Word Problems Involving Money

WORD PROBLEMS Algebra Word Problems Involving Money The five basic steps for solving algebraic word problems can be used for solving word problems involving money. Writing algebraic expressions for these problems depends on the general relationship between the total value and the unit value of money. The total value of a collection of money or a collection of items with a certain monetary value equals the sum of the numbers of items each multiplied by their unit values. Thus, the total value of five pennies, three nickels, four dimes, and two quarters is found by solving the following equation: x = 5($0.01) + 3($0.05) + 4($.10) + 2($0.25) x = $0.05 + $0.15 + $0.40 + $0.50 x = $1.10 The total value of 25 tickets worth $1.50 each and 30 tickets worth $0.75 each is 25($1.50) + 30($0.75) which equals $37.50 + $22.50 or $60.00. 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 $15,000. 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 Page 50 Rev. 0