Difference between revisions of "1985 AJHSME Problem 19"
(→Solution) |
(→Solution) |
||
| Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
Let the length and width of the original rectangle be <math>L</math> and <math>W,</math> respectively. The perimeter of the original rectangle is <math>2L+2W.</math> If we apply the changes as described in the problem, the perimeter of the new rectangle is <math>2 \cdot (\frac{11L}{10}) + 2 \cdot (\frac{11W}{10}) = \frac{11}{10} (2L+2W).</math> This is an increase of 10%, so the answer is <math>\text{(B)}.</math> | Let the length and width of the original rectangle be <math>L</math> and <math>W,</math> respectively. The perimeter of the original rectangle is <math>2L+2W.</math> If we apply the changes as described in the problem, the perimeter of the new rectangle is <math>2 \cdot (\frac{11L}{10}) + 2 \cdot (\frac{11W}{10}) = \frac{11}{10} (2L+2W).</math> This is an increase of 10%, so the answer is <math>\text{(B)}.</math> | ||
| + | |||
| + | |||
| + | Solution 2 (by mathmax12) | ||
| + | proceed by coordbash | ||
Latest revision as of 18:33, 4 July 2023
Problem
If the length and width of a rectangle are each increased by 
, then the perimeter of the rectangle is increased by
Solution
Let the length and width of the original rectangle be 
and 
respectively. The perimeter of the original rectangle is 
If we apply the changes as described in the problem, the perimeter of the new rectangle is 
This is an increase of 10%, so the answer is 
Solution 2 (by mathmax12)
proceed by coordbash
