Aptitude

Before we dive into percentages, profit-loss, or time-speed-distance, we need to be rock solid on the basics — what kinds of numbers exist, and how to quickly tell if one number divides another. This stuff comes up everywhere in aptitude tests, and knowing it cold saves us precious seconds on every single problem.

Types of Numbers

Let’s build up from the simplest to the most complex:

• Natural Numbers (N): 1, 2, 3, 4, … (counting numbers, no zero)
• Whole Numbers (W): 0, 1, 2, 3, … (natural numbers + zero)
• Integers (Z): …, -3, -2, -1, 0, 1, 2, 3, … (whole numbers + negatives)
• Rational Numbers (Q): Any number we can write as p/q where p and q are integers and q ≠ 0. Examples: 1/2, -3/4, 7 (which is 7/1), 0.333… (which is 1/3)
• Irrational Numbers: Numbers that CAN’T be written as p/q. They go on forever without repeating. Examples: √2, √3, π, e
• Real Numbers (R): All rational + irrational numbers combined. Basically every number on the number line.

Quick exam tip: Every natural number is a whole number, every whole number is an integer, every integer is rational, and every rational number is real. It’s like nested boxes.

Prime vs Composite Numbers

• Prime: A number greater than 1 that has exactly 2 factors — 1 and itself. Examples: 2, 3, 5, 7, 11, 13…
• Composite: A number greater than 1 that has more than 2 factors. Examples: 4, 6, 8, 9, 10…
• 1 is neither prime nor composite. This is a classic trap in exams.
• 2 is the only even prime number. Another favorite exam question.

Primes up to 50 (memorize these): 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

How to check if a number is prime: We only need to check divisibility up to √n. For example, to check if 97 is prime, √97 ≈ 9.8, so we only check primes up to 9: 2, 3, 5, 7. None of them divide 97, so it’s prime.

Divisibility Rules — The Gold Standard

These rules are absolute gold in aptitude tests. Memorize them and we’ll save 30+ seconds per question.

The Rule for 7 — Step by Step

The rule for 7 trips people up, so let’s walk through it. Take the number, remove the last digit, double it, and subtract from the remaining number. Keep going until we get a small enough number.

Example: Is 1029 divisible by 7?

• Remove last digit (9), double it → 18
• Remaining number: 102
• 102 - 18 = 84
• Is 84 divisible by 7? → 84 ÷ 7 = 12. Yes!

The Rule for 11 — Step by Step

Example: Is 918082 divisible by 11?

• Digits at odd positions (1st, 3rd, 5th from right): 2, 0, 1 → sum = 3
• Digits at even positions (2nd, 4th, 6th from right): 8, 8, 9 → sum = 25
• Difference: 25 - 3 = 22, which is divisible by 11. Yes!

Factors and Multiples

• Factor: A number that divides another number evenly. Factors of 12: 1, 2, 3, 4, 6, 12
• Multiple: A number obtained by multiplying. Multiples of 5: 5, 10, 15, 20, 25…

Finding Number of Factors

This is a super common exam question. Here’s the trick:

1. Find prime factorization: N = a^p × b^q × c^r
2. Number of factors = (p+1)(q+1)(r+1)

Example: How many factors does 360 have?

• 360 = 2³ × 3² × 5¹
• Number of factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24

Sum of Factors

For N = a^p × b^q:

Sum of factors = [(a^(p+1) - 1)/(a - 1)] × [(b^(q+1) - 1)/(b - 1)]

Example: Sum of factors of 12 (= 2² × 3¹):

• = [(2³ - 1)/(2 - 1)] × [(3² - 1)/(3 - 1)]
• = [7/1] × [8/2]
• = 7 × 4 = 28
• Verify: 1 + 2 + 3 + 4 + 6 + 12 = 28 ✓

Prime Factorization

Breaking a number down into its prime building blocks. We use this everywhere — HCF, LCM, factor counting, you name it.

Method: Keep dividing by the smallest prime that works.

Example: Prime factorize 1260.

• 1260 ÷ 2 = 630
• 630 ÷ 2 = 315
• 315 ÷ 3 = 105
• 105 ÷ 3 = 35
• 35 ÷ 5 = 7
• 7 ÷ 7 = 1

So 1260 = 2² × 3² × 5 × 7

Worked Examples

Example 1: Find the largest 4-digit number divisible by 12.

Largest 4-digit number = 9999. We need it divisible by both 3 and 4.

• 9999 ÷ 12 = 833.25
• 833 × 12 = 9996
• Answer: 9996

Example 2: How many numbers between 1 and 100 are divisible by both 3 and 7?

Divisible by both 3 and 7 means divisible by LCM(3,7) = 21. Multiples of 21 up to 100: 21, 42, 63, 84 → 4 numbers

Example 3: A number when divided by 13 leaves a remainder of 11. What’s the remainder when it’s divided by 7?

The number can be written as 13k + 11. Let’s try values:

• k=0: 11 → 11 ÷ 7 = remainder 4
• k=1: 24 → 24 ÷ 7 = remainder 3
• k=2: 37 → 37 ÷ 7 = remainder 2

Hmm, the remainder depends on k. This is a trick question — we can’t determine a unique answer without more information. Watch out for these in exams!

Example 4: How many factors of 720 are even?

720 = 2⁴ × 3² × 5¹ Total factors = (4+1)(2+1)(1+1) = 30 Odd factors (ignore all 2s, use 3² × 5¹) = (2+1)(1+1) = 6 Even factors = 30 - 6 = 24

Example 5: What is the smallest number that when divided by 5, 6, and 8 leaves remainder 2 in each case?

The number = LCM(5, 6, 8) + 2 = 120 + 2 = 122

Common Exam Patterns

1. “Divisible by both X and Y” → Find LCM(X, Y) and work with that
2. “How many factors” → Prime factorize, then use (p+1)(q+1)(r+1) formula
3. “Leaves remainder r when divided by d” → Number = dk + r for some integer k
4. “Largest/smallest n-digit number divisible by X” → Divide the boundary number by X, then adjust
5. “Is 1 prime?” → No! Trap question. 1 is neither prime nor composite.

Practice Problems

Q1: How many prime numbers are there between 40 and 60?

Q2: Find the total number of factors of 1800.

Q3: What is the smallest 5-digit number divisible by 11?

Answers:

A1: Primes between 40 and 60: 41, 43, 47, 53, 59 → 5 primes

A2: 1800 = 2³ × 3² × 5² → Factors = (3+1)(2+1)(2+1) = 4 × 3 × 3 = 36

A3: Smallest 5-digit number = 10000. 10000 ÷ 11 = 909.09… → 910 × 11 = 10010

HCF and LCM are two of the most fundamental concepts in aptitude. They show up directly in word problems (bells, circular tracks, tiles) and indirectly in fractions, ratios, and number theory questions. Let’s nail them down.

What Are HCF and LCM?

HCF (Highest Common Factor) — also called GCD (Greatest Common Divisor). It’s the largest number that divides two or more numbers evenly.

In simple language, HCF is the biggest number that “fits into” all the given numbers.

LCM (Lowest Common Multiple) — the smallest number that is a multiple of two or more numbers.

In simple language, LCM is the smallest number that all the given numbers “fit into.”

Example: For 12 and 18:

• Factors of 12: 1, 2, 3, 4, 6, 12
• Factors of 18: 1, 2, 3, 6, 9, 18
• Common factors: 1, 2, 3, 6 → HCF = 6
• Multiples of 12: 12, 24, 36, 48…
• Multiples of 18: 18, 36, 54…
• Common multiples: 36, 72… → LCM = 36

Methods to Find HCF

Method 1: Prime Factorization

1. Find the prime factorization of each number
2. Take the lowest power of all common primes

Example: HCF of 48 and 180

• 48 = 2⁴ × 3¹
• 180 = 2² × 3² × 5¹
• Common primes: 2 and 3
• Take lowest powers: 2² × 3¹ = 4 × 3 = 12

Method 2: Long Division (Euclid’s Algorithm)

This is faster for large numbers. Divide the larger number by the smaller. Then divide the divisor by the remainder. Keep going until the remainder is 0. The last divisor is the HCF.

Example: HCF of 462 and 198

• 462 ÷ 198 = 2 remainder 66
• 198 ÷ 66 = 3 remainder 0
• HCF = 66

This method is super clean — no prime factorization needed. Use it when numbers are large.

Methods to Find LCM

Method 1: Prime Factorization

1. Find the prime factorization of each number
2. Take the highest power of all primes (not just common ones)

Example: LCM of 48 and 180

• 48 = 2⁴ × 3¹
• 180 = 2² × 3² × 5¹
• Take highest powers: 2⁴ × 3² × 5¹ = 16 × 9 × 5 = 720

Method 2: Division Method

Write all numbers in a row. Divide by the smallest prime that divides at least one number. Bring down numbers that aren’t divisible. Keep going until all numbers become 1. Multiply all the divisors.

Example: LCM of 12, 15, and 20

LCM = 2 × 2 × 3 × 5 × 5 = 300

The Golden Relationship

Important: The HCF × LCM = Product formula works ONLY for two numbers. For three or more numbers, we can’t directly use this.

HCF and LCM of Fractions

Example: Find HCF and LCM of 2/3, 4/5, and 6/7.

HCF = HCF(2, 4, 6) / LCM(3, 5, 7) = 2 / 105 = 2/105

LCM = LCM(2, 4, 6) / HCF(3, 5, 7) = 12 / 1 = 12

Key Properties

• HCF of any two consecutive numbers is always 1 (they’re co-prime)
• LCM of any two consecutive numbers is always their product
• HCF always divides LCM
• HCF ≤ smaller number ≤ larger number ≤ LCM
• If HCF(a, b) = a, then a divides b (so LCM = b)

Worked Examples

Example 1: Basic application of the golden formula

The HCF of two numbers is 12 and their LCM is 360. If one number is 60, find the other.

HCF × LCM = Product of two numbers 12 × 360 = 60 × other number Other number = (12 × 360) / 60 = 4320 / 60 = 72

Example 2: The bells problem (classic!)

Three bells ring at intervals of 6, 8, and 12 minutes. If they ring together at 12:00, when will they ring together again?

They’ll ring together after LCM(6, 8, 12) minutes.

• 6 = 2 × 3
• 8 = 2³
• 12 = 2² × 3
• LCM = 2³ × 3 = 24 minutes

They ring together again at 12:24 PM.

Example 3: The tiles problem

A room is 12m × 8m. We want to tile it with the largest possible square tiles. What’s the side length of each tile, and how many tiles do we need?

Side of largest square tile = HCF(12, 8) = 4m Number of tiles = (12 × 8) / (4 × 4) = 96 / 16 = 6 tiles

Example 4: Finding numbers from HCF and LCM

The HCF of two numbers is 6. Their LCM is 180. If one number is between 30 and 50, find both numbers.

Since HCF = 6, both numbers are multiples of 6. Let numbers = 6a and 6b where HCF(a, b) = 1.

• LCM = 6 × a × b = 180 → a × b = 30
• Co-prime pairs with product 30: (1, 30), (2, 15), (3, 10), (5, 6)
• Numbers: (6, 180), (12, 90), (18, 60), (30, 36)
• One number between 30 and 50: The pair (30, 36) or (36, 30). Also possible: (18, 60) if we consider 18 isn’t in range but 60 is… re-read the question. If “one number” is between 30-50: 36 and the other is 30.

Example 5: Circular track problem

Two runners start at the same point on a circular track. Runner A completes a lap in 6 minutes, Runner B in 10 minutes. After how many minutes will they meet at the starting point again?

Time = LCM(6, 10) = 30 minutes

Common Exam Variations

1. “Find the largest number that divides X and Y leaving remainders r₁ and r₂” → HCF(X - r₁, Y - r₂)
2. “Find the largest number that divides X, Y, Z leaving the same remainder” → HCF(X-Y, Y-Z, X-Z) — take differences, then find HCF
3. “N bells ring at intervals…” → LCM of all intervals
4. “Largest square tile for a rectangle” → HCF of length and width
5. “Two numbers given HCF and LCM” → Use HCF × LCM = product, and numbers = HCF × co-prime factors

Practice Problems

Q1: Find the greatest number that divides 65 and 117 leaving remainders 1 and 2 respectively.

Q2: Three traffic lights blink at intervals of 48 seconds, 72 seconds, and 108 seconds. If they blink together at 8:00 AM, when will they blink together again?

Q3: The HCF and LCM of two numbers are 8 and 480 respectively. If one number is 96, find the other.

Answers:

A1: We need HCF(65-1, 117-2) = HCF(64, 115). Using Euclid: 115 = 1×64 + 51, 64 = 1×51 + 13, 51 = 3×13 + 12, 13 = 1×12 + 1, 12 = 12×1. So HCF = 1. Hmm, let’s recheck: HCF(64, 115). 64 = 2⁶, 115 = 5 × 23. No common factors. HCF = 1.

A2: LCM(48, 72, 108). 48 = 2⁴ × 3, 72 = 2³ × 3², 108 = 2² × 3³. LCM = 2⁴ × 3³ = 16 × 27 = 432 seconds = 7 minutes 12 seconds. They blink together at 8:07:12 AM.

A3: HCF × LCM = a × b → 8 × 480 = 96 × b → b = 3840/96 = 40

Fractions and decimals seem basic, but in timed exams, the person who has fraction-decimal equivalents memorized crushes the person who has to calculate them every time. Surds (square roots that can’t be simplified to nice numbers) show up in geometry and simplification questions. Let’s build a toolkit we can use instantly.

Fraction-Decimal Equivalents — Memorize This Table

This table is absolutely worth memorizing. It saves us 10-20 seconds per question, and over a full exam, that adds up to 5-10 extra minutes.

Pro tip: Once we know 1/n, we can get any k/n instantly. For example, 3/8 = 3 × 0.125 = 0.375. And 5/6 = 5 × 0.1667 = 0.8333.

Comparing Fractions — Shortcuts

Method 1: Cross Multiplication (Best for 2 fractions)

To compare a/b and c/d: compare a×d with c×b.

Example: Which is larger: 7/11 or 5/8?

• 7 × 8 = 56
• 5 × 11 = 55
• 56 > 55, so 7/11 > 5/8

Method 2: Converting to Same Denominator

When comparing multiple fractions, find LCM of denominators.

Method 3: The Benchmark Method

Compare each fraction to a benchmark like 1/2 or 1.

Example: Arrange 3/7, 4/9, 5/11 in ascending order.

• 3/7 = 0.4286…
• 4/9 = 0.4444…
• 5/11 = 0.4545…
• Ascending: 3/7 < 4/9 < 5/11

Shortcut for fractions of the form (n)/(2n+1): They increase as n increases. So 3/7 < 4/9 < 5/11 < 6/13… This pattern is common in exams.

Recurring Decimals to Fractions

Pure recurring (all digits repeat)

If 0.abcabc… (the block “abc” repeats), the fraction = abc/999

Examples:

• 0.333… = 3/9 = 1/3
• 0.272727… = 27/99 = 3/11
• 0.142857142857… = 142857/999999 = 1/7

The rule: Put the repeating block in the numerator, and as many 9s as digits in the block in the denominator.

Mixed recurring (some digits don’t repeat)

If 0.abc (a doesn’t repeat, bc repeats): fraction = (abc - a) / 990

Example: 0.1666… = 0.16

• Numerator: 16 - 1 = 15
• Denominator: 90 (one 9 for repeating digit, one 0 for non-repeating digit)
• = 15/90 = 1/6 ✓

General rule: Denominator has as many 9s as repeating digits, followed by as many 0s as non-repeating digits after the decimal point.

Surds — The Basics

A surd is a root that can’t be simplified to a rational number. √2, √3, √5, ∛7 are all surds. √4 is NOT a surd because √4 = 2.

Key Values to Memorize

Simplifying Surds

Look for perfect square factors inside the root.

• √18 = √(9 × 2) = 3√2
• √75 = √(25 × 3) = 5√3
• √200 = √(100 × 2) = 10√2

Operations with Surds

We can only add/subtract like surds (same number under the root):

• 3√2 + 5√2 = 8√2 ✓
• 3√2 + 5√3 → can’t simplify ✗

Multiplication works across different surds:

• √2 × √3 = √6
• 3√2 × 4√5 = 12√10

Rationalizing the Denominator

We don’t leave surds in the denominator. To remove them, we multiply by the conjugate.

Simple case: 1/√a

Multiply top and bottom by √a:

• 1/√3 = (1 × √3)/(√3 × √3) = √3/3

Conjugate case: 1/(a + √b)

The conjugate of (a + √b) is (a - √b). Multiply both sides by it.

Example: Rationalize 1/(3 + √2)

• Multiply by (3 - √2)/(3 - √2)
• = (3 - √2)/(9 - 2)
• = (3 - √2)/7

Double surd case: 1/(√a + √b)

Example: Rationalize 1/(√5 + √3)

• Multiply by (√5 - √3)/(√5 - √3)
• = (√5 - √3)/(5 - 3)
• = (√5 - √3)/2

Worked Examples

Example 1: Simplify √50 + √32 - √18

• √50 = √(25×2) = 5√2
• √32 = √(16×2) = 4√2
• √18 = √(9×2) = 3√2
• = 5√2 + 4√2 - 3√2 = 6√2

Example 2: Which is greater: √7 - √6 or √6 - √5?

Trick: Rationalize each by multiplying by the conjugate.

• √7 - √6 = (7-6)/(√7+√6) = 1/(√7+√6) ≈ 1/5.095
• √6 - √5 = (6-5)/(√6+√5) = 1/(√6+√5) ≈ 1/4.685
• Since 1/4.685 > 1/5.095, √6 - √5 is greater

The pattern: for consecutive numbers, the gap between their square roots decreases as numbers get larger.

Example 3: Convert 0.5̄7̄ (0.575757…) to a fraction.

Repeating block = 57, which has 2 digits. Fraction = 57/99 = 19/33

Example 4: Find the value of (√5 + √3)² - (√5 - √3)²

We could expand both, but there’s a shortcut using the identity a² - b² = (a+b)(a-b):

• Let a = √5 + √3 and b = √5 - √3
• a + b = 2√5
• a - b = 2√3
• a² - b² = 2√5 × 2√3 = 4√15

Answer: 4√15

Example 5: Arrange in ascending order: 3/5, 5/8, 7/12, 2/3

Convert to decimals:

• 3/5 = 0.6
• 5/8 = 0.625
• 7/12 = 0.5833…
• 2/3 = 0.6667…

Ascending: 7/12 < 3/5 < 5/8 < 2/3

Common Exam Patterns

1. “Rationalize the denominator” → Multiply by conjugate
2. “Arrange fractions in order” → Cross multiply pairwise, or convert to decimals
3. “Convert recurring decimal to fraction” → Repeating digits/999… formula
4. “Simplify surd expression” → Factor out perfect squares, combine like surds
5. “Which is greater: √a - √b or √c - √d?” → Rationalize and compare

Practice Problems

Q1: Simplify: 3/(√7 - √3)

Q2: Convert 0.23̄ (where only 3 repeats) to a fraction.

Q3: If √3 = 1.732, find the value of 1/(√3 - 1) correct to 3 decimal places.

Answers:

A1: Multiply by (√7 + √3)/(√7 + √3) = 3(√7 + √3)/(7-3) = 3(√7 + √3)/4

A2: 0.2333… Here, 2 doesn’t repeat, 3 repeats. = (23 - 2)/90 = 21/90 = 7/30

A3: 1/(√3 - 1) = (√3 + 1)/((√3)² - 1²) = (√3 + 1)/2 = (1.732 + 1)/2 = 2.732/2 = 1.366

Powers and roots are the backbone of so many aptitude topics — from simplification to number theory to data interpretation. The laws of exponents let us simplify ugly expressions in seconds, and having common squares and cubes memorized means we can work faster than anyone using a calculator. Let’s get these locked in.

Laws of Exponents

The most common mistakes:

• (a + b)² ≠ a² + b² — Don’t forget the middle term! (a + b)² = a² + 2ab + b²
• (aᵐ)ⁿ ≠ a^(m+n) — It’s a^(m×n). Exponents multiply, not add.
• 0⁰ is undefined (or sometimes treated as 1 by convention, but avoid this in exams)

Squares of 1 to 30

Having these memorized means we can do reverse lookups (square roots) instantly.

Quick Squaring Tricks

Trick 1: Squaring numbers ending in 5

For n5 (a number ending in 5): n5² = n×(n+1) followed by 25

• 25² → 2×3 = 6, append 25 → 625
• 35² → 3×4 = 12, append 25 → 1225
• 75² → 7×8 = 56, append 25 → 5625
• 115² → 11×12 = 132, append 25 → 13225

Trick 2: Squaring numbers near 50

For (50 + x)²: Start with (25 + x), append x²

• 52² → 25 + 2 = 27, append 04 → 2704
• 47² → 25 - 3 = 22, append 09 → 2209

Trick 3: Squaring numbers near 100

For (100 + x)²: Start with (100 + 2x), append x²

• 103² → 100 + 6 = 106, append 09 → 10609
• 97² → 100 - 6 = 94, append 09 → 9409

Cubes of 1 to 15

Fun fact for cube roots: The unit digit of a perfect cube uniquely determines the unit digit of its cube root (unlike squares where both 2² and 8² end in 4/6 patterns). If a cube ends in 8, its cube root ends in 2. If it ends in 2, its root ends in 8. Other digits map to themselves.

Square Root Estimation

When we can’t simplify a square root to a whole number, we can estimate it quickly.

Method: Find the two perfect squares the number falls between, then estimate proportionally.

Example: Estimate √50

• 7² = 49 and 8² = 64
• 50 is just 1 above 49, and the gap from 49 to 64 is 15
• Approximate: 7 + 1/15 ≈ 7.07
• Actual: √50 ≈ 7.071 ✓

Example: Estimate √200

• 14² = 196 and 15² = 225
• 200 is 4 above 196, gap = 29
• Approximate: 14 + 4/29 ≈ 14.14
• Actual: √200 ≈ 14.142 ✓

Powers of 2 and 3 — Worth Memorizing

Worked Examples

Example 1: Simplify: (27)^(2/3) × (16)^(3/4) ÷ (8)^(1/3)

• (27)^(2/3) = (3³)^(2/3) = 3² = 9
• (16)^(3/4) = (2⁴)^(3/4) = 2³ = 8
• (8)^(1/3) = (2³)^(1/3) = 2
• = 9 × 8 ÷ 2 = 36

Example 2: If 2^(x+3) = 4^(x-1), find x.

Convert to same base:

• 2^(x+3) = (2²)^(x-1) = 2^(2x-2)
• So x + 3 = 2x - 2
• x = 5

Example 3: Find the value of: 5⁴ × 5⁻² × 5³ ÷ 5⁴

Using aᵐ × aⁿ = aᵐ⁺ⁿ:

• = 5^(4 + (-2) + 3 - 4)
• = 5^(1)
• = 5

Example 4: What is the square root of 17956?

Step 1: The number ends in 6, so the root ends in 4 or 6. Step 2: 130² = 16900 and 140² = 19600. So root is between 130 and 140. Step 3: Try 134² = 17956. Yes! Answer: 134

(Quick check: 134² = (130 + 4)² = 16900 + 2×130×4 + 16 = 16900 + 1040 + 16 = 17956 ✓)

Example 5: If 9^x - 9^(x-1) = 648, find x.

• 9^x - 9^x × 9⁻¹ = 648
• 9^x (1 - 1/9) = 648
• 9^x × (8/9) = 648
• 9^x = 648 × 9/8 = 729
• 9^x = 729 = 9³
• x = 3

Common Exam Patterns

1. “Simplify expressions with exponents” → Convert everything to same base, then add/subtract powers
2. “If aˣ = bʸ, find relation” → Convert to common base or take logarithm
3. “Find cube root of large number” → Use unit digit and range estimation
4. “Compare: 2³⁰⁰ vs 3²⁰⁰” → Raise both to power 1/100 → compare 2³ vs 3² → 8 vs 9, so 3²⁰⁰ > 2³⁰⁰
5. “a^(m/n) problems” → Rewrite as (ⁿ√a)ᵐ or ⁿ√(aᵐ), whichever is easier to compute

Practice Problems

Q1: If 2ˣ = 32 and 3ʸ = 243, find the value of 2^(x-2) × 3^(y-3).

Q2: Find the cube root of 19683.

Q3: Simplify: (125)^(2/3) + (64)^(2/3) - (27)^(2/3)

Answers:

A1: 2ˣ = 32 = 2⁵, so x = 5. 3ʸ = 243 = 3⁵, so y = 5. Then 2^(5-2) × 3^(5-3) = 2³ × 3² = 8 × 9 = 72

A2: 19683 ends in 3, so cube root ends in 7. 20³ = 8000, 30³ = 27000. Root is between 20-30. Try 27: 27³ = 19683 ✓. Answer: 27

A3: (5³)^(2/3) + (4³)^(2/3) - (3³)^(2/3) = 5² + 4² - 3² = 25 + 16 - 9 = 32

Unit digit and remainder questions are among the easiest to score on — once we know the patterns. The entire concept boils down to this: unit digits repeat in cycles, and remainders follow predictable rules. Learn the patterns, and these become free marks.

Unit Digit Cyclicity

Every digit, when raised to successive powers, repeats its unit digit in a cycle. This is called cyclicity.

Key observations:

• Digits 0, 1, 5, 6 → always give the same unit digit (period = 1)
• Digits 4, 9 → alternate between two values (period = 2)
• Digits 2, 3, 7, 8 → cycle through four values (period = 4)

How to Find the Unit Digit of a^n

1. Look at the unit digit of a (only the unit digit matters)
2. Find the cycle length for that digit
3. Divide n by the cycle length and find the remainder
4. Use the remainder to pick the right digit from the cycle

If the remainder is 0, we use the LAST digit in the cycle (not the first).

Step-by-Step Method

Example: Find the unit digit of 7^253.

1. Unit digit of base = 7
2. Cycle of 7: 7, 9, 3, 1 (period = 4)
3. Divide power by cycle: 253 ÷ 4 = 63 remainder 1
4. Remainder 1 → 1st digit in cycle → 7

Example: Find the unit digit of 3^100.

1. Unit digit of base = 3
2. Cycle of 3: 3, 9, 7, 1 (period = 4)
3. 100 ÷ 4 = 25 remainder 0
4. Remainder 0 → last digit in cycle → 1

Example: Find the unit digit of 2^347 + 3^128.

Handle each separately:

• 2^347: Cycle of 2 is (2,4,8,6), period 4. 347 ÷ 4 = remainder 3. Third in cycle → 8
• 3^128: Cycle of 3 is (3,9,7,1), period 4. 128 ÷ 4 = remainder 0. Last in cycle → 1
• Sum’s unit digit: 8 + 1 = 9

Remainder Shortcuts

Remainder when dividing by 2, 5, 10

• ÷ 2: Just look at the last digit
• ÷ 5: Just look at the last digit
• ÷ 10: Just look at the last digit

Remainder when dividing by 3 or 9

• ÷ 3: Sum of all digits, find remainder of that sum ÷ 3
• ÷ 9: Sum of all digits, find remainder of that sum ÷ 9

Example: Remainder when 5847 is divided by 9.

• Sum of digits: 5 + 8 + 4 + 7 = 24
• 24 ÷ 9 = 2 remainder 6

Remainder when dividing by 11

Use the alternating sum trick (same as divisibility by 11):

• Take digits from right, alternately add and subtract

Example: Remainder when 86345 is divided by 11.

• From right: 5 - 4 + 3 - 6 + 8 = 6
• Remainder = 6

If the alternating sum is negative, keep adding 11 until it’s positive.

Remainder of Powers — The Cycle Method

To find remainder of a^n ÷ d:

1. Find the cycle of remainders that a^1, a^2, a^3… give when divided by d
2. Use the cycle to find the answer

Example: Remainder when 2^100 is divided by 7.

• 2¹ ÷ 7 → remainder 2
• 2² ÷ 7 → remainder 4
• 2³ ÷ 7 → remainder 1
• Cycle: (2, 4, 1) with period 3
• 100 ÷ 3 = remainder 1
• So remainder = first in cycle = 2

Fermat’s Little Theorem (Useful Shortcut)

In simple language, Fermat’s theorem tells us: if we raise any number to the power (prime - 1) and divide by that prime, the remainder is always 1.

Example: Remainder when 3^40 is divided by 41.

41 is prime, and 3 is not divisible by 41. By Fermat’s: 3^(41-1) = 3^40 ≡ 1 (mod 41). Remainder = 1.

Worked Examples

Example 1: Find the unit digit of 17^256 × 13^124.

• 17^256: Unit digit of 17 is 7. Cycle of 7: (7,9,3,1), period 4. 256 ÷ 4 = 0 remainder. Last in cycle → 1
• 13^124: Unit digit of 13 is 3. Cycle of 3: (3,9,7,1), period 4. 124 ÷ 4 = 0 remainder. Last in cycle → 1
• Product: 1 × 1 = 1

Example 2: Find the remainder when 37^73 is divided by 9.

Instead of working with 37, find 37 mod 9 first: 37 ÷ 9 = 4 remainder 1. So 37 ≡ 1 (mod 9). Therefore 37^73 ≡ 1^73 ≡ 1 (mod 9). Remainder = 1.

This is a huge shortcut — always reduce the base first!

Example 3: What is the unit digit of 4^4^4?

First, we need to figure out the power: 4^4 = 256. So we need unit digit of 4^256. Cycle of 4: (4, 6), period 2. 256 is even → unit digit = 6

Example 4: Find the remainder when 2^256 is divided by 17.

17 is prime. By Fermat’s: 2^16 ≡ 1 (mod 17). 256 = 16 × 16, so 2^256 = (2^16)^16 ≡ 1^16 = 1 (mod 17). Remainder = 1

Example 5: Find the unit digit of 1! + 2! + 3! + … + 100!

• 1! = 1 (unit digit 1)
• 2! = 2 (unit digit 2)
• 3! = 6 (unit digit 6)
• 4! = 24 (unit digit 4)
• 5! = 120 (unit digit 0)
• 6! = 720 (unit digit 0)
• Every factorial from 5! onwards ends in 0 (because they contain 2×5=10 as a factor)

So unit digit = unit digit of (1 + 2 + 6 + 4 + 0 + 0 + …) = unit digit of 13 = 3

Common Exam Patterns

1. “Unit digit of a^n” → Find cycle, divide power by period
2. “Unit digit of sum/product” → Find unit digit of each term separately, then add/multiply
3. “Remainder when large power ÷ prime” → Use Fermat’s Little Theorem
4. “Remainder when N ÷ d” → Reduce base first (base mod d), then find cycle of remainders
5. “Unit digit of factorial sum” → Only first 4 terms matter (5! onwards ends in 0)

Practice Problems

Q1: Find the unit digit of 8^2048 + 9^2048.

