CAT 2023 Slot 1 · Quant · Question 10

Let each girl score g and each boy score b.

The ten students average 24, so 4g+6b=240, which simplifies to 2g+3b=120.

The stated conditions are g\ge b (a girl's marks are not less than a boy's) and g\le 2b (at most double).

The quantity wanted is T=2g+6b, the total of 2 girls and 6 boys.

From the average relation, 2g=120-3b, so T=120-3b+6b=120+3b.

So T depends only on b, increasing as b increases.

Apply g\ge b: since g=\frac{120-3b}{2}, this gives 120-3b\ge 2b, so b\le 24.

Apply g\le 2b: 120-3b\le 4b, so b\ge \frac{120}{7}, about 17.14.

Therefore T=120+3b runs from 120+3\times \frac{120}{7}=\frac{1200}{7}\approx 171.43 up to 120+72=192.

The integer values of T in that range run from 172 to 192.

Counting, 192-172+1=21 — the second option.

Shortcut. Marks need not be whole numbers; only the total is required to be an integer. So the task is just counting integers in an interval, and writing T in terms of a single variable b makes both endpoints fall straight out of the two inequalities.