Chapter 01

1. Number System, Divisibility & Number Series Patterns

The Number System is the bedrock of quantitative reasoning, covering classifications, divisibility rules, unit digit cyclicity, remainder theorems, and Number Series patterns (Arithmetic/Geometric progressions, second-order differences, alternating operations, cubes/squares, triangular numbers, digit operations, and wrong number identification).

Key Number System & Series Formulas

• Divisibility Rules: 3 & 9 (sum of digits), 4 & 8 (last 2 or 3 digits), 11 (|(Sum_odd) - (Sum_even)| = 0 or 11k).
• Unit Digit Cyclicity: 2, 3, 7, 8 repeat in cycles of 4; 4, 9 repeat in cycles of 2; 0, 1, 5, 6 repeat in cycles of 1.
• Pattern 1 (AP Series): a_n = a + (n - 1)d; S_n = [n/2][2a + (n - 1)d].
• Pattern 2 (Difference Series): Differences between consecutive terms increase/decrease by a constant d_n = d₁ + (n - 1)c.
• Pattern 3 (GP & Multiplier Series): a_n = a × r^n-1 or alternating multipliers ×2, ×1.5, ×2...
• Pattern 4 (Alternating Operations): Alternating operations (+a, ×b, +a, ×b) or (×n + n, ×n - n).
• Pattern 5 & 6 (Powers & Triangular Numbers): T_n = n², n³, n³ - n², and Triangular Number T_n = [n(n + 1)] / 2.
• Pattern 7 & 8 (Digit Ops & Prime Differences): T_n = (Product of digits of T_n-1) + 1; or differences are consecutive primes (2, 3, 5, 7, 11...).
• Pattern 9 & 10 (Percentages & Higher-Order): Alternating percentage shifts (±10%); 3rd order difference Δ³ is constant for cubic series.

Solved Example 1: Multiplier & Product Sum Missing Term

Problem: What number should come in place of question mark (?) in: 3, 4, 10, 33, 136, 685, ?

1

Analyze the progression step-by-step: 3 × 1 + 1 = 4; 4 × 2 + 2 = 10; 10 × 3 + 3 = 33.

2

Continue verifying the pattern: 33 × 4 + 4 = 136; 136 × 5 + 5 = 685.

3

Compute next term: 685 × 6 + 6 = 4110 + 6 = 4116.

Final Answer: 4116

Solved Example 2: Wrong Term Identification Bank PO Pattern

Problem: Find the wrong number in the series: 529, 841, 961, 1296, 1681, 1849, 2209.

1

Express terms as squares: 529 = 23², 841 = 29², 961 = 31², 1296 = 36², 1681 = 41², 1849 = 43², 2209 = 47².

2

Notice the base numbers (23, 29, 31, 37, 41, 43, 47) must all be prime numbers. 36 is composite! The correct term should be 37² = 1369.

Final Answer: 1296 (Should be 1369)

Common Mistake: Jumping to complex exponent formulas before computing first-order and second-order differences. Over 80% of competitive series can be cracked by taking consecutive differences Δ = T_n - T_n-1.

Speed Tip for Wrong Number Series: When one number in a series is incorrect, it will create TWO incorrect differences in the first difference row. The incorrect number lies right between those two faulty differences!

Chapter 02

2. HCF (GCD) and LCM Fundamentals

Highest Common Factor (HCF) is the greatest divisor common to two or more numbers. Lowest Common Multiple (LCM) is the smallest positive integer divisible by all numbers in the set. Essential for bell-ringing problems, circular tracks, and fractional calculations.

Key Formulas & Properties

• Product Relationship: Product of two numbers (A × B) = HCF(A, B) × LCM(A, B).
• HCF of Fractions: HCF(a/b, c/d) = [HCF of numerators (a, c)] / [LCM of denominators (b, d)].
• LCM of Fractions: LCM(a/b, c/d) = [LCM of numerators (a, c)] / [HCF of denominators (b, d)].
• Greatest number dividing x, y, z leaving remainders a, b, c: HCF(x - a, y - b, z - c).

Solved Example Toll Bells

Problem: Three bells toll together at intervals of 9, 12, and 15 minutes respectively. If they toll together now, after how many hours will they toll together next?

1

Find prime factorizations: 9 = 3², 12 = 2² × 3, 15 = 3 × 5.

