CCNA – Subnetting
Here you will find answers to CCNA Subnetting Questions
Note: If you are not sure about subnetting, please read my Subnetting tutorial.
Question 1
Given a subnet mask of 255.255.255.224, which of the following addresses can be assigned to network hosts? (Choose three)
A – 15.234.118.63
B – 92.11.178.93
C – 134.178.18.56
D – 192.168.16.87
E – 201.45.116.159
F – 217.63.12.192
Answer: B C D
Explanation
A subnet mask of 255.255.255.224 has an increment of 32 (the binary form of the last octet is 1110 0000) so we can’t use numbers which are the multiples of 32 because they are sub-network addresses. Besides, we can’t use broadcast addresses of these sub-networks (the broadcast address of the previous subnet is calculated by subtracting 1 from the network address). For example the network address of the 2nd subnet is x.x.x.32 then the broadcast address of the 1st subnet is 32 – 1 = 31 (means x.x.x.31).
By this method we can calculate the unusable addresses, which are (notice that these are the 4th octets of the IP addresses only):
+ Network addresses: 0, 32, 64, 96, 128, 160, 192, 224.
+ Broadcast addresses: 31, 63, 95, 127,159, 191, 223.
Question 2
Which of the following host addresses are members of networks that can be routed across the public Internet? (Choose three)
A – 10.172.13.65
B – 172.16.223.125
C – 172.64.12.29
D – 192.168.23.252
E – 198.234.12.95
F – 212.193.48.254
Answer: C E F
Explanation
Addresses that can be routed accross the public Internet are called public IP addresses. These addresses belong to class A, B or C only and are not private addresses.
Note:
Private class A IP addresses: 10.0.0.0 to 10.255.255.255
Private class B IP addresses: 172.16.0.0 to 172.31.255.255
Private class C IP addresses: 192.168.0.0 to 192.168.255.255
Class D addresses are reserved for IP multicast addresses and can’t be routed across the Internet (their addresses begin with 224.0.0.0 address).
Also we can’t use 127.x.x.x address because the number 127 is reserved for loopback and is used for internal testing on the local machine.
Question 3
A national retail chain needs to design an IP addressing scheme to support a nationwide network. The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet. Working with only one Class B address, which of the following subnet masks will support an appropriate addressing scheme? (Choose two)
A – 255.255.255.0
B – 255.255.255.128
C – 255.255.252.0
D – 255.255.255.224
E – 255.255.255.192
F – 255.255.248.0
Answer: B E
Explanation
We need to remember the default subnet mask of class B is 255.255.0.0. Next, the company requires a minimum of 300 sub-networks so we have to use at least 512 sub-networks (because 512 is the minimum power of 2 and greater than 300). Therefore we need to get 9 bits for network mask (29=512), leaving 7 bits for hosts which is 27-2 = 126 > 50 hosts per subnet.This scheme satisfies the requirement -> B is correct.
We can increase the sub-networks to 1024 ( 1024 = 210), leaving 6 bits for hosts that is 26= 64 > 50 hosts. This scheme satisfies the requirement, too -> E is correct.
Notice: The question asks “The company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet” but this is a typo, you should understand it as “”The company needs a minimum of 300 sub-networks and a minimum of 50 host addresses per subnet”.
Question 4
Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choose three)
A – 115.64.8.32
B – 115.64.7.64
C – 115.64.6.255
D – 115.64.3.255
E – 115.64.5.128
F – 115.64.12.128
Answer: B C E
Explanation
CIDR stands for Classless In4ter-Domain Routing, the difference between CIDR and VLSM is slim and those terms are interchangeable at CCNA level.
To specify which IP addresses fall into the CIDR block of 115.64.4.0/22 we need to write this IP address and its subnet mask in binary form, but we only care 3rd octet of this address because its subnet mask is /22.
(x means “don’t care”)
Next, we have to write the 3rd octets of the above answers in binary form to specify which numbers have the same “prefixes” with 4.
4 = 0000 0100
8 = 0000 1000
7 = 0000 0111
6 = 0000 0110
3 = 0000 0011
5 = 0000 0101
12=0000 1100
We can see only 7, 6 and 5 have the same “prefixes” with 4 so B C E are the correct answers.
