Quantitative Aptitude Topic-wise Practice - Complete SSC CGL Guide
What is Quantitative Aptitude in SSC CGL? This section tests mathematical skills and numerical ability. Covers Arithmetic, Algebra, Geometry, Trigonometry, Mensuration, and Data Interpretation. Essential for SSC CGL Tier I and Tier II exams.
Quantitative Aptitude Topics - Systematic approach for SSC exams
Quantitative Aptitude section analysis
Approximate questions per topic
Quantitative section analysis - Plan your preparation accordingly
Pro Tip – The 3-Step Quantitative Aptitude Mastery Method!
1. Learn Concepts First: Formulas, theorems, methods, shortcuts
2. Practice Topic-wise: Solve 50+ questions per topic with step-by-step solutions
3. Speed & Accuracy: Timed practice with mixed questions
Visit SKY Practice for 5000+ Quantitative Aptitude questions with detailed solutions.
1. Arithmetic
Arithmetic Fundamentals: The foundation of quantitative aptitude covering numbers, percentages, ratios, averages, and basic calculations.
Arithmetic Concepts & Formulas
Understanding Arithmetic Operations
Arithmetic forms the basis of all mathematical calculations. Mastering these concepts is essential for solving complex problems quickly.
Percentage
• Percentage = (Part/Whole) × 100
• Percentage increase = [(New - Original)/Original] × 100
• Percentage decrease = [(Original - New)/Original] × 100
• Successive percentage change: a + b + (ab/100)
Shortcuts:
• 10% = Divide by 10
• 5% = Half of 10%
• 20% = Divide by 5
• 25% = Divide by 4
• 50% = Divide by 2
Applications: Discount, profit/loss, population growth, data interpretation
Profit & Loss
• Profit = Selling Price - Cost Price
• Loss = Cost Price - Selling Price
• Profit % = (Profit/CP) × 100
• Loss % = (Loss/CP) × 100
• Selling Price = CP × (100 ± Profit/Loss %)/100
Discount:
• Marked Price (MP): Original price
• Selling Price = MP × (100 - Discount %)/100
• Successive discounts: Equivalent single discount = a + b - (ab/100)
Applications: Business calculations, market transactions
Time & Work
• Work = Rate × Time
• If A can do work in 'a' days, then A's 1 day work = 1/a
• If A and B work together: 1 day work = 1/a + 1/b
• Time taken together = 1/(1/a + 1/b) = ab/(a+b)
Efficiency:
• If A is twice as efficient as B, then A takes half the time
• Efficiency ratio inversely proportional to time ratio
Work and Wages:
• Wages distributed in ratio of work done
• If A works 'x' days and B works 'y' days, ratio = x/a : y/b
Ratio & Proportion
• Ratio: a:b = a/b
• Proportion: a:b :: c:d means a/b = c/d
• Continued proportion: a:b :: b:c
Operations:
• Duplicate ratio of a:b = a²:b²
• Triplicate ratio of a:b = a³:b³
• Sub-duplicate ratio of a:b = √a:√b
• Sub-triplicate ratio of a:b = ∛a:∛b
Applications:
• Partnership: Profit sharing in investment ratio
• Mixture & Alligation: Finding ratio of mixtures
• Ages: Ratio of ages changing over time
SSC Shortcut: Arithmetic Formulas to Remember
Average: Sum of observations/Number of observations • Weighted Average: (w₁x₁ + w₂x₂)/(w₁ + w₂)
Simple Interest: SI = (P × R × T)/100 • Compound Interest: CI = P[(1 + R/100)^T - 1]
Speed: Distance = Speed × Time • Relative Speed: Same direction = Difference, Opposite direction = Sum
Time & Work: If A in 'a' days, B in 'b' days, together = ab/(a+b) days
Mixture: Alligation rule: (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c)
Important Arithmetic Problems Table
| Topic | Key Concepts | Difficulty Level | Frequency in SSC |
|---|---|---|---|
| Percentage | Basic percentage, successive percentage, percentage change | Easy-Medium | 2-3 questions |
| Profit & Loss | CP, SP, MP, discount, successive discount | Easy-Medium | 2-3 questions |
| Simple & Compound Interest | SI formula, CI formula, difference, installments | Medium | 1-2 questions |
| Ratio & Proportion | Basic ratio, proportion, partnership, ages | Easy-Medium | 2-3 questions |
| Time & Work | Individual work, combined work, efficiency, wages | Medium-Hard | 2-3 questions |
| Time, Speed & Distance | Basic formula, relative speed, trains, boats | Medium | 2-3 questions |
| Average | Simple average, weighted average | Easy | 1-2 questions |
| Number System | Divisibility, remainders, LCM, HCF | Medium-Hard | 2-3 questions |
Solved Example: Arithmetic
Cost Price (CP) = Let's assume ₹100 for easy calculation
Marked Price (MP) = 20% above CP
Discount = 10% on MP
Need to find: Profit percentage
Step 2: Calculate Marked Price
MP = CP + 20% of CP
MP = 100 + (20/100 × 100)
MP = 100 + 20 = ₹120
Step 3: Calculate Selling Price after discount
Discount = 10% of MP
Discount = (10/100 × 120) = ₹12
Selling Price (SP) = MP - Discount
SP = 120 - 12 = ₹108
Step 4: Calculate Profit
Profit = SP - CP
Profit = 108 - 100 = ₹8
Step 5: Calculate Profit Percentage
Profit % = (Profit/CP) × 100
Profit % = (8/100) × 100 = 8%
Step 6: Alternative method (shortcut)
When MP is x% above CP and discount is y% on MP:
Effective profit % = x - y - (xy/100)
Here x = 20, y = 10
Effective profit % = 20 - 10 - (20×10/100)
= 10 - (200/100) = 10 - 2 = 8%
Step 7: Verification
CP = ₹100
MP = ₹120 (20% above)
Discount = ₹12 (10% of 120)
SP = ₹108
Profit = ₹8 (8% of 100)
Check: 8% of 100 = ₹8 ✓
Step 8: Common mistakes to avoid
• Don't calculate discount on CP instead of MP
• Don't add discount to profit
• Remember: Profit % is always on CP, not on MP
Final Answer: 8%
2. Algebra
Algebra Fundamentals: Study of mathematical symbols and rules for manipulating these symbols. Includes equations, polynomials, and sequences.
Algebraic Concepts & Equations
Understanding Algebraic Principles
Algebra provides tools for solving problems involving unknown quantities. Essential for logical reasoning and problem-solving.
Important Algebraic Identities
Algebraic Formulas & Identities
Basic algebraic identities essential for SSC exams
1 Quadratic Equations
Roots: α, β = [-b ± √(b² - 4ac)]/(2a)
Sum and product of roots:
• α + β = -b/a
• αβ = c/a
Nature of roots:
• Discriminant (D) = b² - 4ac
• D > 0: Real and distinct roots
• D = 0: Real and equal roots
• D < 0: Imaginary roots (no real roots)
Forming quadratic equation from roots:
x² - (sum of roots)x + (product of roots) = 0
x² - (α+β)x + αβ = 0
2 Linear Equations & Inequalities
Solution: x = -b/a (a ≠ 0)
Linear equation in two variables: ax + by + c = 0
Infinite solutions
To find unique solution, need two equations
System of linear equations:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
Solutions:
• a₁/a₂ ≠ b₁/b₂: Unique solution
• a₁/a₂ = b₁/b₂ = c₁/c₂: Infinite solutions
