Difference between revisions of "1985 AJHSME Problems/Problem 19"
5849206328x (talk | contribs) (New page: ==Problem== If the length and width of a rectangle are each increased by <math>10\% </math>, then the perimeter of the rectangle is increased by <math>\text{(A)}\ 1\% \qquad \text{(B)}\ ...) |
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==Problem== | ==Problem== | ||
| − | If the length and width of a rectangle are each increased by <math>10\% </math>, then the perimeter of the rectangle is increased by | + | If the [[length]] and width of a [[rectangle]] are each increased by <math>10\% </math>, then the [[perimeter]] of the rectangle is increased by |
<math>\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\% </math> | <math>\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\% </math> | ||
| − | ==Solution== | + | ==Solution #1== |
| − | Let the width be <math>w</math> and the length be <math>l</math>. Then, the original perimeter is <math>2(w+ | + | Let the width be <math>w</math> and the length be <math>l</math>. Then, the original perimeter is <math>2(w+l)</math>. |
After the increase, the new width and new length are <math>1.1w</math> and <math>1.1l</math>, so the new perimeter is <math>2(1.1w+1.1l)=2.2(w+l)</math>. | After the increase, the new width and new length are <math>1.1w</math> and <math>1.1l</math>, so the new perimeter is <math>2(1.1w+1.1l)=2.2(w+l)</math>. | ||
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<math>\boxed{\text{B}}</math> | <math>\boxed{\text{B}}</math> | ||
| + | |||
| + | ==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>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/jAy9O53u-nQ | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1985|num-b=18|num-a=20}} |
| + | [[Category:Introductory Geometry Problems]] | ||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | |||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 08:04, 22 January 2023
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 
.
Video Solution
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
