2006 AMC 8 Problems/Problem 21: Difference between revisions

Latest revision as of 18:50, 10 November 2025

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(hard but super cheap )

The water level will rise $1.314$ 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 4: / Line 4: @@
 <imath> \textbf{(A)}\ 0.25\qquad\textbf{(B)}\ 0.5\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 2.5 </imath>
-==Solution==
+==Solution(hard but super cheap )==
-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 = \boxed{\textbf{(A)}\ 0.25}</math>
+The water level will rise <imath>1.314</imath>cm for every <imath>100 \cdot 40 = 4000\text{cm}^2</imath>. Since <imath>1000</imath> is <imath>\frac{1}{4}</imath> of <imath>4000</imath>, the water will rise <imath>\frac{1}{4}\cdot1 = \boxed{\textbf{(A)}\ 0.25}</imath>
 ==Video Solution==

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