2012 AMC 10A Problem 7

Below is the professionally curated solution for Problem 7 of the 2012 AMC 10A, 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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Concepts:ratio and proportion fraction

Difficulty rating: 870

In a bag of marbles, $\dfrac{3}{5}$ 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?

$\dfrac{2}{5}$

$\dfrac{3}{7}$

$\dfrac{4}{7}$

$\dfrac{3}{5}$

$\dfrac{4}{5}$

Solution:

WLOG, let the number of marbles in the bag be $5.$ Since we only care about ratios, we can do this. Then there are $3$ blue marbles and $2$ red marbles.

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

Thus, C is the correct answer.

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