In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is

Options

A. 42
B. 102
C. 38
D. 1

Answer

C

Explanation

Let A be the set of even numbered students then `n(A)=[(140)/(2)]=70`([.] denotes greatest integer function)
Let B be the set of those students whose number is divisible by 3,
then `n(B) = [(140)/(3)]=46`
([.]denotes greatest integer function)
Let C be the set of those students whose number is divisible by 5,
then `n(C)=[(140)/(5)]=28`
([.] denotes greatest integer function)
Now, `n(A cap B)=[(140)/(6)]=23`
(number divisible by both 2 and 3)
`n (B cap C)=[(140)/(15)]=9`
(numbers divisible by both 3 and 5)
`n (C cap A)=[(140)/(10)]=14`
(numbers divisible by both 2 and 5)
`n (C cap A) =[(140)/(10)]=14`
(numbers divisible by both 2 and 5)
`n(A cap B cap C)=[(140)/(30)]=4`
(numbers divisible by 2, 3 and 5)
and `n (A cup B cup C) `
`=Sigma n(A)-Sigma n(A cap B) + n(A cap B cap C)`
`=(70+46+28)-(23+9+14)+4=102`
`therefore` Number of students who did not opt any of the three courses
`="Total students"-n(A cup B cup C)=140-102=38`