Q2: Find the remainder when 7^99 is divided by 5.

Q3: What is the unit digit of 2^2 × 3^3 × 4^4 × 5^5 × 6^6?

Answers:

A1: 8^2048: Cycle of 8 is (8,4,2,6), period 4. 2048 ÷ 4 = 0 remainder → last in cycle → 6. 9^2048: Cycle of 9 is (9,1), period 2. 2048 ÷ 2 = 0 remainder → last in cycle → 1. Unit digit of sum: 6 + 1 = 7

A2: 7 mod 5 = 2. So we need 2^99 mod 5. Cycle of 2 mod 5: 2, 4, 3, 1 (period 4). 99 ÷ 4 = remainder 3. Third in cycle = 3.

A3: 2² = 4, 3³ = 27 (unit 7), 4⁴ = 256 (unit 6), 5⁵ ends in 5, 6⁶ ends in 6. Product of unit digits: 4 × 7 = 28(unit 8) → 8 × 6 = 48(unit 8) → 8 × 5 = 40(unit 0) → 0 × 6 = 0

Simplification and approximation questions are the bread and butter of banking exams and placement tests. The good news? They’re mostly about speed, not difficulty. If we master BODMAS and learn when to approximate vs compute exactly, we can blitz through these in under a minute each.

BODMAS / PEMDAS

The order of operations everyone learns in school but still gets tripped up by in exams.

Critical point: Division and multiplication have the same priority — we go left to right. Same with addition and subtraction. This is where most people mess up.

Example: 24 ÷ 6 × 2 = ?

• Left to right: (24 ÷ 6) × 2 = 4 × 2 = 8
• NOT: 24 ÷ (6 × 2) = 24 ÷ 12 = 2 ← WRONG

Example: 48 - 12 + 7 = ?

• Left to right: (48 - 12) + 7 = 36 + 7 = 43

Tricky BODMAS Examples

Example: 3 + 4 × 2 - 6 ÷ 3

• First, multiplication and division (left to right): 4 × 2 = 8 and 6 ÷ 3 = 2
• Now: 3 + 8 - 2 = 9

Example: (8 + 2)² ÷ 5 - 3 × 4

• Brackets: (10)² ÷ 5 - 3 × 4
• Orders: 100 ÷ 5 - 3 × 4
• Division/Multiplication: 20 - 12
• Subtraction: 8

The Vinculum (Bar) — Of/Of

In some problems, we see “of” — this means multiplication and has the same priority as multiplication/division.

Example: 1/3 of 27 + 4 × 2 = ?

• 1/3 of 27 = 9
• 4 × 2 = 8
• 9 + 8 = 17

Approximation Strategies

In timed exams, especially banking/placement tests, we often don’t need exact answers. The options are usually spread far enough that a good approximation is enough.

Strategy 1: Round to Nearest Convenient Number

• 498 → round to 500
• 7.89 → round to 8
• 33.3% → round to 1/3
• 24.98% → round to 25% = 1/4

Example: 498.7 × 31.2 ÷ 9.87 = ?

• ≈ 500 × 31 ÷ 10
• = 15500 ÷ 10
• = ≈ 1550 (actual: 1573.6)

If options are 1250, 1425, 1575, 1720 → we’d pick 1575 (closest to our estimate, and we rounded down slightly overall).

Strategy 2: Percentage-Based Estimation

Use the fact that we know our fraction-decimal equivalents.

Example: 37.5% of 4816 = ?

• 37.5% = 3/8
• 4816 ÷ 8 = 602
• 602 × 3 = 1806

No approximation needed here — fraction equivalents gave us the exact answer fast!

Strategy 3: Break Down Complex Expressions

Example: 786 × 764 = ?

• Notice: 786 = 775 + 11, 764 = 775 - 11
• This is (a+b)(a-b) = a² - b²
• = 775² - 11²
• = 600625 - 121
• = 600504

Strategy 4: Use of “Close to” Reference Points

• 19 × 21 = 20² - 1² = 399
• 48 × 52 = 50² - 2² = 2496
• 997 × 1003 = 1000² - 3² = 999991

Digit Sum Method (Casting Out 9s)

This is a quick way to verify if our answer is correct. It doesn’t tell us the exact answer, but it can catch errors.

Method:

1. Find the digit sum of each number (keep adding digits until single digit, treating 9 as 0)
2. Perform the same operation on the digit sums
3. The digit sum of the result should match the digit sum of our answer

Example: Verify: 347 × 28 = 9716

• Digit sum of 347: 3+4+7 = 14 → 1+4 = 5
• Digit sum of 28: 2+8 = 10 → 1+0 = 1
• Product of digit sums: 5 × 1 = 5
• Digit sum of 9716: 9+7+1+6 = 23 → 2+3 = 5
• Both give 5 → likely correct ✓

Limitation: This method can’t detect errors that are multiples of 9. For example, 9720 also has digit sum 18 → 9 → hmm, that’s different. So it would catch that error. But swapping digits (like 9716 vs 9761) might not always be caught.

Simplification of Fractions and Mixed Expressions

Chain Rule Simplification

When we have a chain like a × b / c × d / e, work in pairs to cancel common factors early.

Example: (144 × 25 × 49) / (35 × 18 × 20) = ?

Cancel before multiplying:

• 144/18 = 8
• 25/5 = 5 (taking 5 from 35)
• 49/7 = 7 (taking 7 from the remaining 35/5 = 7)
• Remaining: 8 × 5 × 7 / 20 = 280/20 = 14

In simple language, always cancel common factors between numerator and denominator before multiplying. This keeps numbers small and manageable.

Worked Examples

Example 1: Simplify: 5/8 of 480 + 3/5 of 250 - 7/12 of 360

• 5/8 × 480 = 300
• 3/5 × 250 = 150
• 7/12 × 360 = 210
• 300 + 150 - 210 = 240

Example 2: Approximate: 3987.05 × 11.02 ÷ 39.97

• ≈ 4000 × 11 ÷ 40
• = 44000 ÷ 40
• = 1100 (actual: 1098.1)

Example 3: What value should replace ? in: 55% of 800 + 40% of 350 = ? + 220

• 55% of 800 = 440
• 40% of 350 = 140
• 440 + 140 = ? + 220
• ? = 580 - 220 = 360

Example 4: Simplify: (13.96)² - (5.99)² + (8.01)²

Approximate: 14² - 6² + 8² = 196 - 36 + 64 = 224

(Exact would be very close: 195.0816 - 35.8801 + 64.1601 = 223.3616, so with options like 218, 224, 230, 236 we’d pick 224)

Example 5: Simplify: √(144 × 81) + √(256 × 49) - √(196 × 64)

• √(144 × 81) = 12 × 9 = 108
• √(256 × 49) = 16 × 7 = 112
• √(196 × 64) = 14 × 8 = 112
• 108 + 112 - 112 = 108

When to Approximate vs Compute Exactly

Common Exam Patterns

1. “Simplify the expression” → Follow BODMAS strictly, cancel common factors early
2. “What approximate value should replace ?” → Round everything, pick closest option
3. “What should come in place of ?” → Isolate the unknown, simplify both sides
4. “Which of the following is closest to…” → Approximate, then eliminate wrong options
5. Mixed fraction/percentage expressions → Convert everything to fractions first

Practice Problems

Q1: Simplify: 12.5% of 640 + 37.5% of 480 - 66.67% of 180

Q2: Approximate: 7985 ÷ 401 × 199

Q3: Verify using digit sums: Is 528 × 47 = 24816?

Answers:

A1: 12.5% = 1/8, 37.5% = 3/8, 66.67% = 2/3. So: 640/8 + 3×480/8 - 2×180/3 = 80 + 180 - 120 = 140

A2: ≈ 8000 ÷ 400 × 200 = 20 × 200 = 4000 (actual: 3962.3)

A3: Digit sum of 528: 5+2+8 = 15 → 6. Digit sum of 47: 4+7 = 11 → 2. Product: 6×2 = 12 → 3. Digit sum of 24816: 2+4+8+1+6 = 21 → 3. They match, so it’s likely correct ✓. (Actual: 528 × 47 = 24816 ✓)

Percentages are everywhere in aptitude — profit/loss, interest, data interpretation, you name it. The person who has percentage-fraction equivalents memorized and knows the successive change formula has a massive advantage. Let’s build that toolkit.

What Is a Percentage?

A percentage is just a fraction with denominator 100. “Per cent” literally means “per hundred.”

25% = 25/100 = 1/4

In simple language, if we say “25% of students passed,” we mean 25 out of every 100 students.

Percentage-Fraction Equivalents — The Power Table

This table is the single biggest time-saver in aptitude. Memorize it and we’ll solve problems twice as fast.

How to use these: Instead of calculating “12.5% of 480,” we think “1/8 of 480 = 60.” Instant.

Instead of “find 37.5% of 640,” we think “3/8 of 640 = 3 × 80 = 240.” Done in 3 seconds.

Basic Percentage Calculations

Common mistake alert: “A is what percent more than B” → base is B. “A is what percent less than B” → base is still B. Always figure out what the “compared to” value is — that’s our denominator.

Successive Percentage Changes

When two percentage changes happen one after another, we DON’T just add them.

Formula: If there are two successive changes of a% and b%, the effective change is:

a + b + (ab/100) %

Example: Price increases by 20%, then decreases by 10%. Net change?

• = 20 + (-10) + (20 × -10)/100
• = 20 - 10 - 2
• = 8% increase

Not 10%! The extra -2% is because the 10% decrease applied to the already-increased amount.

Super useful case: If something increases by x% and then decreases by x%, the net result is always a DECREASE of (x²/100)%.

• Increase 10%, decrease 10% → net decrease of 1%
• Increase 20%, decrease 20% → net decrease of 4%

Reverse Percentage Problems

These are the ones that trip people up: “After a 20% increase, the price is 600. What was the original?”

The mistake: People calculate 20% of 600 = 120, then say original = 480. WRONG.

The right way: After 20% increase, new = 120% of original = 1.2 × original.

• 1.2 × original = 600
• Original = 600/1.2 = 500

Shortcut with fractions: 20% increase means the new price is 6/5 of the original.

• Original = 600 × 5/6 = 500

This fraction approach is faster. Here’s the pattern:

• 10% increase → multiply by 11/10 → to reverse, multiply by 10/11
• 20% increase → multiply by 6/5 → to reverse, multiply by 5/6
• 25% increase → multiply by 5/4 → to reverse, multiply by 4/5
• 33.33% increase → multiply by 4/3 → to reverse, multiply by 3/4

Population / Depreciation Problems

These are compound percentage change problems.

Example: A town’s population is 50,000 and grows at 10% per year. What will it be in 3 years?

• = 50000 × (1.1)³
• = 50000 × 1.331
• = 66,550

Worked Examples

Example 1: In an exam, a student scores 280 out of 400. What percentage did they score?

(280/400) × 100 = 70%. Or even faster: 280/4 = 70%

Example 2: A’s salary is 20% more than B’s. B’s salary is what percent less than A’s?

Let B’s salary = 100. Then A’s = 120. B is less than A by: (120 - 100)/120 × 100 = 20/120 × 100 = 16.67% (= 1/6)

Key insight: “20% more” and “16.67% less” describe the same relationship. They’re NOT the same number because the base changes.

Example 3: The price of sugar increases by 25%. By what percent should consumption be reduced so that expenditure remains the same?

Expenditure = Price × Consumption. If price becomes 5/4, consumption must become 4/5. Reduction = 1 - 4/5 = 1/5 = 20%

Shortcut formula: If price increases by r%, reduce consumption by r/(100+r) × 100%. = 25/125 × 100 = 20%

Example 4: In an election, candidate A gets 55% of votes. If A wins by 2400 votes, what’s the total number of votes?

A gets 55%, B gets 45%. Difference = 10% of total = 2400. Total = 2400/0.10 = 24,000

Example 5: The population of a city was 200,000 two years ago. It increased by 10% in the first year and decreased by 5% in the second year. What is the current population?

• After year 1: 200000 × 1.10 = 220000
• After year 2: 220000 × 0.95 = 209,000

Or using successive change: net change = 10 + (-5) + (10×-5)/100 = 10 - 5 - 0.5 = 4.5% increase. 200000 × 1.045 = 209,000 ✓

Common Exam Patterns

1. “A is what % more/less than B?” → Find difference, divide by the BASE (B, not A)
2. “After x% increase, value is Y. Find original” → Divide Y by (1 + x/100), or use fraction equivalents
3. “Two successive changes” → Use a + b + ab/100 formula
4. “Same expenditure, price changes” → New consumption = Old × (old price/new price)
5. “Population/depreciation after n years” → Use (1 ± r/100)ⁿ formula
6. “x% of A = y% of B” → A/B = y/x (the percentages flip!)

Practice Problems

Q1: If the price of an item is reduced by 30%, by what percent must consumption increase so that total expenditure increases by 5%?

Q2: Two successive discounts of 20% and 15% are equivalent to a single discount of what percent?

Q3: In an exam with 80 questions, a student answers 65 correctly. If each correct answer gives 4 marks and each wrong answer deducts 1 mark, what percentage of maximum marks did the student score?

Answers:

A1: New price = 0.7 × Old. New expenditure = 1.05 × Old expenditure. New consumption = New expenditure / New price = (1.05 × Old exp) / (0.7 × Old price) = 1.05/0.7 × Old consumption = 1.5 × Old consumption. Increase = 50%. Answer: 50%

A2: Using successive change: -20 + (-15) + (-20)(-15)/100 = -20 - 15 + 3 = -32%. Equivalent single discount = 32%.

A3: Correct = 65, Wrong = 80-65 = 15. Marks = 65×4 - 15×1 = 260 - 15 = 245. Max marks = 80×4 = 320. Percentage = 245/320 × 100 = 76.5625% ≈ 76.56%

Profit, loss, and discount problems are a staple of every aptitude exam. The concepts are simple, but the questions can get tricky — especially dishonest dealer problems and successive discount questions. Let’s break it all down.

The Core Terms

• Cost Price (CP): What we paid to buy/produce an item
• Selling Price (SP): What we sold it for
• Marked Price (MP): The price tag on the item (also called “list price”)
• Profit: SP - CP (when SP > CP)
• Loss: CP - SP (when CP > SP)

In simple language, think of it like this: we buy at CP, we label it at MP, and we sell at SP (after possibly giving a discount on MP).

The CP → MP → SP Flow

Successive Discounts

Two discounts of 20% and 10% are NOT the same as a single 30% discount.

Method: Apply discounts one after another.

Example: MP = 1000, discounts of 20% and 10%.

• After 20% discount: 1000 × 0.8 = 800
• After 10% discount on 800: 800 × 0.9 = 720
• SP = 720

A single equivalent discount: 1000 - 720 = 280. So 28% discount, not 30%.

Formula (from our percentages chapter): Equivalent discount for a% and b% = a + b - ab/100 = 20 + 10 - (20×10)/100 = 30 - 2 = 28%

Buy X Get Y Free

This is just a discount in disguise.

Example: “Buy 4 get 1 free” means we pay for 4 but get 5 items.

• Effective discount = 1/5 = 20%
• We can also say: CP of 5 items = SP of 4 items

General rule: “Buy X get Y free” → Discount = Y/(X+Y) × 100%

Dishonest Dealer Problems

These are the trickiest and most fun problems. A dealer cheats on weight while claiming to sell at CP or with a small markup.

Type 1: Uses false weight, sells at CP

If a dealer claims to sell at cost price but uses a weight of w grams instead of 1000 grams:

Profit % = [(True weight - False weight) / False weight] × 100

Example: A shopkeeper claims to sell rice at cost price but uses an 800g weight instead of 1kg.

Profit % = (1000 - 800)/800 × 100 = 200/800 × 100 = 25%

In simple language, he’s paying for 800g and selling (claiming) it as 1000g. So for every 800g he buys, he makes a profit on 200g.

Type 2: False weight + markup or discount

Use the multiplier approach: Total SP factor = (True weight/False weight) × (100 ± markup or discount)/100

Example: A dealer uses 900g instead of 1kg and marks up by 10%. What’s his real profit%?

• Weight factor: 1000/900 = 10/9
• Price factor: 110/100 = 11/10
• Total: (10/9) × (11/10) = 11/9
• Profit = 11/9 - 1 = 2/9 = 22.22%

Worked Examples

Example 1: A shopkeeper bought 100 pens at Rs 10 each and sold them at Rs 12 each. Find profit%.

CP = 10, SP = 12. Profit = 2. Profit % = 2/10 × 100 = 20%

Example 2: An article is marked at Rs 800. After giving two successive discounts of 10% and 15%, what is the selling price?

SP = 800 × (90/100) × (85/100) = 800 × 0.9 × 0.85 = 800 × 0.765 = Rs 612

Example 3: A man buys an article for Rs 450 and sells it at a profit of 20%. What is the selling price?

SP = 450 × (120/100) = 450 × 1.2 = Rs 540

Or using fraction: 20% profit = 1/5 on CP. So profit = 450/5 = 90. SP = 450 + 90 = 540.

Example 4: By selling a watch for Rs 1140, a man loses 5%. At what price should he sell it to gain 5%?

At 5% loss: SP = 95% of CP → 1140 = 0.95 × CP → CP = 1140/0.95 = 1200. For 5% gain: SP = 105% of CP = 1.05 × 1200 = Rs 1260

Shortcut: New SP = Old SP × (100 + desired gain%) / (100 - loss%) = 1140 × 105/95 = Rs 1260

Example 5: A trader marks his goods 40% above CP and gives a discount of 25%. Find his profit or loss%.

Let CP = 100. MP = 100 × 1.4 = 140. SP = 140 × 0.75 = 105. Profit = 105 - 100 = 5. Profit % = 5%

Shortcut formula: When markup is m% and discount is d%: Profit/Loss % = m - d - (m×d)/100 = 40 - 25 - (40×25)/100 = 40 - 25 - 10 = 5% profit

The “Same SP” Trick

If two articles are sold at the same selling price, one at a% profit and the other at a% loss, there is ALWAYS a net loss.

Net Loss % = a²/100 %

Example: Two TVs are sold at Rs 10,000 each. One at 20% profit and the other at 20% loss. Net profit/loss?

Net loss % = (20)²/100 = 400/100 = 4% loss

Let’s verify:

• TV1 CP = 10000/1.2 = 8333.33
• TV2 CP = 10000/0.8 = 12500
• Total CP = 20833.33, Total SP = 20000
• Loss = 833.33, Loss% = 833.33/20833.33 × 100 = 4% ✓

Common Exam Patterns

1. “Marked at X% above CP, discount of Y%” → Profit = X - Y - XY/100
2. “Sold at same SP, one at profit, one at loss” → Always a net loss = a²/100%
3. “Buy X get Y free” → Discount = Y/(X+Y) × 100
4. “Dishonest dealer, false weight” → Profit = (True - False)/False × 100
5. “Find SP to get desired profit given a loss/previous SP” → Find CP first, then calculate new SP
6. “CP of X items = SP of Y items” → If X > Y: profit of (X-Y)/Y × 100%. If X < Y: loss.

Practice Problems

Q1: A shopkeeper marks an article at Rs 1200 and gives successive discounts of 20% and 10%. If he still makes a 20% profit, what was the cost price?

Q2: The cost price of 25 articles equals the selling price of 20 articles. Find the profit percentage.

Q3: A dishonest dealer sells goods at cost price but uses a weight of 850 grams instead of 1 kilogram. He also mixes 15% impurities. What is his actual profit percentage?

Answers:

A1: SP = 1200 × 0.8 × 0.9 = Rs 864. If this is 120% of CP: CP = 864/1.2 = Rs 720

A2: CP of 25 = SP of 20. Let CP of each = 1. Total CP of 25 = 25. SP of 20 = 25, so SP of each = 25/20 = 1.25. Profit per item = 0.25. Profit % = 0.25/1 × 100 = 25%

A3: For every 850g he buys, he adds 15% impurities: actual goods he sells = 850 × 1.15 = 977.5g. But he charges for 1000g at cost price. Profit% = (1000 - 850)/850 × 100 is the weight cheat. Adding impurities: he gets 977.5g of mixture but paid for only 850g. He sells 977.5g as 977.5g… wait, let’s think differently. He buys 850g for the cost of 850g. Adds impurities to make it 977.5g. Sells it claiming 1000g at CP per gram. So he charges for 1000g but his cost was for 850g. Profit = (1000-850)/850 × 100 ≈ 17.65%. But actually, the full calculation: he effectively sells 1000g worth at cost, having spent on 850g. So profit% = (1000-850)/850 × 100 = 150/850 × 100 = 17.65%

Interest problems are a guaranteed topic in every aptitude exam. Simple interest is straightforward, but compound interest has a few powerful shortcuts — especially the CI-SI difference formulas — that can turn a 3-minute calculation into a 30-second answer.

Simple Interest (SI)

In simple interest, we earn interest only on the original amount (principal). The interest stays the same every year.

In simple language, if we put Rs 10,000 in a bank at 5% SI for 3 years, we get 5% of 10,000 = Rs 500 every year. After 3 years, total interest = Rs 1,500.

SI Shortcuts

• Doubling time: At R%, the principal doubles in 100/R years (under SI)

  • At 10% → doubles in 10 years
  • At 12.5% → doubles in 8 years

• If SI = Principal, then R × T = 100

Compound Interest (CI)

In compound interest, we earn interest on the interest too. Each year’s interest gets added to the principal, and next year’s interest is calculated on this new, larger amount.

In simple language, instead of “rate R for n years,” half-yearly means “rate R/2 for 2n half-years.” We’re just changing the unit of time.

Quick Values for (1 + R/100)ⁿ

Memorize these common ones:

The CI - SI Difference — The Most Important Shortcut

This is the single most tested concept in CI problems.

Why this works (for 2 years)

SI for 2 years = 2PR/100 CI for 2 years = P(1 + R/100)² - P = P(R/100)² + 2P(R/100) → which is SI + P(R/100)²

The difference P(R/100)² is literally “interest on one year’s interest” — the extra bit CI gives us.

Effective Rate of Interest

When compounding happens more than once a year, the effective annual rate is higher than the stated rate.

Formula: Effective rate = (1 + R/n)ⁿ - 1, where n = number of compounding periods per year.

Example: 10% compounded half-yearly. What’s the effective annual rate?

• = (1 + 0.10/2)² - 1 = (1.05)² - 1 = 1.1025 - 1 = 0.1025 = 10.25%

So 10% compounded half-yearly is equivalent to 10.25% compounded annually.

Installment Problems

When we borrow money and repay in equal installments, each installment covers part of the principal plus interest.

For SI — Equal Installments:

Each installment = Total amount / number of installments (adjusted for interest). The formula is a bit involved, so let’s see it with an example.

For CI — Equal Annual Installments:

If each installment is X, at rate R%, for n years:

P = X/(1 + R/100) + X/(1 + R/100)² + … + X/(1 + R/100)ⁿ

Example: A sum of Rs 10,000 is borrowed at 10% CI and paid back in 2 equal annual installments. Find the installment.

Let installment = X. 10000 = X/1.1 + X/1.21 10000 = X(1/1.1 + 1/1.21) 10000 = X(1.1/1.21 + 1/1.21) 10000 = X(2.1/1.21) X = 10000 × 1.21/2.1 = 12100/2.1 = Rs 5761.90 (approximately)

Worked Examples

Example 1: Find the difference between CI and SI on Rs 8000 for 2 years at 5% per annum.

CI - SI = P × (R/100)² = 8000 × (5/100)² = 8000 × 1/400 = Rs 20

Let’s verify:

• SI = 8000 × 5 × 2 / 100 = 800
• CI: Amount = 8000 × (1.05)² = 8000 × 1.1025 = 8820. CI = 820.
• Difference: 820 - 800 = 20 ✓

Example 2: A sum of money doubles in 5 years at SI. In how many years will it become 5 times?

If it doubles in 5 years, SI in 5 years = P. So PRT/100 = P → RT = 100. With T = 5: R = 20%. For 5 times: Amount = 5P, so SI = 4P. 4P = P × 20 × T / 100 → T = 400/20 = 20 years

Shortcut: If it doubles (2×) in 5 years, for n× it takes (n-1) × 5 years. So 5× takes 4 × 5 = 20 years.

Example 3: The CI on a sum for 2 years is Rs 832 and the SI is Rs 800. Find the rate and the sum.

CI - SI = 832 - 800 = 32. Also, CI - SI for 2 years = P(R/100)². And SI for 2 years = 2PR/100 = 800 → PR = 40000.

From the SI formula: SI for 1 year = 400. CI - SI = R% of (SI for 1 year) = R/100 × 400 = 32. So R = 32 × 100/400 = 8%

From PR = 40000: P = 40000/8 = Rs 5000

Example 4: Find the CI on Rs 15,000 for 1.5 years at 10% compounded half-yearly.

Half-yearly rate = 5%, number of half-years = 3. Amount = 15000 × (1.05)³ = 15000 × 1.157625 = 17364.375. CI = 17364.375 - 15000 = Rs 2364.38 (approximately)

Example 5: A certain sum amounts to Rs 7260 in 2 years and Rs 7986 in 3 years at CI. Find the rate and the sum.

The difference between 3rd year amount and 2nd year amount = interest on the 2nd year amount. 7986 - 7260 = 726 = interest on 7260 for 1 year. Rate = (726/7260) × 100 = 10%

Now: 7260 = P × (1.1)² = 1.21P → P = 7260/1.21 = Rs 6000

Common Exam Patterns

1. “Difference between CI and SI for 2 years” → P(R/100)². Instant answer.
2. “Sum doubles in n years at SI, when does it triple?” → Triple time = 2n (since SI is linear)
3. “Amount after n years at CI, amount after n+1 years” → Difference gives one year’s CI, which reveals the rate
4. “Half-yearly / quarterly compounding” → Adjust rate and time accordingly
5. “CI for 2 years is X and SI is Y” → CI - SI gives a relation, SI gives another. Solve.
6. “Equal installments” → Set up present value equation

Practice Problems

Q1: Find the CI on Rs 12,000 for 3 years at 10% per annum.

Q2: The difference between CI and SI on a certain sum for 2 years at 12% per annum is Rs 90. Find the sum.

Q3: A sum amounts to Rs 2880 in 2 years and Rs 3456 in 3 years at compound interest. Find the principal and the rate.

Answers:

A1: Amount = 12000 × (1.1)³ = 12000 × 1.331 = 15972. CI = 15972 - 12000 = Rs 3972

A2: CI - SI for 2 years = P(R/100)² → 90 = P × (12/100)² = P × 0.0144 → P = 90/0.0144 = Rs 6250

A3: Interest in 3rd year = 3456 - 2880 = 576 = R% of 2880. So R = (576/2880) × 100 = 20%. Amount after 2 years: 2880 = P × (1.2)² = 1.44P → P = 2880/1.44 = Rs 2000

Ratios and proportions are the silent backbone of aptitude. They show up directly in their own questions, and indirectly in mixtures, partnerships, time-work, age problems — basically everywhere. Getting comfortable with ratio manipulation is one of the highest-leverage things we can do for our score.

What Is a Ratio?

A ratio compares two quantities of the same kind. If A has 30 apples and B has 20 apples, the ratio A:B = 30:20 = 3:2.

In simple language, a ratio tells us “for every 3 apples A has, B has 2.”

Key points:

• A ratio has no units (it’s a pure comparison)
• a:b = ka:kb for any non-zero k (we can multiply/divide both sides)
• a:b is NOT the same as b:a (order matters!)
• a:b = a/b (we can treat it as a fraction)

Simplifying Ratios

Always reduce to lowest terms by dividing by HCF.

• 24:36 → divide by 12 → 2:3
• 1.5:2.5 → multiply by 2 → 3:5
• 1/3 : 1/4 → multiply by LCM(3,4) = 12 → 4:3

Ratios of fractions: To compare a/b : c/d, cross multiply → a×d : c×b

Types of Proportion

Direct Proportion

When one quantity increases, the other increases proportionally. If A ∝ B, then A/B = constant.

Example: If 5 pens cost Rs 40, how much do 8 pens cost? → 8 × (40/5) = Rs 64

Inverse Proportion

When one quantity increases, the other decreases proportionally. If A ∝ 1/B, then A × B = constant.

Example: If 6 workers can do a job in 10 days, how many days for 15 workers? → 6×10/15 = 4 days

Continued Proportion

a, b, c are in continued proportion if a:b = b:c, which means b² = ac. Here b is the mean proportional of a and c.

Key Operations on Ratios

Componendo-Dividendo in Action

This is extremely powerful for simplification problems.

Example: If (x + y)/(x - y) = 7/3, find x:y.

Using componendo-dividendo in reverse:

• (x+y)/(x-y) = 7/3
• By componendo-dividendo: [(x+y)+(x-y)] / [(x+y)-(x-y)] = (7+3)/(7-3)
• 2x/2y = 10/4
• x/y = 5/2
• x:y = 5:2

Dividing a Quantity in a Given Ratio

If we need to divide quantity Q in the ratio a:b:c:

• First part = Q × a/(a+b+c)
• Second part = Q × b/(a+b+c)
• Third part = Q × c/(a+b+c)

Example: Divide Rs 1200 among A, B, C in the ratio 2:3:5.

Sum of ratio = 10.

• A = 1200 × 2/10 = Rs 240
• B = 1200 × 3/10 = Rs 360
• C = 1200 × 5/10 = Rs 600

Combining Ratios

When we know A:B and B:C separately, we can find A:B:C.

Method: Make B common in both ratios.

Example: A:B = 2:3 and B:C = 4:5. Find A:B:C.

B is 3 in the first ratio and 4 in the second. LCM(3,4) = 12.

• A:B = 2:3 → multiply by 4 → 8:12
• B:C = 4:5 → multiply by 3 → 12:15
• A:B:C = 8:12:15

Income, Expenditure, and Savings Problems

These are classic ratio problems. The key relationship is:

Income - Expenditure = Savings

Example: A and B’s incomes are in ratio 5:4. Their expenditures are in ratio 3:2. If each saves Rs 2000, find their incomes.

Let incomes = 5x and 4x. Let expenditures = 3y and 2y.

• 5x - 3y = 2000 … (1)
• 4x - 2y = 2000 … (2)

From (2): 2x - y = 1000, so y = 2x - 1000. Sub in (1): 5x - 3(2x - 1000) = 2000 → 5x - 6x + 3000 = 2000 → -x = -1000 → x = 1000.

