Nielit Scientist-B CS 2016 march
Question 1 |
Which of the following logic expression is incorrect?
1 ⊕ 0=1 | |
1 ⊕ 1 ⊕ 0=1 | |
1 ⊕ 1 ⊕ 1=1 | |
1 ⊕ 1 =0 |
Question 1 Explanation:
This ⊕ symbol is nothing but Ex-OR
Option A: Here, 1 ⊕ 0=1 true according to truth table.
Option B: 1⊕1=0⊕0=0 false
Option C: 1 ⊕ 1=0 ⊕ 1=1 true
Option D: 1 ⊕ 1 =0 true.

Option A: Here, 1 ⊕ 0=1 true according to truth table.
Option B: 1⊕1=0⊕0=0 false
Option C: 1 ⊕ 1=0 ⊕ 1=1 true
Option D: 1 ⊕ 1 =0 true.
Question 2 |
In which of the following adder circuits, the carry look ripple delay is eliminated?
Half Adder | |
Full Adder | |
Parallel adder | |
Carry-Look-Ahead adder |
Question 2 Explanation:
A carry-lookahead adder (CLA) or fast adder is a type of adder used in digital logic. A carry-lookahead adder improves speed by reducing the amount of time required to determine carry bits. It can be contrasted with the simpler, but usually slower, ripple-carry adder (RCA), for which the carry bit is calculated alongside the sum bit, and each bit must wait until the previous carry bit have been calculated to begin calculating its own result and carry bits. The carry-lookahead adder calculates one or more carry bits before the sum, which reduces the wait time to calculate the result of the larger-value bits of the adder.
Question 3 |
The output of a sequential circuit depends on
presents inputs only | |
past inputs only | |
both present and past inputs | |
present outputs only |
Question 3 Explanation:
In digital circuit theory, sequential logic is a type of logic circuit whose output depends not only on the present value of its input signals but on the sequence of past inputs, the input history as well. This is in contrast to combinational logic, whose output is a function of only the present input.
Question 4 |
In a ripple counter using edge triggered JK flip-flops, the pulse input is applied to the
clock input of all flip-flops | |
clock input of one flip flops | |
J and K inputs of all flip flops | |
J and K inputs of one flip-flop |
Question 4 Explanation:
In a ripple counter using edge triggered JK flip-flops, the pulse input is applied to the clock input of one flip flops
Question 5 |
A decimal number has 30 digits. Approximately, how many digits would the binary representation have?
30 | |
60 | |
90 | |
120 |
Question 5 Explanation:
Here, 30 digits numbers means 123....30.
1030 -1=1000000000000000000000000000000-1=999999999999999999999999999999
Therefore, it takes approximately above 90 binary numbers. So, 120 is correct answer.
Therefore, it takes approximately above 90 binary numbers. So, 120 is correct answer.
Question 6 |
The result of the subtraction FD16 - 8816 is
75 16 | |
65 16 | |
5E 16 | |
10 16 |
Question 6 Explanation:
Step-1: Convert Hexadecimal numbers into decimal numbers.
(FD) 16</sub> = (253) 10
(88) 16</sub> = (136) 10
Step-2: Perform subtraction 253-136=(117) 10
Step-3: Convert (117) 10 =(75) 16
(FD) 16</sub> = (253) 10
(88) 16</sub> = (136) 10
Step-2: Perform subtraction 253-136=(117) 10
Step-3: Convert (117) 10 =(75) 16
Question 7 |
How many RAM chips of size (256K x 1 bit) are required to build 1M Byte memory?
8 | |
10 | |
24 | |
32 |
Question 7 Explanation:
1 MB = 1024 KB = (1024*8) bits
New RAM would be (1024*8) / (256*1) = 32 .
→ 32 RAM chips of size 256 k x 1 bit are required to build 1 MB of memory.
New RAM would be (1024*8) / (256*1) = 32 .
→ 32 RAM chips of size 256 k x 1 bit are required to build 1 MB of memory.
Question 8 |
When we move from the outermost track to the innermost track in a magnetic disk, then density(bits per linear inch)
increases | |
decreases | |
remains the same | |
either remains constant or decreases |
Question 8 Explanation:
→ we move from the outermost track to the innermost track in a magnetic disk, then density(bits per linear inch) increases
→ The density, in bits per linear inch, increases as we move from the outermost track to the innermost track (this same phenomenon is present in a phonograph record).
→ The density, in bits per linear inch, increases as we move from the outermost track to the innermost track (this same phenomenon is present in a phonograph record).
