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Networking - Subnetting - Discussion

Discussion Forum : Subnetting - Subnetting (Q.No. 11)
Subnetting
11.
You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?
192.168.19.0 255.255.255.0
192.168.19.33 255.255.255.240
192.168.19.26 255.255.255.248
192.168.19.31 255.255.255.248
Answer: Option
Explanation:
A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.
Discussion:
15 comments Page 2 of 2.
Bharti singh said:   8 years ago
The first host address should be :192.168.19.25.

Because up to 29 bits it is fixed part and the remaining part is 3 bits.
192.168.19.00011000 -> subnet block id.
.00011001-> this will be the first host for this subnet.
(1)
Umesh Prajapati said:   6 years ago
As I solved this question :

Nework id :- 192.168.19.24.
First vaild host :- 192.168.19.25.
Last vaild host - 192.168.19.30.
Broadcost IP address :- 192.168.19.31.
Subnet mask :- 255.255.255.248.

Please check it.
(5)
Inuya said:   3 years ago
D is the correct answer.
(1)
Thant Toe Aung said:   2 years ago
C is the correct answer because the first available host 192.168.19.15 has been assigned to the router and 2nd available host 192.168.19.26 can be assigned to server.
So, the correct answer is C 192.168.19.26 255.255.255.248.
Hari said:   1 year ago
Can we provide a broadcast IP address to a server?

Anyone, please clarify.
(1)
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