.
Assume for a minute that all quarters are of the same charge as nickels, i.e. 5 cents (instead of 25 cents).
Then the full charge of 80 coins would be 5*80 = 400 cents = $4.00.
It is $3.60 less than $7.60, and the difference is due to the difference 25c-5c = 20c in charge between quarters and nickels.
So, we need to add/(to compensate) 20c for each quarter. How many times ? = 18 times.
It gives you the
Answer. 18 quarters and 80 - 18 = 62 nickels are there in the collection.
Check. 18*25 + 62*5 = 760 cents. ! Correct !
Solved // mentally.
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For coin problems and their detailed solutions see the lessons in this site:
- 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
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 attentively and become an expert in this field.
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.