On July 7, 2004, Shania Twain was scheduled to perform at the TD Waterhouse Cent

Question

On July 7, 2004, Shania Twain was scheduled to perform at the TD Waterhouse Centre in Orlando, Florida. The center offered 18,039 seats for the concert in three seating classifications: floor, lower, and upper. Based upon their location, tickets were offered at three different prices: $80, $65, and $45.† Suppose the average price of a floor ticket was $80, the average price of a lower ticket was $65, and the average price of an upper ticket was $45. If concert planners expected a total of 7,000 tickets would be sold, including 5 times as many tickets on the lower level as on the floor, how many of each type of ticket would have to be sold to reach $382,500 in ticket revenue?

Explanation / Answer

Let No of Floor tickets sold be x, Lower ticket be y and upper ticket be z.

Since the total tickets sold was 7000, we have x+y+z = 7000------------------------------------(i)

Now it says that Lower tickets were 5 times that of Floor i.e. y = 5x--------------------------(ii)

Now considering the pricing for the tickets as given in the question, we have

80x + 65y + 45z = 382500 ; Divinding this equation by 5 on both sides we have 16x + 13y + 9z = 76500----(iii)

Now multiplying eq (i) by 9 on both sides we get 9x + 9y + 9z = 63000 ---------------------------------------------------(iv)

Subracting eq (iv) from eq (iii), we get

7x + 4y = 13500
Substituting y = 5x from eq(ii) in the above equation we get,

7x + 20x = 13500
27x = 13500
x = 500

Since y = 5x; y = 2500

Substituting the values of x and y in eq (i), we get z = 4000

Ans) The no of floor tickets sold were:
Floor Ticket : 500
   Lower Ticket : 2500
   Upper Ticket : 4000

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