.
Let he has Q quarters; then the number of nickels is (33-Q).
Nickels contribute 5*(33-Q) cents to the total.
Quarters contribute 25*Q cents to the total.
In all, there are
5*(33-Q) + 25Q = 685 cents
5*33 - 5Q + 25Q = 685
20Q = 685 - 5*33 = 520
Q = = 26.
Answer. 26 quarters and 33-26 = 7 nickels.
Check. 26*25 + 7*5 = 685 cents. ! Correct !
Solved.
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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.
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.