Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2021 Fall AMC 10A Problems/Problem 3"
(→‎Solution 1 (Algebra))
Line 4: Line 4:
 
<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7</math>
 
<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7</math>
  
−
== Solution 1 (Algebra) ==
+
== Solution 1 (Inequality) ==
  
 
The volume of the cube is <math>V_{\text{cube}}=6^3=216,</math> and the volume of a clay ball is <math>V_{\text{ball}}=\frac43\cdot\pi\cdot2^3=\frac{32}{3}\pi.</math>
 
The volume of the cube is <math>V_{\text{cube}}=6^3=216,</math> and the volume of a clay ball is <math>V_{\text{ball}}=\frac43\cdot\pi\cdot2^3=\frac{32}{3}\pi.</math>

Revision as of 16:06, 23 November 2021

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 (Inequality)

The volume of the cube is $V_{\text{cube}}=6^3=216,$ and the volume of a clay ball is $V_{\text{ball}}=\frac43\cdot\pi\cdot2^3=\frac{32}{3}\pi.$

Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is \[\left\lfloor\frac{V_{\text{cube}}}{V_{\text{ball}}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.\] Approximating with $\pi\approx3.14,$ we have $\left\lfloor\frac{81}{13}\right\rfloor \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq \left\lfloor\frac{81}{12}\right\rfloor,$ or \[6 \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq 6.\] Clearly, we get $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.$

~NH14 ~MRENTHUSIASM

Solution 2 (Arithmetic)

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

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_3&oldid=165517"