Difference between revisions of "2014 AMC 12A Problems/Problem 2"
| Line 1: | Line 1: | ||
| + | == Solution == | ||
Suppose <math>x</math> is the price of an adult ticket. The price of a child ticket would be <math>\frac{x}{2}</math>. | Suppose <math>x</math> is the price of an adult ticket. The price of a child ticket would be <math>\frac{x}{2}</math>. | ||
| − | < | + | <cmath>\begin{eqnarray*} |
| + | 5x + 4(x/2) = 7x &=& 24.50\\ | ||
| − | + | x &=& 3.50\\ | |
| + | \end{eqnarray*}</cmath> | ||
Plug in for 8 adult tickets and 6 child tickets. | Plug in for 8 adult tickets and 6 child tickets. | ||
| − | < | + | <cmath>\begin{eqnarray*} |
| + | |||
| + | 8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\ | ||
| + | &=&\boxed{\textbf{(B)}\ \ 38.50}\\ | ||
| + | |||
| + | \end{eqnarray*}</cmath> | ||
Revision as of 18:57, 7 February 2014
Solution
Suppose 
is the price of an adult ticket. The price of a child ticket would be 
.
\begin{eqnarray*}
5x + 4(x/2) = 7x &=& 24.50\\
x &=& 3.50\\
\end{eqnarray*} (Error compiling LaTeX. Unknown error_msg)
Plug in for 8 adult tickets and 6 child tickets.
\begin{eqnarray*}
8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\
&=&\boxed{\textbf{(B)}\ \ 38.50}\\
\end{eqnarray*} (Error compiling LaTeX. Unknown error_msg)
