div \text{number of equal angles}

    each angle=360÷15=24\text{each angle} = 360 \div 15 = 24

    Each angle is 24 degrees. Watch out: 15 is the number of spokes, not the size of the angle, so choice C is wrong. So the answer is B, 24°.

Check your answer

Multiply back: 15 angles of 24 degrees make 15×24=36015 \times 24 = 360 degrees, a full turn. Also, a clock’s 12 hour marks are 30 degrees apart. This wheel has more spokes than that, so each angle must be smaller than 30 degrees.

More questions like this

The original DOE question
Question 73 as printed on page 195 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 73. Open page 195 of the handbook (PDF)

Question 17 of 57

2019 Test B Math, question 17

Question

m∙t=72m \bullet t = 72

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

  • A12
  • B72
  • C132
  • D135
Show the answer

Correct answer: C

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

For t=1t = 1:

1m=721m = 72

m=72m = 72

For t=2t = 2:

2m=722m = 72

m=36m = 36

For t=3t = 3:

3m=723m = 72

m=24m = 24

Then, add those three values together:

72+36+24=13272 + 36 + 24 = 132

Question analysis

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

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

Why it is hard

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

  1. Turning words into math minor Understanding the question

    “The sum of the values of m when t = 1, when t = 2, and when t = 3” asks for three separate solutions added together, not one solution at a combined t. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three divisions, 72 ÷ 1, 72 ÷ 2 and 72 ÷ 3, and an addition make four small steps, each easy but each a chance to drop a value. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The equation is m∙t=72m \bullet t = 72. The dot means times. You need m three times: when t is 1, when t is 2 and when t is 3. Then add the three values of m.

  2. Get m by itself

    Since m times t is 72, m is 72 divided by t.

    m=72÷tm = 72 \div t

    Now put in each value of t, one at a time.

  3. Find m for each value of t

    Each time, divide 72 by the value of t. A table keeps the three answers in order.

    Value of tValue of m
    1m=72÷1=72m = 72 \div 1 = 72
    2m=72÷2=36m = 72 \div 2 = 36
    3m=72÷3=24m = 72 \div 3 = 24

  4. Add the three values

    72+36+24=13272 + 36 + 24 = 132

    Watch out: don’t add the t values first. 72÷(1+2+3)=1272 \div (1 + 2 + 3) = 12 is choice A, and it is not the same thing. Also, 72 alone is only the first value, choice B. So the answer is C, 132.

Check your answer

Check each pair by multiplying: 72×1=7272 \times 1 = 72, 36×2=7236 \times 2 = 72 and 24×3=7224 \times 3 = 72. All three pairs make 72. Then add again: 72 + 36 = 108, and 108 + 24 = 132.

More questions like this

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

Question 18 of 57

2019 Test B Math, question 18

Question

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

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

  • A32 sq ft
  • B64 sq ft
  • C96 sq ft
  • D128 sq ft
Show the answer

Correct answer: C

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

16×8=12816 \times 8 = 128

Then, find the area of the triangle:

12×4×16=32\frac{1}{2} \times 4 \times 16 = 32

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

128−32=96 sq ft128 - 32 = 96\text{ sq ft}

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Composite areas built from rectangles and triangles appear in grades 6 and 7. This one is typical, but the triangle lies on its side, so students must see which length is its height rather than read it from a label.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

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

    The triangle lies on its side: its 4-foot base runs along the right edge, so its height is the rectangle’s full 16-foot length, which the figure labels only as the bottom side. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    The rectangle’s area, the triangle’s area and their difference are three separate results, and the first two, 128 and 32, are themselves listed choices. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle
A=l×wA = l \times w
A is the area, the space inside a flat shape, in square units. l is the length and w is the width. It works because the rectangle is filled by w rows of unit squares, with l squares in each row.
Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side you measure from. h is the height, the distance from the base to the opposite corner, at a right angle to the base. You take half, because a triangle is half of a rectangle with the same base and height.

  1. Sort out the figure

    The figure is a rectangle 16 feet long and 8 feet tall. Inside it is a white triangle. Its tip touches the middle of the left side. Its base lies along the right side and is 4 feet long. The shaded part is the gray area around the triangle. You need its area.

    Plan: find the rectangle’s area, then take away the triangle’s area.

  2. Find the area of the rectangle

    A=l×wA = l \times w

    A=16×8=128A = 16 \times 8 = 128

    The rectangle’s area is 128 square feet. Watch out: this includes the white triangle. Stopping here gives choice D.

  3. Find the triangle’s height

    The triangle lies on its side. Its base is the 4-foot piece on the right side, which runs up and down. The height must meet the base at a right angle, a square corner. So the height runs straight across, from the base to the tip on the left side. That is the full length of the rectangle, 16 feet.

  4. Find the area of the triangle

    A=12×b×hA = \frac{1}{2} \times b \times h

    A=12×4×16=32A = \frac{1}{2} \times 4 \times 16 = 32

    The triangle’s area is 32 square feet. Watch out: don’t forget the 12\frac{1}{2}. Using 4×16=644 \times 16 = 64 leaves 64 for the shaded part, choice B.

  5. Subtract to find the shaded area

    128−32=96128 - 32 = 96

    The shaded part is 96 square feet. Watch out: 32 is the white triangle, not the shaded part, so choice A is wrong. So the answer is C, 96 sq ft.

Check your answer

The triangle’s base, 4 feet, is half the rectangle’s height of 8 feet. Its height is the full 16 feet. So the triangle is half of half the rectangle, or one fourth of it. One fourth of 128 is 32, and 128−32=96128 - 32 = 96.

More questions like this

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

Question 19 of 57

2019 Test B Math, question 19

Question

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

  • A50
  • B62.5
  • C100
  • D125
Show the answer

Correct answer: A

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

2x+3x=2502x + 3x = 250

5x=2505x = 250

x=50x = 50

Question analysis

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

NYS standards
  • 7.EE.4a two-step equations
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an equation from a short word problem is standard grade 7 work. This one has a single unknown, since both juices come in the same number of gallons, and only two short steps.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 6, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Orange juice costs $2 a gallon. Grapefruit juice costs $3 a gallon. The shipment has equal amounts of each, so it has the same number of gallons of each juice. It costs $250 in all. You need the number of gallons of orange juice.

  2. Write the cost of each juice

    Let x be the number of gallons of each juice. Use the cost formula for each one.

    cost=price per gallon×number of gallons\text{cost} = \text{price per gallon} \times \text{number of gallons}

    orange juice cost=2×x=2x\text{orange juice cost} = 2 \times x = 2x

    grapefruit juice cost=3×x=3x\text{grapefruit juice cost} = 3 \times x = 3x

  3. Add the costs and solve

    The two costs add up to $250. Two x’s plus three x’s make five x’s.

    2x+3x=2502x + 3x = 250

    5x=2505x = 250

    x=250÷5=50x = 250 \div 5 = 50

    There are 50 gallons of each juice.

  4. Give the answer

    Watch out: equal amounts means equal gallons, not equal dollars. Spending $125 on each juice gives 62.5 gallons, choice B. Also, 100 gallons is both juices together, choice C. So the answer is A, 50.

Check your answer

Find each cost. 50 gallons of orange juice cost 50×2=10050 \times 2 = 100 dollars. 50 gallons of grapefruit juice cost 50×3=15050 \times 3 = 150 dollars. Together, 100+150=250100 + 150 = 250 dollars, the total in the question.

More questions like this

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

Question 20 of 57

2019 Test B Math, question 20

Question

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

  • A11
  • B18
  • C20
  • D22
Show the answer

Correct answer: C

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

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

5n+10=1105n + 10 = 110

5n=1005n = 100

n=20n = 20

Question analysis

Five consecutive numbers add up to 110, and students find the least of them. This number-properties question can be solved with an equation in one unknown, or by noting that the middle of an odd number of consecutive numbers equals their average. The bold word least matters, because the middle number is easier to find.

NYS standards
  • 7.EE.4a two-step equations
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 A sum of consecutive numbers leads to a two-step equation, standard grade 7 work. The average shortcut makes it quicker, and the only extra demand is answering for the least number rather than the middle one.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Dividing 110 by 5 is the natural first move, and it gives the middle number, which is a listed choice. The bold word least signals that one more step is needed, counting down two from the middle. The equation route, 5n + 10 = 110, goes straight to the least number but takes longer to set up.

  1. Turning words into math minor Understanding the question

    “Consecutive numbers” means each is 1 more than the last, and the bold “least” asks for the first of the five, not the middle one that dividing by 5 gives. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    None of the five numbers is given, so they must be written as n through n + 4, or pictured as two on each side of a middle value. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Divide 110 by 5 to find the middle of the five numbers, then count down to the least of them.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive numbers
n, n+1, n+2, n+3, n+4n,\ n + 1,\ n + 2,\ n + 3,\ n + 4
Consecutive numbers come one right after another when you count, like 7, 8, 9. Each is 1 more than the one before. If n is the least number, the next ones are n + 1, n + 2, and so on.

  1. Sort out the question

    Five consecutive numbers add up to 110. Consecutive means each number is 1 more than the one before. You need the least, or smallest, of the five.

  2. Name the five numbers

    Let n be the least number. Then the five numbers are:

    n, n+1, n+2, n+3, n+4n,\ n + 1,\ n + 2,\ n + 3,\ n + 4

    Each one is 1 more than the one before it.

  3. Write the sum as an equation

    Add all five and set the sum equal to 110.

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

    There are five n’s, which make 5n. The extra numbers add up to 1 + 2 + 3 + 4 = 10.

    5n+10=1105n + 10 = 110

  4. Solve for n

    Subtract 10 from both sides. Then divide both sides by 5.

    5n=1005n = 100

    n=100÷5=20n = 100 \div 5 = 20

    So the least number is 20.

  5. Give the answer

    Watch out: 110÷5=22110 \div 5 = 22 is the middle number, not the least. Picking 22 is choice D. The question asks for the least of the five. So the answer is C, 20.

Check your answer

Write out the five numbers and add them: 20+21+22+23+24=11020 + 21 + 22 + 23 + 24 = 110. The sum matches. Also, the middle number, 22, is 110 ÷ 5, and the least is 2 less than the middle.

More questions like this

The original DOE question
Question 77 as printed on page 197 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 197, question 77. Open page 197 of the handbook (PDF)

Question 21 of 57

2019 Test B Math, question 21

Question

−mn-mn

(−m)(−n)(-m)(-n)

|mn||mn|

|−mn||-mn|

|(−m)(−n)||(-m)(-n)|

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

  • A1
  • B2
  • C3
  • D4
Show the answer

Correct answer: D

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

−mn-mn is negative.

(−m)(−n)(-m)(-n) is positive.

|mn||mn| is positive.

|−mn||-mn| is positive.

|(−m)(−n)||(-m)(-n)| is positive.

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

Question analysis

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

NYS standards
  • 7.NS.2a rational number multiplication
  • 6.NS.7c absolute value as distance
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Multiplying signed numbers and absolute value are grade 6 and 7 content. Checking five expressions one by one takes longer than a typical single computation, but each check uses a familiar rule.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
5 steps
Reading level
Question and choices: about grade 10, 23 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    The five expressions use only m and n, so each sign rule has to be applied in general unless the student substitutes values such as m = 2 and n = 3. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    Five expressions are judged one at a time and a count kept, and two of them combine the product rule for negatives with absolute value bars. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Something you must already know minor Working it out

    The rules that two negative factors give a positive product and that absolute value removes a minus sign are assumed, not stated. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    m and n are positive integers. That means whole numbers greater than 0. Writing mn means m times n, so mn is positive. You need to count how many of the five expressions always equal mn.

  2. Pick numbers to test

    Pick easy values: m=2m = 2 and n=3n = 3. Then mn=2×3=6mn = 2 \times 3 = 6. Any expression that equals mn must come out to 6.

  3. Test the first two expressions

    −mn=−(2×3)=−6-mn = -(2 \times 3) = -6

    This is negative, so it does not equal 6.

    (−m)(−n)=(−2)(−3)=6(-m)(-n) = (-2)(-3) = 6

    A negative times a negative is positive, so this one equals 6. Watch out: if you think a negative times a negative stays negative, you reject this one and get 3, choice C.

  4. Test the three absolute values

    |mn|=|6|=6|mn| = |6| = 6

    |−mn|=|−6|=6|-mn| = |-6| = 6

    |(−m)(−n)|=|6|=6|(-m)(-n)| = |6| = 6

    All three equal 6, because absolute value bars remove a minus sign. Watch out: the bars work even when the minus sign is inside them. Rejecting the last two gives 2, choice B.

  5. Count the matches

    Four expressions equal mn: (−m)(−n)(-m)(-n), |mn||mn|, |−mn||-mn| and |(−m)(−n)||(-m)(-n)|. Only −mn-mn does not. The sign rules work the same for any positive m and n, not just 2 and 3. So the answer is D, 4.

Check your answer

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

More questions like this

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

Question 22 of 57

2019 Test B Math, question 22

Question

54×5652=\frac{5^{4} \times 5^{6}}{5^{2}} =

  • A585^{8}
  • B5105^{10}
  • C5125^{12}
  • D5225^{22}
Show the answer

Correct answer: A

54×5652=51052=58\frac{5^{4} \times 5^{6}}{5^{2}} = \frac{5^{10}}{5^{2}} = 5^{8}

Question analysis

This exponents question asks for the value of 5 to the 4th times 5 to the 6th, divided by 5 squared, written as a single power of 5. It tests the rules for multiplying and dividing powers with the same base. Every choice is a power of 5, so the whole question rests on combining the exponents correctly.

NYS standards
  • 8.EE.1 integer exponent properties
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Combining powers of one base by exponent rules is grade 8 content; grades 6 and 7 only evaluate powers. The task itself is short, one product and one quotient of powers of 5.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 16+, 5 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Everything rests on two grade 8 rules, and the choices are built so that each misapplied rule has an answer. Every choice is a power of 5, so the base gives nothing away and only the exponent arithmetic decides. Students who carry the written operations onto the exponents, or who stop after the numerator, find their result listed; working the top first, then the bottom, keeps the two rules apart.

  1. Something you must already know moderate Working it out

    Neither exponent rule is given: multiplying 5⁴ by 5⁶ adds the exponents to 10, and dividing by 5² subtracts 2; working out the powers instead, 625 × 15,625, is hopeless by hand. (Anderson & Krathwohl, 2001)

Strategy: When powers of the same base are multiplied, add their exponents; when one is divided by another, subtract the exponent of the divisor. Apply the top first, then the bottom.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiplying powers with the same base
am×an=am+na^{m} \times a^{n} = a^{m + n}
a is the base, the number being multiplied. m and n are exponents, the small raised numbers. Each exponent counts how many a’s are multiplied. Putting the two groups together makes m + n of them, so you add the exponents.
Dividing powers with the same base
aman=am−n\frac{a^{m}}{a^{n}} = a^{m - n}
a is the base, and m and n are exponents. The top has m factors of a and the bottom has n. Each matching pair cancels, because a ÷ a = 1. So m − n factors are left, and you subtract the exponents.

  1. Sort out the question

    The expression has three powers of 5. A power like 545^{4} means 5 multiplied by itself 4 times: 5×5×5×55 \times 5 \times 5 \times 5. The 5 is the base. The small raised 4 is the exponent. Every choice is a power of 5, so you only need the final exponent.

  2. Multiply the powers on top

    The top multiplies two powers of 5. Use the rule for multiplying powers.

    am×an=am+na^{m} \times a^{n} = a^{m + n}

    54×56=54+6=5105^{4} \times 5^{6} = 5^{4 + 6} = 5^{10}

    Four 5s times six more 5s makes ten 5s in a row.

    Watch out: add the exponents. Don’t multiply them. 4×6=244 \times 6 = 24 leads to 5225^{22}, choice D.

  3. Divide by the power on the bottom

    Now divide by the bottom. Use the rule for dividing powers.

    aman=am−n\frac{a^{m}}{a^{n}} = a^{m - n}

    51052=510−2=58\frac{5^{10}}{5^{2}} = 5^{10 - 2} = 5^{8}

    Two of the ten 5s on top cancel with the two 5s on the bottom. Eight 5s are left.

