SOLUTION: HOw many oz. of a metal containing 66% silver must be combined with 3 oz. of a metal containing 90% sivler to form an alloy containing 85% silver?
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Question 376445: HOw many oz. of a metal containing 66% silver must be combined with 3 oz. of a metal containing 90% sivler to form an alloy containing 85% silver? Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website! How much silver is there in the alloys?
Let x = the number of oz of 66% silver alloy.
This contains (0.66)x oz of silver.
The 3 oz of 90% silver alloy contain (0.9)(3) oz of silver.
The sum of these is to equal (3+x)(0.85) oz of silver.
We can write the equation to solve for x.
0.66x+0.9(3) = 0.85(3+x) Simplify and solve for x.
0.66x+2.7 = 2.55+0.85x Subtract 0.66x from both sides.
2.7 = 2.55+0.19x Subtract 2.55 from both sides.
0.15 = 0.19x Finally, divide both sides by 0.19
x = 0.789 oz of the 66% silver alloy will be required.
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