CAT Quantitative Aptitude Questions | CAT Set theory Questions - Union and Intersection

CAT Questions | Set theory | Union and Intersection

The question is from CAT Set theory. It is about union and intersection of sets. We need to maximize one particular value which is dependent on the other values. CAT exam is known to test on basics rather than high funda ideas. CAT also tests multiple ideas in the same question, and 2IIMs CAT question bank provides you with CAT questions that can help you gear for CAT Exam CAT 2019. Set Theory (especially constructing venn diagrams) is a frequently tested topic. Make sure you know the basics from this chapter.

Question 18: In its annual fest, a college is organizing three events: B-quiz, Finance and Marketing. The college has a strength of 510 students.The students were allowed to participate in any no. of events they liked. While viewing the statistics of the performance, the general secretary noticed:-

1.The number of students who participated in atleast two events were 52% more than those who participated in exactly one game.
2.The no. of students participating in 1,2 or 3 events respectively was atleast equal to 1.
3.The number of students who did not participate in any of the three events was the minimum possible integral value under these conditions.

What can be the maximum no. of students who participated in exactly 3 games?

  1. 200
  2. 300
  3. 303
  4. 304

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Explanatory Answer

Method of solving this CAT Question from Set theory: How do we maximize one particular value?

Let the no. of students who participated in 0,1,2,3 games be A, B, C, D respectively.
Then from the information we have we can conclude that
C + D = 152% 0f B = 1.52 of B -----------------------(1)
Since the total no. of students the college has is 420,
A + B + C + D = 510 ---------------------------------(2)

From (1) and (2), we can conclude that
A + 2.52B = 510
B = \\frac{25}{63}\\) * (510 – A)
For A to be minimum, 510 – A should give us the largest multiple of 63. Since, 63 * 8 = 504 we will get A = 6.
So B = 200 and C + D = 1.52 B = 304
Thus for the students participating in exactly 3 games to be maximum, the number of students participating in exactly 2 games has to be minimized and made equal to 1.
Thus no of students who participated in exactly 3 games = 304 – 1 = 303.

The question is "What can be the maximum no. of students who participated in exactly 3 games?"

Hence, the answer is "303".

Choice C is the correct answer.



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