Question 9 |
A certain processor supports only the immediate and the direct addressing modes. Which of the following programming language features cannot be implemented on this processor?
Pointers | |
Arrays | |
Records | |
All of these |
Question 9 Explanation:
→ Pointer access requires indirect addressing which can be simulated with indexed addressing or register indirect addressing but not with direct and immediate addressing.
→ An array and record access needs a pointer access. So, options (A), (B) and (C) cannot be implemented on such a processor.
→ Now to handle recursive procedure we need to use stack. A local variable inside the stack will be accessed as *(SP+Offset) which is nothing but a pointer access and requires indirect addressing. Usually this is done by moving the SP value to the base register and then using base register addressing to avoid unnecessary memory access for indirect addressing but not possible with just direct and immediate addressing.
→ An array and record access needs a pointer access. So, options (A), (B) and (C) cannot be implemented on such a processor.
→ Now to handle recursive procedure we need to use stack. A local variable inside the stack will be accessed as *(SP+Offset) which is nothing but a pointer access and requires indirect addressing. Usually this is done by moving the SP value to the base register and then using base register addressing to avoid unnecessary memory access for indirect addressing but not possible with just direct and immediate addressing.
Question 10 |
Disadvantage of dynamic RAM over static RAM is
Higher power consumptions | |
Variable speed | |
need to refresh the capacitor charge every once in two milliseconds | |
higher bit density |
Question 10 Explanation:
→ Dynamic random access memory, implies, this form of memory technology is a type of random access memory. It stores each bit of data on a small capacitor within the memory cell. The capacitor can be either charged or discharged and this provides the two states, "1" or "0" for the cell.
→ Since the charge within the capacitor leaks, it is necessary to refresh each memory cell periodically. This refresh requirement gives rise to the term dynamic - static memories do not have a need to be refreshed.
→ Since the charge within the capacitor leaks, it is necessary to refresh each memory cell periodically. This refresh requirement gives rise to the term dynamic - static memories do not have a need to be refreshed.
Question 11 |
What is the correct way to round off x, a float. to an int value?
y=(int)(x+0.5) | |
y=int(x+0.5) | |
y=(int)x+0.5 | |
y=(int)(int) x+0.5) |
Question 11 Explanation:
Rounding off a value means replacing it by a nearest value that is approximately equal or smaller or greater to the given number.
y = (int)(x + 0.5); here x is any float value. To round off, we have to typecast the value of x by using (int)
Example:
#include
int main ()
{
float x = 3.6;
int y = (int)(x + 0.5);
printf ("Result = %d\n", y );
return 0;
}
Output:
Result = 4.
y = (int)(x + 0.5); here x is any float value. To round off, we have to typecast the value of x by using (int)
Example:
#include
int main ()
{
float x = 3.6;
int y = (int)(x + 0.5);
printf ("Result = %d\n", y );
return 0;
}
Output:
Result = 4.
Question 12 |
which of the following sorting algorithms does not have a worst case running time of O(n 2 )
Insertion sort | |
merge sort | |
Quick sort | |
bubble sort |
Question 12 Explanation:

Question 13 |
What error would the following function given on compilation?
f(int a, int b)
{
int a;
a=20;
return a;
}
f(int a, int b)
{
int a;
a=20;
return a;
}
Missing parenthesis is return statement | |
Function should be defined as int f(int a, int b) | |
Redeclaration of a | |
None of these |
Question 13 Explanation:
We are already declared variable name ‘a’ in function. Again we are declared inside the function. So, the compiler will raise a error “Redeclaration of a”
Question 14 |
Prior to using a pointer variable it should be
declared | |
initialized | |
both declared and initialized | |
none of these |
Question 14 Explanation:
● In the programming, Before using any variable we should declare that variable.If require means initialize the variable.
● Pointer is also one variable
● So before using a pointer variable it should be both declared and initialized.
● Pointer is also one variable
● So before using a pointer variable it should be both declared and initialized.
Question 15 |
Output of the following loop is
for(putchar('c');putchar('a');putchar('r'))
putchar('t');
a syntax error | |
cartrt | |
catrat | |
catratratratrat... |
Question 15 Explanation:
Syntax of ‘for loop’
for(Initialization; Condition; value modification/updation)
As per the above question,
→ Initialization part given by putchar('c)
→ Condition part check by putchar('a')
→ Value modification/updation part given by putchar('r')
→ Inside the for loop we are given putchar('t’)
→ The sequence of the for loop first initialization, second condition, executing inside body of for loop and value modification.
