Lesson Santa Claus helps solving coin problem

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Santa Claus helps solving coin problem

Problem 1

In a collection of nickels, dimes and quarters, there are twice as many dimes as nickels and 3 fewer quarters than dimes.
If the total value of the coins is $4.50, how many of each type of coin are there

Solution

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* Santa Claus came to us to help solving this problem *
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He took 3 quarters from his pocket and added it to the collection.
The "value" of the collection became $4.50 + 3*$0.25 = $5.25 = 525 cents.

Now we can combine all the coins in groups in a way that each group contains 2 dimes, 1 nickel and 2 quarters.

Each group worth is 2*10 + 5 + 2*25 = 75 cents.

How much groups do we have now?

 = 7.

OK. Then 7 groups contain 2*7 = 14 dimes, 7 nickels and 2*7 = 14 quarters.

Now we need to return 3 quarters to Santa.

Thank you, Santa !!

Answer. The collection has 14 dimes, 7 nickels and 14-3 = 11 quarters.

In this calculation we can not make a mistake.
Therefore, please check it on your own if the solution is correct.

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    Thank you Santa again !!
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My other lessons on coin word problems in this site are
    - 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
    - Advanced coin problems
    - Solving coin problems mentally by grouping without using equations
    - Non-typical coin problems
    - Swapping dimes and quarters
    - We easily deal with four unknowns
    - OVERVIEW of lessons on coin word problems

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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