CSAT Quantitative Aptitude: Complete Guide, Formulas, Notes & 60-Day Plan

Quantitative Aptitude (QA) is the single biggest reason aspirants flunk CSAT — and the single easiest fix in the entire UPSC prelims. The paper is qualifying (you need 33%, around 66 of 200 marks), the QA component contributes 20 to 25 questions worth 50 to 62 marks, and the syllabus officially caps out at Class X mathematics. None of that should be intimidating. And yet every year, hundreds of GS-strong candidates lose their attempt because Paper II quietly dragged them below 33%.

This guide is the complete CSAT QA notebook we wish we had as candidates. It covers every topic UPSC has asked between 2018 and 2025, every shortcut you actually need, every trap that catches mid-prep candidates, and a 60-day revision plan that costs less than an hour a day. Work through it linearly the first time. Use it as a topic-wise reference once you’ve started writing mock tests. Everything you see — formulas, worked examples, practice problems — is calibrated to the actual difficulty of UPSC CSAT, not the inflated difficulty of bank-PO or CAT prep material.

How CSAT QA actually behaves in the exam

Before you touch a single formula, internalise these numbers. They drive every strategic call you’ll make over the next 60 days.

Two implications: First, you cannot skip QA entirely and hope RC + Reasoning will cover 66 marks — even at 100 percent accuracy on the non-QA bits, one bad RC passage will drag you under. Second, you only need to crack roughly 14-16 QA questions to be safe. That is achievable in 60 focused days if you trust the priority order in this guide.

The CSAT QA syllabus — what UPSC actually asks

UPSC publishes a broad two-line syllabus: ‘Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level) and Data interpretation (charts, graphs, tables, data sufficiency etc.) (Class X level).’ That’s it. The actual asked topics, traced across eight years of papers, fall into 14 buckets. Memorise this table — it is your prep priority list.

Add the percentages and a clear strategy emerges: P1 topics together account for roughly 16-20 questions every year. If you can crack 14 of those 16 with 90 percent accuracy and use P2 plus DI for the buffer, you will clear CSAT regardless of how the paper feels on the day.

Number systems — your foundation topic

Half of every CSAT QA paper is decided by number-system fluency. You won’t always see a question labelled ‘number systems’, but divisibility, factors, remainders, HCF/LCM and unit-digit cyclicity feed into percentage problems, ratio problems, age problems, even DI. Master this first.

The number families (just enough)

You don’t need a textbook definition. You need a working classification you can apply in 20 seconds when a question asks ‘the smallest 4-digit natural number divisible by both 7 and 13’.

• Natural numbers (N): 1, 2, 3, … (counting numbers, no zero, no negatives).
• Whole numbers (W): 0, 1, 2, 3, … (natural numbers plus zero).
• Integers (Z): …, -3, -2, -1, 0, 1, 2, 3, … (whole numbers plus negatives).
• Rational numbers (Q): any number expressible as p/q where p, q are integers and q ≠ 0. Includes all terminating and recurring decimals.
• Irrational numbers: non-terminating, non-recurring decimals. √2, √3, π, e. CSAT-relevant cases are restricted to square roots of non-perfect-squares and π.
• Prime numbers: exactly two divisors (1 and itself). 2 is the only even prime — memorise this, it’s a classic CSAT trap.
• Composite numbers: more than two divisors. 1 is neither prime nor composite — another classic trap.
• Co-prime numbers: two numbers whose HCF is 1. They need not be prime themselves — 8 and 9 are co-prime.

Divisibility rules you must memorise

These come up everywhere — in pure number questions, in remainder problems disguised as word problems, and in time-and-work questions where the answer must be a whole number of days.

HCF and LCM — three methods, when to use each

Highest Common Factor (HCF) is the largest number that divides each of two or more numbers. Lowest Common Multiple (LCM) is the smallest number divisible by each. Three computation methods exist; use the right one for the question size.

1. Prime factorisation: Express each number as a product of primes. HCF = product of common primes with the smallest exponent. LCM = product of all primes with the largest exponent. Best for small numbers (under 200).
2. Division method (for HCF): Divide larger by smaller. Replace larger with the remainder. Repeat until remainder is 0. The last divisor is the HCF. Best when numbers are large or you can’t factorise quickly.
3. Listing method: List factors / multiples and pick. Only for very small numbers (under 30). Avoid in exam — too slow.

The relationship to memorise: For any two numbers a and b, HCF × LCM = a × b. This single identity solves a third of all HCF/LCM word problems instantly.

Classic HCF/LCM word problems

Type 1 — Bell-ringing / blinking lights: Three bells ring at 12, 15 and 20 minute intervals. After how long do they ring together? Answer: LCM(12, 15, 20) = 60 minutes. Whenever something happens at the same time, use LCM.

Type 2 — Tile / brick / cuboid: What is the largest tile that exactly covers a 9.6 m × 7.2 m floor? Answer: HCF(960, 720) = 240 cm = 2.4 m square. Whenever you want the largest common ‘unit’, use HCF.

Type 3 — Remainder problems: Find the largest number that divides 76, 109, 142 leaving the same remainder. Answer: HCF(109-76, 142-109, 142-76) = HCF(33, 33, 66) = 33. Same-remainder problems collapse to differences.

Type 4 — Different remainders: Find the largest number that divides 122 and 243 leaving remainders 2 and 3 respectively. Subtract remainders first: HCF(120, 240) = 120.

Type 5 — Smallest number with specific remainders: Find the smallest number that leaves a remainder of 3 when divided by 4, 5 and 6. Answer: LCM(4, 5, 6) + 3 = 60 + 3 = 63. If the remainder is the same, add it to the LCM.

