Question

The rate of growth of bacteria is proportional to the number present. If initially there were 1000 bacteria and the number doubles in 1 hour, then the number of bacteria after \( 2\frac{1}{2} \) hours are (Given \( \sqrt{2} = 1.414 \))

• A. 4646 approximately
• B. 5056 approximately
• C. 5656 approximately
• D. \( 400 \sqrt{2} \) approximately

Hint:

For exponential growth, use the formula \( N(t) = N_0 e^{kt} \) and apply the doubling condition to solve for \( k \), then substitute into the equation for \( N(t) \) to find the number of bacteria at a later time.

Answer 1

The Correct Option is C

Step 1: Understand the exponential growth formula.
The exponential growth of bacteria can be modeled using the equation: \[ N(t) = N_0 e^{kt}, \] where \( N(t) \) is the number of bacteria at time \( t \), \( N_0 \) is the initial number of bacteria, and \( k \) is the growth constant.
Step 2: Use the doubling condition to find \( k \).
We are told that the number of bacteria doubles in 1 hour. Therefore, we have the equation: \[ 2N_0 = N_0 e^{k \cdot 1}. \] Solving for \( k \), we get: \[ 2 = e^{k} \quad \Rightarrow \quad k = \ln(2) \approx 0.693. \]
Step 3: Find the number of bacteria after \( 2\frac{1}{2} \) hours.
Substitute \( k \approx 0.693 \) and \( N_0 = 1000 \) into the equation for \( N(t) \): \[ N\left(2\frac{1}{2}\right) = 1000 e^{0.693 \cdot 2.5} \approx 1000 \times e^{1.7325} \approx 1000 \times 5.656 = 5656. \]
Step 4: Conclusion.
Thus, the number of bacteria after \( 2\frac{1}{2} \) hours is approximately 5656, which corresponds to option (C).