An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies 75% of the volume of the frozen ice cream. What is the ratio of the cone's height to its radius? (Note: a cone with radius r and height h has volume πr2h/3 and a sphere with radius r has volume 4πr3/3.)
Try it yourself first — the explanation is most useful after you’ve attempted it.
View mode:
Toolkit + CCSS Solution
Understand
Restated: A ball-shaped scoop of ice cream sits on top of a cone. The cone opening has the same diameter as the ball, so they share the same radius $r$. When the ball melts it shrinks to $75\%$ of its frozen volume, and that melted liquid fills the cone exactly to the brim. Find how the cone's height $h$ compares to its radius $r$, written as a ratio $h:r$.
Givens: The frozen scoop is a sphere of radius $r$, volume $\dfrac{4}{3}\pi r^3$; The cone has the same radius $r$ as the sphere and some height $h$, volume $\dfrac{1}{3}\pi r^2 h$; Melted ice cream is $75\%$ of the frozen volume; The melted ice cream exactly fills the cone (no gap, no overflow)
Unknowns: The ratio of the cone's height to its radius, $h:r$
Understand
Restated: A ball-shaped scoop of ice cream sits on top of a cone. The cone opening has the same diameter as the ball, so they share the same radius $r$. When the ball melts it shrinks to $75\%$ of its frozen volume, and that melted liquid fills the cone exactly to the brim. Find how the cone's height $h$ compares to its radius $r$, written as a ratio $h:r$.
Givens: The frozen scoop is a sphere of radius $r$, volume $\dfrac{4}{3}\pi r^3$; The cone has the same radius $r$ as the sphere and some height $h$, volume $\dfrac{1}{3}\pi r^2 h$; Melted ice cream is $75\%$ of the frozen volume; The melted ice cream exactly fills the cone (no gap, no overflow)
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #1 Draw a Diagram
The problem is a chain of volumes with one shared measurement. Tool #4 (Introduce a Variable) is the anchor: name the shared radius $r$ and the cone height $h$, and every quantity becomes an expression in $r$ and $h$. Then Tool #7 (Identify Subproblems) breaks the work into three clean pieces — frozen sphere volume, then the $75\%$ melted amount, then the cone volume — each a single plug-in of a given formula. Setting the melted amount equal to the cone volume gives one equation, and because both sides carry a factor of $\pi r^2$, that factor cancels and $h$ falls out in terms of $r$. Tool #1 (Draw a Diagram) keeps the picture honest: it shows why the cone and sphere must share the same radius.
Execute — Answer: B
#4 Introduce a Variable 6.EE.B.6Step 1
Name the shared radius and height
Because the cone's opening has the same diameter as the ball, they have the same radius.
Call that radius $r$ and call the cone's height $h$.
Now write the two volumes the problem hands you: the frozen sphere is $\dfrac{4}{3}\pi r^3$, and the cone that will catch the melt is $\dfrac{1}{3}\pi r^2 h$.
Only one radius appears, which is the key that lets the two volumes talk to each other.
💡 One shared radius means one letter $r$ ties the sphere and the cone together.
#7 Identify Subproblems 7.RP.A.3Step 2
Melt the scoop down to 75%
Melting keeps only $75\%=\dfrac{3}{4}$ of the frozen volume.
Multiply the sphere's volume by $\dfrac{3}{4}$.
The $\dfrac{3}{4}$ and the $\dfrac{4}{3}$ are reciprocals, so they cancel and the messy fraction disappears — the melted liquid has volume exactly $\pi r^3$.
💡 Undo the one-third by tripling, and the height stands revealed as three radii tall.
[1]
#4 6.EE.B.6Because the cone's opening has the same diameter as the ball, they have the same
[2]
#7 7.RP.A.3Melting keeps only $75\%=\dfrac{3}{4}$ of the frozen volume. Multiply the sphere
[3]
#7 8.G.C.9"Exactly fills" means the melted liquid and the cone hold the same volume. Set t
[4]
#4 8.EE.C.7Multiply both sides of $\dfrac{1}{3}h=r$ by $3$ to get $h=3r$. So the height is
Review
Reasonableness: A tall, skinny cone makes sense here: a round scoop is a fat shape, and a cone only holds $\dfrac{1}{3}$ of the box (cylinder) around it, so it needs extra height to match a ball's volume. Check the numbers directly with $h=3r$: cone volume $=\dfrac{1}{3}\pi r^2(3r)=\pi r^3$, which is exactly the melted volume $\pi r^3$. The two match, so $h:r=3:1$ is right. A quick sanity bound: if $h$ were only $r$ (ratio $1:1$) the cone would hold just $\dfrac{1}{3}\pi r^3$, far too little — so the ratio must be well above $1$, and $3:1$ fits while a giant $6:1$ would overflow.
Alternative: You can reason in "cone units" instead of algebra. A cone whose height equals its radius $r$ holds $\dfrac{1}{3}\pi r^3$. The melted ice cream is $\pi r^3$, which is $3$ times that base cone. Since a cone's volume grows in direct proportion to its height (radius fixed), tripling the volume means tripling the height: $h=3r$. So $h:r=3:1$, choice (B).
CCSS standards used (min grade 8)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the shared radius $r$ and cone height $h$, then writing each volume as an expression in those letters.)
7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems (Taking $75\%$ of the frozen sphere's volume to get the melted volume $\pi r^3$.)
8.G.C.9 Know the formulas for volumes of cones, cylinders, and spheres (Using the sphere volume $\tfrac{4}{3}\pi r^3$ and cone volume $\tfrac{1}{3}\pi r^2 h$, then equating the cone to the melted amount.)
8.EE.C.7 Solve linear equations in one variable (Solving $\tfrac{1}{3}h=r$ for $h$ to get $h=3r$ and the ratio $3:1$.)
⭐ When one melted volume has to exactly fill another shape, set the two volumes equal, cancel the parts they share, and the leftover tells you the missing measurement.
⭐ When one melted volume has to exactly fill another shape, set the two volumes equal, cancel the parts they share, and the leftover tells you the missing measurement.
