Ccna 640-802 Dumps|640-802 Exam Questions

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.

QUESTION 322
You need subnet a Certifyme network segment. How many subnetworks and hosts are available per subnet if
you apply a /28 mask to the 210.10.2.0 class C network?
A. 30 networks and 6 hosts.
B. 6 networks and 30 hosts.
C. 8 networks and 32 hosts.
D. 32 networks and 18 hosts.
E. 16 networks and 14 hosts.
F. None of the above

Answer: E

Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
A 28 bit subnet mask (11111111.11111111.11111111.11110000) applied to a class C network uses a 4 bits for
networks, and leaves 4 bits for hosts. Using the 2n-2 formula, we have 24-2 (or 2x2x2x2-2) which gives us 14
for the number of hosts, and the number of networks is 24 = 16.
Incorrect Answers:
A. This would be the result of a /29 (255.255.255.248) network.
B. This would be the result of a /27 (255.255.255.224) network.
C. This is not possible, as we must subtract two from the subnets and hosts for the network and broadcast
addresses.
D. This is not a possible combination of networks and hosts.

QUESTION 323
The Certifyme network was assigned the Class C network 199.166.131.0 from the ISP. If the administrator at
Certifyme were to subnet this class C network using the 255.255.255.224 subnet mask, how may hosts will they
be able to support on each subnet?
A. 14
B. 16
C. 30
D. 32
E. 62
F. 64

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 subnet mask 255.255.255.224 is a 27 bit mask (11111111.11111111.11111111.11100000). It uses 3 bits
from the last octet for the network ID, leaving 5 bits for host addresses. We can calculate the number of hosts
supported by this subnet by using the 2n-2 formula where n represents the number of host bits. In this case it
will be 5. 25-2 gives us 30.
Incorrect Answers:
A. Subnet mask 255.255.255.240 will give us 14 host addresses.
B. Subnet mask 255.255.255.240 will give us a total of 16 addresses. However, we must still subtract two
addresses (the network address and the broadcast address) to determine the maximum number of hosts the
subnet will support.
D. Subnet mask 255.255.255.224 will give us a total of 32 addresses. However, we must still subtract two
addresses (the network address and the broadcast address) to determine the maximum number of hosts the
subnet will support.
E. Subnet mask 255.255.255.192 will give us 62 host addresses.
F. Subnet mask 255.255.255.192 will give us a total of 64 addresses. However, we must still subtract two
addresses (the network address and the broadcast address) to determine the maximum number of hosts the
subnet will support.

QUESTION 324
What is the subnet for the host IP address 172.16.210.0/22?
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
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:
This question is much easier then it appears when you convert it to binary and do the Boolean operation as
shown below:
IP address 172.16.210.0 = 10101100.00010000.11010010.00000000 /22 mask =
11111111.11111111.11111100.00000000 AND result = 11111111.11111111.11010000.00000000 AND in
decimal= 172 . 16 . 208 . 0

QUESTION 319
Part of the Certifyme network is shown below:
In the Certifyme network shown above the IP address space of 128.107.7.0/24 has been allocated for all
devices. All devices must use the same subnet mask and all subnets are usable. Which subnet mask is
required to apply the allocated address space to the configuration that is shown?
A. 255.255.255.192
B. 255.255.255.128
C. 255.255.255.0
D. 255.255.255.224
E. 255.255.254.0
F. 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:
In this example the requirement is that the company needs 3 subnets and at least 58 hosts per subnet.
Referring to the following formula we see that 6 bits of subnet masking is needed.
With 6 bits used for the subnet portion, we get will get 4 different subnets with 62 usable IP addresses in each.
The subnet mask for this /28 network translates to 255.255.255.192.

QUESTION 320
DRAG DROP Certifyme 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.

A. Answer:

Explanation:
Section 6: Determine the appropriate classless addressing scheme
using VLSM and summarization to satisfy addressing requirements in a LAN/WAN environment (24
questions)
Answer:
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Section 6: Determine the appropriate classless addressing scheme
using VLSM and summarization to satisfy addressing requirements in a LAN/WAN environment (24 questions)

QUESTION 321
Part of the Certifyme network is shown below:
Leading the way in IT testing and certification tools, www.Certifyme.com
Based on the information shown above, what is the most efficient summarization that Router Certifyme1 can
use to advertise its networks to Router Certifyme2?
A. 172.1.4.0/24
172.1.5.0./24
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24
B. 172.1.0.0/22
C. 172.1.0.0/21
D. 172.1.4.0/24
172.1.4.128/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24
E. 172.1.4.0/22
F. None of the above

