CompTIA Network+ N10-009 PBQ Practice

Fill in the subnet plan. The second subnet is the one that starts right after the first.

Fields: Prefix length (or mask) for each subnet; Usable hosts per subnet; Network address of the second subnet; Broadcast address of the second subnet; First usable host in the second subnet

Expected: Prefix length (or mask) for each subnet: /26; Usable hosts per subnet: 62; Network address of the second subnet: 192.168.10.64; Broadcast address of the second subnet: 192.168.10.127; First usable host in the second subnet: 192.168.10.65

Four equal subnets borrow 2 bits from the /24, giving /26, mask 255.255.255.192. The block size is 256 minus 192, which is 64, so the subnets start at .0, .64, .128, and .192. The second runs from network .64 to broadcast .127 (one less than the next multiple), its first usable host is .65, and each holds 64 minus 2, which is 62 usable hosts.

Pick every address a host in the second subnet can use.

1. 192.168.10.64
2. 192.168.10.65
3. 192.168.10.100
4. 192.168.10.126
5. 192.168.10.127
6. 192.168.10.128

Expected: 192.168.10.65; 192.168.10.100; 192.168.10.126

The second subnet's usable hosts run from .65 to .126, everything between its network address and its broadcast. The network .64 and the broadcast .127 can never be given to a host, and .128 is the network address of the third subnet.

A department is assigned the fourth subnet, 192.168.10.192/26. Which headcount fits in it?

1. A department with 60 hosts
2. A department with 70 hosts
3. A department with 64 hosts
4. A department with 126 hosts

Expected: A department with 60 hosts

A /26 has 6 host bits: 2 to the 6th is 64, minus 2 for the network and broadcast addresses, leaves 62 usable hosts. Sixty hosts fit; 64, 70, and 126 do not, because the network and broadcast addresses can never be given to a host.

Put the steps you take on this case in the order of the methodology.

Items: Identify the problem: question the user, read the ticket, and find out what changed; Establish a theory of probable cause; Test the theory to determine the cause; Establish a plan of action and identify its potential effects; Implement the solution or escalate; Verify full system functionality and implement preventive measures; Document findings, actions, outcomes, and lessons learned

Expected: 1. Identify the problem: question the user, read the ticket, and find out what changed; 2. Establish a theory of probable cause; 3. Test the theory to determine the cause; 4. Establish a plan of action and identify its potential effects; 5. Implement the solution or escalate; 6. Verify full system functionality and implement preventive measures; 7. Document findings, actions, outcomes, and lessons learned

The seven steps run identify, theorize, test, plan, implement or escalate, verify and prevent, then document. After a theory is confirmed the next step is the plan of action, never straight to the fix, and you verify before you document.

The user says the drive worked fine yesterday. What do you ask first?

1. What changed since yesterday?
2. Can I restart the file server now?
3. Can I replace your network cable?
4. Should I escalate this to the server team?

Expected: What changed since yesterday?

Identifying the problem includes determining whether anything changed, because most outages follow a change. Restarting a server, swapping a cable, or escalating all jump ahead before you know the cause.

Pick every item that belongs in the ticket's documentation when you close it.

1. The cause you confirmed
2. The fix you applied
3. The time it took to resolve
4. The user's password, so the next technician can sign in as them
5. Lessons learned for the next similar case

Expected: The cause you confirmed; The fix you applied; The time it took to resolve; Lessons learned for the next similar case

Documentation records findings, actions, outcomes, and lessons learned: the cause, the fix, how long it took, and what to do next time. A user's password never goes in a ticket; it is a credential, not a finding.

Configure PC-A with a static address from the office plan, then name the router's peer on the WAN link.

Fields: PC-A IP address; PC-A subnet mask; PC-A default gateway; Provider's address on the WAN link

Expected: PC-A IP address: 10.0.1.10; PC-A subnet mask: 255.255.255.0; PC-A default gateway: 10.0.1.1; Provider's address on the WAN link: 203.0.113.1

PC-A must sit in the router's LAN block, 10.0.1.0/24, so it takes a plan address with mask 255.255.255.0 and the router's LAN interface, 10.0.1.1, as its default gateway; a host with a gateway outside its subnet reaches its neighbors and nothing beyond. The WAN is a /30: block size 4, network .0, broadcast .3, and only .1 and .2 usable, so with the router on .2 the provider is .1.

Why should the printer have a static address or a DHCP reservation?

1. Every PC finds it by its address, so the address must never change
2. Printers cannot request an address from a DHCP server
3. A fixed address makes the printer print faster
4. Only devices with static addresses can be reached across the switch

Expected: Every PC finds it by its address, so the address must never change

Clients reach a shared printer at a known address, so it must not change; a DHCP reservation pins one address to the printer's MAC while keeping it centrally managed. Printers can use DHCP perfectly well, and a fixed address changes neither speed nor reachability across the switch.

Pick every address that cannot be assigned to a PC on this LAN.

1. 10.0.1.0
2. 10.0.1.25
3. 10.0.1.255
4. 10.0.1.1
5. 10.0.1.200

Expected: 10.0.1.0; 10.0.1.255; 10.0.1.1

In 10.0.1.0/24 the network address .0 and the broadcast address .255 can never be given to a host, and .1 is already the router's gateway address, so a PC there would conflict. The .25 and .200 addresses are ordinary usable hosts.