Difference between revisions of "2013 AMC 12A Problems/Problem 3"
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| − | + | ==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? | ||
| − | + | <math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 70 </math> | |
| − | + | ==Solution== | |
| − | + | 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. | |
| − | + | Since <math>\frac{1}{3}</math> of the pink flowers are roses, <math>\frac{2}{3}</math> of the pink flowers are carnations. | |
| + | |||
| + | We are given that <math>\frac{3}{4}</math> of the red flowers are carnations. | ||
| + | |||
| + | 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> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=2vf843cvVzo?t=290 (problem 3 starts at 4:50) | ||
| + | |||
| + | ~sugar_rush | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2013|ab=A|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:31, 14 July 2021
Contents
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?
Solution
We are given that 
of the flowers are pink, so we know 
of the flowers are red.
Since 
of the pink flowers are roses, 
of the pink flowers are carnations.
We are given that 
of the red flowers are carnations.
The percent of flowers that are carnations is
, which is 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
