Difference between revisions of "2008 AMC 12B Problems/Problem 5"
(New page: ==Problem 5== A class collects <math>50</math> dollars to buy flowers for a classmate who is in the hospital. Roses cost <math>3</math> dollars each, and carnations cost <math>2</math> dol...) |
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− | ==Problem | + | ==Problem== |
A class collects <math>50</math> dollars to buy flowers for a classmate who is in the hospital. Roses cost <math>3</math> dollars each, and carnations cost <math>2</math> dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly <math>50</math> dollars? | A class collects <math>50</math> dollars to buy flowers for a classmate who is in the hospital. Roses cost <math>3</math> dollars each, and carnations cost <math>2</math> dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly <math>50</math> dollars? | ||
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==Solution== | ==Solution== | ||
− | The class could send | + | 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== | ||
+ | {{AMC12 box|year=2008|ab=B|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:47, 15 February 2021
Problem
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.