A collection of nickels dimes and quarters consist of 22 coins with the total value of 1.75. If the number of nickels is three times the number of dimes, how many of each coin are contained in the collection
To solve this problem, we can set up a system of equations based on the given information.
Let's assume that the number of nickels is "N", the number of dimes is "D", and the number of quarters is "Q."
From the problem statement, we have three pieces of information:
1. The total number of coins is 22: N + D + Q = 22
2. The total value of the coins is 1.75 dollars: 0.05N + 0.10D + 0.25Q = 1.75
3. The number of nickels is three times the number of dimes: N = 3D
Now, we can substitute equation 3 into equations 1 and 2 to eliminate one variable.
Substituting N = 3D into equation 1:
3D + D + Q = 22
4D + Q = 22 (equation 4)
Substituting N = 3D into equation 2:
0.05(3D) + 0.10D + 0.25Q = 1.75
0.15D + 0.10D + 0.25Q = 1.75
0.25D + 0.25Q = 1.75 (equation 5)
Now we have two equations (equation 4 and equation 5) with two variables (D and Q). We can solve this system to find their values.
Let's simplify equation 4 by subtracting 4D from both sides:
Q = 22 - 4D (equation 6)
Substitute the value of Q from equation 6 into equation 5:
0.25D + 0.25(22 - 4D) = 1.75
0.25D + 5.5 - D = 1.75
0.75D = 1.75 - 5.5
0.75D = -3.75
D = -3.75 / 0.75
D = -5
Uh-oh! It looks like there might be an error in the problem statement or in our calculations. The number of dimes cannot be negative.
Please double-check the information provided in the problem and try again.
n+d+q = 22
5n+10d+25q = 175
n = 3d
now crank it out