The number of students learning Carrom in college C is ________% less than the…

Concept

When two quantities are known only through their sum S and their difference d, the larger of them is (S + d)/2 and the smaller is (S − d)/2. When one category occupies p% of a total, everything outside that category occupies the remaining (100 − p)%, so the total equals that remaining part divided by (100 − p)/100. A quantity X is said to be k% less than a quantity Y when k = ((Y − X)/Y) × 100; the shortfall is always measured against the reference quantity Y that follows the words “less than”, never against X.

Application

1. In college A, Chess and Squash together account for 2100 students. Carrom occupies 30% of A’s strength, so those 2100 students form the remaining 70%. Hence Total(A) = 2100 ÷ 0.70 = 3000.

2. In college A, Chess and Squash have sum 2100 and difference 300, and Chess is the larger of the two. Hence Chess(A) = (2100 + 300)/2 = 1200 and Squash(A) = (2100 − 300)/2 = 900.

3. In college C, Chess and Squash together account for 1260 students. Carrom occupies 40% of C’s strength, so those 1260 students form the remaining 60%. Hence Total(C) = 1260 ÷ 0.60 = 2100.

4. Therefore Carrom(C) = 40% of 2100 = 0.40 × 2100 = 840.

5. The words “less than” make Chess in college A the reference quantity, so the required percentage = ((1200 − 840)/1200) × 100 = (360/1200) × 100 = 30.

Cross-check

Reducing 1200 by 30% gives 1200 × 0.70 = 840, exactly the Carrom count obtained for college C, so the percentage is consistent both ways. College A’s parts also reconcile with its total: Chess 1200 + Squash 900 + Carrom 900 = 3000, and 900 is indeed 30% of 3000.

So the number of students learning Carrom in college C is 30% less than the number of students learning Chess in college A.