Each of the 90 members of a club plays chess or carrom or both. If 54 members play chess and 51 members play carrom, how many members play both games?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 15
Every member plays at least one game, so the members who play chess or carrom number 90 in all. Those who play both are counted twice when the two numbers are added: \(n(A\text{ or }B)=n(A)+n(B)-n(\text{both})\).
\(90=54+51-n(\text{both})\)
\(n(\text{both})=105-90=15\)
Check: chess only \(=54-15=39\), carrom only \(=51-15=36\), and \(39+15+36=90\)
