ISRO-2007

Question 1
The Boolean expression Y = (A + B' + A'B)C' is given by
A
AC'
B
BC'
C
C'
D
A
       Digital-Logic-Design       Boolean-Algebra
Question 1 Explanation: 
Y = (A + B' + A'B)C'
Y = AC' + B'C' + A'BC'
Y = (A + A'B)C' + B'C'
Y = (A + B)C' + B'C'
Y = AC' + BC' + B'C'
Y = AC' + C'(B + B') → B + B' = 1
Y = AC' + C'
Y = C'
Question 2
The circuit shown in the following figure realizes the function
A
(( A + B )’ +C ) ( D’E’ ))
B
(( A + B )’ + C ) ( DE’ ))
C
( A + ( B + C )’ ) ( D’E )
D
( A + B + C’ ) ( D’E’ )
       Digital-Logic-Design       Combinational-Circuit
Question 2 Explanation: 
The given function is equivalent to the following expression:
Y = (((A + B)' + C)' + ((D + E)')')'
Y = ((A + B)' + C)')' . (D + E)'
Y = ((A + B)' + C) (D'E')
Question 3
The circuit shown in the given figure is a
A
full adder
B
full subtractor
C
shift register
D
decade counter
       Digital-Logic-Design       Combinational-Circuit
Question 3 Explanation: 
The above diagram is full subtractor. The equation is D=X⊕Y⊕Bin and Bout=X'Bin+X'Y+YBin
Question 4
When two numbers are added in excess-3 code and the sum is less than 9, then in order to get the correct answer it is necessary to
A
subtract 0011 from the sum
B
add 0011 to the sum
C
subtract 0110 from the sum
D
add 0110 to the sum
       Digital-Logic-Design       Number-Systems
Question 4 Explanation: 
Subtract 0011 if there is no carry otherwise add 0011.
Example:
x+3
y+3
-------
(x+y+6)
Here, sum is excess-6. Hence, subtract 0011 to make it excess-3.
Question 5
The characteristic equation of an SR flip-flop is given by
A
Qn+1 = S + RQn
B
Qn+1= RQn + SQn
C
Qn+1= S + RQn
D
Qn+1 = S + RQn
       Digital-Logic-Design       Sequential-Circuits
Question 5 Explanation: 
The characteristic table of an SR flip-flop is:

So, by simplifying using k-maps:
characteristic equation of an SR flip-flop = Qn+1 = S + RQn
Question 6
A graph with n vertices and n-1 edges that is not a tree, is
A
Connected
B
Disconnected
C
Euler
D
A circuit
       Engineering-Mathematics       Graph-Theory
Question 6 Explanation: 
Consider a graph with two nodes(n1&n2) and number of edges are 1, There may be chance self edge with node n1 then graph is disconnected.
Consider the graph with three nodes(n1,n2&n3) and has 2 edges.
n1-->n2 and n2--->n1 then the graph is disconnected.
Question 7
If a graph requires k different colours for its proper colouring, then the chromatic number of the graph is
A
1
B
k
C
k-1
D
k/2
       Engineering-Mathematics       Graphs
Question 7 Explanation: 
The chromatic number of a graph is the smallest number of colors needed to color the vertices of so that no two adjacent vertices share the same color and if a graph requires k different colours for its proper colouring, then k is the chromatic number of the graph.
Question 8
A read bit can be read
A
and written by CPU
B
and written by peripheral
C
by peripheral and written by CPU
D
by CPU and written by the peripheral
       Computer-Organization       RAID
Question 8 Explanation: 
The read and write functionality depends on the type of microcontroller peripheral. Generally, the status bits have a read only status and can be modified by peripherals only. So, read bit can be read by CPU and written by the peripheral.
Question 9
The term ‘aging’ refers to
A
booting up the priority of the process in multi-level of queue without feedback
B
keeping track of the following a page has been in memory for the purpose of LRU replacement
C
letting job reside in memory for a certain amount of time so that the number of pages required can be estimated accurately
D
gradually increasing the priority of jobs that wait in the system for a long time to remedy infinite blocking
E
In Operating systems, aging is a scheduling technique used to avoid starvation. In this, the priority of the jobs that have a longer waiting time is increased as compared to the newer processes, to avoid the starvation of older processes
       Operating-Systems       Memory-Management
Question 10
Consider a set of n tasks with known runtimes r1, r2….rn to be run on a uniprocessor machine. Which of the following processor scheduling algorithms will result in the maximum throughput?
A
Round Robin
B
Shortest job first
C
Highest response ratio next
D
first come first served
       Operating-Systems       CPU-Scheduling
Question 10 Explanation: 
Throughput means total number of tasks executed per unit time i.e. sum of waiting time and burst time.
Shortest job first scheduling is a scheduling policy that selects the waiting process with the smallest execution time to execute next.
Thus, in shortest job first scheduling, shortest jobs are executed first. This means CPU utilization is maximum. So, maximum number of tasks are completed.
Question 11
Consider a job scheduling problem with 4 jobs J1, J2, J3, J4 and with corresponding deadlines: ( d1, d2, d3, d4) = (4, 2, 4, 2).
Which of the following is not a feasible schedule without violating any job schedule?
A
J2, J4, J1, J3
B
J4, J1, J2, J3.
C
J4, J2, J1, J3.
D
J4, J2, J3, J1
       Operating-Systems       Deadlock
Question 11 Explanation: 
→ Feasible schedule is completing all the jobs within deadline.
