CCNA – Subnetting Questions 2

The network default gateway applying to a host by DHCP is 192.168.5.33/28. Which option is the valid IP address of this host?

A. 192.168.5.55
B. 192.168.5.47
C. 192.168.5.40
D. 192.168.5.32
E. 192.168.5.14

Answer: C
how it gets???///i got it B

@balaji
our IP: 192.168.5.33 255.255.255.240 (/28)
subnetwork ID: 192.168.5.32
broadcast IP: 192.168.5.47

IP range in which the interface could be configured on this subnetwork: 192.168.5.34 ~ 192.168.5.46

only option C is valid.

why not option B? because that is the broadcast IP and you can not assign it to an interface.

i agree with u, xallax. Because the hop are 16….32…..48…..and so on.

would you please explain question 10?

Which two addresses can be assigned to a host with a subnet mask of 255.255.254.0? (Choose two)

A. 113.10.4.0
B. 186.54.3.0
C. 175.33.3.255
D. 26.35.2.255
E. 17.35.36.0

subnet increment is 2, subnet mask is 23
so, host range will be from 2.1 to 3.254, and … so, the answer is B and D.

@angel
increment of 2.0
x.x.0.0
x.x.2.0
x.x.4.0

x.x.2.255 and x.x.3.0 are both in the x.x.2.0 ~ x.x.3.255 range

very confusing but true. especially when /mask goes beyong /24
Thanxs alot

ye its rht

@all

Can someone please explain question 4? I don’t understand the explanation given.

Thanks.

Dear Des… Focus on third oct
DECIMAL BINARY
10.0.0.0 = X.X.0000 0000.0
10.0.1.0 = X.X.0000 0001.0
10.0.2.0 = X.X.0000 0010.0
10.0.3.0 = X.X.0000 0011.0
Only change happened on the last two bits {00,01,10,11}…Right >??
so
= NNNN NNNN.NNNN NNNN.NNNN NNHH.HHHH HHHH =>our Increment is bit number 22 last bit in Network /22
hope that help you

@Z.World

Thanks a bunch, i thought no one would response.

Which two addresses can be assigned to a host with a subnet mask of 255.255.254.0? (Choose two)
A. 113.10.4.0
B. 186.54.3.0
C. 175.33.3.255
D. 26.35.2.255
E. 17.35.36.0

Im missing something here- Q10:- surely with a mask of 255.255.254 ie /23, only gives a host range of 512 addresses and thats it. then youve only got one bit left in the 3rd octet- how can this accommodate any value more than 0 or 1? Can someone please explain?
Many thanks -Rob

@roblightwater
using that mask, /23, we get these subnets:

x.x.0.0 ~ x.x.1.255
x.x.2.0 ~ x.x.3.255
x.x.4.0 ~ x.x.5.255
….
x.x.254.0 ~ x.x.255.255

the increment is 2.0…
there are many ways to find out the increment. my favorite one is by doing the “magic number” subtraction (seen on TestOut videos).
255.255.256.0 -
255.255.254.0
____________
255.255.002.0

let’s look at our options now:
A. 113.10.4.0
subnet address, False

B. 186.54.3.0
valid IP in the x.x.2.0 ~ x.x.3.255 address range, True

C. 175.33.3.255
last IP in the x.x.2.0 ~ x.x.3.255 address range. this is used as broadcast address for 175.33.2.0/23. False

D. 26.35.2.255
valid IP in the x.x.2.0 ~ x.x.3.255 address range, True

E. 17.35.36.0
the increment is 2.0.
x.x.36.0 is a multiple of 2.0 so we can conclude that this is a network address. False

please read a bit here too:
http://www.9tut.com/subnetting-tutorial

Can somebody explain me question 4.

@Manish
Q4 : Ans : B as 10.0.0.0/22 -> Correct

10.0.0.0 = 0000 1010.0000 0000.0000 0000.0000 0000
10.0.1.0 = 0000 1010.0000 0000.0000 0001.0000 0000
10.0.2.0 = 0000 1010.0000 0000.0000 0010.0000 0000
10.0.3.0 = 0000 1010.0000 0000.0000 0011.0000 0000

Default is Class A network, so we can write subnet mask

10.0.0.0 = 255.0.0.0

10.0.3.0 = 0000 1010.0000 0000.0000 0011.0000 0000
255.255.252.0 = 1111 1111.1111 1111.1111 1100.0000 0000 = /22

It may be help to you as above.

