Question 215299: a mixture containing 8% salt is to be mixed with 40mL of a mixture that is 20% salt, in order to obtain a solution that is 12% salt. How much of the first solution must be used?
Found 2 solutions by RAY100, Theo:
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! Let x = added solution
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.08x + .2(40) = .12(x+40),,,,,salt balance
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.08x +8 = .12x +4.8
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3.2 = .04x
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80 = x,,,,,in ml
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check,,,,.08(80) +.2(40)= .12(120) = 14.4
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! X = amount of mixture that is 8% salt
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Let x = amount of salt in the first solution.
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.08 * x = amount of salt in the first solution.
.20 * 40 = amount of salt in the second solution.
.12 * (40+x) = amount of salt in the combined solution.
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.08*x + .2*(40) = .12 * (40+x)
This means that the amount of salt in the first solution plus the amount of salt in the second solution equals the amount of salt in the combined solution.
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Solve for x in the equation:
.08*x + .2*40 = .12 * (40 + x)
remove parentheses to get:
.08*x + .2*40 = .12*40 + .12*x
Simplify to get:
.08*x + 8 = 4.8 + .12*x
Subtract .08*x and subtract 4.8 from both sides of the equation to get:
8 - 4.8 = .12*x - .08*x
Simplify to get:
3.2 = .04*x
Divide both sides of equation by .04 to get:
3.2/.04 = x
Simplify to get:
80 = x which is the same as:
x = 80
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You should have 80 mL of the first solution to mix with 40 mL of the second solution to get 12% salt in the combined solution.
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.08 * 80 + .2 * 40 = 6.4 + 8 = 14.4 / (80 + 40) = 14.4/120 = .12
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Your answer is you need to use 80 mL of the first solution.
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