Jane has six more nickels than quarters and two fewer dimes than nickels. If she has a total of $2.70, how many of each type of coin does she have?
n=q+6
d=n-2
5n+10d+25q = 270
To solve this problem, we'll set up a system of equations based on the given information.
Let's denote the number of quarters as q, the number of dimes as d, and the number of nickels as n.
From the problem, we have the following information:
1. Jane has six more nickels than quarters, which can be written as: n = q + 6.
2. Jane has two fewer dimes than nickels, which can be written as: d = n - 2.
3. The total value of the coins is $2.70. The value of a quarter is $0.25, a dime is $0.10, and a nickel is $0.05. The equation for the total value is: 0.25q + 0.10d + 0.05n = 2.70.
Now we have a system of equations:
n = q + 6
d = n - 2
0.25q + 0.10d + 0.05n = 2.70
To solve this system, we can use substitution.
1. Substitute n in terms of q in equation 3:
0.25q + 0.10d + 0.05(q + 6) = 2.70
2. Simplify and solve for d:
0.25q + 0.10d + 0.05q + 0.30 = 2.70
0.30q + 0.10d = 2.40
3q + d = 24 (multiplied both sides by 10 to eliminate decimals)
3. Substitute d in terms of q in equation 2:
n - 2 = d
n = d + 2
q + 6 = d + 2
q = d - 4
4. Substitute q in terms of d in equation 3:
3(d - 4) + d = 24
3d - 12 + d = 24
4d - 12 = 24
4d = 36
d = 9
5. Substitute d = 9 into q = d - 4:
q = 9 - 4
q = 5
6. Substitute q = 5 into n = q + 6:
n = 5 + 6
n = 11
Therefore, Jane has 5 quarters, 9 dimes, and 11 nickels.