Question

70% of the students who joined XCRI last year play football, 75% play cricket, 80% play basketball and 85% play carrom. The minimum percentage of students who play all four games is:

• A. 5%
• B. 10%
• C. 15%
• D. 20%
• E. None of the above

Hint:

For \(n\) sets with percentages \(p_1,\dots,p_n\) in a \(100%\) universe, a quick bound for the minimum intersection is \(\max\0,\ \sum p_i - (n-1)\cdot 100%\\). This comes from the inclusion–exclusion principle and is often tight for such minimization problems.

Answer 1

The Correct Option is B

Step 1: Let the four sets be \(F,C,B,K\) (football, cricket, basketball, carrom). Their percentages are \(70,75,80,85\) respectively. For any four sets on a \(100%\) universe, a lower bound on the intersection is \[ |F\cap C\cap B\cap K| \ge \big(|F|+|C|+|B|+|K|\big) - 3\cdot 100%, \] since at most \(100%\) students can lie outside each missing-overlap step (generalized pigeonhole/union bound idea). Step 2: Compute the bound: \[ 70+75+80+85 - 300 = 310 - 300 = 10%. \] Thus at least \(10%\) must belong to all four sets. This bound is tight by suitable distribution, so the minimum possible percentage is \(\boxed{10%}\).