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

m (Solution)
m (Solution)
Line 18: Line 18:
  
 
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>.
 +
 +
==See Also==
 +
 +
{{AMC12 box|year=2020|ab=B|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Revision as of 20:13, 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 $A_g=3(14)=\boxed{\textbf{(C) } 42}$.

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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