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
Explanation:
Let the score of 5 toppers be x each.
To maximize the score of toppers we need to minimize the scores of the remaining 20 students.
∴ Score of remaining 25 students will be 30, 31, 32, … , 49.
⇒ 25 × 50 = 30 + 31 + … + 49 + 5x
⇒ 1250 = 790 + 5x
⇒ x = 92
Hence, 92.
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