→ From the dead line, we can deduce that Job J2 & J4 will complete by time “2” whereas remaining two requires time “4”.
→ So the order of completion of Jobs are Either J2 or J4 and followed by either J1 or J3.
From the given options , Option A,C & D gives the solution because after completion of Jobs J2 and J4 then only jobs J1 and J3 is going to complete.
→ But option B , order of completing jobs is J4,J1,J2 ,J3 which is not possible and it is not feasible schedule
Question 12
By using an eight-bit optical encoder the degree of resolution that can be obtained is (approximately)
A
1.8o
B
3.4o
C
2.8o
D
1.4o
       Digital-Logic-Design       Combinational-Circuit
Question 12 Explanation: 
An optical encoder is an electromechanical device which has an electrical output in digital form proportional to the angular position of the input shaft.
Optical encoders enable an angular displacement to be converted directly into a digital form.
Encoder resolution is often referred to in bits, which are binary units: a 16 bit resolution rotary encoder will have 65,536 (216) increments per turn, or PPR.
In the given question, 8-bit optical encoder will have 28 increments Resolution = 360/2n = 360/28 = 1.4o
Question 13
The principal of the locality of reference justifies the use of
A
virtual memory
B
interrupts
C
main memory
D
cache memory
       Computer-Organization       Cache
Question 13 Explanation: 
Spatial Locality of reference – this says that there is chance that element will be present in the close proximity to the reference point and next time if again searched then more close proximity to the point of reference.
Temporal Locality of reference – In this Least recently used algorithm will be used. Whenever there is page fault occurs within word will not only load word in main memory but complete page fault will be loaded because spatial locality of reference rule says that if you are referring any word next word will be referred in
its register that’s why we load complete page table so complete block will be loaded.
Principle of locality of reference justifies the use of cache.
Question 14
Consider the following pseudo-code
x:=1;
i:=1;
while (x <= 1000)
begin
 x:=2^x;
 i:=i+1;
end;
What is the value of i at the end of the pseudo-code?
A
4
B
5
C
6
D
7
       Programming       Control Flow
Question 14 Explanation: 
Initialisation: x = 1, i = 1;
Loop: x i
21 2
22 3
24 4
216 5
After this condition becomes false.
Question 15
The five items: A, B, C, D, and E are pushed in a stack, one after other starting from A. The stack is popped four items and each element is inserted in a queue. The two elements are deleted from the queue and pushed back on the stack. Now one item is popped from the stack. The popped item is
A
A
B
B
C
C
D
D
       Data-Structures       Queues-and-Stacks
Question 15 Explanation: 
When five items: A, B, C, D, and E are pushed in a stack: Order of stack becomes: A, B, C, D, and E (A at the bottom and E at the top.) stack is popped four items and each element is inserted in a queue: Order of queue: B, C, D, E (B at rear and E at the front) Order of stack after pop operations = A. Two elements deleted from the queue and pushed back stack: New order of stack = A, E, D(A at the bottom, D at the top) As D is on the top so when pop operation occurs D will be popped out. So, correct option is (D).
Question 16
Round Robin schedule is essentially the preemptive version of
A
FIFO
B
Shortest job first
C
Shortest remaining time
D
Longest remaining time
       Operating-Systems       CPU-Scheduling
Question 16 Explanation: 
FIFO is when implemented in preemptive version, it acts like round-robin algorithm.
Question 17
The number of digit 1 present in the binary representation of 3 × 512 + 7 × 64 + 5 × 8 + 3
A
8
B
9
C
10
D
12
       Digital-Logic-Design       Number-Systems
Question 17 Explanation: 
3 × 512 + 7 × 64 + 5 × 8 + 3
= (2 + 1)× 512 + (4 + 2 + 1)× 64 + (4 + 1)× 8 + 2 + 1
= 1024 + 512 + 64 x 4 + 64 x 2 + 64 + 32 + 8 + 2 + 1
= 1024 + 512 + 256 + 128 + 64 + 32 + 8 + 2 + 1
As 1024 has ten 0’s followed by 1, 512 has nine 0’s followed by 1 and so on..
So, the expression will contain total nine 1’s and will be be represented as 11111101011.
Question 18
Assume that each character code consists of 8 bits. The number of characters that can be transmitted per second through an synchronous serial line at 2400 baud rate, and with two stop bits is
A
109
B
216
C
300
D
219
       Computer-Organization       Synchronous-and-asynchronous-Communication
Question 18 Explanation: 
Synchronous communication requires that the clocks in the transmitting and receiving devices are synchronized – running at the same rate – so the receiver can sample the signal at the same time intervals used by the transmitter. No start or stop bits are required. For this reason “synchronous communication permits more information to be passed over a circuit per unit time.
2400 baud means that the serial port is capable of transferring a maximum of 2400 bits per second.
Number of 8-bit characters that can be transmitted per second = 2400/8 = 300
Question 19
If the bandwidth of a signal is 5 kHz and the lowest frequency is 52 kHz, what is the highest frequency
A
5 kHz
B
10 kHz
C
47 kHz
D
57 kHz
       Computer-Networks       Bandwidth-Frequency
Question 19 Explanation: 
Bandwidth = Highest frequency - Lowest frequency.
Highest frequency = Bandwidth+ Lowest Frequency= 5kHz + 52kHz = 57kHz.
