CAT 2022 Slot 3 · Quant · Question 13

Write the six numbers in increasing order as {a}_{1}<{a}_{2}<{a}_{3}<{a}_{4}<{a}_{5}<{a}_{6}.

The two smallest average 14, so {a}_{1}+{a}_{2}=28.

The two largest average 28, so {a}_{5}+{a}_{6}=56.

Those two sums are fixed, so maximising the overall average means maximising {a}_{3}+{a}_{4}.

The middle pair is capped by {a}_{5}, since {a}_{3}<{a}_{4}<{a}_{5}.

Now {a}_{5} is itself capped: with {a}_{5}<{a}_{6} and the pair summing to 56, the largest {a}_{5} can be is 27, paired with {a}_{6}=29.

With {a}_{5}=27, the best the middle pair can do is {a}_{4}=26 and {a}_{3}=25.

That gives {a}_{3}+{a}_{4}=51.

The total is then 28+51+56=135.

The average is \frac{135}{6}=22.5 — the first option.

A full example: 13, 15, 25, 26, 27, 29 — all distinct naturals, with the required pair averages of 14 and 28.

Shortcut. Only the middle pair is free, and it is squeezed from above by {a}_{5}. So push {a}_{5} to its ceiling of 27 and slot the middle two just beneath it — the smallest two look flexible but contribute a fixed 28 either way.