The bacteria in a culture grows by 10% in the first hour,
Question:

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.

Solution:

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 .

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