CCNA – Subnetting
Here you will find answers to CCNA Subnetting Questions
Note: If you are not sure about subnetting, please read my Subnetting tutorial.
Question 1
Given a subnet mask of 255.255.255.224, which of the following addresses can be assigned to network hosts? (Choose three)
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: B C D
Explanation
A subnet mask of 255.255.255.224 has an increment of 32 (the binary form of the last octet is 1110 0000) so we can’t use numbers which are the multiples of 32 because they are sub-network addresses. Besides, we can’t use broadcast addresses of these sub-networks (the broadcast address of the previous subnet is calculated by subtracting 1 from the network address). For example the network address of the 2nd subnet is x.x.x.32 then the broadcast address of the 1st subnet is 32 – 1 = 31 (means x.x.x.31).
By this method we can calculate the unusable addresses, which are (notice that these are the 4th octets of the IP addresses only):
+ Network addresses: 0, 32, 64, 96, 128, 160, 192, 224.
+ Broadcast addresses: 31, 63, 95, 127,159, 191, 223.
Question 2
Which of the following host addresses are members of networks that can be routed across the public Internet? (Choose three)
A – 10.172.13.65
B – 172.16.223.125
C – 172.64.12.29
D – 192.168.23.252
E – 198.234.12.95
F – 212.193.48.254
Answer: C E F
Explanation
Addresses that can be routed accross the public Internet are called public IP addresses. These addresses belong to class A, B or C only and are not private addresses.
Note:
Private class A IP addresses: 10.0.0.0 to 10.255.255.255
Private class B IP addresses: 172.16.0.0 to 172.31.255.255
Private class C IP addresses: 192.168.0.0 to 192.168.255.255
Class D addresses are reserved for IP multicast addresses and can’t be routed across the Internet (their addresses begin with 224.0.0.0 address).
Also we can’t use 127.x.x.x address because the number 127 is reserved for loopback and is used for internal testing on the local machine.
Question 3
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 E
Explanation
We need to remember the default subnet mask of class B is 255.255.0.0. Next, the company requires a minimum of 300 sub-networks so we have to use at least 512 sub-networks (because 512 is the minimum power of 2 and greater than 300). Therefore we need to get 9 bits for network mask (29=512), leaving 7 bits for hosts which is 27-2 = 126 > 50 hosts per subnet.This scheme satisfies the requirement -> B is correct.
We can increase the sub-networks to 1024 ( 1024 = 210), leaving 6 bits for hosts that is 26= 64 > 50 hosts. This scheme satisfies the requirement, too -> E is correct.
Notice: The question asks “The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet” but this is a typo, you should understand it as “”The company needs a minimum of 300 sub-networks and a minimum of 50 host addresses per subnet”.
Question 4
Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choose three)
A – 115.64.8.32
B – 115.64.7.64
C – 115.64.6.255
D – 115.64.3.255
E – 115.64.5.128
F – 115.64.12.128
Answer: B C E
Explanation
CIDR stands for Classless In4ter-Domain Routing, the difference between CIDR and VLSM is slim and those terms are interchangeable at CCNA level.
To specify which IP addresses fall into the CIDR block of 115.64.4.0/22 we need to write this IP address and its subnet mask in binary form, but we only care 3rd octet of this address because its subnet mask is /22.
(x means “don’t care”)
Next, we have to write the 3rd octets of the above answers in binary form to specify which numbers have the same “prefixes” with 4.
4 = 0000 0100
8 = 0000 1000
7 = 0000 0111
6 = 0000 0110
3 = 0000 0011
5 = 0000 0101
12=0000 1100
We can see only 7, 6 and 5 have the same “prefixes” with 4 so B C E are the correct answers.
Question 5
Refer to the diagram. All hosts have connectivity with one another. Which statements describe the addressing scheme that is in use in the network? (Choose three)
A – The subnet mask in use is 255.255.255.192.
B – The subnet mask in use is 255.255.255.128.
C – The IP address 172.16.1.25 can be assigned to hosts in VLAN1.
D – The IP address 172.16.1.205 can be assigned to hosts in VLAN1.
E – The LAN interface of the router is configured with one IP address.
F – The LAN interface of the router is configured with multiple IP addresses.
Answer: B C F
Explanation
VLAN 2 has 114 hosts so we need to leave 7 bits 0 for the host addresses (27 – 2 = 126 > 114). Notice that we are working with class B (both Host A and Host B belong to class B) and the default subnet mask of class B is /16 so we need to use 16 – 7 = 9 bits 1 for the sub-network mask, that means the subnet mask should be 255.255.255.128 -> B is correct.
