# A nurse needs to treat a patient with an infection, using a 0.25% saline solution. He prepared the saline solution by diluting 50.0mL of a stock saline solution with a concentration of 3.0%. What volume of dilute 0.25% saline did the nurse prepare. It seems like there is something missing here.  So, we will do what we can. We can set up a system of equations here.  We need two variables.  I am assuming it was water we diluted the saline with.  We don't know how much water we used; that's x.  Also,...

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It seems like there is something missing here.  So, we will do what we can.

We can set up a system of equations here.  We need two variables.  I am assuming it was water we diluted the saline with.  We don't know how much water we used; that's x.  Also, we don't know how much of the 0.25% solution we ended up with; that's y.  But, we do know when we add the water to the 50 ml solution, we get the 0.25% solution.  So:

50 + x = y

Then, we can use a formula for the second equation.  For percent formula problems, trying to make a formula out of it:

Amount#1 * %#1  +  Amount#2 * %#2  +  Amount#3 * %#3 + . . . = Amount(total) * %total

So, filling in the numbers and variables we have:

50*0.03 + x*0 = y*0.0025

Simplifying each equation:

50+x = y

1.5 = 0.0025y

Solving the second equation for y:

y = 600 ml

And, that is the what we are looking for, how much 0.25% saline solution the nurse prepared.  X stood for how much water was added to achieve this.  Substituting the y back into the first equation for it, we would get:

50+x = 600

x = 550 ml

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