Growth and depreciation through repeated percentage changes

Learning outcome

Calculate repeated growth, depreciation and mixed changes, and recover an original value by reversing the full multiplier.

Concepts and assumptions

A percentage change uses the value immediately before that change. Growth by g% retains the old value and adds g/100 of it, giving multiplier 1 + g/100. Depreciation by d% removes d/100 of the current value, giving multiplier 1 - d/100.

For n complete periods at a constant rate:

Final value = original value × (1 + g/100)^n for growth.

Final value = original value × (1 - d/100)^n for depreciation.

These formulas repeat the same multiplication. Equal percentage depreciation produces decreasing monetary reductions because the base shrinks. It is not the same as subtracting a fixed amount every year.

For different rates or directions, multiply the appropriate factors in sequence. To find the net percentage change, divide final value minus original value by the original value and multiply by 100. A negative result indicates a decrease. To work backwards, divide the final value by the complete multiplier.

Assume positive initial values, no separate additions, removals or payments, and changes only at the stated period boundaries. Growth rates are nonnegative; depreciation rates range from 0% to below 100%, keeping values and reverse divisors positive. A 100% depreciation instead gives zero and cannot be reversed uniquely.

These are changes per named period, not nominal annual interest quotes. Do not divide an annual depreciation rate by 12 and assume an equivalent monthly model. Although a year has 12 months, fractional periods or days need an explicit model; no day-count rule is assumed here. Keep intermediate values exact, rounding final money to ₹0.01 only when needed, with half a paise rounded upward.

Worked examples

Example 1 — Repeated growth. A value of ₹18,000 grows by 5% annually for 2 years. First value = 18000 × 1.05 = ₹18,900. Second value = 18900 × 1.05 = ₹19,845. Total growth = 19845 - 18000 = ₹1,845; percentage growth = (1845 ÷ 18000) × 100 = 10.25%, not 10%.

Example 2 — Repeated depreciation. A machine valued at ₹62,500 depreciates 12% annually for 3 years. Successive values are 62500 × 0.88 = ₹55,000, then 55000 × 0.88 = ₹48,400, then 48400 × 0.88 = ₹42,592. Total depreciation = 62500 - 42592 = ₹19,908. Each reduction uses the current value.

Example 3 — Reverse mixed changes. After a 20% increase followed by a 15% decrease, a value is ₹24,480. Combined multiplier = 1.20 × 0.85 = 1.02. Original value = 24480 ÷ 1.02 = ₹24,000. Check: 24000 × 1.20 = ₹28,800, then 28800 × 0.85 = ₹24,480. Net growth is 2%, not 20% - 15% = 5%.

Common mistakes

Equal percentage increases and decreases do not cancel: (1 + x) × (1 - x) = 1 - x^2 for a decimal rate x between 0 and 1. This identity requires two equal opposite rates, each applied successively.

Practice questions

1. ₹26,000 grows by 6% annually for 2 years. Find the final value.
2. ₹45,000 depreciates by 10% annually for 2 years. Find the final value and total depreciation.
3. ₹16,000 rises by 25%, then falls by 25%. Find the final value and net percentage change.
4. After 2 annual depreciations of 8%, a value is ₹33,856. Find its original value.

Worked answers

1. Values are 26000 × 1.06 = ₹27,560 and 27560 × 1.06 = ₹29,213.60.
2. Values are 45000 × 0.90 = ₹40,500 and 40500 × 0.90 = ₹36,450. Total depreciation = 45000 - 36450 = ₹8,550.
3. Values are 16000 × 1.25 = ₹20,000 and 20000 × 0.75 = ₹15,000. Decrease = ₹1,000, or (1000 ÷ 16000) × 100 = 6.25%.
4. Combined multiplier = 0.92^2 = 0.8464. Original value = 33856 ÷ 0.8464 = ₹40,000. Checking gives ₹36,800 after one year and ₹33,856 after two.