Calculate average of three numbers, where each are again average of numbers

Question

If $x$ is the average (arithmetic mean) of $m$ and $9$, $y$ is the average of $2m$ and $15$, and $z$ is the average of $3m$ and $18$, what is the average of $x, y$, and $z$ in terms of $m$?

A) $m+6$

B) $m+7$

C) $2m+14$

D) $3m+21$

I got this online, and I do not know how to solve it. I know how to find $x$, $y$, and $z$, but don't get the last part.

Answer 1

$x=\cfrac{m+9}{2}$

$y=\cfrac{2m+15}{2}$

$z=\cfrac{3m+18}{2}$

So, the average of $x,y,z$ would be...

$$\cfrac{x+y+z}{3}$$

$$\cfrac{\frac{m+9}{2}+\frac{2m+15}{2}+\frac{3m+18}{2}}{3}$$

$$\cfrac{\frac{6m+42}{2}}{3}$$

$$\frac{6m+42}{6}$$

This simplifies to Choice B, $m+7$.

Answer 2

As per the given definition of $x, y$ and $z$, we have

\begin{align} & x = \dfrac{m + 9}{2} \\ & y = \dfrac{2m + 15}{2} \\ & z = \dfrac{3m + 18}{2} \end{align}

Now the average of $x, y$ and $z$ can be given by \begin{align} \mathbb{E}(x + y + z) & = \dfrac{x + y + z}{3} = \dfrac{6m + 42}{2 \times 3} = m + 7. \end{align}