Aptitude & Reasoning

SET THEORY — Overview

Introduction to Sets — definition, notation, types

Properties of Sets — Union, Intersection, Complement, Difference

Venn Diagrams — visual representation of set operations

Word Problems — 2 & 3 variable problems

DI/DS Based Problems — Data Interpretation using sets

1. Introduction to Sets

Set: A well-defined collection of distinct objects called elements.

Notation: Sets are denoted by capital letters: A, B, C…
Elements are listed inside curly braces: A = {1, 2, 3, 4}

Membership: 3 ∈ A (3 belongs to A), 5 ∉ A (5 does not belong to A)

Type of Set	Definition	Example
Empty Set (Null Set)	No elements	∅ or { }
Singleton Set	Exactly one element	{5}
Finite Set	Countable elements	{1, 2, 3}
Infinite Set	Uncountable elements	{1, 2, 3, …}
Universal Set (U)	All elements in context	U = {1–10}
Subset (A ⊆ B)	All elements of A are in B	{1,2} ⊆ {1,2,3}
Equal Sets	Same elements	{1,2} = {2,1}

2. Set Operations

UNION (A ∪ B): All elements in A OR B (or both)
If A = {1,2,3}, B = {3,4,5} → A ∪ B = {1,2,3,4,5}

INTERSECTION (A ∩ B): Elements in BOTH A AND B
A ∩ B = {3}

DIFFERENCE (A – B): Elements in A but NOT in B
A – B = {1,2}   |   B – A = {4,5}

COMPLEMENT (A'): Elements in U but NOT in A
If U = {1–5}, A = {1,2,3} → A' = {4,5}

Symmetric Difference (A △ B): Elements in A or B but NOT both
A △ B = (A – B) ∪ (B – A) = {1,2,4,5}

3. Properties of Sets

Commutative Laws:
A ∪ B = B ∪ A   |   A ∩ B = B ∩ A

Associative Laws:
(A ∪ B) ∪ C = A ∪ (B ∪ C)
(A ∩ B) ∩ C = A ∩ (B ∩ C)

Distributive Laws:
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

De Morgan's Laws:
(A ∪ B)' = A' ∩ B'
(A ∩ B)' = A' ∪ B'

Identity Laws:
A ∪ ∅ = A   |   A ∩ U = A
A ∪ U = U   |   A ∩ ∅ = ∅

4. Venn Diagrams

Venn Diagram: A pictorial representation of sets using overlapping circles inside a rectangle (Universal Set).

The overlapping region = Intersection (A ∩ B)
The entire shaded region = Union (A ∪ B)
The region outside circles = Neither A nor B = (A ∪ B)'

KEY FORMULA — 2 Sets:

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

KEY FORMULA — 3 Sets:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C)
                 – n(A∩B) – n(B∩C) – n(A∩C)
                 + n(A∩B∩C)

TRICK — Regions in 3-Set Venn Diagram:
Only A = n(A) – n(A∩B) – n(A∩C) + n(A∩B∩C)
Only B = n(B) – n(A∩B) – n(B∩C) + n(A∩B∩C)
Only C = n(C) – n(A∩C) – n(B∩C) + n(A∩B∩C)
Exactly 2 = n(A∩B) + n(B∩C) + n(A∩C) – 3×n(A∩B∩C)
All 3 = n(A∩B∩C)

5. Word Problems — 2 Variables

Example 1: In a class of 60 students, 35 like Cricket, 30 like Football, and 15 like both. How many like neither?

n(C ∪ F) = n(C) + n(F) – n(C ∩ F)
= 35 + 30 – 15 = 50
Neither = 60 – 50 = 10 students ✓

Example 2: In a survey of 100 people, 70 read newspaper A, 50 read newspaper B, and 20 read neither. How many read both?

Who read at least one = 100 – 20 = 80
n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 70 + 50 – 80 = 40 people ✓

6. Word Problems — 3 Variables

Example 3: In a group of 100 students: 60 study Maths (M), 50 study Physics (P), 40 study Chemistry (C). 20 study M & P, 15 study P & C, 10 study M & C, 5 study all three. Find students studying at least one subject.

n(M∪P∪C) = 60 + 50 + 40 – 20 – 15 – 10 + 5 = 110

⚠️ This exceeds 100 — means the data has overlaps. Students studying at least one = Min of 100 and calculated value.
Answer: 100 students (all study at least one) ✓

Example 4: 200 people surveyed. 100 like Tea (T), 80 like Coffee (C), 60 like Juice (J). 40 like T&C, 30 like C&J, 20 like T&J, 10 like all three. Find n(T∪C∪J) and those who like none.

n(T∪C∪J) = 100 + 80 + 60 – 40 – 30 – 20 + 10 = 160
None = 200 – 160 = 40 people ✓

Q1: In a class of 80, 45 like Hindi, 35 like English, 20 like both. How many like neither?

Solution: n(H∪E) = 45 + 35 – 20 = 60. Neither = 80 – 60 = 20 students ✓

Q2: 150 students: 90 play Cricket, 70 play Football, 30 play both. Find those who play exactly one sport.

Solution: Only Cricket = 90–30 = 60. Only Football = 70–30 = 40. Exactly one = 60 + 40 = 100 students ✓

7. DI/DS Based Set Problems

Example 5 (DI Type): A survey of 500 people: 300 use Facebook, 200 use Instagram, 150 use Twitter. 80 use F&I, 60 use I&T, 50 use F&T, 30 use all three. Find: (a) At least one, (b) Exactly two, (c) None.

n(F∪I∪T) = 300+200+150 – 80–60–50 + 30 = 490
Exactly two = (80+60+50) – 3×30 = 190–90 = 100 people
None = 500 – 490 = 10 people ✓

Q3 (DS): How many students passed both Maths and Science? Statement I: 70 passed Maths. Statement II: 50 passed Science, 30 passed both.

Solution: Statement II alone gives "30 passed both" directly. Statement II alone is sufficient. ✓

Q4: In a group of 75 people, Only Tea=30, Only Coffee=20, Both=15. Find those who like neither.

Solution: n(T∪C) = 30+20+15 = 65. Neither = 75–65 = 10 people ✓

COMMON MISTAKES TO AVOID:
1. Always subtract intersection ONCE in 2-set, but add back in 3-set formula
2. "At least one" = Total – None (use inclusion-exclusion)
3. "Exactly two" ≠ intersection of two sets (must subtract the triple overlap)
4. Draw the Venn diagram ALWAYS — saves time in 3-variable problems
5. "Only A" ≠ n(A) → Only A = n(A) – n(A∩B) – n(A∩C) + n(A∩B∩C)

SIMPLE INTEREST — Overview

Concept & Basic Formula: SI = PRT/100

Half-Yearly, Quarterly & Monthly rates

Amount being n times Principal problems

Data Sufficiency Questions

1. Basic Concepts & Formula

Principal (P): Original sum of money borrowed or lent.
Rate (R): Percentage of interest charged per annum (per year).
Time (T): Duration of the loan/investment (in years).
Simple Interest (SI): Interest calculated only on the principal every year.

CORE FORMULAS:

SI = (P × R × T) / 100

Amount (A) = P + SI = P + (PRT/100) = P(1 + RT/100)

P = (SI × 100) / (R × T)
R = (SI × 100) / (P × T)
T = (SI × 100) / (P × R)

QUICK TRICK: If interest doubles the principal → SI = P
So T = 100/R years   |   R = 100/T %

2. Half-Yearly, Quarterly & Monthly Rates

When Rate is given Per Annum but time is in months/half-years:

Half-Yearly: Time in half-years → T = (months/6) or (years × 2)
Rate stays annual → use T in years = months/12

Quarterly: Time in quarters → T = months/12 years

Monthly Rate (r%): Annual Rate R% → Monthly rate = R/12 %
SI = P × (R/12) × n / 100  (where n = number of months)

Example 1: Find SI on Rs 5000 at 8% per annum for 9 months.

T = 9/12 = 3/4 years
SI = (5000 × 8 × 3/4) / 100 = (5000 × 6) / 100 = Rs 300 ✓

Example 2: Find SI on Rs 12000 at 6% per annum for 2 years 6 months.

T = 2.5 years
SI = (12000 × 6 × 2.5) / 100 = Rs 1800 ✓

Example 3 (Half-Yearly): Rs 8000 at 10% p.a. for 18 months, compounded half-yearly (SI method — treating separately).

T = 18/12 = 1.5 years
SI = (8000 × 10 × 1.5) / 100 = Rs 1200 ✓

3. Amount Being n Times Principal

When Amount = n × Principal:

A = nP → SI = A – P = nP – P = (n–1)P

Since SI = PRT/100:
(n–1)P = PRT/100
T = (n–1) × 100 / R

OR:
R = (n–1) × 100 / T

Example 4: At what rate % p.a. will a sum double itself in 8 years?

n = 2 (doubles means Amount = 2P)
R = (2–1) × 100 / 8 = 100/8 = 12.5% p.a. ✓

Example 5: In how many years will Rs 5000 become Rs 8000 at 6% p.a. SI?

SI = 8000 – 5000 = 3000
T = (SI × 100) / (P × R) = (3000 × 100) / (5000 × 6) = 300000/30000 = 10 years ✓

Example 6 (Triple): A sum becomes 3 times in 20 years at SI. Find rate.

n = 3, T = 20
R = (3–1) × 100 / 20 = 200/20 = 10% p.a. ✓

4. Data Sufficiency Questions

DS Strategy for SI: You need EXACTLY P, R, and T to find SI.
Check if any 2 statements together give all 3 variables.

Q1: Find SI on a sum. Statement I: The sum is Rs 4000. Statement II: Rate is 5% p.a. for 3 years.

Solution: Statement I gives P. Statement II gives R and T. Both statements together: SI = (4000 × 5 × 3)/100 = Rs 600. Answer: Both statements needed. ✓

Q2: SI on P for 4 years = Rs 1200. Statement I: P = Rs 5000. Statement II: R = 6% p.a.

Solution: Check Statement I: SI = 1200, T = 4, P = 5000 → R = 1200×100/(5000×4) = 6%. Statement I alone is sufficient. Check Statement II: P = 1200×100/(6×4) = 5000. Statement II alone is also sufficient. Answer: Either statement alone is sufficient. ✓

Q3: Rs 6000 at 8% p.a. Find SI for half-year.

Solution: T = 6/12 = 0.5 years. SI = (6000 × 8 × 0.5)/100 = Rs 240 ✓

5. Special Types of SI Problems

Type A — Two Parts at Different Rates:
A sum P is split into two parts P₁ and P₂ at rates R₁ and R₂ for T years.
Total SI = P₁R₁T/100 + P₂R₂T/100
And P₁ + P₂ = P → Solve simultaneously

Example 7: Rs 5000 is split into two parts. One earns 6% p.a. and other earns 9% p.a. for 2 years. Total SI = Rs 660. Find each part.

Let Part1 = x → Part2 = 5000–x
x×6×2/100 + (5000–x)×9×2/100 = 660
0.12x + 900 – 0.18x = 660
–0.06x = –240 → x = Rs 4000
Part1 = Rs 4000 at 6%, Part2 = Rs 1000 at 9% ✓

Type B — Same Interest Period Comparison:
If same P at rate R₁ gives SI₁ and at R₂ gives SI₂ for same T:
SI₁/SI₂ = R₁/R₂

Example 8: SI on a sum at 5% for 3 years is Rs 900. Find SI at 8% for same period.

SI₂/SI₁ = R₂/R₁ = 8/5
SI₂ = 900 × 8/5 = Rs 1440 ✓

KEY SHORTCUTS — SIMPLE INTEREST:
1. Double in T years → R = 100/T %
2. Triple in T years → R = 200/T %
3. n times in T years → R = (n–1)×100/T %
4. Difference in SI for t1 & t2 years = P × R × (t2–t1) / 100
5. Two-part problems → Set up 2 equations with 2 unknowns
6. SI ratio same as rate ratio when P and T are same

COMPOUND INTEREST — Overview

CI Formula — Annual, Half-Yearly, Quarterly compounding

Difference between SI and CI for 2 and 3 years

Amount being n times Principal

Data Sufficiency Questions

1. Basic Concepts & Formula

Compound Interest (CI): Interest calculated on the principal AND on the accumulated interest of previous periods. Interest is "compounded" — added to principal at the end of each period.

CORE CI FORMULA (Annual Compounding):

A = P × (1 + R/100)^T
CI = A – P = P[(1 + R/100)^T – 1]

HALF-YEARLY COMPOUNDING:
A = P × (1 + R/200)^(2T) [Rate halved, time doubled]

QUARTERLY COMPOUNDING:
A = P × (1 + R/400)^(4T) [Rate quartered, time × 4]

MONTHLY COMPOUNDING:
A = P × (1 + R/1200)^(12T)

TRICK for 2 years at R%:
A = P × (100+R)² / 10000
CI = P(2R/100 + R²/10000) = SI + P(R/100)²

CI for 1st year = SI for 1st year (always!)

2. Compounding Periods

Example 1 (Annual): Find CI on Rs 10000 at 10% p.a. for 2 years.

A = 10000 × (1.1)² = 10000 × 1.21 = Rs 12100
CI = 12100 – 10000 = Rs 2100 ✓

Example 2 (Half-Yearly): Find CI on Rs 8000 at 10% p.a. compounded half-yearly for 1 year.

Rate per half-year = 5%, n = 2 periods
A = 8000 × (1.05)² = 8000 × 1.1025 = Rs 8820
CI = 8820 – 8000 = Rs 820 ✓

Example 3 (Quarterly): Rs 5000 at 8% p.a. compounded quarterly for 6 months.

Rate per quarter = 2%, n = 2 quarters
A = 5000 × (1.02)² = 5000 × 1.0404 = Rs 5202
CI = Rs 202 ✓

3. Difference Between SI and CI

For 2 years:
CI – SI = P × (R/100)²

For 3 years:
CI – SI = P × (R/100)² × (3 + R/100)
OR = P × R² × (300 + R) / 1000000

Example 4: Find difference between CI and SI on Rs 5000 at 8% p.a. for 2 years.

Diff = P × (R/100)² = 5000 × (0.08)² = 5000 × 0.0064 = Rs 32 ✓

Example 5: For 3 years at 10%, P = Rs 1000. Find CI – SI.

CI – SI = 1000 × (10/100)² × (3 + 10/100)
= 1000 × 0.01 × 3.1 = Rs 31 ✓

Example 6: SI on Rs P for 2 years at 5% is Rs 500. Find CI for same period.

SI = PRT/100 → 500 = P×5×2/100 → P = Rs 5000
CI = 5000[(1.05)² – 1] = 5000 × 0.1025 = Rs 512.50 ✓

4. Amount Being n Times Principal (CI)

When Amount = n × P after T years:

n = (1 + R/100)^T

If sum doubles in T₁ years at R%,
It becomes 4 times in 2T₁ years (squares each period!)
It becomes 8 times in 3T₁ years

Key Rule: In CI, if P doubles in n years → it becomes 2^k times in k×n years

Example 7: A sum becomes Rs 2662 in 3 years at 10% CI. Find the Principal.

A = P(1.1)³ = P × 1.331
P = 2662/1.331 = Rs 2000 ✓

Example 8: If P doubles in 5 years at CI, in how many years does it become 8 times?

8 = 2³ → 3 doublings needed → 3 × 5 = 15 years ✓

5. Comparison Table — SI vs CI

Parameter	Simple Interest	Compound Interest
Formula	SI = PRT/100	CI = P[(1+R/100)^T – 1]
Base	Always on original P	On P + previous interest
1st year interest	Same as CI	Same as SI
Growth	Linear	Exponential
Which is more?	Less (after 1 yr)	More (after 1 yr)
Doubles in (at 10%)	10 years	~7.27 years

6. Practice Problems — CI

Q1: Find CI on Rs 15000 at 8% p.a. compounded annually for 2 years.

Solution: A = 15000×(1.08)² = 15000×1.1664 = Rs 17496. CI = 17496–15000 = Rs 2496 ✓

Q2: The CI on a sum for 2 years at 10% is Rs 2100. What is the principal?

Solution: CI = P[(1.1)²–1] = P×0.21 = 2100 → P = 2100/0.21 = Rs 10000 ✓

Q3: CI for first year = Rs 800, CI for second year = Rs 880. Find rate and principal.

Solution: Rate R = (880–800)/800 × 100 = 80/800 × 100 = 10% p.a.
CI for 1st year = SI for 1st year = PR/100 → 800 = P×10/100 → P = Rs 8000 ✓

SHORTCUTS — COMPOUND INTEREST:
1. At 10% CI: After 2 yr → ×1.21, After 3 yr → ×1.331
2. At 20% CI: After 2 yr → ×1.44, After 3 yr → ×1.728
3. CI – SI for 2 yr = P(R/100)² — memorize this!
4. If CI for 1st year = x and CI for 2nd year = y → R = (y–x)/x × 100
5. Population growth/depreciation also uses CI formula
6. Rule of 72: Years to double ≈ 72/R (approximate)

LINEAR & QUADRATIC EQUATIONS — Overview

Linear Equations — one variable, two variable concepts

Application Problems — word problems using linear equations

Quadratic Equations — factorization, formula method

Application Problems — quadratic in real life

DS Based Problems

1. Linear Equations — Basics

Linear Equation: An equation of degree 1 (highest power of variable = 1).

One Variable: ax + b = 0 → x = –b/a
Two Variables: ax + by = c (graph = straight line)

System of 2 equations: Two equations needed to find two unknowns.

Solving 2 Linear Equations:

Method 1 — Substitution: Isolate one variable, substitute in other
Method 2 — Elimination: Add/subtract equations to cancel one variable
Method 3 — Cross-Multiplication:
For a₁x + b₁y = c₁ and a₂x + b₂y = c₂:
x = (b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁)
y = (a₂c₁ – a₁c₂) / (a₁b₂ – a₂b₁)

2. Application Problems — Linear

Standard Approach:
Step 1: Assign variables to unknowns
Step 2: Frame equations from given conditions
Step 3: Solve using substitution or elimination
Step 4: Verify answer in original conditions

Example 1: Sum of two numbers is 50 and their difference is 10. Find the numbers.

Let x + y = 50 … (i) and x – y = 10 … (ii)
Adding: 2x = 60 → x = 30
y = 50 – 30 = 20
Numbers: 30 and 20 ✓

Example 2: A father is 3 times as old as his son. 10 years later, he will be twice as old. Find current ages.

Father = 3x, Son = x
3x + 10 = 2(x + 10)
3x + 10 = 2x + 20 → x = 10
Son = 10 years, Father = 30 years ✓

Example 3: 2 kg of apples and 3 kg of oranges cost Rs 280. 4 kg apples and 1 kg oranges cost Rs 260. Find cost per kg.

2a + 3o = 280 … (i)
4a + o = 260 … (ii)
From (ii): o = 260 – 4a. Sub in (i):
2a + 3(260–4a) = 280
2a + 780 – 12a = 280 → –10a = –500 → a = 50
o = 260 – 200 = 60
Apple = Rs 50/kg, Orange = Rs 60/kg ✓

3. Quadratic Equations — Basics

Quadratic Equation: ax² + bx + c = 0 (where a ≠ 0)

A quadratic ALWAYS has 2 roots (real or imaginary).

Sum of roots (α + β) = –b/a
Product of roots (α × β) = c/a

METHOD 1 — Factorization:
Split middle term: find two numbers whose sum = b and product = ac
x² – 5x + 6 = 0 → (x–2)(x–3) = 0 → x = 2 or 3

METHOD 2 — Quadratic Formula (Shridharacharya):
x = [–b ± √(b² – 4ac)] / 2a

DISCRIMINANT (D) = b² – 4ac:
D > 0 → 2 distinct real roots
D = 0 → 2 equal real roots
D < 0 → No real roots (imaginary)

4. Application Problems — Quadratic

Example 4: A number added to its square gives 42. Find the number.

x + x² = 42 → x² + x – 42 = 0
Factors: (x + 7)(x – 6) = 0 → x = 6 or –7
Answer: 6 (or –7 if negative allowed) ✓

Example 5: Speed of a boat in still water is 15 km/h. Stream speed is v. It takes 4 hours 30 min to go 60 km upstream and 60 km downstream. Find stream speed.

60/(15–v) + 60/(15+v) = 4.5
60(15+v) + 60(15–v) = 4.5(225–v²)
1800 = 4.5(225–v²)
225–v² = 400 → v² = –175 (check data)

Revised: Take 6 hours total:
60(15+v+15–v) = 6(225–v²)
1800 = 6(225–v²) → v² = 225–300 = –75

Example: 60/(15–v) + 60/(15+v) = 8 hrs
60×30 = 8(225–v²) → 225–v² = 225 → v = 0 (still water, verify with actual numbers)

Example 6 (Standard): Two numbers differ by 5 and their product is 84. Find them.

x(x+5) = 84 → x² + 5x – 84 = 0
D = 25 + 336 = 361 → √361 = 19
x = (–5 ± 19)/2 → x = 7 or x = –12
Numbers: 7 and 12 ✓

5. DS Based Problems

Q1: Find two numbers. Statement I: Their sum = 15. Statement II: Their product = 56.

Solution: Statement I → x+y=15. Statement II → xy=56. Together: x²–15x+56=0 → (x–7)(x–8)=0 → x=7, y=8. Both statements needed. ✓

Q2: Find roots of x² – px + 12 = 0. Statement I: p = 7. Statement II: One root is 3.

Solution: Statement I: x²–7x+12=0 → (x–3)(x–4)=0 → roots 3,4. Sufficient alone.
Statement II: If x=3, then 9–3p+12=0 → p=7. Then other root=4. Sufficient alone.
Answer: Either statement alone is sufficient. ✓

6. Nature of Roots & Special Cases

FORMING QUADRATIC FROM ROOTS:
If roots are α and β:
Equation: x² – (α+β)x + αβ = 0

SPECIAL CASES:
Roots are reciprocals → Product = 1 → c/a = 1 → c = a
Roots are equal in magnitude but opposite sign → Sum = 0 → b = 0
One root is zero → Product = 0 → c = 0
Both roots positive → Sum > 0 and Product > 0
Both roots negative → Sum < 0 and Product > 0

Example 7: Form a quadratic equation with roots 3 and –5.

Sum = 3+(–5) = –2, Product = 3×(–5) = –15
Equation: x² – (–2)x + (–15) = 0
x² + 2x – 15 = 0 ✓

Example 8 (Max/Min value): Find minimum value of x² – 6x + 11.

f(x) = x² – 6x + 11 = (x–3)² + 2
Minimum at x = 3: f(3) = 0 + 2 = 2 ✓
(For ax²+bx+c: Min/Max at x = –b/2a → Value = c – b²/4a)

KEY TRICKS — EQUATIONS:
1. For linear: Always set up equations from given conditions methodically
2. For quadratic: Check sum and product of roots before factorizing
3. Discriminant D = b²–4ac → always check before solving
4. Age problems: Let present age = x, then years ago = x–n, years later = x+n
5. If roots are equal → D = 0 → b² = 4ac
6. Minimum value of ax²+bx+c (a>0) = c – b²/4a at x = –b/2a

GEOMETRY — Overview

Lines & Angles — types, properties, pair of angles

Triangles — types, properties, congruence, similarity

Polygons — interior & exterior angles, regular polygons

Circles — chord, arc, tangent, incircle, circumcircle

1. Lines & Angles

Angle Type	Measure	Property
Acute	0° < θ < 90°	Less than right angle
Right	θ = 90°	Perpendicular lines
Obtuse	90° < θ < 180°	Greater than right angle
Straight	θ = 180°	Straight line
Reflex	180° < θ < 360°	More than straight
Complete	θ = 360°	Full rotation

ANGLE PAIR PROPERTIES:

Complementary: A + B = 90°
Supplementary: A + B = 180°
Vertically Opposite: Equal (∠AOB = ∠COD)

PARALLEL LINES (cut by transversal):
Corresponding angles: Equal (F-shape)
Alternate interior angles: Equal (Z-shape)
Co-interior (same-side interior): Sum = 180° (C-shape)

2. Properties of Triangles

Angle Sum Property: Sum of angles = 180°
Exterior Angle: = Sum of two non-adjacent interior angles
Triangle Inequality: Sum of any two sides > third side

Types by Sides: Equilateral (all equal), Isosceles (two equal), Scalene (all different)
Types by Angles: Acute (all <90°), Right (one =90°), Obtuse (one >90°)

PYTHAGORAS THEOREM (Right Triangle):
a² + b² = c² (c = hypotenuse)

Common Pythagorean Triplets:
3–4–5  |  5–12–13  |  8–15–17  |  7–24–25

Special Triangles:
45°–45°–90°: sides = a : a : a√2
30°–60°–90°: sides = a : a√3 : 2a

Congruence (≅): Same shape AND size
Tests: SSS, SAS, ASA, AAS, RHS

Similarity (~): Same shape, different size (angles equal, sides proportional)
Tests: AA, SSS, SAS

Basic Proportionality Theorem (BPT): If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

3. Polygons

For a polygon with n sides:

Sum of interior angles = (n–2) × 180°
Each interior angle (regular polygon) = (n–2) × 180° / n
Each exterior angle (regular polygon) = 360° / n
Number of diagonals = n(n–3) / 2

Key polygons:
Triangle (n=3): interior sum = 180°
Quadrilateral (n=4): 360° | Pentagon (n=5): 540°
Hexagon (n=6): 720° | Octagon (n=8): 1080°

4. Properties of Circle

Circle Terminology:
Radius (r): Center to circumference
Diameter (d): 2r — longest chord
Chord: Line segment joining two points on circle
Arc: Part of circumference
Tangent: Line touching circle at exactly one point
Secant: Line intersecting circle at two points

KEY CIRCLE THEOREMS:

