Population Growth & Depreciation Formula

📑 On this page

⚡ The short answer

Growth: A = P × (1 + r/100)ⁿ. Depreciation: A = P × (1 − r/100)ⁿ. Each year's change acts on last year's value, so it is successive change on repeat. Growth beats n × r; depreciation falls short of it and never reaches zero. To go back in time, divide by the factor.

🎬 The 30-second story

The scooter that was "worth nothing"

“Got the e-scooter for ₹60,000! Dealer says it loses 20% a year.”

🛵

Aarav

Aarav buys his first ride.

“20% × 5 years = 100%. Worthless by year five. Why even insure it?”

🧮

Aarav

He does the maths at a red light.

“Year 1 it loses ₹12,000. Year 2 it loses 20% of ₹48,000. That's only ₹9,600.”

🧾

Zoya

Zoya grabs a napkin.

“After five years it's still worth about ₹19,661. It never hits zero.”

📉

Zoya

Aarav quietly renews the insurance.

Tara's maths: Depreciation takes 20% of what is left, so each year's loss is smaller than the last. Value = 60,000 × 0.8⁵ = ₹19,660.80, a total fall of about 67%, not 100%.

What do population growth and depreciation actually mean?

Both are one percentage change applied again and again. A town growing 5% a year adds 5% of this year's population, not the first year's. A car depreciating 15% a year loses 15% of its current value. It is compound interest with people, trees or gadgets instead of rupees.

🕒 Timeline

Aarav's scooter, year by year at −20%

🛵
Day 1

₹60,000

The original value.
×0.8
Year 1

₹48,000

Loses 20% of 60,000 = ₹12,000.
×0.8
Year 2

₹38,400

Loses 20% of 48,000 = ₹9,600. A smaller bite.
×0.8
Year 3

₹30,720

Loses ₹7,680. The rupee loss keeps shrinking.
×0.8
Year 5

₹19,660.80

Years 4 and 5 take ₹6,144 and ₹4,915.20. A third survives.

What is the formula for population growth and depreciation?

A 20% fall means "keep 80%", so one year is × 0.8. Five years is five multiplications in a row: 0.8⁵. Swap 0.8 for 1.05 and you have a town growing 5% a year. The power n just counts the years.

🧮 The formula

The one formula behind both

🌱

value at the start

🔁

(1 ± r/100)ⁿ

+ for growth, − for depreciation; n = number of years

🎯

value after n years

📌 Depreciation is just growth with a minus sign

Write a fall as a negative rate and one formula covers everything: r = −20 gives (1 − 0.2)ⁿ = 0.8ⁿ. If the rate changes from year to year, multiply each year's own factor: 1.1 × 0.95 × 1.08 for +10%, −5%, +8%.

🎛️ Try it yourself — change any number

Try it: Aarav's scooter

Price when new (₹) Change per year (%), negative = depreciation Years

After 5 years

19,660.8

Total change

−67.23%

60,000 × (1 − 20/100)⁵ = 19,660.8

It opens on the scooter: ₹19,660.80 after 5 years, a total change of −67.23%. Try +10% for 7 years: it roughly doubles.

Why isn't 10% a year for two years just 20%?

Because the second 10% is taken on a bigger number when you grow and a smaller one when you shrink. In a town of 1,00,000, simple thinking adds 10,000 people every year. Real growth adds 10% of the latest count: 10,000, then 11,000, then 12,100.

📊 The data

A town of 1,00,000 growing 10% a year

10% of the starting count each year10% of last year's count (real growth)

Year 1

1,10,000

Year 2

1,20,000

1,21,000

Year 3

1,30,000

1,33,100

Year 4

1,40,000

1,46,410

Year 5 · 1.1⁵ = 1.61051

1,50,000

1,61,051

After 5 years the gap is 11,051 people, and it widens every year.

Depreciation bends the other way. Each square is 1% of a laptop's value. Year one takes 10 squares; year two takes 10% of the 90 left, just 9. Total lost: 19%, not 20%.

