AMC 10 Step by Step

2013 AMC 10A Problem 10

P1-10~2 minPrint

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 100100 flowers.

  1. Pink and red. Six tenths are pink: 6060 pink flowers. The other 4040 are red.
  2. Pink carnations. One third of the 6060 pink flowers are roses, so 2020 pink roses. The rest, 60−20=4060 - 20 = 40, are pink carnations.
  3. Red carnations. Three fourths of the 4040 red flowers are carnations: 3030 red carnations (and 1010 red roses).
  4. Put it in a table.
RosesCarnationsTotal
Pink204060
Red103040
Total3070100
  1. Answer. 7070 carnations out of 100100 flowers is 70%70\%.

The answer is (E) 70\boxed{\textbf{(E)}\ 70}.

Why this works

A percent does not change with the size of the bouquet, so you may choose any total you like. Picking 100100 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 100100 flowers.

Six tenths are pink: 6060 pink flowers and 4040 red flowers.

One third of the pink flowers are roses, so 2020 are pink roses and the other 60−20=4060 - 20 = 40 are pink carnations.

Three fourths of the red flowers are carnations: 34⋅40=30\tfrac34 \cdot 40 = 30 red carnations.

Carnations in total: 40+30=7040 + 30 = 70 out of 100100, which is 70%70\%.

The answer is (E) 70\boxed{\textbf{(E)}\ 70}.

Review in the AoPS books

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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.