2012 AMC 10A Problem 7

The problem only gives fractions, so the actual number of marbles does not matter. Pick an easy number.

1. Say the bag has $5$ marbles. Then $\frac35$ of them, $3$ marbles, are blue, and the other $2$ are red.
2. Double the red marbles: $2$ becomes $4$. The blue marbles stay at $3$.
3. The bag now holds $4 + 3 = 7$ marbles, and $4$ of them are red.
4. The fraction that is red is $\frac47$.

The answer is $\boxed{\textbf{(C)}\ \frac{4}{7}}$.

When a problem talks only about fractions and asks for a fraction, you may choose any total that makes the counts whole numbers. The key step is noticing that adding red marbles also makes the whole bag bigger, so the bottom of the fraction changes too.

To see that $5$ was not a special choice, try $10$ marbles: $6$ blue and $4$ red. Doubling red gives $8$ red, and $8$ out of $6 + 8 = 14$ is again $\frac{8}{14} = \frac47$. Any size of bag gives the same answer.

The answer does not depend on the size of the bag, so pick a convenient number of marbles: $5$. Then $3$ are blue and $2$ are red.

Doubling the red marbles gives $4$ red, while the blue count stays at $3$. The bag now holds $4 + 3 = 7$ marbles, of which $4$ are red.

The fraction of red marbles is $\frac{4}{7}$.

The answer is $\boxed{\textbf{(C)}\ \frac{4}{7}}$.