Difference between revisions of "2007 AMC 8 Problems/Problem 20"
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<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>. | ||
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+ | Dis is kewl-sleepypuppy | ||
+ | nice | ||
+ | -yoda | ||
==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. |
+ | |||
+ | ==Solution 3== | ||
+ | First we simplify <math>45</math>% to <math>\frac{9}{20}</math>. Ratio of won to total is <math>\frac{9}{20}</math>, but ratio of total number won to total number played is <math>\frac{9x}{20x}</math> for some <math>x</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>. | ||
− | + | ==Solution 4== | |
+ | Because 45% can be simplified to <math>9/20</math>, and we know that we cannot play a fraction amount of games, we know that the amount of games before district play is divisible by 20. After district play, there was <math>8</math> games, so in total there must be <math>20x+8</math>. The only answer in this format is <math>\boxed{\mathrm{(A)}48}</math>. | ||
− | ==Solution | + | ==Solution 5== |
− | + | Let <math>n</math> be the number of pre-district games. Therefore, we can write the percentage of total games won as a weighted average, namely <math>.45(n)+.75(8)=(n+8)(.5)</math>. Solving this equation for <math>n</math> gives <math>40</math>, but since the problem asked for all games, the answer is <math>n+8=40+8=\boxed{\mathrm{(A)}48}</math> | |
+ | |||
+ | ==Solution 6== | ||
+ | Let the number of games they won be <math>x</math> and the number of games they lost be <math>y</math>. We are told that <math>\frac{x}{x+y}=\frac{9}{20}</math>, which can be manipulated to <math>20x=9x+9y</math> which simplifies down to <math>11x=9y</math>. Then, after district games, we are told <math>\frac{x+6}{x+y+8}=\frac12</math>, which can be changed into <math>2x+12=x+y+8</math> which simplifies down to <math>x+4=y</math>. Then we can solve for <math>x</math> using substitution: | ||
+ | <cmath>11x=9x+36</cmath> | ||
+ | <cmath>2x=36</cmath> | ||
+ | <cmath>x=18</cmath> | ||
+ | Now that we know <math>x=18</math>, we can figure out that <math>y=22</math>. <math>18+22=40</math>. Now we need to add on the district games: <math>40+8=\boxed{\textbf{(A)} 48}</math>. | ||
− | + | ==Video Solution by OmegaLearn== | |
+ | https://youtu.be/rQUwNC0gqdg?t=1993 | ||
+ | ~pi_is_3.14 | ||
==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}} |
Latest revision as of 14:18, 16 January 2023
Contents
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 .
Dis is kewl-sleepypuppy
nice
-yoda
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.
Solution 3
First we simplify % to . Ratio of won to total is , but ratio of total number won to total number played is for some . 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 .
Solution 4
Because 45% can be simplified to , and we know that we cannot play a fraction amount of games, we know that the amount of games before district play is divisible by 20. After district play, there was games, so in total there must be . The only answer in this format is .
Solution 5
Let be the number of pre-district games. Therefore, we can write the percentage of total games won as a weighted average, namely . Solving this equation for gives , but since the problem asked for all games, the answer is
Solution 6
Let the number of games they won be and the number of games they lost be . We are told that , which can be manipulated to which simplifies down to . Then, after district games, we are told , which can be changed into which simplifies down to . Then we can solve for using substitution: Now that we know , we can figure out that . . Now we need to add on the district games: .
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=1993
~pi_is_3.14
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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