Question

In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of $2.5%$ per hour. Find the bacteria at the end of 2 hours, if the count was initially 50600.

Answer 1

Hint: Use the formula for finding compound rate. Here P is the initial count of bacteria, R is rate of interest and $n=2hrs$. Thus, directly substitute all the values and find the count of bacterias at the end of 2 hours.

Complete step-by-step answer:
Here we have been given a laboratory. The count of bacteria in the beginning of the experiment =506000.
It is said that the bacteria is increasing at the rate of 2.5% per hour. Here 2.5% is the compounded rate. Hence we can use the formula, $A=P{{\left( 1+\dfrac{R}{100} \right)}^{n}}$ ……….(1)
Here A= Count of bacteria at the end of 2 hours.
P= initial count of the bacteria=506000.
R= 2.5%.
n= number of hours =2hrs.
Now let us put all these values in eq (1)
$A=506000{{\left[ 1+\dfrac{2.5}{100} \right]}^{2}}$
$\begin{align}
  & =50600{{\left[ 1+0.025 \right]}^{2}} \\
 & =50600\times {{\left( 1.025 \right)}^{2}} \\
 & =506000\times 1.025\times 1.025 \\
 & =506000\times 1.050625 \\
 & =531616.25 \\
\end{align}$
Since the number of bacteria cannot be in decimals, we can round off the value to 531616.
$\therefore $ The count of bacteria at the end of 2hrs=531616.

Note: As we have to find a count in number, where the compound is increasing at a rate of 2.5% per hour. So, it is important that you use the equation (1) formula to calculate compound rate. This is actually the formula for compound interest but we can use it for other purposes like this too.