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Lesson 16 of 46 · Quantitative question types

Numeric entry and select-one-or-more questions

Answer numeric entry questions with no choices to lean on, and test every choice of a select-one-or-more question against one simplified condition.

About 20 minutes

In this lesson

  • Answer a numeric entry question in the quantity, unit and rounding the question asks for, and verify it without answer choices.
  • Reduce a select-one-or-more condition to a simple form, then test each choice against it.
  • Tell "could be" from "must be" and handle the endpoints of an inequality correctly.

The idea

The four quantitative formats

GRE Quantitative Reasoning uses four question types: Quantitative Comparison; multiple choice with one answer from five; multiple choice where you select one or more answer choices; and Numeric Entry. Any of them can stand alone or be part of a Data Interpretation set built around shared data. An on-screen calculator is available throughout the Quantitative Reasoning sections. The previous lesson covered quantitative comparison; this one covers the two formats in which the usual habits of a single-answer multiple-choice question stop working.

Numeric entry: no choices to lean on

In a numeric entry question you produce the number yourself. That removes two familiar safety nets: you cannot work backwards from the choices, and a wrong answer will not be caught by its absence from the list. The work is the same; the checking has to come from you.

Three habits protect you:

  1. Answer the question asked. Underline, in your head, the exact quantity: the percent or the number, the total or the part, the original value or the new one. Many wrong entries are correct answers to a neighbouring question.
  2. Match the form and units. If the question asks for a percent, decide whether the entry is 57.5 (the percent) or 0.575 (the decimal) by reading how the question frames it. If it gives a unit, answer in that unit.
  3. Round once, at the end. When the question asks for an answer to the nearest tenth or whole number, carry full precision through every step and round only the final result. Rounding an intermediate value can move the final answer.

Then check by a second route: substitute the answer back into the situation, or solve it a different way. Without choices, this is the only check you get.

The calculator

The on-screen calculator is useful for awkward arithmetic: dividing 300 by 4.1667, or multiplying several decimals. It does not set up the problem. Write the expression first, estimate its size, and compare the calculator's result with your estimate. A result ten times too large is a keying slip, and an estimate catches it in a second.

Select one or more: one condition, tested many times

These questions give several choices and ask you to select all that are correct; the number of correct choices is not fixed. Decide each choice on its own and do not stop at the first one that works.

The efficient approach is to simplify the condition once, then test each choice against the simple form:

  • An inequality becomes a range: 3<2x−5≤113 < 2x - 5 \le 11 becomes 4<x≤84 < x \le 8.
  • A divisibility condition becomes a list of required factors.
  • A geometric condition becomes a range of possible lengths or angles.

Testing each choice against the original condition repeats the algebra five times and multiplies the chances of a slip.

"Could be" and "must be"

Read which one the question asks.

  • Could be (or "which of the following are possible"): a choice is correct if at least one allowed situation produces it. Find one example.
  • Must be (or "which must be true"): a choice is correct only if every allowed situation produces it. One counter-example rules it out.

The same set of choices can have different answers under the two wordings.

Endpoints

With a range, check each boundary: is it "less than" or "less than or equal to"? In 4<x≤84 < x \le 8, the value 8 is allowed and 4 is not. Choices that sit exactly on a boundary are there to test this.

Worked examples

Example 1: Numeric entry: the right quantity in the right form

A tank holds 240 litres when full. It is 35% full, and then 54 litres are added. What percent of the tank's capacity is now filled? (Enter the percent, without the percent sign.)

  1. Find the starting volume: 0.35×240=840.35 \times 240 = 84 litres.

  2. Add the new water: 84+54=13884 + 54 = 138 litres.

  3. Convert to a percent of capacity, as the question asks: 138240=0.575\frac{138}{240} = 0.575, which is 57.5%57.5\%. The entry is 57.5, not 0.575 and not 138.

  4. Second route: 54 litres is 54240=22.5%\frac{54}{240} = 22.5\% of the tank, and 35%+22.5%=57.5%35\% + 22.5\% = 57.5\%. The two routes agree.

Answer

Enter 57.5. Tempting wrong entries: 138 (the volume, not the percent), 0.575 (the decimal, when the question asked for the percent) and 89 (adding 54 to 35 as if 54 litres were 54%).

Example 2: Select one or more: reduce once, test each

Which of the following could be the value of xx if 3<2x−5≤113 < 2x - 5 \le 11? Indicate all such values.

3, 4, 6, 8, 9

  1. Simplify once. Add 5 to every part: 8<2x≤168 < 2x \le 16. Divide every part by 2 (positive, so the signs stay): 4<x≤84 < x \le 8.

  2. Test each choice against 4<x≤84 < x \le 8: 3 is below the range; 4 is the left endpoint, which is excluded because the inequality is strict; 6 is inside; 8 is the right endpoint, which is included; 9 is above.

  3. Check the endpoints in the original: x=4x = 4 gives 2x−5=32x - 5 = 3, and 3<33 < 3 is false. x=8x = 8 gives 1111, and 11≤1111 \le 11 is true.

Answer

6 and 8. Including 4 (a strict endpoint) or leaving out 8 (an included endpoint) are the two tempting slips.

Example 3: Select one or more: "must be"

nn is a positive integer, and n2n^2 is divisible by 12. Which of the following must be divisors of nn? Indicate all such numbers.

2, 3, 4, 6, 9

  1. Reduce the condition. 12=22×312 = 2^2 \times 3. Since 33 divides n2n^2 and 3 is prime, 3 divides nn. Since 22 divides n2n^2, 2 divides nn. So nn must be a multiple of 6.

