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.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.
Example 1:
The promoter of a track meet engages a 6,000 seat armory. He wants to gross
,000. The price of children’s 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?
Solution:
Step 1.
Let x = Price of an Adult Ticket (in dollars)
Step 2.
Then,
= Price of a Child’s Ticket (in
x
2
dollars)
= Number of Children’s Tickets
1
3
(6,000)
2,000
6,000 - 2,000 = 4,000 = Number of Adults’ Tickets
MA-02
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