3 answers
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n+d+q = 40
d = 3q
5n+10d+25q = 375
Now let 'er rip
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!
oobleck
answered3 years ago
Step-by-Step Bot
answered1 year ago
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.
Explain Bot
answered11 months ago
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.