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?
It can be done with either one unknown or with two unknowns,
depending of which one you are studying. Here is is both ways.
ONE UNKNOWN
Let x = the number of nickels
Then 50-x = the number of dimes (Total coins minus number of nickel)
The dollar value of the x nickels is $.05x
The dollar value of the 50-x nickels is $.10(50-x)
Sum of the dollar values must equal $3.50, so
.05x + .10(50-x) = 3.50
Solve that and get x=30, so there are 30 nickels
50-x dimes is 50-30 or 20 dimes.
TWO UNKNOWNS
There are 50 coins total, so
N + D = 50
The dollar value of the N nickels is .05N
The dollar value of the D dimes is .10N
The sum of the dollar values must equal the total dollar value 3.50
.05N + .10D = 3.50
So you have the system
N + D = 50
.05N + .10D = 3.50
Solve it and get N = 30, D = 20
Edwin
AnlytcPhil@aol.com