    Watch out: stopping at 5105^{10}, choice B, forgets the bottom. Multiplying and dividing the exponents, 4×6÷2=124 \times 6 \div 2 = 12, gives choice C.

    So the answer is A, 585^{8}.

Check your answer

Count the 5s. The top has 4+6=104 + 6 = 10 fives multiplied together. The bottom has 2. Each 5 on the bottom cancels one 5 on top, since 5 ÷ 5 = 1. That leaves 10−2=810 - 2 = 8 fives, which is 5 to the 8th power.

More questions like this

The original DOE question
Question 79 as printed on page 180 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 180, question 79. Open page 180 of the handbook (PDF)

Question 23 of 57

2019 Test B Math, question 23

Question

FRUIT SOLD
AT STORE XYZ

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

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

  • A18%
  • B28%
  • C36%
  • D64%
Show the answer

Correct answer: C

First, find the number of peaches sold:

50−(8+6+12+6)=50 - (8 + 6 + 12 + 6) =

50−32=1850 - 32 = 18

Then, find the percentage:

1850=0.36=36%\frac{18}{50} = 0.36 = 36\%

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a missing value and writing a part as a percent of a whole are grade 6 skills. With a total of 50, the percent step is quick, so the task is lighter than typical grade 7 percent problems.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 34 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

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

    The table gives four counts, a question mark for peaches and a total of 50, so the missing count must be found by subtraction before any percent. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Adding 8, 6, 12 and 6, subtracting from 50, and turning 18 out of 50 into a percent are three steps, and the count of 18 is itself a listed percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The table shows how many of each fruit Store XYZ sold. The number of peaches is missing, shown by a question mark. The total is 50 pieces of fruit. You need the percent of the total that were peaches.

  2. Add the fruits you know

    8+6+12+6=328 + 6 + 12 + 6 = 32

    Apples, bananas, cherries and oranges add up to 32 pieces of fruit. That is every fruit except the peaches.

  3. Find the number of peaches

    The peaches are whatever is left of the 50.

    50−32=1850 - 32 = 18

    So 18 peaches were sold. Watch out: 18 is a count, not a percent. Choosing 18% is choice A.

  4. Turn the count into a percent

    percent=partwhole×100\text{percent} = \frac{\text{part}}{\text{whole}} \times 100

    1850×100=0.36×100=36\frac{18}{50} \times 100 = 0.36 \times 100 = 36

    Peaches were 36% of the fruit sold. Watch out: 32 out of 50 is 64%, choice D. That is the other four fruits, not the peaches. So the answer is C, 36%.

Check your answer

Out of 50 pieces, each piece is 2% of the total, because 100÷50=2100 \div 50 = 2. So 18 peaches are 18×2=3618 \times 2 = 36 percent. Also, 36% for peaches plus 64% for the rest makes 100%.

More questions like this

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

Question 24 of 57

2019 Test B Math, question 24

Question

Carpenters laid a layer of insulation 6 inches thick to cover a rectangular attic floor that was 32 feet long and 20 feet wide. What was the volume, in cubic feet, of the insulation used to cover the floor?

  • A160
  • B320
  • C640
  • D3,840
Show the answer

Correct answer: B

First convert 6 inches to 0.5 feet. Then, calculate the volume, in cubic feet:

0.5×32×20=32×10=3200.5 \times 32 \times 20 = 32 \times 10 = 320

Question analysis

Carpenters spread insulation 6 inches thick over a rectangular attic floor 32 feet long and 20 feet wide, and students find the volume of insulation in cubic feet. It is a volume word problem with mixed units. The thickness is given in inches while the floor is in feet, a mismatch that is easy to overlook.

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 Finding the volume of a box-shaped layer is grade 6 work, with every dimension but the thickness a whole number of feet. Grade 7 volume problems use triangular or composite prisms; this one adds only a change from inches to feet.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 9, 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

Volume of a box is routine, so the question turns on the unit mix: the floor is measured in feet and the layer's thickness in inches. The handbook prints inches in bold, a warning that the plain product 3,840 is listed for those who miss it. The floor's area, 640, is the other ready trap, since it appears partway through the correct work.

  1. Something you must already know minor Working it out

    Neither the volume of a box, length × width × height, nor the 12 inches in a foot is given, and the bold inches marks the thickness as the one dimension to convert. (Anderson & Krathwohl, 2001)

  2. Error-prone arithmetic minor Working it out

    The converted thickness is the fraction 1/2 foot, so the product is 32 × 20 × 1/2, easiest as 32 × 10; kept in inches, 3,840 still needs dividing by 12. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Notice the bold word inches: convert the 6-inch thickness to feet before anything else, then multiply it by the floor's length and width.

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.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Inches to feet
feet=inches12\text{feet} = \frac{\text{inches}}{12}
There are 12 inches in 1 foot. So dividing a number of inches by 12 tells how many feet long it is. Use it when one measure is in inches and the others are in feet.
Volume of a rectangular box
V=l×w×hV = l \times w \times h
V is the volume, the space inside, in cubic units. l is the length, w is the width, and h is the height, here the thickness of the layer. All three must use the same unit. Multiplying counts the unit cubes that fill the box.

  1. Sort out the question

    The insulation is a flat layer shaped like a box. It covers a floor 32 feet long and 20 feet wide. It is 6 inches thick. Notice the bold word: the thickness is in inches, not feet. You need the volume in cubic feet. A cubic foot is a cube 1 foot long on each side.

  2. Change 6 inches to feet

    All three measures must be in feet before you multiply.

    feet=inches12\text{feet} = \frac{\text{inches}}{12}

    feet=612=12=0.5\text{feet} = \frac{6}{12} = \frac{1}{2} = 0.5

    The layer is 0.5 foot thick, half a foot.

    Watch out: using 6 as if it were feet gives 32×20×6=3,84032 \times 20 \times 6 = 3{,}840, choice D. That is 12 times too big.

  3. Multiply to find the volume

    Use the volume formula with the length, width and thickness, all in feet.

    V=l×w×hV = l \times w \times h

    V=32×20×0.5=640×0.5=320V = 32 \times 20 \times 0.5 = 640 \times 0.5 = 320

    The volume is 320 cubic feet.

    Watch out: 640 is the area of the floor, in square feet. Stopping there gives choice C. You still need to multiply by the thickness.

    So the answer is B, 320.

Check your answer

A layer 1 foot thick would fill 32×20=64032 \times 20 = 640 cubic feet. This layer is only half a foot thick. So it holds half as much: 640÷2=320640 \div 2 = 320 cubic feet. That matches.

More questions like this

The original DOE question
Question 81 as printed on page 181 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 181, question 81. Open page 181 of the handbook (PDF)

Question 25 of 57

2019 Test B Math, question 25

Question

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

  • A7,500
  • B8,400
  • C9,000
  • D9,200
Show the answer

Correct answer: C

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

p50=1,500\frac{p}{50} = 1{,}500

Solve for p:

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

The population decrease from 2005 to 2006 is 84,000−75,000=9,00084{,}000 - 75{,}000 = 9{,}000.

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a rate such as people per square mile is grade 6 content. Here it is one step inside a two-step problem with large round numbers, which matches typical grade 7 multi-step rate work.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 8, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    “The average population per square mile was equal to 1,500” is a density, which has to become a total, 1,500 × 50 = 75,000, before it can be compared with 84,000. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    The 2006 population is only a midpoint; subtracting it from 84,000 is a second step, and the question asks for that decrease, not either population. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Waterville covers 50 square miles. In 2005, 84,000 people lived there. In 2006, there were 1,500 people per square mile, on average. You need how many fewer people lived there in 2006 than in 2005.

  2. Find the 2006 population

    Let p be the 2006 population. Use the formula.

    people per square mile=populationarea\text{people per square mile} = \frac{\text{population}}{\text{area}}

    1,500=p501{,}500 = \frac{p}{50}

    To undo dividing by 50, multiply both sides by 50.

    p=1,500×50=75,000p = 1{,}500 \times 50 = 75{,}000

    So 75,000 people lived there in 2006.

  3. Find the decrease

    A decrease is how much a number went down. Subtract the 2006 population from the 2005 population.

    84,000−75,000=9,00084{,}000 - 75{,}000 = 9{,}000

    The population went down by 9,000 people.

  4. Give the answer

    Watch out: 75,000 is the 2006 population, not the decrease. Choice A, 7,500, looks like it but has one zero fewer. The question asks how much the population dropped. So the answer is C, 9,000.

Check your answer

Work with people per square mile instead. In 2005, there were 84,000÷50=1,68084{,}000 \div 50 = 1{,}680 people per square mile. In 2006, there were 1,500. That is 180 fewer per square mile, and 180×50=9,000180 \times 50 = 9{,}000 fewer people.

More questions like this

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

Question 26 of 57

2019 Test B Math, question 26

Question

A number line with arrows at both ends and four tick marks. From left to right, the ticks are labeled 0 (below the line), x and y (above the line), and x + y (above the line) with 24 below it. A bracket above the line marks the distance from x to y as 4.

According to the number line above, what is the value of y?

  • A4
  • B12
  • C14
  • D20
Show the answer

Correct answer: C

According to the number line, y−x=4y - x = 4 and x+y=24x + y = 24. Add the two equations together and solve for y:

y−x=4+x+y=242y=28\begin{array}{l} y - x = 4 \\ + x + y = 24 \\ \hline 2y = 28 \end{array}

y=14y = 14

Question analysis

A number line marks 0, x, y and x + y, with x + y at 24 and a bracket showing that x and y are 4 units apart. Students find the value of y. The question combines reading a number line with solving for two unknowns, and the bracket and the sum have to be used together.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Two unknowns usually call for grade 8 systems, but here one unknown is 4 more than the other, so a single grade 7 equation of the form px + q = r is enough. Reading those facts from a number line adds a step.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 12 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The number line holds all the information but none of it as an equation, and the drawing is not to scale, so the positions must come from the labels. Students who split 24 evenly stop at the point midway between x and y, and those who take the bracket's 4 for a position subtract it from 24. Writing the two facts down and checking the result, 10 + 14 = 24 with a gap of 4, closes off every listed slip.

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

    The two facts needed, y − x = 4 and x + y = 24, are never written; they come from the bracket over x and y and from the 24 under the tick marked x + y. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Variables instead of numbers minor Understanding the question

    Neither x nor y has a value, and the point x + y is itself built from both, so nothing can be placed until the two facts are combined. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. No obvious method minor Working it out

    Two unknowns usually call for a system of equations, which grade 7 students have not met; they must instead write y as x + 4, or see that x and y sit 2 units either side of 12. (Schoenfeld, 1985)

Strategy: Write the two facts the labels give, y − x = 4 and x + y = 24, then add the equations or replace y with x + 4; the drawing is not to scale, so don't measure.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distance on a number line
distance=right point−left point\text{distance} = \text{right point} - \text{left point}
On a number line, numbers get bigger to the right. The distance between two points is how many units apart they are. Subtract the left number from the right one, so the distance is never negative.
Keeping an equation balanced
if a=b, then a−c=b−c and a÷c=b÷c\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c
An equation says two sides are equal, like a balanced scale. a and b are the two sides. c is any number, but not 0 when you divide. Doing the same thing to both sides keeps them equal. Use it to get the letter by itself.

  1. Sort out the question

    The number line shows 0, x, y and x + y. You need the value of y. The drawing is not to scale, though the question doesn’t say so. That means the spacing on the page doesn’t match the numbers. If you measure, y looks like it is near 15. So don’t measure. Work only from the labels.

  2. Write the labels as equations

    An equation is a math sentence that says two amounts are equal. The bracket marked 4 runs from x to y, and y is to the right.

    distance=right point−left point\text{distance} = \text{right point} - \text{left point}

    y−x=4y - x = 4

    So y is 4 more than x, or y = x + 4. The tick marked x + y has 24 under it, so x + y = 24.

    Watch out: the 4 is the distance from x to y. It is not where x is, and it is not y. Taking x as 4 gives 24−4=2024 - 4 = 20, choice D. Taking y as 4 is choice A.

  3. Put x + 4 in place of y

    Two unknown letters are hard to solve at once. y is the same as x + 4, so swap it into x + y = 24. Now there is only one letter.

    x+(x+4)=24x + (x + 4) = 24

    2x+4=242x + 4 = 24

    Here 2x means 2 times x. Two x’s plus 4 make 24.

  4. Solve for x

    Subtract 4 from both sides. Then divide both sides by 2. Doing the same to both sides keeps the equation balanced.

    if a=b, then a−c=b−c and a÷c=b÷c\text{if } a = b \text{, then } a - c = b - c \text{ and } a \div c = b \div c

    2x+4−4=24−42x + 4 - 4 = 24 - 4

    2x=202x = 20

    x=20÷2=10x = 20 \div 2 = 10

    So x is 10.

  5. Find y

    y is 4 more than x.

    y=x+4=10+4=14y = x + 4 = 10 + 4 = 14

    So y is 14.

    Watch out: half of 24 is 12, choice B. That is the point halfway between x and y. y is 2 more than that.

    So the answer is C, 14.

Check your answer

Put both numbers back. 10+14=2410 + 14 = 24, so x + y lands at 24. 14−10=414 - 10 = 4, so x and y are 4 apart. Both labels work. Notice that y is 14, not the 15 or so that the drawing suggests.

More questions like this

The original DOE question
Question 83 as printed on page 181 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 181, question 83. Open page 181 of the handbook (PDF)

Question 27 of 57

2019 Test B Math, question 27

Question

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

Set L (not shown) is a list of 25 consecutive integers. The median of the integers in set N above is equal to the least integer in set L. What is the median of the integers in set L?

  • A25
  • B26
  • C27
  • D28
Show the answer

Correct answer: B

First, find the median of set N. There are 17 integers in set N, so the 9th integer is the median. The 9th integer is 14.

The least integer in set L is equal to the median of set N, so the least integer in set L is 14.

There are 25 integers in set L, so the median is the 13th integer. That integer is 26.

Question analysis

Set N lists the consecutive integers from 6 to 22, and set L is a list of 25 consecutive integers whose least member equals the median of N. Students find the median of set L. It chains two median questions together, and counting positions in a run of consecutive integers invites off-by-one slips.

NYS standards
  • 6.SP.5c range, center, and overall pattern
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Medians of short lists are grade 6 content. This item hides both lists, one behind set notation with an ellipsis and one not shown at all, and links them, so it needs more planning and careful counting than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 5, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains two medians, and both depend on counting members correctly in runs of consecutive integers. The easy slip is between position and distance: the 13th member of L is 12 steps above its first, not 13. With four consecutive choices, every counting error has a home, so writing out the first and last members of L, 14 and 38, and averaging them is the safest check.

  1. A very close wrong answer moderate Choosing the answer

    The choices 25, 26, 27 and 28 are consecutive, so a count that is off by one position lands on a neighboring choice with nothing to flag it. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. No obvious method moderate Working it out

    Set L is never listed; its median has to be found from its least member, 14, and its size, 25, by working out that the middle is the 13th member. (Schoenfeld, 1985)

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

    Set N is written with an ellipsis, {6, 7, …, 21, 22}, so its 17 members must be counted, and 22 − 6 = 16 undercounts by one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find N's median by averaging its least and greatest members. L's median is its middle member, so work out that member's position and count up from L's least member.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting the integers in a list
count=last−first+1\text{count} = \text{last} - \text{first} + 1
This counts the whole numbers in a list that goes up by 1, like 6, 7, 8. Subtracting gives the number of steps from the first number to the last. You add 1 because the first number counts too.
Position of the median
position=count+12\text{position} = \frac{\text{count} + 1}{2}
The count is how many numbers are in a list that is in order. When the count is odd, the median is at this position. It has the same number of values before it and after it.

  1. Sort out the question

    Set N is 6, 7, and so on up to 22. The dots stand for every integer in between. Integers are whole numbers. Set L has 25 consecutive integers, one right after another. Its least number equals the median of N. You need the median of L.

