British sterling silver is a copper-silver alloy that is 7.5% copper by weight. How many grams of pure copper and how many grams of British sterling silver should be used to prepare 170 grams of a copper-silver alloy that is 16% copper by weight?
x = grams of copper
0.075x = value of copper
170 - x = grams of silver
0.925(170 - x) = value of silver (100 - 7.5 = 92.5)
.16(170) = value of mixture
0.075x + 0.925(170 - x ) = 0.16(170)
Solve for x, grams of copper
(170 - x) = grams of silver
To solve this problem, we need to employ the concept of the weighted average.
Let's assume that we need "x" grams of pure copper and "y" grams of British sterling silver to prepare the desired alloy.
Given that British sterling silver is 7.5% copper by weight, we know that:
1. The amount of copper in "y" grams of British sterling silver is 7.5% of "y" grams, which can be expressed as 0.075y grams.
2. The weight of copper in "x" grams of pure copper is simply "x" grams.
Now, we can set up two equations using the information given in the problem:
1. The total weight of the produced alloy is 170 grams:
x + y = 170
2. The weight of copper in the alloy is 16% of the total weight:
0.16 * 170 = 0.075y + x
Simplify the second equation:
27.20 = 0.075y + x
Now, we have a system of two equations with two unknowns. We can solve them simultaneously to find the values of "x" and "y."
One way to solve this system is by substitution. Solving the first equation for "x," we get:
x = 170 - y
Substituting this value of "x" into the second equation:
27.20 = 0.075y + (170 - y)
Combine like terms:
27.20 = 170 - 0.925y
Rearrange the equation:
0.925y = 142.80
Divide both sides by 0.925:
y = 154.59 grams (approx.)
Substitute this value of "y" back into the first equation to find "x":
x + 154.59 = 170
x = 15.41 grams (approx.)
Therefore, to prepare 170 grams of a copper-silver alloy that is 16% copper by weight, we should use approximately 15.41 grams of pure copper and 154.59 grams of British sterling silver.