Guys just take a look at this --

Identify 3 valid host addresses in the
192.168.27.0 network with the mask 255.255.255.240

1. 192.168.27.33
2. 192.168.27.112
3. 192.168.27.119
4. 192.168.27.126
5. 192.168.27.175

I think non of the results are correct as I feel the address range for the network(192.168.27.0) with the mask 255.255.255.240 is from
192.168.27.1 to 192.168.27.14

I would like to know the comments from every one from the group

Regards
Bhaskar

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b_baruah

I thought :

1. 192.168.27.33 - valid host addresses in subnet 192.168.27.32
2. 192.168.27.112 - subnet address
3. 192.168.27.119 - valid host address in subnet 192.168.27.112
4. 192.168.27.126 - valid host address in subnet 192.168.27.112
5. 192.168.27.175 - subnet broadcast addres in subnet 192.168.27.160


Uncle8

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I think that the question is just VERY poorly worded.

After reading this twice, the second answer I listed here was my first choice, but I think what they are asking, is in the network 192.168.27., with subnetmask .240 what are valid host ips ON ANY NETWORK inside the subnetted networks of 192.168.27.
This would be 1,3,4

Explination below after what else the question could be asking:

OR I think they are asking this: imagine that you have the class c network address 192.168.27.
Now with the subnet 255.255.255.240, what three hosts are valid in a subnetted network. Now the only one close to being correct, and I saw close because it is NOT correct is 2,3,4. They are all in the network of 192.168.27.12-26 but they are NOT ALL VALID HOSTS


So in conclusion, they are asking in the network 192.168.27. with subnet of ./28, after removing all the network and broadcast IPS, what are valid hosts, and you would choose what IPS are valid hosts on any subnetted network in 192.168.27.

How to do this quickly, just find the interval number, and start going through the list(interval number is talked about here www.ciscotrack.com/subetting.html)

btw, sorry for any small mistakes, its 4am and I just work up from sleeping a very long time

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Thanks for the answers.

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Reading it quickly, you assume the same network, which would be answers 2,3,4, except that 2 is a network address, not a host. That should prompt you to re-read it and realize they are asking for host addresses in ANY network, which is 1,3,4. 2 is a network address and 5 is a broadcast.

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The first question, How many people were able to complete this in less than two minutes?


This brings up another very good question, are there going to be questions like this on the test? and if so if you only get a minute or so for each question how are you suppose to run through all the numbers to see which is which? Even if you can subnet in your head, which I am working on everyday, how can you go through all of these and and still complete the exam?

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The acuall solving only took a min or so, it was the figureing out what they wanted that took a little time, for the most part the questions, at least in relation to subnetting are fairly strightforward on what they want, and if you can subnet, they are the easiest questions on the test

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Anyone who thinks they are ready for the CCNA should be able to do this Q in under 2 min. Actually you will get more time than 2 min. The OSI and protocal Q's will provide extra time for the math Q's because they take all of 20 seconds to read and answer.

To answer this question in a fast manner, i would simply find my divisor (which is the last bit borrowed). In this case, 240= the "16" position. So just divide 16 into each answer choices. If it divides without leaving a remander then you know it's not a subnetwork (the "line") which can't be a valid host and then take into consideration the last host bit in each range is the broadcast address and also can't be used.

The first and last subnetwork in the network are not valid along with the host in those subnets.

Give me 60 subnetworking questions and I would feel better about my chances come MONDAY @ 4:30.

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Here's my explination!

Identify 3 valid host addresses in the
192.168.27.0 network with the mask 255.255.255.240

1. 192.168.27.33
2. 192.168.27.112
3. 192.168.27.119
4. 192.168.27.126
5. 192.168.27.175

Ask Questions!

How many Subnets? 2^4-2=14
How many Host? 2^4-2=14
What is the valid subnet? 256-240=16

Subnets:
16 32 48 64 80 96 112 128 144 160



176 192 208 224 240

Now you can see that #2 and #7 cannot be the answer. #2 is a subnet address and #5 is a broadcast.

Subnet 192.168.27.160
First valid host address 192.168.27.161
Last valid host address 192.168.27.174
Broadcast address 192.168.27.175

Hope this clears up the confusion.

LOL

FSW

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quote:

---

Originally posted by KeiWarrior
Here's my explination!

Identify 3 valid host addresses in the
192.168.27.0 network with the mask 255.255.255.240

1. 192.168.27.33
2. 192.168.27.112
3. 192.168.27.119
4. 192.168.27.126
5. 192.168.27.175

Ask Questions!

How many Subnets? 2^4-2=14
How many Host? 2^4-2=14
What is the valid subnet? 256-240=16

Subnets:
16 32 48 64 80 96 112 128 144 160



176 192 208 224 240

Now you can see that #2 and #7 cannot be the answer. #2 is a subnet address and #5 is a broadcast.

Subnet 192.168.27.160
First valid host address 192.168.27.161
Last valid host address 192.168.27.174
Broadcast address 192.168.27.175

Hope this clears up the confusion.

LOL

FSW

---

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Life, the only test that counts!!"

djm
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