  2. Count the numbers in set N

    count=last−first+1\text{count} = \text{last} - \text{first} + 1

    22−6+1=1722 - 6 + 1 = 17

    Set N has 17 integers. Watch out: 22−6=1622 - 6 = 16 misses one, because the first number counts too.

  3. Find the median of set N

    position=count+12\text{position} = \frac{\text{count} + 1}{2}

    position=17+12=9\text{position} = \frac{17 + 1}{2} = 9

    The median is the 9th number. 6 is the 1st, so the 9th is 8 steps up: 6 + 8 = 14. The median of N is 14.

  4. Find the middle place in set L

    So set L starts at 14 and has 25 integers. Find the position of its median.

    position=25+12=13\text{position} = \frac{25 + 1}{2} = 13

    The median of L is its 13th number.

  5. Count up to the 13th number

    14 is the 1st number. The 13th is 12 steps after it.

    14+12=2614 + 12 = 26

    Watch out: the 13th number is 12 steps up, not 13. Adding 13 gives 27, choice C. And 25 is how many numbers L has, not its median. So the answer is B, 26.

Check your answer

Set L runs from 14 to 14+24=3814 + 24 = 38. For consecutive integers, the median is halfway between the ends: (14+38)÷2=26(14 + 38) \div 2 = 26. That matches. The same trick gives N’s median: (6+22)÷2=14(6 + 22) \div 2 = 14.

More questions like this

The original DOE question
Question 84 as printed on page 199 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 199, question 84. Open page 199 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 28 of 57

2019 Test B Math, question 28

Question

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

  • A2
  • B4
  • C8
  • D10
Show the answer

Correct answer: D

The ratio of x to y is 1:2, so xy=12\frac{x}{y} = \frac{1}{2} or x=12yx = \frac{1}{2}y. Similarly, the ratio of y to z is 2:5, so yz=25\frac{y}{z} = \frac{2}{5}. Solve for z in terms of y:

5y=2z5y = 2z

52y=z\frac{5}{2}y = z

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

x+y+z=40x + y + z = 40

12y+y+52y=40\frac{1}{2}y + y + \frac{5}{2}y = 40

4y=404y = 40

y=10y = 10

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Most grade 7 ratio tasks split a total using one ratio. Here two ratios linked through y must first become a single three-part ratio; since y's term already matches, that step is short and the problem stays at grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. No obvious method moderate Working it out

    Nothing says to merge the ratios, but y is 2 in both, so x : y : z becomes 1 : 2 : 5 at once; the explanation’s route through fractions of y is slower. (Schoenfeld, 1985)

  2. Variables instead of numbers minor Understanding the question

    x, y and z appear only in two ratios and a sum of 40, so the ratio numbers 1, 2 and 5 are easy to mistake for the values themselves. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Many steps minor Working it out

    Adding the 8 parts, dividing 40 by 8 to get 5 per part, and taking y’s 2 parts are three steps, and the part count is itself a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    x, y and z add up to 40. A ratio compares two amounts. The ratio of x to y is 1:2, so y is twice x. The ratio of y to z is 2:5. You need the actual value of y, not its number in a ratio.

  2. Combine the two ratios

    y is 2 in both ratios. So both ratios count the same size of part, and they join into one.

    x:y:z=1:2:5x : y : z = 1 : 2 : 5

    So x is 1 part, y is 2 parts and z is 5 parts.

  3. Find the total number of parts

    1+2+5=81 + 2 + 5 = 8

    The 40 is split into 8 equal parts. Watch out: 8 counts parts. It is not y, so choice C is wrong.

  4. Find the size of one part

    one part=total÷number of parts\text{one part} = \text{total} \div \text{number of parts}

    one part=40÷8=5\text{one part} = 40 \div 8 = 5

    The 40 is shared equally among the 8 parts. So each part is worth 5.

  5. Find y

    y is 2 parts, so multiply the size of one part by 2.

    y=2×5=10y = 2 \times 5 = 10

    Watch out: the 2 in the ratio counts parts, so y is not 2, choice A. Adding the two 2s to get 4 is wrong too, choice B. So the answer is D, 10.

Check your answer

Find all three: x = 5, y = 10 and z = 25. They add to 5+10+25=405 + 10 + 25 = 40. x to y is 5 to 10, which is 1:2. y to z is 10 to 25, which is 2:5.

More questions like this

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

Question 29 of 57

2019 Test B Math, question 29

Question

Which number could be the sum of 4 consecutive integers?

  • A−12
  • B−10
  • C−8
  • D−4
Show the answer

Correct answer: B

The sum of 4 consecutive integers can be expressed as:

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

=4n+6=4n+6

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

Option A:

4n+6=−124n+6=-12

4n=−184n=-18

n=−184n=-\frac{18}{4} is not an integer.

Option B:

4n+6=−104n+6=-10

4n=−164n=-16

n=−4n=-4 is an integer.

Option C:

4n+6=−84n+6=-8

4n=−144n=-14

n=−144n=-\frac{14}{4} is not an integer.

Option D:

4n+6=−44n+6=-4

4n=−104n=-10

n=−104n=-\frac{10}{4} is not an integer.

The only option for which n is an integer is −10.

Question analysis

A number properties question about consecutive integers: which of four negative numbers could be the sum of 4 consecutive integers? Students write the sum with a variable, or add runs of negative integers, and test each choice. Keeping signs straight with negative numbers, and remembering that the integers must be whole numbers, is the tricky part.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 consecutive-integer problems usually give the sum and ask for the integers. This one reverses the task: students test which negative total is possible at all, using the structure of the sum rather than one equation.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 8, 9 words.
Corrected from the DOE handbook
  • A stray word in the explanation’s last sentence was removed.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Three of the four choices are multiples of 4, and that is the whole trap: a student who pictures the sum as four times some integer finds three choices that seem to work. The sum of four consecutive integers is 4n + 6, so it is always even but never a multiple of 4. Trying the most natural run, −4, −3, −2 and −1, lands on −10 at once and is the fastest check.

  1. Every choice must be checked moderate Choosing the answer

    The stem asks which number “could be” the sum, so no single equation gives the answer; each of −12, −10, −8 and −4 has to be tested against the form a four-integer sum must take. (Embretson & Wetzel, 1987)

  2. No obvious method minor Working it out

    Nothing points to the shortcut that n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 is always 2 more than a multiple of 4, or to simply adding a run such as −1, −2, −3 and −4. (Schoenfeld, 1985)

  3. Error-prone arithmetic minor Working it out

    Every choice is negative, so solving 4n + 6 = −12 or adding a run of integers means subtracting 6 from a negative number and dividing a negative by 4. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the integers as n, n + 1, n + 2 and n + 3, add them, and set the sum equal to each choice to see which one gives a whole-number n.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook ends with “The only option for which n is an integer answer is −10.” The word “answer” does not belong in that sentence; the site reads “The only option for which n is an integer is −10.” The working above it is unchanged: only 4n + 6 = −10 gives an integer, n = −4, and −4 + (−3) + (−2) + (−1) = −10.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sum of 4 consecutive integers
n+(n+1)+(n+2)+(n+3)=4n+6n + (n + 1) + (n + 2) + (n + 3) = 4n + 6
n is the smallest of the four integers. Each next integer is 1 more than the one before, so they are n, n + 1, n + 2 and n + 3. Adding them gives four n’s, plus 1 + 2 + 3, which is 6. Use it to add any four integers in a row.

  1. Sort out the question

    Integers are whole numbers and their negatives, like −2, 0 and 5. They have no fractions or decimals. Consecutive integers come one right after another, like 7, 8, 9, 10 or −3, −2, −1, 0.

    You need the choice that can be the sum of four consecutive integers. The sum is what you get when you add them.

  2. Write the sum with a letter

    You don’t know the integers yet. So call the smallest one n. Each next one is 1 more: n + 1, n + 2 and n + 3. Use the formula. Collect the four n’s, then add the numbers.

    n+n+n+n+1+2+3=4n+6n + n + n + n + 1 + 2 + 3 = 4n + 6

    Any four consecutive integers add up to 4n + 6.

  3. Test each choice

    Set 4n + 6 equal to each choice and solve for n. First subtract 6 from both sides. Then divide both sides by 4.

    Choice4n + 64nn
    A−12−18−4.5
    B−10−16−4
    C−8−14−3.5
    D−4−10−2.5

    n must be an integer. Only choice B gives one: n = −4.

  4. Name the four integers

    Start at n = −4 and count up by 1. The integers are −4, −3, −2 and −1.

    Watch out: Choices A, C and D are multiples of 4. A multiple of 4 is 4 times an integer, like 4 × (−3) = −12. It is tempting to think four integers must add to one of those. But the sum is 4n + 6, and the extra 6 gets in the way. Choice D works only with n = −2.5, and that is not an integer.

    So the answer is B, −10.

Check your answer

Add the integers: −4+(−3)+(−2)+(−1)=−10-4 + (-3) + (-2) + (-1) = -10. That is choice B. Also, 4n + 6 is always 2 more than a multiple of 4. And −10 is 2 more than −12, which is 4 × (−3).

More questions like this

The original DOE question
Question 86 as printed on page 200 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 200, question 86. Open page 200 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 30 of 57

2019 Test B Math, question 30

Question

PRICE OF A VIDEO GAME AT VARIOUS STORES

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

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

  • A$10.99
  • B$11.49
  • C$11.99
  • D$12.32
Show the answer

Correct answer: C

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

Question analysis

A frequency table gives the price of one video game at 12 stores, from $10.99 to $13.99, with the number of stores charging each price. Students find the median price. This statistics question tests the median, the middle of an ordered data set, when the data arrive as a table of counts rather than a list.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.SP.5a number of observations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 The median is a grade 6 measure of center that grade 7 students use regularly. What lifts this slightly is the table of counts: students must work out which rows hold the middle prices of the 12.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 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

Once the middle positions are found there is nothing to compute: the first 5 stores charge $10.99, and the next two, the 6th and 7th, both charge $11.99. The wrong choices each reward a different misuse of a frequency table, taking the most common price, averaging the 5th and 6th values, or working out the mean, which costs time and still misses. Counting stores down the table, 5 and then 7, shows at once that both middle positions fall in the second row.

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

    Each row stands for several stores, 5 at $10.99, 2 at $11.99, 1 at $12.99 and 4 at $13.99, so the ordered list of 12 prices exists only in the student's head. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    The median of an even number of values is the mean of the two middle ones, here the 6th and 7th; the question assumes this, though both turn out to cost $11.99. (Anderson & Krathwohl, 2001)

Strategy: With 12 prices, the median is the mean of the 6th and 7th once they are in order; add up the store counts from the cheapest price until you reach 7.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Middle places of an even count
middle places=n2 and n2+1\text{middle places} = \frac{n}{2} \text{ and } \frac{n}{2} + 1
n is how many values there are, when n is an even number. Line the values up in order. The two values in these places have the same number of values before them as after them. For 10 values, they are the 5th and 6th.
Median of an even number of values
median=first middle value+second middle value2\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}
The median is the middle of a list of values put in order from least to greatest. With an even number of values, two of them share the middle. The median is their mean: add them and divide by 2.

  1. Sort out the question

    The table shows one video game’s price at 12 stores. Each row gives a price and how many stores charge it. For example, 4 stores charge $13.99, so $13.99 is in the list 4 times. You need the median price. That is the middle price when all 12 prices are lined up from least to greatest.

  2. Find the two middle places

    First check the number of stores: 5+2+1+4=125 + 2 + 1 + 4 = 12. That is an even number, so no one price sits in the middle. Two prices share it.

    middle places=n2 and n2+1\text{middle places} = \frac{n}{2} \text{ and } \frac{n}{2} + 1

    122=6 and 6+1=7\frac{12}{2} = 6 \text{ and } 6 + 1 = 7

    The middle prices are the 6th and 7th. Five prices come before the 6th, and five come after the 7th.

  3. Count down the table

    The table already lists the prices from least to greatest. Keep a running count of stores as you go down:

    • $10.99: 5 stores, places 1 to 5
    • $11.99: 2 stores, places 6 and 7
    • $12.99: 1 store, place 8
    • $13.99: 4 stores, places 9 to 12

    Both middle places, the 6th and the 7th, hold $11.99.

  4. Find the median

    Use the median formula with the 6th and 7th prices.

    median=first middle value+second middle value2\text{median} = \frac{\text{first middle value} + \text{second middle value}}{2}

    median=11.99+11.992=23.982=11.99\text{median} = \frac{11.99 + 11.99}{2} = \frac{23.98}{2} = 11.99

    The median price is $11.99.

    Watch out: $10.99 is the price that shows up most often, called the mode. That is choice A. Averaging the 5th and 6th prices gives $11.49, choice B. Adding all 12 prices and dividing by 12 gives the mean, about $12.32, choice D.

    So the answer is C, $11.99.

Check your answer

Look at each side of $11.99. Cheaper: the 5 stores at $10.99. More expensive: 1 store at $12.99 and 4 at $13.99, which is also 5 stores. Five below and five above, with the two $11.99 stores between them. So $11.99 is in the middle.

More questions like this

The original DOE question
Question 87 as printed on page 182 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 182, question 87. Open page 182 of the handbook (PDF)

Question 31 of 57

2019 Test B Math, question 31

Question

If xy=30xy=30, what is the value of 10÷x10\div x in terms of y?

  • A110y\frac{1}{10}y
  • B13y\frac{1}{3}y
  • C3y3y
  • D10y10y
Show the answer

Correct answer: B

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

xy=30xy=30

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

Then:

10÷x=10÷30y=10×y30=13y10\div x=10\div\frac{30}{y}=10\times\frac{y}{30}=\frac{1}{3}y

Question analysis

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

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.1 dividing fractions by fractions
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students rewrite expressions that have numbers and one variable. This one links two variables through their product and asks for a quotient, with no numbers given to check an answer against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 7, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    x and y are tied only by their product, 30, and the answer must be an expression in y, so there is no number to check unless students pick a pair such as x = 3 and y = 10. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Substituting x = 30/y turns 10 ÷ x into 10 divided by a fraction, which must be flipped to 10 × y/30 and then reduced. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. A very close wrong answer minor Choosing the answer

    The answer, (1/3)y, and the choice 3y use the same 3, one dividing y and one multiplying it, so a student unsure whether 10 ÷ 30 or 30 ÷ 10 applies sees both. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    xyxy means x times y. So x times y is 30. You need 10÷x10 \div x “in terms of y.” That means your answer can use y and numbers, but no x.

  2. Solve for x

    To swap x out, first get x by itself. Divide both sides of xy=30xy = 30 by y.

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

    This says x is 30 divided by y.

  3. Put the fraction in place of x

    Replace x in 10÷x10 \div x with this fraction.

    10÷x=10÷30y10 \div x = 10 \div \frac{30}{y}

    Now use the rule for dividing by a fraction. Flip the fraction and multiply.

    10÷30y=10×y30=10y3010 \div \frac{30}{y} = 10 \times \frac{y}{30} = \frac{10y}{30}

  4. Simplify the fraction

    Both 10 and 30 can be divided by 10.

    10y30=y3=13y\frac{10y}{30} = \frac{y}{3} = \frac{1}{3}y

    Watch out: 10 ÷ 30 is not 3. Dividing the bigger number by the smaller one gives 3y, choice C. Also, dividing by x is not the same as multiplying by y. That only works when xy is 1, not 30. It leads to choice D.

    So the answer is B, 13y\frac{1}{3}y.

Check your answer

Pick numbers that multiply to 30, like x = 3 and y = 10. Then 10÷x=10÷3=10310 \div x = 10 \div 3 = \frac{10}{3}. Choice B gives 13×10=103\frac{1}{3} \times 10 = \frac{10}{3}. They match. Choice C would give 30, and choice D would give 100.

