2006 AMC 8 Problems/Problem 21

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 $\frac{1}{4}\cdot1 = \boxed{\textbf{(A)}\ 0.25}$

Video Solution

https://www.youtube.com/watch?v=DNMuW5prOwg ~David

Video Solution by WhyMath

https://youtu.be/lqgpczqRklA

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2006 AMC 8 Problems/Problem 21

Contents

Problem

Solution

Video Solution

Video Solution by WhyMath

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