Difference between revisions of "2013 AMC 12A Problems/Problem 3"
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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> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=2vf843cvVzo?t=290 (problem 3 starts at 4:50) | ||
+ | |||
+ | ~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 22:08, 23 November 2020
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 number of carnations are
, 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.