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Difference between revisions of "2013 AMC 12B Problems/Problem 10"

(→‎See also: Rearranged order for consistency)
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<math>\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103</math>
 
<math>\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103</math>
  
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==Solution 1==
+
== Solution 1==
 +
 
 +
If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first combination Combo 1, and the second combination Combo 2.
 +
 
 +
In other words, Alex can exchange 5 red tokens for 4 silver tokens as long as he has at least 6 red tokens, and Alex can exchange 5 blue tokens for 3 silver tokens as long as he has at least 6 blue tokens. Hence doing as many Combo 1 and Combo 2's as possible, we end up with 5 red, 5 blue, and 98 silver tokens.
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 +
Finally, Alex can visit the blue booth once, then do Combo 1, then visit the blue booth once more to end up with 1 red token, 2 blue tokens, and <math>\boxed{\textbf{(E)}\ 103}</math> silver tokens, at which point it is clear he cannot use the booths anymore.
 +
 
 +
Solution 2==
  
 
We can approach this problem by assuming he goes to the red booth first. You start with <math>75 \text{R}</math> and <math>75 \text{B}</math> and at the end of the first booth, you will have <math>1 \text{R}</math> and <math>112 \text{B}</math> and <math>37 \text{S}</math>. We now move to the blue booth, and working through each booth until we have none left, we will end up with:<math>1 \text{R}</math>, <math>2 \text{B}</math> and <math>103 \text{S}</math>. So, the answer is <math>\boxed{\textbf{(E)}103}</math>
 
We can approach this problem by assuming he goes to the red booth first. You start with <math>75 \text{R}</math> and <math>75 \text{B}</math> and at the end of the first booth, you will have <math>1 \text{R}</math> and <math>112 \text{B}</math> and <math>37 \text{S}</math>. We now move to the blue booth, and working through each booth until we have none left, we will end up with:<math>1 \text{R}</math>, <math>2 \text{B}</math> and <math>103 \text{S}</math>. So, the answer is <math>\boxed{\textbf{(E)}103}</math>
  
  
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==Solution 2==
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==Solution 3==
  
 
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:
 
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:

Revision as of 00:07, 2 January 2019

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

If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first combination Combo 1, and the second combination Combo 2.

In other words, Alex can exchange 5 red tokens for 4 silver tokens as long as he has at least 6 red tokens, and Alex can exchange 5 blue tokens for 3 silver tokens as long as he has at least 6 blue tokens. Hence doing as many Combo 1 and Combo 2's as possible, we end up with 5 red, 5 blue, and 98 silver tokens.

Finally, Alex can visit the blue booth once, then do Combo 1, then visit the blue booth once more to end up with 1 red token, 2 blue tokens, and $\boxed{\textbf{(E)}\ 103}$ silver tokens, at which point it is clear he cannot use the booths anymore.

Solution 2==

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 3

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 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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
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

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