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Practice Test

6.SP.1b samples and populations

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Question 1 of 5

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 2 of 5

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 3 of 5

2024 Test A Math, question 47

Question

In a survey of 200 adults in the town of Waskegon, 45 reported reading the online version of the Waskegon Bulletin the previous day. If 25,000 adults live in Waskegon, which number is the best estimate of the number of adults who read the online version of the Waskegon Bulletin the previous day?

  • A5,600
  • B9,000
  • C11,300
  • D24,800
Show the answer

Correct answer: A

In the survey, 45 out of 200 adults reported reading the online version of the Waskegon Bulletin, and there are 25,000 adults living in Waskegon. Set up a proportion to find the number of adults who read the online version.

45200=x25,000\frac{45}{200} = \frac{x}{25{,}000}

x=(45)(25,000)200=x = \frac{(45)(25{,}000)}{200} =

(45)(125)=5,625(45)(125) = 5{,}625

Rounded to the nearest 100, this number is 5,600.

Question analysis

In a survey of 200 adults in Waskegon, 45 said they read the online Waskegon Bulletin the day before, and the town has 25,000 adults. This sampling question asks for the best estimate of how many adults in town read it. Students scale the sample's rate up to the population with a proportion and round.

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Using a sample to estimate a count for a whole population is typical grade 7 statistics and ratio work. This version takes one proportion with large numbers and a rounding step, about the usual 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 13, 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 idea is a single proportion, and the choices are far apart, so a reasonable estimate is enough. The main trap is treating 45 out of 200 as 45 percent, which doubles the estimate; the other wrong choices combine the survey numbers without a proportion. Noting that 45 is a bit under a quarter of 200, so the town figure must be a bit under a quarter of 25,000, confirms the choice.

  1. Turning words into math minor Understanding the question

    The survey's 45 out of 200 is a rate to scale up to the town's 25,000 adults, a fraction of the sample rather than a percent. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  2. Error-prone arithmetic minor Working it out

    The scale factor 25,000 ÷ 200 = 125 and the product 45 × 125 = 5,625 involve large numbers, and the result must then be rounded to the nearest hundred. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Set the survey's fraction, 45 out of 200, equal to the unknown number out of 25,000, then scale up: 25,000 is 125 times 200.

Traps

  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
  • 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

Proportion
ab=cd\frac{a}{b} = \frac{c}{d}
A proportion says two ratios are equal. A ratio compares two numbers by dividing. Here a out of b in the survey matches c out of d in the whole town. Use it when a small group stands for a bigger group.

  1. Sort out the question

    A survey asked 200 adults, and 45 of them read the online paper. The town has 25,000 adults. You need a good estimate of how many adults in the whole town read it. Assume the town is like the survey. Then the same fraction of adults read it.

  2. Set up a proportion

    A proportion says two fractions are equal. Let x be the number of readers in the town. The fraction of readers is the same in both groups.

    ab=cd\frac{a}{b} = \frac{c}{d}

    45200=x25,000\frac{45}{200} = \frac{x}{25{,}000}

    Watch out: multiplying 45 by 200 gives 9,000, choice B. That mixes the numbers without a plan.

  3. Find the scale factor

    The scale factor is how many times bigger the town is than the survey group.

    25,000÷200=12525{,}000 \div 200 = 125

    The town is 125 times as big. So multiply the top, 45, by 125 too.

  4. Scale up the readers

    x=45×125x = 45 \times 125

    Break 125 into 100 and 25.

    45×100=4,50045 \times 100 = 4{,}500

    45×25=1,12545 \times 25 = 1{,}125

    4,500+1,125=5,6254{,}500 + 1{,}125 = 5{,}625

    About 5,625 adults read it. Watch out: 45 out of 200 is not 45%. A percent is out of 100. Using 45% gives about 11,300, choice C.

  5. Round to the nearest hundred

    The choices are rounded, since this is an estimate. 5,625 is between 5,600 and 5,700. Its last two digits, 25, are less than 50, so round down to 5,600.

    So the answer is A, 5,600.

Check your answer

45 out of 200 is a bit less than 50 out of 200, which is one fourth. One fourth of 25,000 is 6,250. So the answer should be a bit less than 6,250. 5,600 fits, and the other choices are all much bigger.

More questions like this

The original DOE question
Question 104 as printed on page 88 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 88, question 104. 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 4 of 5

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 5 of 5

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)