2

Calculate LCM: Take highest powers of each prime factor → 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes.

3

Convert minutes to hours: 180 / 60 = 3 hours.

Final Answer: 3 Hours

Common Mistake: Applying Product = HCF × LCM for three or more numbers. This formula holds strictly for two numbers only.

Speed Tip: The HCF of any two numbers can never exceed their positive difference. If numbers are close (e.g., 437 and 483), find Difference = 46 = 2 × 23. Test 23 directly!

Chapter 03

3. Percentages & Successive Changes

Percentage expresses a number as a fraction of 100. It is the primary tool for Data Interpretation, Profit & Loss, and Growth calculations. Mastering fractional equivalences eliminates slow division.

Key Formulas & Equivalences

• Percentage Change: Change % = [(Final Value - Initial Value) / Initial Value] × 100.
• Successive Percentage Formula: Net % = a + b + (a × b / 100) (use negative sign for decreases).
• Product Constancy: If price increases by x%, consumption must decrease by [x / (100 + x)] × 100% to keep expenditure constant.
• Fractions: 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%, 1/12 = 8.33%.

Solved Example Successive Change

Problem: The salary of an employee is first increased by 20% and subsequently decreased by 15%. What is the net percentage change in salary?

1

Identify values: a = +20, b = -15.

2

Apply successive formula: Net % = 20 + (-15) + [20 × (-15)] / 100 = 5 - 300 / 100 = 5 - 3 = +2%.

Final Answer: 2% Net Increase

Common Mistake: Switching the base value. If A is 25% more than B, B is NOT 25% less than A. B is (25 / 125) × 100 = 20% less than A.

Chapter 04

4. Profit, Loss, Discount & Markups

Commercial mathematics tests transactions between Cost Price (CP), Selling Price (SP), Marked Price (MP), Discount, and Profit/Loss margins. Also includes dishonest dealer schemes.

Key Formulas

• Profit % & Loss %: Profit % = [(SP - CP) / CP] × 100; Loss % = [(CP - SP) / CP] × 100 (Always on CP).
• Selling Price with Markup & Discount: SP = MP × [1 - (Discount % / 100)] = CP × [1 + (Profit % / 100)].
• Dishonest Dealer (False Weight): Gain % = [Error / (True Value - Error)] × 100 = [(True Weight - False Weight) / False Weight] × 100.
• Equal SP with equal Gain/Loss x%: Net result is always a loss of (x² / 100)%.

Solved Example False Weight

Problem: A dishonest shopkeeper claims to sell goods at Cost Price, but uses a false weight of 900 grams for a 1 kg weight. What is his profit percentage?

1

Identify values: True Weight = 1000g, False Weight = 900g, Error = 100g.

2

Apply formula: Profit % = (100 / 900) × 100 = (1 / 9) × 100 = 11.11%.

Final Answer: 11.11% (or 11 1/9%)

Speed Shortcut: When two items are sold at the same price—one at 20% profit and another at 20% loss—there is ALWAYS an overall loss of (20² / 100) = 4%.

Chapter 05

5. Ratio, Proportion & Alligation Mixtures

Ratios compare quantities of the same kind. Proportions express equality between two ratios (a : b :: c : d → a × d = b × c). The Alligation rule allows instant solving of weighted mixture problems.

Key Formulas

• Compounding & Merging: If A : B = 2 : 3 and B : C = 4 : 5, then A : B : C = (2 × 4) : (3 × 4) : (3 × 5) = 8 : 12 : 15.
• Mean Proportional: Between a and b is √(ab); Third Proportional to a, b is b² / a.
• Alligation Rule: (Quantity of Cheaper / Quantity of Dearer) = (Dearer Price - Mean Price) / (Mean Price - Cheaper Price).
• Repeated Replacement: If a container of V litres has x litres drawn out and replaced by water n times: Liquid Left = V × [1 - (x / V)]ⁿ.

Solved Example Alligation

Problem: In what ratio must tea at ₹62 per kg be mixed with tea at ₹72 per kg so that the mixture is worth ₹65 per kg?

1

Cheaper price C = 62, Dearer price D = 72, Mean price M = 65.

2

Ratio (Cheaper : Dearer) = (D - M) : (M - C) = (72 - 65) : (65 - 62) = 7 : 3.

