ISRO CS 2011

 Question 1
The encoding technique used to transmit the signal in giga ethernet technology over fiber optic medium is
 A Differential Manchester encoding B Non return to zero C 4B/5B encoding D 8B/10B encoding
Computer-Networks       Encoding-Decoding
Question 1 Explanation:
→ In telecommunications, 8b/10b is a line code that maps 8-bit words to 10-bit symbols to achieve DC-balance and bounded disparity, and yet provide enough state changes to allow reasonable clock recovery
→ The FC(Fiber Channel) -0 standard defines what encoding scheme is to be used (8b/10b or 64b/66b) in a Fibre Channel system – higher speed variants typically use 64b/66b to optimize bandwidth efficiency.Thus, 8b/10b encoding is used for 4GFC and 8GFC variants;
→ The Fibre Channel 8b/10b coding scheme is also used in other telecommunications systems. Data is expanded using an algorithm that creates one of two possible 10-bit output values for each input 8-bit value.
 Question 2
Which of the following is an unsupervised neural network?
 A RBS B Hopfield C Back propagation D Kohonen E Incomplete Question
Neural-Networks       Neural-network
Question 2 Explanation:
→ A self-organizing map (SOM) or self-organizing feature map (SOFM) is a type of artificial neural network (ANN) that is trained using unsupervised learning to produce a low-dimensional (typically two-dimensional), discretized representation of the input space of the training samples, called a map, and is therefore a method to do dimensionality reduction.
→ Kohonen map or network is self-organizing map
→ Hopfield nets serve as content-addressable ("associative") memory systems with binary threshold nodes. They are guaranteed to converge to a local minimum, but will sometimes converge to a false pattern (wrong local minimum) rather than the stored pattern (expected local minimum). Hopfield networks also provide a model for understanding human memory.
→ Backpropagation is a method used in artificial neural networks to calculate a gradient that is needed in the calculation of the weights to be used in the network
 Question 3
In compiler terminology reduction in strength means
 A Replacing run time computation by compile time computation B Removing loop invariant computation C Removing common subexpressions D replacing a costly operation by a relatively cheaper one
Compiler-Design       Code-Optimization
Question 3 Explanation:
An optimization method in which an operator is changed to a less-expensive operator;
Example: Exponentiation is replaced by multiplication and multiplication is in return replaced by addition.(x * 2 becomes x + x)
 Question 4
The following table shows the processes in the ready queue and time required for each process for completing its job.

If round-robin scheduling with 5 ms is used what is the average waiting time of the processes in the queue?
 A 27 ms B 26.2 ms C 27.5 ms D 27.2 ms
Operating-Systems       CPU-Scheduling
Question 4 Explanation:
→In the round robin algorithm, time slices (also known as time quanta) are assigned to each process in equal portions and in circular order, handling all processes without priority (also known as cyclic executive).
→Given scheduling algorithm is round robin with time quantum value is 5ms.

→In the given example, every process will execute for 5ms. process P1 will execute for two times , P2 for one time, P3 for four times , P4 for two times and P5 for three times.
→Waiting time of process P1 = 0+(25-20)=5
→Waiting time of process P2 = 5
→Waiting time of process P3 = 10 + (30-15 )+ (43-35) + (53-48)=10 + 15 + 8 + 5 = 38
→Waiting time of process P4 = 15+(35-20)=15 + 15 = 30
→Waiting time of process P5 =20+(38-25)+48-43= 20 + 13 + 5 = 38
→Average waiting time = 20 + 5 + 38 + 30 + 38 =131/5 = 26.2
 Question 5
MOV [BX], AL type of data addressing is called ?
 A register B immediate C register indirect D register relative
Computer-Organization       Microprocessor
Question 5 Explanation:
Register indirect addressing means that the location of an operand is held in a register.
In register addressing mode, a register contains the operand. Depending upon the instruction, the register may be the first operand, the second operand or both.
In Immediate Addressing,an immediate operand has a constant value or an expression. When an instruction with two operands uses immediate addressing, the first operand may be a register or memory location, and the second operand is an immediate constant.
 Question 6
Evaluate (X XOR Y) XOR Y?
 A All 1’s B All 0’s C X D Y
Digital-Logic-Design       Boolean-Expression
Question 6 Explanation:

 Question 7
Which of the following is true about the z-buffer algorithm?
 A It is a depth sort algorithm B No limitation on total number of objects C Comparisons of objects is done D z-buffer is initialized to background colour at start of algorithm
Graphics       Z-buffer
Question 7 Explanation:
The Z-buffer algorithm is a convenient algorithm for rendering images properly according to depth.
To begin with, a buffer containing the closest depth at each pixel location is created parallel to the image buffer. Each location in this depth buffer is initialized to negative infinity.
Since the algorithm processes objects one at a time, the total number of polygons in a picture can be arbitrarily large.
 Question 8
What is the decimal value of the floating-point number C1D00000 (hexadecimal notation)? (Assume 32-bit, single precision floating point IEEE representation)
 A 28 B -15 C -26 D -28
Digital-Logic-Design       Number-Systems
Question 8 Explanation:
Floating Point number in Hexadecimal = C1D00000
Floating Point number in Binary = 1100 0001 1101 0000 0000 0000 0000 0000
In 32-bit, single precision floating point IEEE representation, first MSB represents sign of mantissa: 1 is used to represent a negative mantissa and 0 for a positive value of mantissa, next 8 bits are for exponent value and then 23 bits represents mantissa.
Value of exponent = 131-127 = 4
Mantissa = -1.1010000 0000 0000 0000 0000
Floating point number = -1.1010000 0000 0000 0000 0000
Converting the above one into decimal no -(1*20+1*2-1*0*2-2+1*2-2+0* 2-3 +.....)
= -(1+½+⅛)=-13/8
Decimal value =sign*Exponent*mantissa=1*4*-13/8

= -26
 Question 9
What is the raw throughput of USB 2.0 technology?
 A 480 Mbps B 400 Mbps C 200 Mbps D 12 Mbps
Computer-Networks       USB
Question 9 Explanation:
The USB 3.0 SuperSpeed path operates at a raw bit rate of 5.0 Gbits/s, while the USB 2.0 path operates at 480 Mbits/s (High Speed), 12 Mbits/s (Full Speed), or 1.5 Mbits/s (Low Speed).
 Question 10
Below is the precedence graph for a set of tasks to be executed on a parallel processing system S.

