Difference between revisions of "2007 AMC 8 Problems/Problem 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. 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: | 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> | <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 | + | 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>. |
==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 14:47, 31 January 2015
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 
.
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. 
