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