Subnetting Question.

The network is 192.168.150.0/24. If you have to subnet this address space to cover a 64-port Ethernet switch at each of four remote sites, what is the network mask?

A. /28
B. /27
C. /26
D. /25

The correct answer is: C

Shouldn't the correct answer be D? You can only have 62 usable host addresses and not 64. But than if you answer with questions D, than you only get 2 subnets. Does this make any sense?

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Comments

Mishra

The big thing here is to cover all 4 remote sites when you will need 4 networks to do it with. So the answer has to be C. Other ports on the switch could be used for different functions.

But the question is misleading.

vjsousa

I agree. Very misleading.

shednik

Mishra wrote:

The big thing here is to cover all 4 remote sites when you will need 4 networks to do it with. So the answer has to be C. Other ports on the switch could be used for different functions.

But the question is misleading.

Yea I agree it had to be C for this to work if you want to use that subnet over 4 sites

zenzen

yeah this looks like one of those where you have to think about the best solution with what you have been given even though there is no 100% perfect solution. I have problems with these because I'll often rule out something if I can see right off that mathematically it won't really perfectly fit the question

JC13540

Would someone mind explaining to me why the correct answer is C? I guess I need an explanation...because as said in the other replies...math doesn't bring it to 64....

Thanks!

CiscoCerts

JC13540 wrote:

Would someone mind explaining to me why the correct answer is C? I guess I need an explanation...because as said in the other replies...math doesn't bring it to 64....

Thanks!

It doesn't bring it to 64, and it is misleading your right. It is the only answer that comes anywhere close however. I'll try to explain...

The network is 192.168.150.0/24. If you have to subnet this address space to cover a 64-port Ethernet switch at each of four remote sites, what is the network mask?

so with a class C address we have 255 in the last octet we can play with for subnetmasking

the networks would be:

192.168.150.0-63
192.168.150.64-127
192.168.150.128-191
192.168.150.192-255

Four networks equally split up, that's the the best we can do with the choices given.

If D were chosen, you'd only have 2 networks! 0-127, and 128-255

mamono

The network is 192.168.150.0/24. If you have to subnet this address space to cover a 64-port Ethernet switch at each of four remote sites, what is the network mask?

I read it a little different. I can understand why this question is so confusing because it can be misleading. I read it as there needs to be four networks, so subnet accordingly.

192.168.150.0/24

Binary breakdown for the values in the octet = 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1

4 networks = 00000011 binary

Trick question because 00000011 = binary value for 2 + binary value for 1 = 3, but really you have values 0 through 3 thus four digits for four networks. Zero is also a value. 0 1 2 3

Subnet mask for a Class C network is normally 255.255.255.0

In binary, that is 11111111.11111111.1111111.00000000 ...

To accommodate the subnetting of 4 networks, the binary values are added to the left side of the fourth octet, thus resulting in the following binary values.

11111111.11111111.1111111.11000000

If we count the 1's in the binary sequence, it comes out to /24 + 2 = /26 ...

ccnpninja

I definitely answered C.
It's simple. We need 4 subnets (4 remote sites). In order to have them, the least number of bits we can borrow from the subnet part is 2. The subnet mask becomes: 255.255.255.192 or /26 in CIDR notation. The subnet numbers are:
192.168.150.0
192.168.150.64
192.168.150.128
192.168.150.192