2018 AMC 10A Problems/Problem 6


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 1

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

Let's consider that Sangho has received 100 votes. This means he has received 65 upvotes and 35 downvotes. Part of these upvotes and downvotes cancel out, so Sangho is now left with a total of 30 upvotes, or a score increase of 30. In order for his score to be 90, Sangho must receive three sets of 100 votes. Therefore, the answer is $\boxed{\textbf{(B) } 300}$.


Solution 3

Let $x$ be the amount of votes cast, $35\%$ of $x$ would be dislikes and $65\%$ of $x$ would be likes. Since a like earns the video 1 point and a dislike takes 1 point away from the video, the amount of points Sangho's video will have in terms of $x$, the amount of votes would be $65\%-35\%=30\%$ of $x$, or $\frac{3}{10}$ of $x$. Because Songho's video had a score of 90 points, $\frac{3x}{10} = 90$. After we solve for $x$, we get $x = \boxed{\textbf{(B) } 300}$.


Solution 4

let x be the amount of like votes and y be the amount of dislike votes. Then $x-y=90$. Since $65\%$ of the total votes are like votes, $0.65(x+y)=x$. then $.65y = .35x$ or $13/7$y = $x$. Plugging that in to the original equation, $6/7 y$ = $90$ and $y =105$ Then $x=195$ and total votes is $105+195 = 300$. The answer is B.


Video Solutions





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See Also

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

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