SOLUTION: 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

Question 54362This question is from textbook Algebra Structure and Method
: 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 orignially? This question is from textbook Algebra Structure and Method

Found 2 solutions by josmiceli, ankor@dixie-net.com:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
n = number of Natalies nickels
d = number of Dirks dimes
q = number of Quincys quarters
5n + 10d + 25q = the sum of all their coins
d = q + 5
5n - 5 + 25 = 10d + 5 - 10 = 25q - 25 + 10
5n + 20 = 10d - 5 = 25q - 15
since d = q + 5
10(q + 5) - 5 = 25q - 15
10q + 45 = 25q - 15
15q = 60
q = 4 answer
d = q + 5
d = 9 answer
5n + 20 = 10d - 5
5n = 10*9 - 5 - 20
5n = 65
n = 13 answer
5n + 10d + 25q = the sum of all their coins
5*13 + 10*9 + 25*4 = 255
each person ends up with 255/3 = 85
5n + 20 = 5*13 + 20 = 85
10d - 5 = 10*9 - 5 = 85
25q - 15 = 25*4 - 15 = 85
so they all end up with an equal share

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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?
:
Originally: N = value of the nickels; D = value of the dimes; Q = value of the quarters
:
"Natalie gives Dirk a nickel,"
Resulting values: N-.05, D+.05
:
"Dirk gives Quincy a dime,"
Resulting values: (D+.05) - .10 = (D-.05); (Q + .10)
:
"Quincy gives Natalie a quarter,"
Resulting values: (Q+.10) - .25 = (Q-.15); (N-.05)+ .25 = (N+.20)
:
The final values are: (N+.20), (D-.05), (Q-.15)
:
The final values are the same so let's call that amt x:
:
Using this we put the original values in terms of x:
N = x-.20; D = x + .05; Q = x + .15
:
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
:
We can write an equation in terms of x using the statement:
"Dirk has five more dimes than Qunicy had quarters"
:
(x+.05)/.10 = (x+.15)/.25 + 5
:
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
:
Using: N = x-.20; D = x + .05; Q = x + .15: to find the original values
N = .65; D = .90; Q = 1.00
:
Find the number of each coin:
.65/.05 = 13 nickels; .90/.10 = 9 dimes; 1.00/.25 = 4 quarters
:
Note that we have 5 more dimes than quarters as it says.
:
A long process and there is probably an easier way, but I hope you understood the process I used.

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