It's hilarious that they had their "unlimited" promotion last week, then immediately removed all Cisco content. It's like they knew this was coming, but wanted a cash grab first.
C is the correct answer. Read the question carefully, it's saying what would be the "Summary of the route" meaning summary network address or also known as route summarisation in subnetting. Option C covers all the addresses in the range starting from .16 to .30. Broadcast address on this subnet is 192.168.25.31 which is always one address less from the next block of addresses, which in this case will be a .32 block ending at .47 as it's Broadcast and .48 as the next block. So, 192.168.25.16/28 255.255.255.240 is the summary address for those EIGRP learnt subnets in the routing table showing up in the exhibit.
Another note for those who are confusing it with /30 network: Read the question as it's asking for route summarisation address which will cover .16 to .28 addresses and your /30 prefix for 192.168.25.16 will only go up to .19(Broadcast on this subnet) and then it would be .20 block which does NOT fall in that .16 block as it's the next block of addresses. So, .16 only has .17 and .18 as two allocable host addresses with of course .19 being the broadcast address on this subnet. Hence it's incorrect.
Right answer is C
The correct answer is C.
You need to find which network range covers all the EIGRP learned addresses in the routing table.
The ranges are as follows:
Answer A:
192.168.25.0 - 192.168.25.15
Answer B:
192.168.25.0 - 192.168.25.3
Answer C:
192.168.25.16 - 192.168.25.31
Answer D:
192.168.25.16 - 192.168.25.19
Answer E:
192.168.25.28 - 192.168.25.43
Answer F:
192.168.25.28 - 192.168.25.31
How to find the ranges? Well, there are different ways to find the ranges. Personally, I'm using the 'Seven Seconds Subnetting' technique by Professor Messer. You may search Youtube for that.
To those of you who are confused about how the subnet is decided, all the binary that matches is considered 1, and those that don't match is considered 0. So, for the example below:
0 0 0 1| 0 1 0 0
0 0 0 1| 0 0 0 0
0 0 0 1| 1 0 0 0
0 0 0 1| 1 0 0 0
The final four digits don't match which makes the final four digital all zeros.
Hence, the final subnet would be:
11111111.11111111.11111111.11110000 = 240
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