How can i solve these kinds of questions ?

in CCNA & CCENT
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
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

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
Comments
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)
A+, Network+, CCNA
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.
Security: CCNA [ ]
Virtualization: VCA-DCV [ ]
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).
Security: CCNA [ ]
Virtualization: VCA-DCV [ ]