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Difference between revisions of "2018 AMC 10A Problems/Problem 8"

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==Solution==
 
==Solution==
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Let <math>x</math> be the number of 5-cent stamps that Joe has. Therefore, he must have <math>(x+3)</math> 10-cent stamps and <math>(23-(x+3)-x)</math> 25-cent stamps. Since the toal value of his collection is 320 cents, we can write
+
Let <math>x</math> be the number of 5-cent stamps that Joe has. Therefore, he must have <math>(x+3)</math> 10-cent stamps and <math>(23-(x+3)-x)</math> 25-cent stamps. Since the total value of his collection is 320 cents, we can write
−
<cmath>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</cmath>
+
\begin{align*}
 +
5x+10(x+3)+25(23-(x+3)-x) &=320 \\
 +
5x+10(x+3)+25(20-2x) &=320 \\
 +
5x+10x+30+500-50x &=320 \\
 +
35x &=210 \\
 +
x &=6
 +
\end{align*}
 
Joe has 6 5-cent stamps, 9 10-cent stamps, and 8 25-cent stamps. Thus, our answer is  
 
Joe has 6 5-cent stamps, 9 10-cent stamps, and 8 25-cent stamps. Thus, our answer is  
 
<math>8-6=\boxed{2}</math>
 
<math>8-6=\boxed{2}</math>

Revision as of 17:12, 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?

$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

Solution

Let $x$ be the number of 5-cent stamps that Joe has. Therefore, he must have $(x+3)$ 10-cent stamps and $(23-(x+3)-x)$ 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 \\ 5x+10(x+3)+25(20-2x) &=320 \\ 5x+10x+30+500-50x &=320 \\ 35x &=210 \\ x &=6 \end{align*} Joe has 6 5-cent stamps, 9 10-cent stamps, and 8 25-cent stamps. Thus, our answer is $8-6=\boxed{2}$

~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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