Incomes: A = 5000, B = 4000. A earns Rs 5000, B earns Rs 4000.

Worked Examples

Example 1: If a:b = 3:4, b:c = 5:7, and c:d = 2:3, find a:d.

a/b × b/c × c/d = a/d = 3/4 × 5/7 × 2/3 = 30/84 = 5/14. a:d = 5:14

Example 2: Rs 5600 is divided among A, B, and C such that A gets 2/3 of what B gets and B gets 1/4 of what C gets. Find each person’s share.

B = C/4, so B:C = 1:4. A = 2B/3, so A:B = 2:3. Combine: A:B = 2:3, B:C = 1:4 → make B common → A:B = 2:3, B:C = 3:12. A:B:C = 2:3:12

Sum = 17. A = 5600 × 2/17 ≈ Rs 658.82, B = 5600 × 3/17 ≈ Rs 988.24, C = 5600 × 12/17 ≈ Rs 3952.94.

Example 3: In a mixture of 60 liters, the ratio of milk to water is 2:1. How much water should be added to make the ratio 1:2?

Current: Milk = 40L, Water = 20L. After adding x liters of water: 40/(20+x) = 1/2 → 80 = 20 + x → x = 60 liters

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

Let numbers = 3x and 5x. (3x + 9)/(5x + 9) = 3/4 4(3x + 9) = 3(5x + 9) 12x + 36 = 15x + 27 3x = 9 → x = 3. Numbers: 9 and 15

Example 5: The mean proportional between two numbers is 12. If one number is 8, find the other.

Mean proportional: b² = ac → 12² = 8 × c → 144 = 8c → c = 18

Common Exam Patterns

1. “Divide X in ratio a:b:c” → Each share = X × (own ratio part / total)
2. “Incomes in ratio, expenditures in ratio, each saves Y” → Set up two equations, solve
3. “A:B and B:C given, find A:B:C” → Make B common using LCM
4. “If X is added to both, ratio becomes…” → Let quantities be ax and bx, set up equation
5. “Componendo-dividendo” → If (a+b)/(a-b) = k, use C-D to find a:b directly
6. “Mean proportional” → b = √(ac)

Practice Problems

Q1: A and B share a profit of Rs 9600. If A’s investment is 1.5 times B’s investment, find each person’s share.

Q2: The ratio of boys to girls in a class is 5:3. If 4 boys leave and 4 girls join, the ratio becomes 1:1. How many students were in the class originally?

Q3: If (3a + 5b)/(3a - 5b) = 5/1, find a:b.

Answers:

A1: A:B = 1.5:1 = 3:2. A’s share = 9600 × 3/5 = Rs 5760. B’s share = 9600 × 2/5 = Rs 3840.

A2: Let boys = 5x, girls = 3x. After changes: (5x-4)/(3x+4) = 1/1 → 5x-4 = 3x+4 → 2x = 8 → x = 4. Total originally = 5(4) + 3(4) = 32 students.

A3: By componendo-dividendo: [(3a+5b)+(3a-5b)] / [(3a+5b)-(3a-5b)] = (5+1)/(5-1) → 6a/10b = 6/4 → a/b = 60/40 = 3/2. a:b = 5:3.

Wait, let me redo: 6a/10b = 6/4 → a/b = (6 × 10)/(4 × 6) → that’s wrong. Let me be careful. 6a/10b = 6/4, so a/b = (6 × 10b)/(4 × 6a)… No. Cross multiply: 6a × 4 = 10b × 6 → 24a = 60b → a/b = 60/24 = 5/2. a:b = 5:2.

Averages seem simple — and the formula is — but the questions can get sneaky. The most common traps involve average speed (it’s NOT the simple average of speeds!) and problems where elements are added or removed. Let’s master the patterns.

What Is an Average?

Average (arithmetic mean) = Sum of all values / Number of values

In simple language, the average tells us “if all values were equal, what would each value be?”

Critical warning: Average speed for equal distances is 2ab/(a+b), which is the harmonic mean, NOT the arithmetic mean (a+b)/2. This is the #1 trap in average speed problems.

The “Sum” Approach — The Master Key

Most average problems become easy if we think in terms of sums, not averages.

The core idea: If the average of n numbers is A, then their sum is n×A. When things change (elements added, removed, replaced), we track the change in sum and recalculate.

Average Speed — The Big Trap

Equal Distances at Different Speeds

Example: A person travels from A to B at 40 km/h and returns at 60 km/h. What’s the average speed?

WRONG answer: (40+60)/2 = 50 km/h. This is wrong because they spend MORE time at the slower speed.

RIGHT answer: Average speed = 2 × 40 × 60 / (40 + 60) = 4800/100 = 48 km/h

Let’s verify: Say distance = 120 km each way.

• Time going: 120/40 = 3 hours
• Time returning: 120/60 = 2 hours
• Average speed = 240/5 = 48 km/h ✓

Equal Times at Different Speeds

If a person drives at 40 km/h for 2 hours and 60 km/h for 2 hours, THEN we can use simple average: (40+60)/2 = 50 km/h. Because the time is the same for both speeds.

Three Different Speeds for Equal Distances

Average speed = 3abc / (ab + bc + ca)

Effect of Adding / Removing Elements

Adding a New Element

Example: The average of 10 numbers is 25. When an 11th number is added, the average becomes 27. What is the 11th number?

• Old sum = 10 × 25 = 250
• New sum = 11 × 27 = 297
• 11th number = 297 - 250 = 47

Removing an Element

Example: The average of 20 numbers is 35. If one number (56) is removed, what’s the new average?

• Old sum = 20 × 35 = 700
• New sum = 700 - 56 = 644
• New average = 644/19 = 33.89 (approximately)

Replacing an Element

Example: The average of 8 numbers is 14. If one number 24 is replaced by 8, what’s the new average?

• Change in sum = 8 - 24 = -16
• Change in average = -16/8 = -2
• New average = 14 - 2 = 12

Shortcut: When replacing one element in a group of n: Change in average = (New value - Old value) / n

Weighted Average

When groups of different sizes have different averages, we can’t just average the averages. We need to weight them.

Example: Section A has 30 students with average marks 70. Section B has 20 students with average marks 80. What’s the overall average?

Weighted average = (30×70 + 20×80) / (30+20) = (2100 + 1600)/50 = 3700/50 = 74

Notice it’s closer to 70 than 80, because Section A has more students. Makes sense!

Age-Based Average Problems

These show up ALL the time. The key insight is: when time passes, everyone’s age increases by the same amount.

Example: The average age of a family of 5 members is 30 years. If a baby is born, what’s the new average after 2 years?

• Current sum = 5 × 30 = 150
• After 2 years: each of the 5 existing members is 2 years older → sum increases by 10
• Baby is 2 years old
• New sum = 150 + 10 + 2 = 162
• New average = 162/6 = 27 years

Cricket Run Rate Problems

Example: In a 50-over match, the target is 282. After 40 overs, a team has scored 200. What run rate is needed in the remaining overs?

• Runs needed = 282 - 200 = 82
• Overs remaining = 10
• Required run rate = 82/10 = 8.2 runs per over

Consecutive Numbers

Useful shortcut: The average of consecutive numbers equals the middle number.

• Average of 3, 4, 5, 6, 7 → middle number = 5
• Average of first n natural numbers = (n+1)/2
• Average of first n even numbers = (n+1)
• Average of first n odd numbers = n

Worked Examples

Example 1: The average marks of 30 students is 60. The average of the top 10 is 75, and the average of the bottom 10 is 40. What’s the average of the remaining 10?

• Total sum = 30 × 60 = 1800
• Top 10 sum = 10 × 75 = 750
• Bottom 10 sum = 10 × 40 = 400
• Remaining sum = 1800 - 750 - 400 = 650
• Remaining average = 650/10 = 65

Example 2: A batsman has an average of 45 after 20 innings. How many runs must he score in the 21st innings to increase his average to 48?

• Current sum = 20 × 45 = 900
• Required sum after 21 innings = 21 × 48 = 1008
• Runs in 21st innings = 1008 - 900 = 108

Example 3: The average weight of a class of 24 students is 35 kg. If the teacher’s weight is included, the average increases by 400g. What is the teacher’s weight?

• Students’ total weight = 24 × 35 = 840 kg
• New average = 35.4 kg for 25 people
• New total = 25 × 35.4 = 885 kg
• Teacher’s weight = 885 - 840 = 45 kg

Shortcut: Teacher adds (25 × 0.4) = 10 kg more than the old average. So teacher weighs 35 + 10 = 45 kg.

Example 4: A car travels from city A to B at 60 km/h, B to C at 40 km/h, and C to D at 80 km/h. If AB = BC = CD, find the average speed.

Using the three-equal-distance formula: Average speed = 3 × 60 × 40 × 80 / (60×40 + 40×80 + 80×60) = 3 × 192000 / (2400 + 3200 + 4800) = 576000 / 10400 = 55.38 km/h (approximately)

Example 5: The average of 5 consecutive odd numbers is 41. Find the largest number.

For consecutive odd numbers, the average = middle number = 41. The 5 numbers: 37, 39, 41, 43, 45. Largest = 45

Common Exam Patterns

1. “Average speed for equal distances” → Use 2ab/(a+b), NOT (a+b)/2
2. “Average changes when element added/removed” → Track sums, find the unknown
3. “Average of n consecutive numbers” → It’s always the middle number
4. “Weighted average of two groups” → (n₁×avg₁ + n₂×avg₂) / (n₁ + n₂)
5. “How much to score to change average to X?” → Required sum - Current sum
6. “Average age after n years” → Add n to every person’s age (sum increases by n × count)

Practice Problems

Q1: The average of 11 numbers is 36. If the average of the first 6 is 32 and the average of the last 6 is 37, find the 6th number.

Q2: A person goes to office at 30 km/h and returns at 20 km/h. If the total travel time is 5 hours, find the distance to the office.

Q3: The average weight of 8 people increases by 2.5 kg when a new person replaces one weighing 65 kg. What is the weight of the new person?

Answers:

A1: Total sum = 11 × 36 = 396. First 6 sum = 6 × 32 = 192. Last 6 sum = 6 × 37 = 222. The 6th number is counted in both groups. So: 192 + 222 - 396 = 18. The 6th number is 18.

A2: Let distance = d. Time = d/30 + d/20 = 5. → (2d + 3d)/60 = 5 → 5d = 300 → d = 60 km.

A3: Change in total weight = 8 × 2.5 = 20 kg increase. New person’s weight = 65 + 20 = 85 kg.

Mixtures and alligation is one of those topics that looks scary but has ONE trick that solves almost everything — the alligation (cross) method. Once we get this, we can solve mixing problems, average problems, percentage problems, and even profit/loss problems in seconds.

What Is Alligation?

Alligation is a method to find the ratio in which two ingredients at different prices (or concentrations, or averages) should be mixed to get a mixture at a desired price (or concentration, or average).

In simple language, if we’re mixing cheap and expensive stuff to get something in between, alligation tells us the exact ratio.

The Alligation (Cross) Method

This is the most important diagram in this entire topic.

The trick is: we take the cross differences. The cheaper quantity’s ratio part = (Dearer - Mean), and the dearer quantity’s ratio part = (Mean - Cheaper).

Simple Mixing Example

Example: In what ratio should Rs 40/kg tea be mixed with Rs 60/kg tea to get a mixture worth Rs 45/kg?

Using alligation:

• Cheaper (C) = 40, Dearer (D) = 60, Mean (M) = 45
• Ratio = (D - M) : (M - C) = (60 - 45) : (45 - 40) = 15 : 5 = 3 : 1

So we mix 3 parts of the Rs 40 tea with 1 part of the Rs 60 tea. Done in 10 seconds.

Alligation Beyond Mixing

Here’s the cool part — alligation works for ANY weighted average situation, not just physical mixing.

For Averages

Example: Section A (30 students) has average marks 70. Section B (20 students) has average marks 80. We already know the combined average is 74 (from the averages chapter). But we can also verify the ratio:

Ratio of A:B students = (80 - 74) : (74 - 70) = 6 : 4 = 3 : 2. And indeed, 30:20 = 3:2 ✓

For Percentages

Example: A shopkeeper mixes two types of sugar — one at Rs 20/kg (with 60% purity) and another at Rs 30/kg (with 90% purity). In what ratio should they be mixed to get 70% pure sugar?

Using alligation on purity: Ratio = (90 - 70) : (70 - 60) = 20 : 10 = 2 : 1

For Profit/Loss

Example: A shopkeeper sells some goods at 10% profit and the rest at 20% profit. If overall profit is 14%, find the ratio of cost prices.

Using alligation on profit%: Ratio = (20 - 14) : (14 - 10) = 6 : 4 = 3 : 2

Milk and Water Problems

These are the most classic mixture problems.

Type 1: Finding the Ratio

Example: A container has 40 liters of milk. How much water should be added to make a 4:1 milk-to-water ratio?

Milk = 40L. We want milk:water = 4:1, so water = 40/4 = 10L. Add 10 liters of water.

Type 2: Mixing Two Mixtures

Example: Vessel A contains milk and water in 3:1 ratio. Vessel B contains milk and water in 5:3 ratio. In what ratio should we mix from A and B to get a 7:3 mixture?

Milk concentration in A = 3/4 = 75% Milk concentration in B = 5/8 = 62.5% Desired milk concentration = 7/10 = 70%

Using alligation: Ratio A:B = (70 - 62.5) : (75 - 70) = 7.5 : 5 = 3 : 2

Repeated Dilution (Replacement Problems)

This is a favorite exam type. We start with a full container of substance X. Each time, we remove some and replace with water (or another substance).

Example: A container has 80 liters of milk. 8 liters are drawn out and replaced with water. This is done 3 times. How much milk remains?

Milk remaining = 80 × (1 - 8/80)³ = 80 × (1 - 1/10)³ = 80 × (9/10)³

= 80 × 729/1000 = 58.32 liters

Key insight: Each replacement reduces the milk by the same fraction. After n replacements, the fraction remaining is (1 - x/V)ⁿ.

Worked Examples

Example 1: A milkman mixes water with milk. The cost of pure milk is Rs 24/liter. He sells the mixture at Rs 26/liter and makes a 30% profit. Find the ratio of water to milk.

His effective CP = 26/1.3 = Rs 20 per liter. But pure milk costs Rs 24/liter and water costs Rs 0/liter. Using alligation: Water:Milk = (24 - 20) : (20 - 0) = 4 : 20 = 1 : 5

Example 2: Two vessels contain milk-water mixtures. Vessel 1 has milk:water = 7:3, Vessel 2 has milk:water = 1:2. They are mixed in ratio 2:1. Find the milk:water ratio in the new mixture.

Milk fraction in V1 = 7/10, in V2 = 1/3. Mixed in 2:1 ratio: Average milk fraction = (2 × 7/10 + 1 × 1/3) / (2 + 1) = (14/10 + 1/3) / 3 = (42/30 + 10/30) / 3 = (52/30) / 3 = 52/90 = 26/45

Milk:Water = 26 : (45-26) = 26 : 19

Example 3: 20 liters of a mixture contains milk and water in the ratio 3:1. How much mixture should be removed and replaced with water so that milk and water become equal (1:1)?

Current: Milk = 15L, Water = 5L. After removing x liters and replacing with water:

• Milk removed = x × 3/4 = 3x/4
• Remaining milk = 15 - 3x/4
• Total water = 5 - x/4 + x = 5 + 3x/4

For equal ratio: 15 - 3x/4 = 5 + 3x/4 10 = 6x/4 = 3x/2 x = 20/3 = 6.67 liters

Example 4: A jar contains 100 liters of pure acid. 10 liters are removed and replaced with water. This is done twice more. What percentage of acid remains?

Acid remaining = 100 × (1 - 10/100)³ = 100 × (0.9)³ = 100 × 0.729 = 72.9 liters. Percentage = 72.9%

Example 5: In what ratio must a person mix two kinds of coffee costing Rs 300/kg and Rs 400/kg so that by selling the mix at Rs 420/kg, they make a 20% profit?

SP = Rs 420, Profit = 20%. So CP of mix = 420/1.2 = Rs 350/kg. Alligation: Ratio = (400 - 350) : (350 - 300) = 50 : 50 = 1 : 1

Common Exam Patterns

1. “In what ratio to mix two things” → Alligation cross method
2. “Repeated removal and replacement” → V × (1 - x/V)ⁿ
3. “Mixing two mixtures” → Find concentration of each, then apply alligation
4. “Selling mixture at profit” → Find effective CP of mixture, then apply alligation
5. “How much to add to change ratio from a:b to c:d” → Set up equation using current quantities
6. “Alligation for average/profit%” → Same cross method, just use averages or percentages instead of prices

Practice Problems

Q1: A shopkeeper mixes rice at Rs 50/kg with rice at Rs 65/kg in the ratio 2:3. At what price per kg should he sell the mixture to make a 20% profit?

Q2: A container has 120 liters of pure milk. 30 liters are taken out and replaced with water. This process is repeated twice more. How much milk remains?

Q3: Two solutions have acid concentrations of 30% and 70%. In what ratio should they be mixed to get a 50% solution?

Answers:

A1: CP of mix = (2×50 + 3×65)/(2+3) = (100+195)/5 = 295/5 = Rs 59/kg. SP for 20% profit = 59 × 1.2 = Rs 70.80/kg

A2: Milk = 120 × (1 - 30/120)³ = 120 × (3/4)³ = 120 × 27/64 = 50.625 liters

A3: Alligation: Ratio = (70-50) : (50-30) = 20 : 20 = 1 : 1

Partnership problems are really just ratio problems in a business context. The core idea is simple: profit is shared in the ratio of each partner’s effective investment (capital multiplied by time). Once we get this one concept, partnership problems become straightforward.

What Is a Partnership?

When two or more people invest money together in a business and share profits (or losses), it’s called a partnership.

• Active / Working Partner: Manages the business day-to-day. May receive a salary or commission before profit is distributed.
• Sleeping / Silent Partner: Invests money but doesn’t participate in running the business.

Types of Partnership

Simple Partnership

All partners invest for the same duration. Profit is shared in the ratio of their investments.

Example: A invests Rs 30,000 and B invests Rs 50,000. They run the business for 1 year. Profit = Rs 24,000.

Ratio = 30000 : 50000 = 3 : 5

• A’s share = 24000 × 3/8 = Rs 9,000
• B’s share = 24000 × 5/8 = Rs 15,000

Compound Partnership

Partners invest for different durations. Profit is shared in the ratio of (Capital × Time) for each partner.

In simple language, the Capital × Time product represents the “weight” of each partner’s contribution. Someone who invests Rs 10,000 for 12 months has the same weight as someone who invests Rs 20,000 for 6 months.

How Compound Partnership Works

Example: A starts a business with Rs 40,000. After 3 months, B joins with Rs 60,000. At the end of the year, the total profit is Rs 28,000. How is it divided?

• A’s investment = 40,000 × 12 = 4,80,000 (invested for full year)
• B’s investment = 60,000 × 9 = 5,40,000 (invested for 9 months)
• Ratio = 480000 : 540000 = 8 : 9

A’s share = 28000 × 8/17 = Rs 13,176.47 B’s share = 28000 × 9/17 = Rs 14,823.53

Simplification trick: We don’t need to keep the full numbers. Just work with simplified units.

• A: 40 × 12 = 480
• B: 60 × 9 = 540
• Ratio = 480:540 = 8:9

Working Partner Problems

When one partner actively manages the business, they often get a salary or commission BEFORE the profit is split.

Example: A and B start a business investing Rs 50,000 and Rs 30,000 respectively. A manages the business and receives 10% of the total profit as salary. The remaining profit is divided in the ratio of investments. If total profit is Rs 40,000, find each person’s share.

1. A’s salary = 10% of 40,000 = Rs 4,000
2. Remaining profit = 40,000 - 4,000 = Rs 36,000
3. Investment ratio = 50000 : 30000 = 5 : 3
4. A’s share from remaining = 36,000 × 5/8 = Rs 22,500
5. B’s share = 36,000 × 3/8 = Rs 13,500

Total for A = 4,000 + 22,500 = Rs 26,500 Total for B = Rs 13,500

Partners Joining and Leaving

When partners join at different times or leave mid-way, calculate Capital × Time for each partner based on when they were actually in the business.

Example: A starts with Rs 20,000. B joins after 4 months with Rs 25,000. C joins after 6 months with Rs 30,000. Profit after 12 months?

• A: 20,000 × 12 = 240,000
• B: 25,000 × 8 = 200,000
• C: 30,000 × 6 = 180,000

Ratio = 240 : 200 : 180 = 12 : 10 : 9

Changing Investment Mid-Year

Sometimes a partner increases or decreases their investment during the year. We treat each period separately.

Example: A starts with Rs 30,000. After 6 months, A adds Rs 20,000 more. B invests Rs 40,000 for the full year. Find the profit ratio.

• A: 30,000 × 6 + 50,000 × 6 = 180,000 + 300,000 = 480,000
• B: 40,000 × 12 = 480,000

Ratio = 480000 : 480000 = 1 : 1

Even though A invested less initially, adding more mid-year equalized it.

Worked Examples

Example 1: A and B partner in a business. A invests Rs 40,000 for 8 months and B invests Rs 35,000 for 10 months. If the total profit is Rs 31,500, find each person’s share.

• A’s contribution = 40 × 8 = 320
• B’s contribution = 35 × 10 = 350
• Ratio = 320 : 350 = 32 : 35

A’s share = 31,500 × 32/67 = Rs 15,074 (approximately) B’s share = 31,500 × 35/67 = Rs 16,426 (approximately)

Example 2: A, B, and C start a business. A invests 3 times as much as B, and B invests 2/3 of C’s investment. If the annual profit is Rs 66,000, find C’s share.

Let C’s investment = x. Then B = 2x/3. And A = 3 × 2x/3 = 2x. Ratio = A : B : C = 2x : 2x/3 : x = 6 : 2 : 3

C’s share = 66,000 × 3/11 = Rs 18,000

Example 3: A starts a business with Rs 10,000. B joins after 3 months with Rs 16,000. C joins after 6 months with Rs 20,000. B withdraws Rs 6,000 after 9 months. Find the profit-sharing ratio at the end of the year.

• A: 10,000 × 12 = 120,000
• B: 16,000 × 6 + 10,000 × 3 = 96,000 + 30,000 = 126,000 (B was in from month 4-12, but withdrew 6000 after month 9, so 16000 × 6 months from month 4-9, then 10000 × 3 months from month 10-12)
• C: 20,000 × 6 = 120,000

Ratio = 120 : 126 : 120 = 20 : 21 : 20

Example 4: Two partners A and B invest Rs 1,00,000 and Rs 1,50,000 respectively. A is a working partner and gets 15% of the profit as management fee. If A’s total income is Rs 30,000, what is the total profit?

Let total profit = P. A’s management fee = 0.15P Remaining = 0.85P, split in 100:150 = 2:3. A’s share of remaining = 0.85P × 2/5 = 0.34P. A’s total = 0.15P + 0.34P = 0.49P = 30,000. P = 30,000/0.49 = Rs 61,224.49 (approximately)

Example 5: A and B are partners. A receives 1/3 of the profit and B receives the rest. If A’s income is increased by Rs 600 when the profit rises from 5% to 7%, find the total capital invested.

A gets 1/3 of the profit. Increase in A’s share = 1/3 × increase in total profit. Increase in total profit = (7% - 5%) × Total capital = 2% × Total capital. A’s increase = 1/3 × 2% × Total capital = 600. Total capital = 600 × 3 × 100/2 = Rs 90,000

Common Exam Patterns

1. “Invest different amounts, same time” → Simple ratio of capitals
2. “Join at different times” → Capital × Time for each partner
3. “Working partner gets salary/commission” → Deduct salary first, then divide remainder in investment ratio
4. “Investment changes mid-year” → Split into periods, calculate C×T for each period, sum them
5. “Find investment given profit share” → Set up ratio equation, solve for unknown
6. “Difference between two partners’ shares” → Express difference in terms of total, solve

Practice Problems

Q1: A starts a business with Rs 25,000. After 4 months, B joins with Rs 30,000. At the end of the year, the profit is Rs 47,000. Find the share of each.

Q2: A, B, and C invest Rs 40,000, Rs 50,000, and Rs 60,000 respectively. A is a working partner and receives 12.5% of the profit as salary. If the total profit is Rs 56,000, find A’s total earnings.

Q3: A starts a business with Rs 4,500. After 6 months, B joins with some capital. If the profit at the end of the year is divided between A and B in the ratio 2:1, how much did B invest?

Answers:

A1: A’s contribution = 25 × 12 = 300. B’s contribution = 30 × 8 = 240. Ratio = 300:240 = 5:4. A’s share = 47,000 × 5/9 ≈ Rs 26,111. B’s share = 47,000 × 4/9 ≈ Rs 20,889.

A2: A’s salary = 12.5% of 56,000 = Rs 7,000. Remaining = Rs 49,000. Ratio = 40:50:60 = 4:5:6. A’s share from remaining = 49,000 × 4/15 ≈ Rs 13,067. A’s total = 7,000 + 13,067 = Rs 20,067 (approximately).

A3: A’s contribution = 4500 × 12 = 54,000. B’s contribution = B × 6. Ratio: 54000 : 6B = 2 : 1. So 54000/6B = 2/1 → 54000 = 12B → B = Rs 4,500.

Linear equations are the bread and butter of algebra. If we can solve these fast, we unlock speed in almost every other aptitude topic because so many word problems boil down to “set up an equation, solve it.”

A linear equation is simply an equation where the variable has power 1 — no squares, no cubes, just plain old x. Think of it like a balance scale: whatever we do to one side, we do to the other, and the scale stays balanced.

Single Variable Equations

The goal is simple: get x alone on one side.

Quick trick: When we see an equation like 3x + 7 = 22, just think: “move 7 to the other side, flip its sign.” So 3x = 22 - 7 = 15, then x = 5. We don’t need to formally “subtract 7 from both sides” every time — just move and flip.

Example 1: Basic single variable

Solve: 5x - 3 = 2x + 12

Move all x terms to the left, constants to the right:

• 5x - 2x = 12 + 3
• 3x = 15
• x = 5

Example 2: Fractions involved

Solve: (2x + 1)/3 = (x - 2)/2

Cross-multiply:

• 2(2x + 1) = 3(x - 2)
• 4x + 2 = 3x - 6
• 4x - 3x = -6 - 2
• x = -8

Simultaneous Linear Equations (Two Variables)

When we have two unknowns, we need two equations. There are two main methods: substitution and elimination. Both get us to the same answer — pick whichever feels faster for the problem.

Method 1: Substitution

Best when one equation already has a variable with coefficient 1 (like y = 3x + 2).

Method 2: Elimination

Best when coefficients are easy to match up.

Example 3: Elimination method

Solve: 3x + 2y = 16 and 5x - 2y = 24

Notice the y coefficients are already +2 and -2. Just add the equations:

• (3x + 2y) + (5x - 2y) = 16 + 24
• 8x = 40
• x = 5

Plug back: 3(5) + 2y = 16 → 2y = 1 → y = 0.5

Answer: x = 5, y = 0.5

Example 4: Substitution method

Solve: x + 3y = 10 and 2x - y = 3

From equation 1: x = 10 - 3y

Plug into equation 2: 2(10 - 3y) - y = 3

• 20 - 6y - y = 3
• -7y = -17
• y = 17/7
• x = 10 - 3(17/7) = 10 - 51/7 = 19/7

Answer: x = 19/7, y = 17/7

Word Problem Translation — The Real Skill

This is where most people get stuck. The math is easy; the translation from English to equation is the hard part. Here are the magic translations:

Example 5: Classic word problem

“The sum of two numbers is 50. One number is 14 more than the other. Find the numbers.”

Let the smaller number = x. Then the larger = x + 14.

• x + (x + 14) = 50
• 2x + 14 = 50
• 2x = 36
• x = 18

So the numbers are 18 and 32.

Quick check: 18 + 32 = 50 and 32 - 18 = 14. Done!

Example 6: Age problem

“A father is 3 times as old as his son. 12 years later, he will be twice as old as his son. Find their present ages.”

Let son’s age = x. Father’s age = 3x.

After 12 years: son = x + 12, father = 3x + 12.

Given: 3x + 12 = 2(x + 12)

• 3x + 12 = 2x + 24
• x = 12

Son is 12, father is 36.

Quick check: After 12 years — son is 24, father is 48. Is 48 = 2 × 24? Yes!

Shortcut: The “Assumed Mean” Trick for Symmetric Equations

When we see something like: “The sum of two numbers is 100, and the difference is 20.”

Instead of setting up full equations, just think:

• Average = 100/2 = 50
• Half the difference = 20/2 = 10
• Numbers are 50 + 10 = 60 and 50 - 10 = 40

This works every time and saves a ton of time. In simple language, we’re just splitting the total evenly and then adjusting by half the gap.

Special Cases to Watch For

1. No solution: When we end up with something like 0 = 5 — the equations are parallel lines that never meet (inconsistent system).

2. Infinite solutions: When we get 0 = 0 — both equations are actually the same line (dependent system).

3. Number of solutions test: For a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

   • Unique solution: a₁/a₂ ≠ b₁/b₂
   • No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
   • Infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂

Common Exam Variations

• Digit problems: “A two-digit number…” → number = 10a + b (tens digit × 10 + units digit). If digits are reversed: 10b + a.
• Fraction problems: “If 1 is added to numerator and 2 to denominator, fraction becomes 1/2” → (x+1)/(y+2) = 1/2.
• Speed/distance word problems that reduce to linear equations.
• Ages — present, past, future relationships.

Practice Problems

Problem 1: The sum of two numbers is 84. One is 3 times the other. Find them.

Problem 2: A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Problem 3: Solve: 2x + 3y = 13 and 5x - y = 7.

Answers

Problem 1: Let smaller = x. Larger = 3x. So x + 3x = 84 → 4x = 84 → x = 21. Numbers are 21 and 63.

Problem 2: Let tens digit = a, units digit = b. Number = 10a + b.

• 10a + b = 4(a + b) → 10a + b = 4a + 4b → 6a = 3b → b = 2a
• 10a + b + 18 = 10b + a → 9a - 9b = -18 → a - b = -2
• Substituting b = 2a: a - 2a = -2 → a = 2, b = 4. Number is 24.

Problem 3: From equation 2: y = 5x - 7. Plug into equation 1: 2x + 3(5x - 7) = 13 → 2x + 15x - 21 = 13 → 17x = 34 → x = 2, y = 3. Answer: x = 2, y = 3.