🟦 Picture the percent

10% depreciation for two years

New

100% of the price

After year 1

lost 10 squares

After year 2

lost only 9 more: 19% gone

Growth and depreciation: worked examples

✍️ Worked example Warm-up ⏱ 30 sec

The colony that keeps growing

Rohan's colony has 12,000 residents and grows 5% a year. What will its population be after 2 years?

1
One year is × 1.05, so two years is × 1.05² = × 1.1025.
2
12,000 × 1.1025 = 13,230. Check year by year: 12,000 → 12,600 → 13,230.

Answer: 13,230 residents (not the 13,200 that "5% + 5% = 10%" gives).

⚡ The exam-hall shortcut

Two years at r% = 2r + r²/100 = 10 + 0.25 = 10.25%. And 10.25% of 12,000 is 1,230. Add it on.

✍️ Worked example CLAT level ⏱ 45 sec

The café's coffee machine

Kabir buys an espresso machine for ₹2,50,000. It depreciates 10% a year on its reduced value. What is it worth after 3 years, and how much value has it lost?

1
Factor per year = 0.9, so three years = 0.9³ = 0.729.
2
Value = 2,50,000 × 0.729 = ₹1,82,250.
3
Loss = 2,50,000 − 1,82,250 = ₹67,750, which is 27.1% of the price.

Answer: ₹1,82,250; it has lost ₹67,750 (27.1%, not 30%).

⚡ The exam-hall shortcut

0.9³ = 0.729, and 2,50,000 is a quarter of 10 lakh: a quarter of 7.29 lakh.

✍️ Worked example Tricky ⏱ 40 sec

Backwards: what did the phone cost?

Zoya's phone is worth ₹21,870 after 3 years of 10% yearly depreciation. What did it cost new?

1
Forwards, P × 0.729 = 21,870. Backwards, divide.
2
P = 21,870 ÷ 0.729 = ₹30,000.
3
Trap: adding 30% back gives ₹28,431. Wrong base, wrong rate.

Answer: ₹30,000.

⚡ The exam-hall shortcut

21,870 ÷ 729 = 30, so the answer is 30 × 1,000. Spot 729 inside the number and the division is free.

✍️ Worked example Tricky ⏱ 45 sec

Find the rate

The footfall at Meher's college fest grew from 40,000 to 57,600 in two years at a steady yearly rate. What was the rate?

1
Overall factor = 57,600 ÷ 40,000 = 1.44.
2
Two equal years: (1 + r/100)² = 1.44, so 1 + r/100 = √1.44 = 1.2.
3
Rate = 20% a year. Check: 40,000 → 48,000 → 57,600.

Answer: 20% a year, not the 22% you get by halving the 44% total.

⚡ The exam-hall shortcut

Know your powers: 1.21 = 1.1², 1.44 = 1.2², 1.331 = 1.1³, 1.728 = 1.2³. Most CLAT rates are 5, 10, 20 or 25.

What is the fastest way to solve growth questions?

Don't multiply decimals in the exam hall. CLAT uses a handful of rates, so learn their factors once. The last column shows how far the real answer sits from n × r.

Rate a year	After 2 years	After 3 years	Net change in 3 years
+5%	× 1.1025	× 1.157625	+15.76% (not 15%)
+10%	× 1.21	× 1.331	+33.1% (not 30%)
+20%	× 1.44	× 1.728	+72.8% (not 60%)
+25%	× 1.5625 (25/16)	× 1.953 (125/64)	+95.3% (not 75%)
−10%	× 0.81	× 0.729	−27.1% (not 30%)
−20%	× 0.64	× 0.512	−48.8% (not 60%)

→ scroll to see more

Fractions make the arithmetic vanish. +20% is × 6/5, so three years is × 216/125: 1,25,000 becomes 2,16,000 on sight. Setters pick starting values that cancel, which is itself a clue.

💡 The rule of 72 for "when will it double?"

Years to double ≈ 72 ÷ rate. At 10% a year: about 7.2 years (the exact figure is 7.27). At 8%: about 9 years. At 12%: about 6. Use it to throw out options fast.

How does growth hide in a CLAT caselet?

