SOLUTION: A dehydrated patient needs a 6.8% saline IV. Unfortunately, the hospital only has bags of 2% and 8% saline solutions. How many liters of each of these solutions should be mixed tog
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Question 491069: A dehydrated patient needs a 6.8% saline IV. Unfortunately, the hospital only has bags of 2% and 8% saline solutions. How many liters of each of these solutions should be mixed together to yield 2 liters of the desired concentration? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A dehydrated patient needs a 6.8% saline IV. Unfortunately, the hospital only has bags of 2% and 8% saline solutions.
How many liters of each of these solutions should be mixed together to yield 2 liters of the desired concentration?
:
Let x = amt of 8% solution required
the resulting amt is to be 2 liters, therefore
(2-x) = amt of 2% solution required
:
Write a decimal form of the amt of saline equation
:
.08x + .02(2-x) = .068(2)
.08x + .04 - .02x = .136
.08x - .02x = .136 - .04
.06x = .096
x =
x = 1.6 liters of the 8% solution
then
2 - 1.6 = .4 liters of the 2% solution
:
:
You can check these solutions in the original equation
.08(1.6) + .02(.4) = .068(2)
