aptitude igate test series

Question

A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits one number say ‘ba’ for ‘ab’. Due to this, the average become 1.8 less than the previous one. Find out the difference of the two digits a and b.?

Answer 1

Let the original number be $ab$ and original average be $x$.

Average of 9 numbers $n_1,...,n_9$ and $ab$ is $x$.

(a) $\therefore n_1+...+n_2+10a+b = 10x$.

Average of 9 numbers $n_1,...,n_9$ and $ba$ is $x-1.8$.

(b) $\therefore n_1+...+n_2+10b+a = 10(x-1.8) = 10x - 18$.

Subtracting equation (b) from equation (a),

$9a - 9b = 18$.

$\therefore a - b = 2$.

Answer 2

let the two digits of the first number be x and y

thus average is   (x*10+y*1 + rest 9 numbers sum)/10  =a

after interchanging the digits of the first number we will get average as 

(y*10+x*1+rest 9 numbers sum)/10 = b

and as given a-b= 1.8

after solving we will get

9(x-y)=18

so x-y must be 2.