The average of the three numbers x, y and z is 45. x is greater than the average of y and z by 9. The average of y and z is greater than y by 2. Then the difference of x and z is :

Answer: Option C

According to the question,
$$\begin{array}{rl}& \Rightarrow \frac{x+y+z}{3}=45\\ & \Rightarrow x+y+z=\mathrm{135.....}(i)\\ & \Rightarrow x=\frac{y+z}{2}+9\\ & \Rightarrow 2x-y-z=\mathrm{18.....}(ii)\\ & x+y+z=135\\ & \underline{2x-y-z=18}\\ & 3x=153\\ & x=51\end{array}$$
From (i)
y + z = 135 - 51 = 84.....(iii)
Also,
$$\begin{array}{rl}& \Rightarrow \frac{y+z}{2}=y+2\\ & \Rightarrow y+z=2y+4\\ & \Rightarrow z-y=4\\ & +y+z=84\\ & \underline{-y+z=4}\\ & 2z=88\\ & z=44\end{array}$$
Required difference
= 51 - 44 = 7