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FREE PREVIEW Week 4 4 lessons

Quantitative Aptitude: Advanced and Data

Algebra identities and equations, bank-style quadratic comparisons, geometry and mensuration formulas, trigonometry with heights and distances, and data interpretation with basic statistics.

Lessons in this module

1

Algebra: Identities, Linear and Quadratic Equations

Lesson • 5 quiz questions

Read notes ↓
2

Geometry and Mensuration: 2D and 3D Formulas

Lesson • 5 quiz questions

Read notes ↓
3

Trigonometry Basics, Heights and Distances

Lesson • 5 quiz questions

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4

Data Interpretation and Basic Statistics

Lesson • 5 quiz questions

Read notes ↓

Lesson notes

1. Algebra: Identities, Linear and Quadratic Equations

SSC CGL and CHSL ask many algebra questions based on standard identities. Bank exams (IBPS and SBI PO and Clerk) ask quadratic equation comparisons, where you solve two equations and compare x and y. This lesson covers both, with step-by-step examples.

Important identities

  • (a + b)² = a² + 2ab + b². (a − b)² = a² − 2ab + b². 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).
  • If a + b + c = 0, then a³ + b³ + c³ = 3abc.
  • If a + 1/a = k, then a² + 1/a² = k² − 2 and a³ + 1/a³ = k³ − 3k. Example: a + 1/a = 5 → a² + 1/a² = 25 − 2 = 23.
  • If a − 1/a = k, then a² + 1/a² = k² + 2. Example: a − 1/a = 4 → 16 + 2 = 18.
  • Worked example: x + y = 10 and xy = 21. Then x² + y² = (x + y)² − 2xy = 100 − 42 = 58.

Linear equations in two variables

  • Solve by elimination: make the coefficient of one variable equal, then add or subtract the equations.
  • Worked example: 3x + 2y = 16 and x − y = 2. Step 1: multiply the second equation by 2: 2x − 2y = 4. Step 2: add: 5x = 20, so x = 4. Step 3: y = x − 2 = 2.
  • Check your answer: 3 × 4 + 2 × 2 = 16. Correct.
  • Shortcut: in MCQs, put the options into both equations. Only the correct pair fits both.

Quadratic equations and bank-style comparison

  • For ax² + bx + c = 0: sum of roots = −b/a, product of roots = c/a. Find two numbers whose product is c (when a = 1) and whose sum is −b.
  • Worked example: x² − 7x + 12 = 0. Product 12, sum 7 → 3 and 4. Roots: x = 3, 4.
  • Bank comparison: solve both equations, then compare every x root with every y root. Example: I. x² − 7x + 12 = 0 (x = 3, 4). II. y² − 3y + 2 = 0 (y = 1, 2). Every x is bigger than every y, so x > y.
  • If some x values equal some y values and the rest are bigger, the answer is x ≥ y. If some x are bigger and some are smaller, the answer is 'relationship cannot be established'.
  • Sign rule: if the equation is x² + bx + c with b and c both positive, both roots are negative. If b is negative and c is positive, both roots are positive.
Quick facts
(a + b)²
a² + 2ab + b²
a² − b²
(a + b)(a − b)
a + 1/a = k
a² + 1/a² = k² − 2; a³ + 1/a³ = k³ − 3k
a + b + c = 0
a³ + b³ + c³ = 3abc
Quadratic ax² + bx + c
Sum of roots −b/a, product c/a
Bank comparison
Compare every x root with every y root

💡 In SSC algebra questions, try putting simple values like a = 1 or x = 0 that satisfy the condition. Often this gives the answer without any algebra.

📝 5-question quiz on this lesson: enrol free to take it and track your score.

2. Geometry and Mensuration: 2D and 3D Formulas

Mensuration means measuring area, perimeter, volume and surface area. Geometry gives the rules for triangles, circles and angles. Most questions are direct formula use, so learning the formula table is the key to scoring here. Use π = 22/7 unless the question says otherwise.