What is the efficiency of this precedence graph on S if each of the tasks T1, T2, T3,….T8 takes the same time and the system S has five processors?
 A 25% B 40% C 50% D 90%
Operating-Systems       Concurrency
Question 10 Explanation:
From the precedence graph, we say that the following tasks executed sequentially
I. T1 ,T2
II. T3 and T6
III. T4 and T7
IV. T5 and T8
(T3,T6),(T4,T7) and (T5,T8) will execute parallely.
So total number of processes that can be executed in 4 units time using 5 available processors = 5*4 = 20
Maximum number of tasks are 8
Efficiency = 8/20 * 100 = 40%
 Question 11
How many distinct binary search trees can be created out of 4 distinct keys?
 A 5 B 14 C 24 D 35
Data-Structures       Binary-search-tree
Question 11 Explanation:
The number of distinct BST for n nodes are given as ((2n)Cn)/(n+1)
So, for 4 distinct nodes, we can have (8C4)/5 = 14 distinct BSTs
 Question 12
The network protocol which is used to get MAC address of a node by providing IP address is
 A SMTP B ARP C RIP D BOOTP
Computer-Networks       MAC
Question 12 Explanation:
Address Resolution Protocol(ARP) is used to find the MAC address of a device using its IP address.
Reverse Address Resolution Protocol(RARP) is the viceversa case of ARP as it is a protocol by which a physical machine in a local area network can request to learn its IP address from a gateway server.
 Question 13
Which of the following statements about peephole optimization is False?
 A It is applied to a small part of the code B It can be used to optimize intermediate code C To get the best out of this, it has to be applied repeatedly D It can be applied to the portion of the code that is not contiguous
Compiler-Design       Code-Optimization
Question 13 Explanation:
Peephole optimization is a type of Code Optimization performed on a small part of the code. It is performed on the very small set of instructions in a segment of code.
It basically works on the theory of replacement in which a part of code is replaced by shorter and faster code without change in output.
 Question 14
Which one of the following in place sorting algorithms needs the minimum number of swaps?
 A Quick sort B Insertion sort C Selection sort D Heap sort
Database-Management-System       Sorting
Question 14 Explanation:
Selection sort requires maximum number of swaps i.e O(n).
The algorithm finds the minimum value, swaps it with the value in the first position, and repeats these steps for the remainder of the list. It does no more than n swaps, and thus is useful where swapping is very expensive.
 Question 15
What is the equivalent serial schedule for the following transactions?
 A T1 − T2 − T3 B T3 − T1 − T2 C T2 − T1 − T3 D T1 − T3 − T2
Database-Management-System       Transactions
Question 15 Explanation:
From the following precedence graph, T3 → T1→ T2

