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. If the original count of the bacteria in a sample is 10000, find the bacteria count at the end of 3 hours.

Answer 1

Hint:
We know about the meaning of percent. Percent means out of one hundred, for example: 10% is equal to $\dfrac{{10}}{{100}}$. The problem is related with time so use the relation of rate of time with the percentage.

Complete step by step solution:
The original count of the bacteria in a sample is 10000.
We know the formula of final count is:
${\text{Final count}} = {\text{Initial count}} \pm \left( {{\text{Initial count}} \times \dfrac{{{\text{time}} \times {\text{rate}}}}{{100}}} \right)$
For the first hour, the number increases by 10%, now calculate the total number of bacteria counts by substituting ${\text{Initial count}} = 10000,{\text{ rate}} = 10\% ,{\text{ and time}} = 1{\text{ hour}}$ in ${\text{Final count}} = {\text{Initial count}} \pm \left( {{\text{Initial count}} \times \dfrac{{{\text{time}} \times {\text{rate}}}}{{100}}} \right)$.
${\text{Final count}} = {\text{10000 + }}\left( {10000 \times \dfrac{{1 \times 10}}{{100}}} \right)\\
 = 10000 + 1000\\
 = 11000$
Hence, the bacteria count for the first hour is 11000.
For the second hour, the number decreases by 10%, now calculate the total number of bacteria counts by substituting ${\text{Initial count}} = 11000,{\text{ rate}} = 10\% ,{\text{ and time}} = 1{\text{ hour}}$ in ${\text{Final count}} = {\text{Initial count}} \pm \left( {{\text{Initial count}} \times \dfrac{{{\text{time}} \times {\text{rate}}}}{{100}}} \right)$.
${\text{Final count}} = {\text{11000}} - \left( {11000 \times \dfrac{{1 \times 10}}{{100}}} \right)\\
 = 10000 - 1100\\
 = 9900$
Hence, the bacteria count for the second hour is 9900.
For the third hour, the number increases by 10%, now calculate the total number of bacteria counts by substituting ${\text{Initial count}} = 9900,{\text{ rate}} = 10\% ,{\text{ and time}} = 1{\text{ hour}}$ in ${\text{Final count}} = {\text{Initial count}} \pm \left( {{\text{Initial count}} \times \dfrac{{{\text{time}} \times {\text{rate}}}}{{100}}} \right)$.

${\text{Final count}} = 99{\text{00}} - \left( {9900 \times \dfrac{{1 \times 10}}{{100}}} \right)\\
 = 9900 + 990\\
 = 10890$

Hence, the bacteria count for the third hour is 10890.

Note:
Here you see the calculation, how to get the bacteria count for the nth hours. Here you should know the difference between increase and decrease formula. If the percent rate is increases then use the formula ${\text{Final count}} = {\text{Initial count}} + \left( {{\text{Initial count}} \times \dfrac{{{\text{time}} \times {\text{rate}}}}{{100}}} \right)$ and if the percent rate is decrease then use the formula ${\text{Final count}} = {\text{Initial count}} - \left( {{\text{Initial count}} \times \dfrac{{{\text{time}} \times {\text{rate}}}}{{100}}} \right)$.