By using above scheme, C is correct because the IP 172.16.1.25 belongs to the sub-network of VLAN 1 (172.16.1.0/25) and can be assigned to hosts in VLAN 1.
For communication between VLAN 1 and VLAN 2, the LAN interface of the router should be divided into multiple sub-interfaces with multiple IP addresses -> F is correct.
Question 6
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
Answer: A C D
Explanation
If the “ip subnet-zero” command is configured then the first subnet is 172.25.0.0. Otherwise the first subnet will be 172.25.32.0 (we will learn how to get 32 below).
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).
First subnet: 172.25.0.0/19
Second subnet: 172.25.32.0/19
Third subnet: 172.25.64.0/19
4th subnet: 172.25.96.0/19
5th subnet: 172.25.128.0/19
6th subnet: 172.25.160.0/19
7th subnet: 172.25.192.0/19
8th subnet: 172.25.224.0/19
In fact, we only need to specify the third subnet as the question requested. The third subnet ranges from 172.25.64.0/19 to 172.25.95.255/19 so A C D are the correct answers.
Question 7
Refer to the exhibit. In this VLSM addressing scheme, what summary address would be sent from router A?
A. 172.16.0.0/16
B. 172.16.0.0/20
C. 172.16.0.0/24
D. 172.32.0.0/16
E. 172.32.0.0/17
F. 172.64.0.0/16
Answer: A
Explanation
Router A receives 3 subnets: 172.16.64.0/18, 172.16.32.0/24 and 172.16.128.0/18.
All these 3 subnets have the same form of 172.16.x.x so our summarized subnet must be also in that form -> Only A, B or C is correct.
The smallest subnet mask of these 3 subnets is /18 so our summarized subnet must also have its subnet mask equal or smaller than /18.
-> Only answer A has these 2 conditions -> A is correct.
@Anonymous January 15th, 2011
Please answer this question…
Refer to the exhibit.According to the routing table, where will the router send a packet destined for 10.1.5.65
===============
Exhibit :
Network Interface Next -Hop
10.1.1.0/24 e0 directly connected
10.1.2.0/24 e1 directly connected
10.1.3.0/25 s0 directly connected
10.1.4.0/24 s1 directly connected
10.1.5.0/24 e0 10.1.1.2
10.1.5.64/28 e1 10.1.2.2
10.1.5.64/29 s0 10.1.3.3
10.1.5.64/27 s1 10.1.4.4
Options :
===============
A)10.1.1.2
B)10.1.2.2
C)10.1.3.3
D)10.1.4.4
==============
Please help me solve this
________________________________
The packet address 10.1.5.65 is contained in all the following networks :
1/ 10.1.5.0/24
2/ 10.1.5.64/28
3/ 10.1.5.64/29
4/ 10.1.5.64/27
The router will choise the network with the longer prefix (= /29) (= the smallest network including the address 10.1.5.65).
Thus the router will send the packet in the route:
10.1.5.64/29 s0 10.1.3.3
Bye
@kroute January 10th, 2011
Q.The Ethernet networks connected to router R1 in the graphic have been summarized for router R2 as 192.1.144.0/20. Which of the following packet destination addresses will R2 forward to R1, according to this summary? (Choose two.)
A. 192.1.159.2
B. 192.1.160.11
C. 192.1.138.41
D. 192.1.151.254
E. 192.1.143.145
F. 192.1.1.144
Answer: A, D .but how plz tell me any body
_____________________
All packets with destination addresses in the range 1192.1.144.0/20 (= from 192.1.144.0 to 192.1.159.255) will be forwarded from router R2 to R1.
[ The two addresses 192.1.144.0 (~network address) and 192.1.159.255 (~broadcast address) are usually not used for packet addresses ].
A. 192.1.159.2 . . . . (OK)
B. 192.1.160.11 . . . (out of range)
C. 192.1.138.41 . . . (out of range)
D. 192.1.151.254 . . (OK)
E. 192.1.143.145 . . (out of range)
F. 192.1.1.144 . . . . (out of range)
Bye
@kroute January 7th, 2011
Thanks a lot ……Hoang Nguyen
but in this question still i m so confused plz explain breifly….
Q. Which mask is correct to use for the WAN link between the routers that will provide connectivity while wasting the least amount of addresses?netwok-A=60 host,network-B=310host
i think /30 is corect. u just need 2 IP for the interfaces between 2routers so….
how
plz tell ……..