CLAT rarely says "compound". It narrates a town or a fleet changing over several years, often at different rates. Mini caselet: Sunpur had 2,00,000 residents at the end of 2023. Its population rose 10% in 2024, fell 5% in 2025 when a textile mill shut, and rose 8% in 2026 when a metro line opened.

📊 The data

Sunpur's population, end of each year

2023 · the base

2,00,000

2024 · × 1.1

2,20,000

2025 · × 0.95

2,09,000

2026 · × 1.08

2,25,720

Overall: 1.1 × 0.95 × 1.08 = 1.1286, a rise of 12.86%. Adding the rates (10 − 5 + 8) gives 13%: close enough to fool you, wrong enough to lose the mark.

More questions from the same paragraph: change from 2023 to 2025? 10 − 5 − 0.5 = +4.5%. 2024 from 2026? Divide 2,25,720 by 0.95 × 1.08, not by 1.03. And if a passage gives birth and death rates, growth is the difference: births 3.2%, deaths 1.2%, growth 2%.

What are the traps in growth and depreciation questions?

⚔️ Side by side

Same 10%, opposite bends

📈Growth

× (1 + r/100) every year

•Each year adds more than the last.
•Always beats n × r.

📉Depreciation

× (1 − r/100) every year

•Each year takes less than the last.
•Falls short of n × r; never hits zero.

Test	Growth	Depreciation
2 years at 10%	+21%	−19%
3 years at 10%	+33.1%	−27.1%
Undo one year	÷ 1.1 (−9.09%)	÷ 0.9 (+11.11%)

An option that equals n × r is almost always the bait.

⚠️ Read the last line twice

"Height after 2 years" and "how much it grows in 2 years" are different answers: CLAT 2019 Q44 lists only the first. And "10% of its original cost every year" is straight-line depreciation: subtract the same amount each year, no powers.

How CLAT asked it

📝 Asked in CLAT 2019 · Q44 Growth at a fraction a year

A tree grows at the rate of 1/5th of its height annually. By how much height will it grow after 2 years, if its present height is 75 cms?

A108 cms

B90 cms

C144 cms

D112 cms

Reveal answer + reasoning

Answer: A

Our solution (2019 has no official key). "1/5th of its height" each year is 20% growth on the current height, so each year is × 6/5. 75 → 90 → 108 cm, or 75 × 36/25 = 108. Strictly it grows by 33 cm, but 33 is not offered, so the setter means the height after 2 years. Time: 20 seconds.

Why the others fail

B — The height after one year only.
C — 1.2² = 1.44 read as 144, the factor without the 75.
D — No route reaches it: compound growth gives 108, simple growth (adding 15 cm twice) gives 105.

🧪 CLAT-style practice (not a past question) CLAT-style: fleet depreciation

Riverton's municipal corporation bought 40 electric buses in early 2024 at ₹1.25 crore each. The buses lose 20% of their value every year on the reducing balance. The city's population was 16,22,400 in early 2026, having grown at a steady 4% a year since early 2024.

What is the combined value of the 40 buses in early 2026?

A₹30 crore

B₹32 crore

C₹40 crore

D₹18 crore

Reveal answer + reasoning

Answer: B

Fleet cost = 40 × 1.25 = ₹50 crore. Two years at −20% = × 0.8² = × 0.64. Value = 50 × 0.64 = ₹32 crore. Time: 25 seconds.

Why the others fail

A — Takes 20% of the original cost twice (straight-line). The passage says reducing balance.
C — Stops after one year.
D — That is the value lost (50 − 32), not what is left.

🧪 CLAT-style practice (not a past question) CLAT-style: going back in time

(Same passage as above.)

What was Riverton's population in early 2024?

A15,00,000

B14,92,608

C15,02,222

D14,95,204

Reveal answer + reasoning

Answer: A

Two years at 4% is × 1.04² = × 1.0816. Going back, divide: 16,22,400 ÷ 1.0816 = 15,00,000. Check: 15,00,000 → 15,60,000 → 16,22,400. Time: 35 seconds.

Why the others fail

B — Subtracts 8% from the 2026 figure: wrong base and ignores compounding.
C — Divides by 1.08, treating 4% + 4% as simple growth.
D — Takes 4% off twice. A fall of 4% does not undo a rise of 4%.

