TNPSC2017PolytechnicCS
Question 1 
If there are ‘n’ number of states in NFA, then its equivalent DFA may contain atmost ________ number of states.
2^{n}  
n  
n^{2}  
2^{n+1} 
Question 1 Explanation:
If there are ‘n’ number of states in NFA ,then its equivalent DFA may contain atmost 2^{n} number of states.
Question 2 
To get the PDA, the CFG should be in the form of :
CFG  
GNF  
RE  
CNF 
Question 2 Explanation:
To get the PDA, the CFG should be in the form of GNF.
Question 3 
One of the uses of CNF is to turn parse tree into :
AVL trees  
Binary search trees  
Binary trees  
None of the above 
Question 3 Explanation:
One of the uses of CNF is to turn parse trees into binary trees. – These trees have some convenient properties, one of which we exploit here.
Theorem 7.17: Suppose we have a parse tree according to a Chomsky NormalForm grammar G = (V, T, P, S), and suppose that the yield of the tree is a terminal string w.
Theorem 7.17: Suppose we have a parse tree according to a Chomsky NormalForm grammar G = (V, T, P, S), and suppose that the yield of the tree is a terminal string w.
Question 4 
The grammar A→AA(A)∈ is not suitable for predictive parsing because the grammar is:
Ambiguous  
Left  Recursive  
Right Recursive  
An operator grammar 
Question 4 Explanation:
Let’s draw parse tree for string ‘( )’
Since for the given string two parse trees are possible. So the given grammar is ambiguous.
Since for the given string two parse trees are possible. So the given grammar is ambiguous.
Question 5 
The overall logical structure of a database can be expressed graphically by:
Entity relationship model  
Relation model  
Object based model  
Semi structured model 
Question 5 Explanation:
The overall logical data structure of a database can be expressed graphically by an ER diagram. Which consists of rectangle (entity), ellipse (attribute), diamond (relationship), and lines.
Question 6 
The relation schema describes:
Set of tuples (Records)  
Set of fields (Column heads/Attributes)  
Set of associated values  
Domain of each field 
Question 6 Explanation:
The relation schema describes set of fields (Column heads/Attributes).
Question 7 
“X is not a proper subset of any Key” is a 3NF violation called as:
Partial dependency  
Total dependency  
Transitive dependency  
None of the above 
Question 7 Explanation:
If X is not a proper subset of any key means there is no partial dependency. And if there is a violation then the only possibility left is transitive dependency, means if there is transitive dependencies then there is 3NF violation.
Question 8 
ALL or NONE refers:
Consistency  
Isolation  
Durability  
Atomicity 
Question 8 Explanation:
There are four properties of transactions that must be followed,
AAtomicity
CConsistency
IIsolation
DDurability
And the question is asking all or none refers to which of the property.
All or None refers to Atomicity.
AAtomicity
CConsistency
IIsolation
DDurability
And the question is asking all or none refers to which of the property.
All or None refers to Atomicity.
Question 9 
Inheritance achieved by the key word in SQL is:
of  
sub  
under  
from 
Question 9 Explanation:
Inheritance achieved by the keyword in SQL is ‘under’.
Question 10 
Which of the following is not specified in Abstract data type?
Type  
Set of operations on that type  
How the type is implemented  
(A) and (B) 
Question 10 Explanation:
An ADT does not specify how the data type is implemented.
Question 11 
Backtracking uses __________ node generation _____________ bounding functions :
Breadth first, with  
Breadth first, without  
Depthfirst, with  
Depthfirst, without 
Question 11 Explanation:
Backtracking is a depthfirst search with any bounding function.
Question 12 
Which of the following are not valid IPV4 addresses?
192.10.14.3  
200.172.287.33  
65.92.11.00  
10.34.110.77 
Question 12 Explanation:
Each octet of IP address has the maximum value of 255. Since in option B, third octet exceeds the value 255,hence invalid.
Question 13 
In which kind of communication, the destination address in each packet is the same for all duplicate?
Unicasting  
Multicasting  
Multiple unicasting  
Broad casting 
Question 13 Explanation:
In broadcasting communication ,the destination address of all the packets are same since that packet has to be distributed among all the hosts of the network.
Question 14 
In IPV4, using the classful addressing scheme, the whole address space is divided into how many classes:
8  
16  
24  
5 
Question 14 Explanation:
There are 5 classful addressing scheme, Class A, Class B, Class C, Class D, Class D, Class E.
Question 15 
A functional dependency is a relationship between:
Tables  
Attributes  
Rows  
Relations 
Question 15 Explanation:
A functional dependency is a relationship between attributes.
