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Population Growth/Decay

Introduction

Some aptitude problems deal with population growth or decay (reduction). These are based on compound percentage, because the change happens every year (or time period), not just once.

For growth → population increases each year. For decay → population decreases each year. We solve these by applying the formula step by step.

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

A town has a population of 10,000. It increases at the rate of 10% per year. Find the population after 2 years.

Solution

  1. Step 1: Identify given values.

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

  2. Step 2: Write the formula for growth.

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

  3. Step 3: Simplify the expression.

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

  4. Step 4: Final Answer.

    Population after 2 years = 12,100

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

    After 1 year = 10,000 + 10% of 10,000 = 11,000
    After 2 years = 11,000 + 10% of 11,000 = 12,100 ✅

Quick Variations

Decay case: If the population decreases by 10% per year → P2 = 10,000 × (0.9)2 = 8,100.

Mixed growth & decay: If a population increases by 10% in the first year and decreases by 20% in the next year, apply successive change formula.

Trick to Always Use

  • Growth: Multiply by (1 + r/100) each year.
  • Decay: Multiply by (1 - r/100) each year.
  • n years: Raise the factor to power n.
  • Cross-check with year-by-year calculation if confused.

Summary

Summary

The Population Growth/Decay pattern uses compound percentage to calculate changes over time.

Formula: Pn = P × (1 ± r/100)n

  • “+” sign for growth (increase).
  • “-” sign for decay (decrease).
  • Apply step by step for each year if unsure.

With practice, these problems become quick and accurate to solve.

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.

Mock Test

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