Difference between revisions of "2012 AMC 10A Problems/Problem 7"

(Problem 7)
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<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{4}{7}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{4}{5} </math>
 
<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{4}{7}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{4}{5} </math>
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== Solution ==
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Assume that there are 5 total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling.
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There are 3 blue marbles in the bag and 2 red marbles. If you double the amount of red marbles, there will still be 3 blue marbles but now there will be 4 red marbles. Thus, the answer is <math>\qquad\textbf{(C)} \frac{4}{7}</math>.

Revision as of 22:00, 8 February 2012

Problem 7

In a bag of marbles, $\frac{3}{5}$ of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?

$\textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{3}{7}\qquad\textbf{(C)}\ \frac{4}{7}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{4}{5}$

Solution

Assume that there are 5 total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling.

There are 3 blue marbles in the bag and 2 red marbles. If you double the amount of red marbles, there will still be 3 blue marbles but now there will be 4 red marbles. Thus, the answer is $\qquad\textbf{(C)} \frac{4}{7}$.

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