SOLUTION: 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?

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
.
.08x + .2(40) = .12(x+40),,,,,salt balance
.
.08x +8 = .12x +4.8
.
3.2 = .04x
.
80 = x,,,,,in ml
.
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
-----
Let x = amount of salt in the first solution.
-----
.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.
-----
.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.
-----
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
-----
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.
-----
.08 * 80 + .2 * 40 = 6.4 + 8 = 14.4 / (80 + 40) = 14.4/120 = .12
-----
Your answer is you need to use 80 mL of the first solution.
-----

RELATED QUESTIONS
A mixture containing 8% salt is to be mixed with 40 mL of a mixture that is 20% salt, in... (answered by checkley77)
A mixture containing 8% salt is added to be mixed with 4 ounces of a mixture that is 20%... (answered by josgarithmetic)
chemistry. A mixture containing 8% salt is to be mixed with 4 ounces of mixture that is... (answered by josgarithmetic)
A mixture containing 10% salt is to be mixed with 4 ounces of a mixture that is 20% salt, (answered by josgarithmetic)
A mixture containing 11 % salt is to be mixed with 4 ounces of a mixture that is 20 %... (answered by ewatrrr,josmiceli)
How many quarts of a 20% salt solution must be mixed with 15 quarts of a 35% salt... (answered by CharlesG2)
4 ounce of the mixture containing 6% salt is to be mixed with 9 ounces of a mixture which (answered by josmiceli)
1 ounce of the mixture containing 6% salt is to be mixed with 2 ounces of a mixture which (answered by Fombitz)
a solution containing 8% salt is to be mixed with a solution containing 16% salt to... (answered by ikleyn)