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

Revision as of 08:14, 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 $\boxed{\textbf{(C)}\ 45}.$

2017 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions

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

@@ Line 2: / Line 2: @@
 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\qquad{(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==

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

Revision as of 08:14, 31 March 2018

Problem 17

Solution

See Also