Science:Math Exam Resources/Courses/MATH100/December 2015/Question 06 (a)

MATH100 December 2015

  •  Q1 (i)  •  Q1 (ii)  •  Q1 (iii)  •  Q1 (iv)  •  Q2 (i)  •  Q2 (ii)  •  Q2 (iii)  •  Q2 (iv)  •  Q3 (i)  •  Q3 (ii)  •  Q3 (iii)  •  Q3 (iv)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  Q7  •  Q8 (a)  •  Q8 (b)  •  Q9  •  Q10 (a)  •  Q10 (b)  •  Q10 (c)  •  Q10 (d)  •  Q10 (e)  •  Q10 (f)  •  Q10 (g)  •  Q10 (h)  •  Q11 (a)  •  Q11 (b)  •  Q12 (a)  •  Q12 (b)  •  Q12 (c)  •  

   Other MATH100 Exams

  •  December 2011  •  December 2010  •  December 2012  •  December 2013  •  December 2014  •  December 2015  •  December 2016  •  December 2018  •  

Question 06 (a)
A colony of bacteria doubles every 4 hours. If the colony has 2000 cells after 6 hours, how many cells were present initially? Simplify your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The number $N(t)$ of the cells in $t$ hours has the form of $N(t)=N_{0}e^{ct}$ for some constant $N_{0}$ and $c$.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The number of cells $N(t)$ in $t$ hours obeys the rule: $N(t)=N_{0}e^{ct}$ for some $N_{0}$ and $c$, where $N(0)=N_{0}$ is the initial number of cells at $t=0$, so the question asks for $N_{0}$. Since a colony of bacteria doubles every 4 hours, the number of cells in 4 hours doubles the initial number of the cells: $N(4)=2N_{0}\Rightarrow N_{0}e^{4c}=2N_{0}\Rightarrow e^{4c}=2\Rightarrow (e^{c})^{4}=2\Rightarrow e^{c}=2^{\frac {1}{4}}$ On the other hand, the number of cells in 6 hours is $2000$, so that we have $N(6)=N_{0}e^{6c}=2000$. Therefore, $N_{0}=2000\cdot e^{-6c}=2000\cdot (2^{\frac {1}{4}})^{-6}=\color {blue}{2000\cdot 2^{-{\frac {3}{2}}}}$.