SOLUTION: Roberto changes a $5 bill for dimes and
quarters. He received 8 more dimes than
quarters. How many coins of each type did
he receive?
Algebra.Com
Question 1083843: Roberto changes a $5 bill for dimes and
quarters. He received 8 more dimes than
quarters. How many coins of each type did
he receive?
Found 2 solutions by Boreal, addingup:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
quarters are x
dimes are x+8
.10(x+8)+.25x=5.00
.10x+.80+.25x=5.00
.35x=4.20
x=12 quarters, or $3.00
x+8=20 dimes, or $2.00
add to $5.00
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
q = d-8
0.1d+0.25(d-8) = 5
0.35d-2 = 5
d = 7/0.35 = 20
q = d-8 = 20-8 = 12
----------------------
Check:
20 x 0.1 = 2
12 x 0.25 = 3
2+3 = 5 Correct
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