Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 4 (Answer Choices: Observations): Reformatted.)
(→‎Video Solution by OmegaLearn)
 
(21 intermediate revisions by 5 users not shown)
Line 19: Line 19:
  
 
==Solution 2 (Two Variables)==
 
==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>
  
−
First, let us assign some variables. Let
+
~AnkitAMC
  
−
<cmath>A_w=2x, A_l=x, A_g=3x,</cmath>
+
==Solution 3 (Two Variables)==
−
<cmath>B_w=5y, B_l=3y, B_g=8y,</cmath>
 
  
 +
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:
 
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*}
−
<cmath>B_w=A_w+7\implies 5y=2x+7,</cmath>
+
B_w=A_w+7&\implies 5y=2x+7, \\
−
<cmath>B_l=A_l+7\implies 3y=x+7.</cmath>
+
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 3 (Answer Choices: Substitutions)==
+
==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 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 <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*}
 
<cmath>\begin{align*}
 
\textbf{(A)} &\implies (14, 7) \\
 
\textbf{(A)} &\implies (14, 7) \\
Line 46: Line 54:
 
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>
  
−
==Solution 4 (Answer Choices: Observations)==
+
==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:
Line 59: Line 67:
  
 
~herobrine-india
 
~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==
 
==Video Solution==
  
 
https://youtu.be/WfTty8Fe5Fo
 
https://youtu.be/WfTty8Fe5Fo
−
 
−
~IceMatrix
 
  
 
==See Also==
 
==See Also==

Latest revision as of 15:01, 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)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_5&oldid=194063"