Word problems involving money, Mathematics

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.

Posted Date: 2/9/2013 4:48:22 AM | Location : United States







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