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

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(Problem)
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==Problem==
 
==Problem==
  
What is the largest number of solid <math>2</math>in x<math>2</math>in x<math>1</math>in blocks that can fit in a <math>3</math>-in by <math>2</math> in by <math>3</math>in box?
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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>
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<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math>
  
 
==Solution==
 
==Solution==

Revision as of 10:27, 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

By simply finding the volume of the larger block, we see that its area is $18$. The volume of the smaller block is $4$. Dividing the two, we see that only a maximum of $4$ $2$in x$2$in x$1$in blocks can fit inside a $3$-in by $2$ in by $3$in box. $\qquad\textbf{(B)}\ [4]$


2017 AMC 10b (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions

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