# SOLUTION: A collection of 20 coins made up of only nickels, dimes and quarters has a total value of \$3.35. If the dimes were nickels, the nickels were quarters and the quarters were dimes, t

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Coins -> SOLUTION: A collection of 20 coins made up of only nickels, dimes and quarters has a total value of \$3.35. If the dimes were nickels, the nickels were quarters and the quarters were dimes, t      Log On

 Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Word Problems: Coins Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Word Problems With Coins Question 263282: A collection of 20 coins made up of only nickels, dimes and quarters has a total value of \$3.35. If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of \$2.75. How many quarters are in the original collection?Answer by ankor@dixie-net.com(15746)   (Show Source): You can put this solution on YOUR website!Let the original number of coins be n, d, q : Write an equation for each statement: : "A collection of 20 coins" n + d + q = 20 : "made up of only nickels, dimes and quarters has a total value of \$3.35." .05n + .10d + .25q = 3.35 : "If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of \$2.75." .25n + .05d + .10q = 2.75 : Multiply the above equation by 4, subtract from the 1st equation 1n + 1d + 1q = 20 1n +.2d +.4q = 11 ---------------------subtraction eliminates n .8d + .6q = 9 : Multiply the 2nd equation by 20, subtract the 1st equation 1n + 2d + 5q = 67 1n + 1d + 1q = 20 -----------------------subtraction eliminates n again d + 4q = 47 : Use elimination again on the these two, 2 unknown equations Multiply the above equation by 4, multiply the 1st by 5, subtract it 4d + 16q + 188, 4d + 3q = 45 ----------------------subtraction eliminates d, find q 13q = 143 q = q = 11 : How many quarters are in the original collection? 11 quarters originally