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Question 1 of 100

2026 Practice Test A Math, question 50

Question

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

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

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

Show the answer

Correct answer: 7/5

Solution:

75\frac{7}{5}

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Error-prone arithmetic moderate Working it out

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

  2. Something you must already know minor Working it out

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

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

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

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

  2. Put in the numbers

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

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

    So the expression becomes:

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

  3. Divide the two fractions

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

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

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

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

  4. Add the last fraction

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

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

    The value is 7/5.

  5. Enter it as a fraction

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

    So you would enter 7/5.

Check your answer

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

More questions like this

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

Question 2 of 100

2026 Practice Test B Math, question 2

Question

The list shows the results of a survey. A survey was taken to see whether students like basketball, soccer, both, or neither. These were the results:

  • 25 students like basketball.
  • 19 students like soccer.
  • 7 students like both basketball and soccer.
  • 10 students don’t like either basketball or soccer.

Use the information provided to determine how many students were surveyed and how many like only soccer.

Move the correct answer to each box.

There were students surveyed, and students like only soccer.

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

Show the answer

Correct answer: blank 1: 47; blank 2: 12

There were 47 students surveyed, and 12 students like only soccer.

To determine how many students were surveyed, start with the total number of students who like basketball and soccer separately and subtract the number of students who were counted twice.

There were 25 students who like basketball, 19 students who like soccer, and 7 students who like both basketball and soccer.

So, the total number of students who like basketball or soccer is (25+19)−7=37(25 + 19) - 7 = 37.

Then, add the students who don’t like either basketball or soccer: 10 students don’t like either sport.

Total students surveyed: 37+10=4737 + 10 = 47

To determine how many students like only soccer, subtract the number of students who like both basketball and soccer from the total number of students who like soccer.

Students who like only soccer: 19−7=1219 - 7 = 12

Question analysis

This survey problem lists how many students like basketball, soccer, both sports or neither, and students drag number tiles into two boxes: how many were surveyed and how many like only soccer. It is a Venn diagram counting question. Keeping track of the students who like both sports, who appear in two of the listed counts, is the tricky part.

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 Overlapping groups like these are seldom taught on their own in grade 7, though the arithmetic is simple whole-number addition and subtraction. Two results come from the same four facts, and both boxes must be right to score.
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 6, 78 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light; the work is keeping track of one overlap that touches both answers. Students often remove the 7 for one box and not the other, and the pool holds 54 and 19, the tiles those half-finished methods produce. A sketch of two overlapping circles, with 7 in the middle, 18 and 12 beside it and 10 outside, turns both boxes into reading off the picture.

  1. Much to keep track of moderate Working it out

    The 7 students who like both sports sit inside the 25 and inside the 19, so they come out once for the total and again for only soccer, and both boxes must be right to score. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    The list never says that the 25 basketball fans already include the 7 who like both sports, and only the word “only” in the question separates the 12 from all 19 soccer fans. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Sketch two overlapping circles, write the students who like both sports in the middle, fill each outer part by subtraction, then add every region and the students who like neither.

Traps

  • Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting two groups that overlap
total=B+S−both+neither\text{total} = B + S - \text{both} + \text{neither}
B is the number who like basketball, and S is the number who like soccer. The students who like both are inside B and inside S, so B + S counts them twice. Subtract them once to fix that. Then add the students who like neither sport, since they are in neither group.
Only one group
only soccer=S−both\text{only soccer} = S - \text{both}
S is everyone who likes soccer, and both is the number who like both sports. Only soccer means the soccer fans who do not also like basketball. Take the both group out of the soccer group to find them.

  1. Sort out the question

    The survey gives you four facts:

    • 25 students like basketball.
    • 19 students like soccer.
    • 7 students like both sports.
    • 10 students like neither sport.

    You need two numbers. The first box is how many students were surveyed in all. The second box is how many like only soccer, which means soccer but not basketball. Each tile can be used once.

  2. See who is counted twice

    The 7 students who like both sports are part of the 25 basketball fans. They are also part of the 19 soccer fans. So if you add 25 and 19, those 7 students get counted two times.

    A Venn diagram shows this. It is a picture of two circles that overlap. The 7 go in the middle, where the circles share space.

  3. Count who likes at least one sport

    Use the overlap rule. Add the two groups, then take away the 7 who were counted twice:

    25+19−7=44−7=3725 + 19 - 7 = 44 - 7 = 37

    So 37 students like basketball, soccer or both.

  4. Add the students who like neither

    The 10 students who like neither sport are not in either group. They were still surveyed, so add them:

    37+10=4737 + 10 = 47

    So 47 students were surveyed.

    Watch out: adding 25 + 19 + 10 = 54 counts the 7 students twice. The tile 54 is waiting for that mistake.

  5. Find who likes only soccer

    The 19 soccer fans include the 7 who like basketball too. Take those 7 away:

    19−7=1219 - 7 = 12

    So 12 students like only soccer.

    Watch out: the word only matters. The tile 19 counts every soccer fan, not just the ones who like only soccer.

  6. Fill in both boxes

    On the test, you drag the tile 47 into the first box and the tile 12 into the second box. On this site, you choose each tile from its box’s menu.

    So the boxes get 47 and 12: there were 47 students surveyed, and 12 students like only soccer.

Check your answer

Split the students into four groups that do not overlap. Only basketball is 25−7=1825 - 7 = 18. Both sports is 7. Only soccer is 12. Neither is 10. Add the four groups: 18+7+12+10=4718 + 7 + 12 + 10 = 47. That matches the first box.

More questions like this

The original DOE question
Question 2 as shown on the DOE’s digital test, page 74 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B Math, page 74, question 2.

Question 3 of 100

2026 Practice Test B Math, question 28

Question

A circle has a radius of 5 inches. What is the area of the circle in terms of π\pi?

Use π\pi to express your answer.

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

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

Show the answer

Correct answer: 25π

Solution:

25π25\pi

The area of a circle is its radius squared times π\pi.

A=πr2A = \pi r^2

r=5r = 5

A=π(5)2A = \pi(5)^2

A=25πA = 25\pi

Question analysis

This circle question gives a radius of 5 inches and asks for the area in terms of π, typed into a box rather than chosen. Students recall and apply the formula for the area of a circle. The formula is not printed on the test, and the area and circumference formulas are easy to mix up.

NYS standards
  • 7.G.4 circle area and circumference
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Finding the area of a circle from its radius is the grade 7 circle standard applied directly, in one routine calculation. Typing the answer in terms of π, with no choices to compare, adds a little risk.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is one formula and one square, so the question comes down to recall and to the form of the answer. The area and circumference formulas both pair π with the radius, and 2πr gives a tidy value with π in it that looks just as right. The other losses are about form: replacing π with 3.14, which the instructions rule out twice, or dropping π after squaring 5.

  1. Something you must already know moderate Working it out

    The formula A = πr² is not printed with the question, and the circumference formula 2πr uses the same r and π, so recalling the wrong one gives a value that looks just as finished. (Anderson & Krathwohl, 2001)

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

    The answer is built on a keypad with a π key and must keep π as a symbol; a decimal worked out with 3.14 is not accepted, and no choices show the expected form. (Enright, Morley & Sheehan, 2002)

Strategy: Recall the area formula rather than the circumference formula, square the radius before multiplying, and leave π as a symbol instead of replacing it with 3.14.

Traps

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

Area of a circle
A=πr2A = \pi r^2
A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, a little more than 3. r² means r times r. Square the radius first, then multiply by π. The area is in square units.
Circumference of a circle
C=2πrC = 2 \pi r
C is the circumference, the distance around the circle. r is the radius, and π is the same number as above. This formula measures a length, not the space inside. It is here so you can tell the two formulas apart.

  1. Sort out the question

    The circle has a radius of 5 inches. The radius is the distance from the center to the edge. You need the area, the space inside the circle. In terms of π means π stays in the answer as a symbol. Do not change it to 3.14.

  2. Choose the area formula

    The area formula is A=πr2A = \pi r^2. It uses the radius squared.

    Watch out: the circumference formula, C=2πrC = 2\pi r, also uses π and the radius. It gives 2×π×5=10π2 \times \pi \times 5 = 10\pi, the distance around the circle in inches. That is not the area.

  3. Square the radius

    Squaring a number means multiplying it by itself:

    r2=52=5×5=25r^2 = 5^2 = 5 \times 5 = 25

    Watch out: squaring is not doubling. 5×2=105 \times 2 = 10 would lead to 10π again.

  4. Multiply by π

    Put 25 into the formula in place of r²:

    A=π×25=25πA = \pi \times 25 = 25\pi

    The area is 25π square inches. Write the number first and then π, the way you write 3x for 3 times x.

    Watch out: leave π as π. Using 3.14 gives 78.5, which is not accepted. Dropping π leaves 25, which is not the area.

  5. Enter it with the π key

    On the test, the keypad has a π key in its bottom left corner. Press 2, then 5, then the π key. You do not need a times sign. Enter only the answer, with no A = and no units. On this site, type 25π or 25pi.

    So you would enter 25π.

Check your answer

Estimate. π is a little more than 3, so 25π is a little more than 75 square inches. A square drawn around the circle is 10 inches on each side, so its area is 100 square inches. The circle fits inside that square, so its area must be less than 100. It is.

More questions like this

The original DOE question
Question 28 as shown on the DOE’s digital test, page 100 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test B Math, page 100, question 28.

Question 4 of 100

2026 Practice Test C Math, question 4

Question

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

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

Correct answer: C

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

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

275×36=9,900275 \times 36 = 9{,}900

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Much to keep track of minor Working it out

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

  2. Error-prone arithmetic minor Working it out

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

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

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Change 3 years into months

    Use the months formula:

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

    So the person makes 36 monthly payments.

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

  3. Add up the monthly payments

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

    275×4=1,100275 \times 4 = 1{,}100

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

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

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

  4. Add the down payment once

    Use the total-cost formula:

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

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

    So the answer is C, $11,100.

Check your answer

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

More questions like this

The original DOE question
Question 4 as shown on the DOE’s digital test, page 77 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 77, question 4.

Question 5 of 100

2026 Practice Test C Math, question 13

Question

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

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

What is the result?

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

Correct answer: D

Begin by multiplying 135.28 by 10:

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

Add 0.5:

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

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

Divide by 10:

1,353÷10=135.31{,}353 \div 10 = 135.3

Question analysis

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

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

Why it is hard

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

  1. A very close wrong answer moderate Choosing the answer

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

  2. Many steps minor Working it out

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

  3. Error-prone arithmetic minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Multiply by 10

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

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

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

  3. Add 0.5

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

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

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

  4. Drop the digits after the point

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

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

  5. Divide by 10

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

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

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

Check your answer

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

More questions like this

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

Question 6 of 100

2026 Practice Test C Math, question 15

Question

A grocery store sells 1-gallon containers of milk for $3.99. The store also sells orange juice in a 6-pack of 5.5-fluid-ounce bottles for $1.79. Suppose the store wants to sell its orange juice in gallon containers instead. To the nearest dollar, how much more would a gallon of orange juice cost than a gallon of milk?

Enter your answer in the space.

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

Show the answer

Correct answer: 3

To determine the cost of a gallon of orange juice, begin by finding the total number of ounces in the 6-pack of bottles:

5.5 oz×6=33 oz5.5\text{ oz} \times 6 = 33\text{ oz}

So, 33 oz of orange juice costs $1.79.

There are 128 oz in 1 gallon. Set up a proportion using the $1.79 cost per 33 oz of orange juice and the unknown cost (x) per 1 gallon of orange juice:

33 oz$1.79=128 ozx\frac{33\text{ oz}}{\$1.79} = \frac{128\text{ oz}}{x}

Cross multiply to solve for x in the proportion:

(33)×(x)=($1.79)×(128)(33) \times (x) = (\$1.79) \times (128)

33x=$229.1233x = \$229.12

x≈$6.94x \approx \$6.94

So, the orange juice costs $6.94 per gallon, and the milk costs $3.99 per gallon.

Subtract the cost per gallon of milk from the cost per gallon of orange juice.

$6.94−$3.99=$2.95\$6.94 - \$3.99 = \$2.95

Round the difference to the nearest dollar.

$2.95 rounds to $3.

Question analysis

A store sells milk by the gallon for $3.99 and orange juice in six-packs of 5.5-fluid-ounce bottles for $1.79. Students work out what a gallon of the juice would cost at the same price per ounce and how much more that is than the milk, to the nearest dollar. The ounces in a gallon must be known, and several steps involve decimals.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3d converting measurement units
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Unit prices are grade 6 and 7 work, but this chains five steps with decimals and needs a fact the question does not give, the number of fluid ounces in a gallon. Typing a rounded difference, with no choices, adds risk.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
5 steps
Reading level
Question and choices: about grade 5, 60 words.
Corrected from the DOE handbook
  • A rounded value in the explanation is now marked as approximate (≈ instead of =).
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The calculation is long, and its key fact is missing from the page. Students who know the 128 fluid ounces still face a division by 33 that does not come out evenly, which tempts them to round the price per ounce to 5 cents and lose about 50 cents a gallon. Keeping the exact quotient to the end, or scaling 33 ounces up by about 3.9, keeps the difference near $2.95, which rounds to 3.

  1. Something you must already know moderate Working it out

    The question never says that a gallon holds 128 fluid ounces, yet the whole comparison depends on it; the 6-pack holds only 6 × 5.5 = 33 ounces. (Anderson & Krathwohl, 2001)

  2. Many steps moderate Working it out

    Five moves follow one another: the ounces in a pack, the scale-up to 128 ounces, the division by 33, the subtraction of $3.99 and the rounding to a whole dollar. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    1.79 × 128 = 229.12 and 229.12 ÷ 33 are long decimal calculations without a calculator, and a slip in either moves the price of a gallon. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

    The rounded difference is typed with no choices, so an unrounded $2.95, or a figure rounded too soon, looks just as finished as the whole-dollar answer. (Enright, Morley & Sheehan, 2002)

Strategy: Find the ounces in one six-pack, scale its price up to the 128 fluid ounces in a gallon, subtract the milk's price, and round only at the end.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Rounds too early: Rounds a value partway through, or rounds the wrong way, so the final answer drifts from the exact one.
  • 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 DOE’s explanation goes from 33x = $229.12 to “x = $6.94”. But 229.12 ÷ 33 = 6.9430…, so $6.94 is the cost per gallon rounded to the cent. The site writes x ≈ $6.94. Nothing else changes: $6.94 − $3.99 = $2.95, which rounds to $3, and the unrounded cost gives the same answer.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fluid ounces in a gallon
fluid ounces=128×gallons\text{fluid ounces} = 128 \times \text{gallons}
A gallon holds 128 fluid ounces. This is a fact you must know, because the question does not give it. A fluid ounce measures an amount of liquid. Multiply gallons by 128 to change them into fluid ounces.
Cross-multiplying a proportion
ab=cd means a×d=b×c\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c
A proportion says two ratios are equal. Here each ratio is ounces over dollars. a and c are the tops, and b and d are the bottoms. Multiply each top by the other bottom, and the two products are equal. That turns the proportion into an equation.

  1. Sort out the question

    Milk costs $3.99 for 1 gallon. Orange juice comes in a 6-pack of 5.5-fluid-ounce bottles for $1.79. You need how much more a gallon of juice would cost than a gallon of milk. Round to the nearest dollar.

    The question doesn’t say how big a gallon is. You must know this fact: 1 gallon is 128 fluid ounces.

  2. Find the ounces in a 6-pack

    Each of the 6 bottles holds 5.5 fluid ounces:

    6×5.5=336 \times 5.5 = 33

    So $1.79 buys 33 fluid ounces of juice.

  3. Set up a proportion

    Let x be the cost in dollars of a gallon, 128 fluid ounces, of juice. The price per ounce stays the same. So ounces over dollars is the same for the pack and the gallon:

    331.79=128x\frac{33}{1.79} = \frac{128}{x}

    Cross-multiply:

    33×x=1.79×128=229.1233 \times x = 1.79 \times 128 = 229.12

  4. Find the cost of a gallon

    Divide both sides by 33:

    x=229.12÷33≈6.94x = 229.12 \div 33 \approx 6.94

    The sign ≈ means about equal. The exact value is 6.943…, so a gallon of juice costs about $6.94.

    Watch out: don’t round too early. One ounce costs about $0.054. Rounding that to $0.05 makes a gallon $6.40, and your answer becomes 2.

  5. Subtract the milk’s price

    Find how much more the juice costs:

    6.94−3.99=2.956.94 - 3.99 = 2.95

    A gallon of juice costs $2.95 more than a gallon of milk.

    Watch out: entering 7, the juice’s price, forgets to subtract the milk.

  6. Round and enter it

    The question asks for the nearest dollar. $2.95 is between $2 and $3. Its cents, 95, are 50 or more, so it rounds up to $3. Do not enter 2.95.

    Type only the number in the box. So you would enter 3.

Check your answer

Estimate. 128 ounces is almost 4 packs, since 4×33=1324 \times 33 = 132. Four packs cost 4×1.79=7.164 \times 1.79 = 7.16 dollars, so a gallon costs a bit less than that. Milk costs about $4. About $7 minus $4 is $3, which matches.

More questions like this

The original DOE question
Question 15 as shown on the DOE’s digital test, page 88 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 88, question 15. This is the DOE’s version, before the corrections listed in the analysis.

Question 7 of 100

2026 Practice Test C Math, question 20

Question

Two circles, with centers at A and C, are shown. Line segment AB is congruent to line segment DE.

Two circles. The larger circle has center A, and a radius is drawn from A to point B on the circle, labeled 5. The smaller circle has center C, and a segment is drawn from point D on the circle, through C, to point E on the circle.

Determine the area, A, and the circumference, C, of each circle to complete the table.

Select one answer in each row.

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

Circle ACircle CNeither
A=25πA = 25\pi
C=5πC = 5\pi
Show the answer

Correct answer: A = 25π: Circle A; C = 5π: Circle C

A=25πA = 25\pi: Circle A

C=5πC = 5\pi: Circle C

As shown in the illustration, AB (the radius of Circle A) is 5 units in length. Since DE (the diameter of Circle C) is congruent to AB, we know it is also 5 units in length.

Circle A:

A=πr2A = \pi r^{2}

A=25πA = 25\pi

r=5r = 5

We know the radius of Circle A is 5 so the answer to row 1 is A.

Circle C:

C=π⋅dC = \pi \cdot d

C=5πC = 5\pi

We have to look for the circle with a diameter of 5. That is circle C.

Question analysis

Two circles centered at A and C are shown: radius AB of Circle A is labeled 5, and segment DE, drawn through C, is congruent to AB. Students decide whether an area of 25π and a circumference of 5π each belong to Circle A, Circle C or neither. Keeping radius and diameter apart, and area and circumference, is the challenge.

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 area and circumference formulas is the grade 7 circle standard. This gives a radius for one circle and, for the other, a segment known only through congruence, and both rows of the table must be right.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
4 steps
Reading level
Question and choices: about grade 5, 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

Circle A is direct: its labeled radius of 5 gives an area of 25π. Circle C is where points are lost, since its size comes from a sentence rather than a label, and the segment through its center is a diameter, so its circumference is 5π while its area is only 6.25π. The row labels favor guessers, because the A and C in A = 25π and C = 5π match the circles' names, though they stand for area and circumference.

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

    Circle C's size is shown only by segment DE, drawn through center C with no label, and only the stem's sentence that DE is congruent to AB makes it a diameter of 5. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know moderate Working it out

    Neither A = πr² nor C = πd is given, and one circle is measured by a radius and the other by a diameter, so each formula needs the right length. (Anderson & Krathwohl, 2001)

  3. Much to keep track of minor Working it out

    Four values are in play, areas of 25π and 6.25π and circumferences of 10π and 5π, and both rows must be matched correctly for the item to count. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Write down the radius and the diameter of each circle, compute both circles' areas and circumferences in terms of π, then match each value in the table.

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".
  • 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

Area of a circle
A=πr2A = \pi r^{2}
A is the area, the space inside the circle. r is the radius, the distance from the center to the edge. π (pi) is a fixed number, a little more than 3. Square the radius, then multiply by π.
Circumference of a circle
C=πdC = \pi d
C is the circumference, the distance around the circle. d is the diameter, the distance across the circle through its center. A diameter is twice as long as a radius. Multiply the diameter by π.

  1. Sort out the question

    Circle A has its center at A. Its radius, AB, is 5. Circle C has its center at C. Segment DE goes through C, so it is a diameter. Congruent means the same length, so DE is 5 too.

    Careful: in the table, A means area and C means circumference. Circle A and Circle C are just names.

  2. Find each radius and diameter

    Circle A: the radius is 5, so the diameter is 2×5=102 \times 5 = 10.

    Circle C: the diameter is 5, so the radius is 5÷2=2.55 \div 2 = 2.5.

    Watch out: without the sentence about DE, Circle C seems to have no size. Then you might pick Neither for the second row.

  3. Check the area row

    Use the area formula for each circle. For Circle A:

    A=π×52=25πA = \pi \times 5^{2} = 25\pi

    For Circle C:

    A=π×2.52=6.25πA = \pi \times 2.5^{2} = 6.25\pi

    Only Circle A has an area of 25π.

    Watch out: squaring Circle C’s diameter instead of its radius gives 25π by mistake.

  4. Check the circumference row

    Use the circumference formula for each circle. For Circle A:

    C=π×10=10πC = \pi \times 10 = 10\pi

    For Circle C:

    C=π×5=5πC = \pi \times 5 = 5\pi

    Only Circle C has a circumference of 5π.

    Watch out: π times the radius is not the circumference. It would give Circle A 5π.

  5. Fill in the table

    Choose one button in each row. In the row for A = 25π, choose the Circle A column. In the row for C = 5π, choose the Circle C column. Neither stays empty.

    So the answers are A = 25π: Circle A and C = 5π: Circle C.

Check your answer

Compare sizes. Circle A’s radius, 5, is twice Circle C’s radius, 2.5. So the distance around Circle C should be half of Circle A’s 10π. Half of 10π is 5π, which matches. And the bigger circle has the bigger area, 25π.

More questions like this

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

Question 8 of 100

2026 Practice Test C Math, question 31

Question

NUMBER OF VISITORS
TO BROOKSIDE ZOO

Male Female
Children 360 340
Adults 210 290

The table above shows the number of people who visited Brookside Zoo in one day. The largest of the four groups made up what percentage of all visitors on that day?

  • A30%
  • B28%
  • C24%
  • D17%
Show the answer

Correct answer: A

First, determine the total number of visitors.

360+340+210+290=1200360 + 340 + 210 + 290 = 1200

The largest group is 360 (Male Children).

Find the percent that represents 3601200\frac{360}{1200}.

3601200×1001%=36012×11%=30%\frac{360}{1200} \times \frac{100}{1}\% = \frac{360}{12} \times \frac{1}{1}\% = 30\%

Question analysis

A two-way table counts one day's visitors to Brookside Zoo by age, children or adults, and by male or female. Students find what percent of all visitors the largest of the four groups made up. They must total every cell first, then make sure the count they divide is really the largest one.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding what percent one count is of a total is grade 6 work. A two-way table adds two quick steps, totaling the four groups and spotting the largest, but the numbers divide evenly, so it stays below grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 39 words.
Students, last 30 days
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 friendly, since 360 over 1,200 reduces to 3 over 10, so the work is in reading the table. Two of its cells are close, 340 just below 360, and every wrong choice is the share of one of the other groups. Adding all four counts first, and circling the largest before dividing, keeps the right cell in play.

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

    The two-way table gives four separate counts and no totals, so the 1,200 visitors must be added up from the cells and the largest group picked out of a grid. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Many steps minor Working it out

    Three short steps run in order: total the four cells, pick the largest, and divide; using the wrong cell in the last step still gives a believable percent. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add all four cells for the day's total, pick the biggest single cell, and divide it by the total; simplifying the fraction before dividing keeps the arithmetic short.

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

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 one group you care about. The whole is everyone. The fraction tells what share of everyone the group is. Multiplying by 100 turns that share into a percent, which means out of 100.

  1. Sort out the question

    The table splits one day’s visitors into four groups:

    MaleFemale
    Children360340
    Adults210290

    You need the largest group as a percent of all the visitors. The table gives no total, so you must find it.

  2. Add up all the visitors

    Add all four groups to get the whole:

    360+340+210+290=1,200360 + 340 + 210 + 290 = 1{,}200

    So 1,200 people visited the zoo that day.

  3. Find the largest group

    Compare the four numbers. The biggest is 360, the male children.

    Watch out: 340 is close to 360, but it is smaller. Using 340 gives about 28%, choice B. Using the 290 female adults gives about 24%, choice C. And 210 is the smallest group, which gives 17.5%, near choice D.

  4. Find the percent

    Put the part over the whole. Then simplify by dividing the top and bottom by 120:

    3601,200=310\frac{360}{1{,}200} = \frac{3}{10}

    Now multiply by 100 to get a percent:

    310×100=30\frac{3}{10} \times 100 = 30

    So the male children made up 30% of the visitors.

  5. Give the answer

    The largest group, 360 male children, is 30% of all 1,200 visitors.

    So the answer is A, 30%.

Check your answer

Work backward. 30% of 1,200 is 0.30×1,200=3600.30 \times 1{,}200 = 360, the male children. Also, with four groups an even split would be 25% each. The largest group should be a bit more than 25%, and 30% is.

More questions like this

The original DOE question
Question 31 as shown on the DOE’s digital test, page 104 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 104, question 31.

Question 9 of 100

2026 Practice Test C Math, question 34

Question

A teacher will randomly select 3 students to be on a school committee. He will choose either a student from Classroom A or a student from Classroom B with each selection. This tree diagram shows the selection possibilities.

A tree diagram with three columns of boxes. From a starting point, two arrows lead to boxes A and B. From each of those, two arrows lead to boxes A and B, four boxes in all. From each of those four, two arrows lead to boxes A and B, so the last column has eight boxes: A, B, A, B, A, B, A, B.

What is the probability that the teacher will select 2 students from Classroom A and 1 student from Classroom B (in any order) for the committee?

  • A18\frac{1}{8}
  • B38\frac{3}{8}
  • C12\frac{1}{2}
  • D34\frac{3}{4}
Show the answer

Correct answer: B

The tree diagram shows that each of the 3 randomly selected students is from Classroom A or Classroom B. So the first chosen is either A or B, represented by the left column of the diagram.

The second student is either A or B, so if the first is A, the first two chosen will be either AA or AB. If the first is B, then the first two chosen will be either BA or BB, represented by the middle column.

The third student is either A or B, so if the first 2 were AA, the group of 3 would be either AAA or AAB. These outcomes are represented by the top two boxes of the right column. This pattern is followed so that 8 distinct outcomes can be visualized in order:

AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB

Since the order does not matter, there are 3 groups made up of 2 students from Classroom A and 1 student from Classroom B: AAB, ABA, and BAA.

Out of 8 possible outcomes, 3 outcomes have 2 students from Classroom A and 1 student from Classroom B. Therefore, the probability is 3 out of 8, or 38\frac{3}{8}.

Question analysis

A teacher picks 3 students for a committee, each from Classroom A or Classroom B, and a tree diagram shows every possible sequence of picks. Students find the probability of getting 2 students from A and 1 from B in any order. Reading the tree as a list of outcomes, and counting each order, is the key.

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 Finding a compound probability from a tree diagram is a grade 7 standard, and here the tree is already drawn. Listing its paths and counting the ones that fit the event, in any order, is typical grade 7 classwork.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 7, 65 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The tree does the hard part of the setup, but it shows each outcome only as a path through three columns, not as a list. The parenthesis “in any order” is the key phrase: missing it leaves a single path, while reading the event loosely, as at least two from Classroom A, adds A, A, A. Writing out all eight paths and marking those with exactly two A’s settles the count.

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

    The tree prints only A or B in each box, so the eight three-pick outcomes, such as A, B, A, have to be read by tracing each path across all three columns. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Much to keep track of minor Working it out

    All eight paths must be tracked while counting those with exactly two A’s and one B, and “in any order” means three of them count, not just A, A, B. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Follow each path through the tree to write out every three-letter outcome, mark the ones with exactly two A's and one B, and divide by the total number of outcomes.

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

Probability
P=outcomes you wantall possible outcomesP = \frac{\text{outcomes you want}}{\text{all possible outcomes}}
P is the probability, the chance something happens. The top counts the outcomes that fit the event. The bottom counts every outcome that can happen. This works when every outcome is equally likely, like each path through this tree.

  1. Sort out the question

    The teacher picks 3 students, one at a time. Each pick is from Classroom A or Classroom B.

    You need the chance of getting 2 students from A and 1 from B. The words “in any order” matter. They mean the B can be the first, second or third pick.

  2. List every path in the tree

    Each path from left to right is one outcome, or possible result. The top path is A, then A, then A. Read all of them:

    AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB

    Each pick has 2 choices, so there are 2×2×2=82 \times 2 \times 2 = 8 outcomes in all.

  3. Mark the ones that fit

    Look for exactly two A’s and one B. Three outcomes fit: AAB, ABA and BAA.

    Watch out: counting only AAB misses “in any order” and gives 18\frac{1}{8}, choice A. AAA has three A’s, not two. Counting it too gives 48=12\frac{4}{8} = \frac{1}{2}, choice C. Counting every mixed outcome gives 68=34\frac{6}{8} = \frac{3}{4}, choice D, but three of those have two B’s.

  4. Find the probability

    Use the probability formula. 3 outcomes fit, out of 8:

    P=38P = \frac{3}{8}

    So the answer is B, 38\frac{3}{8}.

Check your answer

Count every outcome by type. Two A’s and one B: 3 outcomes. Two B’s and one A: 3 outcomes (ABB, BAB, BBA). All A’s: 1. All B’s: 1. 3+3+1+1=83 + 3 + 1 + 1 = 8, so each outcome was counted once.

More questions like this

The original DOE question
Question 34 as shown on the DOE’s digital test, page 107 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 107, question 34.

Question 10 of 100

2026 Practice Test C Math, question 39

Question

A shipment contains 170 small boxes of medical supplies. In a random sample of 20 of these boxes, 8 have damage and the others are undamaged. Based on this sample, what is the best prediction of the number of undamaged boxes in the shipment, not including the boxes in the random sample?

  • A60
  • B68
  • C90
  • D102
Show the answer

Correct answer: C

Determine the fraction of undamaged boxes in the sample, given 8 of the 20 boxes are damaged.

20−820=1220=35 or 0.6\frac{20-8}{20} = \frac{12}{20} = \frac{3}{5}\text{ or }0.6

Subtract the number of boxes in the sample from 170.

170−20=150170 - 20 = 150

Basing the prediction of undamaged boxes on the sample, multiply the number of boxes not in the sample by the fraction of undamaged boxes in the sample.

150×0.6=90150 \times 0.6 = 90

Question analysis

A shipment holds 170 boxes of medical supplies, and in a random sample of 20, 8 are damaged. Students predict how many undamaged boxes are in the shipment, leaving out the boxes in the sample. It tests using a sample to make a prediction, and the question's two conditions each lead to a different tempting wrong choice.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.SP.1b samples and populations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Predicting from a random sample is grade 7 proportional reasoning. This version adds two conditions, undamaged rather than damaged and the shipment without the sample, so it takes three steps and a close reading.
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 8, 49 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The proportion itself is routine; the question is built around two conditions that each change the numbers. The stem reports the damaged boxes, but the question asks about undamaged ones, and its bold not removes the 20 sampled boxes from the shipment. Each wrong choice matches one way of mishandling the two conditions, so checking the result against both words, undamaged and not, is what the question rewards.

  1. Much to keep track of moderate Working it out

    Two conditions point away from the obvious numbers: the question wants undamaged boxes while the stem counts the 8 damaged, and it leaves out the 20 sampled, so 150 boxes, not 170. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Many steps minor Working it out

    Three steps follow in order, the undamaged share 12/20, the 150 boxes outside the sample, and their product, and each wrong turn lands on a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the fraction of the sample that is undamaged, count the boxes that were not sampled, and multiply; then check that the answer describes the group of boxes asked about.

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

Predicting from a sample
prediction=part of the samplesample size×group size\text{prediction} = \frac{\text{part of the sample}}{\text{sample size}} \times \text{group size}
A sample is a small group chosen at random from a bigger group. The fraction is the share of the sample with some feature. A random sample is likely to look like the bigger group. So use the same share of the group you are predicting for.

  1. Sort out the question

    The shipment has 170 boxes. A random sample of 20 boxes was checked, and 8 of them were damaged.

    You need to predict the undamaged boxes in the shipment, not counting the 20 in the sample. Two words matter: undamaged and not.

  2. Find the undamaged share

    The sample has 20−8=1220 - 8 = 12 undamaged boxes. Write that as a share of the sample, and simplify:

    1220=35=0.6\frac{12}{20} = \frac{3}{5} = 0.6

    So 0.6, or 60%, of the sample is undamaged.

    Watch out: 820=0.4\frac{8}{20} = 0.4 is the damaged share. Using it leads to choice A or B.

  3. Count the boxes outside the sample

    The question leaves out the 20 boxes in the sample. Take them away from the shipment:

    170−20=150170 - 20 = 150

    So the prediction is about 150 boxes.

  4. Make the prediction

    Use the same undamaged share for the 150 boxes:

    0.6×150=900.6 \times 150 = 90

    About 90 of those boxes should be undamaged.

    Watch out: 0.6×170=1020.6 \times 170 = 102, choice D, forgets the bold not. It counts the sample boxes again.

  5. Give the answer

    The best prediction is 90 undamaged boxes outside the sample.

    So the answer is C, 90.

Check your answer

Predict the damaged boxes outside the sample too: 0.4×150=600.4 \times 150 = 60. Then 90+60=15090 + 60 = 150, every box outside the sample. Also, 12 out of 20 is a bit more than half. And 90 is a bit more than half of 150.

More questions like this

The original DOE question
Question 39 as shown on the DOE’s digital test, page 112 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 112, question 39.

Question 11 of 100

2026 Practice Test C Math, question 41

Question

Elijah is at the school bookstore. He wants to buy a water bottle for $12.50 and some notebooks that cost $4.75 each. He has at most $35.00 to spend. Write an inequality that represents the number of notebooks, n, Elijah can purchase.

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

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

Show the answer

Correct answer: 12.50 + 4.75n ≤ 35.00

The question uses the phrase “at most,” which means “less than or equal to” (≤).

Since notebooks cost $4.75 each and Elijah will buy n notebooks, the total cost just for the notebooks is found by multiplying $4.75 by n, or 4.75n4.75n.

Combining the cost of a water bottle ($12.50) with the cost of n notebooks, the inequality is 12.50+4.75n≤35.0012.50 + 4.75n \leq 35.00.

This may also be written as 4.75n+12.50≤35.004.75n + 12.50 \leq 35.00.

Question analysis

Elijah has at most $35.00 to buy one water bottle and some notebooks at the school bookstore, with a price given for each. Students type an inequality for the number of notebooks, n, he can buy. It tests writing an inequality from a word problem. The two prices play different roles, and the phrase "at most" decides the inequality symbol.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing an inequality for a budget with a one-time cost and a cost per item is typical grade 7 work. The answer is typed, so both the symbol for "at most" and the role of each price must be right.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 50 words.
Students, last 30 days
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 prices are given in the order they are used, so the expression for the total is the easier half. The points are lost on the symbol: “at most” has to be heard as no more than, with equality allowed, and the keypad offers all four inequality signs side by side. Pricing one and two notebooks with the bottle, $17.25 and $22.00, shows which amount repeats and confirms the total.

  1. Turning words into math moderate Understanding the question

    The $12.50 bottle is bought once and the $4.75 is paid for each of n notebooks, and “at most $35.00” has to become ≤, which allows a total of exactly $35.00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The whole inequality, with n, its symbol and both prices, is built on a keypad offering ≤, <, ≥ and >, so a wrong symbol looks as complete as the right one. (Enright, Morley & Sheehan, 2002)

Strategy: Price a few cases by hand, such as 1 and 2 notebooks with the water bottle, to see which amount repeats, then decide whether a total of exactly $35.00 is allowed.

Traps

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

Total cost with a one-time cost
total=one-time cost+(price each)×n\text{total} = \text{one-time cost} + (\text{price each}) \times n
The one-time cost is paid once, no matter what. The price each is paid once for every item. n is the number of items. You multiply the price each by n because you pay it n times.
At most
total≤limit\text{total} \leq \text{limit}
The sign ≤ means less than or equal to. The total can be under the limit or exactly equal to it, but never over it. Use it for words like at most and no more than.

  1. Sort out the question

    Elijah buys one water bottle for $12.50. He also buys n notebooks at $4.75 each. He can spend at most $35.00.

    You need an inequality for n. An inequality compares two amounts with a sign such as ≤ instead of =.

  2. Price a few cases

    Work out the cost for a few numbers of notebooks:

    NotebooksTotal ($)
    112.50+4.75=17.2512.50 + 4.75 = 17.25
    212.50+9.50=22.0012.50 + 9.50 = 22.00
    312.50+14.25=26.7512.50 + 14.25 = 26.75

    The $12.50 shows up once. The $4.75 shows up once for each notebook.

  3. Write the total cost

    Use the total cost formula. The bottle is the one-time cost, and each notebook costs $4.75:

    12.50+4.75n12.50 + 4.75n

    Here 4.75n means 4.75 times n.

    Watch out: do not add the two prices first. 17.25n17.25n charges for a new bottle with every notebook.

  4. Choose the sign

    At most $35.00 means the total can be less than $35.00 or exactly $35.00. It cannot be more. So use ≤:

    12.50+4.75n≤35.0012.50 + 4.75n \leq 35.00

    Watch out: ≥ means at least, the opposite. And < leaves out a total of exactly $35.00, which Elijah is allowed to spend.

  5. Enter it on the keypad

    Type the whole inequality. Nothing is printed before the box. On the test, press 1, 2, ., 5, 0, +, 4, ., 7, 5, the n key, the ≤ key, then 3, 5, ., 0, 0. The ≤ key is in the fourth row. Skip the dollar signs. On this site, type <= for ≤.

    So you would enter 12.50 + 4.75n ≤ 35.00. An equal form, such as 4.75n + 12.50 ≤ 35.00, counts too.

Check your answer

Test some numbers. For 4 notebooks, 12.50+4.75×4=31.5012.50 + 4.75 \times 4 = 31.50, which is not over 35.00. For 5 notebooks, 12.50+4.75×5=36.2512.50 + 4.75 \times 5 = 36.25, which is too much. So Elijah can buy up to 4 notebooks. That makes sense for a budget of $35.00.

More questions like this

The original DOE question
Question 41 as shown on the DOE’s digital test, page 114 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 114, question 41.

Question 12 of 100

2026 Practice Test D Math, question 7

Question

A middle school library has a shelf with 40 books, each labeled with a different number from 201 to 240. One book is selected at random. What is the probability, as a percentage, of selecting book number 223?

Enter your answer in the space provided.

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

Show the answer

Correct answer: 2.5%

The probability that book number 223 will be selected at random from 40 books is 1 out of 40. To find the probability as a percentage, convert the fraction to a decimal.

The chance of selecting one specific book is 140=0.025\frac{1}{40} = 0.025.

Convert the decimal to a percentage: 0.025×100=2.5%0.025 \times 100 = 2.5\%

Question analysis

A library shelf holds 40 books numbered 201 to 240, and one is picked at random. Students find the probability of picking one particular book and type it as a percentage. It tests simple probability and converting a fraction to a percent; the book numbers play no part, and the decimal point needs care.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the probability of one equally likely outcome is grade 6 content. The only further step is rewriting the fraction as a percent, which needs care with place value, so it is lighter than grade 7 compound probability.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 6, 41 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The book numbers, 201 to 240, and the chosen number 223 play no part, since every book is equally likely and the shelf holds 40. The chance is 1 in 40, and the real demand is the form: changing it to a percent means a decimal with a zero in the tenths place, then a shift of two places. Students who know that 1/4 is 25% can check that 1/40 must be a tenth of that.

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

    Only the bold word percentage sets the form, since the empty box shows no % sign, so the probability 1/40 must be turned into 2.5, not entered as the fraction or as 0.025. (Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Dividing 1 by 40 gives 0.025, with a zero right after the decimal point, and losing that zero turns the percent into 25. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the books, write the chance of picking one particular book as a fraction, then turn the fraction into a decimal and multiply by 100 to get the percent.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome
P=outcomes you wantall possible outcomesP = \frac{\text{outcomes you want}}{\text{all possible outcomes}}
P is the probability, the chance of an event. The top counts the outcomes you want. The bottom counts all the equally likely outcomes. Use it when each outcome has the same chance, like picking one book at random.
Changing a fraction to a percent
percent=ab×100\text{percent} = \frac{a}{b} \times 100
a/b is the fraction. Percent means out of 100. Multiplying by 100 tells you how many out of 100 the fraction is. For example, 1/2 × 100 = 50, so 1/2 is 50%.

  1. Sort out the question

    The shelf has 40 books, numbered 201 to 240. One book is picked at random, so every book has the same chance. You need the chance of picking book 223, written as a percentage. Book 223 is on the shelf, since 223 is between 201 and 240.

  2. Write the probability as a fraction

    Only 1 book is number 223, and there are 40 books in all:

    P=140P = \frac{1}{40}

    The numbers on the labels do not matter beyond this. They just give each book its own name.

  3. Change the fraction to a decimal

    Divide 1 by 40. It helps to start from 1/4 = 0.25. Since 40 is ten times 4, 1/40 is one tenth of 1/4:

    1÷40=0.0251 \div 40 = 0.025

    Watch out: there is a 0 right after the decimal point. Losing it gives 0.25 and then 25, which is wrong.

  4. Change the decimal to a percent

    Multiply by 100 to get a percent. That moves the decimal point two places to the right:

    0.025×100=2.50.025 \times 100 = 2.5

    So the chance is 2.5%.

    Watch out: 0.025 is the probability as a decimal, and 1/40 is the same chance as a fraction. The question asks for a percentage, so neither one is the answer here.

  5. Enter the number

    The box shows no % sign. Type 2.5. Adding the sign, as 2.5%, also counts.

    So you would enter 2.5.

Check your answer

2.5% means 2.5 out of every 100. Out of 40 books, that is 40 × 0.025 = 1 book, the one numbered 223. Also, one book out of 40 should be a small chance, and 2.5% is small.

