# GMAT Quant | Set Theory Questions

#### 4 GMAT Sample Questions | Sets Questionbank

You may get one to two questions from sets in the GMAT quant section - in both variants viz., problem solving and data sufficiency. The concepts tested include union and intersection of 2 or 3 sets, subsets, proper subsets, and complimentary sets. Sample GMAT practice questions from set theory are given below. Attempt questions from this questionbank and check whether you have got the correct answer. If you have not got the correct answer, go to the explanatory answer or the video explanations to learn how to solve the question.

## GMAT Set Theory Practice Questions

1. In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, those whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects?

1. 19
2. 41
3. 21
4. 57
5. 26
Approach to solve this 3 Sets Problem

Step 1: Compute the number of students who opt for Physics, the number who opt for Chemistry and the number who opt for Math. Let these be n(A), n(B), and n(C) respectively.
Step 2: Compute numbers who opt for Physics and Chemistry, Chemistry and Math, and Math and Physics. The values will be n(A ∩ B), n(B ∩ C), and n(c ∩ A) respectively.
Step 3: Compute number of students who opt for all three subjects. This value will correspond to n(A ∩ B ∩ C).
Step 4: Compute number of students who opt for at least one of the three subjects - essentially n(A ∪ B ∪ C) using data from steps 1 to 3.
Step 5: Number of students who opt for none = 120 - n(A ∪ B ∪ C).

2. Of the 200 candidates who were interviewed for a position at a call center, 100 had a two-wheeler, 70 had a credit card and 140 had a mobile phone. 40 of them had both, a two-wheeler and a credit card, 30 had both, a credit card and a mobile phone and 60 had both, a two wheeler and mobile phone and 10 had all three. How many candidates had none of the three?

1. 0
2. 20
3. 10
4. 18
5. 25
Steps to solve this Complement of the Union of 3 Sets Question

Approach: Number who have none = Total - number who have at least one.
Step 1: Number who have at least one is n(A ∪ B ∪ C).
Step 2: Compute n(A ∪ B ∪ C) using the formula.
Step 3: Number who have none = 200 - n(A ∪ B ∪ C).

3. In a class of 40 students, 12 enrolled for both English and German. 22 enrolled for German. If the students of the class enrolled for at least one of the two subjects, then how many students enrolled for only English and not German?

1. 30
2. 10
3. 18
4. 28
5. 32
Approach to solve this Union Intersection of 2 Sets Question

Step 1: Let the number who enrolled for English be n(A) and the number who enrolled for German be n(B).
Step 2: Because students of the class enrolled for at least one of the two subjects, n(A ∪ B) = Total number of students = 40.
Step 3: Applying n(A ∪ B) formula and using data given in the question compute n(A).
Step 4: n(Only A) = n(A) - n(A ∩ B). This is the answer to the question.

4. In a class 40% of the students enrolled for Math and 70% enrolled for Economics. If 15% of the students enrolled for both Math and Economics, what % of the students of the class did not enroll for either of the two subjects?

1. 5%
2. 15%
3. 0%
4. 25%
5. None of these
Approach to solve this easy Sets problem

Step 1: Compute n(A ∪ B) using the formula and the data given in the question.
Step 2: % of students who did not enroll for either subject = 100 - n(A ∪ B).

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