Question 8

# 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

Solution

Let sum of marks of students be x
Now therefore x = 25*50 =1250
Now to maximize the marks of the toppers
We will minimize the marks of 20 students
so their scores will be (30,31,32.....49 )
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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