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 18:42, 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 |