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Population Growth Problems

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Introduction

Population Growth problems என்பது depreciation க்கு எதிர்மாறானது - ஒவ்வொரு ஆண்டும் value குறைவதற்கு பதிலாக, population (அல்லது entities எண்ணிக்கை) ஒரு நிலையான percentage வீதத்தில் வருடாந்திரமாக அல்லது குறிப்பிட்ட கால இடைவெளியில் அதிகரிக்கும். இத்தகைய problems, எதிர்காலம் அல்லது கடந்த கால population அளவுகளை கணக்கிடுவதற்கு பரவலாக பயன்படுத்தப்படுகின்றன.

Pattern: Population Growth Problems

Pattern: Population Growth Problems

Key concept: T ஆண்டுகளுக்குப் பிறகு Population = Present Population × (1 + Growth Rate/100)T.

அதேபோல், population குறைந்தால் (decay), (1 + R/100)க்கு பதிலாக (1 - R/100) பயன்படுத்த வேண்டும்.

Step-by-Step Example

Question

ஒரு நகரத்தின் population 1,00,000. அது வருடத்திற்கு 5% வீதத்தில் அதிகரிக்கிறது. 3 ஆண்டுகளுக்குப் பிறகு population ஐ காண்க.

Solution

  1. Step 1: Values ஐ கண்டறிதல்

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

  2. Step 2: Formula பயன்படுத்துதல்

    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:

    3 ஆண்டுகளுக்குப் பிறகு Population = 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. Fractional years க்கு பின் population கணக்கிடுதல்.

2. கொடுக்கப்பட்ட ஆண்டுகளுக்கு முன் population (reverse growth) கண்டறிதல்.

3. Increase + decrease rates சேர்த்து (migration + growth) கணக்கிடுதல்.

4. Multi-stage growth (ஒவ்வொரு ஆண்டுக்கும் வேறுபட்ட rates).

Trick to Always Use

  • Step 1: Future growth க்கு → (1 + R/100)T ஆல் multiply செய்யுங்கள்.
  • Step 2: Past population க்கு → (1 + R/100)T ஆல் divide செய்யுங்கள்.
  • Step 3: Fractional years க்கு → 1.5, 2.25 போன்ற exponent பயன்படுத்துங்கள்.
  • Step 4: Variable rates இருந்தால் stepwise compounding பயன்படுத்துங்கள்.

Summary

  • Population growth க்கு P × (1 + R/100)T மற்றும் population decay க்கு P × (1 - R/100)T பயன்படுத்தவும்.
  • Reverse questions இல் multiply செய்வதற்குப் பதிலாக divide செய்யவும்.
  • Fractional years இருந்தால் exponent பயன்படுத்தவும் (1.5 போன்றது).
  • எப்போதும் forward அல்லது backward quick check மூலம் accuracy ஐ உறுதி செய்யுங்கள்.

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.