AMC 10 Step by Step

2012 AMC 10A Problem 7

P1-10~2 minPrint

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?

Log in to save your progress.
Show solution without answering

Key insight

Pretend the bag holds 5 marbles: 3 blue and 2 red. Doubling the red gives 4 red, and the bag now holds 7 marbles.

Solution

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

  1. Say the bag has 55 marbles. Then 35\frac35 of them, 33 marbles, are blue, and the other 22 are red.
  2. Double the red marbles: 22 becomes 44. The blue marbles stay at 33.
  3. The bag now holds 4+3=74 + 3 = 7 marbles, and 44 of them are red.
  4. The fraction that is red is 47\frac47.

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

Why this works

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.

Alternative approach

To see that 55 was not a special choice, try 1010 marbles: 66 blue and 44 red. Doubling red gives 88 red, and 88 out of 6+8=146 + 8 = 14 is again 814=47\frac{8}{14} = \frac47. Any size of bag gives the same answer.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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}}.

Review in the AoPS books

Similar problems

Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.