Difference between revisions of "2008 AMC 10B Problems/Problem 8"

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

Revision as of 20:04, 15 July 2013

Problem

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

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

Solution

The cost of a rose is odd, hence we need an even number of roses. Let there be $2r$ roses for some $r\geq 0$. Then we have $50-3\cdot 2r = 50-6r$ dollars left. We can always reach the sum exactly $50$ by buying $(50-6r)/2 = 25-3r$ carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality $25-3r \geq 0$, and as $r$ must be an integer, this solves to $r\leq 8$. Hence there are $\boxed{9}$ possible values of $r$, and each gives us one solution.

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AMC 10 Problems and Solutions

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