Difference between revisions of "2009 AMC 8 Problems/Problem 8"
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==Problem== | ==Problem== | ||
| − | The length of a rectangle is increased by <math> 10\%</math> percent and the width is decreased by <math> 10%</math> percent. What percent of the old area is the new area? | + | The length of a rectangle is increased by <math> 10\%</math> percent and the width is decreased by <math> 10\%</math> percent. What percent of the old area is the new area? |
Revision as of 23:41, 12 July 2015
Problem
The length of a rectangle is increased by 
percent and the width is decreased by percent. What percent of the old area is the new area?
Solution
In a rectangle with dimensions 
, the new rectangle would have dimensions 
. The ratio of the old area to the new area is 
.
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
