Difference between revisions of "2017 AMC 8 Problems/Problem 17"

(Problem 17)
(Problem 17)
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Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty.  So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over.  How many gold coins did I have?
 
Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty.  So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over.  How many gold coins did I have?
  
<math>\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45{(D) }63\qquad\textbf{(E) }81</math>
+
<math>\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C)}45\qquad{(D) }63\qquad\textbf{(E) }81</math>
  
 
==Solution==
 
==Solution==

Revision as of 08:13, 31 March 2018

Problem 17

Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?

$\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C)}45\qquad{(D) }63\qquad\textbf{(E) }81$

Solution

We can represent the amount of gold with $g$ and the amount of chests with $c$. We can use the problem to make the following equations: \[9c-18 = g\] \[6c+3 = g\]

Therefore, $6c+3 = 9c-18.$ This implies that $c = 7.$ We therefore have $g = 45.$ So, our answer is $\boxed{\textbf{(C)}\ 45}.$

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
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 AJHSME/AMC 8 Problems and Solutions

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