Difference between revisions of "2003 AMC 12A Problems/Problem 3"

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{{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #3]] and [[2003 AMC 10A Problems|2003 AMC 10A #3]]}}
 
== Problem ==
 
== Problem ==
 
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?  
 
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?  
  
<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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<math> \mathrm{(A) \ } 4.5\%\qquad \mathrm{(B) \ } 9\%\qquad \mathrm{(C) \ } 12\%\qquad \mathrm{(D) \ } 18\%\qquad \mathrm{(E) \ } 24\% </math>
  
 
== Solution ==
 
== Solution ==
The volume of the original box is <math>15cm\cdot10cm\cdot8cm=1200cm^{3}</math>  
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The volume of the original box is <math>15\cdot10\cdot8=1200</math>  
  
The volume of each cube that is removed is <math>3cm\cdot3cm\cdot3cm=27cm^{3}</math>
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The volume of each cube that is removed is <math>3\cdot3\cdot3=27</math>
  
 
Since there are <math>8</math> corners on the box, <math>8</math> cubes are removed.  
 
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>.  
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So the total volume removed is <math>8\cdot27=216</math>.  
  
Therefore, the desired percentage is <math>\frac{216}{1200}\cdot100 = 18 \Rightarrow D</math>
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Therefore, the desired percentage is <math>\frac{216}{1200}\cdot100 = \boxed{\mathrm{(D)}\ 18\%}</math>
  
 
== See Also ==
 
== See Also ==
*[[2003 AMC 12A Problems]]
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{{AMC10 box|year=2003|ab=A|num-b=2|num-a=4}}
 
{{AMC12 box|year=2003|ab=A|num-b=2|num-a=4}}
 
{{AMC12 box|year=2003|ab=A|num-b=2|num-a=4}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 16:08, 29 July 2011

The following problem is from both the 2003 AMC 12A #3 and 2003 AMC 10A #3, so both problems redirect to this page.

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 $15\cdot10\cdot8=1200$

The volume of each cube that is removed is $3\cdot3\cdot3=27$

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

So the total volume removed is $8\cdot27=216$.

Therefore, the desired percentage is $\frac{216}{1200}\cdot100 = \boxed{\mathrm{(D)}\ 18\%}$

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
2003 AMC 12A (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 12 Problems and Solutions
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