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Population Growth Problems

Introduction

Population growth problems are the reverse of depreciation - instead of value decreasing each year, the population (or number of entities) increases at a constant percentage rate annually or periodically. These problems are widely used to estimate future or past population sizes.

Pattern: Population Growth Problems

Pattern

The key concept: Population after ‘T’ years = Present Population × (1 + Growth Rate/100)T.

Similarly, if population decreases (decay), use (1 - R/100) instead of (1 + R/100).

Step-by-Step Example

Question

The population of a town is 1,00,000 and it increases at 5% per annum. Find the population after 3 years.

Solution

  1. Step 1: Identify values

    Present Population (P) = 1,00,000; Growth Rate (R) = 5% p.a.; Time (T) = 3 years.

  2. Step 2: Apply formula

    Future Population = P × (1 + R/100)T = 1,00,000 × (1.05)3.

  3. Step 3: Compute

    (1.05)3 = 1.157625 → Future Population = 1,00,000 × 1.157625 = 1,15,762.5.

  4. Final Answer:

    Population after 3 years = 1,15,763 (approx.)

  5. Quick Check:

    5% yearly growth: 1,00,000 → 1,05,000 → 1,10,250 → 1,15,762.5 ✅

Quick Variations

1. Finding population after fractional years.

2. Finding population before given years (reverse growth).

3. Combining increase and decrease rates (migration + growth).

4. Multi-stage growth (different rates for different years).

Trick to Always Use

  • Step 1: For future growth → multiply by (1 + R/100)T.
  • Step 2: For past population → divide by (1 + R/100)T.
  • Step 3: For fractional years → use exponent (like 1.5 or 2.25).
  • Step 4: For variable growth rates, apply stepwise compounding.

Summary

Summary

  • Use P × (1 + R/100)T for population growth and P × (1 - R/100)T for population decay.
  • For reverse questions, divide instead of multiply.
  • For fractional years, use exponents (like 1.5).
  • Always check with a quick forward or backward computation for accuracy.

Practice

(1/5)
1. The population of a town is 50,000 and it grows at 3% per annum. Find the population after 2 years.
easy
A. 53,045.00
B. 52,000.00
C. 54,000.00
D. 51,500.00

Solution

  1. Step 1: Identify values

    Present population P = 50,000; growth rate R = 3% p.a.; time T = 2 years.

  2. Step 2: Apply growth formula

    Future population = P × (1 + R/100)^T = 50,000 × (1.03)^2.

  3. Step 3: Compute

    (1.03)^2 = 1.0609 → Population = 50,000 × 1.0609 = 53,045.00.

  4. Final Answer:

    Population after 2 years = 53,045.00 → Option A.

  5. Quick Check:

    Yearwise: 50,000 → 51,500 (after 1y) → 53,045 (after 2y) ✅
Hint: Multiply by (1 + R/100)^T directly for future population.
Common Mistakes: Adding percentage points linearly instead of compounding yearly.
2. The population of a city is 1,21,000 after 2 years. If it grows at 10% per annum, find the population two years ago.
easy
A. 1,10,000
B. 1,00,000
C. 1,05,000
D. 1,02,500

Solution

  1. Step 1: Identify values

    Future population A = 1,21,000; growth rate R = 10% p.a.; time T = 2 years.

  2. Step 2: Use reverse formula

    Past population P = A / (1 + R/100)^T = 1,21,000 / (1.1)^2.

  3. Step 3: Compute

    (1.1)^2 = 1.21 → P = 1,21,000 / 1.21 = 1,00,000.

  4. Final Answer:

    Population two years ago = 1,00,000 → Option B.

  5. Quick Check:

    Forward check: 1,00,000 × 1.1 × 1.1 = 1,21,000 ✅
Hint: Divide by (1 + R/100)^T to find past population from a known future value.
Common Mistakes: Subtracting percentages instead of dividing by growth factor.
3. A village has 20,000 people and grows at 2% per annum. Find its population after 1.5 years.
easy
A. 20,602.99
B. 20,400.99
C. 20,800.99
D. 21,000.99

Solution

  1. Step 1: Identify values

    P = 20,000; R = 2% p.a.; T = 1.5 years.

  2. Step 2: Apply formula with fractional exponent

    A = P × (1 + R/100)^T = 20,000 × (1.02)^1.5.

  3. Step 3: Compute

    (1.02)^1.5 = 1.02 × √1.02 ≈ 1.02 × 1.009950 ≈ 1.0301495 → A ≈ 20,000 × 1.0301495 = 20,602.99.

  4. Final Answer:

    Population after 1.5 years ≈ 20,602.99 → Option A.

  5. Quick Check:

    After 1 year: 20,400; additional half-year factor ≈ √1.02 ≈ 1.00995 → 20,400 × 1.00995 ≈ 20,603 ✅
Hint: For fractional years multiply by (1 + R/100) and then by the root for the fractional part (e.g., √ for 0.5).
Common Mistakes: Treating fractional period as full or using simple interest for the fraction.
4. A town has 10,000 inhabitants and its population grows at 8% per annum compounded quarterly. Find the population after 1 year.
medium
A. 10,800.40
B. 10,824.14
C. 10,824.32
D. 11,000.58

Solution

  1. Step 1: Identify values

    P = 10,000; annual R = 8%; compounding quarterly → periods per year = 4; r_per = 8/4 = 2% = 0.02; total periods for 1 year = 4.

  2. Step 2: Apply per-period formula

    A = P × (1 + r_per)^{periods} = 10,000 × (1.02)^4.

  3. Step 3: Compute

    (1.02)^2 = 1.0404; (1.02)^4 = 1.08243216 → A = 10,000 × 1.08243216 = 10,824.32.

  4. Final Answer:

    Population after 1 year = 10,824.32 → Option C.

  5. Quick Check:

    Quarter steps: 10,000 → 10,200 → 10,404 → 10,612.08 → 10,824.32 ✅
Hint: Convert R to per-period rate and raise (1 + r_per) to number of periods.
Common Mistakes: Using annual 8% once instead of applying per-quarter compounding.
5. A population grows from 50,000 to 80,000 at 6% per annum. Approximately how many years does this take?
medium
A. 7.00 years
B. 6.00 years
C. 8.50 years
D. 8.07 years

Solution

  1. Step 1: Identify values

    Initial P = 50,000; Final A = 80,000; annual rate R = 6% → factor per year = 1.06.

  2. Step 2: Use logarithmic formula for time

    T = ln(A/P) / ln(1 + R/100) = ln(80,000/50,000) / ln(1.06).

  3. Step 3: Compute

    80,000/50,000 = 1.6; ln(1.6) ≈ 0.47000363; ln(1.06) ≈ 0.05826891 → T ≈ 0.47000363 / 0.05826891 ≈ 8.07 years.

  4. Final Answer:

    Time ≈ 8.07 years → Option D.

  5. Quick Check:

    Check growth factor: (1.06)^{8.07} ≈ 1.6 → 50,000 × 1.6 = 80,000 ✅
Hint: Use T = ln(A/P) ÷ ln(1 + R/100) when solving for time.
Common Mistakes: Trying linear scaling (A/P ÷ R) instead of using logarithms for exponential growth.

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