QUESTION:
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.
Answer: D
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 :
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: C, D, E
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:
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: A, C, D
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 :
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
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 :
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
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 :
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
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 6 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:
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: E
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 14 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:
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 with 64 hosts
B. 64 networks and 4 hosts
C. 2 networks and 62 hosts
D. 62 networks and 2 hosts
E. 6 network and 30 hosts
Answer: C
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 (22-2 for the network and 26-2 for hosts)
we have 2 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:
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: A, D
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 set 5
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. 14 networks and 14 hosts.
F. None of the above
Answer: E
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 both the number of networks, and the
number of hosts.
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:
The TestKing network was assigned the Class C network 199.166.131.0 from the
ISP. If the administrator at TestKing 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
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:
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
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:
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
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:
3 addresses are shown in binary form below:
A. 01100100.00001010.11101011.00100111
B. 10101100.00010010.10011110.00001111
C. 11000000.10100111.10110010.01000101
below are correct? (Select three)
A. Address C is a public Class C address.
B. Address C is a private Class C address.
C. Address B is a public Class B address.
D. Address A is a public Class A address.
E. Address B is a private Class B address.
F. Address A is a private Class A address.
Answer: A, D, E
Explanation:
A. Address C converts to 192.167.178.69 in decimal, which is a public class C address.
D. Address A converts to 100.10.235.39, which is a public class A IP address.
E. Address B converts to 172.18.158.15, which is a private (RFC 1918) IP address.
QUESTION :
What is the IP address range for the first octet in a class B address, in binary form?
A. 00000111-10001111
B. 00000011-10011111
C. 10000000-10111111
D. 11000000-11011111
E. 11100000-11101111
F. None of the above
Answer: C
Explanation:
The class B address range is 128.0.0.0-191.255.255.255. When looking at the first octet
alone, the range is 128-191. The binary number for 128 is 10000000 and the binary
number for 191 is 10111111, so the value rang is 10000000-10111111.
QUESTION NO:
Which one of the binary bit patterns below denotes a Class B address?
A. 0xxxxxxx
B. 10xxxxxx
C. 110xxxxx
D. 1110xxxx
E. 11110xxx
Answer: B
Explanation:
Class B addresses start with a binary of 10. The valid class B range is
128.0.0.0-191.255.255.255.
Incorrect Answers:
A. Class A addresses start with 0, as they are addresses that are less than 128.
C. Class C addresses start with 110, for a value of 192.0.0.0-223.255.255.255
D. Class D addresses start with 1110. They are reserved for multicast use
E. Class E addresses start with 11110. They are currently reserved for experimental use.
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. 14 networks and 14 hosts.
F. None of the above
Answer: E
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 both the number of networks, and the
number of hosts.
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:
The TestKing network was assigned the Class C network 199.166.131.0 from the
ISP. If the administrator at TestKing 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
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:
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
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:
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
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:
3 addresses are shown in binary form below:
A. 01100100.00001010.11101011.00100111
B. 10101100.00010010.10011110.00001111
C. 11000000.10100111.10110010.01000101
below are correct? (Select three)
A. Address C is a public Class C address.
B. Address C is a private Class C address.
C. Address B is a public Class B address.
D. Address A is a public Class A address.
E. Address B is a private Class B address.
F. Address A is a private Class A address.
Answer: A, D, E
Explanation:
A. Address C converts to 192.167.178.69 in decimal, which is a public class C address.
D. Address A converts to 100.10.235.39, which is a public class A IP address.
E. Address B converts to 172.18.158.15, which is a private (RFC 1918) IP address.
QUESTION :
What is the IP address range for the first octet in a class B address, in binary form?
A. 00000111-10001111
B. 00000011-10011111
C. 10000000-10111111
D. 11000000-11011111
E. 11100000-11101111
F. None of the above
Answer: C
Explanation:
The class B address range is 128.0.0.0-191.255.255.255. When looking at the first octet
alone, the range is 128-191. The binary number for 128 is 10000000 and the binary
number for 191 is 10111111, so the value rang is 10000000-10111111.
QUESTION NO:
Which one of the binary bit patterns below denotes a Class B address?
A. 0xxxxxxx
B. 10xxxxxx
C. 110xxxxx
D. 1110xxxx
E. 11110xxx
Answer: B
Explanation:
Class B addresses start with a binary of 10. The valid class B range is
128.0.0.0-191.255.255.255.
