| Author |
IT'S BACK!! 70-218 QoD Wednesday 6/25
|
|
| mrfixit 2003-06-25, 7:49 am |
| Been kind of busy lately with the new job. Hope to keep these going again. Todays question is:
As the administrator of a small legal firm, you want to maintain certain desktop configuration standards in the organization. Which tool can you use to maintain control?
A)Group policy
B)Remote access control
C)System management
D)Intellimirror
E)Roaming profiles
See you tomorrow with the answer. Good luck!  (It's an "easy" one.) | |
| isles1 2003-06-25, 7:51 am |
| Glad to have you back...
I will take a shot:
A, E | |
| jeff50ho 2003-06-25, 8:32 am |
| Since the question was placed as which tool with a singular tense then i'll go with
A. | |
| B4yaman3 2003-06-25, 9:17 am |
| It asked which tool.
The answer is A | |
| ghaouf 2003-06-25, 9:41 am |
| a | |
| aawmorris 2003-06-25, 11:49 am |
| A. It would be E also, if everyone shared each other's PC. | |
| KiwiPete 2003-06-25, 2:42 pm |
| Yep. Gotta be A
 | |
|
| This QoD was mailed to me, but I think it's wrong. What say you guys?
You have 300 clients that are preparing to deploy onto your network. You want to set up subnets for your new clients so that each subnet will only have about 25 clients each. You have the IP address range of 172.17.10.0 - 172.17.21.0 to work with. What subnet mask will you need to use to accomplish your goal of limiting the subnets to about 25 clients each?
A) 255.255.240.0
B) 255.255.255.240
C) 255.255.255.224
D) 255.255.255.192
E) 255.255.224.0
F) 255.255.192.0
G) 255.224.0.0
H) 255.192.0.0
I) 255.240.0.0
Correct answer: B
Explanation:
From the Managing a Windows 2000 Network Environment Training Kit (page 450), a subnet mask of 255.255.255.240 will provide 31 usable IP addresses. You could then use the following IP address ranges: 172.17.10.0/28 - 172.17.21.0/28, which will provide 12 subnets. If you put 25 clients on each subnet, you would meet the goal of the scenario.
------------------------
I say the correct answer is C, because if the subnet mask is 255.255.255.240, then using 2^n-2 gives you 14 hosts. However, 224, gives you 30.
Am I out in left field on this or was their a mistake on QoD? | |
| mrfixit 2003-06-25, 9:13 pm |
| quote:
Originally posted by bfh
This QoD was mailed to me, but I think it's wrong. What say you guys?
You have 300 clients that are preparing to deploy onto your network. You want to set up subnets for your new clients so that each subnet will only have about 25 clients each. You have the IP address range of 172.17.10.0 - 172.17.21.0 to work with. What subnet mask will you need to use to accomplish your goal of limiting the subnets to about 25 clients each?
First of all, you should have started a new thread for this question, as this is the 218 Q of the Day thread. Secondly, if M$ says it is the right answer, then it must be. But of course, when you consider the source... I have to agree with C as the answer. With a .240 subnet, your first host address available would be 172.17.10.1 and the last 172.17.10.14, giving you 14 host. But with a .224 subnet, it would give you .1 thru .30. Where they get 31 "usable" addresses is beyond me, as IP address 172.17.10.0 is unusable as a host address(network address), and .31 is a broadcast address.
Of course, we all know M$ is NEVER wrong! | |
|
| Sorry, MrFixIt. I didn't mean to hijack your thread. I'm a relatively new user here and saw the title of this thread (QoD 6/25)and since I get QoD from here, I thought it belonged here. No harm meant!
Having said that, Thanks for validating my thoughts on the question. I was pulling my hair out as I knew B couldn't possibly be right. | |
| ghaouf 2003-06-25, 9:31 pm |
| mrfixit could you explain your answer more, how did you calculate that there were 31 ip addresses. | |
| mrfixit 2003-06-26, 12:10 am |
| quote:
Originally posted by ghaouf
mrfixit could you explain your answer more, how did you calculate that there were 31 ip addresses.
quote:
Where they get 31 "usable" addresses is beyond me
quote:
a subnet mask of 255.255.255.240 will provide 31 usable IP addresses.
You actually get 32 (32=2^5) addresses out of the .224 subnet, two of which are used for the network and broadcast nodes. (.0 and .31) So, you have 30 host addresses. (172.17.10.0 and 172.17.10.31 are not usuable as host addresses) Make sense? In the scenario, you only need 25 host per subnet, giving you 5 to spare. (FYI: this is a Class B Subnet.)
Better? | |
| mrfixit 2003-06-26, 11:18 am |
| Time to give you all the answer... I guess.
As the administrator of a small legal firm, you want to maintain certain desktop configuration standards in the organization. Which tool can you use to maintain control?
A)Group policy
That's right! Group Policy. Group policies from Active Directory can help maintain a standard desktop and reduce administration cost. And we all want to reduce administration. (We Sys Admins work hard enough as it is! ) I'll try to post another later today. | |
| ringuete 2003-06-26, 11:43 pm |
| you said "a subnet mask of 255.255.255.240 will provide 31 usable IP addresses" ??
with a subnet of 240, you only have 4 bits for the hosts..(2^4 = 16)..
where did you get 30?.
regards... | |
| Tarzanboy 2003-06-27, 12:03 am |
| You may want to reread his posts again. He stated to use .224 which would be (2^5), which in turn yield 30 useable IPs, rather than .240 (2^4) which yields 14. Because of this, the answer would be C.
Cheers,
TB | |
| adam salam 2003-06-27, 1:19 am |
| quote:
Originally posted by ghaouf
mrfixit could you explain your answer more, how did you calculate that there were 31 ip addresses.
normally the syntax is:
2^n-2=number of ip address
in this example 224 in the last octate gave us 5 bits for hosts (nodes,
2^5= 32-2 = 30 hosts(nodes)
the 2 here one ip address is for network address
the other is for broadcast
for example: one subnet will be 172.16.10.0
first host 172.16.10.1
-
-
-
last host 172.16.10.30
broadcast 172.16.10.31
------------------
the next subnet will be 172.16.10.32
1 172.16.10.0
172.16.10.1 - 172.16.10.30
172.16.10.31
2 172.16.10.32
172.16.10.33 - 172.16.10.62
172.16.10.63 | |
| ringuete 2003-06-27, 4:35 am |
| it was late at night...
and my english is not so well in that hours..
gracias!! | |
| CmptrDude1 2003-06-29, 9:23 pm |
| I've found several of those QODs to be incorrect, including that subnetting one the other day. 255.255.255.224 was undoubtedly the correct answer because it allowed for 5 bits to be assigned to the host ID (or 31 IPs). |
|
|
|