Khawasbnam1 wrote:

If the arithmetic mean of three positive integers is 8, what are the values of the integers?

(1) The largest integer is twice the smallest integer.

(2) One of the integer is 9.

The official answer was A. But I think both statements are sufficient. I need help on this question.

Given,

a+b+c=24

RTF: a=? b=? c=?

Option-1: Let us consider a=largest and c= smallest, then a=2c

So, a+b+c=2c+b+c=24

=> 3c+b=24

Now from here, as we don't know the value of b, we can't determine the value of b either. So not sufficient.

Option-2: One of the integer is 9.

So it can be either a, b or c. So not sufficient.

Option-1 + Option-2:

a=2c

Now, if a=9, then c will not be an integer => a cannot be 9

If c=9, then a=18 => a+c=27 (>24) => c cannot be 9 either

Hence, b can only take the value of 9

So, 2c+9+c=24 => 3c=15 => c =5

a=10, b=9, c=5....Sufficient.

Hence, IMO OA is wrong and ans should be C.