A network engineer must configure two new subnets using the address block 10.70.128.0/19 to meet these requirements:
✑ The first subnet must support 24 hosts.
✑ The second subnet must support 472 hosts.
✑ Both subnets must use the longest subnet mask possible from the address block.
Which two configurations must be used to configure the new subnets and meet a requirement to use the first available address in each subnet for the router interfaces? (Choose two.)
To configure the subnets and use the first available address for the router interfaces, we need to ensure the subnets support the required number of hosts and use the longest possible subnet masks. The first subnet must support 24 hosts, which requires a /27 mask (255.255.255.224) allowing up to 32 IP addresses. The first available IP in this range is .65. The second subnet must support 472 hosts, necessitating a /23 mask (255.255.254.0) which allows for 512 IP addresses. The first available IP in this range is .1. Hence, the valid configurations are 10.70.155.65 255.255.255.224 for the first subnet and 10.70.148.1 255.255.254.0 for the second subnet.
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.
are you saying D is also valid? so A and D are valid but you preferred A because it was the first option?
D is incorrect because it's not using the first ip of the range. the range of D is 10.70.158.1 to 10.70.159.254
How to know the the range of D is 10.70.158.1 to 10.70.159.254?
Selected Answer: AE
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.
/27 needed for 24 addresses -> magical number is 32. The router needs to be the first IP address in the range. => E is right
Yeap, the first available IP address thing got me.
Agreed, key word here being "First"
B cannot be correct because using a /27 subnet, the first legal network address is .32 then .64, a . 16 would work for a /28 but that would not give up to 24 hosts, so the first available network we can choose from is the .64, if we choose the .64 then the first usable address will be a .65 /27
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
I absolutely didn't understand the meaning of this question, mind you, it's not a subnetting issue, but more about the logic behind it. Could you please explain it to me as you would to your 8-year-old nephew? Ty
A & E are correct
Correct Answers: A & E 1. For 472 hosts, CIDR: /23 (2^9 → 32 - 9 = 23) & Subnet Mask: 255.255.255.224 2. For 24 hosts, CIDR: /27 (2^5 → 32 - 5 = 27) & Subnet Mask: 255.255.254 A. Network: 10.70.148.1/23 & First usable IP: 10.70.148.1 B. Network: 10.70.147.17/27 & First usable IP: 10.70.147.1 C. Incorrect because VLAN ranges are 1-1005 and 1006-4094. D. Network: 10.70.159.1/23 & First usable IP: 10.70.158.1 E. Network: 10.70.155.65/27 & First usable IP: 10.70.155.65
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
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)
Could someone explain me how to calculate this step by step please?
Could someone explain me how to calculate this step by step?
Could someone explain me how to calculate this step by step please?
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I think there is a quicker way to do this question. First, you should be able to quickly notice that .17 is very unlikely to be the first useable address by looking at the netmask and the binary of 17. Hence, B and C is out. .1 is very likely to be the first available address to be used, so let's check if .65 is, and .65 in this case is. Then, let's look at the netmask of A, D and E. For answering this question purpose, E must be right, and because the question explicitly said the smaller subnet must be the first one, so only A is possible and D is out.
B is the first subnet. A is the second bigger subnet.