  2. "Must be" needs every allowed nn. The smallest allowed value is n=6n = 6: 36=12×336 = 12 \times 3, so it is allowed. Use it as the counter-example test.

  3. 2, 3 and 6 divide every multiple of 6, so they must divide nn. 4 does not divide 6, and 9 does not divide 6, so neither must divide nn.

Answer

2, 3 and 6. Choosing 4 comes from reading the factor 222^2 in 12 as a factor of nn; but nn needs only one factor of 2 for n2n^2 to have two.

Common traps

  • Answering a neighbouring question

    The volume instead of the percent, the part instead of the total, the new price instead of the change: numeric entry gives no list to show you that your number is not among the expected answers. Reread the last line of the question before you enter anything.

  • Rounding too early

    Rounding 53\frac{5}{3} hours to 1.7 in the middle of a calculation can change the final answer in the first decimal place. Keep full precision, or exact fractions, until the last step.

  • Stopping after the first correct choice

    In a select-one-or-more question, finding one correct choice tells you nothing about the others. Test every choice against the simplified condition.

  • Treating "could be" as "must be", or the reverse

    "Could be" needs one example; "must be" needs every case. Reading one as the other changes which choices are correct, so check the wording before testing.

Check yourself

Answer each question before you open its answer.

Question 1: Numeric entry: a removed value

Numeric Entry

The mean of five numbers is 18. When one of the numbers is removed, the mean of the remaining four numbers is 16. What number was removed?

  1. A.2
  2. B.17
  3. C.18
  4. D.26
  5. E.34
Show the answer

Answer: D. 26

Working

Work with totals, not means. The five numbers total 5×18=905 \times 18 = 90. The remaining four total 4×16=644 \times 16 = 64. The removed number is 90−64=2690 - 64 = 26. Check: putting 26 back, 64+265=905=18\frac{64 + 26}{5} = \frac{90}{5} = 18.

Why the other options are wrong

  • A. This is the difference between the two means, not a number in the list.
  • B. This averages the two means, which has no meaning here.
  • C. This is the original mean; removing a number equal to the mean would leave the mean at 18, not 16.
  • E. This adds the two means; it does not use the number of values at all.

What carries over

When values are added to or removed from a set, convert every mean to a total first.

Question 2: Select one or more: which values are possible?

Multiple-choice: select one or more answer choices

If ∣x−3∣<4|x - 3| < 4, which of the following could be the value of xx? Consider each value separately.

  • I. −2-2
  • II. 00
  • III. 6.56.5
  1. A.I only
  2. B.II only
  3. C.I and II only
  4. D.II and III only
  5. E.I, II and III
Show the answer

Answer: D. II and III only

Working

Reduce the condition once: ∣x−3∣<4|x - 3| < 4 means −4<x−3<4-4 < x - 3 < 4, so −1<x<7-1 < x < 7. Test each value: −2-2 is below −1-1, so it fails; 00 is inside; 6.56.5 is inside. Direct check: ∣0−3∣=3<4|0 - 3| = 3 < 4 and ∣6.5−3∣=3.5<4|6.5 - 3| = 3.5 < 4, but ∣−2−3∣=5|-2 - 3| = 5.

Why the other options are wrong

  • A. −2-2 is outside the range: ∣−2−3∣=5|-2 - 3| = 5, which is not less than 4. This comes from solving the left half as x>−7x > -7, subtracting 3 instead of adding it.
  • B. Stops after finding that 0 works. 6.5 also satisfies the condition, since ∣6.5−3∣=3.5<4|6.5 - 3| = 3.5 < 4.
  • C. Includes −2-2, which gives ∣−5∣=5|-5| = 5, not less than 4, and leaves out 6.5, which works.
  • E. −2-2 fails the condition: its distance from 3 is 5, more than 4.

What carries over

Turn an absolute-value condition into a range once, then test every choice against the range.

Question 3: Numeric entry: average speed for a round trip

Numeric Entry

A car travels 150 km at an average speed of 60 km/h and returns along the same 150 km at an average speed of 90 km/h. What is its average speed for the whole round trip, in kilometres per hour? (Give the answer to the nearest tenth if it is not a whole number.)

  1. A.36
  2. B.60
  3. C.71.4
  4. D.72
  5. E.75
Show the answer

Answer: D. 72

Working

Outward time: 15060=2.5\frac{150}{60} = 2.5 hours. Return time: 15090=53\frac{150}{90} = \frac{5}{3} hours. Total time: 52+53=256\frac{5}{2} + \frac{5}{3} = \frac{25}{6} hours. Average speed: 300÷256=300×625=72300 \div \frac{25}{6} = \frac{300 \times 6}{25} = 72 km/h. Second route: for equal distances, the average speed is 2×60×9060+90=10 800150=72\frac{2 \times 60 \times 90}{60 + 90} = \frac{10\,800}{150} = 72.

Why the other options are wrong

  • A. Divides the one-way distance, 150 km, by the total time; the round trip is 300 km.
  • B. Assumes each leg takes 2.5 hours; the return at 90 km/h takes only 53\frac{5}{3} hours.
  • C. Rounds the return time to 1.7 hours before dividing; 300÷4.2≈71.4300 \div 4.2 \approx 71.4. Rounding only at the end gives a whole number.
  • E. Averages the two speeds; the car spends longer at the slower speed, so the average speed is below 75.

What carries over

Average speed is total distance over total time; keep exact fractions until the last step.

Key points

  • In numeric entry, answer the exact quantity asked, in its form and unit, and check by a second route.

  • Round once, at the end.

  • In select-one-or-more questions, simplify the condition once and test every choice against it.

  • "Could be" needs one example; "must be" needs every case.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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