More questions like this

The original DOE question
Question 7 as shown on the DOE’s digital test, page 80 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 80, question 7.

Question 13 of 100

2026 Practice Test D Math, question 11

Question

In a scale drawing of a triangular park, the sides measure 5 inches, 7 inches, and 8 inches. The perimeter of the actual park is 25 miles. What is the actual length of the shortest side of the park?

  • A6.25 mi
  • B8.75 mi
  • C10 mi
  • D15.625 mi
Show the answer

Correct answer: A

To find the perimeter of a triangle, add the measures of the three sides. The perimeter of the scale drawing is 5+7+8=205 + 7 + 8 = 20 inches.

Now, use the perimeters to find the scale factor for the scale drawing to the actual perimeter.

The scale factor is 20 inches to 25 miles, which reduces to a scale factor of 4 inches to 5 miles. Note that the scale for the perimeter is the same as the scale for the side lengths.

Since the smallest side on the scale drawing is 5 inches, use the scale factor to find the actual side length of the smallest side:

5 inches(5 miles4 inches)=6.25 miles5\text{ inches}\left(\frac{5\text{ miles}}{4\text{ inches}}\right) = 6.25\text{ miles}

Question analysis

A scale drawing of a triangular park has sides of 5, 7 and 8 inches, and the real park's perimeter is 25 miles. Students find the actual length of the park's shortest side. It tests scale drawings and proportional reasoning; the scale is not given directly, so it must be worked out from the two perimeters first.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Scale drawings are a grade 7 topic, and this is a standard scale problem. The one twist is that the scale is not stated: it must come from comparing the two perimeters before any side can be converted.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 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

A scale drawing multiplies every length by one factor, and the perimeter shares that factor, so the work is short once the two perimeters are paired. The common loss is reasoning with differences: the perimeter grows by 5, so a side seems to grow by 5 as well. Converting to miles per inch first, then checking that the three real sides add back to 25 miles, guards against that error and against scaling the wrong side.

  1. Turning words into math minor Understanding the question

    The 25 miles is the park's perimeter, not a side, so it must be matched with the drawing's perimeter, 5 + 7 + 8 = 20 inches, before any side can be scaled. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three operations in a row, adding the sides, dividing 25 by 20 for 1.25 miles per inch, and multiplying the 5-inch side, each produce a number the next step needs. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Add the three drawn sides to get the drawing's perimeter, divide the real perimeter by it to find the miles per inch, then multiply the shortest drawn side by that scale.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a triangle
P=a+b+cP = a + b + c
P is the perimeter, the distance all the way around. a, b and c are the lengths of the three sides. You add them because going once around the triangle covers each side once.
Using a scale
actual length=drawing length×miles per inch\text{actual length} = \text{drawing length} \times \text{miles per inch}
The drawing length is measured on the drawing, in inches. Miles per inch is how many real miles each inch stands for. In a scale drawing, every length is multiplied by the same number, so the perimeter and each side use the same scale.

  1. Sort out the question

    A scale drawing is a smaller copy, with every length shrunk by the same amount. The drawing’s sides are 5, 7 and 8 inches. The real park’s perimeter, the distance all the way around, is 25 miles. You need the real length of the shortest side.

  2. Find the drawing’s perimeter

    Add the three sides of the drawing:

    P=5+7+8=20P = 5 + 7 + 8 = 20

    The drawing’s perimeter is 20 inches. It stands for the real perimeter of 25 miles.

    Watch out: 25 miles is the whole way around, not one side. Matching it with the 8-inch side gives 15.625 miles, choice D.

  3. Find the scale

    20 inches stand for 25 miles. Divide to find how many miles 1 inch stands for:

    25÷20=1.2525 \div 20 = 1.25

    So each inch on the drawing is 1.25 miles in the park.

  4. Scale the shortest side

    The shortest side on the drawing is 5 inches. Multiply it by the scale:

    5×1.25=6.255 \times 1.25 = 6.25

    The real shortest side is 6.25 miles.

    Watch out: the perimeter grew by 5, from 20 to 25, but you cannot just add 5 to a side. That gives 10 miles, choice C. Scaling multiplies. Also use the 5-inch side, not the 7-inch side, which gives 8.75 miles, choice B.

    So the answer is A, 6.25 mi.

Check your answer

Scale all three sides: 5 × 1.25 = 6.25, 7 × 1.25 = 8.75 and 8 × 1.25 = 10. Add them: 6.25 + 8.75 + 10 = 25 miles. That is the park’s perimeter, so the scale is right.

More questions like this

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

Question 14 of 100

2026 Practice Test D Math, question 14

Question

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

  • A35
  • B42
  • C45
  • D56
Show the answer

Correct answer: B

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Many steps moderate Working it out

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

  2. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find the morning pages

    Take 1/4 of 84:

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

    He reads 21 pages in the morning. That leaves:

    84−21=6384 - 21 = 63

    63 pages are left after the morning.

  3. Find the afternoon pages

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

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

    He reads 21 pages in the afternoon.

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

  4. Find the pages still to read

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

    63−21=4263 - 21 = 42

    He still needs to read 42 pages.

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

    So the answer is B, 42.

Check your answer

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

More questions like this

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

Question 15 of 100

2026 Practice Test D Math, question 26

Question

Jordan has a budget of $120 to rent a food truck. The rental company charges $18 per hour to rent the truck, plus a one-time cleaning fee of $30.

Write an inequality that represents this situation, in which x is the number of hours Jordan can rent the truck.

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

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

Show the answer

Correct answer: 30 + 18x ≤ 120

Let x represent the number of hours Jordan can rent the truck.

The hourly rental cost can be represented as $18⋅x\$18 \cdot x, written as 18x18x.

The one-time cleaning fee of $30 is added to the hourly cost to obtain the total cost.

The sum of the cleaning fee and the hourly rental cost is represented as 30+18x30 + 18x.

As Jordan has $120 to spend, the total cost must be less than or equal to $120.

This is represented as 30+18x≤12030 + 18x \leq 120 or 120≥30+18x120 \geq 30 + 18x.

Question analysis

Jordan has $120 to rent a food truck that costs $18 an hour plus a one-time $30 cleaning fee, and students type an inequality for x, the number of hours. It tests writing a two-step inequality from a word problem. Matching each dollar amount to its role and choosing the symbol that fits a spending limit are the tricky parts.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing a two-step inequality from a word problem is the grade 7 standard itself. Nothing is offered to choose from: students type the whole inequality, so the numbers, the variable term and the symbol are all theirs to get right.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 57 words.
Corrected from the DOE handbook
  • A spelling error in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is small, so the points ride on two decisions the keypad leaves open: which symbol, and whether to stop before solving. A budget means Jordan may spend all $120 but no more, so the total cost must stay at or under it, and a strict < leaves out the five-hour rental that costs exactly $120. Testing x = 5 and x = 6 in the typed inequality confirms both the symbol and the setup.

  1. Turning words into math moderate Understanding the question

    The stem never says “at most”: the limit comes only from “a budget of $120”, while “$18 per hour” must multiply x and the “one-time cleaning fee of $30” is added just once. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The whole inequality is typed on a keypad that offers <, ≤, > and ≥ side by side, with no choices to compare, and the solved form x ≤ 5 is not accepted. (Enright, Morley & Sheehan, 2002)

Strategy: Work out what 1 hour and 2 hours would cost Jordan in all, see which amount repeats each hour, build the total for x hours, then compare it with the budget and test a few values.

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation says “The onetime cleaning fee of $30 is added to the hourly cost …”. “Onetime” means “former”; a fee charged once is a “one-time” fee, as the question itself spells it. The site now says “one-time”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a fee and a rate
total cost=fee+rate×x\text{total cost} = \text{fee} + \text{rate} \times x
The fee is a charge paid only once, so you add it once. The rate is the charge for each hour. x is the number of hours. You multiply the rate by x because it is paid again for every hour.
Staying within a budget
total cost≤budget\text{total cost} \leq \text{budget}
The sign ≤ means is less than or equal to. A budget is the most money you can spend. You may spend all of it, or less, but never more. So the total cost must be less than or equal to the budget.

  1. Sort out the question

    You know three facts:

    • Jordan can spend at most $120. That is the budget.
    • The truck costs $18 for each hour.
    • A cleaning fee of $30 is paid one time only.

    x is the number of hours. You need an inequality. That is a math sentence that compares two amounts with a sign such as < or ≤. It must show all three facts.

  2. Work out a few totals

    Find the total cost for a few hours. Each hour adds one more $18:

    HoursHourly charges ($)Total cost ($)
    118×1=1818 \times 1 = 1830+18=4830 + 18 = 48
    218×2=3618 \times 2 = 3630+36=6630 + 36 = 66
    x18x18x30+18x30 + 18x

    The $30 shows up once in every total. The $18 shows up once for each hour. So x hours cost 30 + 18x dollars. Here 18x means 18 times x.

  3. Choose the right sign

    The total cost must fit inside the budget. Jordan may spend exactly $120, or less, but not more. That is less than or equal to, the sign ≤. So:

    30+18x≤12030 + 18x \leq 120

    Watch out: the sign < means less than, but not equal. 30+18x<12030 + 18x < 120 leaves out spending exactly $120. The sign ≥ is backward too. It says Jordan must spend at least $120. Also keep each number in its place. 30x+1830x + 18 charges $30 an hour.

  4. Stop before you solve

    You could solve the inequality. Subtract 30 from both sides to get 18x≤9018x \leq 90. Then divide both sides by 18 to get x≤5x \leq 5. So Jordan can rent the truck for at most 5 hours.

    Watch out: the question asks for an inequality that models the situation. To model means to show the facts of the story in math. The solved form, x ≤ 5, is true, but it hides the $18, the $30 and the $120. So x ≤ 5 is not accepted. Enter the inequality before you solve it.

  5. Enter it on the keypad

    On the test, build the inequality on the keypad. Press 3, 0, +, 1, 8 and the x key. Then press the ≤ key, in the fourth row of keys. Then press 1, 2, 0. You need no dollar signs and no times sign. On this site, type it, with <= for ≤: 30 + 18x <= 120.

    So you would enter 30 + 18x ≤ 120. The same inequality written the other way round, 120 ≥ 30 + 18x, is also right. The solved form, x ≤ 5, is not accepted, because it does not model the situation.

Check your answer

Test two numbers of hours. For 5 hours, 30+18×5=12030 + 18 \times 5 = 120. Is 120 ≤ 120? Yes, so 5 hours fit the budget. For 6 hours, 30+18×6=13830 + 18 \times 6 = 138. Is 138 ≤ 120? No, and $138 is over budget. The inequality gets both right.

More questions like this

The original DOE question
Question 26 as shown on the DOE’s digital test, page 99 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 99, question 26. This is the DOE’s version, before the corrections listed in the analysis.

Question 16 of 100

2026 Practice Test D Math, question 28

Question

A graph shows the relationship between the number of gallons of water, y, that has been added to a tank and the number of hours, x, that water has been added to the tank at a constant rate. What does the ordered pair (3, 24) represent?

  • AThree gallons of water was added in 24 hours.
  • BThree gallons of water was added per hour.
  • CTwenty-four gallons of water was added in 3 hours.
  • DTwenty-four gallons of water was added per hour.
Show the answer

Correct answer: C

It is given that x-values on the graph represent the numbers of hours for which water was added to the tank, and the y-values represent the numbers of gallons of water added in x hours. In the ordered pair (3, 24), 3 represents an x-value and 24 represents a y-value.

Therefore, (3, 24) represents a data point for 24 gallons of water being added to the tank in a 3-hour period.

Question analysis

A graph relates the gallons of water added to a tank, y, to the hours it has been filling at a constant rate, x. Students choose what the ordered pair (3, 24) means in the situation. It tests interpreting coordinates in context, and the four statements mix totals with rates and switch the two numbers around.

NYS standards
  • 7.RP.2d points on proportional graphs
  • 6.EE.9 dependent and independent variables
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Explaining what a point on the graph of a proportional relationship means is named in the grade 7 standards. This is a direct case with no graph to read, though the stem introduces the two quantities in an order that needs care.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
1 step
Reading level
Question and choices: about grade 5, 76 words.
Students, last 30 days
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 decides the question: an ordered pair lists x first. The choices then test two confusions, which number is the gallons and whether a point is a rate. Reading the pair aloud as “after 3 hours, 24 gallons”, and noting that 24 ÷ 3 gives a rate of 8 gallons an hour that no choice states, rules out all three wrong statements.

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

    No graph is drawn, and the stem names y, the gallons, before x, the hours, so the pair (3, 24) has to be matched to its letters, not to the order the quantities were mentioned. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. A very close wrong answer minor Choosing the answer

    The four statements reuse 3 and 24 in two frames, a total added “in” some hours and an amount “per hour”, so the right statement and its swapped twin differ only in which number counts gallons. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Recall that an ordered pair lists x first and y second, write down what x and y measure here, and read the pair as one sentence about hours and gallons.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

An ordered pair
(x,y)(x, y)
An ordered pair names one point on a graph. The first number is always the x-value. The second number is always the y-value. Here x counts hours and y counts gallons. So a pair tells how many gallons were added after that many hours.
Rate from a point
rate=yx\text{rate} = \frac{y}{x}
When water is added at a constant rate, divide the y-value by the x-value. The result is the rate: gallons added in 1 hour. A point itself shows a total, not a rate.

  1. Sort out the question

    The graph links two amounts:

    • x is the number of hours water has been added.
    • y is the number of gallons added.

    You need to say what the point (3, 24) means. No graph is drawn, so you work from the words.

  2. Match each number to its letter

    In an ordered pair, x always comes first and y second. So in (3, 24):

    • x = 3, which means 3 hours.
    • y = 24, which means 24 gallons.

    Watch out: the question names gallons first and hours second. Do not match the numbers to the words in that order. That mistake reads 3 as gallons and 24 as hours, choice A.

  3. Read the pair as a sentence

    Put the two meanings together: after 3 hours, 24 gallons have been added in all. The 24 is a total for the whole 3 hours.

    Watch out: 24 is not the amount for each hour. The rate per hour is 24÷3=824 \div 3 = 8 gallons. So choice D, 24 gallons per hour, is wrong. Choice B, 3 gallons per hour, is also not the rate.

    So the answer is C, twenty-four gallons of water was added in 3 hours.

Check your answer

Use the rate of 8 gallons each hour. In 3 hours, 3×8=243 \times 8 = 24 gallons go in. That matches choice C. No choice says 8 gallons per hour, so both “per hour” choices must be wrong.

More questions like this

The original DOE question
Question 28 as shown on the DOE’s digital test, page 101 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 101, question 28.

Question 17 of 100

2026 Practice Test D Math, question 29

Question

Karl has one red spinner and one blue spinner. Each spinner is divided into 4 equal sections, numbered 1 through 4. He spins each spinner once and writes down the number that each lands on. What is the probability that the two numbers, when multiplied together, will have 4 as a product?

  • A116\frac{1}{16}
  • B316\frac{3}{16}
  • C47\frac{4}{7}
  • D716\frac{7}{16}
Show the answer

Correct answer: B

There are four possible outcomes for each spin of the spinner:

1, 2, 3, 4

The possible outcomes of two spins are:

11121314212223243132333441424344\begin{array}{cccc} 11 & 12 & 13 & \boxed{14} \\ 21 & \boxed{22} & 23 & 24 \\ 31 & 32 & 33 & 34 \\ \boxed{41} & 42 & 43 & 44 \end{array}

Of the 16 possible outcomes, only 3 pairings have a product of 4: 14, 22, and 41.

Therefore, the probability that the two numbers resulting from two consecutive spins have a product of 4 is 316\frac{3}{16}.

Question analysis

Karl spins a red spinner and a blue spinner, each with four equal sections numbered 1 to 4, and multiplies the two results. This compound probability question asks for the chance that the product is 4. Students build the sample space of outcomes, and listing every pair rather than the ones that come to mind is the safe route.

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 a compound probability from a list or table of equally likely outcomes is standard grade 7 work. Two four-section spinners give a small sample space that is quick to write out, and the event needs one careful scan of the products.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 52 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is decided by the count of favorable outcomes, not by any calculation. Students who think of 4 only as 2 × 2 find one outcome, and those who think in factor pairs find two, which is not a listed choice and should send them back to the table. A 4-by-4 grid of products, red down the side and blue across, shows all three cells at once.

  1. Something you must already know moderate Working it out

    Nothing in the stem says that red 1 with blue 4 and red 4 with blue 1 are separate outcomes, or that two four-part spinners give 16 equally likely pairs. (Anderson & Krathwohl, 2001)

  2. Much to keep track of minor Working it out

    All 16 pairs must be scanned for a product of 4, and the two pairs that use 1 and 4 are easy to miss beside the obvious 2 times 2. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Make a 4-by-4 table with the red spinner's numbers down the side and the blue spinner's across the top, write each product in its cell, and count the cells that match.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event
P=number of outcomes you wanttotal number of outcomesP = \frac{\text{number of outcomes you want}}{\text{total number of outcomes}}
P is the probability, the chance that something happens. An outcome is one possible result. This works when every outcome is equally likely, like landing on equal sections of a spinner.
Counting outcomes of two actions
total outcomes=a×b\text{total outcomes} = a \times b
a is the number of results for the first action, and b is the number for the second. Each result of the first can go with each result of the second, so you multiply.

  1. Sort out the question

    Karl spins two spinners, one red and one blue. Each has 4 equal sections, numbered 1 to 4. He multiplies the two numbers. You need the probability that the product, the answer to a multiplication, is 4.

  2. Count all the outcomes

    The red spinner has 4 results, and so does the blue one. Each red result can go with each blue result:

    4×4=164 \times 4 = 16

    So there are 16 outcomes, all equally likely. Red 1 with blue 4 is not the same as red 4 with blue 1. They come from different spinners.

  3. Make a table of products

    Write each product in a table. Red goes down the side and blue goes across the top.

    Red × Blue1234
    11234
    22468
    336912
    4481216

    Three cells show 4: red 1 with blue 4, red 2 with blue 2, and red 4 with blue 1.

    Watch out: thinking of 4 only as 2 × 2 finds one outcome. That gives 1/16, choice A.

  4. Find the probability

    Use the probability rule. Three outcomes give 4, out of 16 in all:

    P=316P = \frac{3}{16}

    Watch out: 7 outcomes have a 4 on at least one spinner. But that counts pairs that show a 4, not pairs whose product is 4. That mistake gives 7/16, choice D.

    So the answer is B, 316\frac{3}{16}.

Check your answer

List the ways to multiply two numbers from 1 to 4 and get 4: 1×41 \times 4, 2×22 \times 2 and 4×14 \times 1. That is 3 ways. The table has 16 cells, and 13 of them are not 4. So a small chance like 3/16 makes sense.

More questions like this

The original DOE question
Question 29 as shown on the DOE’s digital test, page 102 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 102, question 29.

Question 18 of 100

2026 Practice Test D Math, question 31

Question

A box contains an unknown number of green marbles. An experimenter adds 100 red marbles to the box, mixes the marbles thoroughly, and then draws out 10 marbles at random. Assume that this sample is representative of the proportion in the box. If there are 2 red marbles and 8 green marbles in this sample, approximately how many green marbles are in the box?

  • A80
  • B300
  • C400
  • D500
Show the answer

Correct answer: C

In this context, the phrase “this sample is representative” means that the ratio of 2 red marbles to 8 green marbles is the same as the ratio of red to green marbles in the entire box. Use this ratio of 2 to 8 to set up a proportion and solve, where x represents the number of green marbles in the box.

28=100x\frac{2}{8} = \frac{100}{x}

2x=8002x = 800

x=400x = 400

Question analysis

An experimenter adds 100 red marbles to a box of green marbles, mixes them, and draws a sample of 10 with 2 red and 8 green. This sampling question asks students to estimate how many green marbles the box holds. It applies a ratio from a representative sample, worked from a count known for the box rather than for the sample.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a representative sample to reason about a population is grade 6 and 7 work. The twist is that the known count, the 100 red marbles added, belongs to the box rather than the sample, so the usual scaling runs in an unfamiliar direction.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 8, 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

This is the logic of estimating a population by tagging, and the twist is that the known count belongs to the marbles that were added, not to the ones being estimated. Scaling the sample up by 50, so that its 2 red become the box’s 100, turns its 8 green into the estimate in one step. Students who scale by way of the whole box often stop at the box’s total, and those who treat 100 as the whole box scale the wrong amount.

  1. Turning words into math moderate Understanding the question

    The 100 red marbles are counted in the box, while the 2 red and 8 green are counted in a sample of 10, so “representative of the proportion” must become 2 : 8 = 100 : x. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Three ratios can be read from the sample, red to green, red to all and green to all, and only the first two connect to the known 100 red marbles. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Use the sample's ratio of red to green marbles, then scale it up until the red part matches the 100 red marbles known to be in the box.

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

Equal ratios (a proportion)
red in samplegreen in sample=red in boxgreen in box\frac{\text{red in sample}}{\text{green in sample}} = \frac{\text{red in box}}{\text{green in box}}
A ratio compares two amounts. A proportion says two ratios are equal. A sample that is representative has the same mix as the whole box. So the box is the sample scaled up.

  1. Sort out the question

    You know:

    • The box has an unknown number of green marbles.
    • 100 red marbles are added to the box.
    • A sample of 10 marbles has 2 red and 8 green.

    Representative means the sample has the same mix as the box. You need the number of green marbles in the box. Notice that the 100 red marbles are counted in the box, not in the sample.

  2. Set up the proportion

    In the sample, red to green is 2 to 8. In the box, red to green is 100 to x, where x is the number of green marbles. The two ratios are equal:

    28=100x\frac{2}{8} = \frac{100}{x}

  3. Scale up the sample

    How do you get from 2 red to 100 red? Multiply by 50, since 2×50=1002 \times 50 = 100. The green marbles grow by the same factor:

    8×50=4008 \times 50 = 400

    So the box has about 400 green marbles.

  4. Avoid the tempting wrong totals

    Watch out: there is another way that can trip you up. In the sample, 2 of 10 marbles are red. So the 100 red marbles are 2/10 of the box, and the box holds 500 marbles in all. But 100 of those are red. Stopping at 500 gives choice D.

    Taking away the red ones gives 500−100=400500 - 100 = 400 green marbles, the same answer. Treating 100 as the whole box gives 80 green, choice A. Reading 2 to 8 as red to all marbles gives 300, choice B.

    So the answer is C, 400.

Check your answer

Picture the finished box: 400 green and 100 red, so 500 marbles. The red part is 100500=15\frac{100}{500} = \frac{1}{5}. The sample had 210=15\frac{2}{10} = \frac{1}{5} red too. The mix matches, so 400 makes sense.

More questions like this

The original DOE question
Question 31 as shown on the DOE’s digital test, page 104 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 104, question 31.

Question 19 of 100

2026 Practice Test D Math, question 32

Question

There were 36 people at Emily’s party. If each person at the party drank 2 cups of juice, how many gallons of juice did they drink altogether?

Note: 1 gallon = 4 quarts; 1 quart = 4 cups.

  • A2142\frac{1}{4}
  • B4124\frac{1}{2}
  • C9
  • D72
Show the answer

Correct answer: B

Since 36 people each drank 2 cups of juice, the total amount of juice they drank was 36×2=7236 \times 2 = 72 cups.

Convert 72 cups to gallons:

72 cups×1 quart4 cups×1 gallon4 quarts=72 \text{ cups} \times \frac{1 \text{ quart}}{4 \text{ cups}} \times \frac{1 \text{ gallon}}{4 \text{ quarts}} =

7216 gallons=\frac{72}{16} \text{ gallons} =

92 gallons=\frac{9}{2} \text{ gallons} =

412 gallons4\frac{1}{2} \text{ gallons}

Question analysis

At Emily's party 36 people each drank 2 cups of juice, and this unit conversion question asks how many gallons they drank in all, given 4 quarts to a gallon and 4 cups to a quart. Students multiply to get the total in cups, then convert through quarts to gallons, and the bold word gallons sets the unit of the answer.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Converting units with given factors is grade 6 work. This adds a first step, the total number of cups, and a two-stage conversion through quarts, but the numbers are friendly and the path is routine for grade 7 students.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 32 words.
Corrected from the DOE handbook
  • A stray word at the end of the explanation 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

With both conversion facts printed, the question is mainly about keeping the units straight. Students who skip the word each convert 36 cups instead of 72, and those who stop at cups find a total that is right in the wrong unit. Working out that a gallon is 4 × 4 = 16 cups before dividing turns two conversions into one and avoids adding the 4s.

  1. Many steps minor Working it out

    A multiplication, 36 × 2, comes before two divisions by 4, cups to quarts and quarts to gallons, and stopping after the first step leaves a believable 72. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Turning words into math minor Understanding the question

    The word “each” turns 36 people into 72 cups, and the bold word gallons sets the unit of the answer, a change the note supplies in two steps. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the total number of cups first, then divide by 4 to get quarts and by 4 again to get gallons, or work out how many cups make a gallon and divide once.

Traps

  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The DOE’s explanation ends with a line that reads only “lineExtend”, a formatting code left in the document by mistake. The site leaves it out.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total amount
total=number of people×amount each\text{total} = \text{number of people} \times \text{amount each}
When every person has the same amount, multiply the number of people by the amount each one has. The word each tells you to multiply.
Cups in a gallon
cups in 1 gallon=4×4=16\text{cups in 1 gallon} = 4 \times 4 = 16
A gallon holds 4 quarts, and each quart holds 4 cups. So a gallon holds 4 groups of 4 cups. You multiply, not add, because each of the 4 quarts has its own 4 cups.
Changing cups to gallons
gallons=cups÷16\text{gallons} = \text{cups} \div 16
It takes 16 cups to make 1 gallon. Dividing by 16 finds how many groups of 16 cups there are. A gallon is bigger than a cup, so the number of gallons is smaller.

  1. Sort out the question

    You know:

    • 36 people were at the party.
    • Each person drank 2 cups of juice.
    • 1 gallon = 4 quarts, and 1 quart = 4 cups.

    You need the total amount of juice in gallons. The bold word tells you the unit of the answer.

  2. Find the total in cups

    Each of the 36 people drank 2 cups:

    36×2=7236 \times 2 = 72

    They drank 72 cups in all.

    Watch out: skipping the word each uses only 36 cups. Then 36÷16=21436 \div 16 = 2\frac{1}{4} gallons, choice A.

  3. Find the cups in a gallon

    A gallon is 4 quarts, and each quart is 4 cups:

    4×4=164 \times 4 = 16

    So 1 gallon holds 16 cups.

    Watch out: adding the two 4s gives 8 cups in a gallon. Then 72÷8=972 \div 8 = 9, choice C.

  4. Change cups to gallons

    Divide the cups by 16. Simplify the fraction by dividing the top and bottom by 8:

    72÷16=7216=92=41272 \div 16 = \frac{72}{16} = \frac{9}{2} = 4\frac{1}{2}

    They drank 4124\frac{1}{2} gallons of juice.

    Watch out: 72 is the amount in cups. The question asks for gallons, so 72, choice D, is in the wrong unit.

    So the answer is B, 4124\frac{1}{2}.

Check your answer

Work backward. 4124\frac{1}{2} gallons is 412×4=184\frac{1}{2} \times 4 = 18 quarts. 18 quarts is 18×4=7218 \times 4 = 72 cups. Shared among 36 people, that is 72÷36=272 \div 36 = 2 cups each. It matches.

More questions like this

The original DOE question
Question 32 as shown on the DOE’s digital test, page 105 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 105, question 32. This is the DOE’s version, before the corrections listed in the analysis.

Question 20 of 100

2026 Practice Test D Math, question 33

Question

A car costing $24,000 is subject to a sales tax of 8%. If Bindu made a $10,000 down payment on this car, what is the total amount she has left to pay?

  • A$14,000
  • B$15,620
  • C$15,920
  • D$25,920
Show the answer

Correct answer: C

First, find the sales tax. Then, add the sales tax to the cost of the car. Finally, subtract the down payment to find out how much she has left to pay.

(0.08)(24,000)=1,920(0.08)(24{,}000) = 1{,}920

$24,000+$1,920=$25,920\$24{,}000 + \$1{,}920 = \$25{,}920

$25,920−$10,000=$15,920\$25{,}920 - \$10{,}000 = \$15{,}920

Question analysis

Bindu buys a $24,000 car with an 8% sales tax and makes a $10,000 down payment, and students find how much she still has to pay. This percent word problem combines finding a tax, adding it to the price and subtracting a payment. Every step must be done in order, since leaving one out gives an amount that looks just as reasonable.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Adding sales tax to a price is a standard grade 7 percent task. This one also subtracts a down payment at the end, three routine steps with round numbers, much like a typical classroom problem.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 29 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is the standard one for tax on a purchase, so the question checks order and completeness. Forgetting the tax or forgetting the down payment each produces an amount as reasonable-looking as the answer, and a careless 8 × 24 produces one just $300 below it. Multiplying the price by 1.08 in one step, which gives $25,920, is a second route that catches a slip in the tax.

  1. Many steps minor Working it out

    Three operations come in a fixed order, 8% of $24,000, then adding the tax, then subtracting the $10,000, and leaving out the first or the last gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Finding 8% of $24,000 comes down to 8 × 24 = 192 with two zeros added, and a lost carry in that product turns $1,920 into $1,620. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Find 8% of the price and add it, or multiply the price by 1.08, to get the full cost with tax, then subtract what Bindu has already paid.

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}
Percent means out of 100. So 8% is 8/100, or 0.08. To find a percent of a number, change the percent to a decimal and multiply it by the whole.
Amount left to pay
amount left=price+tax−down payment\text{amount left} = \text{price} + \text{tax} - \text{down payment}
The price plus the tax is the full cost. A down payment is money paid at the start. Subtract it, because that part is already paid.

  1. Sort out the question

    You know:

    • The car costs $24,000.
    • The sales tax is 8%. Sales tax is extra money added to a price.
    • Bindu paid $10,000 at the start. That is the down payment.

    You need the amount Bindu still has to pay. Work in order: tax, then full cost, then subtract.

  2. Find the sales tax

    Change 8% to the decimal 0.08. Multiply it by the price:

    0.08×24,000=1,9200.08 \times 24{,}000 = 1{,}920

    A quick way: 1% of 24,000 is 240, so 8% is 8×240=1,9208 \times 240 = 1{,}920. The tax is $1,920.

    Watch out: in 8 × 24 you carry a 3 from 8 × 4 = 32. Dropping it gives 162, so the tax looks like $1,620. That leads to $15,620, choice B.

  3. Find the full cost

    Add the tax to the price:

    24,000+1,920=25,92024{,}000 + 1{,}920 = 25{,}920

    The car costs $25,920 with tax.

    Watch out: this is not the answer yet. Stopping here gives $25,920, choice D. Bindu already paid part of it.

  4. Subtract the down payment

    Bindu already paid $10,000. Take it away:

    25,920−10,000=15,92025{,}920 - 10{,}000 = 15{,}920

    She has $15,920 left to pay.

    Watch out: leaving out the tax gives 24,000−10,000=14,00024{,}000 - 10{,}000 = 14{,}000, choice A.

    So the answer is C, $15,920.

Check your answer

Use a shortcut for the full cost. The price plus 8% more is 108% of the price, or 1.08 times it: 24,000×1.08=25,92024{,}000 \times 1.08 = 25{,}920. Then 25,920−10,000=15,92025{,}920 - 10{,}000 = 15{,}920. Same answer.

More questions like this

The original DOE question
Question 33 as shown on the DOE’s digital test, page 106 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 106, question 33.

Question 21 of 100

2026 Practice Test D Math, question 36

Question

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

DAILY READING LOG

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

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

Enter your answer in the space.

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

Show the answer

Correct answer: 25.6%

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

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

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

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

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

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

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

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

Question analysis

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

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

Why it is hard

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

  1. Something you must already know moderate Working it out

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

  2. Many steps moderate Working it out

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

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

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

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

  2. Count the pages each day

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

    90−45+1=4690 - 45 + 1 = 46

    135−100+1=36135 - 100 + 1 = 36

    190−145+1=46190 - 145 + 1 = 46

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

  3. Add the three days

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

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

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

  4. Change to a percent

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

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

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

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

  5. Enter the answer

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

Check your answer

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

More questions like this

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

Question 22 of 100

2026 Practice Test D Math, question 37

Question

What 3-digit number between 630 and 650 is divisible by both 5 and 8?

Enter your answer in the space.

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

Show the answer

Correct answer: 640

The final digit of the three-digit number is 0. Each of the digits 0 through 9, inclusive, must be considered.

To narrow the choices, first consider that the number between 630 and 650 must be divisible by both 5 and 8.

A number divisible by 5 must end in 0 or 5. Therefore, the only possibilities are 0 and 5.

  • If the number ended in 0, the whole number 640 must also be divisible by 8:

    640÷8=80 and 640÷5=128640 \div 8 = 80 \text{ and } 640 \div 5 = 128

    Both result in a remainder of 0.

  • If the number ended in 5, the whole numbers 635 and 645 must also be divisible by 8:

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

    645÷8=80R5 and 645÷5=129645 \div 8 = 80R5 \text{ and } 645 \div 5 = 129

    Dividing 635 by 5 results in a remainder of 0, but dividing 635 by 8 results in a remainder of 3.

    Dividing 645 by 5 results in a remainder of 0, but dividing 645 by 8 results in a remainder of 5.

Therefore, 640 is the only whole number between 630 and 650 that is divisible by both 5 and 8.

Question analysis

This number question asks for the 3-digit number between 630 and 650 that is divisible by both 5 and 8, typed into a box. Students use divisibility rules or common multiples to find it. The work is short, but the number has to pass both tests, so checking only one divisor is the easy slip.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Common multiples are grade 6 content. Divisibility by 5 limits the last digit, and only a few numbers lie between 630 and 650, so this is a quick check, shorter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 5, 16 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Few numbers need checking, so most students succeed; the risk is testing only one divisor. Students who start from multiples of 8 meet 632 first, and those who start from multiples of 5 can slip on 645 ÷ 8. Using the least common multiple, 40, shows at once that the number must be a multiple of 40, which leaves one candidate in the range.

  1. Much to keep track of minor Working it out

    Each candidate must pass two tests, divisible by 5 and by 8, and both ends of the range, 630 and 650, pass the first test but fail the second. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Something you must already know minor Working it out

    The shortcut that a multiple of 5 ends in 0 or 5 is assumed, and it cuts the search to 635, 640 and 645 before any division by 8. (Anderson & Krathwohl, 2001)

Strategy: Use divisibility by 5 to limit the last digit, then check only those few numbers in the range for divisibility by 8.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Divisible by 5
last digit=0 or 5\text{last digit} = 0 \text{ or } 5
Divisible means it divides evenly, with no remainder. A whole number is divisible by 5 only when its last digit is 0 or 5. That is because counting by 5s always lands on a number ending in 0 or 5.
Least common multiple
LCM(5,8)=5×8=40\text{LCM}(5, 8) = 5 \times 8 = 40
The least common multiple, or LCM, is the smallest number that both 5 and 8 divide evenly. 5 and 8 share no factor except 1, so their LCM is their product. Any number divisible by both is a multiple of 40.

  1. Sort out the question

    You need a 3-digit number between 630 and 650. It must be divisible by 5 and also by 8. Between means 630 and 650 themselves do not count.

  2. Use the rule for 5

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

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

    Only these need checking for 8.

  3. Test each number for 8

    Divide each one by 8 and look at the remainder, the amount left over:

    • 635 ÷ 8 = 79, with 3 left over.
    • 640 ÷ 8 = 80, with nothing left over.
    • 645 ÷ 8 = 80, with 5 left over.

    Only 640 divides evenly by 8.

    Watch out: 645 passes the test for 5 but fails for 8. And 632, which is 8×798 \times 79, passes for 8 but ends in 2, so it fails for 5.

  4. Enter the number

    640 passes both tests. Type the whole number in the box.

    Watch out: 40 is the smallest number divisible by 5 and 8, but it is not between 630 and 650.

    So you would enter 640.

Check your answer

Use the LCM. A number divisible by both 5 and 8 must be a multiple of 40. 40×16=64040 \times 16 = 640. The multiples next to it are 600 and 680, which are outside the range. So 640 is the only one.

More questions like this

The original DOE question
Question 37 as shown on the DOE’s digital test, page 110 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 110, question 37.

Question 23 of 100

2026 Practice Test D Math, question 39

Question

The table shows the probabilities of a hockey team scoring different numbers of goals in a game.

HOCKEY GOAL PROBABILITIES

Number of Goals Probability
0 0.16
1 0.21
2 0.29
3 0.21
4 0.09
5 0.04

What is the probability that the team will score three or more goals in the next game?

  • A0.13
  • B0.21
  • C0.34
  • D0.50
Show the answer

Correct answer: C

To find the probability that any one of several mutually exclusive events occurs, use the addition rule, and add the probabilities of each event:

P(A or B)=P(A)+P(B)\mathrm{P}(\mathrm{A} \text{ or } \mathrm{B}) = \mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B})

Using the data in the table, the probability that the team will score 3 or more goals in the next game is represented by

P(3)+P(4)+P(5)=\mathrm{P}(3) + \mathrm{P}(4) + \mathrm{P}(5) =

0.21+0.09+0.04=0.34.0.21 + 0.09 + 0.04 = 0.34.

Question analysis

A table gives the probabilities of a hockey team scoring 0 to 5 goals in a game, and students find the chance that it scores three or more goals in the next one. This probability question tests adding the probabilities of separate outcomes. Reading exactly which rows the phrase covers is the main point of care.

NYS standards
  • 6.SP.8 probability models for simple events
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding the probability of an event from a given probability model is grade 6 work in New York. Here the model is a table, and the task is choosing the right rows and adding decimals, shorter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 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

Adding three decimals is easy, so the question turns on which rows the words cover. Including the 3-goal row, and not stopping at it, are the two decisions, and each misreading gives a listed choice. Checking the complement, 1 − (0.16 + 0.21 + 0.29), confirms both the sum and the choice of rows.

  1. Turning words into math minor Understanding the question

    The phrase “three or more goals” has to become the three rows for 3, 4 and 5 goals, starting at the 3-goal row itself. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table lists six probabilities, two of them both 0.21, so the rows for 3, 4 and 5 goals must be picked out by their goal counts, not by their values. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide which rows the words "three or more" cover, add their probabilities, and check by subtracting the total of the other rows from 1.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding probabilities
P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)
A and B are two events that cannot both happen at once, like scoring exactly 3 goals and exactly 4 goals. The chance that one or the other happens is the sum of their chances.
The chance it does not happen
P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)
The chances of all possible outcomes add up to 1. So the chance an event does not happen is 1 minus the chance it does. Use it to check an answer from the other side.

  1. Sort out the question

    The table gives the chance of each number of goals, from 0 to 5. You need the chance the team scores three or more goals in the next game.

  2. Pick the right rows

    Three or more means 3 goals, 4 goals or 5 goals. The row for 3 counts too. Read those three rows:

    • 3 goals: 0.21
    • 4 goals: 0.09
    • 5 goals: 0.04

    Watch out: leaving out the 3-goal row gives 0.09 + 0.04 = 0.13, choice A. Using only the 3-goal row gives 0.21, choice B.

  3. Add the chances

    A game can end with only one number of goals. So add the three chances:

    0.21+0.09+0.04=0.340.21 + 0.09 + 0.04 = 0.34

    Watch out: three of the six rows is half the rows, but that does not make the chance 0.50, choice D. The rows have different chances.

    So the answer is C, 0.34.

Check your answer

Find the chance of fewer than 3 goals, the other rows: 0.16+0.21+0.29=0.660.16 + 0.21 + 0.29 = 0.66. Then the chance of 3 or more is 1−0.66=0.341 - 0.66 = 0.34. Both ways give 0.34.

More questions like this

The original DOE question
Question 39 as shown on the DOE’s digital test, page 112 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 112, question 39.

Question 24 of 100

2026 Practice Test D Math, question 40

Question

A team is participating in a contest to collect canned goods for a food drive. The team has 53 points and must have a total of at least 100 points to win a prize. The team earns 5 points for each can of food a member collects. Which inequality can be used to find x, the number of cans of food the team needs to collect to win a prize?

  • A5+53x≤1005 + 53x \leq 100
  • B5+53x≥1005 + 53x \geq 100
  • C53+5x≤10053 + 5x \leq 100
  • D53+5x≥10053 + 5x \geq 100
Show the answer

Correct answer: D

Since each can is worth 5 points, the total number of points earned from all the cans, x, is 5x. The team started with 53 points. That is a constant value and doesn’t change. So the total number of points the team will earn is 53+5x53 + 5x. The team wants to earn at least (≥) 100 points, so they need 53+5x≥10053 + 5x \geq 100 total points.

Question analysis

A team has 53 points in a canned-food drive, earns 5 points for each can, and needs at least 100 points to win a prize. Students choose the inequality for x, the number of cans the team must collect. It tests writing an inequality from words, where the starting points and the points per can must each take the right place.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing an inequality for a word problem with a starting amount and a rate is grade 7 work. The four choices pair two expressions with two symbols, so students make one decision about the terms and one about the phrase "at least".
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 71 words.
Students, last 30 days
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 choices form a two-by-two grid: which number goes with x, and which way the symbol points. Each decision can be made separately, so a student who settles both gets the item, while one mistake lands on a listed choice. Testing two numbers of cans, 5, which leaves the team at 78 points, and 10, which lifts it to 103, rules out every wrong inequality.

  1. Turning words into math moderate Understanding the question

    The 53 points the team “has” are a fixed start, the “5 points for each can” go with x, and “at least 100” calls for ≥, though “least” can sound like less. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    The four choices are two expressions, 5 + 53x and 53 + 5x, each with ≤ and with ≥, so the right one and the strongest wrong one differ only in the symbol. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out the team's total points after 2 cans and then after x cans, translate "at least 100" into a symbol, and compare the result with each choice.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total with a start and a rate
total=start+rate×x\text{total} = \text{start} + \text{rate} \times x
The start is the amount you have now, here 53 points. The rate is what each item adds, here 5 points for each can. x is the number of items. The start is added once, and the rate is multiplied by x.
At least
total≥goal\text{total} \geq \text{goal}
The sign ≥ means is greater than or equal to. At least 100 means 100 or more. So the total must be greater than or equal to 100.

