Question 21

Let $$C$$ be a circle with centre $$P_0$$ and $$AB$$ be a diameter of $$C$$. Suppose $$P_1$$ is the mid point of the line segment $$P_0B$$,$$P_2$$ is the mid point of the line segment $$P_1B$$ and so on. Let $$C_1,C_2,C_3,...$$ be circles with diameters $$P_0P_1, P_1P_2, P_2P_3...$$ respectively. Suppose the circles $$C_1, C_2, C_3,...$$ are all shaded. The ratio of the area of the unshaded portion of $$C$$ to that of the original circle is

Solution

Radius of the circles $$C_1, C_2, C_3,...$$ would be in GP with (R/4),(R/8),(R/16) and so on. Radii of circles are in the ratio 1:4.

Ratio of unshaded region to the ratio of original circle = 1-$$\frac{Ratio\ of\ shaded\ region}{Ratio\ of\ original\ circle}$$

= 1-$$\frac{\pi r^2/16 + \pi r^2/64+.....}{\pi r^2}$$ = 1-$$\frac{1/16}{(1-1/4)}$$ = 1- 1/12 = $$\frac{11}{12}$$ = 11:12

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