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1. 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

Solution:

Since weight of the disc is proportional to its area, we have :

\begin{aligned} π (R^{2} - r^{2}) = \frac{24}{25} π R^{2} \\ \Rightarrow R^{2} - r^{2} = \frac{24}{25} R^{2} \\ \Rightarrow r^{2} = \frac{1}{25} R^{2} \\ \Rightarrow R^{2} = 25 r^{2} \\ \Rightarrow R = 5 r \end{aligned}

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