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2007 AMC 12A Problems/Problem 2: Difference between revisions

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== Problem ==
== 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?
An aquarium has a [[rectangular prism|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?
 
<math>\mathrm{(A)}\ 0.5\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 1.5\qquad \mathrm{(D)}\ 2\qquad \mathrm{(E)}\ 2.5</math>


== Solution ==
== Solution ==
The water has volume 100*40*40=160000. The brick has volume 8000. The water and the brick combined have a volume of 168000. The water rises
The brick has volume <math>8000 cm^3</math>. The base of the aquarium has area <math>4000 cm^2</math>. For every cm 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>
* <math>\frac{168000}{4000}-\frac{160000}{4000}=42-40=2</math> cm.


== See also ==
{{AMC12 box|year=2007|ab=A|num-b=1|num-a=3}}


== See also ==
[[Category:Introductory Algebra Problems]]
* [[2007 AMC 12A Problems/Problem 1 | Previous problem]]
{{MAA Notice}}
* [[2007 AMC 12A Problems/Problem 3 | Next problem]]
* [[2007 AMC 12A Problems]]

Latest revision as of 12:20, 14 October 2025

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 cm 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 (Problems • Answer Key • Resources)
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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