SOLUTION: Number problems. Jill has #3.50 in nickles and dimes. If she has 50 coins, how many of each type of coins does she have?

Question 87873: Number problems. Jill has #3.50 in nickles and dimes. If she has 50 coins, how many of each type of coins does she have?

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
n = number of nickels; d = number of dime
:
Write a simple equation for each statement
:
"Jill has #3.50 in nickels and dimes."
.05n + .10d = 3.50
:
"If she has 50 coins,"
n + d = 50
Rearrange for substitution; subtract d from both sides:
n = (50 - d)
:
how many of each type of coins does she have?
:
Substitute (50-d) for n in the 1st equation, solve for d:
.05n + .10n = 3.50
.05(50-d) + .10d = 3.50
2.5 - .05d + .10d = 3.50
-.05d + .10d = 3.50 - 2.50
.05d = 1.00
d = 1/.05
d = 20 dimes
Then
50 - 20 = 30 nickels:
:
:
Check solutions:
.05(30) + .10(20) = 3.50
:
That's not hard, is it?
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