CCNA Subnetting Questions:

Started by Vijayb, April 02, 2014, 02:59:09 AM

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Vijayb

#15
Question 1

Workstation A has been assigned an IP address of 192.0.10.24/28. Workstation B has been assigned an IP address of 192.0.10.100/28. The two workstations are connected with a straight-through cable. Attempts to ping between the hosts are unsuccessful. What two things can be done to allow communications between the hosts? (Choose two)

A. Replace the straight-through cable with a crossover cable.
B. Change the subnet mask of the hosts to /25.
C. Change the subnet mask of the hosts to /26.
D. Change the address of Workstation A to 192.0.10.15.
E. Change the address of Workstation B to 192.0.10.111.

Answer: A B

Explanation

If you remember the last post:
Group 1: Router, Host, Server
Group 2: Hub, Switch



To connect two hosts of same group above, you use crossover cable : A is correct.
To properly understand subnetting; I suggest you need the subnet cram table.

Bit Value                             128         64          32          16         8          4             2          1

Bit Borrowed                          1          2            3            4          5           6            7          8

Usable host address           126        62          30         14          6           2

Subnet Mask                       128       192       224         240      248        252       255      256

Subnet Prefix/CIDR              /25       /26        /27          /28       /29         /30

With the subnet mask of /28, which mask is 255.255.255.240.

So 192.0.10.24 belongs to subnet 192.0.10.16/28 subnet which offers on 14 usable host IP addresses.

192.0.10.100 will be in different subnet 192.0.10.96 with mask 255.255.255.128/25). This will provide for 126 usable IP addresses.

Therefore both host are in a different subnet and requires a cross over cable to ping each other.


Vijayb

Question 2

You were given the address 223.6.14.6/29 by your ISP to assign to your router's interface. They have also given you the default gateway address of 223.6.14.7. After you have configured the address, the router is unable to ping any remote devices. What is preventing the router from pinging remote devices?

A. The default gateway is not an address on this subnet.
B. The default gateway is the broadcast address for this subnet.
C. The IP address is the broadcast address for this subnet.
D. The IP address is an invalid class D multicast address.

Answer: B

Explanation

Again use the Cram table:

Bit Value                          128         64          32            16         8            4              2         1

Bit Borrowed                    1             2            3              4          5            6              7          8

Usable host address      126         62           30            14         6            2      

Subnet Mask                  128        192         224           240     248        252           255        256

Subnet Prefix/CIDR         /25        /26          /27           /28       /29        /30

For the network 223.6.14.6/29 which mask will be 255.255.255.248. This offers only 6 usable host addresses.

So the Network address: 223.6.14.0

Usable host IP address: 223.6.14.1 - 6
Broadcast address: 223.6.14.7

      

Vijayb

Question 3

Refer to the exhibit below.
According to the routing table, where will the router send a packet destined for 10.1.6.65?

Network                       Interface                    Next-hop

10.1.1.0/24                   e0                            directly connected

10.1.2.0/24                   e1                            directly connected

10.1.3.0/25                   s0                            directly connected

10.1.4.0/24                   s1                            directly connected

10.1.6.0/24                   e0                              10.1.1.2

10.1.6.64/28                 e1                              10.1.2.2

10.1.6.64/29                 s0                              10.1.3.3

10.1.6.64/27                 s1                              10.1.4.4

A. 10.1.1.2
B. 10.1.2.2
C. 10.1.3.3
D. 10.1.4.4

Answer: C

Explanation

If you look at the table, The destination IP address 10.1.6.65 belongs to  10.1.6.64/28, 10.1.6.64/29 & 10.1.6.64/27 subnets. But the subnet with the longest prefix match will be chosen. In this case, the next hop configured on the router will be where the packet will be sent.

Vijayb

#18
Question 4

Refer to the exhibit. The user at Workstation B reports that Server A cannot be reached. What is preventing Workstation B from reaching Server A?



A. The IP address for Server A is a broadcast address.
B. The IP address for Workstation B is a subnet address.
C. The gateway for Workstation B is not on the same subnet.
D. The gateway for Server A is not on the same subnet.

