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.
q 2 in todays exam
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ThankYou ALLAH! clear the paper just now… access list 1, vtp, eigrp lab… I love 9tut :)
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You should not start learning subnetting unless you know the binary to decimal number conversion system.
hello,
qestion 7 – correct answer is 14
“RIP version 1 is the only routing protocol in use on the network” means that we should not use zero subnet and broadcast subnet, hence the formula is 2n-2 –
if n=3 8-2=6 < 7
if n=4 16-2=14 subnet and 14 hosts
http://www.cisco.com/en/US/tech/tk648/tk361/technologies_tech_note09186a0080093f18.shtml
fjre,
So you use 3 bits (11100000) for subnets: 2^3=8 which satisfies the requirement for 7 LANs. Then that leaves with 5 bits for the host (2^5-2=30) The given answer is correct. You are incorrect
how to do ?????
NETWORK ID =10.0.0.0 SUBNET MASK:255.0.0.0 no of networks :6 ????I need first ID,Broad cast DI,valid host IT ???????
How’s this possible ?
let it be the 0 subnet ?
@ Q7
In RIP1 can we do so,Is it support for VLSM??
Qestion 9 could not understand !!!!!!
How summarized network is 192.168.25.16 ?
Just took it yesterday, I PASSED with a 920, thanks to 9tut…woooohhhooooo!!!!
i think Fjre is correct. answer should based on n=4 not on n=3. as subnet 0 not in use
Can anyone explain Q. 2 in a different way. Please and thanks
@Subhsamal
192.168.25.20 = 192.168.25.0001 0100
192.168.25.16 = 192.168.25.0001 0000
192.168.25.24 = 192.168.25.0001 1000
192.168.25.28 = 192.168.25.0001 1100
192.168.25.16 = 192.168.25.0001 xxxx; (xxxx = 4)
These bit are all = 32; so, 32 – 4 = 28 bit ( /28 )
Answer: 192.168.25.16/28 = 192.168.25.16 255.255.255.240
Fjre is not correct, the formula 2n-2 is to determine the amount of HOSTS that can be supported by the network. To determine the number of networks it is just 2n so if we borrow the 4 bits we would have 16 subnets with 14 hosts per subnet. So in total we would have 7 subnets with only 3 bits borrowed. Just because the 0 subnet isn’t going to be used does not mean it’s not a subnet that needs address space as it might be reserved for expansion or perhaps Infrastructure.
Network Address Broadcast Address
Subnet 1: 192.168.0.0 192.168.0.31
Subnet 2: 192.168.0.32 192.168.0.63
Subnet 3: 192.168.0.64 192.168.0.95
In the end you wind up with 7 Subnets with 30 Hosts available per subnet
So to continue on from my last post (damn mouse) Since the subnets requires are less than 8 we don’t need the 4th bit. So in Binary we would have 00000111 which turns into 4+2+1 = 7 which is exactly the number of Subnets we need in this example.
@Adnan W
The question is asking for only the Phoenix Summerized route. the rest of the diagram is flak to confuse you. So if we take just the Phoenix part of the picture we get
Summerize these routes:
10.4.0.0
10.4.1.0
10.4.2.0
10.4.3.0
Since we have only these 4 route to summerize we take a look at them in the bit value
00001010.00000100.00000000.00000000
00001010.00000100.00000001.00000000
00001010.00000100.00000010.00000000
00001010.00000100.00000011.00000000
Next we see where they all line up which is at the 3 bit mark (aka the 4′s bit)
From there we say OK do we need the 4 bit, does anything require it in the given address space, no. So we only need the 2nd bit for the summerized mask value.
so in the end we wind up with 10.4.0.0/22 since none of the 3rd octect lines up at all with each other but we need a proper mask to cover all the subnets contained within without going over.
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@KIJAGATZ
We need 7 LANs from a Class C Address
Already Class C uses 24bits to represent the Network, to create 7 networks, we need three more bits because (2^3) = 8 which is closer to 7. Then we have: 27 bits for the network. the remaining 5 bits are for the hosts in each of the networks.
2^5 = 32;
The usable addresses are: 32 – 2 = 30
so 30 usable hosts per LAN. Actually we will have 8 LANs.
But 8 is most close to 7
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EVERYONE WHO CAN HELP ME TO SOLVE QUESTION ANSWER THAT ASKED FEWEST WASTING ADDRESSES. I FOUND THE TWO CORRECT ANSWER B AND D. BUT B ONLY IS CORRECT. WHO CAN HELP ME TO FIND OUT THE CORRECT ANSWER IS?
I PREFER THE QUESTION 1 ANSWER.
Adnan W
Easy to calculate the Q2.
10.4.0.0…10.4.3.0 = 1022hosts with mask /24, 254+254+254+254 per network.
1022 < 1024
2^10 = 1024 mask /22 will give that.
