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CCNA – Drag and Drop 1September 17th, 2010 Go to comments

Here you will find answers to drag and drop Questions

Question 1:

A dental firm is redesigning the network that connects its three locations. The administrator gave the networking team192.168.164.0 to use for addressing the entire netwok. After subnetting the address, the team is ready to assign theaddresses. The administrator plans to configure ip subnet-zero and use RIP v2 as the routing protocol. As a member ofthe networking team, you must address the network and at the same time conserver unused addresses for future growth.With those goals in mind, drag the host addresses on the left to the correct router interface. Once of the routers is partiallyconfigured. Move your mouse over a router to view its configuration. Not all of the host addresses on the left arenecessary.

Answer:

Explanation:

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In short, we should start calculating from the biggest network (with 16 hosts) to the smallest one using the formula 2n – 2(n is the number of bits we need to borrow).Therefore:

16 hosts < 25 – 2 (we need to borrow 5 bits -> /27)

11 hosts < 24 – 2 (borrow 4 bits -> /28)

5 hosts < 23 – 2 (borrow 3 bits -> /29)

From the available ip addresses, we see that each of them has only one suitable solution (they are192.168.164.149/27,192.168.164.166/28 and 192.168.164.178/29)

The smallest network is the Floss S0/0 which only requires 2 hosts = 22 – 2 (need to borrow 2 bits ->/30). There are 2suitable answers: 192.168.164.189/30 and 192.168.164.188/30 but notice that 192.168.164.188/30 is the networkaddress so we can not use it (because 188 = 4 * 47) -> we have to choose 192.168.164.189 as the correct solution.

In fact, it is not the formal way to solve a VLSM question so I recommend you to review your CCNA book if you haven’tgrasped it well yet.

Question 2:

In order to complete a basic switch configuration, drag each switch IOS command on the left to its purpose on the right

Answer:

1) enable

2) configure terminal

3) hostname

4) Interface vlan 1

5) no shutdown



6) ip address

7) ip default-gateway

Question 3:

The Missouri branch office router is connected through its s0 interface to the Alabama Headquarters router s1 interface.The Alabama router has two LANs. Missouri users obtain Internet access through the Headquarters router. The networkinterfaces in the topology are addressed as follows: Missouri: e0 – 192.168.35.17/28; s0 – 192.168.35.33/28;Alabama: e0 – 192.168.35.49/28; e1 – 192.168.35.65/28; s1 – 192.168.35.34/28. The accounting server has theaddress of 192.168.35.66/28. Match the access list conditions on the left with the goals on the right. (Not all options onthe left are used.)

Answer:

1) deny ip 192.168.35.16 0.0.0.15 host 192.168.35.66

2) deny ip 192.168.35.55 0.0.0.0 host 192.168.35.66

3) permit ip 192.168.35.0 0.0.0.255 host 192.168.35.66

Explanation:

1) The wildcard mask of the command “deny ip 192.168.35.16 0.0.0.15 host 192.16.35.66″ is 0.0.0.15, which is equalto network mask of 255.255.255.240 = /28. So the access list will deny all traffic from network 192.168.35.16/28 fromaccessing host 192.16.35.66, which is the IP address of accounting server.

2) The command “deny ip 192.168.35.55 0.0.0.0 host 192.168.35.66″ will deny host 192.168.35.55, which is a userand belongs to interface e0 of Alabama router (192.168.35.49/28) from accessing accounting server.

3) Because there is an implicit “deny all” command at the end of each access list so the command “permit ip 192.168.35.00.0.0.255 host 192.168.35.66″ will only let network 192.168.35.0/24 access accounting server whilst prevent traffic fromother networks.

Question 4:

A host with the address of 192.168.125.34/27 needs to be denied access to all hosts outside its own subnet. Toaccomplish this, complete the command in brackets, [access-list 100 deny protocol address mask any], by draggingthe appropriate options on the left to their correct placeholders on the right.



Answer:

1) ip

2) 192.168.125.34

3) 0.0.0.0

Full command: access-list 100 deny ip 192.168.125.34 0.0.0.0

Question 5:

Drag and drop the network user application to the appropriate description of its primary use (not all options are used)

Answer:

1) web browser

2) instant message



3) e-mail

4) database

5) collaboration

Question 6:

This topology contains 3 routers and 1 switch. Complete the topology.

Drag the appropriate device icons to the labeled Device

Drag the appropriate connections to the locations labeled Connections.

Drag the appropriate IP addresses to the locations labeled IP address

(Hint: use the given host addresses and Main router information)

To remove a device or connection, drag it away from the topology.

Use information gathered from the Main router to complete the configuration of any additional routers. Nopasswords are required to access the Main router . The config terminal command has been disabled for the HQ router.The router does not require any configuration.

