Difference between revisions of "2017 AMC 8 Problems/Problem 2"
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Let <math>x</math> be the total amount of votes casted. From the chart, Brenda received <math>30\%</math> of the votes and had <math>36</math> votes. We can express this relationship as <math>\frac{30}{100}x=36</math>. Solving for <math>x</math>, we get <math>x=\boxed{\textbf{(E)}\ 120}.</math> | Let <math>x</math> be the total amount of votes casted. From the chart, Brenda received <math>30\%</math> of the votes and had <math>36</math> votes. We can express this relationship as <math>\frac{30}{100}x=36</math>. Solving for <math>x</math>, we get <math>x=\boxed{\textbf{(E)}\ 120}.</math> | ||
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+ | ==Solution 2== | ||
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+ | We're being asked for the total number of votes casted -- that represents <math>100\%</math> of the total number of votes (obviously). Brenda received <math>36</math> votes, which is <math>\frac{30}{100} = \frac{3}{10}</math> of the total number of votes. Multiplying <math>36</math> by <math>\frac{10}{3},</math> we get the total number of votes, which is <math>\boxed{\textbf{(E)}\ 120}.</math> | ||
==See Also== | ==See Also== |
Revision as of 10:29, 19 December 2017
Contents
Problem 2
Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received 36 votes, then how many votes were cast all together?
Solution
Let be the total amount of votes casted. From the chart, Brenda received of the votes and had votes. We can express this relationship as . Solving for , we get
Solution 2
We're being asked for the total number of votes casted -- that represents of the total number of votes (obviously). Brenda received votes, which is of the total number of votes. Multiplying by we get the total number of votes, which is
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.