Question 1
An organization requires a range of IP addresses to assign one to each of its 1500 computers. The organization has approached an Internet Service Provider (ISP) for this task. The ISP uses CIDR and serves the requests from the available IP address space 202.61.0.0/17. The ISP wants to assign an address space to the organization which will minimize the number of routing entries in the ISP’s router using route aggregation. Which of the following address spaces are potential candidates from which the ISP can allot any one to the organization?
I. 202.61.84.0/21
II. 202.61.104.0/21
III. 202.61.64.0/21
IV. 202.61.144.0/21
 A I and II only B III and IV only C II and III only D I and IV only
Question 1 Explanation:
Given CIDR IP is  202.61.0.0/17 and for HID 32 - 17 = 15 bits can be used.
And to Assign an IP address for 1500 computer, we require 11 bit from HID part.
So NID + SID = 17 + 4 = 21 bits and HID = 11 bits
NID     HID
202.61.0 0000 000.00000000
So,  from the given option, possible IP Address is
I. 84 - >0 1010 100 (BCZ in HID bit 1 is not possible)
II. 104 ->0 1101 000
III. 64 -> 0 1000 000
IV. 144->1 0010 000 (BCZ in NID bit 1 is not possible )
 Question 2
Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same net mask N. Which of the values of N given below should not be used if A and B should belong to the same network?
 A 255.255.255.0 B 255.255.255.128 C 255.255.255.192 D 255.255.255.224
Question 2 Explanation:
When we perform bitwise AND operation between IP Address and Subnet Mask, it gives Network ID. If for both IP results is same Network ID. It means, both IP are belong to the same network else it's on different network.
When we perform AND operation between IP address 10.105.1.113 and 255.255.255.224 result is 10.105.1.96 and when we perform AND operation between IP address 10.105.1.91 and 255.255.255.224 result is 10.105.1.64.
Therefore, 10.105.1.96 and 10.105.1.64 are different network, so D is correct answer.
 Question 3
If a class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?
 A 1022 B 1023 C 2046 D 2047
Question 3 Explanation:
255.255.248.0 can be written as 11111111.11111111.11111000.00000000
Number of bits assigned for host id is the number of zeros in subnet mask. Here 11 bits are used.
for host id so maximum possible hosts are= 211 -2=2046
 Question 4
Host X has IP address 192.168.1.97 and is connected through two routers R1 and R2 to an­other host Y with IP address 192.168.1.80. Router R1 has IP addresses 192.168.1.135 and 192.168.1.110. R2 has IP addresses 192.168.1.67 and 192.168.1.155. The netmask used in the network is 255.255.255.224. Given the information above, how many distinct subnets are guaranteed to already exist in the network?
 A 1 B 2 C 3 D 6
Question 4 Explanation:
Simply, Bitwise AND the bits of fourth octet for each of the following IP address with fourth octet of subnet mask. You will get only
XX.XX.XX.96, XX.XX.XX.64 and XX.XX.XX.128.
 Question 5
Host X has IP address 192.168.1.97 and is connected through two routers R1 and R2 to an­other host Y with IP address 192.168.1.80. Router R1 has IP addresses 192.168.1.135 and 192.168.1.110. R2 has IP addresses 192.168.1.67 and 192.168.1.155. The netmask used in the network is 255.255.255.224. Which IP address should X configure its gateway as?
 A 192.168.1.67 B 192.168.1.110 C 192.168.1.135 D 192.168.1.155
Question 5 Explanation:
X must be able to reach the gateway using the net mask.
Subnet no. of host X is,

