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.
Hi all!
Thank you for the mms.
Please answer or clear my thought?
On the net work 131.1.123.0/ 27 What is the last Ip address that can be assigned to a host?
1.131.1.123.30
2.131.1.123.31
3.131.1.123.32
4.131.1.123.33
Answer is 1.
my answer is 4. ??
examinar do not mention any subnets, he asked boldly last ip address is what?
So my answer is 4. ( last, one he provided very last one.
If he mention find the last from the 1st subnet, answer 1. is right.)
answer 2 and 3 are bcast and Nw respectively.
If anybody answer my confusion please?
@Anonymous
Hi all!
Thank you for the mms.
Please answer or clear my thought?
On the net work 131.1.123.0/ 27 What is the last Ip address that can be assigned to a host?
1.131.1.123.30
2.131.1.123.31
3.131.1.123.32
4.131.1.123.33
Answer is 1.
my answer is 4. ??
examinar do not mention any subnets, he asked boldly last ip address is what?
So my answer is 4. ( last, one he provided very last one.
If he mention find the last from the 1st subnet, answer 1. is right.)
answer 2 and 3 are bcast and Nw respectively.
If anybody answer my confusion please?
_________
The network 131.1.123.0 / 27 has a range from 131.1.123.0 to 131.1.123.31 ;
1/ 131.1.123.30 (last host address)
2/ 131.1.123.31 (broadcast address)
3/ 131.1.123.32 (out of range)
4/ 131.1.123.33 (out of range)
How do u divide 202.84.208/21 into 4 equal subnets?
@kumar
How do u divide 202.84.208/21 into 4 equal subnets?
__________
/21 : 11111111 . 11111111 . 11111 00 0.00000000
/23 : 11111111 . 11111111 . 11111 11 0.00000000
202.84.208.0 /21 = 11001010 . 01010100 . 11010 00 0.00000000
To generate 4 (/23) subnets from the /21 network, we need to borrow 2 bits in the host address field.
The 4 (/23) subnets are:
1/ 202.84.208.0 /23 = 11001010 . 01010100 . 11010 00 0.00000000 /23
2/ 202.84.210.0 /23 = 11001010 . 01010100 . 11010 01 0.00000000 /23
3/ 202.84.212.0 /23 = 11001010 . 01010100 . 11010 10 0.00000000 /23
4/ 202.84.214.0 /23 = 11001010 . 01010100 . 11010 11 0.00000000 /23
@tushar14029@!yahoo.co.in
1) Which two addresses can be assigned to host with in subnet mask of 255.255.254.0 ?
a) 113.10.4.0
b) 186.54.3.0
c).175.33.3.255
d) 26.35.2.255
___________________
255.255.254.0 = 11111111.11111111.1111111 0.00000000
a) 113.10.4.0 . . . . . = X.X.0000010 0.00000000 (network address)
b) 186.54.3.0 . . . . . = X.X.0000001 1.00000000 (host address)
c) 175.33.3.255 . . . = X.X.0000001 1.11111111 (broadcast address)
d) 26.35.2.255 . . . . = X.X.0000001 0.11111111 (host address)
wk A with ip 192.0.2.24/28 and wk B with the ip 192.0.2.100/28 are connected with straight-through cable, the ping between the 2 host are unsuccessful what 2 thing can be done to allow communication between A and B
1-use a cross cable
2- change the subnet mask of the hosts to /25
3-change the subnet mask of the hosts to /26
i don’t understand why subnet mask must be changed from 28 to 25 ???
please help!!!
hi friend,
this website is one of the best website that i have ever seen. always update and great explanation. thanks so much.
@ bouhdila
/28 mask means 255.255.255.240
240 gives a network block size of 256-240=16 in the last (4th) octet.
