The question is from CAT Set theory. It is about multiples of three numbers and their union. We need to find out the numbers below 150 which are not divisible by those three numbers. 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 11: A class in college has 150 students numbered from 1 to 150 , in which all the even numbered students are doing CA, whose number are divisible by 5 are doing Actuarial and those whose numbers are divisible by 7 are preparing for MBA. How many of the students are doing nothing?

- 37
- 45
- 51
- 62

51

Starts Sat, April 27th, 2019

Let the total no of students doing CA be n(A), those doing actuarial be n(B) and those doing MBA be n(C).

Now, ‘CA’ is a set of all even numbered students, thus n(A) = \\frac{150}{2}\\) = 75

‘Actuarial’ is a set of all the students whose number are divisible by 5, thus n(B) = \\frac{150}{5}\\) = 30

‘MBA’ is a set of all the students whose number are divisible by 7, thus n(C) = \\frac{150}{7}\\) = 21

The 10^{th}, 20^{th}, 30^{th}…… numbered students would be doing both CA and Actuarial

∴ n(A∩B) = \\frac{150}{10}\\) = 15

The 14^{th}, 28^{th}, 42^{nd}…… numbered students would be doing both CA and MBA

∴ n(A∩C) = \\frac{150}{14}\\) = 10

The 35^{th}, 70^{th}, …… numbered students would be doing both Actuarial and MBA

∴ n(B∩C) = \\frac{150}{35}\\) = 4

And the 70^{th} and 140^{th} students must be doing all the three.

∴n(A∩B∩C) = 2

Now, n(A∪B∪C) = n(A)+n(B)+n(C) – n(A∩B) – n(A∩C) - n(B∩C) + n(A∩B∩C)

= 99

∴ Number of students doing nothing = 150 – 99 = 51

The question is **"How many of the students are doing nothing?"**

Choice C is the correct answer.

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