  1. Sort out the question

    You know:

    • The team already has 53 points.
    • It earns 5 points for each can.
    • It needs at least 100 points to win.

    x is the number of cans. You need the inequality that shows this.

  2. Write the total points

    Try 2 cans first. The team gets 5 points for each, so 53+5×2=6353 + 5 \times 2 = 63 points. For x cans, the total is:

    53+5x53 + 5x

    Watch out: the 53 points do not repeat for each can. 5+53x5 + 53x gives 53 points for each can, as in choices A and B.

  3. Choose the sign

    At least 100 means 100 or more, so use ≥:

    53+5x≥10053 + 5x \geq 100

    Watch out: “least” sounds like “less,” but at least means the total can be 100 or higher. Using ≤ gives choice C.

    So the answer is D, 53+5x≥10053 + 5x \geq 100.

Check your answer

Test two numbers of cans. With 10 cans, 53+5×10=10353 + 5 \times 10 = 103, and 103 ≥ 100 is true. The team wins. With 9 cans, 53+5×9=9853 + 5 \times 9 = 98, and 98 ≥ 100 is false. The team is short. Choice D gets both right.

More questions like this

The original DOE question
Question 40 as shown on the DOE’s digital test, page 113 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 113, question 40.

Question 25 of 100

2026 Practice Test D Math, question 43

Question

Two student clubs are selling snacks as part of a fundraiser. The Art Club sells a 24-pack of granola bars for $12.00. The Science Club sells a 10-pack of granola bars for $6.50. To the nearest cent, what is the difference in cost per bar between the two clubs?

Select the correct answer from each drop-down to complete the sentences.

The cost for the Art Club’s bar is per bar.

The cost for the Science Club’s bar is per bar.

The difference between the cost of the bars is .

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

Show the answer

Correct answer: blank 1: $0.50; blank 2: $0.65; blank 3: $0.15

To determine the cost per bar from the Art Club, divide the total cost by the total number of bars:

$12.0024=$0.50 per bar\frac{\$12.00}{24} = \$0.50 \text{ per bar}

To determine the cost per bar from the Science Club, divide the total cost by the total number of bars:

$6.5010=$0.65 per bar\frac{\$6.50}{10} = \$0.65 \text{ per bar}

To find the difference in price per bar, subtract the cost per bar for the Art Club from the cost per bar for the Science Club:

$0.65 per bar−$0.50 per bar=$0.15 per bar\$0.65 \text{ per bar} - \$0.50 \text{ per bar} = \$0.15 \text{ per bar}

Question analysis

Two clubs sell granola bars for a fundraiser: the Art Club a 24-pack for $12.00 and the Science Club a 10-pack for $6.50. Students complete three sentences with drop-down menus, giving each club's cost per bar and the difference between them. It tests unit prices with money, and all three menus must be right.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Comparing unit prices is grade 6 rate work. The menus split it into three short steps, two divisions of money amounts and one subtraction, which is lighter than typical grade 7 work.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 87 words.
Corrected from the DOE handbook
  • The order of the answers in the DOE’s key was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The menus guide the work, one sentence per step, and every option is a reasonable price, so the question rewards accurate division more than planning. No common misconception leads to the wrong options; they catch slips, such as misreading a pack size or subtracting carelessly. Checking that 24 bars at the first price cost $12.00 and 10 bars at the second cost $6.50 confirms both menus before the subtraction.

  1. Many steps minor Working it out

    Three menus must all be right to score: two divisions for the costs per bar, then a subtraction that uses both, so a slip in either division spoils the third menu. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Error-prone arithmetic minor Working it out

    Each division is with money, $12.00 ÷ 24 and $6.50 ÷ 10, so the decimal point must be placed carefully, though the menus offer only sensible amounts. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Divide each pack's price by the number of bars in it, then subtract the smaller cost per bar from the larger; checking that 24 bars at your first price make $12.00 catches slips.

Traps

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

What was corrected from the DOE handbook, and why

  • Answer key: The DOE’s explanation opens with the key “($0.65; $0.50; $0.15)”, which puts $0.65 in the first drop-down and $0.50 in the second. Neither is possible: the first menu (the Art Club’s cost per bar) offers $0.40, $0.50, $0.60 and $0.75, and the second (the Science Club’s) offers $0.55, $0.65, $0.75 and $0.85. The explanation’s own working gives the Art Club $12.00 ÷ 24 = $0.50 and the Science Club $6.50 ÷ 10 = $0.65, so the key is $0.50, $0.65, $0.15. The site’s key was already in that order; the DOE’s key line is not shown on the site.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Cost of one item
cost per bar=price of the pack÷number of bars\text{cost per bar} = \text{price of the pack} \div \text{number of bars}
The price of the pack is what you pay for all the bars together. Dividing shares that price equally among the bars. The result is the price of one bar, called the unit price.

  1. Sort out the question

    You know:

    • The Art Club sells 24 bars for $12.00.
    • The Science Club sells 10 bars for $6.50.

    There are three menus. The first is the Art Club’s cost per bar. The second is the Science Club’s. The third is the difference, how much more one bar costs than the other.

  2. Find the Art Club’s cost

    Share $12.00 among 24 bars:

    12.00÷24=0.5012.00 \div 24 = 0.50

    Each bar costs $0.50. That makes sense: 24 half-dollars make $12. Pick $0.50 in the first menu.

    Watch out: dividing by 20 instead of 24 gives $0.60, which is also in the menu.

  3. Find the Science Club’s cost

    Share $6.50 among 10 bars. Dividing by 10 moves the decimal point one place to the left:

    6.50÷10=0.656.50 \div 10 = 0.65

    Each bar costs $0.65. Pick $0.65 in the second menu.

  4. Find the difference

    Subtract the smaller cost from the larger one:

    0.65−0.50=0.150.65 - 0.50 = 0.15

    The bars differ by $0.15. Pick $0.15 in the third menu.

    Watch out: $0.10 sits right next to $0.15 in the menu. A slip in the subtraction lands there.

  5. Fill in the three menus

    On the test and on this site, you choose each answer from a drop-down menu.

    So the menus read $0.50, $0.65 and $0.15.

Check your answer

Multiply back. 24×0.50=12.0024 \times 0.50 = 12.00 and 10×0.65=6.5010 \times 0.65 = 6.50, so both costs per bar match the pack prices. And 0.50+0.15=0.650.50 + 0.15 = 0.65, so the difference fits too.

More questions like this

The original DOE question
Question 43 as shown on the DOE’s digital test, page 116 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 116, question 43. This is the DOE’s version, before the corrections listed in the analysis.

Question 26 of 100

2026 Practice Test D Math, question 44

Question

A bakery charges $1.25 for each cupcake and $2.50 for a serving tray. The equation describes p, the price of c cupcakes and a serving tray, not including tax.

p=2.50+1.25cp = 2.50 + 1.25c

If Samantha bought some cupcakes and a serving tray for a total of $22.50, not including tax, how many cupcakes did she buy?

  • A4
  • B9
  • C16
  • D18
Show the answer

Correct answer: C

Using the specified equation p=2.50+1.25cp = 2.50 + 1.25c, determine the number of cupcakes Samantha bought by substituting the total paid ($22.50) for p, and solve the equation for the value of c, the number of cupcakes bought.

22.50=2.50+1.25c22.50 = 2.50 + 1.25c

22.50−2.50=1.25c22.50 - 2.50 = 1.25c

20.00÷1.25=c20.00 \div 1.25 = c

16=c16 = c

Question analysis

A bakery's price for c cupcakes and a serving tray is given by the equation p = 2.50 + 1.25c, and Samantha paid $22.50 before tax. This equation question asks how many cupcakes she bought. Students substitute the total and solve a two-step equation with decimal money amounts, undoing the steps in the right 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 Solving a two-step equation of the form px + q = r in a money context is core grade 7 algebra. The equation is given, so the work is substituting the total and dividing by a decimal, a routine example.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 50 words.
Students, last 30 days
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 supplied, so the question is a two-step solve with money. The tray’s $2.50 has to come off the total before dividing, and students who divide the whole $22.50 by $1.25 get a whole number that matches a choice, so nothing warns them. Substituting the chosen number back into p = 2.50 + 1.25c is a quick, decisive check.

  1. Error-prone arithmetic minor Working it out

    After subtracting the tray, $20 must be divided by $1.25, a decimal divisor that is easiest handled as 2,000 ÷ 125 or by noting that 4 cupcakes cost $5. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Substitute the total for p, subtract the tray's cost from both sides, then divide by the price of one cupcake; check by putting your number back into the equation.

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

Doing the same to both sides
if a=b, then a−n=b−n and an=bn\text{if } a = b \text{, then } a - n = b - n \text{ and } \frac{a}{n} = \frac{b}{n}
a and b are the two sides of an equation, and n is any number (not zero when you divide). An equation is like a balanced scale. Doing the same thing to both sides keeps it level. That is how you get c, the number of cupcakes, by itself.

  1. Sort out the question

    The equation p=2.50+1.25cp = 2.50 + 1.25c gives the price. p is the price in dollars, and c is the number of cupcakes. The 2.50 is the tray, and 1.25 is the price of one cupcake. Samantha paid $22.50. You need c.

  2. Put in the total

    Samantha paid $22.50 in all, so put 22.50 in place of p. Now the equation has only one unknown, c:

    22.50=2.50+1.25c22.50 = 2.50 + 1.25c

  3. Subtract the tray

    Take 2.50 away from both sides. That removes the cost of the tray:

    22.50−2.50=1.25c22.50 - 2.50 = 1.25c

    20.00=1.25c20.00 = 1.25c

    So $20.00 went to cupcakes alone.

    Watch out: dividing all of $22.50 by $1.25 gives 18, choice D. That forgets the tray.

  4. Divide by the price of one cupcake

    Divide both sides by 1.25:

    c=20.00÷1.25=16c = 20.00 \div 1.25 = 16

    An easy way: 4 cupcakes cost 4×1.25=5.004 \times 1.25 = 5.00 dollars. $20 is 4 groups of $5. So Samantha bought 4×4=164 \times 4 = 16 cupcakes.

    Watch out: 4 is the number of $5 groups, not cupcakes. That is choice A.

    So the answer is C, 16.

Check your answer

Put 16 back into the equation. 1.25×16=20.001.25 \times 16 = 20.00, and 2.50+20.00=22.502.50 + 20.00 = 22.50. That is what Samantha paid, so 16 is right.

More questions like this

The original DOE question
Question 44 as shown on the DOE’s digital test, page 117 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 117, question 44.

Question 27 of 100

2026 Practice Test D Math, question 45

Question

Marcus and Isla are saving money to buy concert tickets.

  • They decide to save the same amount each week.
  • They already have $8 saved.
  • After 3 weeks, the two have $26 saved in total.

How much money, in dollars, is being saved by Marcus and Isla each week?

Enter your answer in the space.

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

Show the answer

Correct answer: 6

Define the variable:

Let w represent the amount of money Marcus and Isla save per week.

Construct an equation:

Start with $8.

Marcus and Isla save w each week for 3 weeks, which can be expressed as 3⋅w3 \cdot w or 3w3w.

The total amount saved can be represented by the expression 8+3w8 + 3w.

After 3 weeks, Marcus and Isla have $26. This can be represented by the equation 8+3w=268 + 3w = 26.

Solve the equation:

8+3w=268 + 3w = 26

3w=183w = 18 (subtracting 8 from both sides)

w=6w = 6 (dividing both sides by 3)

Marcus and Isla are saving $6 per week.

Question analysis

Marcus and Isla already have $8 toward concert tickets, save the same amount each week, and have $26 in all after 3 weeks. Students type how many dollars the two of them save each week. It tests writing and solving a two-step equation, keeping the money saved beforehand apart from the weekly savings.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an equation of the form px + q = r from a word problem is the grade 7 standard. This one uses small whole numbers and two steps, a typical classroom example, with no choices to compare.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 51 words.
Students, last 30 days
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 inverse steps solve it, subtracting the starting $8 and dividing by the 3 weeks, and most mistakes drop one of them. The $18 saved over three weeks is the natural stopping place, and dividing $26 by 3 without removing the $8 gives an awkward value that should prompt a second look. Adding three weeks of the answer to $8 and getting $26 confirms it.

  1. Turning words into math minor Understanding the question

    The bullet points hold a start amount, “already have $8 saved”, and a total after three weeks, “$26 saved in total”, which must become 8 + 3w = 26. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The amount is typed with no choices, so stopping at the 3-week total of $18 leaves a whole number that nothing on screen questions. (Enright, Morley & Sheehan, 2002)

Strategy: Take away the amount saved before the weeks began, then share what is left equally among the weeks; check by adding three weeks of savings to the starting amount.

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.
  • 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

Total savings
total=start+weeks×amount each week\text{total} = \text{start} + \text{weeks} \times \text{amount each week}
The start is the money saved before the weeks begin. The amount each week is added once for every week, so you multiply it by the number of weeks. The total is all the money together.
Doing the same to both sides
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 zero when you divide). Doing the same thing to both sides keeps them equal. That is how you get the unknown by itself.

  1. Sort out the question

    You know:

    • Marcus and Isla already have $8.
    • They save the same amount each week.
    • After 3 weeks they have $26 in all.

    You need the amount the two of them save together each week.

  2. Write an equation

    Let w be the dollars saved each week. Three weeks add 3w dollars to the $8 they started with:

    8+3w=268 + 3w = 26

  3. Subtract the starting money

    Take 8 away from both sides:

    3w=26−8=183w = 26 - 8 = 18

    They saved $18 over the 3 weeks.

    Watch out: 18 is the amount for all 3 weeks, not for each week. Do not stop here.

  4. Divide by the number of weeks

    Divide both sides by 3:

    w=18÷3=6w = 18 \div 3 = 6

    They save $6 each week.

    Watch out: dividing all $26 by 3 gives about 8.67. That forgets the $8 saved before. Halving 6 to get 3 is wrong too: the question asks what the two save together.

  5. Enter the answer

    Type only the number in the box, with no dollar sign and no units. So you would enter 6.

Check your answer

Count up week by week. Start at $8. After week 1: 8+6=148 + 6 = 14. After week 2: 14+6=2014 + 6 = 20. After week 3: 20+6=2620 + 6 = 26. That is the $26 in the question.

More questions like this

The original DOE question
Question 45 as shown on the DOE’s digital test, page 118 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 118, question 45.

Question 28 of 100

2026 Practice Test D Math, question 46

Question

A company charges $3.00 to ship a 4-pound package. The cost includes a base fee of $1.50 plus $0.50 for each pound over the first pound. The table below illustrates how the cost depends on the total weight of the package. Complete the table by matching each total weight to its correct total shipping cost.

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

SHIPPING COSTS

Total Weight (pounds) Total Cost to Ship (dollars)
2
2.50
6
5.50

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

Show the answer

Correct answer: blank 1: 2; blank 2: 3; blank 3: 4; blank 4: 9

Total Weight (pounds) Total Cost to Ship (dollars)
2 2
3 2.50
6 4
9 5.50

Let x represent the weight of the package in pounds. The first pound is included in the base fee of $1.50, so (x−1)(x - 1) represents the number of pounds that cost $0.50 each.

The cost equation can be written as Cost=$1.50+$0.50(x−1)\text{Cost} = \$1.50 + \$0.50\,(x - 1)

  • The cost for a 2-pound package is $1.50+$0.50(2−1)=$1.50+$0.50(1)=$2.00\$1.50 + \$0.50(2 - 1) = \$1.50 + \$0.50(1) = \$2.00
  • The cost for a 6-pound package is $1.50+$0.50(6−1)=$1.50+$0.50(5)=$1.50+$2.50=$4.00\$1.50 + \$0.50(6 - 1) = \$1.50 + \$0.50(5) = \$1.50 + \$2.50 = \$4.00

  • The weight of a package that costs $2.50 to ship can be found by solving

    $2.50=$1.50+$0.50(x−1)\$2.50 = \$1.50 + \$0.50(x - 1)

    $1.00=$0.50(x−1)\$1.00 = \$0.50(x - 1)

    2=x−12 = x - 1

    x=3 poundsx = 3 \text{ pounds}

  • The weight of a package that costs $5.50 to ship can be found by solving

    $5.50=$1.50+$0.50(x−1)\$5.50 = \$1.50 + \$0.50(x - 1)

    $4.00=$0.50(x−1)\$4.00 = \$0.50(x - 1)

    8=x−18 = x - 1

    x=9 poundsx = 9 \text{ pounds}

To check:

Given that the total cost is $3.00 for the initial package

$3.00=$1.50+$0.50(x−1)\$3.00 = \$1.50 + \$0.50(x - 1)

$1.50=$0.50(x−1)\$1.50 = \$0.50(x - 1)

3=x−13 = x - 1

x=4x = 4 pounds fits the linear pattern of the table

Question analysis

A shipping company charges a base fee plus 50 cents for each pound over the first, and a 4-pound package costs $3.00. Students drag number tiles into a table to complete two costs and two weights, with some tiles left unused. It tests building and using a linear rule, and the wording about the first pound must be applied in every row.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.9 dependent and independent variables
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students solve word problems with equations like p(x + q) = r. This table needs that rule four times, twice run backward from a cost to a weight, and the phrase about the first pound must be applied every time.
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, 75 words.
Students, last 30 days
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 rule is stated in words and confirmed by the 4-pound example, so the work is applying it consistently in both directions. Charging every pound breaks the $3.00 example at once, since 4 pounds would cost $3.50, and models that ignore the wording leave costs or weights with no matching tile. A shortcut helps: each extra pound adds $0.50 to the $3.00, so the table can be filled by counting up or down from 4 pounds.

  1. Turning words into math moderate Understanding the question

    The phrase “$0.50 for each pound over the first pound” means the first pound is covered by the $1.50 fee, so a package of x pounds pays for only x − 1 of them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Four boxes must all be right, two found forward from a weight and two backward from a cost, and the backward rows need the first pound added back. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Write the rule as the base fee plus $0.50 times the pounds after the first, check it against the 4-pound example, then use it forward for the costs and backward for the weights.

Traps

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

Shipping cost
C=1.50+0.50×(w−1)C = 1.50 + 0.50 \times (w - 1)
C is the total cost in dollars, and w is the weight in pounds. The $1.50 base fee is paid once, and it covers the first pound. Each pound after the first costs $0.50. There are w − 1 of those pounds.

  1. Sort out the question

    The base fee of $1.50 covers the first pound. Each pound over the first adds $0.50. A 4-pound package costs $3.00.

    You fill four boxes: the costs for 2 and 6 pounds, and the weights that cost $2.50 and $5.50. Each tile is used once at most, and four tiles are not used.

  2. Test the rule on 4 pounds

    Check the rule with the example. A 4-pound package has 3 pounds over the first:

    C=1.50+0.50×(4−1)=1.50+1.50=3.00C = 1.50 + 0.50 \times (4 - 1) = 1.50 + 1.50 = 3.00

    That is $3.00, so the rule fits.

    Watch out: charging $0.50 for every pound, the first one too, gives $3.50 for 4 pounds. That does not match.

  3. Find the two costs

    For 2 pounds, 1 pound is over the first:

    1.50+0.50×1=2.001.50 + 0.50 \times 1 = 2.00

    For 6 pounds, 5 pounds are over the first:

    1.50+0.50×5=1.50+2.50=4.001.50 + 0.50 \times 5 = 1.50 + 2.50 = 4.00

    The costs are the tiles 2 and 4.

    Watch out: 6 pounds is 3 times 2 pounds, but the cost is not 3 times $2. The base fee is paid once. Tripling gives the tile 6. Leaving out the base fee gives the tile 2.5.

  4. Work back to the two weights

    For $2.50, take off the base fee: 2.50−1.50=1.002.50 - 1.50 = 1.00. At $0.50 a pound, $1.00 pays for 2 extra pounds. Add the first pound: 2+1=32 + 1 = 3 pounds.

    For $5.50: 5.50−1.50=4.005.50 - 1.50 = 4.00. $4.00 pays for 8 extra pounds. Add the first pound: 8+1=98 + 1 = 9 pounds.

    Watch out: forgetting to add back the first pound gives 2 and 8.

  5. Fill in the four boxes

    From top to bottom, the boxes get 2, 3, 4 and 9. The tiles 2.5, 3.5, 6 and 7 are not used. On the test, you drag each tile into its box. On this site, you choose it from the box’s menu.

Check your answer

Each extra pound adds $0.50, so count from 4 pounds at $3.00. Going down: 3 pounds cost $2.50 and 2 pounds cost $2.00. Going up: 5 pounds $3.50, 6 pounds $4.00, 7 pounds $4.50, 8 pounds $5.00 and 9 pounds $5.50. All four boxes match.

More questions like this

The original DOE question
Question 46 as shown on the DOE’s digital test, page 119 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 119, question 46.

Question 29 of 100

2026 Practice Test D Math, question 47

Question

A tall right rectangular prism standing on end. The bottom front edge is labeled 2 in., the bottom edge running back on the right is labeled 2 in., and the height is labeled 6 in. Right-angle marks are drawn at the lower left corner of the front face and at the front left corner of the top face.

What shape will result from a horizontal slice of the figure above?

  • Aa square that has 2-inch sides
  • Ba triangle that has 2-inch sides
  • Ca rectangle that has a 2-inch side and a 6-inch side
  • Da triangle that has one 2-inch side and two 6-inch sides
Show the answer

Correct answer: A

Based on the diagram, the figure shown is a right rectangular prism with a square base. A horizontal slice of the figure will be a plane parallel to the square base, as in this example.

The same prism, 2 in. by 2 in. by 6 in., drawn with its hidden edges dashed and a horizontal slice partway up: a square parallel to the base and the same size as the base.

Since the horizontal slice is parallel to the base, it will be congruent to the base, which is a square with 2-inch sides.

Question analysis

A figure shows a right rectangular prism measuring 2 inches by 2 inches by 6 inches, standing on end. This geometry question asks what two-dimensional shape a horizontal slice of the prism makes. It tests visualizing cross-sections of solids, using the labeled lengths to name both the shape and its size.

NYS standards
  • 7.G.3 slicing three-dimensional solids
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Describing the shape made by slicing a prism parallel to its base is exactly the grade 7 standard. This is its simplest case, a single slice of a labeled right prism, typical of a classroom example.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
1 step
Reading level
Question and choices: about grade 2, 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

A horizontal cut runs parallel to the ground, and this prism stands on a square base, so the slice copies the base. The main danger is picturing the cut along the tall front face, which gives a 2-by-6 rectangle, the shape a vertical slice makes. The two triangles need a slanted cut through a corner and catch only students guessing.

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

    The prism is drawn standing on end, 6 inches tall on a 2-by-2 base, so the cut has to be pictured in a flat drawing where the tall front face is the most visible shape. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    Students must know that a slice parallel to the base of a prism has the same shape and size as the base, which here is the square at the bottom. (Anderson & Krathwohl, 2001)

Strategy: Picture a flat cut across the prism parallel to the ground, ask which face of the prism that cut runs parallel to, and read that face's edge lengths from the labels.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

A slice parallel to the base
slice parallel to the base≅base\text{slice parallel to the base} \cong \text{base}
The sign ≅ means congruent: the same shape and the same size. A prism has the same shape all the way up, like a stack of matching cards. So a cut parallel to its base, at any height, is a copy of the base.

  1. Sort out the question

    The figure is a right rectangular prism, a box shape whose faces are all rectangles. It stands on end. The two edges along the bottom are labeled 2 in. The tall edge is labeled 6 in.

    You need the shape you get from a horizontal slice. Horizontal means flat, like the floor or a tabletop.

  2. Find the base

    The base is the face the prism stands on. The two bottom edges you can see are both labeled 2 in. The right-angle mark on the top face shows its corners are square. The top and the bottom match. So the base is a square with 2-inch sides.

  3. Picture the slice

    A horizontal slice cuts straight across the prism, parallel to the base. The walls of the prism go straight up, so the slice is a copy of the base. It is a square with 2-inch sides, at whatever height you cut.

    The 6-inch height does not matter. It only tells how tall the stack of squares is.

  4. Rule out the other shapes

    Watch out: a 2-inch by 6-inch rectangle, choice C, comes from a vertical slice. That cut goes straight down, through the 6-inch height. A triangle, choice B or D, needs a slanted cut through a corner.

    So the answer is A, a square that has 2-inch sides.

Check your answer

Picture the slice sliding up and down. At the bottom, it matches the base, 2 in. by 2 in. At the top, it matches the top face, the same square. In between, the walls are straight, so its size never changes.

More questions like this

The original DOE question
Question 47 as shown on the DOE’s digital test, page 120 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 120, question 47.

Question 30 of 100

2026 Practice Test D Math, question 49

Question

Anise has a piece of cloth measuring 45 inches by 75 inches. She wants to cut it into squares of equal size, without any cloth left over. What is the greatest possible side length, in inches, of one of those squares?

  • A3
  • B5
  • C15
  • D45
Show the answer

Correct answer: C

Given that each piece will be a square, the length and width will be the same. The greatest possible length can be determined by finding the greatest common factor of the length and width of the original piece of cloth.

Begin by finding the prime factorization of the length (75 in.) and width (45 in.).

45=3×3×545 = 3 \times 3 \times 5

75=3×5×575 = 3 \times 5 \times 5

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

The greatest possible side length, in inches, will be 3×5=153 \times 5 = 15.

Question analysis

Anise wants to cut a 45-inch by 75-inch piece of cloth into equal squares with nothing left over. This factors question asks for the greatest possible side length of each square, in inches. Students must see that the side has to divide both lengths evenly, then find the largest number that does.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Greatest common factors of two numbers up to 100 are grade 6 content. The cloth setting asks students to recognize that this is what is needed, but the work is one short factor search, below typical grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the side is seen as a common factor, the work is short, since 45 and 75 share both 3 and 5. Students who spot one shared factor and stop reach a listed choice, and those who think only of the largest square that fits use the full 45-inch width. Checking that 45 and 75 both divide by the chosen side, and that no larger number does, closes it.

  1. Turning words into math minor Understanding the question

    The words “squares of equal size, without any cloth left over” mean the side must divide both 45 and 75 exactly; nothing names the greatest common factor. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    Finding the greatest common factor of 45 and 75 calls for prime factorization or a list of factors, a grade 6 skill the question assumes. (Anderson & Krathwohl, 2001)

Strategy: Write each length as a product of primes, multiply the primes they share, and check that both 45 and 75 divide evenly by the result.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor
GCF=product of the prime factors both numbers share\text{GCF} = \text{product of the prime factors both numbers share}
A factor divides a number evenly. The greatest common factor, or GCF, is the biggest number that divides both. Write each number as a product of primes, numbers whose only factors are 1 and themselves. Then multiply the primes they share.

  1. Sort out the question

    The cloth is 45 inches by 75 inches. It is cut into equal squares, with no cloth left over. You need the greatest side the squares can have.

    No cloth left over means the side must fit evenly into 45 and into 75. So the side is a common factor of both. The greatest one is the GCF.

  2. Break each number into primes

    Write each length as a product of prime numbers:

    45=3×3×545 = 3 \times 3 \times 5

    75=3×5×575 = 3 \times 5 \times 5

    The two lists share one 3 and one 5.

  3. Multiply the shared primes

    Multiply the primes the two numbers share:

    GCF=3×5=15\text{GCF} = 3 \times 5 = 15

    The squares can be 15 inches on a side.

    Watch out: 3 and 5 are each common factors, choices A and B. But they are not the greatest. Use every prime the numbers share.

  4. Rule out the full width

    Watch out: a 45-inch square, choice D, fits the width. But 75 ÷ 45 = 1 with 30 left over. That leaves a strip of cloth 45 inches by 30 inches.

    So the answer is C, 15.

Check your answer

Divide each length by 15: 45÷15=345 \div 15 = 3 and 75÷15=575 \div 15 = 5. So the cloth cuts into 3 rows of 5 squares, with nothing left. Both divisions come out even.

More questions like this

The original DOE question
Question 49 as shown on the DOE’s digital test, page 122 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 122, question 49.

Question 31 of 100

2024 Test A Math, question 4

Question

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

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

Show the answer

Correct answer: 5.5

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

8÷2=48 \div 2 = 4

4×2184 \times 2\frac{1}{8}

4×2.125=8.54 \times 2.125 = 8.5

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

8.5+4=12.58.5 + 4 = 12.5

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

8÷2=48 \div 2 = 4

4×1344 \times 1\frac{3}{4}

4×1.75=74 \times 1.75 = 7

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

12.5−7=5.512.5 - 7 = 5.5

Question analysis

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

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

Why it is hard

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

  1. Many steps moderate Working it out

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

  2. Error-prone arithmetic moderate Working it out

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

  3. Turning words into math minor Understanding the question

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

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

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

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

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Count the 2-week periods

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

    8÷2=48 \div 2 = 4

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

  3. Change the mixed numbers to decimals

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

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

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

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

  4. Add up the growing

    The lawn grows 2.125 inches, 4 times.

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

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

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

    4+8.5=12.54 + 8.5 = 12.5

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

  5. Add up the cutting

    Mr. Chan cuts off 1.75 inches, 4 times.

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

    He cuts off 7 inches in all.

  6. Subtract the cutting

    12.5−7=5.512.5 - 7 = 5.5

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

Check your answer

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

More questions like this

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

Question 32 of 100

2024 Test A Math, question 11

Question

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

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

Correct answer: C

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

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

Question analysis

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

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

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

  2. Error-prone arithmetic minor Working it out

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

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Find how much water is poured

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

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

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

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

  3. Make the bottoms match

    Now add the poured water to the 12\frac{1}{2} cup already in Jar 2. The bottoms are 2 and 8, so rewrite the half with 8 on the bottom. Multiply its top and bottom by 4.

    12=1×42×4=48\frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}

  4. Add the water in Jar 2

    ad+bd=a+bd\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}

    48+18=58\frac{4}{8} + \frac{1}{8} = \frac{5}{8}

    Jar 2 now holds 58\frac{5}{8} cup. Watch out: 18\frac{1}{8} cup is only the water poured in. Stopping there gives choice A. So the answer is C, 58\frac{5}{8} cup.

Check your answer

Jar 2 started with a half cup and got a little more. So it should hold just over a half cup, and 58\frac{5}{8} is. Also, the water in both jars still adds up to 1 cup: Jar 1 has 38\frac{3}{8} cup left, and 38+58=1\frac{3}{8} + \frac{5}{8} = 1.

More questions like this

The original DOE question
Question 68 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 68. Open page 78 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 33 of 100

2024 Test A Math, question 13

Question

When asked a certain question in a poll, 72% of the people polled answered yes. If 56 people did not answer yes to that question, what is the total number of people who were polled?

  • A78
  • B128
  • C144
  • D200
Show the answer

Correct answer: D

72% of the people polled answered yes, so 28% of the people did not answer yes.

Therefore, 56 people is 28% of the total, x.

56x=28100\frac{56}{x} = \frac{28}{100}

56x=725\frac{56}{x} = \frac{7}{25}

56(25)=7x56(25) = 7x

56(25)7=x\frac{56(25)}{7} = x

8(25)=x8(25) = x

200=x200 = x

Question analysis

A percent word problem about a poll: 72% of the people answered yes, and 56 people did not. Students find the total number of people polled. They must first match the 56 people with the right percent and then find the whole from a part, which is where the main slip happens.

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 Finding a whole from a part and a percent is grade 6 and 7 work. This adds one short step, finding the complementary percent first, which makes it a standard grade 7 multistep percent problem.
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 6, 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 only percent printed is 72 and the only count is 56, so the first instinct is to pair them, even though the bold not says they describe different groups. Once 56 is matched with the 28% who did not say yes, a single division finds the total. The listed wrong answers come from that mismatch, from adding the two numbers, and from stopping at the yes count.

  1. Turning words into math moderate Understanding the question

    The 56 people are the ones who did not answer yes, so they match 100% − 72% = 28%, not the 72% printed in the sentence before them. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Finding the total means 56 ÷ 0.28, a division by a decimal; scaling 28% up to 100% at 2 people per percent avoids it. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract 72% from 100% to find the percent who did not answer yes, then find the total that makes 56 equal to that percent.

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.
  • 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

Percent of a whole
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
The whole is the total amount, here everyone polled. p is the percent. The part is the amount that p percent stands for. Percent means out of 100, so p% is p hundredths of the whole.

  1. Sort out the question

    In a poll, 72% of the people answered yes. 56 people did not answer yes. You need the total number of people polled. Notice the bold word not. The 56 people are the ones who did not say yes.

  2. Find the percent who did not say yes

    Everyone polled makes 100%. Take away the 72% who said yes.

    100−72=28100 - 72 = 28

    So 28% did not answer yes. That is the group of 56 people. Watch out: do not match the 56 people with 72%. That leads to about 78, choice A.

  3. Set up the percent equation

    Call the total number of people T. The 56 people are 28% of T.

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    56=28100×T56 = \frac{28}{100} \times T

  4. Solve for the total

    First find how many people make 1%. Divide 56 people by 28.

    56÷28=256 \div 28 = 2

    So each 1% is 2 people. All 100% is 100 times that.

    T=2×100=200T = 2 \times 100 = 200

    Watch out: do not multiply 2 by 72. That gives 144, the people who said yes, choice C. So the answer is D, 200.

Check your answer

Find 28% of 200: 0.28×200=560.28 \times 200 = 56. That matches the people who did not say yes. Also, 72% of 200 is 144 people who said yes, and 144+56=200144 + 56 = 200, everyone.

More questions like this

The original DOE question
Question 70 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 79, question 70. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 34 of 100

2024 Test A Math, question 16

Question

A truck rental company charges a one-time fee of $40 plus $1 per mile driven. Dalia rented a truck and used a coupon for 20% off the total rental cost. After the coupon was applied, she spent a total of $60. How many miles did she drive?

  • A8
  • B20
  • C32
  • D35
Show the answer

Correct answer: D

Let m equal the miles driven during Dalia’s rental. The total cost, c, can be expressed as an equation.

c=40+mc = 40 + m

Dalia spent a total of $60, which included a 20% off coupon. Find the cost before the coupon by dividing the final cost ($60) by the percentage paid (80%):

6080%=600.80=75\frac{60}{80\%} = \frac{60}{0.80} = 75

Use $75 as the total cost to find m.

40+m=7540 + m = 75

m=35m = 35

Dalia drove 35 miles.

Question analysis

A multistep word problem about a truck rental that costs a $40 fee plus $1 per mile, with a 20% off coupon applied to the total. Given that Dalia paid $60 after the coupon, students find how many miles she drove. Undoing the discount correctly before removing the fixed fee is the hard part.

NYS standards
  • 7.EE.4a two-step equations
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve two-step equations and work with discounts, usually separately. This combines them and makes students undo a 20% discount starting from the price paid, a reversal many grade 7 students have not practised.
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, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two ordinary skills, undoing a discount and solving a two-step equation, are stacked, and the first runs in reverse. The natural way to undo 20% off is to add 20% back, which gives $72 instead of $75 and a choice three miles from the answer. Students who check by applying the coupon to the full cost of 40 + 35 = 75 dollars see at once that 80% of it is 60.

  1. Turning words into math moderate Understanding the question

    The sentence "After the coupon was applied, she spent a total of $60" makes 60 the price after 20% off, so it is 80% of the full cost, 40 + m. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Combines ideas moderate Working it out

    A percent discount has to be undone before the fee-plus-mileage equation can be solved: 60 ÷ 0.8 = 75, then 75 − 40 = 35 miles. (Embretson, 1998; Daniel & Embretson, 2010)

  3. A very close wrong answer moderate Choosing the answer

    The choices 32 and 35 are three miles apart, too close to tell apart by estimate; only dividing 60 by 0.8, rather than adding 20% of 60, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Since $60 is 80% of the full cost, divide 60 by 0.80 to find the cost before the coupon, then subtract the $40 fee to get the miles.

Traps

  • Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
  • 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

Rental cost
c=40+1×mc = 40 + 1 \times m
c is the full cost in dollars, before any coupon. 40 is the one-time fee, paid once no matter how far you drive. m is the number of miles, and each mile costs 1 dollar.
Price after a discount
price paid=full cost×(1−p100)\text{price paid} = \text{full cost} \times \left(1 - \frac{p}{100}\right)
The full cost is the price before the discount. p is the percent taken off. You pay what is left: 100% minus p%. For 20% off you pay 80%, so you multiply by 0.80.

  1. Sort out the question

    The truck costs a $40 fee, plus $1 for each mile. Dalia’s coupon took 20% off the total. After the coupon, she paid $60. You need how many miles she drove. The $60 is the price after the discount, not the full cost.

  2. Write what she paid

    20% off means she paid what was left of the full cost.

    100%−20%=80%100\% - 20\% = 80\%

    As a decimal, 80% is 0.80.

    price paid=full cost×(1−p100)\text{price paid} = \text{full cost} \times \left(1 - \frac{p}{100}\right)

    60=full cost×0.8060 = \text{full cost} \times 0.80

  3. Undo the discount

    To undo times 0.80, divide by 0.80. Move the decimal point one place in both numbers to make it easier.

    full cost=60÷0.80=600÷8=75\text{full cost} = 60 \div 0.80 = 600 \div 8 = 75

    The full cost was $75. Watch out: the 20% was taken off the full cost, not off $60. Adding 20% of $60 gives $72 and leads to 32 miles, choice C. Taking 20% off $60 again leads to 8 miles, choice A.

  4. Find the miles

    c=40+1×mc = 40 + 1 \times m

    75=40+m75 = 40 + m

    m=75−40=35m = 75 - 40 = 35

    She drove 35 miles. Watch out: do not skip the coupon. Using $60 as the full cost gives 20 miles, choice B. So the answer is D, 35.

Check your answer

Go forward. 35 miles cost $35. Add the $40 fee to get $75. The coupon takes off 20% of $75, which is 0.20×75=150.20 \times 75 = 15 dollars. Then 75−15=6075 - 15 = 60. That is exactly what she paid.

More questions like this

The original DOE question
Question 73 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 79, question 73. Open page 79 of the handbook (PDF)

Question 35 of 100

2024 Test A Math, question 17

Question

The probability of drawing a red candy at random from a bag of 25 candies is 25\frac{2}{5}. After 5 red candies are removed from the bag, what is the probability of randomly drawing a red candy from the bag?

  • A0
  • B110\frac{1}{10}
  • C15\frac{1}{5}
  • D14\frac{1}{4}
Show the answer

Correct answer: D

Start with a proportion to determine the total number of red candies in the bag:

25=r25\frac{2}{5} = \frac{r}{25}

25(25)=r\frac{2}{5}(25) = r

10=r10 = r

There are currently 10 red candies out of 25 candies in the bag. After 5 red candies are removed, there would be 5 red candies left in the bag and 20 total candies.

The probability of randomly drawing a red candy would then be 520=14\frac{5}{20} = \frac{1}{4}.

Question analysis

A probability question about a bag of 25 candies in which the chance of drawing red is 2/5; then 5 red candies are taken out. Students find the new probability of drawing a red candy. They must work out how many red candies there were, then rebuild the fraction from the counts left in the bag.

NYS standards
  • 6.SP.8a uniform probability models
  • 6.RP.3 ratio and rate reasoning
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Simple probability as a fraction of outcomes is grade 6 and 7 content. This adds a step back from the probability to a count and an update after removal, which is typical grade 7 multi-step work.
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 7, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question gives a probability, not a count, so the first job is to rebuild the bag: 10 red candies among 25. Removing candies changes two numbers at once, and the strongest wrong choice comes from updating the red count while leaving the total at 25. A student who writes both new counts before forming the fraction avoids it.

  1. Turning words into math moderate Understanding the question

    The probability 2/5 has to be turned into a count, 10 red of 25, and "after 5 red candies are removed" lowers both the red count and the bag's total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three steps follow in order: 2/5 of 25 = 10, then 5 red among 20 candies left, then 5/20 reduced to lowest terms. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use 2/5 of 25 to find the number of red candies, subtract 5 from both the red count and the total, and write the new fraction in lowest terms.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=red candiesall candiesP = \frac{\text{red candies}}{\text{all candies}}
P is the probability, the chance of drawing a red candy. Red candies is how many red ones are in the bag. All candies is how many candies are in the bag in all. Each candy is equally likely, so the chance is the fraction of the bag that is red.

  1. Sort out the question

    The bag has 25 candies. The chance of drawing red is 25\frac{2}{5}. Then 5 red candies are taken out. You need the new chance of drawing red. Taking out candies changes both the red count and the total.

  2. Find the number of red candies

    A chance of 25\frac{2}{5} means 2 out of every 5 candies are red. Call the red count r.

    P=red candiesall candiesP = \frac{\text{red candies}}{\text{all candies}}

    25=r25\frac{2}{5} = \frac{r}{25}

    r=25×25=2×5=10r = \frac{2}{5} \times 25 = 2 \times 5 = 10

    There are 10 red candies. Watch out: do not stop at 25÷5=525 \div 5 = 5. You must also multiply by 2. Thinking there are 5 red candies leads to 0, choice A.

  3. Take out the 5 red candies

    red: 10−5=5\text{red: } 10 - 5 = 5

    all: 25−5=20\text{all: } 25 - 5 = 20

    Now 5 of the 20 candies are red. Watch out: the total goes down too. Keeping 25 gives 525=15\frac{5}{25} = \frac{1}{5}, choice C.

  4. Find the new probability

    P=520=14P = \frac{5}{20} = \frac{1}{4}

    To simplify, divide the top and bottom by 5. So the answer is D, 14\frac{1}{4}.

Check your answer

At first, 25\frac{2}{5} is 40% of the bag. Taking out only red candies should lower the chance. 14\frac{1}{4} is 25%, which is lower but not zero. That makes sense, because 5 red candies are still in the bag.

More questions like this

The original DOE question
Question 74 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 74. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 36 of 100

2024 Test A Math, question 18

Question

Each number in a sequence is formed by doubling the previous number and then adding 1. If the 9th number in the sequence is 63, what is the 10th number minus the 7th number?

  • A96
  • B111
  • C112
  • D127
Show the answer

Correct answer: C

First, write an equation based on the rule described in the item. Let x be the previous number in the sequence and y be the number in the sequence. The ninth number in the sequence is 63.