A quadratic equation is any equation where the highest power of the variable is 2. The standard form is ax² + bx + c = 0 where a ≠ 0. Think of it like this — linear equations give us straight lines, quadratics give us parabolas (U-shaped curves). And where that parabola crosses the x-axis? Those are our roots (solutions).

Methods to Solve

We have three weapons:

1. Factorization — fastest when it works
2. Quadratic formula — works every single time
3. Completing the square — rarely needed in exams, but good to know

Method 1: Factorization

The idea is to split the middle term bx into two parts that let us factor nicely.

Steps:

1. Multiply a × c
2. Find two numbers that multiply to give a×c and add up to b
3. Split the middle term and factor by grouping

Example 1: Simple factorization

Solve: x² - 7x + 12 = 0

We need two numbers that multiply to 12 and add to -7. That’s -3 and -4.

• x² - 3x - 4x + 12 = 0
• x(x - 3) - 4(x - 3) = 0
• (x - 3)(x - 4) = 0
• x = 3 or x = 4

Shortcut: For x² + bx + c = 0 (when a = 1), the roots are just the two numbers that multiply to c and add to b. We can skip the grouping entirely.

Example 2: When a ≠ 1

Solve: 2x² + 7x + 3 = 0

a × c = 2 × 3 = 6. We need two numbers multiplying to 6 and adding to 7. That’s 6 and 1.

• 2x² + 6x + x + 3 = 0
• 2x(x + 3) + 1(x + 3) = 0
• (2x + 1)(x + 3) = 0
• x = -1/2 or x = -3

Method 2: Quadratic Formula

When factorization isn’t clean (ugly numbers, irrational roots), this is our go-to.

Example 3: Using the formula

Solve: 3x² - 5x + 1 = 0

Here a = 3, b = -5, c = 1.

D = b² - 4ac = 25 - 12 = 13

x = (5 ± √13) / 6

x = (5 + √13)/6 or x = (5 - √13)/6

The Discriminant — Nature of Roots

The discriminant D = b² - 4ac tells us everything about the roots without actually solving.

Example 4: Finding nature of roots

For what values of k does 2x² + kx + 8 = 0 have equal roots?

For equal roots, D = 0:

• k² - 4(2)(8) = 0
• k² = 64
• k = ±8

Sum and Product of Roots

This is a massive shortcut. We can find relationships between roots without actually solving the equation.

If α and β are roots of ax² + bx + c = 0:

• Sum: α + β = -b/a
• Product: αβ = c/a

Example 5: Using sum and product

If the roots of x² - 5x + 6 = 0 are α and β, find α² + β².

Sum = α + β = 5, Product = αβ = 6.

Now the trick: α² + β² = (α + β)² - 2αβ = 25 - 12 = 13

This identity (α² + β²) = (α + β)² - 2αβ comes up ALL the time. Memorize it.

Other useful identities:

• α² + β² = (α + β)² - 2αβ
• (α - β)² = (α + β)² - 4αβ
• α³ + β³ = (α + β)³ - 3αβ(α + β)

Forming Equations from Roots

If we know the roots, we can build the equation: x² - (sum of roots)x + (product of roots) = 0

Example 6: Building an equation

Form a quadratic equation whose roots are 2 and -5.

Sum = 2 + (-5) = -3. Product = 2 × (-5) = -10.

Equation: x² - (-3)x + (-10) = 0 → x² + 3x - 10 = 0

The Sign-Based Comparison Trick

This shows up in banking exams constantly. We get two equations and need to compare x and y.

The method:

1. Solve both equations to find roots of x and roots of y.
2. Compare the roots directly.

In simple language, we just need to figure out which variable has larger values. If all roots of x are greater than all roots of y, then x > y. If there’s overlap, the relationship can’t be determined.

Example 7: Comparison problem

I: x² - 7x + 12 = 0 II: y² - 9y + 20 = 0

Find the relationship between x and y.

Equation I: (x-3)(x-4) = 0 → x = 3 or 4 Equation II: (y-4)(y-5) = 0 → y = 4 or 5

Comparing: x can be 3 or 4, y can be 4 or 5. Since x values (3, 4) are ≤ y values (4, 5):

x ≤ y

Common Exam Variations

• Find the value of k such that the equation has equal roots / no real roots / one root is twice the other.
• Comparison of roots from two different equations (banking exam favorite).
• Sum/product of roots without solving — “find α² + β²” or “find 1/α + 1/β”.
• If one root is known, find the other root and the unknown coefficient.
• Common root between two quadratic equations.

Shortcut: Finding 1/α + 1/β

Instead of solving for individual roots:

1/α + 1/β = (α + β)/(αβ) = (-b/a)/(c/a) = -b/c

Super fast for exam questions!

Practice Problems

Problem 1: Solve 6x² - x - 2 = 0 by factorization.

Problem 2: If the roots of 2x² - 8x + k = 0 are equal, find k.

Problem 3: The roots of x² - px + 12 = 0 are in the ratio 1:3. Find p.

Answers

Problem 1: a×c = -12. Numbers: -4 and 3 (multiply to -12, add to -1). 6x² - 4x + 3x - 2 = 0 → 2x(3x - 2) + 1(3x - 2) = 0 → (2x + 1)(3x - 2) = 0. x = -1/2 or x = 2/3.

Problem 2: For equal roots, D = 0. b² - 4ac = 64 - 8k = 0 → k = 8.

Problem 3: Let roots be α and 3α. Sum: α + 3α = 4α = p → α = p/4. Product: α × 3α = 3α² = 12 → α² = 4 → α = 2. So p = 4α = 4(2) = 8. (We take α = 2 since ratio is 1:3, both positive.)

Inequalities are like equations but instead of “equals,” we have “greater than,” “less than,” or their “or equal to” versions. The solving process is almost identical to equations, with one critical twist that catches everyone: when we multiply or divide by a negative number, the inequality sign FLIPS.

In simple language, think of inequalities as asking “what range of values makes this true?” instead of “what exact value makes this true?”

Linear Inequalities

These work just like linear equations — move terms around, simplify — but keep the sign direction in mind.

Example 1: Basic linear inequality

Solve: 3x - 7 > 2x + 5

• 3x - 2x > 5 + 7
• x > 12

Solution: x > 12 (all numbers greater than 12)

Example 2: The sign-flip trap

Solve: -2x + 6 ≤ 10

• -2x ≤ 4
• Divide by -2 and FLIP the sign: x ≥ -2

Solution: x ≥ -2

If we had forgotten to flip, we’d get x ≤ -2, which is completely wrong. This is the #1 trap in inequality questions.

Example 3: Compound inequality

Solve: -3 < 2x + 1 ≤ 7

Subtract 1 from all three parts:

• -4 < 2x ≤ 6

Divide by 2:

• -2 < x ≤ 3

Solution: x is between -2 (exclusive) and 3 (inclusive).

Quadratic Inequalities — The Wavy Curve Method

This is where things get interesting. We can’t just “solve” a quadratic inequality the way we solve a quadratic equation. We need to find the roots first, then figure out which intervals satisfy the inequality.

The wavy curve method (also called the sign method or number line method) is the fastest approach.

Example 4: Quadratic inequality

Solve: x² - 5x + 6 > 0

Step 1: Factor: (x - 2)(x - 3) > 0

Step 2: Roots are x = 2 and x = 3.

Step 3: Number line with signs:

We need > 0 (positive regions): x < 2 or x > 3

Quick rule of thumb: For (x - a)(x - b) > 0 where a < b, the answer is x < a or x > b (the “outside” regions). For < 0, the answer is a < x < b (the “inside” region).

Example 5: Less than type

Solve: x² - x - 12 ≤ 0

Factor: (x - 4)(x + 3) ≤ 0. Roots: x = -3 and x = 4.

We need ≤ 0 (negative or zero region, which is the “inside”):

-3 ≤ x ≤ 4

The ≤ means we include the endpoints (since equality is allowed).

Modulus (Absolute Value) Inequalities

The modulus |x| just means “distance from zero.” So |x| < 5 means “x is within 5 units of zero” — or simply -5 < x < 5.

In simple language: “less than” means we’re trapped between two values. “Greater than” means we’re outside, far away.

Example 6: Modulus inequality

Solve: |2x - 3| < 7

Using the rule |expression| < a → -a < expression < a:

• -7 < 2x - 3 < 7
• -4 < 2x < 10
• -2 < x < 5

Solution: -2 < x < 5

Example 7: Greater-than modulus

Solve: |x + 1| ≥ 4

This means x + 1 ≤ -4 OR x + 1 ≥ 4:

• x ≤ -5 OR x ≥ 3

Solution: x ≤ -5 or x ≥ 3

The Wavy Curve for Rational Inequalities

The wavy curve works for fractions too. Say we have (x - 1)(x - 4) / (x - 2) > 0.

Same process: find all critical points (1, 2, 4), plot on number line, alternate signs starting from the right with +. The only difference is we never include points where the denominator is zero (x = 2 here, since that makes the expression undefined).

For > 0, we pick the positive regions: 1 < x < 2 or x > 4

Note: x = 2 is excluded (open circle) because the denominator can’t be zero.

Common Traps

1. Forgetting to flip the sign when multiplying/dividing by a negative.
2. Multiplying both sides by a variable without knowing if it’s positive or negative — never do this! Cross-multiply only when we know denominators are positive.
3. Forgetting that |x| ≥ 0 always — so |x| < -3 has NO solution, and |x| > -5 is true for ALL real x.
4. Including undefined points in rational inequalities.

Common Exam Variations

• Basic linear inequalities with the flip-sign trap.
• Quadratic inequalities asking for the range of x.
• Modulus equations (|expression| = value gives two cases).
• Finding integer solutions in a range.
• “For what values of k does the expression have no real solution?”

Practice Problems

Problem 1: Solve: x² - 4x - 5 > 0.

Problem 2: Solve: |3x - 2| ≤ 10.

Problem 3: Solve: (x + 2)/(x - 1) < 0.

Answers

Problem 1: Factor: (x - 5)(x + 1) > 0. Roots at -1 and 5. Positive outside: x < -1 or x > 5.

Problem 2: -10 ≤ 3x - 2 ≤ 10 → -8 ≤ 3x ≤ 12 → -8/3 ≤ x ≤ 4.

Problem 3: Critical points: x = -2 and x = 1. Sign chart from right: +, -, +. We need < 0 (the middle region): -2 < x < 1. Note: x = 1 is excluded (denominator zero), x = -2 is excluded (strict inequality).

Sequences and series show up everywhere in aptitude — salary increments, population growth, bouncing balls, installment payments. The good news is there are really only two types we need to master: Arithmetic Progression (AP) and Geometric Progression (GP).

In simple language, an AP is a sequence where we add the same number each time (like 2, 5, 8, 11…), and a GP is where we multiply by the same number each time (like 3, 6, 12, 24…).

Arithmetic Progression (AP)

An AP has a constant difference between consecutive terms. That difference is called d (common difference).

The alternate sum formula Sₙ = (n/2)(first + last) is often faster because we just need the first term, last term, and count. Think of it like this — the average of an AP is always the average of the first and last terms, and total = average × count.

Example 1: Find the nth term

Find the 20th term of the AP: 3, 7, 11, 15…

a = 3, d = 7 - 3 = 4.

a₂₀ = 3 + (20 - 1)(4) = 3 + 76 = 79

Example 2: Sum of terms

Find the sum of the first 15 terms of the AP: 5, 9, 13…

a = 5, d = 4, n = 15.

S₁₅ = (15/2)[2(5) + (14)(4)] = (15/2)[10 + 56] = (15/2)(66) = 495

The Middle Term Property

This is a neat shortcut. In any three consecutive AP terms, the middle one is the average of the other two. So if we have terms a, b, c in AP, then b = (a + c)/2.

This extends: in an AP with an odd number of terms, the sum = middle term × number of terms.

Example 3: Using middle term

The sum of three numbers in AP is 27, and their product is 648. Find them.

Let the three numbers be (a - d), a, (a + d). This is the standard trick for three AP terms — it automatically centers on a.

Sum: (a - d) + a + (a + d) = 3a = 27 → a = 9

Product: (9 - d)(9)(9 + d) = 648

• 9(81 - d²) = 648
• 81 - d² = 72
• d² = 9 → d = 3

Numbers: 6, 9, 12

Geometric Progression (GP)

A GP has a constant ratio between consecutive terms. That ratio is called r (common ratio).

Example 4: Sum to infinity

A ball is dropped from 80 meters. Each time it bounces, it rises to 3/4 of the height it fell from. Find the total distance traveled.

Going down: 80, 60, 45, … (GP with a = 80, r = 3/4) Going up: 60, 45, … (GP with a = 60, r = 3/4)

Total down = 80 / (1 - 3/4) = 80 / (1/4) = 320 Total up = 60 / (1 - 3/4) = 60 / (1/4) = 240

Total distance = 320 + 240 = 560 meters

Alternative shortcut: Total distance = first drop × (1 + r)/(1 - r) = 80 × (1 + 3/4)/(1 - 3/4) = 80 × (7/4)/(1/4) = 80 × 7 = 560.

Example 5: Finding terms

The 3rd term of a GP is 12 and the 6th term is 96. Find the GP.

a₃ = ar² = 12 a₆ = ar⁵ = 96

Divide: ar⁵/ar² = r³ = 96/12 = 8 → r = 2

From ar² = 12: a(4) = 12 → a = 3

GP: 3, 6, 12, 24, 48, 96…

Harmonic Progression (HP)

An HP is a sequence whose reciprocals form an AP. We rarely solve HP directly — instead, we convert to AP, solve, and convert back.

If a, b, c are in HP, then 1/a, 1/b, 1/c are in AP, which means 1/b = (1/a + 1/c)/2.

The harmonic mean of a and b = 2ab/(a + b).

The relationship between means: AM ≥ GM ≥ HM (with equality when all terms are equal). This inequality is super useful for optimization problems.

Word Problem Patterns

Salary increment (AP)

“Starting salary is 30,000. Annual increment of 2,000. What’s the salary in the 10th year?”

a = 30000, d = 2000, n = 10. a₁₀ = 30000 + 9(2000) = 48,000

Population growth (GP)

“Population of a city is 50,000 and grows at 10% per year. What will it be after 3 years?”

a = 50000, r = 1.10, n = 3. Population = 50000 × (1.10)³ = 50000 × 1.331 = 66,550

Installments

“Total amount is 1000. Paid in 5 installments increasing by 50 each time. Find the first installment.”

AP with n = 5, d = 50, S₅ = 1000. (5/2)[2a + 4(50)] = 1000 2a + 200 = 400 → a = 100

Useful Shortcuts

1. Sum of first n natural numbers: n(n+1)/2
2. Sum of squares of first n natural numbers: n(n+1)(2n+1)/6
3. Sum of cubes of first n natural numbers: [n(n+1)/2]² (yes, it’s the square of the sum of naturals!)
4. Sum of first n even numbers: n(n+1)
5. Sum of first n odd numbers: n²

These come up so often that memorizing them saves a huge amount of time.

Common Exam Variations

• Find the nth term or sum of an AP/GP.
• “How many terms are needed for the sum to exceed X?”
• Three numbers in AP/GP with given sum and product.
• Insert n arithmetic/geometric means between two numbers.
• Mixed problems: “Sum of an AP equals the sum of a GP…”
• Real-world: salary, depreciation, population, bouncing ball.

Practice Problems

Problem 1: The sum of the first 10 terms of an AP is 155, and the first term is 2. Find the common difference.

Problem 2: Find the sum to infinity of the GP: 27, 9, 3, 1, …

Problem 3: The 4th and 7th terms of an AP are 11 and 23 respectively. Find the sum of the first 15 terms.

Answers

Problem 1: S₁₀ = (10/2)[2(2) + 9d] = 155 → 5[4 + 9d] = 155 → 4 + 9d = 31 → 9d = 27 → d = 3.

Problem 2: a = 27, r = 9/27 = 1/3. Since |r| < 1: S∞ = 27/(1 - 1/3) = 27/(2/3) = 40.5 (or 81/2).

Problem 3: a₄ = a + 3d = 11 and a₇ = a + 6d = 23. Subtracting: 3d = 12 → d = 4. So a = 11 - 12 = -1. S₁₅ = (15/2)[2(-1) + 14(4)] = (15/2)[−2 + 56] = (15/2)(54) = 405.

Logarithms look scary at first, but they’re just another way of writing exponents. If we understand that log is the inverse of exponentiation, everything clicks.

In simple language: if 2³ = 8, then log₂(8) = 3. We’re asking “what power do we raise 2 to, to get 8?” The answer is 3. That’s literally all a log is — the exponent finder.

The Log Laws

These are the tools that make simplification problems solvable in seconds. Every log problem in exams boils down to applying these laws.

Think of the product rule like this: multiplication inside the log becomes addition outside. Division becomes subtraction. And exponents come out front as multipliers. Once these three rules are second nature, we can tear through any log problem.

Values to Memorize

These come up constantly. Memorizing them saves precious seconds.

The big trick: We only need to memorize log 2, log 3, and log 7. Everything else can be derived:

• log 4 = 2 × log 2
• log 5 = log(10/2) = 1 - log 2
• log 6 = log 2 + log 3
• log 8 = 3 × log 2
• log 9 = 2 × log 3

So really, just remember: log 2 ≈ 0.301, log 3 ≈ 0.477, log 7 ≈ 0.845.

Worked Examples

Example 1: Basic simplification

Find: log₂(32)

32 = 2⁵, so log₂(32) = 5

Always try to express the number as a power of the base first. That’s the fastest approach.

Example 2: Using log laws

Simplify: log 72

Break 72 into prime factors: 72 = 8 × 9 = 2³ × 3²

log 72 = log(2³ × 3²) = 3 log 2 + 2 log 3 = 3(0.301) + 2(0.477) = 0.903 + 0.954 = 1.857

Example 3: Change of base

Find: log₈(32)

Using change of base: log₈(32) = log(32)/log(8) = log(2⁵)/log(2³) = 5 log 2 / 3 log 2 = 5/3

Shortcut: When both the base and number are powers of the same thing, just use the exponents. 32 = 2⁵ and 8 = 2³, so log₈(32) = 5/3. Done!

Example 4: The flip rule in action

If log₃(5) = x, find log₅(27) in terms of x.

log₅(27) = log₅(3³) = 3 × log₅(3)

Now, log₅(3) = 1/log₃(5) = 1/x (using the flip rule)

So log₅(27) = 3/x

Example 5: Finding number of digits

How many digits are in 2⁶⁴?

The number of digits in N = floor(log₁₀(N)) + 1.

log₁₀(2⁶⁴) = 64 × log₁₀(2) = 64 × 0.3010 = 19.264

Number of digits = floor(19.264) + 1 = 19 + 1 = 20 digits

This is a classic exam question type. The formula works because log tells us the “order of magnitude” of a number.

Common Traps and Gotchas

1. log(A + B) ≠ log A + log B — There is NO rule for log of a sum! This is the most common mistake. The product rule only works with multiplication inside the log.

2. log(0) is undefined — We can’t take log of zero or negative numbers (in real numbers).

3. The base must be positive and not 1 — log₁(anything) is undefined.

4. log(A) - log(B) ≠ log(A - B) — It equals log(A/B).

5. Don’t confuse log and ln — “log” usually means base 10 (common log), “ln” means base e ≈ 2.718 (natural log). In aptitude exams, log almost always means base 10 unless stated otherwise.

Natural Logarithm (ln) — Quick Basics

The natural log uses base e ≈ 2.718 (Euler’s number). It shows up in continuous growth/decay problems.

Key facts:

• ln(e) = 1
• ln(1) = 0
• ln(eˣ) = x
• e^(ln x) = x
• All the same log laws apply (product, quotient, power)

For aptitude exams, natural log questions are rare, but knowing these basics helps if one pops up.

Common Exam Variations

• Simplify log expressions using log laws.
• Find the value of logₐ(b) when both a and b are given as powers.
• “If log x = 1.732, find x” — this means x = 10^1.732.
• Number of digits in large powers.
• Solve equations with logs: “if log(x + 2) + log(x - 1) = log 10” → log[(x+2)(x-1)] = 1 → (x+2)(x-1) = 10.
• Characteristic and mantissa (the integer and decimal parts of a log).

Characteristic and Mantissa

For log₁₀(N):

• Characteristic = the integer part (tells us how many digits minus 1)
• Mantissa = the decimal part (always positive, 0 ≤ mantissa < 1)

For numbers > 1: characteristic = (number of digits - 1). For numbers between 0 and 1: characteristic is negative. E.g., log(0.005) has characteristic -3.

Practice Problems

Problem 1: Simplify: log₅(125) + log₅(1/25).

Problem 2: If log 2 = 0.301 and log 3 = 0.477, find the number of digits in 6²⁰.

Problem 3: Solve for x: log₃(x) + log₃(x - 8) = 2.

Answers

Problem 1: log₅(125) = log₅(5³) = 3. log₅(1/25) = log₅(5⁻²) = -2. Answer: 3 + (-2) = 1.

Problem 2: log(6²⁰) = 20 × log 6 = 20 × (log 2 + log 3) = 20 × (0.301 + 0.477) = 20 × 0.778 = 15.56. Number of digits = 15 + 1 = 16 digits.

Problem 3: log₃(x) + log₃(x - 8) = 2 → log₃[x(x - 8)] = 2 → x(x - 8) = 3² = 9. So x² - 8x - 9 = 0 → (x - 9)(x + 1) = 0. x = 9 or x = -1. Since log requires positive arguments, x must be > 8. x = 9.

Set theory in aptitude is mostly about counting — how many people like tea, how many like coffee, how many like both? The Venn diagram is our best friend here. Once we draw it, the answer practically jumps out.

In simple language, a set is just a collection of things. A set of students who passed math, a set of people who drink tea, a set of numbers divisible by 3 — anything grouped together.

Basic Set Operations

Inclusion-Exclusion Principle — The Core Formula

This is the formula that solves 90% of set theory problems in exams.

Why do we subtract the intersection? Because when we add n(A) and n(B), the people in both sets get counted twice. Subtracting the intersection fixes that double-counting. For three sets, it’s the same idea but we have to be more careful — we subtract the pairwise overlaps, but then the triple overlap gets subtracted too many times, so we add it back once.

Venn Diagram for 2 Sets

The trick with Venn diagrams: always fill in the intersection first, then work outward. If we fill from outside in, numbers don’t add up correctly.

Example 1: Basic 2-set problem

In a class of 60 students, 35 like cricket, 28 like football, and 10 like both. How many like neither?

n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 35 + 28 - 10 = 53

Neither = Total - n(A ∪ B) = 60 - 53 = 7 students

Example 2: Finding the intersection

In a group of 100 people, 72 read newspaper A, 54 read newspaper B. How many read both? (Everyone reads at least one.)

Since everyone reads at least one: n(A ∪ B) = 100.

100 = 72 + 54 - n(A ∩ B) n(A ∩ B) = 126 - 100 = 26 people

Venn Diagram for 3 Sets

The 3-Set Filling Strategy

When solving a 3-set problem, fill in this order:

1. Center first: n(A ∩ B ∩ C) = g
2. Pairwise intersections: subtract center from each. E.g., “A and B only” = n(A ∩ B) - g
3. Individual only: subtract all overlaps. E.g., “Only A” = n(A) - d - e - g
4. Add everything to verify it matches the total.

Example 3: Three-set problem

In a survey of 200 students: 120 like tea, 80 like coffee, 60 like juice. 40 like tea and coffee, 30 like coffee and juice, 25 like tea and juice, and 15 like all three. How many like none of the three?

Using inclusion-exclusion for 3 sets:

n(T ∪ C ∪ J) = 120 + 80 + 60 - 40 - 30 - 25 + 15 = 180

None = 200 - 180 = 20 students

Let’s also find each region:

• All three: 15
• Tea and coffee only: 40 - 15 = 25
• Coffee and juice only: 30 - 15 = 15
• Tea and juice only: 25 - 15 = 10
• Only tea: 120 - 25 - 10 - 15 = 70
• Only coffee: 80 - 25 - 15 - 15 = 25
• Only juice: 60 - 10 - 15 - 15 = 20

Check: 70 + 25 + 20 + 25 + 15 + 10 + 15 + 20 (none) = 200. It works!

Maxima-Minima Problems

These are the trickiest set theory questions. They ask things like “what is the maximum number who could like all three?” or “at least how many like both?”

The Key Insight

Maximum of A ∩ B = min(n(A), n(B)) — the overlap can’t be more than the smaller set.

Minimum of A ∩ B = n(A) + n(B) - n(U) if this is positive, else 0 — this is when we spread things out as much as possible.

Example 4: Min-max problem

In a class of 40, 25 passed Math and 30 passed English. Find the minimum and maximum number who passed both.

Maximum: min(25, 30) = 25 (all Math-passers also passed English)

Minimum: 25 + 30 - 40 = 15 (spread out as much as possible, 15 must overlap)

Example 5: “At least” problem with 3 sets

100 people were surveyed. 60 read Hindu, 50 read Times, 45 read Express. What is the minimum number reading at least two newspapers?

Maximum reading only one paper = minimize overlaps. Each person reading only one paper “uses” 1 from the count. Total sum = 60 + 50 + 45 = 155. Since there are only 100 people, at least 155 - 100 = 55 “extra” readings come from overlaps. Each double-reader accounts for 1 extra, each triple-reader accounts for 2 extras.

To minimize “at least two” readers, maximize triple-readers (each triple-reader absorbs 2 extras). So we need:

• If all extras are triple-readers: 55/2 = 27.5, which isn’t integer. So 27 triple + 1 double = 55 extras. Minimum reading at least two = 27 + 1 = 28.

Wait, let’s think more carefully. Let d = number reading exactly 2, t = number reading exactly 3. The extras are d + 2t = 55. We want to minimize d + t (people reading at least 2). If t = 27, then d = 55 - 54 = 1, so d + t = 28. If t = 0, d = 55, so d + t = 55.

Minimum reading at least two papers = 28.

Common Exam Variations

• Basic 2-set: find union, intersection, or neither.
• 3-set problems: fill the Venn diagram completely.
• “How many like exactly one?” — sum of the three “only” regions.
• “How many like at least two?” — sum of pairwise + triple.
• Max-min problems: find bounds on intersections.
• Complement questions: “How many do NOT like X?”

Important Identities

• n(A only) = n(A) - n(A ∩ B)
• n(exactly one of A, B) = n(A) + n(B) - 2n(A ∩ B)
• n(at least one) = n(A ∪ B) = n(U) - n(neither)
• For 3 sets: n(exactly one) = n(A) + n(B) + n(C) - 2[n(A∩B) + n(B∩C) + n(A∩C)] + 3n(A∩B∩C)
• For 3 sets: n(exactly two) = n(A∩B) + n(B∩C) + n(A∩C) - 3n(A∩B∩C)

Practice Problems

Problem 1: In a group of 80 people, 50 speak Hindi, 40 speak English, and 20 speak both. How many speak neither?

Problem 2: In a class of 150 students: 90 play cricket, 70 play football, 50 play hockey, 30 play both cricket and football, 25 play both football and hockey, 20 play both cricket and hockey, and 10 play all three. How many play none?

Problem 3: In a group of 50, everyone likes at least one of two subjects — Math or Science. 40 like Math and 35 like Science. How many like exactly one subject?

Answers

Problem 1: n(H ∪ E) = 50 + 40 - 20 = 70. Neither = 80 - 70 = 10 people.

Problem 2: n(C ∪ F ∪ H) = 90 + 70 + 50 - 30 - 25 - 20 + 10 = 145. None = 150 - 145 = 5 students.

Problem 3: n(M ∪ S) = 50 (everyone likes at least one). So 50 = 40 + 35 - n(M ∩ S) → n(M ∩ S) = 25. Exactly one subject = n(M ∪ S) - n(M ∩ S) = 50 - 25 = 25 students. (Or: only Math = 40 - 25 = 15, only Science = 35 - 25 = 10, exactly one = 15 + 10 = 25.)

Time and Work problems look complicated, but there’s a single trick that makes almost every problem a breeze: the LCM method. Once we learn it, we’ll wonder why anyone bothers with fractions at all.

The core idea is simple: Work = Rate × Time. If A can do a job in 10 days, then A’s rate is 1/10 of the job per day. When people work together, their rates ADD UP.

The Fraction Method (Traditional)

If A can finish work in a days, A does 1/a of the work per day. If A and B work together, combined rate = 1/a + 1/b per day. Time to finish together = 1/(1/a + 1/b) = ab/(a+b).

This works, but dealing with fractions gets messy fast. Enter the LCM method.

The LCM Method — The Game Changer

Instead of working with fractions, we assume the total work is a concrete number — specifically, the LCM of the given times. This turns everything into whole numbers. WAY faster.

Example 1: Basic LCM method

A can do a job in 12 days, B can do it in 18 days. How long to finish together?

LCM of 12 and 18 = 36 (this is our total work)

• A’s efficiency = 36/12 = 3 units/day
• B’s efficiency = 36/18 = 2 units/day
• Combined = 3 + 2 = 5 units/day
• Time = 36/5 = 7.2 days (or 7 days and 4.8 hours)

Compare with the fraction method: 1/12 + 1/18 = 3/36 + 2/36 = 5/36, time = 36/5 = 7.2 days. Same answer, but the LCM method feels more natural with whole numbers.

Example 2: Three workers

A, B, and C can do a job in 10, 15, and 20 days respectively. How long if all three work together?

LCM of 10, 15, 20 = 60 units.

• A = 60/10 = 6 units/day
• B = 60/15 = 4 units/day
• C = 60/20 = 3 units/day
• Combined = 6 + 4 + 3 = 13 units/day
• Time = 60/13 = 4 and 8/13 days (approximately 4.6 days)

Example 3: One worker leaves midway

A and B together can finish a job in 12 days. They work together for 8 days, then B leaves. A finishes the remaining work in 10 more days. Find how long A and B individually take.

Let A’s time = a days, B’s time = b days.

Together: 1/a + 1/b = 1/12 … (i)

In 8 days together, they finish 8/12 = 2/3 of the work. Remaining = 1/3.

A finishes 1/3 in 10 days, so A’s rate = (1/3)/10 = 1/30. Thus a = 30 days.

From (i): 1/30 + 1/b = 1/12 → 1/b = 1/12 - 1/30 = (5-2)/60 = 3/60 = 1/20. Thus b = 20 days.

Alternate Day Work

Sometimes A works on day 1, B works on day 2, A works on day 3, and so on.

The trick: find the combined work done in a 2-day cycle, then figure out how many complete cycles are needed.

Example 4: Alternate day work

A can do a job in 10 days, B in 15 days. They work on alternate days, starting with A. When will the work be finished?

