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Q.

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

 A.

8

 B.

6

 C.

2

 D.

4

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Solution:
Option(C) is correct

Let the original number be $ab$ i.e., $(10a + b)$.

After interchanging the digits, the new number becomes ba i.e., $(10b + a)$.

The question states that the average of 10 numbers has become 1.8 less than the original average.

Therefore, the sum of the original 10 numbers will be $10\times 1.8$ more than the sum of the 10 numbers with the digits interchanged.

i.e., $10a+b=10b+a+18$

$9a−9b=18$

$a−b$ = 2


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