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300-320 · Question #327

You are given the 192.168.2.0/24 network and you should divide it in 4 subnets. Which statements are true regarding the first subnet? (Choose Three)

The correct answer is A. 62 hosts in subnet B. 255.255.255.192 mask C. 192.168.2.62 last host. To divide 192.168.2.0/24 into 4 equal subnets, you borrow 2 bits from the host portion (2² = 4). This changes the prefix from /24 to /26, giving a subnet mask of 255.255.255.192. Each /26 subnet has 6 host bits, providing 2⁶ − 2 = 62 usable host addresses. The first subnet is…

Advanced Addressing and Routing Solutions

Question

You are given the 192.168.2.0/24 network and you should divide it in 4 subnets. Which statements are true regarding the first subnet? (Choose Three)

Options

  • A62 hosts in subnet
  • B255.255.255.192 mask
  • C192.168.2.62 last host
  • D255.255.255.128 mask
  • E126 hosts in subnet
  • F192.168.2.126 last host

How the community answered

(49 responses)
  • A
    84% (41)
  • D
    4% (2)
  • E
    2% (1)
  • F
    10% (5)

Explanation

To divide 192.168.2.0/24 into 4 equal subnets, you borrow 2 bits from the host portion (2² = 4). This changes the prefix from /24 to /26, giving a subnet mask of 255.255.255.192. Each /26 subnet has 6 host bits, providing 2⁶ − 2 = 62 usable host addresses. The first subnet is 192.168.2.0/26: Network address = 192.168.2.0, First host = 192.168.2.1, Last host = 192.168.2.62, Broadcast = 192.168.2.63. Therefore: A (62 hosts) , B (255.255.255.192 mask) , and C (last host 192.168.2.62) are all correct. Choice D (255.255.255.128 = /25) would only yield 2 subnets, and choices E/F correspond to that incorrect /25 scenario.

Topics

#subnetting#VLSM#IP addressing#prefix calculation

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