Difference between revisions of "2007 AMC 12A Problems/Problem 2"
Baijiangchen (talk | contribs) (→Solution) |
(→See also) |
||
(One intermediate revision by one other user not shown) | |||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
− | The brick has volume <math>8000 cm^3</math>. The base of the aquarium has area <math>4000 cm^2</math>. For every inch the water rises, the volume increases by <math>4000 cm^3</math>; therefore, when the volume increases by | + | The brick has volume <math>8000 cm^3</math>. The base of the aquarium has area <math>4000 cm^2</math>. For every inch the water rises, the volume increases by <math>4000 cm^3</math>; therefore, when the volume increases by <math>8000 cm^3</math>, the water level rises <math>2 cm \Rightarrow\fbox{D}</math> |
== See also == | == See also == | ||
Line 11: | Line 11: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:30, 3 July 2013
Problem
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. It is filled with water to a height of 40 cm. A brick with a rectangular base that measures 40 cm by 20 cm and a height of 10 cm is placed in the aquarium. By how many centimeters does the water rise?
Solution
The brick has volume . The base of the aquarium has area . For every inch the water rises, the volume increases by ; therefore, when the volume increases by , the water level rises
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.