200-301 practice questions 781–790

1. Question 781 · 1334 in bank

   Select 2

   

   Refer to the exhibit. Which two commands, when configured on router R1, fulfill these requirements? (Choose two.) ✑ Packets toward the entire network 2001:db8:23::/64 must be forwarded through router R2. Packets toward host 2001:db8:23::14 preferably must be forwarded through R3.

   

   • Aipv6 route 2001:db8:23::/128 fd00:12::2
   • Bipv6 route 2001:db8:23::14/128 fd00:13::3
   • Cipv6 route 2001:db8:23::/64 fd00:12::2
   • Dipv6 route 2001:db8:23::14/64 fd00:12::2 200
   • Eipv6 route 2001:db8:23::14/64 fd00:12::2

   Show answer

   Answer: BC

   Explanation

   because the destination is a host address( meaning ip address of the PC), not a network address, /128 in ipv6 is like /32 in ipv4

   Source: Adapted from community discussion — studying_1

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2. Question 782 · 1334 in bank

   

   Refer to the exhibit. Traffic from R1 to the 10.10.2.0/24 subnet uses 192.168.1.2 as its next hop. A network engineer wants to update the R1 configuration so that traffic with destination 10.10 2.1 passes through router R3, and all other traffic to the 10.10.2.0/24 subnet passes through R2. Which command must be used?

   • Aip route 10.10.2.1 255.255.255.255 192.168.1.4115
   • Bip route 10.10.2.0 255.255.255.0 192.168.1.4115
   • Cip route 10.10.2.0 255.255.255.0 192.168.1.4100
   • Dip route 10.10.2.1 255.255.255.255192.168.1.4100

   Show answer

   Answer: D

   Explanation

   I dont understand why A and D are different. Host routes (/32) win because they have the longest prefix. Here, both A and D include the destination host, but they differ on AD. If you set the AD to 115, its higher than OSPF (110) but it will still prefer the longest prefix route, so i believe that A and D are both correct in this scenario. Correct me if im wrong.

   Source: Adapted from community discussion — EliasM

   References

   • Cisco documentation

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3. Question 783 · 1334 in bank

   

   Refer to the exhibit. The image server and client A are running an application that transfers an extremely high volume of data between the two. An engineer is configuring a dedicated circuit between R1 and R2. Which set of commands must the engineer apply to the routers so that only traffic between the image server and client A is forces to use the new circuit?

   • AR1(config)#ip route 10.10.13.10 255.255.255.255 10.10.10.6 R2(config)#ip route 192.168.0.100 255.255.255.255 10.10.10.5
   • BR1(config)#ip route 10.10.13.10 255.255.255.128 10.10.10.6 R2(config)#lp route 192.168.0.100 255.255.255.0 10.10.10.5
   • CR1(config)#ip route 10.10.13.10 255.255.255.252 10.10.10.6 R2(config)#tp route 192.168.0.100 255.255.255.252 10.10.10.5
   • DR1(config)#ip route 10.10.13.10 255.255.255.255 10.10.10.2 R2(config)#ip route 192.168.0.100 255.255.255.255 10.10.10.1

   Show answer

   Answer: A

   Explanation

   For real man, I imagine how many poor souls have been confused and even learned things wrongly because of the dumb answers on here. Never do brain dumps if you haven't fully finished studying kids.

   Source: Adapted from community discussion — daddydagoth

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4. Question 784 · 1334 in bank

   

   Refer to the exhibit. An engineer is checking the routing table in the main router to identify the path to a server on the network. Which route does the router use to reach the server at 192.168.2.2?

   • AS 192.168.0.0/20 [1/0] via 10.1.1.1
   • BS 192.168.2.0/29 [1/0] via 10.1.1.1
   • CS 192.168.2.0/28 [1/0] via 10.1.1.1
   • DS 192.168.1.0/30 [1/0] via 10.1.1.1

   Show answer

   Answer: B

   Explanation

   because the range is 192,168.1.0 - 192.168.1.3, destination is 192.168.2.2.... it doesn't cover it, but 192.168.2.0/29 does, 192.168.2.0 - 192.168.2.7, so correct answer is B, hope that helps :)

   Source: Adapted from community discussion — studying_1

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5. Question 785 · 1334 in bank

   

   Refer to the exhibit. What is the prefix length for the route that router1 will use to reach host A?

