Difference between revisions of "2018 AMC 10A Problems/Problem 8"
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==Solution== | ==Solution== | ||
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 | 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 | ||
| − | \begin{align*} | + | <math>\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 \\ | ||
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35x &=210 \\ | 35x &=210 \\ | ||
x &=6 | x &=6 | ||
| − | \end{align*} | + | \end{align*}</math> |
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 17:13, 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
$\begin{align*}
5x+10(x+3)+25(23-(x+3)-x) &=320 \\
5x+10(x+3)+25(20-2x) &=320 \\
5x+10x+30+500-50x &=320 \\
35x &=210 \\
x &=6
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