So, it prints catratratratrat...
for(Initialization; Condition; value modification/updation)
As per the above question,
→ Initialization part given by putchar('c)
→ Condition part check by putchar('a')
→ Value modification/updation part given by putchar('r')
→ Inside the for loop we are given putchar('t’)
→ The sequence of the for loop first initialization, second condition, executing inside body of for loop and value modification.
So, it prints catratratratrat...
Question 16 |
If space occupied by two strings s1 and s2 in 'C' are respectively m and n, then space occupied by string obtained by concatenating s1 and s2 is always
less than m+n | |
equal to m+n | |
greater than m+n | |
none of these |
Question 16 Explanation:
If space occupied by two strings s1 and s2 in 'C' are respectively m and n, then space occupied by string obtained by concatenating s1 and s2 is always less than m+n
Question 17 |
A hash function f defined as f(key)=key mod 7, with linear probing, insert the keys 37,38,72,48,98,11,56 into a table indexed from 11 will be stored in the location
3 | |
4 | |
5 | |
6 |
Question 17 Explanation:
The hash starts from 0th index.
37%7=2
38%7=3
72%7=2. Here, collision happened because already index 2 is already filled by 37.
According to linear probing, increment index value by 1.
2+1=3 but index 3 also filled with 38 element. So, again increment by 1.
72%7=2( getting index position in hash table 4 )
48%7=6
98%7=0
11%7=4. Here, collision happened because already index 4 is already filled by 72.
According to linear probing, increment index value by 1.
11%7=4 (getting index position in hash table 5 )
56%7=0Here, collision happened because already index 0 is already filled by 98.
According to linear probing, increment index value by 1.
56%7=0 (getting index position in hash table 1 )
37%7=2
38%7=3
72%7=2. Here, collision happened because already index 2 is already filled by 37.
According to linear probing, increment index value by 1.
2+1=3 but index 3 also filled with 38 element. So, again increment by 1.
72%7=2( getting index position in hash table 4 )
48%7=6
98%7=0
11%7=4. Here, collision happened because already index 4 is already filled by 72.
According to linear probing, increment index value by 1.
11%7=4 (getting index position in hash table 5 )
56%7=0Here, collision happened because already index 0 is already filled by 98.
According to linear probing, increment index value by 1.
56%7=0 (getting index position in hash table 1 )
Question 18 |
traversing a binary tree first root and then left and right subtree called__ traversalt
postorder | |
preorder | |
inorder | |
none of these |
Question 18 Explanation:
In order: The left subtree is visited first, then the root and later the right subtree.
Pre order: The root node is visited first, then the left subtree and finally the right subtree.
Post order: First we traverse the left subtree, then the right subtree and finally the root node.
Pre order: The root node is visited first, then the left subtree and finally the right subtree.
Post order: First we traverse the left subtree, then the right subtree and finally the root node.
Question 19 |
If there is in NP-Complete language L whose complement is in NP, then complement of any language in NP is in
P | |
NP | |
Both (A) and (B) | |
None of these |
Question 19 Explanation:
If there is in NP-Complete language L whose complement is in NP, then complement of any language in NP is in NP
Question 20 |
A hash table has space for 100 records. then the probability of collision before the table is 10% full, is
0.45 | |
0.5 | |
0.3 | |
0.37(approximately) |
Question 20 Explanation:
Here, we can get the answer using 1-Probability of no collision for first 10 entries.
So, 1- (100*99*98*97*96*95*94*93*92*91)/100 10 ⇒ 0.37(approximately)
So, 1- (100*99*98*97*96*95*94*93*92*91)/100 10 ⇒ 0.37(approximately)
Question 21 |
A polynomial p(x) is such that p(0)=5, p(1)=4, p(2)=9 and p(3)=20 The minimum degree it can have is
1 | |
2 | |
3 | |
4 |
Question 21 Explanation:
Let's take p(x) = ax + b
p(0) = 5 ⇒ b = 5
p(1) = 4 ⇒ a+b = 4 ⇒ a = -1
p(2) = 9 ⇒ 40+b = 9 ⇒ -4+5 = 9, which is false.
So degree 1 is not possible.
Let's take p(x) = ax 2 + bx +c
p(0) = 5 ⇒ c = 5
p(1) = 4 ⇒ a+b+c = 4 ⇒ a+b = -1 -----(1)
p(2) = 9 ⇒ 4a+2b+c = 9 ⇒ 2a+b = 2 -----(2)
(2) - (1)
⇒ a = 3, b = -1-1 = -4
p(3) = 20 ⇒ 9a+3b+c = 20
⇒ 27-12+5 = 20
⇒ 20 = 20, True
Hence, minimum degree it can have.
p(0) = 5 ⇒ b = 5
p(1) = 4 ⇒ a+b = 4 ⇒ a = -1
p(2) = 9 ⇒ 40+b = 9 ⇒ -4+5 = 9, which is false.
