Jane has $4.50 in pennies, nickels and dimes. She has twice as many pennies as she has nickels and dimes combined. How many of each coin does she have?
does it say how many coins total
To solve this problem, let's assign variables to represent the number of each type of coin Jane has.
Let:
P = number of pennies
N = number of nickels
D = number of dimes
From the information given, we can deduce the following equations:
1. Jane has $4.50 in total. We convert this amount to cents since we are dealing with coins:
Total value in cents = 450
2. Jane has twice as many pennies as she has nickels and dimes combined:
P = 2 * (N + D)
3. The value of the pennies, nickels, and dimes combined is equal to the total value:
P + 5N + 10D = 450
Now, we can use these equations to solve for the number of each coin.
First, substitute the value of P from equation 2 into equation 3:
2 * (N + D) + 5N + 10D = 450
Simplify this equation:
2N + 2D + 5N + 10D = 450
7N + 12D = 450
Now, we need to find integer solutions for N and D that satisfy this equation.
There are several ways to do this, such as substitution or trial-and-error. One efficient method is to consider the possible values for N and then calculate the corresponding value of D.
Let's start by assigning a value to N and calculating D:
If N = 0, then 7 * 0 + 12D = 450
12D = 450
D = 37.5
Since D must be an integer, this value does not work. Therefore, N cannot be 0.
If N = 1, then 7 * 1 + 12D = 450
7 + 12D = 450
12D = 443
D ≈ 36.9
Again, D is not an integer, so N cannot be 1.
We can continue this process, incrementing N by 1 each time until we find an integer solution for D.
If N = 2, then 7 * 2 + 12D = 450
14 + 12D = 450
12D = 436
D ≈ 36.3
Still not an integer, so we continue.
If N = 3, then 7 * 3 + 12D = 450
21 + 12D = 450
12D = 429
Finally, we have an integer value for D:
D = 35
Now that we have the values for N and D, we can substitute them back into equation 2 to find P:
P = 2 * (N + D)
P = 2 * (3 + 35)
P = 2 * 38
P = 76
Jane has 76 pennies, 3 nickels, and 35 dimes.