SOLUTION: Mark has 56 coins totalling $7.10. If he has only dimes and quarters, how many of each coin does he have?

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Mark has 56 coins totalling $7.10. If he has only dimes and quarters, how many of each coin does he have?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 189313: Mark has 56 coins totalling $7.10. If he has only dimes and quarters, how many of each coin does he have?
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
Q+D=56 OR Q=56-D
.25Q+.10D=7.10
.25(56-D)+.10D=7.10
14-.25D+.10D=7.10
-.15D=7.10-14
-.15D=-6.90
D=-6.90/-.15
D=46 IS THE NUMBER OF DIMES.
56-46=10 IS THE NUMBER OF QUARTERS.
PROOF:
.25*10+.10*46=7.10
2.50+4.60=7.10
7.10=7.10