Now, the gateway must also have the same subnet number.
Let's take IP 192.168.1.110 of R1,

and hence this can be used by X.
 Question 6
The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet?
 A 62 subnets and 262142 hosts. B 64 subnets and 262142 hosts. C 62 subnets and 1022 hosts. D 64 subnets and 1024 hosts.
Question 6 Explanation:
It is a class B address, so there 16-bits for NID and 16-bits for HID.
From HID, we took 6-bits for subnetting.
Then total subnets possible = ( 26 ) - 2 = 64
Total hosts possible for each subnet = (210) - 2 = 1022
 Question 7
Two computers C1 and C2 are configured as follows. C1 has IP address 203.197.2.53 and netmask 255.255.128.0. C2 has IP address 203.197.75.201 and netmask 255.255.192.0. Which one of the following statements is true?
 A C1 and C2 both assume they are on the same network B C2 assumes C1 is on same network, but C1 assumes C2 is on a different network C C1 assumes C2 is on same network, but C2 assumes C1 is on a different network D C1 and C2 both assume they are on different networks
Question 7 Explanation:
From C1 side,
Subnet mask for C1 is 255.255.128.0.
So it finds the Network ID as,
C1 → 203.197.2.53 AND 255.255.128.0 = 203.197.0.0
C2 → 203.197.75.201 AND 255.255.128.0 = 203.197.0.0
Both same.
From C2 side,
Subnet mask for C2 is 255.255.192.0.
So it finds the network ID as,
C1 → 203.197.2.53 AND 255.255.192.0 = 203.197.0.0
C2 → 203.197.75.201 AND 255.255.192.0 = 203.197.64.0
Both different.
Hence, option 'C' is correct.
 Question 8
A router uses the following routing table: A packet bearing a destination address 144.16.68.117 arrives at the router. On which interface will it be forwarded?
 A eth0 B eth1 C eth2 D eth3
Question 8 Explanation:
144.16.68.117 = 144.16.68.01110101 AND 255.255.255.224 = 255.255.255.11100000
= 144.16.68.96(Not matching with destination)
Now, take 255.255.255.0
144.16.68.117 AND 255.255.255.0
= 144.16.68.0 (matched)
Hence, option (C) is correct.
 Question 9
 A is necessarily 255.255.224.0 B is necessarily 255.255.240.0 C is necessarily 255.255.248.0 D could be any one of 255.255.224.0, 255.255.240.0, 255.255.248.0
Question 9 Explanation:
In the broadcast address for a subnet, all the host bits are set to 1. So as long as all the bits to the right are 1, bits left to it can be taken as possible subnet.
.95.255 or .01011111.11111111
(as in class B, 16 bits each are used for network and host)
So, we can take minimum 3 bits (from left) as subnet and make rest as host bits (as they are 1)
.224.0 → 11100000.00000000 (leftmost 3 bits for subnet)
.240.0 → 11110000.00000000 (leftmost 4 bits for subnet)
.248.0 → 11111000.00000000 (leftmost 5 bits for subnet)
 Question 10
An organization has a class B network and wishes to form subnets for 64 departments. The subnet mask would be:
 A 255.255.0.0 B 255.255.64.0 C 255.255.128.0 D 255.255.252.0
Question 10 Explanation:
Organization have 64 departments, and to assign 64 subnet we need 6 bits for subnet. In Class B network first two octet are reserved for NID, so we take first 6 bit of third octet for subnets and subnet mask would be 255.255.11111100.00000000 = 255.255.252.0
 Question 11

 A Eth1 and Eth2 B Eth0 and Eth2 C Eth0 and Eth3 D Eth1 and Eth3
Question 11 Explanation:
Router decides route for packet by ANDing subnet mask and IP address.
If results of ANDing subnet masks and IP address are same then subnet mask with higher number of 1s is preferred.
IP address 128.75.43.16 is AND with 255.255.255.0 results 128.75.43.0 Net ID which is similar to destination of this mask, but ANDing 128.75.43.16 with 255.255.255.128 also results same destination. So, here, mask with higher number of one is considered and router will forward packet to Eth1.
ANDing 192.12.17.10 with three subnet mask in table does not result in destination Net ID so router will forward this packet to default network via Eth2.
 Question 12
The subnet mask for a particular network is 255.255.31.0. Which of the following pairs of IP addresses could belong to this network?
 A 172.57.88.62 and 172.56.87.233 B 10.35.28.2 and 10.35.29.4 C 191.203.31.87 and 191.234.31.88 D 128.8.129.43 and 128.8.161.55