Tough challenging subnetting questions that seem to have multiple answers?

Hi,

I am doing some subnetting questions from the book, and some of these answers do no make sense to me. I can't understand why the answers are what they are. Please help in advising why the particular option is correct. So , here are the questions:

1. An IP subnetting design requires 200 subnets and 120 hosts/subnet for the largest subnets, and requires that a single mask be used throughout the one private IP network that will be used. The design also requires planning for 20 percent growth in the number of subnets and number of hosts/subnet in the largest subnet. Which of the following answers lists a private IP network and mask that, if chosen, would meet the requirements?
a. 10.0.0.0/25
b. 10.0.0.0/22

c. 172.16.0.0/23

d. 192.168.7.0/24

2. A subnet design uses Class A network 10.0.0.0, and the engineer must choose a single mask to use throughout the network. The design requires 1000 subnets, with the largest subnet needing 200 hosts. Which of the following masks meets the requirements and also maximizes the number of subnets?

a. /18 (why can't it be /18 here, it satisfies both requirements of 1000 subnets and largest needing 200 hosts)

b. /20
c. /22

d. /24

3. Which of the following is not a valid subnet ID for network 10.0.0.0, using mask /25, assuming that this mask is used throughout the network?
a. 10.0.0.0

b. 10.255.255.0
c. 10.255.127.128
d. 10.1.1.192

• Posts: 897Member ■■■■□□□□□□
#3
the looking for the subnet address.
it is a /25 subnet. last 7 bits are for hosts. So, make those all zeros.
the first 24 bits are all part of the subnet address. So the first 3 bytes, you don't care about.
all you need to do is look at the 25 bit which can be a 1 or zero.
that means the last octette is 0 followed by 7 more zeros for the host portion. Or, 0
if it is a one, that followed by the 7 zeros for the hosts, that 128.
so the subnet address has to end in 0 or 128.
so the address ending in 192 is not valid
• Posts: 897Member ■■■■□□□□□□
#2
the operative phase is maximizes the number of subnets.
So, what is the minimum number of host bits does it take to support 200 hosts. 8 bits.
32-8=24
/24 is the answer.
• Posts: 897Member ■■■■□□□□□□
#1
need 200 subnets + 20% = 240 which is < 256 so you need 8 bits for subnets
need 120 hosts + 20% = 144 which is < 256 so you need 8 bits for hosts.

a. /25 only 7 bits for host. incorrect.
b. /22 8 bits for the class A network. 10 bits for hosts good giving 32-18=14 bits for the subnet. right answer
c. /23 16 bits for the class B network. 9 bits for the hosts, good. giving 32-25=7 bits for the subnet incorrect
d. /24 24 bits for the class C network. 8 bits for the hosts, good giving 32-32=0 bits for the subnet incorrect
• Thank you Clarson!

But, if we are to subnet this with /25, wouldn't the subnets go like this:

- 10.0.0.0/25
-10.0.0.128/25

If so, then wouldn't this make options B and C valid addresses? When we subnet 10.0.0.0/25, why are options B and C valid addresses? I am so confused.

Thank you. Appreciate your help! • Posts: 897Member ■■■■□□□□□□
when subnetting you have so many bits for network, then bits for the subnet and then bits for the host portion of the address.
10.0.0.0 is a class A network. so 8 bits for the network.
/25 means 32-25=7 bit for hosts.
32-15= 17 bits for the subnet. with x = 8 bits and y= 1 bit you get a network address like this.
10.x.x.y000 0000

the network address can have anything from 0-255 for x. and 0 or 1 for y.
so the first octet is 10
the second octet can be anything
the third octet can be anything
and the 4th octet can only be 0 or 128.

b and c are valid