CCNA Subnetting Questions

Here you will find answers to CCNA Subnetting Questions

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= 128 > 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.

Comments

Raghu 02-27-2010

Hi All how many questions will be there on core subnetting?

samer 02-27-2010

hi all i passed today with 1000/1000 thx god
thanks alot for 9tut
about question 4 in fr hot spot the answer a: no password
about your answer 9tut telnet its not like this in exam u can modify it to te1net its one not L
im sure cuz i have full mark

i finish my exam in 55 min
All questions are from TI 11.72 and lab sims are from 9tut.
my labsims are NAT, RIPv2 and Frame relay.
tooo easy...im ready to help any one brothers
my email samer_hussan@hotmail.com

khaled 02-28-2010

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

9tut 02-28-2010

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

azzoz 02-28-2010

i took exam on 28 feb , there are number of questions came up with subnetting questions ,,,the type of questions are same like these,,,,I can tell there are 5 question,,,,so be careful guies and good luck ,,,dont be like me i failed because i didnt review these questions ,,,a also there were a lot of questions about switch vlans ,,,vtp ,,,rstp ,,,one question about ipv6 same as "ccna ipv6" section in this website,,,anyway i got frustrated and never ever take it again until i know i will score good,,,so study the questions here very careful good luck for alllllllll.

Sina 02-28-2010

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

niharika 03-01-2010

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

niharika 03-01-2010

please xplain me with question no 5

KHALED 03-01-2010

thanks SINA

it's clear now

Sina 03-01-2010

To Niharika

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

it`ll become /25 mean we have 128 ip address and 126 usabel ip address ( one in network address and another in broadcat 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 renge is

172.16.1.0 to 172.16.1.128 ( we know 128-127 is not usabel )
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 ) .

to 9tut 03-02-2010

thnx.... 9tut... i got it .......

R 03-02-2010

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

@R 03-04-2010

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.

may 03-09-2010

hi all
question number 6

i dont understand this question how we have number 8 to 32 ?

kamran 03-10-2010

Dear Guys,

Please send correct Answer

1) 209.165.202.128/27 - router has last assignable host address in subnet
2) 209.165.202.128/28 - router has last assignable host address in subnet
3) 192.0.2.16/28 - router has the fourth last assignable host address in the subnet.
4) 192.0.2.0/28 - router has the fourth assignable host address in the subnet.
5) 192.0.2.0/27 - router has the fourth assignable host address in the subnet.
6) 192.0.2.16/28 - router the second usable address in the subnet.

kamran 03-11-2010

Kishore 03-12-2010

Hello Kumara... Here are the answers to your questions :

1) 209.165.202.128/27 answer: 209.165.202.158
2) 209.165.202.128/28 answer: 209.165.202.142
3) 192.0.2.16/28 answer: 192.0.2.27
4) 192.0.2.0/28 answer: 192.0.2.4
5) 192.0.2.0/27 answer: 192.0.2.4
6) 192.0.2.16/28 answer: 192.0.2.18

Kamran 03-12-2010

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 ?

My Email adders...........
t9tut@yahoo.com

muthu 03-13-2010

"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 ? "

to ans ths think abt NAT..... u can easilly formed this network

Kishore 03-13-2010

.............202.128

Kishore 03-13-2010

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

Kishore 03-13-2010

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

Kishore 03-13-2010

Hello Kumran,

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 (This is the reqd subnet)
.............202.144

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 to 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 (the 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

benz 03-15-2010

Question# 3 says a maximum of 50 hosts right? why is .128 correct if this will allow 62 hosts? can you explain, i'm confused. thanks

Muhammad Khan 03-16-2010

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

vinoth 03-23-2010

where can i get latest ccna dump.

moha 03-25-2010

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

lele 03-25-2010

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

9tut 03-25-2010

@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.

moha 03-26-2010

9tut,
Thanks a lot for your clarification.

zf88 03-27-2010

to Muhammad Khan
Asslam alikom,

yes ,the correct ans is E and i am going to explain why:
since there are 3 subnets i have to minus 3*2*=6 addresses from the available hosts.

