ISRO CS 2011
Question 1 
The encoding technique used to transmit the signal in giga ethernet technology over fiber optic medium is
Differential Manchester encoding
 
Non return to zero  
4B/5B encoding
 
8B/10B encoding 
Question 1 Explanation:
→ In telecommunications, 8b/10b is a line code that maps 8bit words to 10bit symbols to achieve DCbalance 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 10bit output values for each input 8bit value.
→ 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 10bit output values for each input 8bit value.
Question 2 
Which of the following is an unsupervised neural network?
RBS  
Hopfield  
Back propagation  
Kohonen  
Incomplete Question 
Question 2 Explanation:
→ A selforganizing map (SOM) or selforganizing feature map (SOFM) is a type of artificial neural network (ANN) that is trained using unsupervised learning to produce a lowdimensional (typically twodimensional), 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 selforganizing map
→ Hopfield nets serve as contentaddressable ("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
→ Kohonen map or network is selforganizing map
→ Hopfield nets serve as contentaddressable ("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
Replacing run time computation by compile time computation  
Removing loop invariant computation  
Removing common subexpressions
 
replacing a costly operation by a relatively cheaper one 
Question 3 Explanation:
An optimization method in which an operator is changed to a lessexpensive operator;
Example: Exponentiation is replaced by multiplication and multiplication is in return replaced by addition.(x * 2 becomes x + x)
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 roundrobin scheduling with 5 ms is used what is the average waiting time of the processes in the queue?
If roundrobin scheduling with 5 ms is used what is the average waiting time of the processes in the queue?
27 ms  
26.2 ms  
27.5 ms  
27.2 ms 
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+(2520)=5
→Waiting time of process P2 = 5
→Waiting time of process P3 = 10 + (3015 )+ (4335) + (5348)=10 + 15 + 8 + 5 = 38
→Waiting time of process P4 = 15+(3520)=15 + 15 = 30
→Waiting time of process P5 =20+(3825)+4843= 20 + 13 + 5 = 38
→Average waiting time = 20 + 5 + 38 + 30 + 38 =131/5 = 26.2
→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+(2520)=5
→Waiting time of process P2 = 5
→Waiting time of process P3 = 10 + (3015 )+ (4335) + (5348)=10 + 15 + 8 + 5 = 38
→Waiting time of process P4 = 15+(3520)=15 + 15 = 30
→Waiting time of process P5 =20+(3825)+4843= 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 ?
register  
immediate  
register indirect  
register relative 
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.
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?
All 1’s  
All 0’s  
X  
Y 
Question 6 Explanation:
Question 7 
Which of the following is true about the zbuffer algorithm?
It is a depth sort algorithm  
No limitation on total number of objects  
Comparisons of objects is done  
zbuffer is initialized to background colour at start of algorithm 
Question 7 Explanation:
The Zbuffer 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.
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 floatingpoint number C1D00000 (hexadecimal notation)? (Assume 32bit, single precision floating point IEEE representation)
28  
15  
26  
28 
Question 8 Explanation:
Floating Point number in Hexadecimal = C1D00000
Floating Point number in Binary = 1100 0001 1101 0000 0000 0000 0000 0000
In 32bit, 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 = 131127 = 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*2^{0}+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
Floating Point number in Binary = 1100 0001 1101 0000 0000 0000 0000 0000
In 32bit, 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 = 131127 = 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*2^{0}+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?
480 Mbps
 
400 Mbps  
200 Mbps  
12 Mbps 
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?
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?
25%  
40%  
50%
 
90%

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%
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?
5  
14  
24  
35 
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
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
SMTP  
ARP  
RIP  
BOOTP 
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.
→ 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?
It is applied to a small part of the code  
It can be used to optimize intermediate code  
To get the best out of this, it has to be applied repeatedly  
It can be applied to the portion of the code that is not contiguous 
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.
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?
Quick sort  
Insertion sort  
Selection sort  
Heap sort 
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.
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?
T1 − T2 − T3
 
T3 − T1 − T2  
T2 − T1 − T3  
T1 − T3 − T2 
Question 15 Explanation:
From the following precedence graph, T3 → T1→ T2
Question 16 
Logic family popular for low power dissipation
CMOS  
ECL  
TTL  
DTL 
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
LEVEL0 DFD
 
