Set Theory, Calendars, Clocks and Binomial Theorem

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Sets and Unions

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 54 are doing Actuarial and those whose numbers are divisible by 7 are preparing for MBA. How many of the students are doing nothing?
1. 37
2. 45
3. 51
4. 62

Choice C. 51

Detailed Solution

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 10th, 20th, 30th…… numbered students would be doing both CA and Actuarial
∴ n(A∩B) = ${\frac{150}{10}}$ = 15
The 14th, 28th, 42nd…… numbered students would be doing both CA and MBA
∴ n(A∩C) = ${\frac{150}{14}}$ = 10
The 35th, 70th, …… numbered students would be doing both Actuarial and MBA
∴ n(B∩C) = ${\frac{150}{35}}$ = 4
And the 70th and 140th 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 = (c)

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