Cisco CCNA mock exam questions sample test – Question 197

.A device has an address of 192.168.144.21 and a mask of 255.255.255.240.
What will be the broadcast address for the subnet to which this device is attached?

A. 192.168.144.23
B. 192.168.144.28
C. 192.168.144.31
D. 192.168.144.32


Correct Answer: C

Explanation:
The broadcast address for the subnet to which this device is attached will be 192.168.144.31.
To determine the broadcast address of a network where a specific address resides, you must first determine the network ID of the subnetwork where the address resides. The
network ID can be obtained by determining the interval between subnet IDs. With a 28-bit mask, the decimal equivalent of the mask will be 255.255.255.240. The interval between
subnets can be derived by subtracting the value of the last octet of the mask from 256. In this case, that operation would be 256 – 240. Therefore, the interval is 16.
The first network ID will always be the classful network you started with (in this case 192.168.144.0). Then each subnetwork ID in this network will fall at 16-bit intervals as follows:
192.168.144.0
192.168.144.16
192.168.144.32
192.168.144.48
At 192.168.144.48 we can stop, because the address that we are given as a guide is in the network with a subnet ID of 192.168.144.16. Therefore, since the broadcast address for
this network will be 1 less than the next subnet ID (192.168.144.32), the broadcast address for the subnet to which this device is attached is 192.168.144.31.
All the other options are incorrect because none of these will be the broadcast address for the subnet to which this device is attached.
Objective:
Network Fundamentals
Sub-Objective:
Apply troubleshooting methodologies to resolve problems
References:
Cisco > Technology Support > IP > IP Routing > Design TechNotes > IP Addressing and Subnetting for New Users > Document ID: 13788 > Understanding IP Addresses