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

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We are given that <math>\frac{3}{4}</math> of the red flowers are carnations.
 
We are given that <math>\frac{3}{4}</math> of the red flowers are carnations.
  
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The number of carnations are
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The percent of flowers that are carnations is
  
 
<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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==Video Solution==
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https://www.youtube.com/watch?v=2vf843cvVzo?t=290 (problem 3 starts at 4:50)
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~sugar_rush
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=A|num-b=2|num-a=4}}
 
{{AMC12 box|year=2013|ab=A|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:31, 14 July 2021

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 percent of flowers that are carnations is

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

Video Solution

https://www.youtube.com/watch?v=2vf843cvVzo?t=290 (problem 3 starts at 4:50)

~sugar_rush

See also

2013 AMC 12A (Problems • Answer Key • Resources)
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

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