2^8=256=>>> 256-6=250 host can be made by VLSM on the fourth octet

i hope it is clear now.
*because each subnet has a subnet ip and brodcast ip=2

Hellwarp 03-28-2010

to Muhammad Khan
Asslam alikom

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.

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

Mohammed 03-29-2010

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

Mohammed 03-29-2010

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

Muiez 04-04-2010

first of All thx to Almighty Allah as i passed my CCNa xam on 29 march 2010......
Labs were Eigrp,VTP n access-list..all of these r availible on this gr8 site.....rest questions were frm pass4sure...i gt 940 points......

9tut cheersssss....u make me fell soo happy......

Hellwarp 04-07-2010

Passed with 936
Thanks to 9tut!!! You rock, and a donation on the way!
Labs were Eigrp, VTP and access-list.
Recommend you go through the tutorial in the beginning to understand the different exam interfaces.
Onto CCNP :)
Good luck !

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 ?
who ans plazzzzzzzzzzzzzzz

Zafar 04-11-2010

Need CCNA 640-802 Dumps asap.

MariaXtopher 04-14-2010

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

leo 04-16-2010

Pls help me out of this.

What is the most appropriate summarization for these routes

10.0.0.0
10.0.1.0
10.0.2.0
10.0.3.0

A. 10.0.0.0/21
B. 10.0.0.0/22
C. 10.0.0.0/23
D. 10.0.0.0/24

which one is the right option with explanation please.

thank you guys

leo 04-16-2010

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

purav 04-16-2010

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

ketally luisa 04-17-2010

anyone have latest p4s ??

please send me on ketul84@yahoo.co.uk

thanks you

leo 04-20-2010

purav,

thanks i got it now.

pls can you help me with other the first question above.

thanks

Sean 04-20-2010

Hi, Regarding IP summarization.. im having trouble doing it in reverse.

Where the summarized IP is given and need to now the IP that belong to that summarized IP.

anyone help please. Thanks!

Rahul 04-22-2010

from where to download p4s..........?? plz help.

ciscocert 04-23-2010

hello all ma dear frnds
visit my blog for larest cisco dumps of pass4sure and testinside and many more
get me ur feedback thr too

ciscocert 04-23-2010

hello all ma dear frnds
visit my blog for larest cisco dumps of pass4sure and testinside and many more
get me ur feedback thr too

http://pawanclub.tk

gp 04-23-2010

hi friends..i am gonna take exam on 28th..pls let me know which dumps to prep?some say got frame relay and some say got dhcp..i have ti v17.11 312q...

purav 04-24-2010

leo

your question is not cleared to me, can you double check the question and post it again? with detail

thanks

beginer 04-25-2010

Dear Admin & All
tahnx for the support you are given and really it admires me and I am planning to revise before going to the exam
anyway can you Sina/any one explain me the sixth Q? as I undrstood your explainantion no. 4th Q
R & thnx

leo 04-28-2010

Purav

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

Help subnetting 04-28-2010

which two addresses can be assigned to a host with a subnet mask of 255.255.254.0?

a. 113.10.4.0
b. 186.54.3.0
c. 175.33.3.255
d. 17.35.36.0

Can anyone explain how to solve this please

tinhboss 04-29-2010

thanks so much

pruav 04-29-2010

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 enogh 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 enogh 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

leo 04-30-2010

pruav,

thanks for your effort, you can find this question in Testinside version 17.11, number 72.

regards.

prakash 05-04-2010

hi i have my exam on 6th may . 9tut suggest me with basic tricks to crack any questions
and weight age fo reach topic and which topic has more marks....plz tel me soon

Anonymous 05-04-2010

hi,

I passed my exam with the score of 987/1000. All the questions were from testinside 17.11. My sims were- ACL-2, VTP and EIGRP.

vrnish 05-05-2010

send the latest dumps in testinside and P4S ?

vrnish 05-05-2010

pls say latest version of p4s dumps

Anonymous 05-08-2010

which two addresses can be assigned to a host with a subnet mask of 255.255.254.0?

a. 113.10.4.0 network address
b. 186.54.3.0 hos
c. 175.33.3.255 broadcast
d. 17.35.36.0 network addrress

Can anyone explain how to solve this please

B only is the correct answer

richard 05-09-2010

PLEASE 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

2) subnetmask of 255.255.255.224 which of the following can be assigned to 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

3) which of the following ip address fall into CIDR block of 115.64.4.0/22 ?
a) 115.64.6.255 b) 115.64.7.64 c) 115.64.5.128

4) how to summaized to network 192.1.144.0/20 ?

