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2012 AMC 10A Problem 7

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2012 AMC 10A Problem 7AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 1 of 5 · about 2 min · position P1-10

In a bag of marbles, 35\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?

  1. A)

    25\frac{2}{5}

  2. B)

    37\frac{3}{7}

  3. C)

    47\frac{4}{7}

  4. D)

    35\frac{3}{5}

  5. E)

    45\frac{4}{5}

Answer

C

Key insight

Take 5 marbles: 3 blue, 2 red; doubling red gives 4 red out of 7 total.

Solution

The answer does not depend on the size of the bag, so pick a convenient number of marbles: 55. Then 33 are blue and 22 are red.

Doubling the red marbles gives 44 red, while the blue count stays at 33. The bag now holds 4+3=74 + 3 = 7 marbles, of which 44 are red.

The fraction of red marbles is 47\frac{4}{7}.

The answer is (C) 47\boxed{\textbf{(C)}\ \frac{4}{7}}.

Why this works

When a problem gives only fractions and asks for a fraction, the actual total is irrelevant, so choose one that makes every count an integer. The essential step is that the denominator changes: adding marbles changes the whole, not just the part.

Alternative approach

With 5n5n marbles, 3n3n blue and 2n2n red, doubling red gives 4n4n red out of 3n+4n=7n3n + 4n = 7n, so the fraction is 4n7n=47\frac{4n}{7n} = \frac47 for every nn.

The trap

Doubling the red fraction to 4/5 without recomputing the new total, or answering with the blue fraction 3/7.

Common mistakes

  • Doubling the red fraction to 4/5 without recomputing the new total, or answering with the blue fraction 3/7.
  • Doubling the total number of marbles as well as the reds, which leaves the fraction at 25\frac25.

Techniques

Test small/specific values or special cases to find or verify the answer