IT1352 CRYPTOGRAPHY AND NETWORK SECURITY CURRICULUM AND SYLLABUS – REGULATIONS – 2004 B.E. COMPUTER SCIENCE AND ENGINEERING

ANNA UNIVERSITY CHENNAI: CHENNAI – 600 025
B.E DEGREE PROGRAMME COMPUTER SCIENCE AND ENGINEERING
(Offered in Colleges affiliated to Anna University)
CURRICULUM AND SYLLABUS – REGULATIONS – 2004
B.E. COMPUTER SCIENCE AND ENGINEERING
LIST OF ELECTIVES

IT1352 CRYPTOGRAPHY AND NETWORK SECURITY

AIM
To understand the principles of encryption algorithms; conventional and public key cryptography. To have a detailed knowledge about authentication, hash functions and application level security mechanisms.

OBJECTIVES
• To know the methods of conventional encryption.
• To understand the concepts of public key encryption and number theory
• To understand authentication and Hash functions.
• To know the network security tools and applications.
• To understand the system level security used.

UNIT I INTRODUCTION 10
OSI Security Architecture - Classical Encryption techniques – Cipher Principles – Data Encryption Standard – Block Cipher Design Principles and Modes of Operation - Evaluation criteria for AES – AES Cipher – Triple DES – Placement of Encryption Function – Traffic Confidentiality

UNIT II PUBLIC KEY CRYPTOGRAPHY 10
Key Management - Diffie-Hellman key Exchange – Elliptic Curve Architecture and Cryptography - Introduction to Number Theory – Confidentiality using Symmetric Encryption – Public Key Cryptography and RSA.

UNIT III AUTHENTICATION AND HASH FUNCTION 9
Authentication requirements – Authentication functions – Message Authentication Codes – Hash Functions – Security of Hash Functions and MACs – MD5 message Digest algorithm - Secure Hash Algorithm – RIPEMD – HMAC Digital Signatures – Authentication Protocols – Digital Signature Standard

UNIT IV NETWORK SECURITY 8
Authentication Applications: Kerberos – X.509 Authentication Service – Electronic Mail Security – PGP – S/MIME - IP Security – Web Security.

UNIT V SYSTEM LEVEL SECURITY 8
Intrusion detection – password management – Viruses and related Threats – Virus Counter measures – Firewall Design Principles – Trusted Systems.

TUTORIAL 15
TOTAL : 60
TEXT BOOK
1. William Stallings, “Cryptography And Network Security – Principles and Practices”, Prentice Hall of India, Third Edition, 2003.
REFERENCES
1. Atul Kahate, “Cryptography and Network Security”, Tata McGraw-Hill, 2003.
2. Bruce Schneier, “Applied Cryptography”, John Wiley & Sons Inc, 2001.
3. Charles B. Pfleeger, Shari Lawrence Pfleeger, “Security in Computing”, Third Edition, Pearson Education, 2003.

IT1352 CRYPTOGRAPHY AND NETWORK SECURITY CURRICULUM AND SYLLABUS – REGULATIONS – 2004 B.E. COMPUTER SCIENCE AND ENGINEERING

ANNA UNIVERSITY CHENNAI: CHENNAI – 600 025
B.E DEGREE PROGRAMME COMPUTER SCIENCE AND ENGINEERING
(Offered in Colleges affiliated to Anna University)
CURRICULUM AND SYLLABUS – REGULATIONS – 2004
B.E. COMPUTER SCIENCE AND ENGINEERING
LIST OF ELECTIVES

IT1352 CRYPTOGRAPHY AND NETWORK SECURITY

AIM
To understand the principles of encryption algorithms; conventional and public key cryptography. To have a detailed knowledge about authentication, hash functions and application level security mechanisms.

OBJECTIVES
• To know the methods of conventional encryption.
• To understand the concepts of public key encryption and number theory
• To understand authentication and Hash functions.
• To know the network security tools and applications.
• To understand the system level security used.

UNIT I INTRODUCTION 10
OSI Security Architecture - Classical Encryption techniques – Cipher Principles – Data Encryption Standard – Block Cipher Design Principles and Modes of Operation - Evaluation criteria for AES – AES Cipher – Triple DES – Placement of Encryption Function – Traffic Confidentiality

UNIT II PUBLIC KEY CRYPTOGRAPHY 10
Key Management - Diffie-Hellman key Exchange – Elliptic Curve Architecture and Cryptography - Introduction to Number Theory – Confidentiality using Symmetric Encryption – Public Key Cryptography and RSA.

UNIT III AUTHENTICATION AND HASH FUNCTION 9
Authentication requirements – Authentication functions – Message Authentication Codes – Hash Functions – Security of Hash Functions and MACs – MD5 message Digest algorithm - Secure Hash Algorithm – RIPEMD – HMAC Digital Signatures – Authentication Protocols – Digital Signature Standard

UNIT IV NETWORK SECURITY 8
Authentication Applications: Kerberos – X.509 Authentication Service – Electronic Mail Security – PGP – S/MIME - IP Security – Web Security.

UNIT V SYSTEM LEVEL SECURITY 8
Intrusion detection – password management – Viruses and related Threats – Virus Counter measures – Firewall Design Principles – Trusted Systems.

TUTORIAL 15
TOTAL : 60
TEXT BOOK
1. William Stallings, “Cryptography And Network Security – Principles and Practices”, Prentice Hall of India, Third Edition, 2003.
REFERENCES
1. Atul Kahate, “Cryptography and Network Security”, Tata McGraw-Hill, 2003.
2. Bruce Schneier, “Applied Cryptography”, John Wiley & Sons Inc, 2001.
3. Charles B. Pfleeger, Shari Lawrence Pfleeger, “Security in Computing”, Third Edition, Pearson Education, 2003.

IT1353 EMBEDDED SYSTEMS DESIGN SYLLABUS ELECTRONICS AND INSTRUMENTATION ENGINEERING

ANNA UNIVERSITY CHENNAI: CHENNAI – 600 025
B.E DEGREE PROGRAMME COMPUTER SCIENCE AND ENGINEERING
(Offered in Colleges affiliated to Anna University)
CURRICULUM AND SYLLABUS – REGULATIONS – 2004
B.E. COMPUTER SCIENCE AND ENGINEERING
LIST OF ELECTIVES IT1353 EMBEDDED SYSTEMS

AIM
To give sufficient background for undertaking embedded systems design.

OBJECTIVES
• To introduce students to the embedded systems, its hardware and software.
• To introduce devices and buses used for embedded networking.
• To explain programming concepts and embedded programming in C and C++.
• To explain real time operating systems, inter-task communication and an exemplary case of MUCOS – IIRTOS.

UNIT I INTRODUCTION TO EMBEDDED SYSTEMS 9
Definition and Classification – Overview of Processors and hardware units in an embedded system – Software embedded into the system – Exemplary Embedded Systems – Embedded Systems on a Chip (SoC) and the use of VLSI designed circuits

UNIT II DEVICES AND BUSES FOR DEVICES NETWORK 9
I/O Devices - Device I/O Types and Examples – Synchronous - Iso-synchronous and Asynchronous Communications from Serial Devices - Examples of Internal Serial-Communication Devices - UART and HDLC - Parallel Port Devices - Sophisticated interfacing features in Devices/Ports- Timer and Counting Devices - ‘12C’, ‘USB’, ‘CAN’ and advanced I/O Serial high speed buses- ISA, PCI, PCI-X, cPCI and advanced buses.

UNIT III PROGRAMMING CONCEPTS AND EMBEDDED PROGRAMMING IN C, C++ 9
Programming in assembly language (ALP) vs. High Level Language - C Program Elements, Macros and functions -Use of Pointers - NULL Pointers - Use of Function Calls – Multiple function calls in a Cyclic Order in the Main Function Pointers – Function Queues and Interrupt Service Routines Queues Pointers – Concepts of EMBEDDED PROGRAMMING in C++ - Objected Oriented Programming – Embedded Programming in C++, ‘C’ Program compilers – Cross compiler – Optimization of memory codes.

UNIT IV REAL TIME OPERATING SYSTEMS – PART - 1 9
Definitions of process, tasks and threads – Clear cut distinction between functions – ISRs and tasks by their characteristics – Operating System Services- Goals – Structures- Kernel - Process Management – Memory Management – Device Management – File System Organisation and Implementation – I/O Subsystems – Interrupt Routines Handling in RTOS, REAL TIME OPERATING SYSTEMS : RTOS Task scheduling models - Handling of task scheduling and latency and deadlines as performance metrics – Co-operative Round Robin Scheduling – Cyclic Scheduling with Time Slicing (Rate Monotonics Co-operative Scheduling) – Preemptive Scheduling Model strategy by a Scheduler – Critical Section Service by a Preemptive Scheduler – Fixed (Static) Real time scheduling of tasks - INTER PROCESS COMMUNICATION AND SYNCHRONISATION – Shared data problem – Use of Semaphore(s) – Priority Inversion Problem and Deadlock Situations – Inter Process Communications using Signals – Semaphore Flag or mutex as Resource key – Message Queues – Mailboxes – Pipes – Virtual (Logical) Sockets – Remote Procedure Calls (RPCs).

UNIT V REAL TIME OPERATING SYSTEMS – PART - 2 9
Study of Micro C/OS-II or Vx Works or Any other popular RTOS – RTOS System Level Functions – Task Service Functions – Time Delay Functions – Memory Allocation Related Functions – Semaphore Related Functions – Mailbox Related Functions – Queue Related Functions – Case Studies of Programming with RTOS – Understanding Case Definition – Multiple Tasks and their functions – Creating a list of tasks – Functions and IPCs – Exemplary Coding Steps.

TOTAL: 45
TEXT BOOKS
1. Rajkamal, Embedded Systems Architecture, Programming and Design, TATA McGraw-Hill, First reprint Oct. 2003

REFERENCES
1. Steve Heath, Embedded Systems Design, Second Edition-2003, Newnes,
2. David E.Simon, An Embedded Software Primer, Pearson Education Asia, First Indian Reprint 2000.
3. Wayne Wolf, Computers as Components; Principles of Embedded Computing System Design – Harcourt India, Morgan Kaufman Publishers, First Indian Reprint 2001
4. Frank Vahid and Tony Givargis, Embedded Systems Design – A unified Hardware / Software Introduction, John Wiley, 2002.

IT1353 EMBEDDED SYSTEMS DESIGN SYLLABUS ELECTRONICS AND INSTRUMENTATION ENGINEERING

ANNA UNIVERSITY CHENNAI: CHENNAI – 600 025
B.E DEGREE PROGRAMME COMPUTER SCIENCE AND ENGINEERING
(Offered in Colleges affiliated to Anna University)
CURRICULUM AND SYLLABUS – REGULATIONS – 2004
B.E. COMPUTER SCIENCE AND ENGINEERING
LIST OF ELECTIVES IT1353 EMBEDDED SYSTEMS

AIM
To give sufficient background for undertaking embedded systems design.

OBJECTIVES
• To introduce students to the embedded systems, its hardware and software.
• To introduce devices and buses used for embedded networking.
• To explain programming concepts and embedded programming in C and C++.
• To explain real time operating systems, inter-task communication and an exemplary case of MUCOS – IIRTOS.

UNIT I INTRODUCTION TO EMBEDDED SYSTEMS 9
Definition and Classification – Overview of Processors and hardware units in an embedded system – Software embedded into the system – Exemplary Embedded Systems – Embedded Systems on a Chip (SoC) and the use of VLSI designed circuits

UNIT II DEVICES AND BUSES FOR DEVICES NETWORK 9
I/O Devices - Device I/O Types and Examples – Synchronous - Iso-synchronous and Asynchronous Communications from Serial Devices - Examples of Internal Serial-Communication Devices - UART and HDLC - Parallel Port Devices - Sophisticated interfacing features in Devices/Ports- Timer and Counting Devices - ‘12C’, ‘USB’, ‘CAN’ and advanced I/O Serial high speed buses- ISA, PCI, PCI-X, cPCI and advanced buses.

