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"

(Added page.)
 
(→‎Solution)
Line 12: Line 12:
 
<cmath>B_w=5y, B_l=3y, B_g=8y,</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</math> in <math>{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</math> in <math>{A, B}</math>. Using the given information, we can set up the following two equations:
  
 
<cmath>B_w=A_w+7\implies 5y=2x+7, B_l=A_l+7\implies 3y=x+7.</cmath>
 
<cmath>B_w=A_w+7\implies 5y=2x+7, B_l=A_l+7\implies 3y=x+7.</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>.

Revision as of 20:09, 7 February 2020

Problem 5

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

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:

\[B_w=A_w+7\implies 5y=2x+7, B_l=A_l+7\implies 3y=x+7.\]

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 $A_g=3(14)=\boxed{\textbf{(C) } 42}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_5&oldid=117116"