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> | ||
| + | ==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 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 15:39, 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 
of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
Solution
If of the votes were likes, then 
of the votes were dislikes. 
, so 90 is 
of his total number of votes. Doing quick arithmetic shows that the answer is 
See Also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
