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

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==Problem==
 
==Problem==
  
What is the largest number of solid <math>2\text{ in } \times 2\text{ in } \times 1\text{ in}</math> blocks that can fit in a <math>3\text{ in x } 2\text{ in x } 3\text{ in}</math> box?
+
What is the largest number of solid <math>2\text{-in} \times 2\text{-in} \times 1\text{-in}</math> blocks that can fit in a <math>3\text{-in} \times 2\text{-in}\times3\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 13:49, 16 February 2017

Problem

What is the largest number of solid $2\text{-in} \times 2\text{-in} \times 1\text{-in}$ blocks that can fit in a $3\text{-in} \times 2\text{-in}\times3\text{-in}$ 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 $4$ $2$ by $2$ by $1$ blocks can fit inside the $3$ by $3$ by $2$ block. Therefore, the answer is $\boxed{\textbf{(B) }4}$.


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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