How to approach change in volume when dimensions change?

Question

If the Height of a right Circular Cone is increased by $200\%$ and the radius of the base reduced by $50\%$ then the Volume of the Cone is?

How to approach this Question and Answer

I have tried some Logic But it is Not working, volume of the Cone is 1/3 22/7 rr h, old cone , New cone is height is h1=3h,New Radius of the base is r1=r/2 after that new volume of cone is v1=18.75v is it correct

Answer 1

How to approach questions:

1.First write down what you know:

Volume of the cone $V$ is given by $$V = \frac {1}{3} \pi r^2 h $$

2.The questions says if you increase this and descrease that, so first make something that you can increase and decrease from, make a specific instance that you can start from let's say $V_1$

$$V_1 = \frac {1}{3} \pi r_1^2 h_1 $$ now make the second instance were things are changed call this $V_2$ $$V_2 = \frac {1}{3} \pi r_2^2 h_2 $$

We know height is increase $200\%$ lets write it down $h_1 \frac{200}{100}= h_2$ and radius is decreased by $50\%$ so lets decrease that $r_1 \frac{50}{100}=r_2$

$$V_2 = \frac {1}{3} \pi (r_1 \frac{50}{100})^2 h_1 \frac{200}{100} $$ $$V_2 = \frac {1}{3} \pi r_1^2 \frac{50^2}{100^2} h_1 \frac{200}{100} $$ $$V_2 = \frac {1}{3} \pi r_1^2 h_1\frac{50^2}{100^2} \frac{200}{100} $$ remember $$V_1 = \frac {1}{3} \pi r_1^2 h_1 $$ so $$V_2 = V_1\frac{50^2}{100^2} \frac{200}{100} $$

so the laststep was to somehow make what know related to what initially had. In this case we said remember, but we could have very well looked at $\frac{V_2}{V_1}$ the result would be the same

Answer 2

The volume of the original cone being $$V_1=\frac 13 \pi r^2\,h$$ the volume of the second cone is $$V_2=\frac 13 \pi \left(\frac r 2\right)^2 (3h)=\frac 14 \pi r^2 \,h$$ Now $$\frac{V_2-V_1}{V_1}$$ would tell you the change in volume.