ISRO CS 2014
Question 1 
Consider a 33 MHz CPU based system. What is the number of wait states required if it is interfaced with a 60 ns memory? Assume a maximum of 10 ns delay for additional circuitry like buffering and decoding.
0  
1  
2  
3 
Question 1 Explanation:
A wait state is a situation in which a computer program or processor is waiting for the completion of some event before resuming activity. A program or process in a wait state is inactive for the duration of the wait stat
When a computer processor works at a faster clock speed (expressed in MHz or millions of cycles per second) than the random access memory ( RAM ) that sends it instructions, it is set to go into a wait state for one or more clock cycles so that it is synchronized with RAM speed. In general, the more time a processor spends in wait states, the slower the performance of that processor.
From the given question, we will get total memory access time by combining access time and delay.
CPU frequency = 33 MHz
1 clock time = 1 / (33 MHz) = (1/33)*106 = 30.30 ns.
Total memory access time = 60 ns + 10 ns = 70 ns.
Total number of wait states = Total number of cycle needed
= Total memory access time / CPU frequency
=70 ns / (30.30 ns) = 2.31 (equivalent to 3 cycles)
When a computer processor works at a faster clock speed (expressed in MHz or millions of cycles per second) than the random access memory ( RAM ) that sends it instructions, it is set to go into a wait state for one or more clock cycles so that it is synchronized with RAM speed. In general, the more time a processor spends in wait states, the slower the performance of that processor.
From the given question, we will get total memory access time by combining access time and delay.
CPU frequency = 33 MHz
1 clock time = 1 / (33 MHz) = (1/33)*106 = 30.30 ns.
Total memory access time = 60 ns + 10 ns = 70 ns.
Total number of wait states = Total number of cycle needed
= Total memory access time / CPU frequency
=70 ns / (30.30 ns) = 2.31 (equivalent to 3 cycles)
Question 2 
The number of states required by a Finite State Machine, to simulate the behavior of a computer with a memory capable of storing ‘m’ words, each of length ‘n’ bits is?
m x 2^{n}  
2^{m+n}  
2^{mn}  
m+n 
Question 2 Explanation:
A finitestate machine (FSM) an abstract machine that can be in exactly one of a finite number of states at any given time. The FSM can change from one state to another in response to some external inputs; the change from one state to another is called a transition.
Generally FSM consists of 2^{x} states for a given input x.
But in the question, the input is specified number of words and length of each word.
For a given, ‘m’ words, each of length ‘n’ bits, so total number of bits are m*n
So, total number of states required by a Finite State Machine are 2^{mn}.
Generally FSM consists of 2^{x} states for a given input x.
But in the question, the input is specified number of words and length of each word.
For a given, ‘m’ words, each of length ‘n’ bits, so total number of bits are m*n
So, total number of states required by a Finite State Machine are 2^{mn}.
Question 3 
What is the output of the following C program?
#include<stdio.h>
#define SQR(x) (x*x)
int main()
{
int a;
int b=4;
a=SQR(b+2);
printf("%d\n",a);
return 0;
}
14  
36  
18  
20 
Question 3 Explanation:
In the question, SQR is macro which is preprocessor statement in c language.
Here macro takes one parameter also.
A macro is a fragment of code that is given a name. You can use that fragment of code in your program by using the name. For example,
#define SQR(x) (x*x)
Here, when we use SQR(x) in our program, it's replaced with (x*x)
The movement program control execute the macro, it will replace SQR(b+2) by (b+2*b+2)
The expression (b+2*b+2) consists of addition and multiplication operators.
Between two operators multiplication operator has highest priority later addition .
So the expression evaluation steps are as follows
a = (b + 2*b + 2)
=(4+2*4+2)// first multiplication
=(4+8+2) // addition equal priority , so it will evaluate left to right
=(12+2)
=14
Here macro takes one parameter also.
A macro is a fragment of code that is given a name. You can use that fragment of code in your program by using the name. For example,
#define SQR(x) (x*x)
Here, when we use SQR(x) in our program, it's replaced with (x*x)
The movement program control execute the macro, it will replace SQR(b+2) by (b+2*b+2)
The expression (b+2*b+2) consists of addition and multiplication operators.
Between two operators multiplication operator has highest priority later addition .
So the expression evaluation steps are as follows
a = (b + 2*b + 2)
=(4+2*4+2)// first multiplication
=(4+8+2) // addition equal priority , so it will evaluate left to right
=(12+2)
=14
Question 4 
Consider the following pseudo code
while (m < n)
if (x > y ) and (a < b) then
a=a+1
y=y1
end if
m=m+1
end while
What is cyclomatic complexity of the above pseudo code?
while (m < n)
if (x > y ) and (a < b) then
a=a+1
y=y1
end if
m=m+1
end while
What is cyclomatic complexity of the above pseudo code?
2  
3  
4  
5 
Question 4 Explanation:
Finding cyclomatic complexity using 3 formulas
1. EV+2
2. P+1
3. No.of Regions
According to above code segment.
→ EdgesVertices+2
= 86+2 = 4
→ Total number of regions = 4
→ Predicates+1 = 3+1 = 4
Note: We can use any one of above formulas.
1. EV+2
2. P+1
3. No.of Regions
According to above code segment.
→ EdgesVertices+2
= 86+2 = 4
→ Total number of regions = 4
→ Predicates+1 = 3+1 = 4
Note: We can use any one of above formulas.
Question 5 
What is the number of steps required to derive the string ((() ()) ()) for the following grammar?
S → SS
S → (S)
S → ε
S → SS
S → (S)
S → ε
10  
15  
12  
16 
Question 5 Explanation:
The required string consists of only only matched parenthesis enclosed with parentheses.
To generate the string ((() ()) ()) , we need to perform the steps sequentially.
As the string starts with left parentheses, we will start with the production S → (S).
Depending upon the string we will go for either left or right tree generation.
1) S → (S) // Replacing S by (S)
2) S → (SS) // Replacing by S by SS
3) S → ((S)S) //Replacing S by(S)
4) S → ((SS)S)// Replacing S by SS
5) S → (((S)S))S)
6) S → (((S)(S))S)
7) S → (((S)(S))(S)) [S → ε]
8) S → ((()(S))S) [S → ε]
9) S → ((()())S) [S → ε]
10) S → ((()()))
To generate the string ((() ()) ()) , we need to perform the steps sequentially.
As the string starts with left parentheses, we will start with the production S → (S).
Depending upon the string we will go for either left or right tree generation.
1) S → (S) // Replacing S by (S)
2) S → (SS) // Replacing by S by SS
3) S → ((S)S) //Replacing S by(S)
4) S → ((SS)S)// Replacing S by SS
5) S → (((S)S))S)
6) S → (((S)(S))S)
7) S → (((S)(S))(S)) [S → ε]
8) S → ((()(S))S) [S → ε]
9) S → ((()())S) [S → ε]
10) S → ((()()))
Question 6 
The process of modifying IP address information in IP packet headers while in transit across a traffic routing device is called
Port address translation (PAT)  
Network address translation (NAT)  
Address mapping  
Port mapping 
Question 6 Explanation:
Network address translation (NAT)
It is a method of remapping one IP address space into another by modifying network address information in the IP header of packets while they are in transit across a traffic routing device.The technique was originally used as a shortcut to avoid the need to readdress every host when a network was moved. It has become a popular and essential tool in conserving global address space in the face of IPv4 address exhaustion. One Internetroutable IP address of a NAT gateway can be used for an entire private network.
Port Address Translation (PAT) is an extension to network address translation (NAT) that permits multiple devices on a local area network (LAN) to be mapped to a single public IP address. The goal of PAT is to conserve IP addresses.
Address mapping (also know as pin mapping or geocoding) is the process of assigning map coordinate locations to addresses in a database. The output of address mapping is a point layer attributed with all of the data from the input database.
Port mapping or port forwarding is an application of network address translation (NAT) which redirects a communication request from one address and port number combination to another while the packets are traversing a network gateway, such as a router or firewall.
It is a method of remapping one IP address space into another by modifying network address information in the IP header of packets while they are in transit across a traffic routing device.The technique was originally used as a shortcut to avoid the need to readdress every host when a network was moved. It has become a popular and essential tool in conserving global address space in the face of IPv4 address exhaustion. One Internetroutable IP address of a NAT gateway can be used for an entire private network.
Port Address Translation (PAT) is an extension to network address translation (NAT) that permits multiple devices on a local area network (LAN) to be mapped to a single public IP address. The goal of PAT is to conserve IP addresses.
Address mapping (also know as pin mapping or geocoding) is the process of assigning map coordinate locations to addresses in a database. The output of address mapping is a point layer attributed with all of the data from the input database.
Port mapping or port forwarding is an application of network address translation (NAT) which redirects a communication request from one address and port number combination to another while the packets are traversing a network gateway, such as a router or firewall.
Question 7 
What does a pixel mask mean?
string containing only 1’s  
string containing only 0’s  
string containing two 0’s  
string containing 1’s and 0’s 
Question 7 Explanation:
Pixel mask: when a given image is intended to be placed over a background, the transparent areas can be specified through a binary mask.
→ This way, for each intended image there are actually two bitmaps:
1. Actual image, in which the unused areas are given a pixel value with all bits set to 0s.
2. Additional mask, in which the correspondent image areas are given a pixel value of all bits set to 0s and the surrounding areas a value of all bits set to 1s.
In the sample at right, black pixels have the allzero bits and white pixels have the allone bits.
