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 \$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? | + | 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> | ||
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| + | ==Solution 1== | ||
| + | |||
| + | <math>\$3</math> boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying <math>2</math>, we have <math>\$2</math> left. We cannot buy a third <math>\$3</math> box, so we opt for the <math>\$2</math> box instead (since it has a higher popsicles/dollar ratio than the <math>\$1</math> pack). We're now out of money. We bought <math>5+5+3=13</math> popsicles, so the answer is <math>\boxed{\textbf{(D) }13}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/str7kmcRMY8?feature=shared&t=64 | ||
| + | |||
| + | (TheBeautyofMath) | ||
| + | |||
| + | ==Video Solution 2== | ||
| + | |||
| + | https://youtu.be/9VGPpemq-qg | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2017|ab=A|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:18, 19 September 2024
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 1
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 
.
Video Solution
https://youtu.be/str7kmcRMY8?feature=shared&t=64
(TheBeautyofMath)
Video Solution 2
~savannahsolver
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. 
