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

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==Solution==
 
==Solution==
By simply finding the volume of the larger block, we see that its area is <math>18</math>. The volume of the smaller block is <math>4</math>. Dividing the two, we see that only a maximum of <math>4</math> <math>2</math>in x<math>2</math>in x<math>1</math>in blocks can fit inside a <math>3</math>-in by <math>2</math> in by <math>3</math>in box. <math>\qquad\textbf{(B)}\ 4</math>
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We find that the volume of the larger block is <math>18</math>, and the volume of the smaller block is <math>4</math>. Dividing the two, we see that only a maximum of <math>4</math> <math>2</math> in <math>2</math> in <math>1</math> in blocks can fit inside a <math>3</math>-in by <math>2</math> in by <math>3</math> in box. <math>\qquad\textbf{(B)}\ 4</math>
  
  
 
{{AMC10 box|year=2017|ab=B|num-b=5|num-a=7}}
 
{{AMC10 box|year=2017|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:42, 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 $4$ $2$ in $2$ in $1$ 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
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

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