SOLUTION: Dan opened up his piggy bank and found a total of 35 coins - nickels, dimes, quarters. He has 5 more nickels than dimes and a total of $6. How many of each coin were in his bank?

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Question 1050005: Dan opened up his piggy bank and found a total of 35 coins - nickels, dimes, quarters. He has 5 more nickels than dimes and a total of $6. How many of each coin were in his bank?
dimes = d
nickels = d+5
quarters = 35-d-(d+5)
I know it is .05(d+5); .10d; I think the quarters is .25(35-2d-5),
I know the sum of all 3 coins = $6
I know to multiply by 100 to get rid of the decimals
but I'm not getting it to come out correctly.
thanks for your help!
Found 3 solutions by josmiceli, advanced_Learner, ikleyn:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +n+ = the number of nickels
Let +d+ = the number of dimes
Let +q+ = the number of quarters
----------------------------------
(1) +n+%2B+d+%2B+q+=+35+
(2) +.05n+%2B+.1d+%2B+.25q+=+6+
(3) +n+=+d+%2B+5+
----------------------------------
(2) +5n+%2B+10d+%2B+25q+=+600+
(2) +n+%2B+2d+%2B+5q+=+120+
-----------------------------------
(2) +d+%2B+5+%2B+2d+%2B+5q+=+120+
(2) +3d+%2B+5q+=+115+
-----------------------------------
(1) +d+%2B+5+%2B+d+%2B+q+=+35+
(1) +2d+%2B+q+=+30+
Multiply this result by +5+
and subtract (2) from (1)
----------------------------
(1) +10d+%2B+5q+=+150+
(2) +-3d+-+5q+=+-115+
-------------------------
+7d+=+35+
+d+=+5+
and
(3) +n+=+d+%2B+5+
(3) +n+=+5+%2B+5+
(3) +n+=+10+
and
(1) +n+%2B+d+%2B+q+=+35+
(1) +10+%2B+5+%2B+q+=+35+
(1) +q+=+20+
----------------------------
There are 10 nickels, 5 dimes, 20 quarters
---------------------------------------
(2) +.05n+%2B+.1d+%2B+.25q+=+6+
(2) +.05%2A10+%2B+.1%2A5+%2B+.25%2A20+=+6+
(2) +.5+%2B+.5+%2B+5+=+6+
(2) +6+=+6+
---------------------
I like to use a different variable for each
type of coin. All through the calculations,
I am sure of what I am looking at. Then, at the
end I can put the results all in terms of either
n, d, or q

Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
your equations are
1.N%2BD%2BQ=35
2.N=5%2BD
3.5N%2B10D%2B25Q=600


simplify
2D%2BQ=30
15D%2B25Q=575



Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
+system%28+%0D%0A++++2%5CD+%2B+1%5CQ+=+30%2C%0D%0A++++15%5CD+%2B+25%5CQ+=+575+%29%0D%0A++We'll use substitution. After moving 1*Q to the right, we get:
2%2AD+=+30+-+1%2AQ, or D+=+30%2F2+-+1%2AQ%2F2. Substitute that
into another equation:
15%2A%2830%2F2+-+1%2AQ%2F2%29+%2B+25%5CQ+=+575 and simplify: So, we know that Q=20. Since D+=+30%2F2+-+1%2AQ%2F2, D=5.

Answer: system%28+D=5%2C+Q=20+%29.

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.
Dan opened up his piggy bank and found a total of 35 coins - nickels, dimes, quarters.
He has 5 more nickels than dimes and a total of $6. How many of each coin were in his bank?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Let "d" be the number of dimes.
Then the number of nickels is d+5, according to the condition.
Then the number of quarters is (35 - d - (d+5)) = 35 - 2d -5 = 30 - 2d.

Nickels contribute 5*(d+5) cents to the total.
Dimes contribute 10d cents to the total.
Quarters contribute 25*(30-2d) cents to the total.

An equation for the total is

5*(d+5) + 10d + 25*(30-2d) = 600 cents.

Simplify it:

5d + 25 + 10d + 750 - 50d = 600,

5d + 10d - 50d = 600 - 25 - 750,

-35d = -175  --->  d = %28-175%29%2F%28-35%29 = 5.

There were 5 dimes.

Now you find all remaining unknowns.

There is entire bunch of the lessons on coin problems
    - Coin problems
    - More Coin problems
    - Solving coin problems without using equations
    - Kevin and Randy Muise have a jar containing coins
    - Typical coin problems from the archive
    - More complicated coin problems
    - OVERVIEW of lessons on coin word problems
in this site.

Read them and become an expert in solution of coin problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

It contains many other solved word problems, as well as many other interesting and useful things.