Difference between revisions of "2003 AMC 10A Problems/Problem 3"
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So the total volume removed is <math>8\cdot27cm^{3}=216cm^{3}</math>. | So the total volume removed is <math>8\cdot27cm^{3}=216cm^{3}</math>. | ||
| − | Therefore, the desired percentage is <math>\frac{216}{1200}\cdot100 | + | Therefore, the desired percentage is <math>\frac{216}{1200}\cdot100 = 18\% \rightarrow D</math>. |
== See also == | == See also == | ||
Revision as of 17:42, 24 January 2009
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 | ||
