ipsss

What is the prefix length notation for the subnet mask 255.255.255.224?

• /25
• /26
• /27
• /28

Answer explanation

The binary format for 255.255.255.224 is 11111111.11111111.11111111.11100000. The prefix length is the number of consecutive 1s in the subnet mask. Therefore, the prefix length is /27.

How many valid host addresses are available on an IPv4 subnet that is configured with a /26 mask?

• 254
• 190
• 192
• 62
• 64

Answer explanation

A /26 mask leaves 6 bits for hosts (32 - 26 = 6). The formula is 2 h −2. 2 6 =64; 64−2=62 valid hosts.

Which subnet mask would be used if 5 host bits are available?

• 255.255.255.0
• 255.255.255.128
• 255.255.255.224
• 255.255.255.240

Answer explanation

If 5 bits are used for hosts, 27 bits are used for the network (32−5=27). A /27 mask corresponds to 255.255.255.224.

A network administrator subnets the 192.168.10.0/24 network into subnets with /26 masks. How many equal-sized subnets are created?

• 1
• 2
• 4
• 8
• 16

Answer explanation

The original mask is /24. The new mask is /26. The number of borrowed bits is 26−24=2. The formula for subnets is 2 n (where n is borrowed bits). 2 2 =4 subnets.

Which of the following represents the valid host range for the subnet 192.168.1.32/27?

• 192.168.1.1 – 192.168.1.30
• 192.168.1.33 – 192.168.1.62
• 192.168.1.65 – 192.168.1.94
• 192.168.1.97 – 192.168.1.126

Answer explanation

Subnet 192.168.1.32/27 has a block size of 32. The network address is .32 and the broadcast address is .63. The valid host range is between them (.33 – .62).

An administrator wants to create four subnetworks from the network address 192.168.1.0/24. What is the network address and subnet mask of the second useable subnet?

• Subnetwork 192.168.1.64, Subnet mask 255.255.255.192
• Subnetwork 192.168.1.32, Subnet mask 255.255.255.240
• Subnetwork 192.168.1.64, Subnet mask 255.255.255.240
• Subnetwork 192.168.1.128, Subnet mask 255.255.255.192

Answer explanation

To create 4 subnets, 2 bits are borrowed, making the mask /26 (255.255.255.192). The subnets are .0, .64, .128, and .192. The second subnet starts at .64.

How many bits must be borrowed from the host portion of an address to accommodate a router with five connected networks?

• Two
• Three
• Four
• Five

Answer explanation

To accommodate 5 networks, you need to calculate 2 n >5. If n=2, 2 2 =4 (not enough). If n=3, 2 3 =8 (enough). Therefore, 3 bits must be borrowed.

How many host addresses are available on the 192.168.10.128/26 network?

• 30
• 32
• 60
• 62
• 64

Answer explanation

A /26 prefix gives 6 host bits (32−26=6). 2 6 =64. Subtracting the network and broadcast addresses leaves 62 usable host addresses.

How many host addresses are available on the network 172.16.128.0 with a subnet mask of 255.255.252.0?

• 510
• 512
• 1022
• 1024
• 2046

Answer explanation

A mask of 255.255.252.0 is /22. This leaves 10 host bits (32−22=10). The number of hosts is 2 10 −2=1024−2=1022.

Which of the following is considered a "Test-Net" address range?

• 169.254.0.0/16
• 192.0.2.0/24
• 240.0.0.0/4
• 127.0.0.0/8

Answer explanation

192.0.2.0/24 is reserved for teaching and learning (TEST-NET). 169.254.x.x is Link-Local, 240.x.x.x is Experimental, and 127.x.x.x is Loopback.