Question 979964
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
<b>n = q + 1</b>


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
<b>d = 11 - 3q</b>


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


11 - 3q = q + 3
-4q = -8
<b>q = 2 </b>


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


d = 2 + 3 
<b>d = 5</b>


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


d = n + 2
5 = n + 2
<b>n = 3</b>


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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.