Refer to the exhibit.
which address and mask combination a summary of the routes learned by EIGRP?
A.
192.168.25.0 255.255.255.240
B.
192.168.25.16 255.255.255.252
C.
192.168.25.0 255.255.255.252
D.
192.168.25.28 255.255.255.240
E.
192.168.25.16 255.255.255.240
F.
192.168.25.28 255.255.255.240
The binary version of 20 is 10100.
The binary version of 16 is 10000.
The binary version of 24 is 11000.
The binary version of 28 is 11100.
The subnet mask is /28. The mask is 255.255.255.240.
From the output above, EIGRP learned 4 routes and we need to find out the summary of them: +
192.168.25.16+ 192.168.25.20+ 192.168.25.24+ 192.168.25.28
-> The increment should be: 28 – 16 = 12 but 12 is not an exponentiation of 2 so we must choose
16 (24). Therefore the subnet mask is /28 (=1111 1111.1111 1111.1111 1111.11110000) =
255.255.255.240
So the best answer should be 192.168.25.16 255.255.255.240
From the output above, EIGRP learned 4 routes and we need to find out the summary of them:
+ 192.168.25.16
+ 192.168.25.20
+ 192.168.25.24
+ 192.168.25.28
-> The increment should be: 28 – 16 = 12 but 12 is not an exponentiation of 2 so we must choose 16 (24). Therefore the subnet mask is /28 (=1111 1111.1111 1111.1111 1111.1111 0000) = 255.255.255.240
So the best answer should be 192.168.25.16 255.255.255.240
Correct answer is