The last part is to choose the Network that will fit every network on Phoenix that will be 10.4.0.0
Address: 10.4.0.0
Netmask: 255.255.252.0 = 22
Network: 10.4.0.0/22
Broadcast: 10.4.3.255
HostMin: 10.4.0.1
HostMax: 10.4.3.254
I cleared 200-120
Question 4 & 9 in today’s exam
Almost all questions from 9tut
Thanks everyone :)
At the end of explanation of Q4 it says:
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”.
It should say “maximum of 50 host addresses per subnet” not minimum of 50……….
hi can you explain me the Q.4?
I can’t understand this explanation. Please help
thanks in advance!!!
The debate on question 7 comes down to how you interpret “subnet 0 is not being used.”
A- For those who interpret that to mean only 7 subnets are available, then you will answer ‘E’ (32-2) since 3 bits will still satisfy the requirement of 7 LANs.
B- For those who interpret that to mean that ‘ ip subnet 0 ‘ has not been issued, then you believe that only 6 subnets would be available. Thus, a 4th bit would be needed and the answer would be B (16-2).
I haven’t settled on either.
Because “subnet 0 is not being used” you will lose one subnet. You can calculate this easy by adding one more subnet. so 8 subnets.
With a class c you’ll have 256 addresses. 256 : 8 Lans= 32 addresses.
In each subnet you can’t use the network and the broadcastaddress so you will lose 2. Still all subnet will contain 30 usable Addresses.
(Just remember, you cant use all the addresses in this first Subnet, so the address x.x.x.0-31, the first usable subnet will start with x.x.x.32. The first usable address will be x.x.x.33)
kind regards
Hello to all. I have a question concerning the summarize of subnetworks. Is it possible to summarize 9 subnetworks in one . My ip address is 132.227.0.0/16 and I have divided it on 9 subnetworks so now I want to make a default route of these 9 subnetworks (no 0.0.0.0 cuz it’s too general).
Thanks in advance.
Hi Can someone explain Question 6
How was the subnet mask able to be found its confusion me
Thanks
@Barry re Q7
After researching, I believe the issue with RIP v1 and IP Subnet Zero only affects the all 0′s subnet and not the all 1′s subnet. Early on, we were all taught to discard the bottom and top subnet when using classful routing protocols. The top subnet (all 1′s) was discarded, not because RIP V1 wouldn’t advertise it, but because of the confusion it caused with broadcast address between the subnet and the network. So I’m guessing we throw away the all 0′s subnet but keep the All 1′s.
It irritates me to see the people that have never heard of discarding the top and bottom subnet chime in without understanding the debate.
@Alex
128 64 32 16 8 4 2 1
1 0 0 0 0 0 0 0 to have 84 Hosts u need to take 2^7-2=126 satisfy 84 hosts
1 0 0 0 0 0 0 0 to have 114 Host need to take 2^7-2=126 satisfy 114 hosts
subnetmask will be 255.255.255.128
subnetworks
0–127 1-126 range addresses in VLAN1
128–255 129–254 range addresses in VLAN2
127 stand for broadcast address
255 stand for broadcast address
Hence subnetmask will be 255.255.255.128 correct answer
172.16.1.25 found within the range of VLAN1 correct answer
172.16.1.205 not found within the range of VLAN1 but found in the range of VLAN2
In order for different VLAN to communicate need router that have subinterface of ip addressess or techinical called router on stick that create subinterface or multiple ip addresses correct answer
I think u undersatand and other guys
Thanks
hey
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Q4 and 9 were in test today
Dear iCisco,
Your comment is very informative.
Thanks!
Q5 in yesterday’s exam.
All questions from 9tut and Watson dump.
Sim ACL, ACL2 and EIGRP.
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I just want to ask you that in real exam IP are also same which are mentioned here in questions??
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Hi all,
the correct answer of Question 8 is : A
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 172.1.0.0/22 because we have all 4 sub-networks so we can move left 2 bits (22=4).
Regards
Taking the exam next week. Would appreciate it if someone could share the latest dump for 200-120 for practice purpose. allay_desai@hotmail.com Thank you!
Shady: That is not true.
You are correct in that 172.1.4.0/25 & 172.1.4.128/25 can be grouped into 172.1.4.0/24
However 172.1.0.0/22 covers the following:
172.1.0.0/24
172.1.1.0/24
172.1.2.0/24
172.1.3.0/24
172.1.4.0/22 covers 172.1.4.0/24 -> 172.1.7.0/24
anyone can help me ??
many thx
host A address:192.168.1.78 255.255.255.224
host B address:192.168.1.130 255.255.255.192
which 3 answers are correct? and why?
A. host A IP address: 192.168.1.79
B. host A IP address: 192.168.1.64
C. host A default gateway: 192.168.1.78
D. host B IP address: 192.168.1.128
E. host B default gateway: 192.168.1.129
F. host B IP address: 192.168.1.190
q9 was on the exam Friday aug 8
Hhi everyone,
Q1, why answer B and not de D, ?
/26 is correct to 60 hots,
please help-me.