For a recent play performance, the ticket prices were $25 per adult and $15 per child. A total of 500 tickets were sold for the performance. How many of the tickets sold were for adults?

(1) Revenue from ticket sales for this performance totaled $10,500.

(2) The average (arithmetic mean) price per ticket sold was $21.

Sol: Let A = Total no. of Adult tickets

C: Total no. of Child Tickets

Given A+C=500, we need to find A ?

Price of Adult Ticket: $25

Price of Child Ticket : $15

From St 1, we have 25*A+15*C = 10500

We also know A+C= 500

We have 2 variables and 2 equations and therefore we can solve for A. We can leave it that.

So B C and E ruled out

From St 2 we have Average price is $ 21. Refer attachment

Attachment:

WA.PNG [ 12.23 KiB | Viewed 1420 times ]
Now $ 21 is 6 $ more than Child Ticket price and $4 Less than Adult ticket price. So by Weighted Average principle.

Total difference between Adult and Child Ticket price is $ 10

Number of Adult Tickets will be: 6/10 *500 = 300 -----> A

Just for clarity purpose we can calculate St 1 as well

5A+3C= 2100-------Eq 1

A+C=500-----> 3A+3C= 1500-------> Eq 2

Subtracting 2 from1, we get 2A=600 or A =300.

Ans D

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