So degree 1 is not possible.
Let's take p(x) = ax 2 + bx +c
p(0) = 5 ⇒ c = 5
p(1) = 4 ⇒ a+b+c = 4 ⇒ a+b = -1 -----(1)
p(2) = 9 ⇒ 4a+2b+c = 9 ⇒ 2a+b = 2 -----(2)
(2) - (1)
⇒ a = 3, b = -1-1 = -4
p(3) = 20 ⇒ 9a+3b+c = 20
⇒ 27-12+5 = 20
⇒ 20 = 20, True
Hence, minimum degree it can have.
Question 22 |
Time complexity of an algorithm T(n), where n is the input size is given by
T(n)=T(n-1)+1/n, if n>1
=1, otherwise
The order of this algorithm is
logn | |
n | |
n 2 | |
n n |
Question 22 Explanation:
T (n) = T (n − 1 ) + 1/n


Question 23 |
If S be an infinite set and S 1 ..., S n be sets such that S 1 U S 2 U...U S n =S, then
At least one of the set S i is a finite set | |
not more than one of the sets S i can be finite | |
At least one of the sets S i is an infinite set | |
not more than one of the sets S i can be infinite |
Question 23 Explanation:
Given sets are finite union of sets. One set must be infinite to make whole thing to be infinite.
Question 24 |
Which of the following regular expressions denotes a language comprising all possible strings over the alphabet {a,b}?
a*b* | |
(a | b)* | |
(ab) * | |
(a | b*) |
Question 24 Explanation:
It is clearly saying that all possible strings over the alphabet {a,b} is (a | b)*
Question 25 |
Regarding power of recognition of language, which of the following is false?
Non deterministic finite-state automata are equivalent to deterministic finite-state automata | |
Non deterministic PDA are equivalent to Deterministic PDA | |
Nondeterministic turing machines are equivalent to deterministic PDA | |
Multi tape Turing machines are equivalent to single tape Turing machines |
Question 25 Explanation:
→ NPDA is more powerful than DPDA.
→ NFA and DFA having same power.
→ NTM and DTM having same power.
→ Multi tape Turing machines are equivalent to single tape Turing machines
→ NFA and DFA having same power.
→ NTM and DTM having same power.
→ Multi tape Turing machines are equivalent to single tape Turing machines
Question 26 |
If L1 and L2 are context free language and R a regular set, then which one of the languages below is not necessarily a context free language?
L1.L2 | |
L1 ∩ L2 | |
L1 ∩ R | |
L1 U L2 |
Question 26 Explanation:
→ Intersection and complementation is not closed under context free languages(CFL).
→ L1 ∩ R is closed under CFL
→ L1.L2 and L1 U L2 is closed under CFL
→ L1 ∩ R is closed under CFL
→ L1.L2 and L1 U L2 is closed under CFL
Question 27 |
A packet switching network
is free | |
can reduce the cost of using an information utility | |
allows communications channel to be shared among more than one user | |
boh (B) and (C) |
Question 27 Explanation:
Packet switching is a method of grouping data that is transmitted over a digital network into packets. Packets are made of a header and a payload. Data in the header are used by networking hardware to direct the packet to its destination where the payload is extracted and used by application software.
Question 28 |
Which of the following is not a transceiver function?
Transmission and receipt of data | |
Checking of line voltages | |
Addition and subtraction of headers | |
Collision detection |
Question 28 Explanation:
A transceiver is a device comprising both a transmitter and a receiver that are combined and share common circuitry or a single housing. When no circuitry is common between transmit and receive functions, the device is a transmitter receiver.
Question 29 |
The CFG S → aS|bS|a|b is equivalent to regular expression
(a+b) | |
(a+b)(a+b)* | |
(a+b)(a+b) | |
all of these |
Question 29 Explanation:
● The option A gives either a or b expression which is of length of 1
● The option C gives aa,ab,ba or bb which are of length of 2
● But the CFG will produce strings, which of length of greater than 2 also.
● The option B gives string starts with either a or b followed by any number of a’s or b’s which is equivalent to CFG.
● The option C gives aa,ab,ba or bb which are of length of 2
● But the CFG will produce strings, which of length of greater than 2 also.
● The option B gives string starts with either a or b followed by any number of a’s or b’s which is equivalent to CFG.