Final Answer: 7 : 3

Chapter 06

6. Averages & Weighted Deviations

Average (Arithmetic Mean) is the sum of observations divided by the total number of observations. Advanced questions involve replacements, inclusions, exclusions, and batting/bowling averages.

Key Formulas

• Fundamental Equation: Average = (Total Sum / Count N) → Total Sum = Average × N.
• Replacement Adjustment: New Value = Replaced Value + (Change in Average × N).
• Average of first n natural numbers: (n + 1) / 2; Average of first n even numbers: n + 1; odd numbers: n.
• Consecutive Numbers Average: Always equals the exact middle term or (First Term + Last Term) / 2.

Solved Example Replacement

Problem: The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one weighing 65 kg. What is the weight of the new person?

1

Total increase in weight for 8 persons = 8 × 2.5 kg = 20 kg.

2

Weight of new person = Old Weight + Total Increase = 65 + 20 = 85 kg.

Final Answer: 85 kg

Chapters 07 & 08

7 & 8. Simple Interest (SI) & Compound Interest (CI)

Simple Interest is calculated uniformly on the principal amount throughout the duration. Compound Interest computes interest on both the principal and previously accumulated interest ("interest on interest").

Key Formulas

• Simple Interest: SI = (P × R × T) / 100; Total Amount A = P + SI.
• Compound Interest: A = P × [1 + (R / 100)]ⁿ; CI = A - P.
• Half-Yearly Compounding: Rate becomes (R / 2)%, time becomes 2n.
• Difference between CI and SI for 2 Years: Diff₂ = P × (R / 100)².
• Difference between CI and SI for 3 Years: Diff₃ = P × (R / 100)² × [(300 + R) / 100].

Solved Example CI-SI Difference

Problem: The difference between Simple Interest and Compound Interest on a certain sum for 2 years at 10% per annum is ₹65. Find the principal sum.

1

Use 2-year difference formula: Diff = P × (R / 100)².

2

Substitute values: 65 = P × (10 / 100)² → 65 = P × (1 / 100) → P = 65 × 100 = 6,500.

Final Answer: ₹6,500

Rule of 72: To estimate the years required for money to double under compound interest, divide 72 by the annual interest rate: Years ≈ 72 / R.

Chapter 09

9. Time and Work & Pipes/Cisterns

Time and Work problems revolve around the inverse relationship between work rate efficiency and the time taken. The LCM unit method simplifies complex multi-person and alternating worker scenarios.

Key Formulas & LCM Method

• Unitary Method: If A completes work in n days, A's 1-day work = 1 / n.
• Two Workers Together: Time taken = (x × y) / (x + y) days.
• Man-Day Equation: (M₁ × D₁ × H₁) / W₁ = (M₂ × D₂ × H₂) / W₂.
• Pipes & Cisterns: Inlet pipe rate is positive (+), leak/outlet pipe rate is negative (-).

Solved Example LCM Method

Problem: A can finish a work in 12 days and B in 18 days. They work together for 4 days, then A leaves. How many days will B take to complete the remaining work?

1

Find total work: LCM(12, 18) = 36 units.

2

Find efficiencies: A's rate = 36 / 12 = 3 units/day; B's rate = 36 / 18 = 2 units/day. Combined rate = 5 units/day.

3

Work done in first 4 days = 4 × 5 = 20 units. Remaining work = 36 - 20 = 16 units.

4

Time for B alone = Remaining Work / B's rate = 16 / 2 = 8 days.

Final Answer: 8 Days

Chapter 10

10. Time, Speed, Distance, Trains & Streams

Speed is distance per unit time. Specialized topics include uniform speed conversions, harmonic average velocities, relative speeds of overtaking/crossing trains, and vector summation in water currents.

Key Formulas

• Unit Conversion: 1 km/h = (5 / 18) m/s; 1 m/s = (18 / 5) km/h.
• Average Speed (Equal Distances): Avg Speed = (2 × x × y) / (x + y) (where x, y are speeds).
• Relative Speed: Same Direction = |u - v|; Opposite Direction = u + v.
• Train Crossing Platform: Distance = Length of Train + Length of Platform.
• Boats & Streams: Downstream = u + v, Upstream = u - v, Speed in Still Water u = (Downstream + Upstream) / 2, Stream Speed v = (Downstream - Upstream) / 2.

Solved Example Train Platform

Problem: A train 150 meters long is traveling at 54 km/h. How long will it take to cross a platform 250 meters long?

