SOLUTION: 56) Number problems. Jill has $3.50 in nickels and dimes. If she has 50 coins, how many of each type of coin does she have?

Algebra ->  Linear-equations -> SOLUTION: 56) Number problems. 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


   

Question 89733: 56) Number problems. 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 checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
n+d=50 or n=50-d
5n+10d=350 now substitute (50-d) for n in this equation & solve for d:
5(50-d)+10d=350
250-5d+10d=350
5d=350-250
5d=100
d=20 number of dimes.
50-20=30 number of nickles.
proof
5*30=10*20=350
150+200=350
350=350