SOLUTION: Forty ounces of a 35% gold alloy are mixed with 62 ounces of a 20% alloy. Find the concentration of the resulting gold alloy.

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Question 628905: Forty ounces of a 35% gold alloy are mixed with 62 ounces of a 20% alloy. Find the concentration of the resulting gold alloy.
Found 2 solutions by Maths68, richwmiller:
Answer by Maths68(1474) (Show Source):
You can put this solution on YOUR website!
Alloy-A
========
Amount = 40 ounces
Concentration = 35%=0.35

Alloy-B
========
Amount = 62 ounces
Concentration = 20%=0.20

Resultant Alloy
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Amount = Alloy-A + Alloy-B = 40+62 = 102 ounces
Concentration = x

(Amount of Alloy-A)(Concentration of Alloy-A)+(Amount of Alloy-B)(Concentration of Alloy-B)= (Amount of Resultant Alloy)(Concentration of Resultant Alloy)
(40)(0.35)+(62)(0.20)= (102)(x)
14+12.4=102x
26.40=102x
102x=26.04
102x/102=26.04/102
x=0.2588
Concentration of the resultant alloy will be approximately 0.2588 or approximately 26%.

Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website!
.35*40+.2*60=102x
solve for x