SOLUTION: I need help with this word math problem. In a collection of nickels, dimes, and quarters, there are twice as many dimes as nickels, and fewer quarters than dimes. If the total val

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Question 300978: I need help with this word math problem.
In a collection of nickels, dimes, and quarters, there are twice as many dimes as nickels, and fewer quarters than dimes. If the total value of the coins is $4.50, how many of each type of coin are there?
Found 2 solutions by solver91311, josmiceli:
Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Impossible to derive a unique answer to this one. You left out how many fewer quarters there are than dimes.

The number of quarters ranges from

The number of dimes ranges from and is divisible by 2

And the number of nickels is half of the dimes.

The difference between the number of dimes and number of quarters is a number divisible by 3.

John

Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
Let = number of nickels
Let = number of dimes
Let = number of quarters
given:
(in cents)

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By substitution:

I know that , since quarters must add up to less than $4.50
If then, suppose , could be
That's $2.75 plus $1.00 = $3.75. If there are twice as many dimes as
nickels, , totaling $.25
$3.75 + $.25 = $4.00
My next guess is ,,
You can go on from here, since you get the idea