Incorrect Answers:
A. Class A addresses start with 0, as they are addresses that are less than 128.
C. Class C addresses start with 110, for a value of 192.0.0.0-223.255.255.255
D. Class D addresses start with 1110. They are reserved for multicast use
E. Class E addresses start with 11110. They are currently reserved for experimental use.
QUESTION NO: 9
Which two of the addresses below are available for host addresses on the subnet
192.168.15.19/28? (Select two answer choices)
A. 192.168.15.17
B. 192.168.15.14
C. 192.168.15.29
D. 192.168.15.16
E. 192.168.15.31
F. None of the above
Answer: A, C
Explanation:
The network uses a 28bit subnet (255.255.255.240). This means that 4 bits are used for
the networks and 4 bits for the hosts. This allows for 14 networks and 14 hosts (2n-2).
The last bit used to make 240 is the 4th bit (16) therefore the first network will be
192.168.15.16. The network will have 16 addresses (but remember that the first address is
the network address and the last address is the broadcast address). In other words, the
networks will be in increments of 16 beginning at 192.168.15.16/28. The IP address we
are given is 192.168.15.19. Therefore the other host addresses must also be on this
network. Valid IP addresses for hosts on this network are: 192.168.15.17-192.168.15.30.
Incorrect Answers:
B. This is not a valid address for this particular 28 bit subnet mask. The first network
address should be 192.168.15.16.
D. This is the network address.
E. This is the broadcast address for this particular subnet.
QUESTION NO: 10
You have 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
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 NO: 11
Which of the following is an example of a valid unicast host IP address?
A. 172.31.128.255./18
B. 255.255.255.255
C. 192.168.24.59/30
D. FFFF.FFFF.FFFF
E. 224.1.5.2
F. All of the above
Answer: A
Explanation
The address 172.32.128.255 /18 is 10101100.00100000.10|000000.11111111 in binary,
so this is indeed a valid host address.
Incorrect Answers:
B. This is the all 1's broadcast address.
C. Although at first glance this answer would appear to be a valid IP address, the /30
means the network mask is 255.255.255.252, and the 192.168.24.59 address is the
broadcast address for the 192.168.24.56/30 network.
D. This is the all 1's broadcast MAC address
E. This is a multicast IP address.
Which two of the addresses below are available for host addresses on the subnet
192.168.15.19/28? (Select two answer choices)
A. 192.168.15.17
B. 192.168.15.14
C. 192.168.15.29
D. 192.168.15.16
E. 192.168.15.31
F. None of the above
Answer: A, C
Explanation:
The network uses a 28bit subnet (255.255.255.240). This means that 4 bits are used for
the networks and 4 bits for the hosts. This allows for 14 networks and 14 hosts (2n-2).
The last bit used to make 240 is the 4th bit (16) therefore the first network will be
192.168.15.16. The network will have 16 addresses (but remember that the first address is
the network address and the last address is the broadcast address). In other words, the
networks will be in increments of 16 beginning at 192.168.15.16/28. The IP address we
are given is 192.168.15.19. Therefore the other host addresses must also be on this
network. Valid IP addresses for hosts on this network are: 192.168.15.17-192.168.15.30.
Incorrect Answers:
B. This is not a valid address for this particular 28 bit subnet mask. The first network
address should be 192.168.15.16.
D. This is the network address.
E. This is the broadcast address for this particular subnet.
QUESTION NO: 10
You have 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
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 NO: 11
Which of the following is an example of a valid unicast host IP address?
A. 172.31.128.255./18
B. 255.255.255.255
C. 192.168.24.59/30
D. FFFF.FFFF.FFFF
E. 224.1.5.2
F. All of the above
Answer: A
Explanation
The address 172.32.128.255 /18 is 10101100.00100000.10|000000.11111111 in binary,
so this is indeed a valid host address.
Incorrect Answers:
B. This is the all 1's broadcast address.
C. Although at first glance this answer would appear to be a valid IP address, the /30
means the network mask is 255.255.255.252, and the 192.168.24.59 address is the
broadcast address for the 192.168.24.56/30 network.
D. This is the all 1's broadcast MAC address
E. This is a multicast IP address.
Question Set 4
QUESTION NO: 5
How would the number 172 be expressed in binary form?
A. 10010010
B. 10011001
C. 10101100
D. 10101110
Answer: C
Explanation:
10101100= 128 + 0 + 32 + 0 + 8 + 4 + 0 + 0 = 172
Incorrect Answers:
A. Binary 10010010 = 128+0+0+16+0+0+2+0=146
B. Binary 10011001 = 128+0+0+16+8+0+0+1=153
D. Binary 10101110 = 128+0+32+0+8+4+2+0= 174
QUESTION NO: 6
The MAC address for your PC NIC is: C9-3F-32-B4-DC-19. What is the address of
the OUI portion of this NIC card, expressed as a binary number?