UNIT III PROGRAMMING CONCEPTS AND EMBEDDED PROGRAMMING IN C, C++ 9
Programming in assembly language (ALP) vs. High Level Language - C Program Elements, Macros and functions -Use of Pointers - NULL Pointers - Use of Function Calls – Multiple function calls in a Cyclic Order in the Main Function Pointers – Function Queues and Interrupt Service Routines Queues Pointers – Concepts of EMBEDDED PROGRAMMING in C++ - Objected Oriented Programming – Embedded Programming in C++, ‘C’ Program compilers – Cross compiler – Optimization of memory codes.

UNIT IV REAL TIME OPERATING SYSTEMS – PART - 1 9
Definitions of process, tasks and threads – Clear cut distinction between functions – ISRs and tasks by their characteristics – Operating System Services- Goals – Structures- Kernel - Process Management – Memory Management – Device Management – File System Organisation and Implementation – I/O Subsystems – Interrupt Routines Handling in RTOS, REAL TIME OPERATING SYSTEMS : RTOS Task scheduling models - Handling of task scheduling and latency and deadlines as performance metrics – Co-operative Round Robin Scheduling – Cyclic Scheduling with Time Slicing (Rate Monotonics Co-operative Scheduling) – Preemptive Scheduling Model strategy by a Scheduler – Critical Section Service by a Preemptive Scheduler – Fixed (Static) Real time scheduling of tasks - INTER PROCESS COMMUNICATION AND SYNCHRONISATION – Shared data problem – Use of Semaphore(s) – Priority Inversion Problem and Deadlock Situations – Inter Process Communications using Signals – Semaphore Flag or mutex as Resource key – Message Queues – Mailboxes – Pipes – Virtual (Logical) Sockets – Remote Procedure Calls (RPCs).

UNIT V REAL TIME OPERATING SYSTEMS – PART - 2 9
Study of Micro C/OS-II or Vx Works or Any other popular RTOS – RTOS System Level Functions – Task Service Functions – Time Delay Functions – Memory Allocation Related Functions – Semaphore Related Functions – Mailbox Related Functions – Queue Related Functions – Case Studies of Programming with RTOS – Understanding Case Definition – Multiple Tasks and their functions – Creating a list of tasks – Functions and IPCs – Exemplary Coding Steps.

TOTAL: 45
TEXT BOOKS
1. Rajkamal, Embedded Systems Architecture, Programming and Design, TATA McGraw-Hill, First reprint Oct. 2003

REFERENCES
1. Steve Heath, Embedded Systems Design, Second Edition-2003, Newnes,
2. David E.Simon, An Embedded Software Primer, Pearson Education Asia, First Indian Reprint 2000.
3. Wayne Wolf, Computers as Components; Principles of Embedded Computing System Design – Harcourt India, Morgan Kaufman Publishers, First Indian Reprint 2001
4. Frank Vahid and Tony Givargis, Embedded Systems Design – A unified Hardware / Software Introduction, John Wiley, 2002.

EC 1032 EMBEDDED SYSTEM DESIGN SYLLABUS ELECTRONICS AND INSTRUMENTATION ENGINEERING

ANNA UNIVERSITY CHENNAI: CHENNAI – 600 025
B.E DEGREE PROGRAMME ELECTRONICS AND INSTRUMENTATION ENGINEERING
(Offered in Colleges affiliated to Anna University)
CURRICULUM AND SYLLABUS – REGULATIONS – 2004
B.E ELECTRONICS AND INSTRUMENTATION ENGINEERING
LIST OF ELECTIVES FOR ELECTRONICS AND INSTRUMENTATION ENGINEERING
EC 1032 EMBEDDED SYSTEM DESIGN 3 0 0 100

AIM
To introduce to the functional building blocks of an embedded system for developing a real time system application.

OBJECTIVES
i. Introduce to features that build an embedded system.
ii. To help the understanding of the interaction that the various components within an embedded system have with each other.
iii. Techniques of inter facing between processors & peripheral device related to embedded processing.
iv. To enable writing of efficient programs on any dedicated processor.
v. To present in lucid manner the basic concepts of systems programming like operating system, assembler compliers etc and to understand the management task needed for developing embedded system.

1. INTRODUCTION TO EMBEDDED SYSTEM 9
Introduction to functional building blocks of embedded systems – Register, memory devices, ports, timer, interrupt controllers using circuit block diagram representation for each categories.

2. PROCESSOR AND MEMORY ORGANIZATION 6
Structural units in a processor; selection of processor & memory devices; shared memory; DMA; interfacing processor, memory and I/O units; memory management – Cache mapping techniques, dynamic allocation - Fragmentation.

3. DEVICES & BUSES FOR DEVICES NETWORK 9
I/O devices; timer & counting devices; serial communication using I2C, CAN, USB buses; parallel communication using ISA, PCI, PCI/X buses, arm bus; interfacing with devices/ports, device drivers in a system – Serial port & parallel port.

4. I/O PROGRAMMING SCHEDULE MECHANISM 12
Intel I/O instruction – Transfer rate, latency; interrupt driven I/O - Non-maskable interrupts; software interrupts, writing interrupt service routine in C & assembly languages; preventing interrupt overrun; disability interrupts.

Multi threaded programming – Context switching, premature & non-premature multitasking, semaphores.

Scheduling – Thread states, pending threads, context switching, round robin scheduling, priority based scheduling, assigning priorities, deadlock, watch dog timers.

5. REAL TIME OPERATING SYSTEM (RTOS) 9
Introduction to basic concepts of RTOS, Basics of real time & embedded system operating systems, RTOS – Interrupt handling, task scheduling; embedded system design issues in system development process – Action plan, use of target system, emulator, use of software tools.

L = 45 Total = 45

TEXT BOOKS
1. Rajkamal, ‘Embedded System – Architecture, Programming, Design’, Tata McGraw Hill, 2003.
2. Daniel W. Lewis ‘Fundamentals of Embedded Software’, Prentice Hall of India, 2004.

REFERENCE BOOKS
1. David E. Simon, ‘An Embedded Software Primer’, Pearson Education, 2004.
2. Frank Vahid, ‘Embedded System Design – A Unified hardware & Software Introduction’, John Wiley, 2002.
3. Sriram V. Iyer, Pankaj Gupte, ‘Embedded Real Time Systems Programming’, Tata McGraw Hill, 2004.
4. Steve Heath, ‘Embedded System Design’, II edition, Elsevier, 2003.

EC 1032 EMBEDDED SYSTEM DESIGN SYLLABUS ELECTRONICS AND INSTRUMENTATION ENGINEERING

ANNA UNIVERSITY CHENNAI: CHENNAI – 600 025
B.E DEGREE PROGRAMME ELECTRONICS AND INSTRUMENTATION ENGINEERING
(Offered in Colleges affiliated to Anna University)
CURRICULUM AND SYLLABUS – REGULATIONS – 2004
B.E ELECTRONICS AND INSTRUMENTATION ENGINEERING
LIST OF ELECTIVES FOR ELECTRONICS AND INSTRUMENTATION ENGINEERING
EC 1032 EMBEDDED SYSTEM DESIGN 3 0 0 100

AIM
To introduce to the functional building blocks of an embedded system for developing a real time system application.

OBJECTIVES
i. Introduce to features that build an embedded system.
ii. To help the understanding of the interaction that the various components within an embedded system have with each other.
iii. Techniques of inter facing between processors & peripheral device related to embedded processing.
iv. To enable writing of efficient programs on any dedicated processor.
v. To present in lucid manner the basic concepts of systems programming like operating system, assembler compliers etc and to understand the management task needed for developing embedded system.

1. INTRODUCTION TO EMBEDDED SYSTEM 9
Introduction to functional building blocks of embedded systems – Register, memory devices, ports, timer, interrupt controllers using circuit block diagram representation for each categories.

2. PROCESSOR AND MEMORY ORGANIZATION 6
Structural units in a processor; selection of processor & memory devices; shared memory; DMA; interfacing processor, memory and I/O units; memory management – Cache mapping techniques, dynamic allocation - Fragmentation.

3. DEVICES & BUSES FOR DEVICES NETWORK 9
I/O devices; timer & counting devices; serial communication using I2C, CAN, USB buses; parallel communication using ISA, PCI, PCI/X buses, arm bus; interfacing with devices/ports, device drivers in a system – Serial port & parallel port.

4. I/O PROGRAMMING SCHEDULE MECHANISM 12
Intel I/O instruction – Transfer rate, latency; interrupt driven I/O - Non-maskable interrupts; software interrupts, writing interrupt service routine in C & assembly languages; preventing interrupt overrun; disability interrupts.

Multi threaded programming – Context switching, premature & non-premature multitasking, semaphores.

Scheduling – Thread states, pending threads, context switching, round robin scheduling, priority based scheduling, assigning priorities, deadlock, watch dog timers.

5. REAL TIME OPERATING SYSTEM (RTOS) 9
Introduction to basic concepts of RTOS, Basics of real time & embedded system operating systems, RTOS – Interrupt handling, task scheduling; embedded system design issues in system development process – Action plan, use of target system, emulator, use of software tools.

L = 45 Total = 45

TEXT BOOKS
1. Rajkamal, ‘Embedded System – Architecture, Programming, Design’, Tata McGraw Hill, 2003.
2. Daniel W. Lewis ‘Fundamentals of Embedded Software’, Prentice Hall of India, 2004.

REFERENCE BOOKS
1. David E. Simon, ‘An Embedded Software Primer’, Pearson Education, 2004.
2. Frank Vahid, ‘Embedded System Design – A Unified hardware & Software Introduction’, John Wiley, 2002.
3. Sriram V. Iyer, Pankaj Gupte, ‘Embedded Real Time Systems Programming’, Tata McGraw Hill, 2004.
4. Steve Heath, ‘Embedded System Design’, II edition, Elsevier, 2003.

IC 1005 ROBOTICS AND AUTOMATION ANNA UNIVERSITY CHENNAI ELECTRONICS AND INSTRUMENTATION

ANNA UNIVERSITY CHENNAI: CHENNAI – 600 025
B.E DEGREE PROGRAMME ELECTRONICS AND INSTRUMENTATION ENGINEERING
(Offered in Colleges affiliated to Anna University)
CURRICULUM AND SYLLABUS – REGULATIONS – 2004
B.E ELECTRONICS AND INSTRUMENTATION ENGINEERING
LIST OF ELECTIVES FOR ELECTRONICS AND INSTRUMENTATION ENGINEERING

IC 1005 ROBOTICS AND AUTOMATION

AIM
To provide comprehensive knowledge of robotics in the design, analysis and control point of view.

OBJECTIVES
i. To study the various parts of robots and fields of robotics.
ii. To study the various kinematics and inverse kinematics of robots.
iii. To study the Euler, Lagrangian formulation of Robot dynamics.
iv. To study the trajectory planning for robot.
v. To study the control of robots for some specific applications.

1. INTRODUCTION TO ROBOTICS 9
History of Robots – Classifications – Various fields of Robotics – Actuators – Sensors – Manipulators – End effectors – Application areas – Robot programming languages.

2. ROBOT KINEMATICS 9
Matrix representation – Homogeneous transformation – DH representation of standard robots – Inverse kinematics.

3. ROBOT DYNAMICS 9
Velocity kinematics – Jacobian and inverse Jacobian – Lagrangian formulation – Eulers Lagrangian formulation – Robot equation of motion.

4. TRAJECTORY PLANNING 9
Introduction – Path Vs trajectory – Joint-space Vs Cartesian-space descriptions – Basics of trajectory planning – Joint-space trajectory planning – Cartesian-space trajectories.

5. CONTROL AND APPLICATION OF ROBOTICS 9
Linear control of robot manipulation – Second-order systems – trajectory following control – Modeling and control of single joint – Architecture of industrial robotic controllers – Robot applications.

L = 45 Total = 45

TEXT BOOKS
1. Saced B. Niku, ‘ Introduction to Robotics Analysis, Systems, Applications’, Prentice Hall of India/Pearson Education, Asia, 2001.
2. Craig, ‘Introduction to Robotics Mechanics and Control’, Second edition, Pearson Education, Asia, 2004.

REFERENCE BOOKS
1. K.S. Fu & Co., ‘Robotics Control, Sensing, Vision and Intelligence’, McGraw Hill International Editions, Industrial Engineering Series, 1991.
2. R.D.Klafter, T.A. Chimielewski and M.Negin, ‘Robotic Engineering – An integrated Approach’, Prentice Hall of India, New Delhi, 1994.
3. Mikell P. Groover, Mitchell Weiss, Roger N. Nagel, Nicholas G. Odrey, ‘Industrial Robotics Technology Programming and Application’, McGraw Hill book company, 1986.

IC 1005 ROBOTICS AND AUTOMATION ANNA UNIVERSITY CHENNAI ELECTRONICS AND INSTRUMENTATION

ANNA UNIVERSITY CHENNAI: CHENNAI – 600 025
B.E DEGREE PROGRAMME ELECTRONICS AND INSTRUMENTATION ENGINEERING
(Offered in Colleges affiliated to Anna University)
CURRICULUM AND SYLLABUS – REGULATIONS – 2004
B.E ELECTRONICS AND INSTRUMENTATION ENGINEERING
LIST OF ELECTIVES FOR ELECTRONICS AND INSTRUMENTATION ENGINEERING

IC 1005 ROBOTICS AND AUTOMATION

AIM
To provide comprehensive knowledge of robotics in the design, analysis and control point of view.

OBJECTIVES
i. To study the various parts of robots and fields of robotics.
ii. To study the various kinematics and inverse kinematics of robots.
iii. To study the Euler, Lagrangian formulation of Robot dynamics.
iv. To study the trajectory planning for robot.
v. To study the control of robots for some specific applications.

1. INTRODUCTION TO ROBOTICS 9
History of Robots – Classifications – Various fields of Robotics – Actuators – Sensors – Manipulators – End effectors – Application areas – Robot programming languages.

2. ROBOT KINEMATICS 9
Matrix representation – Homogeneous transformation – DH representation of standard robots – Inverse kinematics.

3. ROBOT DYNAMICS 9
Velocity kinematics – Jacobian and inverse Jacobian – Lagrangian formulation – Eulers Lagrangian formulation – Robot equation of motion.

4. TRAJECTORY PLANNING 9
Introduction – Path Vs trajectory – Joint-space Vs Cartesian-space descriptions – Basics of trajectory planning – Joint-space trajectory planning – Cartesian-space trajectories.

5. CONTROL AND APPLICATION OF ROBOTICS 9
Linear control of robot manipulation – Second-order systems – trajectory following control – Modeling and control of single joint – Architecture of industrial robotic controllers – Robot applications.

L = 45 Total = 45

TEXT BOOKS
1. Saced B. Niku, ‘ Introduction to Robotics Analysis, Systems, Applications’, Prentice Hall of India/Pearson Education, Asia, 2001.
2. Craig, ‘Introduction to Robotics Mechanics and Control’, Second edition, Pearson Education, Asia, 2004.

REFERENCE BOOKS
1. K.S. Fu & Co., ‘Robotics Control, Sensing, Vision and Intelligence’, McGraw Hill International Editions, Industrial Engineering Series, 1991.
2. R.D.Klafter, T.A. Chimielewski and M.Negin, ‘Robotic Engineering – An integrated Approach’, Prentice Hall of India, New Delhi, 1994.
3. Mikell P. Groover, Mitchell Weiss, Roger N. Nagel, Nicholas G. Odrey, ‘Industrial Robotics Technology Programming and Application’, McGraw Hill book company, 1986.

TCS Placement Rounds| TCS Recruitment Rounds |TCS Technical Interview Questions|TCS Placement Papers

TCS Recruitment Rounds

Written Test
Technical Interview
MR ( Managerial)
HR Interview
WRITTEN TEST :( ONLINE TEST)

Contains 3 sections
1) Verbal (Synonyms - Antonyms - Comprehension Passages)
2) Quantitative Aptitude
3) Critical Reasoning

SECTION-1(Verbal- 30 questions - 20 min)
Synonyms (Refer In GRE BARRONS 12th Edition )
Antonyms (Refer In GRE BARRONS 12th Edition (page no -126))
Passage completion

Some of the previous questions in quant .Go through these models and try to solve them. They will give same models but they change the data.

SECTION: 2(QUANT- 38 questions - 40 min )

Section-3 :( Critical Reasoning- 12 questions - 30 min)
Refer GRE BARRONS text book 12th Edition model papers.
Model paper 1: Sec - 5 , Sec- 6
Model paper 2: Sec -1 , Sec-6
Model paper 3: Sec -5 , Sec- 6
Model paper 4: Sec -5 , Sec-6
Model paper 5: Sec -3 , Sec-7
(Try to Mug up those sections .They will give definitely from those model papers only)

The entire recruitment process consisted of:
i) Initial screening(60% in 10th ,12th and B-Tech)(219 were eligible in this phase)
ii) Online Aptitude Test(190 cleared this test)
iii) Personal Interview (Both Technical and HR interview)(130 cleared and selected
finally)

ONLINE APTITUDE TEST:
The aptitude test was online.
1. Verbal-40 Q / Time- 20 mins.
2. Quantitative-38Q / Time- 40 mins.
3. Critical Reasoning-12Q / Time- 30 mins.

Total time for aptitude test was 90 mins. (No ngtv marking & no sectional cut off.)

1.Verbal:
This section consists of 10 synonyms, 20 antonyms, 5 fill in the blanks and 5 mcq type questions from a passage.
For this section you have to prepare from
1. Barron?s GRE high frequency word list.
2. Barron?s GRE antonym questions.
3. Barron?s GRE sentence completion questions.
4. Barron?s GRE reading comprehension questions.

Prepare above all properly because they repeat questions from previous papers. They exactly give the passage from Barron?s GRE (Mainly from 12th edition).Even they do not change the values.

2.Quantitative:
For this section initially u should start ur preparation with R.S.Agarwal or Taira(any one,no need to follow both of them).Then go thru the old papers,just u hv to know the methods to solve each question.
They change the numerical values and names only.
Also u need to practice Data Interpretation, Venn diagram and Pie Charts from Barron?s GRE.

3.Critical Reasoning:
Refer to GRE 12th edition (Go through these Questions given below???.In some Questions names and data were changed)

TCS Placement Rounds| TCS Recruitment Rounds |TCS Technical Interview Questions|TCS Placement Papers

TCS Recruitment Rounds

Written Test
Technical Interview
MR ( Managerial)
HR Interview
WRITTEN TEST :( ONLINE TEST)

Contains 3 sections
1) Verbal (Synonyms - Antonyms - Comprehension Passages)
2) Quantitative Aptitude
3) Critical Reasoning

SECTION-1(Verbal- 30 questions - 20 min)
Synonyms (Refer In GRE BARRONS 12th Edition )
Antonyms (Refer In GRE BARRONS 12th Edition (page no -126))
Passage completion

Some of the previous questions in quant .Go through these models and try to solve them. They will give same models but they change the data.

SECTION: 2(QUANT- 38 questions - 40 min )

Section-3 :( Critical Reasoning- 12 questions - 30 min)
Refer GRE BARRONS text book 12th Edition model papers.
Model paper 1: Sec - 5 , Sec- 6
Model paper 2: Sec -1 , Sec-6
Model paper 3: Sec -5 , Sec- 6
Model paper 4: Sec -5 , Sec-6
Model paper 5: Sec -3 , Sec-7
(Try to Mug up those sections .They will give definitely from those model papers only)

The entire recruitment process consisted of:
i) Initial screening(60% in 10th ,12th and B-Tech)(219 were eligible in this phase)
ii) Online Aptitude Test(190 cleared this test)
iii) Personal Interview (Both Technical and HR interview)(130 cleared and selected
finally)

ONLINE APTITUDE TEST:
The aptitude test was online.
1. Verbal-40 Q / Time- 20 mins.
2. Quantitative-38Q / Time- 40 mins.
3. Critical Reasoning-12Q / Time- 30 mins.

Total time for aptitude test was 90 mins. (No ngtv marking & no sectional cut off.)

1.Verbal:
This section consists of 10 synonyms, 20 antonyms, 5 fill in the blanks and 5 mcq type questions from a passage.
For this section you have to prepare from
1. Barron?s GRE high frequency word list.
2. Barron?s GRE antonym questions.
3. Barron?s GRE sentence completion questions.
4. Barron?s GRE reading comprehension questions.

Prepare above all properly because they repeat questions from previous papers. They exactly give the passage from Barron?s GRE (Mainly from 12th edition).Even they do not change the values.

2.Quantitative:
For this section initially u should start ur preparation with R.S.Agarwal or Taira(any one,no need to follow both of them).Then go thru the old papers,just u hv to know the methods to solve each question.
They change the numerical values and names only.
Also u need to practice Data Interpretation, Venn diagram and Pie Charts from Barron?s GRE.

3.Critical Reasoning:
Refer to GRE 12th edition (Go through these Questions given below???.In some Questions names and data were changed)

TCS Recruitment Rounds |TCS Technical Interview Questions|TCS Placement Papers -IV

TCS Recruitment Rounds PART-IV
1.Written Test
2.Technical Interview
3.MR ( Managerial)
4. HR Interview
WRITTEN TEST :( ONLINE TEST)

Contains 3 sections

1) Verbal (Synonyms - Antonyms - Comprehension Passages)
2) Quantitative Aptitude
3) Critical Reasoning

SECTION-1(Verbal- 30 questions - 20 min)
1.Synonyms (Refer In GRE BARRONS 12th Edition )
2.Antonyms (Refer In GRE BARRONS 12th Edition (page no -126))
3. Passage completion

For more Question,Placement Papers Log on to http://topengineeringcollegesintamilnadu.blogspot.com/

91) Last year pandit was thrice his sister,s age. Next year he is only twice her age. After 5 years what is pandit,s age.
a) 2 b) 12 c) 11 d) 14
Ans: b

92) Jalia is twice older than qurban. If jalia was 4 years younger, qurban was 3 years older their diff. between their ages is 12 years what is the sum of their ages
a) 67 b) 57 c) 36 d) none
Ans: b

93) Fathers age is 5 times his son,s age. 4 years back the father was 9 times older than son. Find the fathers, present age.
Ans. 40 years
94) Joe,s father will be twice his age 6 years from now. His mother was twice his age 2 years before. If Joe will be 24 two years from now, what is the difference between his father,s and mother,s age?
a) 4 b) 6 c) 8 d) 10

95) Anand finishes a work in 7 days; Bittu finishes the same job in 8 days and Chandu in 6 days. They take turns to finish the work. Anand on the first day, Bittu on the second and Chandu on the third day and then Anand again and so on. On which day will the work get over?
a) 3rd
b) 6th
c) 9th
d) 7th
In d 1st day Anand does 1/7th of total work
Similarly,
Bithu does 1/8th work in d 2nd day
Hence at the end of 3 days, work done = 1/7+1/8+1/6=73/168
Remaining work = (168-73)/168 = 95/168
Again after 6 days of work, remaining work is = (95-73)/168 = 22/168
And hence Anand completes the work on 7th day.
Ans is d) 7th day

96) If TAFJHH is coded as RBEKGI then RBDJK can be coded as ---------
Ans: qcckj
97) BFGE CODED AS CEHD THEN CODE PVHDJ
Ans: QUICK
98) Find the no. of Y,s followed by W but that is not followed by Z.
Y W R U D D Y W Z ...............
99) If VXUPLVH is written as SURMISE, what is SHDVD ?
Ans. PEASA
(Hint: in the first word, the alphabets of the jumbled one are three alphabets after the corresponding alphabet in the word SURMISE. S = V-3, similarly find the one for SHDVD)

100) If DDMUQZM is coded as CENTRAL then RBDJK can be coded as -----
Ans. QCEIL
(Hint: Write both the jumbled and the coded word as a table, find the relation between the corresponding words, i.e. C= D-1, N=M+1 & so on)

101) In the word ECONOMETRICS, if the first and second, third and forth, forth and fifth, fifth and sixth words are interchanged up to the last letter, what would be the tenth letter from right?
Ans. word is CENOMOTEIRSC tenth word is R
102) If D_MUQZM is coded as CENTRAL then RBDJK can be coded as

103) In a certain format TUBUJPO is coded as STATION. The code of which string is FILTER?

104) What is the code formed by reversing the First and second letters, the third and fourth letters and so on of the string SIMULTANEOUSLY?

105) In the word ORGANISATIONAL, by reversing if the first and second the third and fourth letters and so on of the string?

106) A power unit is there by the bank of the river of 750 meters width. A cable is made from power unit to power a plant opposite to that of the river and 1500mts away from the power unit. The cost of the cable below water is Rs. 15/- per meter and cost of cable on the bank is Rs.12/- per meter. Find the total of laying the cable.
Ans: 1000 (24725 - cost) 20250
Ans: Rs. 22,500 (hint: the plant is on the other side of the plant i.e. it is not on the same side as the river)

107) The cost of one pencil, two pens and four erasers is Rs.22 while the cost of five pencils, four pens and two erasers is Rs.32.How much will three pencils, three pens and three erasers cost?
Ans: 27

108) A shopkeeper bought a watch for Rs.400 and sold it for Rs.500.What is his profit percentage?
Ans. 25%

109) What percent of 60 is 12?
Ans. 20%

110) Three men goes to a hotel to stay, the clerk says $30 per room/day so all the three plans to stay in one room so each pays $10.After some time the clerk realizes that he made a mistake of collecting $30 but the room cost only $25, there fore he decides to return $5 to them so he calls the room boy and gives him $5 asking him to return. The room boy keeps $2 with him and he returns only $3($1 for each).Now Totally all have paid $9 each($27)+room boy $2 which is equal to $27.where did $1 go, who has made the mistake?

111) Two pencils cost 8 cents. Then 5 pencils cost?
Ans :( 20 cents)

112)Which is more economical of the following
a)2kg -- 30/- b)8kg -- 160/- c)5kg -- 80/-

113)Satish earns 240 weekly.12% of big amount + earning weekly = 540
what is the big amount
a)3200 b)3600 c)2500 d)1000
Ans: c
114) Bhanu spends 30% of his income on petrol on scooter. ¼ of the remaining on house rent and the balance on food. If he spends Rs.300 on petrol then what is the expenditure on house rent? a) Rs.525 b) Rs.1000 c) Rs.675 d) Rs.175
Ans: 175
115) A sporting goods store ordered an equal number of white and yellow balls. The tennis ball company delivered 45 extra white balls, making the ratio of white balls to yellow balls 1/5: 1/6. How many white tennis balls did the store originally order for?
a) 450 b) 270 c) 225 d) None of these
Ans: 180

116) There is a circular pizza with negligible thickness that is cut into ,x, pieces by 4 straight line cuts. What is the maximum and minimum value of ,x, respectively?
a) 12, 6 b) 11, 6 c) 12, 5 d) 11, 5

117) Match the following:
1. Male - Boy ---> a. A type of
2. Square - Polygon ---> b. A part of
3. Roof - Building ---> c. Not a type of
4. Mushroom - Vegetables ---> d. A superset of
Ans: 1- d, 2- a, 3- b, 4- c

118) Match the following.
1. Brother - sister ---> a. Part of
2. Alsatian - dog ---> b. Sibling
3. Sentence - paragraph ---> c. Type of
4. Car - steering ---> d. Not a type of
Ans. 1-b, 2-c, 3-a, 4-d

119) Match the following
1) Scooter --------- Automobile A. A PART OF
2).Oxygen ---------- Water B. A Type of
3).Shop staff ---------- Fitters C. NOT A TYPE OF
4). Bug ---------- Reptile D. A SUPERSET OF
Ans. 1-b, 2-a, 3-d, 4-c

120) What is the largest prime number stored in a-
-----> 6 bit pattern (ANS~2^6=64, so no is 61)
------> 7 bit pattern (ANS~2^7=128, so no is 127)
-------> 8 bit pattern (ANS~2^8=256, so no is 251)
-------->9 bit pattern (ANS~2^9=512, so no is 503)

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TAGS:TCS Recruitment Rounds,Placement Papers,TCS Placement Papers,TCS Question papers,TCS Written Test,TCS Technical Interview,TCS HR Interview,TCS Online Test,TCS Interview Papers,TCS Interview Questions

TCS Recruitment Rounds |TCS Technical Interview Questions|TCS Placement Papers -V

TCS Recruitment Rounds PART-V
1.Written Test
2.Technical Interview
3.MR ( Managerial)
4. HR Interview
WRITTEN TEST :( ONLINE TEST)

Contains 3 sections

1) Verbal (Synonyms - Antonyms - Comprehension Passages)
2) Quantitative Aptitude
3) Critical Reasoning

SECTION-1(Verbal- 30 questions - 20 min)
1.Synonyms (Refer In GRE BARRONS 12th Edition )
2.Antonyms (Refer In GRE BARRONS 12th Edition (page no -126))
3. Passage completion

121) What is the max 3 digit Prime no?
ANS=997

122) G(0)= -1, G(1)=1, G(N)=G(N-1) - G(N-2), G(5)= ?
Ans - 2

123) G (0) =1 G (1) = -1 IF G (N) =2* (G (N-1)) - 3(G (N-2)) Then what is the value of
G (4)?

124) If f (0) =1 and f (n) = f (n-1)*n, find the value of f (4).
Ans: 24

125) If g (0) =g (1) =1 and g (n) = g (n-1) + g (n -2) find g (6);

126) What is the power of 2? a. 2068 b.2048 c.2668
Ans: 2048
127) 8 to the power of x is 32, what is the value of x?

128) Power of 4 Ans-4096

129) Which one will be the exact power of 3?
(i) 2768 (ii) 2678 (iii) 2187
130) Complete the series-
a )3,8,a,24,b,48,63 [ ANS~ a=15, b=35 ]
[HINT~DIFFERENCE IS 5, 7, 9, 11, 13, 15]
B )26,19,17,13,11, ,8,7 [ ANS=9]
[HINT~26,17,11,8 DECREASING LIKE 9,6,3 & 19,13,9,7 DECREASING
LIKE 6, 4, 2]
c)9,10,11,13,15, ,21,28 [ ANS=19 ]
[HINT~9, 11, 15, and 21 INCREASING LIKE 2, 4, 6 & 10,13,19,28 INCRESING
LIKE 3, 6, and 9]
D) 4, -5, 11, -14, 22, --- [ ANS= -27]

131) Number of faces, vertices and edges of a cube
ANS: 6,8,12
132) Find the value of-
a) @@+25-++@16, where @ denotes"square" & + denotes "square root". [ANS=621]
b) $%$6-%$%6, where $ means "tripling" & % means "change of sign". [ANS= -72]
c) % # % 6 + # %# 6, % means "doubling" & # mean "reciprocal".
132) Select odd one out
1) LINUX, WINDOWS 98, SOLARIS, SMTP (ANS: SMTP)
2)MVS
3).JAVA b) LISP c) Smaltalk d)Eiffle Ans: LISP ( All other languages are OOPS)
4)1.http 2.arp 3.snmp 4.sap Ans:sap
5)1.linux 2.windows NT 3.sql server 4.Unix Ans: Sql server
6)1.SAP 2.ARP 3.WAP 4.TCP IP
7)a. Oracle b. Linux c. Ingress d. DB2
a. SMTP b. WAP c. SAP d. ARP Ans:SAP
9)a. WAP b. HTTP c. BAAN d. ARP Ans:Baan
10)a. LINUX b. UNIX c. SOLARIS d. SQL SERVER Ans:SQL SERVER
11)a. SQL b. DB2 c. SYBASE d. HTTP Ans: HTTP
12)a. Oracle b. Linux c. Ingress d. DB2 ANS:LINUX

133) Find the singularity matrix from a given set of matrices? (Hint det (A) ====0)
134) Which of the following are orthogonal pairs?
a. 3i+2j b. i+j c. 2i-3j d. -7i+j
Ans: a, c
135) (a) 2+3i (b) 1+i (c) 3-2i (d) 1-7i .Find which of the above is orthogonal.
Ans: a, c

136) Sum of slopes of 2 perpendicular st. lines is given. Find the pair of lines from the given set of options which satisfy the above condition?

137) If Rs.1260 is divided between A, B and C in the ratio 2:3:4, what is C,s share?
Ans: Rs. 560
138)A sum of money is divided among A, B and C such that for each rupee A gets, B gets 65paise and C gets 35paise. If C,s share is Rs.560, the sum is .
a) 2400 b) 2800 c) 3200 d) 3800
139) Complete the series.
1) 3, 8, --, 24, --, 48, 63. Ans: 15, 35
2) Complete the series. 4, -5, 11, -14, 22, --- Ans - 27
3) SERIES: 2, 7, 24, 77, ------ (238) or (240)
4)77, 49, 36, 18,? Ans: 8 (7*7=49) (4*9=36) (3*6=1 (1*8=
5) series: 5 6 7 8 10 11 14?? Ans.15 or 18
6)15 14 12 11?? 9 8 Ans.10
7) what is the 12th term of the series 2, 5, 8 ... Ans. 35
8)58, 27, 12, x, 2, 1. Find x.
9)7, 9,13,_,27,37. Ans-19
10)2, 5, __, 19, 37, 75 Ans: 9
11) Complete the sequence 9, 10,11,13,15, __, 21, 28.
140) UNITS
1) (Momentum*Velocity)/ (Acceleration * distance) find units. Ans: mass
2) (energy * time * time)/ (mass * dist) = distance
3) (momentum * velocity)/ (force * time) = velocity
4) Find the physical quantity in units from the equation:
(Force*Distance)/ (Velocity*Velocity) Ans. Ns2/m
5) Find the physical quantity represented by
MOMENTUM *VELOCITY] / [LENGTH * ACCELERATION]?

141) Find the result of the following _expression if, M denotes modulus operation, R denotes round-off, T denotes truncation:
M (373, 5) +R (3.4) +T (7.7) +R (5.8)
ANS: 19
142) Which of the following highest Standard deviation
a) 7, - 7, 7,-7, 7,-7 b) 7, 7, 7,7,7,7 c) -7, - 7, -7,-7,-7,-7 d) -7, 7, -7, 7,-7, 7
Ans: d

143) 232 expressed in base-5 is Ans: 1412

144) A building with height D shadow up to G. A neighbor building with what height shadows C feet.
|----|----|----|----|----|----|----|
A B C D E F G H
Sol: B Ft. or CD/G

145) In a fraction, if 1 is added to both the numerator at the denominator, the fraction becomes 1/2. If numerator is subtracted from the denominator, the fraction becomes 3/4. Find the fraction.
Ans. 3/7
146) The sum of the digits of a two digit number is 8. When 18 is added to the number, the digits are reversed. Find the number?
Ans. 35

147) What number should be added to or subtracted from each term of the ratio 17 : 24 so that it becomes equal to 1 : 2.
Ans. 10 should be subtracted

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TAGS:TCS Recruitment Rounds,Placement Papers,TCS Placement Papers,TCS Question papers,TCS Written Test,TCS Technical Interview,TCS HR Interview,TCS Online Test,TCS Interview Papers,TCS Interview Questions

TCS Recruitment Rounds |TCS Technical Interview Questions|TCS Placement Papers -IV

TCS Recruitment Rounds PART-IV
1.Written Test
2.Technical Interview
3.MR ( Managerial)
4. HR Interview
WRITTEN TEST :( ONLINE TEST)

Contains 3 sections

1) Verbal (Synonyms - Antonyms - Comprehension Passages)
2) Quantitative Aptitude
3) Critical Reasoning

SECTION-1(Verbal- 30 questions - 20 min)
1.Synonyms (Refer In GRE BARRONS 12th Edition )
2.Antonyms (Refer In GRE BARRONS 12th Edition (page no -126))
3. Passage completion

For more Question,Placement Papers Log on to http://topengineeringcollegesintamilnadu.blogspot.com/

91) Last year pandit was thrice his sister,s age. Next year he is only twice her age. After 5 years what is pandit,s age.
a) 2 b) 12 c) 11 d) 14
Ans: b

92) Jalia is twice older than qurban. If jalia was 4 years younger, qurban was 3 years older their diff. between their ages is 12 years what is the sum of their ages
a) 67 b) 57 c) 36 d) none
Ans: b

93) Fathers age is 5 times his son,s age. 4 years back the father was 9 times older than son. Find the fathers, present age.
Ans. 40 years
94) Joe,s father will be twice his age 6 years from now. His mother was twice his age 2 years before. If Joe will be 24 two years from now, what is the difference between his father,s and mother,s age?
a) 4 b) 6 c) 8 d) 10

95) Anand finishes a work in 7 days; Bittu finishes the same job in 8 days and Chandu in 6 days. They take turns to finish the work. Anand on the first day, Bittu on the second and Chandu on the third day and then Anand again and so on. On which day will the work get over?
a) 3rd
b) 6th
c) 9th
d) 7th
In d 1st day Anand does 1/7th of total work
Similarly,
Bithu does 1/8th work in d 2nd day
Hence at the end of 3 days, work done = 1/7+1/8+1/6=73/168
Remaining work = (168-73)/168 = 95/168
Again after 6 days of work, remaining work is = (95-73)/168 = 22/168
And hence Anand completes the work on 7th day.
Ans is d) 7th day

96) If TAFJHH is coded as RBEKGI then RBDJK can be coded as ---------
Ans: qcckj
97) BFGE CODED AS CEHD THEN CODE PVHDJ
Ans: QUICK
98) Find the no. of Y,s followed by W but that is not followed by Z.
Y W R U D D Y W Z ...............
99) If VXUPLVH is written as SURMISE, what is SHDVD ?
Ans. PEASA
(Hint: in the first word, the alphabets of the jumbled one are three alphabets after the corresponding alphabet in the word SURMISE. S = V-3, similarly find the one for SHDVD)

100) If DDMUQZM is coded as CENTRAL then RBDJK can be coded as -----
Ans. QCEIL
(Hint: Write both the jumbled and the coded word as a table, find the relation between the corresponding words, i.e. C= D-1, N=M+1 & so on)

101) In the word ECONOMETRICS, if the first and second, third and forth, forth and fifth, fifth and sixth words are interchanged up to the last letter, what would be the tenth letter from right?
Ans. word is CENOMOTEIRSC tenth word is R
102) If D_MUQZM is coded as CENTRAL then RBDJK can be coded as

103) In a certain format TUBUJPO is coded as STATION. The code of which string is FILTER?

104) What is the code formed by reversing the First and second letters, the third and fourth letters and so on of the string SIMULTANEOUSLY?

105) In the word ORGANISATIONAL, by reversing if the first and second the third and fourth letters and so on of the string?

106) A power unit is there by the bank of the river of 750 meters width. A cable is made from power unit to power a plant opposite to that of the river and 1500mts away from the power unit. The cost of the cable below water is Rs. 15/- per meter and cost of cable on the bank is Rs.12/- per meter. Find the total of laying the cable.
Ans: 1000 (24725 - cost) 20250
Ans: Rs. 22,500 (hint: the plant is on the other side of the plant i.e. it is not on the same side as the river)

107) The cost of one pencil, two pens and four erasers is Rs.22 while the cost of five pencils, four pens and two erasers is Rs.32.How much will three pencils, three pens and three erasers cost?
Ans: 27

108) A shopkeeper bought a watch for Rs.400 and sold it for Rs.500.What is his profit percentage?
Ans. 25%

109) What percent of 60 is 12?
Ans. 20%

110) Three men goes to a hotel to stay, the clerk says $30 per room/day so all the three plans to stay in one room so each pays $10.After some time the clerk realizes that he made a mistake of collecting $30 but the room cost only $25, there fore he decides to return $5 to them so he calls the room boy and gives him $5 asking him to return. The room boy keeps $2 with him and he returns only $3($1 for each).Now Totally all have paid $9 each($27)+room boy $2 which is equal to $27.where did $1 go, who has made the mistake?

111) Two pencils cost 8 cents. Then 5 pencils cost?
Ans :( 20 cents)

112)Which is more economical of the following
a)2kg -- 30/- b)8kg -- 160/- c)5kg -- 80/-

113)Satish earns 240 weekly.12% of big amount + earning weekly = 540
what is the big amount
a)3200 b)3600 c)2500 d)1000
Ans: c
114) Bhanu spends 30% of his income on petrol on scooter. ¼ of the remaining on house rent and the balance on food. If he spends Rs.300 on petrol then what is the expenditure on house rent? a) Rs.525 b) Rs.1000 c) Rs.675 d) Rs.175
Ans: 175
115) A sporting goods store ordered an equal number of white and yellow balls. The tennis ball company delivered 45 extra white balls, making the ratio of white balls to yellow balls 1/5: 1/6. How many white tennis balls did the store originally order for?
a) 450 b) 270 c) 225 d) None of these
Ans: 180

116) There is a circular pizza with negligible thickness that is cut into ,x, pieces by 4 straight line cuts. What is the maximum and minimum value of ,x, respectively?
a) 12, 6 b) 11, 6 c) 12, 5 d) 11, 5

117) Match the following:
1. Male - Boy ---> a. A type of
2. Square - Polygon ---> b. A part of
3. Roof - Building ---> c. Not a type of
4. Mushroom - Vegetables ---> d. A superset of
Ans: 1- d, 2- a, 3- b, 4- c

118) Match the following.
1. Brother - sister ---> a. Part of
2. Alsatian - dog ---> b. Sibling
3. Sentence - paragraph ---> c. Type of
4. Car - steering ---> d. Not a type of
Ans. 1-b, 2-c, 3-a, 4-d

119) Match the following
1) Scooter --------- Automobile A. A PART OF
2).Oxygen ---------- Water B. A Type of
3).Shop staff ---------- Fitters C. NOT A TYPE OF
4). Bug ---------- Reptile D. A SUPERSET OF
Ans. 1-b, 2-a, 3-d, 4-c

120) What is the largest prime number stored in a-
-----> 6 bit pattern (ANS~2^6=64, so no is 61)
------> 7 bit pattern (ANS~2^7=128, so no is 127)
-------> 8 bit pattern (ANS~2^8=256, so no is 251)
-------->9 bit pattern (ANS~2^9=512, so no is 503)

For more Question,Placement Papers Log on to http://topengineeringcollegesintamilnadu.blogspot.com/

TAGS:TCS Recruitment Rounds,Placement Papers,TCS Placement Papers,TCS Question papers,TCS Written Test,TCS Technical Interview,TCS HR Interview,TCS Online Test,TCS Interview Papers,TCS Interview Questions

TCS Recruitment Rounds |TCS Technical Interview Questions|TCS Placement Papers -V

TCS Recruitment Rounds PART-V
1.Written Test
2.Technical Interview
3.MR ( Managerial)
4. HR Interview
WRITTEN TEST :( ONLINE TEST)

Contains 3 sections

1) Verbal (Synonyms - Antonyms - Comprehension Passages)
2) Quantitative Aptitude
3) Critical Reasoning

SECTION-1(Verbal- 30 questions - 20 min)
1.Synonyms (Refer In GRE BARRONS 12th Edition )
2.Antonyms (Refer In GRE BARRONS 12th Edition (page no -126))
3. Passage completion

121) What is the max 3 digit Prime no?
ANS=997

122) G(0)= -1, G(1)=1, G(N)=G(N-1) - G(N-2), G(5)= ?
Ans - 2

123) G (0) =1 G (1) = -1 IF G (N) =2* (G (N-1)) - 3(G (N-2)) Then what is the value of
G (4)?

124) If f (0) =1 and f (n) = f (n-1)*n, find the value of f (4).
Ans: 24

125) If g (0) =g (1) =1 and g (n) = g (n-1) + g (n -2) find g (6);

126) What is the power of 2? a. 2068 b.2048 c.2668
Ans: 2048
127) 8 to the power of x is 32, what is the value of x?

128) Power of 4 Ans-4096

129) Which one will be the exact power of 3?
(i) 2768 (ii) 2678 (iii) 2187
130) Complete the series-
a )3,8,a,24,b,48,63 [ ANS~ a=15, b=35 ]
[HINT~DIFFERENCE IS 5, 7, 9, 11, 13, 15]
B )26,19,17,13,11, ,8,7 [ ANS=9]
[HINT~26,17,11,8 DECREASING LIKE 9,6,3 & 19,13,9,7 DECREASING
LIKE 6, 4, 2]
c)9,10,11,13,15, ,21,28 [ ANS=19 ]
[HINT~9, 11, 15, and 21 INCREASING LIKE 2, 4, 6 & 10,13,19,28 INCRESING
LIKE 3, 6, and 9]
D) 4, -5, 11, -14, 22, --- [ ANS= -27]

131) Number of faces, vertices and edges of a cube
ANS: 6,8,12
132) Find the value of-
a) @@+25-++@16, where @ denotes"square" & + denotes "square root". [ANS=621]
b) $%$6-%$%6, where $ means "tripling" & % means "change of sign". [ANS= -72]
c) % # % 6 + # %# 6, % means "doubling" & # mean "reciprocal".
132) Select odd one out
1) LINUX, WINDOWS 98, SOLARIS, SMTP (ANS: SMTP)
2)MVS
3).JAVA b) LISP c) Smaltalk d)Eiffle Ans: LISP ( All other languages are OOPS)
4)1.http 2.arp 3.snmp 4.sap Ans:sap
5)1.linux 2.windows NT 3.sql server 4.Unix Ans: Sql server
6)1.SAP 2.ARP 3.WAP 4.TCP IP
7)a. Oracle b. Linux c. Ingress d. DB2
a. SMTP b. WAP c. SAP d. ARP Ans:SAP
9)a. WAP b. HTTP c. BAAN d. ARP Ans:Baan
10)a. LINUX b. UNIX c. SOLARIS d. SQL SERVER Ans:SQL SERVER
11)a. SQL b. DB2 c. SYBASE d. HTTP Ans: HTTP
12)a. Oracle b. Linux c. Ingress d. DB2 ANS:LINUX

133) Find the singularity matrix from a given set of matrices? (Hint det (A) ====0)
134) Which of the following are orthogonal pairs?
a. 3i+2j b. i+j c. 2i-3j d. -7i+j
Ans: a, c
135) (a) 2+3i (b) 1+i (c) 3-2i (d) 1-7i .Find which of the above is orthogonal.
Ans: a, c

136) Sum of slopes of 2 perpendicular st. lines is given. Find the pair of lines from the given set of options which satisfy the above condition?

137) If Rs.1260 is divided between A, B and C in the ratio 2:3:4, what is C,s share?
Ans: Rs. 560
138)A sum of money is divided among A, B and C such that for each rupee A gets, B gets 65paise and C gets 35paise. If C,s share is Rs.560, the sum is .
a) 2400 b) 2800 c) 3200 d) 3800
139) Complete the series.
1) 3, 8, --, 24, --, 48, 63. Ans: 15, 35
2) Complete the series. 4, -5, 11, -14, 22, --- Ans - 27
3) SERIES: 2, 7, 24, 77, ------ (238) or (240)
4)77, 49, 36, 18,? Ans: 8 (7*7=49) (4*9=36) (3*6=1 (1*8=
5) series: 5 6 7 8 10 11 14?? Ans.15 or 18
6)15 14 12 11?? 9 8 Ans.10
7) what is the 12th term of the series 2, 5, 8 ... Ans. 35
8)58, 27, 12, x, 2, 1. Find x.
9)7, 9,13,_,27,37. Ans-19
10)2, 5, __, 19, 37, 75 Ans: 9
11) Complete the sequence 9, 10,11,13,15, __, 21, 28.
140) UNITS
1) (Momentum*Velocity)/ (Acceleration * distance) find units. Ans: mass
2) (energy * time * time)/ (mass * dist) = distance
3) (momentum * velocity)/ (force * time) = velocity
4) Find the physical quantity in units from the equation:
(Force*Distance)/ (Velocity*Velocity) Ans. Ns2/m
5) Find the physical quantity represented by
MOMENTUM *VELOCITY] / [LENGTH * ACCELERATION]?

141) Find the result of the following _expression if, M denotes modulus operation, R denotes round-off, T denotes truncation:
M (373, 5) +R (3.4) +T (7.7) +R (5.8)
ANS: 19
142) Which of the following highest Standard deviation
a) 7, - 7, 7,-7, 7,-7 b) 7, 7, 7,7,7,7 c) -7, - 7, -7,-7,-7,-7 d) -7, 7, -7, 7,-7, 7
Ans: d

143) 232 expressed in base-5 is Ans: 1412

144) A building with height D shadow up to G. A neighbor building with what height shadows C feet.
|----|----|----|----|----|----|----|
A B C D E F G H
Sol: B Ft. or CD/G

145) In a fraction, if 1 is added to both the numerator at the denominator, the fraction becomes 1/2. If numerator is subtracted from the denominator, the fraction becomes 3/4. Find the fraction.
Ans. 3/7
146) The sum of the digits of a two digit number is 8. When 18 is added to the number, the digits are reversed. Find the number?
Ans. 35

147) What number should be added to or subtracted from each term of the ratio 17 : 24 so that it becomes equal to 1 : 2.
Ans. 10 should be subtracted

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TAGS:TCS Recruitment Rounds,Placement Papers,TCS Placement Papers,TCS Question papers,TCS Written Test,TCS Technical Interview,TCS HR Interview,TCS Online Test,TCS Interview Papers,TCS Interview Questions

TCS Recruitment Rounds |TCS Technical Interview Questions|TCS Placement Papers -III

TCS Recruitment Rounds PART-III
1.Written Test
2.Technical Interview
3.MR ( Managerial)
4. HR Interview
WRITTEN TEST :( ONLINE TEST)

Contains 3 sections

1) Verbal (Synonyms - Antonyms - Comprehension Passages)
2) Quantitative Aptitude
3) Critical Reasoning

SECTION-1(Verbal- 30 questions - 20 min)
1.Synonyms (Refer In GRE BARRONS 12th Edition )
2.Antonyms (Refer In GRE BARRONS 12th Edition (page no -126))
3. Passage completion

For more Question,Placement Papers Log on to http://topengineeringcollegesintamilnadu.blogspot.com/

61) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent?
Ans: 30 milliseconds

62) A person who decided to go weekend trip should not exceed 8 hours driving in a day average speed of forward journey is 40 mph due to traffic in Sundays the return journey average speed is 30 mph. How far he can select a picnic spot.

Ans: Between 120 and 140 miles

63) Car is filled with four and half gallons of oil for full round trip. Fuel is taken 1/4 gallons more in going than coming. What is the fuel consumed in coming up.
Sol: Let fuel consumed in coming up is x.
Thus equation is: x+1.25x=4.5
Ans: 2gallons

64) 40% employees are male if 60% of supervisors are male so for 100% is 26.4%so the probability is ......
Ans: 0.264

65) Gavaskar average in first 50 innings was 50. After the 51st innings his average was 51 how many runs he made in the 51st innings
Sol: first 50 ings.- run= 50*50=2500
51st ings. - Avg 51. So total run =51*51=2601.
So run scored in that ings=2601-2500=101 runs.
Ans: 101 runs

66) Hansie made the following amounts in seven games of cricket in India: Rs.10, Rs.15, Rs.21, Rs.12, Rs.18, Rs.19 and Rs.17 (all figures in crores of course).Find his average earnings.
Ans: Rs.16 crore

67) Average of 5 numbers is -10 sum of 3 numbers is 16, what is the average of other two numbers?
Ans: -33

68) If A, B and C are the mechanisms used separately to reduce the wastage of fuel by 30%, 20% and 10%. What will be the fuel economy if they were used combined. Ans: 20%

69) In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin
Sol: the minimum number of weightings needed is just 5.as shown below

(1) 80->30-30
(2) 15-15
(3) 7-7
(4) 3-3
(5) 1-1
Ans: 5.

70) 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas, and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples?
Sol: 2x+3y+4z=15
3x+2y+z=10
Adding
5x+5y+5z=25
x+y+z=5 that is for 1 orange, 1 banana and 1 apple requires 5Rs.
So for 3 orange, 3 bananas and 3 apples require 15Rs.
i.e. 3x+3y+3z=15
Ans: 15

71) In 8*8 chess board what is the total number of squares refers odele discovered that there are 204 squares on the board .We found that you would add the different squares
= 1 + 4 + 9 + 16+ 25 + 36 + 49 + 64. =204

Also in 3*3 tic tac toe board what is the total no of squares
Ans: 14 i.e. 9+4(bigger ones) +1 (biggest one)

If you get 100*100 board just use the formula the formula for the sum of the first n perfect squares is

n x (n + 1) x (2n + 1)
______________________
6

72) One fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combine in how much time they will finish.
Sol: Faster one can do 1/2 of work in one hour slower one can do 1/3 of work in one hour both they do (1/2+1/3=5/6) the work in one hour. So work will b finished in 6/5=1.2 hour
i e 1 hour 12 min.
Ans: 1 hour 12 min.

73)If Rs20/- is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive?
Sol: Find 42 % of 20 rs with respect to 70 (i.e. 28 + 42)
==> (42 * 20)/70 ==> 12 Rs
Ans: 12 Rs

74) In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held?
Ans: 138 matches
(Explanation: since one player gets a bye in each round, he will reach the finals of the tournament without playing a match.
Therefore 137 matches should be played to determine the second finalist from the remaining 138 players (excluding the 1st player)
Therefore to determine the winner 138 matches should be played.)

75) ONE RECTANGULAR PLATE WITH LENGTH 8INCHES, BREADTH 11 INCHES AND 2 INCHES THICKNESS IS THERE.WHAT IS THE LENGTH OF THE CIRCULAR ROD WITH DIAMETER 8 INCHES AND EQUAL TO VOLUME OF RECTANGULAR PLATE?
Sol: Vol. of rect. plate= 8*11*2=176
Area of rod= (22/7)*(8/2)*(8/2) = (352/7)
Vol. of rod=area*length=vol. of plate
So length of rod= vol of plate/area=176/ (352/7) =3.5
Ans: 3.5
76) One tank will fill in 6 minutes at the rate of 3cu ft /min, length of tank is 4 ft and the width is 1/2 of length, what is the depth of the tank?
Ans: 3 ft 7.5 inches

77) A power unit is there by the bank of the river of 750 meters width. A cable is made from power unit to power a plant opposite to that of the river and 1500mts away from the power unit. The cost of the cable below water is Rs. 15/- per meter and cost of cable on the bank is Rs.12/- per meter. Find the total of laying the cable.
Ans: 1000 (24725 - cost) 20250
Ans: Rs. 22,500 (hint: the plant is on the other side of the plant i.e. it is not on the same side as the river)

78) The cost of one pencil, two pens and four erasers is Rs.22 while the cost of five pencils, four pens and two erasers is Rs.32.How much will three pencils, three pens and three erasers cost?
Sol :( let x b pencil, y b pen and z b eraser... u get x+2y+4z=22 and 5x+4y+2z=32 add 6x+6y+6z=54 div by 2 you get 27)
Ans: 27

79) A man has to get air-mail. He starts to go to airport on his motorbike. Plane comes early and the mail is sent by a horse-cart. The man meets the cart in the middle after half an hour. He takes the mail and returns back, by doing so, he saves twenty minutes. How early did the plane arrive?
Sol: Assume he started at 1:00, so at 1:30 he met cart .He returned home at 2:00.so it took him 1 hour for the total journey. By doing this he saved 20 min. So the actual time if the plane is not late is 1 hour and 20 min. So the actual time of plane is at 1:40.The cart traveled a time of 10 min before it met him. So the plane is 10 min early.
Ans: 10 min
80) Ram singh goes to his office in the city every day from his suburban house. His driver Mangaram drops him at the railway station in the morning and picks him up in the evening. Every evening Ram singh reaches the station at 5 o,clock. Mangaram also reaches at the same time. One day Ram singh started early from his office and came to the station at 4 o,clock. Not wanting to wait for the car he starts walking home. Mangaram starts at normal time, picks him up on the way and takes him back house, half an hour early. How much time did Ram singh walked?

81) 2 trees are there. One grows at 3/5 of the other. In 4 years total growth of the trees is 8 ft. what growth will smaller tree have in 2 years.
Sol: THE BIG TREE GROWS 8FT IN 4 YEARS=>THE BIG TREE GROWS 4FT IN 2 YEARS.WHEN WE DIVIDE 4FT/5=.8*3=>2.4
4 (x+ (3/5) x) =88x/5=2x=5/4
After 2 years x= (3/5)*(5/4)*2 =1.5 (less than 2 feet)

82) There is a six digit code. Its first two digits, multiplied by 3 gives all ones. And the next two digits multiplied by 6 give all twos. Remaining two digits multiplied by 9 gives all threes. Then what is the code?
Sol: Assume the digit xx xx xx (six digits)
First Two digit xx * 3=111
xx =111/3=37
(First two digits of 1 is not divisible by 3 so we can use 111)
Second Two digit xx*6=222
xx=222/6=37
(First two digits of 2 is not divisible by 6 so we can use 222)
Third Two digit xx*9=333
xx=333/9=37
(First two digits of 3 is not divisible by 9 so we can use 333)

83) There are 4 balls and 4 boxes of colors yellow, pink, red and green. Red ball is in a box whose color is same as that of the ball in a yellow box. Red box has green ball. In which box you find the yellow ball?
Sol: Yellow box can have either of pink/yellow balls.
if we put a yellow ball in "yellow" box then it would imply that "yellow" is also the color of the box which has the red ball(because according 2 d question,d box of the red ball n the ball in the yellow box have same color)
Thus this possibility is ruled out...
Therefore the ball in yellow box must be pink, hence the color of box contain in red ball is also pink....
=>the box color left out is "green", which is allotted to the only box left, the one which has yellow ball.
Ans: green

84) A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets at least 2 balls of same color.
Sol: suppose he picks 5 balls of all different colors then when he picks up the sixth one, it must match any on of the previously drawn ball color. Thus he must pick 6 balls
Ans: he should pick 6 balls totally.

85) WHAT IS THE NUMBER OF ZEROS AT THE END OF THE PRODUCT OF THE NUMBERS FROM 1 TO 100?
Sol: For every 5 in unit place one zero is added
so between 1 to 100 there are 10 nos like 5,15,25,..,95 which has 5 in unit place.
Similarly for every no divisible by 10 one zero is added in the answer so between 1 to 100 , 11 zeros are added
For 25, 50, 75 3 extra zeros are added
So total no of zeros are 10+11+3=24

86) There are two numbers in the ratio 8:9. If the smaller of the two numbers is increased by 12 and the larger number is reduced by 19 thee the ratio of the two numbers is 5:9.
Find the larger number?
Sol: 8x: 9x initially
8x+ 12: 9x - 19 = 5x: 9x
8x+12 = 5x
-> x = 4
9x = 36 (NOT SURE ABOUT THE ANSWER)

87) There are three different boxes A, B and C. Difference between weights of A and B is 3 kgs. And between B and C is 5 kgs. Then what is the maximum sum of the differences of all possible combinations when two boxes are taken each time
Sol: A-B = 3
B-c = 5
A-c = 8 so sum of diff = 8+3+5 = 16 kgs

88) A and B are shooters and having their exam. A and B fall short of 10 and 2 shots respectively to the qualifying mark. If each of them fired at least one shot and even by adding their total score together, they fall short of the qualifying mark, what is the qualifying mark?
Sol: Because each had at least 1 shot done so 10 + 1 = 11
And 9 + 2 = 11
So the ans is 11

89) A, B, C, and D tells the following times by looking at their watches. A tells it is 3 to 12. B tells it is 3 past 12. C tells it is 12:2. D tells it is half a dozen too soon to 12. No two watches show the same time. The difference between the watches is 2,3,4,5 respectively. Whose watch shows maximum time?
Sol: A shows 11:57, B shows 12:03, C shows 12:02 and D shows 11:06 therefore,
Max time is for B

90) Falling height is proportional to square of the time. One object falls 64cm in 2sec than in 6sec from how much height the object will fall.
Sol: The falling height is proportional to the square of the time.
Now, the falling height is 64cm at 2sec
So, the proportional constant is=64/ (2*2) =16;
So, at 6sec the object fall maximum (16*6*6) cm=576cm;
Now, the object may be situated at any where.
If it is>576 only that time the object falling 576cm within 6sec .Otherwise if it is situated

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TAGS:TCS Recruitment Rounds,Placement Papers,TCS Placement Papers,TCS Question papers,TCS Written Test,TCS Technical Interview,TCS HR Interview,TCS Online Test,TCS Interview Papers,TCS Interview Questions

TCS Recruitment Rounds |TCS Technical Interview Questions|TCS Placement Papers -III

TCS Recruitment Rounds PART-III
1.Written Test
2.Technical Interview
3.MR ( Managerial)
4. HR Interview
WRITTEN TEST :( ONLINE TEST)

Contains 3 sections

1) Verbal (Synonyms - Antonyms - Comprehension Passages)
2) Quantitative Aptitude
3) Critical Reasoning

SECTION-1(Verbal- 30 questions - 20 min)
1.Synonyms (Refer In GRE BARRONS 12th Edition )
2.Antonyms (Refer In GRE BARRONS 12th Edition (page no -126))
3. Passage completion

For more Question,Placement Papers Log on to http://topengineeringcollegesintamilnadu.blogspot.com/

61) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent?
Ans: 30 milliseconds

62) A person who decided to go weekend trip should not exceed 8 hours driving in a day average speed of forward journey is 40 mph due to traffic in Sundays the return journey average speed is 30 mph. How far he can select a picnic spot.

Ans: Between 120 and 140 miles

63) Car is filled with four and half gallons of oil for full round trip. Fuel is taken 1/4 gallons more in going than coming. What is the fuel consumed in coming up.
Sol: Let fuel consumed in coming up is x.
Thus equation is: x+1.25x=4.5
Ans: 2gallons

64) 40% employees are male if 60% of supervisors are male so for 100% is 26.4%so the probability is ......
Ans: 0.264

65) Gavaskar average in first 50 innings was 50. After the 51st innings his average was 51 how many runs he made in the 51st innings
Sol: first 50 ings.- run= 50*50=2500
51st ings. - Avg 51. So total run =51*51=2601.
So run scored in that ings=2601-2500=101 runs.
Ans: 101 runs

66) Hansie made the following amounts in seven games of cricket in India: Rs.10, Rs.15, Rs.21, Rs.12, Rs.18, Rs.19 and Rs.17 (all figures in crores of course).Find his average earnings.
Ans: Rs.16 crore

67) Average of 5 numbers is -10 sum of 3 numbers is 16, what is the average of other two numbers?
Ans: -33

68) If A, B and C are the mechanisms used separately to reduce the wastage of fuel by 30%, 20% and 10%. What will be the fuel economy if they were used combined. Ans: 20%

69) In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin
Sol: the minimum number of weightings needed is just 5.as shown below

(1) 80->30-30
(2) 15-15
(3) 7-7
(4) 3-3
(5) 1-1
Ans: 5.

70) 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas, and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples?
Sol: 2x+3y+4z=15
3x+2y+z=10
Adding
5x+5y+5z=25
x+y+z=5 that is for 1 orange, 1 banana and 1 apple requires 5Rs.
So for 3 orange, 3 bananas and 3 apples require 15Rs.
i.e. 3x+3y+3z=15
Ans: 15

71) In 8*8 chess board what is the total number of squares refers odele discovered that there are 204 squares on the board .We found that you would add the different squares
= 1 + 4 + 9 + 16+ 25 + 36 + 49 + 64. =204

Also in 3*3 tic tac toe board what is the total no of squares
Ans: 14 i.e. 9+4(bigger ones) +1 (biggest one)

If you get 100*100 board just use the formula the formula for the sum of the first n perfect squares is

n x (n + 1) x (2n + 1)
______________________
6

72) One fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combine in how much time they will finish.
Sol: Faster one can do 1/2 of work in one hour slower one can do 1/3 of work in one hour both they do (1/2+1/3=5/6) the work in one hour. So work will b finished in 6/5=1.2 hour
i e 1 hour 12 min.
Ans: 1 hour 12 min.

73)If Rs20/- is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive?
Sol: Find 42 % of 20 rs with respect to 70 (i.e. 28 + 42)
==> (42 * 20)/70 ==> 12 Rs
Ans: 12 Rs

74) In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held?
Ans: 138 matches
(Explanation: since one player gets a bye in each round, he will reach the finals of the tournament without playing a match.
Therefore 137 matches should be played to determine the second finalist from the remaining 138 players (excluding the 1st player)
Therefore to determine the winner 138 matches should be played.)

75) ONE RECTANGULAR PLATE WITH LENGTH 8INCHES, BREADTH 11 INCHES AND 2 INCHES THICKNESS IS THERE.WHAT IS THE LENGTH OF THE CIRCULAR ROD WITH DIAMETER 8 INCHES AND EQUAL TO VOLUME OF RECTANGULAR PLATE?
Sol: Vol. of rect. plate= 8*11*2=176
Area of rod= (22/7)*(8/2)*(8/2) = (352/7)
Vol. of rod=area*length=vol. of plate
So length of rod= vol of plate/area=176/ (352/7) =3.5
Ans: 3.5
76) One tank will fill in 6 minutes at the rate of 3cu ft /min, length of tank is 4 ft and the width is 1/2 of length, what is the depth of the tank?
Ans: 3 ft 7.5 inches

77) A power unit is there by the bank of the river of 750 meters width. A cable is made from power unit to power a plant opposite to that of the river and 1500mts away from the power unit. The cost of the cable below water is Rs. 15/- per meter and cost of cable on the bank is Rs.12/- per meter. Find the total of laying the cable.
Ans: 1000 (24725 - cost) 20250
Ans: Rs. 22,500 (hint: the plant is on the other side of the plant i.e. it is not on the same side as the river)

78) The cost of one pencil, two pens and four erasers is Rs.22 while the cost of five pencils, four pens and two erasers is Rs.32.How much will three pencils, three pens and three erasers cost?
Sol :( let x b pencil, y b pen and z b eraser... u get x+2y+4z=22 and 5x+4y+2z=32 add 6x+6y+6z=54 div by 2 you get 27)
Ans: 27

79) A man has to get air-mail. He starts to go to airport on his motorbike. Plane comes early and the mail is sent by a horse-cart. The man meets the cart in the middle after half an hour. He takes the mail and returns back, by doing so, he saves twenty minutes. How early did the plane arrive?
Sol: Assume he started at 1:00, so at 1:30 he met cart .He returned home at 2:00.so it took him 1 hour for the total journey. By doing this he saved 20 min. So the actual time if the plane is not late is 1 hour and 20 min. So the actual time of plane is at 1:40.The cart traveled a time of 10 min before it met him. So the plane is 10 min early.
Ans: 10 min
80) Ram singh goes to his office in the city every day from his suburban house. His driver Mangaram drops him at the railway station in the morning and picks him up in the evening. Every evening Ram singh reaches the station at 5 o,clock. Mangaram also reaches at the same time. One day Ram singh started early from his office and came to the station at 4 o,clock. Not wanting to wait for the car he starts walking home. Mangaram starts at normal time, picks him up on the way and takes him back house, half an hour early. How much time did Ram singh walked?

81) 2 trees are there. One grows at 3/5 of the other. In 4 years total growth of the trees is 8 ft. what growth will smaller tree have in 2 years.
Sol: THE BIG TREE GROWS 8FT IN 4 YEARS=>THE BIG TREE GROWS 4FT IN 2 YEARS.WHEN WE DIVIDE 4FT/5=.8*3=>2.4
4 (x+ (3/5) x) =88x/5=2x=5/4
After 2 years x= (3/5)*(5/4)*2 =1.5 (less than 2 feet)

82) There is a six digit code. Its first two digits, multiplied by 3 gives all ones. And the next two digits multiplied by 6 give all twos. Remaining two digits multiplied by 9 gives all threes. Then what is the code?
Sol: Assume the digit xx xx xx (six digits)
First Two digit xx * 3=111
xx =111/3=37
(First two digits of 1 is not divisible by 3 so we can use 111)
Second Two digit xx*6=222
xx=222/6=37
(First two digits of 2 is not divisible by 6 so we can use 222)
Third Two digit xx*9=333
xx=333/9=37
(First two digits of 3 is not divisible by 9 so we can use 333)

83) There are 4 balls and 4 boxes of colors yellow, pink, red and green. Red ball is in a box whose color is same as that of the ball in a yellow box. Red box has green ball. In which box you find the yellow ball?
Sol: Yellow box can have either of pink/yellow balls.
if we put a yellow ball in "yellow" box then it would imply that "yellow" is also the color of the box which has the red ball(because according 2 d question,d box of the red ball n the ball in the yellow box have same color)
Thus this possibility is ruled out...
Therefore the ball in yellow box must be pink, hence the color of box contain in red ball is also pink....
=>the box color left out is "green", which is allotted to the only box left, the one which has yellow ball.
Ans: green

84) A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets at least 2 balls of same color.
Sol: suppose he picks 5 balls of all different colors then when he picks up the sixth one, it must match any on of the previously drawn ball color. Thus he must pick 6 balls
Ans: he should pick 6 balls totally.

