Please HELP. Questions in subnetting

These are questions from MS 291 self paced practice exams. I got them wrong. but....i keep scratching my head....i still can't figure their correct answers....helppppppppppp.....


Question 1)
Your ISP has allocated your organization a Class C IP address range. As your head office has 48 computers, you have split the Class C range using the subnet mask 255.255.255.192. Your organization has several branch offices, each of which has 10 computers. You want to allocate only one address range per branch office. Given that the Class C range has already been divided, if you subnet the remaining addresses from that range in the most efficient manner possible, how many branch offices can you supply with public IP addresses?
A) 3 branch offices
6 branch offices
C) 2 branch offices
D) 12 branch offices
E) 24 branch offices
Question 2)
IP addresses at the local university are allocated in the following manner:
Faculty of Arts: 10.10.10.0/24
Faculty of Engineering: 10.10.12.0/23
Faculty of Science: 10.10.14.0/23
Faculty of Medicine: 10.10.16.0/24
Faculty of Economics: 10.10.17.0/24
Faculty of Law: 10.10.18.0/24
The university administration is responsible for the university’s root DNS server. Each faculty has its own DNS server. Each faculty’s DNS server will directly resolve reverse lookup queries for hosts with IP addresses within the range that it has been allocated. How many zone delegations must be made from the 10.10.in-addr.arpa reverse lookup zone hosted on the university’s root DNS server?
A) 8 delegations
16 delegations
C) 0 delegations
D) 1 delegation
E) 6 delegations
Please explain in details.
Thank you so much for your help.


Question 1)
Your ISP has allocated your organization a Class C IP address range. As your head office has 48 computers, you have split the Class C range using the subnet mask 255.255.255.192. Your organization has several branch offices, each of which has 10 computers. You want to allocate only one address range per branch office. Given that the Class C range has already been divided, if you subnet the remaining addresses from that range in the most efficient manner possible, how many branch offices can you supply with public IP addresses?
A) 3 branch offices

C) 2 branch offices
D) 12 branch offices
E) 24 branch offices
Question 2)
IP addresses at the local university are allocated in the following manner:
Faculty of Arts: 10.10.10.0/24
Faculty of Engineering: 10.10.12.0/23
Faculty of Science: 10.10.14.0/23
Faculty of Medicine: 10.10.16.0/24
Faculty of Economics: 10.10.17.0/24
Faculty of Law: 10.10.18.0/24
The university administration is responsible for the university’s root DNS server. Each faculty has its own DNS server. Each faculty’s DNS server will directly resolve reverse lookup queries for hosts with IP addresses within the range that it has been allocated. How many zone delegations must be made from the 10.10.in-addr.arpa reverse lookup zone hosted on the university’s root DNS server?
A) 8 delegations

C) 0 delegations
D) 1 delegation
E) 6 delegations
Please explain in details.
Thank you so much for your help.
mean people SUCK !!! BACK OFF !!!
The Next Stop is, MCSE 2003 and CCNA.
Bachelors of Technology in 1 More Year.
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The Next Stop is, MCSE 2003 and CCNA.
Bachelors of Technology in 1 More Year.
-Working on CCENT. Thank you my love

