SOLUTION: Joseph has 4 times as many quarters as he does dimes. If he has a total of $2.65, how many of each type of coin does he have?

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Question 62147: Joseph has 4 times as many quarters as he does dimes. If he has a total of $2.65, how many of each type of coin does he have?
Found 2 solutions by joyofmath, stanbon:
Answer by joyofmath(189) About Me  (Show Source):
You can put this solution on YOUR website!
Joseph has 4 times as many quarters as he does dimes. If he has a total of $2.65, how many of each type of coin does he have?
Joseph has Q quarters and D dimes.
We know that Q=4D.
We know Joseph has 265 centers, so 25Q%2B10D+=+265.
Replace Q with 4D, and we get 25%2A4%2AD%2B10D+=+265.
Or, 110D+=+265.
This makes D not a whole number so there's something wrong with the way the problem is worded.
To verify there's something no right with the problem:
Joseph has 4 times as many quarters as dimes.
If he has 4 quarters and 1 dime, he'll have $1.10.
If he has 8 quarters and 2 dimes, he'll have $2.20.
If he has 12 quarters and 3 dimes, he'll have $3.30, which is more than $2.65.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Joseph has 4 times as many quarters as he does dimes. If he has a total of $2.65, how many of each type of coin does he have?
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Let the number of dimes be "x"; Value of these is 10x cents
The number of quarters is "4x"; Value of these is 25(4x)=100x cents
EQUATION:
value + value = 265 cets
10x+100x=265
110x=265
x=2.4...
This is not possible as "x" is a count of coins.
Check your problem data and your posting.
Cheers,
Stan H.