Question 20
An Ethernet hub
A
functions as a repeater
B
connects to a digital PBX
C
connects to a token-ring network
D
functions as a gateway
       Computer-Networks       Ethernet
Question 20 Explanation: 
A Hub is a network hardware device for connecting multiple Ethernet devices together and making them act as a single network segment.
A repeater is an electronic device that receives a signal and re-transmits it. Repeaters are used to extend transmissions so that the signal can cover longer distances or be received on the other side of an obstruction.
An Ethernet Hub or a Repeater Hub also acts as a repeater and regenerates the signal to avoid its loss.
Question 21
Phase transition for each bit are used in
A
Amplitude modulation
B
Carrier modulation
C
Manchester encoding
D
NRZ encoding
       Computer-Networks       Ethernet
Question 21 Explanation: 
In the Manchester encoding , a logic 0 is indicated by a 0 to 1 transition at the centre of the bit and a logic 1 is indicated by a 1 to 0 transition at the centre of the bit. Note that signal transitions do not always occur at the ‘bit boundaries’ (the division between one bit and another), but that there is always a transition at the centre of each bit.
Question 22
Study the following program: //precondition: x>=0 public void demo(int x) { System.out.print(x % 10); if (x % 10 != 0) { demo(x/10); } System.out.print(x%10); } Which of the following is printed as a result of the call demo(1234)?
A
1441
B
3443
C
12344321
D
43211234
       Programming       Functions
Question 22 Explanation: 
demo(1234) ---> First it will print “4” then 4!=10 condition is true again it call demo(123)
demo(123)------> It will print “3” then 3!=10 condition is true again it call demo(12)
demo(12)------> It will print “2” then 2!=10 condition is true again it call demo(1)
demo(1)------> It will print “1” then 1!=10 condition is true again it call demo(0)
It will print again 1,2,3 and 4 as there is another Print statement is in each function call.
Question 23
Bit stuffing refers to
A
inserting a 0 in user stream to differentiate it with a flag
B
inserting a 0 in flag stream to avoid ambiguity
C
appending a nipple to the flag sequence
D
appending a nipple to the use data stream
       Computer-Networks       Ethernet
Question 23 Explanation: 
→ Bit stuffing is most easily described as insertion of a 0 bit after a long run of 1 bits.
→ In SDLC the transmitted bit sequence "01111110" containing six adjacent 1 bits is the Flag byte.
→ Bit stuffing ensures that this pattern can never occur in normal data, so it can be used as a marker for the beginning and end of frame without any possibility of being confused with normal data.
→ Bit stuffing is the insertion of non information bits into data. stuffed bits should not be confused with overhead bits. Overhead bits are non-data bits that are necessary for transmission (usually as part of headers, checksums etc).
Question 24
What is the name of the technique in which the operating system of a computer executes several programs concurrently by switching back and forth between them?
A
Partitioning
B
Multi-tasking
C
Windowing
D
Paging
       Operating-Systems       Multiprogramming
Question 24 Explanation: 
In a multitasking system, a computer executes several programs simultaneously by switching them back and forth to increase the user interactivity. Processes share the CPU and execute in an interleaving manner. This allows the user to run more than one program at a time.
Question 25
If there are five routers and six networks in intranet using link state routing, how many routing tables are there?
A
1
B
5
C
6
D
11
       Computer-Networks       Routing
Question 25 Explanation: 
Routers have routing tables to transmit packets selectively to other networks. These routers apply shortest path algorithms to choose the links on which, a packet is to be forwarded, so that it can reach the destination in covering minimum number of hops.
If there are 5 routers, then there should be 5 routing tables as well.
Question 26
Virtual memory is
A
Part of Main Memory only used for swapping
B
A technique to allow a program, of size more than the size of main memory, to run
C
Part of secondary storage used in program execution
D
None of these
       Operating-Systems       Virtual Memory
Question 26 Explanation: 
A computer can address more memory than the amount physically installed on the system. This extra memory is actually called virtual memory and it is a section of a hard disk that's set up to emulate the computer's RAM.
The main visible advantage of this scheme is that programs can be larger than physical memory. Virtual memory serves two purposes. First, it allows us to extend the use of physical memory by using disk. Second, it allows us to have memory protection, because each virtual address is translated to a physical address
Question 27
The level of aggregation of information required for operational control is
A
Detailed
B
Aggregate
C
Qualitative
D
None of the above
       Algorithms       Aggregation
Question 27 Explanation: 
Question 28
0.75 decimal system is equivalent to ____ in octal system
A
0.60
B
0.52
C
0.54
D
0.50
       Digital-Logic-Design       Number-Systems
Question 28 Explanation: 
0.75 = (0.110)2
= (0.6)8
Option (A) is correct.
Question 29
In an SR latch made by cross-coupling two NAND gates, if both S and R inputs are set to 0, then it will result in
A
Q = 0, Q’ = 1
B
Q = 1, Q’ = 0
C
Q = 1, Q’ = 1
D
Indeterminate states
       Digital-Logic-Design       Sequential-Circuits
Question 29 Explanation: 
Question 30
Identify the correct translation into logical notation of the following assertion. Some boys in the class are taller than all the girls Note: taller (x, y) is true if x is taller than y.
A
(∃x)(boy(x) → (∀y)(girl(y) ∧ taller(x, y)))
B
(∃x)(boy(x) ∧ (∀y)(girl(y) ∧ taller(x, y)))
C
(∃x)(boy(x) → (∀y)(girl(y) → taller(x, y)))
D
(∃x)(boy(x) ∧ (∀y)(girl(y) → taller(x, y)))
       Engineering-Mathematics       Propositional-Logic
Question 30 Explanation: 
Don't confuse with '∧' and '→'
'∧' → predicts statements are always true, no matter the value of x.
'→' → predicts there is no need of left predicate to be true always, but whenever it becomes true, then right predicate must be true.
Option D:
There exists a some boys who are taller than of all girls y.
Question 31
Ring counter is analogous to
A
Toggle Switch
B
Latch
C
Stepping Switch
D
S-R flip flop
       Digital-Logic-Design       Sequential-Circuits
Question 31 Explanation: 
→ A ring counter is a type of counter composed of flip-flops connected into a shift register, with the output of the last flip-flop fed to the input of the first, making a "circular" or "ring" structure.
There are two types of ring counters:
1. A straight ring counter, also known as a one-hot counter, connects the output of the last shift register to the first shift register input and circulates a single one (or zero) bit around the ring.
2. A twisted ring counter, also called switch-tail ring counter, walking ring counter, Johnson counter, or Möbius counter, connects the complement of the output of the last shift register to the input of the first register and circulates a stream of ones followed by zeros around the ring.
Note: Ring counter is analogous to Stepping Switch
Question 32
Digital-Logic-Design
A
0.1 and 5V
B
0.6 and 3.5 V
C
0.9 and 1.75 V
D
-1.75 and 0.9 V
       Digital-Logic-Design       TTL
Question 32 Explanation: 
→ TTL high signal would be 5.00 volts exactly, and a TTL low signal 0.00 volts exactly.
→ However, real TTL gate circuits cannot output such perfect voltage levels, and are designed to accept “high” and “low” signals deviating substantially from these ideal values.
Question 33
Consider a computer system that stores a floating-point numbers with 16-bit mantissa and an 8-bit exponent, each in two’s complement. The smallest and largest positive values which can be stored are
A
1 × 10-128 and 215× 1015
B
1 × 10-256 and 215× 10255
C
1 × 10-128 and 215× 10127
D
1 × 10-128and 215– 1 × 10127
       Digital-Logic-Design       Number-Systems
Question 33 Explanation: 

According to question 16 bit mantissa and 8 bit Exponent.
Since the mantissa is always 1.xxxxxxxxx in the normalised form, no need to represent the leading 1.
Single Precision: mantissa ===> 1 bit + 15 bits
The largest mantissa value value is 215-1 (one bit meant for sign)
The largest exponent value is 27-1=127
The smallest mantissa value is 0000 0000 0000 0000(one bit is always 1) =1
The Smallest (largest negative) exponent value is 1111 1111 (which is 2’s complement form) 2-8=-128
Question 34
In comparison with static RAM memory, the dynamic RAM memory has
A
lower bit density and higher power consumption
B
higher bit density and higher power consumption
C
lower bit density and lower power consumption
D
higher bit density and lower power consumption
       Digital-Logic-Design       RAM
Question 34 Explanation: 
DRAM (Dynamic Random Access Memory)
→ DRAM stands for Dynamic Random Access Memory. It is used in most of the computers. It is the least expensive kind of RAM. It requires an electric current to maintain its electrical state. The electrical charge of DRAM decreases with time that may result in loss of DATA.
→ DRAM is recharged or refreshed again and again to maintain its data. The processor cannot access the data of DRAM when it is being refreshed. That is why it is slow.
SRAM (Static Random Access Memory)
→ SRAM stands for Static Random Access Memory. It can store data without any need of frequent recharging. CPU does not need to wait to access data from SRAM during processing. That is why it is faster than DRAM. It utilizes less power than DRAM.
→ SRAM is more expensive as compared to DRAM. It is normally used to build a very fast memory known as cache memory
Question 35
The Hexadecimal equivalent of 01111100110111100011 is
A
CD73E
B
ABD3F
C
7CDE3
D
FA4CD
       Digital-Logic-Design       Number-Systems
Question 35 Explanation: 
Binary number = 0111 1100 1101 1110 0011
7 C D E 3

(7CDE3)16
Question 36
Disk requests are received by a disk drive for cylinder 5, 25, 18, 3, 39, 8 and 35 in that order. A seek takes 5 msec per cylinder moved. How much seek time is needed to serve these requests for a Shortest Seek First (SSF) algorithm? Assume that the arm is at cylinder 20 when the last of these requests is made with none of the requests yet served
A
125 msec
B
295 msec
C
575 msec
D
750 msec
       Operating-Systems       Disk-Scheduling
Question 36 Explanation: 
The arm is at cylinder 20, so the service order = 18, 25, 35, 39, 8, 5, 3.
Seek time = (20−18) + (25−18) + (35−25) + (39−35) + (39−8) + (8−5) + (5−3)
= 2 + 7 + 10 + 4 + 31 + 3 + 2 = 59
Total seek time = 59 * 5 = 29
Question 37
Consider a system having ‘m’ resources of the same type. The resources are shared by 3 processes A, B, C, which have peak time demands of 3, 4, 6 respectively. The minimum value of ‘m’ that ensures that deadlock will never occur is
A
11
B
12
C
13
D
14
       Operating-Systems       Deadlock
Question 37 Explanation: 
Minimum resources required to avoid deadlock = (m1 – 1) + (m2 – 1) +..+ (my – 1) + 1
where m = resource required by process
y = number of processes
so, Number of resources that ensures that deadlock will never occur is = (3-1) + (4-1) + (6-1) + 1 = 11
Option (A) is correct.
Question 38
A task in a blocked state
A
is executable
B
is running
C
must still be placed in the run queues
D
is waiting for some temporarily unavailable resources
       Operating-Systems       Peocess-state-transition-diagram
Question 38 Explanation: 
Waiting or Blocked state is when a process has requested some input/output and is waiting for the resource.
Question 39
Semaphores
A
synchronize critical resources to prevent deadlock
B
synchronize critical resources to prevent contention
C
are used to do I/O
D
are used for memory management
       Operating-Systems       Process-Synchronization
Question 39 Explanation: 
Semaphore is a variable and is used to solve critical section problem and to achieve process synchronization in the multi processing environment.
Question 40
On a system using non-preemptive scheduling, processes with expected run times of 5, 18, 9 and 12 are in the ready queue. In what order should they be run to minimize wait time?
A
5, 12, 9, 18
B
5, 9, 12, 18
C
12, 18, 9, 5
D
9, 12, 18, 5
       Operating-Systems       CPU-Scheduling
Question 40 Explanation: 
The processes should execute in SJF manner to get the lowest waiting time. So, the order should be 5, 9, 12, 18.
Question 41
The number of page frames that must be allocated to a running process in a virtual memory environment is determined by
A
the instruction set architecture
B
page size
C
number of processes in memory
D
physical memory size
       Operating-Systems       Virtual Memory
Question 41 Explanation: 
There are two important tasks in virtual memory management: a page-replacement strategy and a frame-allocation strategy. Frame allocation strategy says gives the idea of minimum number of frames which should be allocated. The absolute minimum number of frames that a process must be allocated is dependent on system architecture, and corresponds to the number of pages that could be touched by a single (machine) instruction.So, it is instruction set architecture
Question 42
Consider a small 2-way set-associative cache memory, consisting of four blocks. For choosing the block to be replaced, use the least recently (LRU) scheme. The number of cache misses for the following sequence of block addresses is 8, 12, 0, 12, 8
A
2
B
3
C
4
D
5
       Operating-Systems       Page-Replacement-algorithm
Question 42 Explanation: 
8, 12, 0, 12, 8
Miss 8
Miss 12
Miss 0
No miss for 12
Again a miss for 8
Total misses = 4
Question 43
Which commands are used to control access over objects in relational database?
A
CASCADE & MVD
B
GRANT & REVOKE
C
QUE & QUIST
D
None of these
       Database-Management-System       Relational-databases
Question 43 Explanation: 
DCL(Data Control Language) includes commands such as GRANT and REVOKE which mainly deals with the rights, permissions and other controls of the database system.
Examples of DCL commands:
GRANT-gives user’s access privileges to database.
REVOKE-withdraw user’s access privileges given by using the GRANT command.
Question 44
Which of the following is aggregate function in SQL?
A
Avg
B
Select
C
Ordered by
D
distinct
       Database-Management-System       SQL
Question 44 Explanation: 
Avg is one of the aggregate functions. It returns average value after calculating from values in a numeric column.
Syntax:
SELECT AVG(column_name) FROM table_name;
Question 45
One approach to handling fuzzy logic data might be to design a computer using ternary (base-3) logic so that data could be stored as “true,” “false,” and “unknown.” If each ternary logic element is called a flit, how many blits are required to represent at least 256 different values?
A
4
B
5
C
6
D
7
       Digital-Logic-Design       Number-Systems
Question 45 Explanation: 
In binary representation, to represent 256 different values, you need log_2 (256) = 8 bits. Similarly in ternary representation, you would require log_3 (256) which is 5.something. Now rounding off to the upper integer (since number of bits is an integer) and we get 6
Question 46
A view of database that appears to an application program is known as
A
Schema
B
Subschema
C
Virtual table
D
None of these
       Database-Management-System       Schema
Question 46 Explanation: 
A subschema is a subset of the schema and inherits the same property that a schema has. The plan for a view is often called subschema. Subschema refers to an application programmer’s view of the data item types and record types, which he or she uses.
Question 47
Armstrong’s inference rule does not determine
A
Reflexivity
B
Augmentation
C
Transitivity
D
Mutual dependency
Question 47 Explanation: 
Armstrong inference rules refers to a set of inference rules used to infer all the functional dependencies on a relational database. It consists of the following axioms:
Axiom of Reflexivity:
This axiom states: if Y is a subset of X, then X determines Y
Axiom of Augmentation:
The axiom of augmentation, also known as a partial dependency, states if X determines Y, then XZ determines YZ, for any Z
Axiom of Transitivity:
The axiom of transitivity says if X determines Y, and Y determines Z, then X must also determine Z.
Question 48
Which operation is used to extract specific columns from a table?
A
Project
B
Join
C
Extract
D
Substitute
       Database-Management-System       Relational-Algebra
Question 48 Explanation: 
Projection (π)
Projection is used to project required column data from a relation. By Default projection removes duplicate data.
Example :
R(A B C)
----------
1 2 4
2 2 3
3 2 3
4 3 4
π (BC)
B C
-----
2 4
2 3
3 4
Question 49
In the Big-Endian system, the computer stores  
A
MSB of data in the lowest memory address of data unit
B
LSB of data in the lowest memory address of data unit
C
MSB of data in the highest memory address of data unit
D
LSB of data in the highest memory address of data uni
       Computer-Organization       MSB-LSB
Question 49 Explanation: 
Big-endian is an order in which the “big end” i.e. the most significant value in the sequence is stored first at the lowest storage address.
Question 50
BCNF is not used for cases where a relation has
A
Two (or more) candidate keys
B
Two candidate keys and composite
C
The candidate key overlap
D
Two mutually exclusive foreign keys
       Database-Management-System       Normalization
Question 50 Explanation: 
A relation is in Boyce-Codd normal form if all attributes which are determinants are also candidate keys.
Transformation into Boyce-Codd normal form deals with the problem of overlapping keys.
Question 51
Selection sort algorithm design technique is an example of
A
Greedy method
B
Divide-and-conquer
C
Dynamic Programming
D
Backtracking
       Algorithms       Sorting
Question 51 Explanation: 
The selection sort algorithm sorts an array by repeatedly finding the minimum element (considering ascending order) from unsorted part and putting it at the beginning. The algorithm maintains two subarrays in a given array.
1) The subarray which is already sorted.
2) Remaining subarray which is unsorted.
In every iteration of selection sort, the minimum element (considering ascending order) from the unsorted subarray is picked and moved to the sorted subarray. Clearly, it is a greedy approach to sort the array.
Question 52
Which of the following RAID level provides the highest Data Transfer Rate (Read/Write)
A
RAID 1
B
RAID 3
C
RAID 4
D
RAID 5
       Operating-Systems       RAID
Question 52 Explanation: 
Disk mirroring, also known as RAID 1, is the replication of data to two or more disks. Disk mirroring is a good choice for applications that require high performance and high availability, such as transactional applications, email and operating systems.
Question 53
Which of the following programming language(s) provides garbage collection automatically
A
Lisp
B
C++
C
Fortran
D
C
       Programming       Garbage-Collection
Question 53 Explanation: 
Lisp is the first high-level programming language to introduce automatic garbage collection to simplify the concept of manual garbage collection. Rest of these languages- C, C++, FORTRAN do not perform automatic garbage collection but are capable of doing it manually.
Question 54
The average case and worst case complexities for Merge sort algorithm are
A
O(n2), O(n2)
B
O(n2), O(nlog2n)
C
O(n log2n), O(n2)
D
O(n log2n), O(n log2n)
       Algorithms       Sorting
Question 54 Explanation: 
The best case, average case and worst case complexities for Merge sort algorithm are O( nlog2n ).
Question 55
The time taken by binary search algorithm to search a key in a sorted array of n elements is
A
O ( log2n )
B
O ( n )
C
O ( n log2n )
D
O ( n2 )
       Algorithms       Searching
Question 55 Explanation: 
Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty. It takes a maximum of log(n) searches to search an element from the sorted array.
Question 56
Which of the following is correct with respect to Two phase commit protocol?
A
Ensures serializability
B
Prevents Deadlock
C
Detects Deadlock
D
Recover from Deadlock
       Database-Management-System       Transactions
Question 56 Explanation: 
The two phase commit protocol is a distributed algorithm which lets all sites in a distributed system agree to commit or rollback a transaction based upon consensus of all participating sites.If any database server is unable to commit its portion of the transaction, all database servers participating in the transaction must be prevented from committing their work.
It ensures serializability but does not ensures freedom from deadlock.
Question 57
The Fibonacci sequence is the sequence of integers
A
1, 3, 5, 7, 9, 11, 13
B
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55
C
0, 1, 3, 4, 7, 11, 18, 29, 47
D
0, 1, 3, 7, 15
       Algorithms       Fibbonacci-Series
Question 57 Explanation: 
In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation Fn = Fn-1 + Fn-2
with seed values
F0 = 0 and F1 = 1.
The Fibonacci numbers are the numbers in the following integer sequence.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144..
Question 58
Let X be the adjacency matrix of a graph G with no self loops. The entries along the principal diagonal of X are
A
all zeros
B
all ones
C
both zeros and ones
D
different
       Data-Structures       Graph-Theory
Question 58 Explanation: 
In an adjacency matrix of a graph G the entries along the principal diagonal are reflexive, i.e. elements showing connectivity with themselves. Since the GrapH G has no self loops so all these entries should be 0.
Question 59
Which of these is not a feature of WAP 2.0
A
Push and Pull Model
B
Interface to a storage device
C
Multimedia messaging
D
Hashing
       Computer-Networks       Network-protocols
Question 59 Explanation: 
WAP is a Wireless protocol for synchronization of WAP client computers and HTTP server
It is derived from the architecture of the synchronization between HTTP client (browser) and the HTTP server (web server) on the internet. It does the following functions:
SyncML synchronization
WAP push service
MMS service
Question 60
Feedback queues
A
are very simple to implement
B
dispatch tasks according to execution characteristics
C
are used to favour real time tasks
D
require manual intervention to implement properly
       Operating-Systems       CPU-Scheduling
Question 60 Explanation: 
n a multilevel queue-scheduling algorithm, processes are permanently assigned to a queue on entry to the system. Processes do not move between queues. This setup has the advantage of low scheduling overhead, but the disadvantage of being inflexible.

Multilevel feedback queue scheduling, however, allows a process to move between queues. The idea is to separate processes with different CPU-burst characteristics. If a process uses too much CPU time, it will be moved to a lower-priority queue. Similarly, a process that waits too long in a lower-priority queue may be moved to a higher-priority queue. This form of aging prevents starvation.
Question 61
Which of the following is not a UML DIAGRAM?
A
Use Case
B
Class Diagram
C
Analysis Diagram
D
Swimlane Diagram
       Software-Engineering       UML
Question 61 Explanation: 
Use Case, Class Diagram and Swimlane Diagram are all three UML diagrams but Analysis Diagram is not.
Question 62
Silly Window Syndrome is related to
A
Error during transmission
B
File transfer protocol
C
Degrade in TCP performance
D
Interface problem
       Computer-Networks       TCP
Question 62 Explanation: 
Silly window syndrome is a problem in computer networking caused by poorly implemented TCP flow control.
Question 63
To execute all loops at their boundaries and within their operational bounds is an example of
A
Black Box Testing
B
Alpha Testing
C
Recovery Testing
D
White Box Testing
       Software-Engineering       Software-testing
Question 63 Explanation: 
White box testing is a testing technique which evaluates the code and the internal structure of a program. It involves looking at the structure of the code, when a tester already know the internal structure of a product, tests can be conducted to ensure that the internal operations performed according to the specification and all internal components have been adequately exercised.
White Box Testing is coverage of the specification in the code:
1. Code coverage
2. Segment coverage: Ensure that each code statement is executed once.
3. Branch Coverage or Node Testing: Coverage of each code branch in from all possible was.
4. Compound Condition Coverage: For multiple conditions test each condition with multiple paths and combination of the different path to reach that condition.
5. Basis Path Testing: Each independent path in the code is taken for testing.
6. Data Flow Testing (DFT): It is defining the set of intermediate paths through the code.
7. Path Testing: Path testing is where all possible paths through the code are defined and covered.
8. Loop Testing: these strategies relate to testing single loops, concatenated loops, and nested loops.
Question 64
SSL is not responsible for
A
Mutual authentication of client & server
B
Secret communication
C
Data Integrity protection
D
Error detection and correction
       Computer-Networks       Network-Security
Question 64 Explanation: 
SSL (Secure Sockets Layer) is the standard security technology for establishing an encrypted link between a web server and a browser. This link ensures that all data passed between the web server and browsers remain private and integral. SSL is an industry standard and is used by millions of websites in the protection of their online transactions with their customers.
It is utilized to encrypt Web traffic using Hypertext Transfer Protocol (HTTP) and to authenticate Web servers, and to encrypt communications between Web browsers and Web servers etc.
So, other than error detection and correction, all options are correct.
Question 65
A rule in a limited entry decision table is a
A
row of the table consisting of condition entries
B
row of the table consisting of action entries
C
column of the table consisting of condition entries and corresponding action entries
D
columns of the table consisting of conditions of the stub
Question 65 Explanation: 
A limited entry decision table is a structured way to formulate requirements and test cases when dealing with complex business rules. Using a decision table will make it easier to write requirements that cover all alternative conditions. In these tables, the upper half of the table lists the conditions being tested while the lower half lists the possible actions to be taken and each column represents a certain type of condition or rule.
Question 66
The standard for certificates used on internet is
A
X.25
B
X.301
C
X.409
D
X.509
       Computer-Networks       Network-Security
Question 66 Explanation: 
In cryptography, X.509 is a standard that defines the format of public key certificates. X.509 certificates are used in many Internet protocols, including TLS/SSL, which is the basis for HTTPS, the secure protocol for browsing the web.
Question 67
Hashed message is signed by a sender using
A
his public key
B
his private key
C
receiver’s public key
D
receiver’s private key
       Computer-Networks       Network-Security
Question 67 Explanation: 
After the generation of hashed message that needs to be transmitted over a network, it is signed or encrypted using the private key of the sender which also generates the digital signatures. The message is transmitted along with the digital signatures which ensures Authentication and Non-repudiation.
Question 68
An email contains a textual birthday greeting, a picture of a cake and a song. The order is not important. What is the content-type?An email contains a textual birthday greeting, a picture of a cake and a song. The order is not important. What is the content-type?
A
Multipart/mixed
B
Multipart/parallel
C
Multipart/digest
D
Multipart/alternative
Question 68 Explanation: 
The Multipart/parallel subtype
This document defines a "parallel" subtype of the multipart Content- Type. This type is syntactically identical to multipart/mixed, but the semantics are different. In particular, in a parallel entity, the order of body parts is not significant.
Question 69
Range of IP Address from 224.0.0.0 to 239.255.255.255 are
A
Reserved for loopback
B
Reserved for broadcast
C
Used for multicast packets
D
Reserved for future addressing
       Computer-Networks       IPv4-and-Fragmentation
Question 69 Explanation: 
The full range of multicast addresses is from 224.0.0.0 to 239.255.255.255. Multicast addresses represent a group of IP devices that can only be used as the destination of a datagram and not as the source.
Question 70
IEEE 802.11 is standard for
A
Ethernet
B
Bluetooth
C
Broadband Wireless
D
Wireless LANs
       Computer-Networks       LAN
Question 70 Explanation: 
IEEE 802.11 refers to the set of standards that define communication for wireless LANs (wireless local area networks or WLANs).
Question 71
When a host on network A sends a message to a host on network B, which address does the router look at?
A
Port
B
IP
C
Physical
D
Subnet mask
       Computer-Networks       Network-layers
Question 71 Explanation: 
Routing is done on the basis of IP addresses. A host on network A when sends the packet to the host on network B, it checks the IP address of the receiving host and routes the packet to the suitable hop.
