A dehydrated patient needs a 6.4% saline IV.Unfortunately the hospital only has bags of 6% and 8% saline solutions.How many liters of each of these solutions should be mixed together to yield 1 liter of the desired concentration?
let x be the amount of 6% saline
so, 1-x = amount of 8% saline
.06x + .08(1-x) = .064*1
.06x + .08 - .08x = .064
.02x = .016
x = .8
so, .8L of 6% and .2L of 8%
note that 6.4% is 1/5 of the way from 6% to 8%, so the 8% volume is 1/5 of the total volume.
To solve this problem, we can use a simple formula called the "weighted average" formula, which takes into account the concentrations of the two solutions and the desired concentration.
Let's assume:
x = the amount (in liters) of the 6% saline solution to be mixed
y = the amount (in liters) of the 8% saline solution to be mixed
Based on the given information, we can set up two equations:
Equation 1: x + y = 1 (since we want a total of 1 liter of the desired solution)
Equation 2: 0.06x + 0.08y = 0.064 (the weighted average formula)
Let's solve these equations together. First, we'll solve Equation 1 for one of the variables:
x = 1 - y
Next, we'll substitute this value of x into Equation 2:
0.06(1 - y) + 0.08y = 0.064
Simplifying further:
0.06 - 0.06y + 0.08y = 0.064
0.08y - 0.06y = 0.064 - 0.06
0.02y = 0.004
y = 0.004 / 0.02
y = 0.2
So, we found that y (the amount of 8% saline solution) is 0.2 liters.
Now, we can substitute this value of y back into Equation 1 to find x:
x = 1 - y
x = 1 - 0.2
x = 0.8
Therefore, we need to mix 0.8 liters of the 6% saline solution with 0.2 liters of the 8% saline solution to obtain 1 liter of the desired concentration (6.4% saline IV).
Note: This is a mathematical solution. In practice, it is important to consult and follow the guidance of medical professionals or pharmacists for precise and accurate measurements in medical settings.