Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AMC 12A Problems/Problem 2"

m (→‎Solution 2)
(→‎Solution 3: Deleted the repetitive solution. For this problem, I think two solutions are sufficient. Many students will take Sol 2.)
Line 28: Line 28:
  
 
~MRENTHUSIASM (Reconstruction)
 
~MRENTHUSIASM (Reconstruction)
−
 
−
== Solution 3 ==
 
−
 
−
The ratio of dollar per pound is greatest for the <math>5</math> pound rock, then the <math>4</math> pound, lastly the <math>1</math> pound. So we should take two <math>5</math> pound rocks and two <math>4</math> pound rocks. The total value, in dollars, is <math>2\cdot14+2\cdot11=\boxed{\textbf{(C) } 50}.</math>
 
−
 
−
~steakfails
 
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2018|ab=A|num-b=1|num-a=3}}
 
{{AMC12 box|year=2018|ab=A|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:36, 24 August 2021

Problem

While exploring a cave, Carl comes across a collection of $5$-pound rocks worth $$14$ each, $4$-pound rocks worth $$11$ each, and $1$-pound rocks worth $$2$ each. There are at least $20$ of each size. He can carry at most $18$ pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?

$\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52$

Solution 1

Since each rock is worth $1$ dollar less than $3$ times its weight (in pounds), the answer is just $3\cdot 18=54$ minus the minimum number of rocks we need to make $18$ pounds. Note that we need at least $4$ rocks (two $5$-pound rocks and two $4$-pound rocks) to make $18$ pounds, so the answer is $54-4=\boxed{\textbf{(C) } 50}.$

~Kevindujin (Solution)

~MRENTHUSIASM (Revision)

Solution 2

The value of $5$-pound rocks is $$14\div5=$2.80$ per pound, and the value of $4$-pound rocks is $$11\div4=$2.75$ per pound. Intuitively, we wish to maximize the number of $5$-pound rocks and minimize the number of $1$-pound rocks. We have two cases:

  1. We get three $5$-pound rocks and three $1$-pound rocks, for a total value of $$14\cdot3+$2\cdot3=$48.$

  2. We get two $5$-pound rocks and two $4$-pound rocks, for a total value of $$14\cdot2+$11\cdot2=$50.$

Clearly, Case 2 produces the maximum total value. So, the answer is $\boxed{\textbf{(C) } 50}.$

Remark

Note that an upper bound of the total value is $$2.80\cdot18=$50.40,$ from which we can eliminate choices $\textbf{(D)}$ and $\textbf{(E)}.$

~Pyhm2017 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

See Also

2018 AMC 12A (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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_2&oldid=160895"