Difference between revisions of "2017 AMC 8 Problems/Problem 13"

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==Problem 13==
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==Problem==
  
 
Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?
 
Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?
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==Solution==
 
==Solution==
  
Given <math>n</math> games, there must be a total of <math>n</math> wins and <math>n</math> losses. Hence, <math>4 + 3 + K = 2 + 3 + 3</math> where <math>K</math> is Kyler's wins. <math>K = 1.</math>
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Given <math>n</math> games, there must be a total of <math>n</math> wins and <math>n</math> losses. Hence, <math>4 + 3 + K = 2 + 3 + 3</math> where <math>K</math> is Kyler's wins. <math>K = 1</math>, so our final answer is <math>\boxed{\textbf{(B)}\ 1}.</math>
  
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==Video Solution by OmegaLearn==
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https://youtu.be/6xNkyDgIhEE?t=459
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~ pi_is_3.14
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==Video Solution==
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https://youtu.be/AslL9RYPPy4
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 21:55, 20 November 2022

Problem

Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution

Given $n$ games, there must be a total of $n$ wins and $n$ losses. Hence, $4 + 3 + K = 2 + 3 + 3$ where $K$ is Kyler's wins. $K = 1$, so our final answer is $\boxed{\textbf{(B)}\ 1}.$

Video Solution by OmegaLearn

https://youtu.be/6xNkyDgIhEE?t=459

~ pi_is_3.14

Video Solution

https://youtu.be/AslL9RYPPy4

~savannahsolver

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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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