Statistics, Sets & Counting — GMAT Focus Questions

29 GMAT Focus practice questions on Statistics, Sets & Counting, part of Quantitative Reasoning. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. In a survey of 200 people, 80 read magazine A, 90 read magazine B, 70 read magazine C, 30 read both A and B, 25 read both A and C, 20 read both B and C, and 10 read all three. How many read exactly one magazine?

  1. 100
  2. 110
  3. 120
  4. 130
  5. 140

Answer: 120

Use inclusion-exclusion: |A ∪ B ∪ C| = 80+90+70 - (30+25+20) + 10 = 240 - 75 + 10 = 175. Number reading exactly one = |A|+|B|+|C| - 2*(sum of pairwise intersections) + 3*|triple| = 240 - 2*75 + 3*10 = 240 - 150 + 30 = 120. Alternatively, compute each: A only = 80 - (30+25) + 10 = 35; B only = 90 - (30+20) + 10 = 50; C only = 70 - (25+20) + 10 = 35; sum = 120.

2. A box contains 4 red, 3 blue, and 2 green marbles. If 4 marbles are drawn at random without replacement, what is the probability that there is at least one marble of each color?

  1. 4/7
  2. 12/21
  3. 2/3
  4. 5/9
  5. 8/21

Answer: 4/7

Total ways: C(9,4)=126. Favorable: must have at least one of each color. Cases: (2,1,1) distribution. Number of ways: choose which color has 2: 3 choices. For red2: C(4,2)*C(3,1)*C(2,1)=6*3*2=36; blue2: C(4,1)*C(3,2)*C(2,1)=4*3*2=24; green2: C(4,1)*C(3,1)*C(2,2)=4*3*1=12. Total favorable = 36+24+12=72. Probability = 72/126 = 4/7.

3. A committee of 4 people is to be chosen from a group of 6 men and 4 women. How many committees have at least 2 women?

  1. 115
  2. 185
  3. 195
  4. 210
  5. 230

Answer: 185

Total committees: C(10,4)=210. Committees with 0 women: C(6,4)=15. With 1 woman: C(4,1)*C(6,3)=4*20=80. At least 2 women = total - (0+1) = 210 - 95 = 115. Alternatively, sum: 2 women: C(4,2)*C(6,2)=6*15=90; 3 women: C(4,3)*C(6,1)=4*6=24; 4 women: C(4,4)*C(6,0)=1*1=1; total=115.

4. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} be the universal set. Which of the following statements is true?

  1. The set {1, 2, 3} is a subset of U.
  2. The set {1, 2, 3} is an element of U.
  3. The set {1, 2, 3} is the universal set.
  4. The set {1, 2, 3} is equal to U.
  5. The set {1, 2, 3} is not a subset of U.

Answer: The set {1, 2, 3} is a subset of U.

A set A is a subset of set B if every element of A is also an element of B. Here, every element of {1,2,3} is in U, so it is a subset. The other options are false because {1,2,3} is not an element (elements are numbers), not equal to U, and not the universal set.

5. In how many ways can 6 people be seated around a circular table if two particular people must sit next to each other?

  1. 48
  2. 120
  3. 240
  4. 720
  5. 144

Answer: 48

Treat the two people as a block. For circular arrangements, number of ways to arrange n distinct objects is (n-1)!. Here we have 5 objects (block + 4 individuals): (5-1)! = 4! = 24. Within the block, the two can be arranged in 2! = 2 ways. Total = 24 * 2 = 48.

6. How many ways are there to distribute 10 identical candies to 4 distinct children, if each child must receive at least one candy?

  1. 84
  2. 120
  3. 126
  4. 210
  5. 220

Answer: 84

This is a stars and bars problem with the constraint that each child gets at least one. Give each child 1 candy first, leaving 6 candies to distribute among 4 children without restriction. Number of ways = C(6+4-1,4-1) = C(9,3) = 84.

7. In how many ways can 10 identical candies be distributed among 4 distinct children, if each child must receive at least one candy?

  1. 84
  2. 120
  3. 126
  4. 210
  5. 220

Answer: 84

This is combinations with repetition (stars and bars) with lower bound. Give each child 1 candy first, leaving 6 candies. Number of ways to distribute 6 identical candies among 4 children = C(6+4-1,4-1) = C(9,3) = 84.

8. Let the universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 4, 6, 8, 10}. What is the complement of A, denoted A'?

  1. {1, 3, 5, 7, 9}
  2. {2, 4, 6, 8, 10}
  3. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
  4. {}
  5. {1, 2, 3, 4, 5}

Answer: {1, 3, 5, 7, 9}

The complement of a set A relative to the universal set U is the set of all elements in U that are not in A. Since A contains all even numbers from 1 to 10, its complement is the odd numbers: {1,3,5,7,9}.

9. A committee of 4 people is to be selected from a group of 5 men and 4 women. How many committees can be formed if the committee must include at least one man and at least one woman?

  1. 120
  2. 125
  3. 130
  4. 135
  5. 140

Answer: 120

Total committees without restriction: C(9,4)=126. Committees with no men (all women): C(4,4)=1. Committees with no women (all men): C(5,4)=5. So committees with at least one of each = 126 - 1 - 5 = 120.

10. A committee of 4 members is to be formed from a group of 10 people, and then the committee must select a chairperson, a vice-chairperson, a secretary, and a treasurer. How many different ways can this be done?

  1. 5040
  2. 151200
  3. 210
  4. 50400
  5. 30240

Answer: 5040

First, choose 4 people from 10: C(10,4) = 210. Then assign 4 distinct roles to these 4 people: 4! = 24. Total = 210 × 24 = 5040. Alternatively, directly permute 4 from 10: P(10,4) = 10×9×8×7 = 5040.

11. In a survey of 100 students, 60 like math, 50 like science, and 30 like both. How many students like neither math nor science?

  1. 10
  2. 20
  3. 30
  4. 40
  5. 50

Answer: 20

Let M = math, S = science. |M| = 60, |S| = 50, |M ∩ S| = 30. Number liking at least one = |M ∪ S| = |M| + |S| - |M ∩ S| = 60 + 50 - 30 = 80. Total students = 100, so neither = 100 - 80 = 20.

12. A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. If two marbles are drawn without replacement, what is the probability that at least one is red or at least one is blue?

  1. 1/45
  2. 44/45
  3. 7/9
  4. 8/9
  5. 1

Answer: 44/45

Total marbles = 10. P(at least one red or at least one blue) = 1 - P(neither red nor blue) = 1 - P(both green). P(both green) = (2/10)*(1/9) = 2/90 = 1/45. So probability = 1 - 1/45 = 44/45.

More Quantitative Reasoning topics

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