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,
$A=\{2,4,6,\cdots ,140\}$ $\Rightarrow n\left(A\right)=70$,
$B=\{3,6,9,\cdots ,138\}$ $\Rightarrow n\left(B\right)=46$,
$C=\{5,10,15,\cdots ,140\}$ $\Rightarrow n\left(C\right)=28$,
$(A\cap B)=\{6,12,18,\cdots ,138\}\Rightarrow n(A\cap B)=23$,
$(A\cap C)=\{10,20,30,\cdots ,140\}\Rightarrow n(A\cap C)=14$,
$(C\cap B)=\{15,30,45,\cdots ,135\}\Rightarrow n(A\cap B)=9$,
$(A\cap B\cap C)=\{30,60,90,120\}\Rightarrow n(A\cap C)=4$,

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

Hence $n(A\cup B\cup C{)}^{\prime }=140-102=38$