CCNA – Subnetting Questions 3

Thanks guys 9tut can make us reason, hence widening our minds.
olesimbe@yahoo.com

Guys ,,

please explain Q 5 ?? pleeease

mwaaaaah for 9tut

Can someone explain question no.4?
why server A has wrong gw?Thanks!

Can Someone explain questiom 4?
Rian_55 also confused.Can someone please tell why that is the answer.
We will appreciate your explanatiom.Many Thanks in advance.

@rian55 @lindsayss
the server has this IP:
131.1.123.24
it is on this network:
131.1.123.0 ~ 131.1.123.31

its default gateway (marked as GW in the exhibit) is 131.1.123.33

the server and its default gateway are not part of the same subnet. that is the problem

Which of the following IP addresses can be assigned to the host devices? (Choose two)
A. 205.7.8.32/27
B. 191.168.10.2/23
C. 127.0.0.1
D. 224.0.0.10
E. 203.123.45.47/28
F. 10.10.0.0/13
Answer: B F

Is 191.168.10.2/23 a Public Class B IP Address?

Hi can anyone help me on Q5, i am getting the increment of 16 subnetmask 255.255.255.230

@Anas @srinivasan

Q5 Given the address 192.168.20.19/28, which of the following are valid host addresses on this subnet? (Choose two)

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

1. First you need to find the subnet ranges
Given the /28 subnet mask we know that the network increment is 16 so the subnet ranges are as follows:

0 – 15 .0 being the subnet and .15 being the broadcast
16 – 31 .16 being the subnet and .31 being the broadcast
32 – 47 .32 being the subnet and .47 being the broadcast
.
.
and so on

2. Once this is done you can now know the valid ip addresses within this subnet ranges

NOTE: The address given 192.168.20.19/28 IS NOT A SUBNET as it says in the question BUT IT IS AN IP ADDRESS WITHING the 192.168.20.16/28 subnet. That is a mistake.

Thinking this was a mistake and they were actually meaning to find valid ip addresses withing 192.168.20.16/28 subnet we can determine the following:

Subnet Range 16 – 31

A. 192.168.20.29 is correct b/c is within that subnet range
B. 192.168.20.16 is incorrect b/c that’s the subnet address
C. 192.168.20.17 is correct b/c is within that subnet range
D. 192.168.20.31 is incorrect b/s that’s the broadcast address
E. 192.168.20.0 is incorrect b/c that’s the network address of a different subnet

Answers A & C are CORRECT!!!

@Danilin
Thanks for ur explanation

Q7: is not the 192.1.144.159 is the broadcast address of subnet .144?..since .144 subnet, with increment of 16 =144+16=160 (next subnet).?.hence, broadcast 160-1=159???

Best explanation I can give is the following formula:

1- What is increment value? Based on /20 its 16 as following:
Third octet opening: 128 64 32 16 8 4 2 1
How many bits of 3rd octet shifted to network side = 4 bits
What is incremental value = 16 (rightmost bit in this case 4th bit)
Subnet mask? /20 indicates 255.255.240.0 ( sum of all 4 borrowed bit 128+64+32+16=240)

Now let game begin!

2- What is the summarized address: 192.1.144.0/20

3- What is the incremental value : 16 (3rd octet)

4- Write the next range of 192.1.144.0/20

1st range: 192.1.144.0 —————————————– 192.1.159.255
2nd range: 192.1.160.0 —————————————————— 192.1.175.255
3rd range: 192.1.176.0 ————————————————————————

5- Whatever IP addresses given in options if fell within 144.1 ———————– 159.254 are valid addresses that R1 will forward to R2. OPTION A and D are the only valid answers.

if there is still any difficulty grasping the matter ask freely at syedkashifshahab@hotmail.co.uk

@pk

192.1.144.159 is not the broadcast of subnet 192.1.144.0/20. I think that you forgot that /20 means that the changes are done on the 3rd octet not on the 4th octet.

8bits + 8bits + 4bits = /20
11111111 . 11111111 . 1111 | 0000 . 00000000
1st 2nd 3rd 4th

You only work on the 4th octet with masks like /25, /26, /27 or /28 which also has an increment of 16

Yes the increment is 16 but remember that we are working on the 3rd octet

1st range starts with 192.1.144.0
2nd range will start with 192.1.160.0

Try to see it this way….ask yourself what is that last address before we get to 192.1.160.0 ???

