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Difference between revisions of "2020 AMC 12B Problems/Problem 5"

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==Solution 2 (Two Variables)==
 
==Solution 2 (Two Variables)==
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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*}
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\frac{5}{8}x &= \frac{2}{3}y+7, \\
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\frac{3}{8}x &= \frac{1}{3}y+7.
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\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>
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~AnkitAMC
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 +
==Solution 3 (Two Variables)==
  
 
First, let us assign some variables. Let
 
First, let us assign some variables. Let
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~Argonauts16
 
~Argonauts16
  
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==Solution 3 (Answer Choices: Substitutions)==
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==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:
 
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:
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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>
 
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>
  
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==Solution 4 (Answer Choices: Observations)==
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==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:
 
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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~herobrine-india
 
~herobrine-india
  
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==Solution 5 (Two Variables)==
+
==Video Solution (HOW TO THINK CREATIVELY!!!)==
−
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*}
+
https://youtu.be/KXySDYcUnHk
−
\frac{5}{8}x = \frac{2}{3}y+7\\
+
 
−
\frac{3}{8}x = \frac{1}{3}y+7\\
+
~Education, the Study of Everything
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\\
+
 
−
\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)}}</math>
+
 
 +
 
  
−
~AnkitAMC
+
== Video Solution ==
 +
https://youtu.be/00Ngozqw2d0?t=13
  
 
==Video Solution==
 
==Video Solution==
  
 
https://youtu.be/WfTty8Fe5Fo
 
https://youtu.be/WfTty8Fe5Fo
−
 
−
~IceMatrix
 
  
 
==See Also==
 
==See Also==

Latest revision as of 14:01, 8 June 2023

Problem

Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?

$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$

Solution 1 (One Variable)

Suppose team $A$ has played $g$ games in total so that it has won $\frac23g$ games. It follows that team $B$ has played $g+14$ games in total so that it has won $\frac23g+7$ games.

We set up and solve an equation for team $B$'s win ratio: \begin{align*} \frac{\frac23g+7}{g+14}&=\frac58 \\ \frac{16}{3}g+56&=5g+70 \\ \frac13g&=14 \\ g&=\boxed{\textbf{(C) } 42}. \end{align*} ~MRENTHUSIASM

Solution 2 (Two Variables)

If we consider the number of games team $B$ has played as $x$ and the number of games that team $A$ has played as $y$, then we can set up the following system of equations: \begin{align*} \frac{5}{8}x &= \frac{2}{3}y+7, \\ \frac{3}{8}x &= \frac{1}{3}y+7. \end{align*} 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 $2$ and solving the system, we get $y = 42$ or answer choice $\boxed{\textbf{(C) } 42}.$

~AnkitAMC

Solution 3 (Two Variables)

First, let us assign some variables. Let \[A_w=2x, \ A_l=x, \ A_g=3x,\] \[B_w=5y, \ B_l=3y, \ B_g=8y,\] where $X_w$ denotes number of games won, $X_l$ denotes number of games lost, and $X_g$ denotes total games played for $X\in \{A, B\}$. Using the given information, we can set up the following two equations: \begin{align*} B_w=A_w+7&\implies 5y=2x+7, \\ B_l=A_l+7&\implies 3y=x+7. \end{align*} We can solve through substitution, as the second equation can be written as $x=3y-7$, and plugging this into the first equation gives $5y=6y-7\implies y=7$, which means $x=3(7)-7=14$. Finally, we want the total number of games team $A$ has played, which is $A_g=3(14)=\boxed{\textbf{(C) } 42}$.

~Argonauts16

Solution 4 (Answer Choices: Substitutions)

Using the information from the problem, we can note that team $A$ has lost $\frac{1}{3}$ of their matches. Using the answer choices, we can construct the following list of possible win-lose scenarios for $A,$ represented in the form $(w, l)$ for convenience: \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*} Thus, we have $5$ matching $B$ scenarios, simply adding $7$ to $w$ and $l.$ We can then test each of the five $B$ scenarios for $\frac{w}{w+l} = \frac{5}{8}$ and find that $(35, 21)$ fits this description. Then working backwards and subtracting $7$ from $w$ and $l$ gives us the point $(28, 14),$ making the answer $\boxed{\textbf{(C) } 42}.$

Solution 5 (Answer Choices: Observations)

Let's say that team $A$ plays $n$ games in total. Therefore, team $B$ must play $n + 14$ games in total (7 wins, 7 losses) Since the ratio of $A$ is \[\frac{2}{3} \implies n \equiv 0 \pmod{3}\] Similarly, since the ratio of $B$ is \[\frac{5}{8} \implies n + 14 \equiv 0 \pmod{8}\] Now, we can go through the answer choices and see which ones work: \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*} So we can see $\boxed{\textbf{(C) } 42}$ is the only valid answer.

~herobrine-india

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/KXySDYcUnHk

~Education, the Study of Everything



Video Solution

https://youtu.be/00Ngozqw2d0?t=13

Video Solution

https://youtu.be/WfTty8Fe5Fo

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
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Problem 4 		Followed by
Problem 6
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All AMC 12 Problems and Solutions

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