Data Sufficiency

What is the average (arithmetic mean) of x and y?

(1) The average (arithmetic mean) of x, y and k is 7.

(2) The average (arithmetic mean) of x, y and 3k is 13.


OA C

Source: Magoosh

$$Mean=\frac{E\ Fx\ }{Ef}=\frac{x+y}{2}$$
Target question: What is the mean of x and y?

Statement 1: The average (arithmetic mean) of x, y, and k is 7.
$$\frac{x+y+k}{3}=7$$
$$x+y+k=21$$
The value of x and y remains unknown; so, therefore, we cannot the target question at this point.

Statement 2: The average (arithmetic mean) of x, y and 3k is 13.
$$\frac{x+y+3k}{3}=13$$
$$x+y+3k=39$$
The value of x and y remains unknown. Also, the target question cannot be solved.

Combining both statements together:
From statement 1: x + y + k = 21 eqn 1
From statement 2: x + y + 3k = 39 eqn 2
Multiply equation 1 through by 3
(x*3) + (y*3) + (k*3) = 21*3
3x + 3y + 3k = 63 ====> eqn 3
Subtract equation 2 from Eqn. 3
3x + 3y + 3k = 63
x + y + 3k = 39
----------------
2x + 2y = 24
2 (x+y) = 24
x+y = 12
$$Therefore,\ mean\ of\ 'x'\ and\ 'y'=\frac{x+y}{2};\ \ \ where\ x+y=12$$
$$So,\ \frac{12}{2}=6$$
Both statements combined together are SUFFICIENT.
Answer = option C