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.

Let x = 10a + b and assume x is the correct number
Let y = 10b + a then y is the incorrect transposed number
Let z be the sum of the other nine numbers

So we have....

(z + x)/10 + 1.8 = (z + y)/10
z + x + 18 = z + y
x + 18 = y

re-writing this in terms of a & b gives

10a + b + 18 = 10b + a
9b - 9a = 18
b - a = 2

Posted by fwaff on 2003-11-19 09:38:00