Lines, triangles and Pythagoras

  • The angles of a triangle add up to 180°. An exterior angle equals the sum of the two opposite interior angles. The sum of the interior angles of a polygon with n sides = (n − 2) × 180°.
  • Pythagoras theorem (right-angled triangle): hypotenuse² = base² + height². Common triplets: 3-4-5, 5-12-13, 8-15-17, 7-24-25.
  • Area of a triangle = ½ × base × height. Equilateral triangle with side a: area = (√3/4) × a².
  • Heron's formula: s = (a + b + c) ÷ 2, area = √[s(s − a)(s − b)(s − c)]. Worked example: sides 13, 14, 15 → s = 21 → area = √(21 × 8 × 7 × 6) = √7,056 = 84.
  • Trap: Heron's formula needs s, the HALF perimeter, not the full perimeter.

2D shapes: area and perimeter

  • Rectangle: area = l × b, perimeter = 2(l + b), diagonal = √(l² + b²). Square: area = a², diagonal = a√2.
  • Circle: area = πr², circumference = 2πr. Worked example: circumference 88 cm → 2 × 22/7 × r = 88 → r = 14 cm → area = 22/7 × 14 × 14 = 616 cm².
  • Rhombus: area = ½ × d₁ × d₂. Parallelogram: base × height. Trapezium: ½ × (sum of parallel sides) × height.
  • Semicircle perimeter = πr + 2r. Sector area = (θ/360) × πr².
  • If the sides of a square or the radius of a circle increase by r%, the area increases by 2r + r²/100 %. Example: radius up 10% → area up 21%.

3D shapes: volume and surface area

  • Cube with side a: volume = a³, total surface area = 6a². Worked example: volume 512 → a = 8 → TSA = 6 × 64 = 384.
  • Cuboid: volume = l × b × h, TSA = 2(lb + bh + hl). Cylinder: volume = πr²h, curved surface = 2πrh, TSA = 2πr(r + h).
  • Worked example: cylinder with r = 7 cm and h = 10 cm. Volume = 22/7 × 7 × 7 × 10 = 1,540 cm³.
  • Cone: volume = ⅓πr²h, slant height l = √(r² + h²), curved surface = πrl. Example: r = 7, h = 24 → l = 25 → CSA = 22/7 × 7 × 25 = 550.
  • Sphere: volume = (4/3)πr³, surface = 4πr². Hemisphere: volume = (2/3)πr³, curved surface = 2πr², total surface = 3πr².
  • Trap: a cone has one-third the volume of a cylinder with the same base and height.
Quick facts
Triangle
½bh; Heron √[s(s−a)(s−b)(s−c)]
Equilateral triangle
(√3/4)a²
Circle
Area πr², circumference 2πr
Cube / cuboid
a³, 6a² / lbh, 2(lb + bh + hl)
Cylinder / cone
πr²h, 2πrh / ⅓πr²h, πrl
Sphere / hemisphere
(4/3)πr³, 4πr² / (2/3)πr³, 3πr²

💡 When the radius is a multiple of 7, use π = 22/7. The 7 cancels and the numbers become simple whole numbers.

📝 5-question quiz on this lesson: enrol free to take it and track your score.

3. Trigonometry Basics, Heights and Distances

Trigonometry is asked mainly in SSC CGL and CHSL. Questions use the basic ratios, a small table of standard values and three identities. Heights and distances questions use the same ratios to find the height of a tower or the length of a ladder.

Ratios and standard values

  • In a right-angled triangle with angle θ: 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. Also tan θ = sin θ/cos θ.
  • sin values for 0°, 30°, 45°, 60°, 90°: 0, 1/2, 1/√2, √3/2, 1. cos values are the same list in reverse order: 1, √3/2, 1/√2, 1/2, 0.
  • tan values for 0°, 30°, 45°, 60°: 0, 1/√3, 1, √3. tan 90° is not defined.
  • Worked example: if tan θ = 3/4 (θ acute), take opposite = 3, adjacent = 4. Hypotenuse = √(9 + 16) = 5. So sin θ = 3/5 and cos θ = 4/5.

Identities and complementary angles

  • sin²θ + cos²θ = 1. 1 + tan²θ = sec²θ. 1 + cot²θ = cosec²θ.
  • Complementary angles: sin(90° − θ) = cos θ, cos(90° − θ) = sin θ, tan(90° − θ) = cot θ.
  • Worked example: sin²37° + sin²53°. Since 53° = 90° − 37°, sin 53° = cos 37°. So the value = sin²37° + cos²37° = 1.
  • Another use: tan 20° × tan 70° = tan 20° × cot 20° = 1.
  • Trap: sin 2θ is not 2 sin θ. Example: sin 60° = 0.866, but 2 × sin 30° = 1.

