Dilution is the solution to pollution

For all the noobs and non mathematicians out there,
Did you know it takes 4x25% water changes to reduce nasties or increase goodies by 50%

4x25% water change is equivalent to one 68% water change.
 
How do you figure out the math for this?

I want to know much 4x15% water change actually changes.
48% (rounded)

For non-math people I'd suggest thinking about as what percentage is left after each change,,, so after 15% WC, 85% is left,,, and then multiply that out for each change:
0.85^4 = 0.52 (rounded) so 52% of the original water is left
 
Thanks guys! I got it figured out. The calculator is also a helpful tool!

I looked online for the formula and I tested it with Randy’s test example from above.
 
How do you figure out the math for this?

I want to know much 4x15% water change actually changes.

4x15 —>. .85 x .85 x.85 x .85. =. 0.52
 
I often see in freshwater discussions people saying that nitrate will accumulate indefinitely. They sometimes do the math for a few iterations but not enough to see that it doesn't. Then some say you need to periodically do a water change over 50% otherwise it'll accumulate indefinitely.

Sigh.
 
A more generalized formula would be (1 - (1 - x/100)^y) * 100, which should give you the equivalent single water change percentage. Where x is the percent of water being changed and y is the number of times this water change is performed. Or just use the website calculator someone else posted.
 
Depends at what rate pollutants added vs period between change and why I’ve never placed blind faith on those online calculators. Yes you get X change but if topic is dilution then it gets more complex.
 

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