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> | ||
− | ==Solution 1 ( | + | ==Solution 1 (One Variable)== |
+ | Suppose team <math>A</math> has played <math>g</math> games in total so that it has won <math>\frac23g</math> games. | ||
+ | It follows that team <math>B</math> has played <math>g+14</math> games in total so that it has won <math>\frac23g+7</math> games. | ||
− | + | We set up and solve an equation for team <math>B</math>'s win ratio: | |
+ | <cmath>\begin{align*} | ||
+ | \frac{\frac23g+7}{g+14}&=\frac58 \\ | ||
+ | \frac{16}{3}g+56&=5g+70 \\ | ||
+ | \frac13g&=14 \\ | ||
+ | g&=\boxed{\textbf{(C) } 42}. | ||
+ | \end{align*}</cmath> | ||
+ | ~MRENTHUSIASM | ||
− | <cmath> | + | ==Solution 2 (Two Variables)== |
− | < | + | If we consider the number of games team <math>B</math> has played as <math>x</math> and the number of games that team <math>A</math> has played as <math>y</math>, then we can set up the following system of equations: |
+ | <cmath>\begin{align*} | ||
+ | \frac{5}{8}x &= \frac{2}{3}y+7, \\ | ||
+ | \frac{3}{8}x &= \frac{1}{3}y+7. | ||
+ | \end{align*}</cmath> | ||
+ | The first system equated the number of wins of each team, while the second system equates the number of losses by each team. By multiplying the second equation by <math>2</math> and solving the system, we get <math>y = 42</math> or answer choice <math>\boxed{\textbf{(C) } 42}.</math> | ||
− | + | ~AnkitAMC | |
− | + | ==Solution 3 (Two Variables)== | |
− | |||
+ | First, let us assign some variables. Let | ||
+ | <cmath>A_w=2x, \ A_l=x, \ A_g=3x,</cmath> | ||
+ | <cmath>B_w=5y, \ B_l=3y, \ B_g=8y,</cmath> | ||
+ | where <math>X_w</math> denotes number of games won, <math>X_l</math> denotes number of games lost, and <math>X_g</math> denotes total games played for <math>X\in \{A, B\}</math>. Using the given information, we can set up the following two equations: | ||
+ | <cmath>\begin{align*} | ||
+ | B_w=A_w+7&\implies 5y=2x+7, \\ | ||
+ | B_l=A_l+7&\implies 3y=x+7. | ||
+ | \end{align*}</cmath> | ||
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>. | ||
~Argonauts16 | ~Argonauts16 | ||
− | ==Solution | + | ==Solution 4 (Answer Choices: Substitutions)== |
− | + | ||
− | + | Using the information from the problem, we can note that team <math>A</math> has lost <math>\frac{1}{3}</math> of their matches. Using the answer choices, we can construct the following list of possible win-lose scenarios for <math>A,</math> represented in the form <math>(w, l)</math> for convenience: | |
+ | <cmath>\begin{align*} | ||
+ | \textbf{(A)} &\implies (14, 7) \\ | ||
+ | \textbf{(B)} &\implies (18, 9) \\ | ||
+ | \textbf{(C)} &\implies (28, 14) \\ | ||
+ | \textbf{(D)} &\implies (32, 16) \\ | ||
+ | \textbf{(E)} &\implies (42, 21) | ||
+ | \end{align*}</cmath> | ||
+ | Thus, we have <math>5</math> matching <math>B</math> scenarios, simply adding <math>7</math> 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 <math>7</math> from <math>w</math> and <math>l</math> gives us the point <math>(28, 14),</math> making the answer <math>\boxed{\textbf{(C) } 42}.</math> | ||
+ | |||
+ | ==Solution 5 (Answer Choices: Observations)== | ||
− | + | 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: | |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \ | + | \textbf{(A) } 21 &\implies 21 + 14 = 35 \not \equiv 0\pmod{8} \\ |
− | \ | + | \textbf{(B) } 27 &\implies 27 + 14 = 41 \not \equiv 0\pmod{8} \\ |
− | \ | + | \textbf{(C) } 42 &\implies 42 + 14 = 56 \equiv 0\pmod{8} \\ |
− | + | \textbf{(D) } 48 &\implies 48 + 14 = 62 \not \equiv 0\pmod{8} \\ | |
+ | \textbf{(E) } 63 &\implies 63 + 14 = 77 \not \equiv 0\pmod{8} \\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | So we can see <math>\boxed{\textbf{(C) } 42}</math> is the only valid answer. | |
− | + | ~herobrine-india | |
− | + | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | |
− | + | https://youtu.be/KXySDYcUnHk | |
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− | + | ~Education, the Study of Everything | |
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− | + | == Video Solution == | |
+ | https://youtu.be/00Ngozqw2d0?t=13 | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/WfTty8Fe5Fo | https://youtu.be/WfTty8Fe5Fo | ||
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==See Also== | ==See Also== |
Latest revision as of 14:01, 8 June 2023
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 1 (One Variable)
Suppose team has played games in total so that it has won games. It follows that team has played games in total so that it has won games.
We set up and solve an equation for team 's win ratio: ~MRENTHUSIASM
Solution 2 (Two Variables)
If we consider the number of games team has played as and the number of games that team has played as , then we can set up the following system of equations: The first system equated the number of wins of each team, while the second system equates the number of losses by each team. By multiplying the second equation by and solving the system, we get or answer choice
~AnkitAMC
Solution 3 (Two Variables)
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 4 (Answer Choices: Substitutions)
Using the information from the problem, we can note that team has lost of their matches. Using the answer choices, we can construct the following list of possible win-lose scenarios for represented in the form for convenience: Thus, we have matching scenarios, simply adding to and We can then test each of the five scenarios for and find that fits this description. Then working backwards and subtracting from and gives us the point making the answer
Solution 5 (Answer Choices: Observations)
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 (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/00Ngozqw2d0?t=13
Video Solution
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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