Practice Questions - Part 3

Duration: 1 hr 3 min

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Inside: a video lesson and guided study material.

Module outline

  1. About the Course: About the Course
  2. Paper-1 & 2 | Basic Numeracy: Ratio and Proportion (Ratios), Divisibility Rules, Mixture and Alligation, Permutation and Combination, Logarithm, Number System, HCF LCM, Sequence and Series (Series), Speed Time and Distance, Percentage, Average, Mensuration and Geometry, Age Problems, Arithmetic, Statistics, Probability, Profit and Loss, Simple Interest and Compound Interest, Partnership, Speed Mathematics and Basic Mathematics, Algebra, Powers and Exponents (Surds and Indices), Time and Work, Pipe and Cistern, RSSB Specific Videos - Basic Numeracy
  3. Paper-1 & 2 | Data Interpretation: Data Interpretation
  4. Paper-1 & 2 | General Mental Ability: Mental Ability, RSSB Specific Videos - Mental Ability
  5. Paper-1 & 2 | Logical Reasoning and Analytical Ability: Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Data Sufficiency, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Syllogisms, Seating Arrangements, Blood Relations, Directions (Direction Test), Analytical Reasoning (Counting Figures Reasoning), Puzzle Solving (Puzzles), Calendar, Clock, Cubes & Dices, Venn Diagram, Analogy, Ranking, Order and Sequence, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Data Arrangement, Logical Sequence Of Words, Crypt Arithmetic
  6. Paper-1 & 2 | Decision Making and Problem Solving: Selection Decision Table (Decision Making)
  7. Paper-1 | General Science: Physics, Biology, Chemistry, RSSB Specific Videos
  8. Paper-1 | General Awareness: State Wise GK, GK- Current Affairs
  9. Paper-2 | Fundamentals of Computer: Basics of COA, Number System, Floating Point Rep, Number Representation
  10. Paper - 2 | Digital Logic Systems: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits
  11. Paper-2 | Operating Systems: Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Deadlock, Threads & Process Creation, Memory Management, Virtual Memory, Disc Scheduling, File Management, Linux OS, Distributed Systems
  12. Paper-2 | Database Management System: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, Relational Algebra, SQL, Relational Calculus, File Organization & Indexing, Transaction Management, Concurrency Control
  13. Paper-2 | Data Structures: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
  14. Paper-2 | C Programming: C Fundamentals, Control Flow, Functions, Arrays & Pointers, Storage Classes, Structures & Enums, DMA, Macros, Scoping & File Handling
  15. Paper 2 | Oops, C++, Java: Language Design, OOPS with C++, Java, .Net Programming, Java Fundamentals
  16. Paper 2 | Python: Introduction to Python, Tokens, Variables, Data Types, I/O, Operators & Expressions, Control Flow & Looping, Collections & Strings, Functions & Modules, File Handling & Data Analysis
  17. Paper 2 | Emerging Trends and AI: Emerging Tech Trends, Artificial Intelligence
  18. Paper-2 | Algorithms: Algorithm Analysis, Time Complexity Analysis, Sorting Algorithms, Greedy Algorithms, Advanced Algorithms
  19. Paper 2 | Data Communication and Network Security: Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Net Layer:Routing Protocol, Transport Layer Services, TL: Congestion & UDP, Application Layer, Hardware basics, Network Security, Ethical Hacking
  20. Paper-2 I Software Engineering: Fundamentals of Software Engineering, Software Requirements and Quality Assurance, Software Design, Estimation and Metrics, Software Testing, Software Maintenance and Configuration Management
  21. Paper-2 | Computer Organization and Architecture: Cache Memory Organization, Input Output Organisation, Instr Formats & Modes, Control Unit Design, Register Transfer and Microoperations
  22. Paper-2 | Web Development: JavaScript-Basics, HTML Basics, CSS, XML
  23. Pedagogy: Child Development, Concept of Pedagogy, Psychology, Theories, Inclusive Education, Experiential Pedagogy, Curriculum Framework & Reforms, National Education Policy (NEP) 2020
  24. Live Sessions: RSSB Live Sessions
  25. RSSB Paper 1 - Full Mock Tests:
  26. RSSB Paper 2 - Full Mock Tests:
  27. SCI Previous Year Papers:
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This video is a comprehensive lecture on geometry and mensuration, presented by an instructor from Knowledge Gate. The session systematically covers a series of problems involving the calculation of areas, perimeters, and other properties of various shapes. The instructor begins with a word problem about a wall, using algebraic substitution to find the sum of its length and height. He then moves to a problem on percentage change in the area of a rectangle when its dimensions are altered. The lecture continues with practical applications, such as calculating the number of tiles needed to pave a floor and the cost of levelling a circular track. The instructor demonstrates the use of formulas for the area of a rhombus, the area of a shaded region in a circle, and the radius of an incircle in a triangle, using Heron's formula and the inradius formula. The video is structured as a problem-solving session, with the instructor writing equations and diagrams on a whiteboard to guide the viewer through each step.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a title card for 'GEOMETRY & MENSURATION'. The instructor, Yash Jain Sir, introduces the first problem: 'The length of a wall is 5/4 times of its height. If the area of wall be 180 sq. meters, what is the sum of the length and height of the wall?'. He begins to solve it by setting up variables, writing 'l = 5x' and 'h = 4x' on the whiteboard, representing the length and height in terms of a common variable 'x'. He then writes the area formula: 'Area = length x height'.

  2. 2:00 – 5:00 02:00-05:00

    The instructor continues solving the wall problem. He substitutes the variables into the area formula, writing '180 = 5x x 4x'. He simplifies this to '180 = 20x²' and then divides both sides by 20 to get '9 = x²'. He concludes that 'x = 3'. He then calculates the length as '5x = 15' and the height as '4x = 12'. The final answer is the sum: '15 + 12 = 27'. He circles the answer '27' on the board.

  3. 5:00 – 10:00 05:00-10:00

    The instructor moves to the next problem: 'If the length of a rectangular field is doubled and its breadth is halved. What is % change in its area?'. He uses a numerical example, setting the original length to '1 m' and breadth to '2 m', so the original area is '1 x 2 = 2'. He then calculates the new dimensions: length becomes '2 m' and breadth becomes '1 m'. The new area is '2 x 1 = 2'. He concludes that the area remains unchanged, so the percentage change is '0%'. He writes '0% -> no change' on the board.

  4. 10:00 – 15:00 10:00-15:00

    The third problem is introduced: 'How many tiles 20 cm by 40 cm will be required to pave the floor of a prayer hall of a room 16 meters long and 9 meters wide?'. The instructor converts the room dimensions to centimeters: '16 m = 1600 cm' and '9 m = 900 cm'. He calculates the area of the prayer hall as '1600 x 900 = 1,440,000 cm²'. He then calculates the area of one tile as '20 x 40 = 800 cm²'. He sets up the equation 'N x 800 = 1,440,000' and solves for N, finding 'N = 1800' tiles.

  5. 15:00 – 20:00 15:00-20:00

    The fourth problem is presented: 'The outer circumference of a circular track is 220 meters. The track is 7 meters wide everywhere. Calculate the cost of levelling the track at the rate of 50 paise per square meter?'. The instructor draws a diagram of a circular track (an annulus). He uses the formula for circumference, '2πR = 220', to find the outer radius 'R'. He calculates 'R = 220 / (2π) = 35 m'. He then finds the inner radius 'r' as 'R - 7 = 35 - 7 = 28 m'. He calculates the required area as the difference between the areas of the two circles: 'πR² - πr²'.

  6. 20:00 – 25:00 20:00-25:00

    The instructor completes the calculation for the circular track. He uses the formula for the area of an annulus, 'Required Area = π(R² - r²)'. He substitutes the values: 'π(35² - 28²)'. He simplifies this using the difference of squares: 'π(35 + 28)(35 - 28) = π(63)(7)'. He calculates the area as '22/7 x 63 x 7 = 1386 m²'. He then calculates the cost as '1386 x 50 paise = 693 rupees'. He writes the final answer 'Rs. 693' on the board.