So the networks would be:
Network ID (1st Node – Last Node) Broadcast ID
192.0.2.0 (2.1 – 2.14) 192.0.2.15
192.0.2.16 (2.17 – 2.30) 192.0.2.31
192.0.2.32 (2.33 – 2.46) 192.0.2.47
192.0.2.48 (2.49 – 2.62) 192.0.2.63
192.0.2.64 (2.65 – 2.78) 192.0.2.79
192.0.2.80 (2.81 – 2.94) 192.0.2.95
192.0.2.96 (2.97 – 2.110) 192.0.2.111
192.0.2.112 (2.113 – 2.126) 192.0.2.127
192.0.2.128 (2.129 – 2.142) 192.0.2.143
192.0.2.144 (2.145 – 2.158) 192.0.2.159
192.0.2.160 (2.161 – 2.174) 192.0.2.175
192.0.2.176 (2.177 – 2.190) 192.0.2.191
192.0.2.192 (2.193 – 2.206) 192.0.2.207
192.0.2.208 (2.209 – 2.222) 192.0.2.223
192.0.2.224 (2.225 – 2.238) 192.0.2.239
192.0.2.240 (2.241 – 2.254) 192.0.2.255
Since they’re trying to connect both the PCs with a single cable with no router in between, we know that they have to be on the same LAN. But with IP address 192.0.2.24/28, station A falls in subnet # 2 while station B having IP address 192.0.2.100/28 belongs to the 7th subnet.
- – - – - – - – - – - – - – - – - – - – - – - -
/25 mask means 255.255.255.128
128 gives a network block size of 256-128=128 in the last (4th) octet.
So the networks would be:
Network ID (1st Node – Last Node) Broadcast ID
192.0.2.0 (2.1 – 2.126) 192.0.2.127
192.0.2.128 (2.129 – 2.254) 192.0.2.255
Now A and B with IP addresses of 192.0.2.24 and 192.0.2.100 respectively belong to the same subnet, the 1st one. So with a crossover cable in between them instead of a straight-through and a mask of /25, they are good to go.
- – - – - – - – - – - – - – - – - – - – - – - – - – - – - – - – - -
/26 would be wrong again because 255.255.255.192 means a block size of 64.
So the networks would be:
Network ID (1st Node – Last Node) Broadcast ID
192.0.2.0 (2.1 – 2.62) 192.0.2.63
192.0.2.64 (2.65 – 2.126) 192.0.2.127
192.0.2.128 (2.129 – 2.190) 192.0.2.191
192.0.2.192 (2.193 – 2.254) 192.0.2.255
With the addresses stated in the problem, A and B again separate into different subnets if a /26 mask is used.
Cheers
@ Madani
In response to kumar’s question about dividing 202.84.208/21 into 4 equal subnets, how did you figure out straightaway a /23 mask would work. What’s the trick? Can we use any other mask too or /23 is the only option? Could you plz elaborate further?
Thanks in advance
@UthmanHassan December 8th, 2010
-> @ Madani
In response to kumar’s question about dividing 202.84.208/21 into 4 equal subnets, how did you figure out straightaway a /23 mask would work. What’s the trick? Can we use any other mask too or /23 is the only option? Could you plz elaborate further?
Thanks in advance
_________
/21 :. . . . . . . . . . . . . . . . . . . . . 11111111 . 11111111 . 11111 000.00000000
202.84.208.0 /21 = . . . . . . . . . 11001010 . 01010100 . 11010 000.00000000
Size : 2^11 = 2048 ip addresses for this (/21) network
Range : from 202.84.208.0 . . . 11001010 . 01010100 . 11010 000.00000000
. . . .. . . . . to 202.84.215.255 . 11001010 . 01010100 . 11010 111.11111111
By borrowing 1 bit we obtain 2 (/22) networks
/22 : . . . . . . . . . . . . .. . . . . . . . 11111111 . 11111111 . 11111 1 00.00000000
Size : 2^10 = 1024 ip addresses for each (/22) network
The 2 (/22) subnets are :
1/ 202.84.208.0 /22 = . . . . . . . 11001010 . 01010100 . 11010 0 00.00000000 /22
Range : from 202.84.208.0 . .. . 11001010 . 01010100 . 11010 0 00.00000000
. . . . . . . . . to 202.84.211.255 . 11001010 . 01010100 . 11010 0 11.11111111
2/ 202.84.212.0 /22 = . . . . . . . 11001010 . 01010100 . 11010 1 00.00000000 /22
