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In this problem, the numbers are so small that you can easily solve it MENTALLY.
For a moment, put away four nickels from the collection.
Then you have equal number of nickels and dimes, and you can group them in pairs, each containing one nickel and one dime.
Then each pair is worth 5 + 10 = 15 cents, while the entire reduced collection is worth 125 - 4*5 = 105 cents.
Hence, the number of such pairs is = 7.
Thus, you have 7 dimes in the original collection, and 7 + 4 = 11 nickels.
Answer. 11 nickels and 7 dimes: 11*5 + 7*10 = 55 + 70 = 125 cents in all. ! Correct !
Solved.
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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
- 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.