• a₁/a₂ = b₁/b₂ ≠ c₁/c₂: No solution
Inequalities:
• Adding/subtracting same number: Inequality preserved
• Multiplying/dividing by positive number: Inequality preserved
• Multiplying/dividing by negative number: Inequality reversed
SSC Shortcut: Algebra Formulas & Tricks
(a+b)² = a² + b² + 2ab • (a-b)² = a² + b² - 2ab
a² - b² = (a+b)(a-b) • a³ + b³ = (a+b)(a² - ab + b²)
a³ - b³ = (a-b)(a² + ab + b²) • (a+b)³ = a³ + b³ + 3ab(a+b)
(a-b)³ = a³ - b³ - 3ab(a-b) • a³ + b³ + c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)
If a+b+c = 0, then a³ + b³ + c³ = 3abc
Solved Example: Algebra
Given equation: x² - 5x + 6 = 0
Compare with standard form: ax² + bx + c = 0
Here: a = 1, b = -5, c = 6
Step 2: Find roots first (optional)
x² - 5x + 6 = 0
Factorizing: (x - 2)(x - 3) = 0
Roots: α = 2, β = 3 or α = 3, β = 2
α² + β² = 2² + 3² = 4 + 9 = 13
Step 3: Method 1 - Direct calculation
α + β = -b/a = -(-5)/1 = 5
αβ = c/a = 6/1 = 6
We know: α² + β² = (α + β)² - 2αβ
α² + β² = (5)² - 2(6) = 25 - 12 = 13
Step 4: Method 2 - Using formula
For quadratic equation ax² + bx + c = 0:
α² + β² = (b² - 2ac)/a²
Here: b² - 2ac = (-5)² - 2(1)(6) = 25 - 12 = 13
a² = 1² = 1
α² + β² = 13/1 = 13
Step 5: Method 3 - Verification
If α = 2, β = 3: α² + β² = 4 + 9 = 13 ✓
If α = 3, β = 2: α² + β² = 9 + 4 = 13 ✓
Step 6: Related formulas
• α² + β² = (α + β)² - 2αβ
• α³ + β³ = (α + β)³ - 3αβ(α + β)
• α⁴ + β⁴ = (α² + β²)² - 2(αβ)²
• 1/α + 1/β = (α + β)/αβ
• α/β + β/α = (α² + β²)/αβ
Step 7: Application
This method avoids finding actual roots when numbers are complex
Particularly useful when roots are irrational or complex
Saves time in competitive exams
Step 8: Common mistakes
• Don't confuse α² + β² with (α + β)²
• Remember: (α + β)² = α² + β² + 2αβ
• So α² + β² = (α + β)² - 2αβ
Final Answer: 13
3. Geometry
Geometry Fundamentals: Study of shapes, sizes, positions, and properties of space. Includes lines, angles, triangles, circles, and polygons.
Geometric Concepts & Theorems
Understanding Geometric Principles
Geometry deals with spatial relationships and properties of figures. Essential for visualizing and solving measurement problems.
Sum of angles = 180°
Area = ½ × base × height
Pythagoras: a² + b² = c²
Similarity criteria: AAA, SAS, SSS
Circumference = 2πr
Area = πr²
Arc length = (θ/360) × 2πr
Sector area = (θ/360) × πr²
Square: Area = a², Perimeter = 4a
Rectangle: Area = l×b, Perimeter = 2(l+b)
Parallelogram: Area = b×h
Rhombus: Area = ½ × d₁ × d₂
Sum of interior angles = (n-2)×180°
Each interior angle = (n-2)×180°/n
Sum of exterior angles = 360°
Each exterior angle = 360°/n
Distance formula: √[(x₂-x₁)²+(y₂-y₁)²]
Midpoint: [(x₁+x₂)/2, (y₁+y₂)/2]
Slope: m = (y₂-y₁)/(x₂-x₁)
Section formula
Cube: Volume = a³, Surface area = 6a²
Cuboid: Volume = l×b×h, SA = 2(lb+bh+hl)
Sphere: Volume = 4/3πr³, SA = 4πr²
Cylinder: Volume = πr²h, CSA = 2πrh
1 Triangle Properties & Theorems
• By sides: Equilateral, isosceles, scalene
• By angles: Acute, right, obtuse
Important properties:
• Sum of angles = 180°
• Exterior angle = Sum of opposite interior angles
• Triangle inequality: Sum of any two sides > third side
• Area = ½ × base × height
• Area using Heron's formula: √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2
Similar triangles:
• AAA criterion: All angles equal
