Difference between revisions of "2009 AMC 8 Problems/Problem 8"
(→Solution) |
(→Solution) |
||
| Line 11: | Line 11: | ||
==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 13:55, 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. 
