Question 54362
Natalie has some nickels, Dirk has some dimes, and Quincy has some quateres. Dirk has five more dimes than Qunicy had quarters. If Natalie gives Dirk a nickel, Dirk gives Qunciy a dime, and Quincy gives Natalie a quater, they will all have the same amount of money. How many coins did each have originally?
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Originally: N = value of the nickels; D = value of the dimes; Q = value of the quarters
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"Natalie gives Dirk a nickel," 
Resulting values: N-.05, D+.05
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"Dirk gives Quincy a dime,"
Resulting values: (D+.05) - .10 = (D-.05); (Q + .10)
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"Quincy gives Natalie a quarter,"
Resulting values: (Q+.10) - .25 = (Q-.15); (N-.05)+ .25 = (N+.20)
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The final values are: (N+.20), (D-.05), (Q-.15)
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The final values are the same so let's call that amt x:
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Using this we put the original values in terms of x:
          N = x-.20;  D = x + .05;  Q = x + .15
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Now to find the number of coins of each type:
No. of nickels: (x-.20)/.05;
No. of dimes: (x+.05)/.10;
No. of quarters: (x+.15)/.25
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We can write an equation in terms of x using the statement:
"Dirk has five more dimes than Qunicy had quarters"
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(x+.05)/.10 = (x+.15)/.25 + 5
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Get rid of the denominators; mult eq by 5 and we have
50(x+.05) = 20(x+.15) + 25
50x + 2.5 = 20x + 3 + 25
50x - 20x = 28 - 2.5
      30x = 25.5
        x = 25.5/30
        x = .85 is the final value for all three
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Using: N = x-.20;  D = x + .05;  Q = x + .15: to find the original values
       N = .65;   D = .90; Q = 1.00
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Find the number of each coin:
       .65/.05 = 13 nickels; .90/.10 = 9 dimes; 1.00/.25 = 4 quarters
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Note that we have 5 more dimes than quarters as it says.
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A long process and there is probably an easier way, but I hope you understood the process I used.