2017 AMC 10B Problems/Problem 6

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 (Problems • Answer Key • Resources)
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.

Page

Toolbox

Search

2017 AMC 10B Problems/Problem 6

Problem

Solution