One alloy of brass is 70% copper and 30% zinc. Another alloy of brass is 40% copper and 60% zinc. How many grams of each of these alloys need to be melted and combined to obtain 300 grams of a brass alloy that is 60% copper and 40% zinc
Just concentrate on copper only.
let the amount of the 70:30 alloy used be x g
then the amount of the 40:60 alloy used is 300 - x
.7x + .4(300-x) = .6(300)
.7x + 120 - .4x = 180
.3x = 60
x = 200
so 200 g of the 70:30 alloy and
100 g of the 40:60 alloy
checking for the zinc:
.3(200) + .6(100) = 120
and .4(300) = 120, checks out!
By volume, one alloy is 60% copper, 30% zinc and 10% nickel. A second alloy is 50% copper, 30% zinc and 20% nickel. The third alloy is 30% copper and 70% nickel. How much of each alloy is needed to make 100 cm^3 of a new alloy that is 40% copper, 15% zinc and 45% nickel?
To solve this problem, let's assume a total of x grams of the first alloy and y grams of the second alloy are needed to obtain 300 grams of the desired brass alloy.
We are given that the first alloy is 70% copper and 30% zinc. This means that the amount of copper in the first alloy is 70% of x grams, which is 0.7x grams. Similarly, the amount of zinc in the first alloy is 30% of x grams, which is 0.3x grams.
Likewise, the second alloy is 40% copper and 60% zinc. Thus, the amount of copper in the second alloy is 40% of y grams, which is 0.4y grams, while the amount of zinc is 60% of y grams, which is 0.6y grams.
For the final alloy to be 60% copper, the total amount of copper from both alloys should be 60% of 300 grams, which is 0.6 * 300 = 180 grams.
Therefore, we can write the equation: 0.7x + 0.4y = 180
Similarly, for the final alloy to be 40% zinc, the total amount of zinc from both alloys should be 40% of 300 grams, which is 0.4 * 300 = 120 grams.
Therefore, we can write the second equation: 0.3x + 0.6y = 120
We now have a system of equations:
0.7x + 0.4y = 180
0.3x + 0.6y = 120
To solve this system, we can use any method of solving simultaneous equations.
Multiplying the first equation by 10 and the second equation by 5, we can eliminate the decimals:
7x + 4y = 1800
3x + 6y = 600
Now, let's multiply the second equation by 2:
7x + 4y = 1800
6x + 12y = 1200
Next, we can subtract the second equation from the first equation:
(7x + 4y) - (6x + 12y) = 1800 - 1200
x - 8y = 600
From this equation, we can express x in terms of y:
x = 8y + 600
Now, substitute this expression for x in one of the original equations:
0.7(8y + 600) + 0.4y = 180
Simplifying:
5.6y + 420 + 0.4y = 180
6y + 420 = 180
6y = 180 - 420
6y = -240
y = -40
Now, substitute this value of y back into the expression for x:
x = 8(-40) + 600
x = -320 + 600
x =280
Therefore, to obtain 300 grams of the desired brass alloy, you would need 280 grams of the first alloy (70% copper, 30% zinc) and -40 grams of the second alloy (40% copper, 60% zinc). However, it is not practical to have a negative amount of the second alloy.
This result suggests that it is not possible to obtain a brass alloy with 60% copper and 40% zinc by combining these two given alloys.
To solve this problem, let's assume we need x grams of the first alloy (70% copper and 30% zinc) and y grams of the second alloy (40% copper and 60% zinc) to obtain 300 grams of the desired brass alloy.
Now, let's set up the equation based on the copper content:
For the first alloy:
Copper in the first alloy = 0.7x grams
For the second alloy:
Copper in the second alloy = 0.4y grams
For the desired alloy:
Copper in the desired alloy = 0.6 * 300 grams = 180 grams
Since we want to obtain 300 grams of the desired alloy, the sum of the weights of the first and second alloys should be equal to 300 grams:
x + y = 300
Now, let's set up the equation based on the zinc content:
For the first alloy:
Zinc in the first alloy = 0.3x grams
For the second alloy:
Zinc in the second alloy = 0.6y grams
For the desired alloy:
Zinc in the desired alloy = 0.4 * 300 grams = 120 grams
The sum of the weights of the zinc content should also be equal to 120 grams:
0.3x + 0.6y = 120
Now we have a system of two equations with two variables:
x + y = 300 (Equation 1)
0.3x + 0.6y = 120 (Equation 2)
We can solve this system using a variety of methods, such as substitution, elimination, or graphing. In this case, let's use substitution.
Rearrange Equation 1 to solve for y:
y = 300 - x
Substitute this value of y into Equation 2:
0.3x + 0.6(300 - x) = 120
Now, we can solve for x:
0.3x + 180 - 0.6x = 120
-0.3x = -60
x = 200
Substitute this value of x back into Equation 1 to find y:
200 + y = 300
y = 100
Therefore, we need 200 grams of the first alloy (70% copper, 30% zinc), and 100 grams of the second alloy (40% copper, 60% zinc) to obtain 300 grams of the desired brass alloy (60% copper, 40% zinc).