Networking Questions and Answers – Classful Addressing

This set of Network Security Multiple Choice Questions & Answers (MCQs) focuses on “Classful Addressing”.

1. Each block in class A contains _____________ addresses.
a) 216
b) 224
c) 28
d) 214
View Answer

Answer: b
Explanation: Each block in class A contains 224 addresses.

2. Each block in class B contains __________ addresses.
a) 28
b) 224
c) 214
d) 216
View Answer

Answer: d
Explanation: Each block in class B contains 216 addresses.

3. Number of Blocks in class B are –
a) 212
b) 216
c) 214
d) 218
View Answer

Answer: c
Explanation: Number of Blocks in class B are 214.
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3. Number of Blocks in class C are –
a) 212
b) 216
c) 214
d) 218
View Answer

Answer: c
Explanation: Number of Blocks in class C are 214.

4. Percent of addresses occupied by Class D?
a) 50 %
b) 25 %
c) 6.25 %
d) 12.5 %
View Answer

Answer: c
Explanation: Class D has 228 addresses in total.
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5. Which of the following does not have a Net ID and Host ID?
a) Class A
b) Class B
c) Class C
d) Class D
View Answer

Answer: d
Explanation: Class D is the multicast address class. It does not have Net ID and Host ID fields.

6. Which Class is reserved for future use?
a) A
b) B
c) D
d) None of the Mentioned
View Answer

Answer: d
Explanation: Class E is reserved for future use.
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7. Number of Blocks in class C are –
a) 27
b) 28
c) 214
d) 29
View Answer

Answer: a
Explanation: Number of Blocks in class A are 27.

8. What is the size of the Host ID in Class C?
a) 24 bits
b) 16 bits
c) 8 bits
d) 14 bits
View Answer

Answer: c
Explanation: The size of the Host ID in Class C is 8 bits.
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An address in a block is given as 180.8.17.9. Find the number of addresses in the block, the first address, and the last address.

9. Address of the block is –
a) 214
b) 216
c) 28
d) 224
View Answer

Answer: b
Explanation: Since 180 is between 128 and 191, it is class B address n=16.
N = 2(32-n) = 216 = 65,536.

10. First Address is –
a) 180.8.0.0
b) 180.7.64.0
c) 180.8.1.256
d) 180.12.0.0
View Answer

Answer: a
Explanation: Keep the leftmost 16 bits and set the rightmost 16 bits all to 0s. First address = 180.8.0.0, in which 16 is the value of n.

11. Last Address is –
a) 180.8.255.255.
b) 180.8.255.0.
c) 180.12.0.255.
d) 180.9.255.255.
View Answer

Answer: a
Explanation: Keep the leftmost 16 bits and set the rightmost 16 bits all to 1s. Last address = 180.8.255.255.

12. A router receives a packet with the destination address 132.7.21.84. Find the network address of the packet.
a) 1.32.7
b) 132.7
c) 13.27
d) 21.84
View Answer

Answer: b
Explanation: 132 is between 128 and 191, so it is Class B address i.e. n=16-bit. Therefore,
Network address 132.7
Host address 21.84.

Sanfoundry Global Education & Learning Series – Cryptography and Network Security.

To practice all areas of Cryptography and Network Security, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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