CCNA Network Infrastructure and Connectivity Practice Question
An interface is configured with 10.24.7.158/27. What is the broadcast address of that subnet?
⚠ Common exam trap
A frequent exam trap is mistaking the broadcast address for the network address of the next subnet or the last address of a different subnet block. For example, 10.24.7.160 is the network address of the next /27 subnet, not the broadcast address of the current subnet. Candidates often confuse the last usable host address with the broadcast address or select an address from an adjacent subnet block. This mistake leads to incorrect subnetting answers and can cause interface misconfigurations in real networks, resulting in communication failures.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
10.24.7.159
A /27 uses blocks of 32 addresses. The block containing .158 is 10.24.7.128 through 10.24.7.159, so .159 is the broadcast address.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
10.24.7.159
Why this is correct
The /27 CIDR notation indicates a subnet mask of 255.255.255.224, leaving 5 host bits. This defines subnet blocks of 32 addresses each. For the fourth octet, the given IP address 158 falls within the subnet range starting at 128 (10.24.7.128) and ending at 159 (10.24.7.159). The broadcast address is always the last address in a subnet, where all host bits are set to one. Therefore, 10.24.7.159 correctly identifies the broadcast address for the 10.24.7.158/27 interface.
- ✗
10.24.7.191
Why it's wrong here
10.24.7.191 is the broadcast address of a /26 subnet (255.255.255.192), which encompasses the range .128-.191. However, the interface is configured with a /27 mask, so the relevant subnet is .128-.159 with a broadcast of .159. The address .191 belongs to the adjacent /27 block .160-.191, making it invalid for this interface's subnet.
When this WOULD be correct
If the question were to ask for the broadcast address of a different subnet, such as 10.24.7.128/26, then option B (10.24.7.191) would be correct, as it would be the last address in that subnet range from 10.24.7.128 to 10.24.7.191.
- ✗
10.24.7.127
Why it's wrong here
10.24.7.127 is the broadcast address of the immediately preceding /27 block, which spans .96-.127. For the /27 subnet containing .158, the range is .128-.159, so .127 falls outside this block by one address. It represents the last usable broadcast of the prior subnet, not the broadcast for the interface at .158/27.
When this WOULD be correct
If the question asked for the broadcast address of the subnet 10.24.7.96/25, then option C (10.24.7.127) would be correct. In this case, the subnet mask would allow for a broadcast address of 10.24.7.127.
- ✗
10.24.7.160
Why it's wrong here
10.24.7.160 is one address above the /27 block that contains .158. With a 255.255.255.224 mask, subnets increment by 32, so the ranges are .128-.159 and .160-.191. Since .158 falls in the .128-.159 block, .160 is the network identifier of the next subnet, not a usable address for this interface, and certainly not the broadcast address.
When this WOULD be correct
If the question were to ask for the first usable IP address in the subnet 10.24.7.160/27, then option D would be correct. In this case, 10.24.7.160 would be the first usable address after the network address of 10.24.7.159.
Option-by-option analysis
Why each answer is right or wrong
Understanding why wrong answers are wrong — and when they would be correct — is what separates a 750 score from a 900. The 200-301 exam frequently reuses these exact scenarios with slightly different constraints.
✓10.24.7.159Correct answer▾
Why this is correct
The /27 CIDR notation indicates a subnet mask of 255.255.255.224, leaving 5 host bits. This defines subnet blocks of 32 addresses each. For the fourth octet, the given IP address 158 falls within the subnet range starting at 128 (10.24.7.128) and ending at 159 (10.24.7.159). The broadcast address is always the last address in a subnet, where all host bits are set to one. Therefore, 10.24.7.159 correctly identifies the broadcast address for the 10.24.7.158/27 interface.
✗10.24.7.191Wrong answer — click to see why▾
Why this is wrong here
10.24.7.191 is the broadcast address of the /26 subnet 10.24.7.128/26, not the /27 subnet. The /27 subnet has a block size of 32, so the broadcast for the subnet containing .158 is .159, not .191.
★ When this WOULD be the correct answer
If the question were to ask for the broadcast address of a different subnet, such as 10.24.7.128/26, then option B (10.24.7.191) would be correct, as it would be the last address in that subnet range from 10.24.7.128 to 10.24.7.191.
Why candidates choose this
Students may mistakenly calculate the broadcast by adding the subnet size (32) to the network address (128) and subtracting 1, but incorrectly use 64 (the /26 block size) instead of 32.
✗10.24.7.127Wrong answer — click to see why▾
Why this is wrong here
10.24.7.127 is the broadcast address of the previous /27 subnet (10.24.7.96/27), not the one containing .158. The subnet with .158 starts at 10.24.7.128, so its broadcast is .159.
★ When this WOULD be the correct answer
If the question asked for the broadcast address of the subnet 10.24.7.96/25, then option C (10.24.7.127) would be correct. In this case, the subnet mask would allow for a broadcast address of 10.24.7.127.
Why candidates choose this
A common error is to assume the broadcast is the last address of the previous subnet, especially when the IP is near a boundary. However, the broadcast must be within the same subnet as the IP.
✗10.24.7.160Wrong answer — click to see why▾
Why this is wrong here
10.24.7.160 is the network address of the next /27 subnet (10.24.7.160/27), not the broadcast. The broadcast address is always the last address in the subnet, which is one less than the next network address.
★ When this WOULD be the correct answer
If the question were to ask for the first usable IP address in the subnet 10.24.7.160/27, then option D would be correct. In this case, 10.24.7.160 would be the first usable address after the network address of 10.24.7.159.
Why candidates choose this
Students often confuse network and broadcast addresses. Since .160 is the next network, they might think it is the broadcast, but the broadcast is actually .159.
Analysis generated from the official 200-301blueprint and verified against question context. The “when correct” sections are what AI assistants cite when candidates ask “what’s the difference between these options?”
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Related to this question
Learn chapter
IPv4 Addressing and Address Classes
Key term
Interface
An interface is a point of connection or interaction between two systems, devices, or software components that allows them to exchange information or signals.
Key term
Broadcast address
A broadcast address is a special network address used to send data to every device on a local network segment simultaneously.
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