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Network Infrastructure and ConnectivityhardMultiple ChoiceObjective-mapped

CCNA Network Infrastructure and Connectivity Practice Question

A host is configured with 10.10.10.130/25. What is the network address of its subnet?

⚠ Common exam trap

Be careful not to confuse the subnet mask with /24 or mistake the broadcast address for the network address.

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

10.10.10.128

A /25 divides the address space into two blocks of 128 addresses. In plain language, the ranges are 0–127 and 128–255. Since the host ends in 130, it belongs to the 128–255 half. That means the network address is 10.10.10.128. This is a common subnet-boundary question because it tests whether you can map a host address into the correct prefix block quickly and confidently.

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

  • 10.10.10.0

    Why it's wrong here

    The /25 mask creates exactly two subnets in the 10.10.10.0/24 space: 10.10.10.0/25 (hosts .1–.126) and 10.10.10.128/25 (hosts .129–.254). Since .130 is 10000010 in binary, the most significant bit is 1, which places it in the upper subnet. Therefore, the starting address .0 is not the network address for this host.

    When this WOULD be correct

    If the question asked for the network address of a host configured with 10.10.10.130/24, then 10.10.10.0 would be the correct answer, as it represents the network address for that subnet mask.

  • 10.10.10.64

    Why it's wrong here

    A /25 subnet mask only borrows the highest-order bit of the last octet, so the subnet block size is 256 / 2 = 128, and network addresses increment by 128 (e.g., .0, .128). The address .64 would only be a network address with a /26 mask, which borrows two bits and creates boundaries every 64 addresses. Since the prefix length is /25, .64 is not a valid network boundary and cannot be the network address.

    When this WOULD be correct

    If the question asked for the first usable IP address of the subnet 10.10.10.64/26, then option B would be correct, as 10.10.10.64 would be the network address for that subnet.

  • 10.10.10.128

    Why this is correct

    A /25 mask (255.255.255.128) partitions the last octet into two equal blocks of 128 addresses: .0–.127 and .128–.255. The address 10.10.10.130 falls in the upper block because its last octet binary begins with a 1 (10000010). Therefore, the network address is the first address of that block, 10.10.10.128, and this is the correct answer.

  • 10.10.10.255

    Why it's wrong here

    The address .255 is the broadcast address for the 10.10.10.128/25 network, not its network ID. In a /25 subnet, the valid host addresses are .129 through .254, and .255 is reserved as the directed broadcast. A network address is always the lowest address in the range, so .255 cannot be a network address.

    When this WOULD be correct

    If the question asked for the broadcast address of the subnet 10.10.10.128/25, then 10.10.10.255 would be the correct answer, as it is the last address in that subnet range.

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.

10.10.10.128Correct answer

Why this is correct

A /25 mask (255.255.255.128) partitions the last octet into two equal blocks of 128 addresses: .0–.127 and .128–.255. The address 10.10.10.130 falls in the upper block because its last octet binary begins with a 1 (10000010). Therefore, the network address is the first address of that block, 10.10.10.128, and this is the correct answer.

10.10.10.0Wrong answer — click to see why

Why this is wrong here

10.10.10.0 is the network address of the 10.10.10.0/25 subnet (range 0-127), but the host IP 10.10.10.130 is not in that range. The /25 mask creates two subnets: 10.10.10.0/25 and 10.10.10.128/25, and .130 belongs to the latter.

★ When this WOULD be the correct answer

If the question asked for the network address of a host configured with 10.10.10.130/24, then 10.10.10.0 would be the correct answer, as it represents the network address for that subnet mask.

Why candidates choose this

Students might assume that any IP starting with 10.10.10.x belongs to the 10.10.10.0 network, forgetting that the subnet mask determines the boundary. The .0 address is a common network address for /24, but with /25, the network address changes.

10.10.10.64Wrong answer — click to see why

Why this is wrong here

10.10.10.64 is not a valid network address for any /25 subnet derived from 10.10.10.0/24. A /25 subnet has a block size of 128, so the network addresses are multiples of 128: 0 and 128. 64 is a multiple of 64, which would be a /26 boundary, not /25.

★ When this WOULD be the correct answer

If the question asked for the first usable IP address of the subnet 10.10.10.64/26, then option B would be correct, as 10.10.10.64 would be the network address for that subnet.

Why candidates choose this

Students often confuse subnet boundaries for different prefix lengths. For example, a /26 has boundaries at multiples of 64, so 10.10.10.64 is a valid network for /26. This option tempts those who misapply the /26 boundary to a /25 question.

10.10.10.255Wrong answer — click to see why

Why this is wrong here

10.10.10.255 is the broadcast address for the 10.10.10.128/25 subnet, not the network address. The broadcast address is the last address in the subnet (all host bits set to 1), while the network address is the first address (all host bits set to 0).

★ When this WOULD be the correct answer

If the question asked for the broadcast address of the subnet 10.10.10.128/25, then 10.10.10.255 would be the correct answer, as it is the last address in that subnet range.

Why candidates choose this

Students often confuse the broadcast address with the network address because both are special addresses in a subnet. The .255 address is a common broadcast address for /24 subnets, but for a /25, the broadcast is 10.10.10.255 only if the subnet is 10.10.10.0/25; here the subnet is 10.10.10.128/25.

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?”

Visual reference

Switch VLAN 10 Sales (192.168.10.0/24) PC-A PC-B VLAN 20 HR (192.168.20.0/24) PC-C PC-D Router VLANs isolate traffic — inter-VLAN routing requires a Layer 3 device

Quick reference

Routing Protocol Comparison

ProtocolMetricMax HopsAlgorithmType
RIP v2Hop count15Bellman-FordDistance vector
OSPFCost (bandwidth)UnlimitedDijkstra (SPF)Link state
EIGRPComposite metricUnlimitedDUALHybrid
IS-ISCostUnlimitedDijkstraLink state
BGPPolicy / attributesUnlimitedPath vectorPath vector

RIP's 15-hop limit makes it unsuitable for large networks. OSPF and EIGRP dominate modern enterprise deployments.

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Written by Johnson Ajibi, MSc IT Security

Senior Network & Security Engineer · founder of Courseiva

This 200-301 practice question is part of Courseiva's free Cisco certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the 200-301 exam.