Question 16 
___________ operator is used to retain the unmatched rows of relations when they joined.
Outer join  
Inner join  
Natural join  
Self join 
Question 16 Explanation:
Outer join operator is used to retain the unmatched rows of relations when they joined.
Question 17 
The physical location of a record is determined by a mathematical formula that transforms a file key into a record location is:
B – Tree File  
Hashed File  
Indexed File  
Sequential File 
Question 17 Explanation:
The physical location of a record is determined by a mathematical formula, which transforms a file key into a record location in hashed file. Hashing is the transformation of a string of characters into a usually shorter fixedlength value or key that represents the original string.
Question 18 
Which one of the following statements is FALSE?
A relation with two attributes is in BCNF  
Lossless dependency preserving decomposition into BCNF is always possible  
BCNF is stricter than 3NF  
Lossless, dependency preserving decomposition into 3NF is always possible

Question 18 Explanation:
Option B is false because Lossless decomposition into BCNF is always possible but dependency preserving decomposition into BCNF is not always possible.
Question 19 
The primary objective of formal technical reviews is to:
fix errors  
find errors  
clear errors  
modify errors 
Question 19 Explanation:
Primary objective of formal technical reviews:
Find errors during the process so that they do not become defects after release of the software. Benefit  the early discovery of errors so they do not propagate to the next step in the software process
Find errors during the process so that they do not become defects after release of the software. Benefit  the early discovery of errors so they do not propagate to the next step in the software process
Question 20 
___________ is a class which implement lower level business abstractions required to manage the business domain class :
User interface class  
System class  
Business domain class  
Process class 
Question 20 Explanation:
Process class is a class which implement lower level business abstractions required to manage the business domain class.
Question 21 
If there is a Turing machine that enumerates L in canonical order, L is:
ambiguous  
right – recursive  
left – recursive  
recursive 
Question 21 Explanation:
A language L is recursive iff there exists a TM that enumerates L in canonical order.
Question 22 
Which of the following level of storage is not managed by operating system?
Main memory  
Solid state disk  
Cache  
Magnetic disk 
Question 22 Explanation:
Main memory, cache, magnetic disk are managed by operating system but solid state disk is not managed by operating system.
Question 23 
Which of the following is a Pass 1 task in a typical assembler?
Generate data  
Generate instructions  
Look up value of symbols  
Determine length of machine instructions 
Question 23 Explanation:
Generating data,generating instructions and look up values of symbols are done in pass 2 of assembler. But Determining length of machine instructions is done in pass 1.
Question 24 
Using a larger block size in a fixed block size file system leads to :
Better disk throughout but poor disk space utilization  
Better disk throughout and better disk space utilization  
Poor disk throughput but better disk space utilization  
Poor disk throughput and poor disk space utilization 
Question 24 Explanation:
Larger block size means less number of blocks to fetch and hence better throughput. But larger block size also means space is wasted when only small size is required.
Question 25 
Suppose that a process spends a fraction p of its time waiting for I/O to complete. With n processes in memory at once, the probability that all n processes are waiting for I/O is Pn. What is the CPU utilization?
1 – p^{n}  
1 – n^{p}  
1/p^{n}  
1/n^{p} 
Question 25 Explanation:
The CPU will be idle only if all the process are waiting for I/O, and probability of all process waiting for I/O is p^{n}. So therefore probability of CPU utilization is (1p^{n}).
Question 26 
A ___________ can forward packets across different networks that may also use different protocols :
Repeater  
Bridge  
Router  
Gateway 
Question 26 Explanation:
A gateway can forward packets across different networks that may also use different protocol, because gateway is also used as protocol converter.
Question 27 
A circle, if scaled only in one direction becomes a :
Ellipse  
Parabola  
Hyperbola  
Remains as a circle 
Question 27 Explanation:
A circle, if scaled only in one direction becomes a ellipse.
Question 28 
The point at which a set of projected parallel lines appear to converge is called :
Convergence point  
Vanishing point  
Point of illusion  
Point of delusion 
Question 28 Explanation:
A vanishing point is a point on the image plane of a perspective drawing where the twodimensional perspective projections (or drawings) of mutually parallel lines in threedimensional space appear to converge.
Question 29 
The ISO standard for computer graphics is?
Graphics kernel system  
Graphics standard system  
Computer graphics standard  
None of the above 
Question 29 Explanation:
The Graphical Kernel System (GKS) was the first ISO standard for lowlevel computer graphics, introduced in 1977. A draft international standard was circulated for review in September 1983. Final ratification of the standard was achieved in 1985.
Question 30 
A Image compression technique that determines the most frequently occurring pairs of bytes is called:
Run length encoding  
Diatomic encoding  
Huffman encoding  
Arithmetic encoding 
Question 30 Explanation:
Diatomic encoding is a variation based on combinations of two data bytes. This technique determines the most frequently occurring pairs of bytes.
