2010 AMC 8 Problems/Problem 9: Difference between revisions
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==Solution== | ==Solution== | ||
Ryan answered <math>(0.8)(25)=20</math> problems correct on the first test, <math>(0.9)(40)=36</math> on the second, and <math>(0.7)(10)=7</math> on the third. This amounts to a total of <math>20+36+7=63</math> problems correct. The total number of problems is <math>25+40+10=75.</math> Therefore, the percentage is <math>\dfrac{63}{75} *100 \rightarrow \boxed{\textbf{( | Ryan answered <math>(0.8)(25)=20</math> problems correct on the first test, <math>(0.9)(40)=36</math> on the second, and <math>(0.7)(10)=7</math> on the third. This amounts to a total of <math>20+36+7=63</math> problems correct. The total number of problems is <math>25+40+10=75.</math> Therefore, the percentage is <math>\dfrac{63}{75} *100 \rightarrow \boxed{\textbf{(S)}\ 84}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=8|num-a=10}} | {{AMC8 box|year=2010|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:29, 1 October 2023
Problem
Ryan got 
of the problems correct on a 
-problem test, 
on a 
-problem test, and 
on a 
-problem test. What percent of all the problems did Ryan answer correctly?
Solution
Ryan answered 
problems correct on the first test, 
on the second, and 
on the third. This amounts to a total of 
problems correct. The total number of problems is 
Therefore, the percentage is 
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
