Subnetting help
Assassin2005
Banned Posts: 6 ■□□□□□□□□□
in CCNA & CCENT
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.
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.
Comments
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
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
Just abit of friendly advice.
It appears you may not be aware that these are actual exam questions. When in doubt, check at www.certguard.com
Blog >> http://virtual10.com
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.
Blog >> http://virtual10.com
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..
you are concerning the last octet
IP Address 00010011 = 19
Subnet mask 11110000 = 240
Network ID 00010000 = 16
?
Broadcast ID 00011111 = 31
?
/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)
/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
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
/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.
Thanks Once again.
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