The bacteria in a 11-liter container doubles every 4 minutes. After 50 minutes the container is full. How long did it take to fill a quarter of the container?
I do not really know where to start
I disagree with soroban that this is not quite a true math problem. Once you think backward from the endpoint, a very useful trick in math and other pursuits, the answer can be derived from simple logic without using any math but arithmetic. (Notice even soroban had to compute that
\(\displaystyle \dfrac{1}{4} = \dfrac{1}{2} * \dfrac{1}{2}\ and\ 50 - 4 - 4 = 42\), which is arithmetic and so math.
Now in case you do not have soroban's flair for finding elegant solutions, you can do this algebraically.
Start as always by assigning symbols to the relevant variables.
\(\displaystyle Let\ n\ = the\ number\ of\ four\ minute\ periods\ that\ have\ elapsed.\)
\(\displaystyle Let\ I_n = the\ volume\ of\ bacteria\ in\ the\ container\ at\ the\ end\ of\ period\ n.\)
\(\displaystyle Let\ I_0 = the\ volume\ of\ bacteria\ in\ the\ container\ at\ the\ start\ of\ period\ 1.\)
\(\displaystyle Let\ V = the\ volume\ of\ the\ container.\)
Second step is to put the givens in the problem into mathematical form using these symbols.
What is the volume of bacteria after n periods?
What is n when \(\displaystyle I_n = V?\)
So what is n when \(\displaystyle I_n = \dfrac{1}{4}V?\)