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This set of Network Security Interview Questions and Answers for experienced focuses on “IP Addressing”.

1. 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

Explanation: Converting the binary value to their respective decimal values yields 128.11.3.31.

2. Convert the following binary notation to hexadecimal notation –
10000000 00001011 00000011 00011111
a) 0x 80 0B 03 1E
b) 0x 81 0B 04 1E
c) 0x 80 0C 03 1F
d) 0x 80 0B 03 1F

Explanation: Converting the binary value to their respective hex values yields 0x 80 0B 03 1F.

3. 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

Explanation: Converting the binary value to their respective decimal values yields 129.11.11.239.

4. 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

Explanation: Replacing each decimal number with its binary equivalent we get 01101111 00111000 00101101 01001110.

5. 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

Explanation: There should be no leading zeros.

6. 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

Explanation: We cannot have more than 4 bytes in an IPv4 address.

7. 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

Explanation: None.

8. 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

Explanation: The error is that there is a mixture of binary and dotted-decimal notation.

9. The following IPv4 addresses in hexadecimal notation is – 10000001 00001011 00001011 11101111-
a) 0x810B0BEF
b) 0x810D0AFF
c) 0x810B0BFE
d) 0x810C0CEF

Explanation: We replace each group of 4 bits with its hexadecimal equivalent. Note that 0X (or 0x) is added at the beginning or the subscript 16 at the end. 0x810B0BEF or 810B0BEF_16.

10. 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

Explanation: Subtract the first address from the last address in base 256.
The result = 0.0.3.255
Therefore, number of addresses = (3 X 256 + 255) + 1 = 1024.

11. The first address in a range of addresses is 14.11.45.96. If the number of addresses in the range is 32, what is the last address?
a) 14.11.44.64
b) 14.11.44.128
c) 14.12.44.128
d) 14.11.45.127

Explanation: Last Address = (14.11.45.96 + 0.0.0.31)_256 = 14.11.45.127.

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

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