## TNPSC-2017-Polytechnic-CS

 Question 1
If there are ‘n’ number of states in NFA, then its equivalent DFA may contain atmost ________ number of states.
 A 2n B n C n2 D 2n+1
Theory-of-Computation       Finite-Automata
Question 1 Explanation:
If there are ‘n’ number of states in NFA ,then its equivalent DFA may contain atmost 2n number of states.
 Question 2
To get the PDA, the CFG should be in the form of :
 A CFG B GNF C RE D CNF
Theory-of-Computation       Finite-Automata
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 :
 A AVL trees B Binary search trees C Binary trees D None of the above
Theory-of-Computation       Context-Free-Grammar
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- Normal-Form 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:
 A Ambiguous B Left - Recursive C Right Recursive D An operator grammar
Compiler-Design       Parsers
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.
 Question 5
The overall logical structure of a database can be expressed graphically by:
 A Entity relationship model B Relation model C Object based model D Semi structured model
Database-Management-System       ER-Model
Question 5 Explanation:
The overall logical data structure of a database can be expressed graphically by an E-R diagram. Which consists of rectangle (entity), ellipse (attribute), diamond (relationship), and lines.
 Question 6
The relation schema describes:
 A Set of tuples (Records) B Set of fields (Column heads/Attributes) C Set of associated values D Domain of each field
Database-Management-System       Relational Schema
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:
 A Partial dependency B Total dependency C Transitive dependency D None of the above
Database-Management-System       Normalization
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:
 A Consistency B Isolation C Durability D Atomicity
Database-Management-System       Transactions
Question 8 Explanation:
There are four properties of transactions that must be followed,
A-Atomicity
C-Consistency
I-Isolation
D-Durability
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:
 A of B sub C under D from
Database-Management-System       SQL
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?
 A Type B Set of operations on that type C How the type is implemented D (A) and (B)
OOPS       Properties
Question 10 Explanation:
An ADT does not specify how the data type is implemented.
 Question 11
Backtracking uses __________ node generation _____________ bounding functions :
 A Breadth first, with B Breadth first, without C Depth-first, with D Depth-first, without
Algorithms       Back-Tracking
Question 11 Explanation:
Backtracking is a depth-first search with any bounding function.
 Question 12
Which of the following are not valid IPV4 addresses?
 A 192.10.14.3 B 200.172.287.33 C 65.92.11.00 D 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?
 A Unicasting B Multicasting C Multiple unicasting D 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:
 A 8 B 16 C 24 D 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:
 A Tables B Attributes C Rows D Relations
Database-Management-System       Functional-Dependency
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.
 A Outer join B Inner join C Natural join D Self join
Database-Management-System       Relational-Algebra
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:
 A B – Tree File B Hashed File C Indexed File D Sequential File
Database-Management-System       Transactions
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 fixed-length value or key that represents the original string.
 Question 18
Which one of the following statements is FALSE?
 A A relation with two attributes is in BCNF B Lossless dependency preserving decomposition into BCNF is always possible C BCNF is stricter than 3NF D Lossless, dependency preserving decomposition into 3NF is always possible
Database-Management-System       Normalization
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:
 A fix errors B find errors C clear errors D 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
 Question 20
___________ is a class which implement lower level business abstractions required to manage the business domain class :
 A User interface class B System class C Business domain class D Process class
OOPS       Properties
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:
 A ambiguous B right – recursive C left – recursive D recursive
Theory-of-Computation       Turing-machines
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?
 A Main memory B Solid state disk C Cache D Magnetic disk
Operating-Systems       Working-of-operating-system
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?
 A Generate data B Generate instructions C Look up value of symbols D 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 :
 A Better disk throughout but poor disk space utilization B Better disk throughout and better disk space utilization C Poor disk throughput but better disk space utilization D Poor disk throughput and poor disk space utilization
Computer-Organization       Secondary-Memory
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?
