Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "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} 5x + 4(x/2) = 7x = 24.50\\ x = 3.50\\ \e...")
 
m (→‎Solution)
(5 intermediate revisions by 3 users not shown)
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>.\\
+
==Problem==
−
\begin{eqnarray}
+
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?
−
5x + 4(x/2) = 7x = 24.50\\
+
 
−
x = 3.50\\
+
<math>\textbf{(A) }35\qquad
−
\end{eqnarray}
+
\textbf{(B) }38.50\qquad
−
Plug in for 8 adult tickets and 6 child tickets.\\
+
\textbf{(C) }40\qquad
−
\begin{equation}
+
\textbf{(D) }42\qquad
−
8x + 6(x/2) = 8(3.50) + 3(3.50) = \framebox{38.50}
+
\textbf{(E) }42.50</math>
−
\end{equation}
+
 
 +
 
 +
== 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>.
 +
 
 +
<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.
 +
 
 +
<cmath>\begin{eqnarray*}
 +
8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\
 +
&=&\boxed{\textbf{(B)}\ \ 38.50}\\
 +
\end{eqnarray*}</cmath>
 +
 
 +
==See Also==
 +
{{AMC12 box|year=2014|ab=A|num-b=1|num-a=3}}
 +
{{MAA Notice}}

Revision as of 19:02, 10 March 2015

Problem

At the theater children get in for half price. The price for $5$ adult tickets and $4$ child tickets is $24.50$. How much would $8$ adult tickets and $6$ child tickets cost?

$\textbf{(A) }35\qquad \textbf{(B) }38.50\qquad \textbf{(C) }40\qquad \textbf{(D) }42\qquad \textbf{(E) }42.50$


Solution

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

\begin{eqnarray*} 5x + 4(x/2) = 7x &=& 24.50\\ x &=& 3.50\\ \end{eqnarray*}

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*}

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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 12 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_2&oldid=68654"