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I need to clean my fish tank.  However, I can only remove 25% of the water at a time, then refill.  How many times must I remove and refill until I have reduced/diluted the original polluted water to 5% and then 1%.

 Sep 4, 2014

Best Answer 

 #4
avatar+99109 
+5

Hi anonymous,

Maybe I can help

Instead of having the original amount being y0 units why not let it be 1 unit.  Then everything is a little easier.

So

$$\\y_n=\left( \frac{3}{4}\right)^n y_0\\\\
becomes\\
\\y_n=\left( \frac{3}{4}\right)^n \\\\$$

 

remember, $$y_n$$    is the amount of original water in the tank.

 

You want to know when this become 5%.  5% of the original 1 unit is 0.05*1=0.05

so you want to know when

 $$\left( \frac{3}{4}\right)^n=0.05\\\\$$

NOW

If you would like me to explain without changing the $$y_0$$  to 1 ,   I can do that for you.   

 Sep 4, 2014
 #1
avatar+169 
+5

The amount of old water y is found with the equation, with n being the number of refills:

 

 $$y_{n}=\frac{3}{4}*y_{n-1}$$

 

This can be changed to a version where $$y_{n-1}$$ is irrelevant to $$y_{n}$$:

 

$$y_{n}=(\frac{3}{4})^{n}*y_{0}$$

 

With $$y_{0}$$ being the starting amount of water.

 

We are considering percentages hence we divide $$y_{n}$$ by $$y_{0}$$:

 

$$\frac{y_{n}}{y_{0}}=\frac{(\frac{3}{4})^{n}*y_{0}}{y_{0}}=(\frac{3}{4})^{n}$$

 

Now we can equate $$\frac{y_{n}}{y_{0}}$$ with the desired percentage and the amount of required refills:

 

$$\frac{y_{n}}{y_{0}}=(\frac{3}{4})^{n}=0.05$$

 

$$n=\frac{\log{0.05}}{\log{\frac{3}{4}}}\approx10.4$$

 

You'll need to refill 11 times to achieve a pollution level of 5%, you can figure the second percentage out for yourself. :)

 Sep 4, 2014
 #2
avatar
0

how does $${\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}^{{\mathtt{n}}} = {\mathtt{0.05}}$$

.
 Sep 4, 2014
 #3
avatar
0

Thankyou Honga!

 Sep 4, 2014
 #4
avatar+99109 
+5
Best Answer

Hi anonymous,

Maybe I can help

Instead of having the original amount being y0 units why not let it be 1 unit.  Then everything is a little easier.

So

$$\\y_n=\left( \frac{3}{4}\right)^n y_0\\\\
becomes\\
\\y_n=\left( \frac{3}{4}\right)^n \\\\$$

 

remember, $$y_n$$    is the amount of original water in the tank.

 

You want to know when this become 5%.  5% of the original 1 unit is 0.05*1=0.05

so you want to know when

 $$\left( \frac{3}{4}\right)^n=0.05\\\\$$

NOW

If you would like me to explain without changing the $$y_0$$  to 1 ,   I can do that for you.   

Melody Sep 4, 2014

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