➗ CSAT — BASIC NUMERACY (MATHS) · Complete Topic-by-Topic Explainer

Paper: Prelims Paper II (CSAT) · 80 questions · 200 marks · 2 hours · qualifying at 66.67/200 (33%) Marking: +2.5 correct, −0.83 wrong. Level: Class X. Why it matters: CSAT is only qualifying, but failing it ends your attempt no matter how well you did in GS Paper I. Every year candidates with excellent GS scores are eliminated here. Treat it as a guaranteed-clear paper, not an afterthought.

How to use this file: each topic has (1) What it is, (2) Why it matters for CSAT, (3) Formulas, (4) Technique/shortcut, (5) Worked example. Redo the worked examples from scratch — reading them is not practice.

1. NUMBER SYSTEM

What it is: properties of integers — divisibility, factors, remainders, place value. Why it matters: the base layer. Divisibility and remainder tricks save time everywhere else, and 2–4 direct questions appear most years.

Classification

• Natural (1,2,3…), Whole (0,1,2…), Integers (…−2,−1,0,1,2…), Rational (p/q), Irrational (√2, π), Real.
• Prime: exactly two factors (2 is the only even prime). Composite: more than two. 1 is neither.
• Co-prime: HCF = 1 (e.g., 8 and 15).

Divisibility rules (memorize — high payoff)

Key formulas

 HCF × LCM = product of the two numbers
 Number of factors of N = (a+1)(b+1)(c+1)...  where N = p^a · q^b · r^c
 Sum of first n naturals      = n(n+1)/2
 Sum of first n odd numbers   = n²
 Sum of first n even numbers  = n(n+1)
 Sum of squares 1..n          = n(n+1)(2n+1)/6
 Sum of cubes 1..n            = [n(n+1)/2]²

Unit-digit cyclicity (technique)

Powers repeat every 4. Reduce the exponent mod 4.

 2 → 2,4,8,6   3 → 3,9,7,1   7 → 7,9,3,1   8 → 8,4,2,6
 4 → 4,6       9 → 9,1       0,1,5,6 → unchanged

Worked example: Unit digit of 7^103? 103 mod 4 = 3 → third in cycle (7,9,3,1) = 3.

Remainders

• If N leaves remainder r on division by d, write N = dk + r. Substitute to solve. Worked example: N ÷ 7 leaves 4. Remainder when 3N ÷ 7? N = 7k+4 → 3N = 21k+12 → 12 mod 7 = 5.

2. HCF AND LCM

What it is: greatest common divisor / least common multiple. Why it matters: appears directly, and underlies bells-ringing-together, tiles, and race-track problems.

Methods: prime factorization (HCF = lowest powers of common primes; LCM = highest powers of all primes) or division method.

 HCF × LCM = a × b   (only for TWO numbers)
 HCF of fractions = HCF(numerators) / LCM(denominators)
 LCM of fractions = LCM(numerators) / HCF(denominators)

Typical application: "Bells ring at intervals 6, 8, 12 min — when together again?" → LCM = 24 min. Worked example: Largest number dividing 60, 90, 150 exactly → HCF = 30.

3. PERCENTAGES

What it is: a fraction expressed per hundred. Why it matters: the single most-used concept — profit/loss, interest, DI, data comparison all reduce to percentages.

 x% of y = (x/100)·y
 % change = (New − Old)/Old × 100
 If a value increases by r% then decreases by r%:
     net change = −r²/100  (ALWAYS a net loss)
 If A is x% more than B, then B is [x/(100+x)]×100 % less than A
 Successive changes a% then b%: net = a + b + ab/100

Fraction ↔ percentage table (memorize — huge time saver)

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

Technique: convert percentages to fractions to avoid long multiplication. 37.5% of 64 = (3/8)×64 = 24.

Worked example: A salary rises 20% then falls 20%. Net? = −(20²)/100 = −4% (a loss, not zero).

4. PROFIT, LOSS AND DISCOUNT

What it is: commercial arithmetic on cost price (CP), selling price (SP), marked price (MP). Why it matters: 2–3 questions typically; also a favourite in DI caselets.

