submitting examples -dr. senan

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By Dr. Senan Alkaaby

Submitting Examples

5/5/2012 Dr. Senan alkaaby


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Question 1

Given a subnet mask of 255.255.255.224,which of the following addresses can be

assigned to network hosts? (Choose

three)A 15.234.118.63

B 92.11.178.93

C 134.178.18.56D 192.168.16.87

E 201.45.116.159

F 217.63.12.192



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Solution for example 1

Write the network and broadcast, thencompared the results.

+ Network addresses: 0, 32, 64, 96, 128,

160, 192, 224.+ Broadcast addresses: 31, 63, 95,

127,159, 191, 223.

Answer: B C D



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Question 2

Which of the following host addresses aremembers of networks that can be routed

across the public Internet? (Choose three)

A 10.172.13.65B 172.16.223.125

C 172.64.12.29

D 192.168.23.252E 198.234.12.95

F 212.193.48.254



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Solution Question 2

Answer: C E F Addresses that can be routed across the public Internet are called

public IP addresses. These addresses belong to class A, B or C only

and are not private addresses.

Note:

Private class A IP addresses: 10.0.0.0 to 10.255.255.255

Private class B IP addresses: 172.16.0.0 to 172.31.255.255

Private class C IP addresses: 192.168.0.0 to 192.168.255.255

Class D addresses are reserved for IP multicast addresses and

cant be routed across the Internet (their addresses begin with224.0.0.0 address).

Also we cant use 127.x.x.x address because the number 127 is

reserved for loopback and is used for internal testing on the local

machine.



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Question 3

ECE Department in UoD needs to design an IPaddressing scheme to support his lab network. The

Department 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 255.255.252.0D 255.255.255.224

E 255.255.255.192

F 255.255.248.0



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Solution Question 3



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Which of the following IP addresses fallinto the CIDR block of 115.64.4.0/22?

(Choose three)

A 115.64.8.32B 115.64.7.64

C 115.64.6.255

D 115.64.3.255E 115.64.5.128

F 115.64.12.128



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Solution Question 4

/ 22 means 252. So magic no. 256-252 = 4 .

So our Network Subnet will be 0,4,8,12

Our broadcast will be 3.255,7.255,11.255

Let compared with what we have in Q.,

A ,F both are Subletting Network, so we

cannot choose them, So D , is broadcast.

Correct answer is B,C,E



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Question 5

The network 172.25.0.0 has been divided into eightequal subnets. Which of the following IP addresses can

be assigned to hosts in the third subnet if the ip subnet-

zero command is configured on the router? (Choose

three)

A 172.25.78.243

B 172.25.98.16

C 172.25.72.0

D 172.25.94.255

E 172.25.96.17F. 172.25.100.16



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Solution Question 5

If the ip subnet-zero command is configured then the first subnet is172.25.0.0. Otherwise the first subnet will be 172.25.32.0).

The question stated that the network 172.25.0.0 is divided into eight equal

subnets therefore the increment is 256 / 8 = 32

OR 2N=8 =23 , So N = 3 , it means we have 3 of 1s .

its corresponding subnet mask is /19 (1111 1111.1111 1111.1110 0000). First subnet: 172.25.0.0/19

Second subnet: 172.25.32.0/19

Third subnet: 172.25.64.0/19

4th subnet: 172.25.96.0/19

5th subnet: 172.25.128.0/196th subnet: 172.25.160.0/19

7th subnet: 172.25.192.0/19

8th subnet: 172.25.224.0/19

Answer: A C D5/5/2012 Dr. Senan alkaaby

255 255 224

More

easy


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Question 6



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Solution Question 6

Router A receives 3 subnets: 172.16.64.0/18,172.16.32.0/24 and 172.16.128.0/18.

All these 3 subnets have the same form of

172.16.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.



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Question 7

Workstation A has been assigned an IP address of192.0.2.24/28. Workstation B has been assigned an IP

address of 192.0.2.100/28. 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 192.0.2.15.

E. Change the address of Workstation B to 192.0.2.111.5/5/2012 Dr. Senan alkaaby


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Solution Question 7

To specify when we use crossover cable or straight-through cable,we should remember:

Group 1: Router, Host, Server

Group 2: Hub, Switch

One device in group 1 + One device in group 2: use straight-through

cable

Two devices in the same group: use crossover cable

-> To connect two hosts we must use crossover cable -> A is

correct.

With the subnet mask of /28, 192.0.2.24 & 192.0.2.100 will be in

different subnets (192.0.2.24 belongs to subnet 192.0.2.16/28;

192.0.2.100 belongs to subnet 192.0.2.96). To make them in the

same subnet we need more space for host. Because 100 < 128 so

we the suitable subnet should be /25.



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Question 8

Your ISP has given you the address 223.5.14.6/29 toassign to your routers interface. They have also given

you the default gateway address of 223.5.14.7. 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 thissubnet.

D. The IP address is an invalid class D multicast

address.



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Solution Question 8

For the network 223.5.14.6/29: Increment: 8

Network address: 223.5.14.0

Broadcast address: 223.5.14.7 The default gateway IP address is the

broadcast address of this subnet -> B is

correct.

Answer: B



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Question 9



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Solution Question 9

Answer C Make a compare between the Packet

Destination IP send and the Network IP

to find the valid IP. Because /27 is connected to S1 (

SERIAL) , so thats why we cannot used it

, and we have used part C as the correctanswer since it is used e1 ( ethernet).



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Question 10

The network technician is planning to use the255.255.255.224 subnet mask on the network.

Which three valid IP addresses can the

technician use for the hosts? (Choose three)

A. 172.22.243.127B. 172.22.243.191

C. 172.22.243.190

D. 10.16.33.98

E. 10.17.64.34

F. 192.168.1.160



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Solution Question 10

From the subnet mask of 255.255.255.224we learn:

Increment: 32= ( 256-224 )

Network address: In the form ofx.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.5/5/2012 Dr Senan alkaaby

submitting examples -dr. senan

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