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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
Correct Answer: 92
Let sum of marks of students be xΒ
Now therefore x = 25*50 =1250Β
Now to maximize the marks of the toppersΒ
We will minimize the marks of 20 studentsΒ
so their scores will be (30,31,32.....49 )Β
let score of toppers be y
so we get 5y +$$\frac{20}{2}\left(79\right)$$=1250
we get 5y +790=1250
5y=460
y=92
So scores of toppers = 92
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