Tropical Fish Forums

[Marcos Avila]

OK, here's another one that's just a little more complicated. As we all know, Shawna is a BIG fan of adding salt to her pH 7.0 community tank (as stated in another topic below), so much that she got a little carried away and added too much for her liking, and she now wants to do some water changes with distilled water to help solve the problem. Which of the schemes will result in the least amount of salt in the tank water?

Once again, the reply below contains the correct answer so don't read it yet if you still want to find the solution by yourself!
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[Marcos Avila]

Great! 20 people answered, 15 got it right...the least amount of salt will be left in the tank by doing a single 40% water change. A very common mistake made by many people is to think that, for example, two 20% water changes is the same as a 40% water change. This can be proved wrong by simply choosing an extreme example: if the above is true then two 50% water changes should be equivalent to a single 100% water change, right? Wrong!

Try this experminent at home: fill up two glasses with water, add two spoons of salt to each and mix until the salt is completely dissolved. On the first one do a 100% water change, I.e., remove all the water and fill it up with pure tap water. On the second one do two 50% water changes, I.e., remove half of the glass water and fill it up with pure tap water, then repeat the same procedure again on this glass. Now taste them both, does the second glass still taste salty and the first one not?

Who wants to take a shot and explain (mathematically) why they're different?
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Success with a fish/tank is measured in YEARS, not months or weeks...

[Felipe Telles]

Hi!

I'll try Marcos. I will take the glass of water as my example. When you change 100% of the water you remove all the water and the salt. But when you do a 50% change then you do another one with 50% you just actually changed 75% of the water! That happens because when you change 50% of the water you leave 50% salted water and 50% pure water. And of course, they will get mixed. So when you change 50% of the water again you won't just take the salted water but the salted and the pure together. And when you do this you get 25% of the original salted water plus 25% of the pure water. And 25% plus 50% is 75%! I know it's not mathematically but it's the best I can do !Don't know if you understood what I tried to explain because of my (bad) english, but I hope you did.

[Magnus]

It's been a long time since I've taken a math class (the beauty of college, they can't really force you any more, although you'll have to take them later on anyway), but I'll take a jab at the problem.
Let's say Shawna's tank has 10% salt content, and she does 40% water change. That will bring the salt level down to 6% (4% was taken out with the 40% of water [10% * 40% = 4% removed; 10% - 4% = 6% remained]).
If she does a 20% water change, that will lower the salt lever in the tank to 8% (.1 * .2 = 2% removed; .1 - .02 = 8% remained). The second 20% water change will remove 1.6% salt content in the tank (.08 * .2 = 1.6%). This will leave 6.4% salt remaining in the tank. As you can see, the 40% water change removes a little bit more salt than the two 20% water changes.
Using the same method, you can see that by doing four 10% water changes, Shawna's tank will have 6.561% of salt remained.
Once again, it's been a long time since I've done MATH so I'm not really sure, can you confirm it?

[Marcos Avila]

Yes, both of you have the right idea. When we're talking about water changes we usually mention the amount of old water we're REMOVING from the tank, but when doing the math like in this problem it's best to think about the old water that's LEFT in the tank.
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Success with a fish/tank is measured in YEARS, not months or weeks...