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CCNA Exploration2: Routing Protocols and Concepts – Chapter 6 Exam
01. What two advantages does CIDR provide to a network? (Choose two.)
reduced routing table size
dynamic address assignment
automatic route redistribution
reduced routing update traffic
automatic summarization at classful boundaries
02. 
Refer to the exhibit. The network administrator wants to create a subnet for the point-to-point connection between the two routers. Which subnetwork mask would provide enough addresses for the point-to-point link with the least number of wasted addresses?
255.255.255.192
255.255.255.224
255.255.255.240
255.255.255.248
255.255.255.252
03. 
Refer to the exhibit. A network engineer is summarizing the two groups of routes on router R1 shown in the exhibit. Which summarization will work for all the subnets?
192.168.0.0/23
192.168.0.0/22
192.168.0.0/21
192.168.0.0/20
04. Which of the following are contained in the routing updates of classless routing protocols? (Choose two.)
32-bit address
next hop router interface
subnet mask
unicast host address
Layer 2 address
05. Which of the following problems does VLSM help to alleviate?
the shortage of IP addresses
the difficulty of assigning static IP addresses to hosts in large enterprises
the complexity of implementing advanced routing protocols such as OSPF and EIGRP
the shortage of network administrators qualified in the use of RIP v1 and IGRP
06. What does VLSM allow a network administrator to do?
utilize one subnet mask throughout an autonomous system
utilize multiple subnet masks in the same IP address space
utilize IGRP as the routing protocol in an entire autonomous system
utilize multiple routing protocols within an autonomous system
07. 
Refer to the exhibit. What subnet mask will be applied by router B when it receives a RIPv1 update for the network 172.16.1.0?
none
8
16
24
08. 
Refer to the exhibit. The network administrator wants to minimize the number of entries in Router1’s routing table. What should the administrator implement on the network?
VLSM
CIDR
private IP addresses
classful routing
09. A router has a summary route to network 192.168.32.0/20 installed in its routing table. What range of networks are summarized by this route?
192.168.0.0 – 192.168.32.0/24
192.168.0.0 – 192.168.47.0/24
192.168.32.0 – 192.168.47.0/24
192.168.32.0 – 192.168.48.0/24
192.168.32.0 – 192.168.63.0/24
10. A network administrator is tasked with dividing up a class C network among the QA, Sales, and Administration departments. The QA department is made up of 10 people, the Sales is made up of 28 people, and the Administration has 6. Which two subnets masks adequately address the QA and Sales departments? (Choose two.)
255.255.255.252 for QA
255.255.255.224 for Sales
255.255.255.240 for QA
255.255.255.248 for QA
255.255.255.0 for Sales
11. 
In the network shown in the graphic, three bits were borrowed from the host portion of a Class C address. How many valid host addresses will be unused on the three point-to-point links combined if VLSM is not used?
3
4
12
36
84
180
12. 
A Class C address has been assigned for use in the network shown in the graphic. Using VLSM, which bit mask should be used to provide for the number of host addresses required on Router A, while wasting the fewest addresses?
/31
/30
/29
/28
/27
/26
13. 
An additional subnet is required for a new Ethernet link between Router1 and Router2 as shown in the diagram. Which of the following subnet addresses can be configured in this network to provide a maximum of 14 useable addresses for this link while wasting the fewest addresses?
192.1.1.16/26
192.1.1.96/28
192.1.1.160/28
192.1.1.196/27
192.1.1.224/28
192.1.1.240/28
14. Which three interior routing protocols support VLSM? (Choose three.)
OSPF
RIP v1
RIP v2
EIGRP
BGP
STP
15. 
Refer to the exhibit. The number of required host addresses for each subnet in a network is listed in the exhibit. This number includes the host address requirements for all router ports and hosts on that subnet. After all device and router port address assignments are determined, what will be the total number of unused host addresses available?
6
14
29
34
40
62
16. 
Refer to the exhibit. In the network that is shown, the router interfaces are assigned the first address in each subnet. Which IP address would be usable for a host on one of the LANs in this network?
192.168.1.5/30
192.168.2.17/28
192.168.2.63/27
192.168.2.130/25
17. 
Refer to the exhibit. Which address is a broadcast address for one of the subnets that are shown in the exhibit?