2x+1=y2x + 1 = y

Determine the 10th number in the sequence, y, by making x=63x = 63 and solving:

2(63)+1=y2(63) + 1 = y

126+1=y126 + 1 = y

127=y127 = y

Determine the 8th number in the sequence, x, by making y=63y = 63 and solving:

2x+1=632x + 1 = 63

Subtract one from both sides of the equation.

2x+(1−1)=63−12x + (1 - 1) = 63 - 1

2x=622x = 62

2x2=622\frac{2x}{2} = \frac{62}{2}

x=31x = 31

Determine the 7th number in the sequence, x, by making y=31y = 31 and solving:

2x+1=312x + 1 = 31

2x+(1−1)=31−12x + (1 - 1) = 31 - 1

2x=302x = 30

2x2=302\frac{2x}{2} = \frac{30}{2}

x=15x = 15

Subtract the 7th number in the sequence from the 10th number:

127−15=112127 - 15 = 112

The difference is 112.

Question analysis

A sequence question: each number is found by doubling the one before it and adding 1, and the 9th number is 63. Students find the 10th number minus the 7th. That means running the rule forward once and backward twice, while keeping track of which position each number holds.

NYS standards
  • 7.EE.4a two-step equations
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve two-step equations but rarely have to run a rule backward through a sequence. This mixes forward and backward steps with position counting, so it asks for more planning than a typical grade 7 task.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 7, 32 words.
Corrected from the DOE handbook
  • A misplaced parenthesis in the explanation’s working was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Applying the rule forward is routine; running it backward is not, and the question needs both. Students must also count positions carefully, since the 9th number is given and the 7th is two steps back, not one. Each listed wrong answer stops or slips at a different stage, so writing the four numbers in a row, 15, 31, 63 and 127, before subtracting is the reliable way.

  1. No obvious method moderate Working it out

    The rule is stated only forward, double and add 1, but the 7th and 8th numbers come before the given 9th, so the rule must be undone: subtract 1, then halve. (Schoenfeld, 1985)

  2. Many steps moderate Working it out

    One forward step reaches the 10th number, 127, two backward steps reach the 8th and 7th, 31 and 15, and a final subtraction finishes. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Much to keep track of minor Working it out

    Four positions and their values, 7th to 10th, have to be kept straight, and the 8th number is easy to mistake for the 7th. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Apply the rule to 63 to get the 10th number; to go back a position, subtract 1 and halve, doing it twice to reach the 7th; then subtract.

Traps

  • Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: Finding the 7th number, the handbook writes “2x + (1 − 1 = 31 − 1)”, with the closing parenthesis after the whole equation, so the line reads as an equation inside an expression (and 1 − 1 = 31 − 1 is false). The site now reads 2x + (1 − 1) = 31 − 1, the same step the handbook writes correctly for the 8th number (2x + (1 − 1) = 63 − 1). It leads to 2x = 30 and x = 15 as before.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

The sequence rule
y=2x+1y = 2x + 1
A sequence is a list of numbers that follows a rule. x is a number in the list, and y is the next number. To get the next number, double x and then add 1.
Going back one number
x=(y−1)÷2x = (y - 1) \div 2
y is a number in the list, and x is the one just before it. To go back, undo the rule in reverse order. The rule doubles and then adds 1. So first subtract 1, then divide by 2.

  1. Sort out the question

    Each number is “formed by doubling the previous number and then adding 1”. Previous means the one just before. The 9th number is 63. You need the 10th number minus the 7th number. So you go forward one number and back two numbers.

  2. Find the 10th number

    Use the rule with x = 63.

    y=2x+1y = 2x + 1

    y=2×63+1=126+1=127y = 2 \times 63 + 1 = 126 + 1 = 127

    The 10th number is 127. Watch out: remember to add 1. Doubling only gives 126 and leads to 111, choice B.

  3. Plan how to go back

    The 7th number comes before the 9th, so you must undo the rule. The rule doubled, then added 1. Undo the last step first: subtract 1. Then undo the doubling: divide by 2.

    x=(y−1)÷2x = (y - 1) \div 2

  4. Find the 8th number

    x=(63−1)÷2=62÷2=31x = (63 - 1) \div 2 = 62 \div 2 = 31

    The 8th number is 31. Watch out: 31 is the 8th number, not the 7th. Using it gives 127−31=96127 - 31 = 96, choice A.

  5. Find the 7th number

    Step back once more, from the 8th number. Use the same rule with y = 31.

    x=(31−1)÷2=30÷2=15x = (31 - 1) \div 2 = 30 \div 2 = 15

    The 7th number is 15.

  6. Subtract the 7th from the 10th

    127−15=112127 - 15 = 112

    Watch out: 127 alone is choice D. You still have to subtract the 7th number. So the answer is C, 112.

Check your answer

Go forward from 15 using the rule. 2×15+1=312 \times 15 + 1 = 31, then 2×31+1=632 \times 31 + 1 = 63, then 2×63+1=1272 \times 63 + 1 = 127. You land on 63 as the 9th number, so the list is right. And 127−15=112127 - 15 = 112.

More questions like this

The original DOE question
Question 75 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 75. Open page 80 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 37 of 100

2024 Test A Math, question 19

Question

8.9, 8.2, 8.5, 9.0, 8.4, 8.6, 8.8

At a skating championship, there are seven judges who each award a score for each skater’s performance. The highest and lowest scores given to each skater are discarded, and the mean of the remaining scores is then calculated and reported as the skater’s final score. What is the final score for the skater who received the scores shown above from the judges?

  • A8.60
  • B8.62
  • C8.64
  • D8.70
Show the answer

Correct answer: C

The highest and lowest scores are discarded: 8.2 and 9.0.

To find the mean, add the remaining scores and then divide that sum by the total number of remaining scores.

8.9+8.5+8.4+8.6+8.8=43.28.9 + 8.5 + 8.4 + 8.6 + 8.8 = 43.2

43.2÷5=8.6443.2 \div 5 = 8.64

Question analysis

A statistics question about a skating championship with seven judges: the highest and lowest of a skater's seven scores are dropped, and the rest are averaged. Students apply that rule to the listed scores, decimals close together. Following the dropping rule exactly and adding the decimals carefully both matter, since the choices are very close.

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a mean is grade 6 content, and this adds only a rule about which scores to drop. Grade 7 statistics usually compares whole data sets, so this is easier, though the close choices demand exact decimal arithmetic.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 9, 62 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is short, but the choices are spaced so closely that a small slip or a skipped rule lands on a neighbor. Averaging all seven scores, or picking the middle score, gives values within a few hundredths of the answer. The question rewards following the stated rule exactly and adding the five remaining scores with care.

  1. A very close wrong answer moderate Choosing the answer

    The choices 8.60, 8.62 and 8.64 lie within 0.04 of one another, so no estimate separates them; only the exact sum of the five kept scores, 43.2, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Error-prone arithmetic minor Working it out

    Five one-decimal scores must be added and divided by 5; subtracting 8 from each first leaves 0.9, 0.5, 0.4, 0.6 and 0.8, which are easier to add. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Turning words into math minor Understanding the question

    The scoring rule takes two sentences to state: discard the single highest and the single lowest score, then average the five that remain. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Cross out the single highest and the single lowest score, add the five that remain, and divide that sum by 5.

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

Mean, or average
mean=sum of the scoresnumber of scores\text{mean} = \frac{\text{sum of the scores}}{\text{number of scores}}
The mean is the average. The sum is what you get when you add all the scores. Divide it by how many scores you added. It shares the total out evenly, as if every score were the same.

  1. Sort out the rule

    Seven judges gave the seven scores shown. The rule has two parts. First, throw out the highest score and the lowest score. Then find the mean of the scores that are left. The mean is the average. That is the final score you need.

  2. Cross out the highest and lowest

    The highest score is 9.0. The lowest is 8.2. Cross both out. Five scores are left: 8.9, 8.5, 8.4, 8.6 and 8.8. Watch out: drop both. Dropping only 8.2 gives 8.70, choice D. Keeping all seven gives about 8.63, which is closest to 8.62, choice B.

  3. Add the five scores

    8.9+8.5+8.4+8.6+8.8=43.28.9 + 8.5 + 8.4 + 8.6 + 8.8 = 43.2

    A shortcut: each score is 8 plus a bit. The bits are 0.9, 0.5, 0.4, 0.6 and 0.8, and they add to 3.2. Five 8s make 40, and 40+3.2=43.240 + 3.2 = 43.2.

  4. Divide by the number of scores

    mean=sum of the scoresnumber of scores\text{mean} = \frac{\text{sum of the scores}}{\text{number of scores}}

    mean=43.25=8.64\text{mean} = \frac{43.2}{5} = 8.64

    Watch out: do not just take the middle score, 8.6. That is the median, choice A. The question asks for the mean. So the answer is C, 8.64.

Check your answer

Multiply back: 5×8.64=43.25 \times 8.64 = 43.2, the sum. Also, the mean should sit between the lowest kept score, 8.4, and the highest kept score, 8.9. 8.64 does.

More questions like this

The original DOE question
Question 76 as printed on page 80 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 76. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 38 of 100

2024 Test A Math, question 20

Question

A piece of wood that is 4124\frac{1}{2} feet long is cut into 2 pieces of different lengths. The shorter piece has a length of x feet. Which inequality expresses all possible values of x?

  • A0<x<2140 < x < 2\frac{1}{4}
  • B0≤x≤2140 \leq x \leq 2\frac{1}{4}
  • C0<x<4120 < x < 4\frac{1}{2}
  • D214<x<4122\frac{1}{4} < x < 4\frac{1}{2}
Show the answer

Correct answer: A

The original piece of wood is 4124\frac{1}{2} feet long.

The shorter piece is x feet long, and it must be less than half the length of the original piece of wood.

Since half of 4124\frac{1}{2} is 2142\frac{1}{4}, it follows that x<214x < 2\frac{1}{4}.

The shorter piece of wood cannot be 0 feet long, so it follows that 0<x0 < x.

This gives the compound inequality 0<x<2140 < x < 2\frac{1}{4}.

Question analysis

An inequality question: a piece of wood 4½ feet long is cut into two pieces of different lengths, and the shorter piece is x feet long. Students choose the inequality that gives every possible value of x. They must turn each condition in the words into a boundary and decide whether each end of the range is included.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.8 writing simple inequalities
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve and graph simple inequalities. Writing a compound inequality from a description, where the words decide whether each end is strict, is less familiar and calls for careful reading of every condition.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 5, 37 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Halving 4 1/2 feet is the only computation; the rest is deciding whether each end of the range can actually happen. A shorter piece of 2 1/4 feet would make the two pieces equal, and a piece of 0 feet is no piece at all, so both ends are excluded. Students who find the right bounds but never test the ends are drawn to the version with the inclusive symbols.

  1. Turning words into math moderate Understanding the question

    Two phrases set the two ends: "cut into 2 pieces" means x is more than 0, and "of different lengths" with "the shorter piece" means x is less than half of 4 1/2. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer moderate Choosing the answer

    The inequalities 0 < x < 2 1/4 and 0 ≤ x ≤ 2 1/4 share both bounds and differ only in their symbols; only checking whether each end can happen separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Variables instead of numbers minor Understanding the question

    The question asks for every possible length of the shorter piece, a range rather than one length, so students must picture the cut moving along the board. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

Strategy: Picture cutting the board in different places, find the largest and smallest values the shorter piece could take, and check whether each boundary value is actually possible.

Traps

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

Two pieces make the whole board
x+L=412x + L = 4\frac{1}{2}
x is the length of the shorter piece, and L is the length of the longer piece, both in feet. Together they make the whole board. Since x is smaller than L, x must be less than half of the board.
Inequality signs
a<b and a≤ba < b \text{ and } a \leq b
a < b says a is less than b. a ≤ b says a is less than or equal to b, so a can also equal b. A chain like 0 < x < 5 means x is more than 0 and less than 5.

  1. Sort out the question

    A board is 4124\frac{1}{2} feet long. It is cut into 2 pieces of different lengths. The shorter piece is x feet long. You need an inequality for every possible value of x. An inequality uses signs like <, less than, to show a range of numbers.

  2. Find the top end for x

    If you cut exactly in the middle, each piece is half the board.

    412÷2=92×12=94=2144\frac{1}{2} \div 2 = \frac{9}{2} \times \frac{1}{2} = \frac{9}{4} = 2\frac{1}{4}

    But the pieces have different lengths, so the cut is not in the middle. The shorter piece must be less than half.

    x<214x < 2\frac{1}{4}

  3. Find the bottom end for x

    The shorter piece can be very small, like 0.1 foot. But it can’t be 0 feet. Then nothing would be cut off, and there would not be 2 pieces.

    0<x0 < x

  4. Put the two ends together

    0<x<2140 < x < 2\frac{1}{4}

    This says x is more than 0 and less than 2142\frac{1}{4}. Watch out: choice B uses ≤, which allows x = 0 or x=214x = 2\frac{1}{4}. Neither can happen. Choice C lets x be more than half, so it would not be the shorter piece. Choice D describes the longer piece. So the answer is A, 0<x<2140 < x < 2\frac{1}{4}.

Check your answer

Test values. Cut off 1 foot: the pieces are 1 foot and 3123\frac{1}{2} feet. The shorter one is 1, and 1 is between 0 and 2142\frac{1}{4}. If x were 3, the other piece would be 1121\frac{1}{2} feet, so x would not be the shorter one. Choice A leaves 3 out, as it should.

More questions like this

The original DOE question
Question 77 as printed on page 81 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 81, question 77. Open page 81 of the handbook (PDF)

Question 39 of 100

2024 Test A Math, question 23

Question

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

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

Correct answer: B

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

(x−1)+(x+1)=2x(x - 1) + (x + 1) = 2x

Question analysis

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

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

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

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

  2. Turning words into math minor Understanding the question

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Try a real number first

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

    6+8=146 + 8 = 14

    Notice that 14 is 2×72 \times 7.

  3. Write the neighbors with x

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

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

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

  4. Add the two neighbors

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

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

Check your answer

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

More questions like this

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

Question 40 of 100

2024 Test A Math, question 25

Question

Points M, N, P and Q, each marked with a dot, lie in that order from left to right on a line. A dimension arrow from M to N is labeled x cm, and a longer dimension arrow from M to P is labeled y cm. No lengths are marked for NP, PQ or MQ.

In the figure above, N is the midpoint of MQ\overline{MQ}. Which segment has length (2x−y)(2x - y) centimeters?

  • APQ\overline{PQ}
  • BNP\overline{NP}
  • CMQ\overline{MQ}
  • DMP\overline{MP}
Show the answer

Correct answer: A

These facts are given:

  • N is the midpoint of MQ\overline{MQ}.
  • MN\overline{MN} has a length of x centimeters.
  • MP\overline{MP} has a length of y centimeters.

Use the first two facts to determine that MQ\overline{MQ} has a length of 2x2x centimeters. So (2x−y)(2x - y) can be represented by MQ−MPMQ - MP, which is PQ\overline{PQ}.

Question analysis

A geometry and algebra question with a labeled figure: points M, N, P and Q lie on a line, MN is x cm, MP is y cm, and N is the midpoint of MQ. Students find which segment has length 2x − y. They must turn the midpoint fact into an expression and read a difference of lengths on the figure.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.EE.6 variables representing unknown numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students evaluate and rewrite expressions but rarely read one as a length on a figure. Combining the midpoint idea from geometry with a difference of expressions, and no numbers, makes this more abstract than typical grade 7 work.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 9, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question reverses the usual task: instead of finding a length, students match an expression to a segment. The midpoint fact turns into MQ = 2x, and taking away the y of MP leaves the piece from P to Q. The wrong choices reward stopping at 2x, reading only the labeled arrows, or matching the letter y; picking x = 5 and y = 7 lets students check with numbers.

  1. Variables instead of numbers moderate Understanding the question

    Every length is a letter: MN is x, MP is y, and the target is the expression 2x − y, so the matching segment cannot be found by measuring. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

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

    The figure marks only MN and MP with arrows; the length of MQ comes from the stem's midpoint fact, not from the drawing, and has to be written as 2x. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Use the midpoint to write the length of the whole segment in terms of x, or pick sample values for x and y, then see which segment's length matches 2x − y.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Midpoint
MQ=2×MNMQ = 2 \times MN
N is the midpoint of segment MQ, the point exactly in the middle. MN is the half from M to N, and MQ is the whole segment. The whole is two equal halves, so it is 2 times MN.
Adding segments
MP+PQ=MQMP + PQ = MQ
When points lie in order on a line, the short segments add up to the long one. MP runs from M to P, and PQ runs from P to Q. Put end to end, they make MQ.

  1. Sort out the figure

    Points M, N, P and Q lie in that order on a line. Segment MN is x cm long, and segment MP is y cm long. N is the midpoint of MQ\overline{MQ}, the point exactly halfway from M to Q. You need the segment that is 2x−y2x - y cm long.

  2. Find the length of MQ

    MN is half of MQ, so MQ is twice as long.

    MQ=2×MN=2xMQ = 2 \times MN = 2x

    MQ is 2x cm long. Watch out: 2x is only part of the expression. Stopping here gives MQ, choice C.

  3. Take MP away from MQ

    Subtracting y from 2x means taking MP away from MQ. MP covers the part from M to P. What is left of MQ is the piece from P to Q.

    MP+PQ=MQMP + PQ = MQ

    PQ=MQ−MP=2x−yPQ = MQ - MP = 2x - y

    So PQ is 2x−y2x - y cm long. Watch out: do not pick MP just because y appears. MP is y, choice D. NP is y−xy - x, choice B. So the answer is A, PQ\overline{PQ}.

Check your answer

Try numbers. Say x = 5, so MN = 5 and MQ = 10. Say y = 7, so MP = 7. Then PQ is 10−7=310 - 7 = 3. And 2x−y=10−7=32x - y = 10 - 7 = 3. They match. NP would be 7−5=27 - 5 = 2, which does not.

More questions like this

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

Question 41 of 100

2024 Test A Math, question 27

Question

On Wednesday, a baker produced 100 more loaves of bread than were produced on Tuesday. On Thursday, the baker produced 50 fewer loaves than were produced on Tuesday. If the total number of loaves produced on all three days was 230, how many loaves were produced on Wednesday?

  • A60
  • B80
  • C120
  • D160
Show the answer

Correct answer: D

Let t represent the number of loaves of bread produced on Tuesday.

First, determine the number of loaves of bread produced on Tuesday, t, using the number of loaves produced on Wednesday, t+100t + 100, the number produced on Thursday, t−50t - 50, and the total number produced on these three days, 230:

t+t+100+t−50=230t + t + 100 + t - 50 = 230

t+t+t+100−50=230t + t + t + 100 - 50 = 230

3t+50=2303t + 50 = 230

3t+(50−50)=230−503t + (50 - 50) = 230 - 50

3t=1803t = 180

(3÷3)t=180÷3(3 \div 3)t = 180 \div 3

t=60t = 60

Next, determine the number of loaves of bread produced on Wednesday, t+100t + 100, when t=60t = 60:

t+100=60+100=160t + 100 = 60 + 100 = 160

The number of loaves of bread produced on Wednesday is 160.

Question analysis

A word problem about a baker's bread over three days: Wednesday's output is 100 loaves more than Tuesday's, Thursday's is 50 fewer than Tuesday's, and the total is 230. Students find the number of loaves made on Wednesday. Writing every day in terms of Tuesday, and answering for the right day, are the keys.

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 of the form px + q = r from a word problem is a core grade 7 standard. This is a typical three-step version, with one negative adjustment and an extra step after solving.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 7, 45 words.
Corrected from the DOE handbook
  • A misplaced parenthesis in the explanation’s working was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Writing all three days in terms of Tuesday turns the story into 3t + 50 = 230, a routine equation. The trap comes after it is solved: the value found is Tuesday's, and it is a listed choice. Students who label t as Tuesday from the start, and reread the question before choosing, add the 100 loaves and are done.

  1. Turning words into math moderate Understanding the question

    Both comparisons are made with Tuesday, 100 more loaves on Wednesday and 50 fewer on Thursday, so the three days become t, t + 100 and t − 50. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    The equation's unknown is Tuesday's count, while the question asks for Wednesday's, so one more step remains after the equation is solved. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let t be Tuesday's loaves, write Wednesday and Thursday in terms of t, set the sum equal to 230, solve for t, and add 100.

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.
  • 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

What was corrected from the DOE handbook, and why

  • Explanation: The handbook writes “3t + (50 − 50 = 230 − 50)”, with the closing parenthesis after the whole equation, so the line reads as an equation inside an expression (and 50 − 50 = 230 − 50 is false). The site now reads 3t + (50 − 50) = 230 − 50: subtract 50 from both sides, giving 3t = 180 as before.

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\text{if } a = b \text{, then } a - c = b - c
a and b are the two sides of an equation, and c is any number. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. The same is true if you divide both sides by the same number that is not zero.

  1. Sort out the question

    The baker made bread on Tuesday, Wednesday and Thursday. Wednesday had 100 more loaves than Tuesday. Thursday had 50 fewer loaves than Tuesday. All three days add up to 230 loaves. You need the number of loaves made on Wednesday.

  2. Write each day using Tuesday

    Both other days are compared with Tuesday. So call Tuesday’s loaves t.

    DayLoaves
    Tuesdaytt
    Wednesdayt+100t + 100
    Thursdayt−50t - 50

    100 more means add 100. 50 fewer means subtract 50.

  3. Write the equation and combine

    t+(t+100)+(t−50)=230t + (t + 100) + (t - 50) = 230

    Group the t’s together and the plain numbers together. Three t’s make 3t, and 100−50=50100 - 50 = 50.

    3t+50=2303t + 50 = 230

    Watch out: do not just subtract 100 and 50 from 230. That gives 80, choice B, and skips writing each day with t.

  4. Solve for Tuesday

    Subtract 50 from both sides. Then divide both sides by 3.

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

    3t+50−50=230−503t + 50 - 50 = 230 - 50

    3t=1803t = 180

    t=180÷3=60t = 180 \div 3 = 60

    Tuesday had 60 loaves. Watch out: 60 is Tuesday, choice A. The question asks about Wednesday.

  5. Find Wednesday’s loaves

    Wednesday had 100 more loaves than Tuesday, so add 100 to t.

    t+100=60+100=160t + 100 = 60 + 100 = 160

    Wednesday had 160 loaves. So the answer is D, 160.

Check your answer

Find all three days. Tuesday is 60, Wednesday is 160, and Thursday is 60−50=1060 - 50 = 10. Add them: 60+160+10=23060 + 160 + 10 = 230. That is the total in the question, so 160 is right.

More questions like this

The original DOE question
Question 84 as printed on page 83 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 83, question 84. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D). This is the DOE’s version, before the corrections listed in the analysis.

Question 42 of 100

2024 Test A Math, question 42

Question

Lian bought enough oranges to fill 4 bags. Each bag contains 8 oranges. The total cost was $11.52. At that rate, how much would Lian pay for 42 oranges?

  • A$17.28
  • B$15.12
  • C$15.02
  • D$12.52
Show the answer

Correct answer: B

First, find the amount, in dollars, that Lian paid for each orange. There are 4 bags with 8 oranges each.

4×8=324 \times 8 = 32

Divide the total cost by the total number of oranges to get the cost per orange:

11.52÷32=0.3611.52 \div 32 = 0.36

Multiply to find the cost of 42 oranges:

0.36×42=15.120.36 \times 42 = 15.12

Question analysis

This rate word problem says Lian paid $11.52 for 4 bags of 8 oranges each and asks how much 42 oranges would cost at the same rate. Students find the price of one orange and multiply. Since 42 oranges is not a whole number of bags, working by the bag instead of by the orange invites mistakes.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a unit price and scaling it up is grade 6 rate work. The decimals are multi-digit, but the plan is routine, with less to manage than grade 7 problems built on percents or negative numbers.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 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 plan is familiar, total oranges, price per orange, then the new cost, so the work lies in the decimals. Because 42 oranges is five bags and two oranges, students who price by the bag either round up to six bags or must handle the leftover. Estimating, since 42 is about a third more than 32, gives roughly $15 and rules out the far choices but not the near one.

  1. Error-prone arithmetic moderate Working it out

    Two decimal operations are done by hand: 11.52 ÷ 32 = 0.36 for one orange, then 0.36 × 42 = 15.12, where a dropped carry is easy. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. A very close wrong answer minor Choosing the answer

    The choices $15.12 and $15.02 differ by 10 cents, so a rough estimate of about $15 cannot choose between them; only the exact product can. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Multiply to find how many oranges were bought, divide the total cost by that number to get the price of one orange, then multiply by 42.

Traps

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

Unit price
unit price=total costnumber of items\text{unit price} = \frac{\text{total cost}}{\text{number of items}}
The unit price is what one item costs. Divide the total cost by how many items you got. It works because the total is shared equally among the items.
Cost
cost=unit price×number of items\text{cost} = \text{unit price} \times \text{number of items}
The unit price is the cost of one item, here one orange. Multiply it by how many items you buy to get the total cost. This works when every item costs the same.

  1. Sort out the question

    Lian bought 4 bags with 8 oranges in each bag. She paid $11.52 in all. You need the cost of 42 oranges. “At that rate” means each orange costs the same as before. So find the price of one orange first.

  2. Count the oranges

    Each of the 4 bags holds 8 oranges, so multiply to find how many she bought.

    4×8=324 \times 8 = 32

    Lian bought 32 oranges for $11.52.

  3. Find the price of one orange

    unit price=total costnumber of items\text{unit price} = \frac{\text{total cost}}{\text{number of items}}

    unit price=11.5232=0.36\text{unit price} = \frac{11.52}{32} = 0.36

    This is easier in cents: 1,152÷32=361{,}152 \div 32 = 36. So one orange costs $0.36, or 36 cents.

  4. Find the cost of 42 oranges

    cost=unit price×number of items\text{cost} = \text{unit price} \times \text{number of items}

    cost=0.36×42\text{cost} = 0.36 \times 42

    Break 42 into 40 and 2 to make it easier.

    0.36×40=14.400.36 \times 40 = 14.40

    0.36×2=0.720.36 \times 2 = 0.72

    14.40+0.72=15.1214.40 + 0.72 = 15.12

    The 42 oranges cost $15.12.

  5. Avoid the traps and answer

    Watch out: $15.02, choice C, is only 10 cents away. Multiply and add with care. Watch out: Lian pays for each orange, not for full bags. Buying 6 whole bags would cost $17.28, choice A.

    So the answer is B, $15.12.

Check your answer

42 oranges is 10 more than 32. Those 10 oranges cost $3.60, since 10×0.36=3.6010 \times 0.36 = 3.60. Then 11.52+3.60=15.1211.52 + 3.60 = 15.12. Same answer, so $15.12 is right.

More questions like this

The original DOE question
Question 99 as printed on page 87 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 87, question 99. Open page 87 of the handbook (PDF)

Question 43 of 100

2024 Test B Math, question 13

Question

p+2r=r(p+1)+1p+2r=r(p+1)+1

In the equation above, if r=2r=2, what is the value of p?

  • A0
  • B1
  • C2
  • D3
Show the answer

Correct answer: B

First substitute 2 for r in the equation.

p+2r=r(p+1)+1p+2r=r(p+1)+1

p+2(2)=2(p+1)+1p+2(2)=2(p+1)+1

Then solve for p.

p+2(2)=2(p+1)+1p+2(2)=2(p+1)+1

p+4=2p+2+1p+4=2p+2+1

p+4=2p+3p+4=2p+3

p+(4−3)=2p+(3−3)p+(4-3)=2p+(3-3)

p+1=2pp+1=2p

(p−p)+1=2p−p(p-p)+1=2p-p

1=p1=p

The value of p is 1.

Question analysis

An equation question: p + 2r equals r times (p + 1), plus 1, and r is given as 2. Students substitute for r, expand the parentheses and solve the resulting linear equation for p. With p on both sides and a product to expand, distributing is where a slip is most likely.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 6.EE.2c evaluating expressions and order of operations
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 Equations with the variable on both sides are formally grade 8, but after substituting, this is a short solve with small whole numbers. It asks about as much as a typical grade 7 two-step equation.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 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

Substituting r = 2 is immediate; the points are lost in expanding 2(p + 1) + 1. Each wrong choice solves a slightly wrong expansion: forgetting to multiply the 1 inside, folding the outer 1 into the parentheses, or losing a term. Putting the answer back into the original equation, with both r and p, is a quick and complete check.

  1. Error-prone arithmetic moderate Working it out

    After substituting, 2(p + 1) + 1 must become 2p + 3, multiplying both terms in the parentheses before adding the 1 outside, with p left on both sides. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Replace every r with 2, multiply 2 by both terms in the parentheses, gather the p terms on one side, and check by putting the value back into the original equation.

Traps

  • 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

Distributive property
a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c
a is the number outside the parentheses. b and c are the terms inside. A term is a number, a letter, or a number times a letter. To multiply out the parentheses, multiply each term inside by a. It works because a groups of b + c is a groups of b plus a groups of c.
Keeping an equation balanced
if a=b, then a−c=b−c\text{if } a = b \text{, then } a - c = b - c
a and b are the two sides of an equation, and c is any number or term. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. Adding the same thing, or dividing both sides by the same number that is not zero, works too.

  1. Sort out the question

    The equation has two letters, p and r. You are told that r = 2. You need the value of p. Here 2r means 2 times r, and r(p + 1) means r times the group p + 1.

  2. Put 2 in for r

    Replace every r with 2. Keep the parentheses so you can see what gets multiplied.

    p+2(2)=2(p+1)+1p + 2(2) = 2(p + 1) + 1

    On the left, 2(2)=42(2) = 4, so the left side is p+4p + 4.

  3. Multiply out the parentheses

    The 2 multiplies both the p and the 1 inside the parentheses. Then add the 1 that is outside.

    a(b+c)=a×b+a×ca(b + c) = a \times b + a \times c

    2(p+1)+1=2p+2+1=2p+32(p + 1) + 1 = 2p + 2 + 1 = 2p + 3

    Watch out: multiply the 1 inside by 2 as well. Writing 2p + 1 + 1 gives p = 2, choice C. Adding the outside 1 inside first gives p = 0, choice A.

  4. Get p by itself

    Now the equation is p+4=2p+3p + 4 = 2p + 3. Subtract p from both sides. Then subtract 3 from both sides.

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

    p+4−p=2p+3−pp + 4 - p = 2p + 3 - p

    4=p+34 = p + 3

    4−3=p+3−34 - 3 = p + 3 - 3

    1=p1 = p

    So p = 1, and the answer is B, 1.

Check your answer

Put p = 1 and r = 2 back into the first equation. Left side: 1+2×2=51 + 2 \times 2 = 5. Right side: 2×(1+1)+1=52 \times (1 + 1) + 1 = 5. Both sides are 5, so p = 1 works.

More questions like this

The original DOE question
Question 70 as printed on page 202 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 202, question 70. 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 44 of 100

2024 Test B Math, question 20

Question

Centerville will hold a vote on whether to build a new park. The town council surveyed a random sample of 200 residents to determine whether they would be in favor of building this park. The table shows the results of the survey.

SURVEY RESULTS

Opinion Number of Residents
Yes 142
No 38
Undecided 20

The mayor estimates that half the undecided voters will vote in favor of building the park. Based on the survey results, if the mayor is correct, how many of the 24,100 voters will be in favor of building the park?

  • A17,111
  • B18,316
  • C19,401
  • D19,521
Show the answer

Correct answer: B

If half the undecided voters (half of 20) vote in favor of building the park, the total number of “yes” votes from the survey will be 152.

142+10=152142+10=152

There are 24,100 voters, and 200 residents were surveyed. Set up a proportion to solve:

152200=x24,100\frac{152}{200}=\frac{x}{24{,}100}

1925=x24,100\frac{19}{25}=\frac{x}{24{,}100}

(24,100)(1925)=x(24{,}100)\left(\frac{19}{25}\right)=x

18,316=x18{,}316=x

Question analysis

A survey question: a random sample of 200 Centerville residents answered yes, no or undecided about building a park, and the mayor expects half the undecided to vote yes. Students use the sample to predict how many of 24,100 voters will favor the park. It joins a proportion from a sample with large-number arithmetic.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.SP.1b samples and populations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Using a random sample to estimate a count in a whole population is a typical grade 7 proportion task. The extra step for the undecided group and the large numbers add care without going beyond grade 7.
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, 88 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The setup is a sample-to-population proportion, and the trap is the mayor's assumption in the middle of the stem. Students who scale the 142 yes answers directly, or who count every undecided resident, reach believable totals that sit right among the choices. Once the count of 152 is settled, a rough estimate, about three quarters of 24,000, is enough to confirm the exact product.

  1. Error-prone arithmetic moderate Working it out

    Scaling 152 out of 200 up to 24,100 voters means computing 24,100 × 0.76, or 120.5 × 152, by hand with five-digit numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    The mayor's sentence adds a step: “half the undecided voters” turns the 20 undecided residents into 10 more yes votes, giving 152 before any scaling. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table gives Yes, No and Undecided counts, and the half must be taken of the Undecided row, not of the No row just above it. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Count the sample's expected yes votes, including half of the undecided, write them as a fraction of 200, and multiply that fraction by 24,100; estimate first to check the choice.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Scaling up a sample
yes in samplesample size=yes in townvoters in town\frac{\text{yes in sample}}{\text{sample size}} = \frac{\text{yes in town}}{\text{voters in town}}
A random sample is a smaller group picked by chance. Its share of yes votes should be about the same as the whole town’s share. So the two fractions are set equal. Use it to scale a survey up to a whole group.

  1. Sort out what you know

    The town surveyed 200 people. 142 said Yes, 38 said No, and 20 were Undecided (not sure yet). The mayor thinks half of the undecided people will vote yes. There are 24,100 voters in all. You need how many of them will vote yes.

  2. Count the yes votes in the survey

    Half of the 20 undecided people is 20÷2=1020 \div 2 = 10. Add them to the yes votes.

    142+10=152142 + 10 = 152

    So 152 of the 200 people surveyed would vote yes. Watch out: take half of the Undecided row, not the No row. Half of 38 leads to 19,401, choice C.

  3. Find the share who vote yes

    152200=0.76\frac{152}{200} = 0.76

    That is 76% of the people surveyed. The sample was random, so about 76% of all the voters should vote yes too. Watch out: using only the 142 yes votes gives 17,111, choice A. Counting all 20 undecided people gives 19,521, choice D.

  4. Scale up to all the voters

    yes in samplesample size=yes in townvoters in town\frac{\text{yes in sample}}{\text{sample size}} = \frac{\text{yes in town}}{\text{voters in town}}

    152200=x24,100\frac{152}{200} = \frac{x}{24{,}100}

    x=0.76×24,100x = 0.76 \times 24{,}100

    Break it into easy parts. 75% is 34×24,100=18,075\frac{3}{4} \times 24{,}100 = 18{,}075, and 1% is 24,100÷100=24124{,}100 \div 100 = 241.

    x=18,075+241=18,316x = 18{,}075 + 241 = 18{,}316

    So the answer is B, 18,316.

Check your answer

Estimate. 76% is a little more than three fourths. Three fourths of 24,000 is 18,000. So the answer should be a little more than 18,000, and 18,316 fits. Also, 18,316÷24,100=0.7618{,}316 \div 24{,}100 = 0.76, the same share as the survey.

More questions like this

The original DOE question
Question 77 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 2025 Admissions, page 204, question 77. Open page 204 of the handbook (PDF)

Question 45 of 100

2024 Test B Math, question 36

Question

The integers (n−1)(n-1), n, and (n+1)(n+1) are factors of 168. What is the greatest possible value of n?

  • A3
  • B7
  • C14
  • D84
Show the answer

Correct answer: B

First, determine all possible factors of 168. One method of doing this is to use a table to list pairs of factors.

Factors of 168
1 168
2 84
3 56
4 42
6 28
7 24
8 21
12 14

Then determine the group of three sequential factors (n−1)(n-1), n, and (n+1)(n+1) where n is as large as possible. This group is 6, 7, and 8, where n=7n=7.

Question analysis

Three consecutive integers, written as n − 1, n and n + 1, must all be factors of 168, and students find the greatest possible value of n. It tests factors and consecutive integers. The catch is that a number can be a factor while its neighbors are not, so every member of the run needs checking.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Listing factors is grade 6 work, but here students must read n − 1, n and n + 1 as three consecutive integers and search the factor list for the largest such run. Pairing algebraic notation with factors matches grade 7 demand.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 7, 17 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the factors of 168 are listed in order, 1, 2, 3, 4, 6, 7, 8, 12 and up, the runs of three in a row stand out, and the last of them is 6, 7, 8. Students lose points by stopping at the first run they see or by checking only n itself. Testing each choice with its two neighbors, starting from the largest, settles it in under a minute.

  1. Variables instead of numbers minor Understanding the question

    The stem names the numbers only as (n − 1), n and (n + 1), which has to be read as three whole numbers in a row that all divide 168. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    Each candidate must pass three divisibility tests at once, for itself and for both neighbors, while the search continues for the greatest such n. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: List the factor pairs of 168 in order, then look along the list for three whole numbers in a row; or test each choice together with its two neighbors.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pair
a×b=168a \times b = 168
a and b are whole numbers. If a times b equals 168, then a and b are both factors of 168. A factor divides a number with no remainder. Listing every pair, from small a upward, finds every factor.

  1. Sort out the question

    A factor of 168 divides 168 with no remainder. The numbers n − 1, n and n + 1 are three whole numbers in a row, such as 6, 7 and 8. The middle one is n.

    All three must be factors of 168. You need the greatest n that works.

  2. List the factor pairs of 168

    Try each whole number from 1 up. When it divides 168, write it with its partner. Stop when the pairs meet, after 12 and 14.

    Small factorPartner
    1168
    284
    356
    442
    628
    724
    821
    1214

  3. Look for three in a row

    Here are all the factors in order:

    1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168

    The runs of three in a row are 1, 2, 3, then 2, 3, 4, then 6, 7, 8. No other three factors sit side by side.

  4. Pick the greatest middle number

    The middle numbers of the runs are 2, 3 and 7. The greatest is 7, so n = 7.

    Watch out: Stopping at the run 2, 3, 4 gives 3, choice A. Keep looking. Also, 14 and 84 are factors, but their neighbors are not. Neither 13 nor 15 divides 168, so 14, choice C, fails. For 84, choice D, neither 83 nor 85 is a factor.

    So the answer is B, 7.

Check your answer

Divide 168 by each number in the run: 168÷6=28168 \div 6 = 28, 168÷7=24168 \div 7 = 24 and 168÷8=21168 \div 8 = 21. All three are whole numbers, so 6, 7 and 8 are all factors.

More questions like this

The original DOE question
Question 93 as printed on page 209 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 209, question 93. Open page 209 of the handbook (PDF)

Question 46 of 100

2024 Test B Math, question 39

Question

A closed rectangular box drawn in three dimensions. Its height is labeled 20 cm, the bottom edge running back on the left is labeled 40 cm, and the bottom front edge is labeled 50 cm.

What is the total surface area, in square centimeters, of the closed box shown above?

  • A3,800
  • B5,600
  • C7,600
  • D40,000
Show the answer

Correct answer: C

To find surface area, use the formula A=2(hw+hl+lw)A=2(hw+hl+lw):

A=2[(20)(40)+(20)(50)+(50)(40)]A=2[(20)(40)+(20)(50)+(50)(40)]

A=2(800+1,000+2,000)A=2(800+1{,}000+2{,}000)

A=2(3,800)=7,600A=2(3{,}800)=7{,}600

Question analysis

A drawing of a closed box gives its length, width and height in centimeters, and students find its total surface area in square centimeters. It tests surface area of a rectangular prism. The work is short, but every face of the box has to be accounted for, and surface area must be kept apart from volume.

NYS standards
  • 7.G.6 area, surface area, and volume
  • 6.G.4 nets and surface area
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Surface area of a rectangular prism is a grade 7 standard, and this box has whole-number dimensions shown in a drawing. Three face areas, a sum and the opposite faces make it typical practice rather than a stretch.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 8, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Finding the three face areas, 800, 1,000 and 2,000, is quick; the question tests whether students account for all six faces. The wrong choices are what the box would give with only three faces, with the top left off, or with the dimensions multiplied into a volume. Listing the faces in pairs before adding is the safest route.

  1. Something you must already know minor Working it out

    No surface-area formula is given; students must know that a box's six faces come in three matching pairs, 20 × 40, 20 × 50 and 40 × 50. (Anderson & Krathwohl, 2001)

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

    The drawing shows only three faces and labels 20, 40 and 50 cm on separate edges, so the hidden faces and their dimensions have to be pictured. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Find the areas of the three different faces from the labeled length, width and height, add them, and double the sum to cover the opposite faces.

Traps

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

Surface area of a box
SA=2(lw+lh+wh)SA = 2(lw + lh + wh)
SA is the surface area, the total area of all six faces. l is the length, w is the width and h is the height. lw, lh and wh are the areas of the three different faces. You double because each face has a matching face on the opposite side.

  1. Sort out the question

    The box is a rectangular prism, a box shape with six rectangle faces. Its length is 50 cm, its width is 40 cm and its height is 20 cm.

    The box is closed, so it has a top. Surface area is the total area of all the outside faces, in square centimeters.

  2. Find the three different faces

    The faces come in matching pairs. Find the area of one face from each pair. Multiply its two edges.

    FacesEdgesArea of one (sq cm)
    Front and back50 × 201,000
    Left and right40 × 20800
    Top and bottom50 × 402,000

  3. Add the faces and double

    Use the formula. Add the three areas. Then double the sum to cover the matching faces you can’t see.

    SA=2(50×40+50×20+40×20)SA = 2(50 \times 40 + 50 \times 20 + 40 \times 20)

    SA=2(2,000+1,000+800)=2×3,800=7,600SA = 2(2{,}000 + 1{,}000 + 800) = 2 \times 3{,}800 = 7{,}600

    The surface area is 7,600 square centimeters.

  4. Avoid the traps

    Watch out: Stopping at 3,800 counts only three faces, choice A. Leaving off the top, as for an open box, gives 5,600, choice B. Multiplying all three edges gives 40,000, choice D. That is the volume, in cubic centimeters.

    So the answer is C, 7,600.

