Difference between revisions of "2013 AMC 10A Problems/Problem 10"
Countingkg (talk | contribs) |
Countingkg (talk | contribs) |
||
| Line 9: | Line 9: | ||
==Solution== | ==Solution== | ||
| − | Let the total amount of flowers be <math>x</math>. Thus, the number of pink flowers is <math>0.6x</math>, and the number of red flowers is <math>0.4x</math>. The number of pink carnations is <math>\frac{2}{3}(0.6x) = 0.4x</math> and the number of red carnations is <math>\frac{3}{4}(0.4x) = 0.3x</math>. Summing these, the total number of carnations is <math>0.4x+0.3x=0.7x</math>. Dividing, we see that <math>\frac{0.7x}{x} = 0.7 = | + | Let the total amount of flowers be <math>x</math>. Thus, the number of pink flowers is <math>0.6x</math>, and the number of red flowers is <math>0.4x</math>. The number of pink carnations is <math>\frac{2}{3}(0.6x) = 0.4x</math> and the number of red carnations is <math>\frac{3}{4}(0.4x) = 0.3x</math>. Summing these, the total number of carnations is <math>0.4x+0.3x=0.7x</math>. Dividing, we see that <math>\frac{0.7x}{x} = 0.7 = \boxed{\textbf{(C) }70\%}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2013|ab=A|num-b=9|num-a=11}} | ||
{{AMC12 box|year=2013|ab=A|num-b=2|num-a=4}} | {{AMC12 box|year=2013|ab=A|num-b=2|num-a=4}} | ||
Revision as of 14:25, 8 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
Let the total amount of flowers be 
. Thus, the number of pink flowers is 
, and the number of red flowers is 
. The number of pink carnations is 
and the number of red carnations is 
. Summing these, the total number of carnations is 
. Dividing, we see that 
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 10 Problems and Solutions | ||
| 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 | |
