Question

Please explain answer and show steps

Which two statements describe the IP address 10.16.3.65/22? (Choose two)

1. The subnet address is 10.16.3.0 255.255.252.0.
2. The lowest host address in the subnet is 10.16.0.1 255.255.252.0.
3. The last valid host address in the subnet is 10.16.3.255 255.255.252.0
4. The broadcast address of the subnet is 10.16.3.255 255.255.252.0.
5. The network is not subnetted.

Which IP addresses are valid for hosts belonging to the 10.1.160.0/20 subnet? (Choose three)

1. 10.1.169.0
2. 10.1.176.1
3. 10.1.175.254
4. 10.1.160.0
5. 10.1.175.255
6. 10.1.160.255

Which one of the following IP addresses is the last valid host in the subnet using mask 255.255.255.192?

1. 192.168.2.60
2. 192.168.2.61
3. 192.168.2.62
4. 192.168.2.63
5. 192.168.2.32

An administrator is working with the 192.168.4.0 network, which has been subnetted with a /27 mask. Which two addresses can be assigned to hosts within the same subnet? (Choose two)

1. 192.168.4.30
2. 192.168.4.31
3. 192.168.4.32
4. 192.168.4.33

Answer 1

Solution 1:

b. The lowest host address in the subnet is 10.16.0.1 255.255.252.0.

d. The broadcast address of the subnet is 10.16.3.255 255.255.252.0.

Explanation:

The mask 255.255.254.0 (/22) 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.

Solution 2:

a. 10.1.169.0

c. 10.1.175.254

f. 10.1.160.255

Explanation:

From the /20 we can find all information we need:

Increment: 16 (/20 = 11111111.11111111.11110000.00000000). This is applied for the 3rd octet.
Network address: 10.1.160.0 (because 160 = 16 * 10 and 160 = 160 -> the IP address above is also the network address.
Broadcast address: 10.1.175.255 (because 175 = 160 + 16 - 1)

Therefore only 10.1.169.0, 10.1.175.254 and 10.1.160.255 are in this range.

Solution 3:

c. 192.168.2.62

Explanation:

Using a subnet mask of 255.255.255.192, your 192.168.123.0 network then becomes the four networks 192.168.123.0, 192.168.123.64, 192.168.123.128 and 192.168.123.192. These four networks would have as valid host addresses:

192.168.123.1-62

192.168.123.65-126

192.168.123.129-190

192.168.123.193-254

Solution 4:

a. 192.168.4.30

b. 192.168.4.31

Explanation:

Class C

Address: 192.168.4.0 11000000.10101000.00000100.00000000

Netmask: 255.255.255.224 = 27 11111111.11111111.11111111.11100000

Wildcard: 0.0.0.31 00000000.00000000.00000000.00011111

Network: 192.168.4.0/27 11000000.10101000.00000100.00000000

Broadcast: 192.168.4.31 11000000.10101000.00000100.00011111

HostMin: 192.168.4.1 11000000.10101000.00000100.00000001

HostMax: 192.168.4.30 11000000.10101000.00000100.00011110

Hosts/Net: 30 (Private Internet)

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