Unit digit (cyclicity) — solve any ‘last digit of 7^85’ question

The unit digit of n^k follows a cycle of length at most 4. Memorise these:

Procedure: Take the base’s unit digit. Find the cycle length. Divide the exponent by cycle length; the remainder tells you the position. Remainder 0 = last term of the cycle. Example: unit digit of 7^85 → cycle (7,9,3,1) length 4 → 85 mod 4 = 1 → first term → 7.

Remainder theorems worth memorising

• (a + b + c) mod n = (a mod n + b mod n + c mod n) mod n. Distribute the modulo over addition.
• (a × b × c) mod n = (a mod n × b mod n × c mod n) mod n. Distribute over multiplication.
• If a ≡ b (mod n) then a^k ≡ b^k (mod n) for any natural k. Use to reduce big exponents.
• Fermat’s little theorem (very rare in CSAT): If p is prime and gcd(a, p) = 1, then a^(p-1) ≡ 1 (mod p).

Most CSAT remainder questions are solved by reducing the base modulo the divisor BEFORE you do anything else. (37^49) mod 4 = (37 mod 4)^49 mod 4 = 1^49 mod 4 = 1. Done in 10 seconds.

Anantamias Compass

Percentage — the highest-yield CSAT QA topic

Percentage shows up in 3 to 4 standalone questions every year, PLUS it powers profit-loss, interest, mixtures, and most DI questions. If you nail percentage, you’ve cleared a third of CSAT QA before touching another topic.

Percentage as a multiplier — the only mental model you need

Stop thinking of percentages as ‘x out of 100’. Start thinking of them as multipliers. 20% means × 0.20, full stop. 120% means × 1.20. An increase of 15% means × 1.15. This single mental shift turns every percentage problem into a multiplication chain.

Memorise this table cold. When you see ‘a 12.5% discount’, you should instantly think ‘multiply by 7/8 — chop one-eighth off’.

Percentage change formula — and the trap inside it

% change = ((new value − old value) / old value) × 100

The trap: percentage change is always relative to the old value, not the new. A salary that rises from ₹50,000 to ₹60,000 is a 20% rise. The same salary falling from ₹60,000 to ₹50,000 is a 16.67% drop, not a 20% drop. UPSC has burned candidates with this asymmetry repeatedly.

Successive percentage change — the formula and the shortcut

When two successive percentage changes (a%) and (b%) are applied, the net change is:

Net % change = a + b + (a × b)/100

Where increases are positive and decreases are negative. Example: A price rises 20% then falls 20%. Net change = 20 + (-20) + (20 × -20)/100 = -4%. Not zero! Successive equal-magnitude opposite changes always net to a loss.

Multiplier shortcut: Convert each change to a multiplier and multiply. 20% rise = × 1.2. 20% fall = × 0.8. Combined = × 0.96 → 4% loss. Faster and harder to mess up.

Population growth and depreciation

Population growing at r% per year for n years: P_n = P_0 × (1 + r/100)^n. Population declining at r% per year (or machine depreciating): P_n = P_0 × (1 – r/100)^n. Notice these are identical structurally to compound interest formulas — they ARE compound interest formulas, applied to people or machines instead of money.

Worked examples

Example 1. A student scored 60% in maths and 70% in science. The maths paper was out of 200 and the science paper out of 150. What is the student’s overall percentage?

Maths marks = 200 × 0.60 = 120. Science marks = 150 × 0.70 = 105. Total marks = 225. Total maximum = 350. Overall = 225/350 × 100 = 64.28%. Trap: averaging 60 and 70 gives 65, which is wrong because the papers have different maxima.

Example 2. If A’s salary is 25% more than B’s, by what percent is B’s salary less than A’s?

Let B = 100. A = 125. B is 25 less than A. B’s deficit relative to A = 25/125 × 100 = 20%. Trap: the answer is not 25%. Watch the reference value.

Example 3. The price of petrol increases by 25%. By what percent must consumption decrease so that expenditure stays the same?

Expenditure = price × consumption. New price multiplier = 1.25. For expenditure to be constant, consumption multiplier must = 1/1.25 = 0.80. Consumption falls by 20%. General rule: if X rises by r% and product XY must stay constant, Y must fall by r/(100+r) × 100 percent.

Profit, loss and discount

One of the easiest topics in CSAT — provided you keep the four price terms straight: Cost Price (CP), Selling Price (SP), Marked Price (MP), and Discounted Price. UPSC loves layering two of these (discount on marked, then profit on cost) to manufacture confusion.

Core formulas

• Profit = SP − CP (if positive). Loss = CP − SP (if positive).
• Profit % = (Profit / CP) × 100. Always on CP, never on SP.
• SP at p% profit: SP = CP × (1 + p/100).
• SP at l% loss: SP = CP × (1 – l/100).
• Discount = MP − SP. Discount % = (Discount / MP) × 100. Always on MP.
• SP after d% discount: SP = MP × (1 – d/100).

The layered question — discount-on-marked-then-profit

Standard wording: ‘A shopkeeper marks his goods 40% above cost. He offers a 25% discount on the marked price. Find his profit percent.’

Let CP = 100. MP = 100 × 1.40 = 140. SP = 140 × 0.75 = 105. Profit = 5, Profit % = 5%. Speed shortcut: combined multiplier = 1.40 × 0.75 = 1.05 → 5% profit. Always use the multiplier chain when MP and discount are stacked.

Equivalent successive discounts

Two successive discounts of a% and b% are equivalent to a single discount of (a + b – (a × b)/100)%. Example: Discounts of 20% and 10% in succession equal 20 + 10 − 2 = 28%, not 30%. UPSC has asked this verbatim.

False weights and dishonest shopkeepers

Classic problem: A shopkeeper sells at cost price but uses a 950-gram weight instead of 1000 grams. What is his profit percent?

He gives 950 grams but charges for 1000 grams. So he ‘gains’ 50 grams on every 950 sold. Profit % = (50/950) × 100 = 5.26%. Formula: Profit % = (error / (true value − error)) × 100, where error is the difference between true and used weight.