 Question 16
Logic family popular for low power dissipation
 A CMOS B ECL C TTL D DTL
Integrated-Circuits       Circuits
Question 16 Explanation:
CMOS uses almost no power in the static state, i.e. when inputs are not changing. They have low energy requirements for logic transition and hence less power and heat dissipation.
 Question 17
A context model of a software system can be shown by drawing a
 A LEVEL-0 DFD B LEVEL-1 DFD C LEVEL-2 DFD D LEVEL-3 DFD
Software-Engineering       DFD
Question 17 Explanation:
Explanation:
1. A data flow diagram (DFD) illustrates how data is processed by a system in terms of inputs and outputs. As its name indicates its focus is on the flow of information, where data comes from, where it goes and how it gets stored.
2. Context Diagram. A context diagram is a top level (also known as "Level 0") data flow diagram. It only contains one process node ("Process 0") that generalizes the function of the entire system in relationship to external entities.
 Question 18
An example of polyalphabetic substitution is
 A P-box B S-box C Caesar cipher D Vigenere cipher
Computer-Networks       Polyalphabetic
Question 18 Explanation:
A polyalphabetic cipher is any cipher based on substitution, using multiple substitution alphabets. The Vigenère cipher is probably the best-known example of a polyalphabetic cipher, though it is a simplified special case. The Enigma machine is more complex but is still fundamentally a polyalphabetic substitution cipher.
 Question 19
If node A has three siblings and B is a parent of A, what is the degree of A?
 A 0 B 3 C 4 D None of the above
Data-Structures       Graphs
Question 19 Explanation:
The degree of a vertex of a graph is the number of edges incident to the vertex, and in a multigraph, loops are counted twice.
According to question, there is no information regarding children nodes of node “A”. So the degree of A is 0.
 Question 20
The IEEE standard for WiMax technology is
 A IEEE 802.16 B IEEE 802.36 C IEEE 812.16 D IEEE 806.16
Computer-Networks       Wimax
Question 20 Explanation:
WiMAX (Worldwide Interoperability for Microwave Access) is a family of wireless communication standards based on the IEEE 802.16 set of standards, which provide multiple physical layer (PHY) and Media Access Control (MAC) options.
 Question 21
Which type of DBMS provides support for maintaining several versions of the same entity?
 A Relational Database Management System B Hierarchical C Object Oriented Database Management System D Network
Database-Management-System       ER-Model
Question 21 Explanation:
Many object databases, for example Gemstone or VOSS, offer support for versioning.
An object can be viewed as the set of all its versions.
Object versions can be treated as objects in their own right. Some object databases also provide systematic support for triggers and constraints which are the basis of active databases.
 Question 22
A system is having 8 M bytes of video memory for bit-mapped graphics with 64-bit colour. What is the maximum resolution it can support?
 A 800 x 600 B 1024 x 768 C 1280 x 1024 D 1920 x 1440
Graphics       Video-memory
Question 22 Explanation:
Explanation:
Given file size is 8M bytes= 8*1024**1024*8=83,88,608
From the options,
⦁ 800*600*8=34,80,000
⦁ 1024*768*8=62,91,456
⦁ 1280*1024*8=13,10,720
⦁ 1920*1440*8=22,118,400
From the above , option A and B are less than file size.
From that two , maximum one is option B.
 Question 23
What is the meaning of ¯¯RD¯ signal in Intel 8151A?
 A Read (when it is low) B Read (when it is high) C Write (when it is low) D Read and Write (when it is high)
Computer-Peripherals       Intel-8151
Question 23 Explanation:
RD-Read: Read signal, when low, indicates the peripherals that the processor is performing a memory or I/O read operation. RD is active low and shows the state for T2, T3, TW of any read cycle. The signal remains tristated during the 'hold acknowledge'.
 Question 24
If the page size in a 32-bit machine is 4K bytes then the size of the page table is
 A 1 M bytes B 2 M bytes C 4 M bytes D 4 K bytes
Operating-Systems       Memory-Management
Question 24 Explanation:
→Page size is total space taken up by page and Page table entry size is memory taken for indexing the Page in Page Table
→Size of logical address = 32 bits
→Page size = 4K =22210=212 Bytes
→Number of pages = logical address space/ size of each page = 232/ 212= 220
→Page table size = number of pages * size of a page table entry
= 220 * 22
= 222
 Question 25
A problem whose language is recursive is called?
 A Unified problem B Boolean function C Recursive problem D Decidable
Theory-of-Computation       Resursive
Question 25 Explanation:
A formal language (a set of finite sequences of symbols taken from a fixed alphabet) is called recursive if it is a recursive subset of the set of all possible finite sequences over the alphabet of the language.
Equivalently, a formal language is recursive if there exists a total Turing machine (a Turing machine that halts for every given input) that, when given a finite sequence of symbols as input, accepts it if it belongs to the language and rejects it otherwise.
Recursive languages are also called decidable.
 Question 26
In a system using a single processor, a new process arrives at the rate of six processes per minute and each such process requires seven seconds of service time. What is the CPU utilization?
 A 70% B 30% C 60% D 64%
Operating-Systems       Memory-Management
Question 26 Explanation:
From the given question,
The number of new processes will arrive per minute = 6
Each process require to complete its task = 7 secs
CPU utilization time within a minute = 6*7 = 42 secs
The percentage of CPU utilization = time which is spent for utilization / total time * 100
= (42/60) * 100
= 70%
 Question 27
In HTML, which of the following can be considered a container?
 A SELECT B Value C INPUT D BODY
HTML       HTML
Question 27 Explanation:
→ A container tag in HTML is one which has both opening and closing tags.
→ There are some tags in HTML which don't have a closing tag.
→ They end within the same tag.
Examples:
→ is a container tag, it has it's closing tag as .
→Other examples are , ,

etc.
→These are called container tags because they contain something, within the two tags.

