What is the ratio of empty to filled volume of the glass? The base diameter of a glass is $20$% smaller than the diameter at the rim. The glass is filled to half of the height. Then what is the ratio of empty to filled volume of the glass ?
 A: We assume that the glass is continued downwards to make a big cone. Let the height of that big cone be $10$ units.
Then the part of the big cone that got cut off to make the glass has height $8$.  So the cone cut off plus the water is a cone of height $9$. All the cones under discussion are similar. 
The empty part of the glass has volume a constant $k$ times $10^3-9^3$. The volume of water is $k$ times $9^3-8^3$. 
So the ratio is $\dfrac{10^3-9^3}{9^3-8^3}$. 
A: I know it is an old question. But I think my way of solving the problem is different from the others and an easy to understand.
Solution: Let radius of the rim be $10$ unit. 
Now since the base diameter of a glass is $20~\%$ smaller than the diameter at the rim, so the radius. Hence the radius of the base is $10-\left(10\times \frac {20}{100}\right)=8$ unit.
Let the radius of the circular layer of the water in the glass be $x$ unit and the height of the glass is $2h_1$ unit.
Therefore as per the given condition the height of the water is $h_1$ unit and the height of the remaining part is also $h_1$ unit.
If you try to understand the problem graphically, it seems like the following

From the figure, using the formula for the similar triangles, we have
$$\dfrac{2h_1+h}{h}=\dfrac{10}{8}\implies 16h_1+8h=10h\implies h=8h_1$$
Also by the same approach,
$$\dfrac{9h_1}{8h_1}=\dfrac{x}{8}\implies x=9$$
Therefore ratio of empty to filled volume of the glass is
$$\dfrac{\pi/3\left[10^2\times 10h_1-9^2\times 9h_1\right]}{\pi/3\left[9^2\times 9h_1-8^2\times 8h_1\right]}=\dfrac{10^3-9^3}{9^3-8^3}~.$$
A: Let $r$ the radius at the rim, then at the base the radius is $r_b=\frac{4}{5}r$. Supposing the glass has a truncated cone shape, then the radius at the middle of the glass is $r_m=\frac{9}{10}r$. The ratio you are looking for is
\begin{align}
\frac{V_{\text{empty}}}{V_{\text{filled}}}&=\frac{\frac{\pi h}{6}\left(r^2+rr_m+r_m^2\right)}{\frac{\pi h}{6}\left(r_m^2+r_mr_b+r_b^2\right)}\\[3pt]
&=\frac{r^2+\frac{9}{10}r^2+\frac{81}{100}r^2}{\frac{81}{100}r^2+\frac{36}{50}r^2+\frac{16}{25}r^2}\\[3pt]
&=\color{blue}{\frac{271}{217}}
\end{align}
Where $h$ is the height of the glass. The formula for the volume of a truncated cone is here
A: Let $r_1, r_2$ and $r_3$ be the top, bottom and middle radius respectively, and h be the hight of the cone.
Volume of glass= $\frac{1}{3}\pi h (r_1+r_2+r_1 r_2)$.
Let $r_1=10$, then $r_2=8$ and $r_3=9$.
So the ratio of volume of empty portion of glass to volume of filled portion of glass,
$=\frac{\frac{1}{3} \pi\frac{h}{2} (r_1+r_3+r_1 r_3)}{\frac{1}{3} \pi \frac{h}{2}(r_2+r_3+r_2 r_3)}$
$=\frac{\frac{1}{3} \pi (10+9+10.9)}{\frac{1}{3} \pi (8+9+9.8)}$
$=\frac{271}{217}$.
