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

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==Problem==
 
==Problem==
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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?
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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?
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<math>\textbf{(A) }35\qquad
 
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\textbf{(B) }38.50\qquad
 
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\textbf{(C) }40\qquad
 
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\textbf{(D) }42\qquad
 
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\textbf{(E) }42.50</math>
 
  
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<math>\textbf{(A) }\$35\qquad
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\textbf{(B) }\$38.50\qquad
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\textbf{(C) }\$40\qquad
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\textbf{(D) }\$42\qquad
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\textbf{(E) }\$42.50</math>
  
 
== Solution ==
 
== Solution ==

Latest revision as of 17:18, 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
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All AMC 12 Problems and Solutions

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