CCNA – Subnetting

hi 9tut
can you explan from qes 6 this :

The question stated that the network 172.25.0.0 is divided into eight equal subnets therefore the increment is 256 / 8 = 32 and its corresponding subnet mask is /19 (1111 1111.1111 1111.1110 0000).

because i cann’t anderstand how it came /19
thaks

The subnet mask 255.255.224.0 (/19) has an increment of 32. You can read more about the increment at http://www.certprepare.com/forum/index.php?showtopic=3

To khaled

256/8=32

32-256=224

224=128+64+32 means : 1110000 3 bits is on
it`s class B ip address : 11111111.11111111.11100000.00000000
8+8+3 = 19 it means /19 I hope u got it

hi …. c u pls xplain the second part of the 5th question…

To Niharika

We have Vlan 1 : 84 hosts and vlan2: 114 hosts
According to Vlsm table :

it`ll become /25 mean we have 128 ip address and 126 usable ip address ( one in network address and another in broadcast address )

/25 means : 8+8+8+1 (11111111.11111111.11111111.10000000) means

We have one bit on it means 128
256-128 = 128
Netmask : 255.255.255.128 ( means answer B is correct )

Vlan 1 host A ip address is : 172.16.1.126 it mean vlan1 ip address range is

172.16.1.0 to 172.16.1.128 ( we know 128-127 is not usable )
it means The IP address 172.16.1.25 can be assigned to hosts in VLAN1.
(Answer C is correct )

The LAN interface of the router is configured with multiple IP addresses ( F is correct answer too ) .

i want to subnet 192.248.248.0/24 into five subnets each subnet will contain 170 hosts,95 hosts,60 hosts,50 hosts and 50 hosts. how am i suppose to subnet it ? pls help me

my frnd it’s not possible bcoz u have mention no. of host 170+95+60+50+50 => 254 which is total no of host for /24 network….

as u mention network add. is wrong…so,/24 it’s not possible to create sub network for that all host..

Thanks,
Rajdeep.

Hello Kishore,

2) 209.165.202.128/28 answer: 209.165.202.142 this is the correct answer
142 please Example ?
3) 192.0.2.16/28 answer: 192.0.2.27 this is the correct answer
27 please Example ?

Hello Kamran,

2) 209.165.202.128/28, last assignable IP address

Network: 209.165.202.128 (Class C IP)
Subnet : 255.255.255.240
The block size: 16 (128+64+32+16=240)
Hence, the increments: 209.165.202.0
209.165.202.16
209.165.202.32
………….202.48
………….202.64
………….202.80
………….202.96
………….202.112
………….202.128
So, for N/W 209.165.202.128/28 with block size 16

Network Address 209.165.202.128
Valid IP range: 209.165.202.129 209.165.202.142
Broadcast IP: 209.165.202.143

3) 192.0.2.16/28, fourth last-assignable IP

Network: 192.0.2.16
Subnet : 255.255.255.240 (128+64+32+16=240)
Block Size: 16

So, the N/W(s): 192.0.2.0
192.0.2.16 (reqd block)
192.0.2.32
192.0.0.48 and so on…..

Hence for our network 192.0.2.16/28 with block-size 16

Network : 192.0.2.16
Valid IP : 192.0.2.17 to 192.0.2.30
Broadcast IP : 192.0.2.31 (broadcast IP can’t be assigned to hosts)

Now count from last valid IP:

I last : 192.0.2.30
II last : 192.0.2.29
III last : 192.0.2.28
IV last : 192.0.2.27

That’s it: The fourth last assignable IP is: 192.0.2.27

Hi guyz,
I also have a problem with the the answers to Question# 3
The requirements are: more than 300 sub-nets but not more than 50 hosts,
B 255.255.255.128 meets only the first criteria but yields 128 hosts>50 hosts

E 255.255.255.192 meets only the first criteria but yields 64 hosts>50 host again.
In my opinion, there is only one correct answer: D 255.255.255.224 (2048 sub-nets>300 sub-nets and 32 hosts

@Muhammad Khan
I have see this question in testinside, i thing that answer B is correct.

