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what is the subnet for the host ip address 172.16.210.0/22
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Answer / jitendera kumar sinah
the given ip address is
172.16.210.0/22
scince it is class b address the it orignal mask should be 26 but given mask it 22 that is there is 6 bit of sunating
128+64+32+16+8+4=252
block size=256-252=4
now
210/4= 52
and remnder is 2(no matter)
that is
52*4=208
thus
subnet for the given host ip adress must be
172.16.208.0/22
thanks
| Is This Answer Correct ? | 36 Yes | 11 No |
Answer / jitendera kumar sinah
the given ip address is
172.16.210.0/22
scince it is class b address the it orignal mask should be 16 but given mask it 22 that is there is 6 bit of sunating
128+64+32+16+8+4=252
block size=256-252=4
now
210/4= 52
and remnder is 2(no matter)
that is
52*4=208
thus
subnet for the given host ip adress must be
172.16.208.0/22
thanks
| Is This Answer Correct ? | 14 Yes | 6 No |
Answer / jitendera sinha
i have done little typo mistake .in place of 16 i have written 26 in first answer.so i think on that basic i am wrong
| Is This Answer Correct ? | 9 Yes | 3 No |
Answer / venkatesh i patil
no of subnet bits=6
no of subnets=2power6=64
no of host bits=10
no of valid hosts=(2power10)-2
increment=2power(no of host bits in interested octet)[here
interested octet is 3rd octet]=2power2=4
the multiple of 4 nearer to 210 is 208 so the subnet starts
with 172.16.208.0
subnet first valid host last valid host broadcast id
..16.208.0 172.16.209.0 172.16.210.0 172.16.211.0
172.16.208.0 is currect...
| Is This Answer Correct ? | 7 Yes | 4 No |
Answer / efc
The address is a Class B = default subnet mask is 255.255.0.0
/22 = 255.255.252.0
increment = 4
multiple of 4 closest to 210 = 208
network is 172.16.208.0 /22
network address = 172.16.208.0
first valid host= 172.16.208.1
last valid host = 172.16.211.254
broadcast address= 172.16.211.255
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / anantharaman
this ll come under class B. so it has N:N:H:H. 2 network
portions and 2 host portions. 172.16.210.0/22 in dis qn /22
denotes de subnet n it means leaving de 2 octets n count
after tat.. i.e N(8):N(8):H(1+1+1+1+1+1):H(0)
8+8=16+1+1+1+1+1+1=22 so 128+64+32+16+8+4=252 the ans wud
be 255.255.252.0...
| Is This Answer Correct ? | 3 Yes | 4 No |
Answer / s.m.feroz ahmed
Hi,
The Subnet for host ip address 172.16.210.0/22 is
255.255.0.0 as it belongs to classB.
| Is This Answer Correct ? | 2 Yes | 8 No |
Answer / s.m.feroz ahmed
Hi,
255.255.0.0 is the Subnet for the host ip Address
172.16.210.0/22.
| Is This Answer Correct ? | 0 Yes | 9 No |
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CCNA 
