Question

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 :

• A. 102
• B. 42
• C. 1
• D. 38

Answer 1

Answer: (D)

38

Answer 2

Let $n(A)$ = number of students opted
Mathematics = $70$,
$n(B)$ = number of students opted Physics = $46$,
$n(C)$ = number of students opted Chemistry
= 28,
$n(A \cap B) = 23,$
$n(B \cap C) = 9,$
$n(A \cap C) = 14,$
$n(A \cap B \cap C) = 4,$
Now $n(A \cup B \cup C)$
$= n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C)$
$- n(A \cap C) + n(A \cap B \cap C)$
$= 70 + 46 + 28 - 23 - 9 - 14 + 4 = 102$
So number of students not opted for any course
= Total - $n(A \cup B \cup C)$
$= 140 - 102 = 38$