Difference between revisions of "2014 AMC 12A Problems/Problem 2"
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==Problem== | ==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? | + | 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 | ||
+ | \textbf{(B) }\$38.50\qquad | ||
+ | \textbf{(C) }\$40\qquad | ||
+ | \textbf{(D) }\$42\qquad | ||
+ | \textbf{(E) }\$42.50</math> | ||
== Solution == | == Solution == | ||
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<cmath>\begin{eqnarray*} | <cmath>\begin{eqnarray*} | ||
5x + 4(x/2) = 7x &=& 24.50\\ | 5x + 4(x/2) = 7x &=& 24.50\\ | ||
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x &=& 3.50\\ | x &=& 3.50\\ | ||
\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
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<cmath>\begin{eqnarray*} | <cmath>\begin{eqnarray*} | ||
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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}\\ | ||
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\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
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==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 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 |
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