Difference between revisions of "1995 AJHSME Problems/Problem 14"
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==Solution2== | ==Solution2== | ||
| − | Alternatively we can note that they will play a total of <math>40+50=90</math> games and must win <math>0.7(90)=63</math> games. Since they won <math>40</math> games already they need <math>63-40= | + | Alternatively we can note that they will play a total of <math>40+50=90</math> games and must win <math>0.7(90)=63</math> games. Since they won <math>40</math> games already they need <math>63-40=\boxed{(B)23}</math> more games. |
==See Also== | ==See Also== | ||
{{AJHSME box|year=1995|num-b=13|num-a=15}} | {{AJHSME box|year=1995|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:18, 1 August 2013
Contents
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)
Solution2
Alternatively we can note that they will play a total of 
games and must win 
games. Since they won games already they need 
more games.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