@moha: The question should be “more than 300 sub-nets and more than 50 hosts” and we have B and E as the correct answers. In real life, we usually calculate the minimum, not the maximum hosts per subnet (for later expansion).

In conclusion, this is a wrong requirement by mistake.

Refer to the exhibit. 192.1.1.0/24 network address needs to be subnetted over the three Ethernt networks attached to R1. With the use of VLSM what is the largest number of host addresses that can be used on the THREE networks combined.

A. 126
B.186
C. 192
D.224
E 250

Correct Ans in TestInside is E
…Can any one please explain why E is correct

to Muhammad Khan
Asslam alikom

Refer to the exhibit. 192.1.1.0/24 network address needs to be subnetted over the three Ethernet networks attached to R1. With the use of VLSM what is the largest number of host addresses that can be used on the THREE networks combined.

Subnets will be as follows:
192.1.1.0/25 [Hosts: 192.1.1.1 - 192.1.1.126] = 126 hosts
192.1.1.128/26 [Hosts: 192.1.1.129 - 192.1.1.190] = 62 hosts
192.1.1.192/26 [Hosts: 192.1.1.193 - 192.1.1.254] = 62 hosts

Total = 250 hosts

Dear Sir,
I would like to point into the question 6 it is not valid question.

because there is a misunderstanding for Increment concept

The network 172.25.0.0 has been divided into eight equal subnets

The best and easiest way to find the increment number is the lowest network bit

Here is what I mean:
- Given class B address, with 8 subnets
- Original subnet mask: 255.255.0.0
- 8 = 1000 (4 bits) –> we need to reserve 4 bits for network in the subnet mask:
255.255.11110000.00000000
-The increment number is the lowest network bit in the new subnet mask bit which is = 16

Subnets ranges will be the following:

1Dear Sir,
I would like to point into the question 6 it is not valid question.

because there is a misunderstanding for Increment concept

The network 172.25.0.0 has been divided into eight equal subnets

The best and easiest way to find the increment number is the lowest network bit

Here is what I mean:
- Given class B address, with 8 subnets
- Original subnet mask: 255.255.0.0
- 8 = 1000 (4 bits) –> we need to reserve 4 bits for network in the subnet mask:
255.255.11110000.00000000
-The increment number is the lowest network bit in the new subnet mask bit which is = 16

Subnets ranges will be the following:

172.25.0.0 –> 172.25.15.255
172.25.16.0 –> 172.25.31.255
172.25.32.0 –> 172.25.47.255 ==> here is the third subnet range
172.25.48.0 –> 172.25.63.255
.
.
.

Which none of the given options belongs to this range

please reply as fast as possible

Dear Sir,

I think I made a mistake to conclude so fast into finding Question 6 is not valid ..

I’m so sorry .. It is valid

I just mixed between increment concept and the given question

It means the WHOLE given address has been subnetted into 8 EQUAL subnets

The number of borrowed bits is 2^x=8 –> x =3

255.255.11100000.00000000

so, increment number is lowest network bit which is = 32

so the network ranges are divided within range of 32

———

Regarding IP Subnet-zero:
To enable the use of subnet 0 for interface addresses and routing updates, use the ip subnet-zero command in global configuration mode. To restore the default, use the no form of this command.
ip subnet-zero
no ip subnet-zero

————-
Please remove my previous comment, I don’t want people to get confused as I did

I’m so sorry again, and thank you very much for providing such an amazing resources

The network 172.25.0.0 has been divided into eight equal subnets. Which of the following IP addresses can be assigned to hosts in the third subnet ?
who ans plazzzzzzzzzzzzzzz

To 1 04-07-2010
The network 172.25.0.0 has been divided into eight equal subnets. Which of the following IP addresses can be assigned to hosts in the third subnet ?
Your answer:
172.25.0.0
172.25.32.0
172.25.64.0
172.25.96.0
172.25.128.0
172.25.160.0
172.25.192.0
172.25.224.0
The third subnet will be from 172.25.64.0–172.25.95.255
You can see the above question in p4s. under subnetting.G-luck

Which subnet mask will place all hosts on Network B in the same subnet with the least of wasted address?

