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(D) Total number of students $n = 50$.Average marks $\bar{x} = 47.5$.Total marks obtained by all students $= 50 	imes 47.5 = 2375$.Let $k$ be the num

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(D) Total number of students $n = 50$.Average marks $\bar{x} = 47.5$.Total marks obtained by all students $= 50 	imes 47.5 = 2375$.Let $k$ be the number of students who scored more than the average $(47.5)$. Since marks are integers,these students must have scored at least $48$ marks.Let the remaining $(50 - k)$ students score the minimum possible marks,which is $0$.To maximize $k$,we assume these $k$ students scored the minimum possible marks greater than the average,which is $48$.Thus,$48k + (50 - k) 	imes 0 \leq 2375$.$48k \leq 2375$.$k \leq \frac{2375}{48} \approx 49.479$.Since $k$ must be an integer,the maximum value of $k$ is $49$.

$A$ $100$ mark examination was administered to a class of $50$ students. Despite only integer marks being given,the average score of the class was $47.5$. The maximum number of students who could have received marks greater than the class average is:

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