Exam 200-301 topic 1 question 145 discussion

Actual exam question from Cisco's 200-301

Question #: 145
Topic #: 1

The address block 192.168.32.0/24 must be subnetted into smaller networks. The engineer must meet these requirements:
✑ Create 8 new subnets.
✑ Each subnet must accommodate 30 hosts.
✑ Interface VLAN 10 must use the last usable IP in the first new subnet.
✑ A Layer 3 interface is used.
Which configuration must be applied to the interface?

A. no switchport mode trunk ip address 192.168.32.97 255.255.255.224
B. switchport ip address 192.168.32.65 255.255.255.240
C. no switchport ip address 192.168.32.30 255.255.255.224
D. no switchport mode access ip address 192.168.32.62 255.255.255.240

Show Suggested Answer

Suggested Answer: C 🗳️

by HMaw at Oct. 7, 2022, 8:14 a.m.

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HMaw

Highly Voted 2 years, 9 months ago

C is correct. Requirement is 8 networks with 30 hosts 255.255.255.0 = 11111111.11111111.11111111.00000000 8 networks = 1111 with increment of 16 which is less host number than require. 30 hosts = 11100000 with increment of 32 255.255.255.224 or 11111111.11111111.11111111.11100000 8 networks for /27 are 0,32,64,96,128,160,192,224

upvoted 16 times

SVN05

2 years, 5 months ago

Could someone please explain to us what does these 2 lines mean. Thank you. 8 networks = 1111 with increment of 16 which is less host number than require. 30 hosts = 11100000 with increment of 32

upvoted 3 times

AbdullahMohammad251

1 year, 7 months ago

We need to adjust the number of ones and zeros in the last octet to meet our requirements. We have 255.255.255.0 as our subnet mask for the address block which translates to 11111111.11111111.11111111. 00000000 in binary Replacing the first zero with a one in the last octet of our SM divides our network into 2 networks. Replacing the first 2 zeros with 2 ones in the last octet of our SM divides our network into 4 networks. Replacing the first 3 zeros with 3 ones in the last octet of our SM divides our network into 8 networks. Replacing the first 4 zeros with 4 ones in the last octet of our SM divides our network into 16 networks. That's how subnetting works. Having 3 ones in the last octet means we have 2^3 = 8 networks And having 5 zeros in the last octet means we have 2^5=32 hosts in each network.

upvoted 5 times

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1 year, 7 months ago

Correct answer is A, Why? first things first a subnet with 30hosts means a /27 mask which then equals to 255.255.255.224, and the switch SHOULD use the LAST IP in the FIRST subnet, now look at the commands, if you choose C(which is the last ip in the first subnet) this IP will be removed from the interface, focus on NO ip addr 192.168.32.30. so it can't be C, since it removes that ip from the switch

upvoted 1 times

Rick3390

1 year, 7 months ago

You're wrong answer is C! The ‘switchport’ command is used to configure an interface as a Layer 2 interface, and the ‘no switchport’ command is used to configure it as a Layer 3 interface. In the context of the command ‘no switchport ip address 192.168.32.30 255.255.255.224’, the ‘no’ keyword is not negating or removing the IP address. Instead, it’s used to configure the interface as a Layer 3 interface.

upvoted 5 times

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6dcf910

1 year, 6 months ago

Answer A does not comply with the requirement "Interface VLAN 10 must use the last usable IP in the first new subnet."

upvoted 2 times

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raul_kapone

1 year, 11 months ago

192.168.32.0/24 #s = 3 -> 2'3 = 8 subnets #h = 5 -> 2'5 - 2 = 30 host/subnet Subnets: 192.168.32.0/27 192.168.32.32/27 192.168.32.64/27 192.168.32.96/27 1st Subnet - Last useable IP address: 192.168.32.30/27 192.168.32.30 255.255.255.224 So, the answer is "C".

upvoted 5 times

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