 Profit = SP − CP ;  Profit% = (SP−CP)/CP × 100   [always on CP]
 SP = CP(1 + P%/100)  ;  CP = SP / (1 + P%/100)
 Discount is on MP:  SP = MP(1 − d/100)
 Successive discounts a%, b%: net = a + b − ab/100
 If articles bought at x for ₹1 and sold at y for ₹1:
     Profit% = [(x − y)/y] × 100

Technique: for "buy A for ₹X, sell B for ₹X" — compare per-unit prices, or use the formula above directly.

Worked example: Buys 12 apples for ₹100, sells 10 for ₹100. Profit%? CP/apple = 100/12 = 8.33; SP/apple = 10. Profit% = (10−8.33)/8.33 ≈ 20%.

Worked example: Marks goods 40% above cost, gives 25% discount. Profit? 1.40 × 0.75 = 1.05 → 5% profit.

5. SIMPLE AND COMPOUND INTEREST

What it is: growth of money over time; simple = on principal only, compound = on accumulated amount. Why it matters: guaranteed 1–2 questions; also the mathematical basis of economy concepts (inflation, growth rates).

 SI = P·R·T/100
 Amount(CI) = P(1 + R/100)^T   ;  CI = Amount − P
 Half-yearly: rate R/2, time 2T.  Quarterly: R/4, 4T.
 Difference for 2 years:  CI − SI = P(R/100)²
 Difference for 3 years:  CI − SI = P(R/100)²·(3 + R/100)
 Under SI, money doubles in 100/R years; triples in 200/R years
 Under CI, doubling time ≈ 72/R  (rule of 72)

Technique: for CI at small rates over 2 years, use successive-percentage: 10% then 10% = 21% total.

Worked example: CI on ₹10,000 at 10% for 2 years = 10000×(1.21) − 10000 = ₹2,100. Worked example: A sum doubles in 5 years at SI. Rate? 100% in 5 years → 20% p.a.

6. RATIO, PROPORTION AND PARTNERSHIP

What it is: comparison of quantities. Why it matters: underpins mixtures, partnership, ages, and scaling problems.

 a:b = c:d  →  ad = bc  (cross multiplication)
 Combine a:b and b:c → make b common → a:b:c
 If a:b = 2:3 and b:c = 4:5 → a:b:c = 8:12:15 → a:c = 8:15
 Dividing ₹N in ratio a:b:c → each share = N × (part / total parts)
 Partnership profit ∝ (capital × time)

Technique: always convert ratios to "parts," find the value of one part, then answer.

Worked example: Divide 64 in ratio 3:5. Total = 8 parts; 1 part = 8 → shares 24 and 40.

7. AVERAGES, MIXTURES AND ALLIGATION

What it is: central value; and combining two quantities of different concentrations/prices. Why it matters: averages appear directly; alligation is the fastest route to mixture and average-blend questions.

 Average = Sum / n
 Sum = Average × n
 Average of first n naturals = (n+1)/2
 Average of consecutive numbers = (first + last)/2
 If one value replaces another, change in avg = (new − old)/n

 ALLIGATION:
     (Cheaper qty)   Dearer − Mean
     ------------- = ---------------
     (Dearer qty)    Mean − Cheaper

Technique (alligation): put the two rates on the ends, the mean in the middle, cross-subtract.

Worked example: Average of 30 students is 12; adding the teacher makes it 13. Teacher's age? Old sum 360; new sum 31×13 = 403 → 43.

Worked example (alligation): Mix rice at ₹30/kg and ₹40/kg to get ₹34/kg. Ratio? (40−34) : (34−30) = 6:4 = 3:2.

8. TIME, SPEED AND DISTANCE

What it is: motion relationships. Why it matters: consistently 3–5 questions (trains, boats, races).