→ This way, for each intended image there are actually two bitmaps:
1. Actual image, in which the unused areas are given a pixel value with all bits set to 0s.
2. Additional mask, in which the correspondent image areas are given a pixel value of all bits set to 0s and the surrounding areas a value of all bits set to 1s.
In the sample at right, black pixels have the allzero bits and white pixels have the allone bits.
Question 8 
In the standard IEEE 754 single precision floating point representation, there is 1 bit for sign, 23 bits for fraction and 8 bits for exponent. What is the precision in terms of the number of decimal digits?
5  
6  
7  
8 
Question 8 Explanation:
A floatingpoint variable can represent a wider range of numbers than a fixedpoint variable of the same bit width at the cost of precision. A signed 32bit integer variable has a maximum value of 2^{31} − 1 = 2,147,483,647, whereas an IEEE 754 32bit base2 floatingpoint variable has a maximum value of (2 − 2^{−23}) × 2^{127} ≈ 3.402823 × 10^{38}.
In the IEEE 7542008 standard, the 32bit base2 format is officially referred to as binary32; it was called single in IEEE 7541985. IEEE 754 specifies additional floatingpoint types, such as 64bit base2 double precision and, more recently, base10 representations.
We can convert the binary into decimal representation by using the following steps
let the number of digits in decimal digits be ‘x’
2^{23} = 10^{x }
After taking log on both sides
log_{2}10^{x} =log2 2^{23}
x log_{2}10 = 23log_{2}2 (The value of log_{2}2=1)
3.322 x = 23 (The value of log_{2}10 = 3.321928)
x = 6.92
In the IEEE 7542008 standard, the 32bit base2 format is officially referred to as binary32; it was called single in IEEE 7541985. IEEE 754 specifies additional floatingpoint types, such as 64bit base2 double precision and, more recently, base10 representations.
We can convert the binary into decimal representation by using the following steps
let the number of digits in decimal digits be ‘x’
2^{23} = 10^{x }
After taking log on both sides
log_{2}10^{x} =log2 2^{23}
x log_{2}10 = 23log_{2}2 (The value of log_{2}2=1)
3.322 x = 23 (The value of log_{2}10 = 3.321928)
x = 6.92
Question 9 
Let R be the radius of the circle. What is the angle subtended by an arc of length at the center of the circle?
1 degree  
1 radian  
90 degrees  
π radians 
Question 9 Explanation:
A degree (in full, a degree of arc, arc degree, or arc degree), usually denoted by ° (the degree symbol), is a measurement of a plane angle, defined so that a full rotation is 360 degrees.
It is not an SI unit, as the SI unit of angular measure is the radian, but it is mentioned in the SI brochure as an accepted unit.Because a full rotation equals 2π radians.
The radian (SI symbol rad) is the SI unit for measuring angles, and is the standard unit of angular measure used in many areas of mathematics. The length of an arc of a unit circle is numerically equal to the measurement in radians of the angle that it subtends;
It is not an SI unit, as the SI unit of angular measure is the radian, but it is mentioned in the SI brochure as an accepted unit.Because a full rotation equals 2π radians.
The radian (SI symbol rad) is the SI unit for measuring angles, and is the standard unit of angular measure used in many areas of mathematics. The length of an arc of a unit circle is numerically equal to the measurement in radians of the angle that it subtends;
Question 10 
The number of logical CPUs in a computer having two physical quadcore chips with hyperthreading enabled is
1  
2  
8  
16 
Question 10 Explanation:
There is only one processor chip. That chip can have one, two, four, six, or eight cores.
Currently, an 18core processor is the best you can get in consumer PCs.
Each “core” is the part of the chip that does the processing work. Essentially, each core is a central processing unit (CPU).
From the given information , there are two quad core chips are available
So, total number of physical CPUs = 2*4 = 8
Each CPU has two logical CPU in hyper threading
So, 8 physical CPU have 2*8 = 16 logical CPU.
Currently, an 18core processor is the best you can get in consumer PCs.
Each “core” is the part of the chip that does the processing work. Essentially, each core is a central processing unit (CPU).
From the given information , there are two quad core chips are available
So, total number of physical CPUs = 2*4 = 8
Each CPU has two logical CPU in hyper threading
So, 8 physical CPU have 2*8 = 16 logical CPU.
Question 11 
An aggregation, the association is drawn using which symbol?
A line which loops back on to the same table  
A small open diamond at the end of a line connecting two tables  
A small closed diamond at the end of a line connecting two tables  
A small closed triangle at the end of a line connecting two tables 
Question 11 Explanation:
Association: Relationship between 2 objects. It defines the multiplicity between objects like 11, 1many, many1, many to many.
Aggregation: A directional between objects. When an object “hasa” another object, then you have got an aggregation between them, you have got an aggregation between them. Direction between them specified which object contains the other object.
Aggregation: A directional between objects. When an object “hasa” another object, then you have got an aggregation between them, you have got an aggregation between them. Direction between them specified which object contains the other object.
Question 12 
How many states are there in a minimum state deterministic finite automaton accepting the language L = {w  w {0,1} number of 0’s is divisible by 2 and number of 1’s is divisible by 5, respectively} ?
7  
9  
10  
11 
Question 12 Explanation:
From the given data
String which consists of 0’s and 1’s
Language is number of 0’s is divisible by 2 and number of 1’s is divisible by 5
Machine is minimum state deterministic finite automaton
The set of strings generated by {0,1} are ={ε,0,1,00,11,01,10,000,011,010 .. and so on}
The strings which are generated by language which is specified in the questions are
{ ε , 00, 11111, 0011111, 0011111 , 1111100, 1010111 , 000011111,….and so on }
So, strings accepted by the automaton have to be of length 0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 14, 12….and so on, i.e. equation for length will be 2x + 5y (where x,y>=0 )
Number of 0’s divisible by 2 means , the possible remainders are 0 and 1 i.e; 2
Number of 1’s divisible by 5 means , the possible remainders are 4,3,2,1,0 i.e; 5
Total number of states are 2*5 =10
String which consists of 0’s and 1’s
Language is number of 0’s is divisible by 2 and number of 1’s is divisible by 5
Machine is minimum state deterministic finite automaton
The set of strings generated by {0,1} are ={ε,0,1,00,11,01,10,000,011,010 .. and so on}
The strings which are generated by language which is specified in the questions are
{ ε , 00, 11111, 0011111, 0011111 , 1111100, 1010111 , 000011111,….and so on }
So, strings accepted by the automaton have to be of length 0, 2, 4, 5, 6, 7, 8, 9, 10, 11, 14, 12….and so on, i.e. equation for length will be 2x + 5y (where x,y>=0 )
Number of 0’s divisible by 2 means , the possible remainders are 0 and 1 i.e; 2
Number of 1’s divisible by 5 means , the possible remainders are 4,3,2,1,0 i.e; 5
Total number of states are 2*5 =10
Question 13 
Which of the following is true with respect to Reference?
A reference can never be NULL  
A reference needs an explicit dereferencing mechanism  
A reference can be reassigned after it is established  
A reference and pointer are synonymous 
Question 13 Explanation:
In the C++ programming language, a reference is a simple reference data type that is less powerful but safer than the pointer type inherited from C. A reference is a general concept datatype, with pointers and C++ references being specific reference data type implementations.
We can assign NULL to pointers where as we can’t assign NULL to references.
We can reassign pointers as many as number of times where a reference can never be reassigned once it is established.
We can assign NULL to pointers where as we can’t assign NULL to references.
We can reassign pointers as many as number of times where a reference can never be reassigned once it is established.
Question 14 
There are 200 tracks on a disc platter and the pending requests have come in the order – 36, 69, 167, 76, 42, 51, 126, 12 and 199. Assume the arm is located at the 100 track and moving towards track 199. If the sequence of disc access is 126, 167, 199, 12, 36, 42, 51, 69 and 76 then which disc access scheduling policy is used?
Elevator  
Shortest seektime first  
CSCAN  
First Come First Served 
Question 14 Explanation:
Total tracks are :200
Tracks order is 36, 69, 167, 76, 42, 51, 126, 12 and 199
Track position is at 100 and moving towards 199.
sequence of disc access is 126, 167, 199, 12, 36, 42, 51, 69 and 76
From the above sequence , we can observe that disc moving from 100 to 126 from there to moving right side direction until last disc 199
From the last disc(199) to first disc(12) and from there to next disc(36) and so on.
CSCAN sweeps the disk from endtoend, but as soon it reaches one of the end tracks it then moves to the other end track without servicing any requesting location. As soon as it reaches the other end track it then starts servicing and grants requests headed to its direction
Tracks order is 36, 69, 167, 76, 42, 51, 126, 12 and 199
Track position is at 100 and moving towards 199.
sequence of disc access is 126, 167, 199, 12, 36, 42, 51, 69 and 76
From the above sequence , we can observe that disc moving from 100 to 126 from there to moving right side direction until last disc 199
From the last disc(199) to first disc(12) and from there to next disc(36) and so on.
CSCAN sweeps the disk from endtoend, but as soon it reaches one of the end tracks it then moves to the other end track without servicing any requesting location. As soon as it reaches the other end track it then starts servicing and grants requests headed to its direction
Question 15 
Consider the logic circuit given below:
A’C + BC ‘ + CD  
ABC + C’D  
AB + BC’ + BD’  
AB’ + AC’ + C’D 
Question 15 Explanation:
Q = (((CD)’B)'(AB)’)’
Q = ((CD)’B)” + (AB)”
Q = (CD)’B + AB
Q = (C’ + D’)B + AB
Q = C’B + D’B + AB
Q = AB + BC’ + BD’
Q = ((CD)’B)” + (AB)”
Q = (CD)’B + AB
Q = (C’ + D’)B + AB
Q = C’B + D’B + AB
Q = AB + BC’ + BD’
Question 16 
What is routing algorithm used by OSPF routing protocol?
