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Mathematics
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Sixty ounces of a 50% gold alloy are mixed with 40 oz of a 40% gold alloy. Find the percent concentration of the resulting gold alloy.

Apr 21st, 2015

Since there are 50% gold in the first 60 oz alloy, the mass of pure gold in the first alloy is equal to (50%) x (60 oz) = 30 oz.

Similarly, since there are 50% gold in the first 60 oz alloy, the mass of pure gold in the first alloy is equal to (40%) x (40 oz) = 16 oz. 

When the two alloys are mix, the total mass of pure gold is equal to 30 oz + 16 oz = 46 oz.

Since we have 60 oz + 40 oz = 100 oz resulting gold alloy after mixed, the percent concentration of the final gold alloy can be computed by:

46 oz / 100 oz = 46%

Hence, the percent concentration of the resulting gold alloy is 46%.

Answer: 46%

Apr 21st, 2015

" Similarly, since there are 50% gold in the first 60 oz alloy, the mass of pure gold in the first alloy is equal to (40%) x (40 oz) = 16 oz. "

Sorry, there was some typo here. I actually meant " Similarly, since there are 40% gold in the second 40 oz alloy, the mass of pure gold in the second alloy is equal to (40%) x (40 oz) = 16 oz."

Apr 21st, 2015

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