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

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==Problem==
 
==Problem==
Before 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?
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Before 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?
  
 
<math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60 </math>
 
<math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60 </math>

Revision as of 14:48, 18 November 2013

Problem

Before district play, the Unicorns had won $45%$ (Error compiling LaTeX. Unknown error_msg) 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: At the beginning of the problem, the Unicorns had played $y$ games and they had won $x$ of these games. So we can say that $\frac{x}{y}=0.45.$ Then, 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 $0.5.$ 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 (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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