Difference between revisions of "2007 AMC 8 Problems/Problem 20"
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− | First we 45% to <math>\frac{9}{20}</math>. After | + | First we simplify 45% to <math>\frac{9}{20}</math>. After they won 6 more games and lost 2 more games the number of games he 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>. |
-harsha12345 | -harsha12345 |
Revision as of 10:55, 17 July 2018
Problem
Before the 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 1
At the beginning of the problem, the Unicorns had played games and they had won of these games. From the information given in the problem, we can say that Next, 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 of their games. 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 .
Solution 2
Simplifying 45% to , we see that the numbers of games are a multiple of 20. After that the Unicorns 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 , which is 20(2)+8.
-harsha12345
Solution 3
First we simplify 45% to . After they won 6 more games and lost 2 more games the number of games he won is , and the total number of games is . Turning it into a fraction we get , so solving for we get Plugging in 2 for we get .
-harsha12345
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 |
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