New CCNA – Subnetting
Note: If you are not sure about Subnetting, please read our Subnetting Tutorial – Subnetting Made Easy.
Question 1
Refer to the exhibit. Which subnet mask will place all hosts on Network B in the same subnet with the least amount of wasted addresses?
A. 255.255.255.0
B. 255.255.254.0
C. 255.255.252.0
D. 255.255.248.0
Answer: B
Explanation
310 hosts < 512 = 29 -> We need a subnet mask of 9 bits 0 -> 1111 1111.1111 1111.1111 1110.0000 0000 -> 255.255.254.0
Question 2
Refer to the exhibit. All of the routers in the network are configured with the ip subnet-zero command. Which network addresses should be used for Link A and Network A? (Choose two)
A. Network A – 172.16.3.48/26
B. Network A – 172.16.3.128/25
C. Network A – 172.16.3.192/26
D. Link A – 172.16.3.0/30
E. Link A – 172.16.3.40/30
F. Link A – 172.16.3.112/30
Answer: B D
Explanation
Network A needs 120 hosts < 128 = 27 -> Need a subnet mask of 7 bit 0s -> “/25″.
Because the ip subnet-zero command is used, network 172.16.3.0/30 can be used.
Answer E “Link A – 172.16.3.40/30″ is not correct because this subnet belongs to MARKETING subnet (172.16.3.32/27).
Answer F “Link A – 172.16.3.112/30″ is not correct because this subnet belongs to ADMIN subnet (172.16.3.96/27).
Question 3
You have been asked to come up with a subnet mask that will allow all three web servers to be on the same network while providing the maximum number of subnets. Which network address and subnet mask meet this requirement?
A. 192.168.252.0 255.255.255.252
B. 192.168.252.8 255.255.255.248
C. 192.168.252.8 255.255.255.252
D. 192.168.252.16 255.255.255.240
E. 192.168.252.16 255.255.255.252
Answer: B
Question 4
Which subnet mask would be appropriate for a network address range to be subnetted for up to eight LANs, with each LAN containing 5 to 26 hosts?
A. 0.0.0.240
B. 255.255.255.252
C. 255.255.255.0
D. 255.255.255.224
E. 255.255.255.240
Answer: D
Explanation
A is not correct because it is a wildcard mask (not subnet mask).
This question is a bit unclear but we can suppose we have to begin with default subnet mask and “subnet” it. And the default subnet mask here should be class C: 255.255.255.0
For answer B: 252 = 1111 1100 -> with this subnet mask we can subnet up to 26 = 64 subnets but only 22 – 2 = 2 hosts per subnet -> B is not correct.
C is not correct because it is the default subnet mask of class C and that means we don’t “subnet” it.
For answer E: 240 = 1111 0000 -> There are 24 = 16 subnets but only 24 – 2 = 14 hosts per subnet < 26 hosts -> E does not satisfy the second requirement (of 26 hosts per subnet).
For answer D: 224 = 1110 0000 -> There are 23 = 8 subnets and 25 – 2 = 30 hosts > 26 hosts -> This is the correct answer.
Note: The number “5″ in ” with each LAN containing 5 to 26 hosts” is just used to trick you and it does not have any effect on our answer.
Question 5
An administrator must assign static IP addresses to the servers in a network. For network 192.168.20.24/29, the router is assigned the first usable host address while the sales server is given the last usable host address. Which of the following should be entered into the IP properties box for the sales server?
A. IP address: 192.168.20.14
Subnet Mask: 255.255.255.248
Default Gateway: 192.168.20.9
B. IP address: 192.168.20.254
Subnet Mask: 255.255.255.0
Default Gateway: 192.168.20.1
C. IP address: 192.168.20.30
Subnet Mask: 255.255.255.248
Default Gateway: 192.168.20.25
D. IP address: 192.168.20.30
Subnet Mask: 255.255.255.240
Default Gateway: 192.168.20.17
E. IP address: 192.168.20.30
Subnet Mask: 255.255.255.240
Default Gateway: 192.168.20.25
Answer: C
Question 6
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.
