# Simple math problem that is totally screwing with me
**URL:** https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835
**Category:** Factual Questions
**Created:** [November 3, 2008, 8:02am UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835 "2008-11-03T08:02:32Z")
**Posts on this page:** 10
**Page:** 1
### Author:  [@Pullet](https://boards.straightdope.com/u/Pullet)
#### Post date: [November 3, 2008, 8:02am UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835/1 "2008-11-03T08:02:32Z")
You have a solution that is 2% ingredient X in water. You want to make a new solution that is 0.05% ingredient X, and you want 1 liter of this new solution. How much of the 2% solution do you add to your 1 liter of water?
I remember doing problems like this in chemistry, but, for the life of me, I can’t recreate them now, and my google-fu is failing.
Help!
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### Author:  [@Mosier](https://boards.straightdope.com/u/Mosier)
#### Post date: [November 3, 2008, 8:16am UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835/2 "2008-11-03T08:16:09Z")
you want the stuff diluted from 2% to 0.05%, a ratio of 40/1. You add 1 part of 2% solution to 40 parts water to make a solution that is 0.05% stuff. so you add 1/40th of a liter (.025 liters) of 2% solution to 1 liter of water to make the desired 0.05% dilution.
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### Author:  [@Dervorin](https://boards.straightdope.com/u/Dervorin)
#### Post date: [November 3, 2008, 9:56am UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835/3 "2008-11-03T09:56:34Z")
> [@Mosier](#):
>
> you want the stuff diluted from 2% to 0.05%, a ratio of 40/1. You add 1 part of 2% solution to 40 parts water to make a solution that is 0.05% stuff. so you add 1/40th of a liter (.025 liters) of 2% solution to 1 liter of water to make the desired 0.05% dilution.
That’s correct, but that way you end up with 1.025 litres of diluted solution, which is not what was asked for.
Since the solution and water are in the ratio 1:40, for a total of 41 parts of liquid, you add 1/41 litre of solution to 40/41 litres of water, leaving you with the desired 1 litre.
In decimal, that’s approximately 0.02439 litres of solution to 0.97561 litres of water.
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### Author:  [@Washoe](https://boards.straightdope.com/u/Washoe)
#### Post date: [November 3, 2008, 10:05am UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835/4 "2008-11-03T10:05:35Z")
If you’ll pardon the amateurishness of my web page, I give you—[4 different ways](http://www.hamiltonroots.com/Pharmacy/Solutions.htm) to do it.
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### Author:  [@don\_t\_ask](https://boards.straightdope.com/u/don_t_ask)
#### Post date: [November 3, 2008, 10:15am UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835/5 "2008-11-03T10:15:12Z")
> [@Dervorin](#):
>
> That’s correct, but that way you end up with 1.025 litres of diluted solution, which is not what was asked for.
>
> Since the solution and water are in the ratio 1:40, for a total of 41 parts of liquid, you add 1/41 litre of solution to 40/41 litres of water, leaving you with the desired 1 litre.
>
> In decimal, that’s approximately 0.02439 litres of solution to 0.97561 litres of water.
Uh no you just add 39 parts water to one part solution to get a new solution 1/40 th as strong.
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### Author:  [@Sage\_Rat](https://boards.straightdope.com/u/Sage_Rat)
#### Post date: [November 3, 2008, 10:32am UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835/6 "2008-11-03T10:32:32Z")
If you have a container that is 100 units deep and 100 units wide, then you have a total area of 10,000 units[sup]2[/sup]. 200 units[sup]2[/sup] of this is ingredient X, the other 9,800 is water.
Now imagine that the ingredient isn’t mixed in, but is rather lying on the bottom of our square, with the water above it:
```auto
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
```
If we were to slice it in half vertically and replace the one half with water, like so:
```auto
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o
```
Obviously, we’ll have 100 units[sup]2[/sup] of ingredient to 9,900 units[sup]2[/sup] water, giving us a 1% solution.
So this tells us that a vertical cut of 50% (1/2) shrinks the amount of ingredient by 50%. If we wanted a 0.5% solution, we’d replace 75% (3/4) of the solution with water, and if we wanted a 0.25% solution, we’d replace 87.5% (7/8) with water. We don’t want to shrink it by 50% or 25% nor 12.5% though, we want 0.05/2.0 = 0.025 or 2.5% (1/40). If we cut vertically and replace 97.5% (39/40) of the flask with water, we’ll have a 0.05% ingredient solution.
Since the solution is mixed though, we don’t need to cut vertically or horizontally, or any other particular direction. Any 2.5% units[sup]2[/sup] of the solution added to 97.5% units[sup]2[/sup] water, will create a 0.05% solution.
So pour out 2.5% of your liter of water (reducing it to 97.5%), and pour in the solution until you’re back to a liter, and you’re done.