192.168.4.3/29
192.168.4.15/29
192.168.4.65/26
192.168.4.255/24
18. 
Refer to the exhibit. A network administrator needs to create two subnetworks from 10.0.0.0/8 for a router running RIPv2. The Admin subnet requires 120 hosts and the Sales subnet requires 58 hosts. The network administrator assigned 10.0.1.128/25 to the Admin subnet. The Sales subnet is given 10.0.1.192/26. What will be the result of this addressing scheme?
Because RIPv2 does not support VLSM, the subnet masks will not be allowed.
The subnets will not have enough host addresses for the given network requirements.
The subnets overlap and will be rejected by the router.
The router will support the addressing scheme.
19. 
Refer to the exhibit. A network technician enters the static route in R1 needed to reach network 10.1.1.0/24. A ping from R1 to Host B fails. The technician begins testing the network and has the following results:
1. pings from R1 to the S0/0/0 interface on R2....successful 2. pings from R1 to the Fa0/0 interface on R2....successful 3. pings from Host B to hosts on the 10.1.1.0/24 network....successful 4. pings from Host B to the Fa0/0 interface on R2....successful 5. pings from R2 to Host B....successful
What is the likely cause of the failure of the ping from R1 to Host B?
Host B has a defective Ethernet card.
The default gateway on Host B is not correctly set.
There is a Layer 2 problem between R2 and Host B.
R2 does not have routes back to networks connected to R1.
20. What is a supernet?
the network for a default route
a network that contains both private and public addresses
a set of discontiguous networks that are controlled by an ISP
a summarization of serveral IP classful networks into one IP address range
Related posts:
- CCNA Exploration2: Routing Protocols and Concepts – Chapter 7 Exam
- CCNA Exploration2: Routing Protocols and Concepts – Chapter 5 Exam
- CCNA Exploration2: Routing Protocols and Concepts – Chapter 8 Exam
- CCNA Exploration2: Routing Protocols and Concepts – Chapter 11 Exam
- CCNA Exploration2: Routing Protocols and Concepts – Chapter 4 Exam
- CCNA Exploration2: Routing Protocols and Concepts – Chapter 10 Exam
| Print article | This entry was posted by Nathan on 20/04/2010 at 8:38 pm, and is filed under CCNA, Cisco, Routing Protocols and Concepts. Follow any responses to this post through RSS 2.0. You can leave a response or trackback from your own site. |
about 1 year ago
hey could someone explain Q.19 for me please
about 1 year ago
You need to understand the concept of Default gateway then the answer will make sense.
http://en.wikipedia.org/wiki/Default_gateway
Explanation:
1. According to “pings from R2 to Host B….successful”
you can take out the 1st option “Host B has a defective Ethernet card”
2. according to ” pings from R2 to Host B….successful” there is no physical connectivity issues in here since the ping was successful, No layer two problem.
3. according to “1. pings from R1 to the S0/0/0 interface on R2….successful
2. pings from R1 to the Fa0/0 interface on R2….successful
” you can take out the last option “R2 does not have routes back to networks connected to R1″.
At the end you left with the only option that make sense, since the default gateway miscofigured the ICMP echo request packets “ping” won’t get back to the router R1.. you can test this in real world by examining network traffic on the host, you will notice an ICMP packets arriving at NIC card but with no response from the host, or you may get a response destined to another gateway which maybe or maybe not be exists.
about 1 year ago
Hi,
can someone explain no. 3, 16, and 17. I got confuse when the SM of the IP that we want to advertise is not the same (e.g. /26, /28, /29)…
Appreciate if you can explain the concept.
Thanks
about 1 year ago
no. 17 the answer is right 192.168.4.15/29, bec. the question ask for a broadcast address of one of the subnet which is 192.168.4.16/28
it is /29 bec.
0 0 0 0 1 1 1 1=8 host 192.168.4.8 – 192.168.4.14 192.168.4.15 is broadcast address.
about 1 year ago
any updates?
about 1 year ago
No 3 : Put all IP addr in binary one beneth the other and start looking from left to right to a colomn that doesnt have the same bit. When you reach that colomn stop. Now count the number of bits from left to that colomn and you will have the Subnet Mask.
about 1 year ago
hi my name harri
i think Q no 11 the answer is 12, coz they count by /29 not /27..
thanks…
about 11 months ago
#15 please explain^^
about 11 months ago
You gotta have some basic knowledge of VLSM for this one.
