Difference between revisions of "2008 AMC 12B Problems/Problem 5"
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==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 13:27, 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 |