# 1. Consider the network that the host 10.47.124.51/27 is a part of....

## Question

1. Consider the network that the host 10.47.124.51/27 is a part of.
a. How many addresses does this subnet contain?
b. How many usable hosts does the subnet have?
c. What is the subnet mask for the last octet?
d. How many subnets of this size are possible in a class C network?
e. Write the last netmask octet in binary?
2. Consider the subnet that the host 10.18.191.39/23 is a part of.
a. How many addresses does this subnet contain?
b. How many usable hosts does the subnet have?
c. What is the subnet mask for the last octet?
d. How many subnets of this size are possible in a class C network?
e. Write the last netmask octet in binary.
3. What is the last valid host on the network that the host 10.47.191.189/30 is a part of?
4. What is the CIDR shorthand (ie, /XX) corresponding to a subnet mask of 255.255.240.0?
5. Enter the valid host range for the network that the IP address 172.27.61.236 255.255.128.0 is a part of.
a. First Host:
b. Last Host:
6. What subnet mask would you use for the 172.30.0.0 network, such that you can get 50 subnets and 990 hosts per subnet?
7. What subnet does the host 10.185.175.14/20 belong to?
8. What is the first valid host on the network that the host 192.168.121.68/28 is a part of?
9. What is the maximum number of valid subnets and hosts per subnet that you can get from the network 192.168.81.0 255.255.255.224?
10. Enter the valid host range for the network that the IP address 172.28.43.253 255.255.255.192 is a part of:
a. First Host:
b. Last Host:

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1.
a. 10. 47.124.51/27 => 27 mask bits.
We have 32-27=5 host bits in the mask, so the number of addresses is 2^5=32.
b. Among the 32 addresses, the amount of usable hosts is 32-2=30 (we canâ€™t use as host address the subnetwork and broadcast addresses)....

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