   • A/25
   • B/27
   • C/28
   • D/29

   Show answer

   Answer: D

   Explanation

   Answer D /29 is correct. Why /29 cause lets see why 32 bit mask -29 bit mask=3 bits 2 to the power of 3 is 8 per subnet while 6 host in a subnet(2 power of 3 minus 2) So 8 bits in a subnet. Take this questions example. 1st Subnet = 10.10.13.0 - 10.10.13.7 2nd Subnet = 10.10.13.8 - 10.10.13.15 3rd Subnet = 10.10.13.16 - 10.10.13.23 .... ... Some numbered Subnet later =10.10.13.208 - 10.10.13.216 so 10.10.13.214 is in the range so that's the answer right there Reminder Each subnet has a Network and Broadcast Bit so for knowing how many host in a subnet need to ALWAYS MINUS 2. That's why we minus 2(remember just now with 2 power of 3 minus 2 so yup).

   Source: Adapted from community discussion — SVN05

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6. Question 786 · 1334 in bank

   

   Refer to the exhibit. After applying this configuration to router R1, a network engineer is verifying the implementation. If all links are operating normally, and the engineer sends a series of packets from PC1 to PC3, how are the packets routed?

   • AThey are distributed sent round robin to interfaces S0/0/0 and S0/0/1
   • BThey are routed to 10.0.0.2
   • CThey are routed to 192.168.100.2
   • DThey are routed to 172.16.20.2

   Show answer

   Answer: D

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7. Question 787 · 1334 in bank

   

   Refer to the exhibit. When router R1 receives a packet with destination IP address 10.56.0.62, through which interface does it route the packet?

   • AVlan58
   • BNull0
   • CVlan59
   • DVlan60

   Show answer

   Answer: A

   Explanation

   I'm sure A is the correct answer mask /26 seems 2^6 = 64 ,so 10.56.0.0/26 means 10.56.0.0 ~ 10.56.0.63 is the range,the broadcast is 10.56.0.63 and 10.56.0.62 is the final address can be used refer to the routing table, /26 go to vlan 58, so choose A

   Source: Adapted from community discussion — Chris1225

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8. Question 788 · 1334 in bank

   

   Refer to the exhibit. How much OSPF be configured on the GigabitEthernet0/0 interface of the neighbor device to achieve the destined neighbor relationship?

   • ARouter(config)#interface GigabitEthernet 0/0 Router(config-if)#ip ospf cost 5
   • BRouter(config)#interface GigabitEthernet 0/0 Router(config-if)#ip ospf priority 1
   • CRouter(config)#interface GigabitEthernet 0/0 Router(config-if)#ip ospf area 2
   • DRouter(config)#interface GigabitEthernet 0/0 Router(config-if)#ip ospf network point-to-point

   Show answer

   Answer: D

   Explanation

   It took me a While to understand this Question and it would be very helpful to exchange it like: How must OSPF be configured on the GigabitEthernet0/0 interface of the neighbor device to achieve the desired neighbor relationship?

   Source: Adapted from community discussion — GigaGremlin

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9. Question 789 · 1334 in bank

   

   An engineer just installed network 10.120.10.0/24. Which configuration must be applied to the R14 router to add the new network to its OSPF routing table?

   • ARouter ospf 100 Network 10.120.10.0 0.0.0.255 area 0
   • BRouter ospf 120 Network 10.120.10.0 255.255.255.0 area 0 Ip route 10.120.10.0 255.255.255.0 fa0/1
   • CRouter ospf 100 area 0 Network 10.120.10.0 0.0.0.255
   • DRouter ospf 100 Network 10.120.10.0 255.255.255.0 area 0

   Show answer

   Answer: A

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10. Question 790 · 1334 in bank

    DRAG DROP - Drag and drop the IPv4 network subnets from the left onto the correct usable host ranges on the right. Select and Place:

    

    This is an interactive task on the real exam — a drag-and-drop, hotspot or ordering question. There is nothing to select here, so the answer is shown below for reference.

    Show the answer figure

    Answer: See figure below

    Explanation

    This subnet question requires us to grasp how to subnet very well. To quickly find out the subnet range, we have to find out the increment and the network address of each subnet. Let's take an example with the subnet 172.28.228.144/18: From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) - > The network address is 172.28.192.0. So the first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we don't need to calculate the broadcast address because we found the correct answer. Let's take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111 1110 in 3rd octet) -> The 3rd octet of the network address is 228 (because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address is 172.28.228.0 -> The first usable host is 172.28.228.1. It is not necessary but if we want to find out the broadcast address of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or 172.28.230.0 then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet. Therefore the last usable host is 172.28.229.254.

    Source: Adapted from community discussion — paniaguavo

    

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200-301 practice, question 781. 1334 questions available.