Range : from 202.84.212.0 . .. . 11001010 . 01010100 . 11010 1 00.00000000
. . . . . . . . . to 202.84.215.255 . 11001010 . 01010100 . 11010 1 11.11111111
By borrowing 2 bits we obtain 4 (/23) networks
/23 : . . . . . . . . . . . . . . . . .. . . . 11111111 . 11111111 . 11111 11 0.00000000
Size : 2^9 = 512 ip addresses for each (/23) network
The 4 (/23) subnets are:
1/ 202.84.208.0 /23 = . . . . . . . 11001010 . 01010100 . 11010 00 0.00000000 /23
Range : from 202.84.208.0 . .. . 11001010 . 01010100 . 11010 00 0.00000000
. . . . . . . . . to 202.84.209.255 . 11001010 . 01010100 . 11010 00 1.11111111
2/ 202.84.210.0 /23 = . . . . . . . 11001010 . 01010100 . 11010 01 0.00000000 /23
Range : from 202.84.210.0 . .. . 11001010 . 01010100 . 11010 01 0.00000000
. . . . . . . . . to 202.84.211.255 . 11001010 . 01010100 . 11010 01 1.11111111
3/ 202.84.212.0 /23 = . . . . . . . .11001010 . 01010100 . 11010 10 0.00000000 /23
Range : from 202.84.212.0 . .. . 1001010 . 01010100 . 11010 10 0.00000000
. . . . . . . . . to 202.84.213.255 . 11001010 . 01010100 . 11010 10 1.11111111
4/ 202.84.214.0 /23 = . . . . .. . . 11001010 . 01010100 . 11010 11 0.00000000 /23
Range : from 202.84.214.0 . .. . 11001010 . 01010100 . 11010 11 0.00000000
. . . . . . . . . to 202.84.215.255 . 11001010 . 01010100 . 11010 11 1.11111111
So, to generate 4 (/23) equal subnets from the initial (/21) network, we need to borrow 2 bits in the host address field.
As we can see, there is no other option. We must use the /23 mask.
bye
i got my exam yesterday and passed with 974. especial thanks for the 9tut. pass4sure & testinside dumps are still valid. sims are eigrp,vtp & access-list only the ip addresses are changed. in access-list the the pc also changed. again thanks for the 9tut and all of the contributors
if any one need help from me pls email to
hasy_001@yahoo.com
Which command would allow the translations to be created on the router?
*************************************************************************Pro Inside global Inside local outside local Outside global
— 1.1.128.1 10.18.14.90 — —
— 1.1.129.107 10.18.14.91 — —
— 1.1.130.178 10.18.14.92 — —
— 1.1.131.177 10.18.14.89 — —
— 1.1.132.171 10.10.16.204 — —
— 1.1.133.172 10.10.24.210 — —
— 1.1.134.173 10.10.24.216 — —
— 1.1.135.168 10.19.16.95 — —
— 1.1.134.169 10.19.16.96 — —
— 1.1.130.170 10.20.122.234 — —
— 1.1.135.174 10.20.122.240 — —
*************************************************************************
A. ip nat pool mynats 1.1.128.1 1.1.135.254 prefix-length 19
B. ip nat outside mynats 1.1.128.1 1.1.135.254 prefix-length 19
C. ip nat pool mynats 1.1.128.1 1.1.135.254 prefix-length 18
D. ip nat outside mynats 1.1.128.1 1.1.135.254 prefix-length 18
Answer:A
Why is A? Why is prefix-length 19???
@ Oliver
Which command would allow the translations to be created on the router?
Inside global Inside local outside local Outside global
1.1.128.1 10.18.14.90
1.1.129.107 10.18.14.91
1.1.130.178 10.18.14.92
1.1.131.177 10.18.14.89
1.1.132.171 10.10.16.204
1.1.133.172 10.10.24.210
1.1.134.173 10.10.24.216
1.1.135.168 10.19.16.95
1.1.134.169 10.19.16.96
1.1.130.170 10.20.122.234
1.1.135.174 10.20.122.240
A. ip nat pool mynats 1.1.128.1 1.1.135.254 prefix-length 19
B. ip nat outside mynats 1.1.128.1 1.1.135.254 prefix-length 19
C. ip nat pool mynats 1.1.128.1 1.1.135.254 prefix-length 18
D. ip nat outside mynats 1.1.128.1 1.1.135.254 prefix-length 18
Answer:A
Why is A? Why is prefix-length 19???
