Percentage Notes

Last reviewed 16 Sept 2026 · 5 min read

Meaning

Per cent means "per hundred": $x % = \frac{x}{100}$ .

Conversions

Fraction → %: multiply by 100 ( $3/8 = 37.5%$ ).
% → fraction: divide by 100 and simplify ( $12.5% = 1/8$ ).
Decimal ↔ %: move the decimal point two places ( $0.075 = 7.5%$ ).

Key fraction equivalents

%	Fraction	%	Fraction
50	1/2	12.5	1/8
33.33	1/3	11.11	1/9
25	1/4	10	1/10
20	1/5	9.09	1/11
16.67	1/6	8.33	1/12
14.28	1/7	6.25	1/16
66.67	2/3	37.5	3/8
75	3/4	62.5	5/8
40	2/5	87.5	7/8

Basic formulas

FormulaPercentage relations

x% of N $= \frac{x}{100} \times N$
What % is A of B? $= \frac{A}{B} \times 100$
Number whose x% is A $= \frac{A \times 100}{x}$
Percentage change $= \frac{New - Old}{Old} \times 100$
After increase of x%: New = Old × $(1 + \frac{x}{100})$ ; after decrease: Old × $(1 - \frac{x}{100})$

Percentage point — absolute difference between percentages (from 20% to 25% is a 5 percentage point increase, but a 25% increase).

Comparison formulas

Formula"More than / less than"

If A is x% more than B, then B is less than A by $\frac{x}{100 + x} \times 100%$
If A is x% less than B, then B is more than A by $\frac{x}{100 - x} \times 100%$

Successive percentage changes

FormulaNet effect

Two successive changes of a% and b% (use negative for decrease):

Net change = a + b + \frac{ab}{100} %

Increase x% then decrease x% → net loss of $x^{2} /100$ %
Apply repeatedly for three changes (combine two, then the third)

Product constancy (price and consumption)

FormulaExpenditure = price × consumption

If price increases by r%, consumption must decrease by

\frac{r}{100 + r} \times 100%

to keep expenditure constant; if price decreases by r%, consumption can increase by $\frac{r}{100 - r} \times 100%$ .

Area of rectangle with length and breadth changed by a% and b%: net change $= a + b + ab /100$ %; square/circle with side/radius changed by x%: area changes by $2 x + x^{2} /100$ %.

Population and depreciation

FormulaGrowth and depreciation

Population after n years (growth r% p.a.): $P_{n} = P (1 + \frac{r}{100})^{n}$
Population n years ago: $P / (1 + \frac{r}{100})^{n}$
Value after depreciation of r% p.a.: $V_{n} = V (1 - \frac{r}{100})^{n}$

Worked examples

Worked ExampleExample 1 — basic

(a) Find 37.5% of 640. (b) What percent of 250 is 45? (c) 18 is 12% of what number?

Solution. (a) $\frac{3}{8} \times 640 =$ 240; (b) $\frac{45}{250} \times 100 =$ 18%; (c) $\frac{18 \times 100}{12} =$ 150

Worked ExampleExample 2 — more than / less than

A's salary is 25% more than B's. By what percent is B's salary less than A's?

Solution. $\frac{25}{125} \times 100 =$ 20%

Worked ExampleExample 3 — successive change

The price of cement rises by 20% and then falls by 10%. Find the net change.

Solution. $20 - 10 + \frac{20 \times ( - 10 )}{100} =$ +8%

Worked ExampleExample 4 — equal increase and decrease

A number is increased by 10% and then decreased by 10%. Net change?

Solution. $- \frac{1 0 ^{2}}{100} =$ −1% (a decrease of 1%)

Worked ExampleExample 5 — consumption adjustment

The price of steel increases by 25%. By what percent must a builder reduce consumption to keep expenditure unchanged?

Solution. $\frac{25}{125} \times 100 =$ 20%

Worked ExampleExample 6 — area change

The length of a rectangular plot is increased by 20% and breadth decreased by 20%. Find the change in area. What if the side of a square increases by 10%?

Solution. Rectangle: $20 - 20 - 4 =$ −4%; square: $20 + 1 =$ +21%

Worked ExampleExample 7 — population

A town's population of 50 000 grows at 4% per year. Find the population after 2 years.

Solution. $50000 \times 1.0 4^{2} = 50000 \times 1.0816 =$ 54 080

Worked ExampleExample 8 — elections

In an election between two candidates, the winner got 58% of valid votes and won by 4800 votes. Find the total valid votes.

Solution. Margin = 58% − 42% = 16% → $0.16 V = 4800$ → V = 30 000

Worked ExampleExample 9 — examinations

A student needs 40% to pass. She scored 180 marks and failed by 20 marks. Find maximum marks.

Solution. Pass marks = 200 = 40% → maximum 500

Worked ExampleExample 10 — depreciation

A machine worth ₹ 5 lakh depreciates at 10% per year. Find its value after 3 years.

Solution. $5 \times 0. 9^{3} = 5 \times 0.729 =$ ₹ 3.645 lakh

Worked ExampleExample 11 — percentage points

Unemployment rose from 6% to 7.5%. Express the rise in percentage points and percent.

Solution. 1.5 percentage points; $\frac{1.5}{6} \times 100 =$ 25%

Frequently tested points

x% = x/100; know fraction equivalents (1/8 = 12.5%, 1/6 = 16.67%, 1/7 ≈ 14.28%).
% change = (new − old)/old × 100.
A is x% more than B → B is x/(100 + x) × 100% less than A; x% less → x/(100 − x) × 100% more.
Successive changes: a + b + ab/100; equal rise and fall → −x²/100 %.
Constant expenditure: price up r% → consumption down r/(100 + r) × 100%.
Square/circle area change for x% change in side/radius: 2x + x²/100.
Population $P (1 + r /100)^{n}$ ; depreciation $V (1 - r /100)^{n}$ .
Percentage point vs percent.

Common MistakeCommon mistakes

Calculating percentage change on the new value instead of the original.
Adding successive percentages directly (10% up and 10% down is not zero change).
Confusing percentage points with percent change.

Revision SummaryChapter summary

Percentages express parts per hundred and interconvert with fractions and decimals, with key equivalents enabling fast calculation.
Basic relations find a percentage of a quantity, a quantity from its percentage and percentage change.
Comparison formulas convert "more than" into "less than" statements, and successive changes combine as a + b + ab/100.
Product constancy links price and consumption, and area changes follow the same successive-change rule.
Growth and depreciation compound over time, with applications to elections, examinations and budgets.

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Percentage