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Khawasbnam1
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Updated the question to match the OA.

Best,
Gladi

Khawasbnam1
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.
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The answer is A.
As the numbers are distinct and the average of all three is 8.

Let three numbers be a,b,c where c is largest and a is smallest.


Stmt 1: c=2a
Therefore, a+b+c=24
3a+b=24

Lets check with a=1, then c becomes 2 and b becomes 21.
Not correct as 'c' has to be largest

Try with a=2, 3,4 with all the cases b becomes largest.

Only a=5 is the case when c=10 and b=9.

Thus, Stmt 1 is sufficient.
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St 1: 3a+b = 24

check for a = 5, 6

for a= 5, b =9 , c =10

for a = 6, b = 8, c = 12

In both cases a is smallest and c is largest.

We need St. 2 to eliminate option 2.

Ans should be C.

Am I missing something here ?
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Chanda15
St 1: 3a+b = 24

check for a = 5, 6

for a= 5, b =9 , c =10

for a = 6, b = 8, c = 12

In both cases a is smallest and c is largest.

We need St. 2 to eliminate option 2.

Ans should be C.

Am I missing something here ?


6+8+12 = 26 <> 24
You would need 6, 6, and 12. In that case the numbers are not distinct, so only 5, 9, and 10 is possible from S1.

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