Difference between revisions of "1985 AJHSME Problems/Problem 19"
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==Solution #2 (Quick Fakesolve)== | ==Solution #2 (Quick Fakesolve)== | ||
− | Assume WLOG that the rectangle is a square with length <math>10</math> and width <math>10</math>. Thus, the square has a perimeter of <math>40</math>. Increasing the length and width by <math>10\%</math> will increase the dimensions of the square to 11x11. Thus, the new square's perimeter is <math>44</math>, and because <math>44</math> is <math>110\%</math> of <math>40</math>, our answer is <math>\boxed{\text{B} 10\%}</math>. | + | Assume WLOG that the rectangle is a square with length <math>10</math> and width <math>10</math>. Thus, the square has a perimeter of <math>40</math>. Increasing the length and width by <math>10\%</math> will increase the dimensions of the square to 11x11. Thus, the new square's perimeter is <math>44</math>, and because <math>44</math> is <math>110\%</math> of <math>40</math>, our answer is <math>\boxed{\text{B } 10\%}</math>. |
==See Also== | ==See Also== |
Revision as of 15:07, 17 April 2017
Problem
If the length and width of a rectangle are each increased by , then the perimeter of the rectangle is increased by
Solution #1
Let the width be and the length be . Then, the original perimeter is .
After the increase, the new width and new length are and , so the new perimeter is .
Therefore, the percent change is
Solution #2 (Quick Fakesolve)
Assume WLOG that the rectangle is a square with length and width . Thus, the square has a perimeter of . Increasing the length and width by will increase the dimensions of the square to 11x11. Thus, the new square's perimeter is , and because is of , our answer is .
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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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