@xallax.. I am still not clear about your elaboration of Q9. Will you please explain in detail?
thanks

@ahamad

“The network default gateway applying to a host by DHCP is 192.168.5.33/28.”
from the above we find out on which network the host is: 192.168.5.32 with a subnet mask of 255.255.255.240 (or /28)

the increment for this subnet is 16. how to find it out fast? 256 – the value of the last byte (240). so… 256-240 = 16

subnets with this increment are:
x.x.x.0 ~ x.x.x.15
.16 ~ .31
.32 ~ .47
and so on

we’re interested in the 3rd subnet, x.x.x.32 ~ .47
only option C from the list given matches an IP address on this range (C. 192.168.5.40)

this type of question MUST be answered in less than 1 minute.
here’s 9tut’s tutorial on subnetting:
http://www.9tut.com/subnetting-tutorial

Hi …

Can anybody please explain the answer for Q.10 please ?

@rakhy
subnet mask of 255.255.254.0
the subnet will be:
x.x.0.0 ~ x.x.1.255
x.x.2.0 ~ x.x.3.255
……
add multiple of 2.0
……
x.x.254.0 ~ x.x.255.255

options were:
A. 113.10.4.0
wrong, network ID

B. 186.54.3.0
correct, it’s on the x.x.2.0 ~ x.x.3.255 subnet

C. 175.33.3.255
broadcast IP for x.x.2.0 ~ x.x.3.255

D. 26.35.2.255
correct, it’s on the x.x.2.0 ~ x.x.3.255 subnet

E. 17.35.36.0
multiple of 2.0
this is a network ID. wrong

hi al how this can be true any one who can discribe it please ?
Question 10

Which two addresses can be assigned to a host with a subnet mask of 255.255.254.0? (Choose two)

A. 113.10.4.0
B. 186.54.3.0
C. 175.33.3.255
D. 26.35.2.255
E. 17.35.36.0

Answer: B D

@adburehman
have you read my post just above yours? :)

The zero subnet (or subnet zero) is the one subnet in each classful network
that has all binary 0s in the subnet part of the binary version of the subnet number. In
decimal, the zero subnet happens to be the same number as the classful network number.

so, the answer is B, isn’t it?

@Balaji :
192.168.5.47 is broadcast address of 192.168.1.32/28

can anyone explain me Q10 please…………………….. I did’nt understand the comment given by xallax. please………………………..

Please help me analyze Question number 5. I’m a bit confuse. How come A is the right answer. I was thinking of C as the correct answer.

Q5
On the network 131.1.123.0/27, what is the last IP address that can be assigned to a host?

A. 131.1.123.30
B. 131.1.123.31
C. 131.1.123.32
D. 131.1.123.33
for the mast /27, u will have 32 users/subnet. so ur subnet range will be:131.1.123.0 to
131.1.123.31 (131.1.123.0: network address and 131.1.123.31: broadcast address) so the last usable host will be 13.1.123.30. is it clear……………

@Aonlie

Q5
On the network 131.1.123.0/27, what is the last IP address that can be assigned to a host?

A. 131.1.123.30
B. 131.1.123.31
C. 131.1.123.32
D. 131.1.123.33
for the mast /27, u will have 32 users/subnet. so ur subnet range will be:131.1.123.0 to
131.1.123.31 (131.1.123.0: network address and 131.1.123.31: broadcast address) so the last usable host will be 13.1.123.30. is it clear……………

@uday
to start off, please read 9tut’s subnetting tutorial:
http://www.9tut.com/subnetting-tutorial

then please go over my comment once again
cheers

@uday

Thanks. Crystal clear…..

I am very anxious about the exam. Can somebody help me

Hi 9tut… Hi Guys! Can you please help me… I will take exam this Feb. Please send me latest dump so that I will have an idea for the exam.. rico.blake@ymail.com

Thanks Guys!

