In a class of 140 students numbered 1 to 140 , all even numbered students opted Mathematics course,

English

Gujarati

Hindi

1.Set Theory

hard

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

$102$

$42$

$1$

$38$

(JEE MAIN-2019)

Solution

$n(p)\, = \,\left[ {\frac{{140}}{3}} \right]\, = \,46$

$n(C)\, = \,\left[ {\frac{{140}}{5}} \right]\, = \,28$

$n(M)\, = \,\left[ {\frac{{140}}{2}} \right]\, = \,70$

$n(p\, \cup \,C\, \cup \,M)\, = \,n(P)\, + \,n(C)\, + \,n(M)$ $ – \,n(P \cap C) – \,n(C \cap M) – $ $n(M\, \cap \,P)\, + \,n(P \cap M \cap C)$

$ = \,46\, + \,28\, + 70\, – \,\left[ {\frac{{140}}{{15}}} \right]\, – \,\left[ {\frac{{140}}{{10}}} \right]\, – \,\left[ {\frac{{140}}{6}} \right]\, + \,\left[ {\frac{{140}}{{30}}} \right]\,$

$=\,144\,-\,9\,-14\,-\,23+4\,=\,102$

So required number of student $=\,140\,-\,102\,=\,38$

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