Difference between revisions of "1993 AJHSME Problems/Problem 21"
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==Solution== | ==Solution== | ||
If a rectangle had dimensions <math>10 \times 10</math> and area <math>100</math>, then its new dimensions would be <math>12 \times 15</math> and area <math>180</math>. The area is increased by <math>180-100=80</math> or <math>80/100 = \boxed{\text{(D)}\ 80\%}</math>. | If a rectangle had dimensions <math>10 \times 10</math> and area <math>100</math>, then its new dimensions would be <math>12 \times 15</math> and area <math>180</math>. The area is increased by <math>180-100=80</math> or <math>80/100 = \boxed{\text{(D)}\ 80\%}</math>. | ||
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+ | ==Alternate Solution== | ||
+ | Let the dimensions of the rectangle be <math>x \times y</math>. This rectangle has area <math>xy</math>. The new dimensions would be <math>1.2x \times 1.5y</math>, so the area is <math>=1.8xy</math>, which is <math>\boxed{80\%}</math> more than the original area. | ||
==See Also== | ==See Also== | ||
{{AJHSME box|year=1993|num-b=20|num-a=22}} | {{AJHSME box|year=1993|num-b=20|num-a=22}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:32, 10 May 2015
Problem
If the length of a rectangle is increased by and its width is increased by , then the area is increased by
Solution
If a rectangle had dimensions and area , then its new dimensions would be and area . The area is increased by or .
Alternate Solution
Let the dimensions of the rectangle be . This rectangle has area . The new dimensions would be , so the area is , which is more than the original area.
See Also
1993 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
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