The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. Find the bacteria at the end of 3 hours if the count was initially 20000.
Initial count of bacteria, $P=20000$
Rate of increase, $R=10 \%$
Time, $n=3$ hours
Then the count of bacteria at the end of the first hour is given by
Count of bacteria $=P \times\left(1+\frac{10}{100}\right)^{n}$
$=20000 \times\left(1+\frac{10}{100}\right)^{1}$
$=20000 \times\left(\frac{100+10}{100}\right)$
$=20000 \times\left(\frac{110}{100}\right)$
$=20000 \times\left(\frac{11}{10}\right)$
$=2000 \times 11$
$=22000$
Therefore, the count of bacteria at the end of the first hour is 22000 .
The count of bacteria at the end of the second hour is given by Count of bacteria $=P \times\left(1-\frac{10}{100}\right)^{n}$
$=22000 \times\left(1-\frac{10}{100}\right)^{1}$
$=22000 \times\left(\frac{100-10}{100}\right)$
$=22000 \times\left(\frac{90}{100}\right)$
$=22000 \times\left(\frac{9}{10}\right)$
$=2200 \times 9$
$=19800$
Therefore, the count of bacteria at the end of the second hour is 19800 .
Then the count of bacteria at the end of the third hour is is given by Count of bacteria $=\mathrm{P} \times\left(1+\frac{10}{100}\right)^{\mathrm{n}}$
$=19800 \times\left(1+\frac{10}{100}\right)^{1}$
$=19800 \times\left(\frac{100+10}{100}\right)$
$=19800 \times\left(\frac{110}{100}\right)$
$=19800 \times\left(\frac{11}{10}\right)$
$=1980 \times 11$
$=21780$
Therefore, the count of bacteria at the end of the first 3 hours is 21780 .