Subnetting: Validating First Usable Host Addresses

200-301 By islever Editorial Last updated

Configure and verify IPv4 addressing and subnetting Verify IP parameters for Client OS (Windows, Mac OS, Linux)

Short answer

AE Option A correctly uses the first host (.1) of a /23 subnet for 472 hosts, and Option E correctly uses the first host (.65) of a /27 subnet for 24 hosts.

Question

A network engineer must configure two new subnets using the address block 10.70.128.0/19 to meet these requirements:
✑ The first subnet must support 24 hosts.
✑ The second subnet must support 472 hosts.
✑ Both subnets must use the longest subnet mask possible from the address block.
Which two configurations must be used to configure the new subnets and meet a requirement to use the first available address in each subnet for the router
interfaces? (Choose two.)
  1. interface vlan 1148 ip address 10.70.148.1 255.255.254.0 Correct Answer
  2. interface vlan 3002 ip address 10.70.147.17 255.255.255.224
  3. interface vlan 4722 ip address 10.70.133.17 255.255.255.192
  4. interface vlan 1234 ip address 10.70.159.1 255.255.254.0
  5. interface vlan 155 ip address 10.70.155.65 255.255.255.224 Correct Answer

Key Takeaways

The trap lies in assuming any IP within the range is valid; you must calculate the network ID based on the mask and ensure the configured IP is exactly Network ID + 1.

This guide explains how to select the correct subnet configurations by calculating the longest mask for host requirements and verifying that the assigned IP is the first usable address in the network.

Common mistake: Candidates often choose Option D because .1 looks like a standard gateway IP, failing to realize that an odd third octet (159) under a /23 mask places the network at 158.0, making .159.1 the second host, not the first.

Expert Analysis

To solve this, we must determine the correct subnet masks and verify that the provided IP addresses are the first usable host in their respective subnets.

First, calculate the required prefix lengths using the 'longest subnet mask' rule (smallest valid block):

  1. 472 Hosts: A /23 subnet provides 510 usable IPs ($2^9 - 2$), which fits. A /22 would be wasteful. So, mask is 255.255.254.0 (/23).
  2. 24 Hosts: A /27 subnet provides 30 usable IPs ($2^5 - 2$), which fits. A /26 would be wasteful. So, mask is 255.255.255.224 (/27).

Next, validate the options against the 'first available address' constraint:

  • Option A (10.70.148.1/23): The network address is determined by ANDing the IP with the mask. Since the third octet is 148 (binary 10010100) and the mask increments by 2 (11111110), 148 is even. Thus, the network is 10.70.148.0. The first usable host is .1. This matches the option. Correct.
  • Option B (10.70.147.17/27): The mask is /27 (block size 32). $17 / 32 = 0$ remainder 17. The network is 10.70.147.0. The first usable host is .1, not .17. Incorrect.
  • Option C (10.70.133.17/26): Incorrect mask for 24 hosts (wasteful) and.17 is not the first host of a /26 network (network is 10.70.133.0, first host is.1). Incorrect.
  • Option D (10.70.159.1/23): Third octet 159 is odd (10011111). With a /23 mask, the network boundary falls at 158. The network is 10.70.158.0. The first usable host is .158.1, not .159.1. Incorrect.
  • Option E (10.70.155.65/27): Block size 32. $65 / 32 = 2$ remainder 1. Network base is $2 imes 32 = 64$. Network is 10.70.155.64. First usable host is .65. This matches the option. Correct.

The question requires two configurations. Options A and E satisfy all constraints.

Exam Strategy

When validating 'first usable' IPs, quickly check if the last significant octet is divisible by the block size. If it is, the IP is likely the first host. If not, calculate the nearest lower multiple of the block size to find the true network ID. For /23, check if the 3rd octet is even; for /27, check if the 4th octet is a multiple of 32.

Frequently Asked Questions

How do I quickly find the network ID for a /23 subnet?

