The number of bacteria in a certain culture doubles every hour. If there were 50 bacteria present in the culture originally, how many bacteria would be present at the end of
(i) 2nd hour,
(ii) 5th hour and
(iii) nth hour?
To find: The number of bacteria after
(i) $2^{\text {nd }}$ hour
(ii) $5^{\text {th }}$ hour
(iii) nth hour
Given: (i) Initially, there were 50 bacteria
(ii) Rate – 100% per hour
The formula used: $A=P\left(1+\frac{r}{100}\right)^{t}$
(i) For $2^{\text {nd }}$ hour
$\Rightarrow$ No. of bacteria $=50\left(1+\frac{100}{100}\right)^{2}$
$\Rightarrow$ No. of bacteria $=50(1+1)^{2}$
$\Rightarrow$ No. of bacteria $=50(2)^{2}$
$\Rightarrow$ No. of bacteria $=50 \times 4$
$\Rightarrow$ No. of bacteria $=200$
(ii) For $5^{\text {th }}$ hour
$\Rightarrow$ No. of bacteria $=50\left(1+\frac{100}{100}\right)^{5}$
$\Rightarrow$ No. of bacteria $=50(1+1)^{5}$
$\Rightarrow$ No. of bacteria $=50(2)^{5}$
$\Rightarrow$ No. of bacteria $=50 \times 32$
$\Rightarrow$ No. of bacteria $=1600$
(iii) For $\mathrm{n}^{\text {th }}$ hour
$\Rightarrow$ No. of bacteria $=50\left(1+\frac{100}{100}\right)^{n}$
$\Rightarrow$ No. of bacteria $=50(1+1)^{n}$
$\Rightarrow$ No. of bacteria $=50(2)^{\mathrm{n}}$
$\Rightarrow$ No. of bacteria $=2^{n} 50$
Ans) Number of bacteria in a $2^{\text {nd }}$ hour will be 200 , the number of bacteria in a $5^{\text {th }}$ hour will be 1600 and number of bacteria in an $\mathrm{n}^{\text {th }}$ hour will be $2^{n} 50$