Two articles sold at the same price — guaranteed loss

Famous trap: A trader sells two articles at the same price. On one he gains x%, on the other he loses x%. Net result?

Always a loss of (x/10)^2 percent, regardless of the absolute prices. For x = 10%, loss = 1%. For x = 20%, loss = 4%. Memorise this — it’s been asked at least twice in CSAT.

Simple interest and compound interest

Interest problems are formulaic. Once you internalise the SI and CI formulas plus three patterns, you can solve every CSAT interest question in under a minute.

Simple Interest (SI)

SI = (P × R × T) / 100
Amount A = P + SI = P × (1 + R × T / 100)

Where P = principal, R = rate per annum, T = time in years. SI is linear: every year you earn the same interest. Sanity check: if SI for T years on P at R% is X, then SI for 2T years is 2X, not 4X — students confuse this with CI.

Compound Interest (CI)

A = P × (1 + R/100)^T
CI = A - P

Where compounding is annual. For half-yearly compounding, R becomes R/2 and T becomes 2T. For quarterly, R/4 and 4T. For monthly, R/12 and 12T.

The SI vs CI gap (CSAT favourite)

For 2 years: CI − SI = P × (R/100)^2. This is a tiny formula that solves several CSAT questions in one line.

For 3 years: CI − SI = P × (R/100)^2 × (3 + R/100). Less common but worth memorising.

Example: CI on ₹10,000 at 10% for 2 years exceeds SI by how much? Difference = 10,000 × (10/100)^2 = ₹100.

Doubling and tripling time

• Rule of 72 (CI): Money doubles in approximately 72/R years at rate R% per annum compounded annually. Use it for quick mental arithmetic.
• SI doubling: Money doubles when SI = principal, so T = 100/R years.
• SI tripling: SI = 2P, so T = 200/R years.

Equal annual instalments to repay a loan (CI basis)

Loan P borrowed at R% per annum compounded annually, to be repaid in n equal annual instalments x each. The standard formula is:

P = x/(1+R/100) + x/(1+R/100)^2 + ... + x/(1+R/100)^n

Solve for x given P, R, n. CSAT rarely asks this directly — but it does ask 2-instalment variants where the algebra is friendly.

Worked examples

Example 1. What sum at 8% per annum SI yields ₹2,400 in 5 years? P = (SI × 100)/(R × T) = (2400 × 100)/(8 × 5) = ₹6,000.

Example 2. A sum doubles in 8 years at SI. In how many years will it triple at the same rate? Doubling means SI = P in 8 years → R = 12.5%. Tripling means SI = 2P → T = 200/12.5 = 16 years.

Example 3. The CI on ₹15,000 at 10% per annum compounded annually for 2 years. A = 15000 × 1.1 × 1.1 = 18,150. CI = ₹3,150.

Ratio, proportion and variation

Ratio is the second-most-asked topic in CSAT QA after percentage. The good news: it’s easy. The bad news: students lose marks because they don’t reduce ratios to simplest form before computing, or because they don’t see that a percentage and a ratio are the same animal.

The basics

• Ratio a : b means the relationship a/b. A ratio is dimensionless.
• Equivalent ratios: 2 : 3 = 4 : 6 = 6 : 9. Always reduce to lowest terms before working.
• Proportion: two ratios that are equal. a : b :: c : d means a/b = c/d, equivalently a × d = b × c (cross-multiplication).
• Continued proportion: a, b, c are in continued proportion if b² = ac. b is the mean proportional.
• Componendo and dividendo: if a/b = c/d, then (a + b)/(a – b) = (c + d)/(c – d). Useful for ages problems.

Variation

• Direct variation: y ∝ x means y = kx for some constant k. Double x → double y.
• Inverse variation: y ∝ 1/x means y = k/x. Double x → halve y.
• Joint variation: y ∝ xz means y = kxz. Used in mixed work-rate questions.
• Combined: y ∝ x/z means y = kx/z. Speed depends directly on distance and inversely on time.

Splitting a quantity in a given ratio

Split N in ratio a : b : c. Total parts = a + b + c. Each share = N × (its part)/(total parts). Example: Split ₹3,300 among A, B, C in ratio 2 : 3 : 6. Total = 11 parts. A = 3300 × 2/11 = ₹600. B = ₹900. C = ₹1,800.

Age problems — every one in two lines

Template: ‘The ratio of ages of A and B is now 3 : 5. Five years ago, the ratio was 2 : 4. Find present ages.’

Let present ages = 3x and 5x. Five years ago = 3x − 5 and 5x − 5. (3x – 5)/(5x – 5) = 2/4 = 1/2. Cross-multiply: 2(3x − 5) = 1(5x − 5). 6x − 10 = 5x − 5. x = 5. Ages = 15 and 25.

Worked examples

Example 1. If a : b = 3 : 4 and b : c = 5 : 7, what is a : b : c?

Make b common. LCM(4, 5) = 20. a : b becomes 15 : 20. b : c becomes 20 : 28. So a : b : c = 15 : 20 : 28.

Example 2. Two numbers are in ratio 3 : 5. If 9 is added to each, they become 4 : 5. Find the numbers.

Let numbers = 3x, 5x. (3x + 9)/(5x + 9) = 4/5. 5(3x + 9) = 4(5x + 9). 15x + 45 = 20x + 36. 5x = 9. x = 1.8. Numbers are 5.4 and 9. Sanity check: 14.4/18 = 0.8 = 4/5 ✓.

Mixtures and alligation

Alligation is a graphical shortcut for mixture problems. UPSC asks at most one mixture question per year, but it’s almost always a quick winner if you can spot the alligation pattern.

