Q14
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
Answer: 55
Key idea: All the head-counts (initial and remaining) must be whole numbers.
- Let there be \(b\) boys and \(g\) girls. Remaining: \(0.6g\) girls and \(0.4b\) boys, with \(0.6g=0.4b+8\), i.e. \(3g=2b+40\).
- \(0.4b\) and \(0.6g\) are integers, so \(b\) and \(g\) are multiples of 5. Also \(b>10\).
- The total \(b+g=b+\frac{2b+40}{3}\) increases with \(b\), so test the smallest valid \(b\):
- \(b=15\): \(g=\frac{70}{3}\), not an integer.
- \(b=20\): \(g=\frac{80}{3}\), not an integer.
- \(b=25\): \(g=30\), a multiple of 5. Valid.
- Check: 18 girls and 10 boys remain; \(18=10+8\).
Minimum total \(=25+30=55\).
Hence, the answer is 55.