Networking Questions and Answers – Classful Addressing

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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.
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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.

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.

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.
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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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn