|
|
| Line 1: |
Line 1: |
| − | == Problem ==
| + | #REDIRECT[[2003 AMC 12A Problems/Problem 3]] |
| − | A solid [[rectangular prism|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>
| |
| − | | |
| − | == Solution ==
| |
| − | The volume of the original box is <math>15cm\cdot10cm\cdot8cm=1200cm^{3}</math>
| |
| − | | |
| − | The volume of each cube that is removed is <math>3cm\cdot3cm\cdot3cm=27cm^{3}</math>
| |
| − | | |
| − | Since there are <math>8</math> corners on the box, <math>8</math> cubes are removed.
| |
| − | | |
| − | So the total volume removed is <math>8\cdot27cm^{3}=216cm^{3}</math>.
| |
| − | | |
| − | Therefore, the desired percentage is <math>\frac{216}{1200}\cdot100 = 18\% \rightarrow D</math>.
| |
| − | | |
| − | == See also ==
| |
| − | {{AMC10 box|year=2003|ab=A|num-b=2|num-a=4}}
| |
| − | | |
| − | [[Category:Introductory Geometry Problems]]
| |
