2008 AMC 10B Problems/Problem 8

Problem

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

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

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 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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 10 Problems and Solutions

Page

Toolbox

Search

2008 AMC 10B Problems/Problem 8

Problem

Solution

See also