I don´t undestand explanation of question 1:
“Class answer ‘172.9.0.0/16’ private IP address ranges from 10.0.0.0 to 10.255.255.255”????????
9tut
December 10th, 2020
@Jose: It should be “Class A”. Thanks for your detection, we have just fixed it!
Ray
January 27th, 2021
can anyone explain the Answer to Q4
angy
January 31st, 2021
30-40 is not a number and if a client asks you to set 30-40 you do 40 and not 30… so i really dident understand the logic of calling
on your own its ok to cut 10 possible stations.
D
February 1st, 2021
9tut can please check this answer.
i think it should be A as we need 40 user per floor….with answer D we only get 30 User.
An office has 8 floors with approximately 30-40 users per floor. What command must be configured on the router Switched Virtual Interface to use address space efficiently?
A. ip address 192.168.0.0 255.255.254.0
B. ip address 192.168.0.0 255.255.0.0
C. ip address 192.168.0.0 255.255.255.224
D. ip address 192.168.0.0 255.255.255.128
anon
March 4th, 2021
For “Bad mask /28 for address 192.168.16.143” from question “Refer to exhibit. Which statement explains the configuration error message that is received?”, you have the correct answer is It is a broadcast IP address. But for a /28, that would be address intervals of 16 which would mean 192.168.16.143 is actually a usable address in the subnet 192.168.16.129-144. Perhaps it was supposed to be 192.168.16.144?
CarlaColumna
March 5th, 2021
Question 3 please fix the typo. “B. IT is a network IP address” in “B. It is a network IP address”
9tut
March 5th, 2021
@anon: 192.168.16.144 is the network address so 192.168.16.143 must be the broadcast address of the previous subnet.
9tut
March 5th, 2021
@CarlaColumna: Thanks for your detection, we have just fixed Q.3.
Softail
April 22nd, 2021
9tut can please check this answer.
i think it should be A as we need 40 user per floor….with answer D we only get 30 User.
An office has 8 floors with approximately 30-40 users per floor. What command must be configured on the router Switched Virtual Interface to use address space efficiently?
A. ip address 192.168.0.0 255.255.254.0
B. ip address 192.168.0.0 255.255.0.0
C. ip address 192.168.0.0 255.255.255.224
D. ip address 192.168.0.0 255.255.255.128
So what will be the correct answer ?
MalwarE
May 27th, 2021
@Softali
This is just one of these stupid ass questions that they put on that fuking exam :) It says 30-40 APPROXIMATELY so /27 is exactly 30 users per network. This is the closest correct answer if you think about it. You will just have to memorize this shit out…
Cheers
Robbie
June 24th, 2021
Refer to the exhibit. An engineer must add a subnet for a new office that will add 20 users to the network. Which IPv4 network and subnet mask combination does the engineer assign to minimize wasting addresses?
new_subnet_assignment.jpg
A. 10.10.225.32 255.255.255.224correct
B. 10.10.225.48 255.255.255.224
C. 10.10.225.48 255.255.255.240
D. 10.10.225.32 255.255.255.240wrong
Explanation
We need a subnet with 20 users so we need 5 bits 0 in the subnet mask as 25 – 2 = 30 > 20. Therefore the subnet mask should be /27 (with last octet is 1110 0000 in binary). The increment is 32 so the valid network address is 10.10.225.32.
please, 9tut can give explanation of how mask /27 is arrived at? I didn’t get that.
Q3. An office has 8 floors with approximately 30-40 users per floor.
To accommodate for all users, we should use the larger number to ensure there are enough IP addresses for all users, but the answer + explanations answer as if there are only 30 users per floor rather than 30-40.
9tut
June 29th, 2021
@Yonis: It is impossible to assign IP addresses for 40 users per floor so 30 users per floor is an acceptable solution.
Abdullahi Nadif Aden
January 12th, 2022
this is so confusing , specially question 9 , because both 10.10.255.32 and 10.10.255.48 have same subnet , which is /27 , each of them provides 32 addresses with 30 addresses usable and 2(1 for network address and 1 for broadcast IP ) . So both A and D are correct , 10.10.225.32 is network address whith 32 hosts and 10.10.255.48 is also network address which will provide 32 addresses .
Jacob B
March 20th, 2022
Done
Sansal
March 23rd, 2022
An office has 8 floors with approximately 30-40 users per floor. What command must be configured on the router Switched Virtual Interface to use address space efficiently? Should be D
D. ip address 192.168.0.0 255.255.254.0
Ian F
February 16th, 2023
Hi 9tut. Regarding Q3, can you elaborate on the meaning of ”The increment is 32 so the valid network address is 10.10.225.32.” in the answer please ? Thanks.
Dear 9tut.com owner, Thanks for the well-presented post!
Anonymous
March 8th, 2023
“An office has 8 floors with approximately 30-40 users per floor. What command must be configured on the router Switched Virtual Interface to use address space efficiently?”