A. 11001100-00111111-00011000
B. 11000110-11000000-00011111
C. 11001110-00011111-01100000
D. 11001001-00111111-00110010
E. 11001100-01111000-00011000
F. 11111000-01100111-00011001
Answer: D
Explanation:
The first half of the address identifies the manufacturer of the card. This code, which is
assigned to each manufacturer by the IEEE, is called the organizationally unique
identifier (OUI). In this example, the OUI is C9-3F-32. If we take this number and
convert it to decimal form we have:
C9 = (12x16) + 9 = 201
3F = (3x16) + 15 = 63
32 = (3x16) + 2 = 50
So, in decimal we have 201.63.50. If we then convert this to binary, we have:
201 = 11001001
63 = 00111111
50 = 00110010
So the correct answer is D: 11001001-00111111-00110010
QUESTION NO: 7
How do you express the binary number 10110011 in decimal form?
A. 91
B. 155
C. 179
D. 180
E. 201
F. 227
Answer: C
Explanation:
If you arrange the binary number 10110011, against the place value and multiply the
values, and add them up, you get the correct answer.
1 0 1 1 0 0 1 1
128 64 32 16 8 4 2 1
128 + 0 + 32 +16 + 0 + 0 +2 +1 = 179
How would the number 172 be expressed in binary form?
A. 10010010
B. 10011001
C. 10101100
D. 10101110
Answer: C
Explanation:
10101100= 128 + 0 + 32 + 0 + 8 + 4 + 0 + 0 = 172
Incorrect Answers:
A. Binary 10010010 = 128+0+0+16+0+0+2+0=146
B. Binary 10011001 = 128+0+0+16+8+0+0+1=153
D. Binary 10101110 = 128+0+32+0+8+4+2+0= 174
QUESTION NO: 6
The MAC address for your PC NIC is: C9-3F-32-B4-DC-19. What is the address of
the OUI portion of this NIC card, expressed as a binary number?
A. 11001100-00111111-00011000
B. 11000110-11000000-00011111
C. 11001110-00011111-01100000
D. 11001001-00111111-00110010
E. 11001100-01111000-00011000
F. 11111000-01100111-00011001
Answer: D
Explanation:
The first half of the address identifies the manufacturer of the card. This code, which is
assigned to each manufacturer by the IEEE, is called the organizationally unique
identifier (OUI). In this example, the OUI is C9-3F-32. If we take this number and
convert it to decimal form we have:
C9 = (12x16) + 9 = 201
3F = (3x16) + 15 = 63
32 = (3x16) + 2 = 50
So, in decimal we have 201.63.50. If we then convert this to binary, we have:
201 = 11001001
63 = 00111111
50 = 00110010
So the correct answer is D: 11001001-00111111-00110010
QUESTION NO: 7
How do you express the binary number 10110011 in decimal form?
A. 91
B. 155
C. 179
D. 180
E. 201
F. 227
Answer: C
Explanation:
If you arrange the binary number 10110011, against the place value and multiply the
values, and add them up, you get the correct answer.
1 0 1 1 0 0 1 1
128 64 32 16 8 4 2 1
128 + 0 + 32 +16 + 0 + 0 +2 +1 = 179
QUESTION NO: 5
How would the number 172 be expressed in binary form?
A. 10010010
B. 10011001
C. 10101100
D. 10101110
Answer: C
Explanation:
10101100= 128 + 0 + 32 + 0 + 8 + 4 + 0 + 0 = 172
Incorrect Answers:
A. Binary 10010010 = 128+0+0+16+0+0+2+0=146
B. Binary 10011001 = 128+0+0+16+8+0+0+1=153
D. Binary 10101110 = 128+0+32+0+8+4+2+0= 174
QUESTION NO: 6
The MAC address for your PC NIC is: C9-3F-32-B4-DC-19. What is the address of
the OUI portion of this NIC card, expressed as a binary number?
A. 11001100-00111111-00011000
B. 11000110-11000000-00011111
C. 11001110-00011111-01100000
D. 11001001-00111111-00110010
E. 11001100-01111000-00011000
F. 11111000-01100111-00011001
Answer: D
Explanation:
The first half of the address identifies the manufacturer of the card. This code, which is
assigned to each manufacturer by the IEEE, is called the organizationally unique
identifier (OUI). In this example, the OUI is C9-3F-32. If we take this number and
convert it to decimal form we have:
C9 = (12x16) + 9 = 201
3F = (3x16) + 15 = 63
32 = (3x16) + 2 = 50
So, in decimal we have 201.63.50. If we then convert this to binary, we have:
201 = 11001001
63 = 00111111
50 = 00110010
So the correct answer is D: 11001001-00111111-00110010
QUESTION NO: 7
How do you express the binary number 10110011 in decimal form?
A. 91
B. 155
C. 179
D. 180
E. 201
F. 227
Answer: C
Explanation:
If you arrange the binary number 10110011, against the place value and multiply the
values, and add them up, you get the correct answer.
1 0 1 1 0 0 1 1
128 64 32 16 8 4 2 1
128 + 0 + 32 +16 + 0 + 0 +2 +1 = 179
How would the number 172 be expressed in binary form?
A. 10010010
B. 10011001
C. 10101100
D. 10101110
Answer: C
Explanation:
10101100= 128 + 0 + 32 + 0 + 8 + 4 + 0 + 0 = 172
Incorrect Answers:
A. Binary 10010010 = 128+0+0+16+0+0+2+0=146
B. Binary 10011001 = 128+0+0+16+8+0+0+1=153
D. Binary 10101110 = 128+0+32+0+8+4+2+0= 174
QUESTION NO: 6
The MAC address for your PC NIC is: C9-3F-32-B4-DC-19. What is the address of
the OUI portion of this NIC card, expressed as a binary number?
A. 11001100-00111111-00011000
B. 11000110-11000000-00011111
C. 11001110-00011111-01100000
D. 11001001-00111111-00110010
E. 11001100-01111000-00011000
F. 11111000-01100111-00011001
Answer: D
Explanation:
The first half of the address identifies the manufacturer of the card. This code, which is
assigned to each manufacturer by the IEEE, is called the organizationally unique
identifier (OUI). In this example, the OUI is C9-3F-32. If we take this number and
convert it to decimal form we have:
C9 = (12x16) + 9 = 201
3F = (3x16) + 15 = 63
32 = (3x16) + 2 = 50
So, in decimal we have 201.63.50. If we then convert this to binary, we have:
201 = 11001001
63 = 00111111
50 = 00110010
So the correct answer is D: 11001001-00111111-00110010
QUESTION NO: 7
How do you express the binary number 10110011 in decimal form?
A. 91
B. 155
C. 179
D. 180
E. 201
F. 227
Answer: C
Explanation:
If you arrange the binary number 10110011, against the place value and multiply the
values, and add them up, you get the correct answer.
1 0 1 1 0 0 1 1
128 64 32 16 8 4 2 1
128 + 0 + 32 +16 + 0 + 0 +2 +1 = 179
QUESTION NO: 3
Which one of the binary number ranges shown below corresponds to the value of the first octet in Class B address range?
A. 10000000-11101111
B. 11000000-11101111
C. 10000000-10111111
D. 10000000-11111111
E. 11000000-10111111
Answer: C
Explanation:
Class B addresses are in the range 128.0.0.0 through 191.255.255.255.
In binary, the first octet (128 through 191) equates to 10000000-10111111
Incorrect Answers:
A. Binary 10000000 does equate to 128 but binary 11101111 equates to 239
B. Binary 11000000 equates to 192 and binary 11101111 equates to 239
D. Binary 10000000 does equate to 128 but binary 11011111 equates to 223
E. Binary 11000000 equates to 192 but binary 10111111 does equate to 191
QUESTION NO: 4
How would the number 231 be expressed as a binary number?
A. 11011011
B. 11110011
C. 11100111
D. 11111001
E. 11010011
Answer: C
Explanation
Decimal number 231 equates to 11100111 in binary (128+64+32+0+0+4+2+1)
Incorrect Answers:
A: Binary 11011011 equates to 219 (128+64+0+16+8+0+2+1)
B: Binary 11110011 equates to 243 (128+64+32+16+0+0+2+1)
D: Binary 11101011 equates to 249 (128+64+32+16+8+0+0+1)
E: Binary 11010011 equates to 211 (128+64+0+16+0+0+2+1)
Which one of the binary number ranges shown below corresponds to the value of the first octet in Class B address range?
A. 10000000-11101111
B. 11000000-11101111
C. 10000000-10111111
D. 10000000-11111111
E. 11000000-10111111
Answer: C
Explanation:
Class B addresses are in the range 128.0.0.0 through 191.255.255.255.
In binary, the first octet (128 through 191) equates to 10000000-10111111
Incorrect Answers:
A. Binary 10000000 does equate to 128 but binary 11101111 equates to 239
B. Binary 11000000 equates to 192 and binary 11101111 equates to 239
D. Binary 10000000 does equate to 128 but binary 11011111 equates to 223
E. Binary 11000000 equates to 192 but binary 10111111 does equate to 191
QUESTION NO: 4
How would the number 231 be expressed as a binary number?
A. 11011011
B. 11110011
C. 11100111
D. 11111001
E. 11010011
Answer: C
Explanation
Decimal number 231 equates to 11100111 in binary (128+64+32+0+0+4+2+1)
Incorrect Answers:
A: Binary 11011011 equates to 219 (128+64+0+16+8+0+2+1)
B: Binary 11110011 equates to 243 (128+64+32+16+0+0+2+1)
D: Binary 11101011 equates to 249 (128+64+32+16+8+0+0+1)
E: Binary 11010011 equates to 211 (128+64+0+16+0+0+2+1)
QUESTION NO: 12
Which Layer 1 devices can be used to enlarge the area covered by a single LAN
segment? (Select two)
A. Switch
B. Router
C. NIC
D. hub
E. Repeater
F. RJ-45 transceiver
Answer: D, E
Explanation:
Both hub, Repeater, Router and Switch repeat the packet. But only hub and Repeater do
not segment the network. Repeaters and Hubs are contained in layer one of the OSI model
(Physical layer) while a switch lies in layer two and a router is in layer 3.
QUESTION NO: 13
CDP is running between two devices. What information is supplied by
CDP? (Select three)
A. Device identifiers
B. Capabilities list
C. Platform
D. Route identifier
E. Neighbor traffic data
Answer: A, B, C
QUESTION NO: 14
If a host on a network has the address 172.16.45.14/30, what is the address of the
subnetwork to which this host belongs?
A. 172.16.45.0
B. 172.16.45.4
C. 172.16.45.8
D. 172.16.45.12
E. 172.16.45.18
Answer: D
Explanation:
The last octet in binary form is 00001110. Only 6 bits of this octet belong to the subnet
mask. Hence, the subnetwork is 172.16.45.12.
How many broadcast domains are shown in the graphic assuming only the default
VLAN is configured on the switches?
A. One
B. Two
C. Six
D. Twelve
Answer: A
Explanation:
There is only one broadcast domain because switches and hubs do not segment the
broadcast domains when only a single VLAN is configured. Only layer 3 devices can
segment the broadcast domains, or VLAN-capable switches where multiple VLANs are
configured.
QUESTION:
You have the binary number 10011101. Convert it to its decimal and hexadecimal
equivalents. (Select two answer choices)
A. 158
B. 0x9D
C. 156
D. 157
E. 0x19
F. 0x9F
Answer: B, D
Explanation:
10011101 = 128+0+0+16+8+4+0+1 = 157
For hexadecimal, we break up the binary number 10011101 into the 2 parts:
1001 = 9 and 1101 = 13, this is D in hexadecimal, so the number is 0x9D. We can further
verify by taking the hex number 9D and converting it to decimal by taking 16 times 9,
and then adding 13 for D (0x9D = (16x9)+13 = 157).
QUESTION NO:
The subnet mask on the serial interface of a router is expressed in binary as
11111000 for the last octet. How do you express the binary number 11111000 in
decimal?
A. 210
B. 224
C. 240
D. 248
E. 252
Answer: D
Explanation:
128 + 64+32+16+8 = 248. Since this is the last octet of the interface, the subnet mask
would be expressed as a /29.
Reference:
CCNA Self-Study CCNA ICND exam certification Guide
Incorrect Answers:
A. The number 210 would be 11010010 in binary.
B. The number 224 would be 11100000 in binary.
C. The number 240 would be 11110000 in binary
E. The number 252 would be 11111100 in binary. This is known as a /30 and is used
often in point-point links, since there are only 2 available addresses for use in this subnet.
Which Layer 1 devices can be used to enlarge the area covered by a single LAN
segment? (Select two)
A. Switch
B. Router
C. NIC
D. hub
E. Repeater
F. RJ-45 transceiver
Answer: D, E
Explanation:
Both hub, Repeater, Router and Switch repeat the packet. But only hub and Repeater do
not segment the network. Repeaters and Hubs are contained in layer one of the OSI model
(Physical layer) while a switch lies in layer two and a router is in layer 3.
QUESTION NO: 13
CDP is running between two devices. What information is supplied by
CDP? (Select three)
A. Device identifiers
B. Capabilities list
C. Platform
D. Route identifier
E. Neighbor traffic data
Answer: A, B, C
QUESTION NO: 14
If a host on a network has the address 172.16.45.14/30, what is the address of the
subnetwork to which this host belongs?
A. 172.16.45.0
B. 172.16.45.4
C. 172.16.45.8
D. 172.16.45.12
E. 172.16.45.18
Answer: D
Explanation:
The last octet in binary form is 00001110. Only 6 bits of this octet belong to the subnet
mask. Hence, the subnetwork is 172.16.45.12.
How many broadcast domains are shown in the graphic assuming only the default
VLAN is configured on the switches?
A. One
B. Two
C. Six
D. Twelve
Answer: A
Explanation:
There is only one broadcast domain because switches and hubs do not segment the
broadcast domains when only a single VLAN is configured. Only layer 3 devices can
segment the broadcast domains, or VLAN-capable switches where multiple VLANs are
configured.
QUESTION:
You have the binary number 10011101. Convert it to its decimal and hexadecimal
equivalents. (Select two answer choices)
A. 158
B. 0x9D
C. 156
D. 157
E. 0x19
F. 0x9F
Answer: B, D
Explanation:
10011101 = 128+0+0+16+8+4+0+1 = 157
For hexadecimal, we break up the binary number 10011101 into the 2 parts:
1001 = 9 and 1101 = 13, this is D in hexadecimal, so the number is 0x9D. We can further
verify by taking the hex number 9D and converting it to decimal by taking 16 times 9,
and then adding 13 for D (0x9D = (16x9)+13 = 157).
QUESTION NO:
The subnet mask on the serial interface of a router is expressed in binary as
11111000 for the last octet. How do you express the binary number 11111000 in
decimal?
A. 210
B. 224
C. 240
D. 248
E. 252
Answer: D
Explanation:
128 + 64+32+16+8 = 248. Since this is the last octet of the interface, the subnet mask
would be expressed as a /29.
Reference:
CCNA Self-Study CCNA ICND exam certification Guide
Incorrect Answers:
A. The number 210 would be 11010010 in binary.
B. The number 224 would be 11100000 in binary.
C. The number 240 would be 11110000 in binary
E. The number 252 would be 11111100 in binary. This is known as a /30 and is used
often in point-point links, since there are only 2 available addresses for use in this subnet.
Which of the following statements most accurately describes the characteristics of
the above networks broadcast and collision domains? (Select the two best answer
choices)
A. There are two broadcast domains in the network.
B. There are four broadcast domains in the network.
C. There are six broadcast domains in the network.
D. There are four collision domains in the network.
E. There are five collision domains in the network.
F. There are seven collision domains in the network.
Answer: A, F
Explanation:
In this network we have a hub being used in the Sales department, and a switch being
used in the Production department. Based on this, we have two broadcast domains: one
for each network being separated by a router. For the collision domains, we have 5
computers and one port for E1 so we have 6 collision domains total because we use a
switch in the Production Department so 5 are created there, plus one collision domain for
the entire Sales department because a hub is being used.
QUESTION NO: 4
LAN consists of one large flat network. You decide to
segment this LAN into two separate networks with a router. What will be the affect
of this change?
A. The number of broadcast domains will be decreased.
B. It will make the broadcasting of traffic between domains more efficient between
segments.
C. It will increase the number of collisions.
D. It will prevent segment 1's broadcasts from getting to segment 2.
E. It will connect segment 1's broadcasts to segment 2.
Answer: D
Explanation
A router does not forward broadcast traffic. It therefore breaks up a broadcast domain,
reducing unnecessary network traffic. Broadcasts from one segment will not be seen on
the other segment.
Incorrect Answers:
A. This will actually increase the number of broadcast domains from one to two.
B. All link level traffic from segment one to segment two will now need to be routed
between the two interfaces of the router. Although this will reduce the traffic on the LAN
links, it does also provide a less efficient transport between the segments.
C. Since the network size is effectively cut into half, the number of collisions should
decrease dramatically.
E. Broadcasts from one segment will be completely hidden from the other segment.
QUESTION NO: 5
Which of the following are benefits of segmenting a network with a router? (Select
all that apply)
A. Broadcasts are not forwarded across the router.
B. All broadcasts are completely eliminated.
C. Adding a router to the network decreases latency.
D. Filtering can occur based on Layer 3 information.
E. Routers are more efficient than switches and will process the data more quickly.
F. None of the above.
Answer: A, D
Explanation
Routers do not forward broadcast messages and therefore breaks up a broadcast domain.
In addition, routers can be used to filter network information with the use of access lists.
Incorrect Answers:
B. Broadcasts will still be present on the LAN segments. They will be reduced, because
routers will block broadcasts from one network to the other.
C. Adding routers, or hops, to any network will actually increase the latency.
E. The switching process is faster than the routing process. Since routers must do a layer
3 destination based lookup in order to reach destinations, they will process data more
slowly than switches.
QUESTION NO: 10
Given the choices below, which address represents a unicast address?
A. 224.1.5.2
B. FFFF. FFFF. FFFF.
C. 192.168.24.59/30
D. 255.255.255.255
E. 172.31.128.255/18
Answer: E
Explanation:
172.31.128.255 is the only unicast address. It seems to be a broadcast address, because of
255 in the last octet, the broadcast address for this network is 172.31.131.255.
Incorrect Answers:
A: 224.1.5.2 is a multicast address.
B: This is a broadcast layer 2 (data link) address.
C: Using a /30 for the subnet mask, this IP address becomes the broadcast address.
D. This is a broadcast IP address.
QUESTION NO: 11
With regard to bridges and switches, which of the following statements are true?
(Choose three.)
A. Switches are primarily software based while bridges are hardware based.
B. Both bridges and switches forward Layer 2 broadcasts.
C. Bridges are frequently faster than switches.
D. Switches typically have a higher number of ports than bridges.
E. Bridges define broadcast domain while switches define collision domains.
F. Both bridges and switches make forwarding decisions based on Layer 2 addresses.
Answer: B, E, F
Explanation:
B, F: Both are layer 2 (data link) devices designed to forward layer 2 broadcasts and
multicast addresses.
E: All hosts within a bridged network comprise a single broadcast domain, while switches
can be used to segment LANs into separate collision domains. Switches are
VLAN-capable while bridges are typically not capable of this.
the above networks broadcast and collision domains? (Select the two best answer
choices)
A. There are two broadcast domains in the network.
B. There are four broadcast domains in the network.
C. There are six broadcast domains in the network.
D. There are four collision domains in the network.
E. There are five collision domains in the network.
F. There are seven collision domains in the network.
Answer: A, F
Explanation:
In this network we have a hub being used in the Sales department, and a switch being
used in the Production department. Based on this, we have two broadcast domains: one
for each network being separated by a router. For the collision domains, we have 5
computers and one port for E1 so we have 6 collision domains total because we use a
switch in the Production Department so 5 are created there, plus one collision domain for
the entire Sales department because a hub is being used.
QUESTION NO: 4
LAN consists of one large flat network. You decide to
segment this LAN into two separate networks with a router. What will be the affect
of this change?
A. The number of broadcast domains will be decreased.
B. It will make the broadcasting of traffic between domains more efficient between
segments.
C. It will increase the number of collisions.
D. It will prevent segment 1's broadcasts from getting to segment 2.
E. It will connect segment 1's broadcasts to segment 2.
Answer: D
Explanation
A router does not forward broadcast traffic. It therefore breaks up a broadcast domain,
reducing unnecessary network traffic. Broadcasts from one segment will not be seen on
the other segment.
Incorrect Answers:
A. This will actually increase the number of broadcast domains from one to two.
B. All link level traffic from segment one to segment two will now need to be routed
between the two interfaces of the router. Although this will reduce the traffic on the LAN
links, it does also provide a less efficient transport between the segments.
C. Since the network size is effectively cut into half, the number of collisions should
decrease dramatically.
E. Broadcasts from one segment will be completely hidden from the other segment.
QUESTION NO: 5
Which of the following are benefits of segmenting a network with a router? (Select
all that apply)
A. Broadcasts are not forwarded across the router.
B. All broadcasts are completely eliminated.
C. Adding a router to the network decreases latency.
D. Filtering can occur based on Layer 3 information.
E. Routers are more efficient than switches and will process the data more quickly.
F. None of the above.
Answer: A, D
Explanation
Routers do not forward broadcast messages and therefore breaks up a broadcast domain.
In addition, routers can be used to filter network information with the use of access lists.
Incorrect Answers:
B. Broadcasts will still be present on the LAN segments. They will be reduced, because
routers will block broadcasts from one network to the other.
C. Adding routers, or hops, to any network will actually increase the latency.
E. The switching process is faster than the routing process. Since routers must do a layer
3 destination based lookup in order to reach destinations, they will process data more
slowly than switches.
QUESTION NO: 10
Given the choices below, which address represents a unicast address?
A. 224.1.5.2
B. FFFF. FFFF. FFFF.
C. 192.168.24.59/30
D. 255.255.255.255
E. 172.31.128.255/18
Answer: E
Explanation:
172.31.128.255 is the only unicast address. It seems to be a broadcast address, because of
255 in the last octet, the broadcast address for this network is 172.31.131.255.
Incorrect Answers:
A: 224.1.5.2 is a multicast address.
B: This is a broadcast layer 2 (data link) address.
C: Using a /30 for the subnet mask, this IP address becomes the broadcast address.
D. This is a broadcast IP address.
QUESTION NO: 11
With regard to bridges and switches, which of the following statements are true?
(Choose three.)
A. Switches are primarily software based while bridges are hardware based.
B. Both bridges and switches forward Layer 2 broadcasts.
C. Bridges are frequently faster than switches.
D. Switches typically have a higher number of ports than bridges.
E. Bridges define broadcast domain while switches define collision domains.
F. Both bridges and switches make forwarding decisions based on Layer 2 addresses.
Answer: B, E, F
Explanation:
B, F: Both are layer 2 (data link) devices designed to forward layer 2 broadcasts and
multicast addresses.
E: All hosts within a bridged network comprise a single broadcast domain, while switches
can be used to segment LANs into separate collision domains. Switches are
VLAN-capable while bridges are typically not capable of this.
Question Set 1.
QUESTION NO: 1
Which of the following devices can an administrator use to segment their LAN?
(Choose all that apply)
A. Hubs
B. Repeaters
C. Switches
D. Bridges
E. Routers
F. Media Converters
G. All of the above
Answer: C, D, E
Explanation:
Routers, switches, and bridges don't transmit broadcasts. They segment a large
cumbersome network, into multiple efficient networks.
Incorrect Answers:
A. Hubs is incorrect because a hub doesn't segment a network, it only allows more hosts
on one. Hubs operate at layer one, and is used primarily to physically add more stations to
the LAN.
B. This also incorrect because the job of a repeater is to repeat a signal so it can exceed
distance limitations. It also operates at layer one and provides no means for logical LAN
segmentation.
F. This is incorrect because media converters work by converting data from a different
media type to work with the media of a LAN. It also operates at layer one and provides no
means for logical LAN segmentation.
QUESTION NO: 2
Routers perform which of the following functions? (Select three)
A. Packet switching
B. Collision prevention on a LAN segment.
C. Packet filtering
D. Broadcast domain enlargement
E. Broadcast forwarding
F. Internetwork communication
Answer: A, C, F
Explanation:
A. Routers work in Layer 3 of the OSI Model. A major function of the router is to route
packets between networks.
C. Through the use of access lists, routers can permit and deny traffic using layer 3 and
layer 4 packet information.
F. The primary purpose of a router is to route traffic between different networks, allowing
for internetworking.
Incorrect Answers:
B. While routers can be used to segment LANs, which will reduce the amount of
devices on a LAN segment, the possibility of a collision exists, whether a router is used
or not.
D. The broadcast domain of a LAN is often segmented through the use of a router. This
results in reducing the size of the broadcast domain.
E. Routers do not forward broadcast traffic.
Which of the following devices can an administrator use to segment their LAN?
(Choose all that apply)
A. Hubs
B. Repeaters
C. Switches
D. Bridges
E. Routers
F. Media Converters
G. All of the above
Answer: C, D, E
Explanation:
Routers, switches, and bridges don't transmit broadcasts. They segment a large
cumbersome network, into multiple efficient networks.
Incorrect Answers:
A. Hubs is incorrect because a hub doesn't segment a network, it only allows more hosts
on one. Hubs operate at layer one, and is used primarily to physically add more stations to
the LAN.
B. This also incorrect because the job of a repeater is to repeat a signal so it can exceed
distance limitations. It also operates at layer one and provides no means for logical LAN
segmentation.
F. This is incorrect because media converters work by converting data from a different
media type to work with the media of a LAN. It also operates at layer one and provides no
means for logical LAN segmentation.
QUESTION NO: 2
Routers perform which of the following functions? (Select three)
A. Packet switching
B. Collision prevention on a LAN segment.
C. Packet filtering
D. Broadcast domain enlargement
E. Broadcast forwarding
F. Internetwork communication
Answer: A, C, F
Explanation:
A. Routers work in Layer 3 of the OSI Model. A major function of the router is to route
packets between networks.
C. Through the use of access lists, routers can permit and deny traffic using layer 3 and
layer 4 packet information.
F. The primary purpose of a router is to route traffic between different networks, allowing
for internetworking.
Incorrect Answers:
B. While routers can be used to segment LANs, which will reduce the amount of
devices on a LAN segment, the possibility of a collision exists, whether a router is used
or not.
D. The broadcast domain of a LAN is often segmented through the use of a router. This
results in reducing the size of the broadcast domain.
E. Routers do not forward broadcast traffic.
Subscribe to:
Posts (Atom)