Percentages and percentage change

Percent means a comparison per hundred

One percent means one hundredth: 1% = 1/100 = 0.01. In general, p% means p/100. The number p need not be an integer: 37.5% = 37.5/100 = 0.375 = 3/8. Fractions, decimals and percentages can therefore name exactly the same numerical value, using different units of description.

To convert a percentage to a decimal, divide its stated number by 100; to express a decimal as a percentage, multiply its numerical value by 100 and attach the percent sign. Thus 35% = 0.35, while 0.35% = 0.0035. For a fraction, divide its numerator by its denominator to obtain a decimal, or rewrite it over a suitable denominator: 7/20 = 35/100 = 35%. Keep the percent sign with its number so that thirty-five hundredths are not confused with thirty-five whole units.

A percentage needs a named reference amount

Three litres compared with eight litres are 3/8 = 37.5% of that reference amount. The same three litres compared with twelve litres are 3/12 = 25%. The amount being compared stayed the same; the reference changed. Before calculating, say which quantity represents 100%. It might be a whole group, an original value, a target or a benchmark, and these choices answer different questions.

A hundred-cell model divides the reference whole into one hundred equal parts. Each cell stands for 1% of that whole, not necessarily one object or one litre. For a reference of eight litres, one cell represents 0.08 litre. Three litres occupy 37.5 cells: thirty-seven full cells and half of another. The fractional cell describes a fractional measurement unit; it does not claim that a counted person or indivisible item has been cut in half.

Connect the part, the percentage and the base

Let B be the reference amount and S the amount being compared. If the percentage number is p, then S = (p/100) × B. For example, 15% of 80 litres is (15/100) × 80 L = 12 L. The percentage is a dimensionless scale factor, so the amount keeps its litre unit. To find the percentage instead, use p = 100 × S/B when B is nonzero; 3/8 gives the percentage number 37.5.

To recover the reference amount from a known part and nonzero percentage, use B = 100 × S/p. If 18 litres are 30% of the reference, B = 100 × 18 L ÷ 30 = 60 L. At 0%, every base produces a zero part, so a zero part at 0% cannot determine one base. For ordinary nonnegative amounts, use a positive reference when interpreting a share; a zero reference makes S/B undefined even if another multiplication formula can still produce a zero amount.

More than 100% can compare an amount above its benchmark

A comparison of 100 litres with a reference of 80 litres is 100/80 = 5/4 = 125%. It is one whole reference amount plus another quarter of that reference. A hundred-cell model needs one complete whole and twenty-five cells of a second equal whole. The reference size stays fixed; exceeding one whole does not change the definition of a cell.

For an actual part contained in a nonnegative whole, the share cannot exceed 100%. A different amount can exceed a benchmark, so a percentage above 100% is valid in that comparison. Use compatible units: 2 m compared with a 50 cm reference gives 200 cm ÷ 50 cm = 4 = 400%. Saying an amount is 125% of its original means it is 25% greater than the original; an increase of 125% would instead make it 225% of the original.

A percentage change combines the old amount and its change

Increasing an original amount B by p% adds p/100 of B: new = B + (p/100) × B = (1 + p/100) × B. A 25% increase multiplies by 1.25, so 80 litres become 100 litres. The new amount is 125% of the old, while the added amount is 25% of the old. These are different comparisons with the same original reference.

Decreasing B by p% subtracts that share: new = (1 − p/100) × B. A 25% decrease multiplies by 0.75, so 80 litres become 60 litres. A 0% change uses factor 1 and keeps the amount. A 100% decrease uses factor 0 and leaves zero. For an ordinary nonnegative amount, a decrease cannot remove more than the whole available quantity; a negative result would require a different signed interpretation rather than a negative remaining volume or count.

Divide the signed difference by a positive original amount

For a positive original amount O, signed percentage change is c% where c = 100 × (new − O)/O. A rise gives a positive difference and percentage; a fall gives negative ones. Changing from eight litres to three litres gives an absolute change of −5 L and relative change −5/8 = −0.625 = −62.5%. The remaining three litres are 37.5% of the original, while the decrease is 62.5%; do not exchange these two questions.

If the original is zero, the absolute difference new − 0 is still defined, but percentage change relative to that original is undefined. From 0 to 20 units the difference is +20 units; from 0 to 0 it is zero. Neither case supplies a relative percentage through division by zero. With a positive original and a nonnegative new amount, signed percentage change cannot be below −100%: the limiting case is a fall to zero. This lesson uses positive original references for ordinary relative-change comparisons.

Reverse the multiplier to recover the original

If the new amount equals f times the original and f is nonzero, recover the original by dividing the new amount by f. A planned ribbon length increased by 20% to 96 cm has factor 1.2, so its original length was 96 cm ÷ 1.2 = 80 cm. Subtracting 20% of 96 cm would instead use the new length as its reference and give 76.8 cm. It does not undo the original increase.

