Find the solution if possible. If there is not enough information to solve the problem or if it has no solution, say so. If extra information is given, identify it.
14. A collection of 30 coins worth $5.50 consists of nickels, dimes, and quarters. There are twice as many dimes as nickels. How many quarters are there?
--------------------------------------------------------------------------------------------
I'm pretty sure there's enough information to solve this problem, I just can't figure it out. I either get fractions or negative numbers as my answer. Any help is greatly appreciated!! :-)
Q+D+N=30
.25Q+.10D +.05N=5.50
D=2N
Yes, three equations, three unknowns, enought information.
Ah, ok you made everything a lot simpler. Thanks!! :-)
A collection of 30 coins worth $5.50 consists of nickels, dimes, and quarters. There are twice as many dimes as nickels. How many quarters are there?
To solve this problem, we need to set up a system of equations based on the given information.
Let's first represent the number of nickels as "N", the number of dimes as "D", and the number of quarters as "Q".
We are given that there are twice as many dimes as nickels, so we can write the equation:
D = 2N (Equation 1)
We are also given that there are a total of 30 coins, so we can write another equation:
N + D + Q = 30 (Equation 2)
Next, we know the values of each coin. A nickel is worth $0.05, a dime is worth $0.10, and a quarter is worth $0.25. The total value of the coins is $5.50, so we can write another equation based on the values:
0.05N + 0.10D + 0.25Q = 5.50 (Equation 3)
Now, we have a system of three equations with three unknowns. We can solve this system to find the values of N, D, and Q.
To make the calculations easier, let's first eliminate the decimals in Equation 3 by multiplying both sides of the equation by 100:
5N + 10D + 25Q = 550 (Equation 4)
Now, we can solve the system of equations formed by Equations 1, 2, and 4 using any method such as substitution or elimination.
Substituting Equation 1 into Equation 2, we get:
N + 2N + Q = 30
3N + Q = 30 (Equation 5)
Substituting Equation 1 into Equation 4, we get:
5N + 10(2N) + 25Q = 550
5N + 20N + 25Q = 550
25N + 25Q = 550
N + Q = 22 (Equation 6)
Now, we have a system of Equations 5 and 6, which we can solve simultaneously.
Multiplying Equation 5 by -1, we get:
-3N - Q = -30 (Equation 7)
Adding Equations 6 and 7, we eliminate the variable Q:
N + Q + (-3N) - Q = 22 + (-30)
-2N = -8
N = 4
Substituting the value of N back into Equation 6:
4 + Q = 22
Q = 18
Therefore, there are 18 quarters.
Please note that the calculations above assume that all the coins mentioned are included in the given collection. If there is any missing or extra information, this solution might not be valid.