New CCNA – Subnetting 2
Note: If you are not sure about Subnetting, please read our Subnetting Tutorial – Subnetting Made Easy.
Question 1
Refer to the exhibit. A new subnet with 60 hosts has been added to the network. Which subnet address should this network use to provide enough usable addresses while wasting the fewest addresses?
A. 192.168.1.56/27
B. 192.168.1.64/26
C. 192.168.1.64/27
D. 192.168.1.56/26
Answer: B
Explanation
60 hosts < 64 = 26 -> we need a subnet mask of at least 6 bit 0s -> “/26″. The question requires “wasting the fewest addresses” which means we have to allow only 62 hosts-per-subnet -> B is correct.
Question 2
Refer to the exhibit. The Lakeside Company has the internetwork in the exhibit. The Administrator would like to reduce the size of the routing table to the Central Router. Which partial routing table entry in the Central router represents a route summary that represents the LANs in Phoenix but no additional subnets?
A. 10.0.0.0 /22 is subnetted, 1 subnet
D 10.0.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
B. 10.0.0.0 /28 is subnetted, 1 subnet
D 10.2.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
C. 10.0.0.0 /30 is subnetted, 1 subnet
D 10.2.2.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
D. 10.0.0.0 /22 is subnetted, 1 subnet
D 10.4.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
E. 10.0.0.0 /28 is subnetted, 1 subnet
D 10.4.4.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
F. 10.0.0.0 /30 is subnetted, 1 subnet
D 10.4.4.4 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
Answer: D
Explanation
All the above networks can be summarized to 10.0.0.0 network but the question requires to “represent the LANs in Phoenix but no additional subnets” so we must summarized to 10.4.0.0 network. The Phoenix router has 4 subnets so we need to “move left” 2 bits of “/24″-> /22 is the best choice -> D is correct.
Question 3
Refer to the exhibit. What is the most appropriate summarization for these routes?
A. 10.0.0.0/21
B. 10.0.0.0/22
C. 10.0.0.0/23
D. 10.0.0.0/24
Answer: B
Explanation
We need to summarize 4 subnets so we have to move left 2 bits (22 = 4). In this question we can guess the initial subnet mask is /24 because 10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0 belong to different networks. So “/24″ moves left 2 bits -> /22.
Question 4
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= 128 > 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 5
Which address range efficiently summarizes the routing table of the addresses for router main?
A. 172.16.0.0/18
B. 172.16.0.0/16
C. 172.16.0.0/20
D. 172.16.0.0/21
Answer: C
Explanation
To summarize these networks efficiently we need to find out a network that “covers” from 172.16.1.0 -> 172.16.13.0 (including 13 networks < 16). So we need to use 4 bits (24 = 16). Notice that we have to move the borrowed bits to the left (not right) because we are summarizing.
The network 172.16.0.0 belongs to class B with a default subnet mask of /16 but in this case it has been subnetted with a subnet mask of /24 (we can guess because 172.16.1.0, 172.16.2.0, 172.16.3.0… are different networks).
Therefore “move 4 bits to the left” of “/24″ will give us “/20″ -> C is the correct answer.
Question 6
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
First we should notice that different VLANs must use different sub-networks. In this case Host A (172.16.1.126) and Host B (172.16.1.129) are in different VLANs and must use different sub-networks. Therefore the subnet mask in use here should be 255.255.255.128. In particular, it is 172.16.1.0/25 with 2 sub-networks:
+ Sub-network 1: 172.16.1.0 -> 172.16.1.127 (assigned to VLAN 1)
+ Sub-network 2: 172.16.1.128 -> 172.16.1.255 (assigned to VLAN 2)
-> B is correct.
The IP address 172.16.1.25, which is in the same sub-network with host A so it can be assigned to VLAN 1 -> C is correct.
To make different VLANs communicate with each other we can configure sub-interfaces (with a different IP address on each interface) on the LAN interface of the router -> F is correct.
Question 7
The network administrator needs to address seven LANs. RIP version 1 is the only routing protocol in use on the network and subnet 0 is not being used. What is the maximum number of usable IP addresses that can be supported on each LAN if the organization is using one class C address block?
A. 6
B. 8
C. 14
D. 16
E. 30
F. 32
Answer: E
Explanation
“The network administrator needs to address seven LANs” means we have 7 subnets < 8 = 23, so we need to borrow 3 bits from the host part (to create 8 subnets). We are using class C address block which has 8 bits 0 (the default subnet mask of class C is 255.255.255.0), so the number of bit 0 left is 8 – 3 = 5. Therefore the hosts per subnet will be 25 – 2 = 30 -> E is correct.
