.
Take off these extra 9 dimes for a minute.
Then you will have 270-90 = 180 cents in equal number dimes and nickels.
You can group the coins in sets, each containing 1 dime and one nickel.
Each group is worth 10+5 = 15 cents; so, there are = 12 such groups.
It leads you to the answer: the original collection has 12 nickels and 12+9 = 21 dimes.
Solved (mentally).
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On coin problems, see the lessons
- 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
- Non-typical coin problems
- 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.
A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.
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.