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ldbeane
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« on: August 10, 2005, 03:51:05 PM »

I have the mspress book for 70-216. In the subnetting chapter it has a question stated:
You are the new administrator of a 2000 node network.  There is only one router on the entire network, which provides all the computers with Internet access.  The company's ISP has assigned the following eight network addresses to them:
10.24.32.0/24
10.24.33.0/24
10.24.34.0/24
10.30.35.0/24
10.30.36.0/24
10.30.37.0/24
10.30.38.0/24
10.30.39.0/24

What subnet mask could you use to minimize the complexity of the routing tables while maintaining the existing Internet connectivity?
a.) 255.255.252.0
b.) 255.255.255.252
c.) 255.255.255.248
d.) 255.255.248.0

The book says the answer is d.

Could someone please explain this.  I don't see how this is possible with the second byte in the address being 24 and 30.

Thanks.
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Enforcer
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« Reply #1 on: September 05, 2006, 01:14:04 PM »

My guess that there is a misprint and that all the second numbers should be a '30'. check the mspress website and see if there have been any corections that cover this question.
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luisjo
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« Reply #2 on: January 29, 2007, 03:04:16 AM »

There most be 1,900 users on these subnet, at least 11 digits must be use for these 1,900 clients.  since 2**11 is =2,048 and 2**10 = 1,024 these leaves 21 (32-11) bits for the subnet mask.

Subnet mask in binary 11111111,11111111,11111000,00000000

subnet mask in decima
255.255.248.0

:cool:
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« Reply #3 on: February 13, 2007, 06:12:03 PM »

Thats perfect luisjo. This is exactly how to solve the problem because of finding the subnet that will allow for the requested amount of users of 1900. This automatically rules out answers B and C so  now you need to determine between A and D which luisjo has so aptly illustrated. Smiley
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