Difference between revisions of "2010 AMC 8 Problems/Problem 9"
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==Solution== | ==Solution== | ||
| − | Ryan answered (0.8)(25)=20 problems correct on the first test, (0.9)(40)=36 on the second, and (0.7)(10)=7 on the third. This amounts to a total of 20+36+7=63 problems correct. The total number of problems is 25+40+10=75 | + | 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} = 84\% \rightarrow \boxed{\textbf{(D)}\ 84}</math> |
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| + | ==Video Solution by @MathTalks== | ||
| + | https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT | ||
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==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}} | ||
Latest revision as of 22:30, 6 January 2024
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 
Video Solution by @MathTalks
https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
