Free CompTIA Network+ IP Addressing and Subnetting Questions and Answers

Question 1

A network technician is setting up a small office network and needs to select an appropriate private IPv4 address range. Which of the following IP addresses is a valid, non-routable private IP address according to RFC 1918?

Accepted Answer

172.16.50.1

Answer

RFC 1918 defines specific ranges for private, non-routable IP addresses. The Class B private range is 172.16.0.0 to 172.31.255.255. The address 172.16.50.1 falls within this range. 169.254.10.10 is an APIPA address, 172.32.15.2 is outside the private range, and 100.64.20.5 is part of the shared address space for carrier-grade NAT (CGN).

Question 2

A user reports they cannot access the internet, but they can access local file shares. A technician checks the user's IP configuration and finds the IPv4 address is 169.254.100.5. What is the MOST likely cause of this issue?

Accepted Answer

The computer failed to obtain an IP address from a DHCP server.

Answer

The IP address 169.254.100.5 is an Automatic Private IP Addressing (APIPA) address. Windows and other operating systems assign these addresses when a device is configured for DHCP but cannot contact a DHCP server. This allows for local network communication but no internet access.

Question 3

A network administrator needs to create a subnet that can support at least 50 host devices. Which of the following CIDR notations provides the required number of host addresses while minimizing wasted addresses?

Accepted Answer

/26

Answer

A /26 CIDR notation leaves 6 bits for the host portion (32 - 26 = 6). The number of usable hosts is calculated as 2^6 - 2 = 62. This meets the requirement of 50 hosts. A /27 provides only 30 hosts (2^5 - 2), a /28 provides 14 hosts (2^4 - 2), and a /25 provides 126 hosts (2^7 - 2), which is more than necessary and wastes more addresses than /26.

Question 4

Given a host with the IP address 192.168.10.70 and a subnet mask of 255.255.255.224, what is the broadcast address for this subnet?

Accepted Answer

192.168.10.95

Answer

The subnet mask 255.255.255.224 (/27) creates subnets with a block size of 32 in the last octet (256 - 224 = 32). The subnets are 192.168.10.0, 192.168.10.32, 192.168.10.64, etc. The IP 192.168.10.70 falls into the 192.168.10.64 subnet. The next subnet starts at .96, so the broadcast address for the .64 subnet is the address right before it, which is 192.168.10.95.

Question 5

Which of the following represents a valid Class C default subnet mask?

Accepted Answer

255.255.255.0

Answer

In classful networking, a Class C network uses the first three octets (24 bits) for the network portion and the last octet (8 bits) for the host portion. This corresponds to a default subnet mask of 255.255.255.0.

CompTIA Practice Test

CompTIA Practice Test

Free CompTIA Network+ IP Addressing and Subnetting Questions and Answers