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Population Growth Problems

Introduction

Population growth problems, depreciation के उलट होते हैं - यहाँ value हर साल घटने की बजाय population (या entities की संख्या) एक constant percentage rate से हर साल या periodically बढ़ती है। ऐसे problems future या past population size estimate करने में बहुत काम आते हैं।

Pattern: Population Growth Problems

Pattern

मुख्य concept: ‘T’ साल बाद की Population = Present Population × (1 + Growth Rate/100)T.

इसी तरह, अगर population घटती है (decay), तो (1 - R/100) का उपयोग करें, (1 + R/100) की जगह।

Step-by-Step Example

Question

एक town की population 1,00,000 है और यह 5% प्रति वर्ष की दर से बढ़ती है। 3 साल बाद population ज्ञात करें।

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:

    3 साल बाद 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 के लिए population निकालना।

2. दिए गए वर्षों से पहले की population (reverse growth)।

3. Increase और decrease rates को combine करना (migration + growth)।

4. Multi-stage growth (अलग-अलग सालों में अलग rate)।

Trick to Always Use

  • Step 1: Future growth के लिए → (1 + R/100)T से multiply करें।
  • Step 2: Past population के लिए → (1 + R/100)T से divide करें।
  • Step 3: Fractional years के लिए exponent (जैसे 1.5 या 2.25) का उपयोग करें।
  • Step 4: Variable growth rates हों तो stepwise compounding करें।

Summary

Summary

  • P × (1 + R/100)T population growth के लिए और P × (1 - R/100)T population decay के लिए उपयोग करें।
  • Reverse questions में multiply की जगह divide करें।
  • Fractional years में exponent का उपयोग करें (जैसे 1.5)।
  • Accuracy के लिए हमेशा quick forward या backward check करें।

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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