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

Revision as of 12:52, 26 May 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 1 (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:

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

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 (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 3 (Answer Choices)

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 4 (Answer Choices: Modular Arithmetic)

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:

$\textbf{(A) } 21 \implies 21 + 14 = 35 \not \equiv 0\pmod{8}$ $\textbf{(B) } 27 \implies 27 + 14 = 41 \not \equiv 0\pmod{8}$ $\textbf{(C) } 42 \implies 42 + 14 = 56 \equiv 0\pmod{8}$ $\textbf{(D) } 48 \implies 48 + 14 = 62 \not \equiv 0\pmod{8}$ $\textbf{(E) } 63 \implies 63 + 14 = 77 \not \equiv 0\pmod{8}$

So we can see $\boxed{\textbf{(C) } 42} \text{ \tiny nice}$ is the only valid answer.

~herobrine-india

Video Solution

https://youtu.be/WfTty8Fe5Fo

~IceMatrix

@@ Line 56: / Line 56: @@
 <cmath>\textbf{(E) } 63 \implies 63 + 14 = 77 \not \equiv 0\pmod{8}</cmath>
-So we can see <math>\boxed{\textbf{(C) } 42.0} \text{ \tiny nice}</math> is the only valid answer.
+So we can see <math>\boxed{\textbf{(C) } 42} \text{ \tiny nice}</math> is the only valid answer.
 ~herobrine-india

2020 AMC 12B (Problems • Answer Key • Resources)
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