Subnetting help

Assassin2005

Will somebody help me answering this question. Please along with explanation.

Q 1) Which three Ip address can be assigned to host if subnet mask is /27 and subnet zero is useable?

A)10.15.32.17
B)17.15.66.128
C)66.55.128.1
D)135.1.34.64
E)129.33.192.192
F)192.168.5.63

Q2) Given the address 192.168.20.19/28 which are the valid host address on this subnet?

A)192.168.20.19
B)192.168.20.16
C)192.168.20.17
D)192.168.20.31
E)192.168.20.0

Q3)Given the subnet mask 255.255.255.224 which of the following addresses can be assigned to the network host?

A)15.234.118.63
B)92.11.178.93
C)134.178.18.56
D)192.168.16.87
E)201.45.116.159
F)217.63.12.192

Q4)What valid host range is the IP address 10.166.93.224/20 a part of?

will somebody help me with these questions.PLEASE WITH EXPLANATION COZ I M FINDING IT DIFFICULT TO UNDERSTAND.

THANKS IN ADVANCE.

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Comments

hodgey87

Im hoping his isnt homework or the like but ill help you out some what.

1) /27 = increments of 32 ie 0 32 64 96 128 etc
2) /28 = increments of 16 ie 0 16 32 48 64 etc
3) /27
4)16.0 increments

ill leave the rest to you

Assassin2005

Well bro this isn't my home work these are questions from Test inside where they don't give explanation for their answer. that's why i am asking for help as i am studying CCNA and finding it difficult to understand.

hodgey87

Q 1) Which three Ip address can be assigned to host if subnet mask is /27 and subnet zero is useable?

A)10.15.32.17
B)17.15.66.128
C)66.55.128.1
D)135.1.34.64
E)129.33.192.192
F)192.168.5.63

with /27 you are only working with the last subnet.
Network 0 32 64 96 128 160 192 224
1st usable 1 33 65 97 129 161 193 225
last usable 30 62 94 126 158 190 220 254
Broadcast 31 63 95 127 159 191 223 255

B, D, E are network addresses F is a broadcast address. The only ones you assign to a host are A, C unless im missing something.

Just do the same for the rest

mella060

Test inside are exam questions mate. Forget about exam questions until you have mastered subnetting. Doing exam questions is of no use if you do not understand subnetting properly. Read the CCNA books and do the excercises from them until you can do it in your head.

Just abit of friendly advice.

Essendon

Just like mella060 said, the source you mentioned is a braindump, aka actual exam questions. On this site, everyone's advice is to steer clear of such illegitmate material to safeguard your certs and your future.

It appears you may not be aware that these are actual exam questions. When in doubt, check at www.certguard.com

Assassin2005

Thanks for the advice bro but my CCNA course is over and i have to give exam within 45days and subnetting is the area where i lose marks.

Essendon

Here are a few links that would be of great help.

Online IP Subnet Calculator << A subnet calculator

subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online << For unlimited questions

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html << For learning subnetting

Practice Subnet Questions << Practice

Advice: Stick to one method for learning subnetting, there are quite a few though all of them arive at the same answer.

HTH.

Handbrake

Assassin2005 wrote: »

Q2) Given the address 192.168.20.19/28 which are the valid host address on this subnet?

A)192.168.20.19
B)192.168.20.16
C)192.168.20.17
D)192.168.20.31
E)192.168.20.0

I want to take a shot at this question..please don't laugh hehe

and i'll do the rest tomorrow.. it's kinda late tonite .

subnet mask is /28 representing in binary as
11111111.11111111.11111111.11110000 = 240

you are concerning the last octet
IP Address 00010011 = 19
Subnet mask 11110000 = 240
Network ID 00010000 = 16 <--AND IP to Subnet Mask
Broadcast ID 00011111 = 31 <--AND Subnet mask to Network ID except the last 4 bits of subnet mask..u flip it

So ip address range from 192.168.20.16 ~ 31 anything in the middle are for hosts basically 17 ~ 30 are valid for hosts

so A and C are your answer..I hope i'm correct expert please check..

