Word Problems Involving Money:
The promoter of a track meet engages a 6,000 seat armory. He needs to gross $15,000. The price of children's tickets is to be one-half the price of adults' tickets. If there is one-third of the crowd is children, what should be the price of tickets, supposing capacity attendance?
Solution:
Step 1. Let x = Price of an Adult Ticket (in dollars)
Step 2. Then,
x/2 = Price of a child ticket (in dollars)
1/3 (6,000) = 2. 000 = Number of children's Tickets
6,000 - 2, 000 = 4, 000 = Number of Adults' Tickets
Step 3. Gross Income = (Number of Children's Tickets times their Unit Price) + (Number of Adults' Tickets times their Unit Price)
$ 15,000 = 2,000(x/2) + 4, 000(x)
Step 4. Solving for x:
15,000 = 2,000 (x/2) +4,000 (x)
15,000 1,000x 4,000x
15,000 5,000x
x $3.00
Solving for the other unknown:
x/2 = Price of a Child's (in dollars)
x/2 = $3.00/2
x/2 = $1.50
Answers:
Price of Adults' Tickets = $3.00
Price of Children's Tickets = $1.50
Step 5. The price of children's tickets is one-half the price of adults' tickets.
1/2 ($ 3. 00) = $1. 50
The gross is $15,000.
4,000($3.00) + 2,000($1.50) = $12,000 + $3,000 = $15,000
Therefore, the answers check.