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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== See Also == | == See Also == | ||
Revision as of 17:16, 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 
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 
votes is 
of the 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. 
