Difference between revisions of "2004 AMC 8 Problems/Problem 6"
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After Sally takes <math>20</math> shots, she has made <math>55\%</math> of her shots. After she takes <math>5</math> more shots, she raises her percentage to <math>56\%</math>. How many of the last <math>5</math> shots did she make? | After Sally takes <math>20</math> shots, she has made <math>55\%</math> of her shots. After she takes <math>5</math> more shots, she raises her percentage to <math>56\%</math>. How many of the last <math>5</math> shots did she make? | ||
| − | <math> \textbf{(A)}1\qquad\textbf{(B)}2\qquad\textbf{(C)}3\qquad\textbf{(D)}4\qquad\textbf{(E)}5 </math> | + | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math> |
== 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= | + | 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> |
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2004|num-b=5|num-a=7}} | ||
Revision as of 04:33, 24 December 2012
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 
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 | ||
