CSE AMARONS

HR INTERVIEW SOLUTIONS

2009-03-04T10:31:00.001-08:00

Below is a list of frequently asked questions in various interviews...

Click on the questions to view the requisite solutions...

I would love to receive your comments and suggestions regarding this endeavor of mine... Mail me at sampat.honey@gmail.com / patnaik.sampat@gmail.com

01: Tell me about yourself ?

02: What are your greatest strengths ?

03: What are your greatest weaknesses?

04: Why should I hire you?

05. What are your Short Term Goals?

06: What are your Long Term Goals?

07: How do you evaluate success ?

08: What is your Salary Expectation ?

09: How many hours a week do you normally work?

10: Are you willing to travel?

11: How do you feel about working nights and weekends?

12: The “Silent Treatment”

13: What good books have you read lately?

14: Where do you see yourself five years from now?

15: Describe your ideal company, location and job.

16: Who has inspired you in your life and why?

17: Why do you want to work for us?

18: What do you know about this company?

19: Can you work under pressure?

20: What makes you angry?

21: Aren’t you overqualified for this position?

22: Would you lie for the company?

23: Tell me something negative you’ve heard about our company !!!

24 : Demonstrate Your Skills : Sell this stapler,etc or some object on the desk

25: What are your goals?

26: What was the toughest challenge you’ve ever faced?

27: The “Hypothetical Problem” – What would you do if… ?

28: What do you worry about?

29: Give me an example of your creativity (analytical skill…managing ability, etc.)

30: How could you have improved your career progress?

31: What was the toughest decision you ever had to make?

32: Looking back, what would you do differently in your life?

33: How do you feel about reporting to a younger person (minority, woman, etc)?

34: The “Fatal Flaw” question

35: What are your outside interests?

36: What are your career options right now?

37: Do you prefer to work alone or as a team player?

38: What made you choose your major?

39: Why didn’t you pursue a career in your major?

40: What do you seek from a job?

41: Tell me about something you did – or failed to do – that you now feel a little ashamed of.

42: Tell me about a situation when your work was criticized.

43: What changes would you make if you came on board?

44: I’m concerned that you don’t have as much experience as we’d like in…

45: What would you say to your boss if he’s crazy about an idea, but you think it stinks?

46: Why aren’t you earning more money at this stage of your career?

47: Tell me about the most boring job you’ve ever had.

48: What do you see as the proper role/mission of a good (job title you’re seeking) in serving the community

49: What would you do if a colleague wasn’t pulling his/her weight…and this was hurting your department?

50: POSSIBLE FOLLOW-UP QUESTION:

51: What’s the most difficult part of being a (job title)?

52: Why are you leaving (or did you leave) this position?

53: Tell me honestly about the strong points and weak points of your boss (company, management team, etc.)…

54: CONFIDENTIAL ISSUES

55: Could you have done better in your last job?

56: You’ve been with your firm a long time. Won’t it be hard switching to a new company?

57: Why have you had so many jobs?

58: May I contact your present employer for a reference?

59: Have you considered starting your own business?

60: THE ILLEGAL QUESTION

61: THE SECRET ILLEGAL QUESTION

62: What was the toughest part of your last job?

63: Looking back on your last position, have you done your best work?

64: Why should I hire you from the outside when I could promote someone from within?

65: Do you have the stomach to fire people? Have you had experience firing many people?

66: What do you for when you hire people?

67: Have you been absent from work more than a few days in any previous position?

68: Why have you been out of work so long?

69: If you won A $10 million lottery, would you still work?

70:Is there anything else that we should know about you that would impact our decision?

71: During your performance reviews, what criticism do you hear the most?

72: If I tell you, you are an IDIOT, a FOOL, a RASCAL; can you think anything positive about it?

73: Suppose I tell you I don’t like your face? Then why should I select you?

74: What will you do if you are asked to give a bribe?

75: If your boss is taking all the credits for your work, what will you do?

76: Suppose you find yourself in a deserted island. What three things you will need to survive?

77: How do you think you conducted or performed during this interview?

78: On a scale of one to ten, rate me as an interviewer.

OS APTITUDE

2009-03-04T09:59:00.001-08:00

Following are a few basic questions that cover the essentials of OS:

1. Explain the concept of Reentrancy.

It is a useful, memory-saving technique for multiprogrammed timesharing systems. A Reentrant Procedure is one in which multiple users can share a single copy of a program during the same period. Reentrancy has 2 key aspects: The program code cannot modify itself, and the local data for each user process must be stored separately. Thus, the permanent part is the code, and the temporary part is the pointer back to the calling program and local variables used by that program. Each execution instance is called activation. It executes the code in the permanent part, but has its own copy of local variables/parameters. The temporary part associated with each activation is the activation record. Generally, the activation record is kept on the stack.

Note: A reentrant procedure can be interrupted and called by an interrupting program, and still execute correctly on returning to the procedure.

2. Explain Belady's Anomaly.

Also called FIFO anomaly. Usually, on increasing the number of frames allocated to a process' virtual memory, the process execution is faster, because fewer page faults occur. Sometimes, the reverse happens, i.e., the execution time increases even when more frames are allocated to the process. This is Belady's Anomaly. This is true for certain page reference patterns.

3. What is a binary semaphore? What is its use?

A binary semaphore is one, which takes only 0 and 1 as values. They are used to implement mutual exclusion and synchronize concurrent processes.

4. What is thrashing?

It is a phenomenon in virtual memory schemes when the processor spends most of its time swapping pages, rather than executing instructions. This is due to an inordinate number of page faults.

5. List the Coffman's conditions that lead to a deadlock.

Mutual Exclusion: Only one process may use a critical resource at a time.
Hold & Wait: A process may be allocated some resources while waiting for others.
No Pre-emption: No resource can be forcible removed from a process holding it.
Circular Wait: A closed chain of processes exist such that each process holds at least one resource needed by another process in the chain.

6. What are short-, long- and medium-term scheduling?

Long term scheduler determines which programs are admitted to the system for processing. It controls the degree of multiprogramming. Once admitted, a job becomes a process.

Medium term scheduling is part of the swapping function. This relates to processes that are in a blocked or suspended state. They are swapped out of real-memory until they are ready to execute. The swapping-in decision is based on memory-management criteria.

Short term scheduler, also know as a dispatcher executes most frequently, and makes the finest-grained decision of which process should execute next. This scheduler is invoked whenever an event occurs. It may lead to interruption of one process by preemption.

7. What are turnaround time and response time?

Turnaround time is the interval between the submission of a job and its completion. Response time is the interval between submission of a request, and the first response to that request.

8. What are the typical elements of a process image?

User data: Modifiable part of user space. May include program data, user stack area, and programs that may be modified.
User program: The instructions to be executed.
System Stack: Each process has one or more LIFO stacks associated with it. Used to store parameters and calling addresses for procedure and system calls.
Process control Block (PCB): Info needed by the OS to control processes.

9. What is the Translation Lookaside Buffer (TLB)?

In a cached system, the base addresses of the last few referenced pages is maintained in registers called the TLB that aids in faster lookup. TLB contains those page-table entries that have been most recently used. Normally, each virtual memory reference causes 2 physical memory accesses-- one to fetch appropriate page-table entry, and one to fetch the desired data. Using TLB in-between, this is reduced to just one physical memory access in cases of TLB-hit.

10. What is the resident set and working set of a process?

Resident set is that portion of the process image that is actually in real-memory at a particular instant. Working set is that subset of resident set that is actually needed for execution. (Relate this to the variable-window size method for swapping techniques.)

11. When is a system in safe state?

The set of dispatchable processes is in a safe state if there exists at least one temporal order in which all processes can be run to completion without resulting in a deadlock.

12. What is cycle stealing?

We encounter cycle stealing in the context of Direct Memory Access (DMA). Either the DMA controller can use the data bus when the CPU does not need it, or it may force the CPU to temporarily suspend operation. The latter technique is called cycle stealing. Note that cycle stealing can be done only at specific break points in an instruction cycle.

13. What is meant by arm-stickiness?

If one or a few processes have a high access rate to data on one track of a storage disk, then they may monopolize the device by repeated requests to that track. This generally happens with most common device scheduling algorithms (LIFO, SSTF, C-SCAN, etc). High-density multisurface disks are more likely to be affected by this than low density ones.

14. What are the stipulations of C2 level security?

C2 level security provides for:

Discretionary Access Control
Identification and Authentication
Auditing
Resource reuse

15. What is busy waiting?

The repeated execution of a loop of code while waiting for an event to occur is called busy-waiting. The CPU is not engaged in any real productive activity during this period, and the process does not progress toward completion.

16. Explain the popular multiprocessor thread-scheduling strategies.

Load Sharing: Processes are not assigned to a particular processor. A global queue of threads is maintained. Each processor, when idle, selects a thread from this queue. Note that load balancing refers to a scheme where work is allocated to processors on a more permanent basis.
Gang Scheduling: A set of related threads is scheduled to run on a set of processors at the same time, on a 1-to-1 basis. Closely related threads / processes may be scheduled this way to reduce synchronization blocking, and minimize process switching. Group scheduling predated this strategy.
Dedicated processor assignment: Provides implicit scheduling defined by assignment of threads to processors. For the duration of program execution, each program is allocated a set of processors equal in number to the number of threads in the program. Processors are chosen from the available pool.
Dynamic scheduling: The number of thread in a program can be altered during the course of execution.

17. When does the condition 'rendezvous' arise?

In message passing, it is the condition in which, both, the sender and receiver are blocked until the message is delivered.

18. What is a trap and trapdoor?

Trapdoor is a secret undocumented entry point into a program used to grant access without normal methods of access authentication. A trap is a software interrupt, usually the result of an error condition.

19. What are local and global page replacements?

Local replacement means that an incoming page is brought in only to the relevant process' address space. Global replacement policy allows any page frame from any process to be replaced. The latter is applicable to variable partitions model only.

20. Define latency, transfer and seek time with respect to disk I/O.

Seek time is the time required to move the disk arm to the required track. Rotational delay or latency is the time it takes for the beginning of the required sector to reach the head. Sum of seek time (if any) and latency is the access time. Time taken to actually transfer a span of data is transfer time.

21. Describe the Buddy system of memory allocation.

Free memory is maintained in linked lists, each of equal sized blocks. Any such block is of size 2^k. When some memory is required by a process, the block size of next higher order is chosen, and broken into two. Note that the two such pieces differ in address only in their kth bit. Such pieces are called buddies. When any used block is freed, the OS checks to see if its buddy is also free. If so, it is rejoined, and put into the original free-block linked-list.

22. What is time-stamping?

It is a technique proposed by Lamport, used to order events in a distributed system without the use of clocks. This scheme is intended to order events consisting of the transmission of messages. Each system 'i' in the network maintains a counter Ci. Every time a system transmits a message, it increments its counter by 1 and attaches the time-stamp Ti to the message. When a message is received, the receiving system 'j' sets its counter Cj to 1 more than the maximum of its current value and the incoming time-stamp Ti.

23. How are the wait/signal operations for monitor different from those for semaphores?

If a process in a monitor signal and no task is waiting on the condition variable, the signal is lost. So this allows easier program design. Whereas in semaphores, every operation affects the value of the semaphore, so the wait and signal operations should be perfectly balanced in the program.

24. In the context of memory management, what are placement and replacement algorithms?

Placement algorithms determine where in available real-memory to load a program. Common methods are first-fit, next-fit, best-fit. Replacement algorithms are used when memory is full, and one process (or part of a process) needs to be swapped out to accommodate a new program. The replacement algorithm determines which are the partitions to be swapped out.

25. In loading programs into memory, what is the difference between load-time dynamic linking and run-time dynamic linking?

For load-time dynamic linking: Load module to be loaded is read into memory. Any reference to a target external module causes that module to be loaded and the references are updated to a relative address from the start base address of the application module.
With run-time dynamic loading: Some of the linking is postponed until actual reference during execution. Then the correct module is loaded and linked.

26. What are demand- and pre-paging?

With demand paging, a page is brought into memory only when a location on that page is actually referenced during execution.

With pre-paging, pages other than the one demanded by a page fault are brought in. The selection of such pages is done based on common access patterns, especially for secondary memory devices.

27. Paging a memory management function, while multiprogramming a processor management function, are the two interdependent?

Yes.

28. What is page cannibalizing?

Page swapping or page replacements are called page cannibalizing.

29. What has triggered the need for multitasking in PCs?

Increased speed and memory capacity of microprocessors together with the support fir virtual memory and
Growth of client server computing

30. What are the four layers that Windows NT have in order to achieve independence?

Hardware abstraction layer
Kernel
Subsystems
System Services.

31. What is SMP?

To achieve maximum efficiency and reliability a mode of operation known as symmetric multiprocessing is used. In essence, with SMP any process or threads can be assigned to any processor.

32. What are the key object oriented concepts used by Windows NT?

Encapsulation
Object class and instance

33. Is Windows NT a full blown object oriented operating system? Give reasons.

No Windows NT is not so, because its not implemented in object oriented language and the data structures reside within one executive component and are not represented as objects and it does not support object oriented capabilities .

34. What is a drawback of MVT?

It does not have the features like

ability to support multiple processors
virtual storage
source level debugging

35. What is process spawning?

When the OS at the explicit request of another process creates a process, this action is called process spawning.

36. How many jobs can be run concurrently on MVT?

15 jobs

37. List out some reasons for process termination.

Normal completion
Time limit exceeded
Memory unavailable
Bounds violation
Protection error
Arithmetic error
Time overrun
I/O failure
Invalid instruction
Privileged instruction
Data misuse
Operator or OS intervention
Parent termination.

38. What are the reasons for process suspension?

swapping
interactive user request
timing
parent process request

39. What is process migration?

It is the transfer of sufficient amount of the state of process from one machine to the target machine

40. What is mutant?

In Windows NT a mutant provides kernel mode or user mode mutual exclusion with the notion of ownership.

41. What is an idle thread?

The special thread a dispatcher will execute when no ready thread is found.

42. What is FtDisk?

It is a fault tolerance disk driver for Windows NT.

43. What are the possible threads a thread can have?

Ready
Standby
Running
Waiting
Transition
Terminated.

44. What are rings in Windows NT?

Windows NT uses protection mechanism called rings provides by the process to implement separation between the user mode and kernel mode.

45. What is Executive in Windows NT?

In Windows NT, executive refers to the operating system code that runs in kernel mode.

46. What are the sub-components of I/O manager in Windows NT?

Network redirector/ Server
Cache manager.
File systems
Network driver
Device driver

47. What are DDks? Name an operating system that includes this feature.

DDks are device driver kits, which are equivalent to SDKs for writing device drivers. Windows NT includes DDks.

48. What level of security does Windows NT meets?

C2 level security.

TECHNICAL QUESTIONS

2009-03-04T09:57:00.001-08:00

1. A 2MB PCM(pulse code modulation) has

a) 32 channels
b) 30 voice channels & 1 signalling channel.
c) 31 voice channels & 1 signalling channel.
d) 32 channels out of which 30 voice channels, 1 signalling channel, & 1 Synchronizatio channel.
Ans: (c)
2. Time taken for 1 satellite hop in voice communication is

a) 1/2 second
b) 1 seconds
c) 4 seconds
d) 2 seconds
Ans: (a)
3. Max number of satellite hops allowed in voice communication is :

a) only one
b) more han one
c) two hops
d) four hops
Ans: (c)
4. What is the max. decimal number that can be accomodated in a byte.

a) 128
b) 256
c) 255
d) 512
Ans: (c)
5. Conditional results after execution of an instruction in a micro processor is stored in

a) register
b) accumulator
c) flag register
d) flag register part of PSW(Program Status Word)
Ans: (d)

6. Frequency at which VOICE is sampled is

a) 4 Khz
b) 8 Khz
c) 16 Khz
d) 64 Khz
Ans: (a)
7. Line of Sight is

a) Straight Line
b) Parabolic
c) Tx & Rx should be visible to each other
d) none
Ans: (c)
8. Purpose of PC(Program Counter) in a MicroProcessor is

a) To store address of TOS(Top Of Stack)
b) To store address of next instruction to be executed.
c) count the number of instructions.
d) to store base address of the stack.
Ans: (b)
9. What action is taken when the processor under execution is interrupted by a non-maskable interrupt?

a) Processor serves the interrupt request after completing the execution of the current instruction.
b) Processor serves the interupt request after completing the current task.
c) Processor serves the interupt request immediately.
d) Processor serving the interrupt request depends upon the priority of the current task under execution.
Ans: (a)
10. The status of the Kernel is

a) task
b) process
c) not defined.
d) none of the above.
Ans: (b)

11. To send a data packet using datagram , connection will be established

a) before data transmission.
b) connection is not established before data transmission.
c) no connection is required.
d) none of the above.
Ans: (c)

12. Word allignment is

a) alligning the address to the next word boundary of the machine.
b) alligning to even boundary.
c) alligning to word boundary.
d) none of the above.
Ans: (a)
13 When a 'C' function call is made, the order in which parameters passed to the function are pushed into the stack is

a) left to right
b) right to left
c) bigger variables are moved first than the smaller variales.
d) smaller variables are moved first than the bigger ones.
e) none of the above.
Ans: (b)
14 What is the type of signalling used between two exchanges?

a) inband
b) common channel signalling
c) any of the above
d) none of the above.
Ans: (a)
15 Buffering is

a) the process of temporarily storing the data to allow for small variation in device speeds
b) a method to reduce cross talks
c) storage of data within transmitting medium until the receiver is ready to receive.
d) a method to reduce routing overhead.
Ans: (a)

16. Memory allocation of variables declared in a program is

a) allocated in RAM.
b) allocated in ROM.
c) allocated on stack.
d) assigned to registers.
Ans: (c)

17. A software that allows a personal computer to pretend as a computer terminal is

a) terminal adapter
b) bulletin board
c) modem
d) terminal emulation
Ans: (d)
18. Find the output of the following program
int *p,*q;
p=(int *)1000;
q=(int *)2000;
printf("%d",(q-p));
Ans: 500

19. Which addressing mode is used in the following statements:
(a) MVI B,55
(b) MOV B,A
(c) MOV M,A
Ans. (a) Immediate addressing mode.
(b) Register Addressing Mode
(c) Direct addressing mode

20. RS-232C standard is used in _____________.
Ans. Serial I/O

21. Memory. Management in Operating Systems is done by

a) Memory Management Unit
b) Memory management software of the Operating System
c) Kernel
Ans: (b)

22. What is done for a Push opertion?
Ans: SP is decremented and then the value is stored.
23. Binary equivalent of 52

Ans. 110100

24. Hexadecimal equivalent of 3452

Ans. 72A

25. Explain Just In Time Concept ?

Ans. Elimination of waste by purchasing manufacturing exactly when needed

26. A good way of unit testing s/w program is

Ans. User test

27. OOT uses

Ans. Encapsulated of detect methods

28.EDI useful in

Ans. Electronic Transmission

29. MRPII different from MRP

Ans. Modular version of man redundant initials

30. Hard disk time for R/W head to move to correct sector

Ans. Latency Time

31. The percentage of times a page number bound in associate register is called

Ans. Bit ratio

32. Expand MODEM

Ans. Modulator and Demodulator

33. RDBMS file system can be defined as

Ans. Interrelated

34. Super Key is

Ans. Primary key and Attribute

35. Windows 95 supports

(a) Multiuser
(b) n tasks
(c) Both
(d) None

Ans. (a)

36.In the command scanf, h is used for

Ans. Short int

37.A process is defined as

Ans. Program in execution

38.A thread is

Ans. Detachable unit of executable code)

39.What is the advantage of Win NT over Win 95

Ans. Robust and secure

40.How is memory management done in Win95

Ans. Through paging and segmentation

41.What is meant by polymorphism

Ans. Redfinition of a base class method in a derived class

42.What is the essential feature of inheritance

Ans. All properties of existing class are derived

43.What does the protocol FTP do

Ans. Transfer a file b/w stations with user authentification

44.In the transport layer ,TCP is what type of protocol

Ans. Connection oriented

45.Why is a gateway used

Ans. To connect incompatible networks

46.How is linked list implemented

Ans. By referential structures

47.What method is used in Win95 in multitasking

Ans. Non preemptive check

48.What is a semaphore

Ans. A method synchronization of multiple processes

49.What is the precedence order from high to low ,of the symbols ( ) ++ /

Ans.( ) , ++, /

50.Preorder of A*(B+C)/D-G

Ans.*+ABC/-DG

51.What is the efficiency of merge sort

Ans. O(n log n)

52.In which layer are routers used

Ans.In network layer

53.Which of the following sorting algorithem has average sorting behavior --
Bubble sort,merge sort,heap sort,exchange sort

Ans. Heap sort

54.In binary search tree which traversal is used for getting ascending order values--Inorder ,post order,preorder

Ans.Inorder

55.What are device drivers used for

Ans.To provide software for enabling the hardware

56.What is fork command in unix

Ans. System call used to create process

57.What is make command in unix

Ans. Used forcreation of more than one file

58.In unix .profile contains

Ans. Start up program

59.In unix 'ls 'stores contents in

Ans.inode block

60. Which of the following involves context switch,
(a) system call
(b) priviliged instruction
(c) floating poitnt exception
(d) all the above
(e) none of the above
Ans: (a)

61. In OST, terminal emulation is done in
(a) sessions layer
(b) application layer
(c) presentation layer
(d) transport layer
Ans: (b)
62. For 1 MB memory, the number of address lines required,
(a)11
(b)16
(c)22
(d) 24
Ans. (b)
63. Semaphore is used for
(a) synchronization
(b) dead-lock avoidence
(c) box
(d) none
Ans. (a)
64. Which holds true for the following statement
class c: public A, public B
a) 2 member in class A, B should not have same name
b) 2 member in class A, C should not have same name
c) both
d) none
Ans. (a)
65.Preproconia.. does not do which one of the following
(a) macro
(b) conditional compliclation
(c) in type checking
(d) including load file
Ans. (c)

66. Piggy backing is a technique for
a) Flow control
b) Sequence
c) Acknowledgement
d) retransmition
Ans. (c)

67. Which is not a memory management scheme?
a) buddy system
b) swapping
c) monitors
d) paging
Ans : c
68. There was a circuit given using three nand gates with two inputs and one output.
Find the output.
a) OR
b) AND
c) XOR
d) NOT
Ans. (a)

69.Iintegrated check value(ICV) are used as:
Ans. The client computes the ICV and then compares it with the senders value.
70. When applets are downloaded from web sites , a byte verifier performs _________?
Ans. Status check.

71. An IP/IPX packet received by a computer using... having IP/IPX both how the packet is handled.
Ans. Read the, field in the packet header with to send IP or IPX protocol.
72. The UNIX shell ....
a) does not come with the rest of the system
b) forms the interface between the user and the kernal
c) does not give any scope for programming
d) deos not allow calling one program from with in another
e) all of the above
Ans. (b)
73. In UNIX a files i-node ......?
Ans. Is a data structure that defines all specifications of a file like the file size,
number of lines to a file, permissions etc.
74. The very first process created by the kernal that runs till the kernal process is halts is
a) init
b) getty
c) both (a) and (b)
d) none of these
Ans. (a)
75. In the process table entry for the kernel process, the process id value is
(a) 0
(b) 1
(c) 2
(d) 255
(e) it does not have a process table entry
Ans. (a)

76. Which of the following API is used to hide a window
a) ShowWindow
b) EnableWindow
c) MoveWindow
d) SetWindowPlacement
e) None of the above
Ans. (a)
77. Which function is the entry point for a DLL in MS Windows 3.1
a) Main
b) Winmain
c) Dllmain
d) Libmain
e) None
Ans. (b)
78. The standard source for standard input, standard output and standard error is
a) the terminal
b) /dev/null
c) /usr/you/input, /usr/you/output/, /usr/you/error respectively
d) None
Ans. (a)
79. The redirection operators > and >>
a) do the same function
b) differ : > overwrites, while >> appends
c) differ : > is used for input while >> is used for output
d) differ : > write to any file while >> write only to standard output
e) None of these
Ans. (b)
80. The command grep first second third /usr/you/myfile
a) prints lines containing the words first, second or third from the file /usr/you/myfile
b) searches for lines containing the pattern first in the files
second, third, and /usr/you/myfile and prints them
c) searches the files /usr/you/myfiel and third for lines containing the words first or second and prints them
d) replaces the word first with the word second in the files third and /usr/you/myfile
e) None of the above
Ans. (b)
81. You are creating a Index on EMPNO column in the EMPLOYEE table. Which statement will you use?
a) CREATE INdEX emp_empno_idx ON employee, empno;
b) CREATE INdEX emp_empno_idx FOR employee, empno;
c) CREATE INdEX emp_empno_idx ON employee(empno);
d) CREATE emp_empno_idx INdEX ON employee(empno);

Ans. c

82. Which program construct must return a value?
a) Package
b) Function
c) Anonymous block
d) Stored Procedure
e) Application Procedure

Ans. b

83. Which Statement would you use to remove the EMPLOYEE_Id_PK PRIMARY KEY constraint and all depending constraints fromthe EMPLOYEE table?
a) ALTER TABLE employee dROP PRIMARY KEY CASCAdE;
b) ALTER TABLE employee dELETE PRIMARY KEY CASCAdE;
c) MOdIFY TABLE employee dROP CONSTRAINT employee_id_pk CASCAdE;
d) ALTER TABLE employee dROP PRIMARY KEY employee_id_pk CASCAdE;
e) MOdIFY TABLE employee dELETE PRIMARY KEY employee_id_pk CASCAdE;

Ans. a

84. Which three commands cause a transaction to end? (Chosse three)
a) ALTER
b) GRANT
c) DELETE
d) INSERT
e) UPdATE
f) ROLLBACK
Ans. a ,b ,f
85. Under which circumstance should you create an index on a table?
a) The table is small.
b) The table is updated frequently.
c) A columns values are static and contain a narrow range of values
d) Two columns are consistently used in the WHERE clause join condition of SELECT
statements.

Ans.d
86. What was the first name given to Java Programming Language.

a) Oak - Java
b) Small Talk
c) Oak
d) None

Ans.a

87.When a bicycle is in motion,the force of friction exerted by the ground on the two wheels is such that it acts
(a) In the backward direction on the front wheel and in the forward direction on the rear wheel.
(b) In the forward direction on the front wheel and in the backward direction on the rear wheel.
(c) In the backward direction on both the front and rear wheels.
(d) In the backward direction on both the front and rear wheels.
Ans. (d)

88. A certain radioactive element A, has a half life = t seconds.
In (t/2) seconds the fraction of the initial quantity of the element so far decayed is nearly
(a) 29%
(b) 15%
(c) 10%
(d) 45%
Ans. (a)

89. Which of the following plots would be a straight line ?
(a) Logarithm of decay rate against logarithm of time
(b) Logarithm of decay rate against logarithm of number of decaying nuclei
(c) Decay rate against time
(d) Number of decaying nuclei against time
Ans. (b)

90. A radioactive element x has an atomic number of 100.
It decays directly into an element y which decays directly into element z.
In both processes a charged particle is emitted.
Which of the following statements would be true?
(a) y has an atomic number of 102
(b) y has an atomic number of 101
(c) z has an atomic number of 100
(d) z has an atomic number of 101
Ans. (b)

91. If the sum of the roots of the equation ax2 + bx + c=0 is equal to the sum of the squares of their reciprocals
then a/c, b/a, c/b are in
(a) AP
(b) GP
(c) HP
(d) None of these
Ans. (c)

92. A man speaks the truth 3 out of 4 times.
He throws a die and reports it to be a 6.
What is the probability of it being a 6?
(a) 3/8
(b) 5/8
(c) 3/4
(d) None of the above
Ans. (a)

93. If cos2A + cos2B + cos2C = 1 then ABC is a
(a) Right angle triangle
(b) Equilateral triangle
(c) All the angles are acute
(d) None of these
Ans. (a)

94. Image of point (3,8) in the line x + 3y = 7 is
(a) (-1,-4)
(b) (-1,4)
(c) (2,-4)
(d) (-2,-4)
Ans. (a)

95. The mass number of a nucleus is
(a) Always less than its atomic number
(b) Always more than its atomic number
(c) Sometimes more than and sometimes equal to its atomic number
(d) None of the above
Ans. (c)

96. The maximum KE of the photoelectron emitted from a surface is dependent on
(a) The intensity of incident radiation
(b) The potential of the collector electrode
(c) The frequency of incident radiation
(d) The angle of incidence of radiation of the surface
Ans. (c)

97. Which of the following is not an essential condition for interference
(a) The two interfering waves must be propagated in almost the same direction or
the two interfering waves must intersect at a very small angle
(b) The waves must have the same time period and wavelength
(c) Amplitude of the two waves should be the same
(d) The interfering beams of light must originate from the same source
Ans. (c)

98. When X-Ray photons collide with electrons
(a) They slow down
(b) Their mass increases
(c) Their wave length increases
(d) Their energy decreases
Ans. (c)

99. An electron emits energy
(a) Because its in orbit
(b) When it jumps from one energy level to another
(c) Electrons are attracted towards the nucleus
(d) The electrostatic force is insufficient to hold the electrons in orbits
Ans. (b)

100. How many bonds are present in CO2 molecule?
(a) 1
(b) 2
(c) 0
(d) 4
Ans. (d)

APTITUDE

2009-03-04T07:19:00.000-08:00

1.If 2x-y=4 then 6x-3y=?
(a)15
(b)12
(c)18
(d)10

Ans. (b)

2.If x=y=2z and xyz=256 then what is the value of x?

(a)12
(b)8
(c)16
(d)6

Ans. (b)

3. (1/10)18 - (1/10)20 = ?

(a) 99/1020
(b) 99/10
(c) 0.9
(d) none of these

Ans. (a)

4.Pipe A can fill in 20 minutes and Pipe B in 30 mins and Pipe C can empty the same in 40 mins.If all of them work together, find the time taken to fill the tank

(a) 17 1/7 mins
(b) 20 mins
(c) 8 mins
(d) none of these

Ans. (a)

5. Thirty men take 20 days to complete a job working 9 hours a day.How many hour a day should 40 men work to complete the job?

(a) 8 hrs
(b) 7 1/2 hrs
(c) 7 hrs
(d) 9 hrs

Ans. (b)

6. Find the smallest number in a GP whose sum is 38 and product 1728

(a) 12
(b) 20
(c) 8
(d) none of these

Ans. (c)

7. A boat travels 20 kms upstream in 6 hrs and 18 kms downstream in 4 hrs.Find the speed of the boat in still water and the speed of the water current?

(a) 1/2 kmph
(b) 7/12 kmph
(c) 5 kmph
(d) none of these

Ans. (b)

8. A goat is tied to one corner of a square plot of side 12m by a rope 7m long.Find the area it can graze?

(a) 38.5 sq.m
(b) 155 sq.m
(c) 144 sq.m
(d) 19.25 sq.m

Ans. (a)

9. Mr. Shah decided to walk down the escalator of a tube station. He found that if he walks down 26 steps, he requires 30 seconds to reach the bottom. However, if he steps down 34 stairs he would only require 18 seconds to get to the bottom. If the time is measured from the moment the top step begins to descend to the time he steps off the last step at the bottom, find out the height of the stair way in steps?

Ans.46 steps.

10. The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member ?

Ans.40 years.

11. Three containers A, B and C have volumes a, b, and c respectively; and container A is full of water while the other two are empty. If from container A water is poured into container B which becomes 1/3 full, and into container C which becomes 1/2 full, how much water is left in container A?

12. ABCE is an isosceles trapezoid and ACDE is a rectangle. AB = 10 and EC = 20. What is the length of AE?

Ans. AE = 10.

13. In the given figure, PA and PB are tangents to the circle at A and B respectively and the chord BC is parallel to tangent PA. If AC = 6 cm, and length of the tangent AP is 9 cm, then what is the length of the chord BC?

Ans. BC = 4 cm.

15 Three cards are drawn at random from an ordinary pack of cards. Find the probability that they will consist of a king, a queen and an ace.

Ans. 64/2210.

16. A number of cats got together and decided to kill between them 999919 mice. Every cat killed an equal number of mice. Each cat killed more mice than there were cats. How many cats do you think there were ?

Ans. 991.

17. If Log2 x - 5 Log x + 6 = 0, then what would the value / values of x be?

Ans. x = e2 or e3.

18. The square of a two digit number is divided by half the number. After 36 is added to the quotient, this sum is then divided by 2. The digits of the resulting number are the same as those in the original number, but they are in reverse order. The ten's place of the original number is equal to twice the difference between its digits. What is the number?

Ans. 46

19.Can you tender a one rupee note in such a manner that there shall be total 50 coins but none of them would be 2 paise coins.?

Ans. 45 one paisa coins, 2 five paise coins, 2 ten paise coins, and 1 twenty-five paise coins.

20.A monkey starts climbing up a tree 20ft. tall. Each hour, it hops 3ft. and slips back 2ft. How much time would it take the monkey to reach the top?

Ans.18 hours.

21. What is the missing number in this series? 8 2 14 6 11 ? 14 6 18 12

Ans. 9

22. A certain type of mixture is prepared by mixing brand A at Rs.9 a kg. with brand B at Rs.4 a kg. If the mixture is worth Rs.7 a kg., how many kgs. of brand A are needed to make 40kgs. of the mixture?

Ans. Brand A needed is 24kgs.

23. A wizard named Nepo says "I am only three times my son's age. My father is 40 years more than twice my age. Together the three of us are a mere 1240 years old." How old is Nepo?

Ans. 360 years old.

24. One dog tells the other that there are two dogs in front of me. The other one also shouts that he too had two behind him. How many are they?

Ans. Three.

25. A man ate 100 bananas in five days, each day eating 6 more than the previous day. How many bananas did he eat on the first day?

Ans. Eight.

26. If it takes five minutes to boil one egg, how long will it take to boil four eggs?

Ans. Five minutes.

27. The minute hand of a clock overtakes the hour hand at intervals of 64 minutes of correct time. How much a day does the clock gain or lose?

Ans. 32 8/11 minutes.

28. Solve for x and y: 1/x - 1/y = 1/3, 1/x2 + 1/y2 = 5/9.

Ans. x = 3/2 or -3 and y = 3 or -3/2.

29. Daal is now being sold at Rs. 20 a kg. During last month its rate was Rs. 16 per
kg. By how much percent should a family reduce its consumption so as to keep the expenditure fixed?

Ans. 20 %.

30. Find the least value of 3x + 4y if x2y3 = 6.

Ans. 10.

31. Can you find out what day of the week was January 12, 1979?

Ans. Friday.

32. A garrison of 3300 men has provisions for 32 days, when given at a rate of 850 grams per head. At the end of 7 days a reinforcement arrives and it was found that now the provisions will last 8 days less, when given at the rate of 825 grams per head. How, many more men can it feed?

Ans. 1700 men.

33. From 5 different green balls, four different blue balls and three different red balls, how many combinations of balls can be chosen taking at least one green and one blue ball?

Ans. 3720.

34. Three pipes, A, B, & C are attached to a tank. A & B can fill it in 20 & 30 minutes respectively while C can empty it in 15 minutes. If A, B & C are kept open successively for 1 minute each, how soon will the tank be filled?

Ans. 167 minutes.

35. A person walking 5/6 of his usual rate is 40 minutes late. What is his usual time?

Ans. 3 hours 20 minutes.

36.For a motorist there are three ways going from City A to City C. By way of bridge the distance is 20 miles and toll is $0.75. A tunnel between the two cities is a distance of 10 miles and toll is $1.00 for the vehicle and driver and $0.10 for each passenger. A two-lane highway without toll goes east for 30 miles to city B and then 20 miles in a northwest direction to City C.

1. Which is the shortest route from B to C

(a) Directly on toll free highway to City C
(b) The bridge
(c) The Tunnel
(d) The bridge or the tunnel
(e) The bridge only if traffic is heavy on the toll free highway

Ans. (a)

2. The most economical way of going from City A to City B, in terms of toll and distance is to use the

(a) tunnel
(b) bridge
(c) bridge or tunnel
(d) toll free highway
(e) bridge and highway

Ans. (a)

3. Jim usually drives alone from City C to City A every working day. His firm deducts a percentage of employee pay for lateness. Which factor would most influence his choice of the bridge or the tunnel ?

(a) Whether his wife goes with him
(b) scenic beauty on the route
(c) Traffic conditions on the road, bridge and tunnel
(d) saving $0.25 in tolls
(e) price of gasoline consumed in covering additional 10 miles on the bridge

Ans. (a)

4. In choosing between the use of the bridge and the tunnel the chief factor(s) would be:
I. Traffic and road conditions
II. Number of passengers in the car
III. Location of one's homes in the center or outskirts of one of the cities
IV. Desire to save $0.25

(a) I only
(b) II only
(c) II and III only
(d) III and IV only
(e) I and II only

Ans. (a)

37.The letters A, B, C, D, E, F and G, not necessarily in that order, stand for seven consecutive integers from 1 to 10
D is 3 less than A
B is the middle term
F is as much less than B as C is greater than D
G is greater than F

1. The fifth integer is
(a) A
(b) C
(c) D
(d) E
(e) F

Ans. (a)

2. A is as much greater than F as which integer is less than G
(a) A
(b) B
(c) C
(d) D
(e) E

Ans. (a)

3. If A = 7, the sum of E and G is
(a) 8
(b) 10
(c) 12
(d) 14
(e) 16

Ans. (a)

4. A - F = ?
(a) 1
(b) 2
(c) 3
(d) 4
(e) Cannot be determined

Ans. (a)

5. An integer T is as much greater than C as C is greater than E. T can be written as A + E. What is D?
(a) 2
(b) 3
(c) 4
(d) 5
(e) Cannot be determined

Ans. (a)

6. The greatest possible value of C is how much greater than the smallest possible value of D?
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6

Ans. (a)

38.
1. All G's are H's
2. All G's are J's or K's
3. All J's and K's are G's
4. All L's are K's
5. All N's are M's
6. No M's are G's

1. If no P's are K's, which of the following must be true?

(a) All P's are J's
(b) No P is a G
(c) No P is an H
(d) If any P is an H it is a G
(e) If any P is a G it is a J

Ans. (a)

2. Which of the following can be logically deduced from the conditions stated?

(a) No M's are H's
(b) No M's that are not N's are H's
(c) No H's are M's
(d) Some M's are H's
(e) All M's are H's

Ans. (a)

3. Which of the following is inconsistent with one or more of the conditions?

(a) All H's are G's
(b) All H's that are not G's are M's
(c) Some H's are both M's and G's
(d) No M's are H's
(e) All M's are H's

Ans. (a)

4. The statement "No L's are J's" is
I. Logically deducible from the conditions stated
II. Consistent with but not deducible from the conditions stated
III. Deducible from the stated conditions together with the additional statement "No J's are K's"

(a) I only
(b) II only
(c) III only
(d) II and III only
(e) Neither I, II nor III

Ans. (a)

39.In country X, democratic, conservative and justice parties have fought three civil wars in twenty years. TO restore stability an agreement is reached to rotate the top offices President, Prime Minister and Army Chief among the parties so that each party controls one and only one office at all times. The three top office holders must each have two deputies, one from each of the other parties. Each deputy must choose a staff composed of equally members of his or her chiefs party and member of the third party.

1. When Justice party holds one of the top offices, which of the following cannot be true

(a) Some of the staff members within that office are justice party members
(b) Some of the staff members within that office are democratic party members
(c) Two of the deputies within the other offices are justice party members
(d) Two of the deputies within the other offices are conservative party members
(e) Some of the staff members within the other offices are justice party members.

Ans. (a)

2. When the democratic party holds presidency, the staff of the prime minister's deputies are composed
I. One-fourth of democratic party members
II. One-half of justice party members and one-fourth of conservative party members
III. One-half of conservative party members and one-fourth of justice party members.

(a) I only
(b) I and II only
(c) II or III but not both
(d) I and II or I and III
(e) None of these

Ans. (a)

3. Which of the following is allowable under the rules as stated:

(a) More than half of the staff within a given office belonging to a single party
(b) Half of the staff within a given office belonging to a single party
(c) Any person having a member of the same party as his or her immediate superior
(d) Half the total number of staff members in all three offices belonging to a single party
(e) Half the staff members in a given office belonging to parties different from the party of the top office holder in that office.

Ans. (a)

4. The office of the Army Chief passes from Conservative to Justice party. Which of the following must be fired.

(a) The democratic deputy and all staff members belonging to Justice party
(b) Justice party deputy and all his or hers staff members
(c) Justice party deputy and half of his Conservative staff members in the chief of staff office
(d) The Conservative deputy and all of his or her staff members belonging to Conservative party
(e) No deputies and all staff members belonging to conservative parties.

Ans. (a)

40.In recommendations to the board of trustees a tuition increase of $500 per year, the president of the university said "There were no student demonstrations over the previous increases of $300 last year and $200 the year before". If the president's statement is accurate then which of the following can be validly inferred from the information given:
I. Most students in previous years felt that the increases were justified because of increased operating costs.
II. Student apathy was responsible for the failure of students to protest the previous tuition increases.
III. Students are not likely to demonstrate over new tuition increases.

(a) I only
(b) II only
(c) I or II but not both
(d) I, II and III
(e) None

Ans. (a)

41. The office staff of XYZ corporation presently consists of three bookeepers--A, B, C and 5 secretaries D, E, F, G, H. The management is planning to open a new office in another city using 2 bookeepers and 3 secretaries of the present staff . To do so they plan to seperate certain individuals who don't function well together. The following guidelines were established to set up the new office
I. Bookeepers A and C are constantly finding fault with one another and should not be sent together to the new office as a team
II. C and E function well alone but not as a team , they should be seperated
III. D and G have not been on speaking terms and shouldn't go together
IV Since D and F have been competing for promotion they shouldn't be a team

1.If A is to be moved as one of the bookeepers,which of the following cannot be a possible working unit.

A.ABDEH
B.ABDGH
C.ABEFH
D.ABEGH

Ans.B

2.If C and F are moved to the new office,how many combinations are possible

A.1
B.2
C.3
D.4

Ans.A

3.If C is sent to the new office,which member of the staff cannot go with C

A.B
B.D
C.F
D.G

Ans.B

4.Under the guidelines developed,which of the following must go to the new office

A.B
B.D
C.E
D.G

Ans.A

5.If D goes to the new office,which of the following is/are true

I.C cannot go
II.A cannot go
III.H must also go

A.I only
B.II only
C.I and II only
D.I and III only

Ans.D

42.After months of talent searching for an administrative assistant to the president of the college the field of applicants has been narrowed down to 5--A, B, C, D, E .It was announced that the finalist would be chosen after a series of all-day group personal interviews were held.The examining committee agreed upon the following procedure

I.The interviews will be held once a week
II.3 candidates will appear at any all-day interview session
III.Each candidate will appear at least once
IV.If it becomes necessary to call applicants for additonal interviews, no more 1 such applicant should be asked to appear the next week
V.Because of a detail in the written applications,it was agreed that whenever candidate B appears, A should also be present.
VI.Because of travel difficulties it was agreed that C will appear for only 1 interview.

1.At the first interview the following candidates appear A,B,D.Which of the follwing combinations can be called for the interview to be held next week.

A.BCD
B.CDE
C.ABE
D.ABC

Ans.B

2.Which of the following is a possible sequence of combinations for interviews in 2 successive weeks

A.ABC;BDE
B.ABD;ABE
C.ADE;ABC
D.BDE;ACD

Ans.C

3.If A ,B and D appear for the interview and D is called for additional interview the following week,which 2 candidates may be asked to appear with D?

I. A
II B
III.C
IV.E
A.I and II
B.I and III only
C.II and III only
D.III and IV only

Ans.D

4.Which of the following correctly state(s) the procedure followed by the search committee

I.After the second interview all applicants have appeared at least once
II.The committee sees each applicant a second time
III.If a third session,it is possible for all applicants to appear at least twice

A.I only
B.II only
C.III only
D.Both I and II

Ans.A

43. A certain city is served by subway lines A,B and C and numbers 1 2 and 3
When it snows , morning service on B is delayed
When it rains or snows , service on A, 2 and 3 are delayed both in the morning and afternoon
When temp. falls below 30 degrees farenheit afternoon service is cancelled in either the A line or the 3 line,
but not both.
When the temperature rises over 90 degrees farenheit, the afternoon service is cancelled in either the line C or the
3 line but not both.
When the service on the A line is delayed or cancelled, service on the C line which connects the A line, is delayed.
When service on the 3 line is cancelled, service on the B line which connects the 3 line is delayed.

Q1. On Jan 10th, with the temperature at 15 degree farenheit, it snows all day. On how many lines will service be
affected, including both morning and afternoon.
(A) 2
(B) 3
(C) 4
(D) 5
Ans. D

Q2. On Aug 15th with the temperature at 97 degrees farenheit it begins to rain at 1 PM. What is the minimum number
of lines on which service will be affected?
(A) 2
(B) 3
(C) 4
(D) 5
Ans. C

Q3. On which of the following occasions would service be on the greatest number of lines disrupted.
(A) A snowy afternoon with the temperature at 45 degree farenheit
(B) A snowy morning with the temperature at 45 degree farenheit
(C) A rainy afternoon with the temperature at 45 degree farenheit
(D) A rainy afternoon with the temperature at 95 degree farenheit
Ans. B

44. In a certain society, there are two marriage groups, red and brown. No marriage is permitted within a group. On marriage, males become part of their wives groups; women remain in their own group. Children belong to the same group as their parents. Widowers and divorced males revert to the group of their birth. Marriage to more than one person at the same time and marriage to a direct descendant are forbidden

Q1.
A brown female could have had
I. A grandfather born Red
II. A grandmother born Red
III Two grandfathers born Brown

(A) I only
(B) III only
(C) I, II and III
(D) I and II only
Ans. D

Q2. A male born into the brown group may have
(A) An uncle in either group
(B) A brown daughter
(C) A brown son
(D) A son-in-law born into red group
Ans. A

Q3. Which of the following is not permitted under the rules as stated.
(A) A brown male marrying his father's sister
(B) A red female marrying her mother's brother
(C) A widower marrying his wife's sister
(D) A widow marrying her divorced daughter's ex-husband
Ans. B

Q4. If widowers and divorced males retained their group they had upon marrying which of the following would be permissible ( Assume that no previous marriage occurred)
(A) A woman marrying her dead sister's husband
(B) A woman marrying her divorced daughter's ex-husband
(C) A widower marrying his brother's daughter
(D) A woman marrying her mother's brother who is a widower.
Ans. D

Q5. I. All G's are H's
II. All G's are J's or K's
III All J's and K's are G's
IV All L's are K's
V All N's are M's
VI No M's are G's

45. There are six steps that lead from the first to the second floor. No two people can be on the same step
Mr. A is two steps below Mr. C
Mr. B is a step next to Mr. D
Only one step is vacant ( No one standing on that step )
Denote the first step by step 1 and second step by step 2 etc.

1. If Mr. A is on the first step, Which of the following is true?
(a) Mr. B is on the second step
(b) Mr. C is on the fourth step.
(c) A person Mr. E, could be on the third step
(d) Mr. D is on higher step than Mr. C.
Ans: (d)

2. If Mr. E was on the third step & Mr. B was on a higher step than Mr. E which step must be vacant
(a) step 1
(b) step 2
(c) step 4
(d) step 5
(e) step 6
Ans: (a)

3. If Mr. B was on step 1, which step could A be on?
(a) 2&e only
(b) 3&5 only
(c) 3&4 only
(d) 4&5 only
(e) 2&4 only
Ans: (c)

4. If there were two steps between the step that A was standing and the step that B was standing on, and A was on a higher step than D , A must be on step
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6
Ans: (c)

5. Which of the following is false

i. B&D can be both on odd-numbered steps in one configuration
ii. In a particular configuration A and C must either both an odd numbered steps or both an even-numbered steps
iii. A person E can be on a step next to the vacant step.
(a) i only
(b) ii only
(c) iii only
(d) both i and iii
Ans: (c)

46. Six swimmers A, B, C, D, E, F compete in a race. The outcome is as follows.

i. B does not win.
ii. Only two swimmers separate E & D
iii. A is behind D & E
iv. B is ahead of E , with one swimmer intervening
v. F is a head of D

1. Who stood fifth in the race ?
(a) A
(b) B
(c) C
(d) D
(e) E
Ans: (e)

2. How many swimmers seperate A and F ?

(a) 1
(b) 2
(c) 3
(d) 4
(e) cannot be determined

Ans: (d)

3. The swimmer between C & E is

(a) none
(b) F
(c) D
(d) B
(e) A

Ans: (a)

4. If the end of the race, swimmer D is disqualified by the Judges then swimmer B finishes in which place
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5

Ans: (b)

47. Five houses lettered A,B,C,D, & E are built in a row next to each other. The houses are lined up in the order A,B,C,D, & E. Each of the five houses has a colored chimney. The roof and chimney of each housemust be painted as follows.
i. The roof must be painted either green,red ,or yellow.
ii. The chimney must be painted either white, black, or red.
iii. No house may have the same color chimney as the color of roof.
iv. No house may use any of the same colors that the every next house uses.
v. House E has a green roof.
vi. House B has a red roof and a black chimney

1. Which of the following is true ?

(a) At least two houses have black chimney.
(b) At least two houses have red roofs.
(c) At least two houses have white chimneys
(d) At least two houses have green roofs
(e) At least two houses have yellow roofs

Ans: (c)

2. Which must be false ?

(a) House A has a yellow roof
(b) House A & C have different color chimney
(c) House D has a black chimney
(d) House E has a white chimney
(e) House B&D have the same color roof.

Ans: (b)

3. If house C has a yellow roof. Which must be true.

(a) House E has a white chimney
(b) House E has a black chimney
(c) House E has a red chimney
(d) House D has a red chimney
(e) House C has a black chimney

Ans: (a)

4. Which possible combinations of roof & chimney can house

I. A red roof 7 a black chimney
II. A yellow roof & a red chimney
III. A yellow roof & a black chimney

(a) I only
(b) II only
(c) III only
(d) I & II only
(e) I&II&III

Ans: (e)

48. Find x+2y

(i). x+y=10
(ii). 2x+4y=20

Ans: (b)

49. Is angle BAC is a right angle

(i) AB=2BC
(2) BC=1.5AC

Ans: (e)

50. Is x greater than y

(i) x=2k
(ii) k=2y

Ans: (e)

VERBAL APTITUDE

2009-03-04T07:15:00.000-08:00

1. Depreciation: deflation, depression, devaluation, fall, slump

2. Deprecate : feel and express disapproval,

3. Incentive : thing one encourages one to do (stimulus)

4. Echelon : level of authority or responsibility

5. Innovation : make changes or introduce new things

6. Intermittent : externally stopping and then starting

7. Detrimental: harmful

8. Conciliation : make less angry or more friendly

9. Orthodox: conventional or traditional, superstitious

10. Fallible : liable to error

11. Volatile : ever changing

12. Manifest: clear and obvious

13. Connotation : suggest or implied meaning of expression

14. Reciprocal: reverse or opposite

15. Agrarian : related to agriculture

16. Vacillate : undecided or dilemma

17. Expedient : fitting proper, desirable

18. Simulate : produce artificially resembling an existing one.

19. Access : to approach

20. Compensation: salary

21. Truncate : shorten by cutting

22. Adherence : stick

23. Heterogeneous: non similar things

24. Surplus : excessive

25. Assess : determine the amount or value

26. Cognizance : knowledge

27. Retrospective : review

28. Naive : innocent, rustic

29. Equivocate : tallying on both sides, lie, mislead

30. Postulate : frame a theory

31. Latent : dormant, secret

32. Fluctuation : wavering,

33. Eliminate : to reduce

34. Affinity : strong liking

35. Expedite : hasten

36. Console : to show sympathy

37. Adversary : opposition

38. Affable : lovable or approachable

39. Decomposition : rotten

40. Agregious : apart from the crowd, especially bad

41. Conglomeration: group, collection

42. Aberration: deviation

43. Augury : prediction

44. Creditability : ability to common belief, quality of being credible

45. Coincident: incidentally

46. Constituent : accompanying

47. Differential : having or showing or making use of

48. Litigation : engaging in a law suit

49. Moratorium: legally or officially determined period of delay before
fulfillment of the agreement of paying of debts.

50. Negotiate : discuss or bargain

51. Preparation : act of preparing

52. Preponderant : superiority of power or quality

53. Relevance : quality of being relevant

54. Apparatus : appliances

55. Ignorance : blindness, in experience

56. Obsession: complex enthusiasm

57. precipitate : speed, active

58. corroborative: refutable

59. obnoxious : harmless

60. sanction: hinder

61. empirical: experimental

62. aborigine: emigrant

63. corpulent : emaciated

64. officious: pragmate

65. Agitator : Firebrand :: Renegade : Turncoat

66. Burst : Sound :: Tinder : Fire

67. Star : cluster :: Tree : clump

68. Piston : Cylinder :: elevator : shaft

69. Mitigate : punishment :: commute : sentence

70. Erudite : scholar :: illiterate : ignorant

71. Fire : Ashes :: explosion : debris

72. mason : wall :: Author : Book

73. Fire : Ashes :: Event : memories

74. (a) cheerleaders : pompoms
(b) audience:seats
(c) team:goalposts
(d) conductor:podium
(e) referee:decision

Ans. (a)

75. archipelago:islands::

(a) arbor:bower
(b) garden:flower
(c) mountain:valley
(d) sand:dune
(e) constellation:star

Ans. (a)

76. crow:boastful ::

(a) smirk:witty
(b) conceal:s;y
(c) pout:sulky
(d) blush:coarse
(e) bluster:unhappy

Ans. (a)

77. bracket:shelf ::

(a) hammer:anvil
(b) girder:rivet
(c) strut:rafter
(d) valve:pipe
(e) bucket:well

Ans. (a)

78. taxonomy:classification ::

(a) etymology:derivation
(b) autonomy:authorization
(c) economy:rationalization
(d) tautology:justification
(e) ecology:urbanisation

Ans. (a)

79. moderator:debate ::

(a) legislator:election
(b) chef:banquet
(c) auditor:lecture
(d) conspirator:plot
(e) umpire:game

Ans. (a)

80. glossary:words ::

(a) catalogue:dates
(b) atlas:maps
(c) almanac:synonyms
(d) thesaurus:rhymes
(e) lexicon:numbers

Ans. (a)

81. lumber: bear ::

(a) roost:hen
(b) bray:donkey
(c) waddle:goose
(d) swoop:hawk
(e) chirp:sparrow

Ans. (a)

82. celerity:snail ::

(a) indolence:sloth
(b) cunning:weasel
(c) curiosity:cat
(d) humility:peacock
(e) obstinacy:mule

Ans. (a)

83. wood:sand ::

(a) coal:burn
(b) brick:lay
(c) oil:polish
(d) metal:burnish
(e) stone:quarry

Ans. (a)

84. carpenter:saw ::

(a) stenographer:typist
(b) painter:brush
(c) lawyer:brief
(d) runner:sneakers
(e) seamstress:scissors

Ans. (a)

85. horns:bull ::

(a) mane:lion
(b) wattles:turkey
(c) antlers:stag
(d) hooves:horse
(e) wings:eagle

Ans. (a)

86. gullible:duped ::

(a) credible:cheated
(b) careful:cautioned
(c) malleable:moulded
(d) myopic:mislead
(e) articulate:silenced

Ans. (a)

87. marathon:stamina ::

(a) relay:independence
(b) hurdle:perseverance
(c) sprint:celerity
(d) job:weariness
(e) ramble:directness

Ans. (a)

88. Skin:man ::

(a) hide:animal
(b) jump:start
(c) peel:potato
(d) eat:food
(e) wool:cloth

Ans. (a)

89. Bamboo:Shoot ::

(a) Bean:Sprout
(b) Peas:Pod
(c) Potato:Eye
(d) Carrot:Root
(e) Leaf:Stem

Ans. (a)

90. Deflect:Missile ::

(a) Siege:Castle
(b) Distract:Attraction
(c) Protect:Honour
(d) Drop:Catch
(e) Score:Goal

Ans. (a)

91. Editor:magazine ::

(a) captain:ship
(b) actor:movie
(c) director:film
(d) player:team
(e) jockey:horse

Ans. (a)

92. Volcano : Lava ::

(a) Fault:earthquate
(b) crack:wall
(c) tunnel:dig
(d) water:swim
(e) floor:polish

Ans. (a)

93. Disregarded

(a) heed
(b) hopeful
(c) evade
(d) dense

Ans. (a)

94. Obviate

(a) becloud
(b) necessitate
(c) rationalize
(d) execute

Ans. (b)

95. Superficial

(a) profound
(b) exaggerated
(c) subjective
(d) spirited

Ans. (a)

96. chief : tribe :: governer : state

97. epaulette : shoulder :: tiara : head

98. guttural : throat :: gastric : stomach

99. inept : clever :: languid : active

100. Erudite : scholar :: illiterate : ignorant

C APTITUDE

2009-03-04T07:07:00.000-08:00

C Questions

Note : All the programs are tested under Turbo C/C++ compilers.It is assumed that,

Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.

The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).

Predict the output or error(s) for the following:

1. void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}

Answer:
2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

6. main()
{
extern int i;
i=20;
printf("%d",i);
}

Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

7. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

8. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}

Answer:
1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

9. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

10. main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

11. main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error : Type mismatch in redeclaration of function display
Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

12. main()
{
int c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

13. #define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char

14. main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}
Answer:
i=0

Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

15. #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).

16. #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

17. #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration

18. #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed

20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

22. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
 *p that is value at the location currently pointed by p will be taken
 ++*p the retrieved value will be incremented
 when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.

23. #include
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any problem

25. main()
{
printf("%p",main);
}
Answer:
Some address will be printed.
Explanation:
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

27) main()
{
clrscr();
}
clrscr();

Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

28) enum colors {BLACK,BLUE,GREEN}
main()
{

printf("%d..%d..%d",BLACK,BLUE,GREEN);

return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.

29) void main()
{
char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer

30) main()
{
int i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.

31) main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

32) main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

33) main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.

34) void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i

35) void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

36) #include
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.

37) main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

38) #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer:
100

39) main()
{
int i=0;

for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

40) #include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77("M");

41) #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure declaration

42) #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.

43) main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

44) main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

45) main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.

46) main()
{
show();
}
void show()
{
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

47) main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.

2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116 118 120 122

thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a gives the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.

48) main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.

Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

49) main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
a
0 1 2 3 4
100 102 104 106 108
p
100 102 104 106 108
1000 1002 1004 1006 1008
ptr
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.

50) main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf(“%s” ,(q+j));
for (j=0; j<3; j++) printf(“%c” ,*(q+j));
for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as
M O U S E \0
When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.
M T R A C K \0
The third input starts filling from the location 102
M T V I R T U A L \0
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V
The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.

51) main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

52) main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

53) main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i {
printf(“%s\n”,x);
x++;
}
}
Answer:
(blank space)
irl
rl
l

Explanation:
Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.
54) int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5),>,
Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.

55) main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

56) What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).

57) what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);

Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position
of the file.

58) main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it reads all character upto another quotation mark.

59) What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.

60) main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.

61) main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

62) main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

63) main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

64) #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

65) main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

66) main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.

67) #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.

68) int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

69) main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

70) main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

71) #include
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant

72) #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.

73) #include
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.

74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)

76) struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.

77) struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}

Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes

78) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.

79) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to their corresponding data-types.

80) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.

81) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.

82) # include
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.

83) # include
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

85) #include
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against NULL.

86) main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain unchanged and get printed.

87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.

88) int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution
will be,
t i x
4 0 -4
3 1 -2
2 2 0

89) main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.

90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.

91) In the following pgm add a stmt in the function fun such that the address of
'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.

92) What are the following notations of defining functions known as?
i. int abc(int a,float b)
{
/* some code */
}
ii. int abc(a,b)
int a; float b;
{
/* some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation

93) main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.

94) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so bc will be printed.

95) func(a,b)
int a,b;
{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function 'process'.

96) void main()
{
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

97) void main()
{
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.

98) void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.

99) void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
Explanation:

100) void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.

101) void main()
{
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void pointers are normally used for,
1. Passing generic pointers to functions and returning such pointers.
2. As a intermediate pointer type.
3. Used when the exact pointer type will be known at a later point of time.

102) void main()
{
int i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of its declaration.
So expressions such as i = i++ are valid statements. The i, j and k are automatic variables and so they contain some garbage value. Garbage in is garbage out (GIGO).

103) void main()
{
static int i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}
Answer:
i = 1 j = 1 k = 1
Explanation:
Since static variables are initialized to zero by default.

104) void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.

104) main()
{
unsigned int i=10;
while(i-->=0)
printf("%u ",i);

}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
Since i is an unsigned integer it can never become negative. So the expression i-- >=0 will always be true, leading to an infinite loop.

105) #include
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition reduces to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.

106) main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !

107) #define prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
Explanation:
The macro expands and evaluates to as:
x+2*y-1 => x+(2*y)-1 => 10

108) main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
Note the semicolon after the while statement. When the value of i becomes 0 it comes out of while loop. Due to post-increment on i the value of i while printing is 1.

109) main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is, while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

113) main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
Enumeration constants cannot be modified, so you cannot apply ++.
Bit-wise operators and % operators cannot be applied on float values.
fmod() is to find the modulus values for floats as % operator is for ints.

110) main()
{
int i=10;
void pascal f(int,int,int);
f(i++,i++,i++);
printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler error: unknown type integer
Compiler error: undeclared function write
Explanation:
Pascal keyword doesn’t mean that pascal code can be used. It means that the function follows Pascal argument passing mechanism in calling the functions.

111) void pascal f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
void cdecl f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
main()
{
int i=10;
f(i++,i++,i++);
printf(" %d\n",i);
i=10;
f(i++,i++,i++);
printf(" %d",i);
}
Answer:
10 11 12 13
12 11 10 13
Explanation:
Pascal argument passing mechanism forces the arguments to be called from left to right. cdecl is the normal C argument passing mechanism where the arguments are passed from right to left.

112). What is the output of the program given below

main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
-128
Explanation
Notice the semicolon at the end of the for loop. THe initial value of the i is set to 0. The inner loop executes to increment the value from 0 to 127 (the positive range of char) and then it rotates to the negative value of -128. The condition in the for loop fails and so comes out of the for loop. It prints the current value of i that is -128.

113) main()
{
unsigned char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
infinite loop
Explanation
The difference between the previous question and this one is that the char is declared to be unsigned. So the i++ can never yield negative value and i>=0 never becomes false so that it can come out of the for loop.

114) main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);

}
Answer:
Behavior is implementation dependent.
Explanation:
The detail if the char is signed/unsigned by default is implementation dependent. If the implementation treats the char to be signed by default the program will print –128 and terminate. On the other hand if it considers char to be unsigned by default, it goes to infinite loop.
Rule:
You can write programs that have implementation dependent behavior. But dont write programs that depend on such behavior.

115) Is the following statement a declaration/definition. Find what does it mean?
int (*x)[10];
Answer
Definition.
x is a pointer to array of(size 10) integers.

Apply clock-wise rule to find the meaning of this definition.

116). What is the output for the program given below

typedef enum errorType{warning, error, exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer
Compiler error: Multiple declaration for error
Explanation
The name error is used in the two meanings. One means that it is a enumerator constant with value 1. The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the compiler cannot distinguish the meaning of error to know in what sense the error is used:
error g1;
g1=error;
// which error it refers in each case?
When the compiler can distinguish between usages then it will not issue error (in pure technical terms, names can only be overloaded in different namespaces).
Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for programmer’s convenience.

117) typedef struct error{int warning, error, exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}

Answer
1
Explanation
The three usages of name errors can be distinguishable by the compiler at any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the variable name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.

Note
This code is given here to just explain the concept behind. In real programming don’t use such overloading of names. It reduces the readability of the code. Possible doesn’t mean that we should use it!

118) #ifdef something
int some=0;
#endif

main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}

Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The name something is not already known to the compiler making the declaration
int some = 0;
effectively removed from the source code.

119) #if something == 0
int some=0;
#endif

main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}

Answer
0 0
Explanation
This code is to show that preprocessor expressions are not the same as the ordinary expressions. If a name is not known the preprocessor treats it to be equal to zero.

120). What is the output for the following program

main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) );
}
Answer
1
Explanation
This is due to the close relation between the arrays and pointers. N dimensional arrays are made up of (N-1) dimensional arrays.
arr2D is made up of a 3 single arrays that contains 3 integers each .

The name arr2D refers to the beginning of all the 3 arrays. *arr2D refers to the start of the first 1D array (of 3 integers) that is the same address as arr2D. So the expression (arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero doesn’t change the value/meaning. Again arr2D[0] is the another way of telling *(arr2D + 0). So the expression (*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the result is true(1) and the same is printed.

121) void main()
{
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented in memory”);
}
Answer
You can answer this if you know how values are represented in memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on 0 to produce all ones to fill the space for an integer. –1 is represented in unsigned value as all 1’s and so both are equal.

122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help understand this.

123) main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b

124) main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is required.

125) main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just works as a visual clue for the reader to see which expression is first evaluated.

126)
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}

main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and returns the type int. By the assignment ptr[0] = aaa; it means that the first function pointer in the array is initialized with the address of the function aaa. Similarly, the other two array elements also get initialized with the addresses of the functions bbb and ccc. Since ptr[2] contains the address of the function ccc, the call to the function ptr[2]() is same as calling ccc(). So it results in printing "bye".

127)
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}

Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of higher precedence than = operator. In the inner expression, ++i is equal to 6 yielding true(1). Hence the result.

128) main()
{
char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”. Since this string becomes the format string for printf and ASCII value of 65 is ‘A’, the same gets printed.

129) void ( * abc( int, void ( *def) () ) ) ();

Answer::
abc is a ptr to a function which takes 2 parameters .(a). an integer variable.(b). a ptrto a funtion which returns void. the return type of the function is void.
Explanation:
Apply the clock-wise rule to find the result.

130) main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So “some” and “some\0” are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.

131) main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not appended automatically to the string. Since str1 doesn’t have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the result.

132) main()
{
int i = 3;
for (;i++=0;) printf(“%d”,i);
}

Answer:
Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and hence it cannot appear on the left hand side of an assignment operation.

133) void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc returns the allocated memory space initialized to zeros.

134) void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”, i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional operator evaluates to false, executing i--. This continues till the integer value rotates to positive value (32767). The while condition becomes false and hence, comes out of the while loop, printing the i value.

135) main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}

Answer:
10 10
Explanation:
The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;

136) 1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.

Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant char )
*a='F' : illegal
a="Hi" : legal

2. 'const' applies to 'a' rather than to the value of a (constant pointer to char )
*a='F' : legal
a="Hi" : illegal

3. Same as 1.

137) main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d %d",i,j);
}

Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result.

138) main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}

Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth value of the expression is not known. j is not equal to zero itself means that the expression’s truth value is 1. Because it is followed by || and true || (anything) => true where (anything) will not be evaluated. So the remaining expression is not evaluated and so the value of i remains the same.
Similarly when && operator is involved in an expression, when any of the operands become false, the whole expression’s truth value becomes false and hence the remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.

139) main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.

140) main()
{
float i=1.5;
switch(i)
{
case 1: printf("1");
case 2: printf("2");
default : printf("0");
}
}
Answer:
Compiler Error: switch expression not integral
Explanation:
Switch statements can be applied only to integral types.

141) main()
{
extern i;
printf("%d\n",i);
{
int i=20;
printf("%d\n",i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern has no use in resolving it.

142) main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes through f1 and f2 ultimately affects only the value of a.

143) main()
{
char *p="GOOD";
char a[ ]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p), strlen(p));
printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p) => sizeof(char*) => 2
sizeof(*p) => sizeof(char) => 1
Similarly,
sizeof(a) => size of the character array => 5
When sizeof operator is applied to an array it returns the sizeof the array and it is not the same as the sizeof the pointer variable. Here the sizeof(a) where a is the character array and the size of the array is 5 because the space necessary for the terminating NULL character should also be taken into account.

144) #define DIM( array, type) sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer array of 10 elements is 10 * sizeof(int). The macro expands to sizeof(arr)/sizeof(int) => 10 * sizeof(int) / sizeof(int) => 10.

145) int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and only the pointers can be passed. So the argument is equivalent to int * array (this is one of the very few places where [] and * usage are equivalent). The return statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in this case.

146) main()
{
static int a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static *p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),
*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
}
Answer:
1 1 1 1
2 4 2 4
3 7 3 7
4 2 4 2
5 5 5 5
6 8 6 8
7 3 7 3
8 6 8 6
9 9 9 9
Explanation:
*(*(p+i)+j) is equivalent to p[i][j].

147) main()
{
void swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments.
This convention is historically due to pre-ANSI style (referred to as Kernighan and Ritchie style) style of function declaration. In that style, the swap function will be defined as follows,
void swap()
int *a, int *b
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
where the arguments follow the (). So naturally the declaration for swap will look like, void swap() which means the swap can take any number of arguments.

148) main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1 1
Explanation:
The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.

149) main()
{
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2 1
Explanation:
The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are ‘small-endian’ machines. Small-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010.

150) main()
{
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer value 300 in binary notation is: 00000001 00101100. It is stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it is 00000010 00101100 => 556.

151) #include
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = (*ptrprintf("%d",least);
}
Answer:
0
Explanation:
After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘\0’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.

152) Declare an array of N pointers to functions returning pointers to functions returning pointers to characters?
Answer:
(char*(*)( )) (*ptr[N])( );

153) main()
{
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Inside the struct definition of ‘student’ the member of type struct date is given. The compiler doesn’t have the definition of date structure (forward reference is not allowed in C in this case) so it issues an error.

154) main()
{
struct date;
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Only declaration of struct date is available inside the structure definition of ‘student’ but to have a variable of type struct date the definition of the structure is required.

155) There were 10 records stored in “somefile.dat” but the following program printed 11 names. What went wrong?
void main()
{
struct student
{
char name[30], rollno[6];
}stud;
FILE *fp = fopen(“somefile.dat”,”r”);
while(!feof(fp))
{
fread(&stud, sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10 records and prints the names successfully. It will return EOF only when fread tries to read another record and fails reading EOF (and returning EOF). So it prints the last record again. After this only the condition feof(fp) becomes false, hence comes out of the while loop.

156) Is there any difference between the two declarations,
1. int foo(int *arr[]) and
2. int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The numbers that are allowed inside the [] is just for more readability. So there is no difference between the two declarations.

157) What is the subtle error in the following code segment?
void fun(int n, int arr[])
{
int *p=0;
int i=0;
while(i++ p = &arr[i];
*p = 0;
}
Answer & Explanation:
If the body of the loop never executes p is assigned no address. So p remains NULL where *p =0 may result in problem (may rise to runtime error “NULL pointer assignment” and terminate the program).

158) What is wrong with the following code?
int *foo()
{
int *s = malloc(sizeof(int)100);
assert(s != NULL);
return s;
}
Answer & Explanation:
assert macro should be used for debugging and finding out bugs. The check s != NULL is for error/exception handling and for that assert shouldn’t be used. A plain if and the corresponding remedy statement has to be given.

159) What is the hidden bug with the following statement?
assert(val++ != 0);
Answer & Explanation:
Assert macro is used for debugging and removed in release version. In assert, the experssion involves side-effects. So the behavior of the code becomes different in case of debug version and the release version thus leading to a subtle bug.
Rule to Remember:
Don’t use expressions that have side-effects in assert statements.

160) void main()
{
int *i = 0x400; // i points to the address 400
*i = 0; // set the value of memory location pointed by i;
}
Answer:
Undefined behavior
Explanation:
The second statement results in undefined behavior because it points to some location whose value may not be available for modification. This type of pointer in which the non-availability of the implementation of the referenced location is known as 'incomplete type'.

161) #define assert(cond) if(!(cond)) \
(fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort())

void main()
{
int i = 10;
if(i==0)
assert(i < 100);
else
printf("This statement becomes else for if in assert macro");
}
Answer:
No output
Explanation:
The else part in which the printf is there becomes the else for if in the assert macro. Hence nothing is printed.
The solution is to use conditional operator instead of if statement,
#define assert(cond) ((cond)?(0): (fprintf (stderr, "assertion failed: \ %s, file %s, line %d \n",#cond, __FILE__,__LINE__), abort()))

Note:
However this problem of “matching with nearest else” cannot be solved by the usual method of placing the if statement inside a block like this,
#define assert(cond) { \
if(!(cond)) \
(fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\
__FILE__,__LINE__), abort()) \
}

162) Is the following code legal?
struct a
{
int x;
struct a b;
}
Answer:
No
Explanation:
Is it not legal for a structure to contain a member that is of the same
type as in this case. Because this will cause the structure declaration to be recursive without end.

163) Is the following code legal?
struct a
{
int x;
struct a *b;
}
Answer:
Yes.
Explanation:
*b is a pointer to type struct a and so is legal. The compiler knows, the size of the pointer to a structure even before the size of the structure
is determined(as you know the pointer to any type is of same size). This type of structures is known as ‘self-referencing’ structure.

164) Is the following code legal?
typedef struct a
{
int x;
aType *b;
}aType
Answer:
No
Explanation:
The typename aType is not known at the point of declaring the structure (forward references are not made for typedefs).

165) Is the following code legal?
typedef struct a aType;
struct a
{
int x;
aType *b;
};
Answer:
Yes
Explanation:
The typename aType is known at the point of declaring the structure, because it is already typedefined.

166) Is the following code legal?
void main()
{
typedef struct a aType;
aType someVariable;
struct a
{
int x;
aType *b;
};
}
Answer:
No
Explanation:
When the declaration,
typedef struct a aType;
is encountered body of struct a is not known. This is known as ‘incomplete types’.

167) void main()
{
printf(“sizeof (void *) = %d \n“, sizeof( void *));
printf(“sizeof (int *) = %d \n”, sizeof(int *));
printf(“sizeof (double *) = %d \n”, sizeof(double *));
printf(“sizeof(struct unknown *) = %d \n”, sizeof(struct unknown *));
}
Answer :
sizeof (void *) = 2
sizeof (int *) = 2
sizeof (double *) = 2
sizeof(struct unknown *) = 2
Explanation:
The pointer to any type is of same size.

168) char inputString[100] = {0};
To get string input from the keyboard which one of the following is better?
1) gets(inputString)
2) fgets(inputString, sizeof(inputString), fp)
Answer & Explanation:
The second one is better because gets(inputString) doesn't know the size of the string passed and so, if a very big input (here, more than 100 chars) the charactes will be written past the input string. When fgets is used with stdin performs the same operation as gets but is safe.

169) Which version do you prefer of the following two,
1) printf(“%s”,str); // or the more curt one
2) printf(str);
Answer & Explanation:
Prefer the first one. If the str contains any format characters like %d then it will result in a subtle bug.

170) void main()
{
int i=10, j=2;
int *ip= &i, *jp = &j;
int k = *ip/*jp;
printf(“%d”,k);
}
Answer:
Compiler Error: “Unexpected end of file in comment started in line 5”.
Explanation:
The programmer intended to divide two integers, but by the “maximum munch” rule, the compiler treats the operator sequence / and * as /* which happens to be the starting of comment. To force what is intended by the programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.

171) void main()
{
char ch;
for(ch=0;ch<=127;ch++)
printf(“%c %d \n“, ch, ch);
}
Answer:
Implementaion dependent
Explanation:
The char type may be signed or unsigned by default. If it is signed then ch++ is executed after ch reaches 127 and rotates back to -128. Thus ch is always smaller than 127.

172) Is this code legal?
int *ptr;
ptr = (int *) 0x400; Answer:
Yes
Explanation:
The pointer ptr will point at the integer in the memory location 0x400.
173) main()
{
char a[4]="HELLO";
printf("%s",a);
}
Answer:
Compiler error: Too many initializers
Explanation:
The array a is of size 4 but the string constant requires 6 bytes to get stored.

174) main()
{
char a[4]="HELL";
printf("%s",a);
}
Answer:
HELL%@!~@!@???@~~!
Explanation:
The character array has the memory just enough to hold the string “HELL” and doesnt have enough space to store the terminating null character. So it prints the HELL correctly and continues to print garbage values till it accidentally comes across a NULL character.

175) main()
{
int a=10,*j;
void *k;
j=k=&a;
j++;
k++;
printf("\n %u %u ",j,k);
}
Answer:
Compiler error: Cannot increment a void pointer
Explanation:
Void pointers are generic pointers and they can be used only when the type is not known and as an intermediate address storage type. No pointer arithmetic can be done on it and you cannot apply indirection operator (*) on void pointers.

176) main()
{
extern int i;
{ int i=20;
{
const volatile unsigned i=30; printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
int i;

177) Printf can be implemented by using __________ list.
Answer:
Variable length argument lists
178) char *someFun()
{
char *temp = “string constant";
return temp;
}
int main()
{
puts(someFun());
}
Answer:
string constant
Explanation:
The program suffers no problem and gives the output correctly because the character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.

179) char *someFun1()
{
char temp[ ] = “string";
return temp;
}
char *someFun2()
{
char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};
return temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
Answer:
Garbage values.
Explanation:
Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.

NETWORKING APTITUDE

2009-03-04T07:06:00.000-08:00

1. What are the two types of transmission technology available?
(i) Broadcast and (ii) point-to-point

2. What is subnet?
A generic term for section of a large networks usually separated by a bridge or router.

3. Difference between the communication and transmission.
Transmission is a physical movement of information and concern issues like bit polarity, synchronisation, clock etc.
Communication means the meaning full exchange of information between two communication media.

4. What are the possible ways of data exchange?
(i) Simplex (ii) Half-duplex (iii) Full-duplex.

5. What is SAP?
Series of interface points that allow other computers to communicate with the other layers of network protocol stack.

6. What do you meant by "triple X" in Networks?
The function of PAD (Packet Assembler Disassembler) is described in a document known as X.3. The standard protocol has been defined between the terminal and the PAD, called X.28; another standard protocol exists between hte PAD and the network, called X.29. Together, these three recommendations are often called "triple X"

7. What is frame relay, in which layer it comes?
Frame relay is a packet switching technology. It will operate in the data link layer.

8. What is terminal emulation, in which layer it comes?
Telnet is also called as terminal emulation. It belongs to application layer.

9. What is Beaconing?
The process that allows a network to self-repair networks problems. The stations on the network notify the other stations on the ring when they are not receiving the transmissions. Beaconing is used in Token ring and FDDI networks.

10. What is redirector?
Redirector is software that intercepts file or prints I/O requests and translates them into network requests. This comes under presentation layer.

11. What is NETBIOS and NETBEUI?
NETBIOS is a programming interface that allows I/O requests to be sent to and received from a remote computer and it hides the networking hardware from applications.
NETBEUI is NetBIOS extended user interface. A transport protocol designed by microsoft and IBM for the use on small subnets.

12. What is RAID?
A method for providing fault tolerance by using multiple hard disk drives.

13. What is passive topology?
When the computers on the network simply listen and receive the signal, they are referred to as passive because they don’t amplify the signal in any way. Example for passive topology - linear bus.

14. What is Brouter?
Hybrid devices that combine the features of both bridges and routers.

15. What is cladding?
A layer of a glass surrounding the center fiber of glass inside a fiber-optic cable.

16. What is point-to-point protocol
A communications protocol used to connect computers to remote networking services including Internet service providers.

17. How Gateway is different from Routers?
A gateway operates at the upper levels of the OSI model and translates information between two completely different network architectures or data formats

18. What is attenuation?
The degeneration of a signal over distance on a network cable is called attenuation.

19. What is MAC address?
The address for a device as it is identified at the Media Access Control (MAC) layer in the network architecture. MAC address is usually stored in ROM on the network adapter card and is unique.

20. Difference between bit rate and baud rate.
Bit rate is the number of bits transmitted during one second whereas baud rate refers to the number of signal units per second that are required to represent those bits.
baud rate = bit rate / N
where N is no-of-bits represented by each signal shift.

21. What is Bandwidth?
Every line has an upper limit and a lower limit on the frequency of signals it can carry. This limited range is called the bandwidth.

22. What are the types of Transmission media?
Signals are usually transmitted over some transmission media that are broadly classified in to two categories.
a) Guided Media:
These are those that provide a conduit from one device to another that include twisted-pair, coaxial cable and fiber-optic cable. A signal traveling along any of these media is directed and is contained by the physical limits of the medium. Twisted-pair and coaxial cable use metallic that accept and transport signals in the form of electrical current. Optical fiber is a glass or plastic cable that accepts and transports signals in the form of light.
b) Unguided Media:
This is the wireless media that transport electromagnetic waves without using a physical conductor. Signals are broadcast either through air. This is done through radio communication, satellite communication and cellular telephony.

23. What is Project 802?
It is a project started by IEEE to set standards to enable intercommunication between equipment from a variety of manufacturers. It is a way for specifying functions of the physical layer, the data link layer and to some extent the network layer to allow for interconnectivity of major LAN
protocols.
It consists of the following:
 802.1 is an internetworking standard for compatibility of different LANs and MANs across protocols.
 802.2 Logical link control (LLC) is the upper sublayer of the data link layer which is non-architecture-specific, that is remains the same for all IEEE-defined LANs.
 Media access control (MAC) is the lower sublayer of the data link layer that contains some distinct modules each carrying proprietary information specific to the LAN product being used. The modules are Ethernet LAN (802.3), Token ring LAN (802.4), Token bus LAN (802.5).
 802.6 is distributed queue dual bus (DQDB) designed to be used in MANs.

24. What is Protocol Data Unit?
The data unit in the LLC level is called the protocol data unit (PDU). The PDU contains of four fields a destination service access point (DSAP), a source service access point (SSAP), a control field and an information field. DSAP, SSAP are addresses used by the LLC to identify the protocol stacks on the receiving and sending machines that are generating and using the data. The control field specifies whether the PDU frame is a information frame (I - frame) or a supervisory frame (S - frame) or a unnumbered frame (U - frame).

25. What are the different type of networking / internetworking devices?
Repeater:
Also called a regenerator, it is an electronic device that operates only at physical layer. It receives the signal in the network before it becomes weak, regenerates the original bit pattern and puts the refreshed copy back in to the link.
Bridges:
These operate both in the physical and data link layers of LANs of same type. They divide a larger network in to smaller segments. They contain logic that allow them to keep the traffic for each segment separate and thus are repeaters that relay a frame only the side of the segment containing the intended recipent and control congestion.
Routers:
They relay packets among multiple interconnected networks (i.e. LANs of different type). They operate in the physical, data link and network layers. They contain software that enable them to determine which of the several possible paths is the best for a particular transmission.
Gateways:
They relay packets among networks that have different protocols (e.g. between a LAN and a WAN). They accept a packet formatted for one protocol and convert it to a packet formatted for another protocol before forwarding it. They operate in all seven layers of the OSI model.

26. What is ICMP?
ICMP is Internet Control Message Protocol, a network layer protocol of the TCP/IP suite used by hosts and gateways to send notification of datagram problems back to the sender. It uses the echo test / reply to test whether a destination is reachable and responding. It also handles both control and error messages.

27. What are the data units at different layers of the TCP / IP protocol suite?
The data unit created at the application layer is called a message, at the transport layer the data unit created is called either a segment or an user datagram, at the network layer the data unit created is called the datagram, at the data link layer the datagram is encapsulated in to a frame and finally transmitted as signals along the transmission media.

28. What is difference between ARP and RARP?
The address resolution protocol (ARP) is used to associate the 32 bit IP address with the 48 bit physical address, used by a host or a router to find the physical address of another host on its network by sending a ARP query packet that includes the IP address of the receiver.
The reverse address resolution protocol (RARP) allows a host to discover its Internet address when it knows only its physical address.

29. What is the minimum and maximum length of the header in the TCP segment and IP datagram?
The header should have a minimum length of 20 bytes and can have a maximum length of 60 bytes.

30. What is the range of addresses in the classes of internet addresses?
Class A 0.0.0.0 - 127.255.255.255
Class B 128.0.0.0 - 191.255.255.255
Class C 192.0.0.0 - 223.255.255.255
Class D 224.0.0.0 - 239.255.255.255
Class E 240.0.0.0 - 247.255.255.255

31. What is the difference between TFTP and FTP application layer protocols?
The Trivial File Transfer Protocol (TFTP) allows a local host to obtain files from a remote host but does not provide reliability or security. It uses the fundamental packet delivery services offered by UDP.
The File Transfer Protocol (FTP) is the standard mechanism provided by TCP / IP for copying a file from one host to another. It uses the services offer by TCP and so is reliable and secure. It establishes two connections (virtual circuits) between the hosts, one for data transfer and another for control information.

32. What are major types of networks and explain?
 Server-based network
 Peer-to-peer network
Peer-to-peer network, computers can act as both servers sharing resources and as clients using the resources.
Server-based networks provide centralized control of network resources and rely on server computers to provide security and network administration

33. What are the important topologies for networks?
 BUS topology:
In this each computer is directly connected to primary network cable in a single line.
Advantages:
Inexpensive, easy to install, simple to understand, easy to extend.

 STAR topology:
In this all computers are connected using a central hub.
Advantages:
Can be inexpensive, easy to install and reconfigure and easy to trouble shoot physical problems.

 RING topology:
In this all computers are connected in loop.
Advantages:
All computers have equal access to network media, installation can be simple, and signal does not degrade as much as in other topologies because each computer regenerates it.

34. What is mesh network?
A network in which there are multiple network links between computers to provide multiple paths for data to travel.

35. What is difference between baseband and broadband transmission?
In a baseband transmission, the entire bandwidth of the cable is consumed by a single signal. In broadband transmission, signals are sent on multiple frequencies, allowing multiple signals to be sent simultaneously.

36. Explain 5-4-3 rule?
In a Ethernet network, between any two points on the network ,there can be no more than five network segments or four repeaters, and of those five segments only three of segments can be populated.

37. What MAU?
In token Ring , hub is called Multistation Access Unit(MAU).

38. What is the difference between routable and non- routable protocols?
Routable protocols can work with a router and can be used to build large networks. Non-Routable protocols are designed to work on small, local networks and cannot be used with a router

39. Why should you care about the OSI Reference Model?
It provides a framework for discussing network operations and design.

40. What is logical link control?
One of two sublayers of the data link layer of OSI reference model, as defined by the IEEE 802 standard. This sublayer is responsible for maintaining the link between computers when they are sending data across the physical network connection.

41. What is virtual channel?
Virtual channel is normally a connection from one source to one destination, although multicast connections are also permitted. The other name for virtual channel is virtual circuit.

42. What is virtual path?
Along any transmission path from a given source to a given destination, a group of virtual circuits can be grouped together into what is called path.

43. What is packet filter?
Packet filter is a standard router equipped with some extra functionality. The extra functionality allows every incoming or outgoing packet to be inspected. Packets meeting some criterion are forwarded normally. Those that fail the test are dropped.

44. What is traffic shaping?
One of the main causes of congestion is that traffic is often busy. If hosts could be made to transmit at a uniform rate, congestion would be less common. Another open loop method to help manage congestion is forcing the packet to be transmitted at a more predictable rate. This is called traffic shaping.

45. What is multicast routing?
Sending a message to a group is called multicasting, and its routing algorithm is called multicast routing.

46. What is region?
When hierarchical routing is used, the routers are divided into what we will call regions, with each router knowing all the details about how to route packets to destinations within its own region, but knowing nothing about the internal structure of other regions.

47. What is silly window syndrome?
It is a problem that can ruin TCP performance. This problem occurs when data are passed to the sending TCP entity in large blocks, but an interactive application on the receiving side reads 1 byte at a time.

48. What are Digrams and Trigrams?
The most common two letter combinations are called as digrams. e.g. th, in, er, re and an. The most common three letter combinations are called as trigrams. e.g. the, ing, and, and ion.

49. Expand IDEA.
IDEA stands for International Data Encryption Algorithm.

50. What is wide-mouth frog?
Wide-mouth frog is the simplest known key distribution center (KDC) authentication protocol.

51. What is Mail Gateway?
It is a system that performs a protocol translation between different electronic mail delivery protocols.

52. What is IGP (Interior Gateway Protocol)?
It is any routing protocol used within an autonomous system.

53. What is EGP (Exterior Gateway Protocol)?
It is the protocol the routers in neighboring autonomous systems use to identify the set of networks that can be reached within or via each autonomous system.

54. What is autonomous system?
It is a collection of routers under the control of a single administrative authority and that uses a common Interior Gateway Protocol.

55. What is BGP (Border Gateway Protocol)?
It is a protocol used to advertise the set of networks that can be reached with in an autonomous system. BGP enables this information to be shared with the autonomous system. This is newer than EGP (Exterior Gateway Protocol).

56. What is Gateway-to-Gateway protocol?
It is a protocol formerly used to exchange routing information between Internet core routers.

57. What is NVT (Network Virtual Terminal)?
It is a set of rules defining a very simple virtual terminal interaction. The NVT is used in the start of a Telnet session.

58. What is a Multi-homed Host?
It is a host that has a multiple network interfaces and that requires multiple IP addresses is called as a Multi-homed Host.

59. What is Kerberos?
It is an authentication service developed at the Massachusetts Institute of Technology. Kerberos uses encryption to prevent intruders from discovering passwords and gaining unauthorized access to files.

60. What is OSPF?
It is an Internet routing protocol that scales well, can route traffic along multiple paths, and uses knowledge of an Internet's topology to make accurate routing decisions.

61. What is Proxy ARP?
It is using a router to answer ARP requests. This will be done when the originating host believes that a destination is local, when in fact is lies beyond router.

62. What is SLIP (Serial Line Interface Protocol)?
It is a very simple protocol used for transmission of IP datagrams across a serial line.

63. What is RIP (Routing Information Protocol)?
It is a simple protocol used to exchange information between the routers.

64. What is source route?
It is a sequence of IP addresses identifying the route a datagram must follow. A source route may optionally be included in an IP datagram header.

C++ Aptitude Questions

2009-03-04T06:49:00.001-08:00

1. Base class has some virtual method and derived class has a method with the same name. If we initialize the base class pointer with derived object,. calling of that virtual method will result in which method being called?

a. Base method
b. Derived method..

Ans. b

2. For the following C program

#define AREA(x)(3.14*x*x)
main()
{float r1=6.25,r2=2.5,a;
a=AREA(r1);
printf("\n Area of the circle is %f", a);
a=AREA(r2);
printf("\n Area of the circle is %f", a);
}

What is the output?

Ans. Area of the circle is 122.656250
Area of the circle is 19.625000

3. What do the following statements indicate. Explain.
• int(*p)[10]
• int*f()
• int(*pf)()
• int*p[10]

Refer to:
-- Kernighan & Ritchie page no. 122
-- Schaum series page no. 323

4.
void main()
{
int d=5;
printf("%f",d);
}

Ans: Undefined

5.
void main()
{
int i;
for(i=1;i<4,i++)
switch(i)
case 1: printf("%d",i);break;
{
case 2:printf("%d",i);break;
case 3:printf("%d",i);break;
}
switch(i) case 4:printf("%d",i);
}

Ans: 1,2,3,4

6.
void main()
{
char *s="\12345s\n";
printf("%d",sizeof(s));
}

Ans: 6

7.
void main()
{
unsigned i=1; /* unsigned char k= -1 => k=255; */
signed j=-1; /* char k= -1 => k=65535 */
/* unsigned or signed int k= -1 =>k=65535 */
if(iprintf("less");
else
if(i>j)
printf("greater");
else
if(i==j)
printf("equal");
}

Ans: less

8.
void main()
{
float j;
j=1000*1000;
printf("%f",j);
}

1. 1000000
2. Overflow
3. Error
4. None

Ans: 4

9. How do you declare an array of N pointers to functions returning
pointers to functions returning pointers to characters?

Ans: The first part of this question can be answered in at least
three ways:

1. char *(*(*a[N])())();

2. Build the declaration up incrementally, using typedefs:

typedef char *pc; /* pointer to char */
typedef pc fpc(); /* function returning pointer to char */
typedef fpc *pfpc; /* pointer to above */
typedef pfpc fpfpc(); /* function returning... */
typedef fpfpc *pfpfpc; /* pointer to... */
pfpfpc a[N]; /* array of... */

3. Use the cdecl program, which turns English into C and vice versa:

cdecl> declare a as array of pointer to function returning pointer to function returning pointer to char char *(*(*a[])())()

cdecl can also explain complicated declarations, help with casts, and indicate which set of parentheses the arguments go in (for complicated function definitions, like the one above).
Any good book on C should explain how to read these complicated C declarations "inside out" to understand them ("declaration mimics use").
The pointer-to-function declarations in the examples above have not included parameter type information. When the parameters have complicated types, declarations can *really* get messy.
(Modern versions of cdecl can help here, too.)

10. A structure pointer is defined of the type time . With 3 fields min,sec hours having pointers to integers. Write the way to initialize the 2nd element to 10.

11. In the above question an array of pointers is declared.Write the statement to initialize the 3rd element of the 2 element to 10

12.
int f()
void main()
{
f(1);
f(1,2);
f(1,2,3);
}
f(int i,int j,int k)
{
printf("%d %d %d",i,j,k);
}

What are the number of syntax errors in the above?

Ans: None.

13.
void main()
{
int i=7;
printf("%d",i++*i++);
}

Ans: 56

14.
#define one 0
#ifdef one
printf("one is defined ");
#ifndef one
printf("one is not defined ");

Ans: "one is defined"

15.
void main()
{
int count=10,*temp,sum=0;
temp=&count;
*temp=20;
temp=∑
*temp=count;
printf("%d %d %d ",count,*temp,sum);
}

Ans: 20 20 20

16. what is alloca()
Ans : It allocates and frees memory after use/after getting out of scope

17.
main()
{
static i=3;
printf("%d",i--);
return i>0 ? main():0;
}

Ans: 321

18.
char *foo()
{
char result[100]);
strcpy(result,"anything is good");
return(result);
}
void main()
{
char *j;
j=foo()
printf("%s",j);
}

Ans: anything is good.

19.
void main()
{
char *s[]={ "dharma","hewlett-packard","siemens","ibm"};
char **p;
p=s;
printf("%s",++*p);
printf("%s",*p++);
printf("%s",++*p);
}
Ans: "harma" (p->add(dharma) && (*p)->harma)
"harma" (after printing, p->add(hewlett-packard) &&(*p)->harma)
"ewlett-packard"
20. Output of the following program is
main()
{int i=0;
for(i=0;i<20;i++)
{switch(i)
case 0:i+=5;
case 1:i+=2;
case 5:i+=5;
default i+=4;
break;}
printf("%d,",i);
}
}
a) 0,5,9,13,17
b) 5,9,13,17
c) 12,17,22
d) 16,21
e) Syntax error
Ans. (d)
21. What is the ouptut in the following program
main()
{char c=-64;
int i=-32
unsigned int u =-16;
if(c>i)
{printf("pass1,");
if(cprintf("pass2");
else
printf("Fail2");
}
else
printf("Fail1);
if(iprintf("pass2");
else
printf("Fail2")
}
a) Pass1,Pass2
b) Pass1,Fail2
c) Fail1,Pass2
d) Fail1,Fail2
e) None of these
Ans. (c)
22. What will the following program do?
void main()
{
int i;
char a[]="String";
char *p="New Sring";
char *Temp;
Temp=a;
a=malloc(strlen(p) + 1);
strcpy(a,p); //Line number:9//
p = malloc(strlen(Temp) + 1);
strcpy(p,Temp);
printf("(%s, %s)",a,p);
free(p);
free(a);
} //Line number 15//
a) Swap contents of p & a and print:(New string, string)
b) Generate compilation error in line number 8
c) Generate compilation error in line number 5
d) Generate compilation error in line number 7
e) Generate compilation error in line number 1
Ans. (b)
23. In the following code segment what will be the result of the function,
value of x , value of y
{unsigned int x=-1;
int y;
y = ~0;
if(x == y)
printf("same");
else
printf("not same");
}
a) same, MAXINT, -1
b) not same, MAXINT, -MAXINT
c) same , MAXUNIT, -1
d) same, MAXUNIT, MAXUNIT
e) not same, MAXINT, MAXUNIT
Ans. (a)
24. What will be the result of the following program ?
char *gxxx()
{static char xxx[1024];
return xxx;
}
main()
{char *g="string";
strcpy(gxxx(),g);
g = gxxx();
strcpy(g,"oldstring");
printf("The string is : %s",gxxx());
}
a) The string is : string
b) The string is :Oldstring
c) Run time error/Core dump
d) Syntax error during compilation
e) None of these
Ans. (b)
25. Find the output for the following C program
main()
{
char *p1="Name";
char *p2;
p2=(char *)malloc(20);
while(*p2++=*p1++);
printf("%s\n",p2);
}
Ans. An empty string
26. Find the output for the following C program
main()
{
int x=20,y=35;
x = y++ + x++;
y = ++y + ++x;
printf("%d %d\n",x,y);
}
Ans. 57 94
27. Find the output for the following C program
main()
{
int x=5;
printf("%d %d %d\n",x,x<<2,x>>2);
}
Ans. 5 20 1
28 Find the output for the following C program
#define swap1(a,b) a=a+b;b=a-b;a=a-b;
main()
{
int x=5,y=10;
swap1(x,y);
printf("%d %d\n",x,y);
swap2(x,y);
printf("%d %d\n",x,y);
}
int swap2(int a,int b)
{
int temp;
temp=a;
b=a;
a=temp;
return;
}
Ans. 10 5

29 Find the output for the following C program
main()
{
char *ptr = "Ramco Systems";
(*ptr)++;
printf("%s\n",ptr);
ptr++;
printf("%s\n",ptr);
}
Ans. Samco Systems

30 Find the output for the following C program
#include
main()
{
char s1[]="Ramco";
char s2[]="Systems";
s1=s2;
printf("%s",s1);
}
Ans. Compilation error giving it cannot be an modifiable 'lvalue'

31 Find the output for the following C program
#include
main()
{
char *p1;
char *p2;
p1=(char *) malloc(25);
p2=(char *) malloc(25);
strcpy(p1,"Ramco");
strcpy(p2,"Systems");
strcat(p1,p2);
printf("%s",p1);
}
Ans. RamcoSystems

32. Find the output for the following C program given that
[1]. The following variable is available in file1.c
static int average_float;
Ans. All the functions in the file1.c can access the variable

33. Find the output for the following C program
# define TRUE 0
some code
while(TRUE)
{
some code
}
Ans. This won't go into the loop as TRUE is defined as 0
34. struct list{
int x;
struct list *next;
}*head;

the struct head.x =100

Is the above assignment to pointer is correct or wrong ?
Ans. Wrong
35.What is the output of the following ?

int i;
i=1;
i=i+2*i++;
printf(%d,i);
Ans. 4
36. FILE *fp1,*fp2;

fp1=fopen("one","w")
fp2=fopen("one","w")
fputc('A',fp1)
fputc('B',fp2)
fclose(fp1)
fclose(fp2)
}

Find the Error, If Any?
Ans. no error. But It will over writes on same file.
37. What are the output(s) for the following ?
38. #include
char *f()
{char *s=malloc(8);
strcpy(s,"goodbye");
}

main()
{
char *f();
printf("%c",*f()='A'); }

39. #define MAN(x,y) (x)>(y)?(x):(y)
{int i=10;
j=5;
k=0;
k=MAX(i++,++j);
printf(%d %d %d %d,i,j,k);
}
Ans. 10 5 0
40.
void main()
{
int i=7;
printf("%d",i++*i++);
}

Ans: 56

APTITUDE TEST I

2009-03-02T07:43:00.000-08:00

1. The ages of two persons differ by 16 years. If 6 years ago, the elder one be 3 times as old as the younger one, find their present ages

Ans.)14,30

2. A group consists of equal number of men and women. Of them 10% of men and 45% of women are unemployed. If a person is randomly selected from the group, find the probability for the selected person to be an employee.

Ans.)29/40

3. Sandy's chain of used car dealership sold 16,400 cars in 2007. If the chain sold 15,744 cars in 2008, by what percent did the number of cars sold decrease?

Ans.)4%

4. A radio when sold at a certain price gives a gain of 20%. What will be the gain percent, if sold for thrice the price?

Ans.)260

5. Vivek travelled 1200km by air which formed 2/5 of his trip. One third of the whole trip, he travelled by car and the rest of the journey he performed by train. The distance travelled by train was???

Ans.)800

6. Given that on 27th February 2003 is Thursday. What was the day on 27th February 1603?

Ans.)Thursday

7. Finding the wrong term in the given series

7, 28, 63, 124, 215, 342, 511

Ans.)28

8. + means * and * means / and / means + what is the value of these. 2+3*5/7.

Ans.) 8.2

9. 4_4_4_4 = 20 . Replace the blanks by +,-,/,*.

Ans.) / + *

10. (998-1)(998-2)(998-3)…………..(998-n) = _____ when n>1000

Ans.) 0

11. If x=y=2z and xyz=256 then what is the value of x?

Ans.) 8

12. Thirty men take 20 days to complete a job working 9 hours a day. How many hour a day should 40 men work to complete the job?

Ans.) 7.5

13. If O.T.T.F.F.S.S.E.N.T Is the sequence, predict its next sequence of ten letters.

Ans.)E.T.T.F.F.S.S.E.N.T

14. If the operation,^ is defined by the equation x ^ y = 2x + y,what is the value of a in 2 ^ a = a ^ 3

Ans.)1

15. There are 200 questions on a 3 hr examination.Among these questions are 50 mathematics problems.It is suggested that twice as much time be spent on each maths problem as for each other question. How many minutes should be spent on mathematics problems

Ans.) 72

16. A student divided a number by 2/3 when he required to multiply by 3/2. Calculate the percentage of error in his result.

Ans.) 0

17. If the value of x lies between 0 & 1 which of the following is the largest?

Ans.)1/x

18. A man walks east and turns right and then from there to his left and then 45degrees to his right. In which direction did he go ?

Ans.) North West

19. In objective test a correct ans score 4 marks and on a wrong ans 2 marks are ---. a student score 480 marks from 150 question. how many ans were correct?

Ans.)130

20. A man spends half of his salary on household expenses, 1/4th for rent, 1/5th for travel expenses, the man deposits the rest in a bank. If his monthly deposits in the bank amount 50, what is his monthly salary ?

Ans.)Rs.1000