GMAT Quantitative Quiz: Overlapping Sets

What this quiz covers

This quiz focuses on Overlapping Sets, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT Quantitative.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\tfrac14$ (correct answer)
2. $\tfrac{3}{11}$
3. $\tfrac{9}{40}$
4. $\tfrac{1}{3}$

Explanation: The bar shows: Spreadsheet only 80, Presentation only 55, Both 45, Neither 70.
Employees using at least one $=80+55+45=180$.
Probability of using both $=45/180=1/4$.
B: $3/11$ ≈0.27, would require 49 using both.
C: $9/40$ =0.225, would require 40.5 using both.
D: $1/3$ =0.333, would require 60 using both.

Question 2

1. 45
2. 55 (correct answer)
3. 65
4. 75

Explanation: To find those using exactly two devices, we calculate: (laptops and tablets only) + (laptops and smartphones only) + (tablets and smartphones only). Using the inclusion-exclusion principle: Laptops and tablets only = 50 - 15 = 35; Laptops and smartphones only = 35 - 15 = 20; Tablets and smartphones only = 25 - 15 = 10. Total using exactly two devices = 35 + 20 + 10 = 55. Choice A (45) incorrectly subtracts 15 twice from the sum. Choice C (65) adds 15 instead of subtracting it from each pair. Choice D (75) uses the raw intersection values without accounting for triple overlap.

Question 3

1. 60
2. 75
3. 90 (correct answer)
4. 105

Explanation: Let the regions be: only one activity = 150, exactly two activities = y, all three = 30. Using inclusion-exclusion and the constraint that all students participate in at least one activity: Total = 150 + y + 30 = 300, so students in multiple activities = 150. Also, |S ∪ C ∪ V| = 180 + 150 + 120 - (sum of pairwise intersections) + 30 = 300. The sum of pairwise intersections includes contributions from exactly two (y) and all three (3 × 30 = 90), so sum = y + 90. Therefore: 450 - (y + 90) + 30 = 300, giving y = 90. Choice A (60) miscounts the triple intersection contribution. Choice B (75) uses incorrect inclusion-exclusion setup. Choice D (105) adds rather than subtracts in the formula.

Question 4

1. 10 (correct answer)
2. 15
3. 20
4. 25

Explanation: Using inclusion-exclusion: |E ∪ S ∪ F| = |E| + |S| + |F| - |E ∩ S| - |E ∩ F| - |S ∩ F| + |E ∩ S ∩ F|. Since all employees speak at least one language, |E ∪ S ∪ F| = 200. Substituting: 200 = 140 + 80 + 60 - 40 - 30 - 20 + x = 190 + x. Therefore, x = 10. Choice B (15) incorrectly assumes we can exceed the constraint of pairwise intersections. Choice C (20) uses the minimum of the pairwise intersections without considering the union constraint. Choice D (25) ignores the inclusion-exclusion formula entirely.

Question 5

1. 34
2. 36 (correct answer)
3. 37
4. 39

Explanation: Exactly one newspaper means the counts that appear only in a single circle: Financial Times only 15, Wall Street Journal only 12, Economist only 9.
Sum $=15+12+9=36$.
Choices A, C, and D differ by ±1–3 to look plausible but are incorrect.

Question 6

1. 67.6%
2. 70.6%
3. 76.5% (correct answer)
4. 82.4%

Explanation: The bars give: Science only 28, Both 16, Math only 24.
At least one exam $=28+16+24=68$.
Exactly one $=28+24=52$.
Percentage $=52/68 \approx 76.5\%$.
Other choices correspond to treating 48, 50, or 56 students as the numerator—none matches the diagram.

Question 7

1. 76.7%
2. 80.0%
3. 86.7% (correct answer)
4. 90.0%

Explanation: From the pie segments: Car only 65, Bicycle only 25, Both 40, None 20.
At least one vehicle $=65+25+40=130$ out of 150.
Percentage $=130/150=86.7\%$ (rounded to one decimal).
A and B understate by counting only car owners or omitting the overlap; D assumes all but the bicycle-only group.

Question 8

1. 13
2. 16
3. 19 (correct answer)
4. 22

Explanation: Add the six known regions: 18+22+14+9+7+11 = 81.
Let $x$ be the number in all three.
Total $=81+x=100 \Rightarrow x=19$.
A, B, and D correspond to subtracting or adding 3 to the correct total and are therefore incorrect.