A) 102
B) 42
C) 1
D) 38
Correct Answer: D
Solution :
\[n(P)=\left[ \frac{140}{3} \right]=46\] |
\[n(C)=\left[ \frac{140}{5} \right]=28\] |
\[n(M)=\left[ \frac{140}{2} \right]=70\] |
\[n(P\cup C\cup M)\] |
\[=n(P)+n(C)+n(M)-n(P\cap C)-n(C\cap M)\] |
\[-n(M\cap P)+n(P\cap M\cap C)\] |
\[=46+28+70-\left[ \frac{40}{15} \right]-\left[ \frac{140}{10} \right]-\left[ \frac{140}{6} \right]+\left[ \frac{140}{30} \right]\] |
\[=144-9-14-23+4=102\] |
So, required number of students \[=140-102=38.\] |
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