Difference between revisions of "2017 AMC 12A Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | + | We can take two 5-popsicle boxes and one 3-popsicle box with <math>\$8</math>. Note that it is optimal since one popsicle is at the rate of <math>\$1</math> per popsicle, three popsicles at <math>\$\frac{2}{3}</math> per popsicle and finally, five popsicles at <math>\$\frac{3}{5}</math> per popsicle, hence we want as many <math>\$3</math> sets as possible. It is clear that the above is the optimal method. <math>\boxed{\textbf{D}}</math>. | |
==Video Solution (HOW TO THINK CREATIVELY!!!)== | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
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| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=Ro_koxVqd50 | ||
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| + | ~Math4All999 | ||
==See Also== | ==See Also== | ||
Latest revision as of 06:03, 14 September 2024
Contents
Problem
Pablo buys popsicles for his friends. The store sells single popsicles for 
each, 3-popsicle boxes for 
, and 5-popsicle boxes for 
. What is the greatest number of popsicles that Pablo can buy with 
?
Solution
We can take two 5-popsicle boxes and one 3-popsicle box with . Note that it is optimal since one popsicle is at the rate of per popsicle, three popsicles at 
per popsicle and finally, five popsicles at 
per popsicle, hence we want as many sets as possible. It is clear that the above is the optimal method. 
.
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=Ro_koxVqd50
~Math4All999
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