 Question 28

The above figure represents which one of the following UML diagram for a single send session of an online chat system?
 A Package diagram B Activity diagram C Class diagram D Sequence diagram
UML       UML
Question 28 Explanation:
1. Activity diagram is basically a flowchart to represent the flow from one activity to another activity. The activity can be described as an operation of the system.
2. A sequence diagram shows object interactions arranged in time sequence. It depicts the objects and classes involved in the scenario and the sequence of messages exchanged between the objects needed to carry out the functionality of the scenario.
3. A class diagram is a type of static structure diagram that describes the structure of a system by showing the system's classes, their attributes, operations (or methods), and the relationships among objects.
 Question 29
Which normal form is based on the concept of ‘full functional dependency’ is
 A First Normal Form B Second Normal Form C Third Normal Form D Third Normal Form
Database-Management-System       Normalization
Question 29 Explanation:
A full functional dependency is a state of database normalization that equates to the normalization standard of Second Normal Form (2NF).
Second normal form (2NF) is a normal form used in database normalization.
To qualify for second normal form a relation must:
→be in first normal form (1NF)
→not have any non-prime attribute that is dependent on any proper subset of any candidate key of the relation.
A non-prime attribute of a relation is an attribute that is not a part of any candidate key of the relation.
 Question 30
In Boolean algebra, rule (X+Y)(X+Z) =
 A Y+XZ B X+YZ C XY+Z D XZ+Y
Digital-Logic-Design       Boolean-Algebra
Question 30 Explanation:
in Boolean algebra,
(X+Y)(X+Z) = X + XZ + XY + YZ
= X(1 + Z + Y) + YZ // as (1 + A = A)
= X.1 + YZ
= X + YZ
 Question 31
In an RS flip-flop, if the S line (Set line) is set high (1) and the R line (Reset line) is set low (0), then the state of the flip-flop is
 A Set to 1 B Set to 0 C No change in state D Forbidden
Digital-Logic-Design       Flip=flops
Question 31 Explanation:
 Question 32
In which layer of network architecture, the secured socket layer (SSL) is used?
 A physical layer B session layer C application layer D presentation layer
Computer-Networks       OSI-TCP-layers
Question 32 Explanation:
Secure Socket Layer is networking protocol used at transport layer to provide secure connection between client and server over internet. It places itself as and application layer protocol in the TCP/IP reference model and as presentation layer protocol in the OSI model.
 Question 33
What is the bitrate of a video terminal unit with 80 characters/line, 8 bits/character and a horizontal sweep time of 100 μs (including 20 μs of retrace time)?
 A 8 Mbps B 6.4 Mbps C 0.8 Mbps D 0.64 Mbps
Operating-Systems       I/O-Management
Question 33 Explanation:
Bit rate is the number of bits that are conveyed or processed per unit of time
Given data is video terminal unit has 80 lines and each line consists of 8 bits
Total number of bits transmitted = 80 * 8 = 640 bits
Horizontal sweep time = 100 us=100x 10-6 seconds
Bit rate = (640 * 106) / 100 =6400000=6.4 Mbps
 Question 34
Black Box Software Testing method focuses on the
 A Boundary condition of the software B Control structure of the software C Functional requirement of the software D Independent paths of the software
Software-Engineering       Software-testing
Question 34 Explanation:
Black box testing, which is also known as behavioral, opaque-box, closed-box, specification-based or eye-to-eye testing, is a Software Testing method that analyses the functionality of a software/application without knowing much about the internal structure/design of the item that is being tested and compares the input value with the output value
 Question 35
 A Page fault rate is constant even on increasing the number of allocated frames B Page fault rate may increase on increasing the number of allocated frames C Page fault rate may increase on decreasing the number of allocated frames D Page fault rate may decrease on increasing the number of allocated frames
Operating-Systems       CPU-Scheduling
Question 35 Explanation:
Bélády's anomaly is the phenomenon in which increasing the number of page frames results in an increase in the number of page faults for certain memory access patterns.
This phenomenon is commonly experienced when using the first-in first-out (FIFO) page replacement algorithm.
 Question 36
If A and B are square matrices with same order and A is symmetric, then BTAB
 A Skew symmetric B Symmetric C Orthogonal D Idempotent
Engineering-Mathematics       Linear-Algebra
Question 36 Explanation:
For a Symmetric matrix, A’ = A
So, BTAB = B'.A.B
Taking transpose of B’.A.B
(B'.A.B)' = B'.A'.(B')' = B'.A.B // (B')' = B
So, it is a symmetric matrix.
 Question 37
Find the output of the following Java code line System.out.println(math.floor(-7.4))
 A -7 B -8 C -7.4 D -7
Java       Programming
Question 37 Explanation:
The method floor gives the largest integer that is less than or equal to the argument.
But here the argument is negative value.
Floor of -7.4 will return the lower limit, i.e. -8.
 Question 38
One SAN switch has 24 ports. All 24 supports 8 Gbps Fiber Channel technology. What is the aggregate bandwidth of that SAN switch?
 A 96 Gbps B 192 Mbps C 512 Gbps D 192 Gbps
Computer-Networks       SAN
Question 38 Explanation:
SAN switch has 24 ports.
Each port supports 8 Gbps then all 24 ports have capacity of 24*8 Gbps bandwidth
The aggregate bandwidth is 192 Gbps
 Question 39
Two control signals in microprocessor which are related to Direct Memory Access (DMA) are
 A INTR & INTA B RD & WR C S0 & S1 D HOLD & HLDA
Computer-Organization       DMA
Question 39 Explanation:
HOLD (Hold) is a control signal input and HLDA (Hold acknowledge) is a control signal output.
These two signals are used for hand shaked control during DMA operation (Direct Memory Access)
These two signals are used where there is more than one CPU like devices sharing the same system bus
 Question 40
Consider the following pseudocode:
x:=1;
i:=1;
while (x ≤ 500)
begin
x:=2x ;
i:=i+1;
end
What is the value of i at the end of the pseudocode?
 A 4 B 5 C 6 D 7
Data-Structures       Programming
Question 40 Explanation:
After completion of first iteration x and i values are : x = 2 and i = 2
After completion of second iteration x and i values are : x = 4 and i = 3
After completion of third iteration x and i values are : x = 16 and i = 4
After completion of fourth iteration x and i values are : x =256 and i = 5
After completion of fifth iteration x and i values are : x = 65536 and i = 6(Condition is false)
Then the value of “i” is 5
 Question 41
If a microcomputer operates at 5 MHz with an 8-bit bus and a newer version operates at 20 MHz with a 32-bit bus, the maximum speed-up possible approximately will be
 A 2 B 4 C 8 D 16
Computer-Organization       Microprocessor
Question 41 Explanation:
The newer version is the best as compared to old one, but the second version has both high bandwidth as well as increased CPU speed.
20/5 = 4 and 32/8 = 4
Maximum speed up possible is 4
 Question 42
The search concept used in associative memory is
 A Parallel search B Sequential search C Binary search D Selection search
Computer-Organization       Memory-Interfacing
Question 42 Explanation:
A type of computer memory from which items may be retrieved by matching some part of their content, rather than by specifying their address (hence also called associative storage or Content-addressable memory (CAM)
Associative memory makes a parallel search with the stored patterns as data files.
 Question 43
Which variable does not drive a terminal string in grammar?
S → AB
A → a
B → b
B → C
 A A B B C C D S
Compiler-Design       Context-free-grammar
Question 43 Explanation:
C is the useless variable as there is no production rule which replaces C with a terminal. Hence it does not derive any non terminal.
 Question 44
In Java, after executing the following code what are the values of x, y and z?
int x,y=10; z=12;
x=y++ + z++;
 A x=22, y=10, z=12 B x=24, y=10, z=12 C x=24, y=11, z=13 D x=22, y=11, z=13
Java
Question 44 Explanation:
x = y++ + z++;
Post increment operator first perform the action and then it increment the value.
First perform addition of ‘y’ and ‘z’ and later increment ‘y’ and ‘z’ values
So, the value of x = 10 + 12 = 22, y = 11 and z = 13.
 Question 45
 A 172.16.0.255 B 172.16.255.255 C 255.255.255.255 D 172.255.255.255
Computer-Networks       Subnetting
Question 45 Explanation:
Convert the above two addresses into binary format and perform the bitwise AND operation
 Question 46
Which RAID level gives block-level striping with double distributed parity?
 A RAID 10 B RAID 2 C RAID 6 D RAID 5
Operating-Systems       RAID
