Difference between revisions of "2009 AMC 8 Problems/Problem 8"
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==Solution== | ==Solution== | ||
− | In a rectangle with dimensions <math>10 \times 10</math>, the new rectangle would have dimensions <math>11 \times 9</math>. The ratio of the new area to the old area is <math>99/100 = \boxed{\textbf{(B)}\ 99}</math> | + | In a rectangle with dimensions <math>10 \times 10</math>, the new rectangle would have dimensions <math>11 \times 9</math>. The ratio of the new area to the old area is <math>99/100 = \boxed{\textbf{(B)}\ 99}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=7|num-a=9}} | {{AMC8 box|year=2009|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:55, 12 December 2020
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
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.