$\begin{array}{1 1} 12200 \\ 11180 \\ 22680 \\ 12130 \end{array} $

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Let $B(t) $ = number of bacteria present at time $t \rightarrow \large\frac{dB}{dt} $$\propto B = kB$

Integrating we get, $\log B = kt + C \rightarrow B(t) = e^{kt + c}$

At $t=0$, $B=1000 \rightarrow B(0) = 1000 = e^{k\times0+c} \rightarrow e^c = 1000$

$\Rightarrow B(t) = 1000 e^{kt}$

When $t=10$, $B = 5000 \rightarrow B(10) = 5000 = 1000 e^{k\times10} \rightarrow 5 = e^{10k} \rightarrow 5^{\large\frac{1}{10}}$ $=e^k$

Substituting, we get,$ B(t) = 1000\times 5^{\large\frac{t}{10}}$

Whent $t = 15$, $B = 1000 \times 5^{1.5} \approx 11180$ bacteria.

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