Difference between revisions of "2017 AMC 10A Problems/Problem 2"
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==Problem== | ==Problem== | ||
| − | Pablo buys popsicles for his friends. The store sells single popsicles for <math>\$1</math> each, 3-popsicle boxes for <math>\$2</math> each, and <math>5</math>-popsicle boxes for <math>\$3</math>. What is the greatest number of popsicles that Pablo can buy with <math>\$8</math>? | + | Pablo buys popsicles for his friends. The store sells single popsicles for <math>\$1</math> each, <math>3</math>-popsicle boxes for <math>\$2</math> each, and <math>5</math>-popsicle boxes for <math>\$3</math>. What is the greatest number of popsicles that Pablo can buy with <math>\$8</math>? |
<math>\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15</math> | <math>\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15</math> | ||
Revision as of 14:11, 9 February 2017
Problem
Pablo buys popsicles for his friends. The store sells single popsicles for 
each, 
-popsicle boxes for 
each, and 
-popsicle boxes for 
. What is the greatest number of popsicles that Pablo can buy with 
?
Solution
boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying 
, we have left. We cannot buy a third box, so we opt for the box instead (since it has a higher popsicles/dollar ratio than the pack). We're now out of money. We bought 
popsicles, so the answer is 
.
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. 