Net A Net B
66 Host 310 Host

Options

a. 255.255.255.0
b. 255.255.254.0
c. 255.255.252.0
d. 255.255.248.0

Which option is correct with explantion

@leo:

10.0.0.0 –>0000 1010 .0000 0000 .0000 00¦00 .0000 0000
10.0.1.0 –>0000 1010 .0000 0000 .0000 00¦01 .0000 0000
10.0.2.0 –>0000 1010 .0000 0000 .0000 00¦10 .0000 0000
10.0.3.0 –>0000 1010 .0000 0000 .0000 00¦11 .0000 0000
————————————————————————————-
0000 1010 .0000 0000 .0000 00¦00. 0000 0000 (10.0.0.0)

1. convert into binary
2. out of all addresses you can see all the bits on left of the line are common
3. leave all common bits as it is
4. on right of the line turn all bits to 0
5 this is your common subnet number here we got 10.0.0.0
6. count number of common bits which is left of the line it is 8+8+6=22

so answer is 10.0.0.0/22

do not get confuse, this is the method in book i cant explain in detail but if you try same formula on another question you will get answer right.
let me give you another question
172.16.12.0
172.16.13.0
172.16.14.0
172.15.16.0
answer is 172.16.12.0/22

purav,

thanks i got it now.

pls can you help me with other questions:

you can find the question am trying to solve in testinside v17.11, question 70.

the question goes thus: Refer to the exhibit. Which VLSM mask will allow for the appropriate number of host addresses for Network A

Network A Network B

66 Hosts 310 Hosts

Options

A. /25
B. /26
C. /27
D. /28

2. Refer to the exhibit. Which subnet mask will place all hosts on Network B in the same subnet with least amount of wasted addresses?

Network A Network B

66 Hosts 310 Hosts

A. 255.255.255.0
B. 255.255.254.0
C. 255.255.252.0
D. 255.255.248.0

kindly help me out with explanation.

thanks

@leo

networkA 66 hosts
so use formula 2^h-2 to find usable hosts, where h = number of zeros
2^1=2-2=0
2^2=4-2=2
2^3=8-2=6
2^4=16-2=14
2^5=32-2=30
2^6=64-2=62 >>> you cannot use this as it is less than 66
2^7=128-2=126 >>> you can use this despite wasting some address space
remember 1 means subnet and 0 means hosts
2^7=128-2=126 so h=7 will give us enough hosts for our requirement
1111 1111 . 1111 1111 . 1111 1111 . 1000 0000 (7 zeros and rest 1′s)
its equal to /25 (no. of 1′s) and 255.255.255.128

networkB 310 hosts
2^1=2-2=0
2^2=4-2=2
2^3=8-2=6
2^4=16-2=14
2^5=32-2=30
2^6=64-2=62
2^7=128-2=126
2^8=256-2=254 >>> you cannot use this as it is less than 310
2^9=512-2=510 >>> you can use this despite wasting some address space
remember 1 means subnet and 0 means hosts
2^9=512-2=510 so h=9 will give us enough hosts for our requirement
1111 1111 . 1111 1111 . 1111 1110 . 0000 0000 (9 zeros and rest 1′s)
its equal to /23 (no. of 1′s) and 255.255.254.0

for network A –> /25 is the answer and
for network B –> 255.255.254.0 is the answer

I still don’t know the question properly but I hope you will figure it out from this explanation.

goodluck

PLEASE!!!

HELP EXPLAIN BRIEFLY?

1) Which two address can be assigned to host with in subnet mask of 255.255.254.0 ?
a) 113.10.4.0
b) 186.54.3.0
c).175.33.3.255
d) 26.35.2.255
e) 172.35.36.0

u need to focus in which octet the subnetting is taking place. thats the thirth subnet. 255.255.254.0 is equal to /23 so the range is 2. 256-254=2. the range of ip add are xxx.xxx.0.1 to xxx.xxx.1.254. Now we need to see which addresses we could use in this range…those are b and d.

b and d) falls in the subnet of xxx.xxx.2.0 till xx.xxx.3.255

a) is a network address
c) is a broadcast address
e) is a network address

cheers…

blackcisco

Which three ip address can be assign to hosts if the subnet mask is /27 and subnet zero is usable(choose 3)

1. 10.15.32.17
2. 17.15.66.128
3. 66.55.128.1
4. 135.1.64.34
5. 129.32.192.192
6. 192.168.5.63

plz help me to understand to do the answer.

hi alok

i think here you can understand this question

27 subnet mask mean increasing value is 32. (that u can understand )

(and you can not assign broadcast address and subnet address as host)

1st. 10.15.32.17 belong to class A ,

now 10.15.32.0 —– 10.15.32.31

10.15.32.32 —- 10.15.32.63
here is your 1st answer 10.15.32.17 is comes in between 10.15.32.0–10.15.32.31.

2nd.|| belong to class A
17.15.66.0 — 17.15.66.31 | 17.15.66.96 —17.15.66.127
17.15.66.32– 17.15.66.63 | 17.15.66.128—17.15.66.159
17.15.66.64 –17.15.66.95
here is 17.15.66.128 is subnet so you can’t use as host.

3er. 66.55.128.0– 66.55.128.31 | 66.55.128.32 –66.55.128.63
here is your 2nd answer 66.55.128.1 is comes in between 66.55.128.0– 66.55.128.31.

4rd. belong to class B
135.1.64.0 — 135.1.64.31 || 135.1.64.32—135.1.64.63
here is your 3rd answer ,66.55.128.1 is coming between 135.1.64.0 — 135.1.64.31 .

Answer is 1. 10.15.32.17,|| 2. 17.15.66.128, || 3. 66.55.128.1

i thin its help you
any mistake sorry for that
Amit

@alok

Which three ip address can be assign to hosts if the subnet mask is /27 and subnet zero is usable(choose 3)

1. 10.15.32.17
2. 17.15.66.128
3. 66.55.128.1
4. 135.1.64.34
5. 129.32.192.192
6. 192.168.5.63

plz help me to understand to do the answer.
_________

Solution:
For a prefix / 27, we limit our study for the 4th byte.

The first 3 bits are part of the network address, the last 5 bits represent the host portion.
If these last 05 bits are 0, this represents the network address.
If these last 05 bits are one, this represents the broadcast address.
All other combinations, for these last 05 bits represent valid addresses for hosts.
Once again, the examination of these last 05 bits -alone- determines the validity of the resulting host address.

1. 10.15.32.17 – - — – - X.X.X. 000 10001 (OK)
2. 17.15.66.128 – - – - – X.X.X. 100 00000 (network address)
3. 66.55.128.1 – - – - – - X.X.X. 000 00001 (OK)
4. 135.1.64.34 – - – - – - X.X.X. 001 00010 (OK)
5. 129.32.192.192 – - – - X.X.X. 110 00000 (network address)
6. 192.168.5.63 – - — – - X.X.X. 001 11111 (broadcast address)

Bye

I took the exam last week and I think I missed a very simple question. Does anyone know what Cisco is expecting when they ask “how many IP’s can be given to hosts in a /26?”.

There are 64 IP’s and they don’t tell you whether or not IP subnet zero is a given. There are technically 4 answers and all 4 are available to us.

1. 64 (assuming subnet zero is a given)

2. 63 (assuming subnet zero is ok but minus 1 for the ethernet IP)

3. 62 (no subnet zero)

4. 61 (no subnet zero and minus 1 for the ethernet)

I say minus one for the ethernet because no matter what, you will always have an address out of your pool for the router interface. You help is greatly appreciated.

Hi Dm3md ,

Actually if you see there is nothing to do with IP SUB-NET ZERO command.

For /26 you will get 64 ip addresses , in this Network and Broadcast will occupy each an IP address , so 64-2 = 62 host assignable ip addresses will be available.

Hi,

can u help this with simplest way?