Question 8
Refer to the exhibit. What is the most efficient summarization that R1 can use to advertise its networks to R2?
A. 172.1.0.0/22
B. 172.1.0.0/21
C. 172.1.4.0/22
D. 172.1.4.0/24
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24
E. 172.1.4.0/25
172.1.4.128/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24
Answer: C
Explanation
Network 172.1.4.0/25 and network 172.1.4.128/25 can be grouped to a single network 172.1.4.0/24
Network 172.1.4.0/24 + Network 172.1.5.0/24 + Network 172.1.6.0/24 + Network 172.1.7.0/24 can be grouped to a single network 172.1.4.0/22 because we have all 4 subnetworks so we can move left 2 bits (22=4).
Question 9
| Gateway of last resort is not set 192.168.25.0/30 is subnetted, 4 subnets D 192.168.25.20 [90/2681856] via 192.168.15.5, 00:00:10, Serial0/1 D 192.168.25.16 [90/1823638] via 192.168.15.5, 00:00:50, Serial0/1 D 192.168.25.24 [90/3837233] via 192.168.15.5, 00:05:23, Serial0/1 D 192.168.25.28 [90/8127323] via 192.168.15.5, 00:06:45, Serial0/1 C 192.168.15.4/30 is directly connected, Serial0/1 C 192.168.2.0/24 is directly connected, FastEthernet0/0 |
Which address and mask combination a summary of the routes learned by EIGRP?
A. 192.168.25.0 255.255.255.240
B. 192.168.25.16 255.255.255.252
C. 192.168.25.0 255.255.255.252
D. 192.168.25.28 255.255.255.240
E. 192.168.25.16 255.255.255.240
F. 192.168.25.28 255.255.255.240
Answer: E
Explanation
We have 4 routes learned by EIGRP:
D 192.168.25.20 [90/2681856] via 192.168.15.5, 00:00:10, Serial0/1
D 192.168.25.16 [90/1823638] via 192.168.15.5, 00:00:50, Serial0/1
D 192.168.25.24 [90/3837233] via 192.168.15.5, 00:05:23, Serial0/1
D 192.168.25.28 [90/8127323] via 192.168.15.5, 00:06:45, Serial0/1
These subnets are all /30 (as it says “192.168.25.0/30 is subnetted, 4 subnets”. We have 4 successive subnets = 22 so we can go back 2 bits -> the summarized subnet mask is 30 – 2 = 28 and the summarized network is 192.168.25.16.
BV – Answer D. 192.168.1.56 is already being used in another subnet (192.168.1.48/28). 192.168.1.48/28 covers from 192.168.1.48-63. Therefore the next available subnet is 192.168.1.64/26
Tks Sam. best regards
192.168.1.56 = 192.168.1.00111000 not the network address of /26 (01,000000)
/26 network valid ip start from 192.168.1.65~126 (01,000001~01,111110)
Q4 Should be a minimum of 50 host not (a maximum of 50 host) ,coz answer B 128 gives u 128 and answer E 192 gives u 64 hosts and both Greater that 55!!!!!
I would appreciate further explanation of question 9.
Thank you in advance!
@ Jimmy. in regards to Question 9.
we have 4 subnets with the subnet mask of 30 each.
write out all subnets in binary and summarize.
192.168.25.20 /30
192.168.25.16 /30
192.168.25.24 /30
192.168.25.28 /30
summarize this four subnets
192.168.25.20 x.x.x.00010100
192.168.25.16 x.x.x.00010000
192.168.25.24 x.x.x.00011000
192.168.25.28 x.x.x.00011100
so your summary network and subnetmask is
192.168.25.16 255.255.255.240
so answer is very correct.
@ Jimmy. in regards to Question 9.
we have 4 subnets with the subnet mask of 30 each.
write out all subnets in binary and summarize.
192.168.25.20 /30
192.168.25.16 /30
192.168.25.24 /30
192.168.25.28 /30
summarize this four subnets
192.168.25.20 x.x.x.00010100
192.168.25.16 x.x.x.00010000
192.168.25.24 x.x.x.00011000
192.168.25.28 x.x.x.00011100
so your summary network and subnetmask is
192.168.25.16 255.255.255.240
so answer E is very correct.
Q4, Q6 & Q9 in 5th Sep 2014 exam
@ Chris
Thank you very much, now I understand!
