Match the classes of IP addresses with their functional applications :(a) Class A(1) Multicast Services(b) Classs B(2) Reserved for future use(c) Class D(3) Large Organizations(d) Class E(4) Midsize Organizations

The correct matching is (a)-3, (b)-4, (c)-1, (d)-2.Class A (first octet 1-126): very large networks, for large organisations.Class B (128-191): medium-sized networks, for midsize organisations.Class D (224-239): multicast addresses.Class E (240-255): reserved for future and experimental use.Option A makes Class D reserved and Class E multicast, which reverses them, and options B and C misplace Class A as well.

Subnetting Questions and Answers

Q: Which one of the following CIDR prefixes exactly represents the range of IP addresses 10.12.2.0 to 10.12.3.255?

The prefix is 10.12.2.0/23.The range 10.12.2.0 to 10.12.3.255 holds \(2 \times 256 = 512 = 2^{9}\) addresses, so 9 host bits and a prefix length of \(32 - 9 = 23\). A /23 block must start on an even third octet; 2 is even (binary 0000 0010), so 10.12.2.0/23 spans exactly 10.12.2.0 to 10.12.3.255.10.12.2.0/24 covers only 10.12.2.0 to 10.12.2.255. 10.12.0.0/22 covers 10.12.0.0 to 10.12.3.255, twice the range. 10.12.2.0/22 is not a valid block start, since a /22 must begin at a third octet divisible by 4; it denotes the same 10.12.0.0 to 10.12.3.255 block.

Q: Assuming classbased addressing scheme, in an university has 35 (thirty five) departments and uses a class B address. If use 6 (six)-bit subnet number and 10-bit host number then how many Ethernet segments and maximum hosts in each Ethernet segment are possible ?

The answer is 64, 1022.A class B address has 16 host bits, split here into a 6-bit subnet number and a 10-bit host number:\[ \begin{aligned} \text{subnets} &= 2^{6} = 64 \\ \text{hosts} &= 2^{10} - 2 = 1022 \end{aligned} \]Two host addresses in every subnet are reserved, the all-zeros host part for the network and the all-ones host part for broadcast, so each Ethernet segment can have 1022 hosts. 64 segments are enough for the 35 departments. 1024 forgets the two reserved addresses, while 256 and 254 are the figures for an 8-bit host field.

Q: Find the class of the following two addresses :(a) 0000 0001 0000 1011 00001011 11101111(b) 14.23.120.8

Both addresses are Class A, Class A.An IPv4 address's class depends on its first octet: 0 to 127 (first bit 0) is class A, 128 to 191 is B, 192 to 223 is C, and 224 to 239 is D.(a) The first byte 00000001 starts with 0, which is decimal 1, so it is class A. (b) The first byte of 14.23.120.8 is 14, also between 0 and 127, so it is class A too. Hence options naming B, C or D are wrong.

Q: An Internet Service Provider (ISP) has the following chunk of Classless Inter-Domain Routing (CIDR)-based IP addresses available with it : 245.248.128.0/20. The ISP wants to give half of this chunk of addreses 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 valid allocation is 245.248.136.0/21 to A and 245.248.128.0/22 to B.The block 245.248.128.0/20 has \(2^{32-20} = 4096\) addresses, from 245.248.128.0 to 245.248.143.255. Half of it is a /21 (2048 addresses) and a quarter is a /22 (1024 addresses).\[ \begin{aligned} \text{A}: 136.0/21 &= 136.0 \text{ to } 143.255 \\ \text{B}: 128.0/22 &= 128.0 \text{ to } 131.255 \\ \text{ISP}: 132.0/22 &= 132.0 \text{ to } 135.255 \end{aligned} \]These do not overlap and together fill the /20. In option B, B's block 128.0/22 lies inside A's 128.0/21. Options C and D give A the /22 and B the /21, which reverses the half and the quarter.

Q: What is the subnet mask for a Class C network with default subnetting ?

The default Class C mask is 255.255.255.0.A Class C address (first octet 192 to 223) uses the first 24 bits for the network and the last 8 bits for hosts, so its default mask is /24, or 255.255.255.0. That gives 254 usable host addresses per network.255.0.0.0 (/8) is the Class A default and 255.255.0.0 (/16) the Class B default. 255.255.255.255 (/32) leaves no host bits, so it identifies a single host, not a network.

Q: In the IPv4 addressing format, the number of networks allowed under Class C addresses is

Class C allows \(2^{21}\) networks.A Class C address starts with the bits 110, followed by 21 more network bits and 8 host bits. The network part therefore has 21 free bits:\[ 2^{21} = 2{,}097{,}152\ \text{networks} \]\(2^{14}\) is the number of Class B networks and \(2^{7}\) of Class A networks (before reserved values are removed). \(2^{24}\) is roughly the number of hosts per Class A network.

