Average and Means

Question

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...

Answer 1

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

x1 < x2 < x3 < x4 <.......< x10

.

As per the question :-

(x1+x2+x3+....+x9)/9 = 42   Also we can say that x1 <= 42.

==> x1+x2+x3+...+x9 = 378  ----(i)

(x2+x3+x4+...+x10)/9 = 47   Also we can say that x10 >=47

==> x2+x3+x4+....+x10 = 423   ------(ii)

Mean of 10 students = (x1+x2+x3+........+x10)/10

Maximum mean will occur when x1 = 42 :-

[x1 + (x2+x3+x4+.....+x10)]/10 = (42 + 423)/10 = 46.5  From eq.(ii)

Minimum mean will occur when x10 = 47 :-

[(x1+x2+.....x9) + x10]/10 = (378+47)/10 = 42.5   From eq.(i)

Difference = 4 

Please correct me if I am wrong in my approach.