85) WHAT IS THE NUMBER OF ZEROS AT THE END OF THE PRODUCT OF THE NUMBERS FROM 1 TO 100?
Sol: For every 5 in unit place one zero is added
so between 1 to 100 there are 10 nos like 5,15,25,..,95 which has 5 in unit place.
Similarly for every no divisible by 10 one zero is added in the answer so between 1 to 100 , 11 zeros are added
For 25, 50, 75 3 extra zeros are added
So total no of zeros are 10+11+3=24

86) There are two numbers in the ratio 8:9. If the smaller of the two numbers is increased by 12 and the larger number is reduced by 19 thee the ratio of the two numbers is 5:9.
Find the larger number?
Sol: 8x: 9x initially
8x+ 12: 9x - 19 = 5x: 9x
8x+12 = 5x
-> x = 4
9x = 36 (NOT SURE ABOUT THE ANSWER)

87) There are three different boxes A, B and C. Difference between weights of A and B is 3 kgs. And between B and C is 5 kgs. Then what is the maximum sum of the differences of all possible combinations when two boxes are taken each time
Sol: A-B = 3
B-c = 5
A-c = 8 so sum of diff = 8+3+5 = 16 kgs

88) A and B are shooters and having their exam. A and B fall short of 10 and 2 shots respectively to the qualifying mark. If each of them fired at least one shot and even by adding their total score together, they fall short of the qualifying mark, what is the qualifying mark?
Sol: Because each had at least 1 shot done so 10 + 1 = 11
And 9 + 2 = 11
So the ans is 11

89) A, B, C, and D tells the following times by looking at their watches. A tells it is 3 to 12. B tells it is 3 past 12. C tells it is 12:2. D tells it is half a dozen too soon to 12. No two watches show the same time. The difference between the watches is 2,3,4,5 respectively. Whose watch shows maximum time?
Sol: A shows 11:57, B shows 12:03, C shows 12:02 and D shows 11:06 therefore,
Max time is for B

90) Falling height is proportional to square of the time. One object falls 64cm in 2sec than in 6sec from how much height the object will fall.
Sol: The falling height is proportional to the square of the time.
Now, the falling height is 64cm at 2sec
So, the proportional constant is=64/ (2*2) =16;
So, at 6sec the object fall maximum (16*6*6) cm=576cm;
Now, the object may be situated at any where.
If it is>576 only that time the object falling 576cm within 6sec .Otherwise if it is situated

For more Question,Placement Papers Log on to http://topengineeringcollegesintamilnadu.blogspot.com/

TAGS:TCS Recruitment Rounds,Placement Papers,TCS Placement Papers,TCS Question papers,TCS Written Test,TCS Technical Interview,TCS HR Interview,TCS Online Test,TCS Interview Papers,TCS Interview Questions

TCS Recruitment Rounds | Placement Papers -II |TCS Technical Interview Questions

TCS Recruitment Rounds PART-2
1.Written Test
2.Technical Interview
3.MR ( Managerial)
4. HR Interview
WRITTEN TEST :( ONLINE TEST)

Contains 3 sections

1) Verbal (Synonyms - Antonyms - Comprehension Passages)
2) Quantitative Aptitude
3) Critical Reasoning

SECTION-1(Verbal- 30 questions - 20 min)
1.Synonyms (Refer In GRE BARRONS 12th Edition )
2.Antonyms (Refer In GRE BARRONS 12th Edition (page no -126))
3. Passage completion

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31) A =0 0 0 0 1 1 1 1
B =0 0 1 1 0 0 1 1
C =0 1 0 1 0 1 0 1
( A U B ) n C Find the fourth row, having the bit pattern as an integer in an 8-bit computer, and express the answer in its decimal value.
Ans: 29

32) A, B and C are 8 bit nos. They are as follows:
A 1 1 0 1 1 0 1 1
B 0 1 1 1 1 0 1 0
C 0 1 1 0 1 1 0 1
Find ( (A-B) u C )=?
Hint: 109 A-B is {A} - {A n B}
Ans: 0 1 1 1 1 1 1 1 (DB)

33) If A, B and C are the mechanisms used separately to reduce the wastage of fuel by 30%, 20% and 10%. What will be the fuel economy if they were used combined.
Ans: 20%

34) In the class of 40 students, 30 speak Hindi and 20 speak English. What is the lowest possible number of students who speak both the languages?
(a) 5 (b) 20 (c) 15 (d) 10 (e) 30

35) In a two-dimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of
X (8, 5).
Sol: [HINT~ Formula=Base Add + Byte reqd {N (i-1) + (j-1)}
Where,
Base Add=3000;
Byte reqd=4;
N=no of columns in array=7;
i=8; j=5;
IN ROW MAJOR ORDER]
Ans: 3212
36) If the vertex (5, 7) is placed in the memory. First vertex (1, 1),s address is 1245 and then address of (5, 7) is ----------
Ans: 1279

37) A 2D array is declared as A [9, 7] and each element requires 2 byte. If A [1, 1] is stored in 3000. Find the memory of A [8, 5]?
Ans: 3106

38) One circular array is given (means the memory allocation takes place like a circular fashion) dimension (9X7) .starting address is 3000.find the address of (2, 3)
Ans: 555

39) The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied?
Sol: N is increased by 1%
Therefore new value of N=N + (N/100)
=101N/100
M=sqrt (100 * (101N/100))
Hence, we get
M=sqrt (101 * N)
Ans: 0. 5 %( =SQRT 101N)

40) A bus started from bus stand at 8.00a m and after 30 min staying at destination, it returned back to the bus stand. The destination is 27 miles from the bus stand. The speed of the bus 50 percent fast speed. At what time it retur4ns to the bus stand.
Sol: (this is the step by step solution :)
A bus cover 27 mile with 18 mph in =27/18= 1 hour 30 min.
And it wait at stand =30 min.
After this speed of return increase by 50% so 50%of 18 mph=9mph
Total speed of returning=18+9=27
Then in return it take 27/27=1 hour
Then total time in journey=1+1:30+00:30 =3 hour
So it will come at 8+3 hour=11 a.m.
So Ans==11 a.m
41) A Flight takes off at 2 A.M from northeast direction and travels for 11 hours to reach the destination which is in North West direction. Given the latitude and longitude of source and destination. Find the local time of destination when the flight reaches there?
Ans: 7 AM or 1.00 PM

42) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
a) 6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00
(Hint: Every 1 deg longitude is equal to 4 minutes. If west to east add time else subtract time)
Ans: 8:00

43) A moves 3 kms east from his starting point. He then travels 5 kms north. From that point he moves 8 kms to the east. How far is A from his starting point?
Ans: 13 kms

44) A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time.

45) In Madras, temperature at noon varies according to -t^2/2 + 8t + 3, where t is elapsed time. Find how much temperature more or less in 4pm to 9pm. (May be we can solve it by Definite Integration. Check any way}
Ans: at 9 pm 7.5 more or 385.8 (DB)

46) For Temperature a function is given according to time: ((t**2)/6) + 4t +12 what is the temperature rise or fall between 4.AM TO 9 AM
Sol: In equation first put t=9,
We will get 34.5........................... (1)
Now put t=4,
We will get 27.............................. (2)
So Ans=34.5-27
=7.5

47) For Temperature a function is given according to time: ((t**2)/6) + 4t +12 what is the temperature rise or fall between 5 PM to 8 PM

48) Low temperature at the night in a city is 1/3 more than 1/2 high as higher temperature in a day. Sum of the low tem. And highest temp is 100 degrees. Then what is the low temp?
Sol: Let highest temp be x
So low temp=1/3 of x of 1/2 of x plus x/2 i.e. x/6+x/2
Total temp=x+x/6+x/2=100
Therefore, x=60
Lowest temp is 40
Ans :( 40 deg.)

49) A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53and the product went up by 540. What was the raised product?
a) 780
b) 1040
c) 1590
d) 1720
Sol: x*53-x*35=540=> x=30 therefore, 53*30=1590
Ans: 1590

50) How many positive integer solutions does the equation 2x+3y = 100 have?
a) 50
b) 33
c) 16
d) 35
Sol: Given 2x+3y=100, take l.c.m of ,x, coeff and ,y, coeff i.e. l.c.m of 2,3 ==6then divide 100 with 6 , which turns out 16 hence answer is 16short cut formula--- constant / (l.cm of x coeff and y coeff)

51) The total expense of a boarding house is partly fixed and partly variable with the number of boarders. The charge is Rs.70 per head when there are 25 boarders and Rs.60 when there are 50 boarders. Find the charge per head when there are 100 boarders.
a) 65
b) 55
c) 50
d) 45
Sol: let a = fixed cost and
k = variable cost and n = number of boarders
Total cost when 25 boarders c = 25*70 = 1750 i.e. 1750 = a + 25k
Total cost when 50 boarders c = 50*60 = 3000 i.e. 3000 = a + 50k
Solving above 2 eqns, 3000-1750 = 25k i.e. 1250 = 25k i.e. k = 50
Therefore, substituting this value of k in either of above 2 eqns we get
a = 500 (a = 3000-50*50 = 500 or a = 1750 - 25*50 = 500)
So total cost when 100 boarders = c = a + 100k = 500 + 100*50 = 5500
So cost per head = 5500/100 = 55

52) Amal bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more than what Amal had paid. What % of the total amount paid by Amal was paid for pens?
a) 37.5%
b) 62.5%
c) 50%
d) None of these
Sol: Let, 5 pens + 7 pencils + 4 erasers = x rupees
So 10 pens + 14 pencils + 8 erasers = 2*x rupees
Also mentioned, 6 pens + 14 pencils + 8 erasers = 1.5*x rupees
So (10-6) = 4 pens = (2-1.5) x rupees
So 4 pens = 0.5x rupees => 8 pens = x rupees
So 5 pens = 5x/8 rupees = 5/8 of total (note x rupees is total amt paid by
Amal) i.e. 5/8 = 500/8% = 62.5%
Ans: 62.5%

53) I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend lost Rs.4 more than me in the second race. What is the amount lost by my friend in the second race?
Sol: x + x+6 = rs 68
2x + 6 = 68
2x = 68-6
2x = 62
x=31
x is the amt lost in I race
x+ 6 = 31+6=37 is lost in second race
Then my friend lost 37 + 4 = 41 Rs
Ans: 41 Rs

54) A face of the clock is divided into three parts. First part hours total is equal to the sum of the second and third part. What is the total of hours in the bigger part?
Sol: The clock normally has 12 hr
Three parts x, y, z
x+y+z=12
x=y+z
2x=12
x=6
So the largest part is 6 hrs
Ans: 6 hrs

55) (1- 1/6) (1-1/7).... (1- (1/ (n+4))) (1-(1/ (n+5))) = ?
Sol: Leaving the first numerator and last denominator, all the numerator and denominator will cancelled out one another.
Ans: 5/ (n+5)

56) Ten boxes are there. Each ball weighs 100 gms. One ball is weighing 90 gms.
i) If there are 3 balls (n=3) in each box, how many times will it take to find 90 gms ball? ii) Same question with n=10
iii) Same question with n=9
Sol: The chances are
When n=3
(i) nC1= 3C1 =3 for 10 boxes.. 10*3=30
(ii) nC1=10C1=10 for 10 boxes ....10*10=100
(iii) nC1=9C1=9 for 10 boxes.....10*9=90

57) With 4/5 full tank vehicle travels 12 miles, with 1/3 full tank how much distance travels?
Sol: 4/5 full tank= 12 mile
1 full tank= 12/ (4/5)
1/3 full tank= 12/ (4/5)*(1/3) = 5 miles
Ans: 5 miles

58) Wind flows 160 miles in 330min.for 80 miles how much time required
160 miles?
Sol: 1 mile = 330/160
80 miles= (330*80)/160=165 min.
Ans: 165 min.

59) A person was fined for exceeding the speed limit by 10mph.another person was also fined for exceeding the same speed limit by twice the same if the second person was traveling at a speed of 35 mph. find the speed limit
Sol :( x+10) =(x+35)/2
Solving the eqn we get x=15
Ans: 15

60) A sales person multiplied a number and get the answer is 3 instead of that number divided by 3.what is the answer he actually has to get.
Sol: Assume 1
1* 3 = 3
1*1/3=1/3
So he has to got 1/3
Ans: 1/3

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