Question

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

• A. $85.5$
• B. $92.5$
• C. $90.5$
• D. $87.5$

Answer 1

The correct option is D $87.5$

Let $r_a,r_b$ be radius of A & B respectively

Given S.A of B is 300% greater then S.A of A

$\Rightarrow 4\pi r_b^2=4\times 4\pi r_a^2$

$\Rightarrow r_b=2r_a$

Volume of $A\left(V_a\right)=\frac{4}{3}\pi r_a^3$

Volume of $B\left(V_b\right)=\frac{4}{3}\pi (r_b{)}^3=\frac{4}{3}\pi (2r_a{)}^3=8\times \frac{4}{3}\pi r_a^3$

$V_b=8V_a$

$\therefore$ Volume of A is 87.5% lower than B