2007 AMC 8 Problems/Problem 20

Revision as of 12:26, 7 April 2021 by Putu2003 (talk | contribs) (Problem)


Before the district play, the bols 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 bols 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

Simplifying 45% to $\frac{9}{20}$, 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 $\boxed{48}$, which is 20(2)+8.


Solution 3

First we simplify 45% to $\frac{9}{20}$. After they won 6 more games and lost 2 more games the number of games they won is $9x+6$, and the total number of games is $20x+8$. Turning it into a fraction we get $\frac{9x+6}{20x+8}=\frac{1}{2}$, so solving for $x$ we get $x=2.$ Plugging in 2 for $x$ we get $20(2)+8=\boxed{48}$.


See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS