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Question 5411 – Data-Communication
April 30, 2024
Question 11661 – Queues-and-Stacks
April 30, 2024
Question 5411 – Data-Communication
April 30, 2024
Question 11661 – Queues-and-Stacks
April 30, 2024

Question 8430 – Subnetting

In the network 200.10.11.144/27, the fourth octet (in decimal) of the last IP address of the network which can be assigned to a host is ____________.

Correct Answer: A

Question 4 Explanation: 
No. of bit in HID part = 32-27 = 5 bits
Subnet mask is 255.255.255.224
Do AND with given IP and subnet mask then we get NID 200.10.11.128
In fourth octet first three bit will fixed for subnet and remaining 5 bits is for HID, so maximum value as 11111.
The address with all 1s in host part is broadcast address and can’t be assigned to a host.
So the maximum possible last octal in a host IP is 10011110 which is 158.
A
158
B
157
C
156
D
155
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