192.1.160.0
-1
—————–
192.1.159.255

Remember that by definition the broadcast is the last usable ip address in the range.

Can Anyone give the answer for the following question?
Q.If HostA pings to S0/0 on Router3,what will be the TTL value for that ping before it enter Router3?
Diagram is like:
HostA —–> connected to Router1[s0/0]——-[s0/1]connected to couter2[s0/0]——-[s0/0]Connected to Router3 —– HostB

Answer:253

Can anyone explain how??

@Samreen

By default from host to other n/w TTL value is 255 after 2 hop TTL remaining issssssssssss………..gotta

@Samreen
By default from host to other n/w(cisco) TTL value is 256 after 3 hop TTL remaining 256-3

Q12 with a little twist was in exam.

AOA,

can someone please guide me in Question # 5 given on top ?

Question # 5
efer to the exhibit. The user at Workstation B reports that Server A cannot be reached. What is preventing Workstation B from reaching Server A?

BraveAli, here`s the explanation: ip for Server A is 131.1.123.24. The mask is /27. So the ip range for the subnet is 131.1.123.0-131.1.123.31. But the default gateway (131.123.33) is NOT in this subnet, its outside the subnet.

@Samreen

TTL max value is 255 and not 256.It can’t be 256 since it’s 8-bit value
And answer is 253 because each time it passes a router it is decreased by one

So starts with 255 from host A, becomes 254 after passing router1, and after router 2 and before entering router3 becomes 253

@xallax and 9tut…

can you explain q4.

Refer to the exhibit. The user at Workstation B reports that Server A cannot be reached. What is preventing Workstation B from reaching Server A?

… i think the subnet of workstation is the same with the server since the subnet use is /27.

hoping for your response. thanks in advance.

kindly disregard my question.. Thanks..

Hello guys plz help me on this question::A class C is subnetted & the new subnet mask is 255.255.255.224 (a)determine the address of the highest subnet (b) determine the network of the lowest subnet?

@mcintosh
x.x.x.0 ~ x.x.x.31
x.x.x.224 ~ x.x.x.255

anybody explain me Q3. thanks alot

@ronin,thanks buddy,but can you explain to me how u reached to that answer;i can see the increment is 32 derived frm the last significant bit of the sub mask..start from from there,thank you

sorry,@@XALLAX…hanks buddy,but can you explain to me how u reached to that answer;i can see the increment is 32 derived frm the last significant bit of the sub mask..start from from there,thank you

SORRY,@XALLAX,Thanks buddy,but can you explain to me how u reached to that answer;i can see the increment is 32 derived frm the last significant bit of the sub mask..start from from there,thank you

@mcintosh
ok…
class C network is subnetted. the first 3 bytes will remain the same
the subnet mask in use is 255.255.255.224 (/27)
it asks for the first and last subnet. it says nothing about subnet-zero so i assume it is usable (as it is by default)

example network: 200.150.100.0
subnet mask: 255.255.255.224
increment is 32

subnets:
200.150.100.0 ~ 200.150.100.31 — this is the first subnet
200.150.100.32 ~ .63
.64 ~ .95
.96 ~.127
.128 ~ .159
.160 ~ .191
.192 ~ .223
.224 ~ .255 — this is the last subnet

and… that’s it :)

faster way to get the last subnet: 256 – increment = 256 – 32 = 224
the last subnet starts with 224, ends at 255.

@xallax and to all
pease explain to me how to get the /22 in question no.9
and i need explanation in question no.12…

please help..thank you

@me
that questions means:
Which smaller subnets will be included in the big subnet of 172.31.80.0/20?
172.31.80.0/20 is 172.31.80.0 ~ 172.31.95.255
(subnet mask of 255.255.240.0, increment 16, that’s how i got the range above)

A. 172.31.17.4/30
not in the range of 172.31.80.0 ~ 172.31.95.255

B. 172.31.51.16 /30
not in the range of 172.31.80.0 ~ 172.31.95.255

C. 172.31.64.0/18
not in the range of 172.31.80.0 ~ 172.31.95.255

D. 172.31.80.0/22
this is 172.31.80.0 ~ 172.31.83.255 and it is a part of 172.31.80.0 ~ 172.31.95.255

E. 172.31.92.0/22
this is 172.31.92.0 ~ 172.31.95.255. it is the last small subnet of 172.31.80.0 ~ 172.31.95.255

F. 172.31.192.0/18
not in the range of 172.31.80.0 ~ 172.31.95.255

please go over 9tut’s subnetting tutorial. if you do not master this topic then you stand no chance on the exam.
http://www.9tut.com/subnetting-tutorial

