aayushagrawal wrote:

Between 100 and 200, how many numbers are there in which one digit is the average of the other two?

A) 11

B) 12

C) 10

D) 8

E) 9

How can we solve this question without actually listing all possibilities?

First of all we need to notice that if we have three digits, one of them will be the average of the others only in case when those other two digits have the same parity.

In other words in ABC B will be the average only if A + C = even number. Now let’s look at our question. We have 1AB.

Case #1 A will be the average.

1(Average)B In order for A to be the average B must be odd because 1 is odd and we have: 1*1*5 = 5 ways.

Case #2B will be the average.

1A(Average). In this case A must be odd and we have: 1*1*5 = 5 ways

Together 5 + 5 = 10. Right? No, because we have double counting. In each case we can get 111 as a result so we need to subtract one case.

10 – 1 = 9

Case #3 When actual average will be our 1 in front:

(Average)AB. Because our average is 1 this can be achieved only in 2 ways:

102 and 120.

Finally our total result will be: 9 + 2 = 11

Answer A