• SAS criterion: One angle equal and sides proportional
• SSS criterion: All sides proportional
• Ratio of areas = Square of ratio of corresponding sides
Right triangle:
• Pythagoras theorem: a² + b² = c² (c = hypotenuse)
• Area = ½ × product of legs
2 Circle Theorems
• Circumference = 2πr
• Area = πr²
• Diameter = 2r
Chord properties:
• Perpendicular from center bisects chord
• Equal chords are equidistant from center
• Chords equidistant from center are equal
Tangent properties:
• Tangent is perpendicular to radius at point of contact
• Tangents from external point are equal
• Angle between tangent and chord = Angle in alternate segment
Arc and sector:
• Arc length = (θ/360) × 2πr
• Sector area = (θ/360) × πr²
• Segment area = Sector area - Triangle area
Cyclic quadrilateral:
• Sum of opposite angles = 180°
• Exterior angle = Interior opposite angle
SSC Shortcut: Geometry Formulas & Theorems
Triangle area: ½ × b × h • Heron's formula: √[s(s-a)(s-b)(s-c)]
Equilateral triangle: Area = (√3/4)a², Height = (√3/2)a
Right triangle: Pythagoras: a² + b² = c² • 30-60-90 triangle: sides = x, x√3, 2x
45-45-90 triangle: sides = x, x, x√2 • Circle: C = 2πr, A = πr²
Arc length: (θ/360) × 2πr • Sector area: (θ/360) × πr²
Rectangle: A = l×b, P = 2(l+b) • Square: A = a², P = 4a
Parallelogram: A = b×h • Rhombus: A = ½ × d₁ × d₂
Trapezium: A = ½ × (a+b) × h
Solved Example: Geometry
Sides: a = 6 cm, b = 8 cm, c = 10 cm
Check: 6² + 8² = 36 + 64 = 100 = 10²
This satisfies Pythagoras theorem: a² + b² = c²
Therefore, triangle is right-angled with hypotenuse = 10 cm
Step 2: Method 1 - Direct formula for right triangle
For right triangle, area = ½ × (product of legs)
Legs are 6 cm and 8 cm (since 10 cm is hypotenuse)
Area = ½ × 6 × 8 = ½ × 48 = 24 cm²
Step 3: Method 2 - Using Heron's formula
Heron's formula: Area = √[s(s-a)(s-b)(s-c)]
where s = semi-perimeter = (a+b+c)/2
s = (6+8+10)/2 = 24/2 = 12 cm
Area = √[12(12-6)(12-8)(12-10)]
= √[12 × 6 × 4 × 2]
= √[12 × 6 × 8] = √[576] = 24 cm²
Step 4: Method 3 - Using base and height
In right triangle, legs are perpendicular to each other
Take base = 6 cm, then height = 8 cm
Area = ½ × base × height = ½ × 6 × 8 = 24 cm²
Or take base = 8 cm, height = 6 cm: ½ × 8 × 6 = 24 cm²
Step 5: Verification
All three methods give same answer: 24 cm² ✓
Step 6: Additional insights
• This is a Pythagorean triplet: (6, 8, 10)
• Multiples also form right triangles: (3,4,5), (9,12,15), etc.
• Right triangle area is maximum when legs are equal (isosceles right triangle)
• For given perimeter, equilateral triangle has maximum area
Step 7: Common mistakes
• Using wrong sides as base and height (must be perpendicular)
• Forgetting ½ in area formula
• In Heron's formula: s = (a+b+c)/2, not (a+b+c)
• Confusing perimeter with semi-perimeter
Step 8: Applications
• Finding area of land
• Construction calculations
• Design and architecture
• Trigonometry problems
Final Answer: 24 cm²
4. Trigonometry
Trigonometry Fundamentals: Study of relationships between angles and sides of triangles. Essential for heights and distances problems.
Trigonometric Ratios & Identities
Understanding Trigonometric Principles
Trigonometry connects angles with side ratios in triangles. Widely used in navigation, engineering, and physics.
Basic Ratios
sin θ = Opposite/Hypotenuse
cos θ = Adjacent/Hypotenuse
tan θ = Opposite/Adjacent
cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
Important Identities
sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ
sin(A+B) = sinA cosB + cosA sinB
cos(A+B) = cosA cosB - sinA sinB
tan(A+B) = (tanA+tanB)/(1-tanA tanB)
Standard Values
sin 0° = 0, cos 0° = 1
sin 30° = ½, cos 30° = √3/2
sin 45° = 1/√2, cos 45° = 1/√2
sin 60° = √3/2, cos 60° = ½
sin 90° = 1, cos 90° = 0
tan 45° = 1
1 Trigonometric Identities
• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ
Reciprocal identities:
• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• cot θ = 1/tan θ
• tan θ = sin θ/cos θ
• cot θ = cos θ/sin θ
Angle sum and difference:
• sin(A±B) = sinA cosB ± cosA sinB
• cos(A±B) = cosA cosB ∓ sinA sinB
• tan(A±B) = (tanA ± tanB)/(1 ∓ tanA tanB)
Double angle formulas:
• sin 2A = 2 sinA cosA
• cos 2A = cos²A - sin²A = 2cos²A - 1 = 1 - 2sin²A
• tan 2A = 2tanA/(1 - tan²A)
2 Heights and Distances
1. Draw diagram with given information
2. Identify right triangles
3. Apply trigonometric ratios
4. Solve equations
Common terms:
• Angle of elevation: Looking upward from horizontal
• Angle of depression: Looking downward from horizontal
• Line of sight: Straight line from eye to object
Important ratios:
• For angle θ: tan θ = Height/Distance
• If two angles given, form two equations
• Eliminate common variable (usually distance)
Applications:
• Finding height of building/tree
• Finding distance across river
• Navigation problems
• Surveying
Solved Example: Trigonometry
Fundamental identity: sin²θ + cos²θ = 1
Given: sin θ = 3/5
So sin²θ = (3/5)² = 9/25
Step 2: Apply identity
sin²θ + cos²θ = 1
9/25 + cos²θ = 1
cos²θ = 1 - 9/25
cos²θ = (25/25) - (9/25) = 16/25
Step 3: Take square root
cos θ = ±√(16/25) = ±4/5
Step 4: Determine sign
The problem doesn't specify quadrant
• If θ in first quadrant: cos θ = +4/5
• If θ in second quadrant: cos θ = -4/5
Usually in SSC problems, if not specified, take positive value
Step 5: Method 2 - Using right triangle
sin θ = Opposite/Hypotenuse = 3/5
Draw right triangle with opposite = 3, hypotenuse = 5
Using Pythagoras: Adjacent² = Hypotenuse² - Opposite²
Adjacent² = 5² - 3² = 25 - 9 = 16
Adjacent = √16 = 4
cos θ = Adjacent/Hypotenuse = 4/5
Step 6: Find other ratios (if needed)
tan θ = sin θ/cos θ = (3/5)/(4/5) = 3/4
cosec θ = 1/sin θ = 5/3
sec θ = 1/cos θ = 5/4
cot θ = 1/tan θ = 4/3
Step 7: Verification
Check: sin²θ + cos²θ = (9/25) + (16/25) = 25/25 = 1 ✓
1 + tan²θ = 1 + (9/16) = 25/16 = (5/4)² = sec²θ ✓
Step 8: Common mistakes
• Forgetting ± sign when taking square root
• Not considering quadrant information
• Confusing sin²θ with (sin θ)² - they are same
• Using wrong identity
Final Answer: ±4/5 (usually 4/5 in SSC context)
5. Data Interpretation
Data Interpretation Fundamentals: Analysis and interpretation of data presented in various formats like tables, charts, and graphs.
Data Analysis & Interpretation
Understanding Data Patterns
Data Interpretation tests ability to extract information, perform calculations, and draw conclusions from graphical data.
DI Question Types Frequency
Frequency of different DI question types in SSC exams
1 Data Interpretation Techniques
1. Read titles, labels, units carefully
2. Understand what data represents
3. Look for patterns, trends, extremes
4. Perform calculations step by step
5. Verify answers with approximations
Calculation shortcuts:
• Use approximations when exact values not needed
• Look for percentages instead of actual numbers
• Compare ratios rather than absolute values
• Use elimination for multiple choice
Common calculations:
• Percentage increase/decrease
• Ratio and proportion
• Averages (simple, weighted)
• Growth rates
• Share percentages
Time management: 2-3 minutes per DI question
2 Types of Data Presentation
• Most common format
• Read row and column headings carefully
• Look for totals, subtotals
• Compare across rows and columns
Pie Charts:
• Represents parts of whole (100%)
• Angle = (Percentage/100) × 360°
• Compare sector sizes visually
• Calculate percentages from angles
Bar Graphs:
• Compare quantities across categories
• Vertical/horizontal bars
• Look for highest/lowest values
• Calculate differences
Line Graphs:
• Show trends over time
• Identify increasing/decreasing trends
• Calculate rates of change
• Look for peaks and troughs
