Difference between revisions of "2007 AMC 12A Problems/Problem 2"

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== Solution ==
 
== Solution ==
The water has volume <math>100 \cdot 40 \cdot 40=160000</math>. The brick has volume 8000. The water and the brick combined have a volume of 168000. The water rises <math>\frac{168000}{4000}-\frac{160000}{4000}=42-40=2</math> cm.
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The water has volume <math>100 \cdot 40 \cdot 40=160000</math>. The brick has volume 8000. The water and the brick combined have a volume of 168000. The water rises <math>\frac{168000}{4000}-\frac{160000}{4000}=42-40=2</math> cm <math>\Rightarrow\fbox{D}</math>
  
 
== See also ==
 
== See also ==

Revision as of 21:03, 1 September 2008

Problem

An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. It is filled with water to a height of 40 cm. A brick with a rectangular base that measures 40 cm by 20 cm and a height of 10 cm is placed in the aquarium. By how many centimeters does the water rise?

$\mathrm{(A)}\ 0.5\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 1.5\qquad \mathrm{(D)}\ 2\qquad \mathrm{(E)}\ 2.5$

Solution

The water has volume $100 \cdot 40 \cdot 40=160000$. The brick has volume 8000. The water and the brick combined have a volume of 168000. The water rises $\frac{168000}{4000}-\frac{160000}{4000}=42-40=2$ cm $\Rightarrow\fbox{D}$

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 12 Problems and Solutions
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