Difference between revisions of "2007 AMC 8 Problems/Problem 20"
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− | Simplifying 45% to <math>\frac{9}{20}</math>, we see that the numbers of games are a multiple of 20. After that the | + | Simplifying 45% to <math>\frac{9}{20}</math>, we see that the numbers of games are a multiple of 20. After that the avengers 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. |
-harsha12345 | -harsha12345 |
Revision as of 12:27, 7 April 2021
Problem
Before the district play, the avengers 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 avengers 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 avengers 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 they 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.