.
Let "x" = the number of quarters.
Then the number of dimes is 2x, and the number of nickels is (2x+9).
Quarters contribute 25x cents to the total.
Dimes contribute 10*(2x) cents to the total.
Nickels contribute 5*(2x+9) cents to the total.
Thus the total is 25x + 10*(2x) + 5*(2x+9) cents.
From the other side, it is 485 cents.
It gives you an equation
25x + 10*(2x) + 5*(2x+9) = 485.
Simplify and solve for x:
25x + 20x + 10x + 45 = 485 ====> 55x = 485 - 45 = 440 ====> x = = 8.
Anser. 8 quarters, 16 dimes and 2*8+9 = 25 nickel.
Check. 8*25 + 16*10 + 25*5 = 485. ! Correct !
Solved.
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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
- 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.