Check your answer

Add all six faces one by one: 1,000+1,000+800+800+2,000+2,000=7,6001{,}000 + 1{,}000 + 800 + 800 + 2{,}000 + 2{,}000 = 7{,}600. Also check the units. Area multiplies two lengths, so it is in square centimeters. Multiplying three lengths gives cubic units, which is volume.

More questions like this

The original DOE question
Question 96 as printed on page 210 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 96. Open page 210 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 47 of 100

2024 Test B Math, question 40

Question

The value of a particular car decreases at a constant rate. If the car is worth $15,000 three years after its original purchase date and $10,000 five years after its original purchase date, what was the value of the car on its original purchase date?

  • A$18,000
  • B$20,000
  • C$22,500
  • D$25,000
Show the answer

Correct answer: C

The value of the car decreased $5,000 in two years: $15,000 to $10,000.

5,000÷2=2,5005{,}000\div 2=2{,}500

Therefore, the value of the car decreases $2,500 each year. After the first 3 years, the car’s value had decreased $7,500.

So 7,500+15,000=22,5007{,}500+15{,}000=22{,}500.

Question analysis

A car loses value at a constant rate, and its worth three years and five years after purchase is given. Students find what the car was worth on the day it was bought. It tests rate of change and initial value in a linear relationship, and the tricky part is counting the right number of years when working back.

NYS standards
  • 8.F.4 modeling with linear functions
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Finding a starting value from two later values of a steady change is grade 8 linear-function work. A grade 7 student can reason with a yearly rate, but extending it back to the purchase date goes beyond typical grade 7 problems.
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, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is a straight-line pattern read backward, and its numbers are friendly; the difficulty is counting years correctly. The $5,000 drop happens over 2 years, not 5, and the purchase lies 3 years before the $15,000 value, not 2. Each miscount gives a listed answer, so sketching a timeline from year 0 to year 5 before computing pays off.

  1. Turning words into math moderate Understanding the question

    The phrases "three years after its original purchase date" and "five years after" must become points at year 3 and year 5 on a timeline whose start, year 0, holds the unknown value. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps minor Working it out

    Three moves are chained: the $5,000 drop over 2 years, the $2,500 yearly rate, and three years of that rate added back to $15,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Find the yearly drop from the change between the two given years, then add that drop back once for each year between the purchase and year 3.

Traps

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

Rate of change
rate=change in valuechange in time\text{rate} = \frac{\text{change in value}}{\text{change in time}}
change in value is how many dollars the value went up or down. change in time is how many years passed. rate is the change for one year. A constant rate means the change is the same every year.

  1. Sort out the question

    The car’s value “decreases at a constant rate.” That means the car loses the same number of dollars each year. Year 0 is the day it was bought.

    YearValue
    0?
    3$15,000
    5$10,000

    You need the value at year 0.

  2. Find the drop for one year

    From year 3 to year 5 is 2 years. In that time, the value fell by 15,000−10,000=5,00015{,}000 - 10{,}000 = 5{,}000 dollars.

    rate=5,0002=2,500\text{rate} = \frac{5{,}000}{2} = 2{,}500

    The car loses $2,500 each year. Watch out: Divide by the 2 years between the values, not by 5. Dividing by 5 leads to $18,000, choice A.

  3. Go back to year 0

    From year 0 to year 3 is 3 years. The car lost $2,500 in each of them. So add that amount back 3 times.

    15,000+3×2,500=15,000+7,500=22,50015{,}000 + 3 \times 2{,}500 = 15{,}000 + 7{,}500 = 22{,}500

    Watch out: Adding back only one $5,000 gives $20,000, choice B. That is the value at year 1, not year 0.

    So the answer is C, $22,500.

Check your answer

Count forward from $22,500, taking off $2,500 each year. Year 1 is $20,000. Year 2 is $17,500. Year 3 is $15,000. Year 4 is $12,500. Year 5 is $10,000. Both given values match.

More questions like this

The original DOE question
Question 97 as printed on page 210 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 210, question 97. Open page 210 of the handbook (PDF)

Question 48 of 100

2024 Test B Math, question 42

Question

Jevon earns 12%\frac{1}{2}\% per year on the money in his savings account. Which expression is not another way to write 12%\frac{1}{2}\%?

  • A0.005100\frac{0.005}{100}
  • B12100\frac{\frac{1}{2}}{100}
  • C51,000\frac{5}{1{,}000}
  • D1200\frac{1}{200}
Show the answer

Correct answer: A

Rewrite the percent as a decimal:

12%=0.5%=0.005\frac{1}{2}\%=0.5\%=0.005

So the correct answer must not equal 0.005:

Option A

0.005100=0.00005\frac{0.005}{100}=0.00005

Option B

12100=0.5÷100=0.005\frac{\frac{1}{2}}{100}=0.5\div 100=0.005

Option C

51,000=0.005\frac{5}{1{,}000}=0.005

Option D

1200=0.005\frac{1}{200}=0.005

All answer options except Option A equal 0.005. Therefore, Option A is the correct answer.

Question analysis

Jevon's savings account earns one-half percent a year, and students must find the expression that is not another way to write one-half percent. It tests converting a percent less than 1 into fractions and decimals. Reading the word not carefully, and evaluating each stacked or unfamiliar fraction correctly, are what make it tricky.

NYS standards
  • 6.RP.3c percent as a rate per 100
  • 7.NS.2d fractions to decimals by long division
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Converting between percents, fractions and decimals is grade 6 and 7 work. A percent smaller than 1 is less familiar, and the question asks which form is not equal, so all four must be checked, which keeps it at typical grade 7 demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 7, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every expression is built from the same idea, so the question tests whether students can convert a percent smaller than 1 without losing track of decimal places. The odd one out, 0.005 over 100, looks like a correct form, because 0.005 is one-half percent and dividing by 100 seems to turn it into a percent. Converting one-half percent to 0.005 once and comparing each expression with it beats reasoning about each form.

  1. A question in the negative moderate Understanding the question

    The bold word not asks for the one expression unequal to one-half percent, so the three expressions that check out are exactly the ones to reject. (Haladyna, Downing & Rodriguez, 2002)

  2. Every choice must be checked moderate Choosing the answer

    Three of the four forms are equal, so each has to be converted to a decimal and compared with 0.005 before the odd one out can be named. (Embretson & Wetzel, 1987)

  3. Error-prone arithmetic minor Working it out

    One-half percent is 0.5 with the point moved two places, 0.005, and dividing that by 100 moves it two more places, to 0.00005. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write one-half percent as a single decimal by dividing one half by 100, then evaluate each expression as a decimal and compare it with that value.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent as a fraction
p%=p100p\% = \frac{p}{100}
p is any number. The % sign means out of 100. So p percent is p divided by 100. Dividing by 100 moves the decimal point two places to the left.

  1. Sort out the question

    The question asks which expression is not equal to 12%\frac{1}{2}\%. So three choices are equal to it, and one is different. You need the odd one out.

  2. Write the percent as a decimal

    One half is 0.5. A percent means out of 100, so divide by 100.

    12%=0.5%=0.5100=0.005\frac{1}{2}\% = 0.5\% = \frac{0.5}{100} = 0.005

    So 12%\frac{1}{2}\% is 0.005. It is half of one percent, a very small amount.

  3. Test each choice

    Turn each choice into a decimal. Then compare it with 0.005.

    ChoiceWork it outEqual to 0.005?
    A0.005÷100=0.000050.005 \div 100 = 0.00005No
    B0.5÷100=0.0050.5 \div 100 = 0.005Yes
    C5÷1,000=0.0055 \div 1{,}000 = 0.005Yes
    D1÷200=0.0051 \div 200 = 0.005Yes

  4. Pick the one that is different

    Choice A divides by 100 twice. The 0.005 was already divided by 100 once. Dividing again makes it 100 times too small.

    Watch out: Choice B is the plainest way to write 12%\frac{1}{2}\%: one half over 100. Picking it means missing the word “not.” Also, 1 ÷ 200 is 0.005, not 0.05, so D is equal too.

    So the answer is A, 0.005100\frac{0.005}{100}.

Check your answer

Use fractions instead. Half of one hundredth is 12×1100=1200\frac{1}{2} \times \frac{1}{100} = \frac{1}{200}, which is choice D. And 51,000\frac{5}{1{,}000} simplifies to 1200\frac{1}{200} too. Only A is a different amount.

More questions like this

The original DOE question
Question 99 as printed on page 211 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 99. Open page 211 of the handbook (PDF)

Question 49 of 100

2024 Test B Math, question 46

Question

The least positive integer evenly divisible by the first 6 positive integers is 60. What is the least positive integer evenly divisible by the first 8 positive integers?

  • A1,680
  • B840
  • C480
  • D420
Show the answer

Correct answer: B

Each option is divisible by 1, 2, 3, 4, 5, and 6.

Option A:

1,680 is evenly divisible by the first 8 positive integers.

Option B:

840 is evenly divisible by the first 8 positive integers and is less than 1,680.

Option C:

480 is not evenly divisible by 7.

480÷7≈68.57480\div 7\approx 68.57

Option D:

420 is not evenly divisible by 8.

420÷8=52.5420\div 8=52.5

Option B (840) is the correct answer.

Question analysis

Knowing that 60 is the smallest number divisible by each of 1 through 6, students find the smallest positive integer divisible by each of 1 through 8. It tests least common multiples and prime factors. The challenge is adding only the factors that 7 and 8 really need, so the result works for all eight numbers and is still the least.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Least common multiples are grade 6 content, usually for two small numbers. Extending the given result from six numbers to eight means deciding what 7 and 8 add, more reasoning than typical classwork but no new content.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 13, 25 words.
Corrected from the DOE handbook
  • A rounded decimal in the explanation is now marked as approximate (≈ instead of =).
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The given fact, 60 for the first six integers, is a head start, but only for students who see what 7 and 8 add. Multiplying 60 by 7 is needed; multiplying by all of 8 is not, since 60 already supplies two of its three factors of 2. Students who multiply by the new numbers whole get answers that are too large or that miss one divisor, so checking the smaller choices first is the efficient route.

  1. Every choice must be checked moderate Choosing the answer

    Both 1,680 and 840 are divisible by 7 and by 8, so testing the choices is safe only from the smallest up; 420 fails for 8 and 480 for 7. (Embretson & Wetzel, 1987)

  2. Something you must already know minor Working it out

    Building on 60 needs prime factors: 7 adds a new prime, while 8 = 2 × 2 × 2 needs only one more 2, since 60 already holds two. (Anderson & Krathwohl, 2001)

Strategy: Start from 60 and ask which extra prime factors 7 and 8 require, or test the choices from the smallest up for divisibility by both 7 and 8.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: To show that 480 is not evenly divisible by 7, the handbook writes 480 ÷ 7 = 68.57. The quotient is really 68.571428…, a decimal that never ends, so 68.57 is only a rounded value. The site writes 480 ÷ 7 ≈ 68.57, as the handbook itself does for other rounded results in this test. The point of the step is unchanged: the quotient is not a whole number, so 480 is not evenly divisible by 7 and is not divisible by all of the first 8 positive integers.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Evenly divisible
a÷b=a whole numbera \div b = \text{a whole number}
a is the number you test, and b is the number you divide by. If the answer is a whole number, with no remainder, a is evenly divisible by b. Then b is a factor of a.

  1. Sort out the question

    The first 8 positive integers are 1, 2, 3, 4, 5, 6, 7 and 8. “Evenly divisible” means dividing leaves no remainder.

    You need the least, or smallest, number that all 8 of them divide. You are told that 60 is the smallest for 1 to 6. So build on 60.

  2. Bring in the 7

    60 is not divisible by 7. And 7 is prime, which means its only factors are 1 and 7. So the new number needs a factor of 7. Multiply 60 by 7.

    60×7=42060 \times 7 = 420

  3. Bring in the 8

    8 is 2×2×22 \times 2 \times 2, so the number needs three 2s as factors. But 420 has only two: 420=2×2×105420 = 2 \times 2 \times 105, and 105 is odd. Multiply by one more 2.

    420×2=840420 \times 2 = 840

    Now 840÷8=105840 \div 8 = 105 and 840÷7=120840 \div 7 = 120. Both are whole numbers.

  4. Test the choices from smallest up

    Choice D: 420÷8=52.5420 \div 8 = 52.5. That is not a whole number, so 420 fails.

    Choice C: 480÷7≈68.57480 \div 7 \approx 68.57. The ≈ sign means about, since the decimal never ends. It is not a whole number, so 480 fails.

    Choice B: 840 works for 7 and 8. It also works for 1 to 6, since 840=60×14840 = 60 \times 14.

    Watch out: 1,680, choice A, works too. But it is twice 840, so it is not the least.

    So the answer is B, 840.

Check your answer

Divide 840 by each number from 1 to 8. You get 840, 420, 280, 210, 168, 140, 120 and 105. Every answer is a whole number, so 840 works for all eight.

More questions like this

The original DOE question
Question 103 as printed on page 212 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 103. Open page 212 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 50 of 100

2024 Test B Math, question 55

Question

Typically, an amusement park sells 3 times as many tickets on Sunday as on any weekday, and 4 times as many tickets on Saturday as on Sunday. If the number of tickets sold on Saturday and Sunday totaled 30,000, how many tickets are expected to be sold on the following Monday?

  • A2,000
  • B2,500
  • C3,000
  • D4,285
Show the answer

Correct answer: A

First, determine the number of tickets sold on Sunday, x, using the total number of tickets sold on Saturday and Sunday, 30,000, and the number of tickets sold on Saturday, 4x4x.

30,000=4x+x30{,}000=4x+x

30,000=5x30{,}000=5x

6,000=x6{,}000=x

Then determine the number of tickets expected to be sold on Monday, y, using the number of tickets sold on Sunday, 6,000.

6,000=3y6{,}000=3y

2,000=y2{,}000=y

The number of tickets expected to be sold on Monday is 2,000.

Question analysis

An amusement park sells 3 times as many tickets on Sunday as on a weekday and 4 times as many on Saturday as on Sunday. Given the weekend total, students find the expected ticket sales for the following Monday. It tests writing an equation from multiplicative comparisons, and the trap is comparing Saturday with the wrong day.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving a one-variable equation from a word problem is grade 7 work. Two linked multiplicative comparisons, with Saturday measured against Sunday rather than a weekday, add a layer of translation, but the equation itself stays simple.
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 12, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is two comparisons chained through Sunday, and most wrong answers come from linking them incorrectly. Students who read Saturday's 4 times as 4 weekdays, or who leave Sunday out of the 30,000, find listed answers, one of them from a division that does not come out even. Letting a weekday be one share, so Sunday is 3 and Saturday 12, makes the weekend 15 shares and the check easy.

  1. Turning words into math moderate Understanding the question

    "4 times as many tickets on Saturday as on Sunday" compares Saturday with Sunday, not with a weekday, so Saturday is 4 × 3 = 12 weekdays' worth. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Three days have to be held in relation at once, a weekday, Sunday and Saturday, with the 30,000 covering only the last two. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Let Sunday's sales be the unknown, write Saturday as 4 times it, solve the weekend equation for Sunday, then divide Sunday's figure by 3 for a weekday.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing multiplication
a×x=b means x=b÷aa \times x = b \text{ means } x = b \div a
x is the unknown number. a and b are known numbers. Division undoes multiplication. So if a times x is b, you find x by dividing b by a. This is the same as dividing both sides by a.

  1. Sort out the question

    Sunday sells 3 times as many tickets as a weekday. Saturday sells 4 times as many as Sunday. Saturday and Sunday together sell 30,000.

    Monday is a weekday. You need Monday’s tickets. Notice that Saturday is compared with Sunday, not with a weekday.

  2. Write an equation for the weekend

    Let x be Sunday’s tickets. Then Saturday’s tickets are 4x. Together they make 30,000.

    4x+x=30,0004x + x = 30{,}000

    5x=30,0005x = 30{,}000

  3. Solve for Sunday

    Use the rule for undoing multiplication. Divide both sides by 5.

    x=30,000÷5=6,000x = 30{,}000 \div 5 = 6{,}000

    Sunday sells 6,000 tickets. Saturday sells 4×6,000=24,0004 \times 6{,}000 = 24{,}000.

  4. Find Monday’s tickets

    Sunday sells 3 times as many as a weekday. Let y be Monday’s tickets.

    3×y=6,000 means y=6,000÷3=2,0003 \times y = 6{,}000 \text{ means } y = 6{,}000 \div 3 = 2{,}000

    Watch out: If you read Saturday as 4 times a weekday, you split 30,000 into 7 shares. That gives about 4,285, choice D. Dividing 30,000 by 12 leaves out Sunday and gives 2,500, choice B.

    So the answer is A, 2,000.

Check your answer

Think in shares. A weekday is 1 share. Sunday is 3 shares, and Saturday is 4×3=124 \times 3 = 12 shares. The weekend is 15 shares, and 30,000÷15=2,00030{,}000 \div 15 = 2{,}000. It matches.

More questions like this

The original DOE question
Question 112 as printed on page 215 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 112. Open page 215 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 51 of 100

2023 Test A Math, question 5

Question

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

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

Show the answer

Correct answer: 19.2

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

x8=17\frac{x}{8} = 17

x=136x = 136

Next, subtract the 3 discarded numbers from the sum:

136−9−11−20=96136 - 9 - 11 - 20 = 96

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

965=19.2\frac{96}{5} = 19.2

Question analysis

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

NYS standards
  • 6.SP.5c range, center, and overall pattern
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Finding a mean is grade 6 content, but working back from a mean to a total, removing values and averaging again is a typical grade 7 multistep task. As a grid-in, it offers no choices to check a wrong total against.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 3, 21 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Many steps moderate Working it out

    Three operations come in a fixed order: multiply the mean by 8 to get 136, subtract 9, 11 and 20 to leave 96, then divide by the numbers that remain. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Much to keep track of minor Working it out

    The count changes partway through, from 8 numbers to 5, and the divisor at the end must be the new count, not the 8 from the first sentence. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

    The answer, 19.2, is a decimal that must be gridded with its point, and with no choices a total of 96 or a mean over 8 numbers looks complete. (Enright, Morley & Sheehan, 2002)

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

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    There are 8 numbers. Their mean, or average, is 17. You don’t know what the 8 numbers are. Three of them, 9, 11 and 20, are discarded, or thrown away. That leaves 5 numbers. You need the mean of those 5 numbers.

  2. Find the sum of all 8 numbers

    You can’t add the 8 numbers, because you don’t know them. But the mean tells you their sum.

    sum=mean×how many numbers\text{sum} = \text{mean} \times \text{how many numbers}

    sum=17×8=136\text{sum} = 17 \times 8 = 136

    The 8 numbers add up to 136.

  3. Take away the three discarded numbers

    Subtract the three numbers that were thrown away.

    136−9−11−20=136−40=96136 - 9 - 11 - 20 = 136 - 40 = 96

    The 5 numbers left add up to 96. Watch out: 96 is a sum, not a mean, so gridding 96 is wrong. Also, dividing 136 by 5 without subtracting first gives 27.2, which is wrong.

  4. Divide by the 5 numbers left

    Now use the mean formula for the numbers that are left.

    mean=sum of the numbershow many numbers\text{mean} = \frac{\text{sum of the numbers}}{\text{how many numbers}}

    mean=965=19.2\text{mean} = \frac{96}{5} = 19.2

    Watch out: divide by 5, not by 8. Only 5 numbers are left. Dividing by 8 gives 12. So you would grid 19.2, with the decimal point in its own column.

Check your answer

The three numbers thrown away add to 40, so their mean is about 13. That is less than 17. Removing small numbers should make the mean go up, and 19.2 is more than 17. Also, 19.2×5=9619.2 \times 5 = 96, the sum of the 5 numbers left.

More questions like this

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

Question 52 of 100

2023 Test A Math, question 8

Question

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

  • A$20.50/hr
  • B$26.75/hr
  • C$33.00/hr
  • D$37.65/hr
Show the answer

Correct answer: A

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

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

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

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

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

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

Question analysis

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

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 7.EE.4a two-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Commission is one of the percent applications named in the grade 7 standards, and the remaining work is a two-step equation with decimals. That matches a typical grade 7 multistep problem, with the order of the steps as the main demand.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 44 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    The phrase "paid an hourly rate plus a 15% commission on all computer products he sells" makes $802.50 the sum of two parts, 30 hours of pay and 15% of $1,250. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Many steps moderate Working it out

    Three operations in order: 15% of $1,250 is $187.50, $802.50 − $187.50 = $615.00, and $615.00 ÷ 30 = $20.50; dividing before subtracting gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. Error-prone arithmetic minor Working it out

    Every amount carries cents, and 15% of $1,250 and $615 ÷ 30 are worked by hand, so a misplaced decimal point shifts the hourly rate. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

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

Traps

  • Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Pay with commission
pay=rate×hours+commission\text{pay} = \text{rate} \times \text{hours} + \text{commission}
The rate is the money Jamel earns for each hour. Hours is how long he worked. The commission is extra money that is a percent of what he sells. His pay is these two amounts added together.

  1. Sort out the question

    Jamel’s pay has two parts. One part is an hourly rate, the money he earns for each hour. The other part is a commission. A commission is extra pay that is a percent of what he sells. He worked 30 hours and sold $1,250.00 of products. In all he got $802.50. You need his hourly rate.

  2. Find the commission

    His commission is 15% of what he sold. 15% means 15 out of 100, or 0.15.

    part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}

    part=0.15×1,250=187.50\text{part} = 0.15 \times 1{,}250 = 187.50

    His commission was $187.50. Watch out: take 15% of the $1,250 he sold, not of his $802.50 pay. Mixing them up leads to $37.65, choice D.

  3. Take the commission out of his pay

    The rest of his pay came from his hours.

    802.50−187.50=615.00802.50 - 187.50 = 615.00

    He earned $615.00 for 30 hours of work. Watch out: the word plus does not mean add here. Adding the $187.50 to $802.50 counts the commission twice. That leads to $33.00, choice C.

  4. Divide by the hours he worked

    Share the $615.00 equally over the 30 hours.

    615÷30=20.50615 \div 30 = 20.50

    Watch out: don’t divide all $802.50 by 30. That gives $26.75, choice B. That amount still includes the commission. So the answer is A, $20.50 per hour.

Check your answer

Work forward from the answer. 30 hours at $20.50 is 30×20.50=61530 \times 20.50 = 615 dollars. Add the commission: 615+187.50=802.50615 + 187.50 = 802.50. That is exactly Jamel’s pay for the week.

More questions like this

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

Question 53 of 100

2023 Test A Math, question 10

Question

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

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

  • A2
  • B4
  • C6
  • D8
Show the answer

Correct answer: D

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

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

Question analysis

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

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

Why it is hard

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

  1. Many steps minor Working it out

    The two midpoint statements must be used in order: the first places Y at 4, and only then can the second, about segment WZ, place Z. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

    Only W at 0 and X at 2 are marked on the 0-to-10 line, and the overbars name WY and WZ as segments, so Y and Z have to be placed from the statements alone. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The number line shows W at 0 and X at 2. Points Y and Z are not shown. X is the midpoint of segment WY. Y is the midpoint of segment WZ. A midpoint is the point exactly halfway along a segment. You need the number where Z sits.

  2. Place point Y

    X is halfway from W to Y. So X to Y is as long as W to X.

    WX=2−0=2WX = 2 - 0 = 2

    XY=WX=2XY = WX = 2

    So Y is 2 units past X.

    2+2=42 + 2 = 4

    Point Y is at 4. Watch out: this is Y, not Z. Stopping here gives 4, choice B.

  3. Place point Z

    Now Y is halfway from W to Z. W to Y is 4 units, so Y to Z is also 4 units.

    YZ=WY=4−0=4YZ = WY = 4 - 0 = 4

    4+4=84 + 4 = 8

    Watch out: the second midpoint is for segment WZ, not XZ. Taking just 2 more units from Y puts Z at 6, choice C. So the answer is D, 8.

Check your answer

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

More questions like this

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

Question 54 of 100

2023 Test A Math, question 12

Question

If n is an integer and 3n+33n + 3 is an even number, which expression must also represent an even number?

  • A5n+15n + 1
  • B4n+54n + 5
  • C2n+32n + 3
  • Dn+2n + 2
Show the answer

Correct answer: A

The expression 3n+33n + 3 is equal to 3(n+1)3(n + 1), and it is given that this is an even number. The expression (n+1)(n + 1) must also represent an even number, because 3 is an odd number: if (n+1)(n + 1) were odd, 3(n+1)3(n + 1) would be the product of two odd numbers, which is an odd number. If (n+1)(n + 1) is an even number, then n must be an odd number. Choose an odd number to substitute for n in the answer options to find which expression represents an even number:

Let n=3n = 3.

Option A:

5n+1=5(3)+1=165n + 1 = 5(3) + 1 = 16

Option B:

4n+5=4(3)+5=174n + 5 = 4(3) + 5 = 17

Option C:

2n+3=2(3)+3=92n + 3 = 2(3) + 3 = 9

Option D:

n+2=3+2=5n + 2 = 3 + 2 = 5

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

Question analysis

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

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

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    No value of n is given; the only fact is that 3n + 3 is even, and it has to be turned into a fact about n itself, that n is odd, before any choice is judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Something you must already know minor Working it out

    The parity rules are assumed, not stated: an odd number times an odd number is odd, and an even number plus an odd number is odd. (Anderson & Krathwohl, 2001)

  3. Every choice must be checked minor Choosing the answer

    Each expression has to be tested with an odd n or by parity rules; three of the four come out odd whenever n is odd, so none can be dismissed on sight. (Embretson & Wetzel, 1987)

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    n is an integer, a whole number that can be positive, negative or 0. You are told that 3n+33n + 3 is even. You need the expression that must be even too. Must means it is even every time. First find out what the fact tells you about n.

  2. Pull out the 3

    Both terms in 3n+33n + 3 have a factor of 3. Pull it out.

    3n+3=3(n+1)3n + 3 = 3(n + 1)

    So 3 times (n+1)(n + 1) is an even number.

  3. Decide whether n is odd or even

    3 is odd. If n+1n + 1 were odd, 3(n+1)3(n + 1) would be odd times odd. That is always odd. But we are told it is even. So n+1n + 1 must be even. One less than an even number is odd. So n is odd. Watch out: don’t test n = 2. Then 3n+3=93n + 3 = 9, which is odd. With n = 2, only n+2n + 2 comes out even, choice D.

  4. Test each choice with an odd n

    Pick an odd number, n = 3. It fits, since 3×3+3=123 \times 3 + 3 = 12 is even.

    • A: 5n+1=5×3+1=165n + 1 = 5 \times 3 + 1 = 16, even
    • B: 4n+5=4×3+5=174n + 5 = 4 \times 3 + 5 = 17, odd
    • C: 2n+3=2×3+3=92n + 3 = 2 \times 3 + 3 = 9, odd
    • D: n+2=3+2=5n + 2 = 3 + 2 = 5, odd

    Only A is even.

  5. Make sure A always works

    Must means every odd n, not just 3. If n is odd, 5n is odd times odd, so it is odd. Odd plus 1 is even. So 5n+15n + 1 is always even. Watch out: 4n is even, but adding the odd 5 makes 4n+54n + 5 odd, choice B. In the same way, 2n+32n + 3 is always odd, choice C. So the answer is A, 5n+15n + 1.

Check your answer

Try another odd number, n = 1. Then 3n+3=63n + 3 = 6, which is even, so n = 1 fits. And 5n+1=65n + 1 = 6 is even too. The other three choices give 9, 5 and 3, all odd. Choice A wins again.

More questions like this

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

Question 55 of 100

2023 Test A Math, question 14

Question

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

  • A−1
  • B−2
  • C−3
  • D−5
Show the answer

Correct answer: D

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

Question analysis

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

NYS standards
  • 6.NS.7a inequality as relative position
  • 6.NS.6c positioning rational numbers and points
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Ordering negative numbers is grade 6 content, and the question adds only the condition that the integer be odd. It is lighter than typical grade 7 work, though the order of negatives still trips up students who compare their sizes.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 18 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Error-prone arithmetic minor Working it out

    On the negative side, less means farther from 0, so the integers below −3.25 run −4, −5, −6; −3, though its digit is closer to 3.25, lies to the right. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Much to keep track of minor Working it out

    Three conditions apply at once, odd, less than −3.25 and the greatest such value, and checking only two of them leads to a listed choice. (Sweller, 1988; Carpenter, Just & Shell, 1990)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    n has to pass three tests. It is an integer, a whole number with no decimal part. It is odd, so it is not a multiple of 2. And it is less than −3.25. Of all the numbers that pass, you need the greatest one. That is the one farthest to the right on a number line.

  2. Picture −3.25 on a number line

    −3.25 sits between −4 and −3. Less than −3.25 means to the left of it.

    −5<−4<−3.25<−3-5 < -4 < -3.25 < -3

    So −4 is less than −3.25, but −3 is greater. Watch out: −3 is not less than −3.25. It is to the right, closer to 0. For negative numbers, the one closer to 0 is greater. Picking −3 is choice C.

  3. Find the odd integers to the left

    The integers less than −3.25, starting with the closest, are −4, −5, −6 and so on. The greatest one is −4. But −4 is even, since 4 = 2 × 2. So skip it. The next one, −5, is odd. Watch out: −1 is odd, but it is greater than −3.25, so choice A breaks the rule. So the answer is D, −5.

Check your answer

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

More questions like this

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

Question 56 of 100

2023 Test A Math, question 18

Question

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

  • A$212.55
  • B$262.55
  • C$302.55
  • D$312.55
Show the answer

Correct answer: A

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

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

First, subtract 90 from both sides:

0.15×12x=$382.590.15 \times 12x = \$382.59

1.8x=$382.591.8x = \$382.59

Last, divide both sides by 1.8:

x=$212.55x = \$212.55

Question analysis

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

NYS standards
  • 7.EE.4a two-step equations
  • 7.RP.3 multistep ratio and percent problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an equation of the form px + q = r from a word problem is core grade 7 work. Here the coefficient combines a percent with 12 weeks, and the decimals are long, which adds steps but no new idea.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 4, 48 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math moderate Understanding the question

    The total of $472.59 includes the $90.00 she started with, and only 15% of each week's pay is saved, so the equation is 90 + 12 × 0.15x = 472.59. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic moderate Working it out

    Dividing $382.59 by 1.8 by hand means shifting the decimal point to get 3,825.9 ÷ 18, which comes out to exactly $212.55 only if every digit is placed correctly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  3. Many steps minor Working it out

    Subtract the $90, then divide by 12 weeks and by 0.15, or by their product 1.8; doing the division before the subtraction gives a listed choice. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent of a number
part=p100×whole\text{part} = \frac{p}{100} \times \text{whole}
p is the percent. The whole is the amount you take the percent of. The part is the result. Percent means out of 100. So 15% is 15 out of every 100, or 0.15.
Savings over time
total=start+weekly savings×weeks\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}
The start is the money she had at the beginning. Weekly savings is what she adds each week. Weeks is how many weeks pass. Each week adds the same amount, so multiply. Then add what she started with.

  1. Sort out the question

    Sheila starts with $90.00. Each week she earns the same amount and saves 15% of it. After 12 weeks she has $472.59 in all. That total includes the $90 she started with. You need how much she earns each week, not how much she saves.

  2. Write an equation

    Call her weekly earnings x. She saves 15% of x each week. That is 0.15×x0.15 \times x.

    total=start+weekly savings×weeks\text{total} = \text{start} + \text{weekly savings} \times \text{weeks}

    472.59=90+0.15x×12472.59 = 90 + 0.15x \times 12

  3. Take away the starting $90

    Only the money added during the 12 weeks came from her pay. Subtract 90 from both sides.

    0.15x×12=472.59−90=382.590.15x \times 12 = 472.59 - 90 = 382.59

    She saved $382.59 from her pay. Watch out: subtract the $90, don’t add it. Adding leads to $312.55, choice D. Forgetting the $90 leads to $262.55, choice B.

  4. Combine the numbers next to x

    0.15×12=1.80.15 \times 12 = 1.8

    1.8x=382.591.8x = 382.59

    Each week she saves 0.15 of one week’s pay. Over 12 weeks, that adds up to 1.8 times one week’s pay.

  5. Divide by 1.8

    Divide both sides by 1.8. First move the decimal point one place right in both numbers. Then you divide by the whole number 18.

    x=382.59÷1.8=3,825.9÷18=212.55x = 382.59 \div 1.8 = 3{,}825.9 \div 18 = 212.55

    Sheila earns $212.55 each week. So the answer is A, $212.55.

Check your answer

Work forward. 12 weeks of saving 15% is 1.8 weeks of pay: 1.8×212.55=382.591.8 \times 212.55 = 382.59. Add the $90 she started with: 382.59+90=472.59382.59 + 90 = 472.59. That is the total in the question.

More questions like this

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

Question 57 of 100

2023 Test A Math, question 22

Question

The numbers m, n, p, and q are different, and each is equal to one of the numbers 1, 2, 3, 6, or 12. If 2m=6q=12n=p2m = 6q = \frac{1}{2}n = p, what is the value of p?

  • A2
  • B3
  • C6
  • D12
Show the answer

Correct answer: C

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

Option A:

If p=2p = 2, then 12n=2\frac{1}{2}n = 2, so n=4n = 4. This cannot be the answer, because n cannot equal 4.

Option B:

If p=3p = 3, then 6q=36q = 3, so q=0.5q = 0.5. This cannot be the answer, because q cannot equal 0.5.

Option C:

If p=6p = 6, then 2m=62m = 6, so m=3m = 3; 6q=66q = 6, so q=1q = 1; and 12n=6\frac{1}{2}n = 6, so n=12n = 12. This is the correct answer.

Option D:

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

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 The equations involve only doubling, halving and multiplying by 6, all grade 6 skills. What goes beyond grade 7 practice is the chain of four equal expressions, with every letter limited to a short list of different numbers.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 5, 27 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    Four letters are tied by one chain of equalities, 2m = 6q = ½n = p, and the only numbers are the five allowed values, so each letter has to be written in terms of p. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    Each choice for p must be tested against three letters, m = p/2, q = p/6 and n = 2p, and all of them must land on different numbers from 1, 2, 3, 6 and 12. (Embretson & Wetzel, 1987)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    m, n, p and q are four different numbers. Each one is 1, 2, 3, 6 or 12. The chain 2m=6q=12n=p2m = 6q = \frac{1}{2}n = p says that four amounts are all equal to p. You need the value of p.

  2. Write each letter using p

    Each part of the chain equals p. Undo each multiplication.

    2m=p, so m=p÷22m = p \text{, so } m = p \div 2

    6q=p, so q=p÷66q = p \text{, so } q = p \div 6

    12n=p, so n=2×p\frac{1}{2}n = p \text{, so } n = 2 \times p

    Half of n is p, so n is twice p.

  3. Test each choice for p

    pmqnAll on the list?
    2113\frac{1}{3}4No
    31.50.56No
    63112Yes
    126224No

    Only p = 6 puts every letter on the list.

  4. Check that the letters are different

    With p = 6, you get m = 3, q = 1 and n = 12. All four numbers are on the list, and no two are the same. Watch out: p = 12 makes m = 6 and q = 2, which look fine. But n = 24 is not on the list, so choice D fails. Also, the question asks for p, not m. Picking 3 is choice B. So the answer is C, 6.

Check your answer

Put the values into the chain: 2×3=62 \times 3 = 6, 6×1=66 \times 1 = 6, 12×12=6\frac{1}{2} \times 12 = 6, and p = 6. All four parts equal 6. The numbers 3, 1, 12 and 6 are all different and all on the list.

More questions like this

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

Question 58 of 100

2023 Test A Math, question 29

Question

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

  • A2x+12x + 1
  • B3x−53x - 5
  • C4x−14x - 1
  • D4x+64x + 6
Show the answer

Correct answer: D

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

Let x=3x = 3.

Option A:

2x+1=2(3)+1=72x + 1 = 2(3) + 1 = 7

Option B:

3x−5=3(3)−5=43x - 5 = 3(3) - 5 = 4

Option C:

4x−1=4(3)−1=114x - 1 = 4(3) - 1 = 11

Option D:

4x+6=4(3)+6=184x + 6 = 4(3) + 6 = 18

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

Question analysis

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

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.4 common factors and multiples
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students rewrite and evaluate linear expressions, but they seldom ask whether one is divisible by 3 for every allowed value. One well-chosen substitution settles it, so the demand is about typical for grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 13, 18 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Variables instead of numbers moderate Understanding the question

    Nothing pins x down except that it is a multiple of 3, so each expression must be judged for every such x, and any test value has to be a multiple of 3 itself, such as 3 or 6. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked minor Choosing the answer

    The four expressions differ only in their coefficients and constants, from 2x + 1 to 4x + 6, so none can be dismissed at a glance; each needs a test or a look at its terms. (Embretson & Wetzel, 1987)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    “Evenly divisible by 3” means 3 goes into the number with nothing left over. Such a number is called a multiple of 3, like 3, 6, 9 or 12.

    You know that x is a multiple of 3. You need the expression that is a multiple of 3 every time, whatever multiple of 3 x is.

  2. Test x = 3 in every choice

    Pick a number for x that follows the rule. The easiest one is 3. Put it into each expression.

    ChoiceWorkDivisible by 3?
    A2(3)+1=72(3) + 1 = 7No
    B3(3)−5=43(3) - 5 = 4No
    C4(3)−1=114(3) - 1 = 11No
    D4(3)+6=184(3) + 6 = 18Yes

    Watch out: your test number must be a multiple of 3. With x = 1, choice A gives 3 and looks right. That leads to the wrong choice, A.

  3. See why D always works

    One test could be lucky, so see why D works for every x. Since x is a multiple of 3, 4x is one too. For example, 4 × 3 = 12 and 4 × 6 = 24. The number 6 is also a multiple of 3, since 6 = 3 × 2.

    3a+3b=3(a+b)3a + 3b = 3(a + b)

    So 4x + 6 is a multiple of 3 plus a multiple of 3. That is always a multiple of 3.

  4. Rule out the other choices

    Watch out: choice B starts with 3x, but then it subtracts 5. The 5 is not a multiple of 3, so it spoils the result, and 3(3) − 5 = 4. Choices A and C also add or take away a number that is not a multiple of 3.

    So the answer is D, 4x+64x + 6.

Check your answer

Try two more multiples of 3. With x = 6, choice D gives 4(6)+6=304(6) + 6 = 30, and 30 ÷ 3 = 10. With x = 9, it gives 4(9)+6=424(9) + 6 = 42, and 42 ÷ 3 = 14. Both divide evenly, so D keeps working.

More questions like this

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

Question 59 of 100

2023 Test A Math, question 30

Question

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

  • A45\frac{4}{5}
  • B710\frac{7}{10}
  • C310\frac{3}{10}
  • D15\frac{1}{5}
Show the answer

Correct answer: B

There are currently 20 chips in the jar (6+10+46 + 10 + 4). Of those chips, 14 are not red. So the probability of choosing a chip that is not red is 1420=710\frac{14}{20} = \frac{7}{10}.

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Probability of a single event with equally likely outcomes is grade 6 content in New York. Students add three counts and form one fraction, and the only extra care is the bold word not, so it sits below grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Turning words into math minor Understanding the question

    The bold not turns the question around: the chips that count are the 10 blue and 4 yellow ones, 14 in all, or equivalently 1 minus the red share of 6/20. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The count of 14 out of 20 matches no choice until it is reduced to 7/10, so a correct fraction left unsimplified can look missing from the list. (Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    Lamel’s jar holds 6 red chips, 10 blue chips and 4 yellow chips. He takes one chip without looking. You need the probability that it is not red. That means the chip is blue or yellow.

  2. Count all the chips

    6+10+4=206 + 10 + 4 = 20

    There are 20 chips in all. That is the bottom of the fraction, because any of the 20 could come out.

  3. Count the chips that are not red

    The blue and yellow chips are the ones that are not red.

    10+4=1410 + 4 = 14

    So 14 chips are not red. Watch out: the question asks for not red. Counting the 6 red chips gives 620=310\frac{6}{20} = \frac{3}{10}, choice C. Counting only the 4 yellow chips gives 15\frac{1}{5}, choice D.

  4. Find and simplify the probability

    P=outcomes you wantall outcomesP = \frac{\text{outcomes you want}}{\text{all outcomes}}

    P=1420=14÷220÷2=710P = \frac{14}{20} = \frac{14 \div 2}{20 \div 2} = \frac{7}{10}

    Dividing the top and bottom by 2 puts the fraction in lowest terms. So the answer is B, 710\frac{7}{10}.

Check your answer

A chip is either red or not red, so the two chances must add up to 1. The chance of red is 620=310\frac{6}{20} = \frac{3}{10}. Then 310+710=1\frac{3}{10} + \frac{7}{10} = 1. It works.

More questions like this

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

Question 60 of 100

2023 Test A Math, question 46

Question

ESSAY LENGTH

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

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

  • A20
  • B30
  • C100
  • D120
Show the answer

Correct answer: D

The number of students in the sample who wrote an essay with at least 100 words is 24 out of the 30 students. Write it as a fraction: 2430=45\frac{24}{30} = \frac{4}{5}. To predict the number of students in the entire grade who would write an essay with at least 100 words, find 45\frac{4}{5} of 150, which is 45(150)=120\frac{4}{5}(150) = 120.

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a random sample to predict a count for a whole population is typical grade 7 statistics and ratio work. Students total the sample from a table, read at least 100 against the table's ranges, and scale by a whole-number factor.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 6, 65 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

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

    The table never prints the sample size; it has to be built by adding all four rows, 6 + 4 + 11 + 9 = 30, and the qualifying essays sit in three rows, not one. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Turning words into math minor Understanding the question

    The bold “at least 100 words” must be matched to the rows 100–250, 251–500 and >500, keeping the row that begins at 100 and leaving out only the <100 row. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Many steps minor Working it out

    Four steps follow in order: the sample total, the qualifying count of 24, the fraction 24/30 = 4/5, and that fraction of all 150 students. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The table shows a random sample of the essays from a grade of 150 students. You need to predict how many of the 150 students wrote at least 100 words. At least 100 means 100 or more.

  2. Find the sample size

    The table doesn’t give the total, so add all four rows.

    6+4+11+9=306 + 4 + 11 + 9 = 30

    The sample has 30 essays.

  3. Count the essays with enough words

    Three rows count: 100–250, 251–500 and >500. Only the <100 row is left out.

