Difference between revisions of "2013 AMC 10A Problems/Problem 10"
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| − | we have <math>\dfrac15</math> for pink roses,<math>1-\dfrac6{10}=4\dfrac{10}=\dfrac25</math> red flowers, <math>\dfrac6{10}-\dfrac15=\dfrac35-\dfrac15=\dfrac25</math> pink carnations, <math>\dfrac25\cdot \dfrac34=\dfrac6{20}=\dfrac3{10}</math> red carnations we add them up to get <math>\dfrac25+\dfrac3{10}=\dfrac7{10}=70\%</math> so our final answer is [u][b]70%[/b][/u | + | we have <math>\dfrac15</math> for pink roses,<math>1-\dfrac6{10}=4\dfrac{10}=\dfrac25</math> red flowers, <math>\dfrac6{10}-\dfrac15=\dfrac35-\dfrac15=\dfrac25</math> pink carnations, <math>\dfrac25\cdot \dfrac34=\dfrac6{20}=\dfrac3{10}</math> red carnations we add them up to get <math>\dfrac25+\dfrac3{10}=\dfrac7{10}=70\%</math> so our final answer is [u][b]70%[/b][/u] |
~jimkey17 from web2.0calc.com | ~jimkey17 from web2.0calc.com | ||
Revision as of 14:55, 30 July 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 1
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 
Solution 2
we have 
for pink roses,
red flowers, 
pink carnations, 
red carnations we add them up to get 
so our final answer is [u][b]70%[/b][/u]
~jimkey17 from web2.0calc.com
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
