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 00:41, 13 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.