2012 AMC 10A Problem 7

Attempt Problem 7 of the 2012 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 10A solutions, or check the answer key.

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7.

In a bag of marbles, 35\dfrac{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?

25\dfrac{2}{5}

37\dfrac{3}{7}

47\dfrac{4}{7}

35\dfrac{3}{5}

45\dfrac{4}{5}

Answer: C
Concepts:ratio and proportionfraction
Difficulty rating: 870
Solution:

WLOG, let the number of marbles in the bag be 5.5. Since we only care about ratios, we can do this. Then there are 33 blue marbles and 22 red marbles.

Doubling the red marbles gives us 44 of them. Then the fraction of red marbles is 44+3=47. \dfrac{4}{4 + 3} = \dfrac{4}{7}.

Thus, C is the correct answer.

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