Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 8 Problems/Problem 25"

(→‎Solution 1)
(→‎Solution 3)
Line 29: Line 29:
  
 
==Solution 3==
 
==Solution 3==
−
Since each pair of boxes has a sum of <math>3322</math> or <math>2000</math> and a difference of <math>S_2</math>, we see that the answer is <math>\dfrac{3322 - 2000}{2} = \dfrac{1322}{2} = \boxed{(\text{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 - 2000}{2} = \dfrac{1322}{2} = \boxed{(\text{A}) 651}.</math>
  
 
-A_MatheMagician
 
-A_MatheMagician

Revision as of 19:39, 19 November 2020

Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?

[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]

$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$

Solution 1

For each square $S_{k}$, let the sidelength of this square be denoted by $s_{k}$.

As the diagram shows, $s_{1}+s_{2}+s_{3}=3322, s_{1}-s_{2}+s_{3}=2020.$ We subtract the second equation from the first, getting $2s_{2}=1302$, and thus $s_{2}=651$, so the answer is $\boxed{\textbf{(A)}\text{ }651}$ ~icematrix, edits by starrynight7210

Solution 2

WLOG, assume that $S_1=S_3$ and $R_1=R_2$. Let the sum of the lengths of $S_1$ and $S_2$ be $x$ and let the length of $S_2$ be $y$. We have the system \[x+y =3322\] \[x-y=2020\]

which we solve to find that $y=\textbf{(A) }651$.

-franzliszt

Solution 3

Since each pair of boxes has a sum of $3322$ or $2020$ and a difference of $S_2$, we see that the answer is $\dfrac{3322 - 2000}{2} = \dfrac{1322}{2} = \boxed{(\text{A}) 651}.$

-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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_25&oldid=137895"