ISRO2014-75

Question

An organization is granted the block $130.34.12.64/26.$ It needs to have $4$ subnets. Which of the following is not an address of this organization?

• A. $130.34.12.124$
• B. $130.34.12.89$
• C. $130.34.12.70$
• D. $130.34.12.132$

Comments

@Bikram sir please explain..how to approach this question?

Answer 1

The 4 subnets :

130.34.12.64-130.34.12.79

130.34.12.80-130.34.12.95

130.34.12.96-132.34.12.111

130.34.12.112-130.34.12.127

So ans is d

Comments

please explain

Answer 2

The block given to the Organization here is,

130.34.12.64/26 
so,

255.255.255 .1100 0000 (MASK)
130.34.12    .0100 0000 (IP Start Address)
-------------------------------------------------------------

Complementing The Mask ,We will be able to get the range of the IP's available to the Organization.

00000000 00000000 00000000 00111111=64

Dividing it into four subnets will give 

130.34.12.64
130.34.12.65
130.34.12.66
............
130.34.12.127

It is clearly visible that option : D) is the answer.
 

Comments

CIDR /26 has subnet mask 255.255.255.192 and 192 is 11000000 in binary. We used two host bits in network address.

So N = 2
H = 8-2= 6

Now the range of Total subnets ( 2^N ) :- 2² = 4
Block size (256 - subnet mask) :- 256 - 192 = 64
Valid subnets ( Count blocks from 0) :- 0,64,128,192

The organization in given the subnet with starting address 64, i.e. 130.34.12.64

The range of addresses will be: 130.34.12.65 - 130.34.12.126.
Hence the option is D

 

 

---

130 is class b so why we havent taken subnets bits from 3rd octat?

Answer 3

130.	34.	12.	01	000000
NID	NID	NID	NID	HID

26 bit NID. HID is 6 bits.

So hosts possible = 64.

Divide into four subnets => 16 per subnet.

So, last octets: .64 to .79, .80 to .95, .96 to .111, .112 to .127

which makes Option D out of range.