.75) which equals .50 + .50 or .00. Algebraic word problems
involving money are solved using this general relationship following the same five basic steps
for solving any algebraic word problems.
The promoter of a track meet engages a 6,000 seat armory. He wants to gross
,000. The price of childrens tickets is to be one-half the price of adults
tickets. If one-third of the crowd is children, what should be the price of tickets,
assuming capacity attendance?
Let x = Price of an Adult Ticket (in dollars)
= Price of a Childs Ticket (in
= Number of Childrens Tickets
6,000 - 2,000 = 4,000 = Number of Adults Tickets