Question 30 |
Bounded minimization is a technique for
Proving whether a promotivate recursive function is turing computable or not | |
Proving whether a primitive recursive function is a total function or not | |
generating primitive recursive functions | |
generating partial recursive functions |
Question 30 Explanation:
● Primitive recursive functions are a class of functions that are defined using composition and primitive recursion – described below – as central operations. They are a strict subset of those μ-recursive functions (also called partial recursive functions) which are also total functions. Primitive recursive functions form an important building block on the way to a full formalization of computability.
● Most of the functions normally studied in number theory are primitive recursive.
● Most of the functions normally studied in number theory are primitive recursive.
Question 31 |
In a single pass assembler, most of the forward references can be avoided by putting the restriction
on the number of strings/lifereacs | |
that the data segment must be defined after the code segment | |
on unconditional rump | |
that the data segment be defined before the code segment |
Question 31 Explanation:
Data segment − It is represented by .data section and the .bss. The .data section is used to declare the memory region, where data elements are stored for the program. This section cannot be expanded after the data elements are declared, and it remains static throughout the program.
The .bss section is also a static memory section that contains buffers for data to be declared later in the program. This buffer memory is zero-filled.
Code segment − It is represented by .text section. This defines an area in memory that stores the instruction codes. This is also a fixed area.
The .bss section is also a static memory section that contains buffers for data to be declared later in the program. This buffer memory is zero-filled.
Code segment − It is represented by .text section. This defines an area in memory that stores the instruction codes. This is also a fixed area.
Question 32 |
A linker is given object module for a set of programs that were compiled separately. What information need not be included in an object module?
Object code | |
relocation bits | |
Names and locations of all external symbols defined in the object module | |
Absolute addresses of internal symbols |
Question 32 Explanation:
● In object module it includes names and locations of all external symbols defined in the object module.
● To link to external symbols it must know the location of external symbols.
● So object module won’t consists of absolute addresses of internal symbols , Option D is correct.
● To link to external symbols it must know the location of external symbols.
● So object module won’t consists of absolute addresses of internal symbols , Option D is correct.
Question 33 |
In what module multiple instances of execution will yield the same result even if one instance has not terminated before the next one has begun?
Non reusable module | |
serially usable | |
Re-enterable module | |
recursive module |
Question 33 Explanation:
● A reenterable load module does not modify itself. Only one copy of the load module is loaded to satisfy the requirements of any number of tasks in a job step. This means that even though there are several tasks in the job step and each task concurrently uses the load module, the only central storage needed is an area large enough to hold one copy of the load module (plus a few bytes for control blocks).
● A non-reusable module can only be loaded for one task - the code is assumed to be self-modifying (data and code are in the same load module).
● A reusable module can be used by a second task when the first is finished with it (generally the module will acquire its own data space dynamically).
● A non-reusable module can only be loaded for one task - the code is assumed to be self-modifying (data and code are in the same load module).
● A reusable module can be used by a second task when the first is finished with it (generally the module will acquire its own data space dynamically).
Question 34 |
Global locks
synchronize access to local resources | |
synchronize access to global resources | |
are used to avoid local locks | |
prevent access to global resources |
Question 34 Explanation:
● Many of the resources protected by local locks in single-instance Oracle require global exposure in a parallel server database.
● Whenever one of these local locks is needed, a corresponding instance lock must be held as well, to protect the resource across all instances. The instance locks used to protect local locks globally are called global locks
● Whenever one of these local locks is needed, a corresponding instance lock must be held as well, to protect the resource across all instances. The instance locks used to protect local locks globally are called global locks
Question 35 |
The file structure that redefines its first record at a base of zero uses the term
relative organization | |
key fielding | |
dynamic reallocation | |
all of these |
Question 35 Explanation:
● A relative record file contains records ordered by their relative key, a record number that represents the location of the record relative to where the file begins.
● For example, the first record in a file has a relative record number of 1, the tenth record has a relative record number of 10, and so forth. The records can have fixed length or variable length.
● For example, the first record in a file has a relative record number of 1, the tenth record has a relative record number of 10, and so forth. The records can have fixed length or variable length.
Question 36 |
In context of TCP/IP computer network models, which of the following is FALSE?
Besides span of geographical area, the other major difference between LAN and WAN is that the later uses switching element | |
A repeater is used just to forward bits from one network to another one | |
IP layer is connected oriented layer in TCP/IP | |
A gateway is used to connect incompatible networks |
Question 36 Explanation:
→ A transceiver is a device comprising both a transmitter and a receiver that are combined and share common circuitry or a single housing. When no circuitry is common between transmit and receive functions, the device is a transmitter-receiver.
