Difference between revisions of "2008 AMC 10B Problems/Problem 8"
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| − | 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 | + | 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->C}</math> possible values of <math>r</math>, and each gives us one solution. |
{{AMC10 box|year=2008|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2008|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:50, 24 July 2019
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?
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.
| 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 | ||
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