Free Access CompTIA.N10-009.v2026-01-05.q149 Practice Test (Page 16)

Question 71

Which of the following steps in the troubleshooting methodology includes checking logs for recent changes?

A.Identify the problem.

B.Document the findings and outcomes.

C.Test the theory to determine cause.

D.Establish a plan of action.

Question 72

A company upgrades its network and PCs to gigabit speeds. After the upgrade, users are not getting the expected performance. Technicians discover that the speeds of the endpoint NICs are inconsistent. Which of the following should be checked first to troubleshoot the issue?

A.Speed mismatches

B.Load balancer settings

C.Flow control settings

D.Infrastructure cabling grade

Question 73

A virtual machine has the following configuration:
* IPv4 address: 169.254.10.10
* Subnet mask: 255.255.0.0
The virtual machine can reach colocated systems but cannot reach external addresses on the Internet. Which of the following Is most likely the root cause?

A.The subnet mask is incorrect.

B.The DHCP server is offline.

C.The IP address is an RFC1918 private address.

D.The DNS server is unreachable.

Question 74

A network administrator needs to assign IP addresses to a newly installed network. They choose 192.168.1.0
/24 as their network address and need to create three subnets with 30 hosts on each subnet. Which of the following is a valid subnet mask that will meet the requirements?

A.255.255.255.128

B.255.255.255.192

C.255.255.255.224

D.255.255.255.240

Correct Answer: C

Understanding the Requirements
Network Address: 192.168.1.0/24
The /24 notation means a subnet mask of 255.255.255.0, providing 256 total addresses (192.168.1.0-
192.168.1.255).
Usable hosts: 256 - 2 (network and broadcast) = 254.
Goal: Create 3 subnets, each with 30 hosts.
Each subnet needs enough addresses to accommodate 30 hosts, plus 2 reserved addresses (network and broadcast) per subnet.
Total addresses per subnet = 30 (hosts) + 2 (network/broadcast) = 32 addresses.
Subnetting Basics (Networking Fundamentals)
Subnet Mask: Determines how many bits are borrowed from the host portion to create subnets.
Original Mask: /24 (255.255.255.0) = 24 network bits, 8 host bits.
Formulae:
Number of subnets = 2^(number of borrowed bits).
Number of addresses per subnet = 2^(remaining host bits).
Usable hosts per subnet = 2^(remaining host bits) - 2.
We need:
At least 3 subnets.
At least 32 addresses per subnet (to fit 30 hosts + 2 reserved).
Step-by-Step Analysis
Determine Addresses Needed per Subnet:
32 addresses is a power of 2 (2^5 = 32).
This means each subnet requires 5 host bits (since 2^5 = 32 total addresses, and 32 - 2 = 30 usable hosts).
Calculate Remaining Bits:
Original network has 8 host bits (/24).
If 5 bits are left for hosts, we borrow: 8 - 5 = 3 bits for subnetting.
New Subnet Mask:
Original mask: /24 (24 network bits).
Borrow 3 bits: 24 + 3 = /27.
/27 = 255.255.255.224 (binary: 11111111.11111111.11111111.11100000).
Verify Requirements:
Number of Subnets: 2^3 = 8 subnets (meets the requirement of at least 3).
Addresses per Subnet: 2^5 = 32 addresses.
Usable Hosts per Subnet: 32 - 2 = 30 hosts (exactly meets the requirement).
Subnet Breakdown:
Increment: 256 - 224 = 32 (each subnet increments by 32 in the fourth octet).
Subnets:
192.168.1.0-192.168.1.31 (Network: .0, Broadcast: .31, Hosts: .1-.30)
192.168.1.32-192.168.1.63 (Network: .32, Broadcast: .63, Hosts: .33-.62)
192.168.1.64-192.168.1.95 (Network: .64, Broadcast: .95, Hosts: .65-.94) (And 5 more subnets up to 192.168.1.255.) Three subnets fit perfectly with 30 hosts each.
Evaluating the Options
A). 255.255.255.128 (/25):
Borrow 1 bit: 24 + 1 = /25.
Subnets: 2^1 = 2 (not enough, need 3).
Host bits: 7 (2^7 = 128 addresses, 126 hosts).
Why Not: Only 2 subnets, fails the requirement.
B). 255.255.255.192 (/26):
Borrow 2 bits: 24 + 2 = /26.
Subnets: 2^2 = 4 (meets 3).
Host bits: 6 (2^6 = 64 addresses, 62 hosts).
Why Not: 62 hosts exceeds 30, but it's overkill; /27 is more efficient and still valid.
C). 255.255.255.224 (/27):
Borrow 3 bits: 24 + 3 = /27.
Subnets: 2^3 = 8 (meets 3).
Host bits: 5 (2^5 = 32 addresses, 30 hosts).
Why Yes: Perfectly fits 3 subnets with exactly 30 hosts each.
D). 255.255.255.240 (/28):
Borrow 4 bits: 24 + 4 = /28.
Subnets: 2^4 = 16 (meets 3).
Host bits: 4 (2^4 = 16 addresses, 14 hosts).
Why Not: Only 14 hosts per subnet, fails the 30-host requirement.
Why /27 (255.255.255.224) is Best
It provides exactly 30 usable hosts per subnet, avoiding waste while meeting the minimum requirement.
It allows 8 subnets, exceeding the need for 3, ensuring flexibility.
The study guide emphasizes efficient subnet design, and /27 balances host count and subnet availability.
CompTIA Network+ Context
Networking Fundamentals: Subnetting is a core skill, requiring understanding of CIDR, binary conversion, and address allocation.
Example from Study Guide: Similar problems calculate subnet masks for specific host counts, reinforcing /27 as a common solution for ~30 hosts.

Question 75

SIMULATION
A network technician replaced a switch and needs to reconfigure it to allow the connected devices to connect to the correct networks.
INSTRUCTIONS
Click on the appropriate port(s) on Switch 1 and Switch 3 to verify or reconfigure the correct settings:
- Ensure each device accesses only its correctly associated network
- Disable all unused switch ports
- Require fault-tolerant connections between the switches
- Only make necessary changes to complete the above requirements
If at any time you would like to bring back the initial state of the simulation, please click the Reset All button.