Difference between revisions of "2014 AMC 12A Problems/Problem 2"
(added box+MAA notice) |
Mathgeek2006 (talk | contribs) 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}} |
Revision as of 18:02, 10 March 2015
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.