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… · ABC26GN4591
Subject: General Aptitude · Chapter: Area
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 $R=x r$, what is the value of $x$ ?
(a)4
(b)4.5
(c)24
(d)25
(e)None of these
Answer
Answer (as printed): E
Explanation
Since weight of the disc is proportional to its area, we have: $$\begin{aligned} \pi\left(R^{2}-r^{2}\right) & =\frac{24}{25} \pi R^{2} \Rightarrow R^{2}-r^{2}=\frac{24}{25} R^{2} \Rightarrow r^{2}=\frac{1}{25} R^{2} \\ R^{2} & =25 r^{2} \Rightarrow R=5 r \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.