They give you 4 subnets, each one with it’s own host requeriment and ip address/mask. All you have to do is do the subnet math according to the ip address and mask given in order to provide the required number of hosts and see if there’s any unused host address left for each case.
Net A: 10.1.1.0 /29, that means you have a 255.255.255.248 mask (5 active bits from left to right in the host octet). doing to math for host number: 2 power of 3 less 2 = 6 usable hosts. your requirement is 6, that means there’s 0 host address left for net a.
Net B: 10.1.1.32/28, that means you have a 255.255.255.240 mask (4 active bits from left to right in the host octet). doing the math for host number: 2 power of 4 less 2 = 14 usable hosts. your requirement is 14, that means there’s 0 host address left for net b.
Net C: 10.1.1.64/26, that means you have a 255.255.255.192 mask (2 active bits from left to right in the host octet). doing the math for host number: 2 power of 6 less 2 = 62 usable hosts. your requirement is 33, that gives you 29 usable hosts addresses left.
Net D: 10.1.1.128/25, that means you have a 255.255.255.128 mask (1 active bit form left to right in the host octet. doing the math for host number: 2 power of 7 less 2 = 126 usable hosts. your requirement is 126, the means you have 0 host addresses left.
Now just add: 0+0+29+0 = 29 usable host addresses left in total. there you got your answer.
Admin will say if everything is ok with this one, but i guess it is.
about 11 months ago
no 19 too..it say.. 4. pings from Host B to the Fa0/0 interface on R2….successful
so it means it can ping its gateway
about 11 months ago
The fact that the fa0/0 is the gateway for host B, doesn’t mean that the gateway address on host B is correct. Remember that host B and Fa0/0 are directly connected by an ethernet cable and so host B doesn’t use the gateway to contact the fa0/0.
Host B only uses the gateway when it has to foward packages out of it’s network (it’s wired network), that’s why when router 1 pings host B, the data doesn’t get back to router 1, because the gateway is incorrect and host B doesn’t know by which interface foward that data out.
about 11 months ago
could somebody please explain question 11 to me. I would never have guess the correct answer to that
about 11 months ago
First of all DO NOT mix the subnetting in the example with VLSM subnetting.
Class C subnetting means you are going to subnet the last octet of the IPv4 address (using the very last 8 bits of the address).
The Q telling you that we are going to use only 3 bits of the host portion of a class C address for subnetting which will give us a subnet mask of
255.255.255.224 = /27
Using the formula 2 to the power of (n: number of bits used for the networks subnetting) -2 = (2 to the power of 3 -2 ) = 6 subnets
Again using the same forumla to calculate the number of hosts in each subnet = 2 to the power of (5: number of bits reserved for hosts) – 2 = 30 hosts per subnet.
In the Q there are three point to point links which mean we need only three subnets out of the six provided.
In each of these three subnets we are going to use only 2 addresses for each end of the link, that will waste 28 address in each subnet because we are not going to assign them to any host or router interfaces.
each subnet will wast 28 host address multiplied by 3 subnets = 84
28 + 28 + 28 = 84
about 10 months ago
Nathan, the concept of [(2^n)-2] for calculating the available number of subnets is now not valid. This was true with CCNA v3, but with v4 it’s become (2^n).
If the above was true then we wont have /25 /17 and /9 subnets.
Regards.
about 11 months ago
Help in number 16.
about 11 months ago
@ Nathan,i think that you mean at q.11 answer is 84?,bcs /27 gives you 30 hosts per subnet(2^5-2=30),in picture there are 3 subnets and bcs there are point-to-point links,with 3 routers(6 host addreses),3 subnets(with 30 hosts per subnet) are 90 hosts total,and bcs these routers have 6 host addresses(each router 2),the answer would be 90-6=84 host addresses wasted.
about 11 months ago
Indeed, it is 84 my mistake.. :)
fixed..
Thanks for that mate..
about 11 months ago
@ Nathan,but your method is 100% correct.
about 11 months ago
No prob. ;)