Difference between revisions of "2018 AMC 10A Problems/Problem 6"

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<math>\textbf{(A) }  200  \qquad        \textbf{(B) }  300  \qquad    \textbf{(C) }  400  \qquad  \textbf{(D) }  500  \qquad  \textbf{(E) }  600 </math>
 
<math>\textbf{(A) }  200  \qquad        \textbf{(B) }  300  \qquad    \textbf{(C) }  400  \qquad  \textbf{(D) }  500  \qquad  \textbf{(E) }  600 </math>
  
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==Solution==
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If <math>65\%</math> of the votes were likes, then <math>35\%</math> of the votes were dislikes. <math>65-35=30\%</math>, so 90 is <math>30\%</math> of his total number of votes. Doing quick arithmetic shows that the answer is <math>300\Rightarrow\text{\boxed{B}}</math>
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2018|ab=A|num-b=5|num-a=7}}
 
{{AMC10 box|year=2018|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:39, 8 February 2018

Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of 0, and the score increases by 1 for each like vote and decreases by 1 for each dislike vote. At one point Sangho saw that his video had a score of 90, and that $65\%$ of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?

$\textbf{(A) }   200   \qquad        \textbf{(B) }   300   \qquad    \textbf{(C) }   400   \qquad   \textbf{(D) }  500  \qquad  \textbf{(E) }   600$

Solution

If $65\%$ of the votes were likes, then $35\%$ of the votes were dislikes. $65-35=30\%$, so 90 is $30\%$ of his total number of votes. Doing quick arithmetic shows that the answer is $300\Rightarrow\text{\boxed{B}}$

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 AMC 10 Problems and Solutions

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