In a school, there are a total of *Y* students (boys + girls). 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 sports A and B but not C, 13\% like sports B and C but not A, and 4\% like sports 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 is the value of *Y*?

Question

In a school, there are a total of *Y* students (boys + girls). 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 sports A and B but not C, 13\% like sports B and C but not A, and 4\% like sports 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 is the value of *Y*?

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. The total number of students who like at least two sports includes:
   - Only A and B (not C): $0.16Y$
   - Only B and C (not A): $0.13Y$
   - Only A and C (not B): $0.04Y$
   - All three sports: 114

Therefore,
   $$0.16Y + 0.13Y + 0.04Y + 114 = 312$$

4. Combine the percentages:
   $$0.33Y + 114 = 312$$

5. Subtract 114 from both sides:
   $$0.33Y = 312 - 114 = 198$$

6. Solve for $Y$:
   $$Y = \frac{198}{0.33} = 600$$

However, this contradicts the options and the ratio condition. Let's check the ratio condition.

7. Let the number of students who like only A, only B, and only C be $2x$, $3x$, and $x$ respectively.

8. Total students who like only one sport = $2x + 3x + x = 6x$

9. Total students who like at least two sports = 312 (given)

10. Total students $Y$ = students who like only one sport + students who like at least two sports
    $$Y = 6x + 312$$

11. Total students who like sport A:
    - Only A: $2x$
    - A and B only: $0.16Y$
    - A and C only: $0.04Y$
    - All three: 114

12. Total students who like sport B:
    - Only B: $3x$
    - A and B only: $0.16Y$
    - B and C only: $0.13Y$
    - All three: 114

13. Total students who like sport C:
    - Only C: $x$
    - A and C only: $0.04Y$
    - B and C only: $0.13Y$
    - All three: 114

14. Since all students like at least one sport, the sum of all individual groups equals $Y$:
    $$2x + 3x + x + 0.16Y + 0.13Y + 0.04Y + 114 = Y$$
    $$6x + 0.33Y + 114 = Y$$

15. Rearranged:
    $$6x + 114 = Y - 0.33Y = 0.67Y$$

16. From step 10, $Y = 6x + 312$, substitute into above:
    $$6x + 114 = 0.67(6x + 312)$$

17. Expand right side:
    $$6x + 114 = 4.02x + 209.04$$

18. Bring terms to one side:
    $$6x - 4.02x = 209.04 - 114$$
    $$1.98x = 95.04$$

19. Solve for $x$:
    $$x = \frac{95.04}{1.98} = 48$$

20. Calculate $Y$:
    $$Y = 6x + 312 = 6 	imes 48 + 312 = 288 + 312 = 600$$

21. Check total students who like at least two sports:
    $$0.16Y + 0.13Y + 0.04Y + 114 = 0.33 	imes 600 + 114 = 198 + 114 = 312$$ (matches given)

**Answer:** $600$

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