IPMAT Indore 2022 · Quantitative Ability · Question 15

Step 1 — turn the average condition into a sum condition. Let {S}_{H} be the sum of the largest 25 and {S}_{L} the sum of the smallest 25. The averages differ by 50:

\frac{{S}_{H}}{25}-\frac{{S}_{L}}{25}=50 \Rightarrow{S}_{H}-{S}_{L}=1250

Step 2 — write what is to be maximised.

\text{Total}={S}_{H}+{S}_{L}

Using (1), {S}_{L}={S}_{H}-1250, so

\text{Total}=2{S}_{H}-1250

Maximising the total therefore means maximising {S}_{H} — nothing else matters.

Step 3 — maximise {S}_{H}. The 50 integers are distinct and the largest is 100, so the top 25 are at most 100,99,98,\ldots ,76:

{S}_{H}\le 100+99+\cdots +76=\frac{\left( 100+76 \right)\times 25}{2}=2200

Step 4 — check the bottom half can accommodate it. With {S}_{H}=2200, equation (1) gives

{S}_{L}=2200-1250=950

We need 25 distinct positive integers summing to 950, all below 76 (the smallest of the top group). The range available is

minimum possible: 1+2+\cdots +25=325

maximum possible: 51+52+\cdots +75=\frac{\left( 51+75 \right)\times 25}{2}=1575

Since 325\le 950\le 1575, such a set exists ✔

Step 5 — compute the total.

2200+950=3150

Answer: 3150

Shortcut. Reduce to one variable first. The difference {S}_{H}-{S}_{L}=1250 is fixed, so

\text{Total}=2{S}_{H}-1250

and the problem becomes “make the top 25 as large as possible” — a one-line consecutive-block sum, 100 down to 76, giving 2200.

Then do the feasibility check, which is what separates a correct answer from a lucky one: the bottom 25 must sum to 950 using distinct values under 76, and 325\le 950\le 1575 confirms it.

Total =2\left( 2200 \right)-1250=3150. The general lesson: when a difference is fixed, maximising a sum reduces to maximising one side alone.