Question 7
You are working in a data center environment and are assigned the address range 10.188.31.0/23. You are asked to develop an IP addressing plan to allow the maximum number of subnets with as many as 30 hosts each.Which IP address range meets these requirements?
A. 10.188.31.0/27
B. 10.188.31.0/26
C. 10.188.31.0/29
D. 10.188.31.0/28
E. 10.188.31.0/25
Answer: A
Explanation
Each subnet has 30 hosts < 32 = 25 so we need a subnet mask which has at least 5 bit 0s -> /27. Also the question requires the maximum number of subnets (which minimum the number of hosts-per-subnet) so /27 is the best choice -> A is correct.
Question 8
Which two benefits are provided by using a hierarchical addressing network addressing scheme? (Choose two)
A. reduces routing table entries
B. auto-negotiation of media rates
C. efficient utilization of MAC addresses
D. dedicated communications between devices
E. ease of management and troubleshooting
Answer: A E
Question 9
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
Answer: D
Explanation
We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7 bits (because 2^7 = 128).
The network used for point-to-point connection should be /30.
So our initial network should be 30 – 7 = 23.
So 10.10.0.0/23 is the correct answer.
You can understand it more clearly when writing it in binary form:
/23 = 1111 1111.1111 1110.0000 0000
/30 = 1111 1111.1111 1111.1111 1100 (borrow 7 bits)
Question 10
Given an IP address 172.16.28.252 with a subnet mask of 255.255.240.0, what is the correct network address?
A. 172.16.16.0
B. 172.16.24.0
C. 172.16.0.0
D. 172.16.28.0
Answer: A
Explanation
Increment: 16 (of the third octet)
Network address: 172.16.16.0
-> A is correct.
Q3 not understand ?????????
@Anonymous The question asks you to accommodate 3 hosts in the same subnet. We cant do it wit 255.255.255.252 mask because it only provides 2 addresses for hosts, the other two addressees are network id and broadcast witch we can not assign to hosts. So the mask 255.255.255.248 fills the needs with 6 addresses available. As you can see the third mask is too big we would waste to much subnets. Its written a bit tricky with all the network addresses shuffled but they are irrelevant here as they only confuse. You only need to find the right mask and see where it fits.
Question 7. Small correction.
The address range 10.188.31.0/23 is not exactly correct. It looks better: 10.188.30.0/23
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relative to Q4 is sure that the answer is D (255.255.255.224) and not C (255.255.255.0)?
I’m not english so maybe i have problem to understand exactely the meaning of “Which subnet mask would be appropriate for a network address range to be subnetted for up to eight LANs, with each LAN containing 5 to 26 hosts?”
I supposed was the subnet i started to have eight subnet woth a least 26 host…so i need a network with prefix 24 so 255.255.255.0…where i’m worng?
great works 9tut
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Hi I am following this site for quite sometime, all of you have added your expeience, please tell me if i follow this site only with subnetting and some config commands memorizing …..shall it serve my pupose of passing the CCNA 200-120 exam.
Of couse the lab sim i shall follow from 9tut.
hello guys please suggest me if following 9tut is good enough…..ofcourse with some basic understanding
@sandip, it should be, also use watson and/or examtut dumps…u should be good to go then!
Q#9 Explain very good thanks 9tut….
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Passed today with 958
Q5 and Q6 was on test
thanx 9tut
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Q8 in exam today
hello friends……..i have shared whatever i had …. please foward it to the one who need those /// i find it difficult to foward the same so often………
hello friends..,in Q5 i accept the answer as C but why not option A. anybody explain that.
PLEASE ,9tut a lot of people asked about Q4 why the answer is D.we believe that C is correct answer .can you explain? thanks.
@MOHA: We have just added the explanation for Q.4, please read it again.
@MOHA the answer has to be D for Q4. The question asks for an appropriate subnet mask for a network that houses up to 8 lans and each lan having from 5 – 26 hosts. If you take into consideration your formula 2^borrowed bits = subnet and 2^unborrowed – 2 = host, then 2^3= 8 subnets and 2^5 – 2 = 30 hosts, 5-26 hosts falls perfectly in that range. so therefore the subnet mask would have to be 255.255.255.224 which would give u the least amount of wasted space.
