CCNA – Drag and Drop 1

Here you will find answers to drag and drop Questions – Part 1

Question 1

Explanation

In short, we should start calculating from the biggest network (with 16 hosts) to the smallest one using the formula 2ⁿ – 2 (n is the number of bits we need to borrow).Therefore:

16 hosts < 2⁵ – 2 (we need to borrow 5 bits -> /27)

11 hosts < 2⁴ – 2 (borrow 4 bits -> /28)

5 hosts < 2³ – 2 (borrow 3 bits -> /29)

From the available ip addresses, we see that each of them has only one suitable solution (they are 192.168.164.149/27,192.168.164.166/28 and 192.168.164.178/29)

The smallest network is the Floss S0/0 which only requires 2 hosts = 2² – 2 (need to borrow 2 bits ->/30). There are 2 suitable answers: 192.168.164.189/30 and 192.168.164.188/30 but notice that 192.168.164.188/30 is the network address so we can not use it (because 188 = 4 * 47) -> we have to choose 192.168.164.189 as the correct solution.

In fact, it is not the formal way to solve a VLSM question so I recommend you to review your CCNA book if you haven’t grasped it well yet.

Question 2

Question 3

Explanation

1) The wildcard mask of the command “deny ip 192.168.35.16 0.0.0.15 host 192.16.35.66” is 0.0.0.15, which is equal to network mask of 255.255.255.240 = /28. So the access list will deny all traffic from network 192.168.35.16/28 from accessing host 192.16.35.66, which is the IP address of accounting server.

2) The command “deny ip 192.168.35.55 0.0.0.0 host 192.168.35.66” will deny host 192.168.35.55, which is a user and belongs to interface e0 of Alabama router (192.168.35.49/28) from accessing accounting server.

3) Because there is an implicit “deny all” command at the end of each access list so the command “permit ip 192.168.35.0 0.0.0.255 host 192.168.35.66” will only let network 192.168.35.0/24 access accounting server whilst prevent traffic from other networks.

Question 4

Question 5