 Speed = Distance / Time
 1 km/h = 5/18 m/s   ;   1 m/s = 18/5 km/h
 Average speed (equal DISTANCES, speeds x and y) = 2xy/(x+y)   [harmonic mean]
 Average speed (equal TIMES) = (x+y)/2
 Speed ∝ 1/Time (distance constant)

 TRAINS:
   Crossing a pole/man      : distance = length of train
   Crossing platform/bridge : distance = train + platform
   Two trains same direction: relative speed = |a − b|
   Two trains opposite      : relative speed = a + b

 BOATS:
   Downstream = b + s ;  Upstream = b − s
   b = (down + up)/2  ;  s = (down − up)/2

Technique: convert km/h → m/s (×5/18) immediately whenever lengths are in metres.

Worked example: A 150 m train at 54 km/h crosses a 150 m platform in? 54 km/h = 15 m/s; distance = 300 m → 20 seconds.

Worked example: Boat covers 12 km downstream in 2 h and returns in 3 h. Down = 6, Up = 4 → stream = (6−4)/2 = 1 km/h; boat = 5 km/h.

9. TIME AND WORK (and Pipes & Cisterns)

What it is: rate of doing work. Why it matters: 2–4 questions; the "unitary/LCM method" makes them fast.

 If A finishes in 'a' days → A's 1-day work = 1/a
 A + B together: 1/a + 1/b → time = ab/(a+b)
 M₁D₁H₁ / W₁ = M₂D₂H₂ / W₂   (men-days-hours-work)
 Efficiency ∝ 1/Time
 If efficiency increases by r%, time decreases by [r/(100+r)]×100 %
 PIPES: inlet positive, outlet (leak) negative

Technique (LCM method): take total work = LCM of the given days. Then each person's rate is a whole number — no fractions.

Worked example: A in 12 days, B in 18 days. Together? LCM = 36 units. A = 3/day, B = 2/day → 5/day → 36/5 = 7.2 days.

Worked example: A (20 d) and B (30 d) work together 6 days, then A leaves. B finishes in? LCM 60: A=3, B=2. In 6 days → 30 units done; 30 left; B alone → 30/2 = 15 days.

Worked example: Efficiency rises 25% → time falls by 25/125 × 100 = 20%.

10. MENSURATION

What it is: area, perimeter, surface area, volume. Why it matters: 1–3 direct questions; formulas are pure recall — free marks.

2-D

3-D

Scaling technique: if a linear dimension is multiplied by k → area ×k², volume ×k³. Worked example: Radius doubled → area becomes 4×; volume of sphere becomes 8×.

11. ALGEBRA (basics)

What it is: working with unknowns. Why it matters: the language for ages, mixtures, and word problems.

 (a+b)² = a² + 2ab + b²      (a−b)² = a² − 2ab + b²
 a² − b² = (a+b)(a−b)
 (a+b)³ = a³ + b³ + 3ab(a+b)
 a³ + b³ = (a+b)(a² − ab + b²)
 a³ − b³ = (a−b)(a² + ab + b²)
 Quadratic roots: x = [−b ± √(b²−4ac)]/2a
 Sum of roots = −b/a ; Product = c/a

Ages technique: let present age = x; write the condition after/before n years as (x±n). Worked example: Father is 3× son; in 12 years he'll be 2×. Son now? 3s + 12 = 2(s + 12) → 3s + 12 = 2s + 24 → s = 12.

12. PROBABILITY AND COUNTING

What it is: likelihood of events; arrangements and selections. Why it matters: 1–2 questions; easy marks if the basics are solid.

 P(E) = favourable / total ;  0 ≤ P ≤ 1
 P(not E) = 1 − P(E)
 P(A or B) = P(A) + P(B) − P(A and B)
 Independent events: P(A and B) = P(A)·P(B)
 nPr = n!/(n−r)!   (order matters)
 nCr = n!/[r!(n−r)!] (order doesn't)

Reference values: Dice (1 die: 6 outcomes; 2 dice: 36). Deck: 52 cards, 26 red/26 black, 4 suits of 13, 12 face cards, 4 aces. Two-dice sums: 7 → 6 ways (most likely); 6 and 8 → 5 ways each; 2 and 12 → 1 way each.

Worked example: P(sum 7 with two dice) = 6/36 = 1/6. Worked example: P(a king) = 4/52 = 1/13.

