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 <math>\Rightarrow\fbox{D}</math>
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The brick has volume <math>8000 cm^3</math>. The base of the aquarium has area <math>4000 cm^2</math>. For every inch the water rises, the volume increases by <math>4000 cm^3</math>; therefore, when the volume increases by <math>8000 cm^3</math>, the water level rises <math>2 cm \Rightarrow\fbox{D}</math>
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:30, 3 July 2013

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 brick has volume $8000 cm^3$. The base of the aquarium has area $4000 cm^2$. For every inch the water rises, the volume increases by $4000 cm^3$; therefore, when the volume increases by $8000 cm^3$, the water level rises $2 cm \Rightarrow\fbox{D}$

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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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