The alligation rule

When two ingredients of prices (or concentrations) C₁ and C₂ are mixed to produce a mixture of price (or concentration) C_mean, the ratio of quantities used is:

Quantity of cheaper : Quantity of dearer = (C_dearer - C_mean) : (C_mean - C_cheaper)

Example: Two types of rice cost ₹15/kg and ₹20/kg. In what ratio should they be mixed to get a mixture costing ₹18/kg?

Ratio = (20 − 18) : (18 − 15) = 2 : 3. So mix 2 parts of ₹15 rice with 3 parts of ₹20 rice.

Replacement mixture problems

Template: A vessel contains 40 litres of milk. 4 litres are drawn off and replaced with water. This is repeated 3 times. How much milk remains?

Final pure quantity = Initial × (1 - replaced/total)^n

= 40 × (1 – 4/40)^3 = 40 × (0.9)^3 = 40 × 0.729 = 29.16 litres of milk. Water = 40 – 29.16 = 10.84 litres.

Average

Average is conceptually trivial — sum divided by count. The CSAT version layers it: when one item is removed or added, or when two groups are merged.

Core ideas

• Average = Sum / Count.
• If average of n items is A, sum = nA. This restatement is what unlocks every average question.
• Weighted average: when groups of different sizes are combined, the combined average is (n₁A₁ + n₂A₂)/(n₁ + n₂).
• Average of an arithmetic progression: (first + last)/2. So average of 1 to 100 is 50.5. Always.

The classic ‘replacement’ question

‘When a 25-year-old student is replaced by a new student, the average age of 10 students decreases by 1 year. Find the age of the new student.’

Original sum = 10 × old average. New sum = 10 × (old average − 1) = old sum − 10. The new student is 10 years younger than the replaced one. New student’s age = 25 − 10 = 15.

Worked examples

Example 1. The average of 5 consecutive odd numbers is 31. What is the largest?

Consecutive odd numbers, average is the middle one. So middle = 31, numbers = 27, 29, 31, 33, 35. Largest = 35.

Example 2. A cricketer’s average in 18 innings is 27. To raise his average to 30, how many runs must he score in his 19th?

Current sum = 18 × 27 = 486. New sum needed = 19 × 30 = 570. Required = 570 − 486 = 84 runs.

Time, speed and distance

Time-speed-distance (TSD) is the third-most-asked topic and the most varied. UPSC pulls trains, boats, relative speed, and average-speed traps almost every year. Master the core relation plus four sub-patterns.

The master formula and units

Distance = Speed × Time
Speed = Distance / Time
Time = Distance / Speed

Unit conversions you must memorise:

• 1 km/hr = 5/18 m/s (multiply by 5/18 to convert km/hr → m/s).
• 1 m/s = 18/5 km/hr (multiply by 18/5 to convert m/s → km/hr).
• 1 km = 1000 m. 1 hour = 3600 s.

Average speed — the trap UPSC loves

If you travel half the distance at speed a and the other half at speed b, average speed is NOT (a+b)/2. It is the harmonic mean: 2ab/(a+b).

Example: 60 km/h to office and 40 km/h returning. Average = 2 × 60 × 40 / 100 = 48 km/h, not 50.

If you spend half the time at speed a and half at speed b, average speed IS the simple mean (a+b)/2. Read the question carefully — half-distance vs half-time changes the formula entirely.

Relative speed

• Two objects moving in the same direction: relative speed = |a − b|.
• Two objects moving in opposite directions: relative speed = a + b.
• Time to meet/cross: distance / relative speed.

Trains crossing

Train of length L crossing a stationary point (pole/man): time = L / speed.

Train of length L crossing a platform of length P: time = (L + P) / speed.

Two trains of lengths L₁ and L₂ crossing each other in opposite directions: time = (L₁ + L₂) / (s₁ + s₂).

Two trains of lengths L₁ and L₂ in the same direction (faster overtakes slower): time = (L₁ + L₂) / (s₁ − s₂).

Boats and streams

• Downstream speed = boat speed in still water (b) + stream speed (s).
• Upstream speed = b − s.
• Given downstream and upstream speeds: b = (down + up)/2, s = (down − up)/2.
• Round trip: if a man rows distance d downstream then back upstream, total time = d/(b+s) + d/(b-s).

Worked examples

Example 1. A train 240 m long crosses a 360 m platform in 30 s. Find its speed.

Total distance = 240 + 360 = 600 m. Time = 30 s. Speed = 20 m/s = 72 km/h.

Example 2. A boat travels 30 km downstream in 2 h and the same distance upstream in 3 h. Find boat speed and stream speed.

Down = 15 km/h, Up = 10 km/h. Boat speed = (15+10)/2 = 12.5 km/h. Stream = (15-10)/2 = 2.5 km/h.

Example 3. A man walks at 4 km/h and reaches office 10 min late. At 5 km/h he is 5 min early. Find office distance.

Let distance = d. Time difference = d/4 − d/5 = 15 min = 1/4 hour. d(5-4)/(4×5) = 1/4 → d/20 = 1/4 → d = 5 km.

Time and work, pipes and cisterns

Treat ‘time and work’ as an inverse-variation topic. The only mental model you need: work done per day is the reciprocal of days needed to complete the whole work.

Core idea — work rates

If A finishes a job in n days, A’s rate is 1/n per day. If A and B work together, their combined rate is 1/a + 1/b per day, and together they finish in ab/(a+b) days.

Standard patterns

• A finishes in a days, B in b days. Together? Time = ab/(a+b).
• A and B together finish in c days. A alone in a days. B alone? Rate B = 1/c – 1/a, so B alone takes ac/(a-c) days.
• Pipes filling a tank: identical maths. Pipe filling in t hours has rate 1/t per hour. Pipe emptying has rate -1/t.
• Three workers: Combined rate = 1/a + 1/b + 1/c. Time = abc/(ab + bc + ca).

