Difference between revisions of "2017 AMC 10A Problems/Problem 2"

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==Solution==
 
==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 <math>5+5+3=13</math> popsicles, so the answer is <math>\boxed{(\textbf{D})13}</math>.
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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+3=13</math> popsicles, so the answer is <math>\boxed{(\textbf{D })13}</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 22:00, 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?

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

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

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