How many ounces of other metal must be added to 56 ounces of pure silver to make a composition that is 70% silver?
Please help me set this problem up
56/(56+x) = 70/100
silver/(silver + additive)
silver / total weight
is 70% or 70/100
56/(56+x) = .7
Cross multiply and solve for x
Thank you!
You are welcome.
To set up this problem, we need to determine the amount of pure silver that will be in the final composition.
Let's assume that x ounces of other metal will be added to the 56 ounces of pure silver.
Since the resulting composition needs to be 70% silver, the amount of pure silver in the final composition will be 70% of the total weight of the composition.
The amount of pure silver in the final composition can be calculated as (70/100) * (56 + x) ounces.
Now, we can set up an equation to solve for x:
(70/100) * (56 + x) = 56
To solve for x, we can simplify the equation:
0.7 * (56 + x) = 56
By multiplying out and simplifying the equation, we get:
39.2 + 0.7x = 56
Next, we can isolate the variable x by subtracting 39.2 from both sides:
0.7x = 56 - 39.2
Simplifying further:
0.7x = 16.8
Finally, divide both sides by 0.7 to solve for x:
x = 16.8 / 0.7
Evalutating the expression, we find:
x ≈ 24
Therefore, approximately 24 ounces of other metal must be added to 56 ounces of pure silver to make a composition that is 70% silver.