Mixed Graphs: Combination of above types
Solved Example: Data Interpretation
Year 1: 200 units, Year 2: 220 units, Year 3: 240 units, Year 4: 260 units, Year 5: 300 units
Year 1 sales: 200 units
Year 5 sales: 300 units
Need: Percentage increase from Year 1 to Year 5
Step 2: Recall percentage increase formula
Percentage increase = [(New Value - Original Value)/Original Value] × 100
Here: Original Value (Year 1) = 200
New Value (Year 5) = 300
Step 3: Calculate increase
Increase = New Value - Original Value = 300 - 200 = 100 units
Step 4: Calculate percentage increase
Percentage increase = (Increase/Original Value) × 100
= (100/200) × 100 = 0.5 × 100 = 50%
Step 5: Method 2 - Direct calculation
Percentage increase = [(300 - 200)/200] × 100
= (100/200) × 100 = 50%
Step 6: Verification
50% of 200 = 100
200 + 100 = 300 ✓
Step 7: Alternative approach
New Value/Original Value = 300/200 = 1.5
This means 150% of original (which is 100% + 50% increase)
So increase = 50%
Step 8: Common mistakes in DI
• Calculating from wrong years (Year 2 to Year 5 instead of Year 1 to Year 5)
• Using wrong formula (decrease instead of increase)
• Forgetting to multiply by 100 for percentage
• Misreading table values
• Not checking if answer is reasonable (50% increase from 200 to 300 is reasonable)
Step 9: Additional insights
From Year 1 to Year 2: (220-200)/200 × 100 = 10% increase
From Year 2 to Year 3: (240-220)/220 × 100 = 9.09% increase
From Year 3 to Year 4: (260-240)/240 × 100 = 8.33% increase
From Year 4 to Year 5: (300-260)/260 × 100 = 15.38% increase
Maximum yearly increase: Year 4 to Year 5 (15.38%)
Final Answer: 50%
6. Practice Exercises
Hands-on Practice: Apply what you've learned with these SSC-level quantitative aptitude questions.
Interactive Practice Questions
Practice Approach
Time yourself: 60 seconds per question. Apply shortcuts and elimination techniques.
Practice Question 1: Arithmetic
Practice Question 2: Algebra
Practice Question 3: Geometry
SSC Shortcut: Practice Strategy
Daily practice: 20 quantitative questions daily (mix all topics)
Revision cycle: Revise formulas weekly, shortcuts bi-weekly
Speed practice: Timed sets: 25 questions in 60 minutes
Previous papers: Solve last 5 years' SSC quantitative questions
Weak areas: Identify and focus on 2 weak topics weekly
Time management: Arithmetic: 45-60 sec, Algebra: 60-75 sec, Geometry/Trig: 75-90 sec, DI: 90-120 sec
Ready to Master Quantitative Aptitude?
Access 5000+ Quantitative Aptitude questions with detailed solutions, topic-wise practice, speed tests, and smart analytics
Start Quantitative PracticeIncludes Arithmetic, Algebra, Geometry, Trigonometry, Mensuration, Data Interpretation, and previous year questions
Frequently Asked Questions
Q1: How many quantitative aptitude questions in SSC CGL?
Answer: 25 questions in Tier I (50 marks). Time allocation: 60 minutes. Tier II has separate paper with more advanced questions.
Q2: What are the most important formulas to remember?
Answer: Percentage formulas, profit-loss formulas, simple/compound interest, time-speed-distance, area/volume formulas, algebraic identities, trigonometric ratios.
Q3: How to improve calculation speed?
Answer: Practice mental calculations, learn multiplication tables up to 20, use approximation techniques, practice with timer, learn Vedic math shortcuts.
Q4: Which topics have highest weightage?
Answer: Arithmetic (8-10 questions), Algebra (5-7 questions), Geometry/Trigonometry (6-8 questions), Data Interpretation (3-5 questions).
Q5: How to prepare for data interpretation?
Answer: Practice reading different chart types, improve calculation speed, learn percentage shortcuts, practice approximation, solve previous year DI sets.
Q6: Best way to improve quantitative score?
Answer: 1) Strong foundation in basics, 2) Topic-wise practice, 3) Speed and accuracy practice, 4) Regular revision of formulas, 5) Mock tests and analysis, 6) Learning from mistakes.
Final Exam Strategy for Quantitative Aptitude
Time Allocation: Total 60 minutes for 25 questions (average 2.4 minutes per question).
Priority Order: 1) Arithmetic (easy-medium), 2) Algebra (direct formula), 3) Geometry (visual), 4) Data Interpretation (time-consuming).
Accuracy Check: Verify calculations, check units, ensure answer is reasonable, use approximation to verify.
👉 For complete mastery with 5000+ questions, visit SKY Practice!