More questions like this

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

Question 32 of 57

2019 Test B Math, question 32

Question

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

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

  • A16\frac{1}{6}
  • B12\frac{1}{2}
  • C23\frac{2}{3}
  • D3
Show the answer

Correct answer: B

First, find the lengths of the given segments:

QR=516−3=216\mathrm{QR}=5\frac{1}{6}-3=2\frac{1}{6}

PQ=3−113=123\mathrm{PQ}=3-1\frac{1}{3}=1\frac{2}{3}

Then, find the difference:

216−123=2\frac{1}{6}-1\frac{2}{3}=

136−53=\frac{13}{6}-\frac{5}{3}=

136−106=\frac{13}{6}-\frac{10}{6}=

36=12\frac{3}{6}=\frac{1}{2}

Question analysis

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

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding distances between points on a number line is grade 7 work. Here the points are mixed numbers with different denominators, so finding two distances and their difference makes it a typical but careful grade 7 task.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    Both lengths subtract mixed numbers with different denominators, 5 1/6 − 3 and 3 − 1 1/3, and the second needs regrouping before the two results are compared in sixths. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The three coordinates come from the labels under P, Q and R, and the segments QR and PQ must each be measured from Q at 3, the point they share. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Read the number line

    Point P is at 1131\frac{1}{3}. Point Q is at 3. Point R is at 5165\frac{1}{6}.

    The bar over QR means segment QR, the piece of the line from Q to R. You need how much longer QR is than PQ. So find both lengths, then subtract.

  2. Find the length of QR

    R is on the right, and Q is on the left. Use the length rule.

    QR=516−3=216\text{QR} = 5\frac{1}{6} - 3 = 2\frac{1}{6}

    QR is 2162\frac{1}{6} units long.

  3. Find the length of PQ

    Q is on the right, and P is on the left.

    PQ=3−113\text{PQ} = 3 - 1\frac{1}{3}

    Take it in two parts. First, 3 − 1 = 2. Then take away the extra 13\frac{1}{3}. That gives 2−13=1232 - \frac{1}{3} = 1\frac{2}{3}. So PQ is 1231\frac{2}{3} units long.

    Watch out: Don’t just do 3 − 1 and keep the 13\frac{1}{3}. That gives 2132\frac{1}{3}, which is too long. It leads to choice A.

  4. Write both lengths in sixths

    The fractions have different bottoms, 6 and 3. Change both lengths to sixths so you can subtract. Use the mixed number rule.

    216=2×6+16=1362\frac{1}{6} = \frac{2 \times 6 + 1}{6} = \frac{13}{6}

    123=1×3+23=53=1061\frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{5}{3} = \frac{10}{6}

  5. Subtract the lengths

    Take the shorter length away from the longer one.

    136−106=36=12\frac{13}{6} - \frac{10}{6} = \frac{3}{6} = \frac{1}{2}

    QR is 12\frac{1}{2} unit longer than PQ.

    So the answer is B, 12\frac{1}{2}.

Check your answer

Write every point in sixths. P is 86\frac{8}{6}, Q is 186\frac{18}{6} and R is 316\frac{31}{6}. QR is 31 − 18 = 13 sixths. PQ is 18 − 8 = 10 sixths. The difference is 3 sixths, which is one half.

More questions like this

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

Question 33 of 57

2019 Test B Math, question 33

Question

A line segment with point W at its left end and point Y at its right end. Point X lies on the segment between W and Y, closer to Y.

The distance from W to X is 70% of the distance from W to Y. If the distance from X to Y is 15 miles, what is the distance, in miles, from W to X?

  • A10.5
  • B35
  • C45
  • D50
Show the answer

Correct answer: B

Since the distance from W to X is 70% of the distance from W to Y, the distance from X to Y is 30% of the distance from W to Y. Let d represent the distance from W to Y. Then 15 miles is 30% of d.

0.30d=150.30d=15

d=150.30=50d=\frac{15}{0.30}=50

So, the distance from W to X, in miles, is 50−15=3550-15=35.

Question analysis

A percent question with a line segment: point X lies between W and Y, the distance from W to X is 70% of the whole, and the distance from X to Y is 15 miles. Students find the distance from W to X. Matching the 15 miles to the right percent before finding the whole is the step students most often skip.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 percent problems often give a part and its percent and ask for the whole. This one gives the distance for the part the percent does not describe and asks for the other part, so the whole has to be found on the way.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The percent and the given distance describe different pieces of the road, which is the whole difficulty. Students who apply the 70% straight to the 15 miles get a small number fast, and students who solve correctly for the whole of WY must still remember to take XY away. Checking that 35 is 70% of 50 confirms the answer and catches both slips.

  1. Turning words into math moderate Understanding the question

    The 70% describes WX, but the 15 miles given is XY, so the sentence has to be recast as “XY is 30% of WY” before the 15 miles can be used. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Finding the complement, 30%, then the whole, 15 ÷ 0.30 = 50, then the part, 50 − 15, gives three results, and the middle one is itself a choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work out what percent of WY the 15-mile stretch from X to Y must be, use it to find the whole of WY, then find the part from W to X.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}
The whole is the full amount. The part is a piece of it. The percent tells how many hundredths of the whole the part is. For example, 30% means 30 out of every 100, or 0.30. Use it whenever a percent describes a piece of a whole.

  1. Sort out the distances

    Point X sits between W and Y. So the trip from W to Y has two pieces: W to X, then X to Y.

    W to X is 70% of the whole distance from W to Y. X to Y is 15 miles. You need the distance from W to X, in miles.

  2. Find the percent for X to Y

    The whole distance from W to Y is 100%. W to X uses 70% of it. The rest is X to Y.

    100−70=30100 - 70 = 30

    So the 15 miles from X to Y is 30% of the whole distance.

  3. Find the whole distance

    Let d be the distance from W to Y. Use the percent formula. The part is 15, and the percent is 30.

    15=30100×d=0.30×d15 = \frac{30}{100} \times d = 0.30 \times d

    Divide both sides by 0.30 to get d by itself.

    d=15÷0.30=50d = 15 \div 0.30 = 50

    The distance from W to Y is 50 miles.

  4. Find the distance from W to X

    Take the 15 miles from X to Y away from the whole.

    50−15=3550 - 15 = 35

    W to X is 35 miles.

    Watch out: 50 miles is the whole distance from W to Y, choice D. You still have to subtract. Also, the 70% is of the whole distance, not of the 15 miles. Taking 70% of 15 gives 10.5, choice A.

    So the answer is B, 35 miles.

Check your answer

Is 35 really 70% of 50? 0.70×50=350.70 \times 50 = 35. Yes. And 35 + 15 = 50, the whole distance. The figure also shows X closer to Y. So W to X should be longer than 15 miles, and it is.

More questions like this

The original DOE question
Question 90 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 90. Open page 201 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 34 of 57

2019 Test B Math, question 34

Question

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

  • A10
  • B12
  • C24
  • D36
Show the answer

Correct answer: D

Multiply the numbers of highways between each pair of cities:

3×4×3=363\times 4\times 3=36

Question analysis

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

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 7 students build tree diagrams and lists for compound events. This item needs only the counting principle across three stages, with no restriction on the choices, which makes it shorter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
1 step
Reading level
Question and choices: about grade 11, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know minor Working it out

    The counting principle, multiplying the 3, 4 and 3 highway choices because each route uses one highway on every stretch, is assumed; the stem never says to multiply. (Anderson & Krathwohl, 2001)

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

Traps

  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the trip

    A route goes from City W to X, then to Y, then to Z. It uses exactly one highway on each of three stretches:

    • W to X: 3 highways
    • X to Y: 4 highways
    • Y to Z: 3 highways

    You need to count all the different routes.

  2. Count routes from W to Y

    Start small. Pick any of the 3 highways to City X. From there, you can take any of the 4 highways to City Y. So each first highway leads to 4 routes.

    3×4=123 \times 4 = 12

    There are 12 routes from City W to City Y.

  3. Add the last stretch

    Each of those 12 routes can end with any of the 3 highways to City Z. So multiply again. This is the counting principle with all three stretches.

    3×4×3=12×3=363 \times 4 \times 3 = 12 \times 3 = 36

    There are 36 routes in all.

  4. Avoid the common traps

    Watch out: Adding 3 + 4 + 3 = 10 counts highways, not routes. That gives choice A. Stopping at 12 forgets the last stretch, choice B. Adding 12 + 12 = 24 mixes two partial counts, choice C. A route is built by choosing one highway after another, so you multiply.

    So the answer is D, 36.

Check your answer

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

More questions like this

The original DOE question
Question 91 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 91. Open page 201 of the handbook (PDF)

Question 35 of 57

2019 Test B Math, question 35

Question

In Center City, the ratio of people to dogs is 4:14:1, and the ratio of dogs to cats is 3:23:2. What is the ratio of people to cats?

  • A2:12:1
  • B7:37:3
  • C6:16:1
  • D12:112:1
Show the answer

Correct answer: C

Take each ratio and solve for dogs:

pd=41\frac{p}{d}=\frac{4}{1}

4d=p4d=p

d=14pd=\frac{1}{4}p

And

dc=32\frac{d}{c}=\frac{3}{2}

d=32cd=\frac{3}{2}c

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

14p=32c\frac{1}{4}p=\frac{3}{2}c

p=6cp=6c

pc=61\frac{p}{c}=\frac{6}{1}

So, the ratio of people to cats is 6:16:1.

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.1 ratio concepts and language
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 ratio problems usually work with a single ratio. Here two ratios share the dogs but give them different numbers of parts, so one ratio must be rescaled before people and cats can be compared.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 32 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. No obvious method moderate Working it out

    Dogs are 1 part in the first ratio and 3 parts in the second, so the ratios cannot be joined until 4 : 1 is rescaled to 12 : 3; nothing in the stem says to do this. (Schoenfeld, 1985)

  2. Much to keep track of minor Working it out

    Three groups, people, dogs and cats, run through two ratios, and students must keep track of which number belongs to which group while scaling. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    A ratio compares two amounts. People to dogs is 4 : 1. That means 4 people for every 1 dog. Dogs to cats is 3 : 2. That means 3 dogs for every 2 cats.

    You need people to cats. Dogs show up in both ratios, so they are the link.

  2. Make the dog numbers match

    The dogs are 1 in the first ratio and 3 in the second. To join the ratios, the dog numbers must be the same. So multiply both parts of the first ratio by 3.

    4:1=(4×3):(1×3)=12:34 : 1 = (4 \times 3) : (1 \times 3) = 12 : 3

    Now there are 12 people for every 3 dogs.

  3. Join the two ratios

    Both ratios now have 3 dogs. So you can write all three groups in one ratio.

    people:dogs:cats=12:3:2\text{people} : \text{dogs} : \text{cats} = 12 : 3 : 2

    For every 12 people, there are 3 dogs and 2 cats.

  4. Read off and simplify

    People to cats is 12 : 2. Divide both parts by 2.

    12:2=(12÷2):(2÷2)=6:112 : 2 = (12 \div 2) : (2 \div 2) = 6 : 1

    Watch out: Taking the 4 people and the 2 cats straight from the problem gives 4 : 2 = 2 : 1, choice A. That skips matching the dogs. Stopping at 12 : 1, without using the 2 cats, gives choice D.

    So the answer is C, 6 : 1.

Check your answer

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

More questions like this

The original DOE question
Question 92 as printed on page 201 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 201, question 92. Open page 201 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 36 of 57

2019 Test B Math, question 36

Question

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

  • A3
  • B7
  • C8
  • D10
Show the answer

Correct answer: B

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

The area of square S is:

(x−1)(x−1)=9(x-1)(x-1)=9

(x−1)2=9(x-1)^{2}=9

x−1=3x-1=3

x=4x=4

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

The difference between the areas, in square centimeters, is 16−9=716-9=7.

Question analysis

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

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

Why it is hard

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

  1. Many steps minor Working it out

    The work runs from area to side and back: the square root of 9 gives S’s side, adding 1 gives T’s, squaring gives 16, and subtracting gives the difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The phrase “shorter than the side length of square T by 1 centimeter” describes S from T’s point of view, so T’s side is S’s side plus 1, not minus 1. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the two squares

    Square S has an area of 9 square centimeters. Its side is 1 centimeter shorter than the side of square T. You need the area of square T minus the area of square S, in square centimeters.

  2. Find the side of square S

    Use the area formula backward. Which number times itself makes 9?

    3×3=93 \times 3 = 9

    So each side of square S is 3 centimeters. This number is called the square root of 9.

  3. Find the side of square T

    Square T’s side is 1 centimeter longer than square S’s side.

    3+1=43 + 1 = 4

    Each side of square T is 4 centimeters.

  4. Find the area of square T

    Use the area formula with a side of 4 centimeters.

    A=4×4=16A = 4 \times 4 = 16

    The area of square T is 16 square centimeters.

  5. Subtract the areas

    Take the smaller area away from the larger one.

    16−9=716 - 9 = 7

    The difference is 7 square centimeters.

    Watch out: Picturing the new area as two strips, each 4 by 1, gives 8, choice C. But those strips share a 1-by-1 corner, so it gets counted twice. Also, 3 is a side length, not a difference of areas. That is choice A.

    So the answer is B, 7 square centimeters.

Check your answer

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

More questions like this

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

Question 37 of 57

2019 Test B Math, question 37

Question

If x is an odd number, what is the sum of x and the next 2 odd numbers greater than x?

  • Ax+4x+4
  • B2x+62x+6
  • C3x+33x+3
  • D3x+63x+6
Show the answer

Correct answer: D

If x is an odd number, then the next two odd numbers are (x+2)(x+2) and (x+4)(x+4).

Question analysis

If x is an odd number, what is the sum of x and the next two odd numbers greater than x? This algebra question asks students to represent consecutive odd numbers with expressions and add them. The wording packs three numbers into one sum, so students must read exactly which odd numbers to include.

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 routinely add linear expressions. The extra idea is expressing the odd numbers that follow x in terms of x, a familiar setup from consecutive-number problems, so the demand matches typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 24 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The algebra is one addition of three terms, so the work lies in naming the terms correctly. The choices punish each misreading of the phrase: stepping by 1 as with consecutive integers, leaving x out of the sum, or giving the last number instead of the total. A single test value, such as x = 5, checks all four choices in seconds and exposes every one of these.

  1. Variables instead of numbers moderate Understanding the question

    Here x stands for any odd number, so its odd successors have to be written as x + 2 and x + 4; trying x = 5, with 7 and 9, gives a sum of 21 to test the choices against. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    “The sum of x and the next 2 odd numbers greater than x” names three numbers, x itself included, and “next odd” means a step of 2, not 1. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Write the next two odd numbers in terms of x and add all three, then check by trying an odd number such as 5 in the stem and in each choice.

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.
  • 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

An odd number and the next two
x, x+2, x+4x,\ x + 2,\ x + 4
x is an odd number, like 5. Odd numbers are every other whole number: 1, 3, 5, 7. So the next odd number is always 2 more, not 1 more. The odd number after that is 2 more again, which is x + 4.
Combining like terms
ax+bx=(a+b)xax + bx = (a + b)x
a and b are numbers, and x is the letter. Like terms have the same letter part. Add their numbers in front and keep the letter. A plain x means 1x. Plain numbers are added only to plain numbers.

  1. Sort out the question

    x is an odd number. An odd number can’t be split into pairs, like 3, 5 or 7. You need the sum of three numbers: x itself and the next 2 odd numbers after it. The answer is an expression, a math phrase with a letter that stands for a number.

  2. Name the next two odd numbers

    Odd numbers skip every other number: 5, 7, 9, 11. Each odd number is 2 more than the one before. So x and the next two odd numbers are:

    x, x+2, x+4x,\ x + 2,\ x + 4

    For example, if x is 5, the next two are 7 and 9.

    Watch out: x + 1 is the next number, but it is even. Using x, x + 1 and x + 2 adds up to 3x+33x + 3, choice C.

  3. Add all three numbers

    The sum includes x itself. Write all three, then group the x’s together and the plain numbers together.

    ax+bx=(a+b)xax + bx = (a + b)x

    x+(x+2)+(x+4)=(x+x+x)+(2+4)=3x+6x + (x + 2) + (x + 4) = (x + x + x) + (2 + 4) = 3x + 6

    Three x’s make 3x, because 1 + 1 + 1 = 3. And 2 + 4 = 6.

    Watch out: don’t leave out x. Adding only the next two gives 2x+62x + 6, choice B. And x+4x + 4 is just the third number, not the sum. That is choice A.

