SOLUTION: A man has a collection of dimes and quarters with a total of $3.50. If he has 7 more dimes than quarters, how many of each coins does he have?

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Question 1152747: A man has a collection of dimes and quarters with a total of $3.50. If he has 7 more dimes than quarters, how many of each coins does he have?
Found 2 solutions by math_helper, Alan3354:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!


q = number of quarters

d = number of dimes = q+7





25q + 10(q+7) = 350   (cents)

35q + 70 = 350

35q = 280

q = 280/35 = 8  --->  d = 15



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Ans: 8 quarters and 15 dimes

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Check:

 25(8) + 10(15) = 200 + 150 = 350  (= $3.50) (ok)

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A man has a collection of dimes and quarters with a total of $3.50. If he has 7 more dimes than quarters, how many of each coins does he have?
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Set the "7 more dimes" aside.
Now it's pairs of 1Q & 1D worth $2.80
280/35 = 8 pairs.
+ 7 dimes ---> 8Q & 15D