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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
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Let  the sum of scores of students be $x.$

$\Rightarrow \frac{x}{50} = 25$

$\Rightarrow x= 25 \times 50$

$\Rightarrow \boxed{x= 1250}$

For the scores of the top $5$ students to be as high as possible, the score of the bottom $20$ students should be as low as possible.

The minimum score is $30,$ and the scores of the bottom $20$ students are distinct integers.

So, the score of bottom $20$ students must be $30,31,32, \dots, 49.$

Let the score of the topper be $y.$

So, $5y+(30+31+32+ \dots +49) = x$

$\Rightarrow 5y + \frac{20}{2}(30+49) = 1250$

$\Rightarrow 5y+790 = 1250$

$\Rightarrow 5y = 1250-790$

$\Rightarrow 5y = 460$

$\Rightarrow \boxed{y=92}$

$\therefore$ The maximum possible score of the topper is $92.$

Correct Answer $:92$
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