CAT 2023 Slot 2 · Quant · Question 13

Let the price be t times the square of the weight, so a piece weighing w is worth t{w}^{2}.

The four pieces have distinct positive integer weights adding to 18, and their combined price is t times the sum of their squares.

For a fixed sum, a sum of squares is largest when the parts are as unequal as possible.

The most lopsided split into four distinct positive integers is 1,2,3,12.

Its sum of squares is 1+4+9+144=158.

For a fixed sum, a sum of squares is smallest when the parts are as close together as possible.

The most balanced split into four distinct positive integers is 3,4,5,6.

Its sum of squares is 9+16+25+36=86.

The difference in total price is t\left( 158-86 \right)=72t, and this is given as 288000.

So t=\frac{288000}{72}=4000.

The original stone weighs 18 units, so its price is t\times{18}^{2}=4000\times 324.

That is 1296000 — the first option.

Shortcut. Only the gap between the extreme sums of squares matters, and both extremes are found by inspection: pull the weights apart for the maximum, push them together for the minimum. The constant then follows from one division, and the answer is \frac{288000}{72}\times 324.