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A TestKing.com network administrator is adding host TestKing3 to the network shown in the exhibit. Which IP address can be assigned this host on this network?

A.192.1.1.14

B.192.1.1.18

C.192.1.1.20

D.192.1.1.30

E.192.1.1.31

F.192.1.1.36

Answer: B, D

Explanation:

Subnet Mask of 255.255.255.240 means 4-bits of subnetting. When we do 4-bits of subnetting, we have a total of 16 subnets having 16 hosts each. Subnets will be

192.1.1.0 ----- 191.1.1.15 (0-15)

192.1.1.16 ---- 191.1.1.31

(16-31)

192.1.1.32 ---- 191.1.1.47

(32-47)

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192.1.1.240----

192.1.1.255 (240-255)

Only choices B and D are possible as 192.1.1.20 is already used by host TestKing1

QUESTION NO: 12

A diagram depicting a TestKing user is shown below:

Based on the information above, which IP address should be assigned to the host?

A.192.168.5.5

B.192.168.5.32

C.192.168.5.40

D.192.168.5.63

E.192.168.5.75

F.None of the above

Answer: C Explanation:

Host address should be in same subnet of Connected Router's Interface. In exhibit Router's ethernet address is in 192.168.5.33/27 subnet then host address should be also in same subnet.

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27 bits used for network and 5 bits for host. So Network Address=256-224=32

First Subnet 32-64

So Host address should be between 32-64 but 32, 64, 63 can't e sued in a host address, as they are the network and broadcast addresses for the subnet, so only answer C is correct.

QUESTION NO: 13

TestKing is opening a new branch office. Assuming a subnet mask of 255.255.248.0, which three addresses are valid host IP addresses that could be used in this office? (Choose three.)

A.172.16.20.0

B.172.16.24.0

C.172.16.8.0

D.172.16.16.0

E.172.16.31.0

F.172.16.9.0

Answer: A, E, F

Explanation:

For the 255.255.248.0 subnet mask the following is true.

1.2-2=30 subnets

2.2-2=2,046 hosts per subnet

3.256-248=8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, etc.

4.Broadcast for the 8.0 subnet is 15.255. Broadcast for the 16.0 subnet is 23.255, etc.

5.The valid hosts are:

Subnet

8.0

16.0

24.0

32.0

40.0

48.0

56.0

64.0

 

 

 

 

 

 

 

 

 

first

8.1

16.1

24.1

32.1

40.1

48.1

56.1

64.1

host

 

 

 

 

 

 

 

 

last

15.254

23.254

31.254

39.254

47.254

55.254

63.254

71.254

host

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

15.255

23.255

31.255

39.255

47.255

55.255

63.255

71.255

broadcast

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Reference: http://articles.techrepublic.com.com/5100-6350-5033673.html

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QUESTION NO: 14

A small office TestKing network is shown below:

ipconfig exhibit:

The output shown above is from host TestKingA. What value should be displayed for the Default Gateway of the ipconfig output for this host?

A.172.18.14.6

B.192.168.1.11

C.192.168.1.10

D.192.168.1.254

E.192.168.1.250

F.172.18.14.5

G.None of the above

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Answer: D Explanation:

The default gateway setting, which creates the default route in the IP routing table, is a critical part of the configuration of a TCP/IP host. The role of the default gateway is to provide the next-hop IP address and interface for all destinations that are not located on its subnet. Without a default gateway, communication with remote destination is not possible, unless additional routes are added to the IP routing table. The default gateway must be the router's interface that is on the same IP subnet as the hosts. In this case it router TestKing2 is the default gateway router, and it's LAN interface with IP address 192.168.1.254 would be used.

QUESTION NO: 15

The TestKing network administrator has subnetted the 172.16.0.0 network using a subnet mask of 255.255.255.192. A duplicate IP address of 172.16.2.121 has accidentally been configured on workstation TK1 in this network. The technician must assign this workstation a new IP address within that same subnetwork. Which address should be assigned to TK1?

A.172.16.1.64

B.172.16.1.80

C.172.16.2.80

D.172.16.2.64

E.172.16.2.127

F.172.16.2.128

G.None of the above

Answer: C Explanation:

A subnet mask of 255.255.255.192 (/26) will provide us with 4 subnet (2 usable) each with 62 usable hosts per network. So in our example the four networks will be:

172.16.2.1-62

172.16.2.65-126

172.16.2.129-190 172.16.2.193-254

Since we know that the host must be in the same IP subnet as 172.16.2.120, only choice C is correct.

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Section 5: Calculate and apply an addressing scheme including VLSM IP addressing design to a network (20 questions)

QUESTION NO: 1 DRAG DROP

TestKing has three locations and has plans to redesign the network accordingly. The networking team received 192.168.151.0 to use as the addressing for entire network from the administrator. After subnetting the address, the team is ready to assign the address.

The administrator plans to configure ip subnet-zero and use RIP v2 as the routing protocol. As a member of the networking team, you must address the network and at the same time converse unused addresses for future growth.

Being mindful of these goals, drag the host addresses on the left to the correct router interface. One of the routers is partially configured. Move the mouse over a router to view its configuration (** This information is missing**). Not all of the host addresses on the left will be used.

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Answer:

Explanation:

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QUESTION NO: 2 DRAG DROP

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TestKing has three locations and has plans to redesign the network accordingly. The networking team received 192.168.151.0 to use as the addressing for entire network from the administrator. After subnetting the address, the team is ready to assign the address.

The administrator plans to configure ip subnet-zero and use RIP v2 as the routing protocol. As a member of the networking team, you must address the network and at the same time converse unused addresses for future growth.

Being mindful of these goals, drag the host addresses on the left to the correct router interface. One of the routers is partially configured. Move the mouse over a router to view its configuration (** This information is missing**). Not all of the host addresses on the left will be used.

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Answer:

Explanation:

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