———————————————-
The question shows that we are at the step of defining a pool of inside global addresses. we can see that 1st two octects are the same for all the public address we’ve got. Numbers start changing in the 3rd octect. So we need to find out how many matching bits we can find in the 3rd octect after which thing change. We need to look for THE MAXIMUM NUMBER OF COMMON bits in the 3rd octect.
The list can be re-written with 3rd octect in binary form as:
1.1.10000000.1
1.1.10000001.107
1.1.10000010.178
1.1.10000011.177
1.1.10000100.171
1.1.10000101.172
1.1.10000110.173
1.1.10000111.168
1.1.10000110.169
1.1.10000010.170
1.1.10000111.174
Now 8 bits from 1st octect + 8 from 2nd + 5 bits from the third match i.e., 21 bits after which from 22nd bit the variation starts. So a pool could’ve been defined using:
1.1.128.1 1.1.135.254 prefix-length 21
But that’s not an option given in the answers. We’ve to choose from 18 or 19. Which one is THE LONGEST MATCH?
/19
—————-
@ Madani
Thanks mate. Thanks a lot.
plz anyone explain to me why is it so?
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?
Answer is 16 Subnets, 30 hosts.
@ Rayan
/23 = 11111111.11111111.11111110.00000000
/27 = 11111111.11111111.11111111.11100000 (note how many more network bits here)
You have 4 more bit (1s) for the network. Number of subnets = 2 to the power of y where y is the the number of network bit you gained. Thus it’s 16.
Number of host = 2 to the power of x ( x= number of host bits) – 2 = 2 to the power of 5 – 2 = 30.
Hope this help.
thanku hung..
hi ther..
the network default gateway applying to host by DHCP is 192.168.5.33/28. which option is valid ip address 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
Ans is
c)192.168.5.40
plz explain it..
thanks in adv. -:)
and one more thing
what is subnet-zero..?? plz explain..
thanks
@ ryan
hi ther..
the network default gateway applying to host by DHCP is 192.168.5.33/28. which option is valid ip address 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
Ans is
c)192.168.5.40
plz explain it..
thanks in adv. -:)
and one more thing
what is subnet-zero..?? plz explain..
thanks
———————————————
Gateway’s address is 192.168.5.33/28
/28 means a subnet mask of 255.255.255.240
So Gateway is:
192.168.5.33
255.255.255.240
Network block size is 256-240=16 in the last octet. So networks are:
192.168.5.0
192.168.5.16
192.168.5.32
192.168.5.48
192.168.5.64 and so on
Gateway has the 1st assignable IP from the 192.168.5.32 network. Since a host and its default gateway have to be on same subnet, 192.168.5.40 is the only correct option.
————
In the days before CIDR, 1st and last subnets were reserved and not usable.
In the above example, what’s the 1st usable subnet?
On old routers that do not support the subnet zero, 1st usable subnet is 192.168.5.16
On new routers that support using the subnet zero, the 1st usable subnet is 192.168.5.0 meaning you can actually assign addresses from this subnet too i.e., 5.1, 5.2, 5.3 and so on.
The command to enable and disable subnet zero are:
ip subnet-zero (enabled by default)
no ip subnet-zero
respectively.
———————————————
I’ll email you back once I’m done with uploading the dumps. I appreciate your patience.
Thanks and good luck with the exam
thanks utham..
hey hi again..
plz explain abt summarization in ospf..
can any one explain about the question 3
from where i could get those mashti and acme dumps? plz help me
@anirban
We have a Class B address so the default subnet is: 255.255.0.0
and this implies that we have only the last 2 bytes tow make our design, Now from subnets view:
We need 300 subnets so we should use 9 bit or above:
2^8 = 256 –> Not applicable
2^9 = 512 –> yep it is > 300
2^10 = 1024 –> ohhhh nice it is >> 300
so either 9 bits:
11111111.11111111.11111111.10000000 : 255.255.255.128
or 10 bits:
11111111.11111111.11111111.11000000 : 255.255.255.192
are applicable!
