SOLUTION: Sam found a number of nickels, dimes, and quarters in his room. He found three more dimes than nickels but twice as many quarters as dimes. The total value of the coins was $5.05

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Question 120339: Sam found a number of nickels, dimes, and quarters in his room. He found three more dimes than nickels but twice as many quarters as dimes. The total value of the coins was $5.05. How many coins of each type did Sam find?
Found 2 solutions by solver91311, ankor@dixie-net.com:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Let's say he found n nickels, d dimes, and q quarters.

We know the following things:

d=n%2B3 because he found three more dimes than nickels

q=2d because he found twice as many quarters as dimes

The value of the nickels he found is 5n cents, the value of the dimes is 10d cents, and the value of the quarters is 25q cents. And since $5.05 is equal to 505 cents, we can write:

5n%2B10d%2B25q=505

We can substitute n + 3 for d in q=2d to get q=2%28n%2B3%29 so now we have expressions for d and q in terms of n. We can substitute these expressions into the value equation to get a single equation in one variable.

5n%2B10%28n%2B3%29%2B25%282%28n%2B3%29%29=505

Now simplify by distributing and collecting like terms:

5n%2B10n%2B30%2B50n%2B150=505

65n%2B180=505

Add -180 to both sides

65n%2B180-180=505-180
65n=325

Divide by 65

65n%2F65=325%2F65
n=5

Now we know that he found 5 nickels. Three more than 5 is 8, so he found 8 dimes. Twice that is 16, so he found 16 quarters.

Check the answer:

16 quarters = $4.00
8 dimes = $0.80
5 nickels = $0.25

4.00+%2B+0.80+%2B+0.25=5.05, answer checks.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Sam found a number of nickels, dimes, and quarters in his room.
:
He found three more dimes than nickels
n = d - 3
:
but twice as many quarters as dimes.
q = 2d
:
The total value of the coins was $5.05.
.05n + .10d + .25q = 5.05
:
How many coins of each type did Sam find?
:
Using the 1st and 2nd equations, substitute (d-3) for n; and 2d for q
.05(d-3) + .10d + .25(2d) = 5.05
:
.05d - .15 + .10d + .50d = 5.05
:
.05d + .10d + .50d = 5.05 + .15
:
.65d = 5.20
:
d = 5.2/.65
:
d = 8 dimes
:
n = 8 - 3
n = 5 nickels
:
q = 2(8)
q = 16 quarters
:
:
Check solution in the value equation
.05(5) + .10(8) + .25(16) =
.25 + .80 + 4.0 = 5.05; confirms our solutions