In this section you can learn and practice Networking Questions based on "Subnetting" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank Exam, Railway Exam etc.) with full confidence.
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Where can I get Networking Subnetting Interview Questions and Answers (objective type, multiple choice)?
Here you can find objective type Networking Subnetting questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided.
How to solve Networking Subnetting problems?
You can easily solve all kind of Networking questions based on Subnetting by practicing the objective type exercises given below, also get shortcut methods to solve Networking Subnetting problems.
The router's IP address on the E0 interface is 172.16.2.1/23, which is 255.255.254.0. This makes the third octet a block size of 2. The router's interface is in the 2.0 subnet, and the broadcast address is 3.255 because the next subnet is 4.0. The valid host range is 2.1 through 3.254. The router is using the first valid host address in the range.
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.
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?
A crossover cable should be used in place of the straight-through cable.
A rollover cable should be used in place of the straight-through cable.
The subnet masks should be set to 255.255.255.192.
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
A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets, each with 30 hosts. Does it matter if this mask is used with a Class A, B, or C network address? Not at all. The number of host bits would never change.
You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts-this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts. This is the best answer.