Three devices are on three different subnets. Match the network address and the broadcast address with each subnet where these devices are located. (Not all options are used.)
Device 1: IP address 192.168.10.77/28 on subnet 1
Device 2: IP address192.168.10.17/30 on subnet 2
Device 3: IP address 192.168.10.35/29 on subnet 3
Place the options in the following order:
| Subnet 2 network number | 192.168.10.16 |
| Subnet 1 broadcast address | 192.168.10.79 |
| Subnet 3 broadcast address | 192.168.10.39 |
| Subnet 2 broadcast address | 192.168.10.19 |
| Subnet 1 network number | 192.168.10.64 |
| Subnet 3 network number | 192.168.10.32 |
Explanation: To calculate any of these addresses, write the device IP address in binary. Draw a line showing where the subnet mask 1s end. For example, with Device 1, the final octet (77) is 01001101. The line would be drawn between the 0100 and the 1101 because the subnet mask is /28. Change all the bits to the right of the line to 0s to determine the network number (01000000 or 64). Change all the bits to the right of the line to 1s to determine the broadcast address (01001111 or 79).
Related exam: Introduction to Networks (Version 7.0) – ITNv7 Practice Final Exam
