Mensuration and similarity

Question

Cone P has a volume of 108cm^3

Calculate the volume of 2nd come , Q , whose radius is double that of cone P and its height is one-third that of cone p

Here's my working .... $$V_Q=\frac13 \pi (2r)^2 \cdot \frac{h}{3}\\ = \frac{4}{9} \pi r^2 h\\ = \frac{1}{3}\pi r^2 h \cdot \frac{4}{3}$$

I don't understand why must I do this.

Answer 1

It is not clear what is being asked here.

Everything you have done so far is correct, just express the volume of the new cone in terms of that of the old cone and substitute the volume of the old cone to get the answer.

That is nothing but four thirds of the volume of the cone P.

Answer 2

Hint:

In your equation, you already have $$V_Q = \left(\frac{1}{3}\pi r^2h\right) \cdot 43$$

You also know what the volume of cone $P$ is, and you know that it is equal to $\frac{1}{3}\pi r^2 h$