CCNA – Subnetting Questions 4 Flashcards
Question 1
You are working in a data center environment and are assigned the address range 10.188.31.0/23. You are asked to develop an IP addressing plan to allow the maximum number of subnets with as many as 30 hosts each.Which IP address range meets these requirements?
A. 10.188.31.0/27
B. 10.188.31.0/26
C. 10.188.31.0/29
D. 10.188.31.0/28
E. 10.188.31.0/25
Answer: A
Explanation
Each subnet has 30 hosts < 32 = 25 so we need a subnet mask which has at least 5 bit 0s -> /27. Also the question requires the maximum number of subnets (which minimum the number of hosts-per-subnet) so /27 is the best choice -> A is correct.
Question 2
Refer to the exhibit. The Lakeside Company has the internetwork in the exhibit. The Administrator would like to reduce the size of the routing table to the Central Router. Which partial routing table entry in the Central router represents a route summary that represents the LANs in Phoenix but no additional subnets?
A. 10.0.0.0 /22 is subnetted, 1 subnet
D 10.0.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
B. 10.0.0.0 /28 is subnetted, 1 subnet
D 10.2.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
C. 10.0.0.0 /30 is subnetted, 1 subnet
D 10.2.2.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
D. 10.0.0.0 /22 is subnetted, 1 subnet
D 10.4.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
E. 10.0.0.0 /28 is subnetted, 1 subnet
D 10.4.4.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
F. 10.0.0.0 /30 is subnetted, 1 subnet
D 10.4.4.4 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
Answer: D
Explanation
All the above networks can be summarized to 10.0.0.0 network but the question requires to “represent the LANs in Phoenix but no additional subnets” so we must summarized to 10.4.0.0 network. The Phoenix router has 4 subnets so we need to “move left” 2 bits of “/24″-> /22 is the best choice -> D is correct.
Question 3
Which address range efficiently summarizes the routing table of the addresses for router main?
A. 172.16.0.0/18
B. 172.16.0.0/16
C. 172.16.0.0/20
D. 172.16.0.0/21
Answer: C
Explanation
To summarize these networks efficiently we need to find out a network that “covers” from 172.16.1.0 -> 172.16.13.0 (including 13 networks < 16). So we need to use 4 bits (24 = 16). Notice that we have to move the borrowed bits to the left (not right) because we are summarizing.
The network 172.16.0.0 belongs to class B with a default subnet mask of /16 but in this case it has been subnetted with a subnet mask of /24 (we can guess because 172.16.1.0, 172.16.2.0, 172.16.3.0… are different networks).
Therefore “move 4 bits to the left” of “/24″ will give us “/20″ -> C is the correct answer.
Question 4
Refer to the exhibit. A new subnet with 60 hosts has been added to the network. Which subnet address should this network use to provide enough usable addresses while wasting the fewest addresses?
A. 192.168.1.56/27
B. 192.168.1.64/26
C. 192.168.1.64/27
D. 192.168.1.56/26
Answer: B
Explanation
60 hosts < 64 = 26 -> we need a subnet mask of at least 6 bit 0s -> “/26″. The question requires “wasting the fewest addresses” which means we have to allow only 62 hosts-per-subnet -> B is correct.
Question 5
The network technician is planning to use the 255.255.255.224 subnet mask on the network. Which three valid IP addresses can the technician use for the hosts? (Choose three)
A. 172.22.243.127
B. 172.22.243.191
C. 172.22.243.190
D. 10.16.33.98
E. 10.17.64.34
F. 192.168.1.160
Answer: C D E
Explanation
From the subnet mask of 255.255.255.224 we learn:
Increment: 32
Network address: In the form of x.x.x.(0,32, 64, 96, 128, 160, 192, 224)
Broadcast address: In the form of x.x.x.(31,63,95,127,159,191,223)
-> All IP addresses not in the above forms are usable for host -> C D E are correct answers.
Question 6
In the implementation of VLSM techniques on a network using a single Class C IP address, which subnet mask is the most efficient for point-to-point serial links?
A. 255.255.255.240
B. 255.255.255.254
C. 255.255.255.252
D. 255.255.255.0
E. 255.255.255.248
Answer: C
Explanation
The subnet mask of 255.255.255.252 gives only 2 usable host addresses because it has only 2 bit 0s (22 – 2 = 2) so it is the most efficient subnet mask for point-to-point serial links (and you should remember it).
Question 7
Refer to the exhibit. HostA cannot ping HostB. Assuming routing is properly configured, what could be the cause of this problem?
A. HostA is not on the same subnet as its default gateway.
B. The address of SwitchA is a subnet address.
C. The Fa0/0 interface on RouterA is on a subnet that can’t be used.
D. The serial interfaces of the routers are not on the same subnet.
E. The Fa0/0 interface on RouterB is using a broadcast address.
Answer: D
Explanation
Now let’s find out the range of the networks on serial link:
For the network 192.168.1.62/27:
Increment: 32
Network address: 192.168.1.32
Broadcast address: 192.168.1.63
For the network 192.168.1.65/27:
Increment: 32
Network address: 192.168.1.64
Broadcast address: 192.168.1.95
-> These two IP addresses don’t belong to the same network and they can’t see each other -> D is the correct answer.
Question 8
The network administrator is asked to configure 113 point-to-point links. Which IP addressing scheme best defines the address range and subnet mask that meet the requirement and waste the fewest subnet and host addresses?
A. 10.10.0.0/18 subnetted with mask 255.255.255.252
B. 10.10.0.0/25 subnetted with mask 255.255.255.252
C. 10.10.0.0/24 subnetted with mask 255.255.255.252
D. 10.10.0.0/23 subnetted with mask 255.255.255.252
E. 10.10.0.0/16 subnetted with mask 255.255.255.252
Answer: D
Explanation
We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7 bits (because 2^7 = 128).
The network used for point-to-point connection should be /30.
So our initial network should be 30 – 7 = 23.
So 10.10.0.0/23 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 9
If an Ethernet port on a router was assigned an IP address of 172.1.1.1/20, what is the maximum number of hosts allowed on this subnet?
A. 4094
B. 1024
C. 8190
D. 2046
E. 4096
Answer: A
Explanation
In the prefix /20 we have 12 bit 0s so the number of hosts-per-subnet is 212 – 2 = 4094.
Question 10
A network administrator receives an error message while trying to configure the Ethernet interface of a router with IP address 10.24.24.24/29. Which statement explains the reason for it?
A. The address is a broadcast address
B. The Ethernet interface is faulty
C. VLSM-capable routing protocols must be enable first on the router.
D. This address is a network address.
