QUESTION 352
Part of the Certifyme LAN is shown below:
Based on the diagram shown above, which of the following IP addressing schemes could be used for this
Certifyme LAN segment?
A. Host CertifymeA – 192.168.10.22/24; Host CertifymeB – 192.168.11.23/24; Host CertifymeC -
192.168.10.23/24
B. Host CertifymeA – 192.168.10.22/24; Host CertifymeB – 192.168.10.23/24; Host CertifymeC -
192.168.11.23/24
C. Host CertifymeA – 192.168.10.22/24; Host CertifymeB – 192.168.10.23/24; Host CertifymeC -
192.168.10.24/24
D. Host CertifymeA – 192.168.11.22/24; Host CertifymeB – 192.168.11.23/24; Host CertifymeC -
192.168.10.24/24
E. None of the above
Answer: A
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Hosts in the same VLAN must belong to the same IP subnet. In this case, hosts A and C should have IP
addresses in one subnet, and host B should have an IP address in a separate subnet. Only choice A is correct.
Incorrect Answers:
B: In this case, hosts A and B are both in the 192.168.10.0/24 subnet, but they are in different VLANs C: In this
case, all three hosts are in the same 192.168.10.0/24 subnet, even though there are 2 different VLANs D: Here
hosts A and B are both in the 192.168.11.0/24 subnet even though they are in different VLANs.
QUESTION 353
DRAG DROP Network topology exhibit:
You work as a network technician at TestCMing.com. Study the network topology exhibit carefully. The network
is incomplete. Your task is to complete the Certifyme.com network. You need to select the correct router types,
IP addresses and interface types respectively.
Use the CertifymeII PC connected to the Certifyme1 router to access further information that is required to
complete the task.
Note: The console information is missing in this scenario.
A. Answer:
Explanation:
Pending. Send your suggestion to feedback@Certifyme.com
Answer:
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Pending. Send your suggestion to feedback@Certifyme.com
QUESTION 354
After the router interfaces shown in the diagram have been configured, it is discovered that hosts in the
Certifyme1 LAN cannot access the Internet. Further testing reveals additional connectivity issues. What will fix
this problem?
A. Change the subnet mask of the Certifyme2 router LAN interface.
B. Change the address of the Certifyme1router WAN interface.
C. Change the address of the Certifyme2 router interface to the Internet.
D. Change the address of the Certifyme1 router LAN interface.
E. Change the address of the Certifyme2 router LAN interface.
F. Change the subnet mask of the Certifyme2 router interface to the Internet.
G. None of the above
Answer: B
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
In order for two routers to be able to send and receive traffic across a point to point serial link, the IP addresses
of the two serial interfaces across this link needs to be in the same IP subnet. Since we are using a /30 in this
case, only 2 valid IP addresses are usable but as we can see in this example the 192.168.10.82/30 and
192.168.10.85/30 IP addresses reside in two different subnets.
QUESTION 349
While troubleshooting a connectivity issue from a PC you obtain the following information:
Local PC IP address: 190.0.3.35/24 Default Gateway: 190.0.3.1 Remote Server: 190.0.5.250/24
You then conduct the following tests from the local PC:
Ping 127.0.0.1 – Unsuccessful Ping 190.0.3.35 – Successful Ping 190.0.3.1 – Unsuccessful Ping 190.0.5.250 -
Unsuccessful
What is the underlying cause of this problem?
A. TCP/IP not correctly installed
B. Local physical layer problem
C. NIC not functioning
D. Remote physical layer problem
E. None of the above
Answer: A
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Every Windows based PC uses the 127.0.0.1 as the local loopback IP address. Every PC will respond to this
local IP address if the TCP/IP stack is correctly installed and running on the machine. If you cannot ping the
loopback address of 127.0.0.1, then something is wrong with the TCP/IP protocol stack.
QUESTION 350
A Certifyme network is shown below:
The network shown in the exhibit above is experiencing connectivity problems.
Which of the following will correct the problems? (Select two)
A. Configure the gateway on CertifymeA as 10.1.1.1.
B. Configure the gateway on CertifymeB as 10.1.2.254.
C. Configure the IP address of CertifymeA as 10.1.2.2.
D. Configure the IP address of CertifymeB as 10.1.2.2.
E. Configure the masks on both hosts to be 255.255.255.224.
F. Configure the masks on both hosts to be 255.255.255.240.
Answer: BD
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
All devices must have their default gateways set to an IP address that is in the same IP network that the station
is in. Based on the diagram above, Certifyme B is in VLAN2, so the default gateway for this device should be
the IP address of the VLAN 2 interface on the router. In addition, the IP addresses of both devices reside within
the same IP subnet.
