Difference between revisions of "2007 AMC 8 Problems/Problem 20"
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==Solution 3(no decimals)== | ==Solution 3(no decimals)== | ||
| − | First we 45% to <math>\frac{9}{20}</math>. After he 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=48</math>. | + | First we 45% to <math>\frac{9}{20}</math>. After he 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 | ||
==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 15:16, 13 June 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
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 
.
44==Solution 2(under pressure)==
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(no decimals)
First we 45% to . After he 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
