2017 AMC 8 Problems/Problem 17: Difference between revisions

Revision as of 10:14, 5 February 2018

Problem 17

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
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All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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

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2017 AMC 8 Problems/Problem 17: Difference between revisions

Revision as of 10:14, 5 February 2018

Problem 17

Solution

See Also