Question

Amol was asked to calculate the arithmetic mean of 10 positive integers, each of which had 2
digits. By mistake, he interchanged the 2 digits, say a and b, in one of these 10 integers. As a result,
his answer for the arithmetic mean was 1.8 more than what it should have been. Then b – a equals 1

Answer 1

Answer

suppose the number variety which become interchanged is ba, if we interchange the digits then the wide variety becomes ab

Average earlier than interchanging = (ba + x)/10 ,

wherein x is sum of different nine numbers

Average after interchanging = (ab + x)/10

Given that common earlier than interchanging = common after interchanging + 1.8

(ba + x)/10 = (ab+x)/10 + 1.8

or

ba + x = ab + x + 18

ba may be written as 10b + a and ab may be written as 10a + b

10b + a + x = 10a + b + x + 18

or

b-a = 2

#SPJ6

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