.
= = 6 nickels and 8 dimes.
Take off 2 dimes for a minute.
Your total will be 110-2*10 = 90 cents,
and they are comprised of equal number of dimes and nickels.
It is why I divided this difference of 90 cents by (5+10) cents, obtaining 6 groups, each containing 1 nickel and 1 dime.
Solved.
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Surely, you can solve it algebraically, and you must be able to do it !
But when the problem is so simple, the MENTAL solution simply asks and invokes you to use it ! !
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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
- 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.