SOLUTION: Phineas has $1.15 worth of coin in his pocket. He has three more dimes than quarters and two more dimes than nickels. How much does he have in each coin?

Question 979964: Phineas has $1.15 worth of coin in his pocket. He has three more dimes than quarters and two more dimes than nickels. How much does he have in each coin?
Answer by RicaMath(8) (Show Source): You can put this solution on YOUR website!
This problem can be solved by writing a few algebraic equations.

Let's define the following variables:
d = number of dimes
n = number of nickels
q = number of quarters

The problem states that the sum of all of these coins is $1.15. Translating this into an equation, you get 0.1d + 0.05n + 0.25q = 1.15 (multiply each variable by the monetary value of each coin)

Let's convert the rest of the statements into equations.

He has three more dimes than quarters
d = q + 3 (the number of dimes equals 3 more than the number of quarters)

...and two more dimes than nickels.
d = n + 2 (the number of dimes equals 2 more than the number of nickels)

Next, we solve for each variable by doing a series of substitutions. We can equate the two equations above together to temporarily remove d, and then solve for one of the variables. For this, I will solve for n.

q + 3 = n + 2
n = q + 1

Now, I will plug in this value of n into the first equation and solve for d.

0.1d + 0.05n + 0.25q = 1.15
0.1d + 0.05(q + 1) + 0.25q = 1.15
0.1d + 0.05q + 0.05 + 0.25q = 1.15
0.1d + 0.3q = 1.1
0.1d = 1.1 - 0.3q
d = 11 - 3q

I can then plug this into the equation d = q + 3 and solve for q.

11 - 3q = q + 3
-4q = -8
q = 2

Now that I know q = 2, I can plug this into d = q + 3 to solve for d.

d = 2 + 3
d = 5

Last step is to take d = 5 and solve for n.

d = n + 2
5 = n + 2
n = 3

---

To check the answer, we can plug these values into the original equation:

0.1d + 0.05n + 0.25q = 1.15
0.1(5) + 0.05(3) + 0.25(2) = 1.15
0.5 + 0.15 + 0.5 = 1.15
1.15 = 1.15

The equation checks out!

Final answer: Phineas has 2 quarters, 5 dimes, and 3 nickels.
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