Difference between revisions of "2020 AMC 12B Problems/Problem 5"
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We can solve through substitution, as the second equation can be written as <math>x=3y-7</math>, and plugging this into the first equation gives <math>5y=6y-7\implies y=7</math>, which means <math>x=3(7)-7=14</math>. Finally, we want the total number of games team <math>A</math> has played, which is <math>A_g=3(14)=\boxed{\textbf{(C) } 42}</math>. | We can solve through substitution, as the second equation can be written as <math>x=3y-7</math>, and plugging this into the first equation gives <math>5y=6y-7\implies y=7</math>, which means <math>x=3(7)-7=14</math>. Finally, we want the total number of games team <math>A</math> has played, which is <math>A_g=3(14)=\boxed{\textbf{(C) } 42}</math>. | ||
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+ | ==See Also== | ||
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+ | {{AMC12 box|year=2020|ab=B|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Revision as of 20:13, 7 February 2020
Problem 5
Teams and are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team has won of its games and team has won of its games. Also, team has won more games and lost more games than team How many games has team played?
Solution
First, let us assign some variables. Let
where denotes number of games won, denotes number of games lost, and denotes total games played for . Using the given information, we can set up the following two equations:
We can solve through substitution, as the second equation can be written as , and plugging this into the first equation gives , which means . Finally, we want the total number of games team has played, which is .
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 AMC 12 Problems and Solutions |
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