Question

In a certain culture of bacteria, the rate of increase is proportional to the number of bacteria present at that instant. It is found that there are 10,000 bacteria at the end of 3 hours and 40,000 bacteria at the end of 5 hours, then the number of bacteria present in the beginning are:

• A. 1250
• B. 1200
• C. 1350
• D. 1300

Hint:

Whenever a problem says “rate of increase is proportional to the quantity itself”, immediately think of exponential growth or decay: \[ N=N_0e^{kt} \] A quick shortcut here is: \[ \frac{40000}{10000}=4 \] So population became four times in \(2\) hours, meaning it doubles every \(1\) hour: \[ 10000 \rightarrow 5000 \rightarrow 2500 \rightarrow 1250 \] working backwards.

Answer 1

The Correct Option is A

Concept: When the rate of growth of a quantity is directly proportional to the quantity present at that instant, the process follows an exponential growth model. Such problems are modeled using the differential equation: \[ \frac{dN}{dt}=kN \] where:
• \(N\) = population at time \(t\),
• \(k\) = constant of proportionality or growth constant. The solution of this differential equation is: \[ N(t)=N_0 e^{kt} \] where \(N_0\) represents the initial population.

Step 1: Using the first condition given in the problem.
At the end of \(3\) hours, the number of bacteria becomes \(10,000\). Substituting in the exponential growth equation: \[ N_0 e^{3k}=10000 \qquad \cdots (1) \]

Step 2: Using the second condition.
At the end of \(5\) hours, the number of bacteria becomes \(40,000\). Thus: \[ N_0 e^{5k}=40000 \qquad \cdots (2) \]

Step 3: Eliminating the initial population \(N_0\).
To eliminate \(N_0\), divide equation (2) by equation (1): \[ \frac{N_0 e^{5k}}{N_0 e^{3k}}=\frac{40000}{10000} \] \[ e^{2k}=4 \] Taking square root: \[ e^k=2 \] This means the bacteria population doubles every one unit increase in time.

Step 4: Finding the initial population.
Substitute \(e^k=2\) into equation (1): \[ N_0 (2)^3=10000 \] \[ 8N_0=10000 \] \[ N_0=\frac{10000}{8}=1250 \] Hence, the number of bacteria present initially is: \[ \boxed{1250} \]