please help me with this question about subnetting

[bachmyvo]

Hello everybody,
I got stuck with this question. Please help me to answer and give me some explanation. Thanks.

Which three IP addresses can be assigned to hosts if the subnet mask is /27 and subnet zero is usable? (Choose three.)

A. 10.15.32.17
B. 17.15.66.128
C. 66.55.128.1
D. 135.1.64.34
E. 129.33.192.192
F. 192.168.5.63

[Msizi]

Hello everybody,
I got stuck with this question. Please help me to answer and give me some explanation. Thanks.

Which three IP addresses can be assigned to hosts if the subnet mask is /27 and subnet zero is usable? (Choose three.)

A. 10.15.32.17
B. 17.15.66.128
C. 66.55.128.1
D. 135.1.64.34
E. 129.33.192.192
F. 192.168.5.63

Hello everybody,
I got stuck with this question. Please help me to answer and give me some explanation. Thanks.

Which three IP addresses can be assigned to hosts if the subnet mask is /27 and subnet zero is usable? (Choose three.)

A. 10.15.32.17
B. 17.15.66.128
C. 66.55.128.1
D. 135.1.64.34
E. 129.33.192.192
F. 192.168.5.63


Ok firstly a mask of /27 is 255.255.255.224 in decimal form which gives you a block size of 32,how I got that is subtracted 224 from 256. You can then write down all your valid subnets which will be .0, .32, .64, .96, .128, .160, .192, .224. You must then be able to identify valid hosts range and broadcast address for each subnet to answer this question.You must pay your attention to the fourth octet.
Your correct answeres would be A,C,D.

EXPLANATION:

Option A is correct becos 10.15.32.17 is in the 10.15.32.0 subnet (valid hosts range being 10.15.32.1 - 30 .

Option B is incorrect becos 17.15.66.128 is not a valid host address its a network address (check your block size).

Option C is correct becos 66.55.128.1 is in the 66.55.128.0 subnet (valid hosts range being 66.55.128.1 - 30)

Option D is correct becos 135.1.64.34 is in the 135.1.64.32 subnet (valid hosts range being 135.1.64.33 - 62)

Option E is incorrect becos 129.33.192.192 is a subnet address (check your block size)

Option D is incorrect becos 192.168.5.63 is a broadcast address for the 192.168.5.32 subnet.

READ CHAPTER 3!! its an excellent chapter

Hope that helped.

Cheerz!!

[Msizi]

Fuzz also wrote an excellent article about subnetting, follow this link

http://www.lammle.com/discussion/showthread.php?p=6467#post6467

Cheerz!!!

[sukarabi]

You can also have a look at my finger method, it will make you to subnet very very fastly without the need to write on your paper and it's important for the exam.
Cheers.
http://www.lammle.com/discussion/showpost.php?p=6600&postcount=9