After a 20% decrease, 64 units correspond to factor 0.8 of the original; 64 ÷ 0.8 = 80 units. Reversing the decrease multiplies by 1/0.8 = 1.25, a 25% increase from the smaller base. A 100% decrease has factor zero and cannot be reversed by division: a final zero alone cannot tell you which original amount was removed. Keep the original reference and the nonzero inverse-factor condition visible.

Percentage points and relative percentages answer different questions

When a reported rate changes from 20% to 30%, subtracting its percentage numbers gives an increase of 10 percentage points. This measures the difference between two rates on the same per-hundred scale. The relative increase of the rate uses its original value as denominator: (0.30 − 0.20)/0.20 = 0.50, so the rate increased by 50%. Both statements can be correct because they name different comparisons.

The direction and base matter. From 30% down to 20%, the change is −10 percentage points, but the relative change is −10/30 = −1/3, or approximately −33.33%. From 0% to 5%, the change is +5 percentage points, while relative change against the zero original rate is undefined. When comparing rates, keep their definitions clear; a change in a share does not automatically give the same change in the underlying count if the group sizes also changed.

Successive percentages use the amount at each step

Begin with 100 units. A 20% increase adds 20 and gives 120 units. A subsequent 20% decrease removes 20% of 120, which is 24, leaving 96 units. The numerical percentages have equal sizes and opposite directions, but the actual changes do not cancel because their bases differ. The final amount is 96% of the original, a net decrease of 4%.

The same result follows by multiplying the change factors: 1.2 × 0.8 = 0.96. Each factor scales the current amount, so successive factors multiply rather than their percentage changes simply adding. This calculation assumes the stated percentages apply to each new amount and that intermediate quantities remain exact. If a step instead names the original amount as its base, or requires whole-item rounding, use those specific conditions. The next lesson develops repeated growth and comparisons over multiple periods.

Keep the reference fixed while changing the question

Begin with reference 8 and comparison 3 in the explorer. In the comparison view, compare 3 with 8: the exact fraction 3/8 gives 37.5%. In the change view, compare the difference with the same reference: (3 − 8)/8 gives −62.5%. The amounts have not changed; the selected view asks a different question. Read each complete hundred-cell whole as the chosen reference, and treat a partial cell as part of a percentage unit.

Choose 80 and 100: the comparison is 125% of the reference and its change is +25%. Compare with 80 and 60, 80 and 80, and 80 and 0 to distinguish a decrease, no change and a fall to zero. The pair 40 and 50 gives the same relative increase as 80 and 100, but a different absolute difference. Finally, try a zero reference with comparison 20 or 0. The absolute differences remain defined; neither a share percentage nor a relative change can be inferred from that zero reference.

Keep exact percentages distinct from rounded reports and whole counts

Three out of eight items give exactly 37.5%; a noninteger percentage can describe an integer count. One third has a recurring percentage, 33.(3)%, so writing 33.33% is a rounded report and requires ≈ rather than =. Use exact fractions or full stated factors for calculations, then round once to the requested reporting place. Rounding a percentage number to two decimal places introduces at most 0.005 percentage points of rounding error under nearest rounding; it does not establish the accuracy of an underlying measurement.

A requested percentage may not fit indivisible objects exactly. Ten percent of 25 items is 2.5 items, which cannot be selected as a whole-item count. Choosing two gives 8%; choosing three gives 12%. Neither choice is exactly 10%. If the condition requires at least 10%, three is the smallest allowed whole count; use that stated condition rather than silently rounding and preserving the original claim. Continuous lengths or volumes can allow fractional amounts, with their measurement units retained.

Check the comparison, the denominator and the units

Before calculating, identify whether the question asks for an amount, its percentage of a reference, a signed percentage change, or a difference in percentage points. Name the quantity that represents 100%, convert compatible measurements to a shared unit, and check every denominator. A dimensionless percentage scales an amount without removing that amount’s unit. Absolute change retains the original unit; percentage-point change retains its own rate-scale label.

After calculating, compare the result with simple benchmarks: half is 50%, one whole is 100%, and a quarter is 25%. Multiply a recovered base by its forward factor to check an inverse calculation. For successive changes, verify the intermediate base and final factor, and check whether fractional counts or rounding are allowed. Percentages organize comparisons; their meaning comes from the quantities and references you supplied. The next lesson distinguishes simple growth from compound growth and compares their results under stated conditions.

Worked examples

Try it yourself

1. Convert 7/20 to a decimal and percentage, 37.5% to a fraction in lowest terms, and 125% to a fraction. Are 35% and 0.35% equal? Is 33.33% an exact representation of one third?

