A gold ingot in the shape of a cylinder is melted and the resulting molten metal molded into a few identical conical ingots. If the height of each cone is half the height of the original cylinder and the area of the circular base of each cone is one fifth that of the circular base of the cylinder, then how many conical ingots can be made?

Let the radius of the cylinder be $$r$$ and height be $$h$$ (as shown).

Let the dimensions of the cone be $$r_c$$ and $$h_c$$ (as shown).

It is given that $$h_c = \frac{h}{2}$$

Area of base of cone $$A_{c}=\pi \times r_{c}^{2}$$

Area of base of cylinder $$A=\pi \times r^{2}$$

Given, $$A_{c}=\frac{1}{5} \times A $$

$$ \Rightarrow \pi \times r_{c}^{2} =\frac{1}{5} \times \pi \times r^{2} $$

$$  \Rightarrow r_{c}^{2} = \frac{1}{5} \times r^{2} $$

$$  \Rightarrow \frac{r^{2}}{r_{c}^{2}} = 5 $$

Now we know that the volume of cylinder = total volume of cones 

Let the number of cones be n. 

So, volume of cylinder = n x volume of  each cone

$$  \Rightarrow V = n \times V_{c} $$

$$  \Rightarrow \pi \times r^{2} \times h = n \times \frac{1}{3} \times \pi \times r_{c}^{2} \times h_c $$

$$  \Rightarrow r^{2} \times h = n \times \frac{1}{3} \times r_{c}^{2} \times h_c $$

$$  \Rightarrow \frac{r^{2}}{r_{c}^{2}} \times \frac{h}{h_c} \times 3 = n $$

$$  \Rightarrow 5 \times 2 \times 3 = n $$

$$  \Rightarrow 30 = n $$

Thus, 30 conical ingots can be made.