LCM of 10 and 15 = 30 units.

• A’s efficiency = 3 units/day
• B’s efficiency = 2 units/day

In a 2-day cycle (A then B): 3 + 2 = 5 units done.

After 10 days (5 complete cycles): 5 × 5 = 25 units done. Remaining = 30 - 25 = 5 units.

Day 11 (A’s turn): A does 3 units. Remaining = 5 - 3 = 2 units.

Day 12 (B’s turn): B does 2 units. Remaining = 0. Done!

Work finishes on day 12.

Pipes and Cisterns

Think of it like this: inlet pipes do “positive work” (filling the tank), and outlet pipes/leaks do “negative work” (emptying the tank). Everything else is the same as time and work.

Example 5: Pipes problem

Pipe A fills a tank in 12 hours, pipe B fills it in 15 hours, and pipe C empties it in 20 hours. If all three are opened, how long to fill the tank?

LCM of 12, 15, 20 = 60 units.

• A fills: 60/12 = +5 units/hour
• B fills: 60/15 = +4 units/hour
• C empties: 60/20 = -3 units/hour

Net rate = 5 + 4 - 3 = 6 units/hour

Time = 60/6 = 10 hours

Example 6: Leak problem

A pipe fills a tank in 6 hours. Due to a leak, it takes 8 hours. How long would the leak alone take to empty the full tank?

LCM of 6 and 8 = 24 units.

• Pipe rate = 24/6 = 4 units/hour
• Net rate (pipe + leak) = 24/8 = 3 units/hour
• Leak rate = 4 - 3 = 1 unit/hour (draining)

Time for leak to empty = 24/1 = 24 hours

Efficiency-Based Problems

“A is twice as efficient as B” means A’s rate is 2 times B’s rate. If B takes 20 days, A takes 10 days.

“A is 50% more efficient than B” means A’s rate is 1.5 times B’s rate. If B takes 30 days, A takes 30/1.5 = 20 days.

Example 7: Efficiency problem

A is 60% more efficient than B. If B can do a job in 40 days, how long do they take together?

A’s time = 40/1.6 = 25 days.

LCM of 25 and 40 = 200 units.

• A = 200/25 = 8 units/day
• B = 200/40 = 5 units/day
• Combined = 13 units/day
• Time = 200/13 = 15 and 5/13 days (approximately 15.38 days)

Common Exam Variations

• A and B work together, one leaves partway — find individual times.
• Alternate day work — who finishes and on which day?
• Pipes and cisterns with multiple inlets and outlets.
• “A does half the work, then B finishes the rest.”
• Efficiency ratios: “A is x times as efficient as B.”
• “If 5 men can do it in 10 days, how many men needed for 5 days?” (Use: Men × Days = constant)
• MDH formula: M₁ × D₁ × H₁ = M₂ × D₂ × H₂ (Men × Days × Hours = constant work)

The MDH Formula

When the work stays the same:

M₁ × D₁ × H₁ = M₂ × D₂ × H₂

This means: if 5 men working 8 hours/day finish in 10 days, how many men are needed working 10 hours/day to finish in 4 days?

5 × 10 × 8 = M₂ × 4 × 10 → M₂ = 400/40 = 10 men

Practice Problems

Problem 1: A can do a job in 20 days, B in 30 days. They start together, but A leaves after 5 days. How many more days does B take to finish?

Problem 2: Two pipes can fill a tank in 10 and 12 hours. A drain pipe can empty it in 20 hours. If all three are opened, how long to fill the tank?

Problem 3: 12 men can finish a job in 16 days working 8 hours/day. How many men are needed to finish the same job in 12 days working 10 hours/day?

Answers

Problem 1: LCM of 20 and 30 = 60 units. A = 3/day, B = 2/day. In 5 days together: 5 × (3+2) = 25 units done. Remaining = 60 - 25 = 35 units. B alone: 35/2 = 17.5 more days.

Problem 2: LCM of 10, 12, 20 = 60 units. Pipe 1 = +6, Pipe 2 = +5, Drain = -3. Net = 8 units/hour. Time = 60/8 = 7.5 hours.

Problem 3: M₁D₁H₁ = M₂D₂H₂ → 12 × 16 × 8 = M₂ × 12 × 10 → M₂ = 1536/120 = 12.8, so 13 men (round up since we can’t have partial men).

Speed, distance, and time is probably the most common aptitude topic. Almost every exam has at least 2-3 questions from this chapter. The formula is dead simple — Distance = Speed × Time — but the word problems can get creative. Let’s master the patterns.

Unit conversion trick: To go from km/h to m/s, we’re converting bigger units (km, hours) to smaller ones (meters, seconds). Just remember: 1 km/h = 5/18 m/s. So 36 km/h = 36 × 5/18 = 10 m/s. For reverse, multiply by 18/5.

Average Speed — The Classic Trap

This is where exams love to trick people. Average speed is NOT the average of two speeds. It depends on whether we travel equal distances or equal times.

Equal distances at speeds a and b: Average speed = 2ab/(a + b) (harmonic mean)

Equal times at speeds a and b: Average speed = (a + b)/2 (arithmetic mean)

In exams, 90% of the time it’s equal distances, so 2ab/(a+b) is the formula we need.

Example 1: Average speed trap

A person goes from A to B at 40 km/h and returns at 60 km/h. Find the average speed for the round trip.

Since the distances are equal (A to B = B to A):

Average speed = 2 × 40 × 60 / (40 + 60) = 4800/100 = 48 km/h

NOT 50 km/h! The person spends more time at the slower speed, which pulls the average down.

Relative Speed

When two objects are moving, what matters is how fast one is moving relative to the other.

Same direction: Relative speed = S₁ - S₂ (the faster one “gains” on the slower one)

Opposite direction: Relative speed = S₁ + S₂ (they approach each other faster)

Example 2: Chase problem

A thief starts running at 10 km/h. A cop starts from the same point 30 minutes later at 12 km/h. When does the cop catch the thief?

In 30 minutes, the thief covers: 10 × 0.5 = 5 km (head start).

Relative speed (same direction) = 12 - 10 = 2 km/h.

Time for cop to close 5 km gap = 5/2 = 2.5 hours after the cop starts.

Example 3: Meeting problem

Two people start from points 100 km apart and walk toward each other at 6 km/h and 4 km/h. When and where do they meet?

Relative speed (opposite direction) = 6 + 4 = 10 km/h.

Time to meet = 100/10 = 10 hours.

Meeting point from person 1: 6 × 10 = 60 km from person 1 (and 40 km from person 2).

The Proportion Shortcut

Speed and time are inversely proportional (when distance is constant). So if speed increases by 25% (becomes 5/4 of original), time decreases to 4/5 of original.

This is incredibly useful for problems like:

Example 4: Proportion shortcut

If a person walks at 5 km/h, they’re 30 minutes late. If they walk at 6 km/h, they’re 10 minutes early. Find the distance.

Time difference = 30 + 10 = 40 minutes = 2/3 hours.

At 5 km/h, time = D/5. At 6 km/h, time = D/6.

D/5 - D/6 = 2/3 (6D - 5D)/30 = 2/3 D/30 = 2/3 D = 20 km

This pattern — “late at one speed, early at another” — always works the same way. Set up the time difference equation and solve.

Example 5: Speed ratio to time ratio

A car covers a distance in 4 hours at 60 km/h. How long would it take at 80 km/h?

Speed ratio = 60:80 = 3:4. Time ratio = 4:3 (inverse of speed). New time = 4 × (3/4) = 3 hours.

Or just: Distance = 60 × 4 = 240 km. Time = 240/80 = 3 hours. Both work!

Round Trip Problems

For a round trip (going and coming back to the starting point), the total distance is 2D and we can use the average speed formula.

Example 6: Round trip with rest

A person cycles from A to B (30 km) at 15 km/h, rests for 1 hour, then returns at 10 km/h. What is the average speed for the entire journey?

Time going = 30/15 = 2 hours. Time returning = 30/10 = 3 hours. Total time including rest = 2 + 1 + 3 = 6 hours. Total distance = 30 + 30 = 60 km.

Average speed = 60/6 = 10 km/h.

Note: If the problem asks “average speed,” rest time is included. If it asks “average speed while moving,” rest time is excluded (60/5 = 12 km/h).

Multiple Meeting Problems

When two people walk toward each other repeatedly between two points:

First meeting: They together cover 1D (where D is the distance between points). Second meeting: They together cover 3D. Third meeting: They together cover 5D.

So the ratio of distances covered at each successive meeting follows 1:3:5:7… This means the first meeting point can tell us where subsequent meetings happen.

Shortcut for “Reaches X Minutes Late / Early”

When the same distance is covered at two different speeds and we know the time difference:

Distance = (S₁ × S₂ × Time difference) / (S₁ - S₂)

where S₁ < S₂ (slower and faster speeds), and time difference is in the same unit as distance/speed.

Using Example 4: D = (5 × 6 × 2/3) / (6 - 5) = 20/1 = 20 km. Quick!

Common Exam Variations

• Average speed for round trips or multi-leg journeys.
• Relative speed: chase problems, meeting problems.
• “Late at one speed, early at another” — find the distance.
• Speed increased by x%, time reduced by y% — find distance.
• Two people meet and continue — when do they reach each other’s starting points?
• Journey with stops or changing speeds midway.

Practice Problems

Problem 1: A car covers 300 km at 60 km/h and returns at 40 km/h. Find the average speed for the entire trip.

Problem 2: Two trains start from stations 400 km apart at the same time, heading toward each other at 70 km/h and 50 km/h. After how many hours will they meet?

Problem 3: Walking at 4 km/h, a man reaches his office 10 minutes late. Walking at 5 km/h, he reaches 5 minutes early. What is the distance to his office?

Answers

Problem 1: Average speed = 2 × 60 × 40 / (60 + 40) = 4800/100 = 48 km/h.

Problem 2: Relative speed = 70 + 50 = 120 km/h. Time = 400/120 = 10/3 hours = 3 hours 20 minutes.

Problem 3: Time difference = 10 + 5 = 15 minutes = 1/4 hour. D = (4 × 5 × 1/4) / (5 - 4) = 5/1 = 5 km.

Train problems are just speed-distance-time problems with one twist: the train has length. Unlike a car (which we treat as a point), a train takes extra time to fully pass something because of its own size. Once we internalize this, every train problem becomes a simple D = S × T problem.

The one concept that makes it all click: the “distance” a train covers while passing something includes the train’s own length.

Why the Train’s Length Matters

Think of it like this: when a train passes a pole, the front of the train reaches the pole first, but the train hasn’t “fully passed” until the tail clears the pole too. So the total distance traveled = the entire length of the train.

For a platform, the front has to travel the full platform length PLUS the train’s own length to completely clear the platform.

Case 1: Crossing a Pole or Standing Person

The simplest case. Distance = train length. Speed = train’s speed.

Example 1: Crossing a pole

A train 200 meters long crosses a pole in 10 seconds. Find its speed.

Distance = 200 m. Time = 10 s.

Speed = 200/10 = 20 m/s = 20 × 18/5 = 72 km/h

Case 2: Crossing a Platform or Bridge

Distance = train length + platform length. Speed = train’s speed.

Example 2: Crossing a platform

A train 150 m long runs at 54 km/h. How long does it take to cross a platform 300 m long?

Total distance = 150 + 300 = 450 m.

Speed = 54 × 5/18 = 15 m/s.

Time = 450/15 = 30 seconds

Example 3: Finding train length

A train crosses a 200 m platform in 25 seconds and a 500 m platform in 40 seconds. Find the speed and length of the train.

Let train length = L meters and speed = S m/s.

Crossing 200 m platform: (L + 200)/S = 25 → L + 200 = 25S … (i) Crossing 500 m platform: (L + 500)/S = 40 → L + 500 = 40S … (ii)

Subtract (i) from (ii): 300 = 15S → S = 20 m/s = 72 km/h

From (i): L = 25(20) - 200 = 500 - 200 = 300 m

This is a classic pattern — two platform crossings give us two equations with two unknowns.

Case 3: Two Trains Crossing Each Other

This is where relative speed kicks in. The total distance both trains need to cover to fully pass each other is the sum of their lengths (L₁ + L₂).

Opposite directions: Relative speed = S₁ + S₂ (they’re approaching each other).

Same direction: Relative speed = S₁ - S₂ (the faster one overtakes the slower one).

Example 4: Opposite directions

Two trains of lengths 150 m and 200 m are moving toward each other at 40 km/h and 32 km/h. How long do they take to cross each other?

Total distance = 150 + 200 = 350 m.

Relative speed = 40 + 32 = 72 km/h = 72 × 5/18 = 20 m/s.

Time = 350/20 = 17.5 seconds

Example 5: Same direction

A train 250 m long traveling at 60 km/h overtakes a train 300 m long traveling at 42 km/h. How long does it take to completely pass it?

Total distance = 250 + 300 = 550 m.

Relative speed = 60 - 42 = 18 km/h = 18 × 5/18 = 5 m/s.

Time = 550/5 = 110 seconds

Notice how same-direction crossing takes way longer than opposite-direction crossing. That’s because the relative speed is much smaller.

Case 4: Man on a Train vs. Man on a Platform

This is a subtle but important distinction.

Man standing on the platform: He’s stationary, so relative speed = train’s speed. Distance = train length. (Same as crossing a pole.)

Man sitting on another train: He’s moving! So we use relative speed between the trains. Distance = length of the train that’s passing him (not his own train).

Example 6: Man on another train

Train A (200 m long) at 60 km/h passes a man sitting in Train B going in the opposite direction at 40 km/h. How long does it take?

Distance = Train A’s length only = 200 m (the man is a point, we just need Train A to pass him).

Relative speed = 60 + 40 = 100 km/h = 100 × 5/18 = 250/9 m/s.

Time = 200 / (250/9) = 200 × 9/250 = 1800/250 = 7.2 seconds

Shortcut: The Speed Ratio Method

When a train crosses a pole in time t₁ and a platform of length L in time t₂:

Train length = L × t₁ / (t₂ - t₁)

This saves us from setting up simultaneous equations.

Example: A train crosses a pole in 10 seconds and a 300 m platform in 25 seconds.

Train length = 300 × 10 / (25 - 10) = 3000/15 = 200 m

Common Traps

1. Forgetting unit conversion — speeds in km/h, lengths in meters. Always convert to consistent units (m/s is usually best).
2. Using only one train’s length when two trains cross each other — distance is ALWAYS L₁ + L₂.
3. Man on a moving train — don’t forget he has speed too. Use relative speed.
4. Same direction vs. opposite direction — getting the relative speed wrong (add vs. subtract).

Common Exam Variations

• Find speed given crossing time and length.
• Find train length from two different crossing scenarios.
• Two trains crossing each other (same or opposite direction).
• “A train passes a man in 10 seconds and a platform in 20 seconds” — find both length and speed.
• “At what speed should a train travel to cross a bridge in X seconds?”
• Time for trains to completely cross each other.

Practice Problems

Problem 1: A train 300 m long crosses a bridge 200 m long in 25 seconds. Find the speed of the train in km/h.

Problem 2: Two trains of lengths 200 m and 250 m are traveling in the same direction at 72 km/h and 36 km/h. How long does the faster train take to completely pass the slower one?

Problem 3: A train passes a standing man in 8 seconds and a 264 m long platform in 20 seconds. Find the length and speed of the train.

Answers

Problem 1: Distance = 300 + 200 = 500 m. Speed = 500/25 = 20 m/s = 20 × 18/5 = 72 km/h.

Problem 2: Total distance = 200 + 250 = 450 m. Relative speed = 72 - 36 = 36 km/h = 36 × 5/18 = 10 m/s. Time = 450/10 = 45 seconds.

Problem 3: Let train length = L, speed = S. L/S = 8 and (L + 264)/S = 20. From first equation: L = 8S. Substituting: (8S + 264)/S = 20 → 8S + 264 = 20S → 12S = 264 → S = 22 m/s. L = 8 × 22 = 176 m. Speed = 22 × 18/5 = 79.2 km/h.

Boats and streams is really just speed-distance-time with a current that either helps us or fights us. If we’re going with the flow (downstream), the water pushes us faster. If we’re going against it (upstream), it slows us down. That’s the entire concept.

Once we understand this, these problems become pure formula application. And there are only about 4 formulas to know.

Think of it like walking on a moving walkway at an airport. Walking with the walkway (downstream) — we’re faster. Walking against it (upstream) — we’re slower. Our walking speed is b, the walkway speed is s.

The Core Relationships

Let’s say a boat’s speed in still water is b km/h and the stream speed is s km/h.

• Downstream speed = b + s (the current adds to our speed)
• Upstream speed = b - s (the current subtracts from our speed)

And here’s the neat part — if we add these two:

• (b + s) + (b - s) = 2b → b = (downstream + upstream)/2

If we subtract them:

• (b + s) - (b - s) = 2s → s = (downstream - upstream)/2

So just knowing the downstream and upstream speeds gives us everything.

Example 1: Finding still water speed and stream speed

A boat goes 24 km downstream in 3 hours and 18 km upstream in 3 hours. Find the speed in still water and the speed of the stream.

Downstream speed = 24/3 = 8 km/h. Upstream speed = 18/3 = 6 km/h.

Speed in still water (b) = (8 + 6)/2 = 7 km/h Speed of stream (s) = (8 - 6)/2 = 1 km/h

Example 2: Basic downstream/upstream

A man can row at 8 km/h in still water. The stream flows at 2 km/h. How long to go 30 km downstream and return?

Downstream speed = 8 + 2 = 10 km/h. Time = 30/10 = 3 hours. Upstream speed = 8 - 2 = 6 km/h. Time = 30/6 = 5 hours.

Total time = 3 + 5 = 8 hours

Example 3: Finding the stream speed

A boat takes 6 hours to go 36 km downstream and 9 hours to return. Find the speed of the stream.

Downstream speed = 36/6 = 6 km/h. Upstream speed = 36/9 = 4 km/h.

Stream speed = (6 - 4)/2 = 1 km/h

Round Trip Problems

Round trips are very common. We go downstream some distance and come back upstream the same distance.

Total time for round trip = D/(b+s) + D/(b-s)

This can be simplified to: T = 2Db / (b² - s²)

Example 4: Round trip with total time

A boat’s speed in still water is 10 km/h. The stream speed is 2 km/h. If a round trip takes 10 hours, find the one-way distance.

D/(10+2) + D/(10-2) = 10 D/12 + D/8 = 10 (2D + 3D)/24 = 10 5D = 240 D = 48 km

Example 5: Speed of current from round trip

A man rows 40 km upstream and 55 km downstream in 13 hours. He also rows 30 km upstream and 44 km downstream in 10 hours. Find his speed in still water and the speed of the current.

Let upstream speed = u, downstream speed = d.

40/u + 55/d = 13 … (i) 30/u + 44/d = 10 … (ii)

Let’s substitute 1/u = x and 1/d = y: 40x + 55y = 13 … (i) 30x + 44y = 10 … (ii)

Multiply (i) by 3 and (ii) by 4: 120x + 165y = 39 120x + 176y = 40

Subtract: 11y = 1 → y = 1/11 → d = 11 km/h. From (ii): 30x + 44/11 = 10 → 30x + 4 = 10 → x = 1/5 → u = 5 km/h.

Still water speed (b) = (11 + 5)/2 = 8 km/h Stream speed (s) = (11 - 5)/2 = 3 km/h

The “Time Ratio” Shortcut

If a boat takes time T₁ downstream and T₂ upstream for the same distance:

b/s = (T₂ + T₁)/(T₂ - T₁)

This directly gives us the ratio of boat speed to stream speed without finding individual speeds.

Example 6: Using the time ratio

A boat takes 2 hours to go downstream and 4 hours to come back (same distance). Find the ratio of boat speed to stream speed.

b/s = (4 + 2)/(4 - 2) = 6/2 = 3:1

So the boat is 3 times faster than the current. If the stream is at x km/h, the boat is at 3x km/h in still water.

Problems with Multiple Boats or Objects

Sometimes we get problems about a hat or object falling into the water.

Key insight: The object floats at the stream’s speed. If a boat drops something and turns back to find it, the stream carries both the boat and the object. The time to go back and retrieve = the time that has passed (because relative to the water, the boat’s speed is the same in both directions).

Example 7: The floating object

A boat is going upstream. A bottle falls off the boat. After 30 minutes, the boatman notices and turns back. He catches the bottle 2 km from where it fell. Find the speed of the stream.

The bottle floats downstream at stream speed for the entire time (30 min going away + time to come back).

Since the boat’s speed relative to water is the same both ways, the time to return to the bottle = 30 minutes. So total time bottle is in water = 60 minutes = 1 hour.

In 1 hour, the bottle traveled 2 km downstream. Stream speed = 2/1 = 2 km/h

The beauty here: we don’t even need to know the boat’s speed!

Common Exam Variations

• Basic: find downstream/upstream speed, time for a journey.
• Finding still water speed and stream speed from two journey details.
• Round trip problems: total time given, find distance.
• Rate of current problems: “if there were no current, the trip would take X hours less.”
• Floating object problems.
• “A man can row X km downstream in the same time as Y km upstream” — find speed ratio.

Common Traps

1. Upstream speed can’t be negative — if stream speed > boat speed, the boat can’t go upstream at all.
2. Average speed for a round trip is NOT (downstream + upstream)/2. Use 2(b+s)(b-s) / 2b = (b²-s²)/b.
3. Confusing “speed in still water” with “downstream speed” — the problem might give one but we need the other.

Practice Problems

Problem 1: A man can row 6 km/h in still water. If the river flows at 2 km/h, how long does he take to row 16 km upstream and return?

Problem 2: A boat covers 36 km downstream in 4 hours and 24 km upstream in 4 hours. Find the speed of the boat in still water.

Problem 3: A boat takes 3 hours to go 15 km downstream and 5 hours to go 15 km upstream. Find the speed of the current.

Answers

Problem 1: Upstream speed = 6 - 2 = 4 km/h. Time up = 16/4 = 4 hours. Downstream speed = 6 + 2 = 8 km/h. Time down = 16/8 = 2 hours. Total = 6 hours.

Problem 2: Downstream speed = 36/4 = 9 km/h. Upstream speed = 24/4 = 6 km/h. Still water speed = (9 + 6)/2 = 7.5 km/h.

Problem 3: Downstream speed = 15/3 = 5 km/h. Upstream speed = 15/5 = 3 km/h. Current speed = (5 - 3)/2 = 1 km/h.

Races and circular track problems use the same speed-distance-time fundamentals, but with a specific vocabulary and some neat patterns. Once we learn the language of races (“A beats B by 20 meters,” “A gives B a 5 second head start”), the math is straightforward.

Race Terminology

Let’s get the vocabulary down first.

Linear Race Problems

Example 1: Basic race

In a 100 m race, A beats B by 20 m. What is the ratio of their speeds?

When A finishes 100 m, B has run only 80 m. They ran for the same time.

Speed ratio = Distance ratio = A : B = 100 : 80 = 5 : 4

Example 2: Chaining results

In a 100 m race, A beats B by 10 m and B beats C by 20 m. By how much does A beat C?

When A runs 100 m, B runs 90 m. When B runs 100 m, C runs 80 m.

When B runs 90 m, C runs? → 80/100 × 90 = 72 m.

So when A finishes 100 m, C has run 72 m. A beats C by 28 m.

The key insight: we can’t just add 10 + 20 = 30. We have to scale B’s advantage over C to the actual distance B ran.

Example 3: Head start to make it a dead heat

A beats B by 30 m in a 200 m race. What head start should A give B so they finish together?

Speed ratio = 200 : 170 = 20 : 17.

For a dead heat, B needs to cover less distance. If A gives B a head start of x meters, then A runs 200 m while B runs (200 - x) m.

For them to finish together: time must be equal. 200/Speed_A = (200 - x)/Speed_B Since speed ratio is 20:17: 200/20 = (200 - x)/17 10 = (200 - x)/17 200 - x = 170 x = 30 m head start

This makes sense — A beats B by exactly 30 m, so giving B a 30 m head start makes it a tie.

Example 4: Time-based head start

A runs at 10 m/s and B runs at 8 m/s. In a 200 m race, how much head start in time should A give B for a dead heat?

A’s time = 200/10 = 20 seconds. B’s time = 200/8 = 25 seconds.

A needs to give B a head start of 25 - 20 = 5 seconds.

If B starts 5 seconds early, both finish at the same time.

Circular Track Problems

Circular tracks add a new dimension: the runners go around and around, and we need to figure out when and where they meet. This is where relative speed and LCM become our best friends.

Same Direction — The Faster Catches the Slower

In simple language, when two people run in the same direction on a circular track, the faster person “laps” the slower one. The first meeting happens when the faster runner gains exactly one full lap on the slower runner.

Example 5: Same direction on circular track

A and B start at the same point on a 400 m circular track. A runs at 5 m/s and B at 3 m/s, both in the same direction. When do they first meet?

Relative speed = 5 - 3 = 2 m/s.

A needs to gain 400 m on B (one full lap) to meet again.

Time = 400/2 = 200 seconds

Opposite Direction — They Run Toward Each Other

When running in opposite directions, they’re essentially approaching each other on the track. The first meeting happens when they’ve collectively covered one full lap.

Example 6: Opposite direction on circular track

Same setup, but A and B run in opposite directions. When do they first meet?

Relative speed = 5 + 3 = 8 m/s.

Time = 400/8 = 50 seconds

Much faster meeting time — makes sense since they’re running toward each other.

Meeting at the Starting Point

This is different from “when do they meet?” — this asks when both are simultaneously back at the starting point.

Each person returns to the start after completing full laps. A returns every 400/5 = 80 seconds. B returns every 400/3 = 133.33 seconds.

They’re both at the start at time = LCM(80, 400/3).

For clean numbers, let’s use a simpler example.

Example 7: Meeting at starting point

A and B run around a 600 m circular track at 10 m/s and 8 m/s. When do they first meet at the starting point?

Time for A to complete one lap = 600/10 = 60 seconds. Time for B to complete one lap = 600/8 = 75 seconds.

First time both are at the start = LCM(60, 75) = 300 seconds = 5 minutes.

At 300 seconds: A has done 300/60 = 5 laps, B has done 300/75 = 4 laps. Both are at the start!

Three or More Runners

For three runners A, B, C:

First meeting of all three at the same point on the track: Find when each pair meets, then take the LCM of those times. Or more practically, find LCM of individual lap times for “meeting at starting point.”

First time all three meet anywhere: Find when A&B meet (time T₁) and when A&C meet (time T₂). All three meet at LCM(T₁, T₂).

Example 8: Three runners

A, B, C run on a 300 m track at 6 m/s, 4 m/s, and 3 m/s in the same direction. When do all three first meet?

A&B first meet: 300/(6-4) = 150 seconds. A&C first meet: 300/(6-3) = 100 seconds.

All three meet: LCM(150, 100) = 300 seconds.

Verify: In 300 sec, A runs 1800 m (6 laps), B runs 1200 m (4 laps), C runs 900 m (3 laps). A is 2 laps ahead of B (same spot) and 3 laps ahead of C (same spot). All at the same point!

Number of Distinct Meeting Points

When two runners go in opposite directions with speed ratio a:b (in simplest form), they meet at (a + b) distinct points on the track as they keep running.

When going in the same direction with speed ratio a:b, they meet at |a - b| distinct points.

Common Exam Variations

• “A beats B by x meters” — find speed ratio.
• Chaining: A beats B, B beats C, how much does A beat C?
• Head start (time or distance) for a dead heat.
• Circular track: first meeting time (same/opposite direction).
• Circular track: meeting at starting point (LCM method).
• Three runners on a circular track.
• Number of distinct meeting points.

Practice Problems

Problem 1: In a 1 km race, A beats B by 50 m and B beats C by 80 m. By how many meters does A beat C?

Problem 2: Two runners start from the same point on a 500 m circular track running in opposite directions at 4 m/s and 6 m/s. After how many seconds do they meet for the 3rd time?

Problem 3: A and B run around a 400 m track at 8 m/s and 5 m/s. When do they first meet at the starting point together?

Answers

Problem 1: When A finishes 1000 m, B has run 950 m. When B runs 1000 m, C runs 920 m. When B runs 950 m, C runs 920/1000 × 950 = 874 m. A beats C by 1000 - 874 = 126 m.

Problem 2: Relative speed = 4 + 6 = 10 m/s. Time for each meeting = 500/10 = 50 seconds. 3rd meeting = 3 × 50 = 150 seconds.

Problem 3: A’s lap time = 400/8 = 50 seconds. B’s lap time = 400/5 = 80 seconds. LCM(50, 80) = 400 seconds. (A completes 8 laps, B completes 5 laps.)

Clock problems look unique, but they’re secretly just relative speed problems in disguise. Two hands (hour and minute) are moving in the same direction on a circular track (the clock face). The minute hand is faster and keeps lapping the hour hand. Once we see it this way, everything falls into place.

Speed of the Clock Hands

Let’s break this down. The minute hand completes one full circle (360°) in 60 minutes, so it moves at 6° per minute. The hour hand completes one full circle in 12 hours = 720 minutes, so it moves at 0.5° per minute. The minute hand gains on the hour hand at 5.5° per minute — this is the relative speed.

The Clock Angle Diagram

Each number on the clock represents 30° (since 360°/12 = 30°). So at 3 o’clock, the angle is 3 × 30° = 90°. At 6 o’clock, it’s 180°. This is the foundation.

The Master Formula

For any time H hours and M minutes, the angle between the hands is:

Angle = |30H - 5.5M|

If the result is more than 180°, subtract from 360° (since we want the smaller angle).

In simple language: 30H gives us where the hour hand would be if it were exactly on the hour. Then 5.5M adjusts for the fact that the minute hand is at M × 6° and the hour hand has moved an extra M × 0.5° since the hour.

Why does this work? At H:M, the hour hand is at (30H + 0.5M)° and the minute hand is at 6M°. The difference is |30H + 0.5M - 6M| = |30H - 5.5M|.

Example 1: Finding the angle

Find the angle between clock hands at 3:20.

Angle = |30(3) - 5.5(20)| = |90 - 110| = |-20| = 20°

Example 2: Another angle calculation

Find the angle at 7:45.

Angle = |30(7) - 5.5(45)| = |210 - 247.5| = |-37.5| = 37.5°

Since 37.5° < 180°, the answer is 37.5°.

Example 3: Angle greater than 180

Find the angle at 9:10.

Angle = |30(9) - 5.5(10)| = |270 - 55| = 215°

Since 215° > 180°, the smaller angle = 360 - 215 = 145°

“At What Time” Problems

These flip the formula around — we know the angle and need to find the time.