Distance vector  
Flooding  
Path vector  
Link state 
Question 16 Explanation:
→ Open Shortest Path First (OSPF) is a routing protocol for Internet Protocol (IP) networks.
→ It uses a link state routing (LSR) algorithm and falls into the group of interior gateway protocols (IGPs), operating within a single autonomous system (AS).
→ It implements Dijkstra's algorithm, also known as the shortest path first (SPF) algorithm.
→ It uses a link state routing (LSR) algorithm and falls into the group of interior gateway protocols (IGPs), operating within a single autonomous system (AS).
→ It implements Dijkstra's algorithm, also known as the shortest path first (SPF) algorithm.
Question 17 
If each address space represents one byte of storage space, how many address lines are needed to access RAM chips arranged in a 4 x 6 array, where each chip is 8K x 4 bits ?
13  
15  
16  
17 
Question 17 Explanation:
From the given data
Each chip size is = 8K x 4 bits = 2^{3} x 2^{10} x 2^{2} == 2^{15} bits = 2^{12} bits
Given chip array = 6 x 4 = 24 ( 2^{4}=16 and 2^{5}=32)
So the number bits required for the total chips are 5 bits.
So, total number of bits required = 12 + 5 = 17 bits
Each chip size is = 8K x 4 bits = 2^{3} x 2^{10} x 2^{2} == 2^{15} bits = 2^{12} bits
Given chip array = 6 x 4 = 24 ( 2^{4}=16 and 2^{5}=32)
So the number bits required for the total chips are 5 bits.
So, total number of bits required = 12 + 5 = 17 bits
Question 18 
Consider the following segment table in the segmentation scheme:
What happens if the logical address requested is Segment ID 2 and offset 1000?
What happens if the logical address requested is Segment ID 2 and offset 1000?
Fetches the entry at the physical address 2527 for segment Id2  
A trap is generated  
Deadlock  
Fetches the entry at offset 27 in Segment Id 3 
Question 18 Explanation:
From the question we need to find the logical address for segment id2.
From given table,
Segment2 has a base address = 1527 ‘
limit address = 498.
Process can access memory from the location 1527 to 2025(1527+498)
If the process tries to access the memory with offset 1000 then a segmentation fault trap will be generated.
In computing and operating systems, a trap, also known as an exception or a fault, is typically a type of synchronous interrupt caused by an exceptional condition (e.g., breakpoint, division by zero, invalid memory access)
From given table,
Segment2 has a base address = 1527 ‘
limit address = 498.
Process can access memory from the location 1527 to 2025(1527+498)
If the process tries to access the memory with offset 1000 then a segmentation fault trap will be generated.
In computing and operating systems, a trap, also known as an exception or a fault, is typically a type of synchronous interrupt caused by an exceptional condition (e.g., breakpoint, division by zero, invalid memory access)
Question 19 
The number of bit strings of length 8 that will either start with 1 or end with 00 is?
32  
128  
160  
192 
Question 19 Explanation:
→ Number of bit strings of length 8 that start with 1: 27 = 128.
→ Number of bit strings of length 8 that end with 00: 26 = 64.
→ Number of bit strings of length 8 that start with 1 and end with 00: 25 = 32.
→ Applying the subtraction rule, the number is 128+64−32 = 160
→ Number of bit strings of length 8 that end with 00: 26 = 64.
→ Number of bit strings of length 8 that start with 1 and end with 00: 25 = 32.
→ Applying the subtraction rule, the number is 128+64−32 = 160
Question 20 
Which of the following is not a maturity level as per the Capability Maturity Model?
Initial  
Measurable  
Repeatable  
Optimized 
Question 20 Explanation:
Capability Maturity Model levels
1. Initial
2. Repeatable
3. Defined
4. Managed
5. Optimized
1. Initial
2. Repeatable
3. Defined
4. Managed
5. Optimized
Question 21 
Consider the following sequential circuit
What are the values of Q_{0 }and Q_{1} after 4 clock cycles if the initial values are ?
What are the values of Q_{0 }and Q_{1} after 4 clock cycles if the initial values are ?
11  
01  
10  
00 
Question 21 Explanation:
Question 22 
The test suite (set of test input) used to perform unit testing on module could cover 70% of the code. What is the reliability of the module if the probability of success is 0.95 during testing?
0.665 to 0.95  
At the most 0.665  
At the most 0.95  
At least 0.665 
Question 22 Explanation:
Software Reliability is the probability of failurefree software operation for a specified period of time in a specified environment.
From the given question the module code covers 70% of the entire code, so we need to check with reliability for that module only.
Reliability = percentage of code coverage * probability of success = 0.7 * 0.95 = 0.665.
Probability of success is at most 0.665 %
From the given question the module code covers 70% of the entire code, so we need to check with reliability for that module only.
Reliability = percentage of code coverage * probability of success = 0.7 * 0.95 = 0.665.
Probability of success is at most 0.665 %
Question 23 
In a system an RSA algorithm with p=5 and q=11, is implemented for data security. What is the value of the decryption key if the value of the encryption key is 27?
3  
7  
27  
40 
Question 23 Explanation:
The keys for the RSA algorithm are generated the following way:
1. Choose two distinct prime numbers p and q.
2. Compute n = pq.
3. Compute λ(n) = lcm(λ(p), λ(q)) = lcm(p − 1, q − 1), where λ is Carmichael's totient function. Choose an integer e such that 1 < e < λ(n) and gcd(e, λ(n)) = 1; i.e., e and λ(n) are coprime.
4. Determine d as d ≡ e−1 (mod λ(n)); i.e., d is the modular multiplicative inverse of e modulo λ(n). This means: solve for d the equation d⋅e ≡ 1 (mod λ(n)).
Given two prime numbers are p = 5 and q = 11, encryption key, e = 27
n = p * q = 5 * 11 = 55
λ(n)= (p1) * (q1) = 4 * 10 = 40
Let the value of decryption key be ‘d’ such that:
e * d mod λ(n) = 1
27 * d mod 40 = 1
d = 3
1. Choose two distinct prime numbers p and q.
2. Compute n = pq.
3. Compute λ(n) = lcm(λ(p), λ(q)) = lcm(p − 1, q − 1), where λ is Carmichael's totient function. Choose an integer e such that 1 < e < λ(n) and gcd(e, λ(n)) = 1; i.e., e and λ(n) are coprime.
4. Determine d as d ≡ e−1 (mod λ(n)); i.e., d is the modular multiplicative inverse of e modulo λ(n). This means: solve for d the equation d⋅e ≡ 1 (mod λ(n)).
Given two prime numbers are p = 5 and q = 11, encryption key, e = 27
n = p * q = 5 * 11 = 55
λ(n)= (p1) * (q1) = 4 * 10 = 40
Let the value of decryption key be ‘d’ such that:
e * d mod λ(n) = 1
27 * d mod 40 = 1
d = 3
Question 24 
Suppose you want to build a memory with 4byte words with a capacity of 2^{21} bits.
What is the type of decoder required if the memory is built using 2K x 8 RAM chips?
What is the type of decoder required if the memory is built using 2K x 8 RAM chips?
5 to 32  
6 to 64  
4 to 64  
7 to 128 
Question 24 Explanation:
In digital electronics, a binary decoder is a combinational logic circuit that converts binary information from the n coded inputs to a maximum of 2^{n} unique outputs.
We need to built memory with 4 byte words with a capacity of 221bits
So, number of 4byte words memory = total number of bits / number of byte words
= 2^{21}/(4*bytes)
= 2^{21}/(4*8)
=2^{21}/32=2^{21}/2^{5}
= 2^{16} words
Given RAM chips are of size = 2K x 8
Memory to be built using these RAM chips = 2^{16 Required RAM chips = (216 x 32) / (2K x 8) = 32 x 4 So, RAM chips contains 32 rows and each row with 4 columns. A Decoder is required to select a specific row and multiplexer is required to select a particular column. 5 to 32 Decoder will be required to select the desired row. }
We need to built memory with 4 byte words with a capacity of 221bits
So, number of 4byte words memory = total number of bits / number of byte words
= 2^{21}/(4*bytes)
= 2^{21}/(4*8)
=2^{21}/32=2^{21}/2^{5}
= 2^{16} words
Given RAM chips are of size = 2K x 8
Memory to be built using these RAM chips = 2^{16 Required RAM chips = (216 x 32) / (2K x 8) = 32 x 4 So, RAM chips contains 32 rows and each row with 4 columns. A Decoder is required to select a specific row and multiplexer is required to select a particular column. 5 to 32 Decoder will be required to select the desired row. }
Question 25 
The output of a tristate buffer when the enable input in 0 is
Always 0  
Always 1  
Retains the last value when enable input was high  
Disconnected state 
Question 26 
Suppose there are 11 items in sorted order in an array. How many searches are required on the average, if binary search is employed and all searches are successful in finding the item?
3.00  
3.46  
2.81  
3.33 
Question 26 Explanation:
We can arrange 11 items in the four levels(level0,1,2,3).
Each level has 2i nodes where i is level number
Level 0 has one node so one comparison(1)
Level 1 has two nodes so for each node two comparisons and total four comparisons (4)
Level 2 has four nodes and total comparisons are 4x3 =12 comparisons
Level3 has total 8 nodes but in the given question we need to process only 11 items and already we covered 7 nodes so remaining nodes 4 nodes at level 3. Total comparisons at level3 are 4*4=16
So, total number of comparisons at each level are = 1+4+12+16= 33
Average comparisons required for 11 items = 33/11 = 3
Each level has 2i nodes where i is level number
Level 0 has one node so one comparison(1)
Level 1 has two nodes so for each node two comparisons and total four comparisons (4)
Level 2 has four nodes and total comparisons are 4x3 =12 comparisons
Level3 has total 8 nodes but in the given question we need to process only 11 items and already we covered 7 nodes so remaining nodes 4 nodes at level 3. Total comparisons at level3 are 4*4=16
So, total number of comparisons at each level are = 1+4+12+16= 33
Average comparisons required for 11 items = 33/11 = 3
Question 27 
Consider the following Java code fragment:
1 public class While
2 {
3 public void loop()
4 {
5 int x = 0;
6 while(1)
7 {
8 System.out.println("x plus one is" +(x+1));
9 }
10 }
11 }
1 public class While
2 {
3 public void loop()
4 {
5 int x = 0;
6 while(1)
7 {
8 System.out.println("x plus one is" +(x+1));
9 }
10 }
11 }
There is syntax error in line no. 1  
There is syntax errors in line nos. 1 & 6  
There is syntax error in line no. 8  
There is syntax error in line no. 6 
Question 27 Explanation:
Java does not  unlike C  interpret 0 as false and 1 as true.
So we cannot use integers in the while() statement.
So line number 6 will give syntax error.
So we cannot use integers in the while() statement.
So line number 6 will give syntax error.
Question 28 
An IP packet has arrived in which the fragment offset value is 100, the value of HLEN is 5 and the value of total length field is 200. What is the number of last byte in packet.
194  
394  
979  
1179 
Question 28 Explanation:
→ The first byte number is 100×8 =800.
→ The total length is 200 bytes and the header length is 20 bytes (5×4), which means that there are 180 bytes in this datagram.
→ If the first byte number is 800, the last byte number must 979 (800+1801).
→ The total length is 200 bytes and the header length is 20 bytes (5×4), which means that there are 180 bytes in this datagram.
→ If the first byte number is 800, the last byte number must 979 (800+1801).
Question 29 
Write the output of the following C program
#include <stdio.h>
int main (void)
{
int shifty;
shifty = 0570;
shifty = shifty >>4;
shifty = shifty <<6;
printf("the value of shifty is %o",shifty);
}
#include <stdio.h>
int main (void)
{
int shifty;
shifty = 0570;
shifty = shifty >>4;
shifty = shifty <<6;
printf("the value of shifty is %o",shifty);
}
the value of shifty is 1500  
the value of shifty is 4300  
the value of shifty is 5700  
the value of shifty is 2700 
Question 29 Explanation:
Given, shifty = 0570; // number starts with 0 means that number is octal number.
(0570)8 = (000 101 111 000)_{2} (Converting octal number into binary number)
shifty = shifty >>4 ( >> right shift operator where we need to shift the 4 bits towards right side means discard last four bits )
After right shifting 4 bits the shifty consists of the following digits
shifty = (000 000 010 111) _{2}
shifty = shifty <<6( << left shift operator where we need to shift the 6 bits towards left side means add si bits to the end of the binary number. So the binary number becomes as follows shifty = (010 111 000 000) _{2}
= (2700) _{8}
(0570)8 = (000 101 111 000)_{2} (Converting octal number into binary number)
shifty = shifty >>4 ( >> right shift operator where we need to shift the 4 bits towards right side means discard last four bits )
After right shifting 4 bits the shifty consists of the following digits
shifty = (000 000 010 111) _{2}
shifty = shifty <<6( << left shift operator where we need to shift the 6 bits towards left side means add si bits to the end of the binary number. So the binary number becomes as follows shifty = (010 111 000 000) _{2}
= (2700) _{8}
Question 30 
The following Finite Automaton recognizes which of the given languages?
{1, 0}* {01}  
{1, 0}* {1}  
{1}{1, 0}* {1}  
1*0* {0, 1} 
Question 30 Explanation:
Given DFA accepts all string that ending with “01”, so the language should be (0+1)*01.
Question 31 
How much memory is required to implement the zbuffer algorithm for a 512 x 512 x 24 bitplane image?
768 KB  
1 MB  
1.5 MB  
2 MB 
Question 31 Explanation:
In computer graphics, zbuffering, also known as depth buffering, is the management of image depth coordinates in 3D graphics, usually done in hardware, sometimes in software
In a 3drendering engine, when an object is projected on the screen, the depth (zvalue) of a generated pixel in the projected screen image is stored in a buffer (the zbuffer or depth buffer).
A zvalue is the measure of the perpendicular distance from a pixel on the projection plane to its corresponding 3dcoordinate on a polygon in worldspace.
Zbuffer requires 2 type of buffers to be filled: Depth buffer and Frame buffer
The amount of memory required by depth buffer in terms of bits is 512 x 512 x 24 = 6291456
The amount of memory required by frame buffer in terms of bits is 512 x 512 x 24 = 6291456
Total memory is required is sum of both depth and frame buffer memories = 6291456 + 6291456 = 12582912 bits which is equivalent to 1.5 MB(1.5x1024x1024x8)
In a 3drendering engine, when an object is projected on the screen, the depth (zvalue) of a generated pixel in the projected screen image is stored in a buffer (the zbuffer or depth buffer).
A zvalue is the measure of the perpendicular distance from a pixel on the projection plane to its corresponding 3dcoordinate on a polygon in worldspace.
Zbuffer requires 2 type of buffers to be filled: Depth buffer and Frame buffer
The amount of memory required by depth buffer in terms of bits is 512 x 512 x 24 = 6291456
The amount of memory required by frame buffer in terms of bits is 512 x 512 x 24 = 6291456
Total memory is required is sum of both depth and frame buffer memories = 6291456 + 6291456 = 12582912 bits which is equivalent to 1.5 MB(1.5x1024x1024x8)
Question 32 
Consider a standard Circular Queue ‘q’ implementation (which has the same condition for Queue Full and Queue Empty) whose size is 11 and the elements of the queue are q[0], q[1], q[2]…..,q[10].
The front and rear pointers are initialized to point at q[2] . In which position will the ninth element be added?
q[0]  
q[1]  
q[9]  
q[10] 
Question 32 Explanation:
A circular queue is a data structure that uses a single, fixedsize buffer as if it were connected endtoend.
The front and rear pointers are initialized to point at q[2] which means third element.
First element will add at q[3] , second element will add at q[4] and so on eight element will add at q[10].
Q[10] is the end of the queue which is connected to q[0]
So ninth element can be added at q[0] pointer
The front and rear pointers are initialized to point at q[2] which means third element.
First element will add at q[3] , second element will add at q[4] and so on eight element will add at q[10].
Q[10] is the end of the queue which is connected to q[0]
So ninth element can be added at q[0] pointer
Question 33 
If only one memory location is to be reserved for a class variable, no matter how many objects are instantiated, then the variable should be declared as
extern  
static  
volatile  
const 
Question 33 Explanation:
→ The static storage class instructs the compiler to keep a local variable in existence during the lifetime of the program instead of creating and destroying it each time it comes into and goes out of scope. Therefore, making local variables static allows them to maintain their values between function calls.
→ The static modifier may also be applied to global variables. When this is done, it causes that variable's scope to be restricted to the file in which it is declared.
→ In C programming, when static is used on a global variable, it causes only one copy of that member to be shared by all the objects of its class.
→ The static modifier may also be applied to global variables. When this is done, it causes that variable's scope to be restricted to the file in which it is declared.
→ In C programming, when static is used on a global variable, it causes only one copy of that member to be shared by all the objects of its class.
Question 34 
Consider the following binary search tree T given below:
Which node contains the fourth smallest element in T?
Q  
V  
W  
X 
Question 34 Explanation:
Inorder traversal of binary search tree gives the ascending order of the elements.
The inorder traversal of the above tree is UQXWPVZY
So the fourth smallest element is 4th element of the inorder which is W
The inorder traversal of the above tree is UQXWPVZY
So the fourth smallest element is 4th element of the inorder which is W
Question 35 
Let x, y, z, a, b, c be the attributes of an entity set E. If {x}, {x,y}, {a,b}, {a,b,c}, {x,y,z} are superkeys then which of the following are the candidate keys?
{x,y} and {a,b}  
{x} and {a,b}  
{x,y,z} and {a,b,c}  
{z} and {c} 
Question 35 Explanation:
● A candidate key is simply the "shortest" super key. Candidate Key are individual columns in a table that qualifies for uniqueness of each row/tuple.Every table must have at least one candidate key but at the same time can have several.
Question 36 
The five items: A, B, C, D, and E are pushed in a stack, one after other starting from A. The stack is popped four items and each element is inserted in a queue. The two elements are deleted from the queue and pushed back on the stack. Now one item is popped from the stack. The popped item is
A  
B  
C  
D 
Question 36 Explanation:
Stack representation after inserting 5 elements
Question 37 
A computer has 16 pages of virtual address space but the size of main memory is only four frames. Initially the memory is empty. A program references the virtual pages in the order 0, 2, 4, 5, 2, 4, 3, 11, 2, 10. How many page faults occur if LRU page replacement algorithm is used?
3  
5  
7  
8 
Question 37 Explanation:
LRU Page Replacement Algorithm:
when a page fault occurs, throw out the page that has been unused for the longest time.
when a page fault occurs, throw out the page that has been unused for the longest time.