______________________
The WAN link between routers A and B (regardless their size) is a network.
* We need two addresses. One for each router.
* Don’t forget that the first address of the network is a network address and the last address is a broadcast address.
In total we need only 4 addresses. This is provided by a /30 network.
Note: With a /30 prefix, there is “no loss” of addresses.
Bye
Please i need some help from a question in TI
Given the address 192.168.20.19/28, which of the following are valid addresses on this subnet?(choose 2)
a.192.168.20.29
b.192.168.20.16
c.192.168.20.17
d.192.168.20.31
e.192.168.20.0
answer is: a,c
but i think it should be a,d since b,c are outside the network
I’ve read through the many explanations on subnetting but am at a lost for this question. my experience tells me that the answer should be ‘a,d’ but the correct answer listed is ‘a,c’. I just need some explanation as to how that could be.
thanks
can anyone explain q3. . i didn’t understand how 255.255.255.192 is correct?
@salman:
the IP 192.168.29.19/28 is in the subnet beginning from 172.16.20.16 and ends at 172.16.20.31.
D is in the same subnet, right, but it is the BROADCAST-Adress of this subnet and can’t be used… so it is not a valid address of a host.
Hope this helps
@salman
192.168.20.19 /28 is an IP within the subnet of 192.168.20.16 to 192.168.20.31
192.168.20.16 ——-> network (AKA wire) address and cannot be used by a host
192.168.20.17
192.168.20.18
192.168.20.19
192.168.20.20
192.168.20.21
192.168.20.22
192.168.20.23
192.168.20.24
192.168.20.25
192.168.20.26
192.168.20.27
192.168.20.28
192.168.20.29
192.168.20.30
192.168.20.31 ———-> broadcast address and cannot be used by a host
So any address from 17 to 30 is valid
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A nice practice on ip address. It is very useful and fundamental . Thanks 9tut.
Hi guys,
can you help me out how to solve the subnetting question on ccna exam
The network administrator needs to address seven LANs. RIP version 1 is the only routing
protocol in use on the network and subnet 0 is not being used. What is the maximum number of
usable IP addresses that can be supported on each LAN if the organization is using one class C
address block?
A. 6
B. 8
C. 14
D. 16
E. 30
F. 32
What is answer ?
@ Harry
Class C is 255.255.255.0 or 11111111.11111111.11111111.00000000
1s are network bits, and 0s are host bits. In order to create sub-networks you need to “steal” some bits from host bits. In the question they say that “subnet 0 is not being used” and you need 7 LANs and that is giving 1 + 7 = 8 networks. To subnet class C to eight networks you need to “steal” 3 host bits from class C subnet (because 2 on power of 3 is 8). That gives us:
11111111.11111111.11111111.11100000
Now we have just 5 “host bits ” left. 2 on power of 5 is 32 meaning 32 hosts, but remember that first IP is network ID, and last one is broadcast IP … so 32-2 = 30 … the answer is E.
Hope I was clear. Having a glass of red wine …. bit tipsy. If it is not clear, I could do bit better when sober :-)
@ Harry
one more thing …. this “7 LANs” and “subnet 0 is not being used” is just one of those Cisco “confusing questions” … with two network bits (2 on power of 2) you can create just four subnets .. so if you need 5,6,7 or 8 subnets you need to “steal” 3 host bits anyhow.
PS on the glass no three :-D
cheers
Aalhamdulillah,
Hi Everyone,
I M Just here to say thank u so much to 9tut and all who post there valuable comment on this website.
I took mah Exm on 4th Feb’11 and I hv pass d Exm with scoring 923.
All d best to u all who is gonna be appeare dr exm.
Does anyone know where can I get the latest (test Dump) thanks Millions
I don’t understand question 2, how come VLAN 1 & VLAN 2 share the subnet mask of:
255.255.255.128 <——- this mask only gives 126 hosts
AND
VLAN 1 needs 84 hosts with VLAN 2 that needs 114 hosts, that means 84+114=198 hosts for the whole network.
my point is 198 hosts is bigger than 126 hosts how can that mask be valid for this network !!?!!?!?!?!?!?!?!
@ Black Ninja
…. are you sure you understand VLAN concept?
