Difference between revisions of "2014 AMC 12A Problems/Problem 2"

Revision as of 18:43, 7 February 2014

Suppose $x$ is the price of an adult ticket. The price of a child ticket would be $\frac{x}{2}$ .

$5x + 4(x/2) = 7x = 24.50$

Plug in for 8 adult tickets and 6 child tickets.

$8x + 6(x/2) = 8(3.50) + 3(3.50) = 38.50$

@@ Line 1: / Line 1: @@
-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>.
-\begin{eqnarray}
-x + 4(x/2) = 7x = 24.50\\
+<math>5x + 4(x/2) = 7x = 24.50</math>
-x = 3.50\\
-\end{eqnarray}
+                <math>x = 3.50</math>
-Plug in for 8 adult tickets and 6 child tickets.\\
-\begin{equation}
+Plug in for 8 adult tickets and 6 child tickets.
-x + 6(x/2) = 8(3.50) + 3(3.50) = \framebox{38.50}
-\end{equation}
+<math>8x + 6(x/2) = 8(3.50) + 3(3.50) = 38.50</math>