Difference between revisions of "2017 AMC 10A Problems/Problem 2"
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<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> | ||
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| + | ==Solution== | ||
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| + | \$3 boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying 2, we have \$2 left. We cannot buy a third \$3 box, so we opt for the \$2 box instead (since it has a higher popsicles/dollar ratio than the \$1 pack). We're now out of money. We bought <math>5+5+2=12</math> popsicles, so the answer is <math>\boxed{(C)}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=1|num-a=3}} | {{AMC10 box|year=2017|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:47, 8 February 2017
Problem
Pablo buys popsicles for his friends. The store sells single popsicles for $1 each, 3-popsicle boxes for $2 each, and 5-popsicle boxes for $3. What is the greatest number of popsicles that Pablo can buy with $8?
Solution
$3 boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying 2, we have $2 left. We cannot buy a third $3 box, so we opt for the $2 box instead (since it has a higher popsicles/dollar ratio than the $1 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
