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Revenue problem
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didieravoaka
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GMATGuruNY
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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.
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Matt@VeritasPrep
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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
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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ceilidh.erickson
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To Mitch's point, this is a WEIGHTED AVERAGE. One way to easily visualize this is to think of a scale / seesaw:
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:
So, half of 300 is 150.
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:
So, half of 300 is 150.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
