Question 1189913
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            It can be solved using arithmetic only,  without  Algebra.



<pre>
Since the combined cost of one advance ticket and one some-day tickets is $55,

30 pairs of such combined tickets cost  30*55 = 16750 dollars.


It means that remaining 40-30 = 10 same day tickets cost 1900-1650 = 250  dollars.


Hence, one same-day ticket costs 250/10 = 25 dollars.


It means that each advanced ticket costs 55-25 = 30 dollars.
</pre>

Solved &nbsp;(mentally).