Q: A subnet mask of 255.255.255.0 provides how many host bits ?

A 255.255.255.0 mask leaves 8 host bits.An IPv4 address has 32 bits. In the mask, each 255 is eight 1s marking network bits, and the final 0 is eight 0s marking host bits.\[ \begin{aligned} \text{network bits} &= 8 \times 3 = 24 \\ \text{host bits} &= 32 - 24 = 8 \end{aligned} \]That gives \(2^{8} - 2 = 254\) usable host addresses (a /24 network). 24 is the number of network bits, 16 would be a 255.255.0.0 mask, and 32 is the whole address.

Q: What is the maximum number of hosts that can be supported by 193.1.1.64/27 ?

The block supports 30 hosts.A /27 prefix leaves \(32 - 27 = 5\) bits for host numbers. Two of the addresses cannot be given to hosts: the all-zeros address names the network (193.1.1.64) and the all-ones address is the broadcast address (193.1.1.95).\[ \begin{aligned} \text{hosts} &= 2^{5} - 2 \\ &= 32 - 2 = 30 \end{aligned} \]The usable addresses are 193.1.1.65 to 193.1.1.94. 32 would be the block size, which is not offered, and 10, 20 and 40 match no power of two.

Q: What are the first 4 bits used in the prefix for multicast Class E IP addressing ?

The prefix is 1111.In classful IPv4 addressing the leading bits identify the class: Class A starts 0, Class B 10, Class C 110, Class D 1110 and Class E 1111. Class E addresses (240.0.0.0 to 255.255.255.255) are reserved for experimental use.The stem's word 'multicast' belongs to Class D, whose prefix 1110 is not among the options, so the answer follows the class the question names, Class E. 1000, 1100 and 1010 are not class prefixes at all.

Which one of the following CIDR prefixes exactly represents the range of IP addresses 10.12.2.0 to 10.12.3.255?

A.10.12.2.0/23
B.10.12.2.0/24
C.10.12.0.0/22
D.10.12.2.0/22

Exam Year(s)

GATE CS - 2024,

Answer: A

The prefix is 10.12.2.0/23.

The range 10.12.2.0 to 10.12.3.255 holds \(2 \times 256 = 512 = 2^{9}\) addresses, so 9 host bits and a prefix length of \(32 - 9 = 23\). A /23 block must start on an even third octet; 2 is even (binary 0000 0010), so 10.12.2.0/23 spans exactly 10.12.2.0 to 10.12.3.255.

10.12.2.0/24 covers only 10.12.2.0 to 10.12.2.255. 10.12.0.0/22 covers 10.12.0.0 to 10.12.3.255, twice the range. 10.12.2.0/22 is not a valid block start, since a /22 must begin at a third octet divisible by 4; it denotes the same 10.12.0.0 to 10.12.3.255 block.

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Assuming classbased addressing scheme, in an university has 35 (thirty five) departments and uses a class B address. If use 6 (six)-bit subnet number and 10-bit host number then how many Ethernet segments and maximum hosts in each Ethernet segment are possible ?

A.64, 256
B.64, 254
C.64, 1024
D.64, 1022

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The answer is 64, 1022.

A class B address has 16 host bits, split here into a 6-bit subnet number and a 10-bit host number:

\[ \begin{aligned} \text{subnets} &= 2^{6} = 64 \\ \text{hosts} &= 2^{10} - 2 = 1022 \end{aligned} \]

Two host addresses in every subnet are reserved, the all-zeros host part for the network and the all-ones host part for broadcast, so each Ethernet segment can have 1022 hosts. 64 segments are enough for the 35 departments. 1024 forgets the two reserved addresses, while 256 and 254 are the figures for an 8-bit host field.

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Find the class of the following two addresses :

(a) 0000 0001 0000 1011 00001011 11101111
(b) 14.23.120.8

A.Class D, Class D
B.Class C, Class C
C.Class B, Class B
D.Class A, Class A

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

Both addresses are Class A, Class A.

An IPv4 address's class depends on its first octet: 0 to 127 (first bit 0) is class A, 128 to 191 is B, 192 to 223 is C, and 224 to 239 is D.

(a) The first byte 00000001 starts with 0, which is decimal 1, so it is class A. (b) The first byte of 14.23.120.8 is 14, also between 0 and 127, so it is class A too. Hence options naming B, C or D are wrong.

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Match the classes of IP addresses with their functional applications :

(a) Class A	(1) Multicast Services
(b) Classs B	(2) Reserved for future use
(c) Class D	(3) Large Organizations
(d) Class E	(4) Midsize Organizations

A.(a)-4, (b)-3, (c)-2, (d)-1
B.(a)-2, (b)-4, (c)-1, (d)-3
C.(a)-1, (b)-2, (c)-3, (d)-4
D.(a)-3, (b)-4, (c)-1, (d)-2

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The correct matching is (a)-3, (b)-4, (c)-1, (d)-2.