Question 5
Refer to the diagram. All hosts have connectivity with one another. Which statements describe the addressing scheme that is in use in the network? (Choose three)
A – The subnet mask in use is 255.255.255.192.
B – The subnet mask in use is 255.255.255.128.
C – The IP address 172.16.1.25 can be assigned to hosts in VLAN1.
D – The IP address 172.16.1.205 can be assigned to hosts in VLAN1.
E – The LAN interface of the router is configured with one IP address.
F – The LAN interface of the router is configured with multiple IP addresses.
Answer: B C F
Explanation
VLAN 2 has 114 hosts so we need to leave 7 bits 0 for the host addresses (27 – 2 = 126 > 114). Notice that we are working with class B (both Host A and Host B belong to class B) and the default subnet mask of class B is /16 so we need to use 16 – 7 = 9 bits 1 for the sub-network mask, that means the subnet mask should be 255.255.255.128 -> B is correct.
By using above scheme, C is correct because the IP 172.16.1.25 belongs to the sub-network of VLAN 1 (172.16.1.0/25) and can be assigned to hosts in VLAN 1.
For communication between VLAN 1 and VLAN 2, the LAN interface of the router should be divided into multiple sub-interfaces with multiple IP addresses -> F is correct.
Question 6
The network 172.25.0.0 has been divided into eight equal subnets. Which of the following IP addresses can be assigned to hosts in the third subnet if the ip subnet-zero command is configured on the router? (Choose three)
A – 172.25.78.243
B – 172.25.98.16
C – 172.25.72.0
D – 172.25.94.255
E – 172.25.96.17
F. 172.25.100.16
Answer: A C D
Explanation
If the “ip subnet-zero” command is configured then the first subnet is 172.25.0.0. Otherwise the first subnet will be 172.25.32.0 (we will learn how to get 32 below).
The question stated that the network 172.25.0.0 is divided into eight equal subnets therefore the increment is 256 / 8 = 32 and its corresponding subnet mask is /19 (1111 1111.1111 1111.1110 0000).
First subnet: 172.25.0.0/19
Second subnet: 172.25.32.0/19
Third subnet: 172.25.64.0/19
4th subnet: 172.25.96.0/19
5th subnet: 172.25.128.0/19
6th subnet: 172.25.160.0/19
7th subnet: 172.25.192.0/19
8th subnet: 172.25.224.0/19
In fact, we only need to specify the third subnet as the question requested. The third subnet ranges from 172.25.64.0/19 to 172.25.95.255/19 so A C D are the correct answers.
Question 7
Refer to the exhibit. In this VLSM addressing scheme, what summary address would be sent from router A?
A. 172.16.0.0/16
B. 172.16.0.0/20
C. 172.16.0.0/24
D. 172.32.0.0/16
E. 172.32.0.0/17
F. 172.64.0.0/16
Answer: A
Explanation
Router A receives 3 subnets: 172.16.64.0/18, 172.16.32.0/24 and 172.16.128.0/18.
All these 3 subnets have the same form of 172.16.x.x so our summarized subnet must be also in that form -> Only A, B or C is correct.
The smallest subnet mask of these 3 subnets is /18 so our summarized subnet must also have its subnet mask equal or smaller than /18.
-> Only answer A has these 2 conditions -> A is correct.
@marcos
/20 = 11111111.11111111.11110000.00000000
2 power ( number of 1s in 3rd Octet ) = 16 ( network block size)
2 power ( number of 0s ) = 4096 hosts
Subtracting 2 from last result gives 4094
@marcos: There are 32 – 20 = 12 bits for host so we will have 2^12 – 2 = 4094 available hosts. The formula used here is 2^(number of bit 0) – 2
Hi…can u plz explain hw, in Q 5, we come to d conclusion that the IP 172.16.1.25 belongs to the sub-network of VLAN 1 (172.16.1.0/25) ?!
tanks your is site !!!
hi 9tut
There is question on a ccna exam simulator, I don’t get a way to solve it. The exercise gives a routing table as follows
Network Interface Next Hop
10.1.1.0/24 e0 directly connected
10.1.2.0/24 e1 directly connected
10.1.3.0/25 s0 directly connected
10.1.4.0/24 s1 directly connected
10.1.5.0/24 e0 10.1.1.2
10.1.5.64/28 e1 10.1.2.2
10.1.5.64/29 s0 10.1.3.3
10.1.5.64/27 s1 10.1.4.4
Question is Where will the router send a packet destined for 10.1.5.65?
thanks foir your help
sorry, answer given is 10.1.3.3
Q6: what if the network is not configured for ip subnet-zero? what if it is /19 for example and need to get the eight-th netwrok? i got that question in the exam and i couldn’t answer it.
