SOLUTION: Jill has $3.50 in nickels and dimes.If she has 50 coins ,how many of each type of coin does she have?

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Jill has $3.50 in nickels and dimes.If she has 50 coins ,how many of each type of coin does she have?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 123702: Jill has $3.50 in nickels and dimes.If she has 50 coins ,how many of each type of coin does she have?
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
N+D=50 OR N=50-D
.05N+.10D=3.50
.05(50-D)+.10D=3.50
2.5-.05D+.10D=3.50
.05D=3.50-2.50
.05D=2
D=1/.05
D=20 NUMBER OF DIMES.
N+20=50
N=50-20
N=30 NUMBER OF NICKELS.
PROOF
.05*30+.10*20=3.50
1.5+2=3.5
3.5=3.5