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

(Created page with "==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 t...")
 
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==Solution==
 
==Solution==
 
The water level will rise <math>1</math>cm for every <math>100 \cdot 40 = 4000\text{cm}^2</math>. Since <math>1000</math> is <math>\frac{1}{4}</math> of <math>4000</math>, the water will rise <math>\frac{1}{4}\cdot1 = \textbf{(A)}\ 0.25</math>
 
The water level will rise <math>1</math>cm for every <math>100 \cdot 40 = 4000\text{cm}^2</math>. Since <math>1000</math> is <math>\frac{1}{4}</math> of <math>4000</math>, the water will rise <math>\frac{1}{4}\cdot1 = \textbf{(A)}\ 0.25</math>
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{{AMC8 box|year=2006|n=II|num-b=20|num-a=22}}

Revision as of 08:08, 18 December 2011

Problem

An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. The aquarium is tilled 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)
Preceded by
Problem 20
Followed by
Problem 22
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All AJHSME/AMC 8 Problems and Solutions
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