Physical address i.e. the MAC address is used to recognize a unique host over a network on in a LAN. Port number is required to identify a specific process or application to which a message is to be forwarded when it arrives at a server.
Question 72
Which of the following is not an approach to Software Process Assessment?
A
SPICE(ISO/IEC15504)
B
Standard CMMI Assessment Method for process improvement
C
ISO 9001:2000
D
IEEE 2000:2001
       Software-Engineering       Software-process-models
Question 72 Explanation: 
The process assessment leads to process capability determination and process improvement. Process capability determination is an organized assessment, which analyzes the software processes in an organization.The process improvement identifies the changes to be made in the software processes.
Different approaches are used for assessing software process. These approaches are SPICE (ISO/IEC15504), ISO 9001:2000, standard CMMI assessment method for process improvement, CMM-based appraisal for internal process improvement, and Bootstrap.
Question 73
A physical DFD specifies
A
what processes will be used
B
who generates data and who processes it
C
what each person in an organization does
D
which data will be generated
       Software-Engineering       DFD
Question 73 Explanation: 
Physical DFD specifies actual flow of physical documentation and depicts how the system will be implemented.
The processes represent the programs, program modules, and manual procedures.
The data stores represent the physical files and databases, manual files.
It show controls for validating input data, for obtaining a record, for ensuring successful completion of a process, and for system security.
Question 74
In UML diagram of a class
A
state of object cannot be represented
B
state is irrelevant
C
state is represented as an attribute
D
state is represented as a result of an operation
       Software-Engineering       UML
Question 74 Explanation: 
The Unified Modeling Language (UML) is a general-purpose, developmental, modeling language in the field of software engineering, that is intended to provide a standard way to visualize the design of a system.
A class diagram in the Unified Modeling Language (UML) is a type of static structure diagram that describes the structure of a system by showing the system’s classes as attributes, operations and the relationships among objects.
Question 75
Which of the following models used for software reliability
A
Waterfall
B
Musa
C
COCOMO
D
Rayleigh
       Software-Engineering       Software-process-models
Question 75 Explanation: 
Rayleigh Model:
This model predicts fault detection over the life of the software development effort and can be used in conjunction with the other prediction techniques. Software management may use this profile to gauge the defect status. This model assumes that over the life of the project that the faults detected per month will resemble a Raleigh curve
Musa Model:
This prediction technique is used to predict, prior to system testing, what the failure rate will be at the start of system testing. This prediction can then later be used in the reliability growth modelling. For this prediction method, it is assumed that the only thing known about the hypothetical program is a prediction of its size and the processor speed.
COCOMO:
The constructive cost model was developed as a model for estimating effort, cost, and schedule for software projects.
Waterfall Model:
Waterfall is the software development life cycle model which depicts the phases of conception, initiation, analysis, design, construction, testing, deployment and maintenance.
Musa model is also used for software reliability testing but Rayleigh model is very popularly used with higher accuracy.
Question 76
Dijkstra's algorithm is used to
A
Create LSAs
B
Flood an internet with information
C
Calculate the routing tables
D
Create a link state database
       Computer-Networks       Routing
Question 76 Explanation: 
Dijkstra's algorithm finds shortest paths from the source to all other nodes in the graph, producing a shortest-path tree and is used in building the routing tables on a network.
Question 77
Eigenvectors of

are
A
B
C
D
E
None Of the Above
       Engineering-Mathematics       Linear-algebra
Question 77 Explanation: 
Explanation: Excluded for evaluation
Question 78
The set of all Equivalence Classes of a set A of Cardinality C
A
is of cardinality 2c
B
have the same cardinality as A
C
forms a partition of A
D
is of cardinality C2
       Engineering-Mathematics       Set-Theory
Question 78 Explanation: 
The set of ll equivalence class of a set A of cardinality C forms a partition of A
Note: Ambiguous between answer with option D
Question 79
Company X shipped 5 computer chips, 1 of which was defective and company Y shipped 4 computer chips, 2 of which were defective. One computer chip is to be chosen uniformly at a random from the 9 chips shipped by the companies. If the chosen chip is found to be defective, what is the probability that the chip came from the company Y?
A
2/9
B
4/9
C
2/3
D
1/2
       Engineering-Mathematics       Probability
Question 79 Explanation: 
Probability that chip came chosen from company X = 5/9
Probability that chip came chosen from company Y = 4/9
Number of Defective chips from company X = 1
Number of Defective chips from company Y = 2
Probability that chip is defective from company X = 5/9 * 1/5
Probability that chip is defective from company Y = 4/9 * 2/4
Probability that chip is defective = 5/9 * 1/5 + 4/9 * 2/4
Given chip is defective, probability that it came from the company
Y = P(Defective Company Y)/ P(Defective)
= (4/9 * 2/4) / (5/9 * 1/5 + 4/9 * 2/4)
= 2/3
Question 80
A program consists of two modules executed sequentially. Let f1(t) and f2(t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by
A
f1(t) + f2(t)
B
t0 f1(x), f2(x) dx
C
t0 f1(x), f2(t-x) dx
D
max (f1(t), f2(t))
       Engineering-Mathematics       Probability
Question 80 Explanation: 
The two modules executed sequentially. The total runtime of the program is the sum of runtime of the two module. Probability density function of overall time taken
t0 f1(x), f2(t-x) dx
There are 80 questions to complete.
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