Modules 11 – 13: IP Addressing Exam Flashcards
- What is the prefix length notation for the subnet mask 255.255.255.224?
/25
/26
/27
/28
/27
Explanation: The binary format for 255.255.255.224 is 11111111.11111111.11111111.11100000. The prefix length is the number of consecutive 1s in the subnet mask. Therefore, the prefix length is /27.
- How many valid host addresses are available on an IPv4 subnet that is configured with a /26 mask?
254
190
192
62
64
62
- Which subnet mask would be used if 5 host bits are available?
255.255.255.0
255.255.255.128
255.255.255.224
255.255.255.240
255.255.255.224
- A network administrator subnets the 192.168.10.0/24 network into subnets with /26 masks. How many equal-sized subnets are created?
1
2
4
8
16
64
4
- Match the subnetwork to a host address that would be included within the subnetwork. (Not all options are used.)
VER IMAGEM
Explanation: Subnet 192.168.1.32/27 will have a valid host range from 192.168.1.33 – 192.168.1.62 with the broadcast address as 192.168.1.63
Subnet 192.168.1.64/27 will have a valid host range from 192.168.1.65 – 192.168.1.94 with the broadcast address as 192.168.1.95
Subnet 192.168.1.96/27 will have a valid host range from 192.168.1.97 – 192.168.1.126 with the broadcast address as 192.168.1.127
- 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?
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
subnetwork 192.168.1.64
subnet mask 255.255.255.192
Explanation: 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
- How many bits must be borrowed from the host portion of an address to accommodate a router with five connected networks?
two
three
four
five
three
Explanation: Each network that is directly connected to an interface on a router requires its own subnet. The formula 2n, where n is the number of bits borrowed, is used to calculate the available number of subnets when borrowing a specific number of bits.
- How many host addresses are available on the 192.168.10.128/26 network?
30
32
60
62
64
62
Explanation: 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.
- How many host addresses are available on the network 172.16.128.0 with a subnet mask of 255.255.252.0?
510
512
1022
1024
2046
2048
1022
Explanation: A mask of 255.255.252.0 is equal to a prefix of /22. A /22 prefix provides 22 bits for the network portion and leaves 10 bits for the host portion. The 10 bits in the host portion will provide 1022 usable IP addresses (210 – 2 = 1022).
- Match each IPv4 address to the appropriate address category. (Not all options are used.)
VER IMAGEM
- What three blocks of addresses are defined by RFC 1918 for private network use? (Choose three.)
10.0.0.0/8
172.16.0.0/12
192.168.0.0/16
100.64.0.0/14
169.254.0.0/16
239.0.0.0/8
10.0.0.0/8
172.16.0.0/12
192.168.0.0/16
Explanation: RFC 1918, Address Allocation for Private Internets, defines three blocks of IPv4 address for private networks that should not be routable on the public Internet.
10.0.0.0/8
172.16.0.0/12
192.168.0.0/16
- Refer to the exhibit. An administrator must send a message to everyone on the router A network. What is the broadcast address for network 172.16.16.0/22?
CCNA 1 v7 Modules 11 - 13: IP Addressing Exam Answers Full 3
172.16.16.255
172.16.20.255
172.16.19.255
172.16.23.255
172.16.255.255
VER IMAGEM
R: 172.16.19.255
Explanation: The 172.16.16.0/22 network has 22 bits in the network portion and 10 bits in the host portion. Converting the network address to binary yields a subnet mask of 255.255.252.0. The range of addresses in this network will end with the last address available before 172.16.20.0. Valid host addresses for this network range from 172.16.16.1-172.16.19.254, making 172.16.19.255 the broadcast address.
- A site administrator has been told that a particular network at the site must accommodate 126 hosts. Which subnet mask would be used that contains the required number of host bits?
255.255.255.0
255.255.255.128
255.255.255.224
255.255.255.240
255.255.255.128
Explanation: The subnet mask of 255.255.255.0 has 8 host bits. The mask of 255.255.255.128 results in 7 host bits. The mask of 255.255.255.224 has 5 host bits. Finally, 255.255.255.240 represents 4 host bits.
- Refer to the exhibit. Considering the addresses already used and having to remain within the 10.16.10.0/24 network range, which subnet address could be assigned to the network containing 25 hosts?
CCNA 1 v7 Modules 11 - 13: IP Addressing Exam Answers Full 4
10.16.10.160/26
10.16.10.128/28
10.16.10.64/27
10.16.10.224/26
10.16.10.240/27
10.16.10.240/28
VER IMAGEM
R: 10.16.10.64/27
Explanation: Addresses 10.16.10.0 through 10.16.10.63 are taken for the leftmost network. Addresses 10.16.10.192 through 10.16.10.207 are used by the center network.The address space from 208-255 assumes a /28 mask, which does not allow enough host bits to accommodate 25 host addresses.The address ranges that are available include 10.16.10.64/26 and10.16.10.128/26. To accommodate 25 hosts, 5 host bits are needed, so a /27 mask is necessary. Four possible /27 subnets could be created from the available addresses between 10.16.10.64 and 10.16.10.191:
10.16.10.64/27
10.16.10.96/27
10.16.10.128/27
10.16.10.160/27
- What is the usable number of host IP addresses on a network that has a /26 mask?
256
254
64
62
32
16
62
Explanation: A /26 mask is the same as 255.255.255.192. The mask leaves 6 host bits. With 6 host bits, 64 IP addresses are possible. One address represents the subnet number and one address represents the broadcast address, which means that 62 addresses can then be used to assign to network devices.
- Which address prefix range is reserved for IPv4 multicast?
240.0.0.0 – 254.255.255.255
224.0.0.0 – 239.255.255.255
169.254.0.0 – 169.254.255.255
127.0.0.0 – 127.255.255.255
224.0.0.0 – 239.255.255.255
Explanation: Multicast IPv4 addresses use the reserved class D address range of 224.0.0.0 to 239.255.255.255.
- Refer to the exhibit. Match the network with the correct IP address and prefix that will satisfy the usable host addressing requirements for each network.
VER IMAGEM
Explanation: Network A needs to use 192.168.0.128 /25, which yields 128 host addresses.
Network B needs to use 192.168.0.0 /26, which yields 64 host addresses.
Network C needs to use 192.168.0.96 /27, which yields 32 host addresses.
Network D needs to use 192.168.0.80/30, which yields 4 host addresses.
- A high school in New York (school A) is using videoconferencing technology to establish student interactions with another high school (school B) in Russia. The videoconferencing is conducted between two end devices through the Internet. The network administrator of school A configures the end device with the IP address 209.165.201.10. The administrator sends a request for the IP address for the end device in school B and the response is 192.168.25.10. Neither school is using a VPN. The administrator knows immediately that this IP will not work. Why?
This is a loopback address.
This is a link-local address.
This is a private IP address.
There is an IP address conflict.
This is a private IP address.
- Which three addresses are valid public addresses? (Choose three.)
198.133.219.17
192.168.1.245
10.15.250.5
128.107.12.117
172.31.1.25
64.104.78.227
198.133.219.17
128.107.12.117
64.104.78.227
Explanation: The ranges of private IPv4 addresses are as folllows:
10.0.0.0 – 10.255.255.255
172.16.0.0 – 172.31.255.255
192.168.0.0 – 192.168.255.255
- A message is sent to all hosts on a remote network. Which type of message is it?
limited broadcast
multicast
directed broadcast
unicast
directed broadcast
Explanation: A directed broadcast is a message sent to all hosts on a specific network. It is useful for sending a broadcast to all hosts on a nonlocal network. A multicast message is a message sent to a selected group of hosts that are part of a subscribing multicast group. A limited broadcast is used for a communication that is limited to the hosts on the local network. A unicast message is a message sent from one host to another.