Question 38 
Consider a 50 kbps satellite channel with a 500 milliseconds round trip propagation delay. If the sender wants to transmit 1000 bit frames, how much time will it take for the receiver to receive the frame?
250 milliseconds  
20 milliseconds  
520 milliseconds  
270 milliseconds 
Question 38 Explanation:
The propagation delay is the time it takes for the first bit to travel from the sender to the receiver (During this time the receiver is unaware that a message is being transmitted). The propagation speed depends on the physical medium of the link (that is, fiber optics, twistedpair copper wire, etc.),
The roundtrip time or ping time is the time from the start of the transmission from the sending node until a response is received at the same node. It is affected by packet delivery time as well as the data processing delay, which depends on the load on the responding node. If the sent data packet as well as the response packet have the same length, the round trip time can be expressed as:
Round trip time = 2 × Packet delivery time + processing delay
In case of only one physical link, the above expression corresponds to:
Link round trip time = 2 × packet transmission time + 2 × propagation delay + processing delay
Consider there is no effect of propagation delay and processing delay
round trip time ≈ 2 × packet transmission time
packet transmission time=round trip time/2=500/2= 250 ms
Transmission time= Message (bits) / band width (chanel capacity) = 1000 bits/50kbps =20 ms
Total time to receiver to receive the frame =250+20=270ms
The roundtrip time or ping time is the time from the start of the transmission from the sending node until a response is received at the same node. It is affected by packet delivery time as well as the data processing delay, which depends on the load on the responding node. If the sent data packet as well as the response packet have the same length, the round trip time can be expressed as:
Round trip time = 2 × Packet delivery time + processing delay
In case of only one physical link, the above expression corresponds to:
Link round trip time = 2 × packet transmission time + 2 × propagation delay + processing delay
Consider there is no effect of propagation delay and processing delay
round trip time ≈ 2 × packet transmission time
packet transmission time=round trip time/2=500/2= 250 ms
Transmission time= Message (bits) / band width (chanel capacity) = 1000 bits/50kbps =20 ms
Total time to receiver to receive the frame =250+20=270ms
Question 39 
The following three ‘C’ language statements is equivalent to which single statement?
y=y+1;
z=x+y;
x=x+1
y=y+1;
z=x+y;
x=x+1
z = x + y + 2;  
z = (x++) + (++y);  
z = (x++) + (y++);  
z = (x++) + (++y) + 1; 
Question 39 Explanation:
From the below statements
y=y+1;
z=x+y;
x=x+1
First statement: “y” value is incremented by 1
Second statement : that incremented value is added to x and stored into “z”
Third statement : “y” value is incremented by 1
z = (x++) + (++y);
X++ postincrement , so it will perform action which is addition and later the value of “x” is incremented .
++y preincrement , here first value “y” is incremented and updated value is added to value x.
Finally the result will store into z.
The sequence of operations you can find from the below statements
z=x+y; //z = x + (++y)
x=x+1 //z = (x++) + (++y)
y=y+1;
z=x+y;
x=x+1
First statement: “y” value is incremented by 1
Second statement : that incremented value is added to x and stored into “z”
Third statement : “y” value is incremented by 1
z = (x++) + (++y);
X++ postincrement , so it will perform action which is addition and later the value of “x” is incremented .
++y preincrement , here first value “y” is incremented and updated value is added to value x.
Finally the result will store into z.
The sequence of operations you can find from the below statements
z=x+y; //z = x + (++y)
x=x+1 //z = (x++) + (++y)
Question 40 
Consider the logic circuit given below.
The inverter, AND and OR gates have delays of 6, 10 and 11 nanoseconds respectively. Assuming that wire delays are negligible, what is the duration of glitch for Q before it becomes stable?
The inverter, AND and OR gates have delays of 6, 10 and 11 nanoseconds respectively. Assuming that wire delays are negligible, what is the duration of glitch for Q before it becomes stable?
5  
11  
16  
17 
Question 40 Explanation:
In this circuit, inverter and the AND gate will take total 16 nsecs (6 nsecs will be taken by inverter and another 10 nsecs by AND gate) to reach the XOR gate whereas the OR gate will reach to the XOR gate in only 11 nsecs, which will cause a glitch to happen for 5 nsecs.
Question 41 
The conic section that is obtained when a right circular cone is cut through a plane that is parallel to the side of the cone is called
parabola  
hyperbola  
circle  
ellipse 
Question 41 Explanation:
Question 42 
An IP packet has arrived with the first 8 bits as 0100 0010. Which of the following is correct?
The number of hops this packet can travel is 2.  
The total number of bytes in header is 16 bytes  
The upper layer protocol is ICMP  
The receiver rejects the packet 
Question 42 Explanation:
So, for 0100 0010,
First 4 bits represent Version IPV4
And another 4 bits represent header length (/ 4) which should range between 20 to 60 bytes.
here 0010 represents header length , is equal to 2 * 4 = 8
So, receiver will reject the packet.
First 4 bits represent Version IPV4
And another 4 bits represent header length (/ 4) which should range between 20 to 60 bytes.
here 0010 represents header length , is equal to 2 * 4 = 8
So, receiver will reject the packet.
Question 43 
Which of the following is not valid Boolean algebra rule?
X.X = X  
(X + Y).X = X  
X̄ + XY = Y  
(X + Y).(X + Z) = X + YZ 
Question 43 Explanation:
i) X.X = X
ii) (X + Y).X
= X.X + X.Y
= X + X.Y
= X(1 + Y)
= X
iii) X' + XY
= (X' + X)(X' + Y)
= (1)(X' + Y)
= (X' + Y)
iv) (X + Y).(X + Z)
= X.X + X.Z + X.Y + Y.Z
= X(1 + Z + Y) + Y.Z
= X + Y.Z
ii) (X + Y).X
= X.X + X.Y
= X + X.Y
= X(1 + Y)
= X
iii) X' + XY
= (X' + X)(X' + Y)
= (1)(X' + Y)
= (X' + Y)
iv) (X + Y).(X + Z)
= X.X + X.Z + X.Y + Y.Z
= X(1 + Z + Y) + Y.Z
= X + Y.Z
Question 44 
Assume the following information:
Original timestamp value = 46
Receive timestamp value = 59
Transmit timestamp value = 60
Timestamp at the arrival of packet = 69
Which of the following statements is correct?
Receive clock should go back by 3 milliseconds  
Transmit and Receive clocks are synchronized  
Transmit clock should go back by 3 milliseconds  
Receive clock should go ahead by 1 milliseconds 
Question 44 Explanation:
From the given data(original,receive,transmit and packet arrival time) details we can calculate the below details
Sending time =Receive time  stating time= 59 − 46 = 13 milliseconds
Receiving time = Packet arrival time Packet receive time=67 − 60 = 7 milliseconds
Roundtrip time =Sending time+Receiving time= 13 + 7 = 20 milliseconds
Time difference = receive timestamp − ( original timestamp field + oneway time duration)
Time difference = 59 − (46 + 10) = 3
Sending time =Receive time  stating time= 59 − 46 = 13 milliseconds
Receiving time = Packet arrival time Packet receive time=67 − 60 = 7 milliseconds
Roundtrip time =Sending time+Receiving time= 13 + 7 = 20 milliseconds
Time difference = receive timestamp − ( original timestamp field + oneway time duration)
Time difference = 59 − (46 + 10) = 3
Question 45 
Suppose you are browsing the world wide web using a web browser and trying to access the web servers. What is the underlying protocol and port number that are being used?
UDP, 80  
TCP, 80  
TCP, 25  
UDP, 25 
Question 45 Explanation:
The Hypertext Transport Protocol (HTTP) is an application layer protocol that is used to transmit virtually all files and other data on the World Wide Web, whether they're HTML files, image files, query results, or anything else. Usually, HTTP takes place through TCP/IP sockets.
A browser is an HTTP client because it sends requests to an HTTP server (Web server), which then sends responses back to the client. The standard (and default) port for HTTP servers to listen on is 80, though they can use any port.
HTTP is based on the TCP/IP protocols, and is used commonly on the Internet for transmitting webpages from servers to browsers.
A browser is an HTTP client because it sends requests to an HTTP server (Web server), which then sends responses back to the client. The standard (and default) port for HTTP servers to listen on is 80, though they can use any port.
HTTP is based on the TCP/IP protocols, and is used commonly on the Internet for transmitting webpages from servers to browsers.
Question 46 
A mechanism or technology used in Ethernet by which two connected devices choose common transmission parameters such as speed, duplex mode and flow control is called
Autosense  
Synchronization  
Pinging  
Auto negotiation 
Question 46 Explanation:
1. Ping is a computer network administration software utility used to test the reachability of a host on an Internet Protocol (IP) network. It measures the roundtrip time for messages sent from the originating host to a destination computer that are echoed back to the source
2. Autosense refers to a feature found in network adapters that allows them to automatically recognize the current local network's speed and adjust its own setting accordingly. It is often used with Ethernet, fast Ethernet, switches, hubs and network interface cards
3. Process synchronization refers to the idea that multiple processes are to join up or handshake at a certain point, in order to reach an agreement or commit to a certain sequence of action. Data synchronization refers to the idea of keeping multiple copies of a dataset in coherence with one another, or to maintain data integrity. Process synchronization primitives are commonly used to implement data synchronization.
4. Autonegotiation is a signaling mechanism and procedure used by Ethernet over twisted pair by which two connected devices choose common transmission parameters, such as speed, duplex mode, and flow control. Auto negotiation is defined in clause 28 of IEEE 802.3. and was originally an optional component in the Fast Ethernet standard.
