Difference between revisions of "2010 AMC 8 Problems/Problem 9"
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==Solution== | ==Solution== | ||
| − | Ryan answered | + | 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 So the percentage is 63/75 which equals D:84 |
==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 13:44, 15 August 2021
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 (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 So the percentage is 63/75 which equals D:84
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. 