 A 1 – pn B 1 – np C 1/pn D 1/np
Engineering-Mathematics       Probability
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 pn. So therefore probability of CPU utilization is (1-pn).
 Question 26
A ___________ can forward packets across different networks that may also use different protocols :
 A Repeater B Bridge C Router D Gateway
Engineering-Mathematics       Probability
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 :
 A Ellipse B Parabola C Hyperbola D Remains as a circle
Engineering-Mathematics       Co-ordinate-Geometry
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 :
 A Convergence point B Vanishing point C Point of illusion D Point of delusion
Question 28 Explanation:
A vanishing point is a point on the image plane of a perspective drawing where the two-dimensional perspective projections (or drawings) of mutually parallel lines in three-dimensional space appear to converge.
 Question 29
The ISO standard for computer graphics is?
 A Graphics kernel system B Graphics standard system C Computer graphics standard D None of the above
Question 29 Explanation:
The Graphical Kernel System (GKS) was the first ISO standard for low-level 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:
 A Run length encoding B Diatomic encoding C Huffman encoding D 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
Ax2 + By2 + Cxy + Dx + Ey + F = 0 represents a :
 A Line B Conic section C Circle D None of the above
Engineering-Mathematics       Co-ordinate-Geometry
Question 31 Explanation:
Circle equations are often given in the general format of Ax2 + By2 + Cx + Dy + E = 0.
 Question 32
The standard graphics objects are :
 A Line B Point C Polygon D None of the above
Question 32 Explanation:
The standard graphics objects are polygon surfaces.
 Question 33
The wavelength of visible spectrum falls in :
 A 400 nm to 500 nm B 500 nm to 700 nm C 600 nm to 700 nm D 400 nm to 700 nm
Question 33 Explanation:
Visible light falls within the range of 400-700 nm.
 Question 34
JPEG is a :
 A Image compression standard B Image file format C Both (A) and (B) D 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 224 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 :
 A Dialog presentation and control B Dealing with differences in data representation, encryption and compression C Presenting the full and half duplex services to the user D Enabling end to end error control presentation
Computer-Networks       ISO-OSI-layers
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 :
 A Detect d bit errors and correct (d – 1) bit errors. B Detect (d – 1) bit errors only with no error correction. C Detect (d – 1) bit errors and correct (d – 1)/2 bit errors. D Detect and correct all d bit errors.
Digital-Logic-Design       Number-Systems
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.
 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 :
 A 0.5 B 0.05 C 0.005 D 5
Computer-Networks       Stop-and-Wait-ARQ
Question 37 Explanation:
a = Tp/Tt
Tp = RTT/2 = 20/2 = 10ns
Tt = 1000bits/106bits/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 :
 A Symmetric key cryptography is computationally infeasible. B Symmetric key cryptography involves key distribution. C Symmetric key cryptography is unreliable. D Digital signature requires a pair of private – public keys.
Computer-Networks       Network-Security
Question 38 Explanation:
Digital signature requires a pair of private-public 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 :
 A + 3 B + 2 C – 3 D – 2
Computer-Networks       Network-Security
Question 39 Explanation:
C-2 = A
O-2 = M
N - 2 = L
G-2 = E
R-2 = P
A-2 = Y
T-2 = R
S-2 = Q
 Question 40
___________ is a small program that switches the processor from one process to another.
 A Scheduler B Dispatcher C Swapper D Lazy swapper
Operating-Systems       Process-State-Transition-Diagram
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?
 A Interrupts B Signals C I/O devices D Messages
Computer-Organization       I/O-Devices
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 :
 A Multiple of 8 B Power of 2 C Any size depending on operating system D Any size depending on user program
Operating-Systems       Memory-Management
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.
 A FCFS Scheduling B SCAN Scheduling C C – SCAN Scheduling D Look Scheduling
Operating-Systems       Disk-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 :
 A C (5, 3) B 20 C 10 D 5!/3!2!
