Question

Population of towns A and B increases at a rate proportional to population. In 1984, both were 20,000. In 1989, A was 25,000 and B was 28,000. The difference in 1994 was

• A. 5950
• B. 8000
• C. 7950
• D. 6950

Hint:

Population at $2t$ is $P_{0} \times (P_{t}/P_{0})^{2}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Population growth follows the exponential law $P(t) = P_0 e^{kt}$. We need to find the populations at $t=10$ years (from $1984$ to $1994$).

Step 2: Key Formula or Approach:
For a 5-year period ($t=5$), if population grows from $P_0$ to $P_1$, then for the next 5 years (total $t=10$), it grows to $P_2 = P_1 \times (\frac{P_1}{P_0})$.

Step 3: Detailed Explanation:
Initial population (end of 1984, $t=0$): $P_{A0} = 20000$, $P_{B0} = 20000$.
End of 1989 ($t=5$): $P_{A5} = 25000$, $P_{B5} = 28000$.
Growth factor for town A in $5$ years: $r_A = \frac{25000}{20000} = 1.25$.
Growth factor for town B in $5$ years: $r_B = \frac{28000}{20000} = 1.4$.
Population at the end of 1994 ($t=10$):
$P_{A10} = P_{A5} \times r_A = 25000 \times 1.25 = 31250$.
$P_{B10} = P_{B5} \times r_B = 28000 \times 1.4 = 39200$.
Difference $= P_{B10} - P_{A10} = 39200 - 31250 = 7950$.

Step 4: Final Answer:
The difference in populations is $7950$.