Q1. Given IP Address 10.0.5.130/25, Find the Network Address
A /25 prefix leaves:
- 32 − 25 = 7 host bits
Therefore:
2⁷ = 128 addresses per subnet
The subnet boundaries increase in blocks of 128:
- 0–127
- 128–255
The address 10.0.5.130 falls within the range 128–255.
Answer: 10.0.5.128
Q2. Find the Broadcast Address for 10.0.5.130/25
The subnet range is:
10.0.5.128 – 10.0.5.255
The last address of the subnet becomes the broadcast address.
Answer: 10.0.5.255
Q3. Find the Usable Host Range for 10.0.5.130/25
Usable hosts exclude:
Network Address
Broadcast Address
For the subnet:
10.0.5.128 – 10.0.5.255
Therefore:
First Usable Host → 10.0.5.129
Last Usable Host → 10.0.5.254
Answer: 10.0.5.129 – 10.0.5.254
Q4. How Many Usable Hosts Are Available in a /28 Subnet?
A /28 prefix leaves:
- 32 − 28 = 4 host bits
Total addresses:
2⁴ = 16
Usable hosts exclude the network address and broadcast address:
16 − 2 = 14
Answer: 14 usable hosts
Q5. What Is the Subnet Mask for /26?
A /26 mask contains 26 ones followed by 6 zeros.
Binary representation:
11111111.11111111.11111111.11000000
Decimal representation:
255.255.255.192
Answer: 255.255.255.192
Q6. Are 172.16.10.45/20 and 172.16.18.90/20 in the Same Subnet?
A /20 prefix means the subnet mask is:
255.255.240.0
The block size in the third octet is:
256 − 240 = 16
So the third-octet subnet ranges are:
0–15
16–31
32–47
48–63
For 172.16.10.45, the third octet is 10, which falls in 0–15.
For 172.16.18.90, the third octet is 18, which falls in 16–31.
They are in different subnet ranges.
Answer: No, they are not in the same subnet.
Q7. What Prefix Is Needed for at Least 100 Usable Hosts?
Compare common subnet sizes:
- /26 → 64 total → 62 usable
- /25 → 128 total → 126 usable
Since 100 hosts cannot fit into a /26 subnet, the next suitable prefix is /25.
Answer: /25
### Q8. How Many /28 Subnets Can Be Created from a /24 Network?
Borrowed bits:
28 − 24 = 4 bits
Number of subnets:
2⁴ = 16
Answer: 16 subnets
Q9. What Are the Subnet Ranges When Splitting 192.168.10.0/24 into /28?
A /28 subnet contains:
2⁴ = 16 addresses
Therefore, subnet boundaries increase by 16:
192.168.10.0 – 192.168.10.15
192.168.10.16 – 192.168.10.31
192.168.10.32 – 192.168.10.47
192.168.10.48 – 192.168.10.63
192.168.10.64 – 192.168.10.79
192.168.10.80 – 192.168.10.95
192.168.10.96 – 192.168.10.111
192.168.10.112 – 192.168.10.127
192.168.10.128 – 192.168.10.143
192.168.10.144 – 192.168.10.159
192.168.10.160 – 192.168.10.175
192.168.10.176 – 192.168.10.191
192.168.10.192 – 192.168.10.207
192.168.10.208 – 192.168.10.223
192.168.10.224 – 192.168.10.239
192.168.10.240 – 192.168.10.255
Answer: 16 subnets, each containing 16 addresses.
Q10. Can 192.168.10.32/28 Be Assigned to a Host?
The subnet:
192.168.10.32/28
starts at address:
192.168.10.32
This is the network address of that subnet.
Network addresses cannot be assigned to hosts.
Answer: No
Be the first to add a comment.