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How many mg of a metal containing 30%

silver must be combined with 8 mg of a
metal containing 50% silver to form an
alloy containing 38% silver?

User Albeee
by
7.9k points

1 Answer

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Final answer:

To create an alloy containing 38% silver from a metal containing 30% silver and another containing 50% silver, 12 mg of the 30% silver metal must be mixed with 8 mg of the 50% silver metal.

Step-by-step explanation:

To determine how many milligrams (mg) of a metal containing 30% silver must be combined with 8 mg of a metal containing 50% silver to form an alloy containing 38% silver, we can set up an equation based on the mass of silver in each component. Let's denote the mass of the metal containing 30% silver needed as x mg.

The amount of silver from the 30% metal is 0.30x, and the amount of silver from the 8 mg 50% metal is 0.50×8 = 4 mg. The total mass of the alloy is x + 8 mg and the total amount of silver in the alloy should be 0.38(x + 8). Setting up the equation:

0.30x + 4 = 0.38(x + 8)

Solving for x gives:

0.30x + 4 = 0.38x + 3.04

0.08x = 0.96

x = 12 mg

Therefore, 12 mg of the metal containing 30% silver is needed to be combined with the 8 mg of the metal containing 50% silver.

User Blacksun
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