Difference between revisions of "1995 AJHSME Problems/Problem 14"
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==Solution== | ==Solution== | ||
| − | Noting that 70% is the same as <math>\frac{70}{100}=\frac{7}{10}</math>, and that, when x is the amount of wins in the last 40 games, the fraction of games won is <math>\frac{40+x}{50+40}=\frac{40+x}{90}</math>, all we have to do is set them equal: <cmath>\frac{40+x}{90}=\frac{7}{10}</cmath> <cmath>40+x=63</cmath> <cmath>x= | + | Noting that 70% is the same as <math>\frac{70}{100}=\frac{7}{10}</math>, and that, when x is the amount of wins in the last 40 games, the fraction of games won is <math>\frac{40+x}{50+40}=\frac{40+x}{90}</math>, all we have to do is set them equal: <cmath>\frac{40+x}{90}=\frac{7}{10}</cmath> <cmath>40+x=63</cmath> <cmath>x=\boxed{\text{(B)}\ 23}</cmath> |
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| + | ==See Also== | ||
| + | {{AJHSME box|year=1995|num-b=13|num-a=15}} | ||
Revision as of 02:08, 23 December 2012
Problem
A team won 
of its first 
games. How many of the remaining games must this team win so it will have won exactly 
of its games for the season?
Solution
Noting that 70% is the same as 
, and that, when x is the amount of wins in the last 40 games, the fraction of games won is 
, all we have to do is set them equal: ![\[\frac{40+x}{90}=\frac{7}{10}\]](http://latex.artofproblemsolving.com/c/9/c/c9cf5b2c78d072905953eef21e233fd83893ab69.png)
![\[40+x=63\]](http://latex.artofproblemsolving.com/9/a/8/9a8821ded51adcd10d6092091fd227d2d9fa0fac.png)
![\[x=\boxed{\text{(B)}\ 23}\]](http://latex.artofproblemsolving.com/6/a/4/6a48b964a37501e080f1270263d43d5a2184fbbc.png)
See Also
| 1995 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
