Difference between revisions of "2018 AMC 10A Problems/Problem 8"
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\begin{align*} | \begin{align*} | ||
5x+10(x+3)+25(23-(x+3)-x)= 320 \\ | 5x+10(x+3)+25(23-(x+3)-x)= 320 \\ | ||
| − | &\Rightarrow 5x+10(x+3)+25(20-2x)= 320 \\ | + | & \Rightarrow 5x+10(x+3)+25(20-2x)= 320 \\ |
| − | &\Rightarrow 5x+10x+30+500-50x= 320 \\ | + | & \Rightarrow 5x+10x+30+500-50x= 320 \\ |
| − | &\Rightarrow 35x= 210 \\ | + | & \Rightarrow 35x= 210 \\ |
| − | &\Rightarrow x= 6 \ | + | & \Rightarrow x= 6 \ |
\end{align*} | \end{align*} | ||
Revision as of 17:17, 8 February 2018
Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?
Solution
Let 
be the number of 5-cent stamps that Joe has. Therefore, he must have 
10-cent stamps and 
25-cent stamps. Since the total value of his collection is 320 cents, we can write
\begin{align*} 5x+10(x+3)+25(23-(x+3)-x)= 320 \\ & \Rightarrow 5x+10(x+3)+25(20-2x)= 320 \\ & \Rightarrow 5x+10x+30+500-50x= 320 \\ & \Rightarrow 35x= 210 \\ & \Rightarrow x= 6 \ \end{align*}
Joe has 6 5-cent stamps, 9 10-cent stamps, and 8 25-cent stamps. Thus, our answer is
~Nivek
See Also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
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