→ IP is connectionless, in that a data packet can travel from a sender to a recipient without the recipient having to send an acknowledgement.
→ IP is connectionless, in that a data packet can travel from a sender to a recipient without the recipient having to send an acknowledgement.
Question 37 |
A long-term monitor
Should show any immediate performance problems | |
should show I/O, paging and processor activity | |
need show only the I/O and processor activity | |
usually reports only on terminal displays |
Question 37 Explanation:
● Long term monitor metrics, as the name indicates, are the metrics that are collected for a resource for a longer term monitoring perspective. They are useful for continuous ongoing monitoring of the managed system.
● Monitoring the performance of operating systems and processes is essential to debug processes and systems, effectively manage system resources, making system decisions, and evaluating and examining systems.
● Monitoring the performance of operating systems and processes is essential to debug processes and systems, effectively manage system resources, making system decisions, and evaluating and examining systems.
Question 38 |
Determine the number of page faults when references to pages occur in the following order:1,2,4,5,2,1,2,4. Assume that the main memory can accommodate 3 pages and the main memory already has the pages 1 and 2, with page 1 having been brought earlier than page 2.(LRU algorithm is used)
3 | |
5 | |
4 | |
none of these |
Question 38 Explanation:

Here, total 6 page faults but in question, they are clearly mentioned that the main memory already has the pages 1 and 2, with page one having brought earlier than page 2. It means 6-2=4.
Question 39 |
Working set(t,k) at an instant of time,t,is
The set of k future references that the operating system will make | |
The set of future references that the operating system will make in the next 'k' time units | |
The set of k references with high frequency | |
The set of pages that have been referenced in the last k time units |
Question 39 Explanation:
Working set defines the amount of memory that a process requires in a given time interval.
→ It defines “the working set of information W(t,τ) of a process at time t to be the collection of information referenced by the process during the process time interval (t−τ,t)”.
→ Typically the units of information in question are considered to be memory pages. This is suggested to be an approximation of the set of pages that the process will access in the future (say during the next τ time units), and more specifically is suggested to be an indication of what pages ought to be kept in main memory to allow most progress to be made in the execution of that process.
→ It defines “the working set of information W(t,τ) of a process at time t to be the collection of information referenced by the process during the process time interval (t−τ,t)”.
→ Typically the units of information in question are considered to be memory pages. This is suggested to be an approximation of the set of pages that the process will access in the future (say during the next τ time units), and more specifically is suggested to be an indication of what pages ought to be kept in main memory to allow most progress to be made in the execution of that process.
Question 40 |
In a relational schema, each tuple is divided into fields called
relations | |
Domains | |
Queries | |
none of these |
Question 40 Explanation:
● A domain is defined as the set of all unique values permitted for an attribute. For example, a domain of date is the set of all possible valid dates, a domain of integer is all possible whole numbers, a domain of day-of-week is Monday, Tuesday ... Sunday.
● This in effect is defining rules for a particular attribute. If it is determined that an attribute is a date then it should be implemented in the database to prevent invalid dates being entered.
● This in effect is defining rules for a particular attribute. If it is determined that an attribute is a date then it should be implemented in the database to prevent invalid dates being entered.
Question 41 |
The employee salary should not be greater than Rs.2000. This is
integrity constraint | |
referential constraint | |
Over-defined constraint | |
feasible constraint |
Question 41 Explanation:
● Integrity constraints are used to ensure accuracy and consistency of data in a relational database. Data integrity is handled in a relational database through the concept of referential integrity.
● Integrity constraints are primary key constraints,Foreign Key Constraints, Check Constraints and so on.
● Check (CHK) constraints can be utilized to check the validity of data entered into particular table columns. Check constraints are used to provide back-end database edits, although edits are commonly found in the front-end application as well.
● A referential integrity constraint is specified between two tables.
● In the Referential integrity constraints, if a foreign key in Table 1 refers to the Primary Key of Table 2, then every value of the Foreign Key in Table 1 must be null or be available in Table 2.
● Integrity constraints are primary key constraints,Foreign Key Constraints, Check Constraints and so on.
● Check (CHK) constraints can be utilized to check the validity of data entered into particular table columns. Check constraints are used to provide back-end database edits, although edits are commonly found in the front-end application as well.
● A referential integrity constraint is specified between two tables.
● In the Referential integrity constraints, if a foreign key in Table 1 refers to the Primary Key of Table 2, then every value of the Foreign Key in Table 1 must be null or be available in Table 2.