13. SETS AND VENN DIAGRAMS

What it is: grouping and overlap. Why it matters: the "how many read neither newspaper" question is near-guaranteed.

 n(A∪B) = n(A) + n(B) − n(A∩B)
 n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(C∩A) + n(A∩B∩C)
 Neither = Total − n(A∪B)

Worked example: 60 people, 35 read A, 25 read B, 10 read both. Neither? n(A∪B) = 35+25−10 = 50 → Neither = 60−50 = 10.

14. DATA INTERPRETATION (DI)

What it is: extracting and comparing values from tables, bar/line graphs, pie charts, caselets. Why it matters: typically 10–20 marks — the highest-density scoring block in CSAT after comprehension. Requires no new theory, only careful reading + percentage skill.

 PIE CHART: 360° = 100%  →  1% = 3.6°
     Value = (angle/360) × Total
 Growth% = (New − Old)/Old × 100
 Contribution% = (part/whole) × 100
 Ratio comparison: cross-multiply instead of dividing

Technique: 1. Read the title, axis labels, and units before looking at any question (units trip more people than arithmetic). 2. Note whether values are absolute, percentage, or indexed. 3. Approximate first — eliminate options far from your estimate; compute exactly only if two options are close. 4. For "by what percent more/less," always divide by the base being compared to.

Worked example: A pie chart sector is 72°. Its share = 72/360 = 20% of the total.

15. SEQUENCES AND SERIES (numeric patterns)

What it is: ordered numbers following a rule. Why it matters: overlaps with reasoning; 1–3 questions.

 AP: aₙ = a + (n−1)d ;  Sₙ = n/2[2a + (n−1)d] = n/2(first + last)
 GP: aₙ = ar^(n−1)   ;  Sₙ = a(rⁿ−1)/(r−1)
 Common patterns: n², n²±1, n³, n(n+1), Fibonacci (add previous two),
                  ×2+1, alternate series, prime series

Technique: compute first differences; if not constant, compute second differences; then check ratios; then check squares/cubes. Worked example: 2, 6, 12, 20, 30, ? → differences 4,6,8,10 → next difference 12 → 42. Worked example: 1,1,2,3,5,8,13,? → Fibonacci → 21.

16. SQUARES, CUBES AND ROOTS (recall table)

Why it matters: speed. Every second saved on arithmetic is a second for comprehension.

 Squares 1–30: 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,
               256,289,324,361,400,441,484,529,576,625,676,729,784,841,900
 Cubes 1–15:   1,8,27,64,125,216,343,512,729,1000,1331,1728,2197,2744,3375
 Useful roots: √1225=35, √1369=37, √1764=42, √2025=45, √2304=48, √2500=50

Square-root technique: last digit tells you the candidate (…4 → root ends 2 or 8; …9 → 3 or 7; …5 → 5; …6 → 4 or 6). Then bracket using nearby known squares.

17. EXAM STRATEGY FOR CSAT MATHS

• Order of attempt: Comprehension → Reasoning → DI → Maths. Do arithmetic last; it is the most time-hungry.
• Target: ~55–60 attempts at high accuracy comfortably clears 66.67. You do not need all 80.
• Skip discipline: if a question isn't yielding in 90 seconds, leave it. There are no bonus marks for hard questions.
• Negative marking: −0.83 per wrong. Guess only after eliminating at least one option.
• Daily drill (Months 8–10): 20 minutes of mental math — tables to 20, squares to 30, fraction↔% conversions — plus 10 DI/arithmetic questions.

Bottom line: CSAT Maths is a small, closed, completely predictable syllabus. Master these 16 topics and the paper stops being a risk.

📖 SOURCE & PRACTICE MAP

Reading order: learn the topic here → do RS Aggarwal's solved examples → attempt a CSAT paper section timed → log errors. PYQs matter more than any book — CSAT repeats patterns almost exactly.

Exam-day strategy:

 1. First pass: comprehension + easy reasoning + easy arithmetic (secure the 33%)
 2. Second pass: DI sets and medium arithmetic
 3. Skip: long-calculation traps and unfamiliar puzzles
 Target ~30 confident correct = comfortably above the 66.67 cut-off.