Difference between revisions of "2020 AMC 8 Problems/Problem 25"
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For each square <math>S_{k}</math>, let the sidelength of this square be denoted by <math>s_{k}</math>. | For each square <math>S_{k}</math>, let the sidelength of this square be denoted by <math>s_{k}</math>. | ||
− | As the diagram shows, <math>s_{1}+s_{2}+s_{3}=3322, s_{1}-s_{2}+s_{3}=2020.</math> | + | As the diagram shows, <math>s_{1}+s_{2}+s_{3}=3322, s_{1}-s_{2}+s_{3}=2020.</math> We subtract the second equation from the first, getting <math>2s_{2}=1302</math>, and thus <math>s_{2}=651</math>, so the answer is <math>\boxed{\textbf{(A)}\text{ }651}</math> ~icematrix, edits by starrynight7210 |
==Solution 2== | ==Solution 2== |
Revision as of 21:00, 18 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?
Contents
Solution 1
For each square , let the sidelength of this square be denoted by .
As the diagram shows, We subtract the second equation from the first, getting , and thus , so the answer is ~icematrix, edits by starrynight7210
Solution 2
WLOG, assume that and . Let the sum of the lengths of and be and let the length of be . We have the system
which we solve to find that .
-franzliszt
Solution 3
Since each pair of boxes has a sum of or and a difference of , we see that the answer is
-A_MatheMagician
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 |
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