A mining company extracts ore in two mines. In mine A it contains 1%
nickel and 2% copper; in mine B it is 2% nickel and 5% copper.
How much should we extract from each mine to obtain 4 tons of nickel?
and 9 copper?
To calculate the amount of ore to extract from each mine, we can set up a system of equations.
Let's assume x represents the amount of ore (in tons) to extract from mine A and y represents the amount of ore (in tons) to extract from mine B.
From the problem, we know that:
- In mine A, 1% of the ore is nickel. Therefore, the amount of nickel obtained from mine A is 0.01x tons.
- In mine B, 2% of the ore is nickel. Therefore, the amount of nickel obtained from mine B is 0.02y tons.
We want to obtain 4 tons of nickel in total. So, we can set up the equation:
0.01x + 0.02y = 4
Similarly, we can set up an equation for the amount of copper obtained:
- In mine A, 2% of the ore is copper. Therefore, the amount of copper obtained from mine A is 0.02x tons.
- In mine B, 5% of the ore is copper. Therefore, the amount of copper obtained from mine B is 0.05y tons.
We want to obtain 9 tons of copper in total. So, we can set up the equation:
0.02x + 0.05y = 9
Now, we can solve this system of equations to find the values of x and y.
Multiplying the first equation by 100 to eliminate decimals, we get:
x + 2y = 400 (Equation 1)
Multiplying the second equation by 100 to eliminate decimals, we get:
2x + 5y = 900 (Equation 2)
Multiply Equation 1 by 2:
2x + 4y = 800 (Equation 3)
Subtract Equation 3 from Equation 2:
2x + 5y - (2x + 4y) = 900 - 800
y = 100
Substitute y = 100 into Equation 1:
x + 2(100) = 400
x + 200 = 400
x = 200
Therefore, we should extract 200 tons of ore from mine A and 100 tons of ore from mine B to obtain 4 tons of nickel and 9 tons of copper.
To find out how much ore should be extracted from each mine to obtain 4 tons of nickel and 9 tons of copper, we can set up a system of equations based on the metal content in each mine.
Let's assume x represents the amount of ore extracted from mine A, and y represents the amount of ore extracted from mine B.
For nickel content:
From mine A: 1% of x
From mine B: 2% of y
For copper content:
From mine A: 2% of x
From mine B: 5% of y
We can now write two equations based on the metal content:
1) Nickel content equation: 0.01x + 0.02y = 4
2) Copper content equation: 0.02x + 0.05y = 9
Now we can solve this system of equations.
Using the substitution method, let's solve equation (1) for x:
0.01x + 0.02y = 4
0.01x = 4 - 0.02y
x = (4 - 0.02y) / 0.01
Now substitute this value of x into equation (2):
0.02((4 - 0.02y) / 0.01) + 0.05y = 9
0.08 - 0.04y + 0.05y = 9
0.01y = 9 - 0.08
y = (9 - 0.08) / 0.01
y = 892
Now substitute the value of y into equation (1) to find x:
0.01x + 0.02(892) = 4
0.01x + 17.84 = 4
0.01x = 4 - 17.84
0.01x = -13.84
x = -13.84 / 0.01
x = -1384
Since we cannot have a negative amount of ore extracted, this solution is not feasible. Hence, we cannot obtain 4 tons of nickel and 9 tons of copper from the given mines.
To find out how much we should extract from each mine to obtain a specific amount of nickel and copper, we need to set up a system of equations based on the composition of each mine.
Let's say we extract x tons of ore from mine A and y tons of ore from mine B.
To obtain 4 tons of nickel, we can set up the following equation based on the nickel content in each mine:
0.01x + 0.02y = 4 (Equation 1)
To obtain 9 tons of copper, we can set up another equation based on the copper content in each mine:
0.02x + 0.05y = 9 (Equation 2)
Now, we can solve this system of equations to find the values of x and y, which represent the amount of ore to be extracted from each mine.
To do that, we can use substitution or elimination methods. Let's use the elimination method here.
First, let's multiply Equation 1 by 100 to eliminate the decimals:
100(0.01x + 0.02y) = 100(4)
x + 2y = 400 (Equation 3)
Now, we can subtract Equation 2 from Equation 3 to eliminate x variable:
(x + 2y) - (0.02x + 0.05y) = 400 - 9
x + 2y - 0.02x - 0.05y = 391
0.98x + 1.95y = 391 (Equation 4)
Now, we have a new equation without x variable. Let's solve Equation 4 for y:
0.98x + 1.95y = 391
1.95y = 391 - 0.98x
y = (391 - 0.98x) / 1.95 (Equation 5)
Now, we can substitute the value of y from Equation 5 into Equation 1 to solve for x:
0.01x + 0.02[(391 - 0.98x) / 1.95] = 4
Simplifying the equation further, we get:
0.01x + (0.02 * (391 - 0.98x)) / 1.95 = 4
0.01x + (0.02 * (391 - 0.98x)) = 4 * 1.95
0.01x + 0.02 * (391 - 0.98x) = 7.8
0.01x + 7.82 - 0.0196x = 7.8
0.01x - 0.0196x = 7.8 - 7.82
-0.0096x = -0.02
x = (-0.02) / (-0.0096)
x ≈ 2.083
Now, we have the value of x. We can substitute this value back into Equation 5 to find y:
y = (391 - 0.98 * 2.083) / 1.95
y ≈ 198.974
Therefore, to obtain 4 tons of nickel, we should extract approximately 2.083 tons of ore from mine A and approximately 198.974 tons from mine B.
To obtain 9 tons of copper, we should extract approximately 2.083 tons from mine A and approximately 198.974 tons from mine B.