Difference between revisions of "2007 AMC 8 Problems/Problem 20"

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Regular season wins and losses are related in two ways:
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==Problem==
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Before the district play, the Unicorns had won <math>45</math>% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
wins / (wins + losses) = 0.45
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<math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60 </math>
wins + 6 = losses + 2
 
  
So wins + 4 = losses or wins = losses - 4
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==Solution 1==
  
so wins / (wins + wins + 4) = 0.45 or 9/20
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At the beginning of the problem, the Unicorns had played <math>y</math> games and they had won <math>x</math> of these games. From the information given in the problem, we can say that <math>\frac{x}{y}=0.45.</math> Next, the Unicorns win 6 more games and lose 2 more, for a total of <math>6+2=8</math> games played during district play. We are told that they end the season having won half of their games, or <math>0.5 </math> of their games. We can write another equation: <math>\frac{x+6}{y+8}=0.5.</math> This gives us a system of equations:
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<math>\frac{x}{y}=0.45</math> and <math>\frac{x+6}{y+8}=0.5.</math>
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We first multiply both sides of the first equation by <math>y</math> to get <math>x=0.45y.</math> Then, we multiply both sides of the second equation by <math>(y+8)</math> to get <math>x+6=0.5(y+8).</math> Applying the Distributive Property gives yields <math>x+6=0.5y+4.</math> Now we substitute <math>0.45y</math> for <math>x</math> to get <math>0.45y+6=0.5y+4.</math> Solving gives us <math>y=40.</math> Since the problem asks for the total number of games, we add on the last 8 games to get the solution <math>\boxed{\textbf{(A)}\ 48}</math>.
  
so 20 * wins = 18 * wins + 9 * 4
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==Solution 2==
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Simplifying 45% to <math>\frac{9}{20}</math>, we see that the numbers of games before district play are a multiple of 20. After that the Aces played 8 more games to the total number of games is in the form of 20x+8 where x is any positive integer. The only answer choice is <math>\boxed{48}</math>, which is 20(2)+8.
  
so wins = 9 * 2 = 18
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-harsha12345
  
so losses = 22
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==Solution 3==
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First we simplify <math>45</math>% to <math>\frac{9}{20}</math>. Ratio of won to total is <math>\frac{9}{20}</math>, but ratio of total number won to total number played is <math>\frac{9x}{20x}</math> for some <math>x</math>. After they won 6 more games and lost 2 more games the number of games they won is <math>9x+6</math>, and the total number of games is <math>20x+8</math>. Turning it into a fraction we get <math>\frac{9x+6}{20x+8}=\frac{1}{2}</math>, so solving for <math>x</math> we get <math>x=2.</math> Plugging in 2 for <math>x</math> we get <math>20(2)+8=\boxed{48}</math>.
  
so regular season games = 40.
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-harsha12345
  
So the total number of games is 40 + 6 + 2 = 48, or (A).
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==Solution 4==
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Because 45% can be simplified to <math>9/20</math>, and we know that we cannot play a fraction amount of games, we know that the amount of games before district play is divisible by 20. After district play, there was <math>8</math> games, so in total there must be <math>20x+8</math>. The only answer in this format is <math>\boxed{\mathrm{(A)}48}</math>.
  
This is easily checked by finding 45% of 40 = 18 and noticing an 18-22 record + a 6-2 record is a 24-24 record. So another reasonable strategy in this context is to just check each of the answer choices.
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==See Also==
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{{AMC8 box|year=2007|num-b=19|num-a=21}}
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{{MAA Notice}}

Revision as of 02:46, 13 June 2022

Problem

Before the district play, the Unicorns had won $45$% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution 1

At the beginning of the problem, the Unicorns had played $y$ games and they had won $x$ of these games. From the information given in the problem, we can say that $\frac{x}{y}=0.45.$ Next, the Unicorns win 6 more games and lose 2 more, for a total of $6+2=8$ games played during district play. We are told that they end the season having won half of their games, or $0.5$ of their games. We can write another equation: $\frac{x+6}{y+8}=0.5.$ This gives us a system of equations: $\frac{x}{y}=0.45$ and $\frac{x+6}{y+8}=0.5.$ We first multiply both sides of the first equation by $y$ to get $x=0.45y.$ Then, we multiply both sides of the second equation by $(y+8)$ to get $x+6=0.5(y+8).$ Applying the Distributive Property gives yields $x+6=0.5y+4.$ Now we substitute $0.45y$ for $x$ to get $0.45y+6=0.5y+4.$ Solving gives us $y=40.$ Since the problem asks for the total number of games, we add on the last 8 games to get the solution $\boxed{\textbf{(A)}\ 48}$.

Solution 2

Simplifying 45% to $\frac{9}{20}$, we see that the numbers of games before district play are a multiple of 20. After that the Aces played 8 more games to the total number of games is in the form of 20x+8 where x is any positive integer. The only answer choice is $\boxed{48}$, which is 20(2)+8.

-harsha12345

Solution 3

First we simplify $45$% to $\frac{9}{20}$. Ratio of won to total is $\frac{9}{20}$, but ratio of total number won to total number played is $\frac{9x}{20x}$ for some $x$. After they won 6 more games and lost 2 more games the number of games they won is $9x+6$, and the total number of games is $20x+8$. Turning it into a fraction we get $\frac{9x+6}{20x+8}=\frac{1}{2}$, so solving for $x$ we get $x=2.$ Plugging in 2 for $x$ we get $20(2)+8=\boxed{48}$.

-harsha12345

Solution 4

Because 45% can be simplified to $9/20$, and we know that we cannot play a fraction amount of games, we know that the amount of games before district play is divisible by 20. After district play, there was $8$ games, so in total there must be $20x+8$. The only answer in this format is $\boxed{\mathrm{(A)}48}$.

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

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