SOLUTION: A newspaper carrier has ​$4.35 in change. He has six more quarters than dimes but two times as many nickels as quarters. How many coins of each type does he​ have?

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Question 1146739: A newspaper carrier has ​$4.35 in change. He has six more quarters than dimes but two times as many nickels as quarters. How many coins of each type does he​ have?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!






SUBSTITUTE:
----------solve this for q.

Answer by ikleyn(52898)   (Show Source): You can put this solution on YOUR website!
.

            This problem is to be solved using one single equation in one unknown.

            I will show you on how to do it.

            The major step is to choose the leading unknown reasonably.


Let x be the number of quarters.

Then the number of dimes is (x-6), and the number of nickels is 2x.



By knowing it, you can write the "money" equation momentarily


    5*(2x) + 10*(x-6) + 25x = 435   cents.



Simplify


    10x + 10x - 60 + 25x = 435

    45x = 435 + 60 = 495

      x            =  = 11.


ANSWER.  11 quarters;  11-6 = 5 dimes,  and  2*11 = 22 nickels.


CHECK.   5*22 + 5*10 + 25*11 = 435 cents,  which is precisely correct (!)

Solved.

------------------

The lesson to learn from my post is  THIS :

    This problem can be solved using one single equation in one unknown.


    It is designed and intended to be solved in this way.


    It is intended for 6-7 grade students who are not familiar yet with systems of equations.


To see many other similar solved problems,  look into the lessons
    - More complicated coin problems
in this site.



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