SOLUTION: A newspaper carrier has $4.10 in change. He has four more quarters than dimes but two times as many nickels as quarters. How many coins of each type does he have?

Question 1050880: A newspaper carrier has $4.10 in change. He has four more quarters than dimes but two times as many nickels as quarters. How many coins of each type does he have?
Answer by ikleyn(52863) (Show Source): You can put this solution on YOUR website!
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A newspaper carrier has $4.10 in change. He has four more quarters than dimes but two times as many nickels as quarters.
How many coins of each type does he have?
~~~~~~~~~~~~~~~`

Let Q be the number of quarters.
Then the number of dimes is (Q-4)  ("He has four more quarters than dimes")
and the number of nickels is 2Q.

Quarters contribute 25*Q     cents to the total.
Dimes    contribute 10*(Q-4) cents to the total.
Nickels contribute   5*(2Q)  cents to the total.

The "value" equation is 

25Q + 10*(Q-4) + 5*(2Q) = 410 cents,   or

25Q + 10Q - 40 + 10Q = 410,   or

25Q + 10Q + 10Q = 410 + 40,  or

45Q = 450   --->  Q =  = 10.

Answer.  10 quarters, 25-4 = 21 dimes and 20 nickels.

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.

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