In this note, ask: What is the 100% base? The denominator is the reference quantity, not necessarily the larger value.
Assumptions: Assume a positive original, matching units, and no extra fixed adjustments. Successive changes use the immediately preceding value. Recovery requires a nonzero combined multiplier. All calculations are exact.
Choose the method
| Question type | Calculation | Base or safeguard |
|---|---|---|
| A is what percent of B? | A ÷ B × 100 | B is the reference. |
| Percentage change | (new − old) ÷ old × 100 | Use the starting value. |
| Rate: p% to q% | q − p percentage points | Relative change: (q − p) ÷ p × 100, for p > 0. |
| Two successive changes | Multiply their factors | The second base is updated. |
| Recover the original | final ÷ combined multiplier | Divide; do not “subtract back”. |
An r% increase uses 1 + r/100; a decrease uses 1 − r/100. The new value is the original plus or minus the stated fraction of it. For combined multiplier M, net percentage change is (M − 1) × 100; a negative result means a decrease.
Four worked contrasting examples
Example 1 — Same difference, different denominators
Two strips measure 120 cm and 150 cm. Difference = 150 − 120 = 30 cm.
The longer is 30 ÷ 120 × 100 = 25% longer than the shorter. The shorter is 30 ÷ 150 × 100 = 20% shorter than the longer. Meanwhile, 120 ÷ 150 × 100 = 80% describes the shorter length as a share, not a decrease. “Compared with” or “than” identifies the reference.
Example 2 — Percentage points versus percentages
A task-completion rate rises from 40% to 50%.
Difference = 50 − 40 = 10 percentage points. Relative increase = 10 ÷ 40 × 100 = 25%, not 10%. Subtraction measures the gap between rates; division measures growth relative to the starting rate.
Example 3 — Successive changes do not simply add
From 100 units, successive rises of 20% and 10% give 100 × 1.20 = 120, then 120 × 1.10 = 132. Net increase = (132 − 100) ÷ 100 × 100 = 32%, not 30%.
Contrast a 20% rise followed by a 20% fall: 100 × 1.20 = 120; 120 × 0.80 = 96. Net decrease = (100 − 96) ÷ 100 × 100 = 4%. The fall removes 24 units; the rise added only 20. For an r% rise and r% fall, with 0 < r < 100, the multiplier is (1 + r/100)(1 − r/100) = 1 − (r/100)²; net loss = (r²/100)%.
Example 4 — Recover the original by division
After a 25% rise and a 12% fall, a quantity becomes 220 units.
Multiplier = 1.25 × 0.88 = 1.10, a net 10% increase. Original = 220 ÷ 1.10 = 200 units. Check: 200 × 1.25 = 250; 250 × 0.88 = 220. A 10% reduction instead gives 220 × 0.90 = 198: it wrongly uses 220 as the base.
Practice questions
- How much longer is 80 cm than 50 cm, and how much shorter is 50 cm than 80 cm, in percentages?
- A completion rate rises from 60% to 72%. Find the percentage-point increase and relative percentage increase.
- A quantity of 400 units rises by 25%, then falls by 25%. Find the final quantity and net percentage change.
- A quantity falls by 20%, then rises by 10%, becoming 264 units. Find its original value.
Worked answers
- Difference = 80 − 50 = 30 cm. Longer: 30 ÷ 50 × 100 = 60%. Shorter: 30 ÷ 80 × 100 = 37.5%. The reference length changes.
- Increase = 72 − 60 = 12 percentage points. Relative increase = 12 ÷ 60 × 100 = 20%, using the starting rate.
- First: 400 × 1.25 = 500. Then: 500 × 0.75 = 375 units. Change = (375 − 400) ÷ 400 × 100 = −6.25%, a 6.25% decrease.
- Multiplier = 0.80 × 1.10 = 0.88. Original = 264 ÷ 0.88 = 300 units. Check: 300 × 0.80 = 240; 240 × 1.10 = 264.
Revision checklist
Identify the reference; distinguish shares, changes and percentage points; multiply successive factors; divide to recover; check by applying the changes forward.