A jeweler maker has 20 ounces of a gold alloy worth $160 an ounce and 10 ounce of a silver alloy worth $140 an ounce. How many ounces of another alloy at $124 an ounce must she add to these in order to have a mixture of the three that she can use to make 3-ounce jewelry worth $438 a piece?

10 ounces

Step-by-step explanation:

Let's define some variables:

g = ounces of gold per piece

s = ounces of silver per piece

x = ounces of other alloy per piece

There's twice as much gold as there is silver, so:

g = 2s

The total weight of the piece is 3 ounces, so:

g + s + x = 3

The total cost is $438, so:

160g + 140s + 124x = 438

Three equations, three variables.  We can solve the system of equations.  If we substitute the first equation into the second:

2s + s + x = 3

3s + x = 3

x = 3 − 3s

If we substitute this new equation and the first equation into the third equation:

160 (2s) + 140s + 124 (3 − 3s) = 438

320s + 140s + 372 − 372s = 438

88s = 66

s = 0.75

Now solve for g and x:

g = 2s = 1.5

x = 3 − 3s = 0.75

The amount of alloy equals the amount of silver.  The jeweler has 10 ounces of silver, so she needs 10 ounces of alloy.

853.3

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