   Show the answer and reasoning

   7/20 = 0.35 = 35%; 37.5% = 3/8; 125% = 5/4. 35% and 0.35% differ. One third is approximately 33.33%.

   Percent divides the stated number by 100: 37.5/100 = 375/1000 = 3/8, and 125/100 = 5/4. 35% = 0.35, while 0.35% = 0.0035. One third gives the repeating percentage 33.(3)%; a finite 33.33% report changes its value.

2. Find 15% of 80 litres. If 18 litres are 30% of a reference, find that reference. What percentage is 3 litres of 8 litres? Does a zero part at 0% uniquely determine its base?

   Show the answer and reasoning

   12 litres; 60 litres; 37.5%. A zero part at 0% does not determine one base.

   Use 0.15 × 80 L = 12 L; divide 18 L by 0.30 to get 60 L; divide 3 L by 8 L and express 3/8 as 37.5%. Each nonzero denominator is allowed. Every base multiplied by 0% produces zero, so the last information is insufficient for recovery.

3. Express 2 m as a percentage of a 50 cm reference. Find 125% of 80 litres. If an amount changes from 80 litres to that result, is the increase 125% or 25%?

   Show the answer and reasoning

   400%; 100 litres; the increase is 25%, while the new amount is 125% of the original.

   Convert 2 m to 200 cm and compare 200/50 = 4 = 400%. The factor 125/100 = 1.25 gives 1.25 × 80 L = 100 L. The added 20 L are 20/80 = 25% of the original; the entire new 100 L use a different numerator.

4. For changes 80→60, 80→0, 0→20 and 0→0, find the absolute difference and signed percentage change where defined. Assume all amounts are nonnegative and explain the lower bound with a positive original.

   Show the answer and reasoning

   Differences are −20, −80, +20 and 0 units. Relative changes are −25%, −100%, undefined and undefined. With a positive original and nonnegative final amount, relative change cannot be below −100%.

   Divide each signed difference by its original only when that original is positive. −20/80 = −1/4 and −80/80 = −1. The last two original references are zero, so relative division is undefined even though their differences are valid. The smallest allowed new amount is zero, which removes exactly the whole positive original.

5. After a 25% increase an amount is 100 units; after a 25% decrease another amount is 60 units. Recover each original. Can a final zero after a 100% decrease determine its original?

   Show the answer and reasoning

   Both originals are 80 units: 100 ÷ 1.25 = 80 and 60 ÷ 0.75 = 80. A final zero after a 100% decrease cannot determine one original.

   The forward factors are 1.25 and 0.75, both nonzero, so division reverses them. Check 80 × 1.25 = 100 and 80 × 0.75 = 60. A 100% decrease multiplies by zero, losing the original information; division by that factor is not allowed.

6. A rate changes from 20% to 30%, then consider the reverse change from 30% to 20%. Give each difference in percentage points and its relative percentage change. What can be reported for a change from 0% to 5%?

   Show the answer and reasoning

   20%→30%: +10 percentage points and +50% relative change. 30%→20%: −10 percentage points and approximately −33.33% relative change. 0%→5%: +5 percentage points, with undefined relative change.

   Percentage points subtract the displayed percentage numbers. Relative change divides that difference by the original percentage number: 10/20 = 1/2 and −10/30 = −1/3. The reverse comparison has a different base, and its exact relative percentage is −33.(3)%. The final zero original does not allow the relative quotient.

7. Starting with 100 units, apply +20% and then −20% to the current amount. Find the final amount and net change. Separately, can exactly 10% of 25 indivisible items be selected? What is the smallest whole count meeting a condition that requires at least 10%?

   Show the answer and reasoning

   The final amount is 96 units, a net −4%. Exactly 10% of 25 items would be 2.5 items and is impossible as a whole count. The smallest count meeting at least 10% is three items, which is 12%.

   Multiply the successive factors 1.2 and 0.8, obtaining 0.96 of the original. For the item selection, two items give 2/25 = 8%, below the threshold; three give 3/25 = 12%, above it. The rounded whole count satisfies the minimum condition but does not preserve an exact 10% claim.

What to carry forward

• A percentage means a fraction per hundred. Name the amount representing 100%, and retain the units of any amount it scales.
• Find a part by multiplying the base by p/100; find a percentage by dividing by a nonzero reference. Recovering a base requires a nonzero percentage or change factor.
• Signed relative change uses the difference divided by a positive original. A zero original has no defined relative change, while its absolute difference remains valid.
• Percentage points subtract rates on their per-hundred scale; relative percentage change divides by the original rate. Successive changes use their current bases and multiply factors.
• Percentages above 100% can compare amounts above a benchmark. Check nonnegative-amount limits, exact versus rounded reports, and whether a requested share allows whole items.