3.6 Successive Percentage Change

Figure: Successive discounts multiply, they do not add.

When a quantity undergoes two (or more) percentage changes one after another, you cannot simply add the two percentages. The correct combined (net) percentage change is:

Net % change = a + b + (ab)/(100)

where a and b are the individual percentage changes, taken as positive for an increase and negative for a decrease. A positive result means net increase; a negative result means net decrease. This formula extends to three changes by applying it twice (combine a and b first, then combine that result with c), and it applies to any two multiplicative changes — successive discounts, successive price hikes, population growth followed by decline, area change from two side changes, and (later) compound interest over two years.

Worked Example 1: The price of an article is increased by 20% and then decreased by 10%. Find the net percentage change. a = +20, b = −10. Net % change = 20 + (−10) + (20 × −10)/100 = 10 − 2 = 8%. The price is net 8% higher than the original. (Check directly: 100 → 120 → 120 × 0.9 = 108. Yes, +8%.)

Worked Example 2: A shopkeeper offers two successive discounts of 10% and 20% on a marked price of ₹1,000. Find the effective (single) discount percentage and the final price. a = −10, b = −20. Net % change = −10 − 20 + (−10 × −20)/100 = −30 + 2 = −28%. Effective discount = 28%. Final price = 1000 × 0.72 = ₹720. (Check directly: 1000 × 0.9 × 0.8 = 720. Discount = 280 = 28% of 1000. Confirmed.)

Worked Example 3: A number is increased by 30% and then decreased by 30%. Find the net percentage change. a = +30, b = −30. Net % change = 30 − 30 + (30 × −30)/100 = 0 − 9 = −9%. Key insight: whenever a quantity is increased by x% and then decreased by the same x%, the net effect is never zero — it is always a decrease of x²/100 percent. This shortcut is worth memorizing separately (see Shortcuts section).

Worked Example 4 (three successive changes): A trader increases the price of an article by 10%, then by a further 20%, and finally offers a discount of 10% on the twice-increased price. Find the net percentage change. Combine the first two changes: a = +10, b = +20 → 10 + 20 + (10×20)/100 = 30 + 2 = 32%. Now combine this net +32% with the discount of −10%: 32 − 10 + (32 × −10)/100 = 22 − 3.2 = 18.8% net increase. This demonstrates how to chain the formula for any number of successive changes — combine two at a time, left to right.

Worked Example 5 (three successive changes, mixed order): A number is increased by 10%, then decreased by 10%, and finally increased by 20%. Find the net percentage change. Combine the first two changes: a = +10, b = −10 → 10 − 10 + (10 × −10)/100 = 0 − 1 = −1%. Now combine this −1% with the third change of +20%: −1 + 20 + (−1 × 20)/100 = 19 − 0.2 = 18.8% net increase. (Direct check: 1.10 × 0.90 × 1.20 = 0.99 × 1.20 = 1.188, i.e., +18.8%. Confirmed.)

Worked Example 6 (reverse successive change with a zero-net trap): The price of an article, after being increased by 25% and then decreased by 20%, becomes ₹1,200. If instead the same original price had first been decreased by 25% and then increased by 20%, what would the final price be? First find the original price: net factor for +25% then −20% = 1.25 × 0.80 = 1.00 (net change = 25 − 20 + (25×−20)/100 = 5 − 5 = 0%, i.e., no net change at all). Since the net change is 0%, the original price equals the final price here: Original = ₹1,200. Now apply the reversed order of changes (−25% then +20%) to this same ₹1,200: factor = 0.75 × 1.20 = 0.90. Final price = 1,200 × 0.90 = ₹1,080. This shows that while the order of two successive multiplicative changes does not affect the result when the same two percentages are applied (multiplication is commutative), swapping which percentage is the increase and which is the decrease changes the answer completely — do not confuse the two.