Three different numbers are chosen such that when each of the numbers is added to the average of the remaining two, the numbers 65,69 and 76 result. The average of the three original numbers is
(A) 34
(B) 35
(C) 36
(D) 37
(E) 38
Three different numbers are chosen such that when each of the numbers is added to the average of the remaining two, the numbers 65,69 and 76 result. The average of the three original numbers is
(A) 34
(B) 35
(C) 36
(D) 37
(E) 38

## Solution

Let the three numbers be $a, b$ and $c$.

We construct the first equation to be,

$$
a+\frac{b+c}{2}=65
$$

Or, $2 a+b+c=130$.

Similarly we construct the two other equations to be,

$$
\begin{aligned}
a+2 b+c & =138 \\
\text { and } \quad a+b+2 c & =152
\end{aligned}
$$

If we add the three equations we obtain,

$$
4 a+4 b+4 c=420
$$

The average is $\frac{4(a+b+c)}{12}=\frac{420}{12}$.

Or, $\frac{a+b+c}{3}=35$.

ANSWER: (B)