    So the answer is D, 3x+63x + 6.

Check your answer

Try a real odd number. Let x = 5. The next two odd numbers are 7 and 9, and 5+7+9=215 + 7 + 9 = 21. Now put 5 into choice D: 3×5+6=213 \times 5 + 6 = 21. It matches. Choice B gives 16, and choice C gives 18, so they don’t work.

More questions like this

The original DOE question
Question 94 as printed on page 183 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 183, question 94. Open page 183 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

2019 Test B Math, question 38

Question

How is 1x−12x\frac{1}{x}-\frac{1}{2x} written as a single fraction? (Assume x≠0x\neq 0.)

  • A12x\frac{1}{2x}
  • B13x\frac{1}{3x}
  • C−1x-\frac{1}{x}
  • D−12x-\frac{1}{2x}
Show the answer

Correct answer: A

1x−12x=22x−12x=12x\frac{1}{x}-\frac{1}{2x}=\frac{2}{2x}-\frac{1}{2x}=\frac{1}{2x}

Question analysis

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

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 7.EE.2 rewriting expressions to show relationships
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students subtract fractions with unlike denominators and work with linear expressions, but seldom see a variable in a denominator. The steps match those for numerical fractions, so it is a short task once the common denominator is found.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 11, 14 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    The denominators are x and 2x, so the common denominator is built with a letter, and no number checks the result unless students substitute a value such as x = 1. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Only the first fraction changes: 1/x becomes 2/2x, and then the subtraction 2 − 1 must be done in the order written, first numerator minus second. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    You need to write 1x−12x\frac{1}{x} - \frac{1}{2x} as one fraction. The problem says x is not 0. That matters because you can’t divide by 0.

  2. Find a common bottom

    To subtract fractions, their bottoms must match. 2x is 2 times x. So multiply the top and bottom of 1x\frac{1}{x} by 2.

    1x=1×2x×2=22x\frac{1}{x} = \frac{1 \times 2}{x \times 2} = \frac{2}{2x}

    Now both fractions have the bottom 2x.

  3. Subtract the tops

    Subtract the tops in order. Keep the bottom.

    22x−12x=2−12x=12x\frac{2}{2x} - \frac{1}{2x} = \frac{2 - 1}{2x} = \frac{1}{2x}

    Watch out: Never add or subtract the bottoms. Adding them gives 13x\frac{1}{3x}, choice B. Subtracting them gives −1x-\frac{1}{x}, choice C. And keep the order: 2 − 1, not 1 − 2. The wrong order gives choice D.

    So the answer is A, 12x\frac{1}{2x}.

Check your answer

Try x = 1. The problem becomes 1−12=121 - \frac{1}{2} = \frac{1}{2}. Choice A gives 12×1=12\frac{1}{2 \times 1} = \frac{1}{2}. It matches. Choices B, C and D give 13\frac{1}{3}, −1 and −12-\frac{1}{2}.

More questions like this

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

Question 39 of 57

2019 Test B Math, question 39

Question

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

  • A0.0032%
  • B0.032%
  • C0.32%
  • D3.2%
Show the answer

Correct answer: C

The total number of people who participated in the competition was 1+5+10+4,984=5,0001+5+10+4{,}984=5{,}000. The total number who placed first, second, or third was 1+5+10=161+5+10=16.

165,000=3210,000=0.0032=0.32%\frac{16}{5{,}000}=\frac{32}{10{,}000}=0.0032=0.32\%

Question analysis

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

NYS standards
  • 6.RP.3c percent as a rate per 100
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Writing part of a whole as a percent is grade 6 content. This one first totals four groups, then turns a fraction of a large total into a percent, where a slip of one place value is easy to make.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 10, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

    16 ÷ 5,000 = 0.0032 has two zeros after the decimal point, and writing it as a percent moves the point two more places; each move can land one place off. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    The total of 5,000 is not given; it has to be built from 1 + 5 + 10 + 4,984 before the 16 people who placed can be compared with it. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the groups

    Some people placed: 1 in first, 5 in second and 10 in third. The other 4,984 people did not place. You need the percent of all the people who placed.

  2. Count the people who placed

    These are the people who finished first, second or third. Add them.

    1+5+10=161 + 5 + 10 = 16

    So 16 people placed.

  3. Count everyone

    The whole is everyone in the competition: the people who placed plus the people who did not.

    16+4,984=5,00016 + 4{,}984 = 5{,}000

    So 5,000 people took part. Notice that 4,984 is not the whole. It leaves out the 16 who placed.

  4. Turn the share into a percent

    Use the percent formula.

    percent=165,000×100\text{percent} = \frac{16}{5{,}000} \times 100

    First divide: 16÷5,000=0.003216 \div 5{,}000 = 0.0032. Then multiply by 100. That moves the decimal point two places to the right.

    0.0032×100=0.320.0032 \times 100 = 0.32

    So 0.32% of the people placed.

  5. Avoid the decimal traps

    Watch out: 0.0032 is not a percent yet. Putting a % sign on it gives choice A, which is 100 times too small. Moving the point only one place gives choice B. Moving it three places gives choice D.

    So the answer is C, 0.32%.

Check your answer

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

More questions like this

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

Question 40 of 57

2019 Test B Math, question 40

Question

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

  • A45°
  • B67.5°
  • C75°
  • D135°
Show the answer

Correct answer: B

Let A represent the measure of angle A, and let B represent the measure of angle B, in degrees. Since the angles are complementary, A+B=90A+B=90. The question states that B=3AB=3A, or A=B3A=\frac{B}{3}. Substitute that expression for A in the first equation and solve for B:

A+B=90A+B=90

B3+B=90\frac{B}{3}+B=90

4B3=90\frac{4B}{3}=90

4B=2704B=270

B=67.5B=67.5

Question analysis

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

NYS standards
  • 7.G.5 angle relationships and equations
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an equation from complementary angles is a grade 7 standard. This is the familiar version without a figure: one equation from the definition, then a division and a multiplication, typical of grade 7 classwork.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 7, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know moderate Working it out

    The stem never says what complementary means; the 90° total has to be recalled, and a student who confuses it with the supplementary 180° doubles both angles. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The clause “the measure of angle B is three times as great as the measure of angle A” has to become B = 3A, so the 90° splits into four equal parts, not three. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Complementary angles are two angles that add up to 90 degrees. Angle B is three times as big as angle A. You need the measure of angle B, in degrees.

  2. Write the facts as equations

    Let A and B stand for the two measures. The angles are complementary, so use the formula.

    A+B=90A + B = 90

    Angle B is three times angle A, so:

    B=3AB = 3A

  3. Put 3A in place of B

    Replace B in the sum with 3A. Then there is only one letter to find.

    A+3A=90A + 3A = 90

    4A=904A = 90

    Think of it as 4 equal parts. Angle A is 1 part, and angle B is 3 parts.

  4. Find angle A

    Divide both sides by 4.

    A=90÷4=22.5A = 90 \div 4 = 22.5

    Angle A measures 22.5 degrees. That is not the answer yet, because the question asks for angle B.

  5. Find angle B

    Angle B is three times angle A.

    B=3×22.5=67.5B = 3 \times 22.5 = 67.5

    Angle B measures 67.5 degrees.

    Watch out: Angles that add to 180 degrees are called supplementary. Using 180 by mistake gives 45 and 135, choice D. Splitting 90 in half gives 45, choice A. That ignores the fact that B is three times A.

    So the answer is B, 67.5°.

Check your answer

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

More questions like this

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

Question 41 of 57

2019 Test B Math, question 41

Question

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

  • A3313%33\frac{1}{3}\%
  • B100%
  • C200%
  • D300%
Show the answer

Correct answer: C

Let h represent the height of the tree in 2003. Then the height of the tree in 2013 is 3h3h. Calculate the percent increase:

3h−hh=2hh=2=200%\frac{3h-h}{h}=\frac{2h}{h}=2=200\%

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent-increase problems give two amounts to compare. This gives only a percent of the earlier height and no heights at all, so students must work out what that percent means for the change with nothing concrete to calculate with.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
2 steps
Reading level
Question and choices: about grade 3, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The stem hands students a percent and asks for a different percent, and the given 300% looks like a finished answer. The real step is separating “300% of” from “increased by”: a height that triples has grown by twice its original size. Picking a 2003 height of 100 units makes it concrete, since 300 − 100 = 200 units of growth on a base of 100.

  1. Turning words into math moderate Understanding the question

    “300% of the height it was in 2003” describes the new height, three times the old, not the increase; the question then asks for the change, which is a different percent. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers moderate Understanding the question

    No heights are given, only a percent, so students must invent a height such as 100 units or work with h and 3h to measure the change. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Something you must already know minor Working it out

    A percent increase divides the change by the starting amount, (3h − h) ÷ h, a rule the question assumes without stating. (Anderson & Krathwohl, 2001)

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

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the wording

    The 2013 height is 300% of the 2003 height. That means it is 3 times the 2003 height. You need the percent increase. That is how much the tree grew, compared with its 2003 height.

  2. Pick an easy height

    The question gives no real height, so choose one. Say the tree was 100 feet tall in 2003. Then the 2013 height is 300% of 100 feet.

    300100×100=300\frac{300}{100} \times 100 = 300

    In 2013, the tree was 300 feet tall.

  3. Find how much it grew

    The growth is how much taller the tree got. Subtract the old height from the new one.

    300−100=200300 - 100 = 200

    The tree grew 200 feet in ten years.

  4. Find the percent increase

    Use the formula. The original height is the 2003 height.

    percent increase=300−100100×100=2×100=200\text{percent increase} = \frac{300 - 100}{100} \times 100 = 2 \times 100 = 200

    The height went up by 200%.

    Watch out: 300% compares the new height with the old one. It is not the increase. That trap is choice D. Dividing the old height by the new one gives 3313%33\frac{1}{3}\%, choice A.

    So the answer is C, 200%.

Check your answer

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

More questions like this

The original DOE question
Question 98 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 98. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 42 of 57

2019 Test B Math, question 42

Question

1 gallon = 16 cups

1 cup = 8 fluid ounces

If 1 gallon = 2x2^{x} fluid ounces, what is the value of x?

  • A4
  • B7
  • C8
  • D64
Show the answer

Correct answer: B

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

1 gallon=1\text{ gallon}=

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

128 fluid ounces128\text{ fluid ounces}

128=27128=2^{7}

1 gallon=27 fluid ounces1\text{ gallon}=2^{7}\text{ fluid ounces}

So, x=7x=7.

Question analysis

This question gives two facts, 16 cups in a gallon and 8 fluid ounces in a cup, and says a gallon equals 2 to the power x fluid ounces. Students find the exponent x. It combines a unit conversion with writing a number as a power of 2, and the exponent is easy to confuse with a multiplier.

NYS standards
  • 6.RP.3d converting measurement units
  • 6.EE.1 whole-number exponent expressions
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Converting units with given facts is grade 6 work, and this conversion is a single multiplication. The twist is writing the result as a power of 2, which calls for fluency with powers that grade 7 work rarely asks for.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The conversion is one multiplication, so the question really tests whether students can name the exponent for 128. Stopping at the cups, where 16 = 2⁴, or treating 2 to the power x as 2 times x gives a listed choice. Writing 16 as 2⁴ and 8 as 2³ and adding the exponents is the quickest route, and doubling from 1 seven times is a sure check.

  1. Combines ideas minor Working it out

    A unit conversion, 16 cups × 8 fluid ounces = 128 fluid ounces, has to be joined to exponents by recognizing 128 as a power of 2. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Something you must already know minor Working it out

    Students must know or build the powers of 2 up to 128, and see that the exponent counts factors of 2, not the number of terms in a list starting at 1. (Anderson & Krathwohl, 2001)

Strategy: Convert 1 gallon to fluid ounces by multiplying the two facts, then keep halving the result, or keep doubling from 2, counting how many factors of 2 it takes.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Meaning of an exponent
2x=2×2×⋯×22^x = 2 \times 2 \times \cdots \times 2
The 2 is the base, the number being multiplied. x is the exponent, the small raised number. It tells how many 2s are multiplied together. For example, 2 to the third power means 2 × 2 × 2, which is 8.
Multiplying powers of the same base
2a×2b=2a+b2^a \times 2^b = 2^{a + b}
a and b are exponents. The first power has a twos multiplied together. The second has b more twos. Multiplying puts all the twos together, so you add the exponents.

  1. Sort out the question

    A gallon is 16 cups. A cup is 8 fluid ounces. A gallon is also 2x2^x fluid ounces. You need x, the exponent. It tells how many 2s are multiplied together.

  2. Change a gallon to fluid ounces

    Each of the 16 cups holds 8 fluid ounces. So multiply.

    16×8=12816 \times 8 = 128

    One gallon is 128 fluid ounces.

  3. Write 128 as a power of 2

    Now you need 2x=1282^x = 128. Keep multiplying by 2, and count the 2s.

    Number of 2s1234567
    Product248163264128

    It takes seven 2s to make 128.

    27=1282^7 = 128

    So x = 7.

  4. Give the answer

    Watch out: A gallon is 16 cups, and 16=2416 = 2^4. Stopping there gives 4, choice A. And 2x2^x does not mean 2 times x. Solving 2 × x = 128 gives 64, choice D.

    So the answer is B, 7.

Check your answer

Use the exponent rule. 16 is 242^4 and 8 is 232^3. So 16×8=24×23=24+3=2716 \times 8 = 2^4 \times 2^3 = 2^{4 + 3} = 2^7. The exponent is 7 again.

More questions like this

The original DOE question
Question 99 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 99. Open page 203 of the handbook (PDF)

Question 43 of 57

2019 Test B Math, question 43

Question

A certain square puzzle has an area of 64 square inches, and a cubic block has a volume of 64 cubic inches. What is the ratio of the length of a side of the puzzle to the length of a side of the block?

  • A1:31:3
  • B1:21:2
  • C2:32:3
  • D2:12:1
Show the answer

Correct answer: D

The length of one side of the puzzle is 64=8\sqrt{64}=8 inches. The length of one side of the block is 643=4\sqrt[3]{64}=4 inches.

The ratio of the side of the puzzle to the side of the block is 8:48:4, which reduces to 2:12:1.

Question analysis

A square puzzle has an area of 64 square inches, and a cube-shaped block has a volume of 64 cubic inches. Students find the ratio of a side of the puzzle to a side of the block. Both figures are described by the number 64, but turning an area and a volume back into lengths takes different steps, and the order of the ratio matters.

NYS standards
  • 8.EE.2 square and cube roots
  • 6.G.5 perfect squares and perfect cubes
  • 6.RP.1 ratio concepts and language
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 area and volume problems go from side lengths to an area or a volume. Going backward needs a square root and a cube root, and cube roots are grade 8 content, although 64 is both a perfect square and a perfect cube.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 46 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Because 64 is both a perfect square and a perfect cube, the computation is easy once students know to undo squaring and cubing. The risk is at the end: with sides of 8 and 4 inches in hand, writing the block first turns the ratio upside down. Students who never find the side lengths are left with ideas such as two dimensions against three.

  1. Something you must already know moderate Working it out

    Neither formula is given: the puzzle’s side is the number whose square is 64 and the block’s edge the number whose cube is 64, and cube roots are grade 8 work. (Anderson & Krathwohl, 2001)

  2. Combines ideas minor Working it out

    A square and a cube share the number 64 in different units, so two different inverse operations feed one ratio, which must list the puzzle’s side first. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Find the side of the square from its area and the edge of the cube from its volume by testing small whole numbers, then write the puzzle's side first and simplify.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a square
A=s×sA = s \times s
A is the area, in square units. s is the length of one side. All sides of a square are equal, so the area is the side times itself. To find the side from the area, ask which number times itself gives the area.
Volume of a cube
V=s×s×sV = s \times s \times s
V is the volume, the space inside, in cubic units. s is the length of one edge. A cube’s length, width and height are all equal, so the volume is the edge used three times. To go backward, ask which number used three times gives the volume.

  1. Sort out the shapes

    The puzzle is a square with an area of 64 square inches. The block is a cube with a volume of 64 cubic inches. A cube is a box with every edge the same length. You need the ratio of the puzzle’s side to the block’s side, with the puzzle first.

