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

(Solution)
(Solution)
Line 7: Line 7:
  
 
\begin{align*}
 
\begin{align*}
5x+10(x+3)+25(23-(x+3)-x) &=320 \\
+
5x+10(x+3)+25(23-(x+3)-x) &= 320 \\
5x+10(x+3)+25(20-2x) &=320 \\
+
5x+10(x+3)+25(20-2x) &= 320 \\
5x+10x+30+500-50x &=320 \\
+
5x+10x+30+500-50x &= 320 \\
35x &=210 \\
+
35x &= 210 \\
x &=6
+
x &= 6
 
\end{align*}
 
\end{align*}
  

Revision as of 17:14, 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 (ProblemsAnswer KeyResources)
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

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

Invalid username
Login to AoPS