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.