Difference between revisions of "2008 AMC 12B Problems/Problem 5"
(→Solution: fixed solution) |
(→Solution: fixed arrow and box problem) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | The class could send <math>25</math> carnations and no roses, <math>22</math> carnations and <math>2</math> roses, <math>19</math> carnations and <math>4</math> roses, and so on, down to <math>1</math> carnation and <math>16</math> roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), <math>\ | + | The class could send <math>25</math> carnations and no roses, <math>22</math> carnations and <math>2</math> roses, <math>19</math> carnations and <math>4</math> roses, and so on, down to <math>1</math> carnation and <math>16</math> roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), <math>\Rightarrow \boxed{C}</math> |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=4|num-a=6}} | {{AMC12 box|year=2008|ab=B|num-b=4|num-a=6}} | ||
Revision as of 12:27, 30 May 2011
Problem 5
A class collects 
dollars to buy flowers for a classmate who is in the hospital. Roses cost 
dollars each, and carnations cost 
dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly dollars?
Solution
The class could send 
carnations and no roses, 
carnations and roses, 
carnations and 
roses, and so on, down to 
carnation and 
roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), 
See Also
| 2008 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 | |
