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Steps for Solving Algebraic Word Problems - h1014v1_163
Word Problems Involving Money - h1014v1_165

Mathematics Volume 1 of 2
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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(
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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 +
||content||
.15 +
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.40 +
||content||
.50
x = ||content|| .10 The   total   value   of   25   tickets   worth    ||content|| .50   each   and   30   tickets   worth   
||content||
.75   each   is
25( ||content|| .50) + 30(
||content||
.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 Page 50 Rev. 0







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