Question 414817: how many liters of a 3%saline solution must be added to an 8% saline solution to obtain 20 liters of a 6% saline solution Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! how many liters of a 3%saline solution must be added to an 8% saline solution to obtain 20 liters of a 6% saline solution
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Let x = amount (liters) of 3% solution
then
20-x = amount (liters) of 8% solution
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.03x + .08(20-x) = .06(20)
.03x + 1.6 - .08x = 1.2
1.6 - .05x = 1.2
-.05x = -0.4
x = 8 liters (3% solution)
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Amount of 8% solution:
20-x = 20-8 = 12 liters