5) two networks connected to router R2 have been summarized as 192.168.176.0/21 route sent to R1. which two packet destination address will R1 forward to R2

Thank you

richa 05-11-2010

hi,
could you please tell me sim and drag and drop that is coimg now a days in CCNA paper.I have planned to give the exam on 29th May.
Also please provid eme latest dumbs...

leo 05-11-2010

Hello,

I passed my CCNA exam with 910 score yesterday, this exam was easy with this great website. Testinside 17.11 is still valid with about 80% questions from there. The labs i got was ACL, EIGRP and VTP. Please understand the concept and do not cramm the solution because in the ACL the host to access the web server was host D now not host C. Practise all the Labs in this site are you will be good to go.

I want to Thank 9tuts for this great site, you are my Hero. i also want to thank purav for his help.

Wishing everybody best of luck.

Richa 05-12-2010

Hi Iea..

Thank you very much...can u plzss give me testinside 17.11 dumbs ..that would be very helpful....

shakir 05-14-2010

hi richa r u planning to give ur exam on 29th of this month even i too but one day b4 than u..n i hav downloaded testinside 17.11 n 17.12 dumps....just mail me ur e-mail address will forward u those dumps.

ThankYou

shakir 05-14-2010

hi richa

r u planning to give ur exam on 29th of this month even i too but one day b4 than u..n i hav downloaded testinside 17.11 n 17.12 dumps....just mail me ur e-mail address will forward u those dumps.

ThankYou
shakirfpm@gmail.com

dhruv 05-14-2010

richa how can one send u the mail for the dumps....if u want dumps contact me

dhruvgoel007@gmail.com ......i mm giving my exm on last week of may..cheers friends

Yemi 05-15-2010

Pls help me. I have taken CCNA (640-802) 6-times and failed. I had never known about dumps and practice tests. Luckily, I came across this site. Pls, pls, send me the latest dumps out there. The EIGRP is totally different from what is shown on this website. I guess it's changed. But the concepts are the same. My email address is mamakola@hotmail.com and my name is Yemi. Thanks for your help and God bless you all.

JOn 05-15-2010

Hi Yemi,

please go to this link for the latest useful dumps from those who already pass

http://9tut.com/share-your-experience

Jeevak 05-17-2010

can any one provide me ccna 640-802 test inside dumps v17.11 and 17.12

jeevakmukadam30@gmail.com

thanks in advance

Yemi 05-18-2010

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

blackcisco 05-18-2010

u need to focus in which octet the subnetting is taking place. thats the thirth subnet. 255.255.254.0 is equal to /22 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

blackcisco 05-18-2010

sorry prefix must be /23...excuse me

CCNA2B 05-21-2010

Can someone show the subnetting table they used in the exam. Just need to have an idea what people out there are using

Anonymous 05-22-2010

128 64 32 16 8 4 2 1

ALi 05-22-2010

guys question no: 6 :/

can any ine explain how the netwrok ranges were assigned between 64 - 95, because after 65 we can keep incrementing uptill 224..
y did we stop at 65 only as there was a octet containing a 96 and a 100

MUSAWA 05-23-2010

Pls frnds i need d latest CCNA dumbs, my xams is on 14 of june. pls help me.

musawa4u@yahoo.co.uk

God bless u as u help me.

ogard 05-23-2010

About Q6. You said that if the "ip subnet-zero" command is configured then the first subnet is 172.25.0.0.
What if command isn't used? Is than 172.25.0.0 in use at all or we must take 4 bits instead of 3 for 8 subnets?

ogard 05-23-2010

And my e-mail is drazen_85@yahoo.com so if anyone can send me latest dumbs i would appreciate that.

neman 05-29-2010

hi,frnz can any one fwd me wid da latest dumps........

neman 05-29-2010

hi,frnz can any one fwd me wid da latest dumps..plz i'm gvng xam next week my email id is neman.ahamed@yahoo.com

neman 05-29-2010

hi,frnz can any one fwd me wid da latest dumps..plz i'm gvng xam next week my email id is neman.ahamed@yahoo.co.in

eRg 05-29-2010

Hi all

Latest version of Testinside (V17.14)

http://rs168.rapidshare.com/files/390711902/TIS_640-802v17.14_356Q_GB10.rar

Waleed 06-04-2010

Hi All
If any one give a paper of ccna
plz tell me questions of labsims and simulataions
Thanks
waleed_bba_bcs@yahoo.com

NeedHelp 06-06-2010

@gtut
Please explain this sentence from q3. what is the meaning of 'typo'?

