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"

(→‎Solution 2)
Line 4: Line 4:
  
 
<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==
 
==Solution==
Line 20: Line 21:
  
 
~Argonauts16
 
~Argonauts16
 +
  
 
==Solution 2==
 
==Solution 2==
 +
 
Using the information from the problem, we can note that team A 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 A 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:
  
Line 31: Line 34:
  
 
Thus, we have 5 matching <math>B</math> scenarios, simply adding 7 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 7 from <math>w</math> and <math>l</math> gives us the point <math>(28, 14)</math>, making the answer <math>\boxed{\textbf{C}}</math>.
 
Thus, we have 5 matching <math>B</math> scenarios, simply adding 7 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 7 from <math>w</math> and <math>l</math> gives us the point <math>(28, 14)</math>, making the answer <math>\boxed{\textbf{C}}</math>.
 +
 +
 +
==Solution 3 (Answer Choices)==
 +
 +
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>\textbf{(A) } 21 \implies 21 + 14 = 35 \not \equiv \pmod{8}</cmath>
 +
<cmath>\textbf{(B) } 27 \implies 27 + 14 = 41 \not \equiv \pmod{8}</cmath>
 +
<cmath>\textbf{(C) } 42 \implies 42 + 14 = 56  \equiv \pmod{8}</cmath>
 +
<cmath>\textbf{(D) } 48 \implies 48 + 14 = 62 \not \equiv \pmod{8}</cmath>
 +
<cmath>\textbf{(E) } 63 \implies 63 + 14 = 77 \not \equiv \pmod{8}</cmath>
 +
 +
So we can see <math>\boxed{\textbf{(C) } 42.0} \text{ \tiny nice}</math> is the only valid answer.
 +
 +
~herobrine-india
 +
  
 
==Video Solution==
 
==Video Solution==
 +
 
https://youtu.be/WfTty8Fe5Fo
 
https://youtu.be/WfTty8Fe5Fo
  
 
~IceMatrix
 
~IceMatrix
 +
  
 
==See Also==
 
==See Also==

Revision as of 01:39, 23 April 2021

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

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

~Argonauts16


Solution 2

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 find the following list of possible win-lose scenarios for $A$, represented in the form $(w, l)$ for convenience:

\[A \implies (14, 7)\] \[B \implies (18, 9)\] \[C \implies (28, 14)\] \[D \implies (32, 16)\] \[E \implies (42, 21)\]

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 $\boxed{\textbf{C}}$.


Solution 3 (Answer Choices)

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:

\[\textbf{(A) } 21 \implies 21 + 14 = 35 \not \equiv \pmod{8}\] \[\textbf{(B) } 27 \implies 27 + 14 = 41 \not \equiv \pmod{8}\] \[\textbf{(C) } 42 \implies 42 + 14 = 56 \equiv \pmod{8}\] \[\textbf{(D) } 48 \implies 48 + 14 = 62 \not \equiv \pmod{8}\] \[\textbf{(E) } 63 \implies 63 + 14 = 77 \not \equiv \pmod{8}\]

So we can see $\boxed{\textbf{(C) } 42.0} \text{ \tiny nice}$ is the only valid answer.

~herobrine-india


Video Solution

https://youtu.be/WfTty8Fe5Fo

~IceMatrix


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=152477"