Hope that helps.
Hi Fox,
I need a help on the way i could respond on this type of subnetting question in less than 2 minutes: What is the subnet address of 172.16.159.159/22?
I know that this is a class B address and the default mask is 255.255.0.0
Also, I know that /22 is 255.255.252.0 subnet mask and the magic number is 4
what can be the fastest way to found out that 159 on the third and fourth octect is a broadcast address without going by counting the increment 4(0,4,8,12,16….156,160…). Thanks for your help.
Soiree…
You know the magic number from the mask (256-252 = 4)
Take the octet you are interested in and divide this by the magic number I.e. 159 / 4 = 39 (ignore the leftover/remainders)
Then multiply the result by the magic number I.e. 39 x 4 = 156 and this gives you the network number
Add the Magic for the next network number (156 + 4)
Then you can remove 1 from this to give you the broadcast (160 – 1 = 159)
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In Q 9, please verify.
This looks like borrowing 6 bits not 7 bits.
/30 = 1111 1111.1111 1111.1111 1100
borrowing 7 bits will look more like this :
/30 = = 1111 1111.1111 1111.1111 1110
the answer is D, but the way they develop it is a little bit wrong.
.On q no.3 where we not surpose to subnet the number of networks so that we can come up with a network range and it’s subnet mask. Now on this subnet mask 192.168.252.8 255.255.255.248 is this not unusable bcoz it looks like unusable address in the range network ,please explain !
Can anyone provide explanation for Question 3. I didn’t get why the answer is B
to anonymous
the bit is borrowed from slash 23
Question 4 is asking the size of the mask of a network, which subneted will give 8 subnets able to handle 5 to 26 hosts, not the size of the mask of one resulting subnet! The answer is definitively /24.
I agree with will. The question was asking for the whole network’s mask and not for each subnet. please see to this guys. thanks
can someone explain to me better the question 6? i got the answer, but too different in explanation. i assume it as class B, even the /24 mask. then the network address range is 172.16.32.0-172.16.164.0, its more than 128 subnet so i thought thats why we need /16 mask that has 256 subnet. but what i understand in the explanation is that if theres /18, that would be the better answer. :( pls help me with this. thanks
Q 5,7 8 on10th
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Agree with Will and Rod: the right answer for Q4 is C, not D. Sure about this!
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I dont understand Q7, can any explain me please?
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Hi,
I can’t understand it please help me by details explain.
The network default gateway applying to a host by DHCP is 192.168.5.33/28. Which option is the
valid IP address of this host?
A. 192.168.5.55
B. 192.168.5.47
C. 192.168.5.40
D. 192.168.5.32
E. 192.168.5.14
Answer: C
Hi,
I can’t understand it please help me by details explain.
The network default gateway applying to a host by DHCP is 192.168.5.33/28. Which option is the
valid IP address of this host?
A. 192.168.5.55
B. 192.168.5.47
C. 192.168.5.40
D. 192.168.5.32
E. 192.168.5.14
Answer: C
Hi,
I can’t understand it please help me by details explain.
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
@ Md. Rezaul Islam…..
They are asking about gateway 192.168.5.33/28 which means the network is 192.168.5.32/28 because it is a multiple of 16. (16 * 2 = 32) . add 32+16 = 48 hosts can be accommodated in this network. So any host lying between 33-47 will be a part of this network. .47 will be a broadcast address?
isnt for
isnt for Q5, the answer is d ?
even the subnet is .240 and not 248.
do explain if im wrong, thanks much.
@anush…
Question 5 stated that the network is 192.168.20.24/29 .
1) /29 means 255.255.255.248
2) This mask tells you that the block is 8 because (256-248=8), so the range of addresses will be between 24 and 32.
3) Go back to the address 192.168.20.24 and apply what you have found . the first usable address is 192.168.20.25 /29 which will be assigned to the router (the gateway)
4) The last usable address in the range is 192.168.20.30 will be assigned to the server.
Hope that will clear why C is the correct answer.
Q8 and Q9
Yesterday
My accoutn is locked and it says to contact web server administrator. anyone know how or hwom to contact that can unlock an account?
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can you explain q3 please brief