Check the third octet. If it is even, the network ID ends in that number. If it is odd, subtract 1. For example, 10.70.148.1/23 has network 10.70.148.0.

Why is Option B incorrect for 24 hosts?

While the mask /27 fits 24 hosts, the IP 10.70.147.17 is not the first usable address. The network is 10.70.147.0, so the first usable IP is 10.70.147.1.

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Community Votes

A
48%
E
44%
B
4%
D
4%

48% of anonymous learners picked answer A. Votes are pick records left by other test-takers — they are not the verified answer.

Discussion 54 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 30 of 54

  1. Customexit 👍 45
    Adding my answer since there is so much confusion. Remove C because VLAN ranges are 1-1005 and 1006-4094. !The requirement is to use the first available address in each subnet! For A, the network address is .148.0. The first available is 148.1. 255.255.254.0 is /23, we have 512 addresses so we are good on the 472 host requirement. For B, 10.70.147.1 is the first address. So no. For D, (we already decided on A but we'll do this anyway) 10.70.158.1 is our first. Not 159.1. For E, .64 is our network, .65 is our first. This works. .224 is /27 which gives us 32 total addresses, more than we need. I highly recommend watching 'Subnetting Mastery' youtube playlist for learn how to subnet fast.
  2. MTrap 👍 12
    It has to be A & B. The first subnet will require 24 hosts which would put in the subnet mask ending in 224 yes. However since it is the first subnet, it has to come before the second (obviously) so a since B has an IP address of 10.70.147.17 with a subnet mask of 255.255.255.224 and A has an IP address of 10.70.148.1 and a network mask of 255.255.254.0 (512 IP addresses) and it immediately follows the previous address, this should be the answer.
  3. Equiano 👍 5
    These two are the only ones that have the first IP in their range and meet the requirement of 472 hosts (A) and 24 hosts (E) respectively.
  4. Deeztroyer 👍 5
    /27 needed for 24 addresses -> magical number is 32. The router needs to be the first IP address in the range. => E is right
  5. TNgen9 👍 4
    First subnet is /27 which is 255.255.255.224 second subnet is /23 which is 255.255.254.0 the keyword logest subneet and first usable IP is requiment first option in subnet /27 B and E meet the requirment but answer B ip 10.70.147.17 /27 is not the first usable IP so Answer E is correct 10.70.155.65 /27 meet the requiement Second Option in subnet /23 A and D meet the requiment but Answer D ip 10.70.159.1 255.255.254.0 is not the first Usable IP so A is correct meet the requiement and the first usable ip
Show 25 more comments
  1. Customexit 👍 4
    From my understanding, D cannot be correct because a requirement per the question is "to use the first available address in each subnet". If you subnet 10.70.159.1 255.255.254.0, you will see that the network address is 10.70.158.0. The 'next' network starts at 10.70.160.0. The first available would actually be 10.70.158.1. Not 10.70.159.1.
  2. PerryThePlatypus 👍 4
    I absolutely didn't understand the meaning of this question, mind you, it's not a subnetting issue, but more about the logic behind it. Could you please explain it to me as you would to your 8-year-old nephew? Ty
  3. daddydagoth 👍 4
    It's AE, anyone that thinks otherwise need to hold off on taking the CCNA and brush up their subnetting!
  4. ctoklu 👍 3
    vlan numbers shoud be between 0-4095, so C is eliminated at first sight! for the rest please write a detailed calculation. didn't understand the solution.
  5. ZUMY 👍 3
    Going with A & B
  6. Danielki 👍 2
    Guys please read the question carefully! Specifically "Both subnets must use the longest subnet mask possible from the address block." D. interface vlan 1234 ip address 10.70.159.1 255.255.254.0 This is a /23 subnet with 512 addresses, which is sufficient for 472 hosts, and it is at the end of the /19 address block (10.70.128.0/19), so it uses the longest subnet mask possible. So, the correct configurations to meet the requirements are options D and E: D. interface vlan 1234 ip address 10.70.159.1 255.255.254.0 For the second subnet with 472 hosts, using the longest subnet mask possible. E. interface vlan 155 ip address 10.70.155.65 255.255.255.224 For the first subnet with 24 hosts.
  7. ndaniel99 👍 2
    Looking at the host requirements we know we want /23 and /27. Exclude C. Now, we know that we want 1st usable address in each. Let's convert to binary: 1st) /23 A: 10.70.148.1 = 10.70.10010100.1 The mask will be there: 10.1001010/0.00000001 This means this is a right answer. In opposition ,lets check the other /23: 10.70.159.1 = 10.70.1001111/1.00000001 there is more than one "1" after the mask which means it is not the 1st usable address. Repeat for /27 and you will see the same results.
  8. josop 👍 2
    Correct Answers: A & E 1. For 472 hosts, CIDR: /23 (2^9 → 32 - 9 = 23) & Subnet Mask: 255.255.255.224 2. For 24 hosts, CIDR: /27 (2^5 → 32 - 5 = 27) & Subnet Mask: 255.255.254 A. Network: 10.70.148.1/23 & First usable IP: 10.70.148.1 B. Network: 10.70.147.17/27 & First usable IP: 10.70.147.1 C. Incorrect because VLAN ranges are 1-1005 and 1006-4094. D. Network: 10.70.159.1/23 & First usable IP: 10.70.158.1 E. Network: 10.70.155.65/27 & First usable IP: 10.70.155.65
  9. mfaria 👍 2
    AE corrects. A: Range: 10.70.148.0 ~ 10.70.149.255, Host: x.x.148.1 ~ x.x.149.254. It is the first host and the mask is /23 B: .17 is not the first Host as network is .0 and broadcast is .31. Mask is /27 C: VLAN 4722 is not available. VLAN is 1 - 1005, 1006 - 4094 D: x.x.159.1 is not the first IP. Range is x.x.158.1 ~ x.x.159.254/23 E: Range is x.x.155.64 ~ x.x.155.96. First usable x.x.155.65/27
  10. Mikeserj 👍 2
    I just came to tell you that on August 3rd I got my CCNA certification, and I want you to know that at least 90% of the questions I got, I saw here in Exam Topics, study them and understand the reason for the answers and I assure you that you will pass, by the way, you only have until August 15th to schedule an exam for version 7 of CCNA 200-301. I hope you succeed!
  11. picho707 👍 2
    1001 0100 (148) AND 1111 1110 (254) = 148 ( this means that the first address is 148.1) 0100 0001 (65) AND 1110 0000 ( 224) = 64 ( This means that the first address is 255.65)
  12. doribeqiraj 👍 2
    Correct answer for me is A & E Network 10.70.128.0/19 = 10.70.128.1-10.70.159.25 D is incorrect, because is out of range. 10.70.159.1 - 10.70.160.255
  13. splashy 👍 2
    B & C wrong => not the first address D wrong => goes outside of the provided 10.70.128.0/19 block/range Only A & E can meet the requirements.
  14. ptfish 👍 2
    The keyword is the first available address in each subnet
  15. MTrap 👍 2
    Yeap, the first available IP address thing got me.
  16. Amit_R0y 👍 2
    Selected Answer: AE
  17. Djow 👍 2
    it's right
  18. RahmNn 👍 1
    Options Evaluation: A. IP Address: 10.70.148.1 255.255.254.0 (/23) Subnet: 10.70.148.0/23 Usable IPs: 10.70.148.1 to 10.70.149.254 Total IPs: 512 Usable: 510 Valid for 512 hosts. B. IP Address: 10.70.147.17 255.255.255.224 (/27) Subnet: 10.70.147.0/27 Usable IPs: 10.70.147.1 to 10.70.147.30 Total IPs: 32 Usable: 30 Not valid for 512 hosts. C. IP Address: 10.70.133.17 255.255.255.192 (/26) Subnet: 10.70.133.0/26 Usable IPs: 10.70.133.1 to 10.70.133.62 Total IPs: 64 Usable: 62 Not valid for 512 hosts. D. IP Address: 10.70.159.1 255.255.254.0 (/23) Subnet: 10.70.158.0/23 Usable IPs: 10.70.158.1 to 10.70.159.254 Total IPs: 512 Usable: 510 Valid for 512 hosts. E. IP Address: 10.70.155.65 255.255.255.224 (/27) Subnet: 10.70.155.64/27 Usable IPs: 10.70.155.65 to 10.70.155.94 Total IPs: 32 Usable: 30 Not valid for 512 hosts.
  19. Nmk173 👍 1
    i confused between question and answers The FIRST avaible subnet is this. (vlan number is random in avaible vlan numbers) interface vlan 10 ip address 10.70.128.1 255.255.255.224 then SECOND subnet is this(This is avaible after first subnet for max 510 host) interface vlan 11 ip address 10.70.130.1 255.255.254.0 Otherwise question must be one subnet ..... other subnet .... so it is not important first and second. In the avaible answer A and E is right... 10.70.148.0 255.255.254.0 avaible ips 10.70.148.1 - 10.70.149.254 10.70.155.65 255.255.255.224 avaible ips 10.70.155.65 - 10.70.155.94 AND d CANT BE GOOD ANSWER BECAUSE 10.70.154.1 - 10.70.155.254
  20. riteshm42 👍 1
    There is no explanation as to why option B is not correct. The actual answer is both B and A. If the second subnet is using 10.70.148.1 with a subnet mask of 255.255.254.0, then the first subnet is 10.70.147.1 (network address: 10.70.147.0) only. Also, please read the question carefully: the first subnet must support 24 hosts, so option B [10.70.147.1/27] is correct. What is the logic behind using the first and second subnets starting from X.X.158.X? Even though it meets the requirement, the question specifically asks to use the first subnet with 24 hosts, so we should use the first subnet as X.X.147.X and the second subnet as X.X.148.0.
  21. Besxp 👍 1
    For 472 hosts, the subnet must use /23: - A. 10.70.148.1 255.255.254.0 — network is 10.70.148.0/23, first usable address is .148.1. Correct. - D. 10.70.159.1 255.255.254.0 — .159.1 is not the first usable address of a /23 subnet. The network would be 10.70.158.0/23, whose first usable address is .158.1. Incorrect. For 24 hosts, the subnet must use /27: - B. 10.70.147.17 255.255.255.224 — the network is 10.70.147.0/27, whose first usable address is .147.1. Incorrect. - E. 10.70.155.65 255.255.255.224 — the network is 10.70.155.64/27, whose first usable address is .155.65. Correct.
  22. kastorng0718 👍 1
    I think there is a quicker way to do this question. First, you should be able to quickly notice that .17 is very unlikely to be the first useable address by looking at the netmask and the binary of 17. Hence, B and C is out. .1 is very likely to be the first available address to be used, so let's check if .65 is, and .65 in this case is. Then, let's look at the netmask of A, D and E. For answering this question purpose, E must be right, and because the question explicitly said the smaller subnet must be the first one, so only A is possible and D is out.
  23. thomeboy 👍 1
    The correct options are A and B, as they fulfill the requirements for the second and first subnets, respectively. Option E is incorrect because the address range is too high for the first subnet, and Option D is incorrect because the address range is not in the correct order for the first subnet.
  24. ShadyAbdekmalek 👍 1
    For B, C: the x.x.x.17 is not correct For D: this choice means that ip address range will extend between 10.70.159.1 - 10.70.160.254, and this is beyond the allowed range of 10.70.128.0/19 (allowed range is between 10.70.128.1-10.70.159.254). That leave A and E to be the correct answer
  25. nnfordcion 👍 1
    'meet a requirement to use the first available address in each subnet for the router interfaces?' I don't understand the meaning of this sentence at all.What is the difference between this and normal subnetting?

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