Art of Problem Solving

2017 AMC 10A Problems/Problem 2: Difference between revisions

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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 $\$1$ 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 $\$8$?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$

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 $5+5+3=13$ popsicles, so the answer is $\boxed{\textbf{(D) }13}$.

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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