Difference between revisions of "2017 AMC 8 Problems/Problem 13"
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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> | 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> | ||
| + | ~CHECKMATE2021 | ||
==Video Solution (CREATIVE THINKING!!!)== | ==Video Solution (CREATIVE THINKING!!!)== | ||
Revision as of 16:21, 21 January 2024
Contents
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?
Solution
Given 
games, there must be a total of wins and losses. Hence, 
where 
is Kyler's wins. 
, so our final answer is 
~CHECKMATE2021
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=459
~ pi_is_3.14
Video Solution
~hisolver
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
