Difference between revisions of "2007 AMC 8 Problems/Problem 20"
m (→Solution 2) |
m (→Problem) |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Before the district play, the | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Before the district play, the Unicorns had won <math>45</math>% 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?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<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> | ||
Revision as of 15:15, 23 November 2021
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 Aces had played 
games and they had won 
of these games. From the information given in the problem, we can say that 
Next, the Aces 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 before district play are a multiple of 20. After that the Aces 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 . Ratio of won to total is 9/20, but ratio of total number won to total number played is 9x/20x for some x. 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. 
