I don't understand this question..

nzchuen

3 addresses are shown in binary form below:

A. 01100100.00001010.11101011.00100111
B. 10101100.00010010.10011110.00001111
C. 11000000.10100111.10110010.01000101

Regarding these three binary addresses in the above exhibit; which statements below are correct? (Select three)

A. Address C is a public Class C address.
B. Address C is a private Class C address.
C. Address B is a public Class B address.
D. Address A is a public Class A address.
E. Address B is a private Class B address.
F. Address A is a private Class A address.

Can pls explain to me what is public class and private class?

thanks..

Danman32

private address ranges are:

192.168.x.x/24 class C private address range
172.16.x.x/12 (172.16.x.x - 172.31.x.x) class B private address range
10.x.x.x/8 class A private address range
You can't use these addresses as source or destination addresses directly on the internet. The backbone routers on the internet are supposed to filter these addresses. You have to translate them first using NAT or PAT to valid, registered public internet addresses.

With exception to class D and above, the rest are public addresses.

So the task at hand is to convert the binary octets into decimal, determine the class of the address and if it falls into the address ranges designated as private addresses and should not be thought of as: Address C is a private class, C address.
You could get away with only converting the first two octets, unless the first octet is 10, in which you can stop right there.

Looking over your post again, I think you got confused about the modifier for the term class. Class isn't being applied to public or private, but rather class A, B or C address. Try adding a comma between the word private or public and class, that might make the possible answers clearer.

In other words answer A is can be more clearly written: Address C is a private, class C address and should not be thought of as: Address C is a private class, C address.

PawNtheSandman

nzchuen wrote:

3 addresses are shown in binary form below:

A. 01100100.00001010.11101011.00100111
B. 10101100.00010010.10011110.00001111
C. 11000000.10100111.10110010.01000101

Regarding these three binary addresses in the above exhibit; which statements below are correct? (Select three)

A. Address C is a public Class C address.
B. Address C is a private Class C address.
C. Address B is a public Class B address.
D. Address A is a public Class A address.
E. Address B is a private Class B address.
F. Address A is a private Class A address.

Can pls explain to me what is public class and private class?

thanks..

A. 01100100.00001010.11101011.00100111
is 64.10.235.39

B. 10101100.00010010.10011110.00001111
is 172.18.158.15

C. 11000000.10100111.10110010.01000101
is 192.167.178.69


Using what Danman32 stated, none of the above addresses are private. Therefore you can eliminate answers B,E,F.

And since the question said select 3, and you eliminated 3 of your 6 answers, you already have your answers as A,C,D.

BubbaJ

PawNtheSandman wrote:

Using what Danman32 stated, none of the above addresses are private. Therefore you can eliminate answers B,E,F.

And since the question said select 3, and you eliminated 3 of your 6 answers, you already have your answers as A,C,D.

Actually, A starts with 100 not 64, and B is certainly a private Class B address since it falls in the range of 172.16.0.0/12. The answer would be A, D, E.

forbesl

nzchuen wrote:

Can pls explain to me what is public class and private class?

http://www.isi.edu/in-notes/rfc1918.txt

BubbaJ wrote:

Actually, A starts with 100 not 64, and B is certainly a private Class B address since it falls in the range of 172.16.0.0/12. The answer would be A, D, E.

Yup

PawNtheSandman

Then I feel stupid. Guess that is what happens when your teacher is fresh out the MCT. Never knew class B ended at 172.31.255.255.

Danman32

Of course you also have the 169.254 addresses which aren't officially private addresses, but I don't know if they have been excluded from valid registerable internet addresses.

forbesl

Danman32 wrote:

Of course you also have the 169.254 addresses which aren't officially private addresses, but I don't know if they have been excluded from valid registerable internet addresses.

http://www.rfc-editor.org/rfc/rfc3330.txt

leilani

Class B does not end at 172.31.255.255.

I am sure you know this but,

if the first bit in the first octet is 0 then it is a class A (ie.1-126)
if the first bit in the first octet is 10 then it is a class B (ie 128-191)
if the first bit in the first octet is is 110 then it is a class C (ie 192-223)

so just by these calculations A would considered Class A
B would be considered Class B and C would be considered Class C


as for your answer to class B ending at 172.31.255.255 is not true.....
172.32.0.0 is a class B address

BubbaJ

leilani wrote:

as for your answer to class B ending at 172.31.255.255 is not true.....

He was referring to Private addressing.

Danman32

leilani wrote:

Class B does not end at 172.31.255.255.

I am sure you know this but,

if the first bit in the first octet is 0 then it is a class A (ie.1-126)
if the first bit in the first octet is 10 then it is a class B (ie 128-191)
if the first bit in the first octet is is 110 then it is a class C (ie 192-223)

so just by these calculations A would considered Class A
B would be considered Class B and C would be considered Class C


as for your answer to class B ending at 172.31.255.255 is not true.....
172.32.0.0 is a class B address

Class B specification doesn't end at x.31.255.255, but the class B private address range does. 172.32.x.x is a public address, as is 172.15.x.x.