Percentage

high importance~2 Q in Tier 121 formulas⚡ 15 shortcuts5 subtopics

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Subtopic 4 of 5·← 'x% more than' ↔ 'x% less than' & chains·Marks, income–expenditure–savings & price–consumption →

Population growth, depreciation & elections

Population / value growth-decay: value after n years $= P\left(1 \pm \frac{r}{100}\right)^n$ (+ for growth, − for depreciation). Work backwards by dividing.

Two different rates in two years: apply the multipliers one after another (e.g. ×11/10 then ×19/20).

Election problems: total votes polled = valid + invalid. Both candidates' shares must be read off the VALID votes. Margin = difference of shares × valid votes.

Depreciation is the same formula with a negative rate: present $= P\left(\frac{100-r}{100}\right)^n$ .

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

Growth and depreciation over years

Value after n years at r% per annum: $P\left(1 \pm \frac{r}{100}\right)^n$ — plus for growth, minus for depreciation. Different rates in different years: multiply the chips one after another; never average the rates. Example: 40,000 → +10% then −5%: $40000 \times \frac{11}{10} \times \frac{19}{20} = 41{,}800$ . Handy exact chips: 10% → $\frac{11}{10}$ or $\frac{9}{10}$ ; 20% → $\frac{6}{5}$ ; 25% → $\frac{5}{4}$ ; 5% → $\frac{21}{20}$ ; 12.5% → $\frac{9}{8}$ or $\frac{7}{8}$ .

Backwards questions (present given, past asked)

Present = P × chip¹ × chip² …, so divide by the chips to go back. Example: a machine is worth ₹8,100 after two yearly depreciations of 10%: $8100 \div \frac{9}{10} \div \frac{9}{10} = ₹10{,}000$ . Do not divide by the net change (1 − 20% ≠ 0.9 × 0.9 = 0.81).

Watch the walk-through with P = 100

For any multi-year story, take the starting value as 100 and track it year by year: +10% then −5% then +20%: $100 \to 110 \to 104.5 \to 125.4$ — the final value is 125.4% of the start, a net +25.4%. This kills every option at a glance and needs no formula at all. When the answer choices are far apart (they usually are), even rough chips decide the question: "is it above or below 41,000?" is enough.

Elections — the two golden rules

Every candidate's share is a percentage of the valid votes, never of the total polled.
Margin = (winner% − loser%) × valid votes. With invalid votes, chain once more: valid = (100 − invalid%) of the polled. Example: 20% of the polled votes are invalid, the winner polls 60% of the valid votes and wins by 1,200 → margin = 20% of valid → valid = 6,000 = 80% of polled → polled = 7,500.

One-candidate and turnout variants

"x% did not vote / their votes were invalid" → handle the turnout first, then split the rest.
A candidate who withdraws: subtract that share before splitting between the remaining two.
"Winner got 55% of the total votes (no invalid votes)": margin = (2 × 55 − 100)% = 10% of the votes.

Depreciation nuance

"Depreciates 10% p.a. for 2 years" is $\left(\frac{9}{10}\right)^2 = 0.81$ — a total fall of 19%, not 20%. The "20%" option is always planted.

Elections — one more worked shape

"A candidate got 30% of the total valid votes and lost by 4,000": the winner has 70%, so the margin is 40% of the valid votes → valid $= \frac{4000 \times 100}{40} = 10{,}000$ and the loser's votes $= 3{,}000$ . Everything again reduces to one equation on valid votes. If the question instead names "total voters on the roll" with a turnout of 80%, the valid votes sit inside the turnout: roll → turnout → valid → shares, left to right, one chip each.

Quick revision

$P_n = P \times \text{chip}^n$ ; chips multiply; go back by dividing.
Different rates → different chips, applied in order.
Elections: everything sits on valid votes; margin = share gap × valid.
$\left(1 - \frac{r}{100}\right)^2 \ne 1 - \frac{2r}{100}$ .

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Population growth over n yearsvery common2 practice Q

How to spot it:

'A town's population is P and increases at r% per annum — population after n years?'

P_n = P\left(1 + \frac{r}{100}\right)^n

Write the chip $\frac{100 + r}{100}$ .
Multiply P by chipⁿ (or apply chip n times).
Two different rates in two years → multiply the two different chips.

Why: each year's growth sits on the grown value, so the changes compound multiplicatively.

Example: The population of a town is 6,400 and increases at 25% per annum. What will it be after 2 years?

Chip $\frac{5}{4}$ → $6400 \times \frac{25}{16} = 10{,}000$ . (64 hundred → 100 hundred.)

Type 2: Backwards: present value → past value / depreciationvery common2 practice Q

How to spot it:

'The present population is X; it grew at r% p.a. — find the population n years ago' or 'a machine worth M depreciates at r% p.a. — value after n years'.

\text{past} = \text{present} \div \left(1 + \frac{r}{100}\right)^n, \qquad \text{value after n yrs} = P\left(1 - \frac{r}{100}\right)^n

Forward: multiply by the depreciation chipⁿ.
Backward: divide the present value by the growth chipⁿ — never divide by the net %.
Check the answer by growing it forward once.

Why: compounding is not symmetric — undo a 25% rise with ÷(5/4), not with −25%.

Example: The present population of a town is 16,000. If it has been growing at 25% per annum, what was its population 2 years ago?

Past $= \frac{16000}{(5/4)^2} = 16000 \times \frac{16}{25} = 10{,}240$ .

Type 3: Two different rates in consecutive yearscommon3 practice Q

How to spot it:

'Increases by 10% in the first year and decreases by 5% in the second' — rates differ or a sign flips.

P_n = P \times \text{chip}_1 \times \text{chip}_2 \times \cdots

One chip per year, signs included: +5% → 21/20, −10% → 9/10.
Multiply the chips in order (or take P = 100 and walk through).
For 'what was it 3 years ago', divide by all three chips.

Why: each year's rate applies to the value left by the previous year.

Example: The population of a town is 80,000. It increases by 5% in the first year and decreases by 10% in the second year. What is the population at the end of the second year?

$80000 \times \frac{21}{20} \times \frac{9}{10} = 75{,}600$ . (Averaging +5 and −10 = −5% gives 76,000 — wrong.)

Type 4: Election with valid/invalid votes and marginvery common2 practice Q

How to spot it:

'In an election, x% of the votes were invalid; the winner got y% of the valid votes and won by M votes.'

\text{margin} = \frac{W - L}{100} \times \text{valid votes}, \quad \text{valid} = \frac{100 - \text{invalid}}{100} \times \text{polled}

Chain: polled → valid (drop the invalid %) → shares of the winner and loser.
Margin = share gap × valid votes; equate it to the given margin.
Two candidates: L = 100 − W (on valid votes).

Why: candidates' percentages are defined on valid votes, so every equation must sit on that base.

Example: In an election between two candidates, 20% of the votes polled were invalid. The winner got 60% of the valid votes and won by 1,200 votes. How many votes were polled in all?

Valid = 0.8 × polled; margin = 20% of valid = 1200 → valid = 6000 → polled = $\frac{6000}{0.8} = 7{,}500$ .

Formulas

Growth / decay

P_n = P\left(1 \pm \frac{r}{100}\right)^n

Two different rates

P_2 = P\left(1 + \frac{r_1}{100}\right)\left(1 + \frac{r_2}{100}\right)

Depreciated value back

P = \text{present} \times \left(\frac{100}{100-r}\right)^n

Election margin

\text{margin} = (\text{winner share} - \text{loser share}) \times \text{valid votes}

Shortcut tricks

⚡ Multiplier chips for n years

Chips: +10% → 11/10, −5% → 19/20. Multiply chips across years; the final fraction tells everything.

Example: A town's population of 40,000 grows 10% in year 1 and falls 5% in year 2. Find the present population.

$40000 \times \frac{11}{10} \times \frac{19}{20} = 22000 \times \frac{19}{10}... = 41800$ ? Compute: $40000 \times 1.1 = 44000$ , $44000 \times 0.95 = 41800$ .

⚡ Work backwards with division

For 'value n years ago', divide by the chips instead of guessing.

Example: A machine worth ₹8,100 depreciates at 10% p.a. What was its value 2 years ago?

Divide by the chip twice: ₹ $8100 \times \frac{10}{9} = 9000$ , then $9000 \times \frac{10}{9} = 10000$ .

⚡ Elections: everything on valid votes

Convert both candidates' data to fractions of valid votes; the total polled number then follows from the margin.

Example: In an election, 20% of votes polled were invalid. The winner got 60% of valid votes and won by 1200 votes. Total votes polled?

Margin = 20% of valid = $\frac{1}{5}V = 1200$ ⇒ V = 6000 = 80% of total ⇒ total = 7500.

Where students lose marks

Applying $\left(1 + \frac{r}{100}\right)^n$ when the two years have different rates.
Counting invalid votes as a candidate's votes.
Depreciating 2 years as $P - 2r\%$ instead of $\left(\frac{90}{100}\right)^2$ .
In 'population 2 years ago' questions, dividing by (1 + net%) with the averaged rate.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.