PrepAway - Latest Free Exam Questions & Answers

Which command should you run?

Your company has a main office and four branch offices. The main office contains a server named
Server1 that runs Windows Server 2012 R2.
The IP configuration of each office is configured as shown in the following table.

You need to add a single static route on Server1 to ensure that Server1 can communicate with the
hosts on all of the subnets.
Which command should you run?

PrepAway - Latest Free Exam Questions & Answers

A.
route.exe add -p 10.10.0.0 mask 255.255.252.0 10.10.0.1

B.
route.exe add -p 172.16.16.0 mask 255.255.252.0 10.10.0.1

C.
route.exe add -p 10.10.0.0 mask 255.255.252.0 172.16.0.0

D.
route.exe add -p 172.16.18.0 mask 255.255.252.0 10.10.0.1

Explanation:
These parameters will allow communication with all the hosts.

References:
Exam Ref: 70-410: Installing and Configuring Windows Server 2012 R2, Chapter4: Deploying and
configuring core network services, Objective 4.1: Configure IPv4 and IPv6 addressing, p.192, 196

3 Comments on “Which command should you run?”

  1. Joel says:

    YES! Correct answer = B
    route add -p 172.16.16.0 mask 255.255.252.0 10.10.0.1
    Destination^ Mask^ Gateway^

    IP available range on the 172.16.16.0 network =
    172.16.16.1 thru 172.16.19.254 – Will talk to all branch networks
    Broadcast IP = 172.16.19.255




    0



    0
  2. bytezz says:

    Syntax for route in this instance is:

    route add -p DESTINATION_NETWORK mask SUBNET_MASK GATEWAY

    So, analysing the possible solutiuons:

    A: Destination is on the same network as main & router/gateway, so server 1 can’t communicate with ANY subnet listed.
    B: Using the mask 255.255.252.0 with the destination 172.16.16.0 gives the range of addresses 172.16.16.0 > 172.16.19.255, covering all network IDs listed. The gateway is correct, so this is the answer.
    C: As with A, the destination is incorrect
    D: Using the mask 255.255.252.0 with the destination 172.16.18.0 doesn’t work properly. Splitting the third byte of the IP address (18) into binary, you will get 00010010
    The subnet mask third byte (252) in binary is 11111100. This means 172.16.18.0/22 isn’t a network, but a host on the 172.16.16.0/22 network, therefore answer D is incorrect.




    0



    0
  3. bytezz says:

    If the exam throws the same kind of question at you but PowerShell instead; the following is the syntax for New-NetRoute:

    New-NetRoute -DestinationPrefix “172.16.16.0/22” -InterfaceIndex 1 -NextHop “10.10.0.1”




    0



    0

Leave a Reply