Question 46 Explanation:
RAID (Redundant Array of Inexpensive Disks or Drives, or Redundant Array of Independent Disks) is a data storage virtualization technology that combines multiple physical disk drive components into one or more logical units for the purposes of data redundancy, performance improvement, or both.
RAID levels and their corresponding functionality as shown below
RAID 0: Stripping
RAID 1: Mirroring
RAID 2: Hamming code for error detection
RAID 3: Byte-level striping with a dedicated parity disk
RAID 4: Block-level striping with block-level striping with two parity blocks parity
RAID 5: Block-level striping with distributed parity
RAID 6: Block-level striping with double distributed parity.
 Question 47
The output expression of the following gate network is
 A X. Y + X’ Y’ B X. Y + X. Y C X. Y D X + Y
Digital-Logic-Design       Boolean-Function
Question 47 Explanation:
AND operation is represented by ‘.’ and OR operation is representation is represented with ‘+’. According to diagram, option (A) is correct
 Question 48
Number of comparisons required for an unsuccessful search of an element in a sequential search, organized, fixed length, symbol table of length L is
 A L B L/2 C (L+1)/2 D 2L
Algorithms       Searching
Question 48 Explanation:
A linear search or sequential search is a method for finding an element within a list. It sequentially checks each element of the list until a match is found or the whole list has been searched
A linear search runs in at worst linear time and makes at most n comparisons, where n is the length of the list , whether element is found or not
 Question 49
Consider a 32-bit machine where four-level paging scheme is used. If the hit ratio to TLB is 98%, and it takes 20 nanosecond to search the TLB and 100 nanoseconds to access the main memory what is effective memory access time in nanoseconds?
 A 126 B 128 C 122 D 120
Operating-Systems       Memory-Management
Question 49 Explanation:
Hit ratio to TLB(H) is 98%
Searching time of TLB(T) is 20ns
Access time(M) is 100ns and four level paging scheme is used.
Effective Memory access Time, EAT = H* T+ (1 - H)[ T+ 4*M] + M]
EAT = (0.98 *20) + 0.02(20 + 400) + 100
= 19.6 + 8.4 + 100 = 128 ns
 Question 50
Data is transmitted continuously at 2.048 Mbps rate for 10 hours and received 512 bits errors. What is the bit error rate?
 A 6.9 e-9 B 6.9 e-6 C 69 e-9 D 4 e-9
Computer-Networks       Transmission-and-Propagation-Delay
Question 50 Explanation:
Bandwidth = 2.048 Mbps
The amount of time data transferred is = 10 hours = 36000 secs
Total Data transmitted = 2.048 * 106 * 36000 = 2.048 * 36 * 109 bits
The number of Error bits received = 512
Error rate = total number of error bits/ total data transferred per second
= 512 / 73.728 * 109
= 6.944 * 10-9
 Question 51
Warnier Diagram enables the analyst to represent
 A Warnier Diagram enables the analyst to represent B Information Hierarchy C Data Flow D State Transition
Software-Engineering       Warnier-diagram
Question 51 Explanation:
A Warnier/Orr diagram (also known as a logical construction of a program/system) is a kind of hierarchical flow chart that allows the description of the organization of data and procedures.
This method aids the design of program structures by identifying the output and processing results and then working backwards to determine the steps and combinations of input needed to produce them.
The simple graphic method used in Warnier/Orr diagrams makes the levels in the system evident and the movement of the data between them vivid.
 Question 52
Lightweight Directory Access protocol is used for
 A Routing the packets B Authentication C obtaining IP address D domain name resolving
Computer-Networks       Network-protocols
Question 52 Explanation:
The Lightweight Directory Access Protocol (LDAP ) is an open, vendor-neutral, industry standard application protocol for accessing and maintaining distributed directory information services over an Internet Protocol (IP) network
A common use of LDAP is to provide a central place to store usernames and passwords. This allows many different applications and services to connect to the LDAP server to validate users
 Question 53
In functional dependency Armstrong inference rules refers to
 A Reflexive, Augmentation and Decomposition B Transitive, Augmentation and Reflexive C Augmentation, Transitive, Reflexive and Decomposition D Reflexive, Transitive and Decomposition
Database-Management-System       Functional-dependency
Question 53 Explanation:
Armstrong's axioms are a set of axioms (or, more precisely, inference rules) used to infer all the functional dependencies on a relational database.
The axioms are sound in generating only functional dependencies in the closure of a set of functional dependencies when applied to that set
Axiom of Reflexivity:
→If X is a set of attributes and Y is a subset of X, then X holds Y. Hereby, X holds Y ( X -> Y) means that X functionally determines Y.
Axiom of Augmentation:
→If X holds Y and Z is a set of attributes, then XZ holds YZ. It means that attribute in dependencies does not change the basic dependencies.
Axiom of Transitivity:
The axiom of transitivity says if X holds Y, and Y holds Z, then X must also hold Z.
 Question 54
Number of chips (128 x 8 RAM) needed to provide a memory capacity of 2048 bytes
 A 2 B 4 C 8 D 16
Digital-Logic-Design       Memory-interfacing
Question 54 Explanation:
Given memory capacity is 2048 Bytes RAM memory is 128x8
Number of chips needed are = total memory capacity / RAM memory
= 2048 bytes / 128 x 8 =2048x8/128x8
= 16
 Question 55
There are three processes in the ready queue. When the currently running process requests for I/O how many process switches take place?
 A 1 B 2 C 3 D 4
Question 55 Explanation:
Context switching is the procedure of storing the state of an active process for the CPU when it has to start executing a new one.
For example, process A with its address space and stack is currently being executed by the CPU and there is a system call to jump to a higher priority process B;
the CPU needs to remember the current state of the process A so that it can suspend its operation, begin executing the new process B and when done, return to its previously executing process A
Only one context switch process happened
 Question 56
Let T(n) be defined by T(1) = 10 and T(n + 1) = 2n + T(n) and for all integers n ≥ 1 . Which of the following represents the order of growth of T(n) as a function of
 A O(n) B O(n log n) C O(n2) D O(n3)
Algorithms       Recursion
Question 56 Explanation:
Given T(1) is 10 and substitute “n” from 1 in the following equation
T(n + 1) = 2n + T(n)
By substitution method:
n=1, T(2)=2x1+T(1)=2+10=12
n=2,T(3)=2x2+T(2)=4+12=16
n=3,T(4)=2x3+T(3)=6+16=22
n=4,T(5)=2x4+T(4)=8+22=30
n=5,T(6)=2x5+T(5)=10+30=40
n=6,T(7)=2x6+T(6)=12+40=52
n=7,T(8)=2x7+T(7)=14+52=66
The above T(n) value is always less than n2, Then the function is O(n2)
 Question 57
Which of the following UNIX command allows scheduling a program to be executed at the specifies time?
 A cron B nice C date and time D schedule
Linux       UNIX
Question 57 Explanation:
1. Cron is the responsible for executing scheduled and recurring commands at a specific time, week or month. It is a time-based job scheduler in Unix-like computer operating systems.
2. Nice is used to invoke a utility or shell script with a particular priority, thus giving the process more or less CPU time than other processes. A niceness of −20 is the highest priority and 19 is the lowest priority. The default niceness for processes is inherited from its parent process and is usually 0.
3. The date command is used to print out, or change the value of, the system's time and date information.
4. The time command is used to determine how long a given command takes to run. It is useful for testing the performance of your scripts and commands.
 Question 58
In a DMA transfer scheme, the transfer scheme other than burst mode is
 A cycle technique B stealing technique C cycle stealing technique D cycle bypass technique
Computer-Organization       DMA
Question 58 Explanation:
Direct memory access (DMA) is a method that allows an input/output (I/O) device to send or receive data directly to or from the main memory, bypassing the CPU to speed up memory operations.
DMA transfers can transfer either one byte at a time or all at once in burst mode. If they transfer a byte at a time, this can allow the CPU to access memory on alternating bus cycles – this is called cycle stealing since the CPU and either the DMA controller or the bus master contend for memory access.
In burst mode DMA, the CPU can be put on hold while the DMA transfer occurs and a full block of possibly hundreds or thousands of bytes can be moved.
 Question 59
n-th derivative of xn is
 A n xn-1 B nn . n! C nxn ! D n!
Engineering-Mathematics       Calculus
Question 59 Explanation:

 Question 60
A total of 9 units of a resource type available, and given the safe state shown below, which of the following sequence will be a safe state?
 A (P4, P1, P3, P2) B (P4, P2, P1, P3) C (P4, P2, P3, P1) D (P3, P1, P2, P4)
Question 60 Explanation:

 Question 61
Three coins are tossed simultaneously. The probability that they will fall two heads and one tail is
 A 5/8 B 1/8 C 2/3 D 3/8
Engineering-Mathematics       Probability
Question 61 Explanation:
Three coins tossed means , total number of combinations(possibilities) are 23=8
The combinations are (HHH,HHT,HTH,HTT,TTT,TTH,THT,THH)
The number of combinations with two heads and one tail is HHT,HTH,THH
The the probability is the number of combinations of the event/ total combinations of the event = 3/8
 Question 62
The average depth of a binary search tree is:
 A O(n0.5) B O(n) C O(log n) D O(n log n)
Data-Structures       Binary-search-tree
Question 62 Explanation:
A binary search tree is a rooted binary tree, whose internal nodes each store a key (and optionally, an associated value) and each have two distinguished sub-trees, commonly denoted left and right.
The tree additionally satisfies the binary search property, which states that the key in each node must be greater than or equal to any key stored in the left sub-tree, and less than or equal to any key stored in the right sub-tree.
The leaves (final nodes) of the tree contain no key and have no structure to distinguish them from one another.
The average depth of a binary search tree is log(n)
 Question 63

What is the output of the following C code?

