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

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Question 998249: michael has a collection of dimes and quarters with a total value of $ 3.50. if he has 7 more dimes than quarters, how many of each coins does he have?
Found 2 solutions by addingup, ikleyn:
Answer by addingup(3677) (Show Source):
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d= q+7
He has 7 more dimes than quarters, so to make them equal you have to add 7 to the quarters.
In the equation below we will use this value for d.
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0.10d+0.25q= 3.5 Now substitute for d with the value above:
0.10(q+7)+0.25q= 3.5
0.10q+0.70+0.25q= 3.5
0.35q= 2.8
q= 2.8/0.35
q= 8 He has 8 quarters and 8+7= 15 dimes
Proof:
8*0.25= 2
15*0.10= 1.50
2+1.50= 3.50 We have the correct answer

Answer by ikleyn(52800) (Show Source):
You can put this solution on YOUR website!
.
See the lesson Coin problems in this site where similar problem was solved (Problem 2). Read attentively the solution and then substitute your numbers.

Good luck !