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 :$$\eqalign{ & \pi \left( {{R^2} - {r^2}} \right) = \frac{{24}}{{25}}\pi {R^2} \cr & \Rightarrow {R^2} - {r^2} = \frac{{24}}{{25}}{R^2} \cr & \Rightarrow {r^2} = \frac{1}{{25}}{R^2} \cr & \Rightarrow {R^2} = 25{r^2} \cr & \Rightarrow R = 5r \cr} $$
Related Questions on Area
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D. 307200 m2
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B. 18 cm
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D. Data inadequate
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The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
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B. 42%
C. 44%
D. 46%
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