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Exam 200-301 topic 1 question 145 discussion

Actual exam question from Cisco's 200-301
Question #: 145
Topic #: 1
[All 200-301 Questions]

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 Hide Answer
Suggested Answer: C 🗳️
by HMaw at Oct. 7, 2022, 8:14 a.m.

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HMaw
Highly Voted 2 years, 3 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
1 year, 11 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, 2 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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Bhrino
Highly Voted 1 year, 8 months ago
The question ask for a sub that can hold 30 host meaning you would need a /27 which equals .224. The question also said that this must use the last available ip in the first subnet. Because this is a /27 the subnet will be in increments on 32 With this in mind the network address : .0 Broadcast : .31 The range is from .1 to .30 of usable ip addresses
upvoted 9 times
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drdrunk3nst3in
Most Recent 4 weeks ago
Selected Answer: C
First, we need to convert the L2 switch interface into a L3 one through the 'no switchport' command. Then, we need to assign an ip address with the command 'ip address <ip> <mask>'.
upvoted 1 times
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jangri777
1 month, 1 week ago
Selected Answer: C
C correct because : 1. 30 host = (near to) 31 host => 2^5 (=32) (3rd position : (1) 2^7, (2) 2^6, (3)2^5) 2. 24 + 3 = /27 ==> 2^7 + 2^6 + 2^5 = 128 + 64 + 32 = .224 3. VLAN 10 will be the last address used so : a even address so answer C remember : network add = even first add = odd last add = even broadcats = odd
upvoted 1 times
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xtraMiles
6 months ago
Selected Answer: C
Should've put a comma between the two commands so as not to confuse others - 1st command: no switchport 2nd command: ip address [a.b.c.d]
upvoted 3 times
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[Removed]
9 months, 1 week ago
Selected Answer: C
C is correct
upvoted 1 times
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VanessaR05
10 months, 2 weeks ago
Selected Answer: C
C is correct
upvoted 1 times
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MaDota
1 year, 2 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, 1 month 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, 1 month 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, 5 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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rubzal
1 year, 8 months ago
What does layer 3 interface used means in this question?
upvoted 1 times
studying_1
1 year, 7 months ago
it means it is multilayer switch, need to write the subinterface command "no switchport" in order to be able to configure an IP address
upvoted 7 times
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iMo7ed
1 year, 11 months ago
Selected Answer: C
C is correct
upvoted 2 times
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C (25%)
B (20%)
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