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CAT 2021 Slot 3 — QA Question 20

Change in AverageEasy

The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is

Answer & solution

Answer: 92

Solution

Easy

The total of all 2525 scores is fixed by the mean. To maximize the 55 equal topper scores, minimize the other 2020 scores — and since they must be distinct integers starting at 3030, the smallest possible set is 30,31,…,4930,31,\dots,49.

1

Fix the total. Mean 5050 over 2525 students.

Total=25×50=1250\begin{aligned} &\text{Total} = 25\times 50 = 1250 \end{aligned}
2

Minimize the other 2020 scores. Distinct integers, lowest 3030: take 30,31,…,4930,31,\dots,49.

∑k=3049k=(30+49)⋅202=790\begin{aligned} &\sum_{k=30}^{49} k = \frac{(30+49)\cdot 20}{2} = 790 \end{aligned}
3

Solve for the topper score xx. Five toppers each score xx.

790+5x=1250⇒ 5x=460⇒ x=92\begin{aligned} &790 + 5x = 1250\\ &\Rightarrow\ 5x = 460\\ &\Rightarrow\ x = 92 \end{aligned}

(Note 92>4992>49, so the toppers are indeed above the rest — consistent.)

x=92x = 92
CAT 2021 Slot 3 QA Q20: The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examinati — Solution | TheCATExam