Difference between revisions of "2006 AMC 8 Problems/Problem 21"
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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 
cm by 
cm and has a height of 
cm. The aquarium is tilled with water to a depth of 
cm. A rock with volume 
is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
Solution
The water level will rise 
cm for every 
. Since 
is 
of 
, the water will rise 
| 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 | ||
