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. $1$
• B. $102$
• C. $38$
• D. $42$

Answer 1

The correct option is C $38$

Let $A,B$ and $C$ be the set of students opted for Mathematics, Physics and Chemistry respectively then,
$A=\{2,4,6,\cdots ,140\}$
$n\left(A\right)=70$
$B=\{3,6,9,\cdots ,138\}$
$n\left(B\right)=46$
$C=\{5,10,15,\cdots ,140\}$
$n\left(C\right)=28$
$(A\cap B)=\{6,12,18,\cdots ,138\}$
$n(A\cap B)=23$
$(A\cap C)=\{10,20,30,\cdots ,140\}$
$n(A\cap C)=14$
$(C\cap B)=\{15,30,45,\cdots ,135\}$
$n(A\cap B)=9$
$(A\cap B\cap C)=\{30,60,90,120\}$
$n(A\cap C)=4$

So,
$n(A\cup B\cup C{)}^{\prime }=$
Total students $-n(A\cup B\cup C)$
$n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$
$=70+46+28-23-9-14+4=102$
$n(A\cup B\cup C{)}^{\prime }=140-102=38$