Difference between revisions of "2009 AMC 8 Problems/Problem 8"
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==See Also== | ==See Also== | ||
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Latest revision as of 20:39, 26 December 2023
Contents
[hide]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 new area to the old area is .
Solution 2
If you take the length as and the width as then A(OLD)= A(NEW)= =
Video Solution
~savannahsolver
Easier to understand Video Solution
https://www.youtube.com/watch?v=cZUT7lYu474&t=4s
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.