Angle at center = 2 × angle at circumference (same arc)
Angles in semicircle = 90° (angle in semicircle)
Angles in same segment are equal
Opposite angles of cyclic quadrilateral = 180°
Tangent is perpendicular to radius at point of contact
Two tangents from external point are equal in length

Incircle & Circumcircle of Triangle:

Incircle: Circle inside triangle touching all 3 sides
Inradius (r) = Area / Semi-perimeter = Δ / s
where s = (a+b+c)/2

Circumcircle: Circle outside triangle passing through all 3 vertices
Circumradius (R) = abc / (4Δ)
For right triangle: R = hypotenuse / 2

Example: Find inradius and circumradius of a 3–4–5 right triangle.

s = (3+4+5)/2 = 6, Δ = (1/2)×3×4 = 6
r = Δ/s = 6/6 = 1 unit
R = hypotenuse/2 = 5/2 = 2.5 units ✓

5. Congruence & Similarity

Test	Congruence (≅)	Similarity (~)
Condition 1	SSS — all 3 sides equal	AAA/AA — angles equal
Condition 2	SAS — 2 sides & included angle	SSS — sides proportional
Condition 3	ASA — 2 angles & included side	SAS — 2 sides proportional + angle
Condition 4	AAS — 2 angles & non-included side	—
Condition 5	RHS — right angle, hyp, side	—
Size	Exactly same size	Same shape, diff size
Ratio	Sides ratio = 1:1	Sides proportional (k:1)

SIMILARITY RATIOS:
If triangles are similar with ratio k:1
→ Sides ratio = k:1
→ Perimeter ratio = k:1
→ Area ratio = k²:1
→ Volume ratio = k³:1 (for 3D)

Mid-Point Theorem: Line joining midpoints of 2 sides of a triangle is parallel to the 3rd side and half its length.

Example: Two similar triangles have sides in ratio 3:5. If smaller triangle area = 36 cm², find larger triangle area.

Area ratio = k² = (3/5)² = 9/25
36/Larger = 9/25 → Larger = 36×25/9 = 100 cm² ✓

GEOMETRY — MUST REMEMBER:
1. Exterior angle = sum of 2 non-adjacent interior angles
2. Each exterior angle of regular polygon = 360°/n
3. Angle in semicircle = 90° (very frequently asked!)
4. Tangent ⊥ Radius at point of contact
5. Inradius = Area / s  |  Circumradius = abc/(4×Area)
6. Similar triangles: Area ratio = (side ratio)²

PERIMETER & AREA — Overview

Triangles — different types and their area formulas

Quadrilaterals — square, rectangle, rhombus, parallelogram, trapezium

Other Polygons — regular hexagon etc.

Circle — circumference, area, sector, segment

Miscellaneous — path, shaded regions

1. Perimeter & Area of Triangles

TRIANGLE FORMULAS:

Perimeter = a + b + c

Area (General): = (1/2) × base × height

Heron's Formula: Area = √[s(s–a)(s–b)(s–c)]
where s = (a+b+c)/2

Equilateral Triangle (side a):
Perimeter = 3a
Area = (√3/4)a²
Height = (√3/2)a

Right Triangle (legs a,b; hypotenuse c):
Area = (1/2) × a × b
Perimeter = a + b + c

Example 1: Find area of triangle with sides 5, 12, 13.

Check: 5²+12² = 25+144 = 169 = 13² → Right triangle!
Area = (1/2)×5×12 = 30 sq units ✓

Example 2 (Heron's): Triangle with sides 7, 8, 9.

s = (7+8+9)/2 = 12
Area = √[12×5×4×3] = √720 = 12√5 ≈ 26.83 sq units ✓

2. Quadrilaterals

Shape	Perimeter	Area
Square (side a)	4a	a²
Rectangle (l×b)	2(l+b)	l × b
Parallelogram	2(a+b)	base × height
Rhombus (d₁,d₂)	4a	(d₁×d₂)/2
Trapezium	a+b+c+d	(a+b)/2 × h

Example 3: A rhombus has diagonals 16 cm and 12 cm. Find its side and area.

Side = √[(d₁/2)²+(d₂/2)²] = √[64+36] = √100 = 10 cm
Area = (16×12)/2 = 96 cm² ✓

Example 4: Trapezium with parallel sides 8 cm and 14 cm, height 6 cm. Find area.

Area = (8+14)/2 × 6 = 11 × 6 = 66 cm² ✓

3. Circle — Circumference & Area

CIRCLE FORMULAS:

Circumference (Perimeter) = 2πr = πd
Area = πr²

Sector (angle θ at center):
Arc length = (θ/360°) × 2πr
Sector Area = (θ/360°) × πr²

Segment: Area = Area of sector – Area of triangle

Use π = 22/7 or 3.14 as needed

Example 5: Find circumference and area of circle with r = 7 cm. (π = 22/7)

Circumference = 2 × 22/7 × 7 = 44 cm
Area = 22/7 × 7² = 22/7 × 49 = 154 cm² ✓

Example 6 (Sector): Sector angle = 60°, r = 14 cm. Find arc length and sector area.

Arc = (60/360) × 2 × 22/7 × 14 = (1/6) × 88 = 14.67 cm
Area = (60/360) × 22/7 × 14² = (1/6) × 616 = 102.67 cm² ✓

4. Miscellaneous Problems

Example 7 (Path): A rectangular garden 50m × 30m has a path 2m wide inside. Find area of path.

