A problem of mixture and alligations

Question

Suppose a container contains x of liquid from which y units are taken out and replaced by water.After n operations, the quantity of pure liquid is: $x\left(1-\dfrac{y}{x}\right)^n$ I understand the concept but I am unable to derive it. I tried by taking 10 litre of liquid and adding 2 litre water for 2 operations $10:0$

operation 1: $10-2:2$

operation 2:

quantity of liquid removed=$\dfrac{2*8}{10}$

finally $8-\frac{8}{5}$:$2+2-\frac{2}{5}$

which is same as the formula but I am not able to prove $10(1-\frac{2}{10})^2$ from it. I need a general proof of $(x(1-y/x)^n)$.

Answer 1

Hint: In the first step, amount of liquid left in the mixture is $x-y=x\left(1-\dfrac{y}{x}\right)$. Thus ratio of the liquid in the mixture is $\left(1-\dfrac{y}{x}\right)$. Since $y$ units of water is replaced in every step, so total amount of mixture is always $x$ units. So, when we remove $y$ units in the second step, the amount of liquid left is $(x-y)\left(1-\dfrac{y}{x}\right)=x\left(1-\dfrac{y}{x}\right)^2$. You may proceed this way and use induction to prove the result.

Answer 2

I would expect that the amount 'z' of liquid in the unit amount of the mixture:

$0: x,\, z_0 = 1$

$1: x-y\cdot z_0=x(1-\frac{y}{x}) , z_1 = 1-\frac{y}{x}$

$2: x(1-\frac{y}{x})-y\cdot z_1=x(1-\frac{y}{x})-y(1-\frac{y}{x})=x(1-\frac{y}{x})^2 , z_2 = (1-\frac{y}{x})^2$

$3: x(1-\frac{y}{x})^2-y\cdot z_2=x(1-\frac{y}{x})^2-y(1-\frac{y}{x})^2=x(1-\frac{y}{x})^3 , z_3 = (1-\frac{y}{x})^3$

and the same hereinafter, as

$n: x(1-\frac{y}{x})^{n-1}-y\cdot z_{n-1}=x(1-\frac{y}{x})^{n-1}-y(1-\frac{y}{x})^{n-1}=x(1-\frac{y}{x})^n$

Edit: A colleague Debashish was faster, but I'll leave it.

Answer 3

I have solved it a bit differently, The concept is the same.

Thanks

Satish