LEVEL1 DFD  
LEVEL2 DFD  
LEVEL3 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.
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
Pbox  
Sbox  
Caesar cipher  
Vigenere cipher 
Question 18 Explanation:
A polyalphabetic cipher is any cipher based on substitution, using multiple substitution alphabets. The Vigenère cipher is probably the bestknown 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?
0  
3  
4  
None of the above 
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.
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
IEEE 802.16  
IEEE 802.36  
IEEE 812.16  
IEEE 806.16 
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?
Relational Database Management System  
Hierarchical  
Object Oriented Database Management System  
Network 
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.
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 bitmapped graphics with 64bit colour. What is the maximum resolution it can support?
800 x 600  
1024 x 768  
1280 x 1024  
1920 x 1440 
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.
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?
Read (when it is low)  
Read (when it is high)  
Write (when it is low)  
Read and Write (when it is high) 
Question 23 Explanation:
RDRead: 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 32bit machine is 4K bytes then the size of the page table is
1 M bytes  
2 M bytes  
4 M bytes  
4 K bytes

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 =2^{2}2^{10}=2^{12} Bytes
→Number of pages = logical address space/ size of each page = 2^{32}/ 2^{12}= 2^{20}
→Page table size = number of pages * size of a page table entry
= 2^{20} * 2^{2}
= 2^{22 }
→Size of logical address = 32 bits
→Page size = 4K =2^{2}2^{10}=2^{12} Bytes
→Number of pages = logical address space/ size of each page = 2^{32}/ 2^{12}= 2^{20}
→Page table size = number of pages * size of a page table entry
= 2^{20} * 2^{2}
= 2^{22 }
Question 25 
A problem whose language is recursive is called?
Unified problem  
Boolean function  
Recursive problem  
Decidable 
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.
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?
70%
 
30%
 
60%
 
64%

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%
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?
SELECT  
Value  
INPUT  
BODY 
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 , ,
→ 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?
Package diagram
 
Activity diagram  
Class diagram  
Sequence diagram 
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.
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
First Normal Form  
Second Normal Form
 
Third Normal Form  
Third Normal Form 
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 nonprime attribute that is dependent on any proper subset of any candidate key of the relation.
A nonprime attribute of a relation is an attribute that is not a part of any candidate key of the relation.
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 nonprime attribute that is dependent on any proper subset of any candidate key of the relation.
A nonprime 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) =
Y+XZ  
X+YZ  
XY+Z  
XZ+Y 
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
(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 flipflop, 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 flipflop is
Set to 1  
Set to 0
 
No change in state
 
Forbidden 
Question 31 Explanation:
Question 32 
In which layer of network architecture, the secured socket layer (SSL) is used?
physical layer  
session layer  
application layer  
presentation layer 
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)?
8 Mbps  
6.4 Mbps  
0.8 Mbps  
0.64 Mbps 
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 * 10^{6}) / 100 =6400000=6.4 Mbps
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 * 10^{6}) / 100 =6400000=6.4 Mbps
Question 34 
Black Box Software Testing method focuses on the
Boundary condition of the software  
Control structure of the software  
Functional requirement of the software  
Independent paths of the software 
Question 34 Explanation:
Black box testing, which is also known as behavioral, opaquebox, closedbox, specificationbased or eyetoeye 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 
Belady’s anomaly means
Page fault rate is constant even on increasing the number of allocated frames  
Page fault rate may increase on increasing the number of allocated frames  
Page fault rate may increase on decreasing the number of allocated frames  
Page fault rate may decrease on increasing the number of allocated frames 
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 firstin firstout (FIFO) page replacement algorithm.
This phenomenon is commonly experienced when using the firstin firstout (FIFO) page replacement algorithm.
Question 36 
If A and B are square matrices with same order and A is symmetric, then BTAB
Skew symmetric  
Symmetric  
Orthogonal  
Idempotent 
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.
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))
7  
8  
7.4  
7.0 
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.
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?
96 Gbps  
192 Mbps  
512 Gbps  
192 Gbps 
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
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
INTR & INTA
 
RD & WR  
S0 & S1  
HOLD & HLDA 
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
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?
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?
4  
5  
6  
7 
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
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 8bit bus and a newer version operates at 20 MHz with a 32bit bus, the maximum speedup possible approximately will be
2  
4  
8  
16 
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
20/5 = 4 and 32/8 = 4
Maximum speed up possible is 4
Question 42 
The search concept used in associative memory is
Parallel search  
Sequential search  
Binary search  
Selection search 
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 Contentaddressable memory (CAM)
Associative memory makes a parallel search with the stored patterns as data files.
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
S → AB
A → a
B → b
B → C
A  
B  
C  
S 
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++;
int x,y=10; z=12;
x=y++ + z++;
x=22, y=10, z=12  
x=24, y=10, z=12
 