#include

int main()

{

int index;

for(index=1; index<=5; index++)

{

printf("%d", index);

if (index==3)

continue;

}

}

 A 1245 B 12345 C 12245 D 12354
Programming-for-Output-Problems       Programming
Question 63 Explanation:
The continue statement is used inside loops. When a continue statement is encountered inside a loop, control jumps to the beginning of the loop for next iteration, skipping the execution of statements inside the body of loop for the current iteration.
In the given code, there are no statements after continue statement , So it won’t effect on the output.
The loop will executes for five iterations,For each iteration it will print corresponding value i.e; 12345.
 Question 64
When an n-type semiconductor is heated?
 A number of electrons increases while that of holes decreases B number of holes increases while that of electrons decreases C number of electrons and holes remain the same D number of electron and holes increases equally
Semi-Conductor       Semi-conductors
Question 64 Explanation:
An N-Type semiconductor is created by adding pentavalent impurities like phosphorus (P), arsenic (As), antimony (Sb), or bismuth (Bi). A pentavalent impurity is called a donor atom because it is ready to give a free electron to a semiconductor.
In n-type semiconductors the number of electrons is more than the holes, so electrons are measured as majority charge carriers and holes are referred to as minority charge carriers.
If the temperature or heat energy applied on the semiconductor is further increased then even more number of valence electrons gains enough energy to break the bonding with the parent atom and they jump into the conduction band.
If more number of electrons leaves the valence band and jumps into the conduction band then more number of holes (vacancies) are created in the valence band at the electrons position
 Question 65
The Cyclomatic Complexity metric V(G) of the following control flow graph
 A 3 B 4 C 5 D 6
Software-Engineering       Cyclomatic-complexity
Question 65 Explanation:
Cyclomatic complexity is the measurement of a source code complexity.
It is calculated through a control flow graph which is developed on the basis of source code which measures the number of linearly-independent paths through a program module
The Cyclomatic Complexity of a graph = E − N + 2*P, where
where,
E = represents a number of edges in the control flow graph.
N = represents a number of nodes in the control flow graph.
P = represents a number of nodes that have exit points in the control flow graph.
From the given graph has: E = 7, N = 5 and P = 1
Cyclomatic Complexity = 7 – 5 + 2(1) = 4
 Question 66
Which of the following algorithm design technique is used in merge sort?
 A Greedy method B Backtracking C Dynamic programming D Divide and Conquer
Algorithms       Sorting
Question 66 Explanation:
Merge sort is a divide-and-conquer algorithm based on the idea of breaking down a list into several sub-lists until each sublist consists of a single element and merging those sublists in a manner that results into a sorted list.
Conceptually, a merge sort works as follows:
1. Divide the unsorted list into n sublists, each containing one element (a list of one element is considered sorted).
2. Repeatedly merge sublists to produce new sorted sublists until there is only one sublist remaining. This will be the sorted list.
 Question 67
The arithmetic mean of attendance of 49 students of class A is 40% and that of 53 students of class B is 35%. Then the percentage of arithmetic mean of attendance of class A and B is
 A 27.2% B 50.25% C 51.13% D 37.4%
Aptitude       Aptitude
Question 67 Explanation:
The number of class A students are 49
The arithmetic mean of attendance of Class A is 40%
Arithmetic mean of 49 students attendance of Class A is 49x40
The number of class B students are 53
The arithmetic mean of attendance of Class A is 35%
Arithmetic mean of 53 students attendance of Class A is 53x35
The percentage of arithmetic mean of attendance of class A and B is
= (40x49 + 35x53 )/ (49 + 53)=3815/102
= 37.40
 Question 68

Which of the following sentences can be generated by

S -> aS | bA

A -> d | cA

 A bccdd B abbcca C abcabc D abcd
Theory-of-Computation       Context-free-language
Question 68 Explanation:
Given language:
S -> aS | bA
A -> d | cA
From the given productions, it is not possible to derive the sentence bccdd(two consecutive d’s).So option A is not correct.
In the string “abbcca”, there are two consecutive c’s followed by the letter ‘a’, This can’t be derived from the given productions,
From the given productions, it is also not possible to derive string which consists of the letter ‘c’ followed by ‘a’ , So the option (c ) also not correct
From given productions, we can generate the string “abcd”
 Question 69
The hamming distance between the octets of 0xAA and 0x55 is
 A 7 B 5 C 8 D 6
Digital-Logic-Design       Number-system
Question 69 Explanation:
→ The Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.
→ In other words, it measures the minimum number of substitutions required to change one string into the other, or the minimum number of errors that could have transformed one string into the other.
→ Given HexaDecimal numbers are 0xAA and 0x55.
Decimal equivalent of 0xAA is 170
Binary form of 0xAA is 1010 1010
Decimal equivalent of 0x55 is 85
Binary form of 0xAA is 0101 0101
The two numbers binary length length is 8.
→ If you observe all the bits of above two binary numbers, both numbers have different bits in all positions.
→ So according to definition , the number of positions at which the corresponding symbols are different which is 8.
 Question 70
Given X: 0 10 16 Y: 6 16 28 The interpolated value X=4 using piecewise linear interpolation is
 A 11 B 4 C 22 D 10
Interpolation       Interpolation
Question 70 Explanation:
→ linear interpolation is a method of curve fitting using linear polynomials to construct new data points within the range of a discrete set of known data points.
 Question 71
A fast wide SCSI-II disk drive spins at 7200 RPM, has a sector size of 512 bytes, and holds 160 sectors per track. Estimate the sustained transfer rate of this drive
 A 576000 Kilobytes / sec B 9600 Kilobytes / sec C 4800 Kilobytes / sec D 19200 Kilobytes / sec
Operating-Systems       Disk-scheduling
Question 71 Explanation:
Given data,
Disk drive spins=7200 RPM,
Sector size=512 bytes
Disk drive holds= 160 sectors/track
Step-1: First find the number of rotations for 1 second. They given RPM for minute.
1 sec= 7200 / 60 = 120 rotations
Step-2: Disk Transfer rate =120*160*512
= 98,30,400 Bytes per second
Step-3: Equivalent in kilobytes = 9830400 / 1024
= 9600 KB
Note: The disk spins 120 times per second, and each spin transfers a track of 80 KB. Thus, the sustained transfer rate can be approximated as 9600 KB/s.
 Question 72
How many edges are there in a forest with v vertices and k components?
 A (v+1)−k B (v+1)/2 −k C v−k D v+k
Engineering-Mathematics       Graph-theory
Question 72 Explanation:
Method-1:
→ Suppose, if each vertex is a component, then k=v, then there will not be any edges among them. So, v-k= 0 edges.
Method-2:
→ According to pigeonhole principle, every component have v/k vertices.
→ Every component there will be (v/k)-1 edges.
→ Total k components and edges= k*((v/k)-1)
= v–k
 Question 73
How many 3-to-8 line decoders with an enable input are needed to construct a 6-to-64 line decoder without using any other logic gates?
 A 7 B 8 C 9
Digital-Logic-Design       Encpder-decoder
Question 73 Explanation:
Step-1: Level-1 has one 3x8 Decoder and Level-2 has eight 3x8 Decoders. Out of 6 inputs 3 inputs given to level-1 and remaining to level 2.
Step-2: Level-2 has total 8*8=64 output lines. Based on the input, the decoder in level-1 enables one of the Decoders in level-2.
Step-3: The enabled decoder in level-2 selects one of the output lines and keeps that line high and remaining lines low.
 Question 74
The in-order traversal of a tree resulted in FBGADCE. Then the pre-order traversal of that tree would result in
 A FGBDECA B ABFGCDE C BFGCDEA D AFGBDEC
Data-Structures       Tree-traversal
Question 74 Explanation:
→ In order traversal properties are left,root and right.
→ Pre order properties are root,left and right
→ Based on the given sequence, we will get binary tree is