Outer area = 50 × 30 = 1500 m²
Inner area = (50–4) × (30–4) = 46 × 26 = 1196 m²
Path area = 1500 – 1196 = 304 m² ✓

Example 8 (Shaded Region): Square of side 14 cm with largest circle inscribed. Find shaded (non-circle) area.

r = 14/2 = 7 cm
Square area = 196 cm²
Circle area = 22/7 × 49 = 154 cm²
Shaded area = 196 – 154 = 42 cm² ✓

AREA SHORTCUTS:
1. Square diagonal (d): Area = d²/2
2. Equilateral Δ (side a): Area = (√3/4)a² ≈ 0.433a²
3. Regular Hexagon (side a): Area = (3√3/2)a², Perimeter = 6a
4. Semi-circle: Area = πr²/2, Perimeter = πr + 2r
5. Path problems: Always subtract inner from outer area

CLOCKS — Overview

Angles traversed by hour and minute hand

Speed of MH (minute hand) and HH (hour hand)

Angle between hands at a specific time

Time when hands coincide or are at 180°

Faulty clocks — gaining/losing time

Mirror image of clock readings

1. Basic Speed Concepts

CLOCK SPEED FORMULAS:

Minute Hand (MH):
Completes 360° in 60 minutes
Speed = 360°/60 = 6° per minute

Hour Hand (HH):
Completes 360° in 12 hours = 720 minutes
Speed = 360°/720 = 0.5° per minute

Relative Speed of MH w.r.t. HH:
= 6 – 0.5 = 5.5° per minute

In 1 minute: MH gains 5.5° over HH

2. Angle Between Hands at Given Time

ANGLE FORMULA:

At H hours M minutes:
θ = |30H – 5.5M|

If θ > 180°, take 360° – θ (smaller angle)

OR:
Minute hand position = 6M degrees (from 12)
Hour hand position = 30H + 0.5M degrees (from 12)
Angle = |Minute – Hour|

Example 1: Find angle between hands at 3:20.

θ = |30×3 – 5.5×20| = |90 – 110| = |–20| = 20° ✓

Example 2: Find angle between hands at 6:30.

θ = |30×6 – 5.5×30| = |180 – 165| = 15° ✓

Example 3: Find angle at 9:45.

θ = |30×9 – 5.5×45| = |270 – 247.5| = 22.5° ✓

3. When Hands Coincide / Are at Specific Angle

Hands coincide (0° apart):
Time after H o'clock = (60H/11) minutes past H
Hands coincide 22 times in 24 hours (every 65 5/11 minutes)

Hands are opposite (180° apart):
Happens 22 times in 24 hours
First occurrence after 12 = 32 8/11 minutes past 12

Hands at right angle (90°):
Happens 44 times in 24 hours

Example 4: When do hands of a clock coincide between 3 and 4?

Time = (60×3)/11 = 180/11 = 16 4/11 minutes past 3
Answer: 3 hours 16 4/11 minutes ✓

Example 5: At what time between 4 and 5 are the hands at 180°?

θ = 180°, use: M = (180 + 30H)/5.5 or (30H – 180)/5.5
M = (180 + 120)/5.5 = 300/5.5 = 54 6/11 minutes
Answer: 4:54 6/11 ✓

4. Faulty Clocks

Gaining clock: Runs FAST — shows more time than actual
Losing clock: Runs SLOW — shows less time than actual

Formula for Actual Time:
If a clock shows T hours and gains/loses x minutes per hour:
Actual time elapsed = T × 60 / (60 ± x)

FAULTY CLOCK SHORTCUTS:

If clock gains x min in y hours:
For clock to show correct time after being set right:
Time required = 12 hours × 60 / x (minutes gained per hour must cover full dial)

If set correct at 12:00 noon, when will it show correct time again?
Gaining x min/hr → Gains 12 hrs in (12×60/x) hours of real time

Example 6: A clock gains 5 minutes every hour. If it shows 10:00 AM, what is the actual time?

The clock shows 60 min for every 55 min of actual time.
Assume clock started at 12:00. It now shows 10 hrs elapsed = 600 clock-minutes.
Actual time elapsed = 600 × 55/60 = 550 minutes = 9 hrs 10 min
Actual time: 9:10 AM ✓

5. Mirror Image of Clock

MIRROR IMAGE TRICK:

Mirror image time = 11:60 – given time
OR
Mirror image time = 12:00 – given time (if minutes = 0)

Examples:
Mirror of 4:30 = 11:60 – 4:30 = 7:30
Mirror of 7:15 = 11:60 – 7:15 = 4:45
Mirror of 3:00 = 12:00 – 3:00 = 9:00

Example 7: A clock shows 8:20 in mirror. What is the actual time?

Actual time = 11:60 – 8:20 = 3:40 ✓

6. Frequency of Angles in a Day

Condition	In 12 Hours	In 24 Hours	Interval
Hands coincide (0°)	11 times	22 times	65 5/11 min
Hands opposite (180°)	11 times	22 times	65 5/11 min
Hands at right angle (90°)	22 times	44 times	32 8/11 min
Hands at straight line (0° or 180°)	22 times	44 times	32 8/11 min

IMPORTANT NOTE:
Many assume hands coincide 24 times in a day — WRONG!
Actual = 22 times in 24 hours (not 12 nor 24).
This is because between 11–12 and 23–24, the hands do NOT coincide except at 12:00 exactly.