Question 44 Explanation:
⇒ C(5,2) = 5C2 = 5C3 (∴ nCr = nCn-r)
5C2 = ∠5/∠3∠2 (∴ nCr = ∠n/∠r∠n-r)
5C2 = ∠5/∠3∠2 = 5×4/2 = 10
 Question 45
The value of a1 + a1.b1 + b1 + a + 0 is :
 A a1 + b1 B b1 + a C 1 D 0
Digital-Logic-Design       Boolean-Expression
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)
 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 :
 A 0 B 1 C -1 D None
Digital-Logic-Design       Boolean-Algebra
Question 46 Explanation:
 Question 47
The conditional probability of event A given event B is defined as :
 A P(A/B)= P(A∩B)/P(B) B P(A/B)= P(A∩B)/P(A) C P(B/A)= P(A∩B)/P(A) D P(B/A)= P(A∩B)/P(B)
Engineering-Mathematics       Probability
Question 47 Explanation:
The conditional probability of event A given event B is defined as
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 :
 A 0.0038 B 0.246 C 0.68 D 4.35
Computer-Organization       SD-RAM
Question 48 Explanation:
No of cycles require to refresh= 8k ×4 = 215
Amount of time to refresh = 215 × (1/133×106) secs = 0.246 ms
 Question 49
The Booth technique for recording multiply of +13 and –6 [01101 and 11010] is :
 A 1110 0011 01 B 1110 1100 10 C 1110 1010 10 D 1110 0011 00
Digital-Logic-Design       Number-Systems
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
 Question 50
MFLOPS can be abbreviated as :
 A Millions of Floating – Point operations performed per second. B Millions of Fixed – Length operations performed per second. C Millions of Floating – Limited operations performed per second. D Millions of Fixed – Limited operations performed per second.
Question 50 Explanation:
Short for mega floating-point operations per second, MFLOPs are a common measure of the speed of computers used to perform floating-point calculations.
 Question 51
A router must have atleast __________ NICs
 A 3 B 4 C 2 D 5
Computer-Networks       Routing
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.
 A B B C C D 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.
 Question 53
The gateway that stands between the mobile network and the Internet in GPRS is called as _______.
 A CCNS B SGGN C SGSN D CGSN
\"Computer-Networks \"       Hardware-and-various-devices-in-networking
Question 53 Explanation:
A Gateway GPRS Support Node (GGSN) is part of the core network that connects GSM-based 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 IP-based applications.
 Question 54
The value of the expression (a + b) (a + b1) (a +b1b) is :
 A b1 B a1 C a D b
Digital-Logic-Design       Boolean-Expression
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
 Question 55
The disadvantage of write back strategy in cache is that :
 A If generates repeated memory traffic B It creates a write mechanism whenever there is a write operation to cache C Portions of main memory may be invalid D It requires local cache memory attached to every CPU in a multi processor environment
Computer-Organization       Cache
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?
 A In a processor register B In a memory location associated with the call C On a stack D All of the above
Operating-Systems       Memory-Management
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 :
 A ALU B Memory Address Register C Program counter D Instruction memory
Computer-Organization       I/O-Devices
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?
 A θ(n) B θ(1) C θ(log n) D θ(n log log n)
Algorithms       Searching
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?
 A 3 B 4 C 6 D 5
Data-Structures       Binary-Trees
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
 Question 60
Which Sorting method is an external Sort?
 A Heap Sort B Quick Sort C Insertion Sort D None of the above
Algorithms       Sorting
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 sort-merge strategy.
Example of external sorting is merge sort.
 Question 61
Between any two vertices, there exists a path, then the graph is said to be
 A Strongly Connected B Connected C Weakly Connected D All the above
Engineering-Mathematics       Graph-Theory
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?
 A O(n) B O(n2) C O(n2 log n) D O(n log n)
Algorithms       Recurrences
Question 62 Explanation:
T(n) = 2T(n/2) + n
a=2, b=2, k=1, p=0
Now, a=bk 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?
