# Finding the amount of acid removed from a tank after a repititive process.

The main question is :

A tank contains 729 litres of pure acid (Capacity of tank is 729 litres). 3$x$ litres of acid is removed and replaced by water. The two liquids are thoroughly mixed, and 3$x$ litres of solution is withdrawn and replaced by water. This process is repeated for six times. If finally the volume of press avoid left on the tank is 64 litres (rest is water), find $\sqrt{x}$.

My approach :

% of water after step 1 = $\frac{100x}{243}$

Therefore, amount of water in $3x$ litre removed solution = $\frac{3x*100x}{243}$ = $\frac{100x^2}{81}$

Therefore, amount of acid removed = $3x-\frac{100x^2}{81}$

To simplify the notation, let's call the $729 l$ capacity $T$ and the $3x$ spilled/replaced $y$. Also, do not use percentages but unitary ratios to evaluate the mixture content. So we indicate by $R_w$ the ratio q.ty of water/total, and it will be $R_a=1-R_w$. Now:

• at step 0: you have no water, so that $R_w(0)=0$.
• at step 1: you spill $y$ liters of acid and replace with $y$ of water, $R_w(1)=y/T$
• at step 2: you spill $y$ liters of the mixture that contains a fraction $y/T$ of water, so you spill $yy/T$liters and fill back with $y$ liters, so you change the quantity of the water by $y(1-y/T)$ liters, meaning that you change the ratio by $y/T(1-y/T)$ ( which, with the opposite sign, is also the change of ratio of the acid). So $R_w(2)-R_w(1)=y/T-y/TR_w(1)$

We can easily see that this recurrence keeps also for the next steps, and therefore: $$R_{\,w} (n) = \left( {1 - y/T} \right)R_{\,w} (n - 1) + y/T\quad \left| {\,R_{\,w} (0)} \right. = 0$$ For the acid instead , which you spill but do no replacement, it will be: $$R_{\,a} (n) = \left( {1 - y/T} \right)R_{\,a} (n - 1)\quad \left| {\,R_{\,a} (0)} \right. = 1$$ the above readily solve to: $$R_{\,w} (n) = 1 - \left( {1 - y/T} \right)^{\,n} \quad \quad R_{\,a} (n) = \left( {1 - y/T} \right)^{\,n}$$ And, from here, I suppose you can take over.

• @AksharGandhi, glad you appreciate. Sep 18, 2016 at 12:48

It can be formulated as follows: $$\begin{cases} a_{n+1}=a_n-3x\frac{a_n}{a_n+b_n}\\ b_{n+1}=b_n-3x\frac{b_n}{a_n+b_n}+3x \end{cases}$$ where $a_n-$acid, $b_n-$water (after $n$ turns).Initial conditions: $a_0=729, b_0=0$. Notice, that $a_n+b_n=const=a_0+b_0=729$, hence equations in system above are independent. Consider the first one: $$a_{n+1}=a_n-3x\frac{a_n}{729}=a_n\left(1-\frac{x}{243}\right)$$ Which has solution $a_n=C\left(1-\frac{x}{243}\right)^n$, $C-const$. From $a_0 =729$ immediately $C=729$.

It's easy to calculate then $$a_6=729\left(1-\frac{x}{243}\right)^6=64\\ 3^6\left(1-\frac{x}{243}\right)^6=2^6\\ 1-\frac{x}{243}=\frac{2}{3}\\ x=81$$

• Sir, I have a query. I did nit understand how did you get the solution of $a_{n}$ as $C(1-\frac{x}{243})^n$ and how did C become $729$. Please explain sir. Sep 18, 2016 at 3:16
• @AksharGandhi: one can look for solution of $a_{n+1}=a_n\left(1-\frac{x}{243}\right)$ in the form of $a_n=CA^n$. Substitute it in equation, then you will obtain $A=1-\frac{x}{243}$. Regarding $C$, we know volume of acid before first turn $a_0=729$, at the same time $a_0=CA^0=C$ Sep 18, 2016 at 9:24