Assassin2005

yes bro ur are right. but bro how did you figured out 16 and 31.

you are concerning the last octet
IP Address 00010011 = 19
Subnet mask 11110000 = 240
Network ID 00010000 = 16

?
Broadcast ID 00011111 = 31

Forsaken_GA

Assassin2005 wrote: »

Will somebody help me answering this question. Please along with explanation.

Q 1) Which three Ip address can be assigned to host if subnet mask is /27 and subnet zero is useable?

A)10.15.32.17
B)17.15.66.128
C)66.55.128.1
D)135.1.34.64
E)129.33.192.192
F)192.168.5.63

/27 = 255.255.255.224, so subnet ranges will be in increments of 32.

A. 10.15.32.17 - Useable, in subnet 0, within the range of 1 - 30 (0 is the ID, 31 is the broadcast)
B. 17.15.166.128 - Not usable, it's the network ID for subnet 4 (increments of 32, so 0, 32, 64, 96, 128, etc are all network identifier addresses)
C. 66.55.128.1 - Useable, in subnet 0
D. 135.1.34.64 - Not useable, again, network identifier of subnet 2
E. 129.33.192.192 - Not useable, network identified for subnet 6 (again, increments of 32, so 0,32,64,96,128,160,192, etc are identifiers)
F. 192.168.5.63 - Not useable, broadcast address (increments of 32, so 31,63,95, etc are broadcast addresses)

Q2) Given the address 192.168.20.19/28 which are the valid host address on this subnet?

A)192.168.20.19
B)192.168.20.16
C)192.168.20.17
D)192.168.20.31
E)192.168.20.0

/28 = 255.255.255.240, so increments of 16. So the subnet identifier is 192.168.20.16, the broadcast is 192.168.20.31. So .17 to 30 are the useable IP's

Q3)Given the subnet mask 255.255.255.224 which of the following addresses can be assigned to the network host?

A)15.234.118.63
B)92.11.178.93
C)134.178.18.56
D)192.168.16.87
E)201.45.116.159
F)217.63.12.192

Answer gives the subnet mask, so you can get the range off of that. .224 means the subnet increments are in 32. So multiples of 32 are network identifiers, and multiples of 32 - 1 are broadcast addresses. For the questions above -

A. no, broadcast address (32 x 2 = 64 - 1 = 63, so it's a broadcast)
B. Yes (the network ID is 92.11.178.64, the broadcast would be 92.11.178.95, useable IP's fall between .65 and .94, so .93 is assignable)
C. Yes, .32 is the identifier, .63 is the broadcast, so .56 is valid
D. .87 is valid
E. .159 is invalid, broadcast
F. .192 is invalid, network identifier

Q4)What valid host range is the IP address 10.166.93.224/20 a part of?

/20 = 255.255.240.0, subnet increments are 16. The trick here is that they increment on the third octet. So multiples of 16. 0, 16, 32, 48, 64, 80, 96 - 96 is too much, so previous subnet is what we're looking for.

So the subnet range is 10.166.80.0 to 10.166.95.255, with 10.166.80.1 to 10.166.95.254 being usable

This is not really much of an explanation of why those are the answers, but rather how to get them. The why is a basic core fundamental of subnetting that you need to learn, and there are plenty of links in this forum to teach that.

Assassin2005

Thanks you Guyz for all your help. Finally Broken the Subnetting Barrier.
Thanks Once again.

Handbrake

Assassin2005 wrote: »

yes bro ur are right. but bro how did you figured out 16 and 31.

you are concerning the last octet
IP Address 00010011 = 19
Subnet mask 11110000 = 240
Network ID 00010000 = 16
?
Broadcast ID 00011111 = 31
?

IP Address 00010011 = 19
Subnet mask 11110000 = 240
Network ID 00010000 = 16 <--AND IP to Subnet Mask
Broadcast ID 00011111 = 31 <--AND Subnet mask to Network ID except the last 4 bits of subnet mask which are host bits..what u do is..u flip it
so it turns into 1111