Fred has thirteen coins. Together they're worth $1.65. One of them ia a fifty-cent piece. What are the other coins?
1.65 - .50 is 1.15.
If you have 11 dimes you'll get 1.10, then add a nickel to make it 1.15. So the twelve remaining coins are 11 dimes and a nickel.
1-fifty cent piece .50
2-Quarters .50
3-dimes .30
7-nickels .35
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13 coins 1.65
To determine the other coins that Fred has, we can use algebraic equations. Let's assign variables to the value of each type of coin. Let's assume that Fred has x number of coins worth $0.01, y number of coins worth $0.05, z number of coins worth $0.10, and w number of coins worth $0.25.
We know that there is a fifty-cent piece, which is equivalent to 50 cents or $0.50. So, we can write the equation:
0.05y = 0.50 (Equation 1)
The total value of the coins is $1.65, which can be expressed as:
0.01x + 0.05y + 0.10z + 0.25w = 1.65 (Equation 2)
We also know that Fred has thirteen coins, which can be written as:
x + y + z + w = 13 (Equation 3)
Now we can solve these equations to find the values of x, y, z, and w.
Using Equation 1, we can solve for y:
0.05y = 0.50
y = 0.50 / 0.05
y = 10
So, Fred has 10 coins worth $0.05.
Now let's substitute this value into Equations 2 and 3:
0.01x + 0.05(10) + 0.10z + 0.25w = 1.65 (Equation 4)
x + 10 + z + w = 13 (Equation 5)
Simplifying Equation 4, we have:
0.01x + 0.50 + 0.10z + 0.25w = 1.65
Now, solving Equation 5 for x:
x = 13 - (+ 10 + z + w)
x = 3 - z - w
Substituting this value in Equation 4, we have:
0.01(3 - z - w) + 0.50 + 0.10z + 0.25w = 1.65
0.03 - 0.01z - 0.01w + 0.50 + 0.10z + 0.25w = 1.65
0.03 + 0.50 - 0.01z + 0.10z - 0.01w + 0.25w = 1.65
(0.03 + 0.50) + (0.10z - 0.01z) + (0.25w - 0.01w) = 1.65
0.53 + 0.09z + 0.24w = 1.65
Now, let's simplify this equation:
0.09z + 0.24w = 1.65 - 0.53
0.09z + 0.24w = 1.12
Now, we have two equations:
0.09z + 0.24w = 1.12 (Equation 6)
x + y + z + w = 13 (Equation 7)
By solving Equations 6 and 7 simultaneously, you can find the values of z and w, which will give you the remaining types of coins that Fred has.