Top

Subnetting

1.

The network address of 172.16.0.0/19 provides how many subnets and hosts?

Answer: B

A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits, but it provides 13 host bits, or 8 subnets, each with 8,190 hosts.

Enter details here

2.

To test the IP stack on your local host, which IP address would you ping?

Answer: C

To test the local stack on your host, ping the loopback interface of 127.0.0.1.

Enter details here

3.

If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?

Answer: A

A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on-which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.

Enter details here

4.

You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?

Answer: C

A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.

Enter details here

5.

What is the subnetwork number of a host with an IP address of 172.16.66.0/21?

Answer: C

A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.

Enter details here

6.

Using the following illustration, what would be the IP address of E0 if you were using the eighth subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. The zero subnet should not be considered valid for this question.

Answer: A

A /28 is a 255.255.255.240 mask. Let's count to the ninth subnet (we need to find the broadcast address of the eighth subnet, so we need to count to the ninth subnet). Starting at 16 (remember, the question stated that we will not use subnet zero, so we start at 16, not 0), 16, 32, 48, 64, 80, 96, 112, 128, 144. The eighth subnet is 128 and the next subnet is 144, so our broadcast address of the 128 subnet is 143. This makes the host range 129-142. 142 is the last valid host.

Enter details here

7.

A network administrator is connecting hosts A and B directly through their Ethernet interfaces, as shown in the illustration. Ping attempts between the hosts are unsuccessful. What can be done to provide connectivity between the hosts?

1.A crossover cable should be used in place of the straight-through cable.

2.A rollover cable should be used in place of the straight-through cable.

3.The subnet masks should be set to 255.255.255.192.

4.A default gateway needs to be set on each host.

5.The subnet masks should be set to 255.255.255.0.

Answer: D

First, if you have two hosts directly connected, as shown in the graphic, then you need a crossover cable. A straight-through cable won't work. Second, the hosts have different masks, which puts them in different subnets. The easy solution is just to set both masks to 255.255.255.0 (/24).

Enter details here

8.

Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. Again, the zero subnet should not be considered valid for this question.

Answer: C

A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32, so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.

Enter details here

9.

You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts can have IP addresses on the LAN attached to the router interface?

Answer: A

A /29 (255.255.255.248), regardless of the class of address, has only 3 host bits. Six hosts is the maximum number of hosts on this LAN, including the router interface.

Enter details here

10.

Which two statements describe the IP address 10.16.3.65/23?

1.The subnet address is 10.16.3.0 255.255.254.0.

2.The lowest host address in the subnet is 10.16.2.1 255.255.254.0.

3.The last valid host address in the subnet is 10.16.2.254 255.255.254.0.

4.The broadcast address of the subnet is 10.16.3.255 255.255.254.0.

Answer: B

The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254.

Enter details here

Loading…
Tags: Subnetting Questions and Answers || Subnetting MCQ Questions and Answers || Subnetting GK Questions and Answers || Subnetting GK MCQ Questions || Subnetting Multiple Choice Questions and Answers || Subnetting GK || GK on Subnetting || Computer Networking Questions and Answers || Computer Networking MCQ Questions and Answers || Computer Networking GK Questions and Answers || GK on Computer Networking