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Exam 200-301 topic 1 question 398 discussion

Actual exam question from Cisco's 200-301
Question #: 398
Topic #: 1
[All 200-301 Questions]

Refer to the exhibit. A network associate has configured OSPF with the command:
City(config-router)# network 192.168.12.64 0.0.0.63 area 0
After completing the configuration, the associate discovers that not all the interfaces are participating in OSPF. Which three of the interfaces shown in the exhibit will participate in OSPF according to this configuration statement? (Choose three.)

  • A. FastEthernet0 /0
  • B. FastEthernet0 /1
  • C. Serial0/0
  • D. Serial0/1.102
  • E. Serial0/1.103
  • F. Serial0/1.104
Show Suggested Answer Hide Answer
Suggested Answer: BCD 🗳️
The "network 192.168.12.64 0.0.0.63 equals to network 192.168.12.64/26. This network has:
✑ Increment: 64 (/26= 1111 1111.1111 1111.1111 1111.1100 0000) + Network address:
192.168.12.64
✑ Broadcast address: 192.168.12.127
Therefore all interface in the range of this network will join OSPF.
by Chun9 at Jan. 20, 2021, 9:33 a.m.

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chr
Highly Voted 3 years, 5 months ago
OSPF will match IP addresses based on 192.168.12.64 0.0.0.63. 11000000.10101000.00001100.01000000 => 192.168.12.64 00000000.00000000.00000000.00111111 => 0.0.0.63 Matches will be made on the IP only for the 1's not 0's above. We can invert the bits to make it more familiar as a network mask. This becomes: 11000000.10101000.00001100.01000000 => 192.168.12.64 11111111.11111111.11111111.11000000 => /26 or 255.255.255.192 This therefore gives a match of IPs in the network 192.168.12.64 (the next network is 192.168.12.128) so broadcast is 192.168.12.127 and usable IPs are .65 to 126. We now match IPs in this range which are: FastEthernet0/1 (192.168.12.65) - ANSWER B Serial0/0 (192.168.12.121) - ANSWER C Serial0/1/102 (192.168.12.125) - ANSWER D If you are having problems understanding this one the key to write out 0.0.0.63 in binary and then invert the bits.
upvoted 49 times
FALARASTA
1 year, 6 months ago
Thank you
upvoted 1 times
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Carter_Milk
Highly Voted 2 years, 11 months ago
Use the magic number technique. Wildcard mask minus 255 so 255 minus 63 = 192. 192 give block size of 64 (256-192) network 0 -63 network 64 -127 network 128 - 192
upvoted 13 times
liviuml
1 year, 7 months ago
@Carter_Milk why you use right hand to go to the left ear? Is more simple to add 63 (from wildcard mask) to 64 (the ip of network) an will give you 127 (the upper limit of network). Results the range x.x.x.64-127. Regards,
upvoted 2 times
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eb63e5a
Most Recent 6 months, 3 weeks ago
Given answer is correct. I converted wildcard bits to subnet mask and I used the fist subnet 192.168.12.65 to 192.168.12.126 , any Ip address between this interval are valid
upvoted 2 times
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[Removed]
7 months, 3 weeks ago
Selected Answer: BCD
B - C - D are correct
upvoted 2 times
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dmaster42
2 years, 1 month ago
excellent explanation, chr, that the way
upvoted 2 times
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aliwqa777
3 years, 6 months ago
I see that no one knows the explanation
upvoted 3 times
Jonfernz
3 years, 6 months ago
It has already been explained. The network 192.168.12.64 0.0.0.63 covers IP address ranging from .64 to .127 (it's using a /25 subnet mask). So in this case, that's what you're looking for.
upvoted 6 times
Sayeem
3 years, 6 months ago
But why 192.168.12.64 0.0.0.63 --> IP address ranging .64 to .127 (why /25 subnet)
upvoted 1 times
BooleanPizza
3 years, 2 months ago
becuse it's a wildcard mask, which is the inverse of the subnet mask which in this case is 255.255.255.192
upvoted 1 times
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BooleanPizza
3 years, 2 months ago
It's /26 actually
upvoted 3 times
Jonfernz
3 years, 1 month ago
sorry. typo. i meant /26
upvoted 3 times
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Sayeem
3 years, 6 months ago
why network 192.168.12.64 0.0.0.63 equals to network 192.168.12.64/26? can anyone help me explain please
upvoted 2 times
jehangt3
3 years, 5 months ago
In order to identify what a wildcard mask actually represents you must subtract the amount from 255 so in this case 255-63=192. So the subnet mask is actually 255.255.255.192 which is a /26. Use my subnet calculator below to help you master subnetting 1st OCT /1 /2 /3 /4 /5 /6 /7 /8 2nd OCT /9 /10 /11 /12 /13 /14 /15 /16 3rd OCT /17 /18 /19 /20 /21 /22 /23 /24 4th OCT /25 /26 /27 /28 /29 /30 /31 /32 HOSTS 128 64 32 16 8 4 2 1 SUBNET 1 2 4 8 16 32 64 128 PREFIX 128 192 224 240 248 252 254 255
upvoted 6 times
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ttomer
3 years, 9 months ago
Why Serial 1.102 and not Serial 1.103? 192.168.12.128 is the next network, isn't it?
upvoted 1 times
hokieman91
3 years, 9 months ago
Network 192.168.12.64 0.0.0.63 covers the addresses for 192.168.12.64 to 127 192.168.12.64 - Network 192.168.12.65 to 126 - Gives 62 Usable host addresses 192.168.12.127 - Broadcast Only Int F0/1, S0/0 and S0/1.102 fall inside this host range S0/1.103 is in the next network
upvoted 3 times
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Zerotime0
3 years, 9 months ago
So net id’$ Are 0/64/128 pick which ever fall in .64-127.
upvoted 1 times
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Chun9
3 years, 9 months ago
I got it thanks.
upvoted 1 times
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Chun9
3 years, 9 months ago
Can anyone can explain for me please? Thanks
upvoted 1 times
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