Answer:
The chemist should use 410 grams of the first alloy (which is 10% gold and 20% lead) and 40 grams of the second alloy (which is 40% gold and 30% lead) to make an alloy that contains 57 grams of gold and 94 grams of lead.
Explanation:
Let's call the amount of the first alloy used "x" and the amount of the second alloy used "y". We can set up a system of two equations based on the amount of gold and lead needed in the final alloy:
Equation 1: 0.10x + 0.40y = 57 (the amount of gold in the first alloy is 10%, and in the second alloy is 40%)
Equation 2: 0.20x + 0.30y = 94 (the amount of lead in the first alloy is 20%, and in the second alloy is 30%)
We can then solve for x and y using any method of solving systems of equations. One way is to use substitution:
- Solve equation 1 for x: x = (57 - 0.40y)/0.10 = 570 - 4y
- Substitute this expression for x in equation 2: 0.20(570 - 4y) + 0.30y = 94
- Simplify and solve for y: 114 - 0.8y + 0.3y = 94 → -0.5y = -20 → y = 40
- Substitute this value of y into the expression for x: x = 570 - 4y = 410
Therefore, the chemist should use 410 grams of the first alloy (which is 10% gold and 20% lead) and 40 grams of the second alloy (which is 40% gold and 30% lead) to make an alloy that contains 57 grams of gold and 94 grams of lead.