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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

Updated On: Jan 15, 2026
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Solution and Explanation

Provided Information:
1. The average score of 25 students is 50, resulting in a total score of \(25 \times 50 = 1250\).
2. Five students achieved an identical score, denoted as \(T\), contributing \(5T\) to the total score.
3. The remaining 20 students have distinct integer scores, with the minimum score being 30.
Objective: To maximize the toppers' score (\(T\)), the scores of the other 20 students must be minimized.

Calculation of the minimum total score for the 20 students:

The smallest 20 distinct integers starting from 30 are 30, 31, 32, ..., 49.

The sum of these scores is: \(30 + 31 + 32 + ... + 49\)

This forms an arithmetic progression with: \(a_1 = 30\), \(d = 1\), and \(n = 20\).

The sum \(S\) of an arithmetic progression is calculated as: \(S = \frac{n}{2} [2a_1 + (n-1) d]\)

Substituting the values:

\(S = \frac{20}{2} [2(30) + (20-1) 1]\)

\(S = 10 [60 + 19] = 10 \times 79 = 790\)

The minimum combined score for the 20 students is 790.

Therefore, the combined score of the 5 toppers is: \(1250 - 790 = 460\).
The score of each topper is: \(T = \frac{460}{5} = 92\).

The maximum possible score for each topper is 92.

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