Subnet Mask Required for 192.168.25.100 and 192.168.25.128

200-301 By islever Editorial Last updated

Configure and verify IPv4 addressing and subnetting

Short answer

C Configure 255.255.255.0 (/24) on both PCs so 192.168.25.100 and 192.168.25.128 share the 192.168.25.0/24 subnet and can communicate locally.

Question

An engineer must update the configuration on two PCs in two different subnets to communicate locally with each other. One PC is configured with IP address
192.168.25.128/25 and the other with 192.168.25.100/25. Which network mask must the engineer configure on both PCs to enable the communication?
  1. 255.255.255.248
  2. 255.255.255.224
  3. 255.255.255.0 Correct Answer
  4. 255.255.255.252

Key Takeaways

It tests whether you can convert a mask to block boundaries and recognize that 192.168.25.128 is unusable as a host under /25, /27, /29 and /30; the trap is treating .128 as an ordinary address despite being a network ID.

This subnetting question asks which mask lets 192.168.25.100 and 192.168.25.128 communicate locally; the answer is 255.255.255.0 (/24) because it places both addresses in 192.168.25.0/24.

Common mistake: Choosing 255.255.255.224 (B) because its /27 block looks large enough numerically; however, .100 is in 192.168.25.96/27 while .128 is the network address of 192.168.25.128/27, so they still do not share a subnet.

Expert Analysis

Local delivery starts with a mask match

The governing rule is Cisco's IPv4 addressing model: a host treats a destination as local only when the destination's 32-bit address, masked with the host's own subnet mask, equals the host's own masked network address. The mask therefore defines the block; you cannot decide locality from the first three octets alone.

Why the supplied /25 fails

A /25 mask is 255.255.255.128, which splits 192.168.25.0/24 into two blocks: 192.168.25.0–192.168.25.127 and 192.168.25.128–192.168.25.255. The PC at 192.168.25.100 is a host in the first block, but 192.168.25.128 is the all-zero host address (network ID) of the second block, not a usable host. Thus the two PCs cannot communicate locally as currently addressed.

Check each candidate

A 255.255.255.248 (/29) gives blocks of eight;.128 is a network ID and.100 sits in 192.168.25.96/29. B 255.255.255.224 (/27) gives blocks of 32;.100 is in 192.168.25.96/27 and.128 is the network ID of 192.168.25.128/27. D 255.255.255.252 (/30) gives blocks of four;.100 is in 192.168.25.100/30 and.128 is a network ID. Only C, 255.255.255.0 (/24), makes both.100 and.128 host addresses in 192.168.25.0/24. Community explanations make the same point about merging the two /25s into a larger /24 and note that.128 must stop being a network address.

What to configure

The engineer should change both PCs to 255.255.255.0. This is a supernet relative to the original /25 and gives 192.168.25.0/24, where.100 and.128 are ordinary hosts. If strict subnet isolation were required, a router or Layer 3 device would be needed; the question explicitly asks for local communication, so configuring the common /24 mask is the documented fix. Correct answer: C.

Official Reference

https://www.cisco.com/c/en/us/support/docs/ip/routing-information-protocol-rip/13788-3.html

This link opens the official vendor documentation in a new tab.

Exam Strategy

For mask questions, convert each candidate to a block size using 256 minus the last nonzero octet, then test the suspicious address first: 192.168.25.128 is the trap because under /25, /27, /29 and /30 it becomes a network ID. Find the smallest mask whose range contains both endpoints as host addresses, and verify neither endpoint is all-zeros or all-ones. This avoids re-subnetting every candidate from scratch.

Frequently Asked Questions

Why are 192.168.25.100/25 and 192.168.25.128/25 not in the same subnet?

A /25 divides the last octet at 128: 192.168.25.100 is in 192.168.25.0/25, while 192.168.25.128 is the network address of 192.168.25.128/25, not a usable host in that subnet.

Why does 255.255.255.0 allow both addresses to communicate locally?

With /24, the range is 192.168.25.0–192.168.25.255; 192.168.25.100 and 192.168.25.128 are both host addresses inside it, so they use the same Layer 3 subnet.

Why not use 255.255.255.224 or 255.255.255.248?

Under /27, /29 or /30, 192.168.25.128 is a network address for its block, and .100 falls in a different block; only /24 creates a shared subnet containing both.

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Discussion 20 contributions

Compiled from public community threads and reader submissions, lightly edited for clarity. Handles and vote tallies are shown as recorded, and comments reflect the views of individual contributors.

From public community threads 20

  1. Customexit 👍 23
    Just to add more info here: A, .248 has a group size of 8. At a glance that's too small to include both .100 and .128. .252 has a group size of 4. Same as above. .224 seems large enough with a 32 group size, but if you subnet you'll find that .128 is a network address. That leaves us with 255.255.255.0. Which gives us the first usable at .25.1 and the last usable at .25.254.
  2. Anonymous learner 👍 11
    I hate this kind of question because I know the answer is C but to be confirmed and confident with your answer you need to calculate the other answer too hence wasting the time.
  3. MelbourneJin 👍 4
    To enable communication between the two PCs in different subnets, the engineer needs to configure both PCs with the same network mask. In this case, the PCs have IP addresses 192.168.25.128/25 and 192.168.25.100/25, indicating that they are using a subnet mask of 255.255.255.128. Therefore, the engineer must configure both PCs with the subnet mask 255.255.255.128 to ensure they are in the same subnet and can communicate with each other locally.
  4. g_h_97 👍 4
    192.168.25.128/25 is the network address, I guess they meant 192.168.25.129/25
  5. AbdullahMohammad251 👍 3
    we had 2 networks of /25 SM (192.168.25.0/25) and (192.168.25.128/25) they were merged into a larger network (192.168.25.0/24) Note that 192.168.25.128/25 is not a vlaid IP address, since it is the network ID of the 2nd subnet of a /25 SM. After merging the two networks, it becomes valid (usable IP address range: 192.168.25.1-192.168.25.254)
Show 15 more comments
  1. wav2011 👍 3
    One PC is configured with IP address 192.168.25.128/25 - wrong condition, this is network id. incorrect question. Answer is C
  2. bugmenot240410 👍 3
    /25 is 255.255.255.128, only C became larger host bits
  3. AbdullahMohammad251 👍 2 Answered C
    The two pcs cannot communicate with one another because they lie in different subnets. To enable the communication between the two pcs, they must be in the same subnet. The solution is to use what is known as supernetting (merging smaller networks into a larger network.) Option C will yield both addresses in the same subnet, therefore; communication between the two pcs is available. Option C is the only valid option since the SM for the new network has to be less than the SM of the individual small networks (because when you merge, you are reducing the number of networks.)
  4. ccna2025 👍 2 Answered C
    if you subnet everything. 192.168.25.128 - original host IP can only fit into the /24 network because the other masks would cause the .128 to be the network ID. the only way to prevent the network id from being .128 is to use the /24 mask
  5. Trdelnik 👍 2
    i think the implication was the initial config wouldn't work, so what should it be instead...? i thought the same thing myself until i saw that /24 in the answers
  6. AbdullahMohammad251 👍 2
    You don't have to go over each answer. the SM of the new network has to be less than the SM of the individual networks.
  7. AbiZ17 👍 2
    I wonder how 192.168.25.128 is configured to the host coz in a /25 prefix length it is the network. address
  8. THEKYPTONIAN 👍 2
    The subnet must include addresses 100 and 128 so /24 is correct
  9. Iamm 👍 2 Answered C
    only answer could be
  10. Rich_rude 👍 1
    If it helps anybody out you can always use the up range as well like for c 255.255.255.0 that is /24 which has 254 usable ips. Thats should for the range of 192.168.25.100 - 192.168.25.128 provided in the question The range with /24 give us the 192.168.25.0 - 192.168.25.255 which covers those ips in questions just add your usable ips to the last octet to figure out range
  11. oranjo 👍 1
    None of the options listed represent the correct subnet mask for these specific IP addresses (192.168.25.128/25 and 192.168.25.100/25) to communicate within the same subnet. The accurate subnet mask for these IP addresses within the same subnet should be 255.255.255.128 (/25).
  12. MelbourneJin 👍 1
    It makes me so confused. IP address .128 is a network IP, not assignable to a PC.
  13. herbcla 👍 1 Answered B
    It looks as the right one
  14. Da_Costa 👍 1 Answered C
    Use default class C mask
  15. soRwatches 👍 1
    same thought.

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