The answer to question #3 should be D, not B and E. The question states a MAXIMUM of 50 hosts, not minimum. B will give you 2^7 (128) hosts > 50. E gives you 2^6 which is 64. The correct answer is 2^5, so you will have 5 host bits making the subnet needed to answer this question to 255.255.255.224.
I think however it would be easier if you change the question and state a MINIMUM of 50 hosts.
@Xsumi5: The answer can’t be D as you’re using subnet mask of 255.255.255.224 which gives only 2^5-2 = 30 hosts ( D is incorrect.
and plz help me
Ip address-address 10.0.0.249,subnet mask 255.255.255.0,preferred dns server -192.223.129.100
alternate dns server -192.223.129.10
how default gateway is 10.0.0.254??
plz explain me this
and plz help me
Ip address-address 10.0.0.249,subnet mask 255.255.255.0,preferred dns server -192.223.129.100
alternate dns server -192.223.129.10
how default gateway is 10.0.0.254??
plz explain me this
Class C is 255.255.255.0 or 11111111.11111111.11111111.00000000
but To subnet class C to eight networks you need to “steal” 3 host bits from class C subnet (because 2 on power of 3 is 8). That gives us:
11111111.11111111.11111111.11100000
but why we choose 3 host bits??plz explain me this
@ black angel
Gateway Question:
IP address is 10.0.0.249, subnet 255.255.255.0 …. the network is 10.0.0.0 255.255.255.0.
In this network valid host range is 10.0.0.1 – 10.0.0.254 (.0 is network ID, and .255 is broadcast address). The gateway can be on any IP address within valid host range. In most production networks they are using the first,or the last available address …. looks more tidy :-)
For the second question about network and host bits you will have to read the books, and understand the binary. Sorry.
thnx :)
again help me..
Refer to the exhibit what can be determined from the output ?
Router#show ip arp
Protocol address age(min) hardware addr type interface
Internet 19.168.1.1 – ca00.17d0.0008 ARPA fastethernet0/0
Internet 192.168.3.1 – ca00.17d0.0008 ARPA fastethernet0/0
Internet 192.168.1.2 0 ca00.17d0.0008 ARPA fastethernet0/0
A. 192.168.1.2 is local to the router.
B. 192.168.3.1 is local to the router.
C. 192.168.1.2 will age out in less than one minute.
D. 192.168.3.1 has aged out and is marked for deletion.
Ans. B
why the ans is b?because 192.168.3.1 is the highest ip address?
there is a network device A between two switches.what is true concerning Network device A?
A. with a network wide mask of 255.255.254.0, each interface does not requre an IP address.
B. With a network wide mask of 255.255.255.128, each interface does require an IP address on a unique IP subnet.
C. with a network wide mask of 255.255.255.0, must be a layer 2 device for the PCs to communicate with each other.
D. with a network wide mask of 255.255.255.0,must be a layer 3 device for the PCs to communicate with each other.
Correct ans- A B D
please explain me the answers
There are 66 hosts on network A and 310 hosts on Network B. which of the following VLSM masks provides enough host addresses for Network A?
A./25 B. /28 C. /26 D. /27
correct ans. A
why /25 is the correct ans?
there are four routes in a router
10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0
which of the following options is the best summarization for the four routes?
A. 10.0.0.0/23
B. 10.0.0.0/22
C. 10.0.0.0/24
D. 10.0.0.0/21
correct ans-B
why the ans is B?
there is a server A IP:131.1.123.24/27,GW: 131.1.123.33
and a workstation B IP:131.1.123.43/27,GW: 131.1.123.33
and the server & workstation is connected by a router.
why the workstation B can’t reach server A?
A. The IP address for workstation B is a subnet address
B. The gateway for workstation B is not on the same subnet.
C. The IP address for server A is a broadcast address.
D. The gateway for server A is not on the same subnet.
correct ans is D
why d is the correct ans?
@Black Angel,
Dash (-) points to local interface of the router after “show ip arp” command, and places it under “Age” field.
Regarding other questions, it is obvious that you have no idea about subnetting. I strongly encourage you to get your hands on Tod Lammles CCNAQ Study guide, or get the pdf from the net.
cheers
ya i know and worried bout it and came here to take help from you.anyway. thnx
@ black angel
sorry mate … but to write step by step explanation for these would take a few pages. Best way is to get a/m book, and you will nail the subnets for sure. I know it looks hard at the beginning (I was terrified at the time), but it’s so logical and quite easy when you get the basics.
cheers
For your first question, here’s how to solve it but you’ll need to get the theory from someone else:
—
there are four routes in a router
10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0
which of the following options is the best summarization for the four routes?