Question 42 |
the relational algebra expression equivalent to the tuple calculus expression {t|t ∈ r ⋀ (t[A]=10 ⋀ t[B]=20)} is
σ (A=10 v B=20) (r) | |
σ (A=10) (r) Uσ ( B=20) (r) | |
σ (A=10) (r) ∩ σ ( B=20) (r) | |
σ (A=10) (r) - σ ( B=20) (r) |
Question 42 Explanation:
The given relational algebra having the two tuples such as t.a=10 and t⋅B = 20
→ (Tuple having A=10) ∩ (Tuple having B=20) is equal to (Tuples having A=10 and B=20)
→ (Tuple having A=10) ∩ (Tuple having B=20) is equal to (Tuples having A=10 and B=20)
Question 43 |
Let R=(A,B,C,D,E,F) be a relation scheme with the following dependencies:
C→ F, E→ A, EC→ D, A→ B
Which of the following is a key for R?
CD | |
EC | |
AE | |
AC |
Question 43 Explanation:
Here, simple way to solve this question is,
1. First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
2. In right hand side, CE are missing. It means definitely CE should be there in candidate key.
(CE) + =(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.
1. First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
2. In right hand side, CE are missing. It means definitely CE should be there in candidate key.
(CE) + =(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.
Question 44 |
For a database relation R(a,b,c,d) where the domains of a,b,c,d include only atomic values, only the following functional dependencies and those that can be inferred from them hold:
a→ c
b→ d
the relation is in
first normal form but not in second normal form | |
Second normal form but not in third normal form | |
Third normal form | |
None of these |
Question 44 Explanation:
● First normal form (1NF) is a property of a relation in a relational database. A relation is in first normal form if and only if the domain of each attribute contains only atomic (indivisible) values, and the value of each attribute contains only a single value from that domain.
● The Second Normal Form states that it should meet all the rules for 1NF and there must be no partial dependencies of any of the columns on the primary key.
● The questions states that relation consists of automatic values but there no information about prime key.So we don’t about partial dependency.
● The Second Normal Form states that it should meet all the rules for 1NF and there must be no partial dependencies of any of the columns on the primary key.
● The questions states that relation consists of automatic values but there no information about prime key.So we don’t about partial dependency.
Question 45 |
Control Structures include
iteration | |
rendezvous statements | |
exception statements | |
all of these |
Question 45 Explanation:
● A control structure is a block of programming that analyzes variables and chooses a direction in which to go based on given parameters.
● The term flow control details the direction the program takes (which way program control "flows").
● Hence it is the basic decision-making process in computing; flow control determines how a computer will respond when given certain conditions and parameters.
● The term flow control details the direction the program takes (which way program control "flows").
● Hence it is the basic decision-making process in computing; flow control determines how a computer will respond when given certain conditions and parameters.
Question 46 |
According to Brooks, if n is the number of programmers in a project team, then the number of communication path is
n(n-1)/2 | |
nlogn | |
n | |
n(n+1)/2 |
Question 46 Explanation:
● Brooks' law is an observation about software project management according to which "adding human resources to a late software project makes it later.
● According to Brooks, there is an incremental person who, when added to a project, makes it take more, not less time.
● Brooks introduced a formula to measure the volume of communication within the team. It assumes that every team member needs to communicate with every other team member, in order to perform his or her work.
● Group intercommunication formula measures the number of channels of communication within the project team and can be expressed as n*(n-1)/2, where n is the number of people working on a project.
●With 5 software developers working on a project, the number of channels of communication within the project team becomes 5*(5-1)/2 = 10
● With 10 software developers the number of channels of communication within the project team becomes 10*(10-1)/2 = 45
● According to Brooks, there is an incremental person who, when added to a project, makes it take more, not less time.
● Brooks introduced a formula to measure the volume of communication within the team. It assumes that every team member needs to communicate with every other team member, in order to perform his or her work.
● Group intercommunication formula measures the number of channels of communication within the project team and can be expressed as n*(n-1)/2, where n is the number of people working on a project.
●With 5 software developers working on a project, the number of channels of communication within the project team becomes 5*(5-1)/2 = 10
● With 10 software developers the number of channels of communication within the project team becomes 10*(10-1)/2 = 45
Question 47 |
In object oriented design of software, objects have
attributes and name only | |
operations and name only | |
attributes name and operations | |
mutation and permutation property |
Question 47 Explanation:
● Object-oriented design is the process of planning a system of interacting objects for the purpose of solving a software problem.
● An object contains encapsulated data and procedures (operations) grouped together to represent an entity. The 'object interface' defines how the object can be interacted with.
