Difference between revisions of "2003 AMC 12A Problems/Problem 3"
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+ | {{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #3]] and [[2003 AMC 10A Problems|2003 AMC 10A #3]]}} | ||
== Problem == | == Problem == | ||
A solid box is <math>15</math> cm by <math>10</math> cm by <math>8</math> cm. A new solid is formed by removing a cube <math>3</math> cm on a side from each corner of this box. What percent of the original volume is removed? | A solid box is <math>15</math> cm by <math>10</math> cm by <math>8</math> cm. A new solid is formed by removing a cube <math>3</math> cm on a side from each corner of this box. What percent of the original volume is removed? | ||
− | <math> \mathrm{(A) \ } 4.5\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } 24 </math> | + | <math> \mathrm{(A) \ } 4.5\%\qquad \mathrm{(B) \ } 9\%\qquad \mathrm{(C) \ } 12\%\qquad \mathrm{(D) \ } 18\%\qquad \mathrm{(E) \ } 24\% </math> |
== Solution == | == Solution == | ||
− | The volume of the original box is <math> | + | The volume of the original box is <math>15\cdot10\cdot8=1200</math> |
− | The volume of each cube that is removed is <math> | + | The volume of each cube that is removed is <math>3\cdot3\cdot3=27</math> |
Since there are <math>8</math> corners on the box, <math>8</math> cubes are removed. | Since there are <math>8</math> corners on the box, <math>8</math> cubes are removed. | ||
− | So the total volume removed is <math>8\ | + | So the total volume removed is <math>8\cdot27=216</math>. |
− | Therefore, the desired percentage is <math>\frac{216}{1200}\cdot100 = 18 \ | + | Therefore, the desired percentage is <math>\frac{216}{1200}\cdot100 = \boxed{\mathrm{(D)}\ 18\%}</math> |
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2003|ab=A|num-b=2|num-a=4}} | |
{{AMC12 box|year=2003|ab=A|num-b=2|num-a=4}} | {{AMC12 box|year=2003|ab=A|num-b=2|num-a=4}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Revision as of 16:08, 29 July 2011
- The following problem is from both the 2003 AMC 12A #3 and 2003 AMC 10A #3, so both problems redirect to this page.
Problem
A solid box is cm by cm by cm. A new solid is formed by removing a cube cm on a side from each corner of this box. What percent of the original volume is removed?
Solution
The volume of the original box is
The volume of each cube that is removed is
Since there are corners on the box, cubes are removed.
So the total volume removed is .
Therefore, the desired percentage is
See Also
2003 AMC 10A (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 10 Problems and Solutions |
2003 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 |