Question 31 
Ax^{2} + By^{2} + Cxy + Dx + Ey + F = 0 represents a :
Line  
Conic section  
Circle  
None of the above 
Question 31 Explanation:
Circle equations are often given in the general format of Ax^{2} + By^{2} + Cx + Dy + E = 0.
Question 32 
The standard graphics objects are :
Line  
Point  
Polygon  
None of the above 
Question 32 Explanation:
The standard graphics objects are polygon surfaces.
Question 33 
The wavelength of visible spectrum falls in :
400 nm to 500 nm  
500 nm to 700 nm  
600 nm to 700 nm  
400 nm to 700 nm 
Question 33 Explanation:
Visible light falls within the range of 400700 nm.
Question 34 
JPEG is a :
Image compression standard  
Image file format  
Both (A) and (B)  
Video file format 
Question 34 Explanation:
JPEG is a popular image file format. It is commonly used by digital cameras to store photos since it supports 2^{24} or 16,777,216 colors. The format also supports varying levels of compression, which makes it ideal for web graphics.
Question 35 
The major functions of the presentation layer in OSI stack are :
Dialog presentation and control  
Dealing with differences in data representation, encryption and compression  
Presenting the full and half duplex services to the user  
Enabling end to end error control presentation 
Question 35 Explanation:
The main function of presentation layer in OSI are translation, encryption and data compression.
Question 36 
A code with a Hamming distance d can :
Detect d bit errors and correct (d – 1) bit errors.  
Detect (d – 1) bit errors only with no error correction.  
Detect (d – 1) bit errors and correct (d – 1)/2 bit errors.  
Detect and correct all d bit errors. 
Question 36 Explanation:
To detect t bit error hamming distance should be t+1. And to correct t bit error hamming distance should be 2t+1.
So using above solution most appropriate answer is option C.
So using above solution most appropriate answer is option C.
Question 37 
In a stop and wait protocol used across a link of bandwidth of 1Mbps, data packets to 1000 bits are transmitted. The round trip time for a bit is 20ms. The link utilization is :
0.5  
0.05  
0.005  
5.0 
Question 37 Explanation:
link utilization = 1/1+2a
a = T_{p}/T_{t}
T_{p} = RTT/2 = 20/2 = 10ns
T_{t} = 1000bits/10^{6}bits/s = 1ms
∴ a = 10/1 = 10
∴ Link utilization = 1/(1+2×10) = 1/21 ≅ 0.05
a = T_{p}/T_{t}
T_{p} = RTT/2 = 20/2 = 10ns
T_{t} = 1000bits/10^{6}bits/s = 1ms
∴ a = 10/1 = 10
∴ Link utilization = 1/(1+2×10) = 1/21 ≅ 0.05
Question 38 
A cryptographic system that uses only symmetric key cryptography cannot provide digital signature because :
Symmetric key cryptography is computationally infeasible.  
Symmetric key cryptography involves key distribution.  
Symmetric key cryptography is unreliable.  
Digital signature requires a pair of private – public keys. 
Question 38 Explanation:
Digital signature requires a pair of privatepublic keys, due to which symmetric key cryptography cannot provide digital signature.
Question 39 
If a message “CONGRATS” is encoded as “AMLEPYRQ”, the encryption key is :
+ 3  
+ 2  
– 3  
– 2 
Question 39 Explanation:
C2 = A
O2 = M
N  2 = L
G2 = E
R2 = P
A2 = Y
T2 = R
S2 = Q
O2 = M
N  2 = L
G2 = E
R2 = P
A2 = Y
T2 = R
S2 = Q
Question 40 
___________ is a small program that switches the processor from one process to another.
Scheduler  
Dispatcher  
Swapper  
Lazy swapper 
Question 40 Explanation:
Dispatcher is a small program that switches the processor from one process to another. Scheduler only selects the process from ready queue and not do context switching, it is the dispatcher which do context switching.
Question 41 
Which one of the following is not a consumable resources?
Interrupts  
Signals  
I/O devices  
Messages 
Question 41 Explanation:
Examples of consumable resources are interrupts, signals, messages, and information in I/O buffers.
Question 42 
The size of a page is typically a :
Multiple of 8  
Power of 2  
Any size depending on operating system  
Any size depending on user program 
Question 42 Explanation:
The size of a page is typically a power of 2 because the physical addresses and virtual addresses are represented in bits.
