CAT 2022 Slot 2 · Quant · Question 8

An average of 300 over N numbers means a total of 300N.

After the replacement the average is 400, so the new total is 400N.

Only one term changed, and it grew from {a}_{1} to 6{a}_{1}, an increase of 5{a}_{1}.

So 5{a}_{1}=400N-300N=100N.

Therefore {a}_{1}=20N.

The sequence is non-decreasing, so {a}_{1} is the smallest of the numbers and cannot exceed their average.

That gives 20N\le 300, so N\le 15.

At the other end, N=1 would force the single number to be both 300 and 20, which is impossible, so N\ge 2.

So N runs over the integers from 2 to 15, and each gives a distinct {a}_{1}=20N.

Counting, 15-2+1=14 possible values.

A check at the top end: N=15 gives {a}_{1}=300, so every term is 300 — non-decreasing, as required.

Shortcut. The whole question reduces to {a}_{1}=20N plus the single constraint that the smallest term cannot beat the average. Counting the valid N is then counting integers from 2 to 15 — and forgetting to exclude N=1 is the usual slip.