Lesson 29 of 100 | Quantitative Aptitude / Percentages / प्रतिशत
Successive percentage changes
Learning outcome
Combine successive increases and decreases, calculate the net change, and explain why equal percentage increases and decreases do not cancel.
Changing bases require multiplication
Each successive percentage acts on the amount immediately before that step. An increase of p% has factor 1 + p/100; a decrease has factor 1 − p/100. Decreases here are below 100%, keeping amounts positive.
Therefore final amount = original × first factor × second factor. Adding the stated percentages ignores the changed base.
Let F be the product of all factors. If F > 1, the net increase is (F − 1) × 100%. If F < 1, the net decrease is (1 − F) × 100%. If F = 1, there is no net change. Compare the final amount with the original, not with an intermediate amount.
For equal upward and downward changes of p%, the combined factor is (1 + p/100) × (1 − p/100) = 1 − (p/100)². The result is a decrease of (p²/100)%. After the increase, the equal percentage reduction removes more than the first increase added, because its base is larger.
For proportional changes without rounding or fixed additions, reversing factor order preserves the final amount, though intermediate amounts may differ.
Worked example 1
A 40 L stock increases by 25%, then decreases by 10%. Find the final stock and net change.
First, 40 × 1.25 = 50 L. Next, 50 × 0.90 = 45 L. The net increase is 5 L, so (5/40) × 100% = 12.5%. It is not the result of simply subtracting the stated rates.
Worked example 2
A ₹2,500 price rises by 20%, then falls by 20%.
The increased price is 2500 × 1.20 = ₹3000. The decreased price is 3000 × 0.80 = ₹2400. The loss is ₹100, or (100/2500) × 100% = 4%. Equal percentage changes did not cancel.
Worked example 3
An 80 kg stock decreases by 12.5%, then by 20%.
After the first decrease, 80 × 0.875 = 70 kg. After the second, 70 × 0.80 = 56 kg. The combined factor is 0.70, so the net decrease is 30%, not 32.5%.
Common traps
Do not apply both rates independently to the original. Do not add successive decreases. Use the full multiplier, including the retained whole, and avoid premature rounding.
Practice questions
- A ₹1,200 price rises by 10%, then by 15%. Find the final price and net percentage increase.
- A 900 mL stock decreases by 10%, then increases by 20%. Find the final volume and net percentage change.
- A 64 L stock increases by 25%, then decreases by 25%. Find the final volume and net percentage change.
- A ₹2,000 price increases by 10%, decreases by 20%, then increases by 25%. Find its final value and net percentage change.
Worked answers
- The prices are 1200 × 1.10 = ₹1320, then 1320 × 1.15 = ₹1518. The net increase is (318/1200) × 100% = 26.5%, using the original base.
- Calculate 900 × 0.90 = 810 mL, then 810 × 1.20 = 972 mL. The net increase is (72/900) × 100% = 8%.
- The volumes become 64 × 1.25 = 80 L, then 80 × 0.75 = 60 L. The net decrease is (4/64) × 100% = 6.25%.
- The successive prices are ₹2200, ₹1760 and ₹2200. Their combined factor is 1.10 × 0.80 × 1.25 = 1.10, so the net increase is 10%.