Heights and distances

  • Angle of elevation: looking UP from the ground to the top of an object. Angle of depression: looking DOWN from a height.
  • Most questions use tan θ = height ÷ distance from the foot.
  • Worked example: the angle of elevation of a tower's top from a point 20 m from its foot is 60°. Height = 20 × tan 60° = 20√3 ≈ 34.64 m.
  • Worked example: a ladder makes 60° with the ground and its foot is 5 m from the wall. cos 60° = 5 ÷ length → length = 5 ÷ ½ = 10 m.
  • At 45°, height = distance. At 30°, distance = height × √3. At 60°, height = distance × √3.
Quick facts
sin 0°, 30°, 45°, 60°, 90°
0, 1/2, 1/√2, √3/2, 1
tan 0°, 30°, 45°, 60°
0, 1/√3, 1, √3
Identities
sin² + cos² = 1; sec² − tan² = 1; cosec² − cot² = 1
Complementary
sin(90° − θ) = cos θ; tan(90° − θ) = cot θ
Heights
tan θ = height ÷ distance
√ values
√2 ≈ 1.414, √3 ≈ 1.732

💡 Draw a quick right-angled triangle for every heights and distances question and mark the angle, the height and the base. The correct ratio becomes clear at once.

📝 5-question quiz on this lesson: enrol free to take it and track your score.

4. Data Interpretation and Basic Statistics

Data Interpretation (DI) gives you a table or a chart and asks 4 to 5 questions on it. It carries high marks in bank exams and SSC CGL. DI uses percentages, ratios and averages from earlier lessons. Basic statistics (mean, median, mode) is also asked directly in SSC exams.

Tables and bar charts: worked example

  • Data: Sales of a shop (₹ lakh). Year: 2021 | 2022 | 2023 | 2024 | 2025. Sales: 40 | 50 | 45 | 60 | 75.
  • Percentage increase from 2024 to 2025 = (75 − 60) ÷ 60 × 100 = 25%.
  • Average sales over 5 years = (40 + 50 + 45 + 60 + 75) ÷ 5 = 270 ÷ 5 = ₹54 lakh.
  • Ratio of sales in 2022 to 2025 = 50 : 75 = 2 : 3.
  • A bar chart shows the same kind of data as bars. Read each bar's value first, write it down, then calculate.

Pie charts and line graphs

  • A pie chart shows parts of a whole. 100% = 360°, so 1% = 3.6°. Central angle = percentage × 3.6.
  • Data: Monthly family budget ₹60,000. Food 30% | Rent 25% | Education 15% | Transport 10% | Savings 20%.
  • Rent = 25% of 60,000 = ₹15,000. Central angle for rent = 25 × 3.6 = 90°.
  • Food exceeds Transport by 30% − 10% = 20% of 60,000 = ₹12,000. Use the percentage difference first, then convert once.
  • A line graph shows change over time. The steepest upward line segment shows the largest increase in value (not always the largest percentage increase).

Mean, median and mode

  • Mean = sum of values ÷ number of values. Median = the middle value after arranging in order. Mode = the value that occurs most often.
  • For an even number of values, median = average of the two middle values.
  • Worked example: data 4, 7, 7, 9, 12, 15. Mean = 54 ÷ 6 = 9. Median = (7 + 9) ÷ 2 = 8. Mode = 7.
  • Empirical relation (for moderately skewed data): Mode = 3 × Median − 2 × Mean. Example: mean 12, median 13 → mode = 39 − 24 = 15.
  • Trap: always arrange the data in order before finding the median.
Quick facts
% change
(New − Old) ÷ Old × 100
Pie chart
1% = 3.6°; 100% = 360°
Mean
Sum ÷ Count
Median
Middle value of sorted data (even count: average of two middle)
Mode
Most frequent value
Empirical relation
Mode = 3 Median − 2 Mean

💡 In DI, read all the questions in a set first. Then calculate the totals and values you need once, and use them for every question.

📝 5-question quiz on this lesson: enrol free to take it and track your score.

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