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 2023. 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?
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, โCAโ is a set of all even numbered students, thus n(A) = \\frac{150}{2}\\) = 75
โActuarialโ is a set of all the students whose number are divisible by 5, thus n(B) = \\frac{150}{5}\\) = 30
โMBAโ is a set of all the students whose number are divisible by 7, thus n(C) = \\frac{150}{7}\\) = 21
The 10th, 20th, 30thโฆโฆ numbered students would be doing both CA and Actuarial
โด n(AโฉB) = \\frac{150}{10}\\) = 15
The 14th, 28th, 42ndโฆโฆ numbered students would be doing both CA and MBA
โด n(AโฉC) = \\frac{150}{14}\\) = 10
The 35th, 70th, โฆโฆ numbered students would be doing both Actuarial and MBA
โด n(BโฉC) = \\frac{150}{35}\\) = 4
And the 70th and 140th students must be doing all the three.
โดn(AโฉBโฉC) = 2
Now, n(AโชBโชC) = n(A)+n(B)+n(C) โ n(AโฉB) โ n(AโฉC) - n(BโฉC) + n(AโฉBโฉC)
= 99
โด Number of students doing nothing = 150 โ 99 = 51
The question is "How many of the students are doing nothing?"
Choice C is the correct answer.
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