    4+11+9=244 + 11 + 9 = 24

    Watch out: the 100–250 row counts, because 100 is at least 100. Leaving it out gives 100, choice C. Using the 6 short essays instead gives 30, choice B.

  4. Scale up to the whole grade

    expected number=count in samplesample size×whole group\text{expected number} = \frac{\text{count in sample}}{\text{sample size}} \times \text{whole group}

    2430×150=45×150=120\frac{24}{30} \times 150 = \frac{4}{5} \times 150 = 120

    About 4 out of every 5 students write at least 100 words. So the answer is D, 120 students.

Check your answer

The whole grade, 150, is 5 times the sample of 30. So multiply the sample count by 5: 24×5=12024 \times 5 = 120. The short essays scale to 6×5=306 \times 5 = 30, and 120 + 30 = 150, the whole grade.

More questions like this

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

Question 61 of 100

2023 Test B Math, question 1

Question

LAST YEAR’S TOTAL SALES

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

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

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

Show the answer

Correct answer: 250

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

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 62 of 100

2023 Test B Math, question 10

Question

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

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

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

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

Correct answer: B

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

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

If 75% of the students scored less than 82, then 25% scored 82 points or more. Since the third quartile (Q3) indicates 34\frac{3}{4} (75%) of the data is to the left and 14\frac{1}{4} (25%) of the data is to the right, it is reasonable that Q3 would be at or about 82.

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

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

Question analysis

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

NYS standards
  • 7.SP.1 box plots and interquartile range
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students usually read values off a box plot. This item runs the other way, turning a percent statement into a quartile and checking the whiskers against two more facts, but the content stays at grade 7.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 2, 45 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns three sentences into three features of a box plot, and a plot that matches two of them is easy to accept. The upper-quartile fact is the conceptual one: students who do not know that the box’s right edge marks 75 percent put 82 at the end of the whisker. The other two facts are quick visual checks, but only if every plot is actually checked.

  1. Every choice must be checked moderate Choosing the answer

    Each wrong plot fails a different fact, so all four must be tested against all three bullets: the left whisker against 54, the right whisker against 100, and the box’s right edge against 82. (Embretson & Wetzel, 1987)

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

    The axis is labeled every 5 points, and the deciding features fall between labels: one left whisker starts near 51 instead of 54, and box edges sit near 82.5. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  3. Something you must already know minor Working it out

    Nothing explains that “75% of the students scored less than 82” locates the upper quartile, the right edge of the box, rather than the highest score. (Anderson & Krathwohl, 2001)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    24 students took a test. You know three facts. Nobody scored 100. 75% scored less than 82. The lowest score was 54. You need the box plot that fits all three facts.

  2. Turn each fact into a feature

    Match each fact to a part of the box plot.

    • The lowest score is 54, so the left whisker starts at 54.
    • Nobody scored 100, so the right whisker ends before 100.
    • 75% is three quarters. Three quarters of the scores are below Q3. So the box should end at about 82.

  3. Read the number line

    The ticks are 5 points apart: 50, 55, 60 and so on. A score of 54 sits just left of the 55 tick. A score of 82 sits a little less than halfway from 80 to 85. Look closely, because these points fall between labels.

  4. Check plot A

    In plot A, the right whisker ends near 82, and the box ends near 72. That would make 82 the highest score. Watch out: 75% below 82 does not make 82 the top score. It tells you where the box ends. So A is out.

  5. Check plot C

    In plot C, the left whisker starts near 51, just right of the 50 tick. But the lowest score was 54. So C is out.

  6. Check plot D

    Plot D starts at 54, and its box ends near 82. But its right whisker reaches 100. That means someone scored 100, and nobody did. Watch out: D fits two facts, so it is easy to stop early. Check all three. So D is out.

  7. Check plot B and answer

    Plot B starts at 54. Its box ends near 82. Its right whisker stops near 98, below 100. It fits all three facts. So the answer is B.

Check your answer

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

More questions like this

The original DOE question
Question 67 as printed on page 194 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 194, question 67. Open page 194 of the handbook (PDF)

Question 63 of 100

2023 Test B Math, question 12

Question

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

  • 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 64 of 100

2023 Test B Math, question 14

Question

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

  • A92
  • B90
  • C48
  • D45
Show the answer

Correct answer: D

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

n=3+2xn = 3 + 2x

93=3+2x93 = 3 + 2x

90=2x90 = 2x

45=x45 = x

Question analysis

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

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

Why it is hard

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

  1. Turning words into math moderate Understanding the question

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

  2. No obvious method minor Working it out

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

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

  2. Write the rule as an equation

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

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

    93=2x+393 = 2x + 3

  3. Undo the adding of 3

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

    93−3=2x93 - 3 = 2x

    90=2x90 = 2x

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

  4. Undo the doubling

    Now divide both sides by 2.

    x=90÷2=45x = 90 \div 2 = 45

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

Check your answer

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

More questions like this

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

Question 65 of 100

2023 Test B Math, question 19

Question

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

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

Correct answer: A

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

2x+3x=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 66 of 100

2023 Test B Math, question 29

Question

Which number could be the sum of 4 consecutive integers?

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

Correct answer: B

The sum of 4 consecutive integers can be expressed as:

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

=4n+6=4n+6

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

Option A:

4n+6=−124n+6=-12

4n=−184n=-18

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

Option B:

4n+6=−104n+6=-10

4n=−164n=-16

n=−4n=-4 is an integer.

Option C:

4n+6=−84n+6=-8

4n=−144n=-14

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

Option D:

4n+6=−44n+6=-4

4n=−104n=-10

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

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

Question analysis

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

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

Why it is hard

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

  1. Every choice must be checked moderate Choosing the answer

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

  2. No obvious method minor Working it out

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

  3. Error-prone arithmetic minor Working it out

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

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

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

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

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

  2. Write the sum with a letter

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

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

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

  3. Test each choice

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

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

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

  4. Name the four integers

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

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

    So the answer is B, −10.

Check your answer

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

More questions like this

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

Question 67 of 100

2023 Test B Math, question 40

Question

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

  • A45°
  • B67.5°
  • C75°
  • D135°
Show the answer

Correct answer: B

Let A represent the measure of angle A, and let B represent the measure of angle B, in degrees. Since the angles are complementary, A+B=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 68 of 100

2023 Test B Math, question 44

Question

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

  • 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 69 of 100

2022 Test A Math, question 2

Question

A homeowner has a budget of $135 for the installation of a light fixture. An electrician charges $60 per hour plus a one-time trip charge of $45. What is the greatest amount of time, in hours, the electrician can work on the installation while staying within the homeowner’s budget? Write your answer as a decimal.

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

Show the answer

Correct answer: 1.5

Let x represent the number of hours the electrician can work.

The cost for the number of hours worked is $60 times the number of hours worked, x. The sum of the one-time trip charge ($45) and the cost of the number of hours worked (60x60x) must be less than or equal to $135. Express this relationship as an inequality:

45+60x≤13545 + 60x \leq 135

Solve the inequality for x:

45+60x≤13545 + 60x \leq 135

60x≤9060x \leq 90

x≤9060=96=32x \leq \frac{90}{60} = \frac{9}{6} = \frac{3}{2}

Written as a decimal, 32=1.5\frac{3}{2} = 1.5.

Question analysis

A homeowner has $135 to spend, and an electrician charges $60 an hour plus a one-time $45 trip charge. This grid-in asks for the greatest number of hours the electrician can work within the budget, written as a decimal. Students write and solve a two-step inequality, and turning a length of time into decimal hours is the last trap.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Solving a word problem that leads to an inequality of the form px + q ≤ r is the grade 7 standard itself. The numbers are friendly; the only twists are treating the budget as a limit and writing the time in decimal hours.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 8, 52 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light, $135 − $45 = $90 and then $90 ÷ $60, so the question is decided by the setup and by the form of the entry. The trip charge is paid once, not every hour, and has to come off the budget before dividing. The time works out to an hour and a half, and students who think in hours and minutes may write it the way a clock shows it rather than as a decimal.

  1. Turning words into math moderate Understanding the question

    The stem never says “at most”; the limit comes from “a budget of $135”, and the one-time $45 trip charge is added once while the $60 multiplies the hours, giving 45 + 60x ≤ 135. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The grid-in demands a decimal, so an hour and a half must be entered as 1.5, not as 1:30 or 90 minutes, with no choices to confirm the form. (Enright, Morley & Sheehan, 2002)

Strategy: Take the one-time trip charge off the budget first, divide what is left by the hourly charge, and give the time in hours as a decimal rather than in hours and minutes.

Traps

  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.
  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • 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

Total cost with a one-time charge
C=f+r×hC = f + r \times h
C is the total cost. f is the one-time charge, paid once no matter how long the job takes. r is the rate, the cost for each hour. h is the number of hours. The hourly part grows with time, but the one-time charge does not.
Keeping an inequality balanced
if a≤b, then a−c≤b−c and a÷c≤b÷c\text{if } a \leq b \text{, then } a - c \leq b - c \text{ and } a \div c \leq b \div c
The sign ≤ says the left side is less than or equal to the right side. a and b are the two sides. c is a positive number. Subtracting c from both sides, or dividing both sides by c, keeps the statement true.

  1. Sort out the question

    The homeowner can spend $135 at most. That is the budget. The electrician has a one-time trip charge of $45. One-time means it is paid once, not every hour. On top of that, the electrician charges $60 for each hour. You need the greatest number of hours that keeps the total at $135 or less. The question asks for a decimal.

  2. Write an inequality

    Let h be the number of hours. Use the total cost formula with f = 45 and r = 60. The total must be less than or equal to $135. An inequality is like an equation, but it uses a sign such as ≤ instead of =.

    C=f+r×hC = f + r \times h

    45+60×h≤13545 + 60 \times h \leq 135

  3. Take off the trip charge

    Subtract 45 from both sides. This keeps the inequality balanced.

    45+60×h−45≤135−4545 + 60 \times h - 45 \leq 135 - 45

    60×h≤9060 \times h \leq 90

    So $90 is left to pay for the hours. Watch out: don’t divide the whole $135 by $60. That leaves out the trip charge, and you would grid 2.25.

  4. Divide by the hourly rate

    60 × h means 60 times h. Divide both sides by 60 to get h by itself.

    h≤9060=32=1.5h \leq \frac{90}{60} = \frac{3}{2} = 1.5

    The electrician can work at most 1.5 hours. Watch out: 1.5 hours is 1 hour 30 minutes, but don’t grid 1.3. Thirty minutes is half an hour, 0.5, not 0.3. Also don’t grid 90. That is dollars, not hours. So you would grid 1.5.

Check your answer

Put 1.5 hours back in: 45+60×1.5=45+90=13545 + 60 \times 1.5 = 45 + 90 = 135. That is exactly the budget. Now try 2 hours: 45+60×2=16545 + 60 \times 2 = 165. That is over $135. So 1.5 hours is the most.

More questions like this

The original DOE question
Question 59 as printed on page 79 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 79, question 59. Open page 79 of the handbook (PDF)

Question 70 of 100

2022 Test A Math, question 36

Question

When the three-digit whole number 27 ?  is divided by 3 or by 4, the remainder is zero. What is the missing digit?

  • A0
  • B2
  • C3
  • D6
Show the answer

Correct answer: D

Each of the options 0, 2, 3, and 6 must be considered:

Option A: In the case of 0, the three-digit whole number is 270. Divide 270 by both 3 and 4:

270÷3=90 (or 90R0)270 \div 3 = 90 \text{ (or } 90R0\text{)}

270÷4=6712 (or 67R2)270 \div 4 = 67\frac{1}{2} \text{ (or } 67R2\text{)}

Dividing 270 by 3 results in a remainder of 0, but dividing 270 by 4 results in a remainder of 2, so 0 cannot be the missing digit.

Option B: In the case of 2, the three-digit whole number is 272. Divide 272 by both 3 and 4:

272÷3=9023 (or 90R2)272 \div 3 = 90\frac{2}{3} \text{ (or } 90R2\text{)}

272÷4=68 (or 68R0)272 \div 4 = 68 \text{ (or } 68R0\text{)}

Dividing 272 by 4 results in a remainder of 0, but dividing 272 by 3 results in a remainder of 2, so 2 cannot be the missing digit.

Option C: In the case of 3, the three-digit whole number is 273. Divide 273 by both 3 and 4:

273÷3=91 (or 91R0)273 \div 3 = 91 \text{ (or } 91R0\text{)}

273÷4=6814 (or 68R1)273 \div 4 = 68\frac{1}{4} \text{ (or } 68R1\text{)}

Dividing 273 by 3 results in a remainder of 0, but dividing 273 by 4 results in a remainder of 1, so 3 cannot be the missing digit.

Option D: In the case of 6, the three-digit whole number is 276. Divide 276 by both 3 and 4:

276÷3=92 (or 92R0)276 \div 3 = 92 \text{ (or } 92R0\text{)}

276÷4=69 (or 69R0)276 \div 4 = 69 \text{ (or } 69R0\text{)}

Dividing 276 by 3 results in a remainder of 0, and dividing 276 by 4 results in a remainder of 0, so 6 is the missing digit.

Question analysis

A three-digit number begins with 27 and is missing its last digit, and it divides evenly by both 3 and 4. Students find the missing digit. It tests factors, multiples and divisibility rules; a digit that works for one divisor may fail the other, so both tests must be applied.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Divisibility by 3 and by 4 is grade 6 number sense. Testing four possible digits against two conditions is quick, much like a grade 6 exercise, as long as both divisors are actually checked.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 5, 19 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every candidate passes at least one of the two tests, which makes the question a matter of finishing the checks rather than finding a method. Students who test only by 3, or only by 4, find a fitting number and stop. Seeing that a number divisible by both 3 and 4 is divisible by 12, and that 276 = 12 × 23, settles it in one step.

  1. Turning words into math minor Understanding the question

    The words “divided by 3 or by 4, the remainder is zero” mean that both divisions come out even, though or can be read as either one being enough. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Every choice must be checked minor Choosing the answer

    Each candidate, 270, 272, 273 and 276, has to be tested against two divisors, and three of them pass one test, so stopping after one check is costly. (Embretson & Wetzel, 1987)

  3. Something you must already know minor Working it out

    The divisibility tests are not given: a number is divisible by 3 when its digit sum is, and by 4 when the number formed by its last two digits is. (Anderson & Krathwohl, 2001)

Strategy: Use the divisibility tests: a number is divisible by 4 when its last two digits are, and by 3 when its digit sum is. Keep only a digit that passes both.

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

Test for dividing by 3
digit sum divides by 3→number divides by 3\text{digit sum divides by 3} \rightarrow \text{number divides by 3}
The digit sum is what you get when you add up a number’s digits. Divides by 3 means it divides evenly, with a remainder of 0. If the digit sum divides evenly by 3, so does the whole number. Use it to test a big number quickly.
Test for dividing by 4
last two digits divide by 4→number divides by 4\text{last two digits divide by 4} \rightarrow \text{number divides by 4}
Look only at the number made by the last two digits. If it divides evenly by 4, so does the whole number. This works because every 100 divides evenly by 4, so only the last two digits matter.

  1. Sort out the question

    The number is 27 followed by a missing digit. Remainder zero means the division comes out even, with nothing left over. The remainder is zero when the number is “divided by 3 or by 4.” That means both divisions come out even. So the number must pass both tests. The choices give four digits to try: 0, 2, 3 and 6.

  2. Test each number for 3

    Add the digits of each number.

    NumberDigit sumDivides by 3?
    2702+7+0=92 + 7 + 0 = 9yes
    2722+7+2=112 + 7 + 2 = 11no
    2732+7+3=122 + 7 + 3 = 12yes
    2762+7+6=152 + 7 + 6 = 15yes

    So 270, 273 and 276 pass. 272 fails.

  3. Test each number for 4

    Look at the last two digits: 70, 72, 73 and 76.

    70÷4=17 remainder 272÷4=1870 \div 4 = 17 \text{ remainder } 2 \qquad 72 \div 4 = 18

    73÷4=18 remainder 176÷4=1973 \div 4 = 18 \text{ remainder } 1 \qquad 76 \div 4 = 19

    So 272 and 276 pass. 270 and 273 fail.

  4. Keep the digit that passes both

    Only 276 passes both tests. Check by dividing:

    276÷3=92276÷4=69276 \div 3 = 92 \qquad 276 \div 4 = 69

    Watch out: 0 and 3 pass only the test for 3, so A and C fail. 2 passes only the test for 4, so B fails. And 3 is not right just because the question says 3. So the answer is D, 6.

Check your answer

A number that divides evenly by both 3 and 4 also divides evenly by 12, since 3×4=123 \times 4 = 12. And 12×23=27612 \times 23 = 276. So 276 is a multiple of 12. None of 270, 272 or 273 is.

More questions like this

The original DOE question
Question 93 as printed on page 89 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 89, question 93. Open page 89 of the handbook (PDF)

Question 71 of 100

2022 Test B Math, question 12

Question

Julie is at a clothing store. She wants to buy a bag for $18.75 and some T-shirts for $9.25 each. She can spend no more than $50.00. Which inequality can be used to find the number of T-shirts, x, Julie can purchase?

  • A9.25+18.75x≤50.009.25 + 18.75x \leq 50.00
  • B9.25+18.75x≥50.009.25 + 18.75x \geq 50.00
  • C18.75+9.25x≤50.0018.75 + 9.25x \leq 50.00
  • D18.75+9.25x≥50.0018.75 + 9.25x \geq 50.00
Show the answer

Correct answer: C

The question uses the phrase “no more than,” which translates to “less than or equal to” (≤).

Since T-shirts cost $9.25 each and Julie bought x T-shirts, the total cost just for the T-shirts is found by multiplying $9.25 by x, or 9.25x9.25x.

Combining the cost for one bag ($18.75) plus the cost for x T-shirts, the inequality is 18.75+9.25x≤50.0018.75 + 9.25x \leq 50.00.

Question analysis

Julie wants to buy one bag and some T-shirts at a clothing store, with a price for each, and can spend no more than $50.00. Students choose the inequality for the number of T-shirts, x, she can buy. It tests writing an inequality from a word problem: the choices swap the two prices and the direction of the sign.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing an inequality for a budget with one fixed cost and a cost per item is standard grade 7 work. With choices given, the task is to match each price to its role and the phrase to its symbol.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 3, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The choices form a grid of two decisions: which price goes with x, and which way the symbol points. The prices appear in the order they are used, bag first, so the expression is the easier half and most lost points come from the symbol. Pricing the bag with 2 T-shirts, $37.25, and with 3, $46.50, shows that 9.25 is the amount that repeats with each shirt.

  1. Turning words into math moderate Understanding the question

    The $18.75 bag is bought once and the $9.25 is paid for each of x T-shirts, and “no more than $50.00” has to become ≤, which allows spending exactly $50.00. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    The choices 18.75 + 9.25x ≤ 50.00 and 18.75 + 9.25x ≥ 50.00 match term for term and differ only in the direction of the symbol. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Price one bag and 2 T-shirts by hand to see which price is multiplied by x, then translate “no more than” into an inequality symbol.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost
total=one-time cost+price each×number bought\text{total} = \text{one-time cost} + \text{price each} \times \text{number bought}
The one-time cost is paid once, like one bag. The price each is paid again for every item, so it is multiplied by the number bought. Add them to get the total. Use it when one cost happens once and another repeats.
Spending limit
total≤limit\text{total} \leq \text{limit}
The total is what Julie spends, and the limit is $50.00. The sign ≤ means less than or equal to. So the total can be under the limit or exactly equal to it, but never over. That is what the words “no more than” mean.

  1. Sort out the question

    Julie buys one bag for $18.75. She also buys some T-shirts for $9.25 each. x is the number of T-shirts. She can spend no more than $50.00. You need the right inequality. An inequality is a math sentence that uses a sign like ≤ instead of =.

  2. Write the cost of the T-shirts

    Each T-shirt costs $9.25. For x T-shirts, she pays $9.25 x times.

    9.25×x=9.25x9.25 \times x = 9.25x

    The bag is bought only once. So its $18.75 is not multiplied by x.

    Watch out: putting x on 18.75 prices x bags and one T-shirt. That gives choices A and B.

  3. Add to get the total

    Use the total cost formula. The bag is the one-time cost.

    total=18.75+9.25x\text{total} = 18.75 + 9.25x

    Test it with 2 T-shirts: 18.75+9.25×2=18.75+18.50=37.2518.75 + 9.25 \times 2 = 18.75 + 18.50 = 37.25 dollars. That is one bag and two shirts, as it should be.

  4. Choose the sign

    “She can spend no more than $50.00” means she can spend less than $50.00, or exactly $50.00. That is less than or equal to, the sign ≤.

    18.75+9.25x≤50.0018.75 + 9.25x \leq 50.00

    Watch out: ≥ means greater than or equal to. Choice D says she must spend $50.00 or more, the opposite. So the answer is C, 18.75+9.25x≤50.0018.75 + 9.25x \leq 50.00.

Check your answer

Test whole numbers of T-shirts. With 3, the total is 18.75 + 27.75 = 46.50 dollars. That is not over $50.00. With 4, it is 18.75 + 37.00 = 55.75 dollars, too much. Choice C allows 3 T-shirts and rules out 4, which fits.

More questions like this

The original DOE question
Question 69 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 2023 Admissions, page 200, question 69. Open page 200 of the handbook (PDF)

Question 72 of 100

2022 Test B Math, question 49

Question

A machine part is supposed to be 0.38 millimeters thick. The manufacturer allows a variation of −2.5% to +2.5%. Which interval shows the range of acceptable thicknesses, in millimeters, for the part?

  • A0.285–0.475
  • B0.355–0.405
  • C0.3705–0.3895
  • D0.3724–0.3876
Show the answer

Correct answer: C

A variation of −2.5% to +2.5% means that the thickness of the machine part can be from 2.5% less than the standard of 0.38 mm to 2.5% greater than 0.38 mm.

The least allowed measure is

0.38−0.025(0.38)=0.3705.0.38-0.025(0.38)=0.3705.

The greatest allowed measure is

0.38+0.025(0.38)=0.3895.0.38+0.025(0.38)=0.3895.

Therefore, the allowable range of thicknesses is 0.3705 mm to 0.3895 mm.

Question analysis

A machine part should be 0.38 millimeters thick, and the manufacturer accepts a variation from −2.5% to +2.5%. Students choose the interval of acceptable thicknesses. It tests finding a percent of a decimal and applying it as a decrease and an increase; the percent must be turned into millimeters first, and the four-decimal-place choices need exact work.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3c percent as a rate per 100
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Taking a percent of a number and adding or subtracting it is grade 7 percent work. The decimal percent and results to four decimal places make the arithmetic fussier than usual, so careful work matters more than the idea.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 9, 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 question rewards turning the percent into millimeters before anything else. Students who take ±2.5% as ±0.025 mm find a tidy interval, 0.355 to 0.405, and the other listed intervals come from a misplaced decimal point or from 2% in place of 2.5%. Because the right interval and its near twin differ only from the third decimal place, the product 0.38 × 0.025 has to be exact; multiplying 0.38 by 0.975 and by 1.025 is a quick double-check.

  1. Error-prone arithmetic moderate Working it out

    Two and a half percent of 0.38 is 0.025 × 0.38 = 0.0095, a product of two decimals with four decimal places, and the limits 0.3705 and 0.3895 need the same care. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math minor Understanding the question

    “A variation of −2.5% to +2.5%” has to become 2.5% of 0.38 millimeters, a length, before it is subtracted for the least thickness and added for the greatest. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. A very close wrong answer minor Choosing the answer

    The intervals 0.3705–0.3895 and 0.3724–0.3876 both sit about 0.01 mm either side of 0.38, so an estimate cannot separate them and only the exact 0.0095 decides. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write 2.5% as a decimal, multiply it by 0.38 to find the allowed change in millimeters, then subtract it for the least thickness and add it for the greatest.

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.
  • 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

Percent of a number
part=percent100×whole\text{part} = \frac{\text{percent}}{100} \times \text{whole}
The whole is the amount the percent is taken of. The percent tells how many out of every 100. The part is the piece you get. Dividing the percent by 100 turns it into a decimal, so you can multiply.

  1. Sort out the question

    The part should be 0.38 millimeters thick. It may be off by up to 2.5%, thinner or thicker. You need the thinnest and thickest sizes allowed, in millimeters.

    The 2.5% is a percent of 0.38 mm. So first find how many millimeters 2.5% is.

  2. Write 2.5% as a decimal

    Percent means out of 100. So divide by 100, which moves the decimal point two places left.

    2.5%=2.5100=0.0252.5\% = \frac{2.5}{100} = 0.025

    Watch out: 2.5% is 0.025, not 0.25. Using 0.25 makes the interval ten times too wide, choice A.

  3. Find 2.5% of 0.38

    part=0.025×0.38=0.0095\text{part} = 0.025 \times 0.38 = 0.0095

    First, 25 × 38 = 950. The two numbers have 3 + 2 = 5 decimal places in all. So the answer has 5 decimal places: 0.00950, or 0.0095. The size may change by 0.0095 millimeters.

    Watch out: adding and subtracting 0.025 mm itself gives 0.355 to 0.405, choice B.

  4. Find the two limits

    Subtract the change for the thinnest size. Add it for the thickest.

    0.38−0.0095=0.37050.38 - 0.0095 = 0.3705

    0.38+0.0095=0.38950.38 + 0.0095 = 0.3895

    The part may be from 0.3705 to 0.3895 millimeters thick.

    Watch out: using 2% instead of 2.5% gives 0.3724 to 0.3876, choice D. So the answer is C, 0.3705–0.3895.

Check your answer

A part 2.5% thinner is 97.5% of 0.38: 0.975×0.38=0.37050.975 \times 0.38 = 0.3705. A part 2.5% thicker is 102.5% of it: 1.025×0.38=0.38951.025 \times 0.38 = 0.3895. Both match. Also, 0.38 sits exactly in the middle of the interval.

More questions like this

The original DOE question
Question 106 as printed on page 211 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 211, question 106. Open page 211 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 73 of 100

2019 Test A Math, question 51

Question

The greatest common factor of 24 and x is 8. How many possible values for x are greater than 10 and less than 70?

  • A8
  • B7
  • C6
  • D5
Show the answer

Correct answer: D

First find the multiples of 8 that fall between 10 and 70:

16, 24, 32, 40, 48, 56, 64

Since the greatest common factor of 24 and x is 8, x cannot be 24 or 48 (otherwise the GCF would be 24, not 8).

There are 5 remaining values: 16, 32, 40, 56, and 64.

Question analysis

This number properties question gives the greatest common factor of 24 and an unknown number x as 8, and asks how many possible values of x are greater than 10 and less than 70. It runs the usual GCF task in reverse. Each candidate has to meet the range and the GCF condition exactly, so a quick count is risky.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Greatest common factors are grade 6 content, usually found for two given numbers. Here the factor is given and every number producing it in a range must be found and checked, a longer and less familiar task than grade 7 classwork.
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, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question runs the familiar GCF task backward, and the method has to be built: list the multiples of 8 in range, then strike those whose GCF with 24 is larger than 8. The strike-out is where points are lost, because 24 is plainly out but 48, a multiple of 24, looks like any other multiple of 8. Each wrong count is one step of that process left undone, so a student who skips a single check still finds a matching choice.

  1. No obvious method moderate Working it out

    The usual task finds the GCF of two given numbers; here the GCF is given and x is not, so students must see that x is a multiple of 8 and then test each candidate. (Schoenfeld, 1985)

  2. Much to keep track of moderate Working it out

    A candidate must be a multiple of 8, lie strictly between 10 and 70, and share no larger factor with 24, a test that 24 and 48 fail with a GCF of 24. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. A very close wrong answer minor Choosing the answer

    The four choices are the consecutive counts 5 to 8, so keeping a single bad value, most often 48, still lands on a listed answer that looks right. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: List the multiples of 8 between 10 and 70, then test each one: its greatest common factor with 24 must be exactly 8, so cross out any that 24 divides.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greatest common factor
GCF=product of the prime factors both numbers share\text{GCF} = \text{product of the prime factors both numbers share}
A factor divides a number with nothing left over. The greatest common factor, or GCF, is the biggest number that is a factor of both numbers. A prime number has only two factors, 1 and itself, like 2 or 3. Write each number as primes multiplied together, then multiply the primes they share.

  1. Sort out the question

    A factor of a number divides it with nothing left over. The greatest common factor (GCF) of two numbers is the biggest number that is a factor of both. Here the GCF of 24 and x is exactly 8. You need to count the values x can have. “Greater than 10 and less than 70” means 10 and 70 themselves don’t count.

  2. List the multiples of 8

    8 is a factor of x, so x must be a multiple of 8. A multiple of 8 is 8 times a whole number. These multiples of 8 are greater than 10 and less than 70:

    16, 24, 32, 40, 48, 56, 64

    Leave out 8 itself, since it is less than 10. Counting 8 as well makes 8 numbers, choice A.

    Watch out: stopping here gives 7, choice B. Each of these has 8 as a factor, but the GCF must be exactly 8, not more.

  3. Break 24 into prime factors

    Write 24 as primes multiplied together.

    24=2×2×2×3=8×324 = 2 \times 2 \times 2 \times 3 = 8 \times 3

    Use the GCF formula. Every multiple of 8 shares the three 2s with 24. That shared part is 8. If x also has a factor of 3, it shares the 3 too.

    GCF=product of the prime factors both numbers share\text{GCF} = \text{product of the prime factors both numbers share}

    GCF=2×2×2×3=24\text{GCF} = 2 \times 2 \times 2 \times 3 = 24

    Then the GCF is 24, not 8. 24 has no other primes to share. So the GCF is exactly 8 when x is a multiple of 8 but not a multiple of 3.

  4. Test each number for a factor of 3

    A number is a multiple of 3 when its digits add up to a multiple of 3.

    xDigit sumMultiple of 3?GCF with 24
    167no8
    246yes24
    325no8
    404no8
    4812yes24
    5611no8
    6410no8

    Watch out: 48 is easy to miss. It is 24×224 \times 2, so its GCF with 24 is 24. Keeping 48 gives 6, choice C.

  5. Count the numbers left

    Cross out 24 and 48. Five numbers are left: 16, 32, 40, 56 and 64. Each one has a GCF of exactly 8 with 24. So the answer is D, 5.

Check your answer

Test one number the long way. The factors of 32 are 1, 2, 4, 8, 16 and 32. The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. The greatest number in both lists is 8. For 48, the number 24 is in both lists, so 48 is out.

More questions like this

The original DOE question
Question 108 as printed on page 106 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2019–2020, page 106, question 108. Open page 106 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 74 of 100

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 75 of 100

2018 Test A Math, question 5

Question

The sum of two consecutive integers is −15. If 1 is added to the smaller integer and 2 is subtracted from the larger integer, what is the product of the two resulting integers?

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

Show the answer

Correct answer: 63

If x is the smaller consecutive integer, then

x+1x+1

is the larger consecutive integer. Use their sum (−15) to find x:

x+(x+1)=−15x+(x+1)=-15

2x+1=−152x+1=-15

2x=−162x=-16

x=−8x=-8

The two consecutive integers are −8 and −7. One is added to the smaller integer:

−8+1=−7-8+1=-7

and 2 is subtracted from the larger integer:

−7−2=−9-7-2=-9

Find the product:

(−7)(−9)=63(-7)(-9)=63

Question analysis

A grid-in problem about two consecutive integers whose sum is −15. Students find the two integers, add 1 to the smaller and subtract 2 from the larger, then multiply the results. Deciding which negative integer is the smaller one, and keeping signs straight through every step, are what make it tricky.

NYS standards
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 A typical grade 7 task solves one two-step equation or multiplies signed numbers. This links both: set up a consecutive-integer equation, solve it with negatives, adjust each integer, and multiply, with no choices to check against.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 9, 30 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is routine, yet the question collects most of the ways students lose points with negatives. The crucial judgment is which of −8 and −7 is the smaller; a student who picks −7 still produces a tidy, wrong product. Gridding the product of the original pair, or giving a negative product, are the other usual losses. Writing each intermediate pair down, rather than holding them in mind, protects against all three.

  1. Error-prone arithmetic moderate Working it out

    Every number in play is negative: −8 and −7, then −7 and −9, and a product of two negatives that must come out positive. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Turning words into math moderate Understanding the question

    "1 is added to the smaller integer and 2 is subtracted from the larger" depends on knowing that −8, not −7, is the smaller, since it lies farther below zero. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Many steps moderate Working it out

    Four linked steps: set up x + (x + 1) = −15, solve for x, adjust each integer, then multiply; an error early in the chain carries through to the product. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Halve the sum to see which two integers sit on either side of it, remember that the smaller negative is farther from zero, then adjust each and multiply.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Consecutive integers
n,n+1n, \; n + 1
Integers are whole numbers and their opposites, like −2, −1, 0, 1 and 2. Consecutive integers come one right after another. If n is the smaller one, the next one is n + 1, because each step up the number line adds 1.
Multiplying two negative numbers
(−a)×(−b)=a×b(-a) \times (-b) = a \times b
a and b are positive numbers, so −a and −b are negative. When you multiply two negative numbers, the answer is positive. The two minus signs cancel out.

  1. Sort out the question

    Two consecutive integers add up to −15. Integers are whole numbers and their negatives. Consecutive means one comes right after the other, like 4 and 5. You need to do three things:

    1. Find the two integers.
    2. Add 1 to the smaller one. Subtract 2 from the larger one.
    3. Find the product of the two new numbers. The product is the answer when you multiply.

  2. Find the two integers

    Call the smaller integer n. Then the larger one is n + 1. Their sum is −15.

    n+(n+1)=−15n + (n + 1) = -15

    2n+1=−152n + 1 = -15

    Subtract 1 from both sides. Then divide both sides by 2.

    2n=−162n = -16

    n=−8n = -8

    So the integers are −8 and −7.

  3. Decide which one is smaller

    On a number line, −8 is to the left of −7. Numbers to the left are smaller. So −8 is the smaller integer, and −7 is the larger one. Watch out: 8 is bigger than 7, but −8 is smaller than −7. Mixing them up gives −6 and −10, and the wrong answer 60.

  4. Change each integer

    Add 1 to the smaller integer.

    −8+1=−7-8 + 1 = -7

    Subtract 2 from the larger integer.

    −7−2=−9-7 - 2 = -9

    The two new numbers are −7 and −9.

  5. Multiply the new numbers

    Both numbers are negative, so their product is positive.

    (−7)×(−9)=63(-7) \times (-9) = 63

    Watch out: the answer is not −63, because two negatives make a positive. Also, don’t multiply the first pair, −8 and −7. That gives 56 and skips the changes. So you would grid 63.

Check your answer

Check the pair: −8+(−7)=−15-8 + (-7) = -15, and they are one apart. Half of −15 is −7.5, and −8 and −7 sit on either side of it. Then 7×9=637 \times 9 = 63, and the sign is positive.

More questions like this

The original DOE question
Question 62 as printed on page 81 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 62. Open page 81 of the handbook (PDF)

Question 76 of 100

2018 Test A Math, question 9

Question

Jenny starts a game with twice as many marbles as Keiko. Jenny gives Keiko 5 marbles, but she still has 10 more than Keiko. How many marbles did Jenny have to start with?

  • A25
  • B30
  • C35
  • D40
Show the answer

Correct answer: D

Set up some equations. Jenny (J) has twice as many marbles as Keiko (K):

J=2KJ=2K

Jenny gives Keiko 5 marbles, so now they each have: J − 5 and K + 5 marbles. Jenny still has 10 more than Keiko:

J−5=(K+5)+10J-5=(K+5)+10

To find how many marbles Jenny had to start with, solve J = 2K for K and substitute that into the second equation:

In equation J = 2K, solve for K:

K=J2K=\frac{J}{2}

Substitute J2\frac{J}{2} in for K.

J−5=(K+5)+10J-5=(K+5)+10

J−5=(J2+5)+10J-5=(\frac{J}{2}+5)+10

J−5=J2+15J-5=\frac{J}{2}+15

J=J2+20J=\frac{J}{2}+20

J2=20\frac{J}{2}=20

J=40J=40 marbles

Question analysis

A two-variable word problem about marbles: Jenny starts with twice as many as Keiko, gives Keiko 5, and still has 10 more. Students set up equations for before and after the trade and solve for Jenny's starting number. The trade changes both players' counts, and the question asks about the start, not the end.

NYS standards
  • 8.EE.8c systems in real-world problems
  • 7.EE.4 equations and inequalities from problems
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve one equation in one unknown. This needs two related unknowns and a before-and-after change, which leads to a system of equations or a variable on both sides, both grade 8 skills.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 4, 31 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Most of the difficulty is in setting up the equations, not solving them. The gift of 5 marbles changes both counts, so the gap between the girls shrinks by 10, not 5, and students who take the marbles from Jenny without giving them to Keiko reach a consistent but wrong answer. Two other choices are true counts from after the trade, so finishing the algebra is not the same as answering the question. Testing the choices backward is a fast, safe alternative.

  1. Turning words into math major Understanding the question

    Three relations hide in two sentences: "twice as many", the gift of 5 that lowers Jenny's count and raises Keiko's, and "still has 10 more", which compares the counts after the gift. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of moderate Working it out

    Each girl has a count before and after the trade, four quantities in all, and the question asks for only one of them, Jenny's at the start. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Work backward from the choices: halve each one for Keiko's start, move 5 marbles from Jenny to Keiko, and check whether Jenny ends with exactly 10 more.

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

Twice as many
J=2KJ = 2K
J is the number of marbles Jenny starts with, and K is the number Keiko starts with. Twice as many means 2 times as many, so J equals 2 times K. Because J and 2K are equal, you can write 2K anywhere you see J.
Keeping an equation balanced
if a=b, then a−c=b−c\text{if } a = b \text{, then } a - c = b - c
a and b are the two sides of an equation, and c is any number or term. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. The same is true if you add the same thing, or divide both sides by the same number that is not zero.

  1. Sort out the question

    Let K be Keiko’s marbles at the start. Jenny starts with twice as many, so she has 2K. Then Jenny gives Keiko 5 marbles. After that, “she still has 10 more than Keiko”. You need Jenny’s count at the start, before the gift.

  2. Write the counts after the gift

    The gift changes both counts. Jenny loses 5, and Keiko gains 5.

    • Jenny now has 2K−52K - 5.
    • Keiko now has K+5K + 5.

    Jenny has 10 more than Keiko. So Jenny’s count equals Keiko’s count plus 10.

    2K−5=(K+5)+102K - 5 = (K + 5) + 10

    Watch out: Keiko gains the 5 marbles too. Taking 5 from Jenny but not adding 5 to Keiko leads to 30, choice B.

  3. Solve for K

    Add the numbers on the right side first.

    2K−5=K+152K - 5 = K + 15

    Subtract K from both sides. Then add 5 to both sides. Each move keeps the equation balanced.

    K−5=15K - 5 = 15

    K=20K = 20

    Keiko started with 20 marbles.

  4. Find Jenny’s starting count

    Jenny started with twice as many as Keiko.

    J=2K=2×20=40J = 2K = 2 \times 20 = 40

    Watch out: 25 is Keiko’s count after the gift, choice A. 35 is Jenny’s count after the gift, choice C. The question asks about the start. So the answer is D, 40.

Check your answer

Start: Jenny has 40 and Keiko has 20, and 40 is twice 20. After the gift, Jenny has 35 and Keiko has 25. 35−25=1035 - 25 = 10, so Jenny still has 10 more. Every fact fits.

More questions like this

The original DOE question
Question 66 as printed on page 82 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 82, question 66. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 77 of 100

2018 Test A Math, question 11

Question

PEOPLE PER VEHICLE AT CHECKPOINT

Number of People in Vehicle Percent of Vehicles
1 40%
2 35%
3 15%
4 7%
5 or more 3%

A researcher recorded the number of people in each vehicle that passed through a checkpoint. The table above shows the percent distribution for the 420 vehicles that passed through the checkpoint yesterday morning. How many of the 420 vehicles contained at least 3 people?

  • A42
  • B63
  • C105
  • D315
Show the answer

Correct answer: C

First, add the percentage of cars containing 3 people, 4 people, and 5 or more people:

15% + 7% + 3% = 25%

Thus, 25% of the cars contained at least 3 people, so use that to calculate the number of cars:

420 × 0.25 = 105 cars.

Question analysis

A data table question with percents: a table gives the percent of 420 vehicles carrying 1, 2, 3, 4, or 5 or more people. Students find how many vehicles held at least 3 people by adding the right rows and taking that percent of 420. Reading at least correctly decides which rows count.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Taking a percent of a number is grade 6 work, and grade 7 adds multistep percent problems. This two-step table problem sits at that level, with the phrase at least as the careful-reading point.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 7, 56 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is easy, since 25% of 420 is a quarter of it; the points are lost on the phrase "at least 3". Each common misreading, more than 3, exactly 3 or fewer than 3, gives a listed choice, so a wrong reading never leads to a dead end. Adding the percents before converting saves time and avoids three separate multiplications.

  1. Turning words into math moderate Understanding the question

    The bold "at least 3" has to become the rows for 3, 4 and "5 or more", taking in both the 3-person row and the open-ended last row. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The table gives percents, not counts, so the rows have to be added as percents and then taken of the 420 vehicles named only in the text below the table. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add the percents for every row that meets the condition first, then take that single percent of 420 instead of working row by row.

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).
  • 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

Percent of a number
x=p100×Wx = \frac{p}{100} \times W
W is the whole amount. p is the percent, and x is the part you want. Percent means out of 100, so p percent is the fraction p over 100. Multiplying the whole by that fraction gives the part.

  1. Sort out the question

    420 vehicles passed the checkpoint. The table gives the percent of vehicles with each number of people. You need how many vehicles had “at least 3” people. At least 3 means 3 or more. So count vehicles with 3 people, 4 people, and 5 or more.

  2. Pick the right rows

    Three rows fit: 3 people (15%), 4 people (7%), and 5 or more (3%). Watch out: at least 3 includes 3 itself. Leaving out the 3 row gives 42, choice A. Using only the 3 row gives 63, choice B.

  3. Add the percents first

    Adding the percents first saves work. Then you only multiply once.

    15%+7%+3%=25%15\% + 7\% + 3\% = 25\%

    So 25% of the vehicles had at least 3 people.

