2021 Fall AMC 10A Problems/Problem 3

Problem

What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution 1

A sphere with radius $2$ has volume $\frac {32\pi}{3}$ . A cube with side length $6$ has volume $216$ . If $\pi$ was $3$ , it would fit 6.75 times inside. Since $\pi$ is approximately $5$ % larger than $3$ , it is safe to assume that the $3$ balls of clay can fit $6$ times inside. Therefore, our answer is $\boxed {(D)6}$ .

~Arcticturn

Solution 2

The volume of the cube is $6^3 = 216.$ The volume of the sphere is $\frac{4}{3} \pi r^3 = \frac{4}{3} \pi \cdot 8 = \frac{32}{3} \pi.$ Because the balls can be compressed but not reshaped, the greatest number of balls that can fit inside the cube is $\left\lfloor \frac{216}{\frac{32}{3}\pi} \right\rfloor = 6 .$ Thus, the answer is $\boxed{\textbf{(D)}.}$

~NH14

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2021 Fall AMC 10A Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See Also