Difference between revisions of "2004 AMC 8 Problems/Problem 6"
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== Solution == | == Solution == | ||
Sally made <math>0.55*20=11</math> shots originally. Letting <math>x</math> be the number of shots she made, we have <math>\frac{11+x}{25}=0.56</math>. Solving for <math>x</math> gives us <math>x=\boxed{\textbf{(C)}\ 3}</math> | Sally made <math>0.55*20=11</math> shots originally. Letting <math>x</math> be the number of shots she made, we have <math>\frac{11+x}{25}=0.56</math>. Solving for <math>x</math> gives us <math>x=\boxed{\textbf{(C)}\ 3}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/8sDu95w3R2I Soo, DRMS, NM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2004|num-b=5|num-a=7}} | {{AMC8 box|year=2004|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:31, 24 March 2022
Contents
Problem
After Sally takes 
shots, she has made 
of her shots. After she takes 
more shots, she raises her percentage to 
. How many of the last shots did she make?
Solution
Sally made 
shots originally. Letting 
be the number of shots she made, we have 
. Solving for gives us 
Video Solution
https://youtu.be/8sDu95w3R2I Soo, DRMS, NM
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