Answer: E

Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:

QUESTION 316
Part of the Certifyme WAN is shown below:
Leading the way in IT testing and certification tools, www.Certifyme.com
All of the Certifyme routers in this network segment are configured with the “ip subnet-zero” IOS command.
Because of this, which network addresses should be used for Link Certifyme and Network Certifyme shown
above? (Choose two.)
A. Network Certifyme – 172.16.3.128/25
B. Link Certifyme – 172.16.3.40/30
C. Network Certifyme – 172.16.3.192/26
D. Link Certifyme – 172.16.3.112/30
E. Link Certifyme – 172.16.3.0/30
F. Network Certifyme – 172.16.3.48/26

Answer: AE

Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:

QUESTION 317
DRAG DROP
Certifyme 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.

A. Answer
:
Explanation:
Answer:
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:

QUESTION 318
A host on the Certifyme network has been configured with the IP address 10.16.3.66/23. Which two statements
describe this IP address? (Choose two)
A. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.
B. This network is not subnetted.
C. The last valid host address in the subnet is 10.16.2.254 255.255.254.0
D. The subnet address is 10.16.3.0 255.255.254.0.
E. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.

Answer: AE

Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
A subnet mask of /23 translates to 255.255.254.0 and will provide for up to 512 IP addresses.
If we take the 10.16.X.X network using the /23 subnet mask, the first network available is 10.16.0.0/23, which
will provide host address from 10.16.0.1 to 10.16.2.254, with 10.16.2.255 being the broadcast address. The
next available network in the 10.16.X.X covers our example in this question of 10.16.3.66.
In this case, the first useable IP address is (10.16.2.1 choice E), and the broadcast address is 10.16.3.255
(choice A).

QUESTION 313
If an ethernet port on router CM1 was assigned an IP address of 172.16.112.1/20, what is the maximum
number of hosts allowed on this LAN subnet?
A. 2046
B. 1024
C. 4096
D. 8190
E. 4094
F. None of the above

Answer: E

Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
Since a /20 equates to 12 bits used for the subnet mask, 4094 hosts can be uniquely addressed.

QUESTION 314
Part of the Certifyme WAN is shown below:
A new subnet with 12 hosts has been added to the Certifyme network shown above.
Which subnet address should this network use to provide enough useable addresses, while wasting the fewest
number of IP addresses?
A. 192.168.10.80/29
B. 192.168.10.80/28
C. 192.168.10.96/28
D. 192.168.10.96/29
E. 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:

QUESTION 315
DRAG DROP Certifyme 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.

A. Answer:

Explanation:
Answer:
Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:

QUESTION 310
The Certifyme network administrator has designed the IP scheme as shown in the diagram below:
Based on the information shown above, what effect will this addressing scheme have on the network?
A. IP traffic between subnet A and B will be prevented.
B. Routing information will not be exchanged.
C. The addressing scheme will allow all IP traffic between the LANs.
D. IP traffic between all the LANs will be prevented.
E. 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:
This scheme will allow for communication between all networks, and uses all IP addresses in the
192.168.1.0/24 IP network with no overlap. Note that RIPv2 is being used instead of RIPv1. RIPv2 carries
subnet mask information allowing for VLSM networks like the one shown here.

QUESTION 311
The network with the IP address 172.31.0.0/19 is to be configured on the Certifyme router with the partial
configuration shown in the graphic. Which of the following statements describes the number of available
subnets and hosts that will result from this configuration?
Exhibit:
A. There are 7 usable subnets, with 2046 usable host addresses.
B. There are 8 usable subnets, with 30 usable host addresses.
C. There are 7 usable subnets, with 30 usable host addresses.
D. There are 8 usable subnets, with 2046 usable host addresses.
E. There are 7 usable subnets, with 8190 usable host addresses.
F. There are 8 usable subnets, with 8190 usable host addresses.

Answer: F

Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
The 172.31.0.0/19 will have 3 bits in the network portion, and 13 bits in the host portion.
This will allow for 2^3 = 8 networks and 2^13 = 8192 hosts available for each network (8190 usable). Since the
IP subnet-zero command is used the first network is available, making choice F correct.

QUESTION 312
A portion of he Certifyme network is shown in the diagram below:
Consider the 192.1.1.0/24 network in this exhibit. This network uses RIP v2.
Which combination of subnetwork assignments will satisfy the requirements for networks A, B, and C of this
design? (Select three)
A. Network A = 192.1.1.128/25
B. Network A = 192.1.1.0/25
C. Network B = 192.1.1.252/30
D. Network B = 192.1.1.4/30
E. Network C = 192.1.1.64/26
F. Network C = 192.1.1.224/27

Answer: ADE

Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
To properly answer this question, it is best to start from the end, which is network C.
Since network C requires at least 55 host addresses, a /26 network must be used. A network mask of /26 will
provide for 62 usable IP addresses while a /27 network will only provide for 30 so we must choose E. With
choice E taken, hosts within the range of 192.1.1.65-192.1.1.126 will be used.
For network A, both choices A and B are using the correct subnet mask, but we are only limited to choice A
since many of the hosts in choice B are already being used in network C. Finally, for network B we are left with
choice D since hosts in choice C are already being used by network A.

QUESTION 307
What is the maximum number of IP addresses that can be assigned to hosts on a Certifyme subnet that uses
the 255.255.255.224 subnet mask?
A. 14
B. 15
C. 16
D. 30
E. 31
F. 32

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 subnet mask 255.255.255.224 means that there are 27 network bits. The remaining 5 bits are the host
bits. The maximum possible combinations with 5 bits are 25 = 32. As all zero’s and all one’s hosts are not
allowed so, maximum number of valid hosts with the mask 255.255.255.224 are 25 -2 =32-2 = 30 Hosts

QUESTION 308
In a Certifyme network that supports VLSM, which network mask should be used for point-to-point WAN links in
order to reduce waste of IP addresses?
A. /24
B. /30
C. /27
D. /26
E. /32
F. 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:
A 30-bit mask is used to create subnets with two valid host addresses. This is the exact number needed for a
point-to-point connection.

QUESTION 309
The network 172.25.0.0 has been divided into eight equal 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. 172.25.96.17
F. 172.25.100.16

Answer: ACD

Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
If we divide the address 172.25.0.0 in 8 subnets, the resulting subnets will be 1. 172.25.0.0 2. 172.25.32.0 3.
172.25.64.0 This is the third subnet 4. 172.25.96.0 5. 172.25.128.0 6. 172.25.160.0
7. 172.25.192.0 8. 172.25.224.0
Addresses that fall in the 3rd subnet will be from 172.25.64.0 —- 172.25.95.255 Choices A, C and D lie in this
network range.

QUESTION 304
Certifyme has a Class C network and you need ten subnets. You wish to have as many addresses available for
hosts as possible. Which one of the following subnet masks should you use?
A. 255.255.255.192
B. 255.255.255.224
C. 255.255.255.240
D. 255.255.255.248
E. 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:
Using the 2n-2 formula, we will need to use 4 bits for subnetting, as this will provide for 24-2 = 14 subnets. The
subnet mask for 4 bits is then 255.255.255.240.
Incorrect Answers:
A. This will give us only 2 bits for the network mask, which will provide only 2 networks.
B. This will give us 3 bits for the network mask, which will provide for only 6 networks.
D. This will use 5 bits for the network mask, providing 30 networks. However, it will provide for only for 6 host
addresses in each network, so C is a better choice.

QUESTION 305
You have a single Class C IP address and a point-to-point serial link that you want to implement VLSM on.
Which subnet mask is the most efficient for this point to point link?
A. 255.255.255.0
B. 255.255.255.240
C. 255.255.255.248
D. 255.255.255.252
E. 255.255.255.254
F. 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:
For a single point to point link, only 2 IP addresses are required, one for the serial interface of the router at
each end. Therefore, the 255.255.255.252 subnet mask is often used for these types of links, as no IP
addresses are wasted.

QUESTION 306
You have a network that supports VLSM and you need to reduce IP address waste in your point to point WAN
links. Which of the masks below would you use?
A. /38
B. /30
C. /27
D. /23
E. /18
F. /32

Answer: B

Section: IMPLEMENT AN IP ADDRESSING SCHEME AND IP SERVICES TO MEET NETWORK
REQUIREMENTS IN A MEDIUM-SIZE ENTERPRISE BRANCH
Explanation/Reference:
Explanation:
For a single point to point link, only 2 IP addresses are required, one for the serial interface of the router at
each end. Therefore, the 255.255.255.252 subnet mask is often used for these types of links because no IP
addresses are wasted. The subnet mask 255.255.255.252 is a /30, so answer B is correct.
Incorrect Answers:
A. The largest mask that can be used is the single IP host mask, which is /32. It is not possible to use a /38
mask, unless of course IPv6 is being used.
C, D, E. These masks will provide for a larger number of host addresses, and since only 2 IP addresses are
needed for a point to point link, these extra addresses are wasted.
F: No available host addresses with a /32 mask