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Question 1126127: For the opening night at the Opera House, a total of 1000 tickets were sold. Front orchestra seats cost $80 apiece, rear orchestra seats cost $60 apiece, and front balcony seats cost $50 apiece. The combined number of tickets sold for the front orchestra and rear orchestra exceeded twice the number of front balcony tickets sold by 400. The total receipts for the performance were $62,800. Determine how many tickets of each type were sold.
Found 2 solutions by solver91311, ikleyn:
Answer by solver91311(24713)   (Show Source):
Answer by ikleyn(52803)  (Show Source):
You can put this solution on YOUR website! .
From the first glance, this problem is for three equations in three unknowns.
But it can be easily reduced to TWO equations in only TWO unknowns.
The key phrase is this:
The combined number of tickets sold for the front orchestra and rear orchestra exceeded twice the number
of front balcony tickets sold by 400.
Take away for a minute these 400 extra tickets, and then you will have 1000-400 = 600 tickets, in a way, that
the combined number of tickets sold for the front orchestra and rear orchestra is exactly twice the number of front balcony tickets.
Then you momentarily conclude in your mind that the number of front balcony tickets is exactly 200,
while "the reduced" combined number of tickets sold for the front orchestra and rear orchestra is 2*200 = 400.
Returning back that 400 tickets, you obtain that combined number of tickets sold for the front orchestra
and rear orchestra is 2*200 + 400 = 800 = 1000 - 200.
Thus one third of the problem is just solved and one unknown is just found: the number of front balcony tickets is exactly 200.
Regarding the remaining part, you have now only two unknowns: the number of tickets for the front orchestra
and the number of tickets for rear orchestra.
For these two unknowns you have two equations:
- their combined number is 800, and
- the revenue for these two types of tickets is $62800 minus 200 balcony tickets multiplied by their price of $50.
Now, since you are given such a complicated problem for "almost 3 unknowns", I am sure that you can solve this STANDARD
reduced problem for 2 unknowns WITH NO PROBLEMS.
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You may think that my explanation is too long.
It is not so: this chain of logical arguments must go through you mind in 5 - 10 seconds, bringing you to the reduced system.
Moreover, this problem is SPECIALLY DESIGNED to be solved in exactly this way.
Surely, you can go by that blind way that the tutor @solver91311 suggests in his post.
But it is the "blind" way, while my solution will bring you to much higher scores (in any sense).
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There were many similar problems of this type in this forum, and I personally solved 5 - 7 - 10 of them by this method.
I created a special lesson on it in this site
- HOW TO algebreze and solve this problem on 2 equations in 2 unknowns
where you can find other similar solved problems.
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