Question 361817
<font face="Garamond" size="+2">


At first blush, it looks like this cannot be solved with the information given because you can only write two equations where there are three variables.  But once you realize that not only do all of the values have to be positive integers, the number of child tickets must be a multiple of 10.  That is because the total amount taken in is in even dollars.


Now if 120 child tickets were sold, then the total amount taken in would be $12.  Not enough.


If 110 child tickets were sold then the most that could have been taken in was 11 plus 50 = $61.  Again, not enough.


100 child, 10 + 100 = 110.  Not enough again.


90 child, 9 + 150 = 159.  Ah ha...this has possibilities.


Let *[tex \Large x] represent adult tickets at $5 each.  Let *[tex \Large y] represent junior tickets at $2 each.


Then if 90 child tickets were sold:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 30]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ + 2y\ =\ 111]


Multiply the first equation by -2 and add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 0y\ =\ 51]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 17]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 13]


That's one possibility.


Let child tickets equal 80, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 40]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ + 2y\ =\ 112]


Multiply the first equation by -2 and add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 0y\ =\ 32]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 10\frac{2}{3}]


No good because *[tex \Large x] has to be an integer.


Let child tickets equal 70, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 50]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ + 2y\ =\ 113]


Multiply the first equation by -2 and add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 0y\ =\ 13]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 4\frac{1}{3}]


No good because *[tex \Large x] has to be an integer.


Let child tickets equal 60, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 60]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ + 2y\ =\ 114]


Multiply the first equation by -2 and add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 0y\ =\ -6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -2]


No good because *[tex \Large x] has to be a positive integer.


Any smaller number of child tickets will result in a negative number of adult tickets.


Hence the only answer is 17 adult tickets (17 times 5 is $85), 13 Junior tickets for $26, and 90 child tickets for $9.  85 + 26 + 9 = 120 and 17 + 13 + 90 = 120
Checks.





John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>