Mensuration

Mensuration is a branch of mathematics that deals with the measurement of geometric figures. It involves calculating the length, area, and volume of various shapes. For competitive exams, understanding the formulas and their applications is crucial. We will cover 2D (Plane Figures) and 3D (Solid Figures) mensuration.

2D Mensuration (Plane Figures)

These figures lie on a flat surface and have only length and breadth (or height). We measure their perimeter (or circumference) and area.

1. Square

A square is a quadrilateral with four equal sides and four right angles.

• Side length = a
• Perimeter = 4a
• Area = a²
• Diagonal = a√2

Example: If the side of a square is 10 cm, its perimeter is 4 * 10 = 40 cm, and its area is 10² = 100 cm².

2. Rectangle

A rectangle is a quadrilateral with opposite sides equal and four right angles.

• Length = l, Breadth = b
• Perimeter = 2(l + b)
• Area = l × b
• Diagonal = √(l² + b²)

Example: A rectangle has a length of 15 cm and a breadth of 8 cm. Its perimeter is 2(15 + 8) = 2(23) = 46 cm, and its area is 15 * 8 = 120 cm².

3. Triangle

A triangle is a polygon with three sides and three angles.

• Equilateral Triangle: All sides are equal (a).
• Perimeter = 3a
• Area = (√3/4) * a²
• Height = (√3/2) * a
• Isosceles Triangle: Two sides are equal. Let the equal sides be a and the base be b.
• Perimeter = 2a + b
• Area = (b/4) * √(4a² - b²)
• Height (to base b) = (1/2) * √(4a² - b²)
• Scalene Triangle: All sides are different (a, b, c).
• Perimeter = a + b + c
• Area (using Heron's formula): Let s = (a + b + c)/2 (semi-perimeter). Area = √[s(s-a)(s-b)(s-c)]
• Right-Angled Triangle: One angle is 90 degrees. Sides are base (b), perpendicular (p), and hypotenuse (h).
• Perimeter = b + p + h
• Area = (1/2) * b * p
• By Pythagoras theorem: h² = b² + p²

Example: For a triangle with sides 5 cm, 12 cm, and 13 cm (a right-angled triangle), the area is (1/2) * 5 * 12 = 30 cm².

4. Parallelogram

A quadrilateral with opposite sides parallel and equal.

• Adjacent sides = a, b; Height corresponding to base a = h_a
• Perimeter = 2(a + b)
• Area = base × corresponding height = a × h_a = b × h_b

Example: A parallelogram has a base of 10 cm and a corresponding height of 7 cm. Its area is 10 * 7 = 70 cm².

5. Rhombus

A parallelogram with all sides equal. Its diagonals bisect each other at right angles.

• Side length = a
• Diagonals = d₁, d₂
• Perimeter = 4a
• Area = (1/2) * d₁ * d₂
• Also, Area = a * h (where h is the height)
• Relationship between side and diagonals: 4a² = d₁² + d₂²

Example: A rhombus has diagonals of length 6 cm and 8 cm. Its area is (1/2) * 6 * 8 = 24 cm².

6. Trapezium (or Trapezoid)

A quadrilateral with one pair of opposite sides parallel.

• Parallel sides = a, b; Height = h
• Area = (1/2) * (sum of parallel sides) * height = (1/2) * (a + b) * h

Example: A trapezium has parallel sides of 10 cm and 15 cm, with a height of 6 cm. Its area is (1/2) * (10 + 15) * 6 = (1/2) * 25 * 6 = 75 cm².

7. Circle

A set of points equidistant from a central point.

• Radius = r
• Diameter = d = 2r
• Circumference (Perimeter) = 2πr = πd
• Area = πr²
• Area of a Semicircle = (1/2)πr²
• Area of a Sector (with angle θ in degrees) = (θ/360) * πr²
• Length of an Arc (with angle θ in degrees) = (θ/360) * 2πr

Note: Use π ≈ 22/7 or 3.14.

Example: A circle has a radius of 7 cm. Its circumference is 2 * (22/7) * 7 = 44 cm, and its area is (22/7) * 7² = 154 cm².

3D Mensuration (Solid Figures)

These are solid shapes that occupy space and have length, breadth, and height. We measure their surface area (lateral and total) and volume.

1. Cuboid

A rectangular solid with six rectangular faces.