💥 Myth vs fact

Myth or fact?

🙅 Myth

Losing 20% a year, anything is worthless in five years.

Tap to bust it →

✅ Fact

0.8⁵ ≈ 0.328, so about a third of the value survives. Reducing-balance depreciation never reaches zero.

🙅 Myth

10% growth for three years is 30%.

Tap to bust it →

✅ Fact

It is 33.1% (1.1³ = 1.331). Growth always beats n × r.

🙅 Myth

To find last year's population, take the growth rate off this year's.

Tap to bust it →

✅ Fact

Divide by the factor. After 10% growth, last year = this year ÷ 1.1, which is 9.09% less, not 10% less.

🎯 Quick quiz

Five quick ones: growth and depreciation

0 / 5

A hill town of 64,000 people is growing 25% a year as remote workers move in.

Q1. What will its population be after 2 years?

A delivery bike worth ₹1,00,000 depreciates 10% a year on its reduced value.

Q2. What percentage of its value has it lost after 3 years?

After 2 years of 20% yearly depreciation, Rohan's camera is worth ₹32,000.

Q3. What did it cost new?

A meme page grows from 8,000 to 13,824 followers in 3 years at a steady yearly rate.

Q4. What is the yearly growth rate?

A village of 40,000 has a birth rate of 6% and a death rate of 1% a year.

Q5. What will its population be after 2 years?

🎯 Growth and depreciation in a nutshell

Same rate every year = successive change on repeat: multiply by the factor n times.
Growth: P(1 + r/100)ⁿ. Depreciation: P(1 − r/100)ⁿ.
Growth beats n × r; depreciation falls short of it and never hits zero.
Different rates each year: multiply each year's own factor.
Going back in time: divide by the factor, never subtract the percentage.

📎 Formula card

Growth: A = P × (1 + r/100)ⁿ.
Depreciation: A = P × (1 − r/100)ⁿ.
Changing rates: A = P × (1 + r₁/100)(1 + r₂/100)(1 + r₃/100) …, falls negative.
n years ago: P = A ÷ (1 + r/100)ⁿ.
Two years at r%: net = 2r + r²/100 (r negative for depreciation).
Factors: 1.1² = 1.21, 1.1³ = 1.331, 1.2² = 1.44, 1.2³ = 1.728, 0.9² = 0.81, 0.9³ = 0.729, 0.8² = 0.64, 0.8³ = 0.512.
Doubling time: about 72 ÷ r years.
Straight-line depreciation ("r% of the original cost"): value = P × (1 − nr/100).

Frequently asked questions

What is the formula for population growth?

Population after n years = P × (1 + r/100)^n, where P is today's population and r is the yearly growth rate. A town of 12,000 growing 5% a year has 12,000 × 1.1025 = 13,230 people after two years. If the rate changes each year, multiply each year's own factor.

What is the formula for depreciation?

On the reducing balance, value after n years = P × (1 − r/100)^n. A machine bought for ₹2,50,000 losing 10% a year is worth 2,50,000 × 0.729 = ₹1,82,250 after three years. If the question says the fall is a percentage of the original cost every year, it is straight-line depreciation: subtract the same amount each year.

How do I find the population or value n years ago?

Divide the present figure by the growth factor raised to n: P = A ÷ (1 + r/100)^n. A city of 16,22,400 that grew 4% a year for two years had 16,22,400 ÷ 1.0816 = 15,00,000 people. Never subtract the percentage from today's figure; that uses the wrong base.

Is population growth the same as compound interest?

Yes, the maths is identical. Compound interest is money growing at a fixed rate on its latest value; population growth is a headcount doing the same. The formula, the factor table and the shortcuts all carry over. Depreciation is the same idea with a negative rate.

Can depreciation ever make a value zero?

Not on the reducing balance. Each year you keep a fixed share of what is left, such as 80%, and a share of something is never zero. After five years about 33% survives; after ten, about 11%. Only straight-line depreciation, which removes a fixed amount each year, can reach zero.