Question 43 
__________ Algorithm is sometimes called the elevator algorithm.
FCFS Scheduling  
SCAN Scheduling  
C – SCAN Scheduling  
Look Scheduling 
Question 43 Explanation:
Scan Algorithm. It is also called as Elevator Algorithm. In this algorithm, the disk arm moves into a particular direction till the end, satisfying all the requests coming in its path,and then it turns back and moves in the reverse direction satisfying requests coming in its path.
Question 44 
C(5, 2) is not equal to :
C (5, 3)  
20  
10  
5!/3!2! 
Question 44 Explanation:
⇒ C(5,2) = ^{5}C_{2} = ^{5}C_{3} (∴ ^{n}C_{r} = ^{n}C_{nr})
⇒ ^{5}C_{2} = ∠5/∠3∠2 (∴ ^{n}C_{r} = ∠n/∠r∠nr)
⇒ ^{5}C_{2} = ∠5/∠3∠2 = 5×4/2 = 10
⇒ ^{5}C_{2} = ∠5/∠3∠2 (∴ ^{n}C_{r} = ∠n/∠r∠nr)
⇒ ^{5}C_{2} = ∠5/∠3∠2 = 5×4/2 = 10
Question 45 
The value of a^{1} + a^{1}.b^{1} + b^{1} + a + 0 is :
a^{1} + b^{1}  
b^{1} + a  
1  
0 
Question 45 Explanation:
a’ + a’b’ + b’ + a + 0
= a’(a+b’) + b’ + a + 0
= a’ + a + b’ (∵1+x=1)
= 1+b (∵ x’+x=1)
= 1 (∵ 1+x=1)
= a’(a+b’) + b’ + a + 0
= a’ + a + b’ (∵1+x=1)
= 1+b (∵ x’+x=1)
= 1 (∵ 1+x=1)
Question 46 
If p and q are two statements and they take truth values p = 1 and q = 1, then their conjunction p and q written as p ^ q takes truth value :
0  
1  
1  
None 
Question 46 Explanation:
Question 47 
The conditional probability of event A given event B is defined as :
P(A/B)= P(A∩B)/P(B)  
P(A/B)= P(A∩B)/P(A)  
P(B/A)= P(A∩B)/P(A)  
P(B/A)= P(A∩B)/P(B) 
Question 47 Explanation:
The conditional probability of event A given event B is defined as
P(A/B)= P(A∩B)/P(B)
P(A/B)= P(A∩B)/P(B)
Question 48 
An SD RAM has 8K rows, with an access time of 4 clock cycles for each row, and a refresh period of 64 ms. If the clock rate is 133 MHz, the refresh overhead will be :
0.0038  
0.246  
0.68  
4.35 
Question 48 Explanation:
No of cycles require to refresh= 8k ×4 = 2^{15}
Amount of time to refresh = 2^{15} × (1/133×10^{6}) secs = 0.246 ms
Overhead = (0.246/64) = 0.0038
Amount of time to refresh = 2^{15} × (1/133×10^{6}) secs = 0.246 ms
Overhead = (0.246/64) = 0.0038
Question 49 
The Booth technique for recording multiply of +13 and –6 [01101 and 11010] is :
1110 0011 01  
1110 1100 10  
1110 1010 10  
1110 0011 00 
Question 49 Explanation:
Firstly,
13 × 6 = 78
Let’s first find binary value of 78,
Now to get 78 let’s take 2’s complement of above binary no.,
10110010
which can be also written as,
1110110010
13 × 6 = 78
Let’s first find binary value of 78,
Now to get 78 let’s take 2’s complement of above binary no.,
10110010
which can be also written as,
1110110010
Question 50 
MFLOPS can be abbreviated as :
Millions of Floating – Point operations performed per second.  
Millions of Fixed – Length operations performed per second.  
Millions of Floating – Limited operations performed per second.  
Millions of Fixed – Limited operations performed per second. 
Question 50 Explanation:
Short for mega floatingpoint operations per second, MFLOPs are a common measure of the speed of computers used to perform floatingpoint calculations.
Question 51 
A router must have atleast __________ NICs
3  
4  
2  
5 
Question 51 Explanation:
There are multiple network interfaces in a router. Routers route between networks, so there must be at least two, often more, network interfaces in a router.
Question 52 
If an IP address starts with a bit sequence of 11110, it is a class _______ address.
B  
C  
D  
E 
Question 52 Explanation:
Class A starts with 0.
Class B starts with 10.
Class C starts with 110.
Class D starts with 1110.
Class E starts with 11110.
Class B starts with 10.
Class C starts with 110.
Class D starts with 1110.
Class E starts with 11110.
Question 53 
The gateway that stands between the mobile network and the Internet in GPRS is called as _______.
