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- Question
You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
Options- A. 172.16.17.1 255.255.255.252
- B. 172.16.0.1 255.255.240.0
- C. 172.16.20.1 255.255.254.0
- D. 172.16.18.255 255.255.252.0
- Correct Answer
- 172.16.18.255 255.255.252.0
ExplanationA Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask listed, and 172.16.18.255 is a valid host. - 1. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
Options- A. 172.16.36.0
- B. 172.16.48.0
- C. 172.16.64.0
- D. 172.16.0.0 Discuss
- 2. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224?
Options- A. Router(config)#ip classless
- B. Router(config)#no ip classful
- C. Router(config)#ip unnumbered
- D. Router(config)#ip subnet-zero Discuss
- 3. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Options- A. 255.255.255.192
- B. 255.255.255.224
- C. 255.255.255.240
- D. 255.255.255.248 Discuss
- 4. What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?
Options- A. 14
- B. 15
- C. 16
- D. 30 Discuss
- 5. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
Options- A. 200.10.5.56
- B. 200.10.5.32
- C. 200.10.5.64
- D. 200.10.5.0 Discuss
- 6. You have an interface on a router with the IP address of 192.168.192.10/29. What is the broadcast address the hosts will use on this LAN?
Options- A. 192.168.192.15
- B. 192.168.192.31
- C. 192.168.192.63
- D. 192.168.192.127 Discuss
- 7. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?
Options- A. 192.168.19.0 255.255.255.0
- B. 192.168.19.33 255.255.255.240
- C. 192.168.19.26 255.255.255.248
- D. 192.168.19.31 255.255.255.248 Discuss
- 8. If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?
Options- A. 172.16.112.0
- B. 172.16.0.0
- C. 172.16.96.0
- D. 172.16.255.0 Discuss
- 9. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet. How many bits must you borrow from the host field to provide the correct subnet mask?
Options- A. 2
- B. 3
- C. 4
- D. 5 Discuss
- 10. On a VLSM network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
Options- A. /27
- B. /28
- C. /29
- D. /30 Discuss
Subnetting problems
Search Results
Correct Answer: 172.16.64.0
Explanation:
A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.
Correct Answer: Router(config)#ip subnet-zero
Explanation:
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets, each with 30 hosts. However, if the command
ip subnet-zero is not used, then only 6 subnets would be available for use.
Correct Answer: 255.255.255.224
Explanation:
You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts-this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts. This is the best answer.
Correct Answer: 30
Explanation:
A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets, each with 30 hosts. Does it matter if this mask is used with a Class A, B, or C network address? Not at all. The number of host bits would never change.
Correct Answer: 200.10.5.64
Explanation:
This is a pretty simple question. A /28 is 255.255.255.240, which means that our block size is 16 in the fourth octet. 0, 16, 32, 48, 64, 80, etc. The host is in the 64 subnet.
Correct Answer: 192.168.192.15
Explanation:
A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0, 8, 16, 24, etc. 10 is in the 8 subnet. The next subnet is 16, so 15 is the broadcast address.
Correct Answer: 192.168.19.26 255.255.255.248
Explanation:
A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.
Correct Answer: 172.16.112.0
Explanation:
A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on-which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.
Correct Answer: 5
Explanation:
A 240 mask is 4 subnet bits and provides 16 subnets, each with 14 hosts. We need more subnets, so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 host bits (6 hosts per subnet). This is the best answer.
Correct Answer: /30
Explanation:
A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.
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