Difference between revisions of "2020 AMC 8 Problems/Problem 25"
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==Solution 3 (fast)== | ==Solution 3 (fast)== | ||
Since each pair of boxes has a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, we see that the answer is <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}</math>. | Since each pair of boxes has a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, we see that the answer is <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}</math>. | ||
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− | ==Video | + | ==Video Solutions== |
− | https:// | + | https://youtu.be/KN441ecLfKM |
+ | https://youtu.be/LebVAuPkpcg | ||
− | + | ==See also== | |
+ | {{AMC8 box|year=2020|num-b=24|after=Last Problem}} | ||
− | + | [[Category:Introductory Geometry Problems]] | |
+ | {{MAA Notice}} |
Revision as of 15:54, 24 November 2020
Rectangles and and squares and shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of in units?
Solution 1
Let the side length of each square be . Then, from the diagram, we can line up the top horizontal lengths of , , and to cover the top side of the large rectangle, so . Similarly, the short side of will be , and lining this up with the left side of to cover the vertical side of the large rectangle gives . We subtract the second equation from the first to obtain , and thus .
Solution 2
Assuming that the problem is well-posed, it should be true in the particular case where and . Let the sum of the side lengths of and be , and let the length of rectangle be . We then have the system which we solve to find that .
Solution 3 (fast)
Since each pair of boxes has a sum of or and a difference of , we see that the answer is .
Video Solutions
https://youtu.be/KN441ecLfKM https://youtu.be/LebVAuPkpcg
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.