2. Autosense refers to a feature found in network adapters that allows them to automatically recognize the current local network's speed and adjust its own setting accordingly. It is often used with Ethernet, fast Ethernet, switches, hubs and network interface cards
3. Process synchronization refers to the idea that multiple processes are to join up or handshake at a certain point, in order to reach an agreement or commit to a certain sequence of action. Data synchronization refers to the idea of keeping multiple copies of a dataset in coherence with one another, or to maintain data integrity. Process synchronization primitives are commonly used to implement data synchronization.
4. Autonegotiation is a signaling mechanism and procedure used by Ethernet over twisted pair by which two connected devices choose common transmission parameters, such as speed, duplex mode, and flow control. Auto negotiation is defined in clause 28 of IEEE 802.3. and was originally an optional component in the Fast Ethernet standard.
Question 47 
Consider the following sorting algorithms.
 Quicksort
 Heapsort
 Mergesort
1 and 2 only  
2 and 3 only  
3 only  
1, 1 and 3 
Question 47 Explanation:
Question 48 
Consider the following table
The table is in which normal form?
The table is in which normal form?
First Normal Form  
Second Normal Form  
Third Normal Form but not BCNF  
Third Normal Form but BCNF 
Question 48 Explanation:
From the given figure, we can write functional dependencies as
AB→ CDE
C→ B
The candidate keys are "AB" and "AC".
The relation is not in BCNF because the functional dependencies C→ B do not contains the key AB or AC
So, the given relation is in 3NF but not in BCNF
AB→ CDE
C→ B
The candidate keys are "AB" and "AC".
The relation is not in BCNF because the functional dependencies C→ B do not contains the key AB or AC
So, the given relation is in 3NF but not in BCNF
Question 49 
Consider a 13 element hash table for which f(key)=key mod 13 is used with integer keys. Assuming linear probing is used for collision resolution, at which location would the key 103 be inserted, if the keys 661, 182, 24 and 103 are inserted in that order?
0  
1  
11  
12 
Question 49 Explanation:
The hash table size is 13 so the indexes are 0 to 12.So that the elements will store in to any of the 13 locations
661 mod 13 = 11
182 mod 13 = 0
24 mod 13 = 11, already filled, so after linear probing it will get index 12
103 mod 13 = 12, already filled, so after linear probing it will get index 1
661 mod 13 = 11
182 mod 13 = 0
24 mod 13 = 11, already filled, so after linear probing it will get index 12
103 mod 13 = 12, already filled, so after linear probing it will get index 1
Question 50 
Consider a system where each file is associated with a 16bit number. For each file, each user should have the read and write capability. How much memory is needed to store each user’s access data?
16 KB  
32 KB  
64 KB  
128 KB 
Question 50 Explanation:
The user has read and write capability to the file and there are four possible combinations to file read, write,noread,nowrite operations.
We can represent that four options we require 2 bits(2^{2}=4).
So total memory(in bits) to require to access the data is 2^{16}x2=2^{17}bits
Convert the above into bytes 2^{4}2^{10}2^{3}=16KB(2^{10}=1 KB and 2^{3}=8=1Byte)
We can represent that four options we require 2 bits(2^{2}=4).
So total memory(in bits) to require to access the data is 2^{16}x2=2^{17}bits
Convert the above into bytes 2^{4}2^{10}2^{3}=16KB(2^{10}=1 KB and 2^{3}=8=1Byte)
Question 51 
What is the time complexity for the following C module? Assume that n>0 .
int module(int n)
{
if (n == 1)
return 1;
else
return (n + module(n1));
}
int module(int n)
{
if (n == 1)
return 1;
else
return (n + module(n1));
}
O(n)  
O(log n)  
O(n^{2})  
O(n!) 
Question 51 Explanation:
Question 52 
What are the minimum number of resources required to ensure that deadlock will never occur, if there are currently three processes P1, P2 and P3 running in a system whose maximum demand for the resources of the same type are 3, 4, and 5 respectively?
3  
7  
9  
10 
Question 52 Explanation:
let the resources needed by P1 is R1 =3,
the number of resources needed by P2 is R2 = 4 and so on.
. Minimum resources required to ensure that deadlock will never occur = (R11) + (R21) + (R31) + 1
= (31) + (41)+ (51) + 1 = 10.
the number of resources needed by P2 is R2 = 4 and so on.
. Minimum resources required to ensure that deadlock will never occur = (R11) + (R21) + (R31) + 1
= (31) + (41)+ (51) + 1 = 10.
Question 53 
For a software project, the spiral model was employed. When will the spiral stop?
When the software product is retired  
When the software product is released after Beta testing  
When the risk analysis is completed  
After completing five loops 
Question 53 Explanation:
→ The spiral model is a riskdriven software development process model.
→ Based on the unique risk patterns of a given project, the spiral model guides a team to adopt elements of one or more process models, such as incremental, waterfall, or evolutionary prototyping.
→ Spiral model stop, when the software product is retired
→ Based on the unique risk patterns of a given project, the spiral model guides a team to adopt elements of one or more process models, such as incremental, waterfall, or evolutionary prototyping.
→ Spiral model stop, when the software product is retired
Question 54 
Dirty bit is used to indicate which of the following?
A page fault has occurred  
A page has corrupted data  
A page has been modified after being loaded into cache  
An illegal access of page 
Question 54 Explanation:
→ The dirty bit allows for a performance optimization i.e., Dirty bit for a page in a page table helps to avoid unnecessary writes on a paging device
→ When a page is modified inside the cache and the changes need to be stored back in the main memory, the valid bit is set to 1 so as to maintain the record of modified pages.
→ When a page is modified inside the cache and the changes need to be stored back in the main memory, the valid bit is set to 1 so as to maintain the record of modified pages.
Question 55 
Which of the following is not a valid multicast MAC address?
01:00:5E:00:00:00  
01:00:5E:00:00:FF  
01:00:5E:00:FF:FF  
01:00:5E:FF:FF:FF 
Question 55 Explanation:
A multicast addressed frame is either flooded out all ports (if no multicast optimization is configured) or sent out only the ports interested in receiving the traffic.
The range of multicast MAC Address lie between 01005E000000 to 01005E7FFFFF.
Question 56 
The rank of a matrix A =
0  
1  
2  
3 
Question 56 Explanation:
Given matrix
Question 57 
How many different trees are there with four nodes A, B, C and D ?
30  
60  
90  
None of the above 
Question 57 Explanation:
For a given nodes”n” , we can get n^{n2} number of different trees
n=4. Here, n is number of nodes.
4^{42}=16.
Given options are wrong options.
Note: Correct answer is 16
n=4. Here, n is number of nodes.
4^{42}=16.
Given options are wrong options.
Note: Correct answer is 16
Question 58 
What is the median of data if its mode is 15 and the mean is 30?
30  
25  
22.5  
27.5 
Question 58 Explanation:
In a moderately symmetric distribution, the mean, median and mode are connected by the formula:
Mode = 3 Median – 2 Mean
15 = 3 Median  2(30)
Median = 25
Mode = 3 Median – 2 Mean
15 = 3 Median  2(30)
Median = 25
Question 59 
An organization is granted the block 130.34.12.64/26. It needs to have 4 subnets. Which of the following is not an address of this organization?
130.34.12.124  
130.34.12.89  
130.34.12.70  
130.34.12.132 
Question 59 Explanation:
→ The suffix length is 6. This means the total number of addresses in the block is 64 (2^{6}). If we create four subnets, each subnet will have 16 addresses.
→ So, addresses from 130.34.12.64 to 130.34.12.127 will be included in this organization.
→ So, addresses from 130.34.12.64 to 130.34.12.127 will be included in this organization.
Question 60 
A web client sends a request to a web server. The web server transmits a program to that client and is executed at the client. It creates a web document. What are such web documents called?
Active  
Static  
Dynamic  
Passive 
Question 60 Explanation:
→ Static
A static web document resides in a file that it is associated with a web server. The author of a static document determines the contents at the time the document is written. Because the contents do not change, each request for a static document results in exactly the same response.
→ Dynamic:
A dynamic web document does not exist in a predefined form. When a request arrives the web server runs an application program that creates the document. The server returns the output of the program as a response to the browser that requested the document. Because a fresh document is created for each request, the contents of a dynamic document can vary from one request to another.
→ Active:
An active web document consists of a computer program that the server sends to the browser and that the browser must run locally. When it runs, the active document program can interact with the user and change the display continuously.
A static web document resides in a file that it is associated with a web server. The author of a static document determines the contents at the time the document is written. Because the contents do not change, each request for a static document results in exactly the same response.
→ Dynamic:
A dynamic web document does not exist in a predefined form. When a request arrives the web server runs an application program that creates the document. The server returns the output of the program as a response to the browser that requested the document. Because a fresh document is created for each request, the contents of a dynamic document can vary from one request to another.
→ Active:
An active web document consists of a computer program that the server sends to the browser and that the browser must run locally. When it runs, the active document program can interact with the user and change the display continuously.
Question 61 
What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid and invalid bit) and a page size of 512
2^{11}  
2^{15}  
2^{19}  
2^{20} 
Question 61 Explanation:
Size of Physical Address = Paging bits + Offset bits
Paging bits = 11 – 1 = 10 (As 1 valid bit is also included)
Offset bits = log_{2 } (page size) =log_{2} (512) =9
Size of Physical Address = 10 + 9 = 19 bits
Paging bits = 11 – 1 = 10 (As 1 valid bit is also included)
Offset bits = log_{2 } (page size) =log_{2} (512) =9
Size of Physical Address = 10 + 9 = 19 bits
Question 62 
Which of the following is not an optimization criterion in the design of a CPU scheduling algorithm?
