Cisco CCNA 1 Routing and Switching : Subnetting IP Networks Chapter 8 Quiz 8

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Cisco CCNA 1 Routing and Switching : Subnetting IP Networks Chapter 8 Quiz 8

 

Problem 1

Question :

Fill in the blank.
The last host address on the 10.15.25.0/24 network is ______________________ .

Correct Answer :

10.15.25.254
.254
254
10.15.25.254/24

Explanation :

Refer to curriculum topic: 7.1.2
The host portion of the last host address will contain all 1 bits with a 0 bit for the lowest order or rightmost bit. This address is always one less than the broadcast address. The range of addresses for the network 10.15.25.0/24 is 10.15.25.0 (network address) through 10.15.25.255 (broadcast address). So the last host address for this network is 10.15.25.254.

Problem 2

Question :

The network portion of the address 172.16.30.5/16 is _______________________.

Correct Answer :

172.16
172.16
172 .16
172.16
172 .16
172. 16
172. 16
172.16
172.16
172.16
172.16
172 .16
172 .16
172. 16
172. 16
172 .16
172 .16
172. 16
172. 16
172 . 16
172 . 16

Explanation :

Refer to curriculum topic: 7.1.2
A prefix of /16 means that 16 bits are used for the network part of the address. The network portion of the address is therefore 172.16.

 

Problem 3

Question :

What does the IP address 192.168.1.15/29 represent?

Answer Option Choices :

subnetwork address

multicast address

unicast address
broadcast address
Correct Answer :
broadcast address

Explanation :

Refer to curriculum topic: 8.1.2
A broadcast address is the last address of any given network. This address cannot be assigned to a host, and it is used to communicate with all hosts on that network.

 

Problem 4

Question :

A network engineer is subnetting the 10.0.240.0/20 network into smaller subnets. Each new subnet will contain between a minimum of 20 hosts and a maximum of 30 hosts. Which subnet mask will meet these requirements?

Answer Option Choices :

255.255.224.0
255.255.240.0
255.255.255.224
255.255.255.240
Correct Answer :
255.255.255.224

Explanation :

Refer to curriculum topic: 8.1.4
For each new subnet to contain between 20 and 30 hosts, 5 host bits are required. When 5 host bits are being used, 27 network bits are remaining. A /27 prefix provides the subnet mask of 255.255.255.224.

Problem 5

Question :

Three devices are on three different subnets. Match the network address and the broadcast address with each subnet where these devices are located. (Not all options are used.)

Device 1: IP address 192.168.10.77/28 on subnet 1

Device 2: IP address192.168.10.17/30 on subnet 2

Device 3: IP address 192.168.10.35/29 on subnet 3

Answer Option Choices :

Subnet 1 network number
Subnet 1 broadcast address
Subnet 2 network number
Subnet 2 broadcast address
Subnet 3 network number
Subnet 3 broadcast address
192.168.10.64
192.168.10.79
192.168.10.16
192.168.10.19
192.168.10.47
192.168.10.95
192.168.10.48
192.168.10.255
192.168.10.0
Correct Answer :
Subnet 1 network number – 192.168.10.64
Subnet 1 broadcast address – 192.168.10.79
Subnet 2 network number – 192.168.10.16
Subnet 2 broadcast address – 192.168.10.19
Subnet 3 network number – 192.168.10.32
Subnet 3 broadcast address – 192.168.10.39

Other Incorrect Match Options:

  • 192.168.10.47
  • 192.168.10.95
  • 192.168.10.48
  • 192.168.10.255
  • 192.168.10.0

Explanation :

Refer to curriculum topic: 8.1.5
To calculate any of these addresses, write the device IP address in binary. Draw a line showing where the subnet mask 1s end. For example, with Device 1, the final octet (77) is 01001101. The line would be drawn between the 0100 and the 1101 because the subnet mask is /28. Change all the bits to the right of the line to 0s to determine the network number (01000000 or 64). Change all the bits to the right of the line to 1s to determine the broadcast address (01001111 or 79).

Problem 6

Question :

An administrator wants to create four subnetworks from the network address 192.168.1.0/24. What is the network address and subnet mask of the second useable subnet?

Answer Option Choices :

subnetwork 192.168.1.64
subnet mask 255.255.255.192
subnetwork 192.168.1.32
subnet mask 255.255.255.240
subnetwork 192.168.1.64
subnet mask 255.255.255.240
subnetwork 192.168.1.128
subnet mask 255.255.255.192
subnetwork 192.168.1.8
subnet mask 255.255.255.224
Correct Answer :
subnetwork 192.168.1.64
subnet mask 255.255.255.192

Explanation :

Refer to curriculum topic: 8.1.4
The number of bits that are borrowed would be two, thus giving a total of 4 useable subnets:
192.168.1.0
192.168.1.64
192.168.1.128
192.168.1.192
Because 2 bits are borrowed, the new subnet mask would be /26 or 255.255.255.192

 

Problem 7

Question :

Three methods allow IPv6 and IPv4 to co-exist. Match each method with its description. (Not all options are used.)