1) Which two address can be assigned to host with in subnet mask of 255.255.254.0 ?
a) 113.10.4.0
b) 186.54.3.0
c).175.33.3.255
d) 26.35.2.255

regards

Hi jack ,

Yeah they have cleanly stated that /23 is the prefix length.

Block value will be bit ’2′ in the third octet of the IP address right?

so the network range must be like 0-2 , 2-4 ,4-6 etc,

in this question if you see

a) 113.10.4.0(wrong)
b) 186.54.3.0(right)
c).175.33.3.255(wrong)
d) 26.35.2.255(right)

Ans b,d comes under 2-4 range .

Hi,
If you can understand Block values in IP addressing you can capture this explanation.

This is how IP address will look ,

00000000.00000000.00000000.00000000

block values are

128 | 64 |32 | 16 | 8 | 4 | 2 | 1 128 | 64 |32 | 16 | 8 | 4 | 2 | 1
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 . 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 .128 | 64 |32 | 16 | 8 | 4 | 2 | 1 128 | 64 |32 | 16 | 8 | 4 | 2 | 1
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 . 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0

/23 means occupies the first 2 bytes completely and 7 bits in the third octet. 7th bit in the third octet is bit 2.

So the , if the SUB-NET ZERO command is enabled

Network range is 0-2 , 2-4 , 4-6,…

In the first subnet 0.0 – 2.0

First host assignable IP address is 0.1 and last one is 3.254 right ,
So the broadcast address is 3.255 and 0.0 is network address for this subnetwork.

All the IP address which lies under 0.1–3.254 are valid and host-assignable ones.

I think my explanation could be helpful.

Regards,

Hello All,

This question is from Testinside v17.14 for subnetting.

Q. For the network 192.0.2.0/23 which option is a valid IP address that can be assigned to a host?

A. 192.0.2.0
B. 192.0.2.255
C. 192.0.3.255
D. 192.0.4.0

The correct answer is B. Can someone explain why the answer is B and not C?

192.0.2.1 first available address
192.0.3.254 last available address
192.0.2.0 network address
192.0.3.255 broadcast address

@John

Hello All,

This question is from Testinside v17.14 for subnetting.

Q. For the network 192.0.2.0/23 which option is a valid IP address that can be assigned to a host?

A. 192.0.2.0
B. 192.0.2.255
C. 192.0.3.255
D. 192.0.4.0

The correct answer is B. Can someone explain why the answer is B and not C?

__________

Solution:

Subnet mask 255.255.254.0 = (11111111.11111111.1111111 0.00000000) = prefix /23.

192.0.2.0 / 23 ………… [ 11000000 . 00000000 .0000001 ] 0.00000000 (network address)

A) 192.0.2.0 …………… X.X. 0000001 0.00000000 (network address)
B) 192.0.2.255 ………… X.X. 0000001 0.11111111 (OK, valid host address)
C) 192.0.3.255 ………… X.X. 0000001 1.11111111 (broadcast address)
D) 192.0.4.0 …………… X.X. 0000010 0.00000000 (out of range)

Bye

Hi guys anyone knows how to solve this question from TI.

The networks connected to router R2 have been summarized as a 192.168.176.0/21 route and sent to R1. Which two packet destination addresses will R1 forward to R2?

(Choose two)

A. 192.168.194.160
B. 192.168.183.41
C. 192.168.159.2
D. 192.168.183.255
E. 192.168.179.4
F. 192.168.184.45

thanks…

The networks connected to router R2 have been summarized as a 192.168.176.0/21 route and sent to R1. Which two packet destination addresses will R1 forward to R2?

(Choose two)

B. 192.168.183.41
E. 192.168.179.4

/21 therefore subnets of 8 …
this subnet is 192.168.176.0-192.168.183.255

plz anyone explain to me why is it so?

How many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask, and how many usable host addresses will there be per subnet?

Answer is 16 Subnets, 30 hosts.

thnx in advance

172.17.32.0/27 will give 30 hosts and 8 subnets.

