CAT 2025 Slot 1 · Quant · Question 14

Let there be B boys and G girls.

The remaining counts are 0.4B boys and 0.6G girls, and

0.6G=0.4B+8, that is 3G=2B+40.

The hidden condition is that the numbers leaving must be whole: 0.6B and 0.4G are integers, so both B and G must be multiples of 5.

We are told B>10, so try B=15,20,25,\ldots and require G=\frac{2B+40}{3} to be an integer multiple of 5.

B=15 gives G=\frac{70}{3}, not an integer. B=20 gives G=\frac{80}{3}, not an integer.

B=25 gives G=30, which works.

Check:

0.6\left( 30 \right)=18 girls and 0.4\left( 25 \right)=10 boys remain, and 18=10+8.

The minimum total is 25+30=55.