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(52903)   (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.

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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.

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?

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?