9 Apr, 2007
Networking Interview Questions
- You are given the following address: 153.50.6.27/25. Determine the subnet mask, address class, subnet address, and broadcast address.
* 255.255.255.128, B,153.50.6.0, 153.50.6.127
* 255.255.255.128, C,153.50.6.0, 153.50.6.127
* 255.255.255.128, C,153.50.6.127, 153.50.6.0
* 255.255.255.224, C,153.50.6.0, 153.50.6.127
Correct answer: A
- You are given the following address: 128.16.32.13/30. Determine the subnet mask, address class, subnet address,
and broadcast address.
* 255.255.255.252, B,128.16.32.12, 128.16.32.15
* 255.255.255.252, C,128.16.32.12, 128.16.32.15
* 255.255.255.252, B,128.16.32.15, 128.16.32.12
* 255.255.255.248, B,128.16.32.12, 128.16.32.15
Correct answer: A
- You are given the following address: 15.16.193.6/21. Determine the subnet mask, address class, subnet address,
and broadcast address.
* 255.255.248.0, A, 15.16.192.0, 15.16.199.255
* 255.255.248.0, B, 15.16.192.0, 15.16.199.255
* 255.255.248.0, A, 15.16.199.255, 14.15.192.0
* 255.255.242.0, A, 15.16.192.0, 15.16.199.255
Correct answer: A
- You have an IP host address of 201.222.5.121 and a subnet mask of 255.255.255.248. What is the broadcast address?
* 201.222.5.127
* 201.222.5.120
* 201.222.5.121
* 201.222.5.122
Correct answer: A
The easiest way to calculate this is to subtract 255.255.255.248 (subnet mask) from 255.255.255.255, this
equals 7. Convert the address 201.222.5.121 to binary–11001001 11011110 00000101 01111001. Convert the
mask 255.255.255.248 to binary–11111111 11111111 11111111 11111000. AND them together to get: 11001001 11011110
- 01111000 or 201.222.5.120. 201.222.5.120 is the subnet address, add 7 to this address for 201.222.5.127 or
the broadcast address. 201.222.5.121 through 201.222.5.126 are the valid host addresses.
- Given the address 172.16.2.120 and the subnet mask of 255.255.255.0. How many hosts are available?
* 254
* 510
* 126
* 16,372
Correct answer: A
172.16.2 120 is a standard Class B address with a subnet mask that allows 254 hosts. You are a network administrator and have been assigned the IP address of 201.222.5.0. You need to have 20 subnets with 5 hosts per subnet. The subnet mask is 255.255.255.248.
- Which addresses are valid host addresses?
* 201.222.5.17
* 201.222.5.18
* 201.222.5.16
* 201.222.5.19
* 201.222.5.31
Correct answer: A,B & D
Subnet addresses in this situation are all in multiples of 8. In this example, 201.222.5.16 is the subnet, 201.22.5.31 is the broadcast address. The rest are valid host IDs on subnet 201.222.5.16.
- You are a network administrator and have been assigned the IP address of 201.222.5.0. You need to have 20 subnets with
- hosts per subnet. What subnet mask will you use?
* 255.255.255.248
* 255.255.255.128
* 255.255.255.192
* 255.255.255.240
Correct answer: A
By borrowing 5 bits from the last octet, you can. have 30 subnets. If you borrowed only 4 bits you could only have 14 subnets. The formula is (2 to the power of n)-2. By borrowing 4 bits, you have (2×2×2×2)-2=14. By borrowing 5 bits, you have (2×2×2×2×2)-2=30. To get 20 subnets, you would need to borrow 5 bits so the subnet mask would be 255.255.255.248.
- You are given the IP address of 172.16.2.160 with a subnet mask of 255.255.0.0. What is the network address in binary?
* 10101100 00010000
* 00000010 10100000
* 10101100 00000000
* 11100000 11110000
Correct answer: A
To find the network address, convert the IP address to binary–10101100 000100000 00000010 10100000–then ANDed it with the subnet mask–11111111 11111111 00000000 00000000. The rest is 10101100 00010000 00000000 00000000, which is 172.16.0.0 in decimal.
The first octet rule states that the class of an address can be determined by the numerical value of the first octet.
