Subnetting Questions 1
Question 1
A router is connecting with a LAN department with 299 hosts. What subnet mask is best to cover all hosts in one subnet while leaving least amount of wasted hosts?
A. 255.255.255.0
B. 255.255.254.0
C. 255.255.252.0
D. 255.255.248.0
Answer: B
Explanation
299 hosts < 512 = 29 -> We need a subnet mask of 9 bits 0 -> 1111 1111.1111 1111.1111 1110.0000 0000 -> 255.255.254.0
Question 2
Check the topology below. Suppose “ip subnet-zero” command was issued on all routers. What subnets should you configure for the connection A and subnet B?
A. B – subnet 192.168.1.48/26
B. B – subnet 192.168.1.128/25
C. B – subnet 192.168.1.192/26
D. A – 192.168.1.0/30
E. A – 192.168.1.40/30
F. A – 192.168.1.112/30
Answer: B D
Explanation
Subnet B needs 100 hosts < 128 = 27 -> Need a subnet mask of 7 bit 0s -> “/25″.
Because the ip subnet-zero command is used, network 192.168.1.0/30 can be used.
Answer E “A – 192.168.1.40/30″ is not correct because this subnet belongs to R3 subnet (192.168.1.32/27).
Answer F “A – 192.168.1.112/30″ is not correct because this subnet belongs to R5 subnet (192.168.1.96/27).
Question 3
Which network and subnet mask allow three hosts to be on the same subnet while keeping the maximum number of subnets?
A. 192.168.12.0 255.255.255.252
B. 192.168.12.8 255.255.255.248
C. 192.168.12.8 255.255.255.240
D. 192.168.12.0 255.255.255.240
E. 192.168.12.32 255.255.255.224
Answer: B
Explanation
First we should observe the answers, they are all 192.168.12.x so they are in class C with a default subnet mask of 255.255.255.0. We need three web servers (three hosts) in a subnet so we need 3 bits 0 (using the formula 23 – 2 = 6 > 5 hosts) so the last octet of our subnet mask should be 1111 1000 -> 248.
Question 4
Which is the suitable subnet mask for a network to be subnetted into 8 subnets, each contains up to 28 hosts?
A. 0.0.0.224
B. 255.255.255.248
C. 255.255.255.240
D. 255.255.255.224
E. 255.255.255.128
Answer: D
Explanation
A is not correct because it is a wildcard mask (not subnet mask).
This question is a bit unclear but we can suppose we have to begin with default subnet mask and “subnet” it. And the default subnet mask here should be class C: 255.255.255.0
For answer B: 248 = 1111 1000 -> with this subnet mask we can subnet up to 25 = 32 subnets but only 23 – 2 = 6 hosts per subnet -> B is not correct.
For answer C: 240 = 1111 0000 -> with this subnet mask we can subnet up to 24 = 16 subnets but only 24 – 2 = 14 hosts per subnet -> C is not correct.
For answer E: 240 = 1111 0000 -> There are 24 = 16 subnets but only 24 – 2 = 14 hosts per subnet < 28 hosts -> E does not satisfy the second requirement (of 28 hosts per subnet).
For answer D: 224 = 1110 0000 -> There are 23 = 8 subnets and 25 – 2 = 30 hosts > 28 hosts -> This is the correct answer.
Question 5
Your company is using the network 192.168.10.8/29. You are required to manually assign the first usable IP for the router and the last usable IP for the FTP server. What will you enter into the IP box of the FTP server?
A.
IP: 192.168.10.14
Subnet mask: 255.255.255.248
Default Gateway: 192.168.10.10
B.
IP: 192.168.10.13
Subnet mask: 255.255.255.248
Default Gateway: 192.168.10.10
C.
IP: 192.168.10.14
Subnet mask: 255.255.255.248
Default Gateway: 192.168.10.9
D. IP: 192.168.10.14
Subnet mask: 255.255.255.240
Default Gateway: 192.168.10.10
E. IP: 192.168.10.15
Subnet mask: 255.255.255.240
Default Gateway: 192.168.10.10
F. IP: 192.168.10.15
Subnet mask: 255.255.255.248
Default Gateway: 192.168.10.9
Answer: A
Question 6
Check the topology below. After receiving all the subnets, what if the summarized subnet that R1 should send to R2?
A. 10.148.0.0/16
B. 10.148.0.0/20
C. 10.148.0.0/24
D. 10.32.0.0/16
E. 10.32.0.0/17
F. 10.64.0.0/16
Answer: A
Explanation
Router A receives 3 subnets: 10.148.64.0/18, 10.148.32.0/24 and 10.148.128.0/18.
All these 3 subnets have the same form of 10.148.x.x so our summarized subnet must be also in that form -> Only A, B or C is correct.
The smallest subnet mask of these 3 subnets is /18 so our summarized subnet must also have its subnet mask equal or smaller than /18.
-> Only answer A has these 2 conditions -> A is correct.
Question 7
Your company has just been assigned an address range 14.1.6.0/23. You are required to divide this range into the maximum number of subnets but each subnet can have as many as 30 hosts. What subnet mask you will suggest to them?
A. 14.1.6.0/25
B. 14.1.6.0/26
C. 14.1.6.0/27
D. 14.1.6.0/28
E. 14.1.6.0/29
Answer: C
Explanation
Each subnet has 30 hosts < 32 = 25 so we need a subnet mask which has at least 5 bits 0 -> /27. Also the question requires the maximum number of subnets (which minimum the number of hosts-per-subnet) so /27 is the best choice -> C is correct.
Question 8
What advantages does a hierarchical addressing network provide?
A. Reduces MAC address table size
B. Increase the speed of transmission
C. Reduces routing table size
D. The management and troubleshooting are simple
Answer: C D
Question 9
You are working in an Internet Service Provider (ISP) company. You are tasked to connect to 100 customers on point-to-point connections. Which initial IP and subnet mask you can use to complete this task while saving subnet and host addresses for future use?
A. 16.1.0.0/20 and subnetted with subnet mask of /30
B. 16.1.0.0/21 and subnetted with subnet mask of /30
C. 16.1.0.0/22 and subnetted with subnet mask of /30
D. 16.1.0.0/23 and subnetted with subnet mask of /30
E. 16.1.0.0/24 and subnetted with subnet mask of /30
Answer: D
Explanation
We need 100 point-to-point links which equal to 100 sub-networks < 128 so we need to borrow 7 bits (because 27 = 128).
The network used for point-to-point connection should be /30.
So our initial network should be 30 – 7 = 23.
So 16.1.0.0/23 is the correct answer.
You can understand it more clearly when writing it in binary form:
/23 = 1111 1111.1111 1110.0000 0000
/30 = 1111 1111.1111 1111.1111 1100 (borrow 7 bits)
Question 10
Find the network address of the subnet with an IP of 130.16.17.252 and subnet mask 255.255.240.0.
A. 130.16.16.0
B. 130.16.0.1
C. 130.16.32.0
D. 130.16.32.255
E. 130.16.64.0
Answer: A
Explanation
Increment: 16 (of the third octet)
Network address: 130.16.16.0
-> A is correct.