Minimum CPU utilization  
Maximum throughput  
Minimum turnaround time  
Minimum waiting time 
Question 62 Explanation:
Goal of an operating system is “maximum utilization of CPU”.
The following are criterion for designing CPU scheduling
● CPU utilization  Ideally the CPU would be busy 100% of the time, so as to waste 0 CPU cycles. On a real system CPU usage should range from 40% ( lightly loaded ) to 90% ( heavily loaded. )
● Throughput  Number of processes completed per unit time. May range from 10 / second to 1 / hour depending on the specific processes.
● Turnaround time  Time required for a particular process to complete, from submission time to completion. ( Wall clock time. )
● Waiting time  How much time processes spend in the ready queue waiting their turn to get on the CPU.
→ In general one wants to optimize the average value of a criteria ( Maximize CPU utilization and throughput, and minimize all the others.) However sometimes one wants to do something different, such as to minimize the maximum response time.
The following are criterion for designing CPU scheduling
● CPU utilization  Ideally the CPU would be busy 100% of the time, so as to waste 0 CPU cycles. On a real system CPU usage should range from 40% ( lightly loaded ) to 90% ( heavily loaded. )
● Throughput  Number of processes completed per unit time. May range from 10 / second to 1 / hour depending on the specific processes.
● Turnaround time  Time required for a particular process to complete, from submission time to completion. ( Wall clock time. )
● Waiting time  How much time processes spend in the ready queue waiting their turn to get on the CPU.
→ In general one wants to optimize the average value of a criteria ( Maximize CPU utilization and throughput, and minimize all the others.) However sometimes one wants to do something different, such as to minimize the maximum response time.
Question 63 
A computing architecture, which allows the user to use computers from multiple administrative domains to reach a common goal is called as
Grid Computing  
Neutral Networks  
Parallel Processing  
Cluster Computing 
Question 63 Explanation:
→ Grid computing combines computers from multiple administrative domains to reach a common goal, to solve a single task, and may then disappear just as quickly.
→ Grid computing is the use of widely distributed computer resources to reach a common goal.
→ The grid can be thought of as a distributed system with noninteractive workloads that involve a large number of files.
→ Grid computing is distinguished from conventional highperformance computing systems such as cluster computing in that grid computers have each node set to perform a different task/application.
→ Grid computers also tend to be more heterogeneous and geographically dispersed (thus not physically coupled) than cluster computers.
→ Although a single grid can be dedicated to a particular application, commonly a grid is used for a variety of purposes.
→ Grids are often constructed with generalpurpose grid middleware software libraries. Grid sizes can be quite large.
→ Grid computing is the use of widely distributed computer resources to reach a common goal.
→ The grid can be thought of as a distributed system with noninteractive workloads that involve a large number of files.
→ Grid computing is distinguished from conventional highperformance computing systems such as cluster computing in that grid computers have each node set to perform a different task/application.
→ Grid computers also tend to be more heterogeneous and geographically dispersed (thus not physically coupled) than cluster computers.
→ Although a single grid can be dedicated to a particular application, commonly a grid is used for a variety of purposes.
→ Grids are often constructed with generalpurpose grid middleware software libraries. Grid sizes can be quite large.
Question 64 
Which of the following is NOT represented in a subroutine activation record frame for a stackbased programming language?
Values of local variables  
Return address  
Heap area  
Information needed to access non local variables 
Question 64 Explanation:
Stack is used for static memory allocation and Heap for dynamic memory allocation, both stored in the computer's RAM .
Variables allocated on the stack are stored directly to the memory and access to this memory is very fast, and it's allocation is dealt with when the program is compiled
Variables allocated on the heap have their memory allocated at run time and accessing this memory is a bit slower, but the heap size is only limited by the size of virtual memory . Element of the heap have no dependencies with each other and can always be accessed randomly at any time
Variables allocated on the stack are stored directly to the memory and access to this memory is very fast, and it's allocation is dealt with when the program is compiled
Variables allocated on the heap have their memory allocated at run time and accessing this memory is a bit slower, but the heap size is only limited by the size of virtual memory . Element of the heap have no dependencies with each other and can always be accessed randomly at any time
Question 65 
Consider a single linked list where F and L are pointers to the first and last elements respectively of the linked list. The time for performing which of the given operations depends on the length of the linked list?
Delete the first element of the list  
Interchange the first two elements of the list  
Delete the last element of the list  
Add an element at the end of the list 
Question 65 Explanation:
1. Two pointers are pointing to first and last node in the linked list.
2. In order to delete first element , change first pointer to the next element.It won’t require length of the linked list.
3. To interchange first two elements also, We need to work with only first two nodes only.Here also no need of length of linked list.
4. To add an element at the last node, we already has one pointer which is pointing to the last node, simple add new node to last node by storing last pointer next address to new node.
5. But in order to delete last node , we need to traverse the entire list , So it requires length of the linked list. By using the last node pointer , we can’t move to previous node in the single linked list.
2. In order to delete first element , change first pointer to the next element.It won’t require length of the linked list.
3. To interchange first two elements also, We need to work with only first two nodes only.Here also no need of length of linked list.
4. To add an element at the last node, we already has one pointer which is pointing to the last node, simple add new node to last node by storing last pointer next address to new node.
5. But in order to delete last node , we need to traverse the entire list , So it requires length of the linked list. By using the last node pointer , we can’t move to previous node in the single linked list.
Question 66 
Let A be a finite set having x elements and let B be a finite set having y elements. What is the number of distinct functions mapping B into A.
x^{y}  
2^{ (x+y) }  
y^{x}  
y! / (yx)! 
Question 66 Explanation:
A function on a set involves running the function on every element of the set A, each one producing some result in the set B. So, for
the first run, every element of A gets mapped to an element in B. Take this
example, mapping a 2 element set A, to a 3 element set B. There are 9 different
ways, all beginning with both 1 and 2, that result in some different combination of
mappings over to B.
Question 67 
Using the page table shown below, translate the physical address 25 to virtual address. The address length is 16 bits and page size is 2048 words while the size of the physical memory is four frames.


25  
6169  
2073  
4121 
Question 67 Explanation:
Given data,
Virtual address length =16 bits,
Page size=2048 words
= 211 bytes
Step1: Total number of pages = 216/211
= 25
Step2: The physical address is nothing but [number of frames * size of each frame] Physical address= 4*211
= 213
Step3: Given physical address (25)10 = (0000000011001)2 in 13 bits
The 13 bits address, we are representing into
Virtual address length =16 bits,
Page size=2048 words
= 211 bytes
Step1: Total number of pages = 216/211
= 25
Step2: The physical address is nothing but [number of frames * size of each frame] Physical address= 4*211
= 213
Step3: Given physical address (25)10 = (0000000011001)2 in 13 bits
The 13 bits address, we are representing into
Question 68 
The probability that two friends are born in the same month is?
1/6  
1/12  
1/144  
1/24 
Question 68 Explanation:
Probability of a person to be born in a one month out of 12 months is 1/12
Probability of both friends born in a same month is (1/12) * (1/12)
Suppose, they born in january = 1/12 * 1/12
Suppose, they born in february = 1/12 * 1/12
; ; Suppose, they born in December = 1/12 * 1/12
Probability that two friends are born in the same month is= 12*(1/12)*(1/12)
= 1/12
Probability of both friends born in a same month is (1/12) * (1/12)
Suppose, they born in january = 1/12 * 1/12
Suppose, they born in february = 1/12 * 1/12
; ; Suppose, they born in December = 1/12 * 1/12
Probability that two friends are born in the same month is= 12*(1/12)*(1/12)
= 1/12
Question 69 
How many lines of output does the following C code produce?
#include<stdio.h>
float i=2.0;
float j=1.0;
float sum = 0.0;
main()
{
while (i/j > 0.001)
{
j+=j;
sum=sum+(i/j);
printf("%f\n", sum);
}
}
#include<stdio.h>
float i=2.0;
float j=1.0;
float sum = 0.0;
main()
{
while (i/j > 0.001)
{
j+=j;
sum=sum+(i/j);
printf("%f\n", sum);
}
}
8  
9  
10  
11 
Question 69 Explanation:
Iteration1:
while (1.000000 > 0.001)
{
j=2.0
sum=0+1.000000;
printf("%f\n",sum); /* It will print 1.000000 */
}
Iteration2: 1.500000
Iteration3: 1.750000
Iteration4: 1.875000
Iteration5: 1.937500
Iteration6: 1.968750
Iteration7: 1.984375
Iteration8: 1.992188
Iteration9: 1.996094
Iteration10: 1.998047
Iteration11: 1.999023
The program will terminate after 11th iteration. So, it print 11 lines.
while (1.000000 > 0.001)
{
j=2.0
sum=0+1.000000;
printf("%f\n",sum); /* It will print 1.000000 */
}
Iteration2: 1.500000
Iteration3: 1.750000
Iteration4: 1.875000
Iteration5: 1.937500
Iteration6: 1.968750
Iteration7: 1.984375
Iteration8: 1.992188
Iteration9: 1.996094
Iteration10: 1.998047
Iteration11: 1.999023
The program will terminate after 11th iteration. So, it print 11 lines.
