Question

At the start of an experiment there are \[20000\] bacteria. The number of bacteria increases at a rate of \[30\%\] per hour.
\[\left( a \right)\] Work out the number of bacteria after \[4\] hours.

Answer 1

Hint:In order to find the number of bacteria after \[4\] hours, firstly, we would be considering the total number of bacteria given and also the rate at which the bacteria increases. Then we will be applying the formula of compound interest and substituting the required values and solving it will provide us with the required answer.

Complete step-by-step solution:
Now let us have brief information regarding the compound interest. It is basically the interest that is calculated on the principal amount and the interest accumulated over the previous period which means that the interest changes over time upon the same principal amount. The formula of the compound interest is \[A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\].
Now let us start solving the given problem.
We are given that;
Total number of bacteria is \[20000\].
The rate at which the bacteria increases per hour is \[30%\].
The formula we will be considering for calculation is \[A=P{{\left( 1+\dfrac{r}{n} \right)}^{n}}\]
Now let us substitute the values in it. We get,
\[\begin{align}
  & A=P{{\left( 1+\dfrac{r}{n} \right)}^{n}} \\
 & \Rightarrow A=20000{{\left( 1+\dfrac{30}{100} \right)}^{n}}=20000{{\left( 1.3 \right)}^{n}} \\
\end{align}\]
Since we are asked to find the number of bacteria after \[4\] hours, we will be substituting \[n=4\] in the obtained equation.
Upon doing so, we get
\[\begin{align}
  & \Rightarrow 20000{{\left( 1.3 \right)}^{n}} \\
 & \Rightarrow 20000{{\left( 1.3 \right)}^{4}}=20000\times 2.8561=57122 \\
\end{align}\]
\[\therefore \] The total number of bacteria after \[4\] hours is \[57122\].

Note: In this above problem, we have used the concept of compound interest because the initial amount of bacteria would be constant at any hour. We can also apply the concept of compound interest in calculating the increase or decrease in population and also in rise or depreciation in the value of an item.