You are working in a data center environment and are assigned the address range 10.188.31.0/23. You are asked to develop an IP addressing plan to allow the maximum number of subnets with as many as 30 hosts each. Which IP address range meets these requirements?
A. 10.188.31.0/26 B. 10.188.31.0/25 C. 10.188.31.0/28 D. 10.188.31.0/27 E. 10.188.31.0/29
Correct Answer: D
Explanation:
Each subnet has 30 hosts < 32 = 25 so we need a subnet mask which has at least 5 bit 0s -> /27. Also the question requires the maximum number of subnets (which minimum the number of hosts-per-subnet) so /27 is the best choice.
Which two of these statements are true of IPv6 address representation? (Choose two.)
A. There are four types of IPv6 addresses: unicast, multicast, anycast, and broadcast. B. A single interface may be assigned multiple IPv6 addresses of any type. C. Every IPv6 interface contains at least one loopback address. D. The first 64 bits represent the dynamically created interface ID. E. Leading zeros in an IPv6 16 bit hexadecimal field are mandatory.
Correct Answers: B,C
Explanation:
A single interface may be assigned multiple addresses of any type (unicast, anycast, multicast).
Every IPv6-enabled interface must contain at least one loopback and one link-local address.
Optionally, every interface can have multiple unique local and global addresses.
A new subnet with 60 hosts has been added to the network. Which subnet address should this network use to provide enough usable addresses while wasting the fewest addresses?
A. 192.168.1.56/26 B. 192.168.1.56/27 C. 192.168.1.64/26 D. 192.168.1.64/27
Correct Answer: C
Explanation:
A subnet with 60 host is 2*2*2*2*2*2 = 64 -2 == 62
6 bits needed for hosts part. Therefore subnet bits are 2 bits (8-6) in fourth octet.
8bits+ 8bits+ 8bits + 2bits = /26
/26 bits subnet is 24bits + 11000000 = 24bits + 192
256 – 192 = 64
0 -63
64 – 127