CCNA – Subnetting Questions 2
Here you will find answers to Subnetting Questions – Part 2
Question 1
Refer to the exhibit. Which VLSM mask will allow for the appropriate number of host addresses for Network A?
A. /25
B. /26
C. /27
D. /28
Answer: A
Explanation
We need 66 hosts < 128 = 27 -> We need 7 bits 0 -> The subnet mask should be 1111 1111.1111 1111.1111 1111.1000 0000 -> /25
Question 2
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 3
Refer to the exhibit. Which mask is correct to use for the WAN link between the routers that will provide connectivity while wasting the least amount of addresses?
A. /23
B. /24
C. /25
D. /30
Answer: D
Explanation
For WAN link we only need 2 usable host addresses for 2 interfaces on the routers. The subnet mask of /30 gives us 22 – 2 = 2 usable host addresses. Also remember that “/30″ is famous for point-to-point connection because it wastes the least amount of addresses.
Question 4
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 5
On the network 131.1.123.0/27, what is the last IP address that can be assigned to a host?
A. 131.1.123.30
B. 131.1.123.31
C. 131.1.123.32
D. 131.1.123.33
Answer: A
Explanation
Increment: 32
Network address: 131.1.123.0 & 131.1.123.32
Broadcast address: 131.1.123.31
Both 131.1.123.30 & 131.1.123.33 can be assigned to host but the question asks about the “last IP address” so A is the correct answer.
Question 6
The ip subnet zero command is not configured on a router. What would be the IP address of Ethernet0/0 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
Explanation
The “ip subnet zero” is not configured so the first subnet will start at 192.168.8.8 (ignoring 192.168.8.0).
Increment: 8
1st subnet: 192.168.8.8
2nd subnet: 192.168.8.16
3rd subnet: 192.168.8.24
4th subnet: 192.168.8.32
5th subnet: 192.168.8.40
6th subnet: 192.168.8.48 -> The first usable IP address of 6th subnet is 192.168.8.49
Question 7
For the network 192.0.2.0/23, which option is a valid IP address that can be assigned to a host?
A. 192.0.2.0
B. 192.0.2.255
C. 192.0.3.255
D. 192.0.4.0
Answer: B
Explanation
Increment: 2
Network address: 192.0.2.0, 192.0.4.0
Broadcast address: 192.0.3.255
-> 192.0.2.255 is not a broadcast address, it is an usable IP address.
Question 8
How many addresses for hosts will the network 124.12.4.0/22 provide?
A. 510
B. 1022
C. 1024
D. 2048
Answer: B
Explanation
/22 gives us 10 bits 0 -> 210 – 2 = 1022. Notice that the formula to calculate the number of host is: 2k – 2.
Question 9
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
Question 10
Which two addresses can be assigned to a host with a subnet mask of 255.255.254.0? (Choose two)
A. 113.10.4.0
B. 186.54.3.0
C. 175.33.3.255
D. 26.35.2.255
E. 17.35.36.0
Answer: B D
Thanks MH.
Is the Mask 255.254.0.0 or 255.255.254.0 I am confused. Could u please explain me thanks.
Hi Mohamed,
the difference is that, 255.254.0.0. -> .254 is belong to 2nd octet and the it has /15 notation. while the 255.255.254.0 -> .254 is belong to 3rd octet which have a /23 notation.
So how do you find which IP could be assigned?
the mask 255.255.254.0 is for /23
so the network range will be for example 26.35.2.0-26.35.4.0
broadcast mask will be 26.35.3.255 and so the first host will be 26.35.2.255
remember 256-254 you get 2.. the subnets will be 0,2,4,6,8….
@mista, 26.35.2.255? is this the first usable ip.. i think wrong.
the first usable ip should be 26.35.2.1 ; usable ip are from 26.35.2.1 to 26.35.3.254
Q 10;
simple understanding for question 10. the question asked is 255.255.254.0. The play is in the thrid octet ’254′ which has 7 bits having an increment of 2…. like 128–64–32–16–8–4–2–1.
the bits taken from 128 to 2….so the magic number that i call the rightmost bit is 2. Now create the ranges as following..
increment 2:
beginning scope with the increment of 2.
