Question:medium

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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Correct Answer: 92

Solution and Explanation

Given:

  • The average score of 25 students is 50.
  • The aggregate score of all students is $25 \times 50 = 1250$.
  • Five students achieved identical scores, denoted as $T$. Their combined score is $5T$.
  • The scores of the remaining 20 students are unique integers, with the minimum score being 30.

Objective:

To determine the highest possible score for the top five students, it is necessary to assign the lowest possible scores to the other 20 students.

Step 1: Select the 20 lowest unique integers starting from 30:

These integers are: 30, 31, 32, ..., 49

Step 2: Calculate the sum of this arithmetic progression:

First term: $a_1 = 30$
Common difference: $d = 1$
Number of terms: $n = 20$

Using the sum formula for an arithmetic progression: $S = \dfrac{n}{2} \left[2a_1 + (n - 1)d\right]$

Substituting the values: $S = \dfrac{20}{2} \left[2 \times 30 + (20 - 1) \times 1\right] = 10 \left[60 + 19\right] = 10 \times 79 = 790$

Step 3: Compute the aggregate score of the five toppers:

Total score of all students: 1250
Total score of the other 20 students: 790
Therefore, the combined score of the 5 toppers is: $1250 - 790 = 460$

Step 4: Determine the score of each topper:

$T = \dfrac{460}{5} = 92$

Conclusion:

The maximum achievable score for each of the top five students is 92.

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