New CCNA – Subnetting

Question 2 was in today’s exam.

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Can someone explain No.6?

@Wazif
If you translate the three networks in binary you’ll note that the addresses match only up to the second octet. So you would have:

10101100.00010000.100000000.0 (128 subnet)
10101100.00010000.010000000.0 (64 subnet)
10101100.00010000.001000000.0 (32 subnet)
10101100.00010000.000000000.0 (All bits that not match are 0. Result: summarized subnet is 172.16.0.0/16)

This Q9 is not clear to me.

Thanks Yefry for the answer Q.6
But /16 notation giving me bit confusion can you explain it please

What i can see from last line of your answer is
10101100.00010000.00000000.0 i can’t see 16 bit’s are on. though i can see 16 bits are off.

@Imtiaz
As you can see from the binary addresses, matches until the bit 16. Also, you’ve answered yourself: if you can see 16 bits off, so you should see 16 bits on too (32-16=16). Note: /16 are representing subnet mask bits!

@Fezzy
For each point-point network you have /30 subnet mask, so 4 ip addreess. As you need 113 subnet, so simply do these operations:

113×4= 452 (total ip you need)
2^9= 512 (you need to take 9 borrow bits from subnet mask)
11111111.11111111.11111110.00000000 (total bits 32 – borrow bit 9 = 23 bits)
So you can deduce that you need a subnet mask /23.

Could someone plz explain ques 10? Thanks

Thanks Yefry

Q3. Why B?

@cinemaxik
From the question,the main area of interest is ‘maximum number of sub-nets’ we only need to secure three bits for the three servers.therefore our subnet mask will be all ones with only three bits reserved for three servers,the rest of the bits will be 1s so that we can have the maximum number of sub-nets.This will yield:
111111.11111111.11111000/29 subnet mask or 255.255.255.248.
To find the network address,we need to find the increment,which is obtained by looking at what value corresponds to the least significant bit,which is 8 looking at our
111111.11111111.11111000/29 subnet mask or 255.255.255.248.
Ans.B meets both these conditions.

I left out two bits in the first octet,please add them so it has eight bits in total(not 6 bits)

Cinermaxik the answer is B because it requires 3 host. Host = 2^3 – 2 = 8-2 = 6 by using 3 bits for host network have 5 Bit for the computation of the mask.

Mascara=8+8+8+5=/29 o 255.255.255.248

Could someone explain ques 10? Thanks

@Omar – answer toac q 10
172.16.28.252 with a subnet mask of 255.255.240.0

In such case, you need to see the subnet mask..which is 255.255.240.0, so it means we have to work on 3rd octet. 240 means – a block size of 16 network…. so networks will be in this range :

1st network – 172.16.16.0 2nd network – 172.16.32.0 and so on.. it means this IP – 172.16.28.252 lies in network 172.16.16.0 which is your answer…

hope its clear to you..

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Hello.
why in Q4 answer is 255.255.255.224?

we need to define the subnet mask of the lan which:
to be subnetted for up to eight LANs, with each LAN containing 5 to 26 hosts.
This mean that that we need 3 bits (2^3=8) for subnets and 5 bits (2^5=32>26) for hosts.
Thus we need 3+5=8 bits for parent subnet. (255.255.255.0)
Or i’m wrong?

can someone here explain question 4? im really confuse about it!

i got confuse about it too, coz i choose (E:255.255.255.240) which makes 16 networks and 16 host, but the question states that UP TO EIGHT lans and 5-26host, if i choose (C:255.255.255.224) it makes 8 networks BUT 32 HOST! which also conflicts the 5-26 HOST LIMITS!

@efowl ,, dere is no conflict ,,, Slash 28 , will only give u 16 host and 16 networks , , bt d questn requires 8netwoks and 5-26host , since slash 27 will give u 8networks and 32 honest , dis is more closer , u choose slash 27 (255.255.255.224)

i think for ques 7 it should be /28 and not /27 can anyone pls explain ?

/28 would only leave 4 bits for hosts the would be only up to 14 hosts per subnet, the questions asks for as many as 30, so you need 5 bits for hosts which would allow exactly up to 30 hosts per subnet. you need a /27 to leave 5 host bits

Somebody explain Q-6 using total host and no of bits used. Is it 256+512+512 and nearest round up is 2048?

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please any one can help me in question 5 i don’t understand

Can someone explain why the answer is not D for question 3.

@aNONYMOUS,
Q3 For the 3 servers to be in the same network it means that you need 3 valid host addresses, 252 will give 2 valid addresses and 248 will give 14 addresses, 240 will give 30 valid addresses but will give less number of networks compare to 248 as demanded by the question.

Thanks to every one who explained the complexities of subnetting.
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@Anonymous I fail to download help please ma exam is this week

Q3 the 3 networks 192.168.252.0 / 252.8 / 252.16 only falls in the same network when the mask is 255.255.255.248
Magic number is 8 and the subnet address will be multiples of 8. (this is how is figured the answer. don’t whether it is correct)

I am so grateful to all you guys who are sharing these worthy experience and knowledge with us. I Think I can see real clearly now on subnetting. And many thanks and appreciation the the staff of 9tut…..

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Any one pls explain q5?

Jas,, Expln to q5 is for /29 select the subnet 248 in the last octet. and then look for range where the given ip address falls into. As they asked the sales server to be the last one, it is .30 as .31 will be the broadcast address.

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Q9 was in today’s exam.

Thank you Yefry

Plz anyone can help me on question 8? I can’t understand meaning of answers.

Using any IP address class B subnet according to the following requirement 60, 100, 120.

(a) what is the new subnet mask?
(b) How many usable IP can be found in a subnet and what is the host size per subnets.
(c) write out the first 10 IP blocks.
(d) How many subnets are derivable by this.

Please help me with this questions.

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Q7 i think the answer is D not A , am i right ?

Plz. I didnt get Q9. Help me frndzzzzzzzzzzzzzzzz

Q7

10.188.31.0/23 cannot exist

10.188.31.0 would fall in the net 10.188.30.0/23 as the 2nd /24 in the net

Hello guys kindly assist , I don’t understand this question

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.

HI kijaga from tz

Question clearly states that we need to assign Ip address from third sub-network which is
172.25.64.0/19
so valid hosts in this subnet would be
from 172.25.64.1 to 172.25.95.255

only A, C and D ip addresses fall in this range so these would be the correct answer other are from different subnetworks.

Hope this helps…

Q9 where did the /30 come from?? Not mentioned anywhere on the question. Is that a given for point to point links??