CCNS  
SGGN  
SGSN  
CGSN 
Question 53 Explanation:
A Gateway GPRS Support Node (GGSN) is part of the core network that connects GSMbased 3G networks to the Internet. The GGSN, sometimes known as a wireless router, works in tandem with the Serving GPRS Support Node (SGSN) to keep mobile users connected to the Internet and IPbased applications.
Question 54 
The value of the expression (a + b) (a + b^{1}) (a +b^{1}b) is :
b^{1}  
a^{1}  
a  
b 
Question 54 Explanation:
(a + b)(a + b’)(a + b’b)
= (a + b)(a + b’)(a + 0) [x・x’ = 0]
= (a + b)(a + b’)・a
= (a + b・b’)・a
= (a + 0)・a
= a・a
= a
= (a + b)(a + b’)(a + 0) [x・x’ = 0]
= (a + b)(a + b’)・a
= (a + b・b’)・a
= (a + 0)・a
= a・a
= a
Question 55 
The disadvantage of write back strategy in cache is that :
If generates repeated memory traffic  
It creates a write mechanism whenever there is a write operation to cache  
Portions of main memory may be invalid  
It requires local cache memory attached to every CPU in a multi processor environment 
Question 55 Explanation:
There is data availability risk because the cache could fail (and so suffer from data loss) before the data is persisted to the backing store. This result in the data being lost.
Question 56 
Which of the following possibilities for saving the return address of a sub – routine, support sub – routine recursion?
In a processor register  
In a memory location associated with the call  
On a stack  
All of the above 
Question 56 Explanation:
The return address is saved in some specific location of main memory associated with the call.
Question 57 
The unit responsible for tracking the next instruction to be executed in :
ALU  
Memory Address Register  
Program counter  
Instruction memory

Question 57 Explanation:
Program counter keeps the address of the next instruction to be executed.
Question 58 
What is the best case running time of binary search?
θ(n)  
θ(1)  
θ(log n)  
θ(n log log n) 
Question 58 Explanation:
Best case running time complexity of binary search is O(1) and worst case running time complexity of binary search is O(log n).
Question 59 
How many binary trees are possible with three nodes?
3  
4  
6  
5 
Question 59 Explanation:
No. of binary trees possible with n nodes is,
(2n)!/n!(n+1)!
So, for n=3,
6!/3!4! = 6×5/3×2 = 5
(2n)!/n!(n+1)!
So, for n=3,
6!/3!4! = 6×5/3×2 = 5
Question 60 
Which Sorting method is an external Sort?
Heap Sort  
Quick Sort  
Insertion Sort  
None of the above 
Question 60 Explanation:
External sorting is a term for a class of sorting algorithms that can handle massive amounts of data. External sorting is required when the data being sorted do not fit into the main memory of a computing device (usually RAM) and instead they must reside in the slower external memory (usually a hard drive). External sorting typically uses a hybrid sortmerge strategy.
Example of external sorting is merge sort.
Example of external sorting is merge sort.
Question 61 
Between any two vertices, there exists a path, then the graph is said to be
Strongly Connected  
Connected  
Weakly Connected  
All the above 
Question 61 Explanation:
If between any two vertices there exists a path then the graph is said to be strongly connected and every strongly connected graph is weakly connected(vice versa need not be true). And moreover it is connected graph.
Question 62 
What is the asymptotic value for the recurrence equation T(n) = 2T(n/2) + n?
O(n)  
O(n^{2})  
O(n^{2} log n)  
O(n log n) 
Question 62 Explanation:
T(n) = 2T(n/2) + n
a=2, b=2, k=1, p=0
Now, a=b^{k} and p>1
So from Master’s theorem,
a=2, b=2, k=1, p=0
Now, a=b^{k} and p>1
So from Master’s theorem,
Question 63 
In which order the edges of the given graph are chosen while constructing the minimum spanning tree using prim’s algorithm?
(1, 6), (6, 5), (5, 4), (4, 3), (3, 2), (2, 7)  
(1, 6), (3, 4), (2, 7), (2, 3), (7, 4), (5, 4)  
(1, 6), (3, 4), (2, 7), (4, 5), (1, 2), (5, 6)  
(1, 6), (6, 5), (5, 4), (4, 3), (3, 2), (4, 7) 
Question 63 Explanation:
The Prim's algorithm operates by building this tree one vertex at a time, from an arbitrary starting vertex, at each step adding the cheapest possible connection from the tree to another vertex.
B cant be the answer because after edge (1,6) is considered then (3,4) cant be taken because vertex 3 or 4 is not connected directly to the tree having edge (1,6).
C cant be the answer. Same reason as B.
D cant be the answer because after edge (3,2) is considered then next cheapest edge is (2,7) and not (4,7).
