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Difference between revisions of "2008 AMC 12B Problems/Problem 5"
 
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==Problem 5==
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==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==
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The class could send just carnations (25 of them). They could also send 22 carnations and 2 roses, 19 carnations and 4 roses, and so on, down to 1 carnation and 16 roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), which is answer choice C.
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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}}
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{{MAA Notice}}

Latest revision as of 12:47, 15 February 2021

Problem

A class collects $50$ dollars to buy flowers for a classmate who is in the hospital. Roses cost $3$ dollars each, and carnations cost $2$ dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly $50$ dollars?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$

Solution

The class could send $25$ carnations and no roses, $22$ carnations and $2$ roses, $19$ carnations and $4$ roses, and so on, down to $1$ carnation and $16$ roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), $\Rightarrow \boxed{C}$

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

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