x=24, y=11, z=13  
x=22, y=11, z=13 
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.
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 
The broadcast address for IP network 172.16.0.0 with subnet mask 255.255.0.0 is
172.16.0.255  
172.16.255.255  
255.255.255.255  
172.255.255.255 
Question 45 Explanation:
IP address: 172.16.0.0
Sub network address: 255.255.0.0
Convert the above two addresses into binary format and perform the bitwise AND operation
Convert the above two addresses into binary format and perform the bitwise AND operation
Question 46 
Which RAID level gives blocklevel striping with double distributed parity?
RAID 10  
RAID 2  
RAID 6  
RAID 5 
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: Bytelevel striping with a dedicated parity disk
RAID 4: Blocklevel striping with blocklevel striping with two parity blocks parity
RAID 5: Blocklevel striping with distributed parity
RAID 6: Blocklevel striping with double distributed parity.
RAID levels and their corresponding functionality as shown below
RAID 0: Stripping
RAID 1: Mirroring
RAID 2: Hamming code for error detection
RAID 3: Bytelevel striping with a dedicated parity disk
RAID 4: Blocklevel striping with blocklevel striping with two parity blocks parity
RAID 5: Blocklevel striping with distributed parity
RAID 6: Blocklevel striping with double distributed parity.
Question 47 
The output expression of the following gate network is
X. Y + X’ Y’  
X. Y + X. Y  
X. Y
 