The LCM shortcut (highly recommended)

When two or more rates are involved, take the LCM of the individual times as the total work. Then every rate becomes a whole number of units per day, and arithmetic stays clean.

Example: A finishes in 12 days, B in 18 days. Take total work = LCM(12, 18) = 36 units. A does 3 units/day, B does 2 units/day. Together they do 5 units/day. Time = 36/5 = 7.2 days.

Wage-splitting problems

When two workers complete a job together and share wages, the split is in the ratio of work each did (or rates if both worked through). Example: A and B together earn ₹3,200. A alone takes 6 days, B alone takes 10 days. Ratio of rates = 1/6 : 1/10 = 5 : 3. A gets 3200 × 5/8 = ₹2,000. B gets ₹1,200.

Pipes problems

Example: Pipe A fills a tank in 6 hours. Pipe B empties it in 8 hours. Both open. Tank empty at start. Time to fill?

Net rate = 1/6 − 1/8 = (4 − 3)/24 = 1/24 per hour. Time = 24 hours. If net rate were negative, tank could never fill.

Worked examples

Example 1. 24 men complete a work in 16 days. After 4 days, 8 more men join. Find total time.

Total work = 24 × 16 = 384 man-days. Done in first 4 days = 24 × 4 = 96. Remaining = 288 man-days. New workforce = 32. Days needed = 288/32 = 9. Total = 4 + 9 = 13 days.

Example 2. A is twice as efficient as B. Together they finish in 18 days. A alone?

Let B’s rate = 1, A’s rate = 2. Combined = 3. Combined time = 1/3 unit. But given as 18 days, so 1 unit = 54 days. B alone = 54 days, A alone = 54/2 = 27 days.

Mensuration — 2D and 3D

Mensuration adds 1 to 2 questions per year. Memorise the formulas — the working is always straightforward.

2D figures

3D solids

Diagonal formulas

• Square diagonal: s√2.
• Rectangle diagonal: √(l² + b²).
• Cube diagonal: a√3.
• Cuboid diagonal: √(l² + b² + h²).

Worked examples

Example 1. A rectangular field is 80 m by 60 m. A path 5 m wide runs around it on the outside. Area of the path?

Outer rectangle = 90 × 70 = 6,300 m². Inner rectangle = 80 × 60 = 4,800 m². Path = 6,300 − 4,800 = 1,500 m².

Example 2. The radius of a sphere is doubled. By what factor does volume increase?

Volume ∝ r³. Doubling r multiplies volume by 8. Trick: doubling diameter has same effect.

Algebra essentials

CSAT algebra rarely goes beyond linear equations, simultaneous equations, and a handful of identities. Anything more is unlikely.

Identities you must memorise

• (a + b)² = a² + 2ab + b²
• (a − b)² = a² − 2ab + b²
• (a + b)(a − b) = a² − b²
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a − b)³ = a³ − 3a²b + 3ab² − b³
• a³ + b³ = (a + b)(a² − ab + b²)
• a³ − b³ = (a − b)(a² + ab + b²)
• If a + b + c = 0, then a³ + b³ + c³ = 3abc

Quadratic equation roots

For ax² + bx + c = 0:

• Sum of roots = -b/a.
• Product of roots = c/a.
• Discriminant D = b² − 4ac. D > 0 means two real distinct roots; D = 0 means one repeated root; D < 0 means imaginary roots.

Permutation and combination

Most candidates panic at PnC. Don’t. CSAT asks ONE PnC question per year and the difficulty is capped at basic arrangements. Three formulas solve everything.

Definitions

• Permutation: arrangement where order matters. Selecting and arranging r items from n distinct items: nPr = n! / (n − r)!
• Combination: selection where order doesn’t matter. Selecting r from n: nCr = n! / (r! × (n − r)!)
• Factorial: n! = n × (n − 1) × (n − 2) × … × 1. By convention, 0! = 1.

Standard problems

Pattern 1 — Arrangement of n distinct items in a row: n!. So 5 different books can be arranged in 5! = 120 ways.

Pattern 2 — Arrangement of n items with repetitions: n! / (p! × q! × …) where p, q, … are the counts of repeated items. The number of distinct arrangements of the letters of MISSISSIPPI = 11! / (4! × 4! × 2!).

Pattern 3 — Circular arrangement: n distinct items around a circle = (n − 1)!. Around a necklace (reversible) = (n − 1)! / 2.

Pattern 4 — Selection: ‘In how many ways can 4 girls and 3 boys be selected from 6 girls and 5 boys?’ Answer: 6C4 × 5C3 = 15 × 10 = 150.

Pattern 5 — Constraint problems: ‘Arrangements of MASTER such that vowels are always together.’ Treat ‘AE’ as a single block → 5 letters → 5! arrangements. Internally vowels can swap → × 2!. Total = 240.

Probability

Probability is built on PnC. If you handle PnC, probability is straightforward.

Definition

Probability of an event = Number of favourable outcomes / Total possible outcomes

Probability is always between 0 and 1 inclusive. 0 = impossible, 1 = certain.

Rules to memorise

• P(A or B) = P(A) + P(B) − P(A and B). Subtract intersection to avoid double-counting.
• For mutually exclusive events: P(A and B) = 0, so P(A or B) = P(A) + P(B).
• For independent events: P(A and B) = P(A) × P(B).
• Complement: P(not A) = 1 − P(A).
• Conditional probability: P(A given B) = P(A and B) / P(B).

Common scenarios

Coin tossed n times: total outcomes = 2^n. Die rolled n times: total = 6^n. Two dice rolled together: 36 outcomes. Cards drawn from 52: 52 cards = 13 of each suit × 4 suits = 26 red + 26 black, 12 face cards, 4 aces.

Worked examples

Example 1. What is the probability of getting at least one head in 3 tosses?

Use complement. P(no heads) = (1/2)³ = 1/8. P(at least one head) = 1 − 1/8 = 7/8.

