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

Revision as of 08:28, 11 February 2020

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

Video Solution

https://youtu.be/WfTty8Fe5Fo

~IceMatrix

@@ 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>.
+~Argonauts16
+==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:
+<cmath>A \implies (14, 7)</cmath>
+<cmath>B \implies (18, 9)</cmath>
+<cmath>C \implies (28, 14)</cmath>
+<cmath>D \implies (32, 16)</cmath>
+<cmath>E \implies (42, 21)</cmath>
+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>.
+==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
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