Difference between revisions of "2007 AMC 8 Problems/Problem 20"
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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> | <math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60 </math> | ||
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+ | '''SOLUTION:''' At the beginning of the problem, the Unicorns had played <math>y</math> games and they had won <math>x</math> of these games. So we can say that <math>\frac{x}{y}=0.45.</math> Then, 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> We can write another equation: <math>\frac{x+6}{y+8}=0.5.</math> This gives us a system of equations: | ||
+ | <math>\frac{x}{y}=0.45</math> and <math>\frac{x+6}{y+8}=0.5</math> | ||
+ | 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 '''(A) 48'''. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=19|num-a=21}} | {{AMC8 box|year=2007|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:02, 1 October 2013
Problem
Before district play, the Unicorns had won 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?
SOLUTION: At the beginning of the problem, the Unicorns had played games and they had won
of these games. So we can say that
Then, the Unicorns win 6 more games and lose 2 more, for a total of
games played during district play. We are told that they end the season having won half of their games, or
We can write another equation:
This gives us a system of equations:
and
We first multiply both sides of the first equation by
to get
Then, we multiply both sides of the second equation by
to get
Applying the Distributive Property gives yields
Now we substitute
for
to get
Solving gives us
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 (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.