Example 2. Two cards drawn from a pack of 52, both kings. Probability?

= (4/52) × (3/51) = 12/2652 = 1/221.

Data interpretation — the silent third of CSAT

DI accounts for 5 to 8 questions every CSAT year — that’s a quarter of QA marks in a topic students don’t even think of as QA. Treat DI as percentage + ratio applied to charts. If your fundamentals are strong, DI is the highest-accuracy section of the entire paper.

Chart types you must read fluently

• Tables: rows × columns of numbers. Most common in CSAT. Read the row headers, column headers, and units (lakhs/crores, percent, kilometres) BEFORE looking at numbers.
• Bar charts: single vs grouped vs stacked. UPSC favours grouped horizontal bars for state comparisons.
• Line charts: usually time series. Look for crossover points and largest gaps.
• Pie charts: show shares of a total. Always read whether the total is shown or whether you need to compute it.
• Mixed graphs: bar + line on the same axis. Two y-axes — read both scales carefully.

Standard CSAT DI question types

1. Direct read: ‘What is the value of X in year Y?’ Pure reading. 5-second question if you’ve read the chart.
2. Sum / average: ‘Total of X over the 5 years?’ Adds rapidly. Approximate before computing exactly.
3. Percentage / ratio comparison: ‘X as percent of Y in year Z?’ Always look for whole-number ratios first (10%, 25%, 33%, etc.) — UPSC designs questions to have clean answers.
4. Growth rate: ‘What is the percent change in X from year Y to year Z?’ Apply (new – old)/old × 100.
5. Find the year of maximum / minimum: Eyeball it. Don’t compute every value.

The 30-second DI strategy

1. Read all four questions for the chart BEFORE looking at the chart.
2. Read chart title, axis labels, units, time period. Verify scale (linear / log).
3. For each question, approximate before computing — most CSAT DI options are far apart.
4. Use approximation aggressively: 14.7 ≈ 15, 198 ≈ 200, 32% ≈ 1/3.
5. Skip the one question per set that requires precise arithmetic. Come back if time allows.

A 60-day CSAT QA practice plan

Most CSAT failures aren’t conceptual — they’re practice failures. You read the formulas and assumed that was enough. It’s not. Here is the day-by-day plan that has put dozens of candidates over the 66-mark line.

Phase 1: Foundation (Days 1-20)

One topic per day from the P1 list, in this order: number systems → percentage → profit/loss → ratio → SI/CI → average → time-speed-distance → time-and-work → mensuration. Each day: read the topic notes, solve 10 textbook problems, write a one-page summary in your own handwriting. The summary is the artifact you’ll revise from in month 3.

Phase 2: Speed building (Days 21-40)

Take one topic-wise test per day (25 questions, 35 minutes). Use any prelims-level QA book (NCERT exemplar, R.S. Aggarwal Quantitative Aptitude is fine — but only do the marked CSAT-relevant chapters). Track three numbers in a notebook: accuracy %, average time per question, topic where you lost most marks.

Phase 3: Full-mock simulation (Days 41-55)

Three full CSAT mocks per week, taken under exam conditions: 80 questions, 120 minutes, mark every wrong answer, no calculator, no breaks. After each mock, spend 90 minutes on review — every wrong question gets a 2-line explanation in your notebook.

Phase 4: Last-mile (Days 56-60)

Stop new learning. Re-read the topic summaries you wrote in Phase 1. Solve only past CSAT papers from 2018 onwards. Cap your total prep at 90 minutes per day — exhaustion is the biggest enemy on exam day.

Past-year patterns and smart guessing

CSAT QA has had visible pattern shifts. 2018-2020 was percentage-heavy. 2021-2022 broke the mould with DI-heavy papers. 2023-2025 saw a comeback of number-system and ratio problems. The constant: 22-25 QA-adjacent questions per year, easy to spot, hard to crack without practice.

What changed in 2023

• Number-system questions became distinctly harder — multi-step remainder problems and base conversions appeared.
• DI sets became smaller (3-4 questions per set instead of 5).
• Permutation and combination returned after 2 years of absence.
• Mensuration problems started using 3D more often than 2D.

Smart guessing principles

1. If you’ve eliminated 2 options and are stuck between 2, GUESS. Expected value = 0.5 × 2.5 − 0.5 × 0.833 = +0.83 marks. Positive EV.
2. If you can’t eliminate any options, SKIP. Expected value = 0.25 × 2.5 − 0.75 × 0.833 = 0. Zero EV, but stress matters too.
3. If you can eliminate 1 option (3 options left), do NOT guess. EV = -0.27. Negative.
4. If a numerical answer is one of (a) absurd, (b) very clean, (c) ugly — usually clean wins. UPSC designs answers.

Time management on exam day

1. First pass (60 min): Solve everything you can do in under 90 seconds. Skip otherwise.
2. Second pass (40 min): Attempt the trickier QA + DI you skipped.
3. Final pass (20 min): Revisit, verify, fill OMR. Never leave OMR-filling for the end of the last 5 minutes.

Common mistakes that cost CSAT marks

1. Treating CSAT as optional. Even 1% of candidates fail because of CSAT. You don’t want to be one.
2. Reading the question half-way. ‘Profit on cost’ vs ‘profit on SP’ is the single most common trap. Read every word.
3. Computing what you can approximate. If the answer options are 84%, 87%, 91%, 96%, you don’t need decimal precision — eyeball it.
4. Linear thinking on percentage problems. A 20% rise followed by a 20% fall is NOT zero net change. It’s a 4% loss. Always use multipliers.
5. Forgetting units. km/h vs m/s. Hours vs minutes. UPSC has burned candidates by switching units mid-question.
6. Negative marking ignorance. Mark only what you’re 60%+ sure of. ‘I’ll guess this one’ compounds over 80 questions into -15 marks.
7. Skipping DI because it ‘looks long’. DI is the highest-accuracy section once you can read charts fluently. Practice 30 DI sets and your speed doubles.
8. Not reading the question stem of DI sets. Most DI sets have a 2-line introduction that defines units, time periods, and exclusions. Skip it and you’ll fail every question.
9. Carrying a ‘lucky’ calculator-feel into the exam. No calculators. Build arithmetic stamina in Phase 2 or you’ll lose 30 minutes to mental math fatigue.
10. Mixing up SI and CI formulas. They differ in compounding behaviour. SI is linear, CI is exponential. The 2-year CI minus SI = P(R/100)² formula is your fastest way to spot which is which.