Answer: D

Vijayb

Question 5

Given the address 192.168.10.19/28, which of the following are valid host addresses on this subnet? (Choose two)

A. 192.168.10.29
B. 192.168.10.16
C. 192.168.10.17
D. 192.168.10.31
E. 192.168.10.0

Answer: A C

Explanation

192.168.10.19/28 belongs to 192.168.10.16 network with mask of 255.255.255.240. This offers 14 usable ip address range from 192.168.10.17 – 30. Use the cram table above if you are confused.

Vijayb

Question 6

Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choose three)

A. 115.64.8.32
B. 115.64.7.64
C. 115.64.6.255
D. 115.64.3.255
E. 115.64.5.128
F. 115.64.12.128

Answer: B C E

To really understand this question and the provided answers, read How to Subnet Class B tutorial.

Vijayb

#21
Question 7

The Ethernet networks connected to router R1 in the graphic have been summarized for router R2 as 192.1.144.0/20. Which of the following packet destination addresses will R2 forward to R1, according to this summary? (Choose two)


A. 192.1.159.2
B. 192.1.160.11
C. 192.1.138.41
D. 192.1.151.254
E. 192.1.143.145
F. 192.1.1.144

Answer: A D

Vijayb

#22
Question 8

Refer to the exhibit. All of the routers in the network are configured with the ip subnet-zero command. Which network addresses should be used for Link A and Network A? (Choose two)



A. Network A – 172.17.3.48/26
B. Network A – 172.17.3.128/25
C. Network A – 172.17.3.192/26
D. Link A – 172.17.3.0/30
E. Link A – 172.17.3.40/30
F. Link A – 172.17.3.112/30

Answer: B D

Explanation

Network A connects 120 hosts. /25 will be the best because it offers 126 usable host IP addresses (follow the cram table above)

The ip subnet-zero command can be used for link A network 172.16.3.0/30. Other /30 IP addresses in the option above are a waste of ip addresses and not hierarchical in structure.

Vijayb

Question 9

Which two subnetworks would be included in the summarized address of 172.31.80.0/20? (Choose two)

A. 172.31.17.4/30
B. 172.31.51.16 /30
C. 172.31.64.0/18
D. 172.31.80.0/22
E. 172.31.92.0/22
F. 172.31.192.0/18

Answer: D E

Explanation

From the summarized address of 172.31.80.0/20, we find the range of this summarized network:

16 bits were borrowed
Network address: 172.31.80.0
Broadcast address: 172.31.95.255

Read more on IP address summarization

Vijayb

Question 10

Which three IP addresses can be assigned to hosts if the subnet mask is /27 and subnet zero is usable? (Choose three)

A. 10.15.32.17
B. 17.15.66.128
C. 66.55.128.1
D. 135.1.64.34
E. 129.33.192.192
F. 192.168.5.63

Answer: A C D

Explanation

If you follow the cram table,  /27 is mask 255.255.255.224. 3 bits were borrowed with value of 32. Follow the fourth octets of the addresses:

A is a usable host address.

B is classified as a network address

C is a usable host address form the subnet 0

D is a usable host address

E is a network address

F is a broadcast address

Vijayb

#25
Question 3

A national retail chain needs to design an IP addressing scheme to support a nationwide network. The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet. Working with only one Class B address, which of the following subnet masks will support an appropriate addressing scheme? (Choose two)

A – 255.255.255.0
B – 255.255.255.128
C – 255.255.252.0
D – 255.255.255.224
E – 255.255.255.192
F – 255.255.248.0


Answer: B E

Explanation

We need to remember the default subnet mask of class B is 255.255.0.0. Next, the company requires a minimum of 300 sub-networks so we have to use at least 512 sub-networks (because 512 is the minimum power of 2 and greater than 300). Therefore we need to get 9 bits for network mask (29=512), leaving 7 bits for hosts which is 27-2 = 126 > 50 hosts per subnet.This scheme satisfies the requirement -> B is correct.




We can increase the sub-networks to 1024 ( 1024 = 210), leaving 6 bits for hosts that is 26= 64 > 50 hosts. This scheme satisfies the requirement, too -> E is correct.



Notice: The question asks "The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet" but this is a typo, you should understand it as ""The company needs a minimum of 300 sub-networks and a minimum of 50 host addresses per subnet".