Answer Option Choices :

The IPv4 packets and IPv6 packets coexist in the same network.
The IPv6 packet is transported inside an IPv4 packet.
IPv6 packets are converted into IPv4 packets, and vice versa.
dual-stack
tunneling
translation
DHCP
Correct Answer :
The IPv4 packets and IPv6 packets coexist in the same network. – dual-stack
The IPv6 packet is transported inside an IPv4 packet. – tunneling
IPv6 packets are converted into IPv4 packets, and vice versa. – translation

Other Incorrect Match Options:DHCP

Explanation :

Refer to curriculum topic: 7.2.1
The term for the method that allows for the coexistence of the two types of packets on a single network is dual-stack. Tunneling allows for the IPv6 packet to be transported inside IPv4 packets. An IP packet can also be converted from version 6 to version 4 and vice versa. DHCP is a protocol that is used for allocating network parameters to hosts on an IP network.

 

Problem 8

Question :

How many host addresses are available on the 192.168.10.128/26 network?

Answer Option Choices :

30
32
60

62

64
Correct Answer :
62

Explanation :

Refer to curriculum topic: 8.1.2
A /26 prefix gives 6 host bits, which provides a total of 64 addresses, because 26 = 64. Subtracting the network and broadcast addresses leaves 62 usable host addresses.

Problem 9

Question :

A network administrator has received the IPv6 prefix 2001:DB8::/48 for subnetting. Assuming the administrator does not subnet into the interface ID portion of the address space, how many subnets can the administrator create from the /48 prefix?

Answer Option Choices :

16
256
4096
65536
Correct Answer :
65536

Explanation :

Refer to curriculum topic: 8.3.1
With a network prefix of 48, there will be 16 bits available for subnetting because the interface ID starts at bit 64. Sixteen bits will yield 65536 subnets.

Problem 10

Question :

What is the subnet address for the IPv6 address 2001:D12:AA04:B5::1/64?

Answer Option Choices :

2001::/64
2001:D12::/64​
2001:D12:AA04::/64​
2001:D12:AA04:B5::/64​
Correct Answer :
2001:D12:AA04:B5::/64​

Explanation :

Refer to curriculum topic: 8.3.1
The /64 represents the network and subnet IPv6 fields which are the first four groups of hexadecimal digits. The first address within that range is the subnetwork address of 2001: D12:AA04:B5::/64.​

Problem 11

Question :

Cisco CCNA 1 Routing and Switching : Subnetting IP Networks Chapter 8 Quiz 8
Cisco CCNA 1 Routing and Switching : Subnetting IP Networks Chapter 8 Quiz 8

 

 

 

 

 

 

 

 

 

 

 

Refer to the exhibit. Which two network addresses can be assigned to the network containing 10 hosts? Your answers should waste the fewest addresses, not reuse addresses that are already assigned, and stay within the 10.18.10.0/24 range of addresses. (Choose two.)

Answer Option Choices :

10.18.10.200/28
10.18.10.208/28
10.18.10.224/28
10.18.10.200/27
10.18.10.224/27
10.18.10.240/27
Correct Answer :
10.18.10.208/28
10.18.10.224/28

Explanation :

Refer to curriculum topic: 8.1.5
Addresses 10.18.10.0 through 10.18.10.63 are taken for the leftmost network. Addresses 192 through 199 are used by the center network. Because 4 host bits are needed to accommodate 10 hosts, a /28 mask is needed. 10.18.10.200/28 is not a valid network number. Two subnets that can be used are 10.18.10.208/28 and 10.18.10.224/28.

 

Problem 12

Question :

A college has five campuses. Each campus has IP phones installed. Each campus has an assigned IP address range. For example, one campus has IP addresses that start with 10.1.x.x. On another campus the address range is 10.2.x.x. The college has standardized that IP phones are assigned IP addresses that have the number 4X in the third octet. For example, at one campus the address ranges used with phones include 10.1.40.x, 10.1.41.x, 10.1.42.x, etc. Which two groupings were used to create this IP addressing scheme? (Choose two.)

Answer Option Choices :

geographic location
device type
department
personnel type
support model
Correct Answer :
geographic location
device type

Explanation :

Refer to curriculum topic: 8.1.1
The IP address design being used is by geographic location (for example, one campus is 10.1, another campus 10.2, another campus 10.3). The other design criterion is that the next octet number designates IP phones, or a specific device type, with numbers starting with 4, but which can include other numbers. Other devices that might get a designation inside this octet could be printers, PCs, and access points.

 

Problem 13

Question :

What are two benefits of subnetting networks? (Choose two.)

Answer Option Choices :

combining multiple smaller networks into larger networks
reducing the size of broadcast domains
decreasing the number of broadcast domains
grouping devices to improve management and security
increasing the size of collision domains
Correct Answer :
reducing the size of broadcast domains
grouping devices to improve management and security

Explanation :

Refer to curriculum topic: 8.1.1
When a single network is subnetted into multiple networks the following occurs:

  • A new broadcast domain is created for every network that is created through subnetting.
  • The amount of network traffic that crosses the entire network decreases.
  • Devices can be grouped together to improve network management and security.
  • More IP addresses are usable because each network will have a network address and broadcast address.

Problem 14

Question :

A network administrator subnets the 192.168.10.0/24 network into subnets with /26 masks. How many equal-sized subnets are created?

Answer Option Choices :

1
2
4
8
16
64
Correct Answer :
4

Explanation :

Refer to curriculum topic: 8.1.2
The normal mask for 192.168.10.0 is /24. A /26 mask indicates 2 bits have been borrowed for subnetting. With 2 bits, four subnets of equal size could be created.​

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