Difference between revisions of "2011 AMC 8 Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | The area of a rectangle is given by the formula length times width. Karl's garden is <math>20 \times 45 = 900</math> square feet and Makenna's garden is <math>25 \times 40 = 1000</math> square feet. Since <math>1000 > 900,</math> Makenna's garden is larger by <math>1000-900=100</math> square feet. <math>\Rightarrow \boxed{\textbf{(E)}}</math> | + | The area of a rectangle is given by the formula length times width. Karl's garden is <math>20 \times 45 = 900</math> square feet and Makenna's garden is <math>25 \times 40 = 1000</math> square feet. Since <math>1000 > 900,</math> Makenna's garden is larger by <math>1000-900=100</math> square feet. <math>\Rightarrow \boxed{ \textbf{(E)}\ \text{Makenna's garden is larger by 100 square feet.} }</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=1|num-a=3}} | {{AMC8 box|year=2011|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:35, 13 November 2016
Problem
Karl's rectangular vegetable garden is 
feet by 
feet, and Makenna's is 
feet by 
feet. Whose garden is larger in area?
Solution
The area of a rectangle is given by the formula length times width. Karl's garden is 
square feet and Makenna's garden is 
square feet. Since 
Makenna's garden is larger by 
square feet. 
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
