Maximum score of toppers in an examination

Total sum of scores of 25 students = $25 \times 50 = 1250$.

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: $S_{min} = \frac{20}{2} (30 + 49) = 10 \times 79 = 790$

So: $5T + 790 = 1250 \implies 5T = 460 \implies T = 92$.