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]]}}
 
{{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?  
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>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? <!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
 
<math> \mathrm{(A) \ } 4.5\%\qquad \mathrm{(B) \ } 9\%\qquad \mathrm{(C) \ } 12\%\qquad \mathrm{(D) \ } 18\%\qquad \mathrm{(E) \ } 24\% </math>
 
<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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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 16:14, 18 November 2015

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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