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Difference between revisions of "2010 AMC 8 Problems/Problem 9"

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==Solution==
 
==Solution==
  
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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{(D)}\ 84}</math>
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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==
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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 $80\%$ of the problems correct on a $25$-problem test, $90\%$ on a $40$-problem test, and $70\%$ on a $10$-problem test. What percent of all the problems did Ryan answer correctly?

$\textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86$

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 $25+40+10=75.$ Therefore, the percentage is $\dfrac{63}{75} = 84\% \rightarrow \boxed{\textbf{(D)}\ 84}$

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. AMC logo.png

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