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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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.**

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