Averages , Ratio and Proportion Question 1 - CAT 2025
In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was:
Step-by-Step Explanation
Let the initial number of boys be b and the initial number of girls be g.
- Analyze the Remaining Students:
- 40% of the girls left ⟹ 60% of the girls remain = 0.6g
- 60% of the boys left ⟹ 40% of the boys remain = 0.4b
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Set up the Diophantine Equation:
0.6g = 0.4b + 8
Multiply the entire equation by 10 to clear decimals:
6g = 4b + 80 ⟹ 3g = 2b + 40
- Apply Integer Constraints:
Since count variables must be integers, 0.6g (which equals 3g/5) dictates that g must be a multiple of 5. Concurrently, 2b + 40 is even, making 3g even, so g must also be an even number. Thus, g must be a multiple of 10.
- If g = 10 ⟹ 3(10) = 2b + 40 ⟹ 2b = -10 (invalid).
- If g = 20 ⟹ 3(20) = 2b + 40 ⟹ 2b = 20 ⟹ b = 10. However, the problem specifies there are strictly more than 10 boys (b > 10).
- If g = 30 ⟹ 3(30) = 2b + 40 ⟹ 2b = 50 ⟹ b = 25. This perfectly satisfies b > 10.
- Find the Minimum Total:
Minimum initial students = b + g = 25 + 30 = 55
Correct Answer: 55
💡 FundaMakers Tip:
When handling percentage drops on discrete human counts, look for divisibility constraints immediately. The fractional component multiplier (3g/5) locks g into multiples of 5, while the even outcome constraint shifts it to multiples of 10. Testing values along this logical sequence bypasses infinite trial-and-error completely.