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Difference between revisions of "2012 AMC 12A Problems/Problem 3"

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Revision as of 13:21, 12 February 2012

Problem 3

A box $2$ centimeters high, $3$ centimeters wide, and $5$ centimeters long can hold $40$ grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold $n$ grams of clay. What is $n$?

$\textbf{(A)}\ 120\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 200\qquad\textbf{(D)}\ 240\qquad\textbf{(E)}\ 280$

Solution

The first box has volume $2\times3\times5=30\text{ cm}^3$, and the second has volume $(2\times2)\times(3\times3)\times(5)=180\text{ cm}^3$. The second has a volume that is $6$ times greater, so it holds $6\times40=\boxed{\textbf{(D)}\ 240}$ grams.

See Also

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AMC 12 Problems and Solutions
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