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"

(→‎Problem)
m (→‎Problem)
 
(50 intermediate revisions by 22 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 avengers 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 avengers 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 10: Line 10:
 
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 (Answer Choices)==
−
Simplifying 45% to <math>\frac{9}{20}</math>, 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 <math>\boxed{48}</math>, which is 20(2)+8.
+
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>.
  
−
-harsha12345
+
Note: If A didn't work, we would have similarly tested the other choices until we found one that did.
  
−
==Solution 3==
+
~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
−
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>.
 
  
−
-harsha12345
+
==Solution 3 (Quick)==
 +
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>.
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/rQUwNC0gqdg?t=1993
 +
 
 +
~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 19:20, 18 November 2024

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 (Quick)

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 $\boxed{\textbf{(A)} 48}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems/Problem_20&oldid=234950"