Difference between revisions of "2014 AMC 12A Problems/Problem 2"
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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 == | == 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>. | ||
| Line 4: | Line 13: | ||
<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> | ||
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<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> | ||
| − | + | ==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 
adult tickets and 
child tickets is 
. How much would 
adult tickets and 
child tickets cost?
Solution
Suppose 
is the price of an adult ticket. The price of a child ticket would be 
.
Plug in for 8 adult tickets and 6 child tickets.
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. 