  2. Find the puzzle’s side

    Use the area formula backward. Which number times itself makes 64?

    8×8=648 \times 8 = 64

    The puzzle’s side is 8 inches. This number is the square root of 64.

  3. Find the block’s side

    Use the volume formula backward. Which number used three times makes 64? Try small numbers. 3 × 3 × 3 = 27 is too small.

    4×4×4=644 \times 4 \times 4 = 64

    The block’s side is 4 inches. This number is the cube root of 64.

  4. Write and simplify the ratio

    Put the puzzle first, as the question asks. Then divide both parts by 4.

    8:4=(8÷4):(4÷4)=2:18 : 4 = (8 \div 4) : (4 \div 4) = 2 : 1

    Watch out: Writing the block’s side first gives 1 : 2, choice B. The order in a ratio matters.

    So the answer is D, 2 : 1.

Check your answer

Check both sides. 8 × 8 = 64 square inches, and 4 × 4 × 4 = 64 cubic inches. The puzzle’s side is longer, so its number in the ratio must be bigger. Only choice D puts the bigger number first.

More questions like this

The original DOE question
Question 100 as printed on page 203 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 203, question 100. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 44 of 57

2019 Test B Math, question 44

Question

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

  • A76
  • B82
  • C88
  • D90
Show the answer

Correct answer: A

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

81+83+x+y4=85\frac{81+83+x+y}{4}=85

164+x+y=340164+x+y=340

x+y=176x+y=176

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

Question analysis

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

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.EE.8 writing simple inequalities
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 mean problems usually find one missing score. Here two tests remain, and the least possible score depends on how high the other unknown score is allowed to go, a planning step beyond typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 60 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. No obvious method moderate Working it out

    The least third score depends on how high the fourth can go; only the parenthesis “test scores can range from 0 to 100” tells students to set the fourth test at 100. (Schoenfeld, 1985)

  2. Many steps minor Working it out

    The 340 points needed, the 164 already earned, the 176 still needed and the 100 allowed on the last test make four steps, each a chance to lose track. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Rosa takes 4 tests. She scored 81 and 83 on the first two. She wants an average of 85 on all 4. You need the lowest score she can get on test 3 and still reach that average. Each score is from 0 to 100.

  2. Find the total points needed

    Turn the average formula around. The sum is the average times the number of tests.

    sum of the scores=85×4=340\text{sum of the scores} = 85 \times 4 = 340

    Rosa needs 340 points in all.

  3. Find the points still needed

    Subtract the points she already has.

    81+83=16481 + 83 = 164

    340−164=176340 - 164 = 176

    Tests 3 and 4 must add up to 176 points.

  4. Make test 4 as high as possible

    To make test 3 as low as it can be, test 4 must carry as many points as it can. The highest score is 100.

    176−100=76176 - 100 = 76

    So the lowest score on test 3 is 76.

    Watch out: Splitting 176 evenly gives 88 on each test, choice C. That works, but it is not the least score. And 82 is just the average of her first two tests, choice B.

    So the answer is A, 76.

Check your answer

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

More questions like this

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

Question 45 of 57

2019 Test B Math, question 45

Question

The Barnes family and the Ramirez family each have a pizza of equal size. The first pizza is cut into 18 equal slices, and the second pizza is cut into 15 equal slices. If the Barnes family eats 11 slices from the first pizza, what is the greatest number of whole slices from the second pizza that the Ramirez family can eat without eating a greater percentage of a pizza than the Barnes family ate?

  • A6
  • B7
  • C8
  • D9
Show the answer

Correct answer: D

The fraction of the pizza that the Barnes family eats is 1118\frac{11}{18}. Let x represent the number of slices of pizza that the Ramirez family eats. Then the fraction of the pizza that the Ramirez family eats is x15\frac{x}{15}. Write an inequality and solve for x:

x15≤1118\frac{x}{15}\leq\frac{11}{18}

x≤11×1518x\leq\frac{11\times 15}{18}

x≤16518x\leq\frac{165}{18}

x≤916x\leq 9\frac{1}{6}

So, the greatest number of whole slices of pizza the Ramirez family can eat is 9.

Question analysis

Two pizzas of the same size are cut into 18 and 15 equal slices, and one family eats 11 slices of the first. Students find the most whole slices of the second pizza the other family can eat without eating a larger share. It tests comparing fractions with different denominators and rounding to a whole number in the right direction.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 inequality problems usually ask for the greatest whole number that fits a budget. This one wraps that idea in a long stem about shares of two pizzas sliced differently, with several conditions to keep straight at once.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 11, 72 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The math is one inequality between fractions, but the long final sentence hides it, so students who skim fall back on comparing slice counts. Reasoning that the second pizza has 3 fewer slices, so the Ramirez family should eat 3 fewer, is the strongest pull. Comparing 11/18 with x/15, or turning 11/18 into about 61% and taking 61% of 15, keeps the comparison about shares of a pizza.

  1. Complex sentences moderate Understanding the question

    The question is one 42-word sentence ending in a nested condition, “without eating a greater percentage of a pizza than the Barnes family ate”, a reading hurdle rather than more math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

  2. Turning words into math moderate Understanding the question

    The limit of no “greater percentage of a pizza” must become the inequality x/15 ≤ 11/18, comparing fractions of two pizzas cut into different numbers of slices, not slice counts. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Solving gives x ≤ 11 × 15 ÷ 18 = 165/18, a division by 18 that leaves 9 1/6, which must then be cut back to whole slices. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the Barnes family's share as a fraction of their pizza, then find the largest whole number of fifteenths that does not exceed it, by cross-multiplying or by converting to decimals.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a pizza
fraction eaten=slices eatenslices in the pizza\text{fraction eaten} = \frac{\text{slices eaten}}{\text{slices in the pizza}}
The top counts the slices eaten. The bottom counts all the equal slices the pizza was cut into. The fraction tells how much of the whole pizza was eaten. Use fractions to compare pizzas cut into different numbers of slices.

  1. Sort out the question

    The two pizzas are the same size. The Barnes pizza has 18 slices, and they eat 11. The Ramirez pizza has 15 slices, so its slices are bigger. You need the most whole slices the Ramirez family can eat without eating a bigger share of their pizza.

  2. Find the Barnes family’s share

    Their share is the slices they ate out of all the slices in their pizza. Use the fraction formula.

    fraction eaten=1118\text{fraction eaten} = \frac{11}{18}

    The Barnes family ate 1118\frac{11}{18} of their pizza.

  3. Write an inequality

    Let x be the number of Ramirez slices. Their share is x15\frac{x}{15}. It must be no more than the Barnes share. The sign ≤ means less than or equal to.

    x15≤1118\frac{x}{15} \leq \frac{11}{18}

  4. Solve for x

    Multiply both sides by 15 to get x by itself.

    x≤11×1518=16518x \leq \frac{11 \times 15}{18} = \frac{165}{18}

    Now divide. 18 goes into 165 nine times, since 18 × 9 = 162. There are 3 left over.

    x≤9318=916x \leq 9\frac{3}{18} = 9\frac{1}{6}

  5. Pick the whole number

    x must be a whole number no bigger than 9169\frac{1}{6}. So the most is 9 slices. 10 slices would be too many.

    Watch out: The Ramirez pizza has 3 fewer slices, but that does not mean they eat 3 fewer. Their slices are bigger. Eating 11 − 3 = 8 slices is choice C, which is not the most they can eat.

    So the answer is D, 9.

Check your answer

Use decimals. The Barnes share is 11÷18≈0.61111 \div 18 \approx 0.611. Nine Ramirez slices are 9÷15=0.69 \div 15 = 0.6, just under it. Ten slices are 10÷15≈0.66710 \div 15 \approx 0.667, too much. So 9 is the most.

More questions like this

The original DOE question
Question 102 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 102. Open page 204 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 46 of 57

2019 Test B Math, question 46

Question

03+40+2−1+(−1)20^{3}+4^{0}+2^{-1}+(-1)^{2}

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

  • A0
  • B0.5
  • C1.5
  • D2.5
Show the answer

Correct answer: D

03+40+2−1+(−1)20^{3}+4^{0}+2^{-1}+(-1)^{2}

=0+1+12+1=0+1+\frac{1}{2}+1

=212=2.5=2\frac{1}{2}=2.5

Question analysis

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

NYS standards
  • 8.EE.1 integer exponent properties
  • 7.NS.2a rational number multiplication
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 work uses whole-number exponents and signed numbers. Zero and negative exponents are grade 8 content, and this expression tests both alongside a negative base, each with its own common misconception.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
4 steps
Reading level
Question and choices: about grade 6, 12 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know major Working it out

    Three rules the test does not give decide the sum: any nonzero number to the 0 power is 1, 2 to the −1 power is 1/2, and (−1)² is positive 1. (Anderson & Krathwohl, 2001)

  2. Error-prone arithmetic minor Working it out

    The four terms carry a zero base, a zero exponent, a negative exponent and a negative base, so a sign or a stray 0 can slip in at each one. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Zero exponent
a0=1a^0 = 1
a is any number except 0. An exponent of 0 always gives 1. It works because each step down in the exponent divides by a. For example, 4 to the first power is 4, and 4 ÷ 4 = 1.
Negative exponent
a−n=1ana^{-n} = \frac{1}{a^n}
a is the base, and n is a positive whole number. A negative exponent does not make the number negative. It means 1 divided by the base to the positive power. For example, 2 to the −1 power is one half.

  1. Sort out the question

    The expression adds four powers. A power is a base with an exponent, a small raised number. The exponent tells how many times to use the base as a factor, a number you multiply. Find each power, then add.

  2. Find the first two powers

    Zero times itself is still zero.

    03=0×0×0=00^3 = 0 \times 0 \times 0 = 0

    Next, use the zero exponent rule.

    40=14^0 = 1

    Watch out: 404^0 is 1, not 0. Taking it as 0 gives choice C.

  3. Find the negative power

    Use the negative exponent rule.

    2−1=121=12=0.52^{-1} = \frac{1}{2^1} = \frac{1}{2} = 0.5

    Watch out: 2−12^{-1} is not −2. That mistake gives choice A.

  4. Find the last power

    The base is −1, inside the parentheses. Multiply it by itself. A negative times a negative is positive.

    (−1)2=(−1)×(−1)=1(-1)^2 = (-1) \times (-1) = 1

    Watch out: The answer is not −1. That mistake gives choice B.

  5. Add the four values

    Now add them. The question wants a decimal, so write one half as 0.5.

    0+1+0.5+1=2.50 + 1 + 0.5 + 1 = 2.5

    So the answer is D, 2.5.

Check your answer

See why 40=14^0 = 1 with a pattern. 42=164^2 = 16 and 41=44^1 = 4. Each step down divides by 4, so 40=4÷4=14^0 = 4 \div 4 = 1. In the same way, 20=12^0 = 1 and 2−1=1÷2=122^{-1} = 1 \div 2 = \frac{1}{2}.

More questions like this

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

Question 47 of 57

2019 Test B Math, question 47

Question

A company wants to study 6 brands of soap by comparing each brand with every other brand. If each comparison costs $2,000, how much will the company spend altogether?

  • A$13,000
  • B$20,000
  • C$30,000
  • D$60,000
Show the answer

Correct answer: C

First, find the number of comparisons. Let A, B, C, D, E, and F represent the 6 brands of soap. Then the comparisons are AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, and EF. So, there are 15 comparisons. Multiply to find the total cost:

15×$2,000=$30,00015\times \$2{,}000=\$30{,}000

Question analysis

A company will compare each of 6 soap brands with every other brand, at $2,000 per comparison, and students find the total cost. This counting question tests listing combinations of two and then using the count. With six brands, the list is long enough that a pair is easily missed or counted twice.

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. Counting every pair among six brands is that skill, with the extra demand that each pair be counted once whichever brand comes first, followed by one multiplication.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 9, 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 cost step is trivial, so the question is about counting comparisons. The multiplication 6 × 5 feels natural but counts every pair twice, and a listing that starts one short, with 4 partners for the first brand, gives the count for five brands. Halving 6 × 5, or adding 5 + 4 + 3 + 2 + 1, gives 15 comparisons, and multiplying by $2,000 is the last, easy step.

  1. Turning words into math moderate Understanding the question

    “Comparing each brand with every other brand” counts each pair once, since comparing A with B is the same study as comparing B with A, so the count is half of 6 × 5. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    A systematic list, the first brand with 5 others, the second with 4 new ones, and so on, has to be kept going to 5 + 4 + 3 + 2 + 1 without losing a row. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Count the pairs systematically: the first brand pairs with each of the other five, the second with each of the four after it, and so on; then multiply by $2,000.

Traps

  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of pairs in a group
pairs=n×(n−1)2\text{pairs} = \frac{n \times (n - 1)}{2}
n is the number of things in the group, here soap brands. Each one can be paired with the n − 1 others. Multiplying counts every pair twice, once from each side, so you divide by 2.
Total cost
total cost=number of comparisons×cost of each\text{total cost} = \text{number of comparisons} \times \text{cost of each}
When every item costs the same, multiply how many items there are by the cost of one. Here every comparison costs the same amount, $2,000.

  1. Sort out the question

    There are 6 brands of soap. Each brand is compared with every other brand, one pair at a time. Each comparison costs $2,000. You need the total cost, so first count the comparisons. Comparing brand A with brand B is the same study as comparing B with A. It counts once.

  2. List the pairs in order

    Call the brands A, B, C, D, E and F. Pair A with each brand after it. Then do the same for B, and so on. Never pair a brand with one before it, so no pair is counted twice.

    BrandNew pairsCount
    AAB, AC, AD, AE, AF5
    BBC, BD, BE, BF4
    CCD, CE, CF3
    DDE, DF2
    EEF1

  3. Count the comparisons

    Add the counts.

    5+4+3+2+1=155 + 4 + 3 + 2 + 1 = 15

    There are 15 comparisons. The pairs formula agrees. Each of the 6 brands pairs with 5 others, but that counts each pair twice.

    pairs=n×(n−1)2\text{pairs} = \frac{n \times (n - 1)}{2}

    pairs=6×52=302=15\text{pairs} = \frac{6 \times 5}{2} = \frac{30}{2} = 15

    Watch out: 6×5=306 \times 5 = 30 counts AB and BA as two comparisons. That leads to $60,000, choice D. Giving brand A only 4 partners counts 10 pairs, which leads to $20,000, choice B.

  4. Find the total cost

    Each of the 15 comparisons costs $2,000.

    total cost=number of comparisons×cost of each\text{total cost} = \text{number of comparisons} \times \text{cost of each}

    total cost=15×2,000=30,000\text{total cost} = 15 \times 2{,}000 = 30{,}000

    The company will spend $30,000. So the answer is C, $30,000.

Check your answer

Test the counting on a group small enough to picture. With 3 brands, the pairs are AB, AC and BC, which is 3 pairs. The formula gives 3×22=3\frac{3 \times 2}{2} = 3. It works. Also, 30,000÷2,000=1530{,}000 \div 2{,}000 = 15, the number of comparisons.

More questions like this

The original DOE question
Question 104 as printed on page 184 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 184, question 104. Open page 184 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

2019 Test B Math, question 48

Question

What number is halfway between 120\frac{1}{20} and 110\frac{1}{10} on a number line?

  • A15\frac{1}{5}
  • B215\frac{2}{15}
  • C130\frac{1}{30}
  • D340\frac{3}{40}
Show the answer

Correct answer: D

First, subtract the two fractions:

110−120=220−120=2−120=120\frac{1}{10}-\frac{1}{20}=\frac{2}{20}-\frac{1}{20}=\frac{2-1}{20}=\frac{1}{20}

Then, divide the difference by 2 to find half the distance between the two fractions:

120÷2=120×12=140\frac{1}{20}\div 2=\frac{1}{20}\times\frac{1}{2}=\frac{1}{40}

Finally, add the distance to the first point (120)\left(\frac{1}{20}\right) to find the halfway point:

120+140=240+140=2+140=340\frac{1}{20}+\frac{1}{40}=\frac{2}{40}+\frac{1}{40}=\frac{2+1}{40}=\frac{3}{40}

Question analysis

This number line question asks for the number exactly halfway between the fractions 1/20 and 1/10. It tests finding the midpoint of two rational numbers when their denominators differ. Both fractions are small and close together, so a slip in combining them is easy to make.

