IP-Address

Question 1

Consider three machines M, N and P with IP addresses 100.10.5.2, 100.10.5.5, and 100.10.5.6 respectively. The subnet mask is set to 255.255.255.252 for all the three machines. Which one of the following is true?

A
M, N, and P all belong to the same subnet
B
Only M and N belong to the same subnet
C
M, N and P belong to three different subnets
D
Only N and P belong to the same subnet
       Computer-Networks       IP-Address       GATE 2019       Video-Explanation
Question 1 Explanation: 
Take each IP and do bitwise AND with the given Subnet Mask. If we get the same network ID for the given IP'S then it will belong to the same subnet.

Therefore, N and P belong to the same subnet.
Question 2

Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same net mask N. Which of the values of N given below should not be used if A and B should belong to the same network?

A
255.255.255.0
B
255.255.255.128
C
255.255.255.192
D
255.255.255.224
       Computer-Networks       IP-Address       GATE 2010
Question 2 Explanation: 
When we perform bitwise AND operation between IP Address and Subnet Mask, it gives Network ID. If for both IP results is same Network ID. It means, both IP are belong to the same network else it's on different network.
When we perform AND operation between IP address 10.105.1.113 and 255.255.255.224 result is 10.105.1.96 and when we perform AND operation between IP address 10.105.1.91 and 255.255.255.224 result is 10.105.1.64.
Therefore, 10.105.1.96 and 10.105.1.64 are different network, so D is correct answer.
Question 3

If a class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?

A
1022
B
1023
C
2046
D
2047
       Computer-Networks       IP-Address       GATE 2008
Question 3 Explanation: 
255.255.248.0 can be written as 11111111.11111111.11111000.00000000
Number of bits assigned for host id is the number of zeros in subnet mask. Here 11 bits are used.
for host id so maximum possible hosts are= 211 - 2 = 2046
Question 4

The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet?

A
62 subnets and 262142 hosts.
B
64 subnets and 262142 hosts.
C
62 subnets and 1022 hosts.
D
64 subnets and 1024 hosts.
       Computer-Networks       IP-Address       GATE 2007
Question 4 Explanation: 
It is a class B address, so there 16-bits for NID and 16-bits for HID.
From HID, we took 6-bits for subnetting.
Then total subnets possible = ( 26 ) - 2 = 64
Total hosts possible for each subnet = (210) - 2 = 1022
Question 5

Two computers C1 and C2 are configured as follows. C1 has IP address 203.197.2.53 and netmask 255.255.128.0. C2 has IP address 203.197.75.201 and netmask 255.255.192.0. Which one of the following statements is true?

A
C1 and C2 both assume they are on the same network
B
C2 assumes C1 is on same network, but C1 assumes C2 is on a different network
C
C1 assumes C2 is on same network, but C2 assumes C1 is on a different network
D
C1 and C2 both assume they are on different networks
       Computer-Networks       IP-Address       GATE 2006
Question 5 Explanation: 
From C1 side,
Subnet mask for C1 is 255.255.128.0.
So it finds the Network ID as,
C1 → 203.197.2.53 AND 255.255.128.0 = 203.197.0.0
C2 → 203.197.75.201 AND 255.255.128.0 = 203.197.0.0
Both same.
From C2 side,
Subnet mask for C2 is 255.255.192.0.
So it finds the network ID as,
C1 → 203.197.2.53 AND 255.255.192.0 = 203.197.0.0
C2 → 203.197.75.201 AND 255.255.192.0 = 203.197.64.0
Both different.
Hence, option 'C' is correct.
Question 6

An organization has a class B network and wishes to form subnets for 64 departments. The subnet mask would be:

A
255.255.0.0
B
255.255.64.0
C
255.255.128.0
D
255.255.252.0
       Computer-Networks       IP-Address       GATE 2005
Question 6 Explanation: 
Organization have 64 departments, and to assign 64 subnet we need 6 bits for subnet. In Class B network first two octet are reserved for NID, so we take first 6 bit of third octet for subnets and subnet mask would be 255.255.11111100.00000000 = 255.255.252.0
Question 7

The routing table of a router is shown below:

 Destination     Sub net mask 	     Interface
 128.75.43.0 	 255.255.255.0 	        Eth0
 128.75.43.0 	 255.255.255.128 	Eth1
 192.12.17.5 	 255.255.255.255 	Eth3
 Default 	  	                Eth2 

On which interfaces will the router forward packets addressed to destinations 128.75.43.16 and 192.12.17.10 respectively?

