2007 AMC 10A Problems/Problem 3: Difference between revisions
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== Solution == | == Solution == | ||
The volume of the brick is <math>40 \times 20 \times 10 = 8000</math>. Thus the water volume rose <math>8000 = 100 \times 40 \times h \Longrightarrow h = 2\ \mathrm{(D)}</math>. | Disregard the information about the water filling the box. The volume of the brick is <math>40 \times 20 \times 10 = 8000</math>. Thus the water volume rose <math>8000 = 100 \times 40 \times h \Longrightarrow h = 2\ \mathrm{(D)}</math>. | ||
== See also == | == See also == | ||
Revision as of 02:48, 19 June 2017
Problem
An aquarium has a rectangular base that measures 
cm by 
cm and has a height of 
cm. It is filled with water to a height of cm. A brick with a rectangular base that measures cm by 
cm and a height of 
cm is placed in the aquarium. By how many centimeters does the water rise?
Solution
Disregard the information about the water filling the box. The volume of the brick is 
. Thus the water volume rose 
.
See also
| 2007 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 AMC 10 Problems and Solutions | ||
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