subnetting question

i need to know how to work this out the quickest way, possibly with binary as well

which of the folowing ip hosts would be valid for PC users, assuming that a /27 network mask was used
for all of the networks?

a. 15.234.118.63
b. 83.121.178.93
c. 134.178.18.56
d. 192.168.19.37
e. 201.45.116.159
f. 217.63.12.192

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Comments

eurotrash

to me it looks like:

b.
c.
d.

EdTheLad

pally_21 wrote:

i need to know how to work this out the quickest way, possibly with binary as well

which of the folowing ip hosts would be valid for PC users, assuming that a /27 network mask was used
for all of the networks?

a. 15.234.118.63
b. 83.121.178.93
c. 134.178.18.56
d. 192.168.19.37
e. 201.45.116.159
f. 217.63.12.192

/27 = 255.255.255.224
256-224=32 i.e. subnets are multiples of 32
a) 63 is the broadcast address of subnet 32 i.e. invalid
b) 93 is valid in subnet 64
c) 56 is valid in subnet 32
d) 37 is valid in subnet 32
e) 159 is the broadcast address of subnet 128 i.e. invalid
f) 192 is the subnet address i.e invalid