● An object contains encapsulated data and procedures (operations) grouped together to represent an entity. The 'object interface' defines how the object can be interacted with.
Question 48 |
The shell
Accepts command from the user | |
maintains directories of files | |
translates the keyboard character codes | |
none of these |
Question 48 Explanation:
● A Unix shell is a command-line interpreter or shell that provides a command line user interface for Unix-like operating systems.
● The shell is both an interactive command language and a scripting language, and is used by the operating system to control the execution of the system using shell scripts.
● The shell is both an interactive command language and a scripting language, and is used by the operating system to control the execution of the system using shell scripts.
Question 49 |
The extent to which the software can control to operate correctly despite the introduction of invalid input is called as
reliability | |
robustness | |
fault tolerance | |
Portability |
Question 49 Explanation:
● Software Reliability is the probability of failure-free software operation for a specified period of time in a specified environment.
● robustness is the ability of a computer system to cope with errors during execution[1and cope with erroneous input
● Fault tolerance is the property that enables a system to continue operating properly in the event of the failure of some (one or more faults within) of its components.
● Portability in high-level computer programming is the usability of the same software in different environments.
● robustness is the ability of a computer system to cope with errors during execution[1and cope with erroneous input
● Fault tolerance is the property that enables a system to continue operating properly in the event of the failure of some (one or more faults within) of its components.
● Portability in high-level computer programming is the usability of the same software in different environments.
Question 50 |
On an average, the programmer months is given by 3.6 * (KLOC) 1.2 . If so, a project requiring one thousand source instructions will require
3.6PM | |
0.36PM | |
0.0036PM | |
7.23PM |
Question 50 Explanation:
Given programmer months is given by 3.6 * (KLOC) 1.2 and lines of codes are 1000 =1K.
=3.6(1) 1.2 =3.6 Programmer months.
=3.6(1) 1.2 =3.6 Programmer months.
Question 51 |
Relation of COCOMO model is
E=a*(KLOC) 6 | |
E=a*(KLOC) 5 | |
E=a*(KLOC) 7 | |
E=a*(KLOC) 3 | |
None of These |
Question 51 Explanation:
The basic COCOMO equations take the form
Effort Applied (E) = a b (KLOC) b b [ man-months ]
Development Time (D) = c b (Effort Applied) d b [months]
People required (P) = Effort Applied / Development Time [count]
Effort Applied (E) = a b (KLOC) b b [ man-months ]
Development Time (D) = c b (Effort Applied) d b [months]
People required (P) = Effort Applied / Development Time [count]
Question 52 |
Which of the following OS possible in a token passing bus network?
in-service expansion | |
unlimited number of stations | |
both (A) and (B) | |
unlimited distance |
Question 52 Explanation:
Token bus is a network implementing the token ring protocol over a virtual ring on a coaxial cable. A token is passed around the network nodes and only the node possessing the token may transmit. If a node doesn't have anything to send, the token
is passed on to the next node on the virtual ring. Each node must know the address of its neighbour in the ring, so a special protocol is needed to notify the other nodes of connections to, and disconnections from, the ring.
Question 53 |
If L be a language recognizable by a finite automation, then language from {L}={w such that w is prefix of v where v ∈ L}, is a
regular language | |
context free language | |
context sensitive language | |
recursive enumeration language |
Question 54 |
Which of the following statements is correct?
A={anbn |n= 0,1,2,3..} is regular language | |
Set B of all strings of equal number of a's and b's defines a regular language | |
L(A*B*) ∩ B gives the set A | |
None of these |
Question 54 Explanation:
Option-A is not regular language because it require one stack. If they given only an then it belongs to regular language.
Option-B, equal no of a's and equal no of b's is belongs to deterministic context free languages but not regular
Option-C is TRUE. If they given
L(A*B) ∩ B gives the set B but not set A.
L(A*B)= {AB+ AAB + …+AAAAB+B}
L(A*B)∩B=B
Option-B, equal no of a's and equal no of b's is belongs to deterministic context free languages but not regular
Option-C is TRUE. If they given
L(A*B) ∩ B gives the set B but not set A.
L(A*B)= {AB+ AAB + …+AAAAB+B}
L(A*B)∩B=B
Question 55 |
What is the elapsed time of P if records of F are organized using a blocking factor of 2(i.e. each block on D contains two records of F) and P uses one buffer?
12sec | |
14sec | |
17sec | |
21sec | |
None of these |
Question 55 Explanation:
Question and options are wrong. Excluded for evaluation.
There are 55 questions to complete.