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".

9tut 06-06-2010

A typo is a mistake made by accident while typing or transcribing something. I mean they made a mistake while typing "maximum" (should be minimum).

NeedHelp 06-06-2010

@9tut
Thanx for help. By the way what is meant by '9tut'.

Frank 06-09-2010

Can Sum1 do sumthing about QUESTION 4. cos im thinking 7.64 is a broadcast address

Nohelprequired 06-10-2010

@ Frank
How is Question 4 115.64.7.64 a broadcast address? 115.64.4.0/22 subnet address range is 115.64.4.1 to 115.64.7.254, 7.255 is the broadcast for this subnet.

pankaj 06-14-2010

hi, guys
plz tell me ip subnetting & ip addressing

Mooney 06-16-2010

I am taking the ccna in 6 days.
I would be very greatful for any braindump available.
Please reccomend best dump.

e-mail - eastcoastmooney@gmail.com

ketan 06-16-2010

hi all
can any one forward me the latest dumps.
my email id is ketan.lade9@gmail.com

thanks

Pradeeep 06-16-2010

Hi friends,
plz share latest CCNA dump as i am planning to take exam by this month end.
my maild id is mspradeep81@hotmail.com,pradeep_nemom@yahoo.com.
Kindly help.
thanks and regards -MS Pradeep

Suleman 06-17-2010

@NeedHelp
I think 9 is for "network" and tut is for "tutor", so 9tut is "network tutor".

Anonymous 06-17-2010

hiiiiiiii i m nilesh
just now given CCNA exam
& i cleard with 949 marks
thhxxx 9tut
cherrrrssssss

cj2a 06-18-2010

In reference to question 6, where the question states that 172.25.0.0 is to be divided into 8 subnets, I do not understand where the 256 in the formula 256/8 is coming from. Can someone please explain this to me clearly??

Thanks,
cherrrrrssss

alok 06-24-2010

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.

tester 06-25-2010

test

Amit 06-25-2010

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 assine broadcast address and subnet addres 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 beetween 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 beetween 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 beetween 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 mistace sorry for that
Amit

Madani 06-26-2010

@alok 06-24-2010

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

khaled 06-28-2010

latest dumps of 640-802 testinside 14.72

http://www.epda3at.com/vb/showthread.php?t=86

epda3at2010@hotmail.com

Muhammad Nadeem 06-29-2010

Many thanks to to 9TUT for making me clear q6. May GOD bless on every one.

Priya 07-01-2010

Good and perfect explanation which i found really awesome...thanks for the nice article...

keep posting more this kind of question....

my profile : http://www.twitterons.com/Priya

Jack 07-01-2010

Hi Pruav,

There is explanation in previous post for the following but do you have much better explanation for a novice.

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

I am trying to solve it like Question 1 explanation but still I am not able to figure it out.

Jack 07-01-2010

Hi Madani,

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

Dm3md 07-01-2010

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 wether 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.

Vijay0n26 07-04-2010

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.

Vijay0n26 07-04-2010

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 .

Vijay0n26 07-04-2010

Hi prauv,

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,

Kamran 07-04-2010

Hi 9tut,

Please add CCNP 642-902 route testking at 9tut,
or
send me this email address

sherazs_ali@yahoo.com
sherazs_ali@hotmail.com

regard........

Raza, Pakistan 07-05-2010

Cleared CCNA (640-820) on 3rd july 2010 with 1000/1000 marks.

The questions were from TI 17.14 and all the labs were from 9tut with slight modifications

Many Thanks 9tut for such a great website.

