CCNA – Subnetting Questions 2
Here you will find answers to Subnetting Questions – Part 2
Question 1
Refer to the exhibit. Which VLSM mask will allow for the appropriate number of host addresses for Network A?
A. /25
B. /26
C. /27
D. /28
Answer: A
Explanation
We need 66 hosts < 128 = 27 -> We need 7 bits 0 -> The subnet mask should be 1111 1111.1111 1111.1111 1111.1000 0000 -> /25
Question 2
Refer to the exhibit. Which subnet mask will place all hosts on Network B in the same subnet with the least amount of wasted addresses?
A. 255.255.255.0
B. 255.255.254.0
C. 255.255.252.0
D. 255.255.248.0
Answer: B
Explanation
310 hosts < 512 = 29 -> We need a subnet mask of 9 bits 0 -> 1111 1111.1111 1111.1111 1110.0000 0000 -> 255.255.254.0
Question 3
Refer to the exhibit. Which mask is correct to use for the WAN link between the routers that will provide connectivity while wasting the least amount of addresses?
A. /23
B. /24
C. /25
D. /30
Answer: D
Explanation
For WAN link we only need 2 usable host addresses for 2 interfaces on the routers. The subnet mask of /30 gives us 22 – 2 = 2 usable host addresses. Also remember that “/30″ is famous for point-to-point connection because it wastes the least amount of addresses.
Question 4
Refer to the exhibit. What is the most appropriate summarization for these routes?
A. 10.0.0.0/21
B. 10.0.0.0/22
C. 10.0.0.0/23
D. 10.0.0.0/24
Answer: B
Explanation
We need to summarize 4 subnets so we have to move left 2 bits (22 = 4). In this question we can guess the initial subnet mask is /24 because 10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0 belong to different networks. So “/24″ moves left 2 bits -> /22.
Question 5
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
Answer: A
Explanation
Increment: 32
Network address: 131.1.123.0 & 131.1.123.32
Broadcast address: 131.1.123.31
Both 131.1.123.30 & 131.1.123.33 can be assigned to host but the question asks about the “last IP address” so A is the correct answer.
Question 6
The ip subnet zero command is not configured on a router. What would be the IP address of Ethernet0/0 using the first available address from the sixth subnet of the network 192.168.8.0/29?
A. 192.168.8.25
B. 192.168.8.41
C. 192.168.8.49
D. 192.168.8.113
Answer: C
Explanation
The “ip subnet zero” is not configured so the first subnet will start at 192.168.8.8 (ignoring 192.168.8.0).
Increment: 8
1st subnet: 192.168.8.8
2nd subnet: 192.168.8.16
3rd subnet: 192.168.8.24
4th subnet: 192.168.8.32
5th subnet: 192.168.8.40
6th subnet: 192.168.8.48 -> The first usable IP address of 6th subnet is 192.168.8.49
Question 7
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: B
Explanation
Increment: 2
Network address: 192.0.2.0, 192.0.4.0
Broadcast address: 192.0.3.255
-> 192.0.2.255 is not a broadcast address, it is an usable IP address.
Question 8
How many addresses for hosts will the network 124.12.4.0/22 provide?
A. 510
B. 1022
C. 1024
D. 2048
Answer: B
Explanation
/22 gives us 10 bits 0 -> 210 – 2 = 1022. Notice that the formula to calculate the number of host is: 2k – 2.
Question 9
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
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
Hi there, can someone please send me the latest dumps of the 640-802 exam to (bharat.dande@usa.com). taking exam next week.
bharat.dande@usa.com
Hi can anyone explain Q10 please?
Hello,
Does anyone have the latest brain dumps they can send? I am taking the exam 30OCT2012.
Thank you
shire1978@gmail.com
hello ,
If any know about 9th question solution ,please post it
baddigam.vijaykumar@gmail.com
@Vijay Kumar
host add is 192.168.5.33/28. therefore subnet range is 192.168.5.32-47.
32 is the network id (cannot be assigned to host) and 47 is broadcast (cannot be assigned to host) only answer C (192.168.5.40) is the valid host ID. Rest are from other subnets.
Hi,
I worked out first question, for me solution comes out as /27
Please correct me if i am wrong.
Tnx in advance.
Lakshman
Hi,
Apologies for my first comment. Solution was correct /25.
Lakshman
answer to q10 not clear..any idea guys.
thanks.
@vola
Given the subnet mask, you will have /23 and block size of 2.
/23 belongs to 3rd octet. Paying attention to the 3rd octet.. Network addresses are as follows with their valid host addresses.
.0 (0.1 – 1.254)
.2 2.1 – 3.254
.4. 4.1 – 5.254
.
.
.
.36 36.1 – 37.254
First and last cannot be used since those are network and broadcast address.
So only valid hosts are B and D
Can someone please send me the latest dumps of the 640-802 exam to (m.osman@hotmail.co.uk). taking exam next week. Thanks.
Hi can someone please send me the latest dumps (cmaneklal@mweb.co.za) Thanks
I going to take the test next month. Can someone please send me the latest dump.
email is abliej@gmail.com
Please explain to me why is it that B and D are the correct answers for Question # 10.
thanks
@Juvia
Given the subnet mask, you will have /23 and block size of 2.
/23 belongs to 3rd octet. Paying attention to the 3rd octet.. Network addresses are as follows with their valid host addresses.
.0 (0.1 – 1.254)
.2 2.1 – 3.254
.4. 4.1 – 5.254
.
.
.
.36 36.1 – 37.254
First and last cannot be used since those are network and broadcast address.
So only valid hosts are B and D
Hope that was the answer for your question
Thnxs
Any question email me yacoub9308@gmail.com
question 7
255.255.254.0 which is 11111111.11111111.11111110.00000000
network bit =23 host bit = 9
no of hosts (2^9)=512
if you make the subnet table
subnet or network ips are ends with 0.0 , 2.0, 4.0, 6.0,….
broadcast ips ends with 1.255, 3.255, 5.255, 7.255….
so A C E are can’t be hosts
only B D are the hosts
question 9
/28 block sise 16 2^4
so increment in order of 16. Default gateway given is the first valid ip of the network 192.168.5.32.
Can somebody please explain the question number 10. I reckon the answer provided is more or less incorrect.
The answer for question 10 must be C and D
San the answers are B, D:
You have to remember this:
128, 64,32, 16, 8, 4, 2, 1
11111111.11111111.11111110.00000000
11111110 = there are 7 One’s which stop at the 2 bit. See Below:
128, 64,32, 16, 8, 4, 2, 1 series
1 1 1 1 1 1 1 0
so your multiples are:
0 – 2 = 1 – 2.254
2 – 4 = 2.1 – 3.254
4 – 6 = 4.1 – 5.254
etc.
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
question 6 !!
how is the increment 8 occurs? explain. i think its increment 6… !!
Hi,Iam planning to give exam in next week,can any one share the latest dump.My Mail Id is naikare.vijay@gmail.com.Thanks in advance.
TY 9tut.
Today I have passed the CCNA. (860/825)
50 questions 3 labs (VTP, EIGRP, ACL). 35 from 9tut.
Also thanks a lot Brar and Sekhar (still valid from examcollection)
Ty again 9tut
from here Q 10
any new dumps avalible?
b2468@hotmail.com
heyy all.i’m planning for giving ccna exam.i’m very confused please help me and fwd me stuff nd latest dumps at rizwan_shaikh@live.com
In the below question, why isn’t 0-7 the first subnet?
0-7
8-15
16-23
24-31
32-39
40-47 <<<Is what I came up with. Being 192.168.8.41
The ip subnet zero command is not configured on a router. What would be the IP address of Ethernet0/0 using the first available address from the sixth subnet of the network 192.168.8.0/29?
A. 192.168.8.25
B. 192.168.8.41
C. 192.168.8.49
D. 192.168.8.113
I answered my own question. Because the subzero command is not configured. Duh
How many subnets and hosts per subnet can you get from the network 172.28.0.0 255.255.240.0?
