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| − | == Problem ==
| + | #REDIRECT[[2003 AMC 12A Problems/Problem 3]] |
| − | 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?
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| − | <math> \mathrm{(A) \ } 4.5\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } 24 </math>
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| − | == Solution ==
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| − | The volume of the original box is <math>15cm\cdot10cm\cdot8cm=1200cm^{3}</math>
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| − | The volume of each cube that is removed is <math>3cm\cdot3cm\cdot3cm=27cm^{3}</math>
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| − | Since there are <math>8</math> corners on the box, <math>8</math> cubes are removed.
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| − | So the total volume removed is <math>8\cdot27cm^{3}=216cm^{3}</math>.
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| − | Therefore, the desired percentage is <math>\frac{216}{1200}\cdot100% = 18 \Longrightarrow D</math>.
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| − | == See also ==
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| − | {{AMC10 box|year=2003|ab=A|num-b=2|num-a=4}}
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| − | [[Category:Introductory Geometry Problems]] | |
