SOLUTION: Franklin has $3.95 in nickels, dimes, and quarters. If he has three times as many dimes as quarters, and five fewer nickels than dimes, how many coins of each type does he have?

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Question 1065219: Franklin has $3.95 in nickels, dimes, and quarters. If he has three times as many dimes as quarters, and five fewer nickels than dimes, how many coins of each type does he have?
All I have is:
3d=q
d=n-5

Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52812)   (Show Source): You can put this solution on YOUR website!
.
Franklin has $3.95 in nickels, dimes, and quarters. If he has three times as many dimes as quarters, and five fewer nickels than dimes,
how many coins of each type does he have?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let Q be # of quarters.

Then the number of dimes is 3Q, the number of nickels is (3Q-5).

Your "value" equation is

5*(3Q-5) + 10*(3Q) + 25Q = 395.   Or

15Q - 25 + 30Q + 25Q = 395.

Can you complete the solution from this point ?


Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

Franklin has $3.95 in nickels, dimes, and quarters. If he has three times as many dimes as quarters, and five fewer nickels than dimes, how many coins of each type does he have?
All I have is:
3d=q
d=n-5
Your equations are incorrect!
Let the number of quarters be Q
Then number of dimes = 3Q
Number of nickels = 3Q - 5
We then get: .25Q + .1(3Q) + .05(3Q - 5) = 3.95
.25Q + .3Q + .15Q - .25 = 3.95
.7Q = 3.95 + .25
.7Q = 4.2
Q, or number of quarters =
Number of dimes:
Number of nickels:
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