I got that question wrong too, I took for granted that I entered a wrong number but it could be an error in the system. Here is the calculation focusing only on the last octet, the "turned on" relevant bits are in bold:
240 in binary:111100008 in binary:00001000
The network address will be the relevant bits as they are, and the last four bits turned to 0, so it is 10.10.10.050 in binary:00110010The network address will be 10.10.10.48
The CIDR is also correct, a 240 in the last octet is a /28Links relevant to this answer:
If my calculations are correct, then there may be an issue with this question.
Hope this helps,Franco
I am checking this one out. I'll post what I find. Take care.
Thanks for using the forum.
I just ran through the review and after doing subnetting on my end I came up with the same answer you did.
It doesn't seem to like the 10.10.1.40/28 network for some reason. Let me ask around and I'll post what I find.
This is a great opportunity for others in this community to jump in and help as well.
Take care Wilson
Your calculations are correct. I am unable to reproduce the error though. See the screenshot below. Are you trying to add the /28 somewhere?
I worked through this with a colleague of mine and he figured this out. I didn't go far enough in my subnetting before I answered the question but this is correct.
so with a .240 subnet mask there are 16 subnets-- .0 through .15, .16 through .31, .32 through .47, .48 through .63 etc.
the first address in each of those ranges is the network ID
so .50 falls within the .48 network
It's been a while since I subnetted so I am out of shape in that department.
Thanks a lot everyone for working together on this
thats right Jas. i was typing wrong. (ex: 10.10.1.0/28) in the answer you can not type "/28"
Thanks for the support all of you Team.