Difference between revisions of "2012 AMC 12A Problems/Problem 3"
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<math>\textbf{(A)}\ 120\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 200\qquad\textbf{(D)}\ 240\qquad\textbf{(E)}\ 280</math> | <math>\textbf{(A)}\ 120\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 200\qquad\textbf{(D)}\ 240\qquad\textbf{(E)}\ 280</math> | ||
| − | ==Solution== | + | ==Solution 1== |
The first box has volume <math>2\times3\times5=30\text{ cm}^3</math>, and the second has volume <math>(2\times2)\times(3\times3)\times(5)=180\text{ cm}^3</math>. The second has a volume that is <math>6</math> times greater, so it holds <math>6\times40=\boxed{\textbf{(D)}\ 240}</math> grams. | The first box has volume <math>2\times3\times5=30\text{ cm}^3</math>, and the second has volume <math>(2\times2)\times(3\times3)\times(5)=180\text{ cm}^3</math>. The second has a volume that is <math>6</math> times greater, so it holds <math>6\times40=\boxed{\textbf{(D)}\ 240}</math> grams. | ||
| + | |||
| + | ==Solution 2 (faster)== | ||
| + | Note that we don't need to consider the two individual volumes at all, just the ratio between them. We just need to multiply the original mass of clay by this ratio. | ||
| + | The volume of the second box is <math>2\times3\times1 = 6</math> times greater than the first. | ||
| + | Therefore we get that the second box holds a weight of <math>6\times 40=\boxed{\textbf{(D)}\ 240}</math> grams of clay. | ||
| + | |||
| + | ~rawr3507 | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2012|ab=A|num-b=2|num-a=4}} | {{AMC12 box|year=2012|ab=A|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 07:04, 3 November 2024
Problem
A box 
centimeters high, 
centimeters wide, and 
centimeters long can hold 
grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold 
grams of clay. What is ?
Solution 1
The first box has volume 
, and the second has volume 
. The second has a volume that is 
times greater, so it holds 
grams.
Solution 2 (faster)
Note that we don't need to consider the two individual volumes at all, just the ratio between them. We just need to multiply the original mass of clay by this ratio.
The volume of the second box is 
times greater than the first.
Therefore we get that the second box holds a weight of 
grams of clay.
~rawr3507
See Also
| 2012 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
