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% = 18 \Longrightarrow D</math>.
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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 $15$ cm by $10$ cm by $8$ cm. A new solid is formed by removing a cube $3$ cm on a side from each corner of this box. What percent of the original volume is removed?

$\mathrm{(A) \ } 4.5\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } 24$

Solution

The volume of the original box is $15cm\cdot10cm\cdot8cm=1200cm^{3}$

The volume of each cube that is removed is $3cm\cdot3cm\cdot3cm=27cm^{3}$

Since there are $8$ corners on the box, $8$ cubes are removed.

So the total volume removed is $8\cdot27cm^{3}=216cm^{3}$.

Therefore, the desired percentage is $\frac{216}{1200}\cdot100 = 18\% \rightarrow D$.

See also

2003 AMC 10A (ProblemsAnswer KeyResources)
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
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