2006 AMC 8 Problems/Problem 21: Difference between revisions

Revision as of 17:09, 8 November 2024

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
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 AJHSME/AMC 8 Problems and Solutions

@@ Line 10: / Line 10: @@
 https://www.youtube.com/watch?v=DNMuW5prOwg   ~David
+==Video Solution by WhyMath==
+https://youtu.be/lqgpczqRklA
 ==See Also==
 {{AMC8 box|year=2006|n=II|num-b=20|num-a=22}}
 {{MAA Notice}}

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