The answer was given for Q5 war right, but the explanation was confusing…and not perfect, we dont have to care with the next subnet, 131.1.123.0/27 means, cant be 131.1.123.32…
i guess that uday gave the perfect answer

i m not clear in Q7. will somebody clear me

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

answer is B. i understand how it came but my question is the network here is 192.0.2.0/23 it a class c network so definitly the subnet mask will be 255.255.255.0 which is /24 but here it is /23 the solution here is borrowing the 7 bits the increment becomes 2 so the network

continue…. is 192.0.2.0 and 192.0.4.0 but how the class c network is considered to be as class B network… will some body explain?

How many addresses for hosts will the network 124.12.4.0/22 provide?

A. 510
B. 1022
C. 1024
D. 2048

Answer is 1022 because /22 is 10 bits. How did you get 10bits? Thanks.

Hi Everyone,

I am taking CCNA exam pretty soon. Can anyone of you please email me latest dump.

Cheers
Email: jazzyb60@hotmail.com

hello everyone
i am taking CCNA exam in march can any one send me latest dumps plz.@ dileep_shardha@yahoo.com

Hi there, can someone please send me the latest dumps of the 640-802 exam to laurazxz@yahoo.co.uk. taking exam next week.

Thanks

@Joy:
How many addresses for hosts will the network 124.12.4.0/22 provide?
A. 510
B. 1022
C. 1024
D. 2048
Answer is 1022 because /22 is 10 bits. How did you get 10bits? Thanks.

See… 2the power 10 =1024 – 2(broadcast and network IP) so 1022 numbers of addresses you can assign to the host…
Hope u understand…

all questions and answers are 100% correct!

The network default gateway applying to a host by DHCP is 192.168.5.33/28. Which option is the valid IP address of this host?

A. 192.168.5.55
B. 192.168.5.47
C. 192.168.5.40
D. 192.168.5.32
E. 192.168.5.14

Answer :E

@Anonymous

Why E is the right Answer?
My understanding is /28 means Increment will be 16. So default Gateway IP 192.168.5.33/28 is under 192.168.5.32 subnet. Valid host IP range for this Subnet is 192.168.5.33 to 192.168.5.46. So Answer C is within this range. Q9 is the right answer for C.

E is not correct bcz 192.168.5.14 is a host IP under 192.168.5.0/28 subnet not 192.168.5.32/28.

can someone please explain question 4.I still cant understand the explanation

@Joy
How many addresses for hosts will the network 124.12.4.0/22 provide?
Answer is 1022 because /22 is 10 bits. How did you get 10bits?

You have 32 bits for an IPv4 address, you take 22 for network and subnets (32 – 22 = 10) so the rest of the bits (10) are for hosts addresses

@Ritah
Hope this helps
You have networks 10.0.0.0 up to 10.0.3.0, here the octet that you have to check is the third (where the subnetting is occuring) so this means that at least the subnet mask is /16 (255.255.x.x), for the summarization you have to find where the subnets (0,1,2,3) matches: 0 0 0 0 0 0 0 0 (subnet 0)
0 0 0 0 0 0 0 1 (subnet 1)
0 0 0 0 0 0 1 0 (subnet 2)
0 0 0 0 0 0 1 1 (subnet 3)
as we do not have a subnet where all bits are 1s then the network summarized will be 0 10.0.0.0 (remember we are at the third octect)
and for the subnet mask simply add the number of bits that are all 0s (6 in this case) to the subnet that you had before 16 (16+6= 22)
so the answer is 10.0.0.0 /22

any question?

thanks 9tut!! I passed CCNA exam this morning!

thanks Javier dats true, i actually wid u

i m not clear in Q7. will somebody clear me

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

why”/23″

@vipin
the subnet mask (/23) is for a classless network on this scenario.
notice that the IP is of a class C network, the subnet mask is typical for class A or class B.

192.0.2.0/23
/23 is 255.255.254.0

how big is this subnet? just do this simple math:
255.255.255.255-
255.255.254.000
———————-
000.000.001.255

add 1.255 to the value of the subnet at its beginning and that’s it
so the network is 192.0.2.0 ~ 192.0.3.255

now look at the given options and see which of them are correct.

you must learn subnetting on your own if you truly wish to be successful in your cisco career.
http://www.9tut.com/subnetting-tutorial
there’s also a subnetting quiz that comes with the training packs on http://www.ciscovce.com

cheers