A. 10.0.0.0/23
B. 10.0.0.0/22
C. 10.0.0.0/24
D. 10.0.0.0/21
correct ans-B
why the ans is B?
—
Our interesting octet is obviously the third since everything else is the same. So let’s look at the binary:
10.0.0000 0000.0 (10.0.0.0)
10.0.0000 0001.0 (10.0.1.0)
10.0.0000 0010.0 (10.0.2.0)
10.0.0000 0011.0 (10.0.3.0)
They match up to the sixth bit of the the 3rd octet. That equals a /22 network address or 255.255.252.0 which would look like this in binary:
1111 1111.1111 1111.1111 1100.0000 0000
…where the 1′s are your network bits, and the zeros are your hosts bits.
Your second question:
—
there is a server A IP:131.1.123.24/27,GW: 131.1.123.33
and a workstation B IP:131.1.123.43/27,GW: 131.1.123.33
and the server & workstation is connected by a router.
why the workstation B can’t reach server A?
A. The IP address for workstation B is a subnet address
B. The gateway for workstation B is not on the same subnet.
C. The IP address for server A is a broadcast address.
D. The gateway for server A is not on the same subnet.
correct ans is D
why d is the correct ans?
—
A /27 subnet is 255.255.255.224 or 1110 0000 meaning 2^5 (since 5 bits are being used as hosts) so your subnet range is 32.
131.1.123.0 – Subnet 1
131.1.123.32
131.1.123.64
and so forth
You can see immediately that Server A is on subnet 1 but it’s gateway is on subnet 2 (with workstation B).
—
There are 66 hosts on network A and 310 hosts on Network B. which of the following VLSM masks provides enough host addresses for Network A?
A./25 B. /28 C. /26 D. /27
correct ans. A
why /25 is the correct ans?
—
66 hosts needed. What is our formula? That’s right, 2^n – 2 for hosts where n equals number of host bits.
So let’s see…
2^5 – 2 = 30 hosts…nope!
2^6 – 2 = 62 hosts…nope!
2^7 – 2 = 126 hosts…yep!
borrowing 7 bits = 1000 0000 in binary = /25. Yes, we have more hosts than needed, but nothing below that will work.
Don’t worry about the 2nd network since the question only asks about the first.
which of the 2 sub networks would be included in the summarized address of 172.31.80.0/20
A 172.31.92.0/22
B 172.31.51.16/30
C 172.31.80.0/22
D 172.31.17.4/30
ANS for above is A and C
PlZZZZZZZZZZZZ help me with above one plzzzzzzzzzzzzz explain people i hav exam in fw dayssssssssssssss plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
Summarized Address is:
172.31.80.0 /20 = 172.31.80.0 255.255.240.0
Increment is 16 in third octet / network range 172.31.80.0 – 172.31.95.255
A) 172.31.92.0/22 = 172.31.92.0 255.255.252.0
Increment is 4 in third octet / network range is 172.31.92.0 – 172.31.95.255
B) 172.31.51.16/30 = 172.31.51.16 255.255.255.252
Increment is 2 in fourth octet / network range 172.31.51.16 – 172.31.51.20
C) 172.31.80.0 /22 = 172.31.80.0 255.255.252.0
Increment is 4 in third octet / network range is 172.31.80.0 – 172.31.83.255
D) 172.31.17.4/30 = 172.31.17.4 255.255.255.252
Increment is 4 in fourth octet / network range is 172.31.17.4 – 172.31.17.8
Can you see that A and C networks fall in total range of sumarizzed address?
Cheers
Thnx TJ for helping me with this, i hav some douts how 172.31.92.0/22, 172.31.80.0/22 fall in the full network range. i didnt undrerstand wht is the full network range here.
how many hosts and submasks in 172.16.0.0/23 subnetted /27options are
A 8 subnetts 32host
B 8subnets 31hosts
C 16subnets30hosts
D 16subnetts 32hosts
ans C
how many hosts and submasks in 172.16.0.0/23 subnetted /27options are?
16 subnets and 30 host!
How did you get that?
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
i got this one the exam “WITH” the typo…..this is impossible to answer……………… how can cisco judge me if they dont get it right???????
iven a subnet mask of 255.255.255.224, which of the following addresses can be assigned to network hosts? (Choose three)
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
How we calculate ??
TO DAVE. WHY IS THAT Q IMPOSSIBLE.
LOOK MY FRIEND.
