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> and the width is decreased by <math> 10%</math>. 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? |
| 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 | + | 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>. |
| + | |||
| + | ===Solution 2=== | ||
| + | If you take the length as <math>x</math> and the width as <math>y</math> then | ||
| + | A(OLD)= <math>xy</math> | ||
| + | A(NEW)= <math>1.1x\times.9y</math> = <math>.99xy</math> | ||
| + | <math>0.99/1=99\%</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/4io4bjzgQKI | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Easier to understand Video Solution== | ||
| + | https://www.youtube.com/watch?v=cZUT7lYu474&t=4s | ||
==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}} | ||
Latest revision as of 20:39, 26 December 2023
Contents
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. 