——
Now let’s solve it from Hosts view:
We need 50 hosts so, How many bits:
2^5 = 32 —> Nooo we need more
2^6 = 64 —> Good it is > 50
2^7 = 128 —> nice it is >> 50
so at the same manner(hosts’ bits 0s, other bits 1s):
so either 7 bits should be Zero:
11111111.11111111.11111111.10000000 : 255.255.255.128
or 6 bits:
11111111.11111111.11111111.11000000 : 255.255.255.192
——
So u can either solve it from Hosts view or Subnets view.
nice explainations guys
Refer to the exhibit. According to the routing table, where will the router send the packet destined for 10.1.5.65 ?
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
A. 10.1.1.2
B. 10.1.2.2
C. 10.1.3.3
D. 10.1.4.4
Ans. C
email: bouhdila80@hotmail.com please helpppppppppppppppp?
How CCNA can be clear easily. MY paper will be on 31st Jan 2011……………………
plz explain me how it is possible…
Q.Which IP address can be assigned to an Internet interface?
A. 10.180.48.224
B. 9.255.255.10
C. 192.168.20.223
ans=B
Kroute plz look up which IP address ranges are restricted, that is how you know the answer. 10 and 192.168 are restricted, but there are others too.
thanks yousef.
my paper will be on 13 jan 2011…….so help me how i prepare and emphasize on what?
Q. Which mask is correct to use for the WAN link between the routers that will provide connectivity while wasting the least amount of addresses?netwok-A=60 host,network-B=310host
A. /23
B. /24
C. /25
D. /30
Answer: D
so plz briefly explain how it will?????
Qn2.
Q.n.2.plz explin how??????
The ip subnet zero command is not configured on a router. What would be the IP address of Ethernet
0/0 using the first available address from the sixth 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
Answer: C
but how????????????
to kroute:
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
So just only 9.255.255.10 is internet IP.
The ip subnet zero command is not configured on a router. What would be the IP address of Ethernet
0/0 using the first available address from the sixth subnet of the network 192.168.8.0/29?
dear kroute: C is right cause /29 u will have magic number is 8 and the network are:
net 0 : 192.168.8.0
net 1 :192.168.8.8
net 2 :192.168.8.16
net3 : 192.168.8.24
net 4:192.168.8.32
net 5: 192.168.8.40
net 6: 192.168.8.48
first IP :192.168.8.49
second ip:192.168.8.50
…………………………….
…………………..
broadcast ip: 192.168.8.55
Hope u get it now
Q. Which mask is correct to use for the WAN link between the routers that will provide connectivity while wasting the least amount of addresses?netwok-A=60 host,network-B=310host
i think /30 is corect. u just need 2 IP for the interfaces between 2routers so….
I have a question and I don’t know where to start:
Which of the following IP addresses can be assigned to the host dvices (Chose two)
205.7.8.32/27
191.168.10.2/23
127.0.0.1 ( obvious not )
224.0.0.10
203.123.45.47/28
10.10.0.0./13
Without doing each IP address to see if it is broadcast how am I best to figure this out??
I am so confuse! Plz help!
I have a question and I don’t know where to start:
Which of the following IP addresses can be assigned to the host dvices (Chose two)
205.7.8.32/27
191.168.10.2/23
127.0.0.1 ( obvious not )
224.0.0.10
203.123.45.47/28
10.10.0.0./13
Without doing each IP address to see if it is broadcast how am I best to figure this out??
I am so confuse! Plz help!
……………………………………………………………………………………………………………………….
i think everything is based on magic number and we have to know the Class of Ip address
*127.0.0.0 and 224.0.0.0 is not usable
*205.7.8.32/27 32-27=5 Magic number is 2^5=32-> this is the network not an usable IP
*191.168.10.2 /23 24-23 =1 Magic number is 2^1=2 –> net :192.168.5.0 and 7.0 and 9.0 and 11.0 —> 192.168.10.2 is usable Ip address
*203.123.45.47/28 32-23=4 Magic Number is 2^4=16 —> this is broacast Ip
*10.10.0.0/13 is usable
The answer is B F
nice explainations….