Comments
You have Class C which ends in 192. 192 is 11000000. Bit value from left to right is 128, 64, 32, 16, 8, 4, 2, and 1. Since the 2 left-most bits are turned on, that is 128 + 64 = 192. Since you have 2 bits that are turned on, you end up with 2^2 = 4 networks (bits turned on) with 2^6-2 = 62 clients (bits turned off). Since you need to split your Class C into multiple additional subnets, this is known as Virtual Length Subnetting.
So lets be as efficient as possible. Head office has 48 computers and branch offices have 10 machines. For the head office, the minimum amount of bits we need is:
2^2-2 = 2 - Not enough
2^3-2 = 6 - Not enough
2^4-2 = 12 - Enough
Since we are allocated /26 - 255.255.255.192 = 11111111.11111111.11111111.11000000 (26 consistent 1s), we can use /26, /27, and /28. We cannot use /29 because that'd only need 2 bits left (2^3-2=6) which does not meet the 10 client requirement.
So we can use the following subnets:
/26 = 192 = 11111111.11111111.11111111.11000000 = 62 clients - Main office
/27 = 224 = 11111111.11111111.11111111.11100000 = 30 clients at each BO - 2^2 = 8 Branch offices
/28 = 340 = 11111111.11111111.11111111.11110000 = 12 clients at each BO - 2^4 = 16 Branch offices
24 branch offices
I'm pretty sure 0 delegations because any of the addresses created would fall in the 10.10 address space.
Answer: 12 Branch offices
Using the 255.255.255.192 subnet mask at the head office gives you the following options:
Three branch offices using 255.255.255.192, each branch office having 62 available addresses.
Six branch offices using 255.255.255.224, each branch office having 30 available addresses.
Twelve branch offices using 255.255.255.240, each branch office having 14 available addresses.
You cannot use subnet mask 255.255.248 as this will only allow 6 available addresses and each branch office requires 10.
Question 2 Answer:
Answer: 8 delegations
A class-C IP address range is the smallest delegation that can be made for a reverse lookup zone. The faculties of Arts, Medicine, Economics, and Law will require one delegation each. The faculties of Engineering and Science will require two delegations each. Reverse lookup zone can be delegated only in class-A, class-B or class-C lots.
I m trying to figure out why question 1 needs to change to 240. the question says it has the 192 lot. so what i thought was that we will need the 192 to allocate the IP addresses....oh well....i don understand.......whatever.... on question 2 answer is 8. the answer says /24 will make 1 reverse lookup zone, this is understandable. however, why does /23 give you 2 reverse lookup zone thus giving the answer 8 delegations........
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I don't get why you added the two together to get 24. Here's what I'm seeing, I'm new to this so feel free to correct me:
If we make the remaining space:
/26 = 3 additional subnets with 62 clients each.
/27 = 6 additonal subnets with 30 clients each.
/28 = 12 additional subnets with 14 clients each.
The /28 seems to satisfy the requirements here. I agree with the MSPress answer.
That was my thinking exactly Adrian. However, since the subnet mask is /23, wouldn't that put all those 510 in the same logical subnet? If they're in the same subnet, why wouldn't one reverse lookup zone be enough? Can anyone explain this one, I'm lost.
Isp gives you /24, You split this range into 4 networks each with 62 hosts.One network should be assigned to the head office, leaving you three to play with.
So you have 3 /26 networks, each branch requires 10 computers which means you need a block of 16 i.e. /28 for each branch.
So looking at the first /26, this can contain 4 blocks of /28, so 3 networks with /26 can support
12 blocks of /28.Answer D
I was looking at this a little further and tell me if I'm wrong with my thinking here.
Part A - Subnet to /26
2^(32-26) - 2 = 62 Hosts
26-24 = 2
2^2 = 4 Subnets
4 Subnets - 1 reserved for your main office (62 > 48 ) = 3 Subnets for 3 remaining branches.
Part B - Subnet to /27
2^(32-27) - 2 = 30 Hosts
27-24 = 3
2^3 = 8 Subnets
8 Subnets - 2 subnets reserved for your main office (30hosts x 2 > 48 ) = 6 Subnets for remaining branches.
Part C - Subnet /28
2^(32-28 ) - 2 = 14 Hosts
28 - 24 = 4
2^4 = 16 Subnets
16 Subnets - 4 Subnets reserved for your main office (14 Hosts x 4 = 48 ) = 12 Subnets for your remaining branches.
I think the kicker for that question is you need to remember the hosts that the main office requires. Because of the higher requirement, it will demand more subnets. Pretty tricky and interesting question.
Everlife, Notice how the 10.10.13.x and 10.10.15.x subnets are missing???? That is because they are included in the engineering and science address ranges respectively.
Reverse Lookups work via in-addr.arpa zones filled with PTR(Pointer) records. The zone is listed with the subnet address in reverse followed by in-addr.arpa e.g (10.10.12.0 reverse lookup zone would be x.12.10.10.in-addr-arpa)
The PTR record identifies the last octet for the reverse lookup zone...... So if a client was 10.10.12.1 then the PTR record mapping to that client would contain the number 1.
Do you get where I'm coming from??? Only 254 hosts can be listed in the one reverse lookup zone..... Which is why both the engineering and science division will have two reverse lookup zones adding up to 510 clients each because they take up two address ranges not the normal one......
Hope this clarifies things for you
If anyone notices anything wrong in my post please feel free to correct me as I can assure you I am not a expert (Well not yet anyways!
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Awesome thanks APA!
Say they are using a 192.168.0.0 address scheme.
The head office would be 192.168.0.0 /26 (192.168.0.0 - 192.168.0.63)
This leaves you with 3 /26 address blocks left, which you will further subnet into 12 /28 address blocks giving you the following:
Branch 1 -> 192.168.0.0 /28 (192.168.0.64-192.168.0.79)
Branch 2 -> 192.168.0.0 /28 (192.168.0.80-192.168.0.95)
Branch 3 -> 192.168.0.0 /28 (192.168.0.96-192.168.0.111)
Branch 4 -> 192.168.0.0 /28 (192.168.0.112-192.168.0.127)
Branch 5 -> 192.168.0.0 /28 (192.168.0.128-192.168.0.143)
Branch 6 -> 192.168.0.0 /28 (192.168.0.144-192.168.0.159)
Branch 7 -> 192.168.0.0 /28 (192.168.0.160-192.168.0.175)
Branch 8 -> 192.168.0.0 /28 (192.168.0.176-192.168.0.191)
Branch 9 -> 192.168.0.0 /28 (192.168.0.192-192.168.0.207)
Branch 10-> 192.168.0.0 /28 (192.168.0.208-192.168.0.223)
Branch 11-> 192.168.0.0 /28 (192.168.0.224-192.168.0.239)
Branch 12-> 192.168.0.0 /28 (192.168.0.240-192.168.0.255)
Satisfies 48 hosts for the head office and the 10 hosts per branch office.
Am I grasping what you are saying Ed?
Thanks for you help!
No Probs.......
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