Question 9
Gateway of last resort is not set
192.168.25.0/30 is subnetted, 4 subnets
D 192.168.25.20 [90/2681856] via 192.168.15.5, 00:00:10, Serial0/1
D 192.168.25.16 [90/1823638] via 192.168.15.5, 00:00:50, Serial0/1
D 192.168.25.24 [90/3837233] via 192.168.15.5, 00:05:23, Serial0/1
D 192.168.25.28 [90/8127323] via 192.168.15.5, 00:06:45, Serial0/1
C 192.168.15.4/30 is directly connected, Serial0/1
C 192.168.2.0/24 is directly connected, FastEthernet0/0
Which address and mask combination a summary of the routes learned by EIGRP?
A. 192.168.25.0 255.255.255.240
B. 192.168.25.16 255.255.255.252
C. 192.168.25.0 255.255.255.252
D. 192.168.25.28 255.255.255.240
E. 192.168.25.16 255.255.255.240
F. 192.168.25.28 255.255.255.240
Explanation
We have 4 routes learned by EIGRP:
D 192.168.25.20 [90/2681856] via 192.168.15.5, 00:00:10, Serial0/1
D 192.168.25.16 [90/1823638] via 192.168.15.5, 00:00:50, Serial0/1
D 192.168.25.24 [90/3837233] via 192.168.15.5, 00:05:23, Serial0/1
D 192.168.25.28 [90/8127323] via 192.168.15.5, 00:06:45, Serial0/1
These subnets are all /30 (as it says “192.168.25.0/30 is subnetted, 4 subnets”. We have 4 successive subnets = 2^2 so we can go back 2 bits -> the summarized subnet mask is 30 – 2 = 28 and the summarized network is 192.168.25.16.
Can anyone please explain if 4 successive subnets = 2^2, then how about if 6 successive subnets? Or 8 successive subnets? so value to ^2 or 2^ to value?
Can somebody plz explain quest 8? thanks
hello BV both answer are correct
Passed today with 958
Q2 was on test
thanx 9tut
PASS MY EXAM TODAY SEPT 17 1000/1000
Q6 in exam today
anyone can expalain further on Q7.
about Q7, Rip v1 is a classfull routing protocol so how is it correct to subnet a class c ip address to seven subnets.
Tomorrow I have Exam. Any valid suggestion or update or tips..
It would be appreciate .
i too seat for my exam tomorrow, best wishes to us @impartent
@Q7. I thank those who explained about Q7. I forgot that the protocol is RIP v1. the question by it self is not correct. And ,being dependent on 9tut and examtut solutions may not be 100% guaranteed. So study hard to escape from such kinds of problems.
Regarding Question 7
“RIP version 1 is the only routing protocol in use on the network and subnet 0 is not being used, ” are we safe to assume that the router has the no ip suibnet zero command configured? If so then are we not counting the first LAN 0 Subnet and the last broadcast subnet?
Last Octect – 00000000 – Typically if we are looking for 7 subnets then we would need to take 3 bits with 2^3= 8 subnets- 11100000. Total of 8 subnets which meets the LAN requirements but we can’t use the first and last subnets if the no ip subnet zero is applied where the new subnet formula becomes 2^3-2 = 6 subnets which does not meet the LAN requirements.
So I was thinking could the answer really be C. 14 since if we borrow 4 bits then it would 2^4-2 = 14 subnets. Or if it’s not a true no ip subnet zero formula and we are just ignoring the very first 0 subnet then 2^3 = 8 – 1 subnet = 7. Where now we have exactly 7 subnets not counting only the 0 subnet.
This question has really confused me lol, if someone can shed some further light it would be appreciated.
Thank you.
Greetings, I also agree with Aleks. I think that the answer to Question 7 should be C http://www.cisco.com/c/en/us/support/docs/ip/dynamic-address-allocation-resolution/13711-40.html “Problems with Subnet Zero and the All-Ones Subnet
Traditionally, it was strongly recommended that subnet zero and the all-ones subnet not be used for addressing. According to RFC 950 leavingcisco.com, “It is useful to preserve and extend the interpretation of these special (network and broadcast) addresses in subnetted networks. This means the values of all zeros and all ones in the subnet field should not be assigned to actual (physical) subnets.” This is the reason why network engineers required to calculate the number of subnets obtained by borrowing three bits would calculate 23-2 (6) and not 23 (8). The -2 takes into account that subnet zero and the all-ones subnet are not used traditionally.”
m going for exam tommrow. plz pry for me
Bilawal, how did you’re exam go? Hope you cleared the exam.
