SOLUTION: The nickles and dimes all fell on the floor. There were 12 more nickles than dimes, and the total value of the coins was $5.10. How many nickles and how many dimes were on the floo

Question 1079471: The nickles and dimes all fell on the floor. There were 12 more nickles than dimes, and the total value of the coins was $5.10. How many nickles and how many dimes were on the floor?
Answer by ikleyn(52832) (Show Source): You can put this solution on YOUR website!
.

Take off these 12 nickels (mentally).


Then you will have equal number of nickels and dimes and the updated total value of coins is $5.10 - $0.05*12 = $4.50.


Now you can group the remaining coins in pairs containing 1 nickel in 1 dime in each pair.
Each pair worth is 5 + 10 = 15 cents.


How many pairs do you have now ? -  = 30.


So, initially there were 30 dimes and 30 + 12 = 42 nickels.


Answer.  30 dimes and 42 nickels.

Solved.

There is entire bunch of 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
    - Solving coin problems mentally by grouping without using equations
    - Santa Claus helps solving coin problem
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".

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