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This section includes 9 Mcqs, each offering curated multiple-choice questions to sharpen your Networking knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Find the number of addresses in a range if the first address is 146.102.29.0 and last address is 146.102.32.255. |
| A. | 1028 |
| B. | 1024 |
| C. | 578 |
| D. | 512 |
| Answer» C. 578 | |
| 2. |
The following IPv4 addresses in hexadecimal notation is 10000001 00001011 00001011 11101111- |
| A. | 0x810B0BEF |
| B. | 0x810D0AFF |
| C. | 0x810B0BFE |
| D. | 0x810C0CEF |
| Answer» B. 0x810D0AFF | |
| 3. |
What is the error (if any) in the following representation 11100010.23.14.67? |
| A. | There should be no leading zeros |
| B. | We cannot have more than 4 bytes in an IPv4 address |
| C. | Each byte should be less than or equal to 255 |
| D. | None of the mentioned |
| Answer» E. | |
| 4. |
What is the error (if any) in the following representation 75.45.301.14? |
| A. | There should be no leading zeros |
| B. | We cannot have more than 4 bytes in an IPv4 address |
| C. | Each byte should be less than or equal to 255 |
| D. | No error |
| Answer» D. No error | |
| 5. |
What is the error (if any) in the following representation 221.34.7.8.20? |
| A. | There should be no leading zeros |
| B. | Each byte should be less than or equal to 255 |
| C. | We cannot have more than 4 bytes in an IPv4 address |
| D. | No error |
| Answer» D. No error | |
| 6. |
What is the error (if any) in the following representation 111.56.045.78? |
| A. | There should be no leading zeros |
| B. | We cannot have more than 4 bytes in an IPv4 address |
| C. | Each byte should be less than or equal to 255 |
| D. | No error |
| Answer» B. We cannot have more than 4 bytes in an IPv4 address | |
| 7. |
Convert the following dotted-decimal notation to binary notation 111.56.45.78 |
| A. | 01101111 00111000 00101101 01001110 |
| B. | 11101111 00111000 00101101 10001110 |
| C. | 10000000 00001011 00000011 00011111 |
| D. | 10000001 00001011 00001011 11101111 |
| Answer» B. 11101111 00111000 00101101 10001110 | |
| 8. |
Convert the following binary notation to dotted-decimal notation 10000001 00001011 00001011 11101111 |
| A. | 129.11.11.239 |
| B. | 128.11.12.231 |
| C. | 127.11.13.244 |
| D. | 129.12.1.231 |
| Answer» B. 128.11.12.231 | |
| 9. |
Convert the following binary notation to dotted-decimal notation 10000000 00001011 00000011 00011111 |
| A. | 128.11.5.32 |
| B. | 128.11.3.31 |
| C. | 127.11.3.32 |
| D. | 127.12.5.31 |
| Answer» C. 127.11.3.32 | |