## Ice-Cream Cone

An ice-cream cone’s height is three times its diameter. Two spherical scoops of ice cream with the same diameter as the cone are packed into the cone, filling it completely. Some ice cream bulges above the cone’s top. What percent of the scoops remains outside the cone? (assume that there is no dripping and no compression from the packing and that the cone is a standard shape.)
Source: NCTM Mathematics Teacher, February 2006

SOLUTION
Let $r$ represent the radius of the cone and $h$ its height.
Volume of cone = $(1/3)\pi r^2h$
$=(1/3)\pi r^2(6r)$
$=2\pi r^3$
Volume of scoops = $2(4/3)\pi r^3$
$=(8/3)\pi r^3$
Bulge above cone = volume of scoops – volume of cone
$=(8/3)\pi r^3-2\pi r^3$
$=(2/3)\pi r^3$
Percent of scoops above cone = bulge $\div$ volume of scoops
$=(2/3)\pi r^3\div (8/3)\pi r^3$
$=25\%$

Answer: $25\%$