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

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Therefore, <math>6c+3 = 9c-18.</math> This implies that <math>c = 7.</math> We therefore have <math>g = 45.</math> So, our answer is <math>\boxed{\textbf{(C)}\ 45}</math>.
 
Therefore, <math>6c+3 = 9c-18.</math> This implies that <math>c = 7.</math> We therefore have <math>g = 45.</math> So, our answer is <math>\boxed{\textbf{(C)}\ 45}</math>.
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==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
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https://youtu.be/JvgeBrx9Q0U
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~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==

Revision as of 18:01, 1 April 2023

Problem

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?

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

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: \[9c-18 = g\] \[6c+3 = g\]

We do this because for 9 chests there are 2 empty and if 9 were in each 9 multiplied by 2 is 18 left.

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

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/JvgeBrx9Q0U

~Education, the Study of Everything

Video Solution

https://youtu.be/PxO6VxSHD9A

https://youtu.be/vmg51kO7LKg

https://youtu.be/DkVbXdBAYeg

~savannahsolver

See Also

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

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