Networking subjective question for competitive exam(1)

Q1.IP address of host is 182.44.82.16/26.What is first and last address .

subnet mask=255.255.255.192(because 26)
first address=any ip AND network mask
first address=182.44.82.16 AND 255.255.255.192=182.44.82.0/26
last address=any ip OR not(network mask)
last address= 182.44.82.16 OR  0.0.0.63=182.44.82.63/26

Q2 Find subnet mask if 1024 subnets in class A.

In class A(8bit+24bit)8 bit for network and 24 bit for host
number of bits required for 1024 subnets
=n+log(s)
=8+log(1024)
=8+10=18
therefore subnet mask
255.255.192.0

Q3.Find subnet mask if 1024 subnets in class B.

In class B(16bit+16bit) 16 bit for network and 16 bit for host.
number of  bits required for 1024 subnets
=n+log(s)
=16+log(1024)=16+8=24
therefor subnet mask=255.255.255.0

Q4.Find the range of address in blocks 123.56.77.32/29

subnet mask=255.255.255.248(because 29 bits)
first address=123.56.77.32 AND 255.255.255.248=123.56.77.32/29
last address=123.56.77.32 OR 0.0.0.7=123.56.77.39/29
Total address is block=2 power 3=8

Q5.An organization is granted block 16.0.0.0/8.The administrator wants to create 500 fixed length subnets. Find subnet mask,number of address in each subnet,find first and last address.

bits allocation=(8+24)
bits for subnet=n+log(s)
                         =8+log(512)=8+9=17
subnet mask=255.255.128.0(because 17 bit for subnet)
number of address in each subnet=2 pwer 15
first address=16.0.0.0 AND 255.255.128.0=16.0.0.0/17
Last address=16.0.0.0 OR 0.0.63.255=16.0.63.255/17

Q6 An interface on a router with the IP address of 192.168.192.10/29.What is 
the broadcast address.
 

Last address of this block(is broadcast address.)
subnet mask=255.255.255.248(because 29 bits)
Last address=ip address OR not(subnet mask)
last address=192.168.192.10 OR 0.0.0.7=192.168.192.15

Q7 What is the last valid host on the subnetwork 165.21.80.128/26.

subnet mask=255.255.255.192(because 26 bit)
Last valid host is the second last address of block is broadcast address.
Last address=any ip address OR not(network mask)
last address=165.21.80.128 OR  0.0.0.63=165.21.80.191/26
last valid host=165.21.80.190/26

Q8 Find the number of address in a range if first address is 106.92.29.0 and
last address is 106.92.31.255.
 

subtract last address with first address
get
0.0.2.255
number of address=(0*256^3+0*256^2+0*256^1+255*256^0)+1
number of address=768

Q9 Find netid and hostid of classful address 114.34.2.8

that ip 114.34.2.8  in class A.
bits allocation(8+24) 8 bits for network and 24 bits for host.
net id=114
host id=34.2.8
 

Q10 An IP address 1.2.3.4 find class,network ip,limited adress,broadcast address. 

IP address belong to class A.
Network ip address=1.0.0.0(because in class A 8 bit for network and 24 bits for host).
Direct broadcast address=1.255.255.255
limited broadcast address=255.255.255.255