@xallax

thanks for your time to explained it to me.i really appreciate it. i know how to get the range :) .. My questioned is, how to do reverse engineering in the summarized address of 172.31.80.0/20, how to get the subnet mask of /22.. (from the answers D. 172.31.80.0/22, E. 172.31.92.0/22)..

Thank you in advance :)

@me
what is a summary? is a network with a classless mask that comprises several networks
let’s say you had these:

172.31.80.0/22
172.31.84.0/22
172.31.88.0/23
172.31.90.0/23
172.31.92.0/22

if you were to wrap all those subnets up in just one big subnet which one would it be?
starts at .80.0, ends at .95.255… how convenient, it’s a multiple of 16.0 :)
so… guess what subnet mask has this subnet:
172.31.80.0 ~ 172.31.95.255

the subnet mask is 255.255.240.0, which converted to CIDR is 8+8+4+0 = 20

hope this helps you better understand summaries :)

thanks 9tut!=D

Question 4

server A’s subnet will be 131.1.123.24~55 it contain GW.

Why answear is D?

Q1 server’s gateway should be in the range from 1-30

ooops == I meant Q4 on the previous comment

HI everbody, wish the best to all, and thanks to 9tut and everyone who contribute to this gorgeous site…

concerning Q.2, i have a question: Can a Router be configured with a default-gateway (as in the switch case: ip default-gateway) or he means that it is the route of the last resort which is usually connected to the internet through the internet service provider?
hope to be clear..

thanks in advance and best regards….

Q4,and Q11 were in my exam today

Hi all, I am taking CCNA 640-802 exam first time on 30/05/2012. Could anyone please send me latest dumps which are valid for UK? My e-mail address is puneet_gill84@yahoo.co.uk. Many thanks.

is question no 7 is correct as address is from class C and we can not touch the 3rd octet. please tell me

Can someone explain QN 11 pl.

Hi Can anybody send me latest dumps valid for usa to my mail id, basavarajpardi@gmail.com.
thank you

Question 7

Hi, can someone please explain why is the answer for the question 7 on this page is AD? Thank you!

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

I PASSED CCNA EXAM TODAY THANKS TO ALL MIGHTY ALLAH
960/1000

Question 7
192.1.144.0/20 = 192.1.144.0 255.255.240.0; in the third octec the magic number is 256-240 = 16 therefore valid IP host addresses are
192.1.144.1 – 192.1.159.254 so answers A and D are within this range.

Hope this helps
Cheers.

Muhammad
Congratulations!!, don’t forget to praise yourself, you did the hard work with the help of your faith, Allah and God helps those who help themselves!
Cheers.

Q11

A. 205.7.8.32/27 – Net address
B. 191.168.10.2/23 – OK
C. 127.0.0.1 – Loopback address
D. 224.0.0.10 – Multicast address
E. 203.123.45.47/28 – Broadcast address
F. 10.10.0.0/13 – OK

Dear CCNA Passers!!!
I have a question on question # 9!!!

Which two subnetworks would be included in the summarized address of 172.31.80.0/20? (Choose two)

A. 172.31.17.4/30
B. 172.31.51.16 /30
C. 172.31.64.0/18
D. 172.31.80.0/22
E. 172.31.92.0/22
F. 172.31.192.0/18

the given summarized is 172.31.80.0/20, so its 3rd octet, 16 increment
increment 16 so –> 172.31.96.255

so the range for valid ip addreses is 172.31.80.1 up to 172.31.95.254 am i right?
answer letter E is correct but how about letter D?

Please give me a clear explaination why letter D is acceptable to this question. Thanks to all!!! I need to hear from your eexplaination soon..

Hi Gilbert,
(172.31.80.0/22) & (172.31.92.0/22) both of them are Sub Network ip address with 4 increment.
If u summarized both of them then u will get the summarized IP address of 172.31.80.0/20. don’t calculate 16 increment.
I m not CCNA passers !!!