Difference between revisions of "2013 AMC 12A Problems/Problem 3"
| Line 1: | Line 1: | ||
| + | ==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. | 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. | ||
| Line 8: | Line 15: | ||
<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> | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2013|ab=A|num-b=2|num-a=4}} | ||
Revision as of 18:30, 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?
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 
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 | |
