Difference between revisions of "2010 AMC 8 Problems/Problem 9"
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==Problem== | ==Problem== | ||
| − | Ryan got <math>80\%</math> of the problems correct on a <math>25</math>-problem test, <math>90\%</math> on a <math>40</math>-problem test, and <math>70\%</math> on a <math>10</math>-problem test. What percent of all the problems did Ryan answer correctly? | + | Ryan got <math>80\%</math> of the problems correct on a <math>25</math>-problem test, <math>90\%</math> on a <math>40</math>-problem test, and <math>70\%</math> on a <math>10</math>-problem test. What percent of all the problems did Ryan answer correctly? :) |
<math> \textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86 </math> | <math> \textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86 </math> | ||
Revision as of 15:26, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
