SOLVING SYSTEMS:
30 coins having a value of $3.30 consists of nickels, dimes and quarters. If there are twice as many quarters as dimes, how many coins of each kind were there?
-Thank you!
n = nickels
q = quarters
d = dimes
There are a total of 30 coins.
n + d + q = 30
There are twice as many quarters as dimes means:
q = 2 d
n + d + q = 30
n + d + 2 d = 30
n + 3 d = 30
1 nickel = $0.05
1 dime = $0.10
1 quarter = $0.25
A value is $3.30
0.05 ∙ n + 0.1 ∙ d + 0. 25 ∙ q = 3.30
0.05 n + 0.1 d + 0. 25 ∙ 2 d = 3.30
0.05 n + 0.1 d + 0. 50 d = 3.30
0.05 n + 0.60 d = 3.30
Divide both sides by 0.05
n + 12 d = 66
Now you must solve system:
n + 3 d = 30
n + 12 d = 66
___________
First equation:
n + 3 d = 30
Subtract 3 d to both sides
n = 30 - 3 d
Second equation:
n + 12 d = 66
Subtract 12 d to both sides
n = 66 - 12 d
n = n
30 - 3 d = 66 - 12 d
Add 12 d to both sides
30 + 9 d = 66
Subtract 30 to both sides
9 d = 36
d = 36 / 9 = 4
n = 30 - 3 d = 30 - 3 ∙ 4 = 30 - 12 = 18
q = 2 d = 2 ∙ 4 = 8
18 nickels
4 = dimes
8 = quarters
Proof:
q = 2 d
8 = 2 ∙ 4
n + d + q = 18 + 8 + 4 = 26 + 4 = 30
0.05 ∙ n + 0.1 ∙ d + 0. 25 q =
0.05 ∙ 18 + 0.1 ∙ 4 + 0. 25 ∙ 8 =
0.9 + 0.4 + 2 = 1.3 + 2 = 3.30
To solve this problem, let's follow these steps:
Step 1: Let's assume the number of dimes as "x".
Step 2: Since there are twice as many quarters as dimes, the number of quarters would be "2x".
Step 3: Now, let's calculate the value of all the dimes. Since each dime is worth $0.10, the value of dimes would be 0.10x.
Step 4: Similarly, the value of all the quarters would be 0.25 * 2x = 0.50x.
Step 5: The total number of nickels can be found by subtracting the number of dimes and quarters from the total number of coins. Therefore, the number of nickels would be (30 - x - 2x) = (30 - 3x).
Step 6: Since each nickel is worth $0.05, the value of all the nickels would be 0.05 * (30 - 3x) = 1.50 - 0.15x.
Step 7: The total value of all the coins is given as $3.30. Therefore, we can write the equation as follows:
0.10x + 0.50x + 1.50 - 0.15x = 3.30
Step 8: Simplifying the equation, we get:
0.45x + 1.50 = 3.30
Step 9: Now, subtract 1.50 from both sides of the equation:
0.45x = 1.80
Step 10: Divide by 0.45 on both sides of the equation:
x = 4
So, there were 4 dimes, 2x = 2 * 4 = 8 quarters, and the number of nickels would be 30 - x - 2x = 30 - 4 - 8 = 18.
Therefore, there were 4 dimes, 8 quarters, and 18 nickels.
To solve this problem, we can use a system of equations. Let's first assign variables to the unknowns.
Let's say the number of nickels is 'n,' the number of dimes is 'd,' and the number of quarters is 'q.'
We have the following information given in the problem:
1. We have a total of 30 coins: n + d + q = 30
2. The value of all the coins is $3.30: 5n + 10d + 25q = 330 (since a nickel is worth 5 cents, a dime is worth 10 cents, and a quarter is worth 25 cents)
3. There are twice as many quarters as dimes: q = 2d
We can use these equations to solve for the variables 'n,' 'd,' and 'q.'
First, let's simplify the equation q = 2d by substituting it into the first equation:
n + d + 2d = 30
n + 3d = 30
Now, we have two equations with two variables:
n + 3d = 30
5n + 10d + 25q = 330
We can solve this system of equations using various methods, such as substitution or elimination.
Let's use substitution:
1. Rearrange the first equation to solve for n: n = 30 - 3d
2. Substitute this expression for n in the second equation:
5(30 - 3d) + 10d + 25q = 330
150 - 15d + 10d + 25q = 330
-5d + 25q = 180
Now we have two equations with two variables:
-5d + 25q = 180
n + 3d = 30
Solving this system of equations will give us the values of 'n,' 'd,' and 'q,' which represent the number of nickels, dimes, and quarters, respectively.