Completion Time

msu2011 · Mar 3, 2011

I THINK I may have figured this problem out...not sure. But here it goes....

45gtt/min x 60min/hr= 2700gtts/hr

2700gtts/hr divided by 60gtts/mL = 45mL/hr

40mL divided by 40mL/hr = 0.888...

0.888... x 60min/hr = 54.28 or 54 min to complete

Is this correct??? Thanks!

Boog'sCRRN246, RN · Mar 3, 2011

I did it this way, but our answers were pretty close:

60gtt/1mL = 45gtt/XmL X = 0.75mL/min

0.75mL/1min = XmL/60min X = 45mL/60min

45mL/60min = 40mL/Xmin X = 53 minutes

solneeshka, BSN, RN · Mar 4, 2011

You're both right. In the interest of teaching you to fish (instead of giving you a fish :)), here's an easy way to set up ANY IVF calc problem.

1) On the left side of the equal sign, put what it is that you're looking for. It's easiest if you do this in the form of a fraction. You're looking for a length of time, so the left side of your equal sign looks like this:

x minutes / 1

2) The very first thing on the right side of the equal sign will be whatever ratio you have that has the same kind of variable in the numerator of the ratio. In this case, we're looking for minutes, so the first thing on the right side of the equal sign will be whatever ratio has "minutes" in the numerator:

x minutes / 1 = 1 minute / 45 gtt ...

In order for this to work for you, you have to have enough of a grasp of mathematics to understand that in a ratio of variables, saying "there are 45 gtts per 1 minute of time" is the same as saying "there is 1 minute of time to each 45 gtt of fluid." So the ratio presented was 45 gtt/minute, which is the same as saying there is 1 minute/45 gtt of fluid. You have to stick the ratio into the equation with the right number on top, so this part is important! The numerator of the ratio on the right has to have the same kind of variable the numerator of the first ratio on the left (that is, whatever it is you're looking for).

3) From here, you just keep multiplying on the right side by whatever ratios will cancel out the variables that are *not* the variable you are looking for. In this case, you want to cancel out any variables that are not minutes, because you're looking for minutes:

x minutes / 1 = 1 minute / 45 gtt * 60 gtt / 1 mL * 40 mL / 1

This is easier to see when you are writing all this out by hand and are using horizontal (instead of diagonal) lines to separate your numerators from your denominators. Try it! With every ratio you're adding to the equation, the next numerator should be whatever unit is in the previous denominator. In this case, the 1st ratio to the right has "gtt" in the denominator so I know the next ratio I'm going to plug in will have gtt in the *numerator*. I do that and the next denominator is "mL" so I know I'm going to plug in whatever has mL in the *numerator* (in this case, I had to "create" a ratio via dividing by 1 -- again, this is all easier to see when you write it out by hand). You know you're done when you've canceled out all of the units on the right hand side that are *not* the kind of unit you are looking for (in this case, "minutes").

4) The gtt in the 1st denominator cancels out out the gtt in the 2nd numerator. The mL in the 2nd denominator cancels out the mL in the 3rd numerator, leaving you with this:

x minutes / 1 = 1 minute / 45 * 60 / 1 * 40 / 1

5) Run the math and you're left with x = 53.33333. This would still have worked if you had defined time in hours instead of minutes. The difference is that you either would have ended up with a fraction of an hour (in this case, .88888889), or you would have had to have added another ratio to the righthand side to convert minutes into hours.