Example 4: When are hands at 90°?

At what time between 4 and 5 will the hands be at a right angle?

We need |30(4) - 5.5M| = 90.

Case 1: 120 - 5.5M = 90 → 5.5M = 30 → M = 30/5.5 = 60/11 = 5 and 5/11 minutes past 4.

Case 2: 120 - 5.5M = -90 → 5.5M = 210 → M = 210/5.5 = 420/11 = 38 and 2/11 minutes past 4.

So the hands are at right angles at 4:05 5/11 and 4:38 2/11.

Example 5: When do hands coincide?

At what time between 7 and 8 do the hands overlap?

We need |30(7) - 5.5M| = 0.

210 - 5.5M = 0 → M = 210/5.5 = 420/11 = 38 and 2/11 minutes past 7.

So the hands coincide at 7:38 2/11 (approximately 7:38:11).

Important Counts

How many times do the hands form certain angles in 12 hours?

Why 11 and not 12? Because between 11 and 1, the hands coincide only once (at 12:00), not twice. Same for the straight line — between 5 and 7, the hands form 180° only once (at 6:00), not twice.

In 24 hours: Double everything. Coincidences = 22 times. Right angles = 44 times.

Interval between consecutive coincidences: 12 hours / 11 = 12/11 hours = 1 hour 5 minutes 27.27 seconds ≈ 65 5/11 minutes.

Faulty Clock Problems

A faulty clock runs too fast or too slow. We need to figure out what the actual time is when the faulty clock shows a certain time, or vice versa.

Key concept: If a clock gains 5 minutes per hour, then in a “true” hour, it shows 65 minutes. If it loses 5 minutes per hour, it shows 55 minutes.

Example 6: Fast clock

A clock gains 5 minutes every hour. If it was set correctly at 12 noon, what is the true time when it shows 6:00 PM?

The clock shows 6 hours = 360 minutes have passed (on the clock).

In every real 60 minutes, the clock shows 65 minutes.

Real time elapsed = 360 × (60/65) = 360 × 12/13 = 4320/13 ≈ 332.3 minutes ≈ 5 hours 32 minutes 18 seconds.

True time ≈ 5:32 PM (when the clock shows 6:00 PM).

Example 7: Slow clock

A clock loses 3 minutes every hour. It is set right at 8 AM. What does it show when the actual time is 8 PM?

Real time elapsed = 12 hours = 720 minutes.

In every 60 real minutes, the clock shows 57 minutes.

Clock shows: 720 × (57/60) = 720 × 19/20 = 684 minutes = 11 hours 24 minutes.

Clock shows 7:24 PM when actual time is 8 PM.

Mirror/Reflection Problems

When we see a clock in a mirror, the time appears reversed. To find the actual time from a mirror image:

Actual time = 11:60 - Mirror time (or equivalently, 12:00 - mirror time)

If mirror shows 2:25, actual time = 11:60 - 2:25 = 9:35. If mirror shows 8:40, actual time = 11:60 - 8:40 = 3:20.

This works because a mirror reversal on a clock is like subtracting from 12:00.

Common Exam Variations

• Find the angle at a given time.
• Find the time when hands form a specific angle.
• How many times do hands coincide / form right angles in X hours?
• Faulty clock: find true time or clock time.
• Mirror reflection: find actual time from mirror image.
• “Between X and Y o’clock, at what time are the hands at angle θ?”

Practice Problems

Problem 1: Find the angle between the clock hands at 5:30.

Problem 2: At what time between 3 and 4 o’clock do the minute and hour hands coincide?

Problem 3: A clock gains 4 minutes every hour. It is set correctly at 10 AM. What is the true time when the clock shows 4 PM?

Answers

Problem 1: Angle = |30(5) - 5.5(30)| = |150 - 165| = 15°. The answer is 15°.

Problem 2: We need |30(3) - 5.5M| = 0 → 90 = 5.5M → M = 90/5.5 = 180/11 = 16 and 4/11 minutes. The hands coincide at 3:16 4/11 (approximately 3:16:22).

Problem 3: The clock shows 6 hours = 360 minutes. In every real 60 minutes, the clock advances 64 minutes. Real time = 360 × 60/64 = 360 × 15/16 = 337.5 minutes = 5 hours 37.5 minutes. True time = 3:37:30 PM (about 3:37 PM).

Welcome to geometry — this is where aptitude gets visual. Lines, angles, and triangles form the backbone of every geometry question we’ll ever see. The good news? Most of it is pattern recognition. Once we know the properties, we can spot the answer almost instantly. Let’s build up from the basics.

Types of Angles

• Acute angle: Less than 90° (think: a cute little angle)
• Right angle: Exactly 90° (the square corner)
• Obtuse angle: Between 90° and 180° (the fat angle)
• Straight angle: Exactly 180° (a straight line)
• Reflex angle: Between 180° and 360° (the “wrap around” angle)

Quick fact: Complementary angles add up to 90°. Supplementary angles add up to 180°. These definitions come up all the time.

Parallel Lines and a Transversal

When a line (transversal) cuts two parallel lines, it creates 8 angles — but we really only need to know 2 values, because angles come in matching sets.

Here are the angle relationships to memorize:

• Corresponding angles are equal: a = e, b = f, c = g, d = h (same position at each intersection)
• Alternate interior angles are equal: c = e, d = f (opposite sides, between the lines)
• Alternate exterior angles are equal: a = h, b = g (opposite sides, outside the lines)
• Co-interior (same-side interior) angles are supplementary: c + f = 180°, d + e = 180°

The shortcut: If we know just ONE angle, we know all eight. Angles are either equal to it or supplementary (add up to 180°).

Triangle Basics

Angle Sum Property

The angles inside any triangle always add up to exactly 180°. No exceptions. This is probably the single most useful fact in geometry.

Exterior Angle Theorem

An exterior angle of a triangle equals the sum of the two non-adjacent interior angles.

In simple language: if we extend one side of a triangle, the angle formed outside equals the sum of the two “far away” interior angles. This is a massive time-saver.

Triangle Inequality

For three sides to form a triangle, the sum of any two sides must be greater than the third side. In simple language: the two shorter sides together must be longer than the longest side. If not, no triangle.

Types of Triangles

By sides:

• Equilateral: All 3 sides equal, all angles = 60°
• Isosceles: 2 sides equal, base angles are equal
• Scalene: All sides different, all angles different

By angles:

• Acute: All angles < 90°
• Right-angled: One angle = 90°
• Obtuse: One angle > 90°

Pythagoras Theorem

The Pythagoras theorem only works for right-angled triangles. The hypotenuse (the side opposite the right angle) is always the longest side.

Pythagorean Triplets — Memorize These!

These sets of three numbers always form a right triangle. Knowing them saves us from doing any calculation:

Trick: Any multiple of a Pythagorean triplet is also a Pythagorean triplet. So if we see sides 15, 20, 25 — that’s just 3-4-5 multiplied by 5. Instant right triangle!

Identifying the triangle type from sides:

• If c² = a² + b² → Right triangle
• If c² < a² + b² → Acute triangle
• If c² > a² + b² → Obtuse triangle

Congruence of Triangles

Two triangles are congruent when they’re exactly the same shape AND size — one is a perfect copy of the other. Think of it like photocopying a triangle.

The conditions for congruence:

• SSS (Side-Side-Side): All three sides are equal
• SAS (Side-Angle-Side): Two sides and the included angle are equal
• ASA (Angle-Side-Angle): Two angles and the included side are equal
• AAS (Angle-Angle-Side): Two angles and a non-included side are equal
• RHS (Right-Hypotenuse-Side): For right triangles — hypotenuse and one side are equal

Watch out: SSA (Side-Side-Angle) is NOT a valid congruence rule (the ambiguous case). AAA is also not valid — it gives similarity, not congruence.

Similarity of Triangles

Two triangles are similar when they have the same shape but can be different sizes. Think of it like zooming in or out on a triangle.

The conditions for similarity:

• AA (Angle-Angle): Two angles are equal (third is automatically equal since angles sum to 180°)
• SAS (Side-Angle-Side): Two pairs of sides are proportional and the included angle is equal
• SSS (Side-Side-Side): All three pairs of sides are proportional

Key property: If two triangles are similar with sides in ratio k, then:

• Corresponding altitudes, medians, angle bisectors are also in ratio k
• Areas are in ratio k²
• This area ratio (k²) is the most tested property in exams!

Area of a Triangle

We have three main formulas — use whichever fits the given information:

1. Base-Height formula: Area = ½ × base × height (most common)

2. Heron’s formula (when all 3 sides are known):

• First find semi-perimeter: s = (a + b + c) / 2
• Then Area = √[s(s-a)(s-b)(s-c)]

3. Two sides and included angle: Area = ½ × a × b × sin(C)

Special Triangle Areas

• Equilateral triangle with side a: Area = (√3/4)a², Height = (√3/2)a
• Right triangle with legs a and b: Area = ½ × a × b (legs are base and height!)
• Isosceles triangle with equal sides a and base b: Height = √(a² - b²/4)

Worked Examples

Example 1: Angle Finding

In a triangle, two angles are 45° and 72°. Find the third angle.

Sum of angles = 180° Third angle = 180° - 45° - 72° = 63°

Example 2: Exterior Angle

One exterior angle of a triangle is 120°. If one of the non-adjacent interior angles is 50°, find the other non-adjacent interior angle.

Exterior angle = sum of non-adjacent interior angles 120° = 50° + other angle Other angle = 120° - 50° = 70°

And the adjacent interior angle = 180° - 120° = 60°. Quick check: 50° + 70° + 60° = 180° ✓

Example 3: Pythagoras

A ladder 13 m long is placed against a wall. The foot of the ladder is 5 m from the wall. How high up the wall does the ladder reach?

We recognize 5 and 13 — this is the 5-12-13 triplet! Height = 12 m

Without recognizing the triplet: h² = 13² - 5² = 169 - 25 = 144, so h = 12 m.

Example 4: Heron’s Formula

Find the area of a triangle with sides 7, 8, and 9.

Semi-perimeter s = (7 + 8 + 9) / 2 = 12 Area = √[12 × (12-7) × (12-8) × (12-9)] Area = √[12 × 5 × 4 × 3] Area = √720 Area = √(144 × 5) Area = 12√5 ≈ 26.83 sq units

Example 5: Similar Triangles

Two similar triangles have sides in the ratio 3:5. If the area of the smaller triangle is 36 cm², find the area of the larger triangle.

For similar triangles, area ratio = (side ratio)² Area ratio = (3/5)² = 9/25 36/Area₂ = 9/25 Area₂ = 36 × 25/9 = 100 cm²

Common Exam Variations

1. Finding missing angles using angle sum property + parallel line properties combined
2. Identifying right triangles from given sides (check if they form a Pythagorean triplet or its multiple)
3. Area of shaded regions — usually involves subtracting one triangle’s area from another
4. Similar triangles to find unknown sides when two triangles share an angle (very common with overlapping triangles)
5. Median/centroid problems — the centroid divides each median in ratio 2:1 from vertex

Practice Problems

Q1: In a triangle PQR, the exterior angle at R is 110°. If ∠P = 60°, find ∠Q.

Q2: The sides of a triangle are 6, 8, and 10. Find its area and determine what type of triangle it is.

Q3: Two similar triangles have areas 64 cm² and 100 cm². If the longest side of the smaller triangle is 16 cm, find the longest side of the larger triangle.

Answers

A1: Exterior angle at R = ∠P + ∠Q → 110° = 60° + ∠Q → ∠Q = 50°. Check: ∠R = 180° - 110° = 70°, and 60° + 50° + 70° = 180° ✓

A2: Check: 6² + 8² = 36 + 64 = 100 = 10². It’s a right triangle (the 3-4-5 triplet × 2). Area = ½ × 6 × 8 = 24 cm².

A3: Area ratio = 64/100 = (side ratio)². Side ratio = √(64/100) = 8/10 = 4/5. Longest side of larger = 16 × 5/4 = 20 cm.

Circles are everywhere in aptitude exams — from finding areas of shaded regions to tangent-based problems. The trick with circles is that there are a handful of theorems, and once we nail those down, every problem becomes a plug-and-solve situation. Let’s get into it.

Circle Basics

A quick refresher on the terminology:

• Radius (r): Distance from center to any point on the circle
• Diameter (d): Distance across the circle through the center (d = 2r)
• Chord: A line segment connecting any two points on the circle (diameter is the longest chord)
• Arc: A part of the circumference (minor arc < semicircle, major arc > semicircle)
• Sector: The “pizza slice” area between two radii and an arc
• Segment: The area between a chord and the arc it cuts off

Key Formulas

In simple language: for arc length and sector area, we just take that fraction of the full circle. If the angle is 60°, that’s 60/360 = 1/6 of the full circumference or area.

Circle Theorems — The Must-Know Set

1. Perpendicular from Center to Chord

A perpendicular drawn from the center of a circle to a chord bisects the chord (cuts it in half). This is incredibly useful — whenever we see a chord problem, we drop a perpendicular from the center and form a right triangle.

2. Equal Chords

Equal chords are equidistant from the center. The only difference is, the converse is also true — chords equidistant from the center are equal.

3. Tangent Properties

• A tangent is always perpendicular to the radius at the point of contact (this creates a right angle — use Pythagoras!)
• Two tangents drawn from an external point are equal in length
• The angle between two tangents from an external point and the angle subtended at the center are supplementary (add up to 180°)

4. Inscribed Angle Theorem (The Star Theorem)

The angle subtended by an arc at the center is twice the angle subtended at any point on the remaining circle.

In simple language: if we draw an angle from an arc to the center, and another angle from the same arc to a point on the circle, the center angle is exactly double. So the inscribed angle = ½ × central angle.

Special case: An angle in a semicircle is always 90° (since the central angle is 180°, and half of that is 90°). This is a classic exam question.

5. Cyclic Quadrilateral

A cyclic quadrilateral is a four-sided figure where all four vertices lie on a circle.

Key property: Opposite angles of a cyclic quadrilateral add up to 180°.

So if ∠A + ∠C = 180° and ∠B + ∠D = 180°, it’s cyclic.

6. Angles in the Same Segment

Angles subtended by the same arc (or chord) at points on the same side of the chord are equal. Think of it like this — no matter where we put the point on the same arc, the angle stays the same.

Shortcut Methods and Tricks

Trick 1: The right triangle in chord problems Whenever a chord problem gives us the chord length and distance from center (or radius), immediately draw the perpendicular from center to chord. We get a right triangle with:

• Hypotenuse = radius
• One leg = half the chord length
• Other leg = distance from center to chord

Trick 2: Two tangent problems When two tangents are drawn from an external point, the figure formed (two tangents + two radii) creates a kite shape. The tangent lengths are equal, and we can use this symmetry to solve quickly.

Trick 3: Angle in semicircle = 90° If a question says “diameter subtends an angle at a point on the circle,” the answer is always 90°. Don’t even calculate.

Trick 4: Quick sector calculations For common angles, memorize these fractions:

• 60° = 1/6 of circle
• 90° = 1/4 of circle
• 120° = 1/3 of circle
• 180° = 1/2 of circle

Worked Examples

Example 1: Arc Length and Sector Area

Find the arc length and sector area for a circle with radius 14 cm and central angle 90°.

Arc length = (90/360) × 2π(14) = (1/4) × 28π = 7π = 22 cm (using π = 22/7)

Sector area = (90/360) × π(14)² = (1/4) × 196π = 49π = 154 cm²

Example 2: Chord and Distance

A chord of length 24 cm is drawn in a circle of radius 13 cm. Find the distance of the chord from the center.

Drop a perpendicular from center to chord. It bisects the chord, so half-chord = 12 cm.

Now we have a right triangle: radius = 13, half-chord = 12, distance = ?

distance² = 13² - 12² = 169 - 144 = 25

Distance = 5 cm

Notice: 5-12-13 is a Pythagorean triplet! Spotting these saves calculation time.

Example 3: Tangent Length

From a point 17 cm away from the center of a circle of radius 8 cm, a tangent is drawn. Find the tangent length.

The tangent is perpendicular to the radius at the point of contact, so we have a right triangle:

• Hypotenuse = distance from external point to center = 17
• One leg = radius = 8
• Other leg = tangent length = ?

Tangent² = 17² - 8² = 289 - 64 = 225

Tangent length = 15 cm

(Another triplet: 8-15-17!)

Example 4: Inscribed Angle

A chord subtends an angle of 50° at a point on the major arc. Find the angle it subtends at the center.

By the inscribed angle theorem, the central angle is twice the inscribed angle.

Central angle = 2 × 50° = 100°

Example 5: Cyclic Quadrilateral

In a cyclic quadrilateral ABCD, ∠A = 70° and ∠B = 110°. Find ∠C and ∠D.

Opposite angles sum to 180°:

• ∠C = 180° - ∠A = 180° - 70° = 110°
• ∠D = 180° - ∠B = 180° - 110° = 70°

Common Exam Variations

1. Finding the area of a segment — sector area minus triangle area. The triangle is formed by two radii and the chord.
2. Two circles touching — externally (distance between centers = R + r) or internally (distance = R - r)
3. Common tangent problems — direct common tangents (don’t cross the line joining centers) vs transverse tangents (do cross it)
4. Shaded region between two concentric circles (ring/annulus area = π(R² - r²))
5. Semicircle problems — using the “angle in semicircle = 90°” property

Practice Problems

Q1: A sector has a central angle of 120° and radius 21 cm. Find its area and the length of the arc. (Use π = 22/7)

Q2: Two tangents PA and PB are drawn from a point P to a circle with center O and radius 5 cm. If OP = 13 cm, find the length of each tangent and the angle ∠APB.

Q3: In a circle of radius 10 cm, a chord is at a distance of 6 cm from the center. Find the length of the chord.

Answers

A1: Area = (120/360) × (22/7) × 21² = (1/3) × (22/7) × 441 = (1/3) × 1386 = 462 cm². Arc length = (120/360) × 2 × (22/7) × 21 = (1/3) × 132 = 44 cm.

A2: Tangent is perpendicular to radius, so PA² = OP² - OA² = 169 - 25 = 144. PA = 12 cm. In triangle OAP: sin(∠OPA) = 5/13, so ∠OPA = sin⁻¹(5/13) ≈ 22.6°. Since the figure is symmetric, ∠APB = 2 × 22.6° ≈ 45.2°. (Or we can note: ∠AOB + ∠APB = 180°, and tan(∠AOB/2) = 12/5, giving the same result.)

A3: Half-chord² = radius² - distance² = 100 - 36 = 64. Half-chord = 8 cm. Full chord = 16 cm.

Quadrilaterals (four-sided figures) and polygons show up constantly in aptitude exams. The secret here is understanding the hierarchy — a square is a special rectangle, which is a special parallelogram. Once we see these relationships, properties cascade down and we don’t need to memorize everything separately.

The Quadrilateral Family Tree

The key insight: as we go down the tree, each shape has all the properties of its parent PLUS some extra ones. A square has every property of a rectangle AND every property of a rhombus.

Properties of Each Quadrilateral

Parallelogram

• Opposite sides are equal AND parallel
• Opposite angles are equal
• Adjacent angles are supplementary (add to 180°)
• Diagonals bisect each other (cut each other in half)
• Area = base × height

Rectangle (Parallelogram + all right angles)

• All angles = 90°
• Diagonals are equal and bisect each other
• Area = length × breadth
• Diagonal = √(l² + b²)

Rhombus (Parallelogram + all sides equal)

• All sides are equal
• Diagonals bisect each other at right angles (90°)
• Diagonals bisect the vertex angles
• Area = ½ × d₁ × d₂ (half the product of diagonals)

Square (Rectangle + Rhombus = everything)

• All sides equal, all angles = 90°
• Diagonals are equal, bisect at right angles, bisect vertex angles
• Area = side² = ½ × diagonal²
• Diagonal = side × √2

Trapezium (only one pair of parallel sides)

• One pair of parallel sides (called parallel sides or bases)
• Area = ½ × (sum of parallel sides) × height = ½(a + b) × h
• Isosceles trapezium: non-parallel sides are equal, diagonals are equal

Kite

• Two pairs of adjacent sides are equal
• One diagonal bisects the other at right angles
• Area = ½ × d₁ × d₂

Key Formulas

Diagonal Properties Quick Reference

Memory trick: As we go from parallelogram to square, diagonals gain more and more properties. Rectangle adds “equal,” rhombus adds “perpendicular,” and square has both.

Polygons

A polygon is any closed figure with straight sides. Quadrilaterals are 4-sided polygons. Let’s look at the general rules.

Angle Sum of a Polygon

Interior angle sum = (n - 2) × 180°, where n is the number of sides.

This makes sense: a triangle (3 sides) has 180°, a quadrilateral (4 sides) has 360°, a pentagon has 540°, and so on. We’re basically dividing the polygon into (n-2) triangles.

Regular Polygons

A regular polygon has all sides equal AND all angles equal.

• Each interior angle = (n - 2) × 180° / n
• Each exterior angle = 360° / n
• Number of diagonals = n(n - 3) / 2

Important fact: The sum of exterior angles of ANY convex polygon is always 360°, regardless of the number of sides.

Shortcut Methods and Tricks

Trick 1: Finding shape from diagonals If a question describes a quadrilateral by its diagonal properties, use the table above. “Diagonals bisect at right angles but aren’t equal” → Rhombus. “Diagonals are equal and bisect each other” → Rectangle.

Trick 2: Rhombus area shortcut If we know the diagonals of a rhombus, we can find the side using: side = ½√(d₁² + d₂²). This is because the diagonals form 4 right triangles with legs d₁/2 and d₂/2.

Trick 3: Finding exterior angle first For regular polygons, always find the exterior angle first (360°/n) — it’s simpler. Then interior = 180° - exterior.

Trick 4: Reverse problem — finding n from angle If given an interior angle and asked to find the polygon: exterior angle = 180° - interior angle, then n = 360° / exterior angle.

Worked Examples

Example 1: Parallelogram Area

A parallelogram has a base of 12 cm and a height of 8 cm. Find its area.

Area = base × height = 12 × 8 = 96 cm²

Note: the slant side is NOT the height! The height is the perpendicular distance between the parallel sides.

Example 2: Rhombus from Diagonals

The diagonals of a rhombus are 16 cm and 12 cm. Find the area and the side length.

Area = ½ × d₁ × d₂ = ½ × 16 × 12 = 96 cm²

Side = ½√(16² + 12²) = ½√(256 + 144) = ½√400 = ½ × 20 = 10 cm

(We used the 3-4-5 triplet here — half-diagonals are 8 and 6, so side = 10. Quick!)

Example 3: Trapezium Area

A trapezium has parallel sides of 10 cm and 16 cm, and the height between them is 9 cm. Find the area.

Area = ½ × (a + b) × h = ½ × (10 + 16) × 9 = ½ × 26 × 9 = 117 cm²

Example 4: Finding the Polygon

Each interior angle of a regular polygon is 140°. How many sides does it have?

Exterior angle = 180° - 140° = 40° Number of sides = 360° / 40° = 9 sides (nonagon)

Example 5: Number of Diagonals

How many diagonals does a decagon (10-sided polygon) have?

Diagonals = n(n - 3)/2 = 10 × 7 / 2 = 35 diagonals

Common Exam Variations

1. “Which quadrilateral…” — identifying from properties (very common in SSC/Banking exams)
2. Finding area from diagonals — especially rhombus and kite
3. Perimeter of regular polygon — given one side or the interior angle
4. Square inscribed in circle (diagonal = diameter) or circle inscribed in square (diameter = side)
5. Angle-based problems — combining polygon angle formulas with parallel line properties

Practice Problems

Q1: The diagonals of a rhombus are 24 cm and 10 cm. Find its side and area.

Q2: Each exterior angle of a regular polygon is 30°. How many sides does it have, and how many diagonals?

Q3: A trapezium has an area of 150 cm². If the parallel sides are in the ratio 3:7 and the height is 12 cm, find the lengths of the parallel sides.

Answers

A1: Half-diagonals are 12 and 5. Side = √(12² + 5²) = √(144 + 25) = √169 = 13 cm (5-12-13 triplet!). Area = ½ × 24 × 10 = 120 cm².

A2: Sides = 360°/30° = 12 sides (dodecagon). Diagonals = 12 × 9 / 2 = 54 diagonals.

A3: Let parallel sides = 3x and 7x. Area = ½(3x + 7x) × 12 = 150. So 60x = 150, x = 2.5. Parallel sides = 7.5 cm and 17.5 cm.

We’ve covered individual shapes — now let’s put them together. Real exam problems rarely give us a clean rectangle or circle. Instead, they give us shaded regions, pathways, tiling problems, and composite figures. This is where mensuration gets practical and fun. It’s all about knowing which formula to pull out and when to add or subtract areas.

The Complete 2D Formula Reference

Important note on semicircle perimeter: It’s πr + 2r (the curved part + the diameter), NOT just πr. This is a common trap.

Key Formulas

Composite Shapes — The Add/Subtract Strategy

The golden rule for composite shapes: break the figure into basic shapes, then add or subtract their areas.

• Shaded region inside a larger shape: Subtract the smaller area from the larger area
• Combined figure: Add the areas of the parts
• Overlapping figures: Add both areas and subtract the overlap

In simple language: if we can see rectangles, triangles, circles, or semicircles hiding inside the figure, we calculate each one and combine them.

Pathway Problems

These come up all the time. A path runs around (or inside) a rectangular garden/field/room.

Path Outside a Rectangle

If a rectangle (l × b) has a path of width w running outside it:

• Outer dimensions: (l + 2w) × (b + 2w)
• Path area = (l + 2w)(b + 2w) − l × b = 2w(l + b + 2w)

Path Inside a Rectangle

If the path runs inside the rectangle:

• Inner dimensions: (l − 2w) × (b − 2w)
• Path area = l × b − (l − 2w)(b − 2w) = 2w(l + b − 2w)

Path Around a Circle

If a circular path of width w surrounds a circle of radius r:

• Outer radius R = r + w
• Path area = πR² − πr² = π(R² − r²)

Trick: This can also be written as π(R + r)(R − r) = π(2r + w)(w).

Carpet and Tiling Problems

These are essentially “how many small shapes fit in a big shape” problems.

Number of tiles = Area of floor / Area of one tile

The catch: make sure both areas are in the same units! Convert everything to the same unit before dividing.

Cost of carpeting/painting = Area × Rate per unit area

Wire Bending Problems

When a wire is bent from one shape to another:

• The perimeter stays the same (the wire length doesn’t change)
• The area changes (different shapes with the same perimeter have different areas)

Among all shapes with the same perimeter, a circle has the maximum area. Among rectangles with the same perimeter, a square has the maximum area.

Worked Examples

Example 1: Pathway Problem

A rectangular garden is 40 m long and 30 m wide. A path 2 m wide runs outside around it. Find the area of the path.

Path area = 2w(l + b + 2w) = 2 × 2 × (40 + 30 + 2 × 2) = 4 × (70 + 4) = 4 × 74 = 296 m²

Let’s verify: Outer dimensions = 44 × 34 = 1496 m². Garden = 40 × 30 = 1200 m². Path = 1496 - 1200 = 296 m² ✓

Example 2: Composite Shape

A rectangular sheet of paper is 20 cm × 14 cm. A semicircle is cut out from each of the shorter sides (diameter = 14 cm). Find the area of the remaining paper.

Area of rectangle = 20 × 14 = 280 cm²

Two semicircles with diameter 14 (radius 7) = one full circle = π × 7² = 49π = 154 cm² (using π = 22/7)

Remaining area = 280 - 154 = 126 cm²

Example 3: Tiling Problem

A room is 6 m × 4 m. How many tiles of size 25 cm × 20 cm are needed to cover the floor?

First, convert to same units. Room = 600 cm × 400 cm.

Number of tiles = (600 × 400) / (25 × 20) = 240000 / 500 = 480 tiles

Example 4: Wire Bending

A wire is bent into a square of side 11 cm. If the same wire is bent into a circle, find the radius of the circle.

Wire length (perimeter of square) = 4 × 11 = 44 cm

This becomes the circumference: 2πr = 44 r = 44 / (2 × 22/7) = 44 × 7 / 44 = 7 cm

Bonus: Area of square = 121 cm². Area of circle = 22/7 × 49 = 154 cm². See? The circle has more area for the same perimeter!

Example 5: Painting Cost

A rectangular room is 12 m long, 8 m wide, and 4 m high. It has 2 doors (2 m × 1.5 m each) and 3 windows (1.5 m × 1 m each). Find the cost of painting the walls at Rs. 15 per m².

Total wall area = 2 × (l + b) × h = 2 × (12 + 8) × 4 = 160 m²

Door area = 2 × (2 × 1.5) = 6 m² Window area = 3 × (1.5 × 1) = 4.5 m² Total to subtract = 10.5 m²

Paintable area = 160 - 10.5 = 149.5 m² Cost = 149.5 × 15 = Rs. 2242.50

Common Exam Variations

1. Shaded region between a circle and inscribed/circumscribed square
2. Four circles placed inside a square (touching each other and the sides) — find the uncovered area
3. Path around a rectangular pool/garden — inside vs outside path (read the question carefully!)
4. Wire bent into different shapes — comparing areas
5. Fencing problems — perimeter-based with one side being a wall (so only 3 sides need fencing)

Shortcut: Fencing with One Side as Wall

When one side of a rectangle is against a wall:

• Fencing needed = l + 2b (one length + two breadths)
• For maximum area with given fencing F: l = F/2 and b = F/4 (the optimal shape is when l = 2b)

Practice Problems

Q1: A circular garden of radius 21 m has a 3.5 m wide path running outside around it. Find the area of the path. (Use π = 22/7)

Q2: How many square tiles of side 20 cm are needed to pave a rectangular courtyard 3 m long and 2 m wide? What would be the cost at Rs. 5 per tile?

Q3: A wire shaped as a rectangle 16 cm × 12 cm is reshaped into a square. Find the side of the square and compare the areas.

Answers

A1: Outer radius = 21 + 3.5 = 24.5 m. Path area = π(24.5² - 21²) = (22/7)(600.25 - 441) = (22/7)(159.25) = 22 × 22.75 = 500.5 m².

A2: Floor area = 300 × 200 = 60000 cm². Tile area = 20 × 20 = 400 cm². Tiles needed = 60000/400 = 150 tiles. Cost = 150 × 5 = Rs. 750.

A3: Perimeter of rectangle = 2(16 + 12) = 56 cm. Side of square = 56/4 = 14 cm. Rectangle area = 192 cm², Square area = 196 cm². The square has 4 cm² more area — same perimeter but greater area.

