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This section includes 13 Mcqs, each offering curated multiple-choice questions to sharpen your Networking knowledge and support exam preparation. Choose a topic below to get started.
1. |
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 |
Answer» B. No | |
2. |
An ISP has requested a block of 1000 addresses. What is its prefix length? |
A. | 28 |
B. | 22 |
C. | 26 |
D. | 24 |
Answer» C. 26 | |
3. |
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 |
Answer» D. None of the mentioned | |
4. |
ICANN stands for Internet Corporation for Assigned Names and Addresses. |
A. | True |
B. | False |
Answer» B. False | |
5. |
No. of address in the block= N= NOT(mask)+1. |
A. | True |
B. | False |
Answer» B. False | |
6. |
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 |
Answer» B. 167.199.170.256 | |
7. |
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 |
Answer» D. 167.199.170.78 | |
8. |
A classless address is given as 167.199.170.82/27. Find the number of addresses. |
A. | 128 |
B. | 64 |
C. | 32 |
D. | 16 |
Answer» D. 16 | |
9. |
The slash notation in classless addressing is referred to as |
A. | NIFT |
B. | PITF |
C. | CIDR |
D. | TRS |
Answer» D. TRS | |
10. |
Which of the following values can n not take? |
A. | 16 |
B. | 8 |
C. | 20 |
D. | 24 |
Answer» D. 24 | |
11. |
The value n in classless addressing is referred to as |
A. | prefix length |
B. | suffix length |
C. | intermediate length |
D. | none of the mentioned |
Answer» B. suffix length | |
12. |
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 |
Answer» E. | |
13. |
Classless Addressing overcomes the problem of |
A. | address completion |
B. | address depletion |
C. | address extraction |
D. | all of the mentioned |
Answer» C. address extraction | |