0.0——-
2.0——-
4.0——
6.0——
8.0——-
The last IP is 0.255-1=254 usable ip i.e. x.x.x.254 likewise The last IP is 2.255-1=254 usable ip i.e. x.x.x.254 so we will write like this..
0.0——-1.255
2.0——-3.255
4.0——5.255
6.0——7.255
8.0——-9.255
36.0—–37.255
Now check from the given ip list which comes under these subnets. remember not to use subnet id or broadcast id as they are not assignable ips…i.e. for example for subnet 2.0——-3.255 2.1 >>>>>>2.1 —– 3.254 are usable IPs.
Question 4 in my exam 640-802
Can someone please send me the latest dumps to kimms12@gmail.com. Thank you in advance!!
A network administrator is configuring ACLs on a Cisco router, to allow IP access form the 192.168.146.0/24,192.168.147.0/24,192.168.148.0/2,. and 192.168.149.0/24 networks only. Which two ACLs, when combined, should be used.
A.access-list 10 permit ip 192.168.146.0 0.0.0.255
B.access-list 10 permit ip 192.168.146.0 255 255.255.0
C.access-list 10 permit ip 192.168.147.0 0.0.255 255
D.access-list 10 permit ip 192.168.149.0 0.0.255.255.0
E.access-list 10 permit ip 192.168.148.0 0.0.1.255
F.access-list 10 permit ip 192.168.146.0 0.0.1.255
answer :E F
can anybody explain to me this one? Thanks!
@Yaser
I dont think answer is E and F because /24 will not have wildcard mask of 0.0.1.255
Wild card mask of /24 is 0.0.0.255
@Yaser
Wildcard of 0.0.1.255 = subnet of 255.255.254.0 = CIDR /23 which covers 2 times /24 ranges.
Only necessary because you are being limited to 2 lines in the ACL instead of 4.
Can check with binary calculation: host address AND subnet mask = network address
e.g. 147 = 10010011, 254 = 11111110, do an AND operation and you get 146 i.e. 192.168.146.0 so you see you are covered for 192.168.146.0-192.168.147.255
PLS SOME EXPLAIN Q10
@Raj
Question 10.
A. 113.10.4.0 (Its a network number whose broadcast address is 113.10.7.255. So, it cant be assigned as host)
B. 186.54.3.0(This is a valid host whose network number is 186.54.0.0 and broadcast is 186.54.3.255)
C. 175.33.3.255(It’s not a valid host because it’s a broadcast number for the network 175.33.0.0)
D. 26.35.2.255(It’s a valid host whose network number is 26.35.0.0 and broadcast is 26.35.3.255)
E. 17.35.36.0(It’s not a valid host because it’s a network number whose broadcast is 17.35.39.255)
I hope this will help.
Please correct me if I am wrong somewhere.
PLS SOME EXPLAIN Q-9
Q-9 solved
Can some send the latest dumbs on below gven address:
kumar_vikash@hotmail.com
Thanks
@ yaser
Those are the correct answers.
E. Access-list 10 permit ip 192.168.148.0 0.0.1.255
Lets change the wild card mask to a subnet mask.
0.0.1.255 is the same as 255.255.254.0 = (510 host)
This ACL covers 192.168.148.1 – 192.168.149.255
Meaning it cover both 192.168.148.0 & 192.168.149.0 networks.
——————————————————————————————-
F. Access-list 10 permit ip 192.168.146.0 0.0.1.255
Lets change the wild card mask to a subnet mask.
0.0.1.255 is the same as 255.255.254.0 = (510 host)
This ACL covers 192.168.146.1 – 192.168.147.255
Meaning it cover both 192.168.146.0 & 192.168.147.0 networks.
——————————————————————————————-
Using both these two ACL will cover all 4 networks.
Can someone please send me the latest dumps to sg980321@hotmail.com. Thank you in advance!!
so easy…
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