Network Security Interview Subcategories

Freshers	Network Security Interview
Network Hardware	Experienced
Classful Addressing

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.

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.

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. Step-by-Step ConversionFor each octet, we multiply each bit by $2^n$ where $n$ is its position (from 7 down to 0).First Octet: 10000001$1 times 2^7 = 128$$0 times 2^6 = 0$$0 times 2^5 = 0$$0 times 2^4 = 0$$0 times 2^3 = 0$$0 times 2^2 = 0$$0 times 2^1 = 0$$1 times 2^0 = 1$Total: $128 + 1 = mathbf{129}$Second Octet: 00001011$0 times 128 = 0$$0 times 64 = 0$$0 times 32 = 0$$0 times 16 = 0$$1 times 8 = 8$$0 times 4 = 0$$1 times 2 = 2$$1 times 1 = 1$Total: $8 + 2 + 1 = mathbf{11}$Third Octet: 00001011This is identical to the second octet.Total: 11Fourth Octet: 11101111$1 times 128 = 128$$1 times 64 = 64$$1 times 32 = 32$$0 times 16 = 0$$1 times 8 = 8$$1 times 4 = 4$$1 times 2 = 2$$1 times 1 = 1$Total: $128 + 64 + 32 + 8 + 4 + 2 + 1 = mathbf{239}$Final ResultBy combining the four decimal values with periods, we get:Dotted-Decimal Notation: 129.11.11.239

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.

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.

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.

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

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.

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.

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.