Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 AMC 8 Problems/Problem 20"

(→‎Solution 3 (Quick))
 
(7 intermediate revisions by 6 users not shown)
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 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>
+
<!-- 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>
Line 11: Line 11:
  
 
==Solution 2 (Answer Choices)==
 
==Solution 2 (Answer Choices)==
−
We can check each answer choice from left to right to see which one is correct. Suppose the Unicorns played <math>48</math> games in total. Then, after district play, they would have won <math>24</math> games. Now, consider the situation before district play. The Unicorns would have won <math>18</math> games out of <math>40</math>. Converting to a percentage, <math>18/40 = 45</math>%. Thus, the answer is <math>\boxed{\textbf{(A)} 48}</math>.
+
We can check each answer choice from left to right to see which one is correct. Suppose the Unicorns played <math>48</math> games in total. Then, after district play, they would have won <math>24</math> games. Now, consider the situation before district play. The Unicorns would have won <math>18</math> games out of <math>40</math>. Converting to a percentage, <math>18/40 = 45\%</math>. Thus, the answer is <math>\boxed{\textbf{(A)} 48}</math>.
  
 
Note: If A didn't work, we would have similarly tested the other choices until we found one that did.
 
Note: If A didn't work, we would have similarly tested the other choices until we found one that did.
Line 17: Line 17:
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
  
−
==Solution 3 (Quick)==
+
==Solution 3 (Cheese)==
−
We know that <math>45\%=\frac9{20}</math>. Therefore, the number of games before district play must be a multiple of <math>20</math> in order for the number of games won to be an integer. The Unicorns played <math>6+2=8</math> more games during district play. The only answer choice that is <math>8</math> more than a multiple of <math>20</math> is <math>\boxed{\textbf{(A)}48}</math>.
+
We know that <math>45\%=\frac9{20}</math>. Therefore, the number of games before district play must be a multiple of <math>20</math> in order for the number of games won to be an integer. The Unicorns played <math>6+2=8</math> more games during district play. The only answer choice that is <math>8</math> more than a multiple of <math>20</math> is <math>\boxed{\textbf{(A)} 48}</math>.
 +
 
 +
==Solution 4==
 +
At the start of the problem we figure out the Unicorns have won <math>45\%</math> of their games, which can be represented by <math>\frac9{20}</math>. Since this is a simplified fraction and not necessarily the amount of games they have played, we can add a variable (in this case I am using <math>x</math>) to the numerator and denominator to get <math>\frac{9x}{20x}</math>. We then know that during the district play they have won 6 more games and lost 2 of them for a total of 6+2=8 games played. We then can add it to our fraction we have built to make <math>\frac{9x+6}{20x+8}</math>. We also know that the fraction is equal to <math>\frac1{2}</math>. Putting these equations together we get  <math>\frac{9x+6}{20x+8}=\frac1{2}</math>. Now we can cross multiply to come out with <math>18x+12=20x+8</math>. We now solve for <math>x</math>!<math>18x+12=20x+8</math> ----> <math>4=2x</math> ----> <math>2=x</math>. After we get <math>x</math> which comes out to be 2 we then plug it back into the denominator to get <math>20(2)+8</math> which is equal to 48, so the answer must be <math>\boxed{\textbf{(A)} 48}</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Am24 AM24]
  
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==
Line 24: Line 29:
  
 
~pi_is_3.14
 
~pi_is_3.14
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/1CAxNXM8TWo
 +
 
==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 16:50, 14 June 2025

Problem

Before the district play, the Unicorns had won $45\%$ 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?

$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution 1

At the beginning of the problem, the Unicorns had played $y$ games and they had won $x$ of these games. From the information given in the problem, we can say that $\frac{x}{y}=0.45.$ Next, the Unicorns win 6 more games and lose 2 more, for a total of $6+2=8$ games played during district play. We are told that they end the season having won half of their games, or $0.5$ of their games. We can write another equation: $\frac{x+6}{y+8}=0.5.$ This gives us a system of equations: $\frac{x}{y}=0.45$ and $\frac{x+6}{y+8}=0.5.$ We first multiply both sides of the first equation by $y$ to get $x=0.45y.$ Then, we multiply both sides of the second equation by $(y+8)$ to get $x+6=0.5(y+8).$ Applying the Distributive Property gives yields $x+6=0.5y+4.$ Now we substitute $0.45y$ for $x$ to get $0.45y+6=0.5y+4.$ Solving gives us $y=40.$ Since the problem asks for the total number of games, we add on the last 8 games to get the solution $\boxed{\textbf{(A)}\ 48}$.

Solution 2 (Answer Choices)

We can check each answer choice from left to right to see which one is correct. Suppose the Unicorns played $48$ games in total. Then, after district play, they would have won $24$ games. Now, consider the situation before district play. The Unicorns would have won $18$ games out of $40$. Converting to a percentage, $18/40 = 45\%$. Thus, the answer is $\boxed{\textbf{(A)} 48}$.

Note: If A didn't work, we would have similarly tested the other choices until we found one that did.

~cxsmi

Solution 3 (Cheese)

We know that $45\%=\frac9{20}$. Therefore, the number of games before district play must be a multiple of $20$ in order for the number of games won to be an integer. The Unicorns played $6+2=8$ more games during district play. The only answer choice that is $8$ more than a multiple of $20$ is $\boxed{\textbf{(A)} 48}$.

Solution 4

At the start of the problem we figure out the Unicorns have won $45\%$ of their games, which can be represented by $\frac9{20}$. Since this is a simplified fraction and not necessarily the amount of games they have played, we can add a variable (in this case I am using $x$) to the numerator and denominator to get $\frac{9x}{20x}$. We then know that during the district play they have won 6 more games and lost 2 of them for a total of 6+2=8 games played. We then can add it to our fraction we have built to make $\frac{9x+6}{20x+8}$. We also know that the fraction is equal to $\frac1{2}$. Putting these equations together we get $\frac{9x+6}{20x+8}=\frac1{2}$. Now we can cross multiply to come out with $18x+12=20x+8$. We now solve for $x$!$18x+12=20x+8$ ----> $4=2x$ ----> $2=x$. After we get $x$ which comes out to be 2 we then plug it back into the denominator to get $20(2)+8$ which is equal to 48, so the answer must be $\boxed{\textbf{(A)} 48}$.

~AM24

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=1993

~pi_is_3.14

Video Solution by WhyMath

https://youtu.be/1CAxNXM8TWo

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_20&oldid=251163"