If this method is confusing for you, then just go with something that seems more intuitive. But if you can try this out a few times and get the hang of it, you will be able to solve any and every IV fluid calc that comes your way, and you won't have to memorize a single formula to do it. The equation just writes itself. If you get this theory down, you can apply it to any practical situation.

msu2011 · Mar 4, 2011

I did it this way, but our answers were pretty close:
60gtt/1mL = 45gtt/XmL X = 0.75mL/min
0.75mL/1min = XmL/60min X = 45mL/60min
45mL/60min = 40mL/Xmin X = 53 minutes

Yes! You are correct! :yeah: I re-calculated, and came up with 53min as well. I noticed I accidently stated 40mL divided by 40mL/hr = 0.888... I meant to type 40Ml/hr divided by 45mL/hr=0.888... That was a typo :rolleyes:

Thank you so much for your helpful reply!

msu2011 · Mar 4, 2011

You're both right. In the interest of teaching you to fish (instead of giving you a fish :)), here's an easy way to set up ANY IVF calc problem.
1) On the left side of the equal sign, put what it is that you're looking for. It's easiest if you do this in the form of a fraction. You're looking for a length of time, so the left side of your equal sign looks like this:
x minutes / 1
2) The very first thing on the right side of the equal sign will be whatever ratio you have that has the same kind of variable in the numerator of the ratio. In this case, we're looking for minutes, so the first thing on the right side of the equal sign will be whatever ratio has "minutes" in the numerator:
x minutes / 1 = 1 minute / 45 gtt ...
In order for this to work for you, you have to have enough of a grasp of mathematics to understand that in a ratio of variables, saying "there are 45 gtts per 1 minute of time" is the same as saying "there is 1 minute of time to each 45 gtt of fluid." So the ratio presented was 45 gtt/minute, which is the same as saying there is 1 minute/45 gtt of fluid. You have to stick the ratio into the equation with the right number on top, so this part is important! The numerator of the ratio on the right has to have the same kind of variable the numerator of the first ratio on the left (that is, whatever it is you're looking for).
3) From here, you just keep multiplying on the right side by whatever ratios will cancel out the variables that are *not* the variable you are looking for. In this case, you want to cancel out any variables that are not minutes, because you're looking for minutes:
x minutes / 1 = 1 minute / 45 gtt * 60 gtt / 1 mL * 40 mL / 1
This is easier to see when you are writing all this out by hand and are using horizontal (instead of diagonal) lines to separate your numerators from your denominators. Try it! With every ratio you're adding to the equation, the next numerator should be whatever unit is in the previous denominator. In this case, the 1st ratio to the right has "gtt" in the denominator so I know the next ratio I'm going to plug in will have gtt in the *numerator*. I do that and the next denominator is "mL" so I know I'm going to plug in whatever has mL in the *numerator* (in this case, I had to "create" a ratio via dividing by 1 -- again, this is all easier to see when you write it out by hand). You know you're done when you've canceled out all of the units on the right hand side that are *not* the kind of unit you are looking for (in this case, "minutes").
4) The gtt in the 1st denominator cancels out out the gtt in the 2nd numerator. The mL in the 2nd denominator cancels out the mL in the 3rd numerator, leaving you with this:
x minutes / 1 = 1 minute / 45 * 60 / 1 * 40 / 1
5) Run the math and you're left with x = 53.33333. This would still have worked if you had defined time in hours instead of minutes. The difference is that you either would have ended up with a fraction of an hour (in this case, .88888889), or you would have had to have added another ratio to the righthand side to convert minutes into hours.
If this method is confusing for you, then just go with something that seems more intuitive. But if you can try this out a few times and get the hang of it, you will be able to solve any and every IV fluid calc that comes your way, and you won't have to memorize a single formula to do it. The equation just writes itself. If you get this theory down, you can apply it to any practical situation.

WOW! Thank you so much for taking your time to type all this information! :up: Ill take this same problem and go through this equation you posted. Mabey I can get the hang of it by practicing over and over. I made another post on here about the kind of word math problems where a number is thrown in there just to throw you off. I miss these problems everytime on tests. I assume this information you gave would help for those type of tricky math problems? Thanks!

solneeshka, BSN, RN · Mar 4, 2011

It does help with that, msu2011. If a given figure or ratio doesn't cancel out a unit that you need to have canceled out on the right side of the equal sign, then you don't need it for the problem. It might be good to try it out on a few problems where you already know what the right answer is. That would give you confidence in the process when you follow it and also come up with the right answer.