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

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==Solution==
 
==Solution==
 
If <math>65\%</math> of the votes were likes, then <math>35\%</math> of the votes were dislikes. <math>65\%-35\%=30\%</math>, so <math>90</math> votes is <math>30\%</math> of the total number of votes. Doing quick arithmetic shows that the answer is <math>\boxed{\textbf{(B) } 300}</math>
 
If <math>65\%</math> of the votes were likes, then <math>35\%</math> of the votes were dislikes. <math>65\%-35\%=30\%</math>, so <math>90</math> votes is <math>30\%</math> of the total number of votes. Doing quick arithmetic shows that the answer is <math>\boxed{\textbf{(B) } 300}</math>
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==Solution 2==
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== See Also ==
 
== See Also ==

Revision as of 18:16, 30 January 2019

Problem

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 $90$ votes is $30\%$ of the total number of votes. Doing quick arithmetic shows that the answer is $\boxed{\textbf{(B) } 300}$

Solution 2

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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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