SHSAT Sunset

Practice Test

NOT, LEAST or EXCEPT

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Question 1 of 2

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 2 of 2

2018 Test B Math, question 53

Question

At North High School, a survey asked two questions, Question A and Question B. For each question, students could answer either “yes” or “no.” Of the 800 students who responded to the survey, 720 answered “yes” to Question A, and 640 answered “yes” to Question B. What is the least possible number of these students who could have answered “yes” to both questions?

  • A80
  • B160
  • C560
  • D640
Show the answer

Correct answer: C

None of the 80 students (800 − 720) who answered “no” to Question A could have answered “yes” to both questions.

Therefore, the least possible number of students who could have answered “yes” to both questions can be found by subtracting the 80 who answered “no” to Question A from the 640 who answered “yes” to Question B or 640 − 80 = 560.

Question analysis

In a survey of 800 students at North High School, 720 said yes to the first question and 640 said yes to the second. Students must find the smallest possible number who said yes to both. It tests reasoning about overlapping groups, and asking for the least possible overlap rather than an exact count makes it unusual.

NYS standards
  • 7.SP.8b sample spaces for compound events
  • 7.EE.4 equations and inequalities from problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students work with Venn diagrams and two-way tables when all counts are given. Finding the smallest possible overlap requires reasoning about extremes rather than computing, which is rare in grade 7 and new to most students.
Estimated difficulty
Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Solution
2 steps
Reading level
Question and choices: about grade 8, 60 words.
Corrected from the DOE handbook
  • A repeated phrase and a stray comma in the explanation were 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

Most students have never been asked for a smallest overlap, so they fall back on a picture of nested groups and give the largest one. The key move is to push the two yes groups apart as far as possible: only the 80 students who said no to Question A can absorb some of Question B's yes answers. Once that is seen, the arithmetic is one subtraction, so the question rewards reasoning far more than computation.

  1. No obvious method major Working it out

    Nothing in the wording points to the method: the smallest overlap appears only when all 80 students who said no to Question A are imagined saying yes to Question B. (Schoenfeld, 1985)

  2. Turning words into math moderate Understanding the question

    "The least possible number of these students who could have answered “yes” to both questions" asks for an extreme case, which has to be turned into an arrangement of the 800 students. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

  3. A very close wrong answer minor Choosing the answer

    640 is only 80 above the right 560, so a rough sense of size cannot choose between them; only asking whether the overlap is the least or the greatest separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Add the two yes counts and compare the sum with the number of students; any amount above the total must come from students counted in both groups.

Traps

The wrong answers, classified

What was corrected from the DOE handbook, and why

  • Explanation: The handbook prints “(800 − 720)” twice in its first sentence, after “80 students” and again after “Question A,” and puts a comma between subject and verb in “the least possible number of students who could have answered “yes” to both questions, can be found.” The site keeps the first “(800 − 720)” and drops the comma.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Least overlap of two groups
least overlap=a+b−t\text{least overlap} = a + b - t
a is the size of the first group and b is the size of the second. t is the total number of people. If a + b is more than t, some people must be in both groups. The extra amount is the smallest number that can be in both.

  1. Sort out what you know

    There were 800 students. Of them, 720 said yes to Question A and 640 said yes to Question B. You need the least number who could have said yes to both. The overlap is the group in both yes groups. You want the smallest overlap.

  2. Find who said no to A

    Students who said no to Question A cannot be in the overlap.

    800−720=80800 - 720 = 80

    So 80 students said no to Question A.

    Watch out: 80 is not the answer. It is only a step, and it is choice A.

  3. Keep the overlap small

    To make the overlap as small as you can, put as many B yes answers outside it as possible. The only students outside it are the 80 who said no to A. So pretend all 80 of them said yes to B.

  4. Count the rest of the yes group

    The other yes answers to B must come from students who also said yes to A.

    640−80=560640 - 80 = 560

    So at least 560 students said yes to both.

    Watch out: 640 is the most that could say yes to both, not the least. That is choice D.

  5. Give the answer

    The fewest students who could say yes to both questions is 560. So the answer is C, 560.

Check your answer

Use the overlap formula: 720+640−800=560720 + 640 - 800 = 560. There are 1,360 yes answers but only 800 students, so 560 students must be counted twice. You can also check the groups. 560 said yes to both, 160 to A only and 80 to B only. That makes 560+160+80=800560 + 160 + 80 = 800.

More questions like this

The original DOE question
Question 110 as printed on page 163 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 110. Open page 163 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.