SOLUTION: A pile of 55 coins, consisting of nickels and dimes, is worth $3.90. Find the number of each.

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Question 1093877: A pile of 55 coins, consisting of nickels and dimes, is worth $3.90. Find the number of each.

Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Been done at least 100's of times.
Look it up.

Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.
Solution   (using system of 2 equations)

 N +   D =  55   coins   (1)   (N stands for nickels,
5N + 10D = 390   cents   (2)    D stands for dimes)


Simplify (2) by canceling the factor 5 in both sides:

 N  + 2D =  78           (3)


Subtract eq(1) from eq(3). You will get

D = 78 - 55 = 23.


Answer.  23 dimes and 55-23 = 32 nickels.

Check.   23*10 + 32*5 = 390 cents.  ! correct ! 


For coin problems and their detailed solutions see the lessons in this site:
    - 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
    - Solving coin problems mentally by grouping without using equations
    - Santa Claus helps solving coin problem

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 attentively and become an expert in this field.

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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