Practice question

Duration: 2 min

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Inside: a video lesson and guided study material.

Module outline

  1. English Comprehension/Verbal Ability: Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Vocabulary, Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Subject Verb Agreement (Verb Noun Agreement), Tenses, Verb, Articles, Preposition, Conjunction, Pronoun, Noun, Adjectives, Adverb, Modals, Cloze Test (Cloze Passage), Active Passive Voice (Voice), Direct and Indirect Speech, Sentence Construction
  2. Logical Ability: Coding Decoding, Directions (Direction Test), Analogy, Inequality, Data Sufficiency, Series (Number and Letter Series) (Numerical Relations and Reasoning), Syllogisms, Seating Arrangements, Puzzle Solving (Puzzles), Blood Relations, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Ranking, Order and Sequence, Logical Sequence Of Words, Data Arrangement, Venn Diagram, Calendar, Clock
  3. Quantitative Ability: Logarithm, Permutation and Combination, Probability, Profit and Loss, Speed Time and Distance, Mixture and Alligation, Partnership, Simple Interest and Compound Interest, Percentage, Ratio and Proportion (Ratios), Average, Time and Work, Pipe and Cistern, Number System, Divisibility Rules, HCF LCM, Algebra, Powers and Exponents (Surds and Indices), Mensuration and Geometry, Age Problems, Data Interpretation, Sequence and Series (Series)
  4. Basic Analytical Ability: Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Analytical Reasoning (Counting Figures Reasoning), Cubes & Dices, Selection Decision Table (Decision Making)
  5. Computer Programming: Pseudo Codes, Coding Questions by Yash Sir, C Programming by Yash Sir, Programming in C++, Java, Python for Placements by Yash Sir
  6. DBMS (Computer Science): Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Transaction Management, Concurrency Control, Database Security & Authorization
  7. SQL (Computer Science): SQL
  8. OS (Computer Science): Introduction to OS, Process Management, CPU Scheduling, Deadlock, Process Synchronization, Threads & Process Creation, Memory Management, Virtual Memory, File Management
  9. Computer Networks (Computer Science): Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, DLL: Framing, 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
  10. Data Structures (Computer Science): Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
  11. OOPs (Computer Science): Object Oriented Programming (OOPS)
  12. Software Engineering (Computer Science): Fundamentals of Software Engineering, Software Requirements and Quality Assurance, Software Design, Estimation and Metrics, Software Testing, Software Maintenance and Configuration Management
  13. SVAR Conversational - Spoken English: Communication Assessment
  14. HR Interview: MR & HR Interview Preparation
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This short practice-question video demonstrates how to compute the memory address of a specific element in a one-dimensional array. The on-screen question states that the base address of the first element is 250, each element occupies 3 bytes, and asks for the address of the fifth element in array a[10]. The instructor first identifies the given values, writing B = 250 and w = 3B in red handwriting. A diagram of the array is drawn as a rectangle with indices 0 through 9 written above it, emphasizing that the fifth element corresponds to index 4 because indexing starts at zero. The solution is then computed manually as 250 + (4 × 3) = 262. A formal array addressing formula, a[k] = B + w * (k - Lb), is also shown and applied with k = 4, B = 250, w = 3, and Lb = 0 to verify the same result. The final answer, 262 bytes, is underlined in red.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The slide displays the orange question: "Let the base address of the first element of the array is 250 and each element of the array occupies 3 bytes in the memory, then address of the fifth element of a one-dimensional array a[10]?" Red handwriting appears showing "B = 250" and "w = 3B", with a red underline under "first element". An empty red rectangle is drawn, then index numbers "0 1 2 3 4 5 6 7 8 9" are written above it to show the fifth element is at index 4. The working expands to "=250 + 4*3B", then "250 + 12", and finally "= 262B" underlined in red. A formal formula, a[k] = B + w * (k - Lb), is shown and applied with k=4, B=250, w=3 to yield 262.

The central concept is array element address calculation using the formula a[k] = B + w * (k - Lb), where B is the base address, w is the word size in bytes, k is the target index, and Lb is the lower bound (starting index). The key teaching point is that "fifth element" means index 4 in a zero-based array, not index 5. The instructor demonstrates both a manual offset calculation (base + index × word size) and the formal formula, arriving at 262 bytes. This reinforces that address computation depends on the index offset from the lower bound multiplied by element size, added to the base address.

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