Elliot has nickels and dimes in his pocket. He has twice as many dimes as nickels. If the total value of the coins in his pocket is $1.00, how many of each coin does he have?
assign variables to each kind of coin:
n = # of nickels
d = # of dimes
A nickel has a value of $0.05 and a dime has a value of $0.10
Since the total value of the coins is $1.00, then
(value of nickel * # of nickels) + (value of dime * # of dimes) = 1.00
(0.05 * n) + (0.10 * d) = 1.00
0.05n + 0.10d = 1.00
Since there are twice as many dimes as nickels, then the # of dimes is equal to two times the # of nickels. That is, d = 2n.
Substitute 2n for d in the equation above:
0.05n + 0.10(2n) = 1.00
0.05n + 0.20n = 1.00
Combine like term:
0.25n = 1.00
Divide both sides of the equation by 0.25 to solve for n:
0.25n/0.25 = 1.00/0.25
n = 4
Since we found that the # of dimes equals two times the # of nickels, then we solve for d as follows:
d = 2n
d = 2(4)
d = 8
Therefore, we have 4 nickels and 8 dimes.
We can check the answer by finding the total amount of each kind of coin we have them adding the two values up:
4 nickels worth $0.05 per nickel: (0.05)(4) = 0.20 ==> $0.20
8 dimes worth $0.10 per dime: (0.10)(8) = 0.80 ==> $0.80
$0.20 + $0.80 = $1.00
