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 ?

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,
$=\frac{\frac{1}{3} \pi\frac{h}{2} (r_1+r_3+r_1 r_3)}{\frac{1}{3} \pi \frac{h}{2}(r_2+r_3+r_2 r_3)}$
$=\frac{\frac{1}{3} \pi (10+9+10.9)}{\frac{1}{3} \pi (8+9+9.8)}$
$=\frac{271}{217}$.
