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200-301 · Question #820

200-301 Question #820: Real Exam Question with Answer & Explanation

The correct answer is A: 192.168.4.67. To identify valid host addresses within a 192.168.4.0 network subnetted with a /26 mask, one must determine the network and broadcast addresses for each subnet.

Submitted by fernanda_arg· Mar 5, 2026[DOMAIN_LIST_MISSING_IN_PROMPT]

Question

An administrator is working with the 192.168.4.0 netwrok, which has been subnetted with a /26 mask. Which two addresses can be assigned to hosts within the same subnet? (Choose two.)

Options

  • A192.168.4.67
  • B192.168.4.61
  • C192.168.4.128
  • D192.168.4.132
  • E192.168.4.125
  • F192.168.4.63

Explanation

To identify valid host addresses within a 192.168.4.0 network subnetted with a /26 mask, one must determine the network and broadcast addresses for each subnet.

Common mistakes.

  • B. 192.168.4.61 is a valid host address but belongs to the first subnet (192.168.4.0/26), not the same subnet as options A and E.
  • C. 192.168.4.128 is the network address for the third subnet (192.168.4.128/26) and cannot be assigned to a host.
  • D. 192.168.4.132 is a valid host address but belongs to the third subnet (192.168.4.128/26), not the same subnet as options A and E.
  • F. 192.168.4.63 is the broadcast address for the first subnet (192.168.4.0/26) and cannot be assigned to a host.

Concept tested. IPv4 Subnetting and Host Address Assignment

Reference. https://learn.microsoft.com/en-us/windows-server/networking/technologies/ip-addressing/ipv4-subnetting

Topics

#Subnetting#IP Addressing#CIDR notation#Host addresses

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