 A (1, 6), (6, 5), (5, 4), (4, 3), (3, 2), (2, 7) B (1, 6), (3, 4), (2, 7), (2, 3), (7, 4), (5, 4) C (1, 6), (3, 4), (2, 7), (4, 5), (1, 2), (5, 6) D (1, 6), (6, 5), (5, 4), (4, 3), (3, 2), (4, 7)
Algorithms       Greedy-approach
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).
 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 __________
 A Dynamic programming, O (n2) B Dynamic programming, O (log n) C Greedy, O (n2) D Greedy, O (log n) E None of the above
Algorithms       Greedy-approach
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 _________.
 A 16 bits B 18 bits C 20 bits D 22 bits
Operating-Systems       Memory-Management
Question 65 Explanation:
No. of bits required to indicate no. of pages = log2 64 = 6
No. of bits required to indicate page size = log2 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.
 A RARP B IPV6 C TFTP D 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)  ```
 A (i) (ii) (iii) B (i) (iii) (ii) C (iii) (ii) (i) D (ii) (iii) (i)
Computer-Networks       Network-Security
Question 67 Explanation:
Masquerading-One pretends to be someone else.
Phishing-Phishing 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.
Hijacking-Hijacking 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 → A `
Which one of the following options is false ?
 A BG+={ABCDG} B CF+={ACDEFG} C AB+={ABCDG} D AF+={ACDEFG}
Database-Management-System       Functional-Dependency
Question 68 Explanation:
(BG)+ = BGACD
(AB)+ = ABCDG
(AF)+ = AFDE
Hence option (D) is False.
 Question 69
Consider the following transactions with data items X and y initialized to zero :
```T1: read x;
if x=0 then y≔y+1;
write y;
write x;
if y=0 then x≔x+1
write x; ```
The concurrent execution of T1 and T2 leads to
 A Serializable schedule B A schedule that is not conflict serializable C A conflict serializable schedule D A schedule for which a precedence graph cannot be drawn
Database-Management-System       Transactions
Question 69 Explanation:
One of the schedule that is possible is,

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.
 A Free variable and bound variable B Bound variable and free variable C Free variable and free variable D Bound variable and bound variable
Database-Management-System       Relational-Calculus
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) :
 A π, ∞, σ B π, σ, ∞ C σ, π, ∞ D σ, ∞, π
Database-Management-System       Relational-Algebra
Question 71 Explanation:
In SQL we write
SELECT
FROM
WHERE
in a sequence.
So it is equivalent to
SELECT-π
FROM- ∞
WHERE-σ
 Question 72
If T1 and T2 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 :
 A T1 + (1-H) × T2 B (1 - H) × T1 + T2 C T1 + (1 - H) D T1 + (1 + H)
Computer-Organization       Memory-hierarchy
Question 72 Explanation:
Average access time
= hit rate × T1 + miss rate × (T1 + T2)
= H × T1 + (1 - H)(T1 + T2)
= H × T1 + T1 - T1 H + T2 - T2H
= T1 + (1 - H)T2
 Question 73
Which of the following is the recurrence relation for binary search?
 A T(n) = T(n/2) + 1 B T(n) = T(n/2) + n C T(n) = 2T(n - 1) + 1 D T(n) = T(n - 1) + 1
Algorithms       Searching
Question 73 Explanation:
A-applying 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(2n) which is not the time complexity for binary search.
D-Using 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?
 A Leftist Heap B Binary Heap C AVL trees D Red – Black trees
Data-Structures       Queues-and-Stacks
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 = at2; y = 2at is the parametric equation of:
 A Circle B Parabola C Rectangular hyperbola D Ellipse
Engineering-Mathematics       Co-ordinate-Geometry
Question 75 Explanation:
x=at2, y=2at is the parametric equation of y2 = 4ax which is parabola.
 Question 76
An example for a pattern – scanning language is :
 A lex B awk C bison D yacc
Compiler-Design       Phases-of-Compilers
Question 76 Explanation:
An example for a pattern - scanning language is lex.
