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 "2013 AMC 12B Problems/Problem 10"

(→‎Solution 2)
m (→‎Solution 2)
Line 13: Line 13:
 
==Solution 2==
 
==Solution 2==
  
−
Let <math>\text{x}</math> denote the number of visits to the first booth and <math>\text{y}</math> denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows:
+
Let <math>x</math> denote the number of visits to the first booth and <math>y</math> denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows:
−
 
+
<cmath>R(x,y)=-2x+y+75</cmath>
−
<math>\text{R(x,y)=-2x+y+75}</math>
+
<cmath>B(x,y)=x-3y+75</cmath>
−
<math>\text{B(x,y)=x-3y+75}</math>
+
There are no legal exchanges when he has fewer than <math>2</math> red coins and fewer than <math>3</math> blue coins, namely when he has <math>1</math> red coin and <math>2</math> blue coins. We can then create a system of equations:
−
 
+
<cmath>1=-2x+y+75</cmath>
−
There are no legal exchanges when he has fewer than 2 red coins and fewer than 3 blue coins, namely when he has 1 red coin and 2 blue coins. We can then create a system of equations:
+
<cmath>2=x-3y+75</cmath>
−
 
+
Solving yields <math>x=59</math> and <math>y=44</math>. Since he gains one silver coin per visit to each booth, he has <math>x+y=44+59=\boxed{\textbf{(E)}103}</math> silver coins in total.
−
<math>\text{1=-2x+y+75}</math>
 
−
<math>\text{2=x-3y+75}</math>
 
−
 
 
−
Solving yields <math>\text{x=59}</math> and <math>\text{y=44}</math>. Since he gains one silver coin per visit to each booth, he has <math>\text{x+y=44+59=103}</math> silver coins in total. <math>\boxed{\textbf{(E)}103}</math>
 
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=9|num-a=11}}
 
{{AMC12 box|year=2013|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:15, 27 January 2015

The following problem is from both the 2013 AMC 12B #10 and 2013 AMC 10B #17, so both problems redirect to this page.

Problem

Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?

$\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103$

Solution 1

We can approach this problem by assuming he goes to the red booth first. You start with $75 \text{R}$ and $75 \text{B}$ and at the end of the first booth, you will have $1 \text{R}$ and $112 \text{B}$ and $37 \text{S}$. We now move to the blue booth, and working through each booth until we have none left, we will end up with:$1 \text{R}$, $2 \text{B}$ and $103 \text{S}$. So, the answer is $\boxed{\textbf{(E)}103}$


Solution 2

Let $x$ denote the number of visits to the first booth and $y$ denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows: \[R(x,y)=-2x+y+75\] \[B(x,y)=x-3y+75\] There are no legal exchanges when he has fewer than $2$ red coins and fewer than $3$ blue coins, namely when he has $1$ red coin and $2$ blue coins. We can then create a system of equations: \[1=-2x+y+75\] \[2=x-3y+75\] Solving yields $x=59$ and $y=44$. Since he gains one silver coin per visit to each booth, he has $x+y=44+59=\boxed{\textbf{(E)}103}$ silver coins in total.

See also

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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=2013_AMC_12B_Problems/Problem_10&oldid=67066"