In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by
How I have approached this question till now :-
Let x1,x2,x3,……,x10 be the marks of 10 students such that x1<x2<x3<….<x10.
(x1+x2+x3+….+x9)/9 = 42
==> x1+x2+x3+…+x9 = 378
(x2+x3+x4+…+x10)/9 = 47
==> x2+x3+x4+….+x10 = 423
From the above two equations subtraction :-
x10 – x1 = 45
from the above two equations addition :-
x1 + 2(x2+x3+....+x9) + x10 = 801
==> x2+x3+...+x9 = [801 - (x10+x1)]/2
After this I tried using this relation in the average of 10 students but couldn't work that out towards a solution. Need some help…
Best Answer
I solved the question with exact answer of 4. Below is my solution approach:-
Let the marks of 10 students be
x1,x2,x3,...........,x10
.Suppose
.
As per the question :-
Please correct me if I am wrong in my approach.