FIRSTLY WE NEED TO CALCULATE THE SUBNET MASK FOR THE DESIRED N/W, OK.
AND FOR 300 SUB-NETWORKS WE SHOULD RESERV 9-BITS IN THE CLASS B SUBNETMASK (255.255.0.0)——> WE SHOULD THEN GET 255.255.255.128 , AND IT´S /25 N/W
THEN WE DEAL WITH THE ALTERNATIVES THAT WE HAVE
A) 255.255.255.0 NO! WE CANT HAVE THIS IF WE WANT TO HAVE 300 SUB-NETWORKS, BECAUSE WE THERE IS NOT ENOUGH OF “BITS LEFT” FOR 300 N/W
B) 255.255.255.128, YES! THIS MATCHES EXACT OUR DESIRED SUBNETMASK , AND WE COULD HAVE 126 HOST ADDRESSES (WICH IS MORE THAN THE 50 HOSTS WE NEED, REMEMBER THERE IS NO ONE WHO IS TELING US WE CAN’T HAVE MORE THAN 50HOSTS, ONLY THAT WE NEED MAXIMUM 50)
C) 255.255.252.0 , NO! WE CAN’T HAVE THIS ALSO BECAUSE THERE IS NOT “ENOUGH OF BITS LEFT” FOR 300 SUB-NETWORKS (SINCE WE NEED TO RESERV A MINIMUM OF 9-BITS FOR THE 300 SUB-NETWORKS, SO WE NEED TO RESRV ALL 8-BITS(255) IN THE 3TH OCT. OF THE CLASS B SUBNETMASK)
D) 255.255.255.224 , NO! NOT ENOUGH OF BITS LEFT FOR 50 HOSTS
E) 255.255.255.192 , YES! WE CAN HAVE THIS ALSO SINCE WE CAN HAVE MORE THAN 300 N/W AND 62 HOST
F)255.255.255.248.0 , NO! WE CAN’T HAVE 300 WITH THIS, WE WOULD END UP HAVING LESS.
iven a subnet mask of 255.255.255.224, which of the following addresses can be assigned to network hosts? (Choose three)
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
How we calculate ??
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1255.255.255.224 ——-> WE SOULD LOOK INTO 4TH OCT ONLY (.224)
255.255.255.224 IN BINARY = 11111111.11111111.11111111.11100000
THE INCREMETN IS 32 AS WE CAN SEE —–> SO LAST OCT(4TH) SHOULD INCREASE BY 32 FOR EVERY STEP. SO WE CAN HAVE IP’S THAT LOOK LIKE THIS
XXXX.XXXX.XXXX.0 – XXXX.XXXX.XXXX.31
XXXX.XXXX.XXXX.32 – XXXX.XXXX.XXXX.63 ( 15.234.118.63, BROADCASTING, NOT VALID)
(134.178.18.56 , VALID IP)
XXXX.XXXX.XXXX.64 – XXXX.XXXX.XXXX.95 (92.11.178.93, VALID IP)
(192.168.16.87, VALID IP)
XXXX.XXXX.XXXX.96 – XXXX.XXXX.XXXX.127 (
XXXX.XXXX.XXXX.128 – XXXX.XXXX.XXXX.159 (201.45.116.159, BROADCAST, NOT VALID)
XXXX.XXXX.XXXX.160 – XXXX.XXXX.XXXX.191
XXXX.XXXX.XXXX.192 – XXXX.XXXX.XXXX.223 (217.63.12.192, N/W ID, NO VALID IP)
SO THIS MEANS THAT THE VALID IP’S WOULD BE THOSE IN B, C AND D
SPELLING ERROR , SOULD = SHOULD AND INCREMETN = INCREMENT , BUT I GUESE YOU UNDERSTAND WHAT I MEANT…LOL
Can any one give me answer and explanation for following question ?
The ip Subnet Zero command is not configured on a router . What would be the Ip address of Ethernet0/0 using first available address from the SIXTH subnet of the network 192.168.8.0/29
A 192.268.8.25
B 192.268.8.41
C 192.168.8.49
D 192.168.8.113
A 192.268.8.25
B 192.268.8.41
C 192.168.8.49
D 192.168.8.113
How many subnet can be gained by sub netting 172.17.32.0/23 in to a /27 mask , and hoe many usable host address will be there be per sunet ?
8 subnets , 31 hots
8 subnets, 32 hots
16 subnest , 30 hots
16 subnets, 32 hots