  4. Find 25% of 420

    Use the percent formula with p = 25 and W = 420.

    x=p100×Wx = \frac{p}{100} \times W

    x=25100×420=105x = \frac{25}{100} \times 420 = 105

    25% is one quarter, so this is the same as 420÷4420 \div 4. Watch out: 315 is the number of vehicles with fewer than 3 people, choice D. So the answer is C, 105.

Check your answer

The other rows, 1 and 2 people, make 40%+35%=75%40\% + 35\% = 75\%. 75% of 420 is 315. Then 105+315=420105 + 315 = 420, which is all the vehicles. The two groups fit together.

More questions like this

The original DOE question
Question 68 as printed on page 83 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 83, question 68. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 78 of 100

2018 Test A Math, question 12

Question

Square pyramid with apex R over the square base MNPQ; corner N is at the back, and its edges are dashed as hidden. Segment RS runs down face QRP from R to point S on base edge QP, with a right-angle mark at S.

In the pyramid above, each triangular face has the same area, and the base MNPQ is a square that measures 8 centimeters on each side. If the length of RS\overline{\mathrm{RS}} = 6 centimeters, what is the surface area of the pyramid excluding the base?

  • A48 sq cm
  • B96 sq cm
  • C128 sq cm
  • D160 sq cm
Show the answer

Correct answer: B

Line segment RS\overline{\mathrm{RS}} is the altitude, or height, of triangle QRP. The length of QP\overline{\mathrm{QP}} is 8 cm. Use that information to find the area of triangle QRP:

A=12bh=12(8)(6)=24A=\frac{1}{2}bh=\frac{1}{2}(8)(6)=24 sq cm.

There are 4 congruent triangles in the pyramid, so the surface area is

4 × 24 = 96 sq cm

Question analysis

A geometry question about the surface area of a square pyramid, leaving out its base. The base edges are 8 cm and segment RS, the height of one triangular face drawn from the top to a base edge, is 6 cm. Students find one face's area and multiply by the number of faces, reading excluding the base carefully.

NYS standards
  • 7.G.6 area, surface area, and volume
  • 6.G.4 nets and surface area
  • 6.G.1 area of polygons
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Surface area of a right pyramid from its faces is named in the grade 7 standards, and this is a direct two-step case. It matches typical grade 7 work, though the drawing's slant height must be read correctly.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 5, 49 words.
Students, last 30 days
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 RS is read as the height of a face, the computation is short: one triangle, then four of them. Points go to the figure and to one word. Students who count only the faces drawn in front, or who include the square base, reach listed choices, and the bold "excluding" is all that separates the lateral area from the total. The statement that every face has the same area is what makes multiplying by four valid.

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

    The drawing shows two lateral faces in front and two behind dashed edges, and RS, with its right-angle mark at S, is the height of face QRP, not of the pyramid. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

  2. Something you must already know minor Working it out

    The triangle area formula, one half times base times height, and the fact that a square pyramid has four lateral faces are both assumed, not given. (Anderson & Krathwohl, 2001)

Strategy: Find the area of one triangular face with one half times base times height, using RS as the height, then multiply by the four congruent faces.

Traps

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

Area of a triangle
A=12×b×hA = \frac{1}{2} \times b \times h
A is the area. b is the base, the side the triangle sits on. h is the height: the straight-up distance from the base to the top corner, 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 pyramid

    The pyramid sits on a square base, MNPQ. Each side of the square is 8 cm. Four triangle faces rise from the base and meet at the top point, R. Every face has the same area. Segment RS is 6 cm. You need the area of the four faces, “excluding the base”. Excluding means leaving it out.

  2. See what RS measures

    RS runs from R down face QRP to point S on edge QP. The small square at S marks a right angle. So RS is the height of triangle QRP. The base of that triangle is QP, one side of the square, so it is 8 cm.

  3. Find the area of one face

    Use the triangle formula with base 8 cm and height 6 cm.

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

    A=12×8×6=24A = \frac{1}{2} \times 8 \times 6 = 24

    One face has an area of 24 sq cm.

  4. Add the four faces

    All four faces are the same, so multiply by 4.

    4×24=964 \times 24 = 96

    Watch out: only two faces are drawn in front. The other two hide behind the dashed lines. Counting just two faces gives 48, choice A. Watch out: don’t add the base, 8×8=648 \times 8 = 64. That gives 160, choice D. So the answer is B, 96 sq cm.

Check your answer

Each face fits inside an 8 by 6 rectangle, with an area of 48 sq cm. The triangle is half of that, 24 sq cm, and four faces make 96 sq cm. That is less than 160, which would include the base.

More questions like this

The original DOE question
Question 69 as printed on page 83 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 83, question 69. Open page 83 of the handbook (PDF)

Question 79 of 100

2018 Test A Math, question 13

Question

The perimeter of a rectangle is 510 centimeters. The ratio of the length to the width is 3:2. What are the dimensions of this rectangle?

  • A150 cm by 105 cm
  • B153 cm by 102 cm
  • C158 cm by 97 cm
  • D165 cm by 90 cm
Show the answer

Correct answer: B

Let 2x2x = the width and 3x3x = the length.

Draw the rectangle to help visualize.

Rectangle with its top and bottom sides labeled 3x and its left and right sides labeled 2x.

Since 2 times width + 2 times length = perimeter, we get

2(2x)+2(3x)=5102(2x)+2(3x)=510

4x+6x=5104x+6x=510

10x=51010x=510

x=51x=51

2x=1022x=102 cm and 3x=1533x=153 cm

Question analysis

A rectangle word problem that combines perimeter with a ratio: the perimeter is 510 centimeters and the length and width are in the ratio 3 to 2. Students find the rectangle's dimensions. Every choice has the correct perimeter, so only the ratio separates them.

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 Perimeter and ratio are each familiar from grades 6 and 7, and combining them with a five-part split is a typical grade 7 multistep ratio problem. It asks about the same as classroom work.
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, 35 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is built so that the perimeter, the easier condition, rules nothing out: every choice adds to 255. A student who tests the choices against 510 alone will accept whichever comes first, and the first is 150 by 105. Solving with ratio parts, or checking each pair's ratio, avoids that; 3:2 means the length is exactly one and a half times the width.

  1. A very close wrong answer moderate Choosing the answer

    153 by 102 and 150 by 105 differ by just 3 cm in each side and give the same 510 cm perimeter; only the ratio, exactly 3:2 against 10:7, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Turning words into math minor Understanding the question

    "The ratio of the length to the width is 3:2" has to become 3 parts and 2 parts of the half-perimeter, 255, not of the full 510. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Halve the perimeter to get length plus width, split that sum into 5 equal parts, and give the length 3 parts and the width 2.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Perimeter of a rectangle
P=2×(l+w)P = 2 \times (l + w)
P is the perimeter, the distance all the way around the shape. l is the length and w is the width. A rectangle has two lengths and two widths. So you add one length and one width, then double it.
Sharing a total in a ratio
one part=total÷number of parts\text{one part} = \text{total} \div \text{number of parts}
In a ratio like 3:2, the amounts are made of equal-sized parts. Add the numbers in the ratio to count the parts. Divide the total by that count to get the size of one part. Then multiply by how many parts each amount has.

  1. Sort out the question

    The perimeter of the rectangle is 510 cm. The perimeter is the distance all the way around. The ratio of length to width is 3:2. That means the length is 3 equal parts, and the width is 2 of the same parts. You need the length and the width.

  2. Find length plus width

    The perimeter counts each side twice, once for each matching side. So halve it to get one length plus one width.

    P=2×(l+w)P = 2 \times (l + w)

    510=2×(l+w)510 = 2 \times (l + w)

    l+w=510÷2=255l + w = 510 \div 2 = 255

    Length plus width is 255 cm.

  3. Find the size of one part

    Length and width together are 3+2=53 + 2 = 5 parts. Share the 255 cm equally among them.

    one part=255÷5=51\text{one part} = 255 \div 5 = 51

    Each part is 51 cm.

  4. Find the length and width

    The length is 3 parts and the width is 2 parts.

    l=3×51=153l = 3 \times 51 = 153

    w=2×51=102w = 2 \times 51 = 102

    The length is 153 cm and the width is 102 cm.

  5. Check the ratio, then answer

    Watch out: every choice has a perimeter of 510 cm. For example, 150+105=255150 + 105 = 255 too. So checking only the perimeter can trick you into choice A. The ratio decides it. 153 is exactly 1.5 times 102, but 150 is not 1.5 times 105. So the answer is B, 153 cm by 102 cm.

Check your answer

Check both facts. Perimeter: 2×(153+102)=2×255=5102 \times (153 + 102) = 2 \times 255 = 510. Ratio: 153÷51=3153 \div 51 = 3 and 102÷51=2102 \div 51 = 2, so 153 to 102 is 3:2.

More questions like this

The original DOE question
Question 70 as printed on page 83 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 83, question 70. Open page 83 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 80 of 100

2018 Test A Math, question 19

Question

In Centerville, 45% of the population is female, and 60% of the population commutes to work daily. Of the total Centerville population, 21% are females who commute to work daily. What percentage of the total Centerville population are males who do not commute to work daily?

  • A15%
  • B16%
  • C24%
  • D39%
Show the answer

Correct answer: B

Create a table with the information provided in the problem and use subtraction to fill in the rest of the table:

Female Male TOTAL
Commutes to work 21% 39%
(60 − 21)		 60%
Does not commute to work 24%
(45 − 21)		 16%
(40 − 24)		 40%
(100 − 60)
TOTAL 45% 100%

16% of the population is male and does not commute to work.

Question analysis

A percent problem about overlapping groups in a town's population: the percents who are female, who commute daily, and who are female commuters are given. Students find the percent who are male and do not commute. A two-way table helps, since several of its cells are easy to confuse with the one asked for.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 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 percent problems usually take a percent of one quantity. This needs two overlapping categories organized in a two-way table, which is not in New York's grade 7 standards, with three subtractions and two tempting cells.
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 12, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question gives three overlapping percents and asks for the one group none of them describes directly. Without a table, students tend to subtract the given numbers in some order, and several orders yield a listed choice. The last trap is the word "not", since the males who do commute are one cell away. Filling every cell of a two-by-two table before answering takes under a minute and leaves nothing to guess.

  1. No obvious method moderate Working it out

    Nothing in the wording points to a two-way table, yet the four groups, female or male by commuting or not, fall into place only once one is drawn. (Schoenfeld, 1985)

  2. Much to keep track of moderate Working it out

    Three given percents, 45, 60 and 21, plus their complements, 55 and 40, have to be kept straight while the cells are filled in by subtraction. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  3. Turning words into math moderate Understanding the question

    "Of the total Centerville population, 21% are females who commute" is a share of everyone, not of the females, and belongs in a single cell of the table. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Make a two-by-two table of female and male against commutes and does not commute, fill in the totals, and complete each cell by subtraction.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Parts add up to the whole
a+b=Wa + b = W
In a two-way table, every row and every column splits a group into two parts. a and b are the two parts, and W is the whole group. If you know the whole and one part, subtract to find the other part.

  1. Sort out the question

    Every number is a percent of the whole town. 45% of people are female. 60% commute, which means they travel to work every day. 21% are females who commute. You need the percent who are “males who do not commute to work daily”.

  2. Make a two-way table

    A two-way table sorts people two ways at once: female or male, and commutes or not. Put in what you know. The whole town is 100%.

    FemaleMaleTotal
    Commute21%?60%
    Do not commute???
    Total45%?100%

  3. Fill in the two totals

    Everyone is female or male, so the males are what is left.

    100%−45%=55%100\% - 45\% = 55\%

    Everyone either commutes or does not.

    100%−60%=40%100\% - 60\% = 40\%

    So 55% are male, and 40% do not commute.

  4. Find the males who commute

    Use the commute row. All commuters are 60%, and the female commuters are 21%.

    60%−21%=39%60\% - 21\% = 39\%

    So 39% are males who commute. Watch out: the question says do not commute. Picking 39% is choice D.

  5. Find the males who do not commute

    Use the male column. All males are 55%, and the males who commute are 39%.

    55%−39%=16%55\% - 39\% = 16\%

    Watch out: 24% is the females who do not commute, 45%−21%45\% - 21\%. That is choice C. So the answer is B, 16%.

Check your answer

Find the same cell another way. Females who do not commute: 45%−21%=24%45\% - 21\% = 24\%. All who do not commute: 40%. So males who do not commute: 40%−24%=16%40\% - 24\% = 16\%. The answers match.

More questions like this

The original DOE question
Question 76 as printed on page 84 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 76. Open page 84 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 81 of 100

2018 Test A Math, question 20

Question

Mrs. Cranston bought five bottles of water for $0.90 each and 8 pounds of meat. She paid a total of $26.90 for these items, not including tax. What was the price per pound of the meat?

  • A$2.80
  • B$3.25
  • C$14.40
  • D$22.40
Show the answer

Correct answer: A

Let xx be the price per pound for the meat. Set up an equation to show what Mrs. Cranston spent:

5(0.90)+8x=26.905(0.90)+8x=26.90

4.50+8x=26.904.50+8x=26.90

8x=22.408x=22.40

x=2.80x=2.80

The price per pound for the meat is $2.80.

Question analysis

This money word problem has a shopper buying five bottles of water at a set price plus 8 pounds of meat, for a known total. Students write and solve a two-step equation to find the price per pound of the meat. The decimals call for care, and the question wants a price per pound rather than a total.

NYS standards
  • 7.EE.4a two-step equations
  • 7.NS.3 four operations with rational numbers
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Word problems that lead to an equation like px + q = r are the core of grade 7 algebra. This is a standard example with money decimals and no extra twist.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 2, 33 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a two-step money problem: take away the water, then divide by the pounds. The listed wrong answers come from doing one of those steps wrongly or not at all, especially stopping at the total cost of the meat, which looks like a finished dollar amount. Checking that 8 pounds at the chosen price plus $4.50 makes $26.90 confirms the answer.

  1. Turning words into math minor Understanding the question

    "five bottles of water for $0.90 each" has to become 5 × 0.90 before the total is split, and the unknown is a price per pound, not the meat's total cost. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    Every amount is a decimal of dollars, so 26.90 − 4.50 and 22.40 ÷ 8 must be lined up and worked carefully without a calculator. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Subtract the cost of all five bottles from the total, then divide what is left by the 8 pounds of meat.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost
C=p×qC = p \times q
C is the total cost. p is the price of one item, or of one pound. q is the quantity, how many items or pounds you buy. Each one costs p, so q of them cost p times q.

  1. Sort out the question

    Mrs. Cranston bought five bottles of water “for $0.90 each”. She also bought 8 pounds of meat. Everything cost $26.90 in all. You need the price per pound of the meat, the cost of just 1 pound.

  2. Find the cost of the water

    Each bottle costs $0.90, and there are 5 bottles. Use the cost formula.

    C=p×qC = p \times q

    C=0.90×5=4.50C = 0.90 \times 5 = 4.50

    The water cost $4.50. Watch out: each means every bottle costs $0.90. Taking away just one $0.90 leads to $3.25, choice B.

  3. Find the cost of the meat

    Take the water away from the total. What is left paid for the meat.

    26.90−4.50=22.4026.90 - 4.50 = 22.40

    All 8 pounds of meat cost $22.40. Watch out: this is the cost of all the meat, not of one pound. Stopping here gives choice D.

  4. Find the price per pound

    Use the cost formula again. Now C is 22.40 and q is 8 pounds. Divide to find p.

    22.40=p×822.40 = p \times 8

    p=22.40÷8=2.80p = 22.40 \div 8 = 2.80

    Each pound costs $2.80. So the answer is A, $2.80.

Check your answer

Put it all back together. Meat: 8×2.80=22.408 \times 2.80 = 22.40. Water: 4.50. Total: 22.40+4.50=26.9022.40 + 4.50 = 26.90. That is exactly what she paid.

More questions like this

The original DOE question
Question 77 as printed on page 85 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 77. Open page 85 of the handbook (PDF)

Question 82 of 100

2018 Test A Math, question 21

Question

In a sample of 10 cards, 4 are red and 6 are blue. If 2 cards are selected at random from the sample, one at a time without replacement, what is the probability that both cards are not blue?

  • A215\frac{2}{15}
  • B425\frac{4}{25}
  • C310\frac{3}{10}
  • D13\frac{1}{3}
Show the answer

Correct answer: A

The probability that both cards are not blue is the same as the probability that both cards are red. There are 4 red cards out of the 10, so the probability of the first card being red is

410\frac{4}{10}

Now there are 9 cards left, and 3 of those are red, so the probability of the second card being red is

39\frac{3}{9}

Multiply the two probabilities to find the probability that both cards are red (not blue):

(410)(39)=1290=215\left (\frac{4}{10} \right )\left ( \frac{3}{9} \right )=\frac{12}{90}=\frac{2}{15}

Question analysis

This probability question draws 2 cards, one at a time and without replacement, from 10 cards: 4 red and 6 blue. Students find the chance that neither card drawn is blue. The second draw has one fewer card, and the word not changes which color to count.

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
Above grade 7 Grade 7 compound probability is usually found with a list or tree of equally likely outcomes. Dependent draws without replacement, plus a negative phrasing to translate, go beyond that typical 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

Two ideas stack here: restating not blue as red, and changing the second draw's numbers after a card is removed. Students who know to multiply but forget the removal reach a listed choice, and so do students who compute both blue. Writing the second fraction explicitly as 3 red out of 9 cards, before multiplying, is what separates right from nearly right.

  1. Something you must already know moderate Working it out

    Multiplying the first draw's chance by the second draw's, with the second adjusted for the card already taken, is a rule the test assumes and does not give. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    2/15 and 4/25 are about 0.13 and 0.16, too close to tell apart by estimate; only the second draw's fraction, 3 red of 9 cards rather than 4 of 10, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Turning words into math minor Understanding the question

    "Both cards are not blue" has to be restated as both cards red, and the bold not also invites computing the chance that both are blue. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The product comes out as 12/90, and it has to be reduced to lowest terms before it matches any of the listed fractions. (Enright, Morley & Sheehan, 2002)

Strategy: Restate not blue as red, then multiply the chance of red on the first draw by the chance of red on the second, with one card fewer.

Traps

  • Without replacement: In a second pick there is one fewer item, which changes the probability.
  • 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

Probability
P=favorabletotalP = \frac{\text{favorable}}{\text{total}}
P is the probability, the chance that something happens. Favorable is the number of ways to get what you want, here a red card. Total is the number of all possible picks, here all the cards you can pick from. It works when every pick is equally likely.
Two events in a row
P(A and B)=P(A)×P(B after A)P(\text{A and B}) = P(\text{A}) \times P(\text{B after A})
A is the first event and B is the second. P( ) means the probability of what is inside. The chance that both happen is the first chance times the second. For the second chance, count only what is left after the first event.

  1. Sort out the question

    There are 10 cards: 4 red and 6 blue. You pick 2 cards “one at a time without replacement”. That means the first card is not put back. “Both cards are not blue” means both are red, since red is the only other color. You need the probability, or chance, of that.

  2. Find the chance the first card is red

    4 of the 10 cards are red. Each card is equally likely to be picked. Use the probability formula.

    P(red first)=410P(\text{red first}) = \frac{4}{10}

  3. Find the chance the second card is red

    One red card is gone now, and it is not put back. So 9 cards are left, and 3 of them are red.

    P(red second)=39P(\text{red second}) = \frac{3}{9}

    Watch out: using 4/10 again acts as if the card went back. That gives 4/25, choice B.

  4. Multiply the two chances

    Both cards must be red. So multiply the first chance by the second. To multiply fractions, multiply the tops, then multiply the bottoms.

    410×39=4×310×9=1290\frac{4}{10} \times \frac{3}{9} = \frac{4 \times 3}{10 \times 9} = \frac{12}{90}

  5. Simplify the fraction

    None of the choices is 12/90, so simplify it. Divide the top and the bottom by 6, the biggest number that goes into both.

    12÷690÷6=215\frac{12 \div 6}{90 \div 6} = \frac{2}{15}

    Watch out: the chance that both are blue is 610×59=13\frac{6}{10} \times \frac{5}{9} = \frac{1}{3}, choice D. The question says not blue. So the answer is A, 215\frac{2}{15}.

Check your answer

Count the ways. There are 10×9=9010 \times 9 = 90 ways to pick two cards in order. Red then red has 4×3=124 \times 3 = 12 ways. That gives 12/90, or 2/15. The answer is small, which makes sense, because there are fewer red cards than blue ones.

More questions like this

The original DOE question
Question 78 as printed on page 85 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 78. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 83 of 100

2018 Test A Math, question 25

Question

Which of the following numbers has factors that include the smallest factor (other than 1) of 91?

  • A30
  • B35
  • C39
  • D44
Show the answer

Correct answer: B

To find the smallest factor of 91, list the factors:

1, 7, 13, and 91

The smallest factor (other than 1) is 7.

Of the options listed (30, 35, 39, and 44), only 35 is a multiple of 7.

Question analysis

A number properties question about factors: students find the smallest factor of 91 other than 1, then pick the number that shares it. Because 91 looks like a prime number, many students skip the factoring, and the choices include numbers built from factors that 91 might seem to have.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Factors and divisibility are grade 4 to 6 skills, and this takes two quick checks. It is easier than typical grade 7 work, though recognizing that 91 is not prime catches many students.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 8, 15 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question tests a grade 6 skill inside a hard-to-parse sentence. Once the stem is unpacked, the work is finding the smallest prime factor of 91 and checking the choices for it. Students lose points in two ways: 91 looks prime or divisible by 3, so they guess at its factors, or they find 13 before 7 and accept the choice that shares it.

  1. Complex sentences moderate Understanding the question

    The stem nests three ideas, "has factors that include the smallest factor (other than 1) of 91", so it has to be unpacked before any math; this is reading difficulty, not number difficulty. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

  2. Something you must already know moderate Working it out

    Students must find that 91 = 7 × 13, and 91 is a number many students wrongly take to be prime or a multiple of 3. (Anderson & Krathwohl, 2001)

Strategy: Test 91 against the small primes in order, starting with 2, until one divides it evenly, then check which choice is a multiple of that prime.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

What a factor is
b=a×cb = a \times c
a, b and c are whole numbers. If b = a × c, then a and c are factors of b. That means a divides b evenly, with no remainder. Every whole number has 1 and itself as factors.

  1. Unpack the long sentence

    A factor of a number divides it evenly, with nothing left over. For example, 3 is a factor of 12, because 12÷3=412 \div 3 = 4. The question has two steps. First, find “the smallest factor (other than 1) of 91”. Then find the choice that also has that factor.

  2. Test small numbers on 91

    Try dividing 91 by 2, 3, 5 and so on, until one works.

    • 2: no. 91 is odd. That also rules out 4 and 6, which are even.
    • 3: no. A number is a multiple of 3 only if its digits add to a multiple of 3. Here 9 + 1 = 10, which is not.
    • 5: no. 91 does not end in 0 or 5.
    • 7: yes!

    91=7×1391 = 7 \times 13

    So 7 is the smallest factor of 91 other than 1.

  3. Avoid the 3 trap

    Watch out: 91 is close to 90, and 3 goes into 90. But 91÷391 \div 3 is 30 with 1 left over. So 3 is not a factor of 91. That trap leads to 30, choice A.

  4. Check each choice for 7

    Now look for a multiple of 7 among the choices.

    • 30: no. 7×4=287 \times 4 = 28 and 7×5=357 \times 5 = 35.
    • 35: yes. 35=7×535 = 7 \times 5.
    • 39: no. 39=3×1339 = 3 \times 13.
    • 44: no. 7×6=427 \times 6 = 42.

    Watch out: 39 shares 13 with 91, but 13 is the larger factor, not the smallest. That is choice C. So the answer is B, 35.

Check your answer

Multiply to be sure: 7×13=917 \times 13 = 91 and 7×5=357 \times 5 = 35. So 7 goes into both numbers evenly. No number from 2 to 6 goes into 91, so 7 really is the smallest.

More questions like this

The original DOE question
Question 82 as printed on page 86 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 86, question 82. Open page 86 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 84 of 100

2018 Test A Math, question 31

Question

How many positive even factors of 48 are greater than 24 and less than 48?

  • A0
  • B1
  • C2
  • D12
Show the answer

Correct answer: A

List the factors of 48:

1 and 48, 2 and 24, 3 and 16, 4 and 12, 6 and 8

There are no factors greater than 24 and less than 48.

Question analysis

This number properties question asks how many positive even factors of 48 are greater than 24 and less than 48. Students list the factors of 48 and test which ones fall strictly inside that range. Reading both conditions precisely, and listing every factor without skipping any, is what matters.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Listing the factors of a two-digit number is grade 6 work, and here it takes one short list. The only demand beyond routine is reading both conditions exactly, which grade 7 students already do regularly.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 7, 12 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The factor list is short, so the question is decided by the conditions. The pairs 2 and 24 and 1 and 48 put the only candidates exactly on the boundaries, and the strict words "greater than" and "less than" exclude them. Students also tend to distrust an answer of zero, which nudges them toward counting an endpoint.

  1. A very close wrong answer moderate Choosing the answer

    The choices 0 and 1 differ by a single factor, so the count cannot be sized up; only checking whether 48 itself, or 24, falls strictly inside the range separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Much to keep track of minor Working it out

    Each factor must pass three tests at once, even, greater than 24 and less than 48, and the factor list itself must be complete. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: List the factors of 48 in pairs, starting from 1 × 48, then check each one against both conditions: it must be even and lie strictly between 24 and 48.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Factor pairs
a×b=na \times b = n
n is the number you start with, here 48. If two whole numbers a and b multiply to make n, both are factors of n. A factor divides n with nothing left over. Listing factors in pairs makes sure you skip none.

  1. Sort out what the question asks

    A factor of 48 is a whole number that divides 48 with nothing left over. You need factors that pass three tests. They must be even. They must be greater than 24. They must be less than 48.

    “Less than 48” means 48 itself does not count. “Greater than 24” means 24 does not count either.

  2. List the factors of 48 in pairs

    Start with 1 and work up. Each pair multiplies to 48:

    1×482×243×164×126×81 \times 48 \quad 2 \times 24 \quad 3 \times 16 \quad 4 \times 12 \quad 6 \times 8

    The next number, 7, does not divide 48. After 6 and 8 the pairs repeat. So the factors are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

  3. Look between 24 and 48

    In the list, the factor right after 24 is 48. Nothing falls between them.

    Here is why. A factor bigger than 24 must pair with a number smaller than 2. The only such number is 1, and it pairs with 48. So no factor of 48 is between 24 and 48, even or odd.

  4. Give the answer

    Watch out: 24 and 48 are both factors, but they sit on the edges. Counting 48 gives 1, which is choice B. Counting both gives 2, which is choice C.

    So the answer is A, 0.

Check your answer

Try the even numbers 26, 28, and so on up to 46. Dividing 48 by any of them gives more than 1 but less than 2. That is never a whole number, so none of them is a factor. Choice D, 12, counts even numbers without checking that they are factors.

More questions like this

The original DOE question
Question 88 as printed on page 87 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 88. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 85 of 100

2018 Test A Math, question 33

Question

Johan leased a car for three years. He paid a one-time fee of $1,000, and an additional $300 per month for the full three years. At the end of the three years, what is the total amount Johan paid for leasing this car?

  • A$1,900
  • B$4,600
  • C$10,800
  • D$11,800
Show the answer

Correct answer: D

Three years is 36 months (12 × 3).

Set up an expression to find the total amount Johan paid:

1,000+300(36)=$11,8001{,}000+300(36)=\$11{,}800

Question analysis

This word problem about leasing a car combines a one-time fee with a monthly payment over three years and asks for the total amount paid. Students must convert years to months, multiply by the monthly payment and add the fee. Leaving out one piece of the cost, or mixing years with months, is the usual slip.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 6.RP.3d converting measurement units
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 This is a short total-cost calculation with whole numbers, easier than typical grade 7 word problems, which bring in percents, negative numbers or fractions. The only care needed is turning years into months.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
3 steps
Reading level
Question and choices: about grade 4, 41 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is the easiest kind of multi-step word problem: whole numbers, a familiar money situation and a direct method. The only places to slip are changing three years into 36 months and remembering the one-time fee after the long multiplication. Each slip has its own choice, so a quick check that both costs are in the total is worth the seconds.

  1. Much to keep track of minor Working it out

    The $1,000 fee is paid once while the $300 repeats monthly, so it must be held aside while 36 × 300 is computed. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. A very close wrong answer minor Choosing the answer

    The choices $11,800 and $10,800 differ by only $1,000 out of about $11,000, so a rough total cannot separate them; only including the one-time fee does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Count the months in the lease first, multiply the monthly payment by that number, and then add the one-time fee.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost with a fee
T=f+m×nT = f + m \times n
T is the total paid. f is the one-time fee, paid only once. m is the monthly payment. n is the number of months. You multiply first because the monthly payment repeats every month. Then you add the fee once.

  1. Sort out what you know

    Johan paid $1,000 one time. He also paid $300 every month for 3 years. You need the total he paid.

    The payments are counted by the month, but the time is given in years. So first change the years to months.

  2. Count the months

    Each year has 12 months:

    3×12=363 \times 12 = 36

    So Johan made 36 monthly payments. Watch out: multiplying $300 by 3 years instead of 36 months gives $900. Adding the fee then gives $1,900, which is choice A.

  3. Add up the monthly payments

    Multiply the monthly payment by the number of months:

    300×36=10,800300 \times 36 = 10{,}800

    The monthly payments add up to $10,800.

  4. Add the one-time fee

    Use the formula. The fee goes in once:

    T=1,000+300×36=1,000+10,800=11,800T = 1{,}000 + 300 \times 36 = 1{,}000 + 10{,}800 = 11{,}800

    Watch out: $10,800 is choice C. It leaves out the $1,000 fee. So the answer is D, $11,800.

Check your answer

Round to check. $300 a month for 36 months is a bit more than $10,000. Add the $1,000 fee and you get a bit more than $11,000. $11,800 fits. Choices A and B are far too small for three years of payments.

More questions like this

The original DOE question
Question 90 as printed on page 87 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 90. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 86 of 100

2018 Test A Math, question 35

Question

For a presentation, Deion can create 5 slides in 20 minutes, working at a constant rate. Kyra can create 3 slides in 10 minutes, working at her own constant rate. What is the total number of slides the two of them can create in one hour?

  • A16
  • B30
  • C33
  • D55
Show the answer

Correct answer: C

Set up proportions to figure out how many slides Deion and Kyra can create in 1 hour:

Deion

520=x60\frac{5}{20}=\frac{x}{60}

20x=30020x=300

x=15x=15

Deion can create 15 slides in 1 hour.

Kyra

310=x60\frac{3}{10}=\frac{x}{60}

10x=18010x=180

x=18x=18

Kyra can create 18 slides in 1 hour.

Add Deion and Kyra to figure out how many slides they can create together in 1 hour:

15 + 18 = 33.

Question analysis

This rate word problem says how many slides Deion makes in 20 minutes and Kyra makes in 10 minutes, then asks how many slides they make together in one hour. Students must scale each person's rate to an hour before combining. Because their time periods differ, the two rates cannot simply be added as given.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 rate problems often ask for a single unit rate. This one needs two rates over different time periods, each scaled to an hour and then combined, which is typical grade 7 work built on grade 6 content.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 5, 42 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is easy once each person is scaled to an hour on their own; the trouble is the urge to combine first. Adding the slides and the minutes into one rate treats two different rates like fractions added across the top and bottom, and it gives a listed answer. Keeping the two workers separate until the last step avoids every trap.

  1. Much to keep track of moderate Working it out

    Deion's rate is given per 20 minutes and Kyra's per 10, so each needs its own scale factor to an hour, 3 and 6, before the totals can be added. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Many steps minor Working it out

    Two scalings and an addition: 5 × 3 = 15 for Deion, 3 × 6 = 18 for Kyra, then the sum, each a simple step. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The choices 33 and 30 differ by only 3 slides, so a rough sense of the total cannot separate them; only scaling Kyra's own rate to 18 an hour, not 15, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Find how many 20-minute blocks and how many 10-minute blocks fit in 60 minutes, scale each person's slides by that number, and then add the two totals.

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.
  • 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

Amount of work
a=r×ta = r \times t
a is the amount of work done, here slides. r is the rate: how much gets done in one block of time. t is the number of those blocks. Use it when someone works at a steady speed.

  1. Sort out what you know

    Deion makes 5 slides every 20 minutes. Kyra makes 3 slides every 10 minutes. Each works at a constant rate, which means a steady speed that does not change.

    One hour is 60 minutes. You need how many slides they make together in 60 minutes.

  2. Find Deion’s slides in an hour

    Deion’s rate is 5 slides per 20-minute block. An hour holds 3 of those blocks, since 60 ÷ 20 = 3:

    a=5×3=15a = 5 \times 3 = 15

    Deion makes 15 slides in an hour.

  3. Find Kyra’s slides in an hour

    Kyra’s rate is 3 slides per 10-minute block. An hour holds 6 of those blocks, since 60 ÷ 10 = 6:

    a=3×6=18a = 3 \times 6 = 18

    Kyra makes 18 slides in an hour. Watch out: Kyra has her own rate. Doubling Deion’s 15 to get 30, choice B, treats her as if she worked at his speed.

  4. Add the two amounts

    15+18=3315 + 18 = 33

    Together they make 33 slides. Watch out: do not add the slides and the minutes first. Mixing 5 + 3 slides with 20 + 10 minutes blends two different rates and leads to choice A, 16.

    So the answer is C, 33.

Check your answer

Compare per minute. Deion makes 5 ÷ 20 = 0.25 slide a minute, and Kyra makes 3 ÷ 10 = 0.3. Together that is 0.55 slide a minute. Times 60 minutes gives 33. Choice D, 55, comes from multiplying by 100 instead of 60.

More questions like this

The original DOE question
Question 92 as printed on page 87 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 92. Open page 87 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 87 of 100

2018 Test A Math, question 37

Question

An unmarked straight stick will be laid end over end to measure a distance of exactly 72 feet. The same stick will be used in the same way to measure a distance of exactly 30 feet. What is the length of the longest possible stick that can be used for both measurements?

  • A3 ft
  • B4 ft
  • C6 ft
  • D8 ft
Show the answer

Correct answer: C

The length of the stick must be the greatest common factor of 72 and 30.

The factors of 30 are

1, 2, 3, 5, 6, 10, 15, and 30.

Of those, only 1, 2, 3, and 6 are also factors of 72. The greatest of these is 6.

Question analysis

This word problem asks for the longest stick that can be laid end over end to measure exactly 72 feet and exactly 30 feet. Students must recognize that the stick's length has to divide both distances evenly, so the task is finding a greatest common factor. The story wording hides the factor idea.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Finding a greatest common factor is a sixth-grade skill, and the arithmetic is light. The question is a little less routine than a bare factor problem only because students must see that the story calls for one.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 3, 54 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the story is recognized as a greatest common factor problem, the arithmetic is light and the choices can simply be tested. The trap is checking only one of the two distances, because the longest choice works for 72 feet but not for 30. A student who lists the factors of 30 first, the smaller number, sees the answer quickly.

  1. No obvious method moderate Working it out

    Nothing in the stick story says "common factor"; students must see that laying the stick a whole number of times along 72 and 30 feet means it divides both. (Schoenfeld, 1985)

  2. Hard words minor Understanding the question

    The phrases "unmarked straight stick" and "laid end over end" are unusual wording for a math item; they describe whole-stick lengths, a reading hurdle rather than more math. (Perfetti, 2007; Freedle & Kostin, 1993)

Strategy: Translate 'measures both distances exactly' into 'divides both numbers evenly', list the factors of the smaller number, 30, and take the largest that also divides 72.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of sticks
n=dsn = \frac{d}{s}
d is the distance to measure. s is the length of the stick. n is how many times you lay the stick down. For an exact measure, n must be a whole number. So s must divide d evenly.

  1. Sort out the story

    You lay a stick down again and again, end to end, like taking steps. The stick has no marks, so you can only count whole stick lengths.

    It must fit exactly into 72 feet and exactly into 30 feet. You need the longest stick that does both.

  2. Turn it into factors

    Fitting exactly means no piece is left over. So the stick length must divide 72 evenly and divide 30 evenly.

    A number that divides both is a common factor. You want the greatest common factor: the biggest number that divides both.

  3. List the factors of 30

    Start with the smaller number, 30, since its list is shorter. Its factors are 1, 2, 3, 5, 6, 10, 15 and 30.

    Now test them on 72, starting from the biggest. 30, 15, 10 and 5 leave a remainder. But 6 works:

    72÷6=1230÷6=572 \div 6 = 12 \qquad 30 \div 6 = 5

    So 6 is the greatest common factor.

  4. Check the other choices

    Watch out: 8 feet fits 72 feet, 9 times. But 30 ÷ 8 = 3.75, not a whole number, so choice D fails. Choice B fails too, since 30 ÷ 4 = 7.5. A 3-foot stick works, but it is not the longest, so choice A is wrong.

    So the answer is C, 6 ft.

Check your answer

Use the stick. A 6-foot stick goes down 12 times for 72 feet, since 12 × 6 = 72. It goes down 5 times for 30 feet, since 5 × 6 = 30. Both counts are whole numbers, so 6 feet works.

More questions like this

The original DOE question
Question 94 as printed on page 88 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 88, question 94. Open page 88 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 88 of 100

2018 Test A Math, question 40

Question

In the set of consecutive integers from 12 to 30, inclusive, there are four integers that are multiples of both 2 and 3. How many integers in this set are multiples of neither 2 nor 3?

  • A5
  • B6
  • C13
  • D15
Show the answer

Correct answer: B

The question asks for integers from 12 to 30 that are not divisible by 2 or 3. The set of consecutive integers is

{12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}

Since all even numbers are divisible by 2, eliminate all even numbers, leaving the odd numbers in the set:

{13, 15, 17, 19, 21, 23, 25, 27, 29}

Eliminate those integers that are multiples of 3 (15, 21, and 27). The remaining integers are:

{13, 17, 19, 23, 25, 29}

Therefore, there are 6 numbers in the set that are multiples of neither 2 nor 3.

Question analysis

This number properties question takes the integers from 12 to 30, inclusive, and asks how many are multiples of neither 2 nor 3. Students must count the whole set, remove multiples of 2 and of 3 without removing any number twice, and count what remains. Inclusive endpoints and the overlap make it easy to miscount.

NYS standards
  • 6.NS.4 common factors and multiples
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Above grade 7 Multiples are sixth-grade content, but grade 7 problems rarely ask students to count a set under two overlapping conditions. The inclusive range and the numbers that are multiples of both make this longer and trickier than usual.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 9, 30 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question gives the overlap, four multiples of both 2 and 3, which invites a subtraction method with three places to go wrong: the count of the set, the overlap, and the final complement. Each of those errors has its own choice. Listing the 19 numbers and crossing out evens and then odd multiples of 3 is slower to describe but faster and safer to do.

  1. Much to keep track of moderate Working it out

    Each of the 19 integers must be checked against two conditions, and the four multiples of 6 meet both, so they must be removed once, not twice. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. A very close wrong answer moderate Choosing the answer

    The choices 5 and 6 differ by one, the result of counting 30 − 12 = 18 integers instead of 19, so only counting the inclusive set exactly separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Cross out every even number first, then cross out the odd multiples of 3 that remain, and count the survivors; listing all nineteen numbers is quick enough.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting whole numbers in a list
n=b−a+1n = b - a + 1
a is the first number and b is the last. n is how many whole numbers there are, counting both ends. You add 1 because subtracting counts the gaps, and there is one more number than gaps.

  1. Sort out the question

    The set is every integer (whole number) from 12 to 30. “Inclusive” means 12 and 30 are both in.

    A multiple of 2 is an even number. A multiple of 3 is a number in the 3 times table. You need the numbers that are neither: not even, and not in the 3 times table.

  2. Count the numbers in the set

    n=30−12+1=19n = 30 - 12 + 1 = 19

    There are 19 integers. Watch out: 30 − 12 = 18 forgets one end. That slip leads to choice A, 5.

  3. Cross out the even numbers

    Every even number is a multiple of 2, so remove them all. The odd numbers are left:

    13, 15, 17, 19, 21, 23, 25, 27, 29

    That is 9 numbers.

  4. Cross out the multiples of 3

    Check each odd number for 3. A quick test: add its digits. If the sum is in the 3 times table, so is the number. That removes 15, 21 and 27. These are left:

    13, 17, 19, 23, 25, 29

    Watch out: only 12, 18, 24 and 30 are multiples of both. Removing just those gives 19 − 4 = 15, choice D. “Neither” means you remove every multiple of 2 and every multiple of 3.

  5. Count what is left

    Six numbers are left. Watch out: 13 is how many you crossed out, choice C. The question asks for the ones left. So the answer is B, 6.

Check your answer

Count another way. Of the 19 numbers, 10 are even. Of the 9 odd ones, 3 are multiples of 3. So 10 + 3 = 13 are crossed out, and 19 − 13 = 6 are left. Same answer.

More questions like this

The original DOE question
Question 97 as printed on page 88 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 88, question 97. Open page 88 of the handbook (PDF)

Question 89 of 100

2018 Test A Math, question 42

Question

A box contains 5 strawberry candies, 3 banana candies, and 2 orange candies. If Braden selects 2 candies at random from this box, without replacement, what is the probability that both candies are not banana?

  • A115\frac{1}{15}
  • B9100\frac{9}{100}
  • C715\frac{7}{15}
  • D49100\frac{49}{100}
Show the answer

Correct answer: C

The total number of candies in the box is

5 + 3 + 2 = 10

The number of candies that are not banana is

5 + 2 = 7

The probability of the first candy not being banana is

710\frac{7}{10}

Now, out of 9 candies, there are 6 candies left that are not banana. The probability of the second candy not being banana is

69\frac{6}{9}

Multiply these two probabilities to get the solution:

(710)(69)=4290=715(\frac{7}{10})(\frac{6}{9})=\frac{42}{90}=\frac{7}{15}

Question analysis

This probability question has a box of strawberry, banana and orange candies and asks the chance that two candies drawn without replacement are both not banana. Students must count the non-banana candies, find the chance for each draw, and multiply. The second draw changes because one candy is already gone.

