A boy was asked to calculate the arithmetic mean of ten positive integers each of which had 2 digits.
By mistake, he interchanged the two digits, a & b, in one of these ten integers. As a result, his answer for the arithmetic mean was 1.8 more than what it should have been.
Find the value of b-a.
Since there are ten numbers to begin with, it follows that by interchanging the digits, the total must have been appreciated by 1.8*10 = 18.
(10b+a) - (10a+b) = 18, giving:
9(b-a) = 18, so that;
b-a = 2
Consequently, the required difference is 2.
Edited on October 16, 2007, 12:50 am