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Revenue problem

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by GMATGuruNY » Thu Mar 10, 2016 2:38 pm
Average ticket price = (total revenue)/(total number of tickets) = 4500/300 = $15.

Since the average ticket price is HALFWAY between the price of a child ticket ($10) and the price of an adult ticket ($20), an EQUAL NUMBER of each ticket type must have been sold.
To illustrate:
If 1 child ticket and 1 adult ticket are sold, the average price for the 2 tickets = (10 + 20)/2 = 30/2 = 15.
If 2 child tickets and 2 adult tickets are sold, the average price for the 4 tickets = (2*10 + 2*20)/4 = 60/4 = 15.
If 3 child tickets and 3 adult tickets are sold, the average price for the 6 tickets = (3*10 + 3*20)/6 = 90/6 = 15.

Thus:
Of the 300 tickets, 150 were adult tickets, and 150 were child tickets.

The correct answer is C.
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by Matt@VeritasPrep » Thu Mar 17, 2016 9:41 pm
Algebra works too:

a + c = 300

20a + 10c = 4500

To find c, multiply the top equation by 20: 20a + 20c = 6000

then subtract the second from the first:

(20a + 20c = 6000) - (20a + 10c = 4500) =>

10c = 1500 =>

c = 150
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by ceilidh.erickson » Fri Mar 18, 2016 9:09 am
To Mitch's point, this is a WEIGHTED AVERAGE. One way to easily visualize this is to think of a scale / seesaw:

Image

If all 300 tickets sold were for children, the total would be $3000. If all 300 tickets were for adults, the total would be $6000. We can tell how many of each ticket we sell by seeing where the total falls between 3000 and 6000. If closer to 3000, more children's were sold; if closer to 6000, more adults' were sold.

Since the actual total of $4500 is exactly halfway between 3000 and 6000, there must have been an equal number of each sold:

Image

So, half of 300 is 150.
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