.
For a minute, take off 8 dimes from the collection.
Then your updated collection consists of equal number of dimes and quarters and is worth 850-80 = 770 cents.
But now you can combine your coins in pairs containing (1 dime + 1 quarter) each.
Each pair is worth 35 cents (= 10 + 25), so the number of such pairs is = 22.
Answer. There are 22 quarters and 22+8 = 30 dimes in the original collection.
Solved.
As you see, I did not forget to return back that 8 dimes I took off at the beginning.
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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.