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

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