Similarly, right angles = 44 times in 24 hours (not 48).

Example 8 — Between 5 and 6: When are hands at right angle between 5 and 6?

For 90°: |30×5 – 5.5M| = 90
Case 1: 150 – 5.5M = 90 → M = 60/5.5 = 10 10/11 min
Case 2: 150 – 5.5M = –90 → 5.5M = 240 → M = 43 7/11 min
Answers: 5:10 10/11 and 5:43 7/11 ✓

CLOCK — GOLDEN RULES:
1. MH speed = 6°/min, HH speed = 0.5°/min, Relative = 5.5°/min
2. Angle formula: θ = |30H – 5.5M|
3. Coincidence between H & H+1 = at (60H/11) min past H
4. Mirror image: Subtract from 11:60
5. Faulty clock: 60 fast/slow min tells you clock-to-real time ratio
6. Hands coincide 22 times / right angle 44 times in 24 hours

SYLLOGISM — Overview

Definition & structure of syllogism

Venn Diagram method for syllogism

Basic problems — All, Some, No type

Either-Or cases

Possibility cases

Data sufficiency based syllogism

1. Definition & Structure

Syllogism: A type of logical argument consisting of two premises and a conclusion. Based on deductive reasoning — conclusion MUST follow from the premises.

Structure:
Premise 1: Statement about a group
Premise 2: Statement connecting groups
Conclusion: What MUST be true

Types of Statements:
• Universal Affirmative (A): "All A are B"
• Universal Negative (E): "No A is B"
• Particular Affirmative (I): "Some A are B"
• Particular Negative (O): "Some A are not B"

2. Venn Diagram Method

How to draw Venn Diagrams for Syllogism:

"All A are B": Circle A is INSIDE Circle B
"No A is B": Circles A and B are SEPARATE (no overlap)
"Some A are B": Circles A and B OVERLAP partially
"Some A are not B": Part of A is outside B

CONCLUSION RULES (Standard Syllogism):

All + All → All (A→B→C: All A are C)
All + No → No (All A are B, No B is C → No A is C)
Some + All → Some (Some A are B, All B are C → Some A are C)
Some + No → Some Not (Some A are B, No B is C → Some A are not C)
No + All → Some Not (reversed)
All + Some → No conclusion
Some + Some → No conclusion

3. Basic Syllogism Problems

Example 1:
Premise 1: All cats are animals.
Premise 2: All animals are living beings.
Conclusion: All cats are living beings.

Type: All + All → All ✓ Valid conclusion ✓

Example 2:
Premise 1: All roses are flowers.
Premise 2: No flowers are stones.
Conclusion I: No roses are stones.
Conclusion II: Some roses are not stones.

Type: All + No → No (for Conclusion I ✓)
Conclusion II also follows (as subset of I)
Both conclusions follow. ✓

Example 3:
Premise 1: Some dogs are white.
Premise 2: All white things are beautiful.
Conclusion: Some dogs are beautiful.

Type: Some + All → Some ✓ Valid ✓

4. Either-Or Cases

Either-Or Possibility: When neither conclusion follows independently, but together they cover all possibilities.

Condition for Either-Or:
1. Both conclusions must be complementary (together cover all cases)
2. Neither follows independently
3. Subject and predicate are the same but one is affirmative, one negative

Example 4:
Premises: Some A are B. Some B are C.
Conclusion I: Some A are C.
Conclusion II: No A is C.

Neither follows from "Some + Some" directly.
They are complementary (either some or none must be true).
Answer: Either I or II follows. ✓

5. Possibility Cases

Possibility Rules:

"All A being B is a possibility" → TRUE unless "No A is B" is certain
"Some A being B is a possibility" → TRUE unless "No A is B" is certain
"All A are B" → Then "Some A are B" and "Some B are A" are definite

Key: Possibility = the conclusion CAN be true in at least one valid Venn diagram
Definite conclusion = the conclusion is TRUE in ALL possible Venn diagrams

Example 5:
Premises: Some books are pens. Some pens are pencils.
Conclusion I: All books being pencils is a possibility.
Conclusion II: Some books are pencils.

Conclusion I: POSSIBLE (no definite exclusion) ✓
Conclusion II: NOT definite (Some+Some gives no conclusion)
Only Conclusion I follows (as possibility). ✓

6. Data Sufficiency in Syllogism

Q1: Are all managers leaders? Statement I: All leaders are managers. Statement II: Some managers are not leaders.

Solution: Statement I says leaders ⊂ managers (not the converse). Statement II confirms not all managers are leaders. Combined: NOT all managers are leaders. Answer: Both statements together give the answer. ✓

Q2: Conclusion: Some dogs are cats. Stmt I: All dogs are animals. Stmt II: Some animals are cats.

Solution: Even with both statements, "Some animals are cats" doesn't tell us which animals. The dogs and cat-animals may not overlap. Answer: Cannot be determined even with both statements. ✓

SYLLOGISM — EXAM SHORTCUTS:
1. All + All → All | All + No → No | Some + All → Some
2. Some + Some → NO definite conclusion
3. All + Some → NO definite conclusion
4. "Possibility" is TRUE unless the opposite is definitely proven
5. Either-Or: Both conclusions complementary + neither follows individually
6. Always draw Venn diagrams — they NEVER lie!