B cant be the answer because after edge (1,6) is considered then (3,4) cant be taken because vertex 3 or 4 is not connected directly to the tree having edge (1,6).
C cant be the answer. Same reason as B.
D cant be the answer because after edge (3,2) is considered then next cheapest edge is (2,7) and not (4,7).
Question 64 
Dijkstra’s algorithm follows ___________ method of algorithm design. The complexity of the algorithm to find the shortest path from a vertex to all other vertices in a graph is __________
Dynamic programming, O (n^{2})  
Dynamic programming, O (log n)  
Greedy, O (n^{2})  
Greedy, O (log n)  
None of the above 
Question 64 Explanation:
Dijkstra's algorithm follows greedy method of algorithm design. The complexity of algorithm to find the shortest path from a vertex to all other vertices in a graph is O(ElogV).
Question 65 
If there are 64 pages, and the page size is 4096 words, the length of the logical address is _________.
16 bits  
18 bits  
20 bits  
22 bits 
Question 65 Explanation:
No. of bits required to indicate no. of pages = log_{2} 64 = 6
No. of bits required to indicate page size = log_{2} 4096 = 12
∴ Length of logical address is,
6 + 12 = 18
No. of bits required to indicate page size = log_{2} 4096 = 12
∴ Length of logical address is,
6 + 12 = 18
Question 66 
__________ is used to obtain the IP address of a host based on its physical address.
RARP  
IPV6  
TFTP  
TELNET 
Question 66 Explanation:
RARP is used to obtain the IP address of a host based on its physical address.
Question 67 
Match the terms with the definition.
(a) Masquerading (i) Session is intercepted (b) Phishing (ii) One pretends to be someone else (c) Hijacking (iii) A email misleads a user into entering confidential information Codes: (a) (b) (c)
(i) (ii) (iii)
 
(i) (iii) (ii)
 
(iii) (ii) (i)  
(ii) (iii) (i) 
Question 67 Explanation:
MasqueradingOne pretends to be someone else.
PhishingPhishing is a cybercrime in which a target or targets are contacted by email, telephone or text message by someone posing as a legitimate institution to lure individuals into providing sensitive data such as personally identifiable information, banking and credit card details, and passwords.
HijackingHijacking is a type of network security attack in which the attacker takes control of a communication  just as an airplane hijacker takes control of a flight  between two entities and masquerades as one of them. So interception between two entities is done.
PhishingPhishing is a cybercrime in which a target or targets are contacted by email, telephone or text message by someone posing as a legitimate institution to lure individuals into providing sensitive data such as personally identifiable information, banking and credit card details, and passwords.
HijackingHijacking is a type of network security attack in which the attacker takes control of a communication  just as an airplane hijacker takes control of a flight  between two entities and masquerades as one of them. So interception between two entities is done.
Question 68 
Consider the following dependencies :
AB → CD, AF → D, DE → F, C → G, F → E, G → AWhich one of the following options is false ?
BG^{+}={ABCDG}  
CF^{+}={ACDEFG}  
AB^{+}={ABCDG}  
AF^{+}={ACDEFG} 
Question 68 Explanation:
(BG)^{+} = BGACD
(CF)^{+} = CFGEAD
(AB)^{+} = ABCDG
(AF)^{+} = AFDE
Hence option (D) is False.
(CF)^{+} = CFGEAD
(AB)^{+} = ABCDG
(AF)^{+} = AFDE
Hence option (D) is False.
Question 69 
Consider the following transactions with data items X and y initialized to zero :
T_{1}: read x; read (y) if x=0 then y≔y+1; write y; T_{2}: read y; write x; if y=0 then x≔x+1 write x;The concurrent execution of T_{1} and T_{2} leads to
Serializable schedule  
A schedule that is not conflict serializable  
A conflict serializable schedule  
A schedule for which a precedence graph cannot be drawn 
Question 69 Explanation:
One of the schedule that is possible is,
Let’s check for conflict serializability,
Let’s check for conflict serializability,
Question 70 
onsider the expression t ε instructor ⋏∃ s ∈ department (t [dept_name] = s [dept_name])
The variables t and s are ______________ respectively.
The variables t and s are ______________ respectively.
Free variable and bound variable  
Bound variable and free variable  
Free variable and free variable  
Bound variable and bound variable 
Question 70 Explanation:
If the variables is used along with quantifiers ∃ and ∀ then that variable is bounded variable otherwise it is a free variable.
Question 71 
The materialization approach of query evaluation includes (from root to leaf) :
π, ∞, σ  
π, σ, ∞  
σ, π, ∞  
σ, ∞, π 
Question 71 Explanation:
In SQL we write
SELECT
FROM
WHERE
in a sequence.
