How can i solve these kinds of questions ?

mustafa86m

interface FastEthernet1/1.3
encapsulation dot1Q 3
ip address 192.168.0.129 255.255.255.192
interface FastEthernet1/1.4
encapsulation dot1Q 4
ip address 192.168.0.193 255.255.255.224
You plan to add two new VLANS: VLAN 5 and VLAN 6. VLAN 5 should support 19 hosts, while VLAN 6 should support 3 hosts. Which of the following answers makes sense. For your answer simply give the letter of the correct choice (e.g. 'A').
A) 192.168.1.1 255.255.255.248 on Fa1/1.6
192.168.0.198 255.255.255.224 on Fa1/1.5
C) 192.168.0.208 255.255.255.248 on Fa1/1.6
D) 192.168.0.169 255.255.255.248 on Fa1/1.6

Priston

This really isn't a good question because they don't tell you what address space you can use to subnet. So the only way to get an answer is to know what the wrong answers are.

So currently you have
vlan 3 which uses 192.168.0.128/26 (192.168.0.128 - 192.168.0.191)
vlan 4 which uses 192.168.0.192/27 (192.168.0.192 - 192.168.0.223)

B is wrong because it's the same subnet as vlan 4 (192.168.0.192 - 192.168.0.223)
C is wrong because it uses address space inside vlan 4 (192.168.0.208 - 192.168.0.215)
D is wrong because it uses address space inside vlan 3 (192.168.0.168 - 192.168.0.175)

satishtech

Agrees with Priston.
Simple Subnetting with interVLAN subinterfaces.

seems like a wrong question with over lapping subnets....

Answer A which is a /30 will give the required 3 IP's
for VLAN 6

192.168.1.1 ----to

---

192.168.1.6
network number 192.168.1.0
broadcast 192.168.1.7

The other answers as Priston said are wrong / overlapping subnets.

theodoxa

At first glance, it appears B, C, and D would overlap the existing subnets. So, A is the only possible answer.

theodoxa

satishtech wrote: »

Answer A which is a /30 will give the required 3 IP's
for VLAN 6

A is a /29, which would give you 8 IP Addresses of which 6 are usable. A /30 (255.255.255.252) would not work as it would only provide 2 usable IP Addresses (out of 4 Total IP Addresses).

VLAN 5 -- 19 hosts. 30 > 19 > 14. A /28 would provide 14 usable addresses, which would not be enough. The next step up is a /27, which would give you 30 usable addresses


where n = the number of network bits (e.g. /27 = 27 network bits).

Dr169

I don't think there is any problem with this question. You just need to look for non overlapping space, which would clearly result in Option A.