@Santi:
First notice that 10.1.5.65 belongs to network 10.1.5.64/29.
From the output, we learned that the network 10.1.5.64/29 is learned from 10.1.3.3 (10.1.5.64/29 s0 10.1.3.3) so the router can reach 10.1.5.65 through 10.1.3.3 -> the router will send that packet to 10.1.3.33 (on s0 interface).
@ef: If the network is not configured for ip subnet-zero, it will not use the first sub-network. For example network 192.168.1.0/25, the router will not use the first sub-network 192.168.1.0 -> 192.168.1.127.
Por que en la explicacion me dice que la mascara es de /19 No entiendo como saca que la mascara es de 19, yo en un principio pense que era de /20
ANSWER ME PLEASE
Why in the explanation tells me that the mask is / 19. Do not understand how the mask takes is /19? I initially thought I was / 20
I had to go tow a auto that had sat for months.
thanks a lot for your help and response 9tut
just one more question
¿how did you notice 10.1.5.65 belongs to 10.1.5.64/29 subnet? from my point it also can be considered belongs to 10.1.5.64/27 or /28
Can someone please explain Question 2? I am little confused, because answer B.
Thanks in advance.
@Santi: 10.1.5.65 belong to 10.1.5.64/29, 10.1.5.64/27 and 10.1.5.64/28 but 10.1.5.64/29 will be selected to route this packet. This is called the “longest prefix match” rule, which selects the most specific network to route the packet. Because /29 is the most specific (longest) prefix comparing to /27 and /28 prefixes -> it will be used to route the packet.
hello everyone,
thanks for this site
@9tut now that makes sense, just like what you’ve said
10.1.5.65 is a valid address under subnet 10.1.5.65 /27;subnet 10.1.5.65 /28 and subnet 10.1.5.65 /29 so the router will use the one with the longest prefix.
correct me if i’m wrong so if there is an address of 10.1.5.64 /30, then the router will use this address?
Which two subnetworks would be included in the summarized address of 172.31.80.0/20 (choose 2)
a. 172.31.17.4/30
b.172.31.51.16/30
c.172.31.64.0/18
d.172.31.80.0/22
e.172.31.92.0/22
f.172.31.192.0/18
Ans. D n E
Hw do v find d answer? Is it by converting every 3rd octet no into binary and then checking to c which comes under d summarized add or is there an easier method?
Which 3 IP addresses can be assigned to hosts if the subnet mask is /27 and subnet zero is usable? (choose 3)
a.10.15.32.17
b.17.15.66.128
c.66.55.128.1
d.135.1.64.34
e.129.33.192.192
f.192.168.5.63
Answer A, C, D
How do v find this answer?
@lanc3l0t: Yes, the router will route to 10.1.5.64/30 if it is available.
@Ans: First you need to find out the range of 172.31.80.0/20 and find the IPs belong to this range.
thanks a lot @9tut, pretty clear to me know
what is the subnetmask for 172.16.0.1 and 192.168.0.1 i know as 255.255.255.0 and 255.255.255.0 but give me the explanation plz guys
Can any one pls explain this question?
1. the ip subnet zero commend is **NOT* configured on a router. What would be the ip address of E0/0 using the first available address from the 6th subnet of the network 192.168.8.0/29 ?
a.192.168.8.25
b.192.168.8.41
c.192.168.8.49
d.192.168.8.113
ans is C.
but my question is, if subnet zero commend is * configured on a router what would be the answer ????????
