# Networking Questions and Answers – Classless Addressing

This set of Network Security Test focuses on “Classless Addressing”.

1. Classless Addressing overcomes the problem of –
a) address completion
b) address depletion
c) address extraction
d) all of the mentioned
View Answer

Answer: b
Explanation: Classless Addressing overcomes the problem of address depletion.

2. Which of the following is true for classless addressing?
a) The addresses are contiguous
b) The number of addresses is a power of 2 (16 = 24), and the first address is divisible by 16
c) The first address, when converted to a decimal number, is 3,440,387,360, which when divided by 16 results in 215,024,210
d) All of the mentioned
View Answer

Answer: d
Explanation: All the statements are true.

3. The value ‘n’ in classless addressing is referred to as –
a) prefix length
b) suffix length
c) intermediate length
d) none of the mentioned
View Answer

Answer: a
Explanation: ‘n’ is referred to as prefix length.
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4. Which of the following values can ‘n’ not take?
a) 16
b) 8
c) 20
d) 24
View Answer

Answer: c
Explanation: ‘n’ can only have values 8, 16, or 24.

5. The slash notation in classless addressing is referred to as –
a) NIFT
b) PITF
c) CIDR
d) TRS
View Answer

Answer: c
Explanation: It is referred to as CIDR.
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6. CIDR stands for –
a) Class In Domain Range
b) Classless Inter Domain Routing
c) Classless In Domain Range
d) None of the mentioned.
View Answer

Answer: b
Explanation: CIDR stands for Classless Inter Domain Routing.

7. A classless address is given as 167.199.170.82/27. Find the number of addresses.
a) 128
b) 64
c) 32
d) 16
View Answer

Answer: c
Explanation: n =27
Therefore number of addresses is 2(32-n) = 32.
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8. A classless address is given as 167.199.170.82/27. Find the first address.
a) 167.199.170.32
b) 167.199.170.82
c) 167.199.170.64
d) 167.199.170.78
View Answer

Answer: c
Explanation: First address = (any address) AND (network mask).

9. A classless address is given as 167.199.170.82/27. Find the last address.
a) 167.199.170.95
b) 167.199.170.256
c) 167.199.170.128
d) 167.199.170.88
View Answer

Answer: a
Explanation: Last address = (any address) OR [NOT (network mask)].
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10. No. of address in the block= N= NOT(mask)+1.
a) True
b) False
View Answer

Answer: a
Explanation: The statement is true.

11. ICANN stands for Internet Corporation for Assigned Names and Addresses.
a) True
b) False
View Answer

Answer: a
Explanation: The address block allocation is managed by a global authority called the ICANN.

12. An ISP has requested a block of 1000 addresses. How many blocks are granted to it?
a) 512
b) 1000
c) 1024
d) None of the mentioned
View Answer

Answer: c
Explanation: The number of blocks allotted can only be in the powers of 2.

13. An ISP has requested a block of 1000 addresses. What is its prefix length?
a) 28
b) 22
c) 26
d) 24
View Answer

Answer: b
Explanation: n =32- log_2 (1024) = 22.

14. An ISP has requested a block of 1000 addresses. Can 18.14.12.0 be its first address?
a) Yes
b) No
c) Can’t Say
d) Insufficient Data
View Answer

Answer: a
Explanation: 18.14.12.0 is divisible by 1024 hence it can be taken as first address.

15. If the first address is First address = 18.14.12.0/22. What is the last address?
a) 18.14.15.128/22
b) 18.14.15.64/22
c) 18.14.15.32/22
d) 18.14.15.255/22
View Answer

Answer: d
Explanation: Last address = (any address) OR [NOT (network mask)].
The Last Address = 18.14.15.255/22.

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

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

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