A
Eth1 and Eth2
B
Eth0 and Eth2
C
Eth0 and Eth3
D
Eth1 and Eth3
       Computer-Networks       IP-Address       GATE 2004
Question 7 Explanation: 
Router decides route for packet by ANDing subnet mask and IP address.
If results of ANDing subnet masks and IP address are same then subnet mask with higher number of 1s is preferred.
IP address 128.75.43.16 is AND with 255.255.255.0 results 128.75.43.0 Net ID which is similar to destination of this mask, but ANDing 128.75.43.16 with 255.255.255.128 also results same destination. So, here, mask with higher number of one is considered and router will forward packet to Eth1.
ANDing 192.12.17.10 with three subnet mask in table does not result in destination Net ID so router will forward this packet to default network via Eth2.
Question 8

The subnet mask for a particular network is 255.255.31.0. Which of the following pairs of IP addresses could belong to this network?

A
172.57.88.62 and 172.56.87.233
B
10.35.28.2 and 10.35.29.4
C
191.203.31.87 and 191.234.31.88
D
128.8.129.43 and 128.8.161.55
       Computer-Networks       IP-Address       GATE 2003
Question 8 Explanation: 
To find whether hosts belong to same network or not , we have to find their net id, if net id is same then hosts belong to same network and net id can be find by ANDing subnet mask and IP address.
128.8.129.43 (Bitwise AND) 255.255.31.0 = 128.8.1.0
128.8.161.55 (Bitwise AND) 255.255.31.0 = 128.8.1.0
Question 9

An organization requires a range of IP addresses to assign one to each of its 1500 computers. The organization has approached an Internet Service Provider (ISP) for this task. The ISP uses CIDR and serves the requests from the available IP address space 202.61.0.0/17. The ISP wants to assign an address space to the organization which will minimize the number of routing entries in the ISP’s router using route aggregation. Which of the following address spaces are potential candidates from which the ISP can allot any one to the organization?

I. 202.61.84.0/21
II. 202.61.104.0/21 
III. 202.61.64.0/21
IV. 202.61.144.0/21 
A
I and II only
B
III and IV only
C
II and III only
D
I and IV only
       Computer-Networks       IP-Address       GATE 2020
Question 9 Explanation: 
Given CIDR IP is 202.61.0.0/17 and for HID 32 - 17 = 15 bits can be used.
And to Assign an IP address for 1500 computer, we require 11 bit from HID part.
So NID + SID = 17 + 4 = 21 bits and HID = 11 bits
NID HID
202.61.0 0000 000.00000000
So, from the given option, possible IP Address is
I. 84 -> 0 1010 100 (Because in HID bit 1 is not possible)
II. 104 -> 0 1101 000
III. 64 -> 0 1000 000
IV. 144 -> 1 0010 000 (Because in NID bit 1 is not possible )
Question 10

Host X has IP address 192.168.1.97 and is connected through two routers R1 and R2 to an­other host Y with IP address 192.168.1.80. Router R1 has IP addresses 192.168.1.135 and 192.168.1.110. R2 has IP addresses 192.168.1.67 and 192.168.1.155. The netmask used in the network is 255.255.255.224. Given the information above, how many distinct subnets are guaranteed to already exist in the network?

A
1
B
2
C
3
D
6
       Computer-Networks       IP-Address       GATE 2008-IT
Question 10 Explanation: 
Simply, Bitwise AND the bits of fourth octet for each of the following IP address with fourth octet of subnet mask. You will get only
XX.XX.XX.96, XX.XX.XX.64 and XX.XX.XX.128.
Question 11

Host X has IP address 192.168.1.97 and is connected through two routers R1 and R2 to an­other host Y with IP address 192.168.1.80. Router R1 has IP addresses 192.168.1.135 and 192.168.1.110. R2 has IP addresses 192.168.1.67 and 192.168.1.155. The netmask used in the network is 255.255.255.224. Which IP address should X configure its gateway as?

A
192.168.1.67
B
192.168.1.110
C
192.168.1.135
D
192.168.1.155
       Computer-Networks       IP-Address       GATE 2008-IT
Question 11 Explanation: 
X must be able to reach the gateway using the net mask.
Subnet no. of host X is,

Now, the gateway must also have the same subnet number.
Let's take IP 192.168.1.110 of R1,

and hence this can be used by X.
Question 12

A router uses the following routing table:

A packet bearing a destination address 144.16.68.117 arrives at the router. On which interface will it be forwarded?

A
eth0
B
eth1
C
eth2
D
eth3
       Computer-Networks       IP-Address       GATE 2006-IT
Question 12 Explanation: 
Firstly start with longest mask.
144.16.68.117 = 144.16.68.01110101 AND 255.255.255.224 = 255.255.255.11100000
= 144.16.68.96(Not matching with destination)
Now, take 255.255.255.0
144.16.68.117 AND 255.255.255.0
= 144.16.68.0 (matched)
Hence, option (C) is correct.
Question 13

A subnetted Class B network has the following broadcast address: 144.16.95.255. Its subnet mask

A
is necessarily 255.255.224.0
B
is necessarily 255.255.240.0
C
is necessarily 255.255.248.0
D
could be any one of 255.255.224.0, 255.255.240.0, 255.255.248.0
       Computer-Networks       IP-Address       GATE 2006-IT
Question 13 Explanation: 
In the broadcast address for a subnet, all the host bits are set to 1. So as long as all the bits to the right are 1, bits left to it can be taken as possible subnet.
Broadcast address for subnet is
.95.255 or .01011111.11111111
(as in class B, 16 bits each are used for network and host)
So, we can take minimum 3 bits (from left) as subnet and make rest as host bits (as they are 1)
.224.0 → 11100000.00000000 (leftmost 3 bits for subnet)
.240.0 → 11110000.00000000 (leftmost 4 bits for subnet)
.248.0 → 11111000.00000000 (leftmost 5 bits for subnet)
Question 14
Which protocol suite designed by IETF to provide security for a packet at the Internet layer?
A
IPsec
B
NetSec
C
PacketSec
D
SSL
       Computer-Networks       IP-Address       ISRO-2017 May
Question 14 Explanation: 
→ The Internet Engineering Task Force(IETF) IP Security Working Group formed to standardize these efforts as an open, freely available set of security extensions called IPsec.
Question 15
Which network protocol allows hosts to dynamically get a unique IP number on each bootup
A
DHCP
B
BOOTP
C
RARP
D
ARP
       Computer-Networks       IP-Address       ISRO-2016
Question 15 Explanation: 
→ A router(or) a residential gateway can be enabled to act as a DHCP server.
→ Most residential network routers receive a globally unique IP address within the ISP network.
→ Within a local network, a DHCP server assigns a local IP address to each device connected to the network.
Question 16
The subnet mask for a particular network is 255.255.31.0. Which of the following pairs of IP addresses could belong to this network?
A
172.57.88.62 and 172.56.87.23
B
10.35.28.2 and 10.35.29.4
C
191.203.31.87 and 191.234.31.88
D
128.8.129.43 and 128.8.161.55
       Computer-Networks       IP-address       ISRO CS 2009
Question 16 Explanation: 
To find whether hosts belong to same network or not , we have to find their net id, if net id is same then hosts belong to same network and net id can be find by ANDing subnet mask and IP address.
128.8.129.43 (Bitwise AND) 255.255.31.0 = 128.8.1.0
128.8.161.55 (Bitwise AND) 255.255.31.0 = 128.8.1.0
Question 17
What is IP class and number of sub-networks if the subnet mask is 255.224.0.0?
A
Class A, 3
B
Class A, 8
C
Class B, 3
D
Class B, 32
       Computer-Networks       IP-address       ISRO CS 2013
Question 17 Explanation: 
The binary number equivalent to 255 is 1111 1111
The binary number equivalent to 224 is 1110 0000
We can write the subnet mask 255.224.0.0 in binary form as follows
11111111.11100000.00000000.00000000
In the binary representation, the first eight bits represent Class A network address and next three bits( ones) used for represent the number of subnets.
Then the total number of subnets are 23=8.
Question 18
The process of modifying IP address information in IP packet headers while in transit across a traffic routing device is called
A
Port address translation (PAT)
B
Network address translation (NAT)
C
Address mapping
D
Port mapping
       Computer-Networks       IP-Address       ISRO CS 2014
Question 18 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 Internet-routable 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 19
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?
A
130.34.12.124
B
130.34.12.89
C
130.34.12.70
D
130.34.12.132
       Computer-Networks       IP-Address       ISRO CS 2014
Question 19 Explanation: 
→ The suffix length is 6. This means the total number of addresses in the block is 64 (26). 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.
Question 20

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?

A
205.16.42.56
B
205.17.32.76
C
205.16.31.10
D
205.16.39.44
       Computer-Networks       IP-Address       ISRO CS 2014
Question 20 Explanation: 
Given data,
Supernet has a first address=205.16.32.0
Supernet Mask = 255.255.248.0
IP address=?
Step-1: Perform AND operation between supernet mask and IP address. Let us take IP address is 205.16.39.44.
Step-2: AND operation
Question 21

An internet service provider (ISP) has following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20 . The ISP want to give half of this chunk of addresses to organization A and a quarter to Organization B while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?

A
245.248.132.0/22 and 245.248.132.0/21
B
245.248.136.0/21 and 245.248.128.0/22
C
245.248.136.0/24 and 245.248.132.0/21
D
245.248.128.0/21 and 245.248.128.0/22
       Computer-Networks       IP-Address       UGC-NET CS 2018 DEC Paper-2
Question 21 Explanation: 
Question 22

The four byte IP Address consists of

A
Neither network nor Host Address
B
Network Address
C
Both Network and Host Address
D
Host Address
       Computer-Networks       IP-Address       UGC-NET CS 2018 DEC Paper-2
Question 22 Explanation: 
The IP address of 32 bit (4 byte) consists of both network and host address.
Question 23
The DNS maps the IP address to
A
A binary address as strings
B
An alphanumeric address
C
A hierarchy of domain names
D
A hexadecimal address
       Computer-Networks       IP-Address       ISRO CS 2015
Question 23 Explanation: 
→ DNS maps the IP addresses into a hierarchy of domain names.
→ DNS is a host name to IP address translation service.
→ DNS is a distributed database implemented in a hierarchy of name servers.
→ It is an application layer protocol for message exchange between clients and servers.
Question 24
In a class B subnet, we know the IP address of one host and the mask as given below:
IP address: 125.134.112.66
Mask: 255.255.224.0
What is the first address(Network address)?
A
125.134.96.0
B
125.134.112.0
C
125.134.112.66
D
125.134.0.0
       Computer-Networks       IP-Address       ISRO CS 2015
Question 24 Explanation: 
Given
IP address: 125.134.112.66
Subnet Mask: 255.255.224.0
IP address in binary Form = 01111101 10000110 01110000 01000010
Subnet Mask in binary Form = 11111111 11111111 11100000 00000000
Every IP address has Network id and Host id,
We will get Network id by doing bitwise -AND (&) operation with IP address and Subnet Mask

Question 25
A router uses the following routing table:

A packet bearing a destination address 144.16.68.117 arrives at the router. On which interface will it be forwarded?
A
eth0
B
eth1
C
eth2
D
eth3
       Computer-Networks       IP-Address       ISRO CS 2015
Question 25 Explanation: 
Question 26
In a classful addressing, first four bits in Class-A IP address is
A
1010
B
1100
C
1011
D
1110
E
None of the above
       \"Computer-Networks        IP-Address       UGC NET CS 2014 Dec-Paper-2
Question 26 Explanation: 
The class-A address starts with 0 to 127 but we never used 0 and 127 numbers. The numbers in binary format is 00000001 to 01111110. So, no option is starts with 0.
Class-A is 0
Class-B is 10
Class-C is 110
Class-D is 1110
Class-E is 1111
Question 27
​An internet service provider (ISP) has following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20 . The ISP want to give half of this chunk of addresses to organization A and a quarter to Organization B while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?
A
245.248.132.0/22 and 245.248.132.0/21
B
245.248.136.0/21 and 245.248.128.0/22
C
245.248.136.0/24 and 245.248.132.0/21
D
245.248.128.0/21 and 245.248.128.0/22
       Computer-Networks       IP-Address       UGC NET CS 2018-DEC Paper-2
Question 27 Explanation: 
Question 28
The four byte IP Address consists of
A
Neither network nor Host Address
B
Network Address
C
Both Network and Host Address
D
Host Address
       Computer-Networks       IP-Address       UGC NET CS 2018-DEC Paper-2
Question 28 Explanation: 
The IP address of 32 bit (4 byte) consists of both network and host address.
Question 29
What percentage(%) of the IPv4, IP address space do all class C addresses consume?
A
12.5%
B
25%
C
37.5%
D
50%
       Computer-Networks       IP-Address       UGC NET June-2019 CS Paper-2
Question 29 Explanation: 
Total possible IP numbers numbers of all classes 0 to 255.
Class-A: 0 to 127. It means 50%
Class-B: 128 to 191. It means 25%
Class-C: 192 to 223. It means 12.5 %
Class-D: 224 to 239. It means 6.25%
Class-E: 240 to 255. It means 6.25%
Question 30
What is the name of the protocol that allows a client to send a broadcast message with its MAC address and receive an IP address in reply?
A
ARP
B
DNS
C
RARP
D
ICMP
       Computer-Networks       IP-Address       UGC NET June-2019 CS Paper-2
Question 30 Explanation: 
ARP maps 32-bit logical (IP) address to 48-bit physical address.
RARP maps 48-bit physical address to 32-bit logical (IP) address.
Question 31
Consider the following two statements with respect to IPv4 in computer networking:
P: The loopback(IP) address is a member of class B network
Q: The loopback(IP) address is used to send a packet from host to itself
What can you say about the statements P and Q?
A
P-True: Q-False
B
P-False; Q-True
C
P-True; Q-True
D
P-False; Q-false
       Computer-Networks       IP-Address       UGC NET June-2019 CS Paper-2
Question 31 Explanation: 
Explanation: The loopback(IP) address is a member of class A network. Any address between the range 127 . 0. 0. 0 to 127.255.255.255 can be used a loopback address but we can’t use 127 . 0. 0. 0 and 127.255.255.255 as loopback addresses.
The loopback(IP) address is used to send a packet from host to itself
Question 32
You need 500 subnets, each with about 100 usable host addresses  per subnet. What network mass will you assign using a class B network address?
A
255.255.255.252
B
255.255.255.128
C
255.255.255.0
D
255.255.254.0
       Computer-Networks       IP-Address       UGC NET June-2019 CS Paper-2
Question 32 Explanation: 
Explanation: We know that in class B addressing we have 16-bit network ID and !6-bits for host ID.
Since in question 500 subnets are required. Since 29 = 512, So we need to take 9 bits from host id part of class B addressing in order to provide 500 subnets and with remaining 7 bits of host id we can easily 100 usable hosts per subnet because 27 = 128.

Since we know that Subnet mask contains 1’s in subnet id part and the 0's in host ID part. Hence subnet mask is:
11111111 . 11111111 . 11111111 . 10000000
In decimal form it can be represented as:
255 . 255 . 255 . 128
c
Question 33
Which of the following class of IP address has the last address as 223.255.255.255?
A
Class A
B
Class B
C
Class C
D
Class D
       Computer-Networks       IP-Address       UGC-NET DEC-2019 Part-2
Question 33 Explanation: 
Class A → 0.0.0.0 to 127.255.255.255
Class B → 128.0.0.0 to 191.255.255.255
Class C → 192.0.0.0 to 223.255.255.255
Class D → 224.0.0.0 to 239.255.255.255
Class E → 240.0.0.0 to 254.255.255.254
Question 34
An organization has a Class B network and wishes to form subnets for 60 departments. The subnet mask would be:
A
255.255.64.0
B
255.255.0.0
C
255.255.252.0
D
255.255.255.0
       Computer-Networks       IP-Address       CIL Part - B
Question 34 Explanation: 
Organization have 60 departments, and to assign 60 subnet we need 6 bits for subnet. In Class B network first two octet are reserved for NID, so we take first 6 bit of third octet for subnets and subnet mask would be 255.255.11111100.00000000 = 255.255.252.0
Question 35

A block of addresses is granted to a small organization, If one of the addresses is 210.32.64.79/26, then what will be the values of the following?

(i) First address

(ii) Last address

(iii) Total number of addresses
A
(i) 210.32.64.64, (ii) 210.32.64.127, (iii)64
B
(i) 210.32.64.64, (ii) 210.32.64.255, (iii)32
C
(i) 210.32.64.79, (ii) 210.32.64.255, (iii)64
D
(i) 210.32.64.64, (ii) 210.32.64.79, (iii)128
       Computer-Networks       IP-Address       CIL Part - B
Question 35 Explanation: 
→Given address is CIDR network address 210.32.64.79/26.
→Classless addressing treats the IP address as a 32 bit stream of ones and zeroes, where the boundary between network and host portions can fall anywhere between bit 0 and bit 31.Classless addressing system is also known as CIDR(Classless Inter-Domain Routing).
→The first address in the block can be found by setting the rightmost 32 − n bits to 0s.
→The last address in the block can be found by setting the rightmost 32 − n bits to 1s.
→The number of addresses in the block can be found by using the formula 232−n.
→Here “n” is 26.
Question 36
Which of the following classes of languages can validate an IPv4 address in dotted decimal format? It is to be ensured that the decimal values lie between 0 and 255.
A
RE and higher
B
CFG and higher
C
CSG and higher
D
Recursively enumerable language
       Computer-Networks       IP-Address       ISRO CS 2020       Video-Explanation
Question 36 Explanation: 
To ensure that decimal value lie between 0 and 255 we can make finite automata as number of decimal values between 0 and 255 is finite. So RE (regular) and higher is correct.
Question 37

In the IPv-4 addressing format, the number of networks allowed under Class C addresses is

A
214
B
221
C
27
D
224
       Computer-Networks       IP-Address       APPSC-2016-DL-CA
Question 37 Explanation: 
The number of networks allowed under Class C addresses are 221 and each network contains 28 IP addresses.
Question 38

Which protocol will be used to automate the IP configuration mechanism which includes IP address, subnet mask, default gateway, and DNS information?

A
SMTP
B
DHCP
C
ARP
D
TCP/IP
       Computer-Networks       IP-Address       APPSC-2016-DL-CA
Question 39

Consider the classful addressing, the IP address 128.252.144.84 denotes:

A
0.0.0.0 as network ID and 128.252.252.84 as node ID
B
128.0.0.0 as network ID and 128.252.127.84 as node ID
C
128.252.144.0 as network ID and 128.252.144.84 as node ID
D
128.252.0.0 as network ID and 128.252.144.84 as node ID
       Computer-Networks       IP-Address       CIL 2020
Question 39 Explanation: 
The range of class B IP addresses is 128-191. So first octet of given IP address is 128 ,hence the given IP address is of class B in which first two octet or first 16 bit is Network ID which remains same for all IP address of that network and rest bits are part of host ID which will be zero for Network ID of the given network.
So the Network ID of given IP address will be 128.252.0.0. And the node ID will be the same as the given IP address which is 128.252.144.84.
Question 40