Dear 9tut,
Could you PLEASE make a tutorial for subneting and VLSM too, like other brilliant, useful and wonder tutorials that you have been made, and I deeply appreciate your wonderful serving…
Question 3 indicates that maximum number of hosts should not exceed 50 right?
If I’m doing it correctly, I only see one answer and that’s “D”
D – 255.255.255.224
Hello…! 9tut
How many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask, and how many
usable host addresses will there be per subnet?
A. 8 subnets, 31 hosts
B. 8 subnets, 32 hosts
C. 16 subnets, 30 hosts
D. A Class B address can’t be subnetted into the fourth octet.
E. 16 subnets, 32 hosts
As the network administrator of your company, you have been assigned the task of designing a new Office internetwork. So you need to consider IP addressing scheme, Which two subnetworks would be included in the summarized address of 172.31.80.0 /20? (Choose two.)
A. 172.31.92.0 /22
B. 172.31.51.16 /30
C. 172.31.80.0 /22
D. 172.31.17.4 /30
Hello…! 9tut
I understand /30 give us only 2 host it is small for company. so /22 is correct? can you clear me pl…!
@Swati: The summarized address of 172.31.80.0/20 ranges from 172.32.80.0 to 172.32.96.0 (increment:16) so we have
A. 172.31.92.0 /22
C. 172.31.80.0 /22 are the correct answers.
Hello…! 9tut
Hello…! 9tut
Thank you for help. pl… give me this question ans.
————————————————————————
How many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask, and how many usable host addresses will there be per subnet?
A. 8 subnets, 31 hosts
B. 8 subnets, 32 hosts
C. 16 subnets, 30 hosts
D. A Class B address can’t be subnetted into the fourth octet.
E. 16 subnets, 32 hosts
@Swati: We have
2^(27-23) = 16 subnets
2^(32 – 27) – 2 = 30 usable host addresses per subnet.
So C is the correct answer.
Hello…! 9tut
Thank you for help. Pretty clear to me know. This side really help to me. Thank you for this side.
Thank u 9tut..was refreshing!
@ Santi
You have a destination network with 4 possible paths. To choose the best path, the router will select the lowest Administrative Distance, if those are equal (same routing protocol) then lowest cost is chosen ( you would see these in the numbers in brackets I.E [120|1] for RIP, AD is 120, second number is cost, RIP uses hop count, and is not shown in your example). If those are equal then the longest matching subnet mask would be used. So the router will choose 10.1.5.65/29
@ jila boy
the answer w’ll be ‘ B ‘ coz sub net zero active >Router>>> w’ll start from 0
the new ranges : (usable IP’s)
1. (1-7)
2. (9-15)
3.(17-23)
4.(25-31)
5.(33-39)
6.(41-47) >>> it shows B is the correct answer
“””” i want to thank 9tut for the helpful website , it’s great .
HELLO 9tut,
PLEASE make a tutorial for subneting and VLSM too, like other brilliant, useful and wonder tutorials that you have been made, as subneting is the major topic of CCNA exam, we will really appreciate it.
Which two subnetworks would be included in the summarized address of 172.31.80.0/20?
A. 172.31.17.4/30
B. 172.31.51.16/30
C. 172.31.64.0/18
D. 172.31.80.0/22
E. 172.31.92.0/22
F. 172.31.192.0/18
please tell me What is answer and how?
D, E
172.31.80.0/20 is 172.31.80.0 ~ 172.31.95.255
the only 2 networks that belong to this network range are those at options D and E.
you get a script that generates subnetting questions with any of the VCEs posted here:
http://www.tinyurl.com/xallaxvce
Im trying to do a lab on CCNA exploration Chapter 6. Lab 6.7.5. Please help with the addressing as it seems the lab is getting the incorrect ip-addresses. Given an 192.168.1.0/24 address space.
The network attached to R1 will require enough IP addresses to support 15 hosts.
R2 will require enough IP addresses to support 30 hosts.
link between router R1 and router R2 will require IP addresses at each end of the link.
Assign subnet 1 to the network attached to R1.
Assign subnet 2 to the link between R1 and R2.
Assign subnet 3 to the network attached to R2.
