From the above network you need to create 7subnets What will

From the above network you need to create 7-subnets. What will be your new subnet mask? 255.255.255.240 255.255.255.252 255.255.255.224 255.255.255.192 Given the following subnet mask: 198.0.0.0/24 How many host can you get? 60-hosts 62-hosts 64-hosts 128-hosts An ISP leases you the following network: 198.10.10/24 You need to create 26-subnetworks from this network What will be the new subnet mask (dotted-decimal)? How many hosts will be supported In each subnet? What is the subnet address of the third subnet? What Is the broadcast address of the fourth subnet?. An ISP leases you the following network: 138.10.8.0/21 You need to create 60-subnetworks from this network What will be the new sublet (dotted - decimal)? How many hosts will be supported in each subnet? What is the subnet address of the third subnet? What is the broadcast address of the fourth subnet?

Solution

1. The answer is option c. 255.255.255.224
   this is because the given network is class C address which has 256 addresses from 0-255. therefore it is only the last octet which occupies both host bits and network bits.As we need 7 subnets, the last and filling octet should have 3 ones in the MSB and remaining with 0s. 3 ones because 2³=8 which is the least possible number of subnets that match the requirement.MSB three ones whens summed results in 224 value. therefore the answer is C.

2. The answer is: 256 hosts as the slash notation specified /24 which is 255.255.255.0. the last octet consists of 8 bits. therefore the number of hosts is 2⁸ which is 256. If we consider usable hosts then it is 256-2=254. Both the answers are not specified in the answer.So, cross check the question.

3. i) the new subnet mask will be:255.255.255.248 (look question 1 for the explanation)
   ii)2³=8 will be the number of hosts in each network. usabke hosts are 8-2=6 hosts.(1 address is used for net ID and the other is used for broadcast ID)
  iii)198.10.10.16(1st -198.10.10.0-198.10.10.7, 2nd- 198.10.10.8-198.10.10.15, 3rd-198.10.10.16-198.10.10.23)
  iv)198.10.10.31 is the broadcast of the fourth subnet.

4.i)the new subnet mask would be 255.255.255.224
   ii) 32 host will be supported by each network 32-2=30 usable hosts.
  iii) the subnet address of the third 138.10.8.64
  iv) the broadcast address of the 138.10.8.127

for further queries kindly get back