IP address range for E class network is

A
1.0.0.0 to 127.0.0.0
B
128.0.0.0 to 191.255.0.0
C
224.0.0 to 239.255.255.0
D
240.0.0.0 to 255.255.255
       Computer-Networks       IP-Address       APPSC-2012-DL CA
Question 40 Explanation: 
IP addresses of class E ranges from 240.0. 0.0 – 255.255. 255.255.
Question 41
IP address can be used to specify a broadcast and map to hardware broadcast, if available. By conversion broadcast address has hosted with all bits is
A
0
B
1
C
Both (A) and (B)
D
None of these
       Computer-Networks       IP-Address       TNPSC-2012-Polytechnic-CS
Question 41 Explanation: 
For limited broadcast all bits are 1 in destination IP address.
Question 42
Which of the following are not valid IPV4 addresses?
A
192.10.14.3
B
200.172.287.33
C
65.92.11.00
D
10.34.110.77
       Computer-Networks       IP-Address       TNPSC-2017-Polytechnic-CS
Question 42 Explanation: 
Each octet of IP address has the maximum value of 255. Since in option B, third octet exceeds the value 255,hence invalid.
Question 43
In which kind of communication, the destination address in each packet is the same for all duplicate?
A
Unicasting
B
Multicasting
C
Multiple unicasting
D
Broad casting
       Computer-Networks       IP-Address       TNPSC-2017-Polytechnic-CS
Question 43 Explanation: 
In broadcasting communication ,the destination address of all the packets are same since that packet has to be distributed among all the hosts of the network.
Question 44
In IPV4, using the classful addressing scheme, the whole address space is divided into how many classes:
A
8
B
16
C
24
D
5
       Computer-Networks       IP-Address       TNPSC-2017-Polytechnic-CS
Question 44 Explanation: 
There are 5 classful addressing scheme, Class A, Class B, Class C, Class D, Class D, Class E.
Question 45
If an IP address starts with a bit sequence of 11110, it is a class _______ address.
A
B
B
C
C
D
D
E
       Computer-Networks       IP-Address       TNPSC-2017-Polytechnic-CS
Question 45 Explanation: 
Class A starts with 0.
Class B starts with 10.
Class C starts with 110.
Class D starts with 1110.
Class E starts with 11110.
Question 46
__________ is used to obtain the IP address of a host based on its physical address.
A
RARP
B
IPV6
C
TFTP
D
TELNET
       Computer-Networks       IP-Address       TNPSC-2017-Polytechnic-CS
Question 46 Explanation: 
RARP is used to obtain the IP address of a host based on its physical address.
Question 47
How many host interfaces may be addressed in the subnet 123.224.00.00/11?
A
2048
B
2,097,150
C
1,000,192
D
2,097,152
       Computer-Networks       IP-Address       TNPSC-2017-Polytechnic-CS
Question 47 Explanation: 
No. of host id bits is 32-11 = 21
So, no. of host interfaces that can be addressed in the subnet is 221 - 2 = 2097150.
Question 48
Which of the following is used in the options field of IPv4 ?
A
Strict source routing
B
Loose source routing
C
time stamp
D
All of the above
       Computer-Networks       IP-Address       UGC NET CS 2014 June-paper-3
Question 48 Explanation: 
IPv4 option field used to provide:
Source routing: also called path addressing, allows a sender of a packet to partially or completely specify the route the packet takes through the network..
Source routing can be of two types:
1. Strict source routing
2. Loose source routing
Timestamp may also be included in option field.
Question 49
A network on the Internet has a subnet mask of 255.255.240.0. What is the maximum number of hosts it can handle ?
A
1024
B
2048
C
4096
D
8192
E
none of the above
       Computer-Networks       IP-Address       UGC NET CS 2014 June-paper-3
Question 49 Explanation: 
Subnet mask: It contains 1’s in the network ID part and 0’s in the host ID part.
255.255.240.0 = 11111111 . 11111111 .11110000 . 00000000
maximum number of hosts = 212 - 2
= 4094
Question 50