Madani 07-09-2010

@Jack 07-01-2010

Hi Madani,

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
______________
Jack 07-01-2010

Hi Pruav,

There is explanation in previous post for the following but do you have much better explanation for a novice.

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

I am trying to solve it like Question 1 explanation but still I am not able to figure it out.
________

Solution:

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

The first 23 bits represent the network address and the last 09 bits represent the host address.
If these last 09 bits are 0, this represents the network address.
If these last 09 bits are 1, this represents the broadcast address.
All other combinations, for these last 09 bits represent valid addresses for hosts.
We limit our study for the last 09 bits.

a) 113.10.4.0 ............... X.X.0000010 0.00000000 (network address)
b) 186.54.3.0 ............... X.X.0000001 1.00000000 (OK)
c).175.33.3.255 ............ X.X.0000001 1.11111111 (broadcast address)
d) 26.35.2.255 .............. X.X.0000001 0.11111111 (OK)

Bye

Willis 07-18-2010

Hi guys! please give me the link or the latest p4s and testinside dumps im planning to exam thi last week of july thanks!

Have a nice day!

willis 07-21-2010

william 07-21-2010

Is it like this

Since /24 is equal to 256 hosts(with Netw + Broadcast address)

The first Subnet is to divide 256/2 which = 128-2=126 host

/25 =126 host

Second divide 128/2=64 or 62 hosts /26 = 62 hosts

Third get the remaining 64 hosts which is also equals to /26 = 62 hosts

Add: 126+62+62=250 hosts

Hope this will help:

Mercenary :)

rahulrawat_2 07-24-2010

guys i m supposed to give my xam on 7th august..and the dumps havnt been updated on pass4sure since 15th may..sum1 plz help me out..

KHALED 07-27-2010

latest dumps of 640-802 testinside 14.72

http://www.epda3at.com/vb/t86.html

epda3at2010@hotmail.com

John 07-31-2010

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?

Alexander 08-03-2010

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

jarich 08-03-2010

hi guys.. i'm hoping to take the exam this august 12.. just want to know if this is updated for this month's exam... If not can i ask a big favor? can i ask for sites or reviewers that would really help me for the exam.. thanks in advance!

jarich_jarich@yahoo.com

Madani 08-03-2010

@John 07-31-2010

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

Angel 08-04-2010

@Madani

Thank you for explaining the concept.

jarich 08-06-2010

hi guys... is this updated for this month's exam?

ccnaboy 08-07-2010

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...

ccnaboy 08-08-2010

9tut pls help...

jarich 08-08-2010

hi 9tut...are the questions here updated for the ccna exam?

which is more updated? http://9tut.com/ccna-access-list or http://9tut.net/ccna-access-list-questions?
pls help... thnx in advance... i really want to pass the exam... It means alot to me.. thnx..

Anonymous 08-09-2010

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

ccnaboy 08-09-2010

thank you for the explanation anonymou.. i understand now...

SANTHOSH 08-10-2010

ARAWERYERGTG

ccna ben 08-13-2010

Hi all i pass my ccna exam on 11th aug. got 936/1000..
i got eigrp, acl and vtp.. questions all came out from Ti ver 17.14...
thanks to this website (9tut) and ppl here for helping.. =) thanks...

imaginative 08-13-2010

@ all

Hey does any1 know the .rar password to Testinside V17.14?? i downloaded it but when i try to extract it asks for a password! :(

NO ONE SEEMS TO KNOW THE PASSWORD TO IT?? help out plz

didiTUT 08-25-2010

Can 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.

shree 08-26-2010

hi
i have CCNA exam on monday but still i dont know from where should i start any one help me , give the question from which i could practice and get gud marks
thanks
shree

Fas 08-31-2010

Done

edun morris 09-03-2010

am planing to write my CCNA exam by monday next week am looking for lab question that i can pratices

divyaprakash 09-05-2010

hi im doing ccna pls can any 1 help me to get the dumps and training materials... catch me at divyaprakasht@gmail.com

CCNA Training with Real Questions & Answers!

CCNA Subnetting Questions

3.26 Copyright (C) 2008 Compojoom.com / Copyright (C) 2007 Alain Georgette / Copyright (C) 2006 Frantisek Hliva. All rights reserved."

CCNA Exam 640-802

Network Resources