Difference between revisions of "2014 AMC 12A Problems/Problem 2"
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
| + | At the theater children get in for half price. The price for <math>5</math> adult tickets and <math>4</math> child tickets is <math>24.50</math>. How much would <math>8</math> adult tickets and <math>6</math> child tickets cost? | ||
| + | |||
| + | <math>\textbf{(A) }35\qquad | ||
| + | \textbf{(B) }38.50\qquad | ||
| + | \textbf{(C) }40\qquad | ||
| + | \textbf{(D) }42\qquad | ||
| + | \textbf{(E) }42.50</math> | ||
| + | |||
| + | |||
== Solution == | == 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>. | ||
Revision as of 20:03, 7 February 2014
Problem
At the theater children get in for half price. The price for 
adult tickets and 
child tickets is 
. How much would 
adult tickets and 
child tickets cost?
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)