So it is equivalent to
SELECTπ
FROM ∞
WHEREσ
SELECT
FROM
WHERE
in a sequence.
So it is equivalent to
SELECTπ
FROM ∞
WHEREσ
Question 72 
If T_{1} and T_{2} are average access times of upper level memory M1 and lower level memory M2 in a 2 – level memory hierarchy and H is the hit rate in M1, then the overall average access time is given by ________, assuming that in case of a miss in M1, a block is first copied from M2 to M1 and then accessed from M1 :
T_{1} + (1H) × T_{2}  
(1  H) × T_{1} + T_{2}  
T_{1} + (1  H)  
T_{1} + (1 + H) 
Question 72 Explanation:
Average access time
= hit rate × T_{1} + miss rate × (T_{1} + T_{2})
= H × T_{1} + (1  H)(T_{1} + T_{2})
= H × T_{1} + T_{1}  T_{1} H + T_{2}  T_{2}H
= T_{1} + (1  H)T_{2}
= hit rate × T_{1} + miss rate × (T_{1} + T_{2})
= H × T_{1} + (1  H)(T_{1} + T_{2})
= H × T_{1} + T_{1}  T_{1} H + T_{2}  T_{2}H
= T_{1} + (1  H)T_{2}
Question 73 
Which of the following is the recurrence relation for binary search?
T(n) = T(n/2) + 1  
T(n) = T(n/2) + n  
T(n) = 2T(n  1) + 1  
T(n) = T(n  1) + 1 
Question 73 Explanation:
Aapplying masters theorem we get O(logn) which is the time complexity of binary search.
B applying masters theorem we get O(n) which is not the time complexity for binary search.
C Using back substitution we get O(2^{n}) which is not the time complexity for binary search.
DUsing back substitution we get time complexity as O(n) which is not of binary search.
B applying masters theorem we get O(n) which is not the time complexity for binary search.
C Using back substitution we get O(2^{n}) which is not the time complexity for binary search.
DUsing back substitution we get time complexity as O(n) which is not of binary search.
Question 74 
Which one of the following is a Meldable priority queue?
Leftist Heap  
Binary Heap  
AVL trees  
Red – Black trees 
Question 74 Explanation:
The classic way to implement a priority queue is using a data structure called a binary heap. A binary heap will allow us to enqueue or dequeue items in O ( log n ).
Question 75 
x = at^{2}; y = 2at is the parametric equation of:
Circle  
Parabola  
Rectangular hyperbola  
Ellipse 
Question 75 Explanation:
x=at^{2}, y=2at is the parametric equation of y^{2} = 4ax which is parabola.
Question 76 
An example for a pattern – scanning language is :
lex  
awk  
bison  
yacc 
Question 76 Explanation:
An example for a pattern  scanning language is lex.
Question 77 
Consider the following grammar :
S → La L → LSSAfter the elimination of left – recursion, we get the following:
S→La L→SA A→SAε  
S→LA A→SAL L→aSLε  
S→Lε L→SA A→SAε  
S→La A→aSAε L→aSA 
Question 77 Explanation:
S → (L)a
L → LSS //contains left recursion
Now eliminating left recursion we get,
S → (L)a
L → SA
A → SA∊
Using formula,
A → Aαβ
⇓
A → βA’
A’ → αA’∊
L → LSS //contains left recursion
Now eliminating left recursion we get,
S → (L)a
L → SA
A → SA∊
Using formula,
A → Aαβ
⇓
A → βA’
A’ → αA’∊
Question 78 
The Language L = {a^{p} p is prime } is :
regular  
not regular  
accepted by NFA with ε  
None 
Question 78 Explanation:
The given language is not regular because because p is not in arithmetic sequence.
Question 79 
Consider the grammar
S→AS/b A→SA / a then Closure (S'→.S,$) is:
S→AS/b A→SA / a then Closure (S'→.S,$) is:
S^{1}→.S,$ S→.AS,$ / a/b S→.b,$/a/b A→.SA, a/b A→.a,a /b  
S^{1}→.S,$ S→.AS,$ s/ b S→.b,$ / b  
S^{1}→.S,$ S→.AS,$ / a/b S→.AS,$ / a/b S→.b,$ / a/b  
S^{1}→.S,$ S→.AS,$ S→.b,$ 
Question 79 Explanation:
The given grammar is
S → ASb
A → SAa
Now closure for S →.S, $,
S’ → .S, $
S → .AS.b, $
A → .SA.a, ab
S → .AS.b, ab
⇓
S → .S, $
S → .AS.b, $ab
A → .SA.a, ab
S → ASb
A → SAa
Now closure for S →.S, $,
S’ → .S, $
S → .AS.b, $
A → .SA.a, ab
S → .AS.b, ab
⇓
S → .S, $
S → .AS.b, $ab
A → .SA.a, ab
Question 80 
How many host interfaces may be addressed in the subnet 123.224.00.00/11?