Answer: 16 subnets and 4094 hosts
EXplain how
here is all vbout reverse engineering,this is a classB address,convert 255.255.240.0=11111111.1111111.11110000.00000000 if we save the host then we have 12 zeros 2 pow 12=4096 minus 2=4094 and the networ portion in third octed is 4 ones 2 power 4= 16
Thank u renolph
all question are well treated no doubts please could some one help me with an exam collection software that can enable me take 30 to 40 exam question at a time mine is just 5 coming for test thanks (hardcorepadiman@gmail.com)
Hi ,
in your 6th question you miss the subnet from 192.168.8.0 – 192.168.8.7 which is the 1st subnet. so answer will be ” B ”
please explain if i’m wrong.
Balaji,
No “subnet Zero” means ignore first subnet
.0, .8, .16, .24, .32, .40, (.48,) .56, .64
1, 2, 3, 4, 5, 6
I am confuse wid the 10 question. need Help ?
How do i get to open vce files.sme1 help
Hello friends,
Pls explain quesion 10. How can two addresses can be assigned to a host with a subnet mask of 255.255.254.0.
anita94587@yahoo.com
Thanks,
Anita
Question 10
Subnet mask is 255.255.254.0..It meanz that we have 9 host bits => 512 hosts per subnet.
So
A) 113.10.4.0
network ip = 113.10.0.0 , 113.10.2.0 , 113.10.4.0 and so on
Broadcast ip = 113.10.1.255 , 113.10.3.255 , 113.10.5.255 and so on
So Option A is Wrong bcz it is network ip
B) 186.54.3.0
n/w ip = 186.54.0.0 , 186.54.2.0 , 186.54.4.0 and so on
B/c ip = 186.54.1.255 , 186.54.3.255 , 186.54.5.255 and so on
So Option B is Correct bcz it is neither n/w ip nor b/c ip
C) 175.33.3.255
n/w ip = 175.33.0.0 , 175.33.2.0 , 175.33.4.0 and so on
b/c ip = 175.33.1.255 , 175.33.3.255 , 175.33.5.255 and so on
So C is wrong
D) 26.35.2.255
n/w ip = 26.35.0.0 , 26.35.2.0 , 26.35.4.0 and so on
b/c ip = 26.35.1.255 , 26.35.3.255 , 26.35.5.255 and so on
So D is correct
Similarly for E which is n/w ip and wrong.
Appreciate your help Muhammad. Crystal clear now
in Q9,
y is not the options a & c correct ????
they also come under 1st,3rd & 4th subnet of /28 n/w
I am taking the exam on March 1,2013
Could anybody please send me the latest dumps on gauriwalinjkar@gmail.com
Could anybody explain question 9 please.
The network default address is 192.168.5.33/28, so hence 5.0,5.8,5.16,5.24,5.32,5.40,5.48,5.56 will be the address assigned to subnets and 5.7, 5.15, 5.23, 5.31, 5.39, 5.47, 5.55 will be their coreesponding broadcast address.
So a valid IP address will be in the range 192.168.5.32-5.39, which is not given in the options.
Could anyone explain if I am missing any point here. How is the answer C ?
@Gauri /28 mean increment of 16 not 8 valid networks will be 5.16,5.32,5.48…so on
accordingly the valid ip address are 5.33-5.46 (5.33 already in use by DHCP server and 5.47 is the broadcast address of this range) so 5.40 is correct choice..
question 9 makes no sense. it needs to be re-worded
please send new dumps…juminda98@yahoo.com
@Big Dr 74 where does q9 not make sense my friend? Show us where it is not making so that others can help you because that question is simple and straight forward, very very simple i can tell you.
kindly give me some web site for sub netting practice
Passed CCNA, question 6 from here.
que 6 was in the exam today
Hi All,
Can anyone help me to do a Subnet a 100 Subnets?
IP address is 150.X.x.x can we subnet into 100 networks? Thanks in advance…
@Mohammad
U need 7 bits i.e. 2^7 = 128 Sub-Nets therefore Mask will be 255.254.0.0.