A clown blows up a spherical balloon such that its...

Hello AAPL.

This is an interesting question. I will try to explain how to solve it.

It takes 3 seconds for the radius of the balloon to increase from 1 inch to 2 inches, so the volume increases from $$V_1=\frac{4\pi\left(1\right)^3}{3}=\frac{4\pi}{3}\ \ \ to\ \ V_2=\frac{4\pi\left(2\right)^3}{3}=\frac{32\pi}{3}$$ in 3 seconds.

We know that the volume increases at a constant rate. So it increase at $$\frac{28\pi}{3}\ each\ 3\ \sec onds,\ or\ \frac{28\pi}{9}\ per\ \sec ond.$$ Now, $$V_3=\frac{4\pi\left(3\right)^3}{3}=36\pi\ \ and\ V_5=\frac{4\pi\left(5\right)^3}{3}=\frac{500\pi}{3}.$$ So $$V_5-V_4=\frac{500\pi}{3}-36\pi=\frac{392\pi}{3}.$$ Finally, $$\frac{\frac{392\pi}{3}}{\frac{28\pi}{9}}=\frac{3528\pi}{84\pi}=42\ .$$ It implies that it takes 42 seconds to increase the radius from 3 inches to 5 inches. The correct answer is E .

I hope it can help you.

Feel free to ask me again if you have a doubt.

Regards.