Question 70 
If the maximum output voltage of a DAC is V volts and if the resolution is R bits then the weight of the most significant bit is
V/ (2^{R} 1)  
(2^{R1}).V/(2^{R} 1)  
(2^{R1}).V  
V/(2^{R1}) 
Question 70 Explanation:
→ If the maximum output voltage of a Digital to Analogue Converter is V volts and if the resolution is R bits then the weight of the most significant bit is (2^{R1}).V/(2^{R}1)
Question 71 
A frame buffer array is addressed in row major order for a monitor with pixel locations starting from (0,0) and ending with (100,100). What is address of the pixel(6,10)? Assume one bit storage per pixel and starting pixel location is at 0.
1016  
1006  
610  
616 
Question 71 Explanation:
Given data,
Pixel location starts from =(0,0)
Pixel location ends from =(100,100)
Address of the pixel= ?
Hint: Frame buffer array is addressed in row major order
. Step1: Row major order is= 0+1((60)+101(100))
= 0+6+1010
= 1016
Pixel location starts from =(0,0)
Pixel location ends from =(100,100)
Address of the pixel= ?
Hint: Frame buffer array is addressed in row major order
. Step1: Row major order is= 0+1((60)+101(100))
= 0+6+1010
= 1016
Question 72 
Consider the following Deterministic Finite Automaton.
Let denote the set of eight bit strings whose second, third, sixth and seventh bits are 1. The number of strings in that are accepted by M is
Let denote the set of eight bit strings whose second, third, sixth and seventh bits are 1. The number of strings in that are accepted by M is
0  
1  
2  
3 
Question 72 Explanation:
According to DFA transition diagram, the possible strings accepted by M is
01110110
01110111
Above two strings are are having constraint that 2nd, 3rd, 6th and 7th bits are 1. So, it satisfied above condition.
01110110
01110111
Above two strings are are having constraint that 2nd, 3rd, 6th and 7th bits are 1. So, it satisfied above condition.
Question 73 
Consider the schema R(A,B,C,D) and the functional dependencies A→ B and C→ D. If the decomposition is made as R1(A,B) and R2(C,D), then which of the following is TRUE?
Preserves dependency but cannot perform lossless join  
Preserves dependency and performs lossless join  
Does not perform dependency and cannot perform lossless join  
Does not preserve dependency but perform lossless join 
Question 73 Explanation:
Given Data,
Schema R(A, B, C, D)
Functional dependencies are A→ B and C→ D
Decomposed Schema is R1(A,B) and R2(C,D)
Dependency preservation decomposition:
It is another property of decomposed relational database schema D in which each functional dependency X → Y specified in F either appeared directly in one of the relation schemas Ri in the decomposed D or could be inferred from the dependencies that appear in some Ri.
Decomposition D={ R1 , R2, R3,,.., ,Rm} of R is said to be dependencypreserving with respect to F if the union of the projections of F on each Ri , in D is equivalent to F.
In other words, R ⊂ join of R1, R1 over X. The dependencies are preserved because each dependency in F represents a constraint on the database. If decomposition is not dependencypreserving, some dependency is lost in the decomposition.
→ R1(A,B) is covered A→ B
→ R2(C,D) is covered C→ D
It is Functional Dependency preserving because both the functional dependencies are covered.
Lossless join:
The decomposition is a losslessjoin decomposition of R if at least one of the following functional dependencies are in F+ (where F+ stands for the closure for every attribute or attribute sets in F):
R1 ∩ R2 → R1
R1 ∩ R2 → R2
According to functional dependency R1(A,B) ∩ R2(C,D) = null. There is no common key in both the tables. So, it is not lossless join.
Schema R(A, B, C, D)
Functional dependencies are A→ B and C→ D
Decomposed Schema is R1(A,B) and R2(C,D)
Dependency preservation decomposition:
It is another property of decomposed relational database schema D in which each functional dependency X → Y specified in F either appeared directly in one of the relation schemas Ri in the decomposed D or could be inferred from the dependencies that appear in some Ri.
Decomposition D={ R1 , R2, R3,,.., ,Rm} of R is said to be dependencypreserving with respect to F if the union of the projections of F on each Ri , in D is equivalent to F.
In other words, R ⊂ join of R1, R1 over X. The dependencies are preserved because each dependency in F represents a constraint on the database. If decomposition is not dependencypreserving, some dependency is lost in the decomposition.
→ R1(A,B) is covered A→ B
→ R2(C,D) is covered C→ D
It is Functional Dependency preserving because both the functional dependencies are covered.
Lossless join:
The decomposition is a losslessjoin decomposition of R if at least one of the following functional dependencies are in F+ (where F+ stands for the closure for every attribute or attribute sets in F):
R1 ∩ R2 → R1
R1 ∩ R2 → R2
According to functional dependency R1(A,B) ∩ R2(C,D) = null. There is no common key in both the tables. So, it is not lossless join.
Question 74 
Every time the attribute A appears, it is matched with the same value of attribute B but not the same value of attribute C. Which of the following is true?
A→ (B,C)  
A → B, A→→ C  
A→ B, C→→A  
A→→B, B→ C 
Question 74 Explanation:
→ represents functional dependency and
→→ is multivalued dependency.
Functional Dependency:
A → B means that the values of B are determined by the values of A. Two tuples sharing the same values of A will necessarily have the same values of B.
Multivalued dependency: It is a special case of a join dependency, with only two sets of values involved, i.e. it is a binary join dependency.
A multivalued dependency exists when there are at least three attributes (like A,B and C) in a relation and for a value of A there is a well defined set of values of B and a well defined set of values of C. However, the set of values of B is independent of set C and vice versa.
→→ is multivalued dependency.
Functional Dependency:
A → B means that the values of B are determined by the values of A. Two tuples sharing the same values of A will necessarily have the same values of B.
Multivalued dependency: It is a special case of a join dependency, with only two sets of values involved, i.e. it is a binary join dependency.
A multivalued dependency exists when there are at least three attributes (like A,B and C) in a relation and for a value of A there is a well defined set of values of B and a well defined set of values of C. However, the set of values of B is independent of set C and vice versa.
Question 75 
Consider the following grammar.
S → AB
A → a
A → BaB
B → bbA
Which of the following statements is FALSE?
S → AB
A → a
A → BaB
B → bbA
Which of the following statements is FALSE?
The length of every string produced by this grammar is even  
No string produced by this grammar has three consecutive a’s  
The length of substring produced by B is always odd  
No string produced by this grammar has four consecutive b’s 
Question 75 Explanation:
Question 76 
A cube of side 1 unit is placed in such a way that the origin coincides with one of its top vertices and the three axes along three of its edges. What are the coordinates of the vertex which is diagonally opposite to the vertex whose coordinates are (1,0,1)?
(0, 0, 0)  
(0, 1, 0)  
(0, 1, 0)  
(1, 1, 1) 
Question 76 Explanation:
Question 77 
How many different BCD numbers can be stored in 12 switches? (Assume two position or onoff switches)
2^{12}  
2^{12}1  
10^{12}  
10^{3} 
Question 77 Explanation:
Step1: A binarycoded decimal (BCD) is a class of binary encodings of decimal numbers where each decimal digit is represented by a fixed number of bits, usually four (or) eight.
Step2: Decimal number 0 can be represented 0000 and 9 can be represented by using 1001.
Step3: A switch can store maximum 1 bit data that may be either 0 (or) 1. In switch terminology, 0 means “off” and 1 means “on”. With 4 bit we can represent 10 BCD numbers.
Step4: A BCD digit can be from 0 to 9 (total 10 possibility).
Step5: Different possible BCD numbers in 12 switches are = 10*10*10
= 1000
= 10^{3}
Step2: Decimal number 0 can be represented 0000 and 9 can be represented by using 1001.
Step3: A switch can store maximum 1 bit data that may be either 0 (or) 1. In switch terminology, 0 means “off” and 1 means “on”. With 4 bit we can represent 10 BCD numbers.
Step4: A BCD digit can be from 0 to 9 (total 10 possibility).
Step5: Different possible BCD numbers in 12 switches are = 10*10*10
= 1000
= 10^{3}
Question 78 
Which of the following is FALSE with respect to possible outcomes of executing a Turing Machine over a given input?
it may halt and accept the input  
it may halt by changing the input  
it may halt and reject the input  
it may never halt 
Question 78 Explanation:
→ The halting problem is the problem of determining, from a description of an arbitrary computer program and an input, whether the program will finish running (i.e., halt) or continue to run forever.
→ Halting of Turing machine is an undecidable problem. A general algorithm to solve the halting problem for all possible programinput pairs cannot exist.
→ Possible outcomes are
1. It may halt and accept the input.
2. It may halt and reject the input.
3. It may never halt.
→ Halting of Turing machine is an undecidable problem. A general algorithm to solve the halting problem for all possible programinput pairs cannot exist.
→ Possible outcomes are
1. It may halt and accept the input.
2. It may halt and reject the input.
3. It may never halt.
Question 79 
A supernet has a first address of 205.16.32.0 and a supernet mask of 255.255.248.0. A router receives 4 packets with the following destination addresses. Which packet belongs to this supernet?
205.16.42.56  
205.17.32.76  
205.16.31.10  
205.16.39.44 
Question 79 Explanation:
Given data,
Supernet has a first address=205.16.32.0
Supernet Mask = 255.255.248.0
IP address=?
Step1: Perform AND operation between supernet mask and IP address. Let us take IP address is 205.16.39.44.
Step2: AND operation
Supernet has a first address=205.16.32.0
Supernet Mask = 255.255.248.0
IP address=?
Step1: Perform AND operation between supernet mask and IP address. Let us take IP address is 205.16.39.44.
Step2: AND operation
There are 79 questions to complete.