CompTIA Network+ IP Addressing and Subnetting Questions and Answers — Questions and Answers
Question 1: A network technician is configuring a workstation and observes its IP address is 169.254.100.51. The user reports they cannot access the internet or the company's file server. What is the MOST likely cause of this issue?
- The workstation has a duplicate IP address on the network.
- The workstation is configured with a static IP address in the wrong subnet.
- The workstation failed to obtain an IP address from the DHCP server. (Correct answer)
- The router's default gateway is configured incorrectly.
Correct answer: The workstation failed to obtain an IP address from the DHCP server.
The IP address 169.254.100.51 is an Automatic Private IP Addressing (APIPA) address. Operating systems assign APIPA addresses when a device is configured for DHCP but cannot contact a DHCP server. This allows for communication on the local subnet but not for routed traffic to other networks or the internet.
Question 2: A network administrator needs to create a subnet that can support up to 25 usable host addresses. Which of the following CIDR notations provides the required number of hosts while minimizing wasted IP addresses?
- /26
- /27 (Correct answer)
- /28
- /25
Correct answer: /27
To find the number of usable hosts, use the formula 2^h - 2, where 'h' is the number of host bits. A /27 prefix leaves 5 host bits (32 - 27 = 5). Calculating 2^5 - 2 gives 30 usable hosts. A /28 (14 hosts) is too small, and a /26 (62 hosts) wastes more addresses.
Question 3: Which of the following IP addresses is a non-routable, private address according to RFC 1918?
- 11.200.50.15
- 172.32.10.25
- 192.168.20.1 (Correct answer)
- 169.254.1.10
Correct answer: 192.168.20.1
RFC 1918 defines specific ranges for private IP addressing. The range 192.168.0.0 to 192.168.255.255 is one of these private ranges. The 172.16.0.0/12 range is also private, but 172.32.x.x falls outside of it. The 11.x.x.x address is public, and 169.254.x.x is reserved for APIPA.
Question 4: A network administrator is given the IP address 192.168.50.133 and the subnet mask 255.255.255.240. What is the subnet address for this host?
- 192.168.50.0
- 192.168.50.128 (Correct answer)
- 192.168.50.132
- 192.168.50.143
Correct answer: 192.168.50.128
The subnet mask 255.255.255.240 (/28) creates subnets in increments of 16 in the fourth octet (256-240=16). The subnets are .0, .16, .32, etc. The address 192.168.50.133 falls into the subnet that starts at 192.168.50.128 and ends with the broadcast address 192.168.50.143.
Question 5: A subnet mask of 255.255.252.0 is equivalent to which CIDR prefix notation?
- /21
- /23
- /22 (Correct answer)
- /24
Correct answer: /22
CIDR notation represents the number of leading '1's in the subnet mask's binary representation. 255.255.252.0 in binary is 11111111.11111111.11111100.00000000. Counting the number of '1's gives a total of 8 + 8 + 6 = 22. Therefore, the CIDR notation is /22.
Question 6: A company is using the network address 10.10.0.0/22. How many total IP addresses are available within this network range?
- 256
- 512
- 1024 (Correct answer)
- 2048
Correct answer: 1024
A /22 CIDR prefix means that 22 bits are used for the network portion, leaving 10 bits for the host portion (32 - 22 = 10). The total number of addresses in a subnet is calculated as 2^h, where 'h' is the number of host bits. Therefore, 2^10 equals 1024 total addresses.
CompTIA Network+ IP Addressing and Subnetting Questions and Answers
A network technician is configuring a workstation and observes its IP address is 169.254.100.51.
The user reports they cannot access the internet or the company's file server.
What is the MOST likely cause of this issue?