Percentage
🔒 Log in to trackThe language of the whole arithmetic section — profit-loss, interest, discount and DI are all percentage questions in disguise. Tier 1 asks 1–2 direct questions per shift (successive change, more/less flips, election and population problems) and Tier 2 leans on it heavily.
One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.
Every formula on one printable page, grouped by subtopic.
5 exam-level questions worked step by step.
66 questions — untimed practice or a timed test with analysis.
Track record in the exam
Questions per shift in recent SSC CGL papers.
Test difficulty mix (66 questions)
Question patterns exams keep repeating
Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.
x% of a number / fraction conversion
very common'What is 35% of 480', '1/8 as a percentage', 'if p% of N is k find N'.
How to solve: Fraction table + one multiplication or division; recover the whole with N = 100k/p. Under 30 seconds — never long-multiply.
Example: What is 35% of 480?
Table route: 35% = 7/20 → 480 × 7/20 = 168. Or 10% is 48, so 35% = 3.5 × 48 = 168.
Successive percentage change
very commonTwo or three changes applied one after another (price/population/marks), asking net change or final value.
How to solve: Multiplier chips (×6/5, ×3/4 …) or a + b + ab/100 for two changes. Up-then-down by the same rate → net −x²/100%. Three changes: chain the chips.
Example: The price of an article rises by 20% and then falls by 20%. What is the net change?
Net = −20²/100 = −4%. Chips: 6/5 × 4/5 = 24/25 → a 4% decrease.
'More than / less than' flip
very common'A's income is 25% more than B's; B's income is what % less than A's?' — the base changes mid-question.
How to solve: Flip formulas: 100x/(100+x) (more→less) and 100x/(100−x) (less→more). Or set the unmentioned quantity to 100. Chains of percentages: multiply fractions.
Example: A's income is 20% more than B's. B's income is what per cent less than A's?
B = 100, A = 120 → gap 20 on base 120 → 100 × 20/120 = 16⅔% less.
Population / depreciation over years
commonPopulation grows or falls at stated yearly rates (same or different); find present, past, or future value.
How to solve: Chain the chips forward (×1.1, ×0.9 …); divide by them to go back. Two different rates: never average — multiply chips. (1 − r/100)² ≠ 1 − 2r/100.
Example: The population of a town is 40,000. It increases by 10% in the first year and decreases by 5% in the second. Population at the end of the second year?
40000 × 11/10 × 19/20 = 44,000 × 0.95 = 41,800.
Election / voting percentages
commonTwo candidates, invalid votes given, winner's share of valid votes and margin given; find total votes or votes got.
How to solve: Everything sits on VALID votes: valid = (100 − invalid)% of polled; margin = share gap × valid; then unmask the total.
Example: In an election, 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?
Margin = 20% of valid → valid = 6,000 = 80% of polled → polled = 7,500.
Pass marks with two candidates
commonOne candidate fails by f marks at p%, another exceeds pass by e at q%; find maximum marks or the pass mark.
How to solve: (q − p)% of M = f + e. Solve M, then pass mark = pM/100 + f; verify with the second candidate.
Example: A candidate scoring 33% fails by 45 marks; another scoring 45% gets 15 marks more than the pass marks. Find the maximum marks.
12% of M = 45 + 15 = 60 → M = 500; pass mark = 165 + 45 = 210.
Price rise vs consumption / expenditure
common'Price rises 20%, by what % reduce consumption to keep expenditure same' or the reverse (k kg more for a fixed outlay).
How to solve: Quantity factor = expenditure factor ÷ price factor; standard cut = 100r/(100+r)%. Outlay version: r% of the money buys the extra quantity at the reduced price.
Example: A 20% cut in the price of oil lets a family buy 5 kg more for ₹400. Find the original price per kg.
Saving = 20% of 400 = ₹80 → reduced price = 80/5 = ₹16 → original = 16 × 5/4 = ₹20.
Income–expenditure–savings change
occasionalSavings % given; income and expenditure change by given %; find the % change in savings.
How to solve: Set income = 100, track all three quantities with chips, and compute the new savings directly — the % change in savings is never guessed.
Example: A person spends 80% of his income. Income rises 15% and expenditure rises 10%. Find the % increase in savings.
100/80/20 → 115/88/27 → savings rise 7/20 × 100 = 35%.