CAT 2025 Slot 3 · Quant · Question 21

Write S=p+q=900-r, so everything depends on r through S.

The upper condition p\le 0.7q with q=S-p gives p\le 0.7\left( S-p \right), so 1.7p\le 0.7S and p\le \frac{7S}{17}.

The lower condition p\ge 0.3q gives p\ge 0.3\left( S-p \right), so 1.3p\ge 0.3S and p\ge \frac{3S}{13}.

Both bounds rise with S, so the largest p comes from the largest S, that is the smallest r, and the smallest p from the largest r.

The perfect squares strictly between 150 and 500 run from {13}^{2}=169 to {22}^{2}=484.

Taking r=169 gives S=731 and p\le \frac{7\times 731}{17}=301, which is a whole number, so p=301 with q=430.

Taking r=484 gives S=416 and p\ge \frac{3\times 416}{13}=96, again a whole number, so p=96 with q=320.

Both are attainable, since 0.3\times 430=129\le 301\le 0.7\times 430=301 and 0.3\times 320=96\le 96\le 0.7\times 320=224.

The sum is 301+96=397.

Shortcut. Convert each condition on p and q into a bound on p alone using q=S-p. Both bounds then increase with S, so only the two extreme perfect squares need testing — {13}^{2} and {22}^{2} — rather than all ten.