Now we step into the third dimension. 3D mensuration is about volume (how much stuff fits inside) and surface area (how much material covers the outside). The formulas might look like a lot, but once we see the pattern and practice a few problems, they become second nature. Let’s build our 3D toolkit.

The Complete 3D Formula Reference

Where l = slant height for cone and frustum. For a cone: l = √(r² + h²).

Key Formulas

Understanding the Relationships

Some beautiful patterns to remember:

• Cone volume = ⅓ × Cylinder volume (with same radius and height). In simple language: we need 3 cones to fill 1 cylinder.
• Hemisphere volume = ⅔ × Cylinder volume (with same radius, and cylinder height = radius)
• Sphere = 2 Hemispheres (obvious, but people forget this in TSA calculations)
• Hemisphere TSA ≠ Half of sphere’s SA. We need to add the flat circular face: TSA = 2πr² + πr² = 3πr².

Cube and Cuboid

Cube

All edges are equal (side = a).

• Volume = a³
• Longest diagonal (space diagonal) = a√3
• CSA = 4a² (4 faces, since top and bottom are excluded)
• TSA = 6a² (all 6 faces)

Cuboid

Three different dimensions (l, b, h).

• Volume = l × b × h
• Longest diagonal = √(l² + b² + h²)
• CSA = 2h(l + b) (the 4 vertical faces)
• TSA = 2(lb + bh + hl) (all 6 faces)

Cylinder

Think of it as a rectangle wrapped around to form a tube, with circles on top and bottom.

• CSA = 2πrh (the “label” around the can)
• TSA = 2πrh + 2πr² = 2πr(r + h) (label + two circular caps)
• Volume = πr²h (base area × height)

Hollow Cylinder (like a pipe)

Outer radius R, inner radius r, height h.

• Volume = πh(R² − r²) = πh(R + r)(R − r)
• CSA = 2πh(R + r) (both inner and outer curved surfaces)

Cone

A cone is essentially a cylinder that tapers to a point. That’s why its volume is exactly one-third of a cylinder.

• Slant height l = √(r² + h²) — always check if the question gives height or slant height!
• CSA = πrl (the curved surface, like an ice cream cone)
• TSA = πr(r + l) (curved surface + circular base)
• Volume = ⅓πr²h

Sphere and Hemisphere

Sphere

• SA = 4πr² (a sphere has no “curved” vs “flat” — it’s all curved)
• Volume = 4/3 πr³

Hemisphere

• CSA = 2πr² (just the dome part)
• TSA = 3πr² (dome + flat circular base)
• Volume = 2/3 πr³

Frustum (Truncated Cone)

When we cut a cone parallel to its base, the remaining piece is a frustum. Think of it like a bucket shape — wider at the top, narrower at the bottom.

• R = larger radius (top/bottom), r = smaller radius
• Slant height l = √(h² + (R − r)²)
• Volume = ⅓πh(R² + r² + Rr)
• CSA = π(R + r)l

Melting and Recasting Problems

These are the most common 3D mensuration problems in exams. The concept is simple:

When a solid is melted and recast into another shape, the volume stays the same.

So we just set Volume₁ = Volume₂ and solve for the unknown.

Common Variations:

• Sphere melted into smaller spheres → (4/3)πR³ = n × (4/3)πr³ → n = (R/r)³
• Cylinder melted into cone → πr₁²h₁ = ⅓πr₂²h₂
• Multiple small objects melted into one large object → sum of volumes = large volume

Water Flow Problems

These combine volume with rate:

• Water flowing through a pipe at speed v, with pipe cross-section area A:
• Volume per unit time = A × v
• For a circular pipe: Volume per second = πr² × v

Largest Shape Inscribed Problems

• Largest sphere in a cube (side a): diameter = a, so r = a/2
• Largest cube in a sphere (radius r): side = 2r/√3
• Largest cone in a cylinder (r, h): same r and h (volume = ⅓ of cylinder)
• Largest sphere in a cylinder (r, h): radius = min(r, h/2)
• Largest cylinder in a sphere (radius R): when h = 2R/√3 and r = R√(2/3) for maximum volume

Worked Examples

Example 1: Cube Problem

Find the volume and total surface area of a cube whose diagonal is 6√3 cm.

Diagonal = a√3 = 6√3, so a = 6 cm Volume = 6³ = 216 cm³ TSA = 6 × 6² = 216 cm²

(Fun coincidence — the numbers match for a cube with side 6!)

Example 2: Cylinder

A cylindrical tank has a radius of 7 m and height 14 m. Find the volume of water it can hold and the cost of painting its outer surface at Rs. 10/m². (Use π = 22/7)

Volume = πr²h = 22/7 × 49 × 14 = 22 × 7 × 14 = 2156 m³

For painting the outer surface (CSA + top, assuming open-top tank): CSA = 2πrh = 2 × 22/7 × 7 × 14 = 616 m² Top = πr² = 22/7 × 49 = 154 m²

If we paint only the curved surface: Cost = 616 × 10 = Rs. 6160 If we include the top: Cost = (616 + 154) × 10 = Rs. 7700

Example 3: Melting/Recasting

A metallic sphere of radius 6 cm is melted and recast into 27 smaller spheres. Find the radius of each small sphere.

Volume of large sphere = Volume of 27 small spheres

4/3 × π × 6³ = 27 × 4/3 × π × r³

6³ = 27 × r³

216 = 27r³

r³ = 8

r = 2 cm

Shortcut: n = (R/r)³, so 27 = (6/r)³, meaning 6/r = 3, so r = 2. Much faster!

Example 4: Water Flow

Water flows through a cylindrical pipe of diameter 14 cm at the rate of 5 m per second. How long will it take to fill a rectangular tank 50 m × 44 m × 7 m?

Pipe radius = 7 cm = 0.07 m Volume flow per second = πr²v = 22/7 × 0.07² × 5 = 22/7 × 0.0049 × 5 = 0.077 m³/s

Tank volume = 50 × 44 × 7 = 15400 m³

Time = 15400 / 0.077 = 200000 seconds = 55 hours 33 minutes 20 seconds

Example 5: Frustum

A bucket is in the shape of a frustum with top radius 20 cm, bottom radius 12 cm, and height 15 cm. Find its volume.

Volume = ⅓πh(R² + r² + Rr) = ⅓ × 22/7 × 15 × (400 + 144 + 240) = ⅓ × 22/7 × 15 × 784 = 22/7 × 5 × 784 = 22 × 5 × 112 = 12320 cm³ ≈ 12.32 liters

Common Exam Variations

1. Melting a cylinder into cones (or vice versa) — equate volumes
2. Water rise problems — object immersed in water raises the level (volume of object = l × b × rise in water level)
3. Painting/wrapping problems — need surface area, not volume
4. Hollow shapes — subtract inner volume from outer volume
5. Compound shapes — hemisphere on top of cylinder, cone on top of cylinder, etc. Add volumes and adjust surface areas (remove the common circular face)

Practice Problems

Q1: A cone has a radius of 7 cm and height of 24 cm. Find its slant height, volume, and curved surface area. (Use π = 22/7)

Q2: A solid metallic cylinder of radius 3 cm and height 12 cm is melted and recast into a cone of radius 6 cm. Find the height of the cone.

Q3: A hemispherical bowl has an inner radius of 10 cm. Find the volume of milk it can hold and the cost of painting the inner surface at Rs. 2 per cm². (Use π = 3.14)

Answers

A1: Slant height l = √(7² + 24²) = √(49 + 576) = √625 = 25 cm (7-24-25 triplet!). Volume = ⅓ × 22/7 × 49 × 24 = 22 × 7 × 24/3 = 1232 cm³. CSA = 22/7 × 7 × 25 = 550 cm².

A2: Volume of cylinder = Volume of cone. π × 9 × 12 = ⅓ × π × 36 × h. 108 = 12h. h = 9 cm.

A3: Volume = 2/3 × 3.14 × 1000 = 2093.33 cm³ ≈ 2.09 liters. Inner surface (CSA of hemisphere) = 2 × 3.14 × 100 = 628 cm². Cost = 628 × 2 = Rs. 1256.

Coordinate geometry is where algebra meets geometry. Instead of drawing shapes and eyeballing, we use numbers — every point has an address (x, y), and we can calculate distances, areas, and slopes with formulas. It’s one of the most formula-driven topics in aptitude, which is great news because formulas mean guaranteed marks once we memorize them.

The Coordinate Plane

Quick refresher: the plane is divided into 4 quadrants by two axes.

• Q1: x > 0, y > 0 (both positive)
• Q2: x < 0, y > 0 (x negative, y positive)
• Q3: x < 0, y < 0 (both negative)
• Q4: x > 0, y < 0 (x positive, y negative)

Points on axes have one coordinate as zero. The origin is (0, 0).

Key Formulas

Distance Formula

The distance between two points (x₁, y₁) and (x₂, y₂) is:

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

In simple language: it’s just the Pythagoras theorem applied to coordinates. The horizontal difference is one leg, the vertical difference is the other, and the distance is the hypotenuse.

Special cases:

• Distance from origin: d = √(x² + y²)
• Horizontal line (y₁ = y₂): d = |x₂ − x₁|
• Vertical line (x₁ = x₂): d = |y₂ − y₁|

Midpoint and Section Formula

Midpoint Formula

The midpoint of (x₁, y₁) and (x₂, y₂) is simply the average of the coordinates:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Section Formula (Internal Division)

If point P divides the line joining A(x₁, y₁) and B(x₂, y₂) in the ratio m:n internally:

P = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))

Memory trick: The point closer to B gets the weight of m (associated with B’s coordinates), and the point closer to A gets the weight of n.

The midpoint is just a special case where m:n = 1:1.

Centroid of a Triangle

The centroid (intersection of medians) of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is:

G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)

In simple language: just average all three x-coordinates and all three y-coordinates.

Slope of a Line

The slope tells us how steep a line is — how much it rises (or falls) for every unit it moves right.

Slope m = (y₂ − y₁)/(x₂ − x₁) = rise / run

• Positive slope: line goes up from left to right (↗)
• Negative slope: line goes down from left to right (↘)
• Zero slope: horizontal line (m = 0)
• Undefined slope: vertical line (division by zero — the run is 0)

Parallel and Perpendicular Lines

• Parallel lines have the same slope: m₁ = m₂
• Perpendicular lines have slopes that multiply to −1: m₁ × m₂ = −1

In simple language: for perpendicular lines, flip the fraction and change the sign. If one line has slope 2/3, the perpendicular line has slope −3/2.

Equation of a Line

There are several forms — use whichever fits the given information:

1. Slope-intercept form: y = mx + c

• m = slope, c = y-intercept (where the line crosses the y-axis)
• Best when we know the slope and y-intercept

2. Point-slope form: y − y₁ = m(x − x₁)

• Best when we know the slope and one point

3. Two-point form: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁)

• Best when we know two points on the line

4. Intercept form: x/a + y/b = 1

• a = x-intercept, b = y-intercept
• Best when we know both intercepts

Area of Triangle from Coordinates

Given three vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):

Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

The absolute value is important — area can’t be negative! If the formula gives us zero, the three points are collinear (they lie on the same line).

Collinearity check: Three points are collinear if and only if the area of the triangle they form is zero.

Shortcut Methods and Tricks

Trick 1: Quick distance check Before applying the distance formula, check if the differences form a Pythagorean triplet. E.g., if Δx = 3 and Δy = 4, the distance is immediately 5.

Trick 2: Vertical/horizontal sides If a triangle or quadrilateral has any vertical or horizontal sides, calculate those distances by simple subtraction — no need for the full formula.

Trick 3: Area using the “shoelace” pattern For the triangle area formula, arrange the coordinates in a column and cross-multiply diagonally:

Write the vertices in order and repeat the first:

x₁  y₁
x₂  y₂
x₃  y₃
x₁  y₁

Sum of (down-right diagonals) − Sum of (down-left diagonals), divide by 2, take absolute value.

Trick 4: When given the equation ax + by + c = 0

• Slope = −a/b
• x-intercept = −c/a
• y-intercept = −c/b

Worked Examples

Example 1: Distance

Find the distance between (3, −2) and (−1, 1).

d = √[(−1 − 3)² + (1 − (−2))²] d = √[(−4)² + (3)²] d = √[16 + 9] d = √25 = 5 units

(We spotted the 3-4-5 triplet! Δx = 4, Δy = 3.)

Example 2: Section Formula

Find the point that divides the line joining A(2, 3) and B(8, −1) in the ratio 1:3.

P = ((1×8 + 3×2)/(1+3), (1×(−1) + 3×3)/(1+3)) P = ((8 + 6)/4, (−1 + 9)/4) P = (14/4, 8/4) P = (3.5, 2)

Example 3: Equation of Line

Find the equation of a line passing through (2, 5) with slope 3.

Using point-slope form: y − 5 = 3(x − 2) y − 5 = 3x − 6 y = 3x − 1

In general form: 3x − y − 1 = 0

Example 4: Area of Triangle

Find the area of the triangle with vertices A(1, 2), B(4, 6), C(7, 2).

Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| = ½ |1(6 − 2) + 4(2 − 2) + 7(2 − 6)| = ½ |1(4) + 4(0) + 7(−4)| = ½ |4 + 0 − 28| = ½ |−24| = ½ × 24 = 12 sq units

Example 5: Perpendicular Lines

Line L₁ passes through (1, 3) and (4, 9). Find the slope of a line perpendicular to L₁.

Slope of L₁ = (9 − 3)/(4 − 1) = 6/3 = 2

For perpendicular: m₁ × m₂ = −1 2 × m₂ = −1 m₂ = −1/2

Common Exam Variations

1. Collinearity check — show that area = 0, or show that slope between any two pairs is the same
2. Finding the fourth vertex of a parallelogram (use midpoint property — diagonals bisect each other)
3. Circumcenter/Incenter — usually avoided in aptitude, but centroid is commonly tested
4. Line through origin — always has the form y = mx (no c term, since it passes through (0,0))
5. Distance of a point from a line — d = |ax₁ + by₁ + c| / √(a² + b²) for line ax + by + c = 0

Practice Problems

Q1: Show that the points (1, 1), (3, 5), and (5, 9) are collinear.

Q2: Find the equation of the line passing through (−2, 3) and (4, −1). Also find the x-intercept and y-intercept.

Q3: The vertices of a triangle are P(0, 0), Q(8, 0), and R(4, 6). Find the area of the triangle and the coordinates of the centroid.

Answers

A1: Area = ½|1(5−9) + 3(9−1) + 5(1−5)| = ½|−4 + 24 − 20| = ½|0| = 0. Since the area is zero, the points are collinear. (Alternatively: slope of AB = (5−1)/(3−1) = 2, slope of BC = (9−5)/(5−3) = 2. Same slope, so collinear.)

A2: Slope = (−1−3)/(4−(−2)) = −4/6 = −2/3. Using point-slope: y − 3 = −2/3(x + 2) → 3y − 9 = −2x − 4 → 2x + 3y − 5 = 0. x-intercept: set y = 0 → x = 5/2 = 2.5. y-intercept: set x = 0 → y = 5/3 ≈ 1.67.

A3: Area = ½|0(0−6) + 8(6−0) + 4(0−0)| = ½|0 + 48 + 0| = 24 sq units. Centroid = ((0+8+4)/3, (0+0+6)/3) = (4, 2).

Permutations are all about arrangements — how many ways can we arrange things when the order matters. Choosing a president, then a vice-president, then a secretary from a group is a permutation problem because who gets which role matters. Let’s build this up step by step.

The Fundamental Counting Principle

Before we even touch formulas, we need this one idea:

If task 1 can be done in m ways, and task 2 can be done in n ways, then both tasks together can be done in m × n ways.

In simple language: multiply the choices at each step.

Example: We have 3 shirts and 4 pants. Total outfits = 3 × 4 = 12.

This extends to any number of tasks. If we have a 3-step process with 5, 3, and 2 choices at each step, total ways = 5 × 3 × 2 = 30.

Factorial

The factorial of n (written n!) is the product of all positive integers from 1 to n:

• 0! = 1 (by definition — this trips people up, just memorize it)
• 1! = 1
• 2! = 2
• 3! = 6
• 4! = 24
• 5! = 120
• 6! = 720
• 7! = 5040
• 8! = 40320
• 9! = 362880
• 10! = 3628800

Memorize up to 7! at least. These come up constantly.

Useful property: n! = n × (n-1)!. So 8! = 8 × 7! = 8 × 5040 = 40320.

Key Formulas

nPr — The Permutation Formula

nPr = n! / (n-r)! gives us the number of ways to arrange r items chosen from n distinct items.

• nPn = n! (arrange all n items — the (n-n)! = 0! = 1 in the denominator)
• nP1 = n (just picking 1 item from n — n ways)
• nP0 = 1 (doing nothing — there’s exactly 1 way to do nothing)

Example: How many 3-digit numbers can be formed from digits 1-5 (no repetition)?

5P3 = 5!/(5-3)! = 120/2 = 60

Think of it as: 5 choices for first digit × 4 for second × 3 for third = 60.

Arrangements with Repeated Items

When some items are identical, we divide by the factorial of each group of repeats:

Arrangements = n! / (p! × q! × r!…)

where p, q, r… are the frequencies of the repeated items.

Why? Because swapping identical items doesn’t create a new arrangement. If we have 3 identical A’s, those 3! = 6 swaps among themselves are all the same arrangement, so we divide by 3!.

The MISSISSIPPI Problem

This is the classic example. How many ways can we arrange the letters of MISSISSIPPI?

Letters: M(1), I(4), S(4), P(2) — total 11 letters

Arrangements = 11! / (1! × 4! × 4! × 2!) = 39916800 / (1 × 24 × 24 × 2) = 39916800 / 1152 = 34650

Circular Permutations

When we arrange items in a circle, we fix one item’s position (since rotations of the same arrangement are identical) and arrange the rest.

Circular permutations of n items = (n-1)!

In simple language: in a line, ABCD and BCDA are different. In a circle, they’re the same (just rotated). So we divide by n, giving n!/n = (n-1)!.

Necklace/Bracelet Problem

For a necklace or bracelet, clockwise and counterclockwise arrangements look the same (we can flip it). So we divide by 2 as well:

Necklace arrangements = (n-1)! / 2

Restricted Permutations

”Always Together” Problems

When certain items must always be next to each other:

1. Bundle them into a single unit (treat the group as one item)
2. Arrange the units (including the bundle)
3. Arrange the items within the bundle
4. Multiply the results

”Never Together” Problems

When certain items must never be adjacent:

Never together = Total arrangements − Always together

Or use the gap method: arrange the other items first, then place the restricted items in the gaps between them.

Fixed Position Problems

• Starts with / ends with a specific item: Fix that item, arrange the rest
• Even/odd positions only: Count the available positions, fill them, then fill the rest

Shortcut Methods and Tricks

Trick 1: Building numbers step by step For “how many n-digit numbers” problems, think step by step. For a 4-digit number:

• First digit: can’t be 0 (so count carefully)
• Remaining digits: fill one by one based on constraints

Trick 2: At least one / at most one “At least one” = Total - None “At most one” = None + Exactly one

Trick 3: Odd/Even number formations If the number must be even, fix the last digit first (it must be even), then fill the rest. If the number must be odd, fix the last digit first (it must be odd).

Trick 4: Division shortcut When calculating nPr, we don’t need the full factorial. 10P3 = 10 × 9 × 8 = 720. Just multiply from n going down, r times.

Worked Examples

Example 1: Basic Permutation

How many 4-letter words (with or without meaning) can be formed from the letters of REACT, without repeating any letter?

We have 5 distinct letters and need to arrange 4 of them.

5P4 = 5 × 4 × 3 × 2 = 120 words

Example 2: Number Formation

How many 3-digit even numbers can be formed using digits 1, 2, 3, 4, 5 (no repetition)?

For an even number, the last digit must be even: 2 or 4 → 2 choices First digit: any of the remaining 4 digits → 4 choices Second digit: any of the remaining 3 digits → 3 choices

Total = 4 × 3 × 2 = 24 numbers

Key point: We filled the most restricted position (last digit) first. Always handle constraints first!

Example 3: Word Arrangement with Repeats

How many ways can the letters of COMMITTEE be arranged?

Letters: C(1), O(1), M(2), I(1), T(2), E(2) — total 9 letters

Arrangements = 9! / (2! × 2! × 2!) = 362880 / 8 = 45360

Example 4: Always Together

In how many ways can the letters of GARDEN be arranged so that the vowels (A, E) are always together?

Treat the vowels as one unit: [AE], G, R, D, N → 5 units Arrange 5 units = 5! = 120 Arrange vowels within the unit = 2! = 2

Total = 120 × 2 = 240 ways

Example 5: Circular Permutation

In how many ways can 8 people sit around a circular table?

Circular permutation = (8-1)! = 7! = 5040 ways

If the table is replaced by a bracelet of 8 beads: 7!/2 = 5040/2 = 2520 arrangements.

Common Exam Variations

1. Number formation with/without repetition, with constraints (even, odd, greater than X, divisible by 5)
2. Word arrangements from words like INDEPENDENCE, MATHEMATICS, PERMUTATION (repeating letters)
3. Circular seating — sometimes with constraints like “A and B must sit together” or “A and B must not sit adjacent”
4. Books on a shelf — sometimes with “all math books together” or “no two science books adjacent”
5. Flag signals — using colored flags in different arrangements

Practice Problems

Q1: How many 4-digit numbers greater than 5000 can be formed using digits 0, 1, 3, 5, 7 (no repetition)?

Q2: In how many ways can the letters of BANANA be arranged?

Q3: 6 people are to sit around a circular table. In how many ways can they sit such that two specific people (A and B) always sit next to each other?

Answers

A1: First digit must be 5 or 7 (to be > 5000) → 2 choices. Remaining 3 positions from 4 remaining digits = 4 × 3 × 2 = 24. Total = 2 × 24 = 48 numbers. But wait — check if any start with 0 in remaining positions. Since 0 isn’t in the first position, all are valid 4-digit numbers. Answer: 48.

A2: Letters: B(1), A(3), N(2) — total 6 letters. Arrangements = 6! / (3! × 2!) = 720 / 12 = 60.

A3: Treat A and B as one unit → 5 units around a circle = (5-1)! = 24. A and B can swap within the unit = 2! = 2. Total = 24 × 2 = 48 ways.

Combinations are about selection — how many ways can we choose things when the order doesn’t matter. Picking 3 people for a team is a combination problem because we just care about WHO is on the team, not the order we picked them. The only difference from permutations is: in combinations, ABC = ACB = BAC = BCA = CAB = CBA — they’re all the same selection.

Permutation vs Combination — The Key Difference

Permutation: Arrangement matters → choosing a president, VP, and secretary → nPr

Combination: Only selection matters → choosing a 3-person committee → nCr

In simple language: if we’re assigning roles or positions, it’s a permutation. If we’re just picking a group, it’s a combination.

Quick test: After selecting, does rearranging the chosen items give something different? If yes → permutation. If no → combination.

Key Formulas

Understanding nCr

nCr = n! / [r! × (n-r)!]

This gives us the number of ways to choose r items from n distinct items, where order doesn’t matter.

The relationship: nCr = nPr / r!. We take the permutation (which overcounts because order matters) and divide by r! (the number of ways to arrange the r chosen items) to remove the ordering.

Key Properties

• nC0 = 1 — there’s exactly one way to choose nothing
• nCn = 1 — there’s exactly one way to choose everything
• nC1 = n — choosing 1 item from n gives n choices
• nCr = nC(n-r) — choosing r items to include is the same as choosing (n-r) items to exclude. So 10C7 = 10C3 (use this to simplify calculations!)

Quick Calculation Trick

Always use the smaller of r and (n-r) to minimize computation.

10C3 = (10 × 9 × 8) / (3 × 2 × 1) = 720/6 = 120

We only need to multiply r terms going down from n, then divide by r!. Much faster than computing full factorials.

Useful Values to Know

Committee / Team Selection

This is the bread-and-butter application of combinations.

Basic Selection

“Choose r people from n” = nCr. That’s it.

Selection with Constraints

These are where exams get tricky:

“At least k” type: Calculate for each valid value and add up, OR use complement: nCr(at least k) = Total − nCr(fewer than k)

“At most k” type: Calculate for 0, 1, 2, …, k and add up.

“Exactly k from group A and exactly j from group B”: Multiply the individual combinations.

The “Must Include / Must Exclude” Approach

• Person X must be included: Fix X in the team, choose remaining (r-1) from remaining (n-1). Answer: (n-1)C(r-1)
• Person X must be excluded: Choose all r from remaining (n-1). Answer: (n-1)Cr
• Both X and Y must be included: Fix both, choose (r-2) from (n-2)
• At least one of X or Y: Total − Neither = nCr − (n-2)Cr

Handshakes and Diagonals

Handshakes

If n people are in a room and each shakes hands with everyone else:

Number of handshakes = nC2 = n(n-1)/2

Each handshake involves 2 people, and order doesn’t matter (A shaking B’s hand = B shaking A’s hand).

Diagonals of a Polygon

Each diagonal connects 2 non-adjacent vertices.

Diagonals = nC2 − n = n(n-1)/2 − n = n(n-3)/2

We subtract n because nC2 counts all connections (including the n sides, which are not diagonals).

Distribution Problems

Identical Objects into Distinct Groups

Distributing n identical objects into r distinct groups (each group can get 0 or more):

Ways = (n+r-1)C(r-1)

This is the “stars and bars” theorem. Think of it like: we have n stars (objects) and need (r-1) bars (dividers) to create r groups.

Distinct Objects into Distinct Groups

Each object has r choices → r^n ways (each of the n objects independently goes to one of r groups).

Shortcut Methods and Tricks

Trick 1: Symmetry saves time 20C17 = 20C3 = (20 × 19 × 18) / 6 = 1140. Always reduce to the smaller side.

Trick 2: Committee with groups “Select 5 from 6 men and 4 women with at least 2 women” → don’t calculate “at least” directly. Calculate: (2W, 3M) + (3W, 2M) + (4W, 1M). It’s usually faster.

Trick 3: Complementary counting “At least 1” = Total − “none” is almost always faster than direct calculation.

Trick 4: Points and lines/triangles

• Lines from n points (no 3 collinear): nC2
• Triangles from n points (no 3 collinear): nC3
• If k points are collinear: subtract kC2 from lines (or kC3 from triangles) and add back what the collinear points contribute (1 line for collinear points, 0 triangles)

Worked Examples

Example 1: Basic Combination

A class has 12 students. In how many ways can a team of 4 be chosen?

12C4 = (12 × 11 × 10 × 9) / (4 × 3 × 2 × 1) = 11880 / 24 = 495 ways

Example 2: Committee with Constraints

From 5 men and 4 women, a committee of 4 is to be formed with at least 2 women. How many ways?

We need: (2W, 2M) or (3W, 1M) or (4W, 0M)

• 2W, 2M: 4C2 × 5C2 = 6 × 10 = 60
• 3W, 1M: 4C3 × 5C1 = 4 × 5 = 20
• 4W, 0M: 4C4 × 5C0 = 1 × 1 = 1

Total = 60 + 20 + 1 = 81 ways

Example 3: Handshakes

In a party of 10 people, everyone shakes hands with everyone else. How many handshakes?

10C2 = 10 × 9 / 2 = 45 handshakes

Example 4: Must Include / Must Exclude

From 8 students, a team of 5 is chosen. In how many ways if (a) a particular student must be in the team, and (b) two particular students cannot both be in the team?

(a) Fix that student. Choose remaining 4 from 7: 7C4 = 35. 35 ways.

(b) Total − Both in = 8C5 − 6C3. Total = 8C5 = 56. Both in (fix both, choose 3 from 6) = 6C3 = 20. Answer = 56 − 20 = 36 ways.

Example 5: Points and Triangles

There are 10 points in a plane, no 3 collinear. How many triangles can be formed? How many straight lines?

Triangles = 10C3 = (10 × 9 × 8) / 6 = 120 triangles

Lines = 10C2 = (10 × 9) / 2 = 45 lines

Follow-up: If 4 of the 10 points are collinear, triangles = 10C3 − 4C3 = 120 − 4 = 116 triangles. Lines = 10C2 − 4C2 + 1 = 45 − 6 + 1 = 40 lines (we remove the 6 lines those 4 points would have made and add back the 1 line they actually form).

Common Exam Variations

1. Committee formation — with constraints on gender, age group, or department representation
2. Handshakes and diagonals — sometimes combined (e.g., at a conference, each table has n people)
3. Selection from groups — “choose 3 books from 5 math and 4 science, with at least 1 from each”
4. Identical objects — distributing chocolates, fruits, or balls into groups
5. Cards — choosing hands from a deck of 52 (very common in probability problems too)

Practice Problems

Q1: How many diagonals does a dodecagon (12-sided polygon) have?

Q2: A box contains 5 red and 4 blue balls. In how many ways can 3 balls be drawn such that at least 1 is red?

Q3: From a group of 7 men and 6 women, a committee of 5 is to be formed such that there are at least 3 men. Find the number of ways.

Answers

A1: Diagonals = 12 × (12-3) / 2 = 12 × 9 / 2 = 54 diagonals.

A2: Total ways to draw 3 from 9 = 9C3 = 84. Ways with NO red (all blue) = 4C3 = 4. At least 1 red = 84 − 4 = 80 ways. (Complement method — much faster than calculating 1 red + 2 red + 3 red separately.)

A3: Cases: (3M, 2W) + (4M, 1W) + (5M, 0W) = 7C3 × 6C2 + 7C4 × 6C1 + 7C5 × 6C0 = 35 × 15 + 35 × 6 + 21 × 1 = 525 + 210 + 21 = 756 ways.

Probability tells us how likely something is to happen. It’s a number between 0 (impossible) and 1 (certain). If we can count the favorable outcomes and the total outcomes, we can find the probability. Most aptitude probability questions follow a handful of patterns — once we learn those, we’re golden.

The Basics

P(E) = Number of favorable outcomes / Total number of outcomes

This is the classical definition, and it works when all outcomes are equally likely.

• P(E) = 0 means the event is impossible
• P(E) = 1 means the event is certain
• P(not E) = 1 − P(E) — this is called the complement

The complement rule is probably the most useful trick in probability. When “at least one” is hard to calculate directly, we flip it: P(at least one) = 1 − P(none).

Mutually Exclusive vs Independent Events

These are two completely different concepts, and exams love to test whether we know the difference.

Mutually Exclusive Events

Two events are mutually exclusive if they cannot happen at the same time. Think of it like: if A happens, B can’t, and vice versa.

• P(A ∩ B) = 0 (no overlap)
• P(A ∪ B) = P(A) + P(B) (just add — no double counting needed)

Example: Rolling a 3 and rolling a 5 on the same die roll — can’t happen simultaneously.