Since they belong to different VLANs, the best method to ensure proper connectivity would be to give
CertifymeB an IP address within the same IP range as the VLAN that it belongs to, which is VLAN2 in this
example.
QUESTION 351
Part of the Certifyme network is shown below:
A junior network associate has created the network design shown above. The goal of this network design is to
provide the most efficient use of IP address space in a network expansion. Each circle defines a network
segment and the number of users required on that segment. An IP subnet number and default gateway
address are shown for each segment.
Based on the information shown above, what are three problems with the network design as shown? (Choose
three)
A. The IP subnet 10.1.1.0/30 is invalid for a segment with a single server.
B. Network 10.1.3.128/25 requires more user address space.
C. Network 10.1.2.0/25 requires more user address space.
D. Interface fa0/2 has an invalid IP address for the subnet on which it resides.
E. Interface fa0/1 has an invalid IP address for the subnet on which it resides.
F. Interface fa0/3 has an incorrect IP address
Answer: CEF
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
C: A /25 will provide for 128 IP addresses (126 usable) so a larger subnet mask should be used. A /24 will
provide for up to 254 usable addresses.
E: The 10.1.1.0/30 network will allow for only 2 usable IP addresses. In this network, 10.1.1.0 is reserved for
the network address, and 10.1.1.3 is the broadcast address, so only 10.1.1.1 or 10.1.1.2 can be used for the
fa0/1 IP address.
F: The 10.1.3.1 IP address is in the 10.1.3.0/25 network, not the 10.1.3.0/25 network. A more appropriate IP
address for this interface would be 10.1.3.129
QUESTION 346
The Certifyme network topology exhibit is shown below:
Configuration exhibit:
The Certifyme network administrator is adding two new hosts to switch CertifymeA.
Of the following choices, which values could be used for the configuration of these hosts? (Choose three.)
A. Host CertifymeB IP Address: 192.168.1.128
B. Host CertifymeA default gateway: 192.168.1.78
C. Host CertifymeA IP Address: 192.168.1.64
D. Host CertifymeB IP Address: 192.168.1.190
E. Host CertifymeA IP address: 192.168.1.79
F. Host CertifymeB default gateway: 192.168.1.129
Answer: BDE
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
QUESTION 347
The new Certifyme location is displayed below:
A new PC is installed on the LAN of the Certifyme1 router as shown above. This PC is unable to connect to the
CertifymeA server located on the Ethernet 1 network.
What is the cause of this?
A. IP address of the Ethernet 0 router interface is wrong
B. Server is using an invalid IP address
C. Workstation default gateway is set incorrectly
D. Workstation subnet mask is incorrect
E. Workstation IP address is invalid
F. None of the above
Answer: C
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
The default gateway of the host (192.168.10.64) is wrong. 192.168.10.64 is the network address of the host’s
network in this question. The default gateway should be the address of the local interface of the router. In this
case: 192.168.10.65.
Incorrect Answers:
A: The IP address of the Eternet0 interface is valid.
B: The IP address of the server is valid.
D: The network uses a 27 bit subnet mask which equates to 255.255.255.224.
E: The IP host address 192.168.10.66 is a valid host address on the subnet.
QUESTION 348
A new LAN is being implemented on the Certifyme1 network as shown below:
The local host CM1 can’t access any of the resources on the other networks. The configuration of CM1 is as
follows:
host address: …..192.168.166.45
subnet mask: ……255.255.255.240
default gateway: ..192.168.166.32
What is the underlying cause of this problem?
A. The default gateway is a network address.
B. The default gateway is on a different subnet address as the host.
C. The IP address of the host is on a different subnet.
D. The host subnet mask is incompatible to the subnet mask of the attached router interface.
Answer: A
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
The range of the subnet used in this question is 192.168.166.32 to 192.168.166.47.
192.168.166.32 is the network address and 192.168.166.47 is the broadcast. This leaves the usable host
address range of 192.168.166.33 to 192.168.166.46.
The default gateway for the host should be 192.168.166.33.
Incorrect Answers:
B: The default gateway is on the same network but it is a network address.
C: The host address is correct.
D: The subnet mask 255.255.255.240 uses 28 bits and is therefore correct.
QUESTION 343
Refer to the following exhibit:
In this network segment, the IP subnets connected to router Certifyme2 have been summarized as a
192.18.176.0/21 route and sent to Certifyme1. Based on this information, which two packet destination
addresses will Certifyme1 forward to Certifyme2? (Choose two)
A. 192.18.183.255
B. 192.18.159.2
C. 192.18.194.160
D. 192.18.179.4
E. 192.18.183.41
F. 192.18.184.45
Answer: DE
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
QUESTION 344
What is the subnet address of the host with an IP address of 172.16.159.159/22?
A. 172.16.128.0
B. 172.16.156.0
C. 172.16.159.128
D. 172.16.159.0
E. 172.16.192.0
F. 172.16.0.0
G. None of the above
Answer: B
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
In this question there are a total of 6 bits borrowed from the host portion with the subnet mask 255.255.252.0.
This will leave us 64 networks with 1022 and the IP subnets for each will be a multiple of 4. Some of the
networks are as follows:
Section 7: Describe the technological requirements for running IPv6 in conjunction with IPv4 (including:
protocols, dual stack, tunneling, etc). (0 questions)
Section 8: Describe IPv6 addresses (1 question)
QUESTION 345
Certifyme is migrating to an IPv6 addressing scheme. Identify the four valid IPv6 addresses below that could be
used in this network. (Choose four)
A. ::192:168:0:1
B. 2002:c0a8:101::42
C. 2000::
D. ::
E. 2001:3452:4952:2837::
F. 2003:dead:beef:4dad:23:46:bb:101
Answer: ABDF
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
From RFC 1884: IP Version 6 Addressing Architecture
There are three conventional forms for representing IPv6 addresses as text strings:
1. The preferred form is x:x:x:x:x:x:x:x, where the ‘x’s are the hexadecimal values of the eight 16-bit pieces of
the address.
Examples:
FEDC:BA98:7654:3210:FEDC:BA98:7654:3210
1080:0:0:0:8:800:200C:417A
Note that it is not necessary to write the leading zeros in an individual field, but there must be at least one
numeral in every field (except for the case described in 2.).
2. Due to the method of allocating certain styles of IPv6 addresses, it will be common for addresses to contain
long strings of zero bits. In order to make writing addresses
containing zero bits easier a special syntax is available to compress the zeros. The use of “::” indicates
multiple groups of 16-bits of zeros. The “::” can only appear once in an address. The “::” can also be used to
compress the leading and/or trailing zeros in an address.
For example the following addresses:
1080:0:0:0:8:800:200C:417A a unicast address FF01:0:0:0:0:0:0:43 a multicast address 0:0:0:0:0:0:0:1 the
loopback address 0:0:0:0:0:0:0:0 the unspecified addresses
may be represented as:
1080::8:800:200C:417A a unicast address FF01::43 a multicast address ::1 the loopback address :: the
unspecified addresses
3. An alternative form that is sometimes more convenient when dealing with a mixed environment of IPv4 and
IPv6 nodes is x:x:x:x:x:x:d.d.d.d, where the ‘x’s are the hexadecimal values of the six high-order 16-bit pieces
of the address, and the ‘d’s are the decimal values of the four low-order 8-bit pieces of the address (standard
IPv4 representation). Examples:
0:0:0:0:0:0:13.1.68.3
0:0:0:0:0:FFFF:129.144.52.38
or in compressed form:
::13.1.68.3
::FFFF:129.144.52.38
QUESTION 340
On the topic of VLSM, which one of the following statements best describes the concept of the route
aggregation?
A. Deleting unusable addresses through the creation of many subnets.
B. Combining routes to multiple networks into one supernet.
C. Reclaiming unused space by means of changing the subnet size.
D. Calculating the available host addresses in the AS.
E. None of the above
Answer: B
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
In the networking world route aggregate means combining routes to multiple networks into one. This is also
known as route summarization or supernetting. It is normally used to reduce the number of route entries in the
routing table by advertising numerous routes into one larger route.
QUESTION 341
In the Certifyme network shown below, what is the most efficient summarization that CM1 can use to advertise
its networks to CM2?
A. 172.1.4.0/24172.1.5.0/24172.1.6.0/24172.1.7.0/24
B. 172.1.0.0/22
C. 172.1.4.0/25172.1.4.128/25172.1.5.0/24172.1.6.0/24172.1.7.0/24
D. 172.1.0.0/21
E. 172.1.4.0/22
Answer: E
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Route Summarization Overview:
In large internetworks, hundreds, or even thousands, of network addresses can exist. It is often problematic for
routers to maintain this volume of routes in their routing tables.
Route summarization (also called route aggregation or supernetting) can reduce the number of routes that a
router must maintain, because it is a method of representing a series of network numbers in a single summary
address.
For example, in the figure above, router D can either send four routing update entries or summarize the four
addresses into a single network number. If router D summarizes the information into asingle network number
entry, the following things happen:
1. Bandwidth is saved on the link between routers D and E.
2. Router E needs to maintain only one route and therefore saves memory.
3. Router E also saves CPU resources, because it evaluates packets against fewer entries
in its routing table.
A summary route is announced by the summarizing router as long as at least one specific route in its routing
table matches the summary route.
QUESTION 342
Part of the Certifyme network is shown below:
The five Ethernet networks connected to router Certifyme1 in the graphic have been summarized for router
Certifyme2 as 192.1.144.0/20. Based on this information, which of the following packet destination addresses
will Certifyme2 forward to Certifyme1, according to this summary? (Choose two.)
A. 192.1.1.144
B. 192.1.159.2
C. 192.1.138.41
D. 192.1.151.254
E. 192.1.160.11
F. 192.1.143.145
G. 0.0.0.0
Answer: BD
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
QUESTION 337
A Certifyme PC has the IP address 172.16.209.10 /22. What is the subnet of this address?
A. 172.16.42.0
B. 172.16.107.0
C. 172.16.208.0
D. 172.16.252.0
E. 172.16.254.0
Answer: C
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
172.16.209.10/22 translates to 10101100.00010000.11010001.00001010 in binary form.
The network portion is 22 bits, so after the logical AND comparison the network address translates
to10101100.00010000.110100001.00001010. Converting the network portion to decimal results in the address
172.16.208.0/22
QUESTION 338
You’ve been assigned the CIDR (classless inter domain routing) block of 115.64.4.0/22 from your ISP. Which
of the IP addresses below can you use for a host? (Select all valid answers)
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: BCE
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
115.64.4.0 = 01110011.01000000.00000100.00000000 Subnet mask =
11111111.11111111.11111100.00000000= 255.255.252.0 Subnet number =
01110011.01000000.00000100.00000000= 115.64.4.0 Broadcast =
01110011.01000000.00000111.11111111= 115.64.7.255
Valid address range = 115.64.4.1 – 115.64.7.254
QUESTION 339
A Certifyme remote office branch is set up as shown in the diagram below:
All of the hosts in the above exhibit are connected with each other via the single Catalyst switch. Which of the
following statements correctly describe the addressing scheme of this network? (Select three)
A. The subnet mask in use is 255.255.255.192.
B. The subnet mask in use is 255.255.255.128.
C. The IP address 172.16.1.25 can be assigned to hosts in VLAN1.
D. The IP address 172.16.1.205 can be assigned to hosts in VLAN1
E. The LAN interface of the router is configured with one IP address.
F. The LAN interface of the router is configured with multiple IP addresses.
Answer: BCF
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Based on the diagram above, the subnet mask used for each VLAN is 255.255.255.128.
This means that hosts in VLAN 1 will be addressed 172.16.1.1-172.16.1.126, with 172.16.1.127 being used as
the broadcast address. Hosts in VLAN 2 will be addressed 172.16.1.129-172.16.1.254. Because there is only
one LAN interface on the router, sub interfaces will be used, so the router’s LAN interface will be configured
with 2 IP addresses, one for VLAN 1 and 1 for VLAN 2.
Incorrect Answers:
A. This subnet mask will only provide 62 host IP addresses, and the diagram shows that as many as 114 host
IP addresses are needed.
D. This IP address can be used in VLAN 2, not VLAN 1.
E. Since there are 2 subnets in this network, each separate network will require a distinct default gateway IP
address, so 2 IP addresses will be required on the LAN interface of the router.
QUESTION 334
You have a class B network with a 255.255.255.0 mask. Which of the statements below are true of this
network? (Select all valid answers)
A. There are 254 usable subnets.
B. There are 256 usable hosts per subnet.
C. There are 50 usable subnets.
D. There are 254 usable hosts per subnet.
E. There are 24 usable hosts per subnet.
F. There is one usable network.
Answer: AD
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
The default subnet mask for Class B is 255.255.0.0. Thus an extra 8 bits have been used for the network
portion, leaving 8 for hosts. The 2n – 2 formula (28 – 2 in this case for both the network and IP hosts) gives us
254 networks and 254 hosts per network.
Incorrect Answers:
B. We must remember to always subtract 2 (one for the network, and one for the broadcast) so the result is
254, not 256.
C, E: No possible network mask would give us this exact number of subnets or hosts.
F. This would be true if this were a class C network, not a class B.
QUESTION 335
How many usable IP addresses can you get from a conventional Class C address?
A. 128
B. 192
C. 254
D. 256
E. 510
Answer: C
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Class Caddresses range from 192.0.0.0 through 223.225.225.225 and default subnet maskof 255.255.255.0.
In Class C addresses, the first 24 bits are used as for the network IDwhile only the last 8 bits is used for the
host ID. Using the 2n-2 formula, we can calculate that Class C addresses can support a maximum of 254 (28-
2) hosts.
Incorrect Answers:
D. Note that the question asked for the number of usable addresses, and not the total number of all addresses.
We must subtract 2 for the network and broadcast addresses to calculate the number of usable addresses in
any subnet.
QUESTION 336
Your ISP assigned you a full class B address space. From this, you need at least 300 sub-networks that can
support at least 50 hosts each. Which of the subnet masks below are capable of satisfying your needs? (Select
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: BE
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Requirement in the question is that the company needs 300 subnets and 50 hosts per subnet.
With 9 bits used for the subnet portion, we get 510 subnets and using the remaining 7 bits for the hosts gives
us 126 hosts per subnet. The subnet mask will be 255.255.255.128
With 10 bits used for the subnet portion, we get 1022 subnets and then using the remaining 6 bits for hosts
provides 62 hosts per subnet. The subnet mask will be 255.255.255.192 in this case which will also fulfill the
requirement.
QUESTION 331
You have a class C network, and you need to design it for 5 usable subnets with each subnet handling a
minimum of 18 hosts each. Which of the following network masks should you use?
A. 225.225.224.0.
B. 225.225.240.0.
C. 225.225.255.0.
D. 255.255.255.224
E. 225.225.255.240
Answer: D
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
The default subnet mask for class C network is 255.255.255.0. If one has to create 5 subnets, then 3 bits are
required. With 3 bits we can create 8 subnets. The remaining 5 bits are used for Hosts. One can create 30
hosts using 5 bits in host field. This matches with the requirement.
Incorrect Answers:
A, B: This is an illegal subnet mask for a class C network, as the third octet can not be divided when using a
class C network.
C. This is the default subnet mask for a class C network. It provides for one network, with 254 usable host IP
addresses.
E. This subnet mask will provide for 14 separate networks with 14 hosts each. This does not meet the
requirement of a minimum of 18 hosts.
QUESTION 332
The 213.115.77.0 network was subnetted using a /28 subnet mask. How many usable subnets and host
addresses per subnet were created as a result of this?
A. 2 networks with 62 hosts
B. 6 networks with 30 hosts
C. 16 networks and 16 hosts
D. 62 networks and 2 hosts
E. 14 networks and 14 hosts
F. None of the above
Answer: F
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
A class C subnet with a 28 bit mask requires 4 bits for the network address, leaving 4 bits for host addresses.
Using the 2n-2 formula (24-2 in this case) we have 14 host addresses and 16 network addresses.
Incorrect Answers:
A. This would be the result of a /26 network mask B. This would be the result of a /27 network mask C.
Remember we need to always subtract two for the network and broadcast addresses, so this answer is
incorrect.
D. This would be the result of a /30 network mask.
QUESTION 333
The 201.145.32.0 network is subnetted using a /26 mask. How many networks and IP hosts per network exists
using this subnet mask?
A. 4 networks and 64 hosts
B. 64 networks and 4 hosts
C. 4 networks and 62 hosts
D. 62 networks and 2 hosts
E. 6 network and 30 hosts
Answer: C
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
A class C network with a 26 bit mask requires 2 bits for the network address, leaving 6 bits for host addresses.
Using the 2n-2 formula (22 for the network and 26-2for hosts) we have 4 network addresses and 62 host
addresses.
Incorrect Answers:
A, B: This is not a possible combination. No network mask will provide for 64 usable hosts, because we must
always subtract 2 for the network and broadcast address.
D. This would be the result of a /30 mask.
E. This would be the result of a /27 network mask.
QUESTION 328
Your ISP has assigned you the following IP address and subnet mask:
IP address: 199.141.27.0 Subnet mask: 255.255.255.240
Which of the following addresses can be allocated to hosts on the resulting subnet? (Select all that apply)
A. 199.141.27.2
B. 199.141.27.175
C. 199.141.27.13
D. 199.141.27.11
E. 199.141.27.208
F. 199.141.27.112
Answer: ACD
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
IP address = 11001000.10001101.00011011.00000000 = 199.141.27.0 Subnet mask =
11111111.11111111.11111111.11110000 = 255.255.255.240 Subnet # =
11001000.10001101.00011011.00000000 = 199.141.27.0 Broadcast =
11001000.10001101.00011011.00001111 = 199.141.27.15
The valid IP address range = 199.141.27.1 – 199.141.27.14
QUESTION 329
The IP network 210.106.14.0 is subnetted using a /24 mask. How many usable networks and host addresses
can be obtained from this?
A. 1 network with 254 hosts
B. 4 networks with 128 hosts
C. 2 networks with 24 hosts
D. 6 networks with 64 hosts
E. 8 networks with 36 hosts
Answer: A
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
A subnet with 24 bits on would be 255.255.255.0. Since this is a class C network, this subnet can have only 1
network and 254 usable hosts.
QUESTION 330
Given that you have a class B IP address network range, which of the subnet masks below will allow for 100
subnets with 500 usable host addresses per subnet?
A. 255.255.0.0
B. 255.255.224.0
C. 255.255.254.0
D. 255.255.255.0
E. 255.255.255.224
Answer: C
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Using the 2n-2 formula for host addresses, 29-2 = 510 host address, so a 9-bit subnet mask will provide the
required number of host addresses. If these 9 bits are used for the hosts in a class B network, then the
remaining 7 bits are used for the number of networks.
Again using the 2n-2 formula, we have 2n-2 = 126 networks that are available.
Incorrect Answers:
A. This will provide for only 1 network with 216-2 = 65534 hosts B. This will provide for 6 networks with 8190
host addresses.
D. This will provide 254 networks and 254 hosts.
E. This will provide 2046 different networks, but each network will have only 30 hosts.
QUESTION 325
What is the subnet for the host IP address 201.100.5.68/28?
A. 201.100.5.0
B. 201.100.5.32
C. 201.100.5.64
D. 201.100.5.65
E. 201.100.5.31
F. 201.100.5.1
Answer: C
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
This question is much easier then it appears when you convert it to binary and do the Boolean operation as
shown below:
IP address 201.100.5.68 = 11001001.01100100.00000101.01000100
/28 mask = 11111111.11111111.11111111.11000000 AND result = 11001001.01100100.00000101.01000000
AND in decimal= 200 . 100 . 5 . 64
QUESTION 326
Your network uses the172.12.0.0 class B address. You need to support 459 hosts per subnet, while
accommodating the maximum number of subnets. Which mask would you use?
A. 255.255.0.0.
B. 255.255.128.0.
C. 255.255.224.0.
D. 255.255.254.0.
E. None of the above
Answer: D
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
To obtain 459 hosts the number of host bits will be 9. This can support a maximum of 510 hosts. To keep 9 bits
for hosts means the last bit in the 3rd octet will be 0. This gives 255.255.254.0 as the subnet mask.
QUESTION 327
Using a subnet mask of 255.255.255.224, which of the IP addresses below can you assign to the hosts on this
subnet? (Select all that apply)
A. 16.23.118.63
B. 87.45.16.159
C. 92.11.178.93
D. 134.178.18.56
E. 192.168.16.87
F. 217.168.166.192
Answer: CDE
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Since the subnet mask is 255.255.255.224, the number of network hosts that is available is 30. Every network
boundary will be a multiple of 32. This means that every subnet will be a multiple (0, 32, 64, 96, 128, 160, 192,
224) and the broadcast address for each of these subnets will be one less this number (31, 63, 95, 127, 159,
191, 223). Therefore, any IP address that does not end in one of these numbers will be a valid host IP address.
C. Valid Host in subnetwork 2 (92.11.178.64 to 92.11.178.95) D. Valid Host in subnetwork 1 (134.178.18.32 to
134.178.18.63) E. Valid Host in subnetwork 2 (192.168.16.64 to 192.168.16.95) Incorrect Answers:
A. This will be the broadcast address for the 16.23.118.32/27 network.
B. This will be the broadcast address for the 87.45.16.128/27 network F. This will be the network address for
the 217.168.166.192/27 network.