• Length = l, Breadth = b, Height = h
• Surface Area (Lateral) = 2(l + b) * h (Area of the four walls)
• Surface Area (Total) = 2(lb + bh + hl) (Area of all six faces)
• Volume = l × b × h
• Diagonal = √(l² + b² + h²)

Example: A cuboid has dimensions 10 cm × 5 cm × 3 cm. Its total surface area is 2((10*5) + (5*3) + (3*10)) = 2(50 + 15 + 30) = 2(95) = 190 cm². Its volume is 10 * 5 * 3 = 150 cm³.

2. Cube

A cuboid with all edges equal.

• Edge length = a
• Surface Area (Lateral) = 4a²
• Surface Area (Total) = 6a²
• Volume = a³
• Diagonal = a√3

Example: A cube has an edge of 4 cm. Its total surface area is 6 * 4² = 6 * 16 = 96 cm². Its volume is 4³ = 64 cm³.

3. Cylinder

A solid with two parallel circular bases connected by a curved surface.

• Radius of base = r, Height = h
• Curved Surface Area (Lateral) = 2πrh
• Total Surface Area = 2πrh + 2πr² = 2πr(h + r)
• Volume = Area of base × height = πr²h

Example: A cylinder has a radius of 7 cm and a height of 10 cm. Its curved surface area is 2 * (22/7) * 7 * 10 = 440 cm². Its total surface area is 440 + 2 * (22/7) * 7² = 440 + 308 = 748 cm². Its volume is (22/7) * 7² * 10 = 1540 cm³.

4. Cone

A solid with a circular base and a vertex opposite to the base.

• Radius of base = r, Height = h, Slant height = l
• Relationship: l² = r² + h²
• Curved Surface Area (Lateral) = πrl
• Total Surface Area = πrl + πr² = πr(l + r)
• Volume = (1/3) * Area of base * height = (1/3)πr²h

Example: A cone has a radius of 3 cm and a height of 4 cm. Its slant height is √(3² + 4²) = √25 = 5 cm. Its curved surface area is π * 3 * 5 = 15π cm². Its volume is (1/3)π * 3² * 4 = 12π cm³.

5. Sphere

A perfectly round geometrical object in three-dimensional space.

• Radius = r
• Surface Area (Total) = 4πr²
• Volume = (4/3)πr³

Example: A sphere has a radius of 3 cm. Its surface area is 4π * 3² = 36π cm². Its volume is (4/3)π * 3³ = 36π cm³.

6. Hemisphere

Half of a sphere.

• Radius = r
• Curved Surface Area (Lateral) = 2πr²
• Total Surface Area = 2πr² (curved) + πr² (base) = 3πr²
• Volume = (1/2) * Volume of sphere = (1/2) * (4/3)πr³ = (2/3)πr³

Example: A hemisphere has a radius of 7 cm. Its total surface area is 3 * (22/7) * 7² = 462 cm². Its volume is (2/3) * (22/7) * 7³ = 718.67 cm³ (approx).

Statistics

Statistics is the science of collecting, analyzing, interpreting, presenting, and organizing data. In competitive exams, this topic primarily focuses on understanding and interpreting data presented in various formats like tables, charts, and graphs. Key areas include measures of central tendency, dispersion, and data interpretation.

1. Data Interpretation (DI)

This is a major part of the statistics section. You will be presented with data in the form of tables, line graphs, bar graphs, pie charts, etc., and asked to answer questions based on this data. The key is to read the data accurately and perform calculations quickly.

Types of Data Presentation:

• Tables: Data organized in rows and columns.
• Bar Graphs: Uses rectangular bars to represent data, comparing different categories.
• Line Graphs: Uses points connected by lines to show trends over time.
• Pie Charts: Represents data as sectors of a circle, showing proportions.
• Mixed Graphs: Combinations of the above.

Common Calculations in DI:

• Percentage: (Part / Whole) * 100
• Ratio: Comparing two quantities (e.g., A:B)
• Average: Sum of values / Number of values
• Difference: Finding the variation between two values.
• Sum: Adding up values.

Example DI Problem: A pie chart shows the distribution of students in 5 different colleges (A, B, C, D, E). College A has 20% of the total students.

• If the total number of students is 5000, then the number of students in College A is 20% of 5000 = (20/100) * 5000 = 1000.
• If College B has 15% of students and College C has 25%, what is the ratio of students in B to C? Ratio = 15% : 25% = 3 : 5.
• What is the average number of students in Colleges A, B, and C if their percentages are 20%, 15%, and 25% respectively? Average % = (20+15+25)/3 = 60/3 = 20%. If total students are 5000, average students = 20% of 5000 = 1000.