Please update when you get a chance. Is 9tut enough ?
to Polaris
subnet zero means we are not using the 1st subnet out of the 8 subnets( that does not affect the number of hosts on each LAN) leaving us with 7 subnets.
2^3=8 subnet (3 subnets bits borrowed) leaving us with 5 hosts bits.
2^5 -2 =30 usable hosts .
in the reference you gave you should notice that each LAN have the same amount of host addresses.;;…
I hope that helps
the Q7 is not correct, because RIPv1 is a classfull IP routing protocol which means when it send it routing update it don’t include the subnet mask as result the neighbor or remote router will place the route with a default mask for this exemple /24. a part from that the IP subnet-zero is not in use, it means the first and last subnet won’t be use we’ll have just 6 subnet. 9 if I’m making mistake please make us understand.
Cedric:
Don’t forget that you are told that you need to configure this for 7 LANs; which means you need 3 bits for the network portion, which leaves 5 bits for the host portion. 2^3 = 8, and 2^5 = 32. Since you don’t have IP subnet zero configured, you would subtract 2 from 32; which leaves you with 30 configurable hosts per subnet.
q5 today
I don’t understand Q5. pls help. why did u use 4 bits? to get 16
Q 1 and 2 were on 10th
To “proxil”
But from example I gave I see that we need to exclude 2 subnets. The one with all “0″ and all “1″ thus we need to subtract number 2 from total number of subnets of 8…that gives us 6 subnets, which means we need to borrow 1 more bit. Simply put I think we need to exclude first network subnet AND broadcast subnet. Can you explain it further as I still don’t understand what you meant by: “in the reference you gave you should notice that each LAN have the same amount of host addresses”
I believe Q7 is a bit of a trick question as it does not say ip subnet-zero has not been configured, it just says ‘subnet 0 is not being used’, meaning ignore the first subnet – which turns out to be an irrelevant part of the question, along with the mention of RIPv1.
By the way, as it seems a lot of people are not too familiar with the ip subnet-zero command, it applies to the network and not the hosts. You always minus 2 for host calculation. ip subnet-zero (which is enabled by default) allows the use of the first and last subnets.
Q5 is confusing for me. can someone xplain it pls.
and also Q8 please. so when summarizing an net address there diff methods use?
Q8
Answer: C
Explanation
Network 172.1.4.0/25 and network 172.1.4.128/25 can be grouped to a single network 172.1.4.0/24
Network 172.1.4.0/24 + Network 172.1.5.0/24 + Network 172.1.6.0/24 + Network 172.1.7.0/24 can be grouped to a single network 172.1.4.0/22 because we have all 4 subnetworks so we can move left 2 bits (22=4).
++++++++++++++++++++++++++++++
why did it become /22????? if the 4 networkds are subnetted to /24
172.1.[4/5/6/7].0 subnets use first 3 octets for their individual network mask – 255.255.255.0
To summarise you have to include all subnets (it’s like subnetting a subnet)
The octet position and block size (increment) of combined networks now become important.
Now the third octet has to be subnetted because it’s a summary of all networks.
A block or increment of 4 in the third octet is a /22 (252)
For Q8 this covers 172.1 .[4-8].0
help pls i dnt understand these questions and can’t find answers.
for the following IP adress 130.22.40.1 and 194.174.3.2
a) maximum hosts is 60 for network( subnet) ?
b) Bits number of hosts?
c) Bits numbers of network ( subnet)?
d) Broadcast and network (subnet) ID?
in the exam do they change the ip’s / diagram of the questions? (not talking about the lab sims)
Q5 – It is a little confusing but all they are asking for is a summarization of all the networks going into the Main Router
Hi Guys,
Going to sit CCNA exams by next Monday. If you have any new Dumbs please mail me to rifamym@ymail.com
Thanks.
q6…Dec 18
q5 on 11 dec
hai guys
took my exam recently scored 1000/1000
acl1,acl2 modified ,eigrp
all question from the dumps
Q2 today
Uday, how was the exam? were all the questions from this website only?
hi
whic one is eaiser ICND! or CCNA
thanks
tara, ICND is easier than CCNA. However, CCNA has less question than the first because you have to take ICND1 and ICND2, which means more questions for you to deal with. Let us say if CCNA has 45 question (excluding the Labs), then ICND 1 & 2 may have double that. Got it?
can someone please explain Q2 ??
Sir i Hear from one person that he said the C class private Musk is 17 i can’t understand that How it could?
Q4 & Q9 in my exam today.
RE: Q9 Can someone please explain how 255.255.255.240 came in answer? Why answer is E why not B?