shubhangi wrote:

A boy finds an average of 10 positive integers. Each integeres contain two digits .By mistake the boy interchanges the digits of one number say ba for ab. Due to this the average become 1.8 less than the previous one.What was the difference of the two digit a and b?

a)4

b)2

c)6

d)8

Any easy & understandable solution?

let the number be represented by 10a + b

a = tens digits

b = unit digit

we know that.

Orginal Average - Mistaken Average = 1.8

[(Sum of 9 nos.) + (10 a+b) / 10] - [Sum of 9 nos +(10b +a) / 10]= 1.8

that is just...

[(10a + b ) - (10b +a)] /10 =1.8 ( 10 is the common denominator..so

we can simplify)

10a + b -10b -a = 18

9a - 9b =18

thus,

a- b =2

recall a = tenths digits, b =units digits

Answer B