WORD PROBLEMSAlgebraWordProblemsInvolvingMoneyThe five basic steps for solving algebraic word problems can be used for solving word problemsinvolving money. Writing algebraic expressions for these problems depends on the generalrelationship between the total value and the unit value of money. The total value of a collectionof money or a collection of items with a certain monetary value equals the sum of the numbersof 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(
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.01) + 3(
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.05) + 4($.10) + 2(
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.25)x =
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.05 +
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.15 +
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.40 +
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.50x =
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.10The total value of 25 tickets worth
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.50 each and 30 tickets worth
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.75 each is25(
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.50) + 30(
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.75) which equals .50 + .50 or .00. Algebraic word problemsinvolving money are solved using this general relationship following the same five basic stepsfor 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 (inx2dollars)= Number of Children’s Tickets13(6,000)2,0006,000 - 2,000 = 4,000 = Number of Adults’ TicketsMA-02 Page 50 Rev. 0
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