Difference between revisions of "2007 AMC 8 Problems/Problem 20"
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==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 Aces 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 Aces 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> | ||
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==Solution 1== | ==Solution 1== | ||
− | At the beginning of the problem, the | + | At the beginning of the problem, the Aces had played <math>y</math> games and they had won <math>x</math> of these games. From the information given in the problem, we can say that <math>\frac{x}{y}=0.45.</math> Next, the Aces 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> of their games. 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 <math>\boxed{\textbf{(A)}\ 48}</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 <math>\boxed{\textbf{(A)}\ 48}</math>. | ||
==Solution 2== | ==Solution 2== | ||
− | 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 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 <math>\boxed{48}</math>, which is 20(2)+8. |
-harsha12345 | -harsha12345 | ||
==Solution 3== | ==Solution 3== | ||
− | 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 they 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>. | + | First we simplify 45% to <math>\frac{9}{20}</math>. 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 <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 |
Latest revision as of 12:29, 8 July 2021
Problem
Before the district play, the Aces 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 Aces 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.