Question 57
Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be
[CAT 2003 leaked]
Solution
Surface area of sphere A (of radius a) is $$4\pi*a^2$$
Surface area of sphere B (of radius b) is $$4\pi*b^2$$
=> $$4\pi*a^2$$/$$4\pi*b^2$$ = 1/4 => a:b = 1:2
Their volumes would be in the ratio 1:8
Therefore, k = 7/8 * 100% = 87.5%
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