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The bacteria in culture grows by 5% in the first hour .gets destroyed by 5% in the second hour and again grows by 5 % in the third hour .If the count of the bacteria at the end the third hour is $8.379\times 10^{8}$, find the original count of bacteria in the sample?

Answer:

Explanation:

Let p be the original count of bacteria in the sample

R1 = 5%,R2 = -5% ,R3 = 5% ,t = 3 hours

$8.379\times 10^{8}$= $P \left[\left(1+\frac{5}{100}\right)\left(1-\frac{5}{100}\right)\left(1+\frac{5}{100}\right)\right]$

$8.379\times 10^{8}$= $P\left[\left(\frac{21}{20}\right)\left(\frac{19}{20}\right)\left(\frac{21}{20}\right)\right]$

P= $8\times 10^{8}$