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Network Infrastructure and ConnectivityhardMultiple ChoiceObjective-mapped

CCNA Network Infrastructure and Connectivity Practice Question

A host is configured with IP address 192.168.70.18/30. Which addresses belong to the same subnet block?

⚠ Common exam trap

Be careful not to confuse adjacent subnet blocks or miscalculate the block size of a /30 subnet.

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

192.168.70.16 through 192.168.70.19

A /30 subnet has a block size of 4. In practical terms, the relevant blocks in the last octet are 0–3, 4–7, 8–11, 12–15, 16–19, and so on. Because 18 falls inside the 16–19 block, the subnet includes network address .16, usable hosts .17 and .18, and broadcast .19. This question checks whether you can identify the correct /30 block and understand all addresses that fall inside it.

Answer analysis

Option-by-option breakdown

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

  • 192.168.70.16 through 192.168.70.19

    Why this is correct

    A /30 prefix (255.255.255.252) defines a block of exactly 4 addresses aligned to multiples of 4. The IP 192.168.70.18 falls within the block 192.168.70.16/30, where .16 is the network address, .17 and .18 are usable host addresses, and .19 is the broadcast address. Therefore the entire subnet range is 192.168.70.16 through 192.168.70.19.

  • 192.168.70.18 through 192.168.70.21

    Why it's wrong here

    Starting the range at .18 would imply a network boundary at .18, but /30 subnets must begin at a multiple of 4 (i.e., .0, .4, .8, .12, .16, .20). The block 192.168.70.18 through .21 arbitrarily splits the real .16/.30 subnet and the following .20/.30 subnet, so it cannot represent a valid /30 addressing scheme. The host .18 belongs exclusively to the .16-.19 block.

    When this WOULD be correct

    If the question asked for the range of addresses in a /29 subnet instead of /30, then option B would be correct, as a /29 subnet allows for IP addresses from 192.168.70.16 to 192.168.70.23.

  • 192.168.70.12 through 192.168.70.15

    Why it's wrong here

    192.168.70.12 through .15 is the /30 subnet immediately preceding the one that holds .18, with network address .12 and broadcast .15. A /30 boundary occurs every four addresses, so the block containing .18 must start at .16, not at .12. Thus this range is too early to include 192.168.70.18.

    When this WOULD be correct

    If the question were to ask for the range of addresses in the subnet defined by a different CIDR notation, such as 192.168.70.12/30, then the range 192.168.70.12 through 192.168.70.15 would be correct, as it would represent the valid addresses in that subnet.

  • 192.168.70.20 through 192.168.70.23

    Why it's wrong here

    The range 192.168.70.20 through .23 is the next sequential /30 subnet after the one containing .18. Since the network .16/.30 spans .16-.19, the address .18 is not part of the .20/.30 block. Choosing this range confuses the next network boundary with the actual subnet for this host.

    When this WOULD be correct

    If the question asked for the addresses in the subnet of 192.168.70.20/30, then option D would be correct, as it would include the range from 192.168.70.20 to 192.168.70.23, which belongs to that subnet.

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.

192.168.70.16 through 192.168.70.19Correct answer

Why this is correct

A /30 prefix (255.255.255.252) defines a block of exactly 4 addresses aligned to multiples of 4. The IP 192.168.70.18 falls within the block 192.168.70.16/30, where .16 is the network address, .17 and .18 are usable host addresses, and .19 is the broadcast address. Therefore the entire subnet range is 192.168.70.16 through 192.168.70.19.

192.168.70.18 through 192.168.70.21Wrong answer — click to see why

Why this is wrong here

A /30 subnet always has a block size of 4 addresses, starting at multiples of 4. The block starting at .18 would be 192.168.70.16–.19, not .18–.21. The range .18–.21 crosses a subnet boundary and includes addresses from two different subnets.

★ When this WOULD be the correct answer

If the question asked for the range of addresses in a /29 subnet instead of /30, then option B would be correct, as a /29 subnet allows for IP addresses from 192.168.70.16 to 192.168.70.23.

Why candidates choose this

Students might think that since .18 is the host address, the subnet includes .18 and the next three addresses, not realizing that subnet boundaries are fixed at multiples of the block size.

192.168.70.12 through 192.168.70.15Wrong answer — click to see why

Why this is wrong here

The block 192.168.70.12–.15 is a different /30 subnet (network .12, broadcast .15). The host .18 belongs to the subnet .16–.19, not .12–.15.

★ When this WOULD be the correct answer

If the question were to ask for the range of addresses in the subnet defined by a different CIDR notation, such as 192.168.70.12/30, then the range 192.168.70.12 through 192.168.70.15 would be correct, as it would represent the valid addresses in that subnet.

Why candidates choose this

A student might miscalculate the block size or confuse the subnet boundaries, thinking that .18 falls into the previous block because of incorrect arithmetic.

192.168.70.20 through 192.168.70.23Wrong answer — click to see why

Why this is wrong here

The block 192.168.70.20–.23 is the next /30 subnet (network .20, broadcast .23). The host .18 is not in this range; it is in the .16–.19 subnet.

★ When this WOULD be the correct answer

If the question asked for the addresses in the subnet of 192.168.70.20/30, then option D would be correct, as it would include the range from 192.168.70.20 to 192.168.70.23, which belongs to that subnet.

Why candidates choose this

Students might incorrectly add the block size to the host address (18+4=22) and assume the subnet starts at .20, forgetting that subnet boundaries are based on the network address, not the host address.

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

192.168.1.0 /24 256 addresses (254 usable) 192.168.1.0 /25 Subnet A 128 addr (126 usable) 192.168.1.128 /25 Subnet B 128 addr (126 usable) Borrowing 1 bit from host portion creates 2 subnets (/25)

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.