Hello Hoang,
Thanks for being quick about my question but if you please your final solution
10.10.0.0/13
This would be 8-13?
or 2^-5?
Also
191.168.10.2 /23
24-23 = 1
Magic number 2^1 =2
Therfore 191.168.0
.2
.4
Unless magic number is diff than incrementing value.
Thanks a lot ……Hoang Nguyen
but in this question still i m so confused plz explain breifly….
Q. Which mask is correct to use for the WAN link between the routers that will provide connectivity while wasting the least amount of addresses?netwok-A=60 host,network-B=310host
i think /30 is corect. u just need 2 IP for the interfaces between 2routers so….
how
plz tell ……..
Q.The Ethernet networks connected to router R1 in the graphic have been summarized for router R2 as 192.1.144.0/20. Which of the following packet destination addresses will R2 forward to R1, according to this summary? (Choose two.)
A. 192.1.159.2
A. 192.1.159.2
B. 192.1.160.11
C. 192.1.138.41
D. 192.1.151.254
E. 192.1.143.145
F. 192.1.1.144
Answer: AD .but how plz tell me any body
can i get crack for visual cert exam manager keygen please
give me different ospf and eirgp
Please answer this question…
Refer to the exhibit.According to the routing table, where will the router send a packet destined for 10.1.5.65
===============
Exhibit :
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
Options :
===============
A)10.1.1.2
B)10.1.2.2
C)10.1.3.3
D)10.1.4.4
==============
Please help me solve this
The question stated that the network 172.25.0.0 is divided into eight equal subnets therefore the increment is 256 / 8 = 32
172 mans class B, so 255.255.224.0(block add 256-224=32)
that means /19 = 225.225.224.0
so we got /19
Refer to the exhibt. The networks connected to router R2 have been summarized as 192.168.176.0/21 route and sent to R1. which two packet destination addresses will R1 forward to R2 ? (Choose two).
A. 192.168.194.160
B. 192.168.183.41
C. 192.168.159.2
D. 192.168.183.255
E. 192.168.179.4
F. 192.168.184.45
Correct Answer is B,E
Can anyone explain it to me. i am new in subnetting.
regards…
This blog seems to get a large ammount of visitors. How do you get traffic to it? It gives a nice unique spin on things. I guess having something authentic or substantial to post about is the most important factor.
The ip subnet zero command is not configured on a router. What would be the IP address of Ethernet
0/0 using the first available address from the sixth 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
Answer: C
kindly explain
response to Zulfiqar
if the ip subnet-zero command is not used this means you cant use the first and last subnet since /29 is a block size of 8 we start counting from 8 without the 0 subnet
so
the rabge will be 8 16 24 32 40 48 the 48 subnet is the sixth subnet its range is 49-55 and the first available address is 192.168.8.49
@Butt G January 15th, 2011
Refer to the exhibit. The networks connected to router R2 have been summarized as 192.168.176.0/21 route and sent to R1. Which two packet destination addresses will R1 forward to R2 ? (Choose two).
A. 192.168.194.160
B. 192.168.183.41
C. 192.168.159.2
D. 192.168.183.255
E. 192.168.179.4
F. 192.168.184.45
Correct Answer is B,E
Can anyone explain it to me. i am new in subnetting.
regards…
______________________________
All packets with destination addresses in the range 192.168.176.0/21 (= from 192.168.176.0 to 192.168.183.255) will be forwarded from router R1 to R2.
The two addresses 192.168.176.0 (~network address) and 192.168.183.255 (~broadcast address) are usually not used for packet addresses.
A. 192.168.194.160 . . . (out of range)
B. 192.168.183.41 . . . . (OK)
C. 192.168.159.2 . . . . . (out of range)
D. 192.168.183.255 . . . (? …usually not used)
E. 192.168.179.4 . . . . . (OK)
F. 192.168.184.45 . . . . (out of range)
Bye