RRB Non Technical Model Papers 10 with solutions

(3) Since, average of x, y and z is 45 and x is greater than average of y and z by 9,
According to the question,
$\frac{\mathit{y}+\mathit{z}}{2}+9=\mathit{x}$
Sum of three numbers is
$\frac{\mathit{y}+\mathit{z}}{2}+9+\mathit{y}+\mathit{z}=45\times 3\Rightarrow \mathit{y}+\mathit{z}=84$
Now, average of y and z is greater than y by 2, we get:
$\frac{\mathit{y}+\mathit{z}}{2}=\mathit{y}+2\Rightarrow \mathit{z}=4$
Therefore, z = 44 and y = 40
Now,$\mathit{x}=\frac{40+44}{2}+9=51$
So, difference between x and z is 51 – 44 = 7