Question

In a class of $120$ students numbered from $1$ to $120$, all even numbered students opt for Physics, those whose numbers are divisible by $5$, opt for Chemistry and those whose numbers are divisible by $7$, opt for Maths. How many opt for none of the three subjects?

• A. $19$
• B. $41$
• C. $21$
• D. $26$

Answer 1

The correct option is B $41$

$P=\{2,4,6,\dots ,118,120\}$
$C=\{5,10,15,\dots ,115,120\}$
$M=\{7,14,21,\dots ,112,119\}$
$n\left(P\right)=60,n\left(C\right)=24,n\left(M\right)=17$
(Here, $P=$ Physics, $C=$ Chemistry, $M=$ Maths)

The elements in $P\cap C$ will be the multiples of $\text{LCM}(2,5)$.
So, $n(P\cap C)=12$
Similarly, $n(C\cap M)=3,n(P\cap M)=8$
and $n(P\cap C\cap M)=1$


$\therefore n(P\cup C\cup M)$
$=n\left(P\right)+n\left(C\right)+n\left(M\right)-n(P\cap C)-n(C\cap M)-n(P\cap M)+n(P\cap C\cap M)$
$=60+24+17-(12+3+8)+1$
$=79$

$\therefore$ Number of students who opt for none of the subjects $=120-79=41$