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

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Suppose <math>x</math> is the price of an adult ticket. The price of a child ticket would be <math>\frac{x}{2}</math>.\\
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==Problem==
\begin{eqnarray}
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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\\
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x = 3.50\\
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<math>\textbf{(A) }\$35\qquad
\end{eqnarray}
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\textbf{(B) }\$38.50\qquad
Plug in for 8 adult tickets and 6 child tickets.\\
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\textbf{(C) }\$40\qquad
\begin{equation}
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\textbf{(D) }\$42\qquad
8x + 6(x/2) = 8(3.50) + 3(3.50) = \framebox{38.50}
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\textbf{(E) }\$42.50</math>
\end{equation}
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 +
== 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\\
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x &=& 3.50\\
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\end{eqnarray*}</cmath>
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 +
Plug in for 8 adult tickets and 6 child tickets.
 +
 
 +
<cmath>\begin{eqnarray*}
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8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\
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&=&\boxed{\textbf{(B)}\ \ 38.50}\\
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\end{eqnarray*}</cmath>
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==See Also==
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{{AMC12 box|year=2014|ab=A|num-b=1|num-a=3}}
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{{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 (ProblemsAnswer KeyResources)
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

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