NYS standards
  • 7.NS.1c rational number subtraction
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Locating a point between two rational numbers on a number line is grade 6 and 7 work. Here the endpoints are small fractions with related denominators, so once they share a denominator the task is quick, matching typical grade 7 practice.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is short once the midpoint is seen as an average, but small fractions with different denominators invite the straight-across habit. Adding numerators and denominators separately gives 2/30, and halving that either way lands on a listed choice outside the two endpoints. Checking that the answer sits between 0.05 and 0.1, which 0.075 does, rules out every wrong choice.

  1. Error-prone arithmetic minor Working it out

    The fractions 1/20 and 1/10 have different denominators, so 1/10 must become 2/20 before adding, and halving 3/20 then needs a denominator of 40. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Something you must already know minor Working it out

    The stem does not say that the point halfway between two numbers is their mean; students must bring that, or the idea of adding half the gap to the smaller number. (Anderson & Krathwohl, 2001)

Strategy: Find the average of the two fractions: rewrite them with a common denominator, add them, and divide the sum by 2.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Equivalent fractions
ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}
a is the top number and b is the bottom number, the denominator. k is any whole number except 0. Multiplying the top and bottom by the same number gives an equal fraction. Use it to give two fractions the same denominator so you can add them.
Midpoint on a number line
midpoint=a+b2\text{midpoint} = \frac{a + b}{2}
a and b are the numbers at the two ends. The midpoint is the number exactly halfway between them. Adding the ends and dividing by 2 finds their mean, or average, which sits in the middle.

  1. Sort out the question

    You need the number exactly halfway between 120\frac{1}{20} and 110\frac{1}{10} on a number line. That number is the midpoint. It is the same distance from both ends. Notice that one tenth is the bigger number, because tenths are bigger pieces than twentieths.

  2. Give both fractions the same denominator

    The denominator is the bottom number. To add fractions, they need the same one. 20 is a multiple of 10, so use 20.

    ab=a×kb×k\frac{a}{b} = \frac{a \times k}{b \times k}

    110=1×210×2=220\frac{1}{10} = \frac{1 \times 2}{10 \times 2} = \frac{2}{20}

    Now the two ends are 120\frac{1}{20} and 220\frac{2}{20}.

  3. Average the two ends

    Use the midpoint formula. First add the ends. Then divide by 2.

    midpoint=a+b2\text{midpoint} = \frac{a + b}{2}

    a+b=120+220=320a + b = \frac{1}{20} + \frac{2}{20} = \frac{3}{20}

    midpoint=320÷2=320×12=340\text{midpoint} = \frac{3}{20} \div 2 = \frac{3}{20} \times \frac{1}{2} = \frac{3}{40}

    Dividing by 2 is the same as multiplying by 12\frac{1}{2}. Multiply the tops and the bottoms.

    Watch out: don’t add straight across, top plus top and bottom plus bottom. That gives 230\frac{2}{30}. Halving it gives 130\frac{1}{30}, choice C, which is less than both ends. Halving only the bottom gives 215\frac{2}{15}, choice B, which is past both ends. Choice A, 15\frac{1}{5}, is also past one tenth.

    So the answer is D, 340\frac{3}{40}.

Check your answer

Write both ends in fortieths: 120=240\frac{1}{20} = \frac{2}{40} and 110=440\frac{1}{10} = \frac{4}{40}. Halfway between 2 and 4 is 3, so halfway between the ends is three fortieths. As decimals, 0.075 sits right between 0.05 and 0.1.

More questions like this

The original DOE question
Question 105 as printed on page 184 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 184, question 105. Open page 184 of the handbook (PDF)

Question 49 of 57

2019 Test B Math, question 49

Question

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

  • A40
  • B286
  • C560
  • D1,680
Show the answer

Correct answer: C

First, change 6 feet to yards:

6 feet×1 yard3 feet=2 yards6\text{ feet}\times\frac{1\text{ yard}}{3\text{ feet}}=2\text{ yards}

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

20×14×2=56020\times 14\times 2=560

Question analysis

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

NYS standards
  • 6.G.2 volume with fractional edge lengths
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 The volume of a rectangular prism is grade 5 and 6 content, and the conversion here gives a whole number. Grade 7 volume work is usually longer, with composite or triangular prisms; here the only extra step is one unit conversion.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Combines ideas minor Working it out

    The length and width are in yards but the depth is in feet, so a conversion, 6 feet = 2 yards, must come before the volume formula is used. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Something you must already know minor Working it out

    Neither 3 feet = 1 yard nor volume = length × width × depth is given; both are assumed in a stem that asks for cubic yards. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the measurements

    The pool is shaped like a box. Uniform depth means the water is equally deep everywhere. The pool is 20 yards long and 14 yards wide. The water is 6 feet deep. You need the volume, the space the water fills, in cubic yards.

  2. Change the depth to yards

    Two sides are in yards, but the depth is in feet. All three must use the same unit before you multiply. The answer is in cubic yards, so change feet to yards.

    6÷3=26 \div 3 = 2

    The water is 2 yards deep.

  3. Find the volume

    Use the volume formula with all three measurements in yards.

    V=20×14×2=560V = 20 \times 14 \times 2 = 560

    The pool holds 560 cubic yards of water.

  4. Give the answer

    Watch out: Multiplying 20 × 14 × 6 mixes yards with feet. It gives 1,680, choice D, which is three times too big.

    So the answer is C, 560 cubic yards.

Check your answer

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

More questions like this

The original DOE question
Question 106 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 106. Open page 205 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 50 of 57

2019 Test B Math, question 50

Question

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

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

  • A320\frac{3}{20}
  • B310\frac{3}{10}
  • C12\frac{1}{2}
  • D35\frac{3}{5}
Show the answer

Correct answer: B

First, list the products:

3×4=123\times 4=12

3×5=153\times 5=15

3×6=183\times 6=18

3×7=213\times 7=21

4×5=204\times 5=20

4×6=244\times 6=24

4×7=284\times 7=28

5×6=305\times 6=30

5×7=355\times 7=35

6×7=426\times 7=42

Of the 10 products, 3 are odd. The fraction is 310\frac{3}{10}.

Question analysis

A student lists every product of two different numbers from the set 3, 4, 5, 6, 7. Students find what fraction of those products are odd. It tests listing outcomes systematically without counting a pair twice, and reasoning about odd and even numbers.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.SP.8a compound event as fraction of outcomes
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Organized lists of outcomes are grade 7 work. Here students list the products of pairs from five numbers, without repeating a pair or pairing a number with itself, then count the ones that fit.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 28 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The count of odd products is easy, since only 3, 5 and 7 can make one; the total is where students go wrong. Counting ordered pairs, 5 × 4 = 20, while listing odd products once mixes two kinds of counting. Writing the 10 pairs in order, 3 with each larger number, then 4, and so on, keeps the total honest.

  1. Much to keep track of moderate Working it out

    Listing the products of 2 different numbers from five means tracking 10 pairs, with no pair repeated in reverse order and no number paired with itself. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Something you must already know minor Working it out

    The shortcut, that a product is odd only when both factors are odd, cuts the work to counting pairs from 3, 5 and 7, but it must be known. (Anderson & Krathwohl, 2001)

Strategy: List each pair of different numbers once, then find each product, or use the fact that a product is odd only when both factors are odd, and count.

Traps

  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a list
fraction=odd productsall products\text{fraction} = \frac{\text{odd products}}{\text{all products}}
The top counts the products that are odd. The bottom counts all the products in the list. A product is the answer when you multiply. The fraction tells what share of the list is odd.
Odd times odd
odd×odd=odd\text{odd} \times \text{odd} = \text{odd}
An odd number can’t be split into pairs, like 3 or 7. An even number can, like 4. A product is odd only when both numbers are odd. If either number is even, the product is even, because it splits into pairs.

  1. Sort out the question

    The set has 3, 4, 5, 6 and 7. The student multiplies 2 different numbers, every possible way. A product is the answer to a multiplication. Each pair counts once, since 3 × 4 and 4 × 3 give the same product. You need the fraction of the products that are odd.

  2. List every pair once

    Pair 3 with each bigger number. Then pair 4 with each bigger number, and so on. This way no pair is counted twice.

    3 × 4 = 123 × 5 = 153 × 6 = 183 × 7 = 21
    4 × 5 = 204 × 6 = 244 × 7 = 28
    5 × 6 = 305 × 7 = 35
    6 × 7 = 42

    There are 10 products.

  3. Count the odd products

    The odd products are 15, 21 and 35. That makes 3. They come from the only pairs where both numbers are odd: 3 and 5, 3 and 7, and 5 and 7.

  4. Write the fraction

    Use the fraction formula.

    fraction=310\text{fraction} = \frac{3}{10}

    Watch out: Counting 4 × 3 as well as 3 × 4 makes 20 pairs. Then 320\frac{3}{20}, choice A, mixes two ways of counting. And 35\frac{3}{5}, choice D, is the share of the numbers that are odd, not the products.

    So the answer is B, 310\frac{3}{10}.

Check your answer

Count without listing. Each of the 5 numbers can pair with 4 others, which makes 20. But that counts every pair twice, so there are 20 ÷ 2 = 10 pairs. Only the odd numbers 3, 5 and 7 make odd products, and they form 3 pairs.

More questions like this

The original DOE question
Question 107 as printed on page 205 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 107. Open page 205 of the handbook (PDF)

Question 51 of 57

2019 Test B Math, question 51

Question

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

  • A10.20
  • B10.25
  • C10.50
  • D10.75
Show the answer

Correct answer: A

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

10+525=1015=10.2010+\frac{5}{25}=10\frac{1}{5}=10.20

Question analysis

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

NYS standards
  • 6.NS.2 multi-digit division
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Division with remainders and decimal quotients are grade 6 skills. This takes two short steps, with no word problem to translate, though students must know what a remainder means once a quotient is written as a decimal.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 22 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know moderate Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the words

    n is divided by 25. So 25 is the divisor, the number you divide by. The quotient is the answer to a division. Here it is 10 with a remainder of 5. The remainder is the amount left over. You need the exact quotient as a decimal.

  2. Turn the remainder into a fraction

    The 5 left over still has to be divided by 25. So it becomes the fraction 525\frac{5}{25}. Use the formula.

    quotient=10+525\text{quotient} = 10 + \frac{5}{25}

  3. Change the fraction to a decimal

    Simplify first. Divide the top and bottom by 5. Then write the fraction in hundredths.

    525=15=20100=0.20\frac{5}{25} = \frac{1}{5} = \frac{20}{100} = 0.20

    So the quotient is 10 + 0.20 = 10.20.

  4. Give the answer

    Watch out: A remainder is not the digits after the decimal point. Writing the 5 after the point gives 10.50, choice C. Writing the 25 after it gives 10.25, choice B.

    So the answer is A, 10.20.

Check your answer

Rebuild n. Ten 25s plus the 5 left over is 25×10+5=25525 \times 10 + 5 = 255. Now divide: 255÷25=10.2255 \div 25 = 10.2. That matches 10.20.

More questions like this

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

Question 52 of 57

2019 Test B Math, question 52

Question

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

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

  • A6
  • B12
  • C3π3\pi
  • D6π6\pi
Show the answer

Correct answer: C

The perimeter (or circumference) of one of the three semicircles is 12(diameter)π=12(2π)=π\frac{1}{2}(\text{diameter})\pi=\frac{1}{2}(2\pi)=\pi. Since there are 3 semicircles that make up the perimeter, the perimeter is 3π3\pi.

Question analysis

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

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Applying the circumference formula is a grade 7 standard. Here it is used for a semicircle and repeated three times, a short extension of routine classwork, with a dashed triangle inside the figure to look past.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 8, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

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

    The dashed triangle lies inside the garden, so its three 2-meter sides are not part of the perimeter; only the three curved arcs bound the garden. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    The circumference formula is not given, and each arc is half of a circle with a 2-meter diameter, so students must take half of π × 2. (Anderson & Krathwohl, 2001)

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

Traps

  • Wrong formula: Uses a related but wrong formula: perimeter for area, diameter for radius, area for volume, or a misremembered one.
  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Circumference of a circle
C=π×dC = \pi \times d
C is the circumference, the distance around a circle. d is the diameter, the straight line across the circle through its center. π is a number a little more than 3.14. The distance around any circle is π times as long as the distance across it.
Arc of a semicircle
arc=12×π×d\text{arc} = \frac{1}{2} \times \pi \times d
A semicircle is half of a circle. Its arc is the curved part. d is the diameter, the straight side. The arc is half of the full circle’s circumference, so take half of π times d.

  1. Sort out the figure

    The garden’s edge is made of three semicircles, or half circles. Each one sits on a side of the dashed triangle. Each side is a diameter, 2 meters long. The perimeter is the distance around the outside. Here that is only the three arcs, the curved parts. The dashed triangle is inside, so it does not count.

  2. Find a full circle’s circumference

    Picture the whole circle that one semicircle comes from. Its diameter is 2 meters. Use the circumference formula.

    C=π×2=2πC = \pi \times 2 = 2\pi

    The full circle would measure 2π2\pi meters around.

  3. Find the length of one arc

    Each semicircle’s arc is half of that circle. Use the arc formula.

    arc=12×π×2=π\text{arc} = \frac{1}{2} \times \pi \times 2 = \pi

    One arc is π\pi meters long.

    Watch out: Using the full circle’s 2π2\pi for each arc doubles the answer. That gives 6π6\pi, choice D.

  4. Add the three arcs

    The garden’s edge is three equal arcs.

    π+π+π=3π\pi + \pi + \pi = 3\pi

    The perimeter is 3π3\pi meters.

    Watch out: The dashed sides add to 2 + 2 + 2 = 6 meters, choice A. That is the triangle, not the curved edge.

    So the answer is C, 3π3\pi meters.

Check your answer

Estimate. 3π3\pi is about 3 × 3.14 = 9.42 meters. A curved path is longer than the straight line across it. So the edge should be longer than the triangle’s 6 meters, and 9.42 is.

More questions like this

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

Question 53 of 57

2019 Test B Math, question 53

Question

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

  • A1125\frac{1}{125}
  • B3125\frac{3}{125}
  • C115\frac{1}{15}
  • D35\frac{3}{5}
Show the answer

Correct answer: A

Each time Micha draws a card, the probability of drawing the number 5 is 15\frac{1}{5}. To find the probability of drawing the number 5 three times, multiply:

15×15×15=1125\frac{1}{5}\times\frac{1}{5}\times\frac{1}{5}=\frac{1}{125}

Question analysis

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

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the probability of a compound event is a grade 7 standard, and repeated draws with replacement are a standard case. Three equal factors make this a short, typical grade 7 task.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 3, 59 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Something you must already know minor Working it out

    The rule that the chance of independent events all happening is the product of their chances is assumed, as is that replacing the card keeps each draw at 1/5. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The bag has 5 cards, numbered 1 to 5. Micha draws a card three times. He puts it back each time, so every draw starts with all 5 cards. He wins only if he draws a 5 all three times. You need the probability, or chance, of that.

  2. Find the chance for one draw

    One card out of 5 is a 5. Use the probability formula.

    P(5)=15P(5) = \frac{1}{5}

    Each draw has a 1 in 5 chance of being a 5.

  3. Multiply for three draws

    He needs a 5 on the first draw, and the second, and the third. The card goes back each time, so the draws don’t change each other. Use the rule for events in a row.

    P(5 and 5 and 5)=15×15×15=15×5×5=1125P(\text{5 and 5 and 5}) = \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = \frac{1}{5 \times 5 \times 5} = \frac{1}{125}

  4. Give the answer

    Watch out: Adding the chances gives 35\frac{3}{5}, choice D. That would make winning likely, which makes no sense. Multiplying 5 × 3 instead of 5 × 5 × 5 gives 115\frac{1}{15}, choice C.

    So the answer is A, 1125\frac{1}{125}.

