Difference between revisions of "2017 AMC 10B Problems/Problem 6"

m (Solution)
m (Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
What is the largest number of solid <math>2\text{in}</math> by <math>2\text{in}</math> by <math>1\text{in}</math> blocks that can fit in a <math>3\text{in}</math> by <math>2\text{in}</math> by <math>3\text{in}</math> box?
+
What is the largest number of solid <math>2\text{ in}</math> by <math>2\text{ in}</math> by <math>1\text{ in}</math> blocks that can fit in a <math>3\text{ in}</math> by <math>2\text{ in}</math> by <math>3\text{ in}</math> box?
  
 
<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>

Revision as of 12:46, 16 February 2017

Problem

What is the largest number of solid $2\text{ in}$ by $2\text{ in}$ by $1\text{ in}$ blocks that can fit in a $3\text{ in}$ by $2\text{ in}$ by $3\text{ in}$ box?

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

Solution

We find that the volume of the larger block is $18$, and the volume of the smaller block is $4$. Dividing the two, we see that only a maximum of $4$ $2$x$2$x$1$ blocks can fit inside the $3$x$3$x$2$ block. $\boxed{\textbf{(B) }4}$


2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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. AMC logo.png

Invalid username
Login to AoPS