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Subnetting Questions and Answers

11.

In classful addressing, an IP address 123.23.156.4 belongs to ______ class format.

Exam Year(s)

UGC NET - 2012,

Answer: A

In classful addressing the class is decided by the first octet. The range 1 to 126 is class A, 128 to 191 class B, 192 to 223 class C and 224 to 239 class D. The address 123.23.156.4 has first octet 123, which lies in 1 to 126, so it belongs to class A. It is not class B, which would need a first octet of 128 or more, nor class C or D. The keyed answer is A.

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12.

In IPV 4, the IP address 200.200.200.200 belongs to

Exam Year(s)

UGC NET - 2013,

Answer: C

The address 200.200.200.200 belongs to Class C. In IPv4 the class is decided by the first octet: 0 to 127 is Class A, 128 to 191 is Class B, 192 to 223 is Class C, and 224 to 239 is Class D (multicast). The first octet here is 200, which lies in the Class C range.

It is not Class A or B because 200 is above 191, and not Class D because 200 is below 224.

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13.

In classful addressing, the IP address 190.255.254.254 belongs to

Exam Year(s)

UGC NET - 2013,

Answer: B

The address 190.255.254.254 belongs to Class B, option (B). In classful addressing the class is read from the first octet: 0 to 127 is Class A, 128 to 191 is Class B, 192 to 223 is Class C, 224 to 239 is Class D (multicast) and 240 to 255 is Class E. The first octet here is 190, which falls in 128 to 191, and in binary it begins with the bits 10. Class A would need the first octet below 128, Class C at least 192 and Class D at least 224, so (A), (C) and (D) are wrong.

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14.

Which of the following is the size of Network bits and Host bits of Class A of IP address ?

Exam Year(s)

UGC NET - 2013,

Answer: A

A Class A address has 7 network bits and 24 host bits. The first bit of the first octet is 0, which leaves 7 bits for the network number in the first octet, and the other three octets, 24 bits, are for the host.

Fourteen network bits and 16 host bits is the split of Class B, which starts with the bits 10. Fifteen and sixteen, and sixteen and sixteen, are not splits of any address class, so the other options are wrong.

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15.

Given the IP address 201.14.78.65 and the subnet mask 255.255.255.224. What is the subnet address ?

Exam Year(s)

UGC NET - 2013,

Answer: B

The subnet address is 201.14.78.64. It is obtained by a bitwise AND of the address and the mask. Only the last octet is affected, since the first three octets of the mask are 255.

\[ 65 = 01000001_{2} \]

\[ 224 = 11100000_{2} \]

\[ 01000001 \wedge 11100000 = 01000000 = 64 \]

So the subnet address is 201.14.78.64. The value 201.14.78.32 would need a last octet in the range 32 to 63, and .65 and .224 are the host address and the mask, which are not the network address.

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16.

A network on the Internet has a subnet mask of 255.255.240.0. What is the maximum number of hosts it can handle ?

Exam Year(s)

UGC NET - 2014,

Answer: C

The network can handle 4096 hosts.

The mask 255.255.240.0 has \(8 + 8 + 4 = 20\) network bits, so 12 bits remain for hosts and \(2^{12} = 4096\) addresses are available.

1024, 2048 and 8192 correspond to 10, 11 and 13 host bits. The count of usable hosts is often given as 4094 because the all-zero and all-ones host addresses are reserved, but 4096 is the number of addresses in the block and is the only option near that value.

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17.

Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host IDs on the LAN interface attached to the router?

1.172.16.1.100

2.172.16.1.198

3.172.16.2.255

4.172.16.3.0

Answer: C

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.

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18.

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.

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19.

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).

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20.

What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?

Answer: D

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.

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