# 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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