Consider a router that interconnects three subnets A, B, and C that are required

Question

Consider a router that interconnects three subnets A, B, and C that are required to support 100, 50, and 25 interfaces, respectively. The overall network mask for the overall subnet (all three subnets) is 10.10.10.024. A possible network address for subnet A would be Oa. 10.10.10.0/26 Ob. 10.10.10.0/27 O10.10.10.0/25 d. 10.10.10.0/24 A possible network address for subnet B would be O a. 10.10.10. 12825 b.10.10.10.128/26 OC.10.10.10.128/24 Od. 10.10.10.128/27 A possible network address for subnet C would be a. 10.10.10.192/27 Ob. 10.10.10.192/26 OG. 10.10.10.192/25 Od. 10.10.10.192/24

Explanation / Answer

Subnet A requires 100 interfaces + 2 (network + broadcast) => so 128 addresses

Subnet B requires 50 interfaces + 2 (network + broadcast) => so 64 addresses

Subnet C requires 25 interfaces + 2 (network + broadcast) => so 32 addresses

So Solution -

Subnet A 10.10.10.0/25 -> 10.10.10.0 to 10.10.10.127 = 128

(10.10.10.00000000 to 10.10.10.01111111) SO OPTION C IS TRUE

Subnet B 10.10.10.128/26 -> 10.10.10.128 to 10.10.10.191 =64

(10.10.10.10000000 to 10.10.10.10111111) SO OPTION B IS TRUE

Subnet C 10.10.10.192/27 -> 10.10.10.192 to 10.10.10.223 = 32

(10.10.10.11000000 to 10.10.10.11011111) SO OPTION A IS TRUE

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