2012 AMC 10A Problem 7
In a bag of marbles, 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?
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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.
- Say the bag has marbles. Then of them, marbles, are blue, and the other are red.
- Double the red marbles: becomes . The blue marbles stay at .
- The bag now holds marbles, and of them are red.
- The fraction that is red is .
The answer is .
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 was not a special choice, try marbles: blue and red. Doubling red gives red, and out of is again . 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: . Then are blue and are red.
Doubling the red marbles gives red, while the blue count stays at . The bag now holds marbles, of which are red.
The fraction of red marbles is .
The answer is .
Review in the AoPS books
- Introduction to Algebra, Chapter 6: Ratios and Percents
- Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
- Prealgebra, Chapter 13: Data and Statistics (§13.1)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.