Subnetting

Question 1
A
i-a, ii-c, iii-e, iv-d
B
i-a, ii-d, iii-b, iv-e
C
i-b, ii-c, iii-d, iv-e
D
i-b, ii-c, iii-e, iv-d
       Computer-Networks       Subnetting       GATE 2015 -(Set-2)
Question 1 Explanation: 
Do the AND operation of group I IP address with Netmask and compare the result with network number. if it match with network number, forward packet through that interface. Ex: 128.96.167.151 AND 255.255.254.0 = 128.96.166.0 Therefore packet will forward through R2
Question 2
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 ____________.
A
158
B
157
C
156
D
155
       Computer-Networks       Subnetting       GATE 2015(Set-03)
Question 2 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.
There are 2 questions to complete.