If $x=3, y=2 x$, and $z=3 y$, then the average of $x, y$ and $z$ is
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Since $x=3$ and $y=2 x$, then $y=2 \times 3=6$.

Since $y=6$ and $z=3 y$, then $z=3 \times 6=18$.

Therefore, the average of $x, y$ and $z$ is $\frac{x+y+z}{3}=\frac{3+6+18}{3}=9$.

ANSWER: (D)