I take 900ml of a liquid and dilute it with 100ml of water. Then I take 900ml of the mixture and dilute it with 100ml of water. I repeat this process twenty times.
What proportion of the original liquid remains in the mixture at the end?
How much mixture should I take if I want it to contain 20ml of the original liquid?
Sorry I didn't know what to tag this question as and I forgot to write in a part of the problem.
On
I take 900ml of a liquid and dilute it with 100ml of water.
Let $a_n$ be the original liquid left in the mix after $n$ steps, with $a_1=900\,$.
I repeat this process twenty times.
On the assumption that "repeat this process" literally means "take 900ml of a liquid and dilute it with 100ml of water": $\;a_{n+1} = 0.9 \,a_n\,$, since at each step $100 \,\text{ml}$ of the previous mix are discarded to make room for the new $100 \,\text{ml}$ of water, and what's discarded includes $\,\frac{100}{1000}=0.1\,$ of the original liquid that still existed in the previous mix. Therefore $a_n=0.9^{n-1} a_1\,$, and for $n= 20 \ldots$
Hint: You started with $900$ ml and added $2000$ ml of water. The fact that you did it in steps does not matter at all. What proportion is the original solution now?