SOLUTION: John has 100 coins made up of nickels and dimes in his pocket that equal $7.85 how many nickels and dimes does he have

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Question 1095347: John has 100 coins made up of nickels and dimes in his pocket that equal $7.85 how many nickels and dimes does he have
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52802) About Me  (Show Source):
You can put this solution on YOUR website!
.
5N + 10*(100-N) = 785  cents  ====>

5N + 1000 - 10N = 785  ====>  -5N = 785 - 1000 = -215  ====>  N = %28-215%29%2F%28-5%29 = 43.


Answer.  43 nickels and 100 - 43 = 57 dimes.

Check.   43*5 + 57*10 = 785 cents.   ! Correct !

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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.



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!

You can solve this kind of problem informally, with logical analysis. The actual calculations you will need to do are nearly identical to what you do in the formal algebraic solution, as provided by another tutor.

The method I use is first to suppose what the total amount would have been if all 100 coins had been dimes. That's easy: $10.00.

But the actual total was only $7.85; simple arithmetic shows that is $2.15 short of $10.00.

But the difference between the value of a dime and a nickel is 5 cents. So the number of nickels there must be the $2.15, divided by 5 cents:
2.15%2F0.05+=+215%2F5+=+43

So there are 43 nickels, leaving 57 for the number of dimes.