Difference between revisions of "2020 AMC 12B Problems/Problem 5"
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<math>\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63</math> | <math>\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63</math> | ||
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==Solution== | ==Solution== | ||
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~Argonauts16 | ~Argonauts16 | ||
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==Solution 2== | ==Solution 2== | ||
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Using the information from the problem, we can note that team A has lost <math>\frac{1}{3}</math> of their matches. Using the answer choices, we can find the following list of possible win-lose scenarios for <math>A</math>, represented in the form <math>(w, l)</math> for convenience: | Using the information from the problem, we can note that team A has lost <math>\frac{1}{3}</math> of their matches. Using the answer choices, we can find the following list of possible win-lose scenarios for <math>A</math>, represented in the form <math>(w, l)</math> for convenience: | ||
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Thus, we have 5 matching <math>B</math> scenarios, simply adding 7 to <math>w</math> and <math>l</math>. We can then test each of the five <math>B</math> scenarios for <math>\frac{w}{w+l} = \frac{5}{8}</math> and find that <math>(35, 21)</math> fits this description. Then working backwards and subtracting 7 from <math>w</math> and <math>l</math> gives us the point <math>(28, 14)</math>, making the answer <math>\boxed{\textbf{C}}</math>. | Thus, we have 5 matching <math>B</math> scenarios, simply adding 7 to <math>w</math> and <math>l</math>. We can then test each of the five <math>B</math> scenarios for <math>\frac{w}{w+l} = \frac{5}{8}</math> and find that <math>(35, 21)</math> fits this description. Then working backwards and subtracting 7 from <math>w</math> and <math>l</math> gives us the point <math>(28, 14)</math>, making the answer <math>\boxed{\textbf{C}}</math>. | ||
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+ | ==Solution 3 (Answer Choices)== | ||
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+ | Let's say that team <math>A</math> plays <math>n</math> games in total. Therefore, team <math>B</math> must play <math>n + 14</math> games in total (7 wins, 7 losses) Since the ratio of <math>A</math> is <cmath>\frac{2}{3} \implies n \equiv 0 \pmod{3}</cmath> Similarly, since the ratio of <math>B</math> is <cmath>\frac{5}{8} \implies n + 14 \equiv 0 \pmod{8}</cmath> Now, we can go through the answer choices and see which ones work: | ||
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+ | <cmath>\textbf{(A) } 21 \implies 21 + 14 = 35 \not \equiv \pmod{8}</cmath> | ||
+ | <cmath>\textbf{(B) } 27 \implies 27 + 14 = 41 \not \equiv \pmod{8}</cmath> | ||
+ | <cmath>\textbf{(C) } 42 \implies 42 + 14 = 56 \equiv \pmod{8}</cmath> | ||
+ | <cmath>\textbf{(D) } 48 \implies 48 + 14 = 62 \not \equiv \pmod{8}</cmath> | ||
+ | <cmath>\textbf{(E) } 63 \implies 63 + 14 = 77 \not \equiv \pmod{8}</cmath> | ||
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+ | So we can see <math>\boxed{\textbf{(C) } 42.0} \text{ \tiny nice}</math> is the only valid answer. | ||
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+ | ~herobrine-india | ||
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==Video Solution== | ==Video Solution== | ||
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https://youtu.be/WfTty8Fe5Fo | https://youtu.be/WfTty8Fe5Fo | ||
~IceMatrix | ~IceMatrix | ||
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==See Also== | ==See Also== |
Revision as of 02:39, 23 April 2021
Contents
Problem
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 .
~Argonauts16
Solution 2
Using the information from the problem, we can note that team A has lost of their matches. Using the answer choices, we can find the following list of possible win-lose scenarios for , represented in the form for convenience:
Thus, we have 5 matching scenarios, simply adding 7 to and . We can then test each of the five scenarios for and find that fits this description. Then working backwards and subtracting 7 from and gives us the point , making the answer .
Solution 3 (Answer Choices)
Let's say that team plays games in total. Therefore, team must play games in total (7 wins, 7 losses) Since the ratio of is Similarly, since the ratio of is Now, we can go through the answer choices and see which ones work:
So we can see is the only valid answer.
~herobrine-india
Video Solution
~IceMatrix
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
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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