/27=255.255.255.224
8 8 8 3 = 27

256-224=32, hence increment is 32

172.17.32.0 – 172.17.32. 31——1 Ntw
172.17.32.32 – 172.17.32.63 —— 2 Ntw
172.17.32.64 – 95 —— 3 Ntw
96 127—– 4 Ntw
128 159—– 5 Ntw
160 191—— 6 Ntw
192 223—— 7 Ntw
224 255—— 8 Ntw

Total client IP 32
No of valid host: 32-2=30( where 2 ips are used for network IP and broadcast IP )

“i want to subnet 192.248.248.0/24 into five subnets each subnet will contain 170 hosts,95 hosts,60 hosts,50 hosts and 50 hosts. how am i suppose to subnet it ? ”

can any 1 explain this plz

@apple

170 hosts need a /24 subnet. So, there is no suffisant ip space for another subnets…
Please, check the question. Thanks.

i’ll try.

/23= IIIIIII.IIIIIII.IIIIIII0.00000000
/27= IIIIIII.IIIIIII.IIIIIII.III00000

the difference would be:

I.III00000

you have 4 borrowed bit.

getting the subnet: 2 xy 4=16
getting the host: 2 xy 5 – 2= 30

Hope it helps.

Hi, on question # 6

8 subnet is needed so 8 = 00001000 ( 4 bits)
4 bits should be borrowed on the host portion

11111111.1111111.11110000.00000
|
16

why is it that it became 32 or 3 bits??

@rhuna

Here’s the question:
The network 172.25.0.0 has been divided into eight equal subnets. Which of the following IP addresses can be assigned to hosts in the third subnet if the ip subnet-zero command is configured on the router? (Choose three)

A – 172.25.78.243
B – 172.25.98.16
C – 172.25.72.0
D – 172.25.94.255
E – 172.25.96.17
F. 172.25.100.16

Focus first on the subnet.. 8 equal subnet is the clue…

it goes like this:
IIIIIIII.IIIIIII.III00000.00000000
3 borrowed bits in the 3rd octet…
2 xy 3= 8… that satisfies the question.
the subnet mask is 224…
increment of 32…
IP subnet zero is in effect…
therefore the 3rd subnet is 172.25.64.0/19…
the range would be 65-94…

A, C and D are the correct hosts…

hope that helps…

I completely agree with closetfan!!!

The best way to read that question is that you have created “8″ equal subnets…well..if we have created 8 equal subnets…how many host bits did we borrow to create at least 8 subnets???? We need at least 3 which gives us a subnet of 255.255.224.0 since this is a class B address with a default subnet of 255.255.0.0. The block size of 32 is incremented in the third octet. :-)

@closetfan

the 3rd subnet is 172.25.64.0/19 as you say,
but the range would be 172.25.64.0 to 172.25.95.255

so :
* 172.25.64.0 . . . . . . . . . . . . . . . . (network address)
* 172.25.64.1 to 172.25.95.254 . . . . (host addresses)
* 172.25.95.255 . . . . . . . . . . . . . . . (broadcast address)

Hi I have a question about creating a translation on a router, with 11 inside global address 1.1.128.1-1.1.135.174 but they are not consecutive, 11 inside local 10.18.14.90 – 10.20.122.240 also not conservative, now the question asks to create a translation on the router. With answer as follow

A ip nat pool mynats 1.1.128.1 1.1.1.135.254 prefix-length 19 correct

B ip nat pool mynats 1.1.128.1 1.1.1.135.254 prefix-length 18 wrong

can some one please explain what the 1 and 135 254 mean, and the prefex i dont understand.

second question refer to the diagram below

———–e0 router1 s0———–s0 192.1.1.5/30 router2 e0 192.1.1.65/26

to explain router1 is connected to router 2 through s0 to s0 on router 2, router 2 ‘s s0 address is 192.1.1.5/30 and router 2 ‘s e0 address is 192.1.1.65/26 now the question is what configuration would be done on router 1 is correct.

