Test by Bikram | Computer Networks | Test 1 | Question: 11

Question

An organization with a class C IP address of $195.37.5.0$ wish to divide the network into different sub networks, the subnet mask is 255.255.255.192. Which one among these is not a broadcast address in any of the sub network?

• A. $195.37.5.127$
• B. $195.37.5.191$
• C. $195.37.5.31$
• D. $195.37.5.255$

Comments

Can anyone explain how it is option (c)?

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as subnet mask is 192 which is 11 000 000 so 2 bits we take for subnet bits from HID and total 4 subnets 2^2=4 are there and first subnet 00 second subnet 01 third subnet 10 and fourth subnet 11 .

now for broadcast address all Host bits are 1 that means

first broadcast address is 00 111 111 which is 63 

second broadcast address is 01 111 111 which is 127 

third broadcast address is 10 111 111 which is 191  and

fourth broadcast address is 11 111 111 which is 255 .

last octet ::: 63(00) , 127(01), 191(10)  , and 255(11) for 195.37.5.127,    195.37.5.191,    195.37.5.255,     195.37.5.63 broadcast addresses possible  for 4 subnets  .

As given option c says 31 so c) is not broadcast address. 

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Thanx for the answer.

Answer 1

the subnet mask is 192 that means 11 000000 , so 2 bits from HID taken for subnets there are 4 subnets possible like 00, 01, 10 and 11 so for 00 broadcast address is 195.37.5. 63 , for 01 broadcast address is 195.37.5. 127, for 10 broadcast address is 195.37.5. 191, for 11 broadcast address is 195.37.5. 255. but option c says 195.37.5.31 which not match with anyone hence option C is not a broadcast address in any of the sub network .

Comments

I think there is problem with the test, My wrong answer it says right. Problem with lot of question please check @Bikram
 

Test name : Computer Networks-12