Exam 200-301 topic 1 question 166 discussion

I think that the answer is good. we are looking at 31 VLANs ( a sin 31 Subnets) and each one will have 3 hosts. So /29 will give us 32 subnets each with 8 hosts (6 usable). /29 means 248 mask, so B is correct.

30 ADDITIONAL SUBNETS , IT MUST BE 32 = 5 BITS CLASS C /24+5BITS = /29 = 225.255.255.248

30 more vlans = 30 more subnets

B is correct

2^5 = 32 subnet-

B is wrong, in the mark is missing the point! 255 here 255.255.248, so D s the only option

I will choose B

A is wrong. Mask # can not be 265 only goes to 255

couldn't be /30 because you only could configure one PC and the subinterface router's, you can't grow up, and instead of 30 plus networks you achieved 126 so exceeds the actual requirement

Because we are looking at 31 vlans we 31 different subnets each with at least 3 host address the closet one to that would be /29 or .248 giving us 8 total per subnet and 6 available for use in each one making the answer b

need 31 Subnet

The correct command set that fulfills the given request while reserving address space for the expected growth is: B. Switch(config)#interface vlan 10 Switch(config-if)#ip address 192.168.0.1 255.255.255.248 The scenario mentions that a Class C subnet needs to be used, which means we have a default subnet mask of 255.255.255.0. With the requirement to implement 30 additional VLANs, we need a subnet mask that will provide enough IP addresses for all these VLANs with the same Class C network . By allocating a /29 subnet to each VLAN, it will provide 6 bit host addresses (2^6 - 2, where 2 is subtracted for the network address and broadcast address) and will provide enough IP addresses for all 30 additional VLANs.

you need /29 for 3 hosts, /30 is only for 2 and would be used in point-to-point connection

With .248 you get 8-2= 6 hosts to receive IPs per subnet With .252 you get 4-2=2 hosts to rececive IPs per subnet. As you need 3 hosts for addressing, Answer B is the correct

B is correct

255 255.255.248 allows for 32 subnets while 255.255.255.252 allows for 64 subnets. Since we need 33 subnets(3 PCs and 30 additional vlans) I think A is answer. Correct me if I am wrong.