→ Based upon the inorder traversal, we will get preorder sequence is ABFGCDE.
 Question 75
Which one of the following is true?
 A R ∩ S = (R ∪ S) − [(R − S) ∪ (S − R)] B R ∪ S = (R ∩ S) − [(R − S) ∪ (S − R)] C R ∩ S = (R ∪ S) − [(R − S) ∩ (S − R)] D R ∩ S = (R ∪ S) ∪ (R − S)
Engineering-Mathematics       Set-Theory
Question 75 Explanation:
This is direct formula for set theory.
R∩S = (R ∪ S) − [(R − S) ∪ (S − R)]
 Question 76
A symbol table of length 152 is processing 25 entries at any instant. What is occupation density?
 A 0.164 B 127 C 8.06 D 6.08
Data-Structures
Question 76 Explanation:
Given data,
Symbol table length=152,
Number of entries=25,
Occupation density=?
Step-1: To find Occupation density require number of entries and length of symbol table.
Occupation Density = Number of entries/ Length of symbol table
= 25/152
= 0.164
 Question 77
Consider a direct mapped cache with 64 blocks and a block size of 16 bytes. To what block number does the byte address 1206 map to
 A does not map B 6 C 11 D 54
Computer-Organization       I/O-mapping
Question 77 Explanation:
Given data,
Direct memory cache have = 64 block
Block size = 16 Bytes
Block number does the byte address of 1206 map=?
Step-1: To find block number = Byte Address / Block size
= 1206/16
= 75.3
Step-2: Byte address 1206 map to 75th block.
Step-3: We have to find the cache block number.
Cache block number = (Block number) mod (Block size in cache)
= 75 mod 16
= 11
 Question 78
What is the matrix that represents the rotation of an object by θ degree about the origin in 2D?
 A B C D
Engineering-Mathematics       Linear-algebra
Question 78 Explanation:
→The matrix representation of a counter-clockwise rotation by θ degrees about the origin.
 Question 79
A processor takes 12 cycles to complete an instruction I. The corresponding pipelined processor uses 6 stages with the execution times of 3, 2, 5, 4, 6 and 2 cycles respectively. What is the asymptotic speedup assuming that a very large number of instructions are to be executed?
 A 1.83 B 2 C 3 D 6
Computer-Organization       Microprocessor
Question 79 Explanation:
Step-1: Speed Up= Time without Pipeline / Time with Pipeline
Step-2: It is given that without pipeline it takes 12ns to execute one instruction. Assuming there are n-instructions, time without pipeline = (12*n) ns.
Step-3: For time with pipeline, when there are k-stages in the pipeline, time taken to execute n-instructions is = (k+n-1) clock cycles.
Step-4: There are six stages in the pipeline, so k=6.
Time with pipeline = (6+n-1) clock cycles
= (n+5) clock cycles.
Step-5: It is also given that very large number of instructions are to be executed. So in the time with pipeline, (n+5) clock cycles, we can ignore 5. So time with pipeline for running large number of instructions = n clock cycles.
Step-6: 1 clock cycle time in pipeline = max. of all stage delays
= max(3, 2, 5, 4, 6, 2)
= 6ns
Now, time with pipeline= (n*6) ns
Asymptotic speedup = (12*n) / (6*n)
= 2
 Question 80
Find the memory address of the next instruction executed by the microprocessor (8086), when operated in real mode for CS=1000 and IP=E000
 A 10E00 B 1E000 C F000 D 1000E
Computer-Organization       Microprocessor
There are 80 questions to complete.