Class A (first octet 1-126): very large networks, for large organisations.
Class B (128-191): medium-sized networks, for midsize organisations.
Class D (224-239): multicast addresses.
Class E (240-255): reserved for future and experimental use.

Option A makes Class D reserved and Class E multicast, which reverses them, and options B and C misplace Class A as well.

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An Internet Service Provider (ISP) has the following chunk of Classless Inter-Domain Routing (CIDR)-based IP addresses available with it : 245.248.128.0/20. The ISP wants to give half of this chunk of addreses 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.245.248.136.0/21 and 245.248.128.0/22
B.245.248.128.0/21 and 245.248.128.0/22
C.245.248.132.0/22 and 245.248.132.0/21
D.245.248.136.0/22 and 245.248.132.0/21

Exam Year(s)

Tripura Public Service Commission - 2019, Tripura Public Service Commission - 2025,

Answer: A

A valid allocation is 245.248.136.0/21 to A and 245.248.128.0/22 to B.

The block 245.248.128.0/20 has \(2^{32-20} = 4096\) addresses, from 245.248.128.0 to 245.248.143.255. Half of it is a /21 (2048 addresses) and a quarter is a /22 (1024 addresses).

\[ \begin{aligned} \text{A}: 136.0/21 &= 136.0 \text{ to } 143.255 \\ \text{B}: 128.0/22 &= 128.0 \text{ to } 131.255 \\ \text{ISP}: 132.0/22 &= 132.0 \text{ to } 135.255 \end{aligned} \]

These do not overlap and together fill the /20. In option B, B's block 128.0/22 lies inside A's 128.0/21. Options C and D give A the /22 and B the /21, which reverses the half and the quarter.

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What is the subnet mask for a Class C network with default subnetting ?

A.255.0.0.0
B.255.255.0.0
C.255.255.255.0
D.255.255.255.255

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: C

The default Class C mask is 255.255.255.0.

A Class C address (first octet 192 to 223) uses the first 24 bits for the network and the last 8 bits for hosts, so its default mask is /24, or 255.255.255.0. That gives 254 usable host addresses per network.

255.0.0.0 (/8) is the Class A default and 255.255.0.0 (/16) the Class B default. 255.255.255.255 (/32) leaves no host bits, so it identifies a single host, not a network.

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In the IPv4 addressing format, the number of networks allowed under Class C addresses is

A.2^14
B.2^7
C.2^21
D.2^24

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: C

Class C allows \(2^{21}\) networks.

A Class C address starts with the bits 110, followed by 21 more network bits and 8 host bits. The network part therefore has 21 free bits:

\[ 2^{21} = 2{,}097{,}152\ \text{networks} \]

\(2^{14}\) is the number of Class B networks and \(2^{7}\) of Class A networks (before reserved values are removed). \(2^{24}\) is roughly the number of hosts per Class A network.

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A subnet mask of 255.255.255.0 provides how many host bits ?

A.24
B.8
C.16
D.32

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: B

A 255.255.255.0 mask leaves 8 host bits.

An IPv4 address has 32 bits. In the mask, each 255 is eight 1s marking network bits, and the final 0 is eight 0s marking host bits.

\[ \begin{aligned} \text{network bits} &= 8 \times 3 = 24 \\ \text{host bits} &= 32 - 24 = 8 \end{aligned} \]

That gives \(2^{8} - 2 = 254\) usable host addresses (a /24 network). 24 is the number of network bits, 16 would be a 255.255.0.0 mask, and 32 is the whole address.

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What is the maximum number of hosts that can be supported by 193.1.1.64/27 ?

A.10
B.20
C.30
D.40

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: C

The block supports 30 hosts.

A /27 prefix leaves \(32 - 27 = 5\) bits for host numbers. Two of the addresses cannot be given to hosts: the all-zeros address names the network (193.1.1.64) and the all-ones address is the broadcast address (193.1.1.95).

\[ \begin{aligned} \text{hosts} &= 2^{5} - 2 \\ &= 32 - 2 = 30 \end{aligned} \]

The usable addresses are 193.1.1.65 to 193.1.1.94. 32 would be the block size, which is not offered, and 10, 20 and 40 match no power of two.

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10.

What are the first 4 bits used in the prefix for multicast Class E IP addressing ?

A.1000
B.1100
C.1010
D.1111

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: D

The prefix is 1111.

In classful IPv4 addressing the leading bits identify the class: Class A starts 0, Class B 10, Class C 110, Class D 1110 and Class E 1111. Class E addresses (240.0.0.0 to 255.255.255.255) are reserved for experimental use.

The stem's word 'multicast' belongs to Class D, whose prefix 1110 is not among the options, so the answer follows the class the question names, Class E. 1000, 1100 and 1010 are not class prefixes at all.

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