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How to Solve Percentage Population Growth Problems

Solve population growth aptitude problems with the compounding formula P(1+r/100)^n, mixed growth/decline years, and practice questions.

mediumQ103 of 225 in Aptitude Est. time: 5 minsLast updated:
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Expected Interview Answer

Population growth at a fixed percentage rate compounds exactly like compound interest, so a population P growing at r% per year for n years becomes P×(1+r/100)^n, and decline uses (1−r/100)^n instead.

Because each year’s growth is calculated on the already-grown population from the previous year, not the original population, percentage population growth is multiplicative, not additive: you cannot simply add r% of the original population n times. When growth and decline alternate across years, apply each year’s factor in sequence: multiply by (1+r/100) for a growth year and (1−r/100) for a decline year, in the order given. To find the population n years ago from today’s population, divide by (1+r/100)^n rather than multiplying, since you are reversing the compounding.

  • One compounding formula, P(1+r/100)^n, handles growth, decline, and mixed years
  • Prevents the common error of adding percentages instead of compounding them
  • The reverse (divide) formula finds past population from a present figure

AI Mentor Explanation

A stadium’s season-ticket holder base growing at 10% every year is like compound interest, not simple addition: 10,000 holders becoming 11,000 after year one, then 12,100 after year two, since the second year’s 10% is computed on 11,000, not the original 10,000. This is exactly why population growth problems use P×(1+r/100)^n rather than P + n×(r% of P) — each period’s increase builds on the already-grown total.

Worked example (mixed growth and decline)

Step-by-Step Explanation

  1. Step 1

    Identify the compounding rate and period

    Extract the yearly (or periodic) growth/decline rate r and number of periods n.

  2. Step 2

    Apply the compound formula

    Growth: P×(1+r/100)^n; decline: P×(1−r/100)^n.

  3. Step 3

    Chain mixed years in sequence

    For alternating growth/decline years, multiply by each year’s factor in order given.

  4. Step 4

    Reverse for past population

    Divide today’s population by (1+r/100)^n to find the population n years ago.

What Interviewer Expects

  • Recognition that population growth compounds, mirroring compound interest
  • Correct use of (1+r/100)^n for growth and (1−r/100)^n for decline
  • Correct sequencing when growth and decline years alternate
  • Correct reverse calculation for past population from a present figure

Common Mistakes

  • Adding r% of the original population every year instead of compounding
  • Using the same growth factor for a decline year
  • Multiplying instead of dividing when working backward to a past population
  • Misordering growth and decline years when they alternate

Best Answer (HR Friendly)

Population growth at a fixed percentage compounds exactly like compound interest, so I use P times one plus r over 100, raised to the number of years, for growth, and one minus r over 100 for decline. If growth and decline years alternate, I chain each year’s factor in the given order rather than averaging the rates. To go backward and find a past population, I divide by the same compounding factor instead of multiplying.

Follow-up Questions

  • How would you find the annual growth rate given population figures for two years?
  • How does population growth differ when the rate itself changes each year?
  • How would you model population growth with a fixed net migration number added each year?
  • Why can’t you simply add the percentage growth rates across multiple years?

MCQ Practice

1. A town of 40,000 grows at 5% per year. Its population after 2 years is closest to?

40000×(1.05)² = 40000×1.1025 = 44,100.

2. A population of 10,000 declines by 10% per year for 2 years. The population after 2 years is?

10000×(0.9)² = 10000×0.81 = 8,100.

3. A city’s population today is 33,275 after growing 10% and 9% in two successive years. The population before these two years (approximately) is?

33275 ÷ (1.10×1.09) = 33275 ÷ 1.199 ≈ 27,750.

Flash Cards

Population growth formula?P×(1+r/100)^n for growth over n periods at rate r%.

Population decline formula?P×(1−r/100)^n for decline over n periods.

Why not add percentages across years?Each year’s growth compounds on the already-grown population, not the original.

How to find past population from present?Divide by (1+r/100)^n instead of multiplying.

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