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) About Me  (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) About Me  (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 = highlight_green%28matrix%281%2C3%2C+4.2%2F.7%2C+or%2C+6%29%29

Number of dimes: highlight_green%28matrix%281%2C3%2C+3%286%29%2C+or%2C+18%29%29

Number of nickels: highlight_green%28matrix%281%2C5%2C+3%286%29+-+5%2C+or%2C+18+-+5%2C+or%2C+13%29%29