- A company has a network address of 192.168.1.64 with a subnet mask of 255.255.255.192. The company wants to create two subnetworks that would contain 10 hosts and 18 hosts respectively. Which two networks would achieve that? (Choose two.)
192.168.1.16/28
192.168.1.64/27
192.168.1.128/27
192.168.1.96/28
192.168.1.192/28
192.168.1.64/27
192.168.1.96/28
Explanation: Subnet 192.168.1.64 /27 has 5 bits that are allocated for host addresses and therefore will be able to support 32 addresses, but only 30 valid host IP addresses. Subnet 192.168.1.96/28 has 4 bits for host addresses and will be able to support 16 addresses, but only 14 valid host IP addresses.
- Which address is a valid IPv6 link-local unicast address?
FEC8:1::FFFF
FD80::1:1234
FE80::1:4545:6578:ABC1
FE0A::100:7788:998F
FC90:5678:4251:FFFF
FE80::1:4545:6578:ABC1
Explanation: IPv6 LLAs are in the fe80::/10 range. The /10 indicates that the first 10 bits are 1111 1110 10xx xxxx. The first hextet has a range of 1111 1110 1000 0000 (fe80) to 1111 1110 1011 1111 (febf).
- Which of these addresses is the shortest abbreviation for the IP address:
3FFE:1044:0000:0000:00AB:0000:0000:0057?
3FFE:1044::AB::57
3FFE:1044::00AB::0057
3FFE:1044:0:0:AB::57
3FFE:1044:0:0:00AB::0057
3FFE:1044:0000:0000:00AB::57
3FFE:1044:0000:0000:00AB::0057
3FFE:1044:0:0:AB::57
Explanation: The rules for reducing the notation of IPv6 addresses are:
1. Omit any leading 0s (zeros) in any hextet.
2. Replace any single, contiguous string of one or more 16-bit hextets consisting of all zeros with a double colon (::) .
3. The double colon (::) can only be used once within an address.
- 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?
16
256
4096
65536
65536
Explanation: 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.
- Given IPv6 address prefix 2001:db8::/48, what will be the last subnet that is created if the subnet prefix is changed to /52?
2001:db8:0:f00::/52
2001:db8:0:8000::/52
2001:db8:0:f::/52
2001:db8:0:f000::/52
2001:db8:0:f000::/52
Explanation: Prefix 2001:db8::/48 has 48 network bits. If we subnet to a /52, we are moving the network boundary four bits to the right and creating 16 subnets. The first subnet is 2001:db8::/52 the last subnet is 2001:db8:0:f000::/52.
- Consider the following range of addresses:
2001:0DB8:BC15:00A0:0000::
2001:0DB8:BC15:00A1:0000::
2001:0DB8:BC15:00A2:0000::
…
2001:0DB8:BC15:00AF:0000::
The prefix-length for the range of addresses is /60 .
Explanation: All the addresses have the part 2001:0DB8:BC15:00A in common. Each number or letter in the address represents 4 bits, so the prefix-length is /60.
- What type of IPv6 address is FE80::1?
loopback
link-local
multicast
global unicast
link-local
Explanation: Link-local IPv6 addresses start with FE80::/10, which is any address from FE80:: to FEBF::. Link-local addresses are used extensively in IPv6 and allow directly connected devices to communicate with each other on the link they share
Refer to the exhibit. A company is deploying an IPv6 addressing scheme for its network. The company design document indicates that the subnet portion of the IPv6 addresses is used for the new hierarchical network design, with the s ite subsection to represent multiple geographical sites of the company, the s ub-site section to represent multiple campuses at each site, and the s ubnet section to indicate each network segment separated by routers. With such a scheme, what is the maximum number of subnets achieved per sub-site ?
0
4
16
256
VER IMAGEM
R: 16
Explanation: Because only one hexadecimal character is used to represent the subnet, that one character can represent 16 different values 0 through F.