1

Total distance to cover = 150 m + 250 m = 400 m.

2

Convert speed to m/s: 54 × (5 / 18) = 3 × 5 = 15 m/s.

3

Time = Distance / Speed = 400 / 15 = 80 / 3 = 26.67 seconds.

Final Answer: 26.67 seconds (or 26 2/3 s)

Chapter 11

11. Problems on Ages

Age word problems translate real-world scenarios into linear equations in one or two variables. Key invariant: the age difference between two people remains constant across all points in time.

Key Principles

• Timeline Shifts: If present age is x, age n years ago was (x - n), age n years hence will be (x + n).
• Constant Difference: If Father is 26 years older than Son today, he will still be 26 years older after 20 years.

Solved Example Age Ratios

Problem: The ratio of the ages of Father and Son at present is 7 : 2. After 5 years, the ratio becomes 4 : 1. Find the present age of the Father.

1

Let present ages be 7x and 2x.

2

After 5 years: (7x + 5) / (2x + 5) = 4 / 1 → 7x + 5 = 4(2x + 5) → x = 5.

3

Present age of Father = 7 × 5 = 35 years.

Final Answer: 35 Years

Chapter 12

12. Probability & Independent Events

Probability measures the likelihood of an outcome occurring within a defined sample space (0 ≤ P(E) ≤ 1). Core exam patterns include playing cards (52-card deck), dice throws, ball urns, and coin tosses.

Key Formulas

• Classical Formula: P(E) = n(E) / n(S) = (Number of Favorable Outcomes) / (Total Possible Outcomes).
• Complementary Event: P(not E) = 1 - P(E).
• Addition Rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B). For mutually exclusive events, P(A ∩ B) = 0.
• Independent Events: P(A ∩ B) = P(A) × P(B).
• 52-Card Deck Breakdown: 4 Suits (13 each), 26 Red, 26 Black, 12 Face Cards (4 Kings, 4 Queens, 4 Jacks), 4 Aces.

Solved Example Dice Pair

Problem: In a single throw of two unbiased dice, what is the probability of getting a total sum of 8?

1

Total outcomes when rolling two dice = 6 × 6 = 36.

2

List pairs summing to 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 favorable outcomes.

3

P(Sum = 8) = 5 / 36.

Final Answer: 5/36 (≈ 0.1389)

Chapter 13

13. Permutation & Combination (P&C)

Permutation concerns the arrangement and ordering of items where sequence matters (ⁿP_r). Combination deals strictly with selection and grouping where order does NOT matter (ⁿC_r).

Key Formulas

• Factorial: n! = n × (n - 1) × … × 1; 0! = 1.
• Permutation (Arrangement): ⁿP_r = n! / (n - r)!.
• Combination (Selection): ⁿC_r = n! / [r!(n - r)!] = ⁿP_r / r!.
• Circular Arrangements: For n distinct objects, circular arrangements = (n - 1)!.
• Repetitive Letters in Word: Arrangements = n! / (p! × q! × r!) where p, q, r are repeat counts.

Solved Example Team Selection

Problem: In how many ways can a committee of 5 members be formed from 6 men and 4 women, such that it contains exactly 3 men and 2 women?

1

Selecting 3 men out of 6: ⁶C₃ = (6 × 5 × 4) / (3 × 2 × 1) = 20.

2

Selecting 2 women out of 4: ⁴C₂ = (4 × 3) / (2 × 1) = 6.

3

By Fundamental Counting Principle: Total Ways = 20 × 6 = 120.

Final Answer: 120 Ways

Chapter 14

14. Data Interpretation (DI)

Data Interpretation tests your ability to extract, analyze, calculate, and infer conclusions from graphical representations such as Bar Charts, Pie Charts, Line Graphs, and Complex Tables.

Key Calculation Formulas in DI

• Percentage Distribution in Pie Chart: Angle in Degrees = (Percentage / 100) × 360°.
• Percentage Growth Rate: Growth % = [(Value in Year 2 - Value in Year 1) / Value in Year 1] × 100.
• Ratio Comparison Shortcut: To compare (a / b) vs (c / d), cross-multiply: if a × d > b × c, then a / b > c / d.

DI Strategy Tip: In placement exams, DI questions are often approximation friendly. Round large numbers to the nearest hundreds or use percentage fractions to eliminate wrong options instantly without calculating exact 4-decimal values.