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Question 1199176: A biologist has two brine solutions, one containing 2% salt and another containing 8% salt. How many milliliters of each solution should she mix to obtain 1 L of a solution that contains 5.6% salt?
2% solution mL
8% solution mL
Found 4 solutions by Theo, MathTherapy, josgarithmetic, greenestamps:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = the number of milliliters of 2% salt solution.
y = the number of milliliters of 8% salt solution.
total solution needs to be 1 liter = 1000 milliliters
your equations are:
x + y = 1000.
.02 * x + .08 * y = .056 * 1000 = 56
these two equations need to be solved simultaneously
multiply both sides of the first equation by .02 and leave the second equation as is to get:
.02 * x + .02 * y = 20
.02 * x + .08 * y = 56
subtract the first equation from the second to get:
.06 * y = 36
divide both sides of the equation by .06 to get:
y = 36 / .06
solve for y to get:
y = 600
in the equation x + y = 1000, solve for x to get:
x = 400
when x = 400 and y = 600, .....
x + y = 400 + 600 = 1000
.02 * 400 + .08 * 600 = 56
56 / 1000 = .056
your solution is that 400 milliliters of 2% solution needs to be mixed with 600 milliliters of 8% solutopm to get 1000 milliliters = 1 liter of 5.6% solution.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
A biologist has two brine solutions, one containing 2% salt and another containing 8% salt. How many milliliters of each solution should she mix to obtain 1 L of a solution that contains 5.6% salt?
2% solution mL
8% solution mL
Let amount of 2% salt be T
Then amount of 8% salt = 1 - T
We then get: .02T + .08(1 - T) = .056(1)
.02T + .08 - .08T = .056
.02T - .08T = .056 - .08
- .06T = - .024
Amount of 2% solution, or 
Amount of 8% solution = .6 L = .6(1,000) = 600 mL
Answer by josgarithmetic(39618)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
You have received three responses, all showing variations of the standard formal algebraic method for solving 2-part mixture problems like this.
Here is an informal method that can be used to solve this kind of problem quickly -- if a formal algebraic solution is not required.
(1) Look at the three percentages -- 2, 5.6, and 8 -- (on a number line, if it helps)
(2) observe/calculate that 5.6 is 3.6/6 = 0.6 = 3/5 of the way from 2 to 8
(3) that means 3/5 of the mixture must be the solution with the higher percentage
3/5 of the total 1 liter (1000 ml) is 600ml
ANSWER: 600 ml of the 8% solution and 400 ml of the 2% solution
CHECK:
.08(600)+.02(400) = 48+8 = 56
.056(1000) = 56
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