2. Measures of Central Tendency

These are statistical measures used to represent a dataset by a single value that indicates the center of the distribution. The most common measures are Mean, Median, and Mode.

a. Mean (Average)

The sum of all observations divided by the number of observations.

• For ungrouped data (x₁, x₂, ..., x_n): Mean (X̄) = (Σx_i) / n
• For grouped data (class intervals with frequencies): Mean = (Σf_ix_i) / Σf_i, where x_i is the midpoint of the class interval.

Example: Data: 2, 4, 6, 8, 10. Mean = (2+4+6+8+10)/5 = 30/5 = 6.

b. Median

The middle value of a dataset when arranged in ascending or descending order.

• If the number of observations (n) is odd: Median = Value of the ((n+1)/2)^th term.
• If the number of observations (n) is even: Median = Average of the (n/2)^th term and the ((n/2)+1)^th term.

Example 1 (Odd n): Data: 2, 4, 6, 8, 10. Arranged: 2, 4, 6, 8, 10. n=5. Median = value of ((5+1)/2)^th term = 3rd term = 6.

Example 2 (Even n): Data: 2, 4, 6, 8, 10, 12. Arranged: 2, 4, 6, 8, 10, 12. n=6. Median = Average of (6/2)^th and ((6/2)+1)^th terms = Average of 3rd and 4th terms = (6+8)/2 = 7.

c. Mode

The value that appears most frequently in a dataset.

• A dataset can have one mode (unimodal), two modes (bimodal), or no mode.

Example: Data: 2, 4, 4, 6, 8, 8, 8, 10. The mode is 8 because it appears most frequently (3 times).

3. Measures of Dispersion

These measures indicate the spread or variability of the data points around the central value.

a. Range

The difference between the maximum and minimum values in a dataset.

• Range = Maximum Value - Minimum Value

Example: Data: 2, 4, 6, 8, 10. Range = 10 - 2 = 8.

b. Standard Deviation (σ)

A measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

• For ungrouped data: σ = √[Σ(x_i - X̄)² / n]
• For grouped data: σ = √[Σf_i(x_i - X̄)² / Σf_i]

Calculation can be tedious, but understanding its concept is important for interpreting data.

c. Variance (σ²)

The square of the standard deviation.

• Variance = σ²

4. Probability Basics (often linked with Statistics)

Probability is the measure of the likelihood that an event will occur.

• Probability of an event E = P(E) = (Number of favorable outcomes) / (Total number of possible outcomes)
• Probabilities range from 0 (impossible event) to 1 (certain event).

Example: What is the probability of drawing an ace from a standard deck of 52 cards? There are 4 aces. Total cards = 52. P(Ace) = 4/52 = 1/13.

Algebra

Algebra is a fundamental branch of mathematics that uses symbols, usually letters, to represent unknown quantities or numbers. It allows us to generalize arithmetic operations and solve equations.

1. Basic Concepts

Variable: A symbol (usually a letter like x, y, a, b) that can take any value.

Constant: A value that does not change (e.g., 5, -3, π).

Term: A combination of constants and variables (e.g., 5x, -3y², 7).

Expression: A combination of terms using addition, subtraction, multiplication, or division (e.g., 2x + 5, 3y² - 7).

Equation: A statement that two expressions are equal (e.g., 2x + 5 = 11).

Polynomial: An expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.

• Monomial: One term (e.g., 7x²)
• Binomial: Two terms (e.g., 3x + 4)
• Trinomial: Three terms (e.g., x² + 2x + 1)

2. Algebraic Identities

These are equations that are true for all values of the variables. They are extremely useful for simplifying expressions and solving equations.

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• (a + b)(a - b) = a² - b²
• (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
• (a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b)
• (a - b)³ = a³ - 3a²b + 3ab² - b³ = a³ - b³ - 3ab(a - b)
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)
• a² + b² = (a + b)² - 2ab
• a² + b² = (a - b)² + 2ab
• 2(a² + b²) = (a + b)² + (a - b)²

Example: Simplify (x + 2y)². Using the identity (a + b)² = a² + 2ab + b², where a=x and b=2y:

(x + 2y)² = x² + 2(x)(2y) + (2y)² = x² + 4xy + 4y².

