Practice Question

Duration: 7 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 | Digital Logic Systems: Digital Systems & Boolean Basics, Logic Gates & Hardware, Combinational Circuit, Sequential Circuits
  10. Paper-2 | Database Management System: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, Relational Calculus, Relational Algebra, SQL, NoSQL
  11. Paper-2 | Data Structures & Algorithms: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Algorithm Analysis, Sorting Algorithms, Intro to Compilers, Hashing
  12. Paper 2 | Data Communication and Network Security: Introduction to CN, Data Communication, Data Link Layer - Ethernet, DLL: Error Control, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Application Layer, Hardware basics, Network Security, Ethical Hacking
  13. Paper-2 | Operating Systems: Introduction to OS, Process Management, Memory Management, File Management, Deadlock
  14. Paper-2 | Computer Organization and Architecture: Basics of COA, Floating Point Rep, Cache Memory Organization, Input Output Organisation, Instr Formats & Modes, Control Unit Design, Register Transfer and Microoperations, Programming the Basic Computer, Number System, Number Representation
  15. Paper 2 | System Analysis and Design: Fundamentals of Systems and Information Systems, Fundamentals of Software Engineering, SDLC and Development Approaches, Software Documentation and Software Requirements Specification (SRS), SDD and Quality Assurance, Security, Risk Management & Maintenance, Software Testing
  16. Paper-2 | C Programming: C Fundamentals, Control Flow, Functions, Arrays & Pointers, Storage Classes, Structures & Enums, DMA, Macros, Scoping & File Handling
  17. Paper 2 | OOPS , C++ , JAVA: Language Design, OOPS with C++, Java, .Net Programming, Java Fundamentals
  18. 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
  19. Emerging Trends and AI: Emerging Tech Trends, Artificial Intelligence
  20. Paper-2 | Data Processing (Ms-Office): MS Office Applications
  21. Paper-2 | Internet and Web Technologies: HTML Basics, CSS, XML, JavaScript-Basics, Internet Fundamentals, Protocols & Services, Web Publishing & Hosting, Server Side & Security, Emerging Technologies, Terms & Abbreviations, Cyber Threats and Malware Attacks, Role of Internet and Web Services, Security in Mobile and Wireless Computing
  22. Paper 2 | Pedagogy: Child Development, Theories, Concept of Pedagogy, Inclusive Education, Experiential Pedagogy, Assessment & Evaluation, Psychology, Curriculum Framework & Reforms, National Education Policy (NEP) 2020
  23. Live Classes: RSSB Live Sessions
  24. Paper 1 - Full Mock Tests:
  25. Paper 2 - Full Mock Tests:
  26. BCI Previous Year Papers:
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This educational video segment focuses on solving a practice problem regarding the memory location calculation of two-dimensional array elements. The instructor presents an array VAL with dimensions ranging from 1 to 15 for rows and 1 to 10 for columns. Each element occupies 4 bytes of storage, with a base address of 1500. The core task is to determine the memory location of the specific element VAL[12][9] using two distinct storage methods: Row Major Order (RMO) and Column Major Order (CMO). The session begins by defining the problem parameters, including lower bounds L1 and L2, which are both 1 in this context. The instructor systematically derives the general formula for Row Major Order, Loc(A[i][j]) = BA + W[(i - L1) * #cols + (j - L2)], and applies it to the specific indices. Subsequently, the video transitions to Column Major Order, utilizing a similar but structurally different formula that prioritizes column indices over row indices in the offset calculation. The final output provides precise memory addresses for both storage schemes, demonstrating the arithmetic progression required to solve such data structure problems.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with the instructor displaying a practice question on screen regarding an array VAL[1...15][1...10] stored in memory. The problem statement specifies that each element requires 4 bytes of storage and the base address is 1500. The instructor identifies the task as finding the location of VAL[12][9] using both Row Major Order (RMO) and Column Major Order. Visible text on screen includes the array dimensions, base address value, and element size. The instructor begins writing 'Row wise (RMA)' on the board to initiate the solution process for the first storage method. The setup phase involves identifying the lower bounds L1 and L2 as 1, and calculating the total number of columns as U2 - L2 + 1 = 10. The instructor writes the general formula for Row Major Order: Loc(A[i][j]) = BA + W [(i - L1) x #cols + j - L2]. This section establishes the foundational parameters required for subsequent calculations.

  2. 2:00 – 5:00 02:00-05:00

    The instructor proceeds to solve the Row Major Order (RMO) calculation step-by-step. The screen shows the substitution of specific values into the previously written formula: Loc(A[i][j]) = BA + w[(i-L1)*#cols + (j-L2)]. The target indices are identified as i=12 and j=9. The calculation begins with the substitution: 1500 + 4[(12-1)*10 + (9-1)]. The instructor simplifies the terms inside the brackets, calculating 12 minus 1 to get 11, and 9 minus 1 to get 8. The intermediate step shows the multiplication of 11 by the number of columns (10), resulting in 110, which is then added to 8. The expression becomes 1500 + 4[11*10 + 8], which simplifies to 1500 + 4[118]. The final arithmetic yields 1500 + 472, resulting in a memory address of 1972. The instructor writes the final result clearly on the screen, confirming the location of VAL[12][9] under row-wise storage.

  3. 5:00 – 7:13 05:00-07:13

    The video transitions to the second part of the problem, calculating the memory location using Column Major Order (CMO). The instructor writes the specific formula for column-wise storage: Loc(A[i][j]) = BA + W[(i-Lr) * #rows + (j-Lc)]. The visible text indicates the base address remains 1500 and element size W is 4. The instructor substitutes the indices i=12 and j=9 into this new formula, along with the number of rows (15) instead of columns. The calculation proceeds as 1500 + 4[(9-1) * 15 + (12-1)]. The instructor simplifies the bracketed terms, calculating 9 minus 1 to get 8, and multiplying by 15 rows to obtain 120. The row index offset is calculated as 12 minus 1, which equals 11. Adding these gives 120 + 11 = 131. The final step multiplies 4 by 131 to get 524, which is added to the base address. The screen displays the final result as 1500 + 4 * 131 = 2024, completing the solution for column-wise storage.

The lecture segment effectively demonstrates the application of memory addressing formulas for two-dimensional arrays. The key distinction lies in how the offset is calculated: Row Major Order multiplies the row index difference by the number of columns, while Column Major Order multiplies the column index difference by the number of rows. The problem uses 1-based indexing, requiring the subtraction of lower bounds (L1=1, L2=1) from the target indices before calculation. The arithmetic progression is clearly shown, moving from general formulas to specific substitutions and finally to numerical results. The final addresses differ significantly (1972 vs 2024), highlighting the impact of storage order on memory layout. The instructor's methodical approach ensures clarity in deriving both results from the same initial problem statement.

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