Check your answer

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

More questions like this

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

Question 54 of 57

2019 Test B Math, question 54

Question

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

  • A10,800
  • B12,000
  • C36,000
  • D120,000
Show the answer

Correct answer: D

10 m1 sec×3,600 sec1 h×1 ft0.3 m=\frac{10\text{ m}}{1\text{ sec}}\times\frac{3{,}600\text{ sec}}{1\text{ h}}\times\frac{1\text{ ft}}{0.3\text{ m}}=

36,000 ft0.3 h=120,000fthr\frac{36{,}000\text{ ft}}{0.3\text{ h}}=120{,}000\,\frac{\text{ft}}{\text{hr}}

Question analysis

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

NYS standards
  • 6.RP.3d converting measurement units
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students usually convert one unit with a given factor. This converts both parts of a rate, seconds to hours and meters to feet, with a decimal conversion factor and large numbers to keep track of.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 18 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Many steps moderate Working it out

    Both parts of the rate change: seconds become hours by multiplying by 3,600, and meters become feet by dividing by 0.3, two conversions that each have a wrong direction. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic moderate Working it out

    Dividing 36,000 by 0.3 means moving the decimal point in the divisor, and a slip of one place gives 12,000 instead of 120,000. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Something you must already know minor Working it out

    The stem gives 1 foot = 0.3 meter but not the 3,600 seconds in an hour, which students must build from 60 × 60. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The speed is 10 meters per second. That means it moves 10 meters every second. You need the speed in feet per hour. Two units change: seconds become hours, and meters become feet. Change one at a time. Use 1 foot = 0.3 meter.

  2. Find the seconds in an hour

    An hour is much longer than a second. So in one hour, the object moves much farther. First find how many seconds are in an hour.

    60×60=3,60060 \times 60 = 3{,}600

    One hour is 3,600 seconds.

  3. Find the meters per hour

    Each second it moves 10 meters. In 3,600 seconds it moves 3,600 times as far.

    10×3,600=36,00010 \times 3{,}600 = 36{,}000

    The speed is 36,000 meters per hour.

    Watch out: Stopping here gives 36,000, choice C. That number is still in meters, not feet.

  4. Change meters to feet

    Each foot is 0.3 meter. To find how many feet fit in 36,000 meters, divide by 0.3. Multiplying both numbers by 10 makes the division easier.

    36,000÷0.3=360,000÷3=120,00036{,}000 \div 0.3 = 360{,}000 \div 3 = 120{,}000

    So 36,000 meters is 120,000 feet. The speed is 120,000 feet per hour.

  5. Give the answer

    Watch out: Multiplying by 0.3 instead of dividing gives 10,800, choice A. Dividing by 3 instead of 0.3 gives 12,000, choice B.

    So the answer is D, 120,000 feet per hour.

Check your answer

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

More questions like this

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

Question 55 of 57

2019 Test B Math, question 55

Question

6.34−4.126.\overline{34}-4.\overline{12}

What is the value of the expression above, written as an improper fraction?

  • A2411\frac{24}{11}
  • B224101\frac{224}{101}
  • C11150\frac{111}{50}
  • D209\frac{20}{9}
Show the answer

Correct answer: D

6.34−4.12=2.22=2.26.\overline{34}-4.\overline{12}=2.\overline{22}=2.\overline{2}

x=2.2222...x=2.2222... Set x equal to the repeating decimal.
10x=22.2222...10x=22.2222... Multiply both sides by 10.
−x=−2.2222...-x=-2.2222... Subtract 10x and x.
9x=209x=20 Apply the multiplicative inverse property; divide both sides by 9.
x=209x=\frac{20}{9}

Question analysis

Two repeating decimals are subtracted, and students write the difference as an improper fraction. It tests subtracting repeating decimals and converting a repeating decimal into a fraction, a grade 8 skill. The repeating part has to be handled exactly, so a careful method matters more than speed.

NYS standards
  • 8.NS.1 decimal expansions and irrational numbers
  • 7.NS.2d fractions to decimals by long division
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students change fractions into repeating decimals by long division. Going the other way, from a repeating decimal to a fraction, is grade 8 work, though the subtraction that comes first here is simple.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 18 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The subtraction is easy, since the repeating digits line up and give 2.222…; the conversion is the test. Every choice sits within a few hundredths of the answer, so rounding or estimating cannot separate them, and the exact 2.22 of 111/50 is the strongest pull. Students who know that 0.222… is 2/9 can write 2 2/9 = 20/9 directly.

  1. A very close wrong answer moderate Choosing the answer

    All four choices lie between 2.18 and 2.23: 111/50 is exactly 2.22 and 224/101 is about 2.218, so only 20/9 equals the repeating 2.222… exactly. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Something you must already know moderate Working it out

    Turning 2.222… into a fraction, by setting x equal to it and subtracting x from 10x, is grade 8 work that the question does not explain. (Anderson & Krathwohl, 2001)

Strategy: Subtract the decimals digit by digit to see which digits repeat, then turn that repeating decimal into a fraction, or divide each choice's numerator by its denominator and compare.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Repeating decimal to fraction
10x−x=9x10x - x = 9x
x is the repeating decimal. Multiplying by 10 moves the decimal point one place to the right. The endless repeating digits still line up after the point. So when you subtract x from 10x, they cancel, and only whole numbers are left.

  1. Sort out the bars

    A bar over digits means they repeat forever. So 6.346.\overline{34} means 6.343434… and 4.124.\overline{12} means 4.121212… with no end. You need the difference as an improper fraction. That is a fraction whose top is bigger than its bottom.

  2. Subtract the decimals

    Line up the digits. Each block of 34 minus 12 gives 22. The whole numbers give 6 − 4 = 2.

    6.343434…−4.121212…=2.222222…6.343434\ldots - 4.121212\ldots = 2.222222\ldots

    The difference is 2.222…, with the 2 repeating forever. You can write it 2.22.\overline{2}.

  3. Name the decimal x

    Let x be the repeating decimal. Then multiply both sides by 10. This moves the decimal point one place to the right.

    x=2.222…x = 2.222\ldots

    10x=22.222…10x = 22.222\ldots

  4. Subtract to remove the repeats

    Subtract the first line from the second. The repeating 2s after the point match up, so they cancel.

    10x−x=22.222…−2.222…10x - x = 22.222\ldots - 2.222\ldots

    9x=209x = 20

  5. Divide to find x

    Divide both sides by 9.

    x=209x = \frac{20}{9}

    Watch out: 11150\frac{111}{50} is exactly 2.22, which stops. The real difference never stops. That trap is choice C. Every choice is close to 2.2, so estimating won’t help.

    So the answer is D, 209\frac{20}{9}.

Check your answer

Divide 20 by 9. 9 goes into 20 two times, with 2 left over. Each next step leaves 2 again, so you get 2.222… forever. That matches the difference. Choice A, 2411\frac{24}{11}, is 2.1818… instead.

More questions like this

The original DOE question
Question 112 as printed on page 207 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 207, question 112. Open page 207 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 56 of 57

2019 Test B Math, question 56

Question

A coordinate grid with the origin O, the x-axis marked 1, 2, and 3, and the y-axis marked 1 through 5. Line t passes through the point (0, 1) and line s passes through the point (1, 0); they intersect at P (3, 5), and dashed lines run from P across to 5 on the y-axis and down to 3 on the x-axis. The shaded region is the four-sided figure with vertices O, (0, 1), P (3, 5), and (1, 0).

Lines s and t intersect at point P (3, 5). What is the area, in square units, of the shaded region in the graph above?

  • A2
  • B3
  • C4
  • D7
Show the answer

Correct answer: C

First, find the area of the rectangle in the graph:

3×5=153\times 5=15

Then, find the areas of the two right triangles that are not shaded:

Upper triangle:

12×4×3=6\frac{1}{2}\times 4\times 3=6

Lower triangle:

12×2×5=5\frac{1}{2}\times 2\times 5=5

The area of the shaded region, in square units, is calculated by subtracting the total area of the two right triangles from the area of the rectangle:

15−(6+5)=15−11=415-(6+5)=15-11=4

Question analysis

Two lines meet at the point (3, 5) on a coordinate grid, and the shaded region between them and the axes is a four-sided shape. Students find its area in square units. The shape is not a standard figure, so no single area formula applies to it directly.

NYS standards
  • 6.G.1 area of polygons
  • 6.G.3 polygons in the coordinate plane
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Areas of triangles and rectangles are grade 6 content, and grade 7 students often find composite areas. This region is an irregular four-sided shape with two slanted sides, so students need a plan, and the lengths come from coordinates rather than labels.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 4, 23 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question rewards a plan more than computation: the region has no name, so students must see it inside the dashed rectangle from O to P, with two right triangles cut away. Reading each triangle’s legs from the axes is the second hurdle, since the 1-unit intercepts shorten one leg of each, to 4 instead of 5 and to 2 instead of 3. Only one wrong choice, from using one coordinate as both heights, has a clear working behind it, so most errors here are guesses after a failed plan.

  1. No obvious method moderate Working it out

    The shaded shape has four sides, two of them slanted, and no formula of its own; it must be boxed in the 3-by-5 rectangle with two triangles removed, or cut along OP. (Schoenfeld, 1985)

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

    No lengths are labeled; every one comes from the coordinates of P (3, 5), the y-intercept 1 of line s and the x-intercept 1 of line t. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Many steps minor Working it out

    The rectangle, two triangle areas and a subtraction make four computations, 15, 6 and 5, then 15 − 11, each built from different coordinates. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Draw the rectangle from O to P, find the areas of the two unshaded right triangles inside it from the coordinates, and subtract both from the rectangle's area.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle
A=l×wA = l \times w
A is the area, in square units. l is the length and w is the width. Multiplying them counts the unit squares that fill the rectangle.
Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, one side. h is the height: the distance from the base straight up to the far corner. A right triangle has a square corner, and the two sides that meet there are the base and height. A triangle is half of a rectangle with the same base and height.

  1. Sort out the shaded region

    The shaded region has four corners: O at (0, 0), (0, 1) on the y-axis, P at (3, 5), and (1, 0) on the x-axis. Line t meets the y-axis at (0, 1), and line s meets the x-axis at (1, 0). There is no easy formula for this shape. But the dashed lines make a rectangle around it. So find the rectangle’s area and take away the white parts.

  2. Find the rectangle’s area

    The rectangle has corners O, (3, 0), P and (0, 5). It is 3 units wide and 5 units tall.

    A=3×5=15A = 3 \times 5 = 15

    The rectangle’s area is 15 square units.

  3. Find the upper triangle’s area

    The white triangle at the top left has corners (0, 1), (0, 5) and P. Its square corner, or right angle, is at (0, 5). One side runs up the y-axis from 1 to 5, so it is 4 units. The other runs across from 0 to 3, so it is 3 units.

    A=12×4×3=6A = \frac{1}{2} \times 4 \times 3 = 6

    Its area is 6 square units.

  4. Find the lower triangle’s area

    The white triangle at the bottom right has corners (1, 0), (3, 0) and P. Its square corner is at (3, 0). One side runs along the x-axis from 1 to 3, so it is 2 units. The other goes up from 0 to 5, so it is 5 units.

    A=12×2×5=5A = \frac{1}{2} \times 2 \times 5 = 5

    Its area is 5 square units.

    Watch out: The lines cross the axes at 1, not at O. So these sides are 4 and 2, not 5 and 3.

  5. Subtract the white triangles

    Take both triangles away from the rectangle.

    15−(6+5)=15−11=415 - (6 + 5) = 15 - 11 = 4

    The shaded area is 4 square units.

    So the answer is C, 4.

Check your answer

Cut the shaded region along a line from O to P. One piece has a base of 1 on the x-axis and a height of 5, since P is 5 units up. Its area is 2.5. The other has a base of 1 on the y-axis and a height of 3, since P is 3 units across. Its area is 1.5. Then 2.5 + 1.5 = 4. Using 3 for both heights gives 3, choice B.

More questions like this

The original DOE question
Question 113 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 113. Open page 208 of the handbook (PDF)

Question 57 of 57

2019 Test B Math, question 57

Question

A box contains 14 red marbles, 10 black marbles, and 16 white marbles. How many more white marbles need to be added to the box so that the probability of randomly drawing a white marble is 35\frac{3}{5}?

  • A8
  • B10
  • C14
  • D20
Show the answer

Correct answer: D

Let w represent the number of additional white marbles.

Set the probability of drawing a white marble equal to 35\frac{3}{5} and solve the equation for w:

16+w14+10+16+w=35\frac{16+w}{14+10+16+w}=\frac{3}{5}

16+w40+w=35\frac{16+w}{40+w}=\frac{3}{5}

5(16+w)=3(40+w)5(16+w)=3(40+w)

80+5w=120+3w80+5w=120+3w

2w=402w=40

w=20w=20

Question analysis

A box holds 14 red, 10 black and 16 white marbles, and students find how many white marbles must be added so that the chance of drawing a white one is 3/5. Working backward from a target probability to a number of marbles is the unusual step.

NYS standards
  • 6.SP.8a uniform probability models
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 probability questions usually give the counts and ask for a probability. This runs the other way: the target probability is given, and the number to add changes both the white count and the total.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 8, 34 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The tempting shortcut is to take 3/5 of the 40 marbles already in the box, which forgets that the new white marbles join the total. Students who set up the probability with the unknown in both the top and the bottom solve it in a few lines, and testing choices works too: adding 20 gives 36 white of 60, exactly 3/5. The other two choices are counts copied from the stem.

  1. Turning words into math moderate Understanding the question

    “How many more white marbles need to be added” puts the unknown in two places, (16 + w)/(40 + w), because each white marble added also raises the total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Variables instead of numbers minor Understanding the question

    The unknown number of added marbles has to be carried as a letter through a proportion, 5(16 + w) = 3(40 + w), before it can be found. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  3. Many steps minor Working it out

    Totaling the marbles, setting up the proportion, cross-multiplying and solving 80 + 5w = 120 + 3w make four steps, unless the choices are tested one by one. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Let w be the number of white marbles added, write the new probability with w in both the top and the bottom, set it equal to 3/5 and solve; or test each choice.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P(white)=white marblesall marblesP(\text{white}) = \frac{\text{white marbles}}{\text{all marbles}}
P(white) is the probability, or chance, of drawing a white marble. The top counts the white marbles. The bottom counts every marble in the box. It works because each marble is equally likely to be drawn.
Cross multiplication
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
a and c are the tops. b and d are the bottoms. When two fractions are equal, each top times the other bottom gives the same product. It works because you multiply both sides by both bottoms. Use it to clear fractions from an equation.

  1. Sort out the box

    The box has 14 red, 10 black and 16 white marbles.

    14+10+16=4014 + 10 + 16 = 40

    That is 40 marbles. You will add some white marbles. Call that number w. You need w so that the chance of drawing white is 35\frac{3}{5}.

  2. Write the new probability

    Adding w white marbles changes both counts. There will be 16 + w white marbles and 40 + w marbles in all. Use the probability formula.

    16+w40+w=35\frac{16 + w}{40 + w} = \frac{3}{5}

    Watch out: Taking 35\frac{3}{5} of 40 gives 24, and 24 − 16 = 8, choice A. That forgets that the new marbles also make the box bigger.

  3. Cross multiply

    Fractions are hard to solve with. Clear them with cross multiplication.

    5×(16+w)=3×(40+w)5 \times (16 + w) = 3 \times (40 + w)

    Then multiply each outside number by both terms inside the parentheses.

    80+5w=120+3w80 + 5w = 120 + 3w

  4. Solve for w

    Subtract 3w from both sides. Then subtract 80 from both sides.

    2w=402w = 40

    Divide both sides by 2.

    w=20w = 20

    You need to add 20 white marbles.

    So the answer is D, 20.

Check your answer

Add 20 white marbles. Then there are 16 + 20 = 36 white marbles and 40 + 20 = 60 marbles in all. Divide the top and bottom by 12: 3660=35\frac{36}{60} = \frac{3}{5}. It works.

More questions like this

The original DOE question
Question 114 as printed on page 208 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 114. Open page 208 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).