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### Author:  [@robby](https://boards.straightdope.com/u/robby)
#### Post date: [November 3, 2008, 1:38pm UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835/7 "2008-11-03T13:38:12Z")
> [@Pullet](#):
>
> You have a solution that is 2% ingredient X in water. You want to make a new solution that is 0.05% ingredient X, and you want 1 liter of this new solution. How much of the 2% solution do you add to your 1 liter of water?
>
> I remember doing problems like this in chemistry, but, for the life of me, I can’t recreate them now, and my google-fu is failing.
>
> Help!
I actually taught this back when I was teaching chemistry.
Note that your problem is worded in a conflicting manner. You state that you want 1 L of the dilute solution, but you then ask how much concentrated solution should be added to 1 L of water (which would cause you to have more than 1 L of dilute solution). I will assume you mean the former (which is how these problems are usually worded), and that you want 1 L of dilute solution.
You use the equation:
_C_[sub]i[/sub]_V_[sub]i[/sub] = _C_[sub]f[/sub]_V_[sub]f[/sub]
where:
_C_[sub]i[/sub] = initial concentration
_V_[sub]i[/sub] = initial volume of concentrated solution
_C_[sub]f[/sub] = final concentration
_V_[sub]f[/sub] = final volume of dilute solution
For this problem:
_C_[sub]i[/sub] = 2%
_V_[sub]i[/sub] = initial volume of concentrated solution
_C_[sub]f[/sub] = 0.05%
_V_[sub]f[/sub] = 1 L
Substituting these values into the equation above:
(2%)_V_[sub]i[/sub] = (0.05%)(1 L)
Solving for _V_[sub]i[/sub], you get _V_[sub]i[/sub] = 0.025 L
Therefore if you put 0.025 L of concentrated solution in a flask, then fill the flask up to the 1 L mark, you will have 1 L of 0.05% solution. (You will have to add 0.975 L of water to the 0.025 L of concentrated solution.)
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### Author:  [@robby](https://boards.straightdope.com/u/robby)
#### Post date: [November 3, 2008, 1:55pm UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835/8 "2008-11-03T13:55:40Z")
> [@Mosier](#):
>
> you want the stuff diluted from 2% to 0.05%, a ratio of 40/1. You add 1 part of 2% solution to 40 parts water to make a solution that is 0.05% stuff. so you add 1/40th of a liter (.025 liters) of 2% solution to 1 liter of water to make the desired 0.05% dilution.
Wrong. As **don’t ask** states, you have to add 1 part of concentrated solution to 39 parts of water. Therefore you should add 1/40th of a liter (0.025 L) of 2% solution to 39/40 liters (0.975 L) of water to get the desired dilution.
> [@Dervorin](#):
>
> …Since the solution and water are in the ratio 1:40, for a total of 41 parts of liquid, you add 1/41 litre of solution to 40/41 litres of water, leaving you with the desired 1 litre.
Wrong. If you add 1/41 liter of concentrated solution to 40/41 liters of water, the final ratio comes out to be a dilution ratio of 1:41 instead of the desired 1:40.
> [@don\_t\_ask](#):
>
> Uh no you just add 39 parts water to one part solution to get a new solution 1/40 th as strong.
Correct.
> [@Sage\_Rat](#):
>
> …So pour out 2.5% of your liter of water (reducing it to 97.5%), and pour in the solution until you’re back to a liter, and you’re done.
Correct.
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### Author:  [@Dervorin](https://boards.straightdope.com/u/Dervorin)
#### Post date: [November 3, 2008, 2:55pm UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835/9 "2008-11-03T14:55:12Z")
> [@robby](#):
>
> Wrong. If you add 1/41 liter of concentrated solution to 40/41 liters of water, the final ratio comes out to be a dilution ratio of 1:41 instead of the desired 1:40.
I stand corrected. This is why I shouldn’t try to re-hash high-school chemistry early in the morning after a late night party. :smack:
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### Author:  [@Pullet](https://boards.straightdope.com/u/Pullet)
#### Post date: [November 3, 2008, 6:31pm UTC](https://boards.straightdope.com/t/simple-math-problem-that-is-totally-screwing-with-me/470835/10 "2008-11-03T18:31:59Z")
Thanks!
To clarify: the solution given in the original probelm was actually a detergent that one intends to use for flushing out a wound, so having exactly 1 liter of it isn’t strictly necessary. But, I knew the chemists on the board would throw a fit with that kind of slipshod math 🙂
**Robby** Your technique is now what I remember being taught in school. Thanks! I think I can pack that one back onto its neuron for my upcomming exam.
**Sage Rat and Washoe** thanks for the alternate methods. Though, I have to admit that it’ll take me a while to sit down and think about them to make sure I can follow the logic all the way through. I never had such a hard time with math before. I guess I’m out of practice. ☹
Thanks everyone!