1. Assign the first valid host address in subnet 1 to the LAN interface on R1.
2. Assign the last valid host address in subnet 1 to PC1.
3. Assign the first valid host address in subnet 2 to the WAN interface on R1.
4. Assign the last valid host address in subnet 2 to the WAN interface on R2.
5. Assign the first valid host address in subnet 3 to the LAN interface of R2.
6. Assign the last valid host address in subnet 3 to PC2.
Please show me how to work it out. I understand that 4 subnets (2 power of 2 = 4) are required so we will borrow two host bits. With subnetting do u start counting subnets from o or 1 as below
subnet 0 – 0 0
subnet 1 – 0 1
subnet 2 – 1 0
subnet 3 – 1 1
pls anyone give the solution for solving the following question:
The network administrator is asked to configure 113 point-to-point links.Which IP addressing scheme
best defines the address range and subnet mask that meet the requirement and waste the fewest subnet
and host addresses?
A. 10.10.0.0/18 subnetted with mask 255.255.255.252
B. 10.10.0.0/25 subnetted with mask 255.255.255.252
C. 10.10.0.0/24 subnetted with mask 255.255.255.252
D. 10.10.0.0/23 subnetted with mask 255.255.255.252
E. 10.10.0.0/16 subnetted with mask 255.255.255.252
@anonymous
each p2p link takes 4 IPs. 1 is the subnet ID, 1 is the broadcast IP and 2 are legal to use for device interfaces.
113 p2p links = 113*4 IPs= 442 IPs.
256 < 442 < 512
512 = 2^9. 9 bits for hosts.
32-9 = 23 bits for subnetting.
the answer is D, 10.10.0.0/23.
452 IPs*, my bad, it doesn’t change the answer. was rushing :)
hi 9tut!
Q3 is fcking crazy.. if that statement minimum of 500 and maximum of 50.. den only D could be the possible answer.. but it requires to choose two and the only two that can fit is your current answer B and E…
damnnn cisco!
hi, pls any latest ccna dumps,
yongtillblisz@yahoo.com
@Santi
10.1.5.65 can be reached the following routing path.
10.1.5.0/24
10.1.5.64/28
10.1.5.64/29
10.1.5.64/27
If there are many routing paths….we choose the longest prefix length…
Thats why …choose this path 10.1.5.64/29 s0 10.1.3.3.
Hi Guys. Thanks for all the good work.
I need help in clarifying this subnetting question:
Which addresses are valid host IP addresses given the subnet mask 255.255.255.248 (choose three)
1. 192.168.200.87
2. 194.10.10.104
3. 223.168.210.100
4. 220.100.100.154
5. 200.152.2.160
6. 196.123.142.190
The answers are option 3,4,6. Can anyone please help varify. Thanks.
mask 255.255.255.248
the IPs are in blocks of 8:
0~7
8~15
16~23
so on…
the first and last IPs can not be assigned.
192.168.200.87 this is a broadcast IP (.80~.87)
194.10.10.104 this is a subnet ID (.104~.111)
223.168.210.100 valid (.104~.111)
220.100.100.154 valid (.152~.159)
200.152.2.160 subnet ID (.160~.167)
196.123.142.190 valid (.184~.191)
Thanks Xallax for pointing me in the right direction. Yes, with increment of 8. The answer is verifiable.
I need your help one more time on subnetting.
What are the three valid host adresses given the subnet mask 255.255.254.0 ( Choose three)
1. 64.100.2.255
2. 172.55.33.255
3. 64.104.129.0
4. 192.168.162.255
5. 10.165.55.255
6. 128.107.62.0
The answers are option 1,3,4. Can you kindly share some light on it.Thanks.
Is ther any way that I could find out mac address by using only IP address but not using subnet mask?
for an example 192.168.1.97 /28, by using subnet mask /28 I know the mac address would be 255.255.255.240 but if subnet mask is not given then how I’ll find out Mac address by using only IP address?
If any one knows then help me out with this doubt.
Thanks
I will like to help in any way I can. But your question is not clear. Mac address are addresses of devices, and not related to subnet mask. While IP address are logical network addresses. This two are on different layers of the OSI -TCP/IP. Layer2 and layer3. I guess I can’t quite get at what you are driving at. Only if you can reword your question differently………