Two systems S1 and S2 are configured with the following IP address:

S1: 203.197.2.53; netmask 255.255.128.0

S2: 203. 197.75.201;netmask 255.255.192.0

Which one of the following statements is true?
A
S1 assumes S2 is on same network, but S2 assumes S1 is on a different network
B
S2 assumes S1 is on same network, but S1 assumes S2 is on a different network
C
S1 and S2 both assume they are on the same network
D
S1 and S2 both assume they are on different networks
       Computer-Networks       IP-Address       HCU PHD CS MAY 2019
Question 50 Explanation: 
For S1,
IP address - 203.197.2.53
So Netmask - 255.255.128.0
For S2,
IP address - 203.197.75.201
Netmask - 255.255.192.0
Now first let’s check for S1,
Network ID according to S1 for itself,
203.197.2.53
AND 255.255.128.0
-------------------------------
203.197.0.0
Network ID according to S1 for S2,
203.197.75.201
255.255.128.0
---------------------
203.197.0.0
Since according to S1 NID of both S1 and S2 are the same, it assumes S2 to be in the same network.
Now let’s check for S2,
Network ID according to S2 for itself,
203.197.75.201
AND 255.255.192.0
-------------------------------
203.197.264.0
Network ID according to S2 for S1,
203.197.2.53
255.255.192.0
---------------------
203.197.0.0
Since according to S2 NID of both S1 and S2 are different, so it assumes S1 to be in a different network.
Question 51
If a host has an IP address 201.40.67.31/25, what are the network ID and broadcast address of the network to which this host belongs:
A
201.40.67.31, 201.40.67.63
B
201.40.67.0, 201.40.67.255
C
201.40.67.0, 201.40.67.63
D
201.40.67.31, 201.40.67.255
E
None of the above
       Computer-Networks       IP-Address       HCU PHD CS MAY 2015
Question 51 Explanation: 
Correct answer is 201.40.67.0, 201.40.67.127
Question 52

IP addresses are generally written in the dotted-decimal notation. Given the IP address in binary is

1000000010000 1 1 10100010000000101

Using dotted decimal notation it is written as
A
100.000001.00001.100
B
80.87.66.05
C
135.128.68.5
D
128.135.68.5
       Computer-Networks       IP-Address       HCU PHD CS MAY 2016
Question 52 Explanation: 
There are four octets in IP address each of 8 bits.And each 8 bits can be represented in decimal. So the IP address using dotted decimal is,
Question 53
All IP addresses in the range 186.220.64.0 to 186.220.71.254 are kept in a VLAN. The correct netmask so that messages are broadcast only within this VLAN is
A
186.220.255.255
B
186.220.248.0
C
186.220.248.0
D
186.220.8.0
       Computer-Networks       IP-Address       HCU PHD CS MAY 2014
Question 53 Explanation: 
We have to select such subnet mask by which ANDing the both IP address we get the same network ID. And option D gives same network ID after ANDing .The network ID we get is 186.220.0.0
Question 54
All IP addresses in the range 186.220 .64.0 to 186.220.71.254 arc kept in a VLAN. The correct netmask so that messages are broadcast only within this VLAN is
A
186.220.255.255
B
186.220.248.0
C
186.220.248.255
D
186.220.8.0
       Computer-Networks       IP-Address       HCU PHD CS MAY 2012
Question 54 Explanation: 
Let check option wise,




Question 55
In classful addressing, a large part of the available addresses are ______.
A
Dispersed
B
Blocked
C
Wasted
D
Reserved
       Computer-Networks       IP-Address       NIC-NIELIT Scientist-B 2020
Question 55 Explanation: 
In classful addressing, a large part of the available addresses are Wasted.
There are 55 questions to complete.