3. Solving Linear Equations

An equation where the highest power of the variable is 1.

• Steps:
1. Simplify both sides of the equation by removing parentheses and combining like terms.
2. Move all terms containing the variable to one side of the equation and all constant terms to the other side using addition or subtraction.
3. Isolate the variable by dividing both sides by the coefficient of the variable.

Example: Solve for x: 3(x - 2) + 5 = 2x + 7

1. Distribute: 3x - 6 + 5 = 2x + 7
2. Combine like terms: 3x - 1 = 2x + 7
3. Move x terms to the left: 3x - 2x = 7 + 1
4. Simplify: x = 8

4. Solving Quadratic Equations

An equation where the highest power of the variable is 2. The standard form is ax² + bx + c = 0.

• Methods to Solve:
• Factorization: Split the middle term (bx) into two terms such that their product is ac and their sum is b.
• Quadratic Formula: x = [-b ± √(b² - 4ac)] / 2a
• Completing the Square: (Less common in exams but a valid method).
• Discriminant (Δ): Δ = b² - 4ac. This determines the nature of the roots:
• If Δ > 0: Two distinct real roots.
• If Δ = 0: Two equal real roots (or one real root).
• If Δ < 0: No real roots (two complex roots).

Example: Solve x² - 5x + 6 = 0

• Using factorization: We need two numbers whose product is 6 (a*c = 1*6) and sum is -5 (b). These numbers are -2 and -3.
• Split the middle term: x² - 2x - 3x + 6 = 0
• Factor by grouping: x(x - 2) - 3(x - 2) = 0
• (x - 2)(x - 3) = 0
• Therefore, x = 2 or x = 3.

Example using Quadratic Formula: For x² - 5x + 6 = 0, a=1, b=-5, c=6.

x = [ -(-5) ± √((-5)² - 4*1*6) ] / (2*1)

x = [ 5 ± √(25 - 24) ] / 2

x = [ 5 ± √1 ] / 2

x = (5 + 1) / 2 = 6 / 2 = 3

or x = (5 - 1) / 2 = 4 / 2 = 2

5. Inequalities

Statements comparing two expressions using symbols like < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to).

• Solving inequalities is similar to solving equations, with one crucial difference: if you multiply or divide both sides by a negative number, you must reverse the inequality sign.

Example: Solve for x: 2x - 3 < 7

1. Add 3 to both sides: 2x < 10
2. Divide by 2 (which is positive, so sign remains): x < 5

Example: Solve for x: -3x + 1 > 10

1. Subtract 1 from both sides: -3x > 9
2. Divide by -3 (which is negative, so reverse sign): x < -3

6. Problems on Ages

These problems involve setting up algebraic equations based on the ages of people now, in the past, or in the future.

• Strategy:
1. Define variables for the current ages.
2. Express past or future ages in terms of these variables.
3. Formulate equations based on the given conditions.
4. Solve the equations.

Example: The sum of the ages of a father and son is 60 years. 5 years ago, the father's age was 7 times the son's age. Find their present ages.

• Let the present age of the son be s and the father be f.
• Given: f + s = 60 (Equation 1)
• 5 years ago: Son's age was s - 5, Father's age was f - 5.
• Given: f - 5 = 7(s - 5)
• Simplify: f - 5 = 7s - 35
• f = 7s - 30 (Equation 2)
• Substitute Equation 2 into Equation 1: (7s - 30) + s = 60
• 8s - 30 = 60
• 8s = 90
• s = 90/8 = 11.25 years (Son's present age)
• From Equation 1: f = 60 - 11.25 = 48.75 years (Father's present age)

7. Ratio and Proportion

Ratio: Compares two quantities. For example, a:b.

Proportion: States that two ratios are equal. For example, a:b :: c:d, which means a/b = c/d.

• In a proportion a:b :: c:d, 'a' and 'd' are called extremes, and 'b' and 'c' are called means.
• Product of extremes = Product of means (ad = bc).

Example: If 3 kg of sugar costs Rs. 90, what is the cost of 5 kg of sugar?

• Let the cost of 5 kg be x.
• Set up the proportion: 3 kg : Rs. 90 :: 5 kg : x
• 3 / 90 = 5 / x
• 3x = 90 * 5
• 3x = 450
• x = 150. So, 5 kg costs Rs. 150.