CAT 2024 Slot 1 · Quant · Question 11

Call the numbers p, q, r and s, and work through the conditions one at a time.

The average of the first two exceeds p by 1: \frac{p+q}{2}=p+1, so q=p+2.

That average is p+1, and the average of the first three is 2 more, namely p+3.

So \frac{p+q+r}{3}=p+3, giving p+\left( p+2 \right)+r=3p+9 and r=p+7.

The average of the first four is 3 more than p+3, namely p+6.

So \frac{p+q+r+s}{4}=p+6, giving 3p+9+s=4p+24 and s=p+15.

The four numbers are p, p+2, p+7 and p+15, which increase throughout.

The difference between the largest and smallest is \left( p+15 \right)-p=15.

Shortcut. Each condition names the new average outright, so the next number follows immediately from (new count \times new average) minus (old total). The starting value p never has to be pinned down, because the question asks only for a difference.