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Let $\text{A}$ and $\text{B}$ two solid spheres such that the surface area of $\text{B}$ is $300\%$ higher than the surface area of $\text{A}.$ The volume of $\text{A}$ is found to be $k\%$ lower than the volume of $\text{B}.$ The value of $k$ must be ________

- $85.5$
- $92.5$
- $90.5$
- $87.5$

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Let radius of sphere $A$ = $r_a$

Let radius of sphere $B$ = $r_b$

Surface area of sphere = $4\pi r^2$

Volume of sphere = $\frac{4}{3} \pi r^3$

Let surface area of sphere $A = x$

Let surface area of sphere $B = x + 3x = 4x$ [Since surface area of B is 300% higher than A]

$\implies \large{\frac{\text{Surface area of sphere A}}{\text{Surface area of sphere B}} = \frac{4 \pi r_a^2}{4 \pi r_b ^2}}$

$\implies \large{\frac{x}{4x}} = (\frac{r_a}{r_b})^2$

$\implies \large{\frac{r_a}{r_b} = \frac{1}{2}}$

Let volume of sphere $B = y$

Let volume of sphere $A = y – \frac{k}{100}y = \frac{(100-k)y}{100}$

$\implies \large{\frac{\text{Volume of sphere A}}{\text{Volume of sphere B}} = \frac{\frac{4}{3} \pi r_a^3}{\frac{4}{3} \pi r_b^3}}$

$\implies \large{\frac{\frac{(100-k)y}{100}}{y} = (\frac{r_a}{r_b}})^3$

$\implies \large{\frac{100-k}{100} = \frac{1}{8}}$

$\implies 800-8k = 100 \implies 8k =700 \implies k = 87.5$