Exam 200-301 topic 1 question 267 discussion

Adding my answer since there is so much confusion. Remove C because VLAN ranges are 1-1005 and 1006-4094. !The requirement is to use the first available address in each subnet! For A, the network address is .148.0. The first available is 148.1. 255.255.254.0 is /23, we have 512 addresses so we are good on the 472 host requirement. For B, 10.70.147.1 is the first address. So no. For D, (we already decided on A but we'll do this anyway) 10.70.158.1 is our first. Not 159.1. For E, .64 is our network, .65 is our first. This works. .224 is /27 which gives us 32 total addresses, more than we need. I highly recommend watching 'Subnetting Mastery' youtube playlist for learn how to subnet fast.

It has to be A & B. The first subnet will require 24 hosts which would put in the subnet mask ending in 224 yes. However since it is the first subnet, it has to come before the second (obviously) so a since B has an IP address of 10.70.147.17 with a subnet mask of 255.255.255.224 and A has an IP address of 10.70.148.1 and a network mask of 255.255.254.0 (512 IP addresses) and it immediately follows the previous address, this should be the answer.

There is no explanation as to why option B is not correct. The actual answer is both B and A. If the second subnet is using 10.70.148.1 with a subnet mask of 255.255.254.0, then the first subnet is 10.70.147.1 (network address: 10.70.147.0) only. Also, please read the question carefully: the first subnet must support 24 hosts, so option B [10.70.147.1/27] is correct. What is the logic behind using the first and second subnets starting from X.X.158.X? Even though it meets the requirement, the question specifically asks to use the first subnet with 24 hosts, so we should use the first subnet as X.X.147.X and the second subnet as X.X.148.0.

If you want to find out the number of subnets from /19, then you need /32-19 which equals 13 -8 (for 256 hosts you always need 8). That gives 5 and 2 to the power of 5 = 32 subnets.

Definitely right answers with no doubt.

The correct options are A and B, as they fulfill the requirements for the second and first subnets, respectively. Option E is incorrect because the address range is too high for the first subnet, and Option D is incorrect because the address range is not in the correct order for the first subnet.

Options Evaluation: A. IP Address: 10.70.148.1 255.255.254.0 (/23) Subnet: 10.70.148.0/23 Usable IPs: 10.70.148.1 to 10.70.149.254 Total IPs: 512 Usable: 510 Valid for 512 hosts. B. IP Address: 10.70.147.17 255.255.255.224 (/27) Subnet: 10.70.147.0/27 Usable IPs: 10.70.147.1 to 10.70.147.30 Total IPs: 32 Usable: 30 Not valid for 512 hosts. C. IP Address: 10.70.133.17 255.255.255.192 (/26) Subnet: 10.70.133.0/26 Usable IPs: 10.70.133.1 to 10.70.133.62 Total IPs: 64 Usable: 62 Not valid for 512 hosts. D. IP Address: 10.70.159.1 255.255.254.0 (/23) Subnet: 10.70.158.0/23 Usable IPs: 10.70.158.1 to 10.70.159.254 Total IPs: 512 Usable: 510 Valid for 512 hosts. E. IP Address: 10.70.155.65 255.255.255.224 (/27) Subnet: 10.70.155.64/27 Usable IPs: 10.70.155.65 to 10.70.155.94 Total IPs: 32 Usable: 30 Not valid for 512 hosts.

Looking at the host requirements we know we want /23 and /27. Exclude C. Now, we know that we want 1st usable address in each. Let's convert to binary: 1st) /23 A: 10.70.148.1 = 10.70.10010100.1 The mask will be there: 10.1001010/0.00000001 This means this is a right answer. In opposition ,lets check the other /23: 10.70.159.1 = 10.70.1001111/1.00000001 there is more than one "1" after the mask which means it is not the 1st usable address. Repeat for /27 and you will see the same results.

I just came to tell you that on August 3rd I got my CCNA certification, and I want you to know that at least 90% of the questions I got, I saw here in Exam Topics, study them and understand the reason for the answers and I assure you that you will pass, by the way, you only have until August 15th to schedule an exam for version 7 of CCNA 200-301. I hope you succeed!

It must be a and e, the first usable ip in 10.70.158.0/23 is 10.70.158.1 not 159.1

32=1 31=2 30=4 29=8 28=16 27=32 26=64 25=128 24=256 23=512 21=1024 20=2048 19=4096

DE is correct according to question.

i confused between question and answers The FIRST avaible subnet is this. (vlan number is random in avaible vlan numbers) interface vlan 10 ip address 10.70.128.1 255.255.255.224 then SECOND subnet is this(This is avaible after first subnet for max 510 host) interface vlan 11 ip address 10.70.130.1  255.255.254.0 Otherwise question must be one subnet ..... other subnet .... so it is not important first and second. In the avaible answer A and E is right... 10.70.148.0 255.255.254.0 avaible ips 10.70.148.1 - 10.70.149.254 10.70.155.65 255.255.255.224 avaible ips 10.70.155.65 - 10.70.155.94 AND d CANT BE GOOD ANSWER BECAUSE 10.70.154.1 - 10.70.155.254

First subnet is /27 which is 255.255.255.224 second subnet is /23 which is 255.255.254.0 the keyword logest subneet and first usable IP is requiment first option in subnet /27 B and E meet the requirment but answer B ip 10.70.147.17 /27 is not the first usable IP so Answer E is correct 10.70.155.65 /27 meet the requiement Second Option in subnet /23 A and D meet the requiment but Answer D ip 10.70.159.1 255.255.254.0 is not the first Usable IP so A is correct meet the requiement and the first usable ip

A & E are correct

focus on the main keyword here : "FIRST AVAILABLE ADDRESS"!

1001 0100 (148) AND 1111 1110 (254) = 148 ( this means that the first address is 148.1) 0100 0001 (65) AND 1110 0000 ( 224) = 64 ( This means that the first address is 255.65)