CAT 2022 Slot 2 · Quant · Question 16

An average of 38 across five students means a total of 38\times 5=190.

Write the five distinct marks in increasing order as {a}_{1}<{a}_{2}<{a}_{3}<{a}_{4}<{a}_{5}, with Amit holding {a}_{1}.

Exactly three students scored above 32, and those must be the top three, so {a}_{3}, {a}_{4} and {a}_{5} all exceed 32 while {a}_{2} does not.

To push Amit's mark as high as possible, make the two lowest marks as large as the rules allow.

Since {a}_{2}\le 32 and the marks are distinct, the best pair is {a}_{1}=31 and {a}_{2}=32.

That leaves {a}_{3}+{a}_{4}+{a}_{5}=190-63=127, which is achievable with marks above 32, for example 33, 44 and 50.

So Amit's highest possible mark is 31.

To push Amit's mark as low as possible, make everyone else's marks as large as possible.

The top three are capped at 50, so the largest they can be is 48, 49 and 50, totalling 147.

That leaves {a}_{1}+{a}_{2}=190-147=43, and {a}_{2} can be at most 32.

Taking {a}_{2}=32 gives {a}_{1}=11, which is still below {a}_{2} as required.

So Amit's lowest possible mark is 11, and the difference is 31-11=20 — the third option.

Shortcut. The phrase "exactly three above 32" is what pins {a}_{2}\le 32, and that single cap drives both extremes. Push the top three to their minimum for Amit's maximum, and to their maximum (48, 49, 50) for his minimum.