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2014 AMC 12A Problems/Problem 2: Difference between revisions

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==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 ==
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>.


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


<math>x &= 3.50</math>
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>


<math>8x + 6(x/2) = 8(3.50) + 3(3.50) = 38.50</math>
==See Also==
{{AMC12 box|year=2014|ab=A|num-b=1|num-a=3}}
{{MAA Notice}}

Latest revision as of 16:18, 25 July 2023

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

These problems are copyrighted © by the Mathematical Association of America.

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