This is a question with a stupid answer. We need 30-40 user and we subnetting for 30. So if in a subnet would be 34 user 4 user would be whitout IP address just because the answer itself is stupid. A more professional work would be:
just use “two” network… This question is missleading, would be better if the question would be changed to only 30 user, or the answer updated. I wish to know what is inside the head of the Cisco employee who made the questions.
Nick
January 14th, 2024
Hello guys
About question n°6, are you really really sure that the correct answer is C?
If so, can you please provide a new explanation cause i think lots of people don’t get it from the one above; i would really appreciate it.
Btw, even chatgpt agree that the correct answer should be A.
Here’s its answer if you mind:
To efficiently use address space in a scenario where there are approximately 30-40 users per floor on 8 floors, you need to choose a subnet mask that allows for enough host addresses per subnet while minimizing waste.
Let’s calculate the number of hosts required per floor:
30 users/floor * 8 floors = 240 users
40 users/floor * 8 floors = 320 users
We need a subnet size that accommodates at least 320 hosts, so we need at least a /23 subnet (2^9 = 512, which is the smallest power of 2 greater than 320).
Among the options:
A. ip address 192.168.0.0 255.255.0.0 – This is a /16 subnet, providing more than enough hosts but wasting a lot of address space.
B. ip address 192.168.0.0 255.255.254.0 – This is a /23 subnet, which is suitable for the scenario. It allows for 510 hosts per subnet.
C. ip address 192.168.0.0 255.255.255.224 – This is a /27 subnet, which provides only 30 usable hosts per subnet, insufficient for the requirement.
D. ip address 192.168.0.0 255.255.255.128 – This is a /25 subnet, providing 126 hosts per subnet, which might be enough but is smaller than the /23 subnet in option B.
Therefore, the most appropriate choice for efficient address space utilization in this scenario is option B:
B. ip address 192.168.0.0 255.255.254.0
Let us know!
Thank you very much for your work guys!
Nick
CCNP in progress
June 19th, 2024
The 8 floors should be C -> 255.255.254.0
@Nick & CCNP in progress
June 21st, 2024
When you change a bit in the default subnet you are about to supernet or summarize it (it is in the explanation as well). Because a 192.168.0.0 is a class C address with a default subnet mask /24.
NA
I don´t undestand explanation of question 1:
“Class answer ‘172.9.0.0/16’ private IP address ranges from 10.0.0.0 to 10.255.255.255”????????
@Jose: It should be “Class A”. Thanks for your detection, we have just fixed it!
can anyone explain the Answer to Q4
30-40 is not a number and if a client asks you to set 30-40 you do 40 and not 30… so i really dident understand the logic of calling
on your own its ok to cut 10 possible stations.
9tut can please check this answer.
i think it should be A as we need 40 user per floor….with answer D we only get 30 User.
An office has 8 floors with approximately 30-40 users per floor. What command must be configured on the router Switched Virtual Interface to use address space efficiently?
A. ip address 192.168.0.0 255.255.254.0
B. ip address 192.168.0.0 255.255.0.0
C. ip address 192.168.0.0 255.255.255.224
D. ip address 192.168.0.0 255.255.255.128
For “Bad mask /28 for address 192.168.16.143” from question “Refer to exhibit. Which statement explains the configuration error message that is received?”, you have the correct answer is It is a broadcast IP address. But for a /28, that would be address intervals of 16 which would mean 192.168.16.143 is actually a usable address in the subnet 192.168.16.129-144. Perhaps it was supposed to be 192.168.16.144?
Question 3 please fix the typo. “B. IT is a network IP address” in “B. It is a network IP address”
@anon: 192.168.16.144 is the network address so 192.168.16.143 must be the broadcast address of the previous subnet.
@CarlaColumna: Thanks for your detection, we have just fixed Q.3.
9tut can please check this answer.
i think it should be A as we need 40 user per floor….with answer D we only get 30 User.
An office has 8 floors with approximately 30-40 users per floor. What command must be configured on the router Switched Virtual Interface to use address space efficiently?
A. ip address 192.168.0.0 255.255.254.0
B. ip address 192.168.0.0 255.255.0.0
C. ip address 192.168.0.0 255.255.255.224
D. ip address 192.168.0.0 255.255.255.128
So what will be the correct answer ?
@Softali
This is just one of these stupid ass questions that they put on that fuking exam :) It says 30-40 APPROXIMATELY so /27 is exactly 30 users per network. This is the closest correct answer if you think about it. You will just have to memorize this shit out…
Cheers
Refer to the exhibit. An engineer must add a subnet for a new office that will add 20 users to the network. Which IPv4 network and subnet mask combination does the engineer assign to minimize wasting addresses?
new_subnet_assignment.jpg
A. 10.10.225.32 255.255.255.224correct
B. 10.10.225.48 255.255.255.224
C. 10.10.225.48 255.255.255.240
D. 10.10.225.32 255.255.255.240wrong
Explanation
We need a subnet with 20 users so we need 5 bits 0 in the subnet mask as 25 – 2 = 30 > 20. Therefore the subnet mask should be /27 (with last octet is 1110 0000 in binary). The increment is 32 so the valid network address is 10.10.225.32.
please, 9tut can give explanation of how mask /27 is arrived at? I didn’t get that.
@Robbie: Please read our Subnetting Made Easy tutorial at https://www.9tut.com/subnetting-tutorial. We have some examples which are identical to this question.
Q3. An office has 8 floors with approximately 30-40 users per floor.
To accommodate for all users, we should use the larger number to ensure there are enough IP addresses for all users, but the answer + explanations answer as if there are only 30 users per floor rather than 30-40.
@Yonis: It is impossible to assign IP addresses for 40 users per floor so 30 users per floor is an acceptable solution.
this is so confusing , specially question 9 , because both 10.10.255.32 and 10.10.255.48 have same subnet , which is /27 , each of them provides 32 addresses with 30 addresses usable and 2(1 for network address and 1 for broadcast IP ) . So both A and D are correct , 10.10.225.32 is network address whith 32 hosts and 10.10.255.48 is also network address which will provide 32 addresses .
Done
An office has 8 floors with approximately 30-40 users per floor. What command must be configured on the router Switched Virtual Interface to use address space efficiently? Should be D
D. ip address 192.168.0.0 255.255.254.0
Hi 9tut. Regarding Q3, can you elaborate on the meaning of ”The increment is 32 so the valid network address is 10.10.225.32.” in the answer please ? Thanks.
Address: 192.168.16.143 11000000.10101000.00010000.1000 1111
Netmask: 255.255.255.240 = 28 11111111.11111111.11111111.1111 0000
Wildcard: 0.0.0.15 00000000.00000000.00000000.0000 1111
Network: 192.168.16.128/28 11000000.10101000.00010000.1000 0000 (Class C)
Broadcast: 192.168.16.143 11000000.10101000.00010000.1000 1111
HostMin: 192.168.16.129 11000000.10101000.00010000.1000 0001
HostMax: 192.168.16.142 11000000.10101000.00010000.1000 1110
Hosts/Net: 14 (Private Internet)
ok
Dear 9tut.com owner, Thanks for the well-presented post!
“An office has 8 floors with approximately 30-40 users per floor. What command must be configured on the router Switched Virtual Interface to use address space efficiently?”
This is a question with a stupid answer. We need 30-40 user and we subnetting for 30. So if in a subnet would be 34 user 4 user would be whitout IP address just because the answer itself is stupid. A more professional work would be:
192.168.0.0/26
192.168.0.64/26
192.168.0.128/26
192.168.0.192/26
192.168.1.0/26
192.168.1.64/26
192.168.1.128/26
192.168.1.192/26
just use “two” network… This question is missleading, would be better if the question would be changed to only 30 user, or the answer updated. I wish to know what is inside the head of the Cisco employee who made the questions.
Hello guys
About question n°6, are you really really sure that the correct answer is C?
If so, can you please provide a new explanation cause i think lots of people don’t get it from the one above; i would really appreciate it.
Btw, even chatgpt agree that the correct answer should be A.
Here’s its answer if you mind:
To efficiently use address space in a scenario where there are approximately 30-40 users per floor on 8 floors, you need to choose a subnet mask that allows for enough host addresses per subnet while minimizing waste.
Let’s calculate the number of hosts required per floor:
30 users/floor * 8 floors = 240 users
40 users/floor * 8 floors = 320 users
We need a subnet size that accommodates at least 320 hosts, so we need at least a /23 subnet (2^9 = 512, which is the smallest power of 2 greater than 320).
Among the options:
A. ip address 192.168.0.0 255.255.0.0 – This is a /16 subnet, providing more than enough hosts but wasting a lot of address space.
B. ip address 192.168.0.0 255.255.254.0 – This is a /23 subnet, which is suitable for the scenario. It allows for 510 hosts per subnet.
C. ip address 192.168.0.0 255.255.255.224 – This is a /27 subnet, which provides only 30 usable hosts per subnet, insufficient for the requirement.
D. ip address 192.168.0.0 255.255.255.128 – This is a /25 subnet, providing 126 hosts per subnet, which might be enough but is smaller than the /23 subnet in option B.
Therefore, the most appropriate choice for efficient address space utilization in this scenario is option B:
B. ip address 192.168.0.0 255.255.254.0
Let us know!
Thank you very much for your work guys!
Nick
The 8 floors should be C -> 255.255.254.0
When you change a bit in the default subnet you are about to supernet or summarize it (it is in the explanation as well). Because a 192.168.0.0 is a class C address with a default subnet mask /24.
255.255.254.0 = /23 = 1111 1111 . 1111 1111 . 1111 111>01commandInterface<…"
I think it would just throw the error: "Bad mask /23 for address 192.168.0.0"
I agreed with the answers