5 Written questions
5 Matching questions
- Given 192.168.10.0/28, What are the valid hosts?
- Unable to Ping Remote Destination
- What subnet and broadcast address is the IP address 172.16.50.10 255.255.224.0 (/19) a member of?
- Given 172.16.0.0/18, How many subnets?
- a The subnet is 172.16.32.0, and the broadcast must be 172.16.63.25
- b .1 - .126 & .129 - .254
- c /30 is 255.255.255.252. The valid subnet is 192.168.100.24, broadcast is 192.168.100.27, and valid hosts are 192.168.100.25 and 26
- d 4
- e Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
5 Multiple choice questions
- 32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address
- A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.
- 256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
- This subnet address must be in the 172.16.32.0 subnet, and the broadcast must be 172.16.47.255
- 11111100 = 252
5 True/False questions
Additional Windows Troubleshooting → Tracert, Show ip arp
Given 192.168.10.0/28, What's the broadcast address for each subnet? → 0 & 128
Given 172.16.0.0/18, How many hosts per subnet? → 16,382
Classful Routing Protocols → RIPv1 and IGRP
Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. Again, the zero subnet should not be considered valid for this question → A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32, so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.