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Population Growth/Decay

Introduction

कुछ aptitude problems में population growth या decay (reduction) से जुड़े सवाल होते हैं। ये compound percentage पर based होते हैं, क्योंकि change हर साल (या time period) में होता है, सिर्फ एक बार नहीं।

Growth में population हर साल बढ़ती है। Decay में population हर साल घटती है। इनको step-by-step formula लगाकर solve किया जाता है।

Pattern: Population Growth/Decay

Pattern

Growth: Pn = P × (1 + r/100)n

Decay: Pn = P × (1 - r/100)n

P = initial population, r = rate %, n = number of years

Step-by-Step Example

Question

एक town की population 10,000 है। यह 10% per year की दर से बढ़ती है। 2 years बाद population निकालें।

Solution

  1. Step 1: दिए गए values पहचानें।

    Initial population = P = 10,000
    Growth rate = r = 10%
    Years = n = 2

  2. Step 2: Growth का formula लिखें।

    Sentence: Population after n years = P × (1 + r/100)n.
    Math: P2 = 10,000 × (1 + 10/100)2

  3. Step 3: Expression simplify करें।

    = 10,000 × (1.1)2
    = 10,000 × 1.21

  4. Step 4: Final Answer.

    2 साल बाद population = 12,100

  5. Step 5: Quick Check (year-by-year)।

    1st year: 10,000 + 10% of 10,000 = 11,000
    2nd year: 11,000 + 10% of 11,000 = 12,100 ✅

Quick Variations

Decay case: Population हर साल 10% घटे → P2 = 10,000 × (0.9)2 = 8,100.

Mixed growth & decay: अगर पहले साल 10% बढ़े और अगले साल 20% घटे, तो successive change formula apply करें।

Trick to Always Use

  • Growth: हर साल (1 + r/100) से multiply करें।
  • Decay: हर साल (1 - r/100) से multiply करें।
  • n years: Factor को power n तक raise करें।
  • Confusion हो तो year-by-year calculation से cross-check करें।

Summary

Summary

Population Growth/Decay pattern में time के साथ होने वाले changes को compound percentage से calculate किया जाता है।

Formula: Pn = P × (1 ± r/100)n

  • “+” → growth (increase)
  • “-” → decay (decrease)
  • हर साल step-by-step apply करने से mistake नहीं होगी।

Practice करने पर ये problems बहुत जल्दी और accurately solve होने लगते हैं।

Practice

(1/5)
1. The population of a town is 10,000. It increases at 10% per year. What will be the population after 1 year?
easy
A. 11,000
B. 10,800
C. 11,200
D. 10,900

Solution

  1. Step 1: Identify base and growth rate

    Initial population = 10,000, growth rate = 10%.

  2. Step 2: Apply the growth formula

    Population after 1 year = 10,000 × (1 + 10/100) = 10,000 × 1.1.

  3. Step 3: Compute

    10,000 × 1.1 = 11,000.

  4. Final Answer:

    11,000 → Option A

  5. Quick Check:

    10% of 10,000 = 1,000 → 10,000 + 1,000 = 11,000 ✅
Hint: Multiply by 1.1 for +10% growth in one year.
Common Mistakes: Adding 10% twice instead of multiplying correctly.
2. A town has 20,000 people. The population increases 5% per year. Find the population after 2 years.
easy
A. 22,000
B. 22,050
C. 21,500
D. 21,800

Solution

  1. Step 1: Note initial, rate and period

    Initial = 20,000, rate = 5%, years = 2.

  2. Step 2: Use compound growth formula

    Population = 20,000 × (1.05)^2.

  3. Step 3: Compute

    (1.05)^2 = 1.1025 → 20,000 × 1.1025 = 22,050.

  4. Final Answer:

    22,050 → Option B

  5. Quick Check:

    Year 1 = 20,000 × 1.05 = 21,000; Year 2 = 21,000 × 1.05 = 22,050 ✅
Hint: Use (1 + r/100)^n to avoid step-by-step repetition.
Common Mistakes: Multiplying by 1.05 only once instead of squaring for 2 years.
3. The population of a city is 50,000. It decreases at 4% per year. Find the population after 1 year.
medium
A. 47,500
B. 48,000
C. 48,500
D. 49,000

Solution

  1. Step 1: Identify base and decay rate

    Initial = 50,000, decay = 4%.

  2. Step 2: Apply decay multiplier

    Population after 1 year = 50,000 × (1 - 4/100) = 50,000 × 0.96.

  3. Step 3: Compute

    50,000 × 0.96 = 48,000.

  4. Final Answer:

    48,000 → Option B

  5. Quick Check:

    4% of 50,000 = 2,000 → 50,000 - 2,000 = 48,000 ✅
Hint: Multiply by (1 - r/100) for decay.
Common Mistakes: Subtracting 4 instead of 4% from total.
4. A village has a population of 8,000. It increases 8% annually. Find the population after 2 years.
medium
A. 9,200
B. 9,300
C. 9,331
D. 9,250

Solution

  1. Step 1: Record base, rate and period

    Initial = 8,000, rate = 8%, years = 2.

  2. Step 2: Use compound growth

    Population = 8,000 × (1.08)^2.

  3. Step 3: Compute

    (1.08)^2 = 1.1664 → 8,000 × 1.1664 = 9,331.

  4. Final Answer:

    9,331 → Option C

  5. Quick Check:

    Year 1 = 8,640; Year 2 = 8,640 × 1.08 = 9,331 ✅
Hint: Use (1.08)^2 directly instead of yearly steps.
Common Mistakes: Forgetting to square the factor for 2 years.
5. The population of a town is 12,000. It decreases 10% annually for 2 years. Find the population after 2 years.
medium
A. 9,700
B. 9,600
C. 9,800
D. 9,720

Solution

  1. Step 1: Note initial, decay rate and period

    Initial = 12,000, decay = 10%, years = 2.

  2. Step 2: Apply compound decay

    Population after 2 years = 12,000 × (0.9)^2.

  3. Step 3: Compute

    (0.9)^2 = 0.81 → 12,000 × 0.81 = 9,720.

  4. Final Answer:

    9,720 → Option D

  5. Quick Check:

    Year 1 = 10,800; Year 2 = 10,800 - 10% of 10,800 = 9,720 ✅
Hint: Multiply by (0.9)^n for repeated 10% decay.
Common Mistakes: Subtracting 20% directly for 2 years instead of compounding.

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