2048  
2,097,150  
1,000,192  
2,097,152 
Question 80 Explanation:
No. of host id bits is 3211 = 21
So, no. of host interfaces that can be addressed in the subnet is 2^{21}  2 = 2097150.
So, no. of host interfaces that can be addressed in the subnet is 2^{21}  2 = 2097150.
Question 81 
Which of the following make(s) filtering decisions based on application payload?
packet filter  
deep inspection firewall  
reverse proxy  
stateful packet inspection firewall 
Question 81 Explanation:
Filtering decisions based on application payload requires all the five layers till application layer .And deep inspection firewall have all the five layers.
Question 82 
If the data frame is 1101011011 and the divisor is 10011 in a CRC error detection process, a burst error 0000010011 occurs in transmission. Justify whether it will be detected:
With a high probability, it will be detected.  
It will not be detected, as the burst error is exactly identical to the divisor.  
With a very small probability, it will be detected.  
It depends on the data rate in the channel. So it may or may not be detected.

Question 82 Explanation:
CRC can detect all burst error of length less than the degree of the polynomial. But in the given question the burst error is exactly identical to the divisor, i.e., 4, so cant be detected.
Question 83 
If X is uniformly distributed in (–2,3), then its variance is __________.
15/12  
35/12  
25/12  
17/12 
Question 83 Explanation:
If X is uniformly distributed in (2,3), then its variance is
(3(2))^{2}/12 = 25/12
(3(2))^{2}/12 = 25/12
Question 84 
The lifetime (in years) of a radio has an exponential distribution with parameter λ = 1/10. If we buy a five – year – old radio, what is the probability that it will work for less than 10 additional years?
e^{1}  
e^{2}  
1  e^{1}  
1  e^{2} 
Question 84 Explanation:
Question 85 
The Product of Sums form of the expression a.b + a^{1}.c is :
(a+b).(a+c^{1})  
(a+c).(a^{1}+b)  
(a^{1}+b^{1}).(a+c^{1})  
(a^{1}+b).(a^{1}+c) 
Question 85 Explanation:
Let's draw kmap for given equation,
ab + a’c
Now let's find product of sum,
(a + c)(c’ + b)
ab + a’c
Now let's find product of sum,
(a + c)(c’ + b)
Question 86 
In a class of 45 students, a boy is ranked 20^{th}. When two boys joined, his rank was dropped by one. What is his new rank from the end ?
25^{th}  
26^{th}  
27^{th}  
28^{th} 
Question 86 Explanation:
Initially the no. of students in the class is 45, and the rank of boy was 20.
Now two boys joined, so now the total no. of students in the class is 47. Also the rank of boy now is 21.
So the new rank of boy from the end is,
47  21 + 1 = 26 + 1 = 27
Now two boys joined, so now the total no. of students in the class is 47. Also the rank of boy now is 21.
So the new rank of boy from the end is,
47  21 + 1 = 26 + 1 = 27
Question 87 
Who won the gold both in the 5,000 and 10,000 metres event in 2017 Asian Athletics Championship?
Lakshmanan  
Gopy Thonkanal  
Jinson Johnson  
Neeraj Chopra 
Question 87 Explanation:
G lakshamanan won the gold both in the 5000 and 10000 metres event in 2017 asian athletics championship.
Question 88 
What temperature are Fahrenheit and Celsius equal?
–40°  
574.59  
40  
–574.59 
Question 88 Explanation:
The Fahrenheit and Celsius scales have one point at which they intersect. They are equal at 40 °C and 40 °F.
Question 89 
First state to fix minimum education qualification for cooperative body poll:
Rajasthan  
West Bengal  
Tamil Nadu  
Karnataka 
Question 89 Explanation:
Rajasthan, First State to fix minimum education qualification for cooperative body poll.
Question 90 
Who wrote the novel – ‘KavalKottam’?
Vannadasan  
S. Venkatesan  
Joe D Cruz  
Puviarasan 
Question 90 Explanation:
S. Venkatesan wrote the novel ’ KavalKottam’.
Question 91 
Article 21A and the RTE Act came into effect:
On 1st April 2010  
On 1st April 2009  
On 1st April 2017  
On 1st April 2005 
Question 91 Explanation:
Article 21A and the RTE Act came into effect on 1 April 2010. The title of the RTE Act incorporates the words 'free and compulsory'. ... The RTE Act provides for the: Right of children to free and compulsory education till completion of elementary education in a neighbourhood school.
There are 91 questions to complete.