Sample DI set — solved walkthrough

If you’ve never solved a CSAT DI set, here’s a representative one with full reasoning. Cover the answer, attempt yourself, then read the explanation.

Set: The table below shows the production (in lakh tonnes) of four crops in five states for 2024-25.

Q1. The state with the highest combined production of rice and wheat: A) Punjab B) UP C) MP D) Haryana.

Approach: Add rice + wheat for each. Punjab = 300. Haryana = 180. UP = 470. Bihar = 140. MP = 220. Answer: B) UP.

Q2. Wheat production in UP is approximately what percent of total wheat in all 5 states? A) 30% B) 35% C) 38% D) 42%.

Approach: Total wheat = 180+120+320+65+190 = 875. UP share = 320/875. Approximate: 320/875 ≈ 320/900 ≈ 35.6%. Answer: C) 38% (precise = 36.6%, closest option).

Q3. Which state’s total agricultural production exceeds 400 lakh tonnes? A) Punjab only B) UP only C) UP and MP D) Punjab and UP.

Approach: Punjab total = 370. UP = 605. Bihar = 200. Haryana = 245. MP = 375. Answer: B) UP only.

Q4. Ratio of maize production in MP to maize production in Punjab? A) 19 : 9 B) 2 : 1 C) 9 : 19 D) 21 : 10.

Approach: 95 : 45 = 19 : 9. Answer: A) 19 : 9.

One-page cheat sheet — every CSAT formula

Print this section. Pin it above your desk. Revise it every morning for the last 30 days. It is everything you need to remember on exam day.

Percentage and profit

• % change = ((new − old) / old) × 100
• Successive change = a + b + ab/100 (a, b signed)
• If X rises r%, consumption must fall by r/(100+r) × 100 to keep expenditure constant
• Profit % = (Profit / CP) × 100 — always on CP, never on SP
• SP at p% profit = CP × (1 + p/100)
• Two articles at same SP, one at +x%, other at -x% → net loss = (x/10)² %
• False weight profit = (error / (true − error)) × 100

Interest

• SI = P × R × T / 100
• CI: A = P × (1 + R/100)^T, CI = A − P
• Half-yearly compounding: R → R/2, T → 2T
• CI − SI for 2 years = P × (R/100)²
• Rule of 72 for CI doubling: T ≈ 72/R years
• SI doubling: T = 100/R years

Ratio and average

• Componendo: a/b = c/d → (a+b)/(a-b) = (c+d)/(c-d)
• Direct variation: y = kx; inverse: y = k/x
• Average = sum/count; sum = count × average
• Weighted average = (n₁A₁ + n₂A₂) / (n₁ + n₂)
• Average of AP = (first + last) / 2

Speed, work, mixtures

• 1 km/h = 5/18 m/s
• Average speed (half-distance trips) = 2ab/(a+b), harmonic mean
• Average speed (half-time trips) = (a+b)/2, arithmetic mean
• Relative speed: same direction = |a−b|, opposite = a+b
• Train + platform: time = (L_train + L_platform) / speed
• Boat: downstream = b+s, upstream = b−s
• Combined work time: 1/a + 1/b → ab/(a+b)
• LCM trick: total work = LCM of individual days; each rate = LCM / its days
• Alligation: cheaper : dearer = (dearer − mean) : (mean − cheaper)
• Replacement formula: final pure = initial × (1 − replaced/total)^n

Number system

• HCF × LCM = product of two numbers
• Same-remainder problem → take HCF of differences
• Smallest number with same remainder R from divisors a, b, c → LCM(a,b,c) + R
• Unit-digit cycle of 4: 2, 3, 7, 8
• Unit-digit cycle of 2: 4, 9
• Unit-digit cycle of 1 (constant): 0, 1, 5, 6
• (a + b) mod n = ((a mod n) + (b mod n)) mod n
• Sum of first n natural numbers = n(n+1)/2
• Sum of first n squares = n(n+1)(2n+1)/6
• Sum of first n cubes = (n(n+1)/2)²

Mensuration

• Square: A = s², P = 4s, diagonal = s√2
• Rectangle: A = lb, P = 2(l+b), diagonal = √(l² + b²)
• Triangle: A = ½bh; equilateral: A = (√3/4)s²
• Circle: A = πr², C = 2πr
• Cube: V = a³, TSA = 6a², diagonal = a√3
• Cuboid: V = lbh, TSA = 2(lb + bh + hl)
• Cylinder: V = πr²h, TSA = 2πr(h+r)
• Cone: V = (1/3)πr²h, l = √(r²+h²)
• Sphere: V = (4/3)πr³, SA = 4πr²
• Hemisphere: V = (2/3)πr³, TSA = 3πr²

PnC and probability

• nPr = n!/(n−r)! (order matters)
• nCr = n!/(r!(n−r)!) (order doesn’t)
• Distinct arrangements with repetitions = n!/(p! × q! × …)
• Circular arrangements of n = (n−1)!
• P(A or B) = P(A) + P(B) − P(A and B)
• Independent: P(A and B) = P(A) × P(B)
• P(at least one) = 1 − P(none)

Standard-problem library — 10 solved patterns

Ten more problem patterns that recur in CSAT, each solved fully. If you can replicate the reasoning on a fresh problem, you’re ready.

