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Difference between revisions of "2006 AMC 8 Problems/Problem 21"

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==Problem==
 
==Problem==
An aquarium has a rectangular base that measures <math>100</math> cm by <math>40</math> cm and has a height of <math>50</math> cm. The aquarium is tilled with water to a  depth of <math>37</math> cm. A rock with volume <math>1000\text{cm}^3</math> is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?  
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An aquarium has a rectangular base that measures <math>100</math> cm by <math>40</math> cm and has a height of <math>50</math> cm. The aquarium is filled with water to a  depth of <math>37</math> cm. A rock with volume <math>1000\text{cm}^3</math> is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?  
  
 
<math> \textbf{(A)}\ 0.25\qquad\textbf{(B)}\ 0.5\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 2.5 </math>
 
<math> \textbf{(A)}\ 0.25\qquad\textbf{(B)}\ 0.5\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 2.5 </math>

Revision as of 23:48, 12 November 2012

Problem

An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. The aquarium is filled with water to a depth of $37$ cm. A rock with volume $1000\text{cm}^3$ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?

$\textbf{(A)}\ 0.25\qquad\textbf{(B)}\ 0.5\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 2.5$

Solution

The water level will rise $1$cm for every $100 \cdot 40 = 4000\text{cm}^2$. Since $1000$ is $\frac{1}{4}$ of $4000$, the water will rise $\frac{1}{4}\cdot1 = \textbf{(A)}\ 0.25$

2006 AMC 8 (ProblemsAnswer KeyResources)
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Problem 20
Followed by
Problem 22
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All AJHSME/AMC 8 Problems and Solutions
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