QUESTION 41:
Which one of the binary bit patterns below denotes a Class B address?
A. 0xxxxxxx
B. 10xxxxxx
C. 110xxxxx
D. 1110xxxx
E. 11110xxx
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 42:
The Certkiller network consists of 5 different departments as shown below:
You are a systems administrator at Certkiller and you've just acquired a new Class
C IP network. Which of one of the subnet masks below is capable of providing one useful subnet for each of the above departments (support, financial, sales & development) while still allowing enough usable host addresses to meet the needs of each department?
A. 255.255.255.128
B. 255.255.255.192
C. 255.255.255.224
D. 255.255.255.240
E. 255.255.255.248
F. 255.255.255.252
Explanation:
The network currently consists of 5 subnets. We need to subnet the Class C network into at least 5 subnets. This requires that we use 3 bits for the network address. Using the formula 2n-2 we get 6. This also leaves us with 5 bits for hosts, which gives us 30 hosts.
Incorrect Answers:
A. Only 1 bit is required to give us 128 but 1 bit gives us 0 subnets.
2 bits are required to give us 192 but 2 bits gives us only 2 subnets. This is too few.
D. 4 bits are required to give us 240. This gives us 14 subnets. However we are left with 4 bits for hosts leaving us with 14 host addresses. Two of the networks require more than 14 hosts so this will not do.
E. 5 bits are required to give us 248. This gives us 30 subnets. However we are left with 3 bits for hosts leaving us with 6 host addresses. All the networks require more than 6 hosts so this will not do.
F. 6 bits are required to give us 252. This gives us 62 subnets. However we are left with 2 bits for hosts leaving us with 2 host addresses. This is too few.
QUESTION 43:
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.
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 44:
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
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 45:
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
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 46:
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
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 47:
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
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 48:
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
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 49:
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
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 50:
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 62 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
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.
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