SOLVING SYSTEMS:
A bag contains nickels, dimes and quarters having a value of $3.75. If there are 40 coins in all and 3 times as many dimes as quarters, how many coins of each kind were there?
-Thank you!
n+d+q = 40
d = 3q
5n+10d+25q = 375
Now let 'er rip
To solve this problem, we'll set up a system of equations based on the given information.
Let's assume:
N = number of nickels
D = number of dimes
Q = number of quarters
1. The total value of the coins is $3.75:
0.05N + 0.10D + 0.25Q = 3.75
2. The total number of coins is 40:
N + D + Q = 40
3. There are three times as many dimes as quarters:
D = 3Q
Now, we can solve this system of equations using substitution or elimination method.
Using substitution method:
From equation 3, we can substitute D with 3Q in equations 1 and 2.
0.05N + 0.10(3Q) + 0.25Q = 3.75
N + (3Q) + Q = 40
Simplifying these equations:
0.05N + 0.30Q + 0.25Q = 3.75
N + 4Q = 40
Combining like terms:
0.05N + 0.55Q = 3.75
N + 4Q = 40
To eliminate decimals, we can multiply the first equation by 100:
5N + 55Q = 375
N + 4Q = 40
Now we can solve this system of equations using elimination or substitution method.
Let's use elimination method:
Multiply the second equation by 5:
5N + 20Q = 200
Now we can subtract the equation (N + 4Q = 40) from the equation (5N + 20Q = 200):
(5N + 20Q) - (N + 4Q) = (200 - 40)
4N + 16Q = 160
Divide this equation by 4 to simplify it further:
N + 4Q = 40
N + 4Q = 40
N = 40 - 4Q
Substitute N = 40 - 4Q into the equation 4N + 16Q = 160:
4(40 - 4Q) + 16Q = 160
160 - 16Q + 16Q = 160
160 = 160
This means that N can be any value since 160 = 160. So, the number of nickels (N) is not determined.
However, we still can calculate the number of dimes (D) and quarters (Q) using the equation D = 3Q.
Therefore, the number of dimes would be 3Q and the number of quarters would be Q.
To solve this problem, let's start by assigning variables to the unknown quantities.
Let's say:
- The number of nickels is N
- The number of dimes is D
- The number of quarters is Q
Now, let's create the equations based on the given information.
1. The total value of the coins is $3.75:
The value of each nickel is $0.05, so the total value of the nickels is 0.05N.
The value of each dime is $0.10, so the total value of the dimes is 0.10D.
The value of each quarter is $0.25, so the total value of the quarters is 0.25Q.
Therefore, we can write the equation: 0.05N + 0.10D + 0.25Q = 3.75
2. There are 40 coins in total:
Therefore, we can write the equation: N + D + Q = 40
3. There are 3 times as many dimes as quarters:
This means D = 3Q
Now we have a system of three equations with three variables:
0.05N + 0.10D + 0.25Q = 3.75 (Equation 1)
N + D + Q = 40 (Equation 2)
D = 3Q (Equation 3)
To solve this system, we can use the method of substitution or elimination. Let's use substitution here:
From Equation 3, we know that D = 3Q. We can substitute this value into Equation 2:
N + 3Q + Q = 40
N + 4Q = 40
Rearranging Equation 2, we have:
N = 40 - 4Q
Now substitute the values of N and D (from Equations 3 and 2) into Equation 1:
0.05(40 - 4Q) + 0.10(3Q) + 0.25Q = 3.75
2 - 0.20Q + 0.30Q + 0.25Q = 3.75
0.35Q = 3.75 - 2
0.35Q = 1.75
Q = 1.75 / 0.35
Q = 5
Now substitute the value of Q back into Equation 3 to find the value of D:
D = 3 * 5
D = 15
Finally, substitute the values of D and Q into Equation 2 to find N:
N + 15 + 5 = 40
N = 40 - 20
N = 20
Therefore, there were 20 nickels, 15 dimes, and 5 quarters in the bag.