NYS standards
  • 7.SP.8a compound event as fraction of outcomes
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 compound probability usually uses lists or tree diagrams for independent events such as coins and number cubes. Here the draws are dependent, and students must also work with the complement, 'not banana'.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 9, 35 words.
Corrected from the DOE handbook
  • A stray punctuation mark in the explanation 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

Two phrases in the stem change the calculation, "not" and "without replacement", and each has a choice for the student who misses it. The cleanest approach treats strawberry and orange together as one group of 7 and then works through the two draws in order. Students who know independent probability well are the most likely to square a fraction out of habit.

  1. Combines ideas moderate Working it out

    Two ideas meet: counting the 7 non-banana candies as the favorable group, and reducing both counts by one for the dependent second draw. (Embretson, 1998; Daniel & Embretson, 2010)

  2. A very close wrong answer moderate Choosing the answer

    The choices 49/100 and 7/15 are about 0.49 and 0.47, so estimating cannot separate them; only noticing that the second draw is 6 of 9, not 7 of 10, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Error-prone arithmetic minor Working it out

    The product 7/10 × 6/9 = 42/90 must be simplified to 7/15 to match a choice, a reduction by 6 that is easy to stall on. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the candies that are not banana, find the chance the first draw is one of them, reduce both counts by one for the second draw, and multiply.

Traps

  • Without replacement: In a second pick there is one fewer item, which changes the probability.
  • 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 reads “The probability of the second candy not being banana is. 6/9.” The period after “is” was removed.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=ftP = \frac{f}{t}
P is the probability, the chance that something happens. f is the number of favorable outcomes, the ones you want. t is the total number of outcomes. It works when every outcome is equally likely, like picking one item at random.
Two picks in a row
P(both)=P(first)×P(second)P(\text{both}) = P(\text{first}) \times P(\text{second})
To find the chance that two things both happen, multiply their chances. For the second chance, use what is left after the first pick. You multiply because you want a fraction of a fraction: the second pick must work in the share of times the first one did.

  1. Sort out what you know

    The box has 5 strawberry, 3 banana and 2 orange candies. Braden takes 2 candies. “Without replacement” means he keeps the first candy out. He does not put it back.

    You need the chance that neither candy is banana. Watch out: the bold “not” matters. The chance that both are banana is 115\frac{1}{15}, choice A.

  2. Count the candies

    5+3+2=105 + 3 + 2 = 10

    There are 10 candies. The ones that are not banana are the strawberry and orange ones:

    5+2=75 + 2 = 7

  3. Find the chance for the first candy

    7 of the 10 candies are not banana. Use the probability formula:

    P=710P = \frac{7}{10}

    So the first candy is not banana 7 times out of 10.

  4. Find the chance for the second candy

    Say the first candy was not banana. Now 9 candies are left, and 6 of them are not banana:

    P=69P = \frac{6}{9}

    Watch out: the first candy is not put back. Using 710\frac{7}{10} again gives 49100\frac{49}{100}, choice D.

  5. Multiply the two chances

    710×69=4290\frac{7}{10} \times \frac{6}{9} = \frac{42}{90}

    Simplify by dividing the top and bottom by 6:

    42÷690÷6=715\frac{42 \div 6}{90 \div 6} = \frac{7}{15}

    So the answer is C, 715\frac{7}{15}.

Check your answer

Count pairs instead. From 10 candies there are 10 × 9 ÷ 2 = 45 different pairs. Pairs with no banana come from the 7 others: 7 × 6 ÷ 2 = 21. And 21 out of 45 is 7/15. It matches.

More questions like this

The original DOE question
Question 99 as printed on page 89 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 89, question 99. Open page 89 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 90 of 100

2018 Test B Math, question 6

Question

If x=9x=9 and y=−7y=-7, what is the value of x(x−2y)x(x-2y)?

  • A18
  • B45
  • C144
  • D207
Show the answer

Correct answer: D

x(x−2y)=9[9−2(−7)]=9(9+14)=9(23)=207x(x-2y)=9[9-2(-7)]=9(9+14)=9(23)=207

Question analysis

This algebra question asks students to evaluate the expression x(x − 2y) when x is 9 and y is negative 7. It practices substitution and the order of operations with integers. The risk is the sign, since subtracting twice a negative number turns into addition, and dropping that sign or the factor of 2 is easy.

NYS standards
  • 6.EE.2c evaluating expressions and order of operations
  • 7.NS.2a rational number multiplication
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Substituting values, including a negative one, into an expression is everyday grade 7 work with integers. This has three short steps and one sign decision, matching a typical classroom exercise.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 8, 10 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Substitution itself is routine, so the question tests one thing: subtracting a negative inside the parentheses. A student who turns 9 − 2(−7) into 9 − 14 gets −45 and may then choose 45, since it is the only choice that looks close. Working the parentheses first and writing −(−14) as +14 before multiplying by 9 avoids every wrong choice.

  1. Error-prone arithmetic moderate Working it out

    The middle step is 9 − 2(−7): the product 2(−7) is −14, and subtracting it means adding 14, so two negatives must be handled in one short move. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Work inside the parentheses first: find 2y, subtract it from x by rewriting minus a negative as plus, and multiply that result by x last.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a negative
a−(−b)=a+ba - (-b) = a + b
a and b are numbers. Taking away a negative number is the same as adding the positive number. For example, 9 − (−14) = 9 + 14. Think of taking away a debt: you end up richer.
Multiplying by a negative
positive×negative=negative\text{positive} \times \text{negative} = \text{negative}
When one number is positive and the other is negative, their product is negative. The product is the answer to a multiplication. So 2 × (−7) = −14.

  1. Write down what you know

    You know x = 9 and y = −7. You need the value of x(x−2y)x(x - 2y).

    A number written right next to parentheses means multiply. So this is x times the amount inside the parentheses.

  2. Put in the numbers

    Replace each letter with its value. Put −7 in parentheses so you do not lose its minus sign.

    x(x−2y)=9×(9−2×(−7))x(x - 2y) = 9 \times (9 - 2 \times (-7))

  3. Work inside the parentheses

    Always work inside parentheses first. Multiply before you subtract. A positive times a negative is negative, so 2×(−7)=−142 \times (-7) = -14.

    9−(−14)=9+14=239 - (-14) = 9 + 14 = 23

    Taking away a negative is the same as adding.

    Watch out: If you write 9 − 14 = −5, you end with −45. Then choice B, 45, looks close, but it is wrong.

  4. Multiply by x last

    Now multiply 9 by the 23 you found inside the parentheses.

    9×23=2079 \times 23 = 207

    Watch out: Do not drop the 2. Using 9 − (−7) = 16 gives 144, choice C.

    So the answer is D, 207.

Check your answer

Check the inside another way. 9−2×(−7)9 - 2 \times (-7) means start at 9 and move 14 to the right on the number line. You land on 23. Then 9×23=180+27=2079 \times 23 = 180 + 27 = 207.

More questions like this

The original DOE question
Question 63 as printed on page 154 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 63. Open page 154 of the handbook (PDF)

Question 91 of 100

2018 Test B Math, question 10

Question

To make party invitations, Macie could buy a package of paper for $10.50, or she could buy x individual sheets of the same paper for $0.15 each. What is the largest value of x that would make buying the individual sheets less expensive than buying the package?

  • A60
  • B65
  • C69
  • D70
Show the answer

Correct answer: C

Set up an inequality to compare the costs:

0.15x < 10.50
x < 70

Therefore, 70 individual sheets of paper would cost $10.50, so 69 is the greatest number of individual sheets of paper that Macie can buy that would be less expensive than the package.

Question analysis

This inequality word problem compares buying a package of paper with buying single sheets at 15 cents each, and asks for the largest number of sheets for which the single sheets cost less. Students write and solve an inequality with decimals. The tricky part is interpreting the solution as a whole number of sheets.

NYS standards
  • 7.EE.4b two-step inequalities
  • 6.NS.3 multi-digit decimal operations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing and solving an inequality from a real-world situation is a grade 7 standard. This one is short, but it asks students to interpret a strict inequality as a count of sheets, the kind of interpretation grade 7 tests expect.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
3 steps
Reading level
Question and choices: about grade 10, 45 words.
Corrected from the DOE handbook
  • An inequality sign in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The computation leads straight to 70, and the whole question is about what to do with that number. Students who treat the boundary as the answer forget that equal cost is not less cost, while the other choices only catch those who misdivide or ignore the word largest. The question rewards checking the boundary case in the story: 70 sheets cost exactly the package price.

  1. A very close wrong answer moderate Choosing the answer

    70 and 69 sheets differ by one, and 70 is exactly where the two costs meet at $10.50; only the strict “less expensive” separates them, ruling out 70. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. Turning words into math minor Understanding the question

    The words “less expensive than buying the package” must become a strict inequality, 0.15x < 10.50, and “largest value” must then pick the biggest whole number below the boundary. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. Error-prone arithmetic minor Working it out

    Dividing 10.50 by 0.15 means shifting the decimal point in both numbers, a step where a misplaced digit gives a wrong boundary. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Divide the package price by the price of one sheet to see where the two costs meet, then find the largest whole number of sheets that still costs less than $10.50.

Traps

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

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook sets up 0.15x ≤ 10.50 and x ≤ 70, which would allow 70 sheets, and then says that 70 sheets cost exactly $10.50. Buying the sheets has to be less expensive than the package, so the site uses strict inequalities, 0.15x < 10.50 and x < 70, which make 69 the largest number of sheets.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Total cost
cost=price of one×number bought\text{cost} = \text{price of one} \times \text{number bought}
Price of one is what a single sheet costs, here $0.15. Number bought is how many sheets, here x. Multiplying gives the total you pay, because you pay the same price for every sheet.

  1. Sort out the question

    The package costs $10.50. Single sheets cost $0.15 each. Macie buys x sheets.

    You need the largest x that makes the sheets cost less than the package. Costing the same is not enough.

  2. Write the inequality

    An inequality is a math sentence that compares two amounts. The sign < means is less than.

    The cost of x sheets is 0.15 times x. It must be less than 10.50.

    cost=price of one×number bought\text{cost} = \text{price of one} \times \text{number bought}

    0.15×x<10.500.15 \times x < 10.50

  3. Find where the costs are equal

    First find how many sheets cost exactly $10.50. Divide the package price by the price of one sheet. Multiply both numbers by 100 to clear the decimals.

    10.50÷0.15=1050÷15=7010.50 \div 0.15 = 1050 \div 15 = 70

    So 70 sheets cost exactly $10.50.

  4. Pick the largest number below 70

    x must be less than 70. You can only buy whole sheets. The largest whole number below 70 is 69.

    Watch out: 70 sheets cost the same as the package, not less. That is why choice D is wrong. Choices A and B are cheaper too, but they are not the largest.

    So the answer is C, 69.

Check your answer

Test both numbers. 69×0.15=10.3569 \times 0.15 = 10.35, and $10.35 is less than $10.50. But 70×0.15=10.5070 \times 0.15 = 10.50, which is not less. So 69 is the largest number that works.

More questions like this

The original DOE question
Question 67 as printed on page 154 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 154, question 67. Open page 154 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 92 of 100

2018 Test B Math, question 14

Question

What is the least common multiple of 24, 6, and 18?

  • A36
  • B48
  • C72
  • D144
Show the answer

Correct answer: C

Since 18 and 24 are both multiples of 6, find the least common multiple of only 18 and 24.

Multiples of 18:

18, 36, 54, 72…

Multiples of 24:

24, 48, 72…

The least common multiple of 6, 18, and 24 is 72.

Question analysis

This number properties question asks for the least common multiple of three numbers: 24, 6 and 18. Students can list multiples or use prime factors. Each candidate has to be checked against all three numbers, and it must be the smallest number that works, not just any shared 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 Least common multiples are grade 6 content, where New York's standard uses two numbers up to 12. This uses three larger numbers, but it is still a short factor task below typical grade 7 work.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 4, 8 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Finding a least common multiple is routine, and listing multiples of 24 reaches the answer by the third one. The wrong choices are built to catch incomplete checking: each passes the test for some of the numbers but fails for one, or passes for all and is simply not the smallest. Testing every candidate against all three numbers is the whole skill.

  1. Every choice must be checked minor Choosing the answer

    Each wrong choice passes part of the test: 36 is a multiple of 6 and 18, 48 of 6 and 24, and 144 of all three, so every candidate needs checking against all three numbers. (Embretson & Wetzel, 1987)

  2. Something you must already know minor Working it out

    Students must know what “least common multiple” means and see that 6 can be dropped, since it already divides both 24 and 18. (Anderson & Krathwohl, 2001)

Strategy: Note that 6 divides both 24 and 18, so only those two matter; list multiples of 24 and stop at the first one that 18 also divides.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Multiples of a number
n×1, n×2, n×3, …n \times 1,\ n \times 2,\ n \times 3,\ \ldots
n is any whole number. Its multiples are what you get when you multiply n by 1, 2, 3 and so on. The least common multiple of some numbers is the smallest number that is a multiple of every one of them.

  1. Understand the question

    A multiple of a number is that number times 1, 2, 3, and so on. A common multiple is a multiple of every number in the list.

    You need the least, or smallest, common multiple of 24, 6 and 18.

  2. Drop the 6

    6 goes evenly into 24, since 6×4=246 \times 4 = 24. It also goes into 18, since 6×3=186 \times 3 = 18.

    So any multiple of 24 is already a multiple of 6. You only need to work with 24 and 18.

  3. List the multiples of 24

    Start with the bigger number, since its list grows faster. Check each multiple to see if 18 divides it evenly.

    • 24×1=2424 \times 1 = 24, and 24 ÷ 18 is not a whole number.
    • 24×2=4824 \times 2 = 48, and 48 ÷ 18 is not a whole number.
    • 24×3=7224 \times 3 = 72, and 72÷18=472 \div 18 = 4. It works.

  4. Give the answer

    72 is the first multiple of 24 that 18 also divides. It is a multiple of 6 too, since 72÷6=1272 \div 6 = 12.

    Watch out: 36 works for 6 and 18, but not for 24, so choice A is wrong. 144 works for all three, but it is not the smallest.

    So the answer is C, 72.

Check your answer

Test 72 with all three numbers: 72÷24=372 \div 24 = 3, 72÷6=1272 \div 6 = 12 and 72÷18=472 \div 18 = 4. All are whole numbers. The smaller choices, 36 and 48, each fail for one of the numbers.

More questions like this

The original DOE question
Question 71 as printed on page 155 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 155, question 71. Open page 155 of the handbook (PDF)

Question 93 of 100

2018 Test B Math, question 16

Question

A cooler contains three types of beverages: 5 bottles of apple juice, 3 bottles of grape juice, and 6 bottles of orange juice. What is the probability that a bottle chosen at random from this cooler is not apple juice?

  • A19\frac{1}{9}
  • B514\frac{5}{14}
  • C914\frac{9}{14}
  • D23\frac{2}{3}
Show the answer

Correct answer: C

The total number of bottles of juice in the cooler is

5 + 3 + 6 = 14

The number of bottles of juice that are not apple juice (grape juice and orange juice) is

3 + 6 = 9

So the probability is 914\frac{9}{14} .

Question analysis

This probability question describes a cooler holding apple, grape and orange juice bottles in different numbers and asks for the chance that a randomly chosen bottle is not apple juice. Students count favorable bottles over all bottles. The word not is the trap, along with counting kinds of drink instead of bottles.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Simple probability with equally likely outcomes is grade 6 content in New York. This needs one count and one fraction, and the only extra demand is reading the word not, so it sits below grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 6, 41 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The counting is easy, 14 bottles in all, so the question mainly tests careful reading. The word “not” turns the event around, and the probability of the opposite event is waiting among the choices. The other wrong choices catch students who count kinds of juice instead of bottles, or who build a fraction with the wrong total.

  1. Turning words into math minor Understanding the question

    The event “not apple juice” must be turned into grape plus orange, 3 + 6 bottles over all 14, so the word “not” decides which counts go in the numerator. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add all the bottles for the denominator, then count only the grape and orange bottles for the numerator, or subtract the chance of apple juice from 1.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=outcomes you wantall outcomesP = \frac{\text{outcomes you want}}{\text{all outcomes}}
P is the probability, the chance that something happens. The top counts the outcomes you want, here bottles that are not apple. The bottom counts all outcomes, here every bottle. It works when every bottle has the same chance.
Probability of “not”
P(not A)=1−P(A)P(\text{not } A) = 1 - P(A)
A is an event, such as picking apple juice. P(A) is its chance. Every bottle either is apple or is not, so the two chances add to 1. Subtract from 1 to get the chance that A does not happen.

  1. Sort out the question

    The cooler has 5 apple, 3 grape and 6 orange bottles. One bottle is picked at random. That means every bottle has the same chance.

    You need the chance that it is not apple juice.

  2. Count all the bottles

    5+3+6=145 + 3 + 6 = 14

    There are 14 bottles. This is the bottom of the probability fraction, since any of them could be picked.

  3. Count the bottles that are not apple

    The bottles that are not apple are the grape and orange ones.

    3+6=93 + 6 = 9

    There are 9 such bottles. This is the top of the fraction.

  4. Write the probability

    P=outcomes you wantall outcomes=914P = \frac{\text{outcomes you want}}{\text{all outcomes}} = \frac{9}{14}

    This fraction cannot be simplified. 9 and 14 share no factor except 1.

    Watch out: 514\frac{5}{14} is the chance of picking apple juice. That is choice B. Read the word “not” carefully.

    So the answer is C, 914\frac{9}{14}.

Check your answer

Use the “not” rule. The chance of apple is 514\frac{5}{14}. So the chance of not apple is 1−514=9141 - \frac{5}{14} = \frac{9}{14}. Both ways give the same answer.

More questions like this

The original DOE question
Question 73 as printed on page 156 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 73. Open page 156 of the handbook (PDF)

Question 94 of 100

2018 Test B Math, question 27

Question

Nicole’s age now is three times Carmen’s age. If Carmen will be 17 in two years, how old was Nicole 5 years ago?

  • A38 yr
  • B40 yr
  • C45 yr
  • D50 yr
Show the answer

Correct answer: B

First, find out how old Nicole and Carmen are now.

Let NN = Nicole’s age now. Let CC = Carmen’s age now.

C+2=17C+2=17

C=15C=15 (Carmen’s age now)

N=3CN=3C

N=3(15)=45N=3(15)=45 (Nicole’s age now)

N−5=45−5=40N-5=45-5=40 (Nicole’s age 5 years ago)

Question analysis

This age word problem says Nicole is three times as old as Carmen now, and that Carmen will be 17 in two years. Students work out both present ages and then go back five years for Nicole. Keeping the times straight is the tricky part: two years ahead for Carmen, five years back for Nicole.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.7 one-step equations
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students write simple equations from word problems. Age problems like this one take three short steps, and the shifts in time are the only real hurdle, so the demand is typical of grade 7.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 1, 25 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is short, but the question moves through three points in time, and only one of them carries the three-times relationship. Students who find Nicole's present age often stop there, since it is a choice. Labeling each age with its time, now or then, keeps the two-year and five-year shifts with the right person and in the right direction.

  1. Turning words into math moderate Understanding the question

    Three points in time appear, now, in two years and 5 years ago, and “three times Carmen's age” holds only for now, so each shift must be applied to the right person. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Much to keep track of minor Working it out

    Students must hold Carmen's future age 17, her present age 15, Nicole's present age 45 and the 5-year shift while deciding which to report. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Find Carmen's age now by going back two years from 17, triple it for Nicole's current age, then subtract five.

Traps

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

Age at another time
age in n years=age now+n\text{age in } n \text{ years} = \text{age now} + n
n is a number of years. To find an age in the future, add n to the age now. To find an age in the past, subtract n instead. Everyone gets older at the same speed, one year per year.

  1. Sort out the times

    The question talks about three times: now, two years from now, and 5 years ago. Nicole is 3 times as old as Carmen now.

    Carmen will be 17 in two years. You need Nicole’s age 5 years ago.

  2. Find Carmen's age now

    In two years Carmen will be 17. So right now she is 2 years younger.

    17−2=1517 - 2 = 15

    Carmen is 15 now.

  3. Find Nicole's age now

    Nicole’s age now is three times Carmen’s age now.

    3×15=453 \times 15 = 45

    Nicole is 45 now.

    Watch out: 45 is her age now, choice C. The question asks about 5 years ago.

  4. Go back 5 years

    Going back in time means subtracting.

    45−5=4045 - 5 = 40

    Nicole was 40 five years ago.

    Watch out: Adding 5 gives 50, choice D. Also, the 2 years only apply to Carmen. Taking 2 more years off Nicole gives 38, choice A.

    So the answer is B, 40 yr.

Check your answer

Work forward from your answer. Nicole was 40 five years ago, so she is 45 now. Carmen is a third of that, 15. In two years Carmen will be 17. That matches the question.

More questions like this

The original DOE question
Question 84 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 84. Open page 158 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 95 of 100

2018 Test B Math, question 29

Question

If w−1w-1 is an odd integer, which one of the following must be an even integer?

  • Aw+1w+1
  • B2w−12w-1
  • C2w−22w-2
  • D2w+12w+1
Show the answer

Correct answer: C

One more than an odd integer must be even. One more than w−1w-1 is ww, therefore ww must be even.

Two times an even integer must be even, therefore 2w2w is even. An even integer decreased by 2 must be even.

Therefore, 2w−22w-2 must be even.

Question analysis

This algebra question says that w − 1 is an odd integer and asks which of four expressions must be even. It tests reasoning about odd and even numbers written with a variable. The difficulty is that "must be" means even for every allowed value of w, so one example that happens to work is not enough.

NYS standards
  • 7.EE.2 rewriting expressions to show relationships
  • 6.NS.4 common factors and multiples
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 work evaluates and rewrites expressions but rarely asks whether one is odd or even for every value. The algebra here is light; the demand is reasoning about parity and reading "must be" carefully.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 8, 20 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns on one early inference: since w − 1 is odd, w itself is even. Students who skip that step, or quietly treat w as the odd number, carry the wrong parity into every choice. After that the work is mechanical, because each expression adds an even or an odd amount to an even number. Trying one value, such as w = 4, confirms the choice in seconds.

  1. Variables instead of numbers moderate Understanding the question

    The stem gives no number at all: w − 1 is only known to be odd, so the parity of w itself has to be worked out before any choice can be judged. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Every choice must be checked moderate Choosing the answer

    All four choices are built from w or 2w plus or minus 1 or 2, so each one has to be tested for evenness; three of them come out odd for every allowed w. (Embretson & Wetzel, 1987)

  3. Something you must already know minor Working it out

    The parity rules are assumed, not stated: one more than an odd number is even, any multiple of 2 is even, and an even number plus or minus 1 is odd. (Anderson & Krathwohl, 2001)

Strategy: Choose a value of w that makes w − 1 odd, evaluate all four expressions, and keep the one that comes out even; then confirm it with a second value.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding 1 to an odd number
odd+1=even\text{odd} + 1 = \text{even}
Odd and even numbers take turns on the number line: 1, 2, 3, 4 and so on. Adding 1 moves you to the next whole number. So the number right after an odd number is always even, and the one after an even number is always odd.
Doubling makes an even number
2×n=even2 \times n = \text{even}
n is any whole number. An even number is one you can split into two equal whole groups. Two times n is always two equal groups of n, so it is always even.

  1. Sort out what you know

    You know that w−1w-1 is an odd integer. An integer is a whole number, like 3, 0 or −5. An even number can be split into two equal groups, like 4 or 10. An odd number is 1 more than an even number, like 3 or 11.

    You need the choice that must be even. Must means it is even for every possible value of ww, not just one.

  2. Decide whether w is odd or even

    The number ww is 1 more than w−1w-1. One more than an odd number is always even. For example, if w−1w-1 is 3, then ww is 4. So ww is even.

    Watch out: do not treat ww itself as the odd number. That mistake makes w+1w+1 look even and leads to choice A.

  3. Show that 2w − 2 is even

    Two times any whole number is even, so 2w2w is even. Taking 2 away from an even number moves you back two spots. You land on another even number. So 2w−22w-2 is even.

    Here is another way to see it. 2w−22w-2 is the same as 2×(w−1)2 \times (w-1). That is 2 times a whole number, so it must be even.

  4. Check the other three choices

    Each one adds or takes away 1 from an even number. That always gives an odd number.

    • A: w+1w+1 is even plus 1, so it is odd.
    • B: 2w−12w-1 is even minus 1, so it is odd.
    • D: 2w+12w+1 is even plus 1, so it is odd.

    Watch out: starting with 2w2w does not make a number even. The +1 in choice D makes it odd.

  5. Give the answer

    Only one choice is even for every possible ww. So the answer is C, 2w−22w-2.

Check your answer

Pick a number. Let w=4w = 4, so w−1=3w-1 = 3, which is odd. Then A is 5, B is 7, C is 6 and D is 9. Only C is even. Try w=10w = 10 too. You get 11, 19, 18 and 21. C is even again.

More questions like this

The original DOE question
Question 86 as printed on page 158 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 158, question 86. Open page 158 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 96 of 100

2018 Test B Math, question 32

Question

PET SURVEY

Number of Pets Number of Students
0 12
1 16
2 7
3 or more 5

Amy surveyed students at her school about the number of pets they have. What is the probability that a student who participated in the survey has at least 2 pets?

  • A740\frac{7}{40}
  • B112\frac{1}{12}
  • C18\frac{1}{8}
  • D310\frac{3}{10}
Show the answer

Correct answer: D

First, add the number of students for each category to find out how many total students were in the survey:

12 + 16 + 7 + 5 = 40

The number of students who had at least 2 pets are the ones who have 2 pets (7) plus the ones who have 3 or more (5). The total number of students with at least 2 pets is

7 + 5 = 12

The probability of a student in the survey having at least two pets is:

1240=310\frac{12}{40}=\frac{3}{10}

Question analysis

Amy's pet survey table shows how many pets each student has, and the question asks for the probability that a randomly chosen student has at least 2 pets. It tests finding a probability from survey data. The tricky part is reading "at least 2" correctly and using the right total as the denominator.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Simple probability from a frequency table is grade 6 work in New York. The only extra demand is interpreting "at least" and adding a column, two short steps a grade 7 student should handle easily.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
3 steps
Reading level
Question and choices: about grade 5, 43 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light, so the question is decided by the phrase "at least 2" and by remembering that the denominator is everyone surveyed. Students who include only one of the two qualifying rows reach an answer with the right denominator, which makes the mistake hard to notice. Rereading the condition before choosing is the main safeguard.

  1. Turning words into math minor Understanding the question

    "At least 2 pets" has to be turned into two rows of the table, the 2-pet row and the 3-or-more row, not one. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    The denominator is not printed anywhere; it has to be built by adding all four rows of the table, 12 + 16 + 7 + 5 = 40. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Total the students in every row to get the denominator, then add the counts for each row that satisfies "at least 2" and simplify the fraction.

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number you wanttotal numberP = \frac{\text{number you want}}{\text{total number}}
P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

  1. Sort out the question

    The table shows how many students have each number of pets. You pick one student who took the survey. You need the chance that the student has “at least 2 pets”.

    At least 2 means 2 or more. So two rows count: the 2 pets row and the 3 or more row.

  2. Find the total number of students

    The bottom of the fraction is everyone in the survey. Add all four rows.

    12+16+7+5=4012 + 16 + 7 + 5 = 40

    There are 40 students in all.

  3. Count the students you want

    Add the 2 pets row and the 3 or more row.

    7+5=127 + 5 = 12

    So 12 students have at least 2 pets.

    Watch out: using only the 7 students with exactly 2 pets gives 740\frac{7}{40}, choice A. Using only the 5 with 3 or more gives 540=18\frac{5}{40} = \frac{1}{8}, choice C.

  4. Write and simplify the probability

    Use the probability formula.

    P=1240P = \frac{12}{40}

    Divide the top and bottom by 4, since 4 goes into both.

    1240=12÷440÷4=310\frac{12}{40} = \frac{12 \div 4}{40 \div 4} = \frac{3}{10}

    So the answer is D, 310\frac{3}{10}.

Check your answer

The students with fewer than 2 pets are 12+16=2812 + 16 = 28. Then 28+12=4028 + 12 = 40, so every student is counted once. Also, 12 out of 40 is a bit less than a third. So is 310\frac{3}{10}.

More questions like this

The original DOE question
Question 89 as printed on page 159 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 89. Open page 159 of the handbook (PDF)

Question 97 of 100

2018 Test B Math, question 39

Question

A wooden box has a square base. The height of this box is 3 times the length of one side of the base. If one side of the base is 3 feet long, what is the volume of this box?

  • A9 cu ft
  • B27 cu ft
  • C36 cu ft
  • D81 cu ft
Show the answer

Correct answer: D

One side of the square base is 3 ft long. Since the height of the box is 3 times the length, then the height is

3 × 3 = 9 ft.

The volume of a rectangular prism is length × height × width. The volume of the wooden box is

V= 3 × 3 × 9 = 81 cu ft.

Question analysis

A wooden box has a square base, and its height is 3 times the length of a side of the base; students find its volume in cubic feet. It tests the volume of a rectangular prism with dimensions described in words. The work is short, but the height must be built from the side length before multiplying.

NYS standards
  • 6.G.2 volume with fractional edge lengths
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Volume of a box is grade 5 and 6 content, and this is a two-step version with whole numbers. Grade 7 volume problems usually involve composite or triangular prisms, so this sits below.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 1, 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

This is a routine volume question once all three dimensions are written down. Errors come from treating the box as a 3-foot cube, from stopping at the base area, or from mixing perimeter into the product. Listing the dimensions 3, 3 and 9 before multiplying prevents each of them.

  1. Turning words into math minor Understanding the question

    "The height of this box is 3 times the length of one side of the base" must become a height of 9 feet before the volume formula can be used. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Something you must already know minor Working it out

    The volume of a box, length times width times height, is not given, and the square base means the length and width are both 3 feet. (Anderson & Krathwohl, 2001)

Strategy: Write down all three dimensions first, computing the height from the base side, then multiply length, width and height.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Volume of a box
V=l×w×hV = l \times w \times h
V is the volume, the space inside the box. l is the length, w is the width and h is the height. l times w is the area of the bottom. Multiplying by h stacks that many layers of cubes.

  1. Sort out what you know

    The box has a square base with sides of 3 feet. A square has equal sides, so the length and width are both 3 feet. The height is “3 times the length of one side of the base”. You need the volume, the space inside the box.

  2. Find the height

    The height is 3 times the 3-foot side.

    h=3×3=9h = 3 \times 3 = 9

    The height is 9 feet.

    Watch out: the height is not 3 feet. If you use 3 for the height, you get 3×3×3=273 \times 3 \times 3 = 27. That is choice B.

  3. Find the volume

    Use the volume formula with all three sizes.

    V=3×3×9=81V = 3 \times 3 \times 9 = 81

    The volume is 81 cubic feet.

    Watch out: 3×3=93 \times 3 = 9 is only the area of the bottom. Stopping there gives choice A.

  4. Give the answer

    The box is 3 feet long, 3 feet wide and 9 feet tall. So the answer is D, 81 cu ft.

Check your answer

Think in layers. The bottom layer holds 3×3=93 \times 3 = 9 cubes, each 1 foot on a side. The box is 9 feet tall, so it holds 9 layers. Then 9×9=819 \times 9 = 81 cubic feet. The units are cubic feet because the answer is a volume.

More questions like this

The original DOE question
Question 96 as printed on page 160 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 160, question 96. Open page 160 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 98 of 100

2018 Test B Math, question 46

Question

In a sample of 50 cars at a local dealership, there are 12 red cars and 10 cars with backup cameras. Of the 12 red cars, 4 have backup cameras. If a car is selected at random from the given sample, what is the probability that both of the following are true: the car is not red and does not have a backup camera?

  • A35\frac{3}{5}
  • B1625\frac{16}{25}
  • C1925\frac{19}{25}
  • D45\frac{4}{5}
Show the answer

Correct answer: B

Create a chart using the given information and use subtraction to figure out how many cars are not red and do not have a back-up camera:

Red Not Red TOTAL
Back-up Camera 4 6
(10-4)		 10
No back-up Camera 32
(38-6)
TOTAL 12 38
(50-12)		 50

The probability of selecting a car that meets both conditions from the total of 50 cars at the dealership is 3250=1625\frac{32}{50}=\frac{16}{25} .

Question analysis

In a sample of 50 cars, 12 are red, 10 have backup cameras, and 4 of the red cars have cameras. Students find the probability that a randomly chosen car is neither red nor equipped with a camera. It tests compound probability with overlapping groups, and the overlap between red cars and camera cars is what makes it tricky.

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
Above grade 7 Grade 7 compound probability usually uses a ready-made table or tree. Here students must build the two-way table from words, handle an overlap, and use a "not this and not that" condition, several steps beyond typical work.
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, 63 words.
Corrected from the DOE handbook
  • A grammar slip in the explanation was corrected.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is an overlap problem dressed as probability: the hard part is counting the cars that avoid both features, not the division by 50. A two-way table makes it routine, but students working in their heads often keep only one of the two conditions. The question rewards placing the 4 red cars with cameras in the table before doing any subtraction.

  1. Much to keep track of major Working it out

    Three overlapping counts, 12 red, 10 with cameras and 4 that are both, must be sorted into four groups before the wanted one, neither red nor with a camera, can be found. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Many steps moderate Working it out

    Find the non-red cars with cameras, 10 − 4 = 6, and the non-red cars, 50 − 12 = 38, subtract to get 32, then simplify 32/50. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer moderate Choosing the answer

    19/25, or 38/50, has the same denominator as the right 16/25 and sits just above it; only removing the 6 non-red cars that have cameras from the 38 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  4. Complex sentences minor Understanding the question

    The last sentence stacks "both of the following are true", a colon and two negatives, "not red" and "does not have"; untangling it is reading work, not extra math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Draw a two-way table with red and not red across, camera and no camera down, fill in the given counts, and complete the missing cells by subtraction.

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

What was corrected from the DOE handbook, and why

  • Explanation: The handbook says “a car that meet both conditions”; the site says “a car that meets both conditions.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number you wanttotal numberP = \frac{\text{number you want}}{\text{total number}}
P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

  1. Sort out what you know

    There are 50 cars. Of these, 12 are red and 10 have backup cameras. Of the red cars, 4 have cameras. You pick one car at random. You need the chance that “both of the following are true”: the car is not red, and it has no camera.

  2. Set up a two-way table

    A two-way table sorts each car by two things at once: its color and its camera. Put in the numbers you know.

    RedNot redTotal
    Camera4?10
    No camera?
    Total12?50

  3. Fill in the missing numbers

    Not red with a camera: 10−4=610 - 4 = 6 cars.

    All cars that are not red: 50−12=3850 - 12 = 38 cars.

    Not red with no camera: 38−6=3238 - 6 = 32 cars.

    Watch out: the 38 cars that are not red include 6 with cameras. Using 38 gives 3850=1925\frac{38}{50} = \frac{19}{25}, choice C. Using all 40 cars with no camera gives 4050=45\frac{40}{50} = \frac{4}{5}, choice D.

  4. Write the probability

    Use the probability formula. There are 32 cars you want out of 50 cars in all.

    P=3250P = \frac{32}{50}

    Divide the top and bottom by 2.

    3250=1625\frac{32}{50} = \frac{16}{25}

    So the answer is B, 1625\frac{16}{25}.

Check your answer

Finish the table. Red cars with no camera: 12−4=812 - 4 = 8. Now add the four inside boxes: 4+6+8+32=504 + 6 + 8 + 32 = 50. That is every car, so the numbers fit. Also, the cars with no camera are 8+32=408 + 32 = 40, and 50−10=4050 - 10 = 40 too.

More questions like this

The original DOE question
Question 103 as printed on page 162 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 103. Open page 162 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 99 of 100

2018 Test B Math, question 54

Question

Raoul is at least 3 years older than Vahn. Which of the following inequalities gives the relationship between Raoul’s age (rr ) and Vahn’s age ( vv )?

  • Ar−v≥3r-v\geq 3
  • Br−v≤3r-v\leq 3
  • C3−v≤r3-v\leq r
  • D3−r≤v3-r\leq v
Show the answer

Correct answer: A

Raoul is at least 3 years older than Vahn, which can be written as

r ≥ v + 3

Rewrite this inequality to match the answer options:

r − v ≥ 3

Question analysis

Raoul is at least 3 years older than Vahn, and students choose the inequality that links their ages, r and v. It tests translating a phrase like "at least" into an inequality. The correct form may not match the one students write first, so they need to rearrange carefully and keep the direction of the inequality.

NYS standards
  • 7.EE.4 equations and inequalities from problems
  • 6.EE.8 writing simple inequalities
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Writing inequalities from words is grade 6 and 7 work. The extra demand is that "at least" must become the right symbol and the choices are rearranged forms, so students must manipulate inequalities rather than just recognize one.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 8, 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

Two skills are tested at once: translating "at least … older than" into an inequality, and recognizing that inequality after a rearrangement. The two choices that begin with 3 − v and 3 − r look like legitimate moves but say only that the two ages add up to at least 3. Testing one pair of ages that fits the sentence and one that does not, such as 20 and 12 against 13 and 12, sorts the choices without algebra.

  1. Turning words into math major Understanding the question

    "Raoul is at least 3 years older than Vahn" has to become r ≥ v + 3; "at least" calls for ≥, though the word "least" seems to point the other way. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

    No choice is written as r ≥ v + 3, so the natural translation has to be rearranged to r − v ≥ 3 before it can be matched. (Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer moderate Choosing the answer

    r − v ≤ 3 is the right r − v ≥ 3 with the symbol reversed; only deciding that "at least" means greater than or equal to separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the sentence directly as Raoul's age compared with Vahn's age plus 3, then rearrange each choice into that form to see which one matches.

Traps

  • Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Answer in another form: The right answer is written differently from the way you worked it out.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Greater than or equal to
a≥ba \geq b
a and b are two amounts. The sign ≥ means a is greater than or equal to b, so a is b or more. The words at least mean the same thing. The sign ≤ means less than or equal to, or at most.

  1. Sort out the question

    The letter rr is Raoul’s age, and vv is Vahn’s age. Raoul is “at least 3 years older than Vahn”. That means the gap between their ages is 3 years or more. An inequality is a math sentence that uses a sign like ≥\geq instead of =. You need the one that says this.

  2. Write the age gap

    Raoul is older, so the gap is his age minus Vahn’s age. The gap is r−vr - v. For example, ages 15 and 10 have a gap of 5.

  3. Turn at least into a sign

    At least 3 means 3 or more. So use ≥\geq, which means greater than or equal to.

    r−v≥3r - v \geq 3

    This is choice A.

    Watch out: the word least does not mean less. Choice B, r−v≤3r - v \leq 3, says the gap is 3 or less. Then Raoul could even be younger.

  4. Check choices C and D

    Add vv to both sides of choice C. You get 3≤r+v3 \leq r + v. That is about the sum of their ages, not the gap. Choice D works the same way. Neither one says Raoul is older.

  5. Give the answer

    Only one choice says the gap is 3 years or more. So the answer is A, r−v≥3r-v \geq 3.

Check your answer

Test ages that do not fit: Raoul 11 and Vahn 10. He is only 1 year older. Choice A gives 11−10=111 - 10 = 1, and 1 is not 3 or more. So A rightly says no. Choice B says yes, so B is wrong.

More questions like this

The original DOE question
Question 111 as printed on page 164 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 111. Open page 164 of the handbook (PDF)

Question 100 of 100

2018 Test B Math, question 56

Question

There are now xx cans stacked on a shelf that holds 36 cans when full. If 4 of these cans were removed, the shelf would be half full. What is the value of xx ?

  • A14
  • B16
  • C18
  • D22
Show the answer

Correct answer: D

The shelf, when full, holds 36 cans. When the shelf is half full, it holds 18 cans.

x − 4 = 18

x = 22

Question analysis

A shelf holds 36 cans when full, and removing 4 of the cans now on it would leave it half full. Students find how many cans are on the shelf now. It tests writing and solving a one-step equation from words, and the risk is undoing the removal in the wrong direction.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
Below grade 7 Grade 6 students solve equations like x − p = q, and this one needs only that and halving 36. It is a quick, routine word problem, easier than the two-step equations typical of grade 7.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 3, 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

Short and routine, this question still punishes a quick reading. Students must keep three amounts apart, the 36-can capacity, the 18 cans on a half-full shelf and the x cans there now, and then undo the removal by adding. Writing x − 4 = 18 before computing prevents every wrong choice.

  1. Turning words into math minor Understanding the question

    The 36 cans are the shelf's capacity, not its contents, and "if 4 of these cans were removed" describes a hypothetical, so the equation is x − 4 = 18. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. A very close wrong answer minor Choosing the answer

    18 sits just below the right 22, and both are near half of 36; only remembering that x counts the cans before the 4 were removed separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Work out how many cans make a half-full shelf, then undo the removal of the 4 cans; write it as an equation in x if that helps.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Undoing subtraction
x−a=b means x=b+ax - a = b \text{ means } x = b + a
x is the number you want. a is the amount taken away, and b is what is left. To undo taking away a, add a back. Adding the same number to both sides keeps the equation balanced.

  1. Sort out what you know

    The shelf holds 36 cans when it is full. That is its size, not how many cans are on it now. Right now it has xx cans. If 4 cans were taken off, the shelf would be half full. You need xx.

  2. Find half of a full shelf

    A full shelf holds 36 cans, so half full is half of that.

    36÷2=1836 \div 2 = 18

    A half-full shelf has 18 cans.

  3. Write an equation

    Taking 4 cans away from xx leaves 18 cans.

    x−4=18x - 4 = 18

    Watch out: 18 is the number after the cans are taken off. Picking 18 gives choice C.

  4. Add the 4 cans back

    Undo the subtraction by adding 4 to both sides.

    x=18+4=22x = 18 + 4 = 22

    Watch out: subtracting the 4 again gives 18−4=1418 - 4 = 14, choice A. The cans were taken from xx, so xx is more than 18.

  5. Give the answer

    There are 22 cans on the shelf now, before any are taken off. So the answer is D, 22.

Check your answer

Start with 22 cans and take off 4. That leaves 22−4=1822 - 4 = 18 cans. Half of 36 is 18, so the shelf is half full. The answer works.

More questions like this

The original DOE question
Question 113 as printed on page 164 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 113. Open page 164 of the handbook (PDF)