Applyo - College Application Platform

CAT 2020 Slot 2 QA Question & Solution

ArithmeticHard

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

Options

5
4
3
6

Solution

Let $x(1)$ be the least number and $x(10)$ be the largest number. Now from the condition given in the question , we can say that

$x(2)+x(3)+x(4)+........x(10)$ = 47*9 = 423 $\quad$...................(1)

Similarly
$x(1)+x(2)+x(3)+x(4)................+x(9)$= 42*9=378 $\quad$...............(2)

Subtracting both the equations we get $x(10)-x(1)=45$

Now, the sum of the 10 observations from equation (1) is $423+x(1)$

Now the minimum value of $x(10)$ will be 47 and the minimum value of $x(1)$ will be 2.

Hence minimum average 425/10 = 42.5

Maximum value of x(1) is 42.

Hence maximum average will be 465/10 = 46.5

Hence difference in average will be $46.5-42.5 = \boxed 4$ which is the correct answer