2013 AMC 10A Problem 10
A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?
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Key insight
Pretend there are 100 flowers: 60 pink (40 of them carnations) and 40 red (30 of them carnations). That is 70 carnations.
Solution
Each fraction talks about a different group (all flowers, pink flowers, red flowers). Using real numbers keeps them straight. The answer is a percent, so pick flowers.
- Pink and red. Six tenths are pink: pink flowers. The other are red.
- Pink carnations. One third of the pink flowers are roses, so pink roses. The rest, , are pink carnations.
- Red carnations. Three fourths of the red flowers are carnations: red carnations (and red roses).
- Put it in a table.
| Roses | Carnations | Total | |
|---|---|---|---|
| Pink | 20 | 40 | 60 |
| Red | 10 | 30 | 40 |
| Total | 30 | 70 | 100 |
- Answer. carnations out of flowers is .
The answer is .
Why this works
A percent does not change with the size of the bouquet, so you may choose any total you like. Picking turns every fraction into a whole number of flowers. A small table with rows (color) and columns (type) makes sure each fraction is used on the right group.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Percentages do not depend on the size of the bouquet, so suppose there are flowers.
Six tenths are pink: pink flowers and red flowers.
One third of the pink flowers are roses, so are pink roses and the other are pink carnations.
Three fourths of the red flowers are carnations: red carnations.
Carnations in total: out of , which 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.