 Question 77
Consider the following grammar :
`S → L|a     L → LS|S  `
After the elimination of left – recursion, we get the following:
 A S→L|a L→SA A→SA|ε B S→L|A A→SAL L→aSL|ε C S→L|ε L→SA A→SA|ε D S→L|a A→aSA|ε L→aSA
Compiler-Design       Grammars
Question 77 Explanation:
S → (L)|a
L → LS|S //contains left recursion
Now eliminating left recursion we get,
S → (L)|a
L → SA
A → SA|∊
Using formula,
A → Aα|β

A → βA’
A’ → αA’|∊
 Question 78
The Language L = {ap| p is prime } is :
 A regular B not regular C accepted by NFA with ε D None
Theory-of-Computation       Regular-Language
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:
 A S1→.S,\$ S→.AS,\$ / a/b S→.b,\$/a/b A→.SA, a/b A→.a,a /b B S1→.S,\$ S→.AS,\$ s/ b S→.b,\$ / b C S1→.S,\$ S→.AS,\$ / a/b S→.AS,\$ / a/b S→.b,\$ / a/b D S1→.S,\$ S→.AS,\$ S→.b,\$
Compiler-Design       Parsers
Question 79 Explanation:
The given grammar is
S → AS|b
A → SA|a
Now closure for S →.S, \$,
S’ → .S, \$
S → .AS|.b, \$
A → .SA|.a, a|b
S → .AS|.b, a|b

S → .S, \$
S → .AS|.b, \$|a|b
A → .SA|.a, a|b
 Question 80
How many host interfaces may be addressed in the subnet 123.224.00.00/11?
 A 2048 B 2,097,150 C 1,000,192 D 2,097,152
Question 80 Explanation:
No. of host id bits is 32-11 = 21
So, no. of host interfaces that can be addressed in the subnet is 221 - 2 = 2097150.
 Question 81
Which of the following make(s) filtering decisions based on application payload?
 A packet filter B deep inspection firewall C reverse proxy D stateful packet inspection firewall
Computer-Networks       Network-Security
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:
 A With a high probability, it will be detected. B It will not be detected, as the burst error is exactly identical to the divisor. C With a very small probability, it will be detected. D It depends on the data rate in the channel. So it may or may not be detected.
Computer-Networks       CRC
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 __________.
 A 15/12 B 35/12 C 25/12 D 17/12
Engineering-Mathematics       Probability
Question 83 Explanation:
If X is uniformly distributed in (-2,3), then its variance is
(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?
 A e-1 B e-2 C 1 - e-1 D 1 - e-2
Engineering-Mathematics       Probability
Question 84 Explanation:
 Question 85
The Product of Sums form of the expression a.b + a1.c is :
 A (a+b).(a+c1) B (a+c).(a1+b) C (a1+b1).(a+c1) D (a1+b).(a1+c)
Digital-Logic-Design       Boolean-Expression
Question 85 Explanation:
Let's draw k-map for given equation,
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 20th. When two boys joined, his rank was dropped by one. What is his new rank from the end ?
 A 25th B 26th C 27th D 28th
Aptitude       Numerical
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
 Question 87
Who won the gold both in the 5,000 and 10,000 metres event in 2017 Asian Athletics Championship?
 A Lakshmanan B Gopy Thonkanal C Jinson Johnson D 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?
 A –40° B 574.59 C 40 D –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:
 A Rajasthan B West Bengal C Tamil Nadu D Karnataka
Question 89 Explanation:
Rajasthan, First State to fix minimum education qualification for cooperative body poll.
 Question 90
Who wrote the novel – ‘KavalKottam’?
 A Vannadasan B S. Venkatesan C Joe D Cruz D Puviarasan
Question 90 Explanation:
S. Venkatesan wrote the novel --’ KavalKottam’.
 Question 91
Article 21-A and the RTE Act came into effect:
 A On 1st April 2010 B On 1st April 2009 C On 1st April 2017 D On 1st April 2005
Question 91 Explanation:
Article 21-A 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.