Difference between revisions of "2008 AMC 10B Problems/Problem 8"
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| − | + | ==Problem== | |
| + | |||
| + | A class collects <math>\</math><math>50</math> to buy flowers for a classmate who is in the hospital. Roses cost <math>\</math><math>3</math> each, and carnations cost <math>\</math><math>2</math> each. No other flowers are to be used. How many different bouquets could be purchased for exactly <math>\</math><math>50</math>? | ||
| − | + | <math> | |
| − | {{ | + | \mathrm{(A)}\ 1 |
| + | \qquad | ||
| + | \mathrm{(B)}\ 7 | ||
| + | \qquad | ||
| + | \mathrm{(C)}\ 9 | ||
| + | \qquad | ||
| + | \mathrm{(D)}\ 16 | ||
| + | \qquad | ||
| + | \mathrm{(E)}\ 17 | ||
| + | </math> | ||
==Solution== | ==Solution== | ||
| + | |||
| + | The cost of a rose is odd, hence we need an even number of roses. Let there be <math>2r</math> roses for some <math>r\geq 0</math>. Then we have <math>50-3\cdot 2r = 50-6r</math> dollars left. We can always reach the sum exactly <math>50</math> by buying <math>(50-6r)/2 = 25-3r</math> carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality <math>25-3r \geq 0</math>, and as <math>r</math> must be an integer, this solves to <math>r\leq 8</math>. Hence there are <math>\boxed{9}</math> possible values of <math>r</math>, and each gives us one solution. | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2008|ab=B|num-b=7|num-a=9}} | ||
Revision as of 14:21, 11 February 2009
Problem
A class collects 
to buy flowers for a classmate who is in the hospital. Roses cost 
each, and carnations cost 
each. No other flowers are to be used. How many different bouquets could be purchased for exactly ?
Solution
The cost of a rose is odd, hence we need an even number of roses. Let there be 
roses for some 
. Then we have 
dollars left. We can always reach the sum exactly by buying 
carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality 
, and as 
must be an integer, this solves to 
. Hence there are 
possible values of , and each gives us one solution.
See also
| 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 | ||
