Question

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

Accepted Answer

92. Total sum of scores of 25 students = $25 	imes 50 = 1250$. Let the score of each of the 5 toppers be $T$. To maximize $T$, we must minimize the sum of the remaining $20$ students' scores. The minimum possible distinct integer scores starting from $30$ for $20$ students are $30, 31, 32, \dots, 49$. Sum of these 20 scores: $$S_{min} = \frac{20}{2} (30 + 49) = 10 	imes 79 = 790$$ So: $$5T + 790 = 1250 \implies 5T = 460 \implies T = 92$$.

Maximum score of toppers in an examination

Explanation

Practise this under exam conditions