In a school with a total of $Y$ students, each student likes at least one of the three sports: A, B, and C. The number of students who like all three sports is 114. Additionally, 16\% of the total students like both A and B but not C, 13\% like both B and C but not A, and 4\% like both A and C but not B. The ratio of students who like only A, only B, and only C is 2:3:1 respectively. The total number of students who like at least two sports is 312.

What percentage less is the number of students who like all three sports compared to the total number of students in the school?

Question

In a school with a total of $Y$ students, each student likes at least one of the three sports: A, B, and C. The number of students who like all three sports is 114. Additionally, 16\% of the total students like both A and B but not C, 13\% like both B and C but not A, and 4\% like both A and C but not B. The ratio of students who like only A, only B, and only C is 2:3:1 respectively. The total number of students who like at least two sports is 312.

What percentage less is the number of students who like all three sports compared to the total number of students in the school?

Accepted Answer

**Solution:**

1. Let the total number of students be $Y$.

2. Given:
   - Number of students who like all three sports (A, B, C) = 114
   - Students who like A and B but not C = 16\% of $Y$ = $0.16Y$
   - Students who like B and C but not A = 13\% of $Y$ = $0.13Y$
   - Students who like A and C but not B = 4\% of $Y$ = $0.04Y$
   - Total students who like at least two sports = 312

3. Sum of students who like exactly two sports:
   $$0.16Y + 0.13Y + 0.04Y = 0.33Y$$

4. Since total students who like at least two sports is 312, and this includes those who like exactly two sports plus those who like all three:
   $$0.33Y + 114 = 312$$

5. Solve for $Y$:
   $$0.33Y = 312 - 114 = 198$$
   $$Y = \frac{198}{0.33} = 600$$

6. Now, find the percentage less of students who like all three sports compared to total students:
   $$	ext{Percentage less} = \left(1 - \frac{114}{600}ight) 	imes 100 = \left(1 - 0.19ight) 	imes 100 = 81\%$$

7. The question asks how much percentage less the number of students who like all three sports is compared to the total number of students. Since 114 is 19\% of 600, it is 81\% less than the total.

**Answer:** 81\%

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