Difference between revisions of "2013 AMC 12A Problems/Problem 3"

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==Problem==
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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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<math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 70 </math>
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==Solution==
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We are given that <math>\frac{6}{10} = \frac{3}{5}</math> of the flowers are pink, so we know <math>\frac{2}{5}</math> of the flowers are red.
 
We are given that <math>\frac{6}{10} = \frac{3}{5}</math> of the flowers are pink, so we know <math>\frac{2}{5}</math> of the flowers are red.
  
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<math>\frac{3}{5} * \frac{2}{3} + \frac{2}{5} * \frac{3}{4} = \frac{2}{5} + \frac{3}{10} = \frac{7}{10} = 70\%</math>, which is <math>E</math>
 
<math>\frac{3}{5} * \frac{2}{3} + \frac{2}{5} * \frac{3}{4} = \frac{2}{5} + \frac{3}{10} = \frac{7}{10} = 70\%</math>, which is <math>E</math>
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=2|num-a=4}}

Revision as of 17:30, 22 February 2013

Problem

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?

$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 70$

Solution

We are given that $\frac{6}{10} = \frac{3}{5}$ of the flowers are pink, so we know $\frac{2}{5}$ of the flowers are red.

Since $\frac{1}{3}$ of the pink flowers are roses, $\frac{2}{3}$ of the pink flowers are carnations.

We are given that $\frac{3}{4}$ of the red flowers are carnations.

The number of carnations are

$\frac{3}{5} * \frac{2}{3} + \frac{2}{5} * \frac{3}{4} = \frac{2}{5} + \frac{3}{10} = \frac{7}{10} = 70\%$, which is $E$

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions