READ Free Dumps For Cisco- 200-125
| Question ID 16099 | Refer to the exhibit.
The company uses EIGRP as the routing protocol. What path will packets take from a host
on the 192.168.10.192/26 network to a host on the LAN attached to router R1?
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| Option A | The path of the packets will be R3 to R2 to R1.
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| Option B | The path of the packets will be R3 to R1 to R2.
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| Option C | The path of the packets will be both R3 to R2 to R1 AND R3 to R1.
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| Option D | The path of the packets will be R3 to R1.
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| Correct Answer | D |
Explanation Explanation: Host on the LAN attached to router R1 belongs to 192.168.10.64/26 subnet. From the output of the routing table of R3 we learn this network can be reach via 192.168.10.9, which is an IP address in 192.168.10.8/30 network (the network between R1 & R3) -> packets destined for 192.168.10.64 will be routed from R3 -> R1 -> LAN on R1.
| Question ID 16100 | Refer to the exhibit.
Which address and mask combination represents a summary of the routes learned by
EIGRP?
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| Option A | 192.168.25.0 255.255.255.240
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| Option B | 192.168.25.0 255.255.255.252
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| Option C | 192.168.25.16 255.255.255.240
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| Option D | 192.168.25.16 255.255.255.252
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| Option E | 192.168.25.28 255.255.255.240
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| Option F | 192.168.25.28 255.255.255.252
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| Correct Answer | C |
Explanation Explanation: 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. Note: 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.