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

(added box+MAA notice)
m (Solution)
 
Line 14: Line 14:
 
<cmath>\begin{eqnarray*}
 
<cmath>\begin{eqnarray*}
 
5x + 4(x/2) = 7x &=& 24.50\\
 
5x + 4(x/2) = 7x &=& 24.50\\
 
 
x &=& 3.50\\
 
x &=& 3.50\\
 
\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
Line 21: Line 20:
  
 
<cmath>\begin{eqnarray*}
 
<cmath>\begin{eqnarray*}
 
 
8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\  
 
8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\  
 
&=&\boxed{\textbf{(B)}\ \ 38.50}\\
 
&=&\boxed{\textbf{(B)}\ \ 38.50}\\
 
 
\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2014|ab=A|num-b=1|num-a=3}}
 
{{AMC12 box|year=2014|ab=A|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest 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 (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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