A town has population 2,65,000 at the beginning of 1986. If the annual growth rate is 52 per thousand (i.e., 5.2% per year), find the population at the beginning of 1991.

Difficulty: Medium

Correct Answer: 3,41,400

Explanation:


Introduction / Context:
Year-on-year percentage growth compounds. Over 5 years (from the start of 1986 to the start of 1991), the population multiplies by (1 + r)^5 with r = 5.2% = 0.052.


Given Data / Assumptions:

  • Initial population P0 = 2,65,000.
  • Annual growth rate r = 0.052 (52 per thousand).
  • Years n = 5.


Concept / Approach:

  • Population after n years: Pn = P0(1 + r)^n.
  • Round to the nearest hundred as per common exam convention if needed.


Step-by-Step Solution:

(1.052)^2 ≈ 1.1067; (1.052)^4 ≈ 1.1067^2 ≈ 1.2248; multiply by 1.052 ⇒ (1.052)^5 ≈ 1.289P5 ≈ 2,65,000 × 1.289 ≈ 3,41,585Closest listed population ≈ 3,41,400


Verification / Alternative check:
Using a more precise value (1.052)^5 ≈ 1.28936 gives ≈ 3,41,690; among the options, 3,41,400 is the nearest standard rounded choice.


Why Other Options Are Wrong:

  • 3,40,400 and 3,42,400 are farther from the precise computation; 3,43,400 overshoots.


Common Pitfalls:

  • Treating the growth as linear (simple increase) instead of compound.


Final Answer:
3,41,400

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