TheCATExam
Sign in

CAT 2022 Slot 3 — QA Question 15

Basics of AverageEasy

Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is

Answer & solution

  • A

    23

  • 22.5

  • C

    24

  • D

    23.5

Solution

Easy

The two smallest sum to 2828 and the two largest sum to 5656 — both fixed. To maximise the overall average, push the two middle numbers as high as possible, limited only by distinctness from the largest pair.

1

Fix the locked sums.

two smallest: average 14⇒sum=28two largest: average 28⇒sum=56\begin{aligned} &\text{two smallest: average }14\Rightarrow\text{sum}=28\\ &\text{two largest: average }28\Rightarrow\text{sum}=56 \end{aligned}
2

Bound the middle pair. The smaller of the two largest must be below the larger; the largest pair sums to 5656, so the two largest can be 2929 and 2727. The middle two must be distinct integers below 2727, so at most 2626 and 2525:

middle two≤26+25=51\text{middle two}\le 26+25=51
3

Maximise the total and divide by 66:

sum=28+(25+26)+56=135average=1356=22.5\begin{aligned} &\text{sum}=28+(25+26)+56=135\\ &\text{average}=\frac{135}{6}=22.5 \end{aligned}

The maximum possible average is 22.5\mathbf{22.5}.

A valid arrangement: 12,16,25,26,27,2912,16,25,26,27,29 — smallest pair 12,1612,16 (avg 1414), largest pair 27,2927,29 (avg 2828), all distinct, sum 135135.

CAT 2022 Slot 3 QA Q15: Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average — Solution | TheCATExam