SOLUTION: A newspaper carrier has $12.00 in change. He has two more quarters than dimes, but three times as many nickels as dimes. How many coins of each type does he have.

Question 1126810: A newspaper carrier has $12.00 in change. He has two more quarters than dimes, but three times as many nickels as dimes. How many coins of each type does he have.
Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
.

Nickels + dimes + quarters   = 1200   cents  total


5*(3D)  + 10D   + (D + 2)*25 = 1200   <<<---=== It is the money equation;  "D" stands for dimes.


15D + 10D + 25D + 50 = 1200


50D = 1200 - 50 = 1150


D = 1150/50 = 23.


Answer.  23 dimes;  3*23 = 69 nickels  and  23+2 = 25 quarters.


Check.   23*10 + 5*69 + 25*25 = 1200 cents = $12.00.   ! Correct !

Solved.

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On coin problems, see the lessons
    - 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
    - Three methods for solving standard (typical) coin word problems
    - More complicated coin problems
    - Solving coin problems mentally by grouping without using equations
    - Santa Claus helps solving coin problem
    - OVERVIEW of lessons on coin word problems
in this site.

You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.

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.

The referred lessons are the part of this online textbook under the topic "Coin problems".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

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