A small disc of radius r is cut out from a disc of radius R. The weight of the disc which now has a hole in it, is reduced to $$\frac{{24}}{{25}}$$ of the original weight. If R = xr, what is the value of x ?

A. 4

B. 4.5

C. 24

D. 25

E. None of these

Answer: Option E

Solution(By Examveda Team)

Since weight of the disc is proportional to its area, we have :
$$\eqalign{ & \pi \left( {{R^2} - {r^2}} \right) = \frac{{24}}{{25}}\pi {R^2} \cr & \Rightarrow {R^2} - {r^2} = \frac{{24}}{{25}}{R^2} \cr & \Rightarrow {r^2} = \frac{1}{{25}}{R^2} \cr & \Rightarrow {R^2} = 25{r^2} \cr & \Rightarrow R = 5r \cr} $$