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Exam 200-301 All Questions

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Exam 200-301 topic 1 question 37 discussion

Actual exam question from Cisco's 200-301
Question #: 37
Topic #: 1
[All 200-301 Questions]

An office has 8 floors with approximately 30-40 users per floor. One subnet must be used. Which command must be configured on the router Switched Virtual
Interface to use address space efficiently?

  • A. ip address 192.168.0.0 255.255.0.0
  • B. ip address 192.168.0.0 255.255.254.0
  • C. ip address 192.168.0.0 255.255.255.128
  • D. ip address 192.168.0.0 255.255.255.224
Show Suggested Answer Hide Answer
Suggested Answer: B 🗳️

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Chosen Answer:
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[Removed]
Highly Voted 2 years, 11 months ago
8 floors and 40 user per floor means 320 users (approx.). How many bits do you need to have 320 IP addresses? 8 bits = (2^8 – 2) =254 IP addresses, and it’s not enough. 9 bits = (2^9 – 2) = 510 IP addresses, and this is enough. You have a class C subnet (192.168.0.0). This means a subnet mask like this: 255.255.255.0 But you need 9 bits for the hosts, so you’ve got left with a subnet mask like this: 255.255.1111111x.xxxxxxxx =255.255.254.0 This means you will use VLSM subnetting. Answer B is correct.
upvoted 82 times
re__roy
2 years, 8 months ago
Well explained brother
upvoted 3 times
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HMaw
2 years, 7 months ago
Reading your explain make me hearing Jeremy voice saying "Save the hosts". Nicely done.
upvoted 12 times
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bikila123
1 year, 9 months ago
very good explanation!!!
upvoted 6 times
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AKA1987
Highly Voted 3 years, 8 months ago
40*8<=2^H-2, will give H=9 which is a /23 OR 255.255.254.0 = Answer B
upvoted 10 times
HakamCnna
2 years, 12 months ago
how you give H=9 .?
upvoted 2 times
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snowy23
Most Recent 2 months ago
Selected Answer: B
The answer is B since 8x40=320 For supporting 320, you'd need to use the remaining host bits formula, 2^9 - 2 = 510 this will give you enough hosts, any less than that is not enough. then you will skip the 4th octet since you borrowed 9 host bits and not any less than 8.. that means you'd go for the 3rd octet, only 1 answer has changes made to 3rd octet, so B is your answer.
upvoted 1 times
snowy23
2 months ago
sorry, option A has changes made to 3rd octet too but it goes 0 and only B goes for 254 which is exactly leaving the last bit off for the 3rd octet and that would be borrowing 9 bits :D
upvoted 1 times
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liptechBR
6 months, 1 week ago
Selected Answer: C
The correct answer is: C. ip address 192.168.0.0 255.255.255.128 With a subnet mask of 255.255.255.128 (or /25), we have 128 addresses per subnet, of which 126 can be assigned to devices. This mask is enough to accommodate between 30 and 40 users on each of the 8 floors, totaling up to 320 users, and also saves address space.
upvoted 1 times
c49555c
6 months ago
your chatgpt generated answer is wrong, you would have realized it yourself if you just read what you copy pasted
upvoted 3 times
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Ineedicecreamdaily
8 months ago
40 cannot fit in 32 so it must fit in the 64, therefor you know that you need (64 addresses) * 8 (floors) = 512 addresses. You know that /23 contains 512 addresses and the subnet mask is 255.255.254.0. Thank you, and sorry for my english.
upvoted 2 times
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Agribaba
11 months, 3 weeks ago
Calculating Total Users: Minimum number of users: 8 floors * 30 users = 240 users Maximum number of users: 8 floors * 40 users = 320 users To ensure there is enough space for growth and network devices (such as printers, IP phones, and other equipment), it is best to choose a subnet that can accommodate more than the maximum number of users. Choosing the Subnet Mask: A /23 subnet provides 512 IP addresses (510 usable). This will accommodate up to 510 devices, which is sufficient for the current and future needs of the office.
upvoted 4 times
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[Removed]
1 year, 2 months ago
Selected Answer: B
B. ip address 192.168.0.0 255.255.254.0 is correct
upvoted 4 times
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KnackerTopf1
1 year, 2 months ago
literally none of these commands would actually work
upvoted 1 times
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T0ga
1 year, 4 months ago
Selected Answer: C
Convert IP address and subnet mask to binary: IP address: 10.10.225.48 becomes 00001010.00001010.11111111.00110000 Subnet mask: 255.255.255.224 becomes 11111111.11111111.11111111.11100000 Compare the first 27 bits: Align the two binary strings and check if the first 27 bits (from left to right) are identical: 00001010.00001010.11111111.00110000 (IP address) 11111111.11111111.11111111.11100000 (Subnet mask) The first 27 bits (highlighted) match perfectly.
upvoted 1 times
2488fcf
1 year ago
I would reread your book, mate. This is the weirdest gibberish I have ever read. You need 8 floors times 40 people that is 320 in total and the question is to use one subnet for the whole building so 320 hosts are needed in the subnet. /23 is 510 hosts /24 is 254 hosts so answer is /23 or B.
upvoted 2 times
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ciscopue
1 year, 5 months ago
The option is B. ip address 192.168.0.0 255.255.254.0. Explanation: An office has 8 floors with approximately 30 to 40 users per floor. They make a total of 320 users, a mask of 255.55.255.0 only fits 254 users, which is not enough, so the next largest mask would be: 255.255.254.0.
upvoted 2 times
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duongccna
1 year, 7 months ago
B correct, ~40 x 8 = 320 user 2^9 -2 =510 >320 (ok) 2^8-2=254 < 320 therefore borrow 9 bit, choose B
upvoted 1 times
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Freitas512
1 year, 8 months ago
Selected Answer: B
To get a subnet to support 40 users we need at least 6 bits. 32-6 = /26 but we want to be able to get 8 different subnets so we need 3 more bits. /26-3 = /23 . We just need to find the subnet mask of /23
upvoted 3 times
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blackcisco
1 year, 9 months ago
B is correct because i quote "One subnet must be used" one subnet is needed for all 8 floors and only A and B provide this with B uses must less address space
upvoted 1 times
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Naetan0809
2 years ago
8 FLOORS = 8 SUBNET "APPROX" 30-40 USERS = HOST NEEDED 192.168.0.0 = Old Subnet Mask =/24 8 SUBNET= 2,4,8 = 3 BITS BORROWED NewSubnetMask= OldSubnetMask + BITS BORROWED = 24+3 = 27 USABLE HOST = 2^(32-NSM)- 2 =2^(32-27) - 2 = 30 USABLE HOST With “192.168.0.0 255.255.255.224” we have 8 subnets:  • + First subnet: 192.168.0.0 to 192.168.0.31  • + Second subnet: 192.168.0.32 to 192.168.0.63  • + Third subnet: 192.168.0.64 to 192.168.0.95  • + Fourth subnet: 192.168.0.96 to 192.168.0.127  • + Fifth subnet: 192.168.0.128 to 192.168.0.159  • + Sixth subnet: 192.168.0.160 to 192.168.0.191  • + Seventh subnet: 192.168.0.192 to 192.168.0.223  • + Eighth subnet: 192.168.0.224 to 192.168.0.255
upvoted 1 times
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timtgh
2 years, 3 months ago
They most likely meant one subnet per floor, not just one big subnet for the whole office. It's a judgement call. They didn't say "per floor," but they often word things poorly, and that is probably what they meant. If the whole office is one subnet, then it's a flat network and no subnetting is needed at all, and the mask doesn't matter.
upvoted 1 times
oatmealturkey
2 years, 2 months ago
But subnetting is still needed if they want only one subnet, because with their allocated Class C address that we can see in the answer choices, they do not have enough space for that many hosts on a single subnet.
upvoted 2 times
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Maycao
2 years, 4 months ago
Correct
upvoted 1 times
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Vile_Yogabear
2 years, 6 months ago
I managed to solve it by I didn't like the way I did it. 8 x 40 = 320 hosts I manually tried to find which subnet can support this many hosts. 512, 254, 128, 64, 32, 16,8 , 4, 2, 1 If a /24 can support 254 - 2 hosts then /23 can support 512 - 2 hosts so its must be a /23 network. I just counted the subnet my memorizing the numbers 128, 192, 224,240,248, 252,254, 255 /17, /18, /19, /20, /21, /22, /23, /24 So the subnet mask for /23 is 255.255.254.0
upvoted 3 times
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C (25%)
B (20%)
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