X + Y 
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
L  
L/2  
(L+1)/2  
2L 
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
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 32bit machine where fourlevel 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?
126  
128  
122  
120 
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
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?
6.9 e^{9}  
6.9 e^{6}  
69 e^{9}  
4 e^{9} 
Question 50 Explanation:
Bandwidth = 2.048 Mbps
The amount of time data transferred is = 10 hours = 36000 secs
Total Data transmitted = 2.048 * 10^{6} * 36000 = 2.048 * 36 * 10^{9} bits
The number of Error bits received = 512
Error rate = total number of error bits/ total data transferred per second
= 512 / 73.728 * 10^{9}
= 6.944 * 10^{9}
The amount of time data transferred is = 10 hours = 36000 secs
Total Data transmitted = 2.048 * 10^{6} * 36000 = 2.048 * 36 * 10^{9} bits
The number of Error bits received = 512
Error rate = total number of error bits/ total data transferred per second
= 512 / 73.728 * 10^{9}
= 6.944 * 10^{9}
Question 51 
Warnier Diagram enables the analyst to represent
Warnier Diagram enables the analyst to represent  
Information Hierarchy  
Data Flow  
State Transition 
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.
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
Routing the packets  
Authentication  
obtaining IP address  
domain name resolving 
Question 52 Explanation:
The Lightweight Directory Access Protocol (LDAP ) is an open, vendorneutral, 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
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
Reflexive, Augmentation and Decomposition  
Transitive, Augmentation and Reflexive  
Augmentation, Transitive, Reflexive and Decomposition  
Reflexive, Transitive and Decomposition 
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.
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
2  
4  
8  
16 
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
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?
1  
2  
3  
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
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
O(n)  
O(n log n)  
O(n^{2})  
O(n^{3}) 
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(n^{2})
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(n^{2})
Question 57 
Which of the following UNIX command allows scheduling a program to be executed at the specifies time?
cron  
nice  
date and time  
schedule 
Question 57 Explanation:
1. Cron is the responsible for executing scheduled and recurring commands at a specific time, week or month. It is a timebased job scheduler in Unixlike 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.
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
cycle technique  
stealing technique  
cycle stealing technique  
cycle bypass technique 
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.
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 
nth derivative of x^{n} is
n x^{n1}  
n^{n} . n!  
nx^{n} !  
n! 
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?
(P4, P1, P3, P2)  
(P4, P2, P1, P3)  
(P4, P2, P3, P1)  
(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
5/8  
1/8  
2/3  
3/8 
Question 61 Explanation:
Three coins tossed means , total number of combinations(possibilities) are 2^{3}=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
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:
O(n^{0.5})  
O(n)  
O(log n)  
O(n log n) 
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 subtrees, 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 subtree, and less than or equal to any key stored in the right subtree.
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)
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 subtree, and less than or equal to any key stored in the right subtree.
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;
}
}
1245  
12345  
12245  
12354 
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.
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 ntype semiconductor is heated?
number of electrons increases while that of holes decreases  
number of holes increases while that of electrons decreases  
number of electrons and holes remain the same  
number of electron and holes increases equally 
Question 64 Explanation:
An NType 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 ntype 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
In ntype 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
3  
4  
5  
6 
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 linearlyindependent 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
It is calculated through a control flow graph which is developed on the basis of source code which measures the number of linearlyindependent 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?
Greedy method  
Backtracking  
Dynamic programming  
Divide and Conquer 
Question 66 Explanation:
Merge sort is a divideandconquer algorithm based on the idea of breaking down a list into several sublists 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.
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
27.2%  
50.25%  
51.13%  
37.4% 
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
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
bccdd  
abbcca  
abcabc  
abcd 
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”
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
7  
5  
8  
6 
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.
→ 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
11  
4  
22  
10 
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 SCSIII 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
576000 Kilobytes / sec  
9600 Kilobytes / sec  
4800 Kilobytes / sec  
19200 Kilobytes / sec 
Question 71 Explanation:
Given data,
Disk drive spins=7200 RPM,
Sector size=512 bytes
Disk drive holds= 160 sectors/track
Step1: First find the number of rotations for 1 second. They given RPM for minute.
1 sec= 7200 / 60 = 120 rotations
Step2: Disk Transfer rate =120*160*512
= 98,30,400 Bytes per second
Step3: 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.
Disk drive spins=7200 RPM,
Sector size=512 bytes
Disk drive holds= 160 sectors/track
Step1: First find the number of rotations for 1 second. They given RPM for minute.
1 sec= 7200 / 60 = 120 rotations
Step2: Disk Transfer rate =120*160*512
= 98,30,400 Bytes per second
Step3: 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?
(v+1)−k  
(v+1)/2 −k  
v−k  
v+k 
Question 72 Explanation:
Method1:
→ Suppose, if each vertex is a component, then k=v, then there will not be any edges among them. So, vk= 0 edges.
Method2:
→ 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
→ Suppose, if each vertex is a component, then k=v, then there will not be any edges among them. So, vk= 0 edges.
Method2:
→ 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 3to8 line decoders with an enable input are needed to construct a 6to64 line decoder without using any other logic gates?
7  
8  
9 
Question 73 Explanation:
Step1: Level1 has one 3x8 Decoder and Level2 has eight 3x8 Decoders. Out of 6 inputs 3
inputs given to level1 and remaining to level 2.
Step2: Level2 has total 8*8=64 output lines. Based on the input, the decoder in level1 enables one of the Decoders in level2.
Step3: The enabled decoder in level2 selects one of the output lines and keeps that line high and remaining lines low.
Step2: Level2 has total 8*8=64 output lines. Based on the input, the decoder in level1 enables one of the Decoders in level2.
Step3: The enabled decoder in level2 selects one of the output lines and keeps that line high and remaining lines low.
Question 74 
The inorder traversal of a tree resulted in FBGADCE. Then the preorder traversal of that tree would result in
FGBDECA  
ABFGCDE  
BFGCDEA  
AFGBDEC 
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.
→ 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?
R ∩ S = (R ∪ S) − [(R − S) ∪ (S − R)]  
R ∪ S = (R ∩ S) − [(R − S) ∪ (S − R)]  
R ∩ S = (R ∪ S) − [(R − S) ∩ (S − R)]  
R ∩ S = (R ∪ S) ∪ (R − S) 
Question 75 Explanation:
This is direct formula for set theory.
R∩S = (R ∪ S) − [(R − S) ∪ (S − R)]
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?
0.164  
127  
8.06  
6.08 
Question 76 Explanation:
Given data,
Symbol table length=152,
Number of entries=25,
Occupation density=?
Step1: 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
Symbol table length=152,
Number of entries=25,
Occupation density=?
Step1: 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
does not map  
6  
11  
54 
Question 77 Explanation:
Given data,
Direct memory cache have = 64 block
Block size = 16 Bytes
Block number does the byte address of 1206 map=?
Step1: To find block number = Byte Address / Block size
= 1206/16
= 75.3
Step2: Byte address 1206 map to 75th block.
Step3: We have to find the cache block number.
Cache block number = (Block number) mod (Block size in cache)
= 75 mod 16
= 11
Direct memory cache have = 64 block
Block size = 16 Bytes
Block number does the byte address of 1206 map=?
Step1: To find block number = Byte Address / Block size
= 1206/16
= 75.3
Step2: Byte address 1206 map to 75th block.
Step3: 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?
Question 78 Explanation:
→The matrix representation of a counterclockwise 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?
1.83  
2  
3  
6 
Question 79 Explanation:
Step1: Speed Up= Time without Pipeline / Time with Pipeline
Step2: It is given that without pipeline it takes 12ns to execute one instruction. Assuming there are ninstructions, time without pipeline = (12*n) ns.
Step3: For time with pipeline, when there are kstages in the pipeline, time taken to execute ninstructions is = (k+n1) clock cycles.
Step4: There are six stages in the pipeline, so k=6.
Time with pipeline = (6+n1) clock cycles
= (n+5) clock cycles.
Step5: 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.
Step6: 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
Step2: It is given that without pipeline it takes 12ns to execute one instruction. Assuming there are ninstructions, time without pipeline = (12*n) ns.
Step3: For time with pipeline, when there are kstages in the pipeline, time taken to execute ninstructions is = (k+n1) clock cycles.
Step4: There are six stages in the pipeline, so k=6.
Time with pipeline = (6+n1) clock cycles
= (n+5) clock cycles.
Step5: 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.
Step6: 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
10E00  
1E000  
F000  
1000E 
There are 80 questions to complete.