Independent Events

Two events are independent if one happening doesn’t affect the probability of the other.

• P(A ∩ B) = P(A) × P(B) (just multiply)
• P(A|B) = P(A) (knowing B happened doesn’t change A’s probability)

Example: Flipping a coin and rolling a die — the coin result doesn’t affect the die.

Key distinction: Mutually exclusive events are about overlap (or lack thereof). Independent events are about influence (or lack thereof). Mutually exclusive events with non-zero probabilities are actually DEPENDENT (if A happened, P(B) = 0, which is different from P(B) in general).

Addition Rule (OR)

When we want P(A or B), we use:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

We subtract P(A ∩ B) to avoid double-counting the overlap.

If A and B are mutually exclusive (no overlap), the formula simplifies to: P(A ∪ B) = P(A) + P(B).

Multiplication Rule (AND)

When we want P(A and B):

P(A ∩ B) = P(A) × P(B|A)

If A and B are independent: P(A ∩ B) = P(A) × P(B)

In simple language: for “AND” problems, we multiply. But we need to be careful about whether events are independent or not.

Conditional Probability

The probability of A happening given that B has already happened:

P(A|B) = P(A ∩ B) / P(B)

In simple language: we’re restricting our sample space to only the outcomes where B is true, and then asking “of those, how many also have A?”

Example: In a class, 60% study math, 40% study science, and 20% study both. What’s the probability that a student studies math given they study science?

P(Math|Science) = P(Math ∩ Science) / P(Science) = 0.20 / 0.40 = 0.50 or 50%

Bayes’ Theorem

Bayes’ theorem lets us reverse conditional probability — if we know P(B|A), we can find P(A|B).

P(A|B) = P(B|A) × P(A) / P(B)

This is incredibly powerful. Let’s see it with a practical example.

Bayes’ Theorem — A Practical Example

A factory has 3 machines: Machine A produces 50% of items, Machine B produces 30%, Machine C produces 20%. The defective rates are: A = 3%, B = 4%, C = 5%. If a randomly selected item is defective, what’s the probability it came from Machine A?

We want P(A | Defective).

P(Defective) = P(D|A)×P(A) + P(D|B)×P(B) + P(D|C)×P(C) = 0.03 × 0.50 + 0.04 × 0.30 + 0.05 × 0.20 = 0.015 + 0.012 + 0.010 = 0.037

P(A | Defective) = P(D|A) × P(A) / P(D) = 0.015 / 0.037 = 15/37 ≈ 0.405 or 40.5%

Even though Machine A has the lowest defective rate, it produces the most items, so it still contributes the most to the defective pool.

Shortcut Methods and Tricks

Trick 1: The complement is your best friend P(at least 1) = 1 − P(none). This saves enormous calculation in most “at least” problems.

Trick 2: Sequential probability For events happening one after another (like drawing cards without replacement), multiply the probabilities at each step. Just update the total and favorable counts after each step.

Trick 3: OR means add, AND means multiply This simple mnemonic covers 90% of cases. Just remember to check for independence/mutual exclusivity.

Trick 4: Odds to probability If odds in favor are a:b, then P = a/(a+b). If odds against are a:b, then P = b/(a+b).

Worked Examples

Example 1: Basic Probability

A bag contains 5 red, 3 blue, and 2 green balls. One ball is drawn at random. Find P(red) and P(not green).

Total = 5 + 3 + 2 = 10

P(red) = 5/10 = 1/2

P(not green) = 1 − P(green) = 1 − 2/10 = 4/5

Example 2: Addition Rule

A card is drawn from a standard deck of 52 cards. Find the probability that it is a King or a Heart.

P(King) = 4/52, P(Heart) = 13/52, P(King AND Heart) = 1/52 (King of Hearts)

P(King or Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13

Example 3: Independent Events

A coin is tossed and a die is rolled. What is the probability of getting a Head and a number greater than 4?

P(Head) = 1/2 P(greater than 4) = P(5 or 6) = 2/6 = 1/3

These are independent events, so: P(Head AND > 4) = 1/2 × 1/3 = 1/6

Example 4: Without Replacement

A bag has 4 red and 6 blue balls. Two balls are drawn without replacement. Find P(both red).

P(1st red) = 4/10 P(2nd red | 1st red) = 3/9 (one red ball is gone)

P(both red) = 4/10 × 3/9 = 12/90 = 2/15

Example 5: At Least One (Complement)

Three coins are tossed. Find P(at least one head).

The hard way: P(1H) + P(2H) + P(3H) = 3/8 + 3/8 + 1/8 = 7/8

The smart way: P(at least 1 head) = 1 − P(no heads) = 1 − P(all tails) = 1 − (1/2)³ = 1 − 1/8 = 7/8

Same answer, way less work!

Common Exam Variations

1. Drawing balls from bags — with and without replacement
2. Card problems — drawing specific suits, ranks, or combinations
3. Dice problems — sum of two dice, specific outcomes
4. “At least one” problems — almost always use the complement
5. Conditional probability in real-world scenarios — disease testing, defective items, exam pass rates

Practice Problems

Q1: Two dice are rolled. Find the probability that the sum is 7 or 11.

Q2: A box has 3 defective and 7 non-defective items. Two items are drawn without replacement. Find the probability that both are non-defective.

Q3: The probability that A solves a problem is 2/3, and the probability that B solves it is 3/4. If both try independently, what’s the probability that the problem is solved?

Answers

A1: Total outcomes = 36. Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 ways. Sum 11: (5,6), (6,5) = 2 ways. These are mutually exclusive. P = (6+2)/36 = 8/36 = 2/9.

A2: P(1st non-defective) = 7/10. P(2nd non-defective | 1st non-defective) = 6/9 = 2/3. P(both) = 7/10 × 2/3 = 7/15.

A3: P(problem solved) = 1 − P(neither solves it) = 1 − P(A fails) × P(B fails) = 1 − (1/3)(1/4) = 1 − 1/12 = 11/12.

Dice, coins, and playing cards — these three objects make up about 80% of all probability questions in aptitude exams. The reason they’re so popular is that they have clean, well-defined sample spaces. Once we know the sample spaces inside out, these problems become almost mechanical. Let’s master each one.

Coins — The Simplest Probability

Sample Space

• 1 coin: {H, T} → 2 outcomes
• 2 coins: {HH, HT, TH, TT} → 4 outcomes
• 3 coins: 8 outcomes
• n coins: 2ⁿ outcomes

The number of ways to get exactly r heads in n tosses = nCr. This is because we’re choosing which r of the n tosses show heads.

Common Coin Probabilities (3 coins)

The power of complement: P(at least 1 head in n tosses) = 1 − P(all tails) = 1 − (1/2)ⁿ. No counting needed!

Dice — The King of Probability Questions

Sample Space

• 1 die: {1, 2, 3, 4, 5, 6} → 6 outcomes
• 2 dice: 6 × 6 = 36 outcomes
• n dice: 6ⁿ outcomes

Single Die Probabilities

Two Dice — Sum Probabilities

This is the most tested topic. There are 36 total outcomes for two dice. Here’s how many ways each sum can occur:

Pattern to memorize: The ways go 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 — a symmetric pyramid peaking at sum = 7. Sum = 7 is the most likely outcome with 6 ways.

The combinations for sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

Useful Two-Dice Facts

• P(sum = 7) = 6/36 = 1/6 (the highest for any specific sum)
• P(doubles) = 6/36 = 1/6 → (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
• P(sum is even) = 18/36 = 1/2
• P(sum is odd) = 18/36 = 1/2
• P(at least one 6) = 1 − P(no 6) = 1 − (5/6)² = 1 − 25/36 = 11/36

Playing Cards — The Most Complex Sample Space

A standard deck has 52 cards organized as follows:

Single Card Drawn

Two Cards Drawn

When drawing 2 cards from 52 without replacement:

Total ways = 52C2 = 1326

Both Kings: 4C2 / 52C2 = 6/1326 = 1/221

Both from same suit: 4 × 13C2 / 52C2 = 4 × 78 / 1326 = 312/1326 = 4/17

One King and one Queen: (4C1 × 4C1) / 52C2 = 16/1326 = 8/663

”At Least” and “At Most” Problems

These are the trickiest question types, but the complement rule makes them easy.

”At Least One” = 1 − P(None)

Example: Two cards drawn from 52. P(at least one ace)?

P(at least one ace) = 1 − P(no ace) = 1 − 48C2/52C2 = 1 − 1128/1326 = 198/1326 = 33/221

”At Most One” = P(None) + P(Exactly One)

Example: Three coins tossed. P(at most one head)?

P(at most 1 head) = P(0 heads) + P(1 head) = 1/8 + 3/8 = 4/8 = 1/2

Shortcut Methods and Tricks

Trick 1: Dice sum = 7 shortcut For any single die value x on one die, the other die needs (7-x). Since x can be 1 through 6, there are always 6 ways. This is why P(sum=7) = 1/6.

Trick 2: Card counting by elimination Instead of counting favorable, sometimes it’s faster to count unfavorable and subtract. “Not a face card” = 52 − 12 = 40 cards.

Trick 3: “At least” always means complement Whenever we see “at least,” our first instinct should be: 1 − P(complement). It’s faster in almost every case.

Trick 4: With/Without replacement matters hugely With replacement: probabilities stay the same each draw. Without replacement: we must update the denominator AND numerator after each draw.

Trick 5: Drawing simultaneously vs one by one Drawing 2 cards simultaneously = drawing 2 cards one by one without replacement. The math is the same. Use combinations for simultaneous, sequential multiplication for one-by-one.

Worked Examples

Example 1: Coin Problem

4 coins are tossed. Find P(exactly 2 heads).

Total outcomes = 2⁴ = 16 Favorable (exactly 2 heads) = 4C2 = 6

P = 6/16 = 3/8

Example 2: Two Dice Problem

Two dice are rolled. Find P(sum > 9).

Sum = 10: (4,6), (5,5), (6,4) → 3 ways Sum = 11: (5,6), (6,5) → 2 ways Sum = 12: (6,6) → 1 way

Total favorable = 6

P = 6/36 = 1/6

Example 3: Card Problem

Two cards are drawn from a deck. Find P(both are red).

Red cards = 26

P = 26C2 / 52C2 = (26 × 25/2) / (52 × 51/2) = 325/1326 = 25/102

Step by step (sequential): P(1st red) × P(2nd red | 1st red) = 26/52 × 25/51 = 25/102. Same answer!

Example 4: At Least One Problem

A die is rolled 3 times. Find P(at least one 6).

P(no 6 in a single roll) = 5/6

P(no 6 in 3 rolls) = (5/6)³ = 125/216

P(at least one 6) = 1 − 125/216 = 91/216

Example 5: Mixed Problem

From a well-shuffled deck, 3 cards are drawn. What is the probability that all 3 are face cards?

Face cards = 12

P = 12C3 / 52C3 = 220 / 22100 = 1/100.45…

Let’s compute exactly: 12C3 = (12 × 11 × 10)/6 = 220. 52C3 = (52 × 51 × 50)/6 = 22100.

P = 220/22100 = 22/2210 = 11/1105

The Birthday Problem (Fun Bonus)

This is a famous probability puzzle: In a group of 23 people, there’s a greater than 50% chance that two people share a birthday. Sounds wild, right?

Here’s the logic: P(all different birthdays) = 365/365 × 364/365 × 363/365 × … × 343/365

For 23 people, this product works out to about 0.493, so P(at least one shared birthday) = 1 − 0.493 ≈ 0.507 or 50.7%.

At 50 people, the probability jumps to about 97%. Our intuition is terrible at this — which is exactly why probability is so important to study!

Common Exam Variations

1. Two dice sum problems — P(sum = k), P(sum > k), P(sum is prime)
2. Card drawing — with specific suit/rank constraints, with or without replacement
3. Coin sequences — P(exactly r heads in n tosses), P(at least r heads)
4. Bag problems — balls of different colors, drawing with/without replacement
5. Conditional dice — “given that the sum is even, find P(both are same)“

Practice Problems

Q1: Two dice are rolled. Find the probability that the product of the numbers is 12.

Q2: 5 coins are tossed simultaneously. Find P(at least 4 heads).

Q3: From a deck of 52 cards, two cards are drawn without replacement. Find the probability that one is a spade and the other is a heart.

Answers

A1: Pairs giving product 12: (2,6), (3,4), (4,3), (6,2) → 4 favorable outcomes. P = 4/36 = 1/9.

A2: P(at least 4 heads) = P(4 heads) + P(5 heads) = 5C4/2⁵ + 5C5/2⁵ = 5/32 + 1/32 = 6/32 = 3/16.

A3: Ways to pick 1 spade and 1 heart = 13C1 × 13C1 = 169. Total ways = 52C2 = 1326. P = 169/1326 = 13/102.

Data Interpretation (DI) is less about formulas and more about speed and accuracy. We’re given a chart, table, or graph, and we need to extract information fast. The math itself is usually percentages, ratios, and averages — stuff we already know. The challenge is doing it quickly while reading data correctly. Let’s build our DI toolkit.

The Types of Data Presentations

Exams typically use five formats:

1. Tables — rows and columns of numbers (most common)
2. Bar charts — comparing values across categories
3. Line charts — showing trends over time
4. Pie charts — showing parts of a whole (percentages)
5. Mixed/Caselets — text-based data that we have to organize ourselves

Each has its own reading strategy. Let’s go through them.

Tables — The Foundation

Tables are the most straightforward. We have rows (usually categories or time periods) and columns (different metrics).

Reading Strategy:

1. Read the title first — what does the table represent?
2. Check the units — are values in lakhs, crores, percentages, thousands?
3. Scan for patterns — increasing, decreasing, or irregular?
4. Don’t read every cell — go straight to the question and find only the cells we need

Common Calculations from Tables:

• Percentage change = [(New − Old) / Old] × 100
• Ratio between two values
• Average of a row or column
• Percentage contribution = (Part / Total) × 100

Bar Charts

Bar charts compare values visually. The height (or length) of each bar represents the value.

Reading Strategy:

1. Check both axes — what does X-axis represent? What about Y-axis? What’s the scale?
2. Look at the legend — if there are grouped/stacked bars, know what each color represents
3. Estimate before calculating — the visual gives us a quick approximation
4. Watch for broken axes — some charts don’t start at 0, which can be misleading

Types of Bar Charts:

• Simple bar chart — one bar per category
• Grouped bar chart — multiple bars per category (for comparison)
• Stacked bar chart — bars divided into segments (showing composition)

Trick for stacked bars: To find one segment’s value, subtract the bottom of the segment from the top. Don’t eyeball it — read the scale carefully.

Line Charts

Line charts are best for showing trends over time. The slope of the line tells us the rate of change.

Reading Strategy:

1. Steep upward slope = rapid increase
2. Gentle upward slope = slow increase
3. Flat line = no change
4. Downward slope = decrease
5. Point of intersection of two lines = they have the same value at that point

Common Questions:

• “In which year was the growth rate highest?” → Look for the steepest positive slope
• “When did X overtake Y?” → Look for where lines cross
• “What’s the CAGR?” → (Final/Initial)^(1/n) − 1 (rarely tested in aptitude, more for MBA)

Pie Charts

Pie charts show composition — what fraction of the total does each category represent.

Reading Strategy:

1. Check if values are in percentages or degrees — convert to one form if needed
2. Identify the total — sometimes given, sometimes we need to calculate
3. For comparisons: percentages work fine. For actual values: we need the total.

Degree-to-Percentage Quick Reference:

Speed Techniques — The Real Differentiator

In DI, speed matters more than in any other topic. Here are the techniques that save us the most time:

1. Approximate First, Then Refine

We don’t need exact values most of the time. If the options are 23%, 31%, 47%, and 52%, we can round aggressively.

Example: 4873 / 15234 ≈ 5000 / 15000 = 1/3 ≈ 33%. Only option close to this is 31%. Done!

2. Percentage Change Shortcuts

• Increase from 200 to 250: Difference = 50. 50/200 = 1/4 = 25%. Don’t reach for a calculator.
• Fraction to percentage: Know your common fractions: 1/8 = 12.5%, 1/6 ≈ 16.7%, 1/5 = 20%, 1/4 = 25%, 1/3 ≈ 33.3%, 2/5 = 40%, 3/8 = 37.5%

3. Compare Ratios Without Dividing

To compare a/b and c/d, cross multiply: if ad > bc, then a/b > c/d. This avoids division entirely.

4. Use Differences, Not Absolutes

“How much more does A have than B?” → Just find the difference. No need to calculate both values individually if they share a common part.

5. Anchor to Nice Numbers

If we need 17% of 5400: 10% = 540, 5% = 270, 2% = 108, so 17% = 540 + 270 + 108 = 918. Much faster than multiplying 0.17 × 5400.

Worked Examples

Example 1: Table Interpretation

A company’s revenue (in lakhs) over 5 years:

Q: What was the percentage growth from 2022 to 2024?

Growth = (210 − 150) / 150 × 100 = 60/150 × 100 = 40%

Q: In which year was the year-on-year growth highest?

2020→2021: 18/120 = 15% 2021→2022: 12/138 ≈ 8.7% 2022→2023: 30/150 = 20% 2023→2024: 30/180 ≈ 16.7%

Highest growth: 2022 to 2023 (20%)

Example 2: Pie Chart

A monthly budget of Rs. 36,000 is divided as: Food 30%, Rent 25%, Transport 15%, Savings 20%, Others 10%.

Q: How much is spent on Food and Transport together?

Food + Transport = 30% + 15% = 45% of 36,000 = 0.45 × 36,000 = Rs. 16,200

Q: What is the central angle for Rent in a pie chart?

Angle = 25% × 360° = 90°

Q: What’s the ratio of Savings to Others?

20% : 10% = 2:1 (we don’t even need the actual amounts — percentages give us the ratio directly)

Example 3: Bar Chart Reading

Production (in thousand units) at a factory:

Q: In which month was the combined production highest?

Jan: 70, Feb: 80, Mar: 82, Apr: 80, May: 95

Highest: May (95 thousand units)

Q: What is the average production of Product B?

Average = (30 + 35 + 32 + 38 + 40) / 5 = 175 / 5 = 35 thousand units

Example 4: Percentage Contribution

Using the same data, what percentage of total production in March was Product A?

A’s share in March = 50 / (50 + 32) × 100 = 50/82 × 100 ≈ 61%

(Quick approximation: 50/80 = 62.5%, close enough for most MCQ options.)

Example 5: Trend Analysis

Exports of a country (in $ billion): 2019: 280, 2020: 250, 2021: 310, 2022: 340, 2023: 320.

Q: Between which two consecutive years was the absolute change maximum?

2019→2020: −30 (decrease) 2020→2021: +60 (increase) 2021→2022: +30 2022→2023: −20

Maximum absolute change: 2020 to 2021 (+60 billion)

Common Exam Patterns

1. “What is the ratio of X to Y?” — straightforward division, but watch the order
2. “What is the percentage increase/decrease?” — always divide by the OLD value, not the new
3. “In which year was the growth rate highest?” — calculate percentage change for each period
4. “What fraction of the total is X?” — part divided by whole
5. “If the total is T, what is the value of sector X?” — percentage × T

Common Traps to Avoid

• Confusing absolute change with percentage change — a jump from 10 to 20 (100%) is more significant percentage-wise than 100 to 120 (20%), even though both are a change of 10-20 units
• Misreading the scale — always check if the Y-axis starts at 0 or some other value
• Wrong base for percentage — “X is what percent of Y” means (X/Y) × 100, not (Y/X)
• Unit confusion — lakhs vs thousands vs crores. A factor of 10 or 100 error is a guaranteed wrong answer

Practice Problems

Q1: A pie chart shows the following distribution of students in a college (total 4000): Engineering 35%, Commerce 25%, Arts 20%, Science 15%, Others 5%. How many more students are in Engineering than Science?

Q2: Revenue (in crores): 2021: 450, 2022: 540, 2023: 594. Find the percentage growth in 2022 over 2021, and 2023 over 2022. In which year was the growth rate higher?

Q3: A company spends: Salaries 40%, Raw materials 30%, Marketing 15%, R&D 10%, Misc 5%. If total expenditure is Rs. 80 lakhs, and Salaries increase by 20% next year (all else same), what’s the new total expenditure?

Answers

A1: Engineering = 35% of 4000 = 1400. Science = 15% of 4000 = 600. Difference = 800 students.

A2: 2022 growth = (540−450)/450 × 100 = 90/450 × 100 = 20%. 2023 growth = (594−540)/540 × 100 = 54/540 × 100 = 10%. Growth rate was higher in 2022.

A3: Current salaries = 40% of 80 = 32 lakhs. New salaries = 32 × 1.2 = 38.4 lakhs. Increase = 6.4 lakhs. New total = 80 + 6.4 = Rs. 86.4 lakhs.

Data Sufficiency (DS) is a completely different beast from regular math. We’re NOT asked to solve the problem — we’re asked whether the given information is enough to solve it. This subtle difference catches people off guard. Many waste time actually calculating the answer when all they needed to do was check if an answer is determinable. Let’s master the framework.

The Standard DS Framework

Every DS question has:

• A question asking for some value or a yes/no answer
• Two statements providing additional information

We need to determine:

Memory trick for the order: The options spell out “A B C D E” naturally, but think of it as: Alone 1, Balone 2, Combined, Dual (either), Either insufficient.

The Systematic Approach

Here’s a step-by-step process that works every time:

Step 1: Understand the Question

Read the question carefully. What exactly do we need to find? Is it a specific value, or is it a yes/no question?

Critical: Don’t start looking at the statements yet. First, understand what would constitute a sufficient answer.

Step 2: Test Statement 1 Alone

Forget Statement 2 exists. Using ONLY Statement 1 plus any information given in the question stem:

• Can we determine a unique answer?
• If we get a single definite value → Statement 1 is sufficient
• If we get multiple possible values → Statement 1 is insufficient

Step 3: Test Statement 2 Alone

Now forget Statement 1. Using ONLY Statement 2 plus the question stem:

• Can we determine a unique answer?

Step 4: Combine if Needed

If neither statement alone was sufficient, combine both. Together, can we find a unique answer?

The Elimination Flowchart

”Sufficient” Doesn’t Mean “Calculate”

This is the biggest mindset shift. We don’t need to find the actual answer. We just need to determine IF we COULD find it. Often, we can tell that the information is sufficient without doing any heavy math.

Example: “What is x? Statement 1: x² = 25”

We don’t need to say x = 5 or x = −5. We just need to notice: x has TWO possible values, so this statement is NOT sufficient (unless the question restricts x to positive integers or something).

Common Traps

Trap 1: Assuming Additional Information

DS statements are all we get. Don’t assume x is positive, x is an integer, or any other constraint unless explicitly stated in the question or statements.

Example: “What is x? Statement 1: x² = 4”

• x could be 2 or −2 → NOT sufficient
• BUT if the question says “x is a positive integer,” then x = 2 → Sufficient

Trap 2: Not Testing Edge Cases

Always ask: “Is there another value that also satisfies this condition?”

Example: “Is x > 5? Statement 1: x > 3”

• x could be 4 (answer: NO) or x could be 10 (answer: YES)
• Since we get different answers, the statement is NOT sufficient

Trap 3: Confusing “Sufficient for a Unique Value” with “Sufficient to Answer Yes/No”

For yes/no questions, a statement is sufficient if it ALWAYS gives YES or ALWAYS gives NO. A consistent “NO” is still sufficient!

Example: “Is x even? Statement 1: x is divisible by 3”

• x = 3 → No, x = 6 → Yes → NOT sufficient (inconsistent)

Example: “Is x > 0? Statement 1: x² + 1 > 0”

• This is always true for any x → Always YES → Sufficient (but the answer to the question is trivially yes)

Wait — actually x² + 1 > 0 tells us nothing special about x (it’s true for all real numbers). We still don’t know if x > 0. Let’s reconsider: we need to know if x > 0, and knowing x² + 1 > 0 doesn’t help because that’s always true. So it’s NOT sufficient — it gives us no new information about the sign of x.

This illustrates why we need to think carefully rather than rush!

Trap 4: Using Both Statements When Only One is Needed

Test each statement independently first. A common mistake is jumping to “Combined” without checking if one alone works.

Trap 5: Forgetting Information from the Question Stem

The question stem often contains hidden constraints (like “x is a positive integer” or “the triangle is equilateral”). These apply to ALL evaluations.

Shortcut Strategies

Strategy 1: Number Plugging For each statement, try 2-3 different values that satisfy it. If they all give the same answer to the question, the statement is likely sufficient. If they give different answers, it’s definitely not sufficient.

Strategy 2: Count the Unknowns If a question has 2 unknowns and a statement gives 1 equation, we usually can’t solve it (not sufficient). Two independent equations for 2 unknowns → sufficient. But be careful — this is a guideline, not a rule. Sometimes 1 equation is enough (e.g., x² = 0 gives x = 0 uniquely).

Strategy 3: For Yes/No Questions Find one case where the answer is YES and one where it’s NO. If we can find both, the statement is NOT sufficient. If we can only get one type of answer, it IS sufficient.

Worked Examples

Example 1: Value Question

What is the value of x?

• Statement 1: 2x + 3 = 11
• Statement 2: x is a prime number less than 6

S1 alone: 2x + 3 = 11 → 2x = 8 → x = 4. Unique value. Sufficient.

S2 alone: Primes less than 6 are {2, 3, 5}. Three possible values. Not sufficient.

Answer: (A) — Statement 1 alone is sufficient.

Example 2: Yes/No Question

Is x positive?

• Statement 1: x² − x > 0
• Statement 2: x > −2

S1 alone: x² − x > 0 → x(x − 1) > 0. This is true when x < 0 OR x > 1.

• If x = −1 → x is negative (NO)
• If x = 2 → x is positive (YES) Both cases satisfy S1 but give different answers. Not sufficient.

S2 alone: x > −2 allows x = −1 (NO) and x = 5 (YES). Not sufficient.

Combined: x(x − 1) > 0 AND x > −2. From S1, x < 0 or x > 1. From S2, x > −2. Combined: −2 < x < 0 or x > 1.

• If x = −1 (which satisfies both) → NO
• If x = 2 (which satisfies both) → YES Still inconsistent. Not sufficient.

Answer: (E) — Neither statement, even combined, is sufficient.

Example 3: Geometric Question

What is the area of rectangle ABCD?

• Statement 1: The perimeter is 24 cm
• Statement 2: The length is twice the breadth

S1 alone: 2(l + b) = 24 → l + b = 12. Infinite possibilities (11×1, 10×2, etc.). Not sufficient.

S2 alone: l = 2b. We know the ratio but not the actual dimensions. Not sufficient.

Combined: l + b = 12 and l = 2b → 2b + b = 12 → b = 4, l = 8. Area = 32 cm². Unique value. Sufficient.

Answer: (C) — Both statements together are sufficient.

Example 4: Each Alone Sufficient

What is the value of |x|?

• Statement 1: x = −5
• Statement 2: x² = 25

S1 alone: x = −5, so |x| = 5. Unique value. Sufficient.

S2 alone: x² = 25, so x = 5 or x = −5. But |x| = 5 in BOTH cases. Unique value for |x|. Sufficient.

Answer: (D) — Each statement alone is sufficient.

This is a great example of why we test what the question actually asks. The question asks for |x|, not x. Even though S2 gives two values of x, it gives only one value of |x|.

Common Question Types in DS

1. Number properties — Is x odd/even/prime/positive? What is the value of x?
2. Geometry — Area, perimeter, angle of a shape given partial information
3. Ratio/Proportion — Finding actual values when ratios are given
4. Inequalities — Is x > y? Is x positive?
5. Averages and sets — Finding mean, median, or range with partial data

Practice Problems

Q1: What is the value of y?

• Statement 1: 3y − 7 = 8
• Statement 2: y² = 25

Q2: Is n divisible by 6?

• Statement 1: n is divisible by 3
• Statement 2: n is divisible by 2

Q3: What is the area of triangle ABC?

• Statement 1: AB = 10 cm, BC = 8 cm
• Statement 2: Angle B = 90°

Answers

A1: S1: 3y = 15, y = 5. Sufficient. S2: y = 5 or y = −5. Two values. Not sufficient. Answer: (A).

A2: S1: n divisible by 3. Could be 3 (not div by 6) or 6 (div by 6). Not sufficient. S2: n divisible by 2. Could be 2 (not div by 6) or 6 (div by 6). Not sufficient. Combined: n is divisible by both 2 and 3, which means n IS divisible by 6 (since 2 and 3 are coprime). Always YES. Answer: (C).

A3: S1: Two sides known, but without the included angle or the third side, we can’t find the area (the triangle could have different shapes). Not sufficient. S2: Angle B = 90° alone tells us nothing about size. Not sufficient. Combined: Right triangle with legs AB = 10 and BC = 8 (since B is the right angle). Area = ½ × 10 × 8 = 40 cm². Answer: (C).

Series problems are the bread and butter of logical reasoning. We’re given a sequence of numbers or letters, and we need to find the pattern and predict the next term (or find the wrong term). The good news is there are only about 10-12 pattern types that cover almost every question we’ll ever see. Once we learn to spot them, these become free marks.

The Alphabet Position Table

Before we dive into letter series, let’s get this memorized. Every letter series problem becomes easy once we see letters as numbers.

The trick is to memorize EJOTY — E(5), J(10), O(15), T(20), Y(25). Any letter’s position can be found quickly by counting from the nearest landmark.

Common Number Series Patterns

The Difference Method — Our Main Weapon

This is the systematic approach that works for 80% of number series problems.

Step 1: Write out the differences between consecutive terms (first differences).

Step 2: If the first differences don’t show a pattern, take differences of the differences (second differences).

Step 3: Keep going until we see a pattern (constant, arithmetic, geometric, etc.).

Example 1: Find the next term — 2, 5, 10, 17, 26, ?

Let’s take differences:

• 5-2=3, 10-5=5, 17-10=7, 26-17=9
• First differences: 3, 5, 7, 9 — they’re increasing by 2!

So the next first difference = 11. Answer: 26 + 11 = 37

We can also see: these are n² + 1 → 1+1=2, 4+1=5, 9+1=10, 16+1=17, 25+1=26, 36+1=37 ✓

Example 2: Find the next term — 1, 2, 6, 24, 120, ?

Let’s look at ratios instead of differences:

• 2/1=2, 6/2=3, 24/6=4, 120/24=5

The pattern is ×2, ×3, ×4, ×5… so next is ×6. Answer: 120 × 6 = 720

This is the factorial series! 1!, 2!, 3!, 4!, 5!, 6! = 720