A interface ethernet 0
ip adddress 192.1.1.129 255.255.255.192
no shut down A IS CORRECT

B interface ethernet 0
ip adddress 192.1.1.97 255.255.255.192
no shut down B is wrong

my question is they did not give what network ethernet 0 of router 1 is on so how should the diagram help me in knowing that the ip address of e0 is 192.1.1.129 with subnet mask 255.255.255.192 and not 192.1.1.97 with subnet mask 255.255.255.192
router 1 has e0 ip address not given and s0

3RD QUESTION
3rd question there is a switch with 2 vlans vlan 1 host a 172.16.1.126 and 84 hosts, vlan 2 host b 172.16.1.129 with 114 hosts.

now the question say which is correct

C ip address 172.16.1.25 can be assigned to hosts in vlan 1 is CORRECT
D ip address 172.16.1.205 can be assigned to hosts in vlan 1 is WRONG,

how does one calculate what ip address ranges fall in vlan 1 and vlan 2 i dont understand both C and D.

4TH QUESTION refer to the diagram below

host 10.1.0.36—–switch———-network device A—–switch—-host 10.1.1.70

two hosts connected to 2 different switches and the 2 switches connected together by a network device A

now question asks which is right
A. With a network wide mask of 255.255.255.128 each interface does not require an ip address WRONG
B. With a network wide mask of 255.255.255.128 each interface require a unique ip subnet CORRECT
E. With a network wide mask of 255.255.254.0 each interface does not require an ip address CORRECT

B,E are correct, A is wrong, how does a subnet mask tell you if each interface require a unique ip address or not, i dont understand this

Thank you anyone. i tried drafting the problem and explaining also, m going to write my ccna next week and i need help on those questions please

A national retail chain needs to design an IP addressing scheme to support a nationwide network. The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet. Working with only one Class B address, which of the following subnet masks will support an appropriate addressing scheme? (Choose two)

A – 255.255.255.0
*B – 255.255.255.128
C – 255.255.252.0
D – 255.255.255.224
*E – 255.255.255.192
F – 255.255.248.0

Answer: B(300) E (50)
Explanations
From 300 subnets so
Subnetting when given a required number of networks 300 thus this belong to a
Class B (255.255.0.0)

512 256 |128 64 32 16 8 4 2 1
1 |0 0 1 0 1 0 0 0 = 9 bits @ 256
The binary representation of 300 = 00000001.00101000.
binary representation of 300 = 00000001.00101000 = 9 BITS.
The full binary representation of the subnet mask is as follows:
255.255.0.0 = 11111111 .11111111 . 00000000 .00000000
- We must “convert” 9 of the client bits (0) to network bits (1) in order to satisfy the requirements-
New Mask = 11111111 .11111111 .111111111 .1000000
- If we convert the mask back to decimal, we now have the subnet mask that will be used on all -
512 256 |128 64 32 16 8 4 2 1
1 |1 0 0 0 0 0 0 0 so, the
the new networks – 255.255.255.128

From 50 subnets so
Subnetting when given a required number of networks 300 thus this belong to a
Class C (255.255.255.0)
512 256 |128 64 32 16 8 4 2 1
|0 0 1 1 0 0 1 0 = 6 bits @ 32
@ 50 = .00110010
From 255.255.255.0, the
11111111 .11111111 . 11111111 .00000000
512 256 |128 64 32 16 8 4 2 1
|1 1 0 0 0 0 0 0 = 192
the new networks – 255.255.255.192

Hope that it can help.

to ccnaman answer to Q5

The 2 VLANs are both using same mask /25 : whcih allow to create 2 subnetwork

1-172.16.1.0 (range 1-126) .127 Bcast for this subnet –VLAN 1
2-172.16.1.128 (range 129-254), .255 bcast for this subnet ….VLAN 2

ip :172.16.1.205 can be in VLAN 2 , but the question (stated Vlan 1) which is not true

so the answer is wrong (and the question asked to choose the corrects answer )
and 172.16.25 is correct , since in range of vlan 1 1 (0-126)

I hope this will clarify things for you

Thanks,

1) Which two address can be assigned to host with in subnet mask of 255.255.254.0 ?
a) 113.10.4.0
b) 186.54.3.0
c).175.33.3.255
d) 26.35.2.255

my email id
tushar14029@!yahoo.co.in