Problem 1. The price of sugar increases by 25%. A family wants to keep its sugar expenditure unchanged. By what percent should it reduce sugar consumption?

Solution. Price multiplier = 1.25. Consumption multiplier needed = 1/1.25 = 0.80. Reduction = 20%. (Reciprocal trick: r/(100+r) × 100 = 25/125 × 100 = 20%.)

Problem 2. If 60 men can complete a job in 25 days, how many men are required to finish the same job in 15 days?

Solution. Man-days = 60 × 25 = 1500. New days = 15. Required men = 1500/15 = 100. So 40 additional men needed.

Problem 3. The simple interest on a sum at 8% per annum for 6 years is ₹4,800 less than the principal. Find the principal.

Solution. SI = P × 8 × 6 / 100 = 0.48P. P − SI = 4800. P − 0.48P = 4800. 0.52P = 4800. P = ₹9,230.77.

Problem 4. A boat goes 12 km downstream in 2 hours and returns in 4 hours. Find speed of boat in still water and speed of current.

Solution. Down speed = 6 km/h. Up speed = 3 km/h. Boat = (6+3)/2 = 4.5 km/h. Current = (6-3)/2 = 1.5 km/h.

Problem 5. A bag contains 5 red and 4 blue balls. Two balls are drawn at random. Probability both are red?

Solution. Favourable = 5C2 = 10. Total = 9C2 = 36. P = 10/36 = 5/18.

Problem 6. The ratio of milk to water in a 60-litre mixture is 7 : 3. How much water must be added to make the ratio 3 : 2?

Solution. Milk = 42 L, Water = 18 L. After adding x L water: 42/(18+x) = 3/2 → 84 = 54 + 3x → x = 10 L.

Problem 7. A man buys two horses for ₹30,000. He sells one at a 15% gain and the other at a 10% loss, and makes neither profit nor loss overall. Find the cost price of each.

Solution. Let CP of horse 1 = x. Then CP of horse 2 = 30,000 − x. Combined SP = x × 1.15 + (30000 − x) × 0.90 = 30000. Solve: 1.15x + 27000 − 0.9x = 30000 → 0.25x = 3000 → x = 12,000. Other = 18,000.

Problem 8. The radius and height of a cylinder are in ratio 3 : 5. Its volume is 4,851 cm³. Find its total surface area. (Take π = 22/7.)

Solution. Let r = 3k, h = 5k. V = π × 9k² × 5k = 45πk³ = 4851. k³ = 4851/(45 × 22/7) = 4851 × 7/990 = 34.3. So k = 3.25 ≈ 3. Actually let me solve cleanly: 45πk³ = 4851 → k³ = 4851/(45 × 22/7) = 4851 × 7 / 990 = 34.3. k = ∛34.3 ≈ 3.25. (CSAT would round to k = 3 for clean numbers.) Use k = 3: r = 9, h = 15. TSA = 2πr(r+h) = 2 × 22/7 × 9 × 24 = 1357.7 cm².

Problem 9. In how many ways can the letters of ‘COMPUTER’ be arranged so that vowels never come together?

Solution. Total arrangements of 8 distinct letters = 8!. Vowels are O, U, E (3 vowels). Treat vowels together = 6 units → 6! × 3! (internal vowel arrangements). Vowels never together = 8! − 6! × 3! = 40320 − 4320 = 36,000.

Problem 10. A clock gains 5 minutes every hour. If it shows the correct time at 12 noon, what is the actual time when the clock shows 6 PM?

Solution. Faulty clock runs 65 min for every actual 60 min. So 6 hours on faulty clock = 6 × 60/65 actual hours = 360/65 = 5.538 hours = 5 hours 32 minutes. Actual time when faulty shows 6 PM = 12:00 + 5h 32m = 5:32 PM.

Exam day checklist

1. Carry 3 pens (black ball-point), 2 pencils, eraser, sharpener. No mechanical pencil — UPSC sometimes objects.
2. Reach the centre 90 minutes early. CSAT is in the afternoon (2:30-4:30 PM). Eat a light lunch by 12:30.
3. First read the question paper for 2 minutes. Don’t solve. Just note: how many DI sets, how many RC passages, where the heavy QA is.
4. Pick your starting section based on the read. If RC looks easy, start there. If QA looks light, start there. Confidence in the first 20 questions is everything.
5. Mark answers on OMR after every 10 questions. Don’t leave it for the end — students who do this every year forget and lose 30+ marks.
6. Use the rough sheet ruthlessly. Don’t try to be mental-math hero. Write down the percentage multiplier; you’ll save 30 seconds of re-reading.
7. Fill OMR completely. A half-filled bubble is treated as unattempted. Use a sharpened pencil; bubble fully.
8. If stuck on a question for more than 90 seconds, skip. One stuck question is six skipped easy ones.
9. Don’t panic if the paper feels hard. Cutoff is 66 marks. Even if the paper feels brutal, you only need 25-30 correct answers out of 80. Focus on what you can do.
10. Verify OMR before you submit. Match the question number on OMR to the question paper. One mis-aligned bubble shifts your entire answer key.

Frequently asked questions

Bottom line

CSAT QA is not a barrier. It’s a tax. Pay it once with 60 disciplined days of practice, and you’ll never think about it again for the rest of your prep. The candidates who fail CSAT are almost never the ones who couldn’t learn the math — they’re the ones who didn’t take it seriously enough to practice. Don’t be that candidate. Start with number systems and percentage today. Solve 10 problems before you close this tab.

If you found this guide useful, the same level of structured prep is what we deliver in our UPSC Foundation programme for GS, optionals, and CSAT — daily classes, weekly tests, individual mentor reviews, and a current-affairs Compass that maps every story to syllabus. Book a free demo class to see how a single week’s prep unfolds.