  7. 25:00 – 30:00 25:00-30:00

    The fifth problem is introduced: 'In the following figure, find the area of the shaded region.'. The figure shows a rectangle inscribed in a circle. The rectangle has sides 8 and 6. The instructor draws the figure and identifies the diagonal of the rectangle as the diameter of the circle. He uses the Pythagorean theorem: 'h² = 8² + 6² = 64 + 36 = 100', so the diagonal 'h = 10'. He concludes the diameter is 10, so the radius is 5.

  8. 30:00 – 35:00 30:00-35:00

    The instructor calculates the area of the shaded region. He finds the area of the circle: 'πr² = 3.14 x 5² = 78.5'. He finds the area of the rectangle: '8 x 6 = 48'. He subtracts the area of the rectangle from the area of the circle to get the shaded area: '78.5 - 48 = 30.5'. He writes the final answer '30.5' on the board.

  9. 35:00 – 40:00 35:00-40:00

    The sixth problem is presented: 'The diagonals of a rhombus are 16 cm and 12 cm. Its perimeter is?'. The instructor draws a rhombus and its diagonals, which bisect each other at right angles. He states that the diagonals are 16 cm and 12 cm, so each half is 8 cm and 6 cm. He uses the Pythagorean theorem to find the side length: 'side² = 8² + 6² = 64 + 36 = 100', so the side length is 10 cm.

  10. 40:00 – 45:00 40:00-45:00

    The instructor calculates the perimeter of the rhombus. He states that all four sides are equal, so the perimeter is '4 x side = 4 x 10 = 40 cm'. He writes the final answer '40 cm' on the board. He also mentions the formula for the area of a rhombus: '1/2 x d1 x d2'.

  11. 45:00 – 50:00 45:00-50:00

    The seventh problem is introduced: 'The area of the square garden is 625 sq. meters. What is the area of a path of width 2.5 meters around it if the path is outside the garden?'. The instructor draws a square garden with a path around it. He calculates the side of the square garden as 'a = √625 = 25 m'. He then calculates the side of the larger square (garden + path) as '25 + 2.5 + 2.5 = 30 m'.

  12. 50:00 – 55:00 50:00-55:00

    The instructor calculates the area of the path. He finds the area of the larger square: '30² = 900 m²'. He finds the area of the garden: '25² = 625 m²'. He subtracts the garden area from the larger square area to get the path area: '900 - 625 = 275 m²'. He writes the final answer '275 m²' on the board.

  13. 55:00 – 60:00 55:00-60:00

    The eighth problem is presented: 'What is the measure of the radius of the circle inscribed in a triangle whose sides measure 8, 15 and 17 units?'. The instructor draws a triangle with an incircle. He identifies the triangle as a right-angled triangle because '17² = 15² + 8²' (289 = 225 + 64). He states the formula for the inradius: 'r = A / s', where A is the area and s is the semi-perimeter.

  14. 60:00 – 63:14 60:00-63:14

    The instructor calculates the inradius. He calculates the semi-perimeter 's = (8 + 15 + 17) / 2 = 20'. He calculates the area 'A = 1/2 x 8 x 15 = 60'. He then calculates the inradius 'r = A / s = 60 / 20 = 3'. He writes the final answer '3' on the board. The video ends with a logo for Knowledge Gate.

This video is a structured problem-solving session on geometry and mensuration. The instructor systematically works through a series of eight distinct problems, each focusing on a different shape or concept. The progression moves from basic algebraic word problems (wall) to more complex applications involving circles, rectangles, rhombuses, and triangles. Key concepts demonstrated include the use of variables to represent unknowns, the application of fundamental formulas for area and perimeter (e.g., Area = length x breadth, Area of a circle = πr², Area of a rhombus = 1/2 x d1 x d2), and the use of geometric theorems (e.g., Pythagorean theorem, properties of a rhombus). The instructor consistently uses a step-by-step approach, writing out equations and diagrams on a whiteboard to clearly illustrate the solution process for each problem, making it a comprehensive review for students preparing for competitive exams.

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