Monica Collects quarters, dimes, nickels and pennies. This afternoon she took $ 1.00 from her collection. She counted the coins and found that she had twice as many nickels as dimes. What coins could Monica have?
Let's solve this problem step-by-step.
Step 1: Assume the number of dimes
Let's assume that Monica has x number of dimes.
Step 2: Calculate the number of nickels
We know that Monica has twice as many nickels as dimes. So, the number of nickels would be 2x.
Step 3: Calculate the value of dimes
Each dime is worth $0.10. So the value of x dimes would be 0.10x.
Step 4: Calculate the value of nickels
Each nickel is worth $0.05. So the value of 2x nickels would be 0.05 * 2x = 0.10x.
Step 5: Calculate the total value
We know that Monica took $1.00 from her collection. So, the total value would be $1.00.
Step 6: Set up the equation
The total value of the coins (0.10x + 0.10x) should equal $1.00. So the equation would be:
0.10x + 0.10x = 1.00
Step 7: Solve the equation
Combining like terms, the equation becomes:
0.20x = 1.00
Dividing both sides by 0.20 to isolate x gives:
x = 5
Step 8: Calculate the number of each type of coin
With x = 5, Monica would have 5 dimes and 2x = 2 * 5 = 10 nickels.
Step 9: Check the total value
The value of 5 dimes is 0.10 * 5 = $0.50, and the value of 10 nickels is 0.05 * 10 = $0.50. Adding these together, we get $0.50 + $0.50 = $1.00.
So, Monica could have 5 dimes and 10 nickels.
To solve this problem, let's use a system of equations. Let's denote the number of quarters, dimes, nickels, and pennies Monica has as q, d, n, and p, respectively.
Given that Monica took $1.00 from her collection, we can write the equation:
0.25q + 0.10d + 0.05n + 0.01p = 1.00 (equation 1)
Next, we know that Monica had twice as many nickels as dimes. Mathematically, we can express this as:
n = 2d (equation 2)
Now, since the problem states that Monica has a certain number of coins, we also have the equation:
q + d + n + p = total coins (equation 3)
To find the possible coin combinations, we need integer values for q, d, n, and p that satisfy all the equations. We can start by checking different values to see which solutions work.
The total coins can range from 4 to a maximum of around 100, as having more coins will make it difficult to meet the equation for the total dollar amount.
For each value of the total coins, we can substitute eq. 2 into eq. 3 and then substitute the result into eq. 1 to solve for the possible values of q, d, and p.
Let's start with a small possibility where the total coins is 6.
Substituting eq. 2 into eq. 3, we get:
q + d + 2d + p = 6
q + 3d + p = 6
Next, we can substitute this equation into eq. 1:
0.25q + 0.10d + 0.05(2d) + 0.01p = 1.00
0.25q + 0.10d + 0.10d + 0.01p = 1.00
0.25q + 0.20d + 0.01p = 1.00
We can then proceed to solve this system of equations using various methods, such as substitution or elimination. By substituting the values of d and p from eq. 3 into eq. 1, and resolving the equations, we can find the possible values of q, d, n, and p that satisfy the given conditions.
Continuing this process, we can check other values of the total coins to find additional possible coin combinations. By iteratively solving the system of equations, we can list all the valid solutions.
Note that there are multiple valid combinations of coins that Monica could have, so this process will help identify all possible cases.