Configure each additional router with the following

Configure the interfaces with the correct IP address and enable the interfaces.

Set the password to allow console access to consolepw

Set the password to allow telnet access to telnetpw

Set the password to allow privilege mode access to privpw

Note: Because routes are not being added to the configurations, you will not be able to ping through the internetwork.

All devices have cable autosensing capabilities disabled.

All hosts are PC’s



Answer:



View full explanation of this question here

Comments

1. TunBKKOctober 7th, 2008

Hi! Question 6 appear in my exam today. I used this website but didn’t pay good attention to this paticular questionand I was almost paying the prcie for it. However, i did manage to get back on track and scored well. I didn’t scrolldown the questions and as a result didn’t see the Notes which say I won’t be able to ping between the host andcable sensing not there. As I am not able to test by pinging I started to nervous but luckily I found these lines andcontinued next questions. It is a lab question, not drag drop.

The question states it is 3 routers and a switch set up. Since Main router is already there, we need to drop tworouters and a switch.

It is obvious to start dragging a router and drop it on the right “device” box since there is serial connection. Then, ifyou look carefully, bottom “device” box has two interfaces which are Fa0/4 and Fa0/2. That fits perfectly for aswtich than a router. Thus, needless to say, router is a must for “device” box on the left side of Main router. Agree!!!

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(When you drop routers in the box, host connected for console session automatically show up)

Then you will need to start dropping cable connections between them. (1) Crossover cable for Host A to router’sFa0/1 interface.(2) Stright-through cable for switch’s Fa0/4 interface to Fa0/0 interface of routerFor connection between the routers, “show run” command on the Main router. Then you will see there is twointerface one is serial and another one is Fast Ethernet. This will be a lead for step 3 and 4.(3) Router you drop on the right side has serial interface and thus it will have to connect serial interface of Mainrouter. Drop serial cable here.(4) Router you drop on the left side has Fast Ethernet port and thus it will have to connect with Fast Ethernet portof Main router. Drop crossover cable here.(In the screenshot above, there is Fast Ethernet and Serial inteface given beside Main router. They will not be therein real exam. You need to use “show run” to find out.)

Then it is all about calculation of IP and subnet and drop the correct ones. I don’t think explanation is needed hereas IP address in the exam will not be the same. Moreover, you need to use the IPs found on the interfaces of themain router, which you can obtain by show run, for calculation.

After that, you will need to start console session of each routers and assigning the IP address which you havecalculated in the above step. Please do not forget, “”no shut” once after you have assign an address.

Then you will need to provide passwords for privillege mode, console session and telnet with the passwords givenin the qeustion on each router.

Finally, do not foget to use “copy run start” command on each router.

Hope this helps. please let me know if you need further explanation.

Cheers !!!

2. WRXedDecember 2nd, 2010

Can you used “show cdp neighbor” on the main router to find more details to complete this sim??

3. TunBKKDecember 2nd, 2010

WRXed, that was where I spent 50 minutes for this lab. show CDP neighbor command works between Router onthe Left side and Main router. However, it didn’t work with Router on the right. I was so exhausted with that andthought I was wrong. The first line in Notes says “Because routes are not being added to the configurations, you willnot be able to ping through the internetwork.” Then second lines in the Notes says “All devices have cableautosensing capabilities disabled.”. Means your pinging between router interface, “show cdp neighbor” are notguaranteed to work.

4. ii tr00f iiDecember 2nd, 2010

I tell you what. It feels GOOD to look at subnetting questions like question one and solve it in 1 minute for theranges. There was a part in there says “look at the router” but I knew I could just figure it out anyway and I knewthe WAN link there were only 2 options….the /30 subnet masks.



That’s the point, People. If you just MEMORIZE questions, then you are WORTHLESS. Know the concept, atthis point I am looking at these questions and I Know them and know them well without even seeing them. I don’tlike tossing 250 bucks out the window so sure I want to be safe, but I knew the material before coming here. Iactually found this site 2 weeks ago, but didn’t look at it until yesterday to see what’s on here.

People on here say “On the exam the ip add’s were different and the interfaces were not the same and bla bla blabla” BOO HOO! If you knew the concept, it wouldn’t matter at all! Read, breathe, love the material. Lammle 6thedition sybex CCNA, CBT nuggets, Train signal and …………….a REAL lab at home to practice with……..yesthat’s what I said. You can pass the CCNA without one, but I tell you what when you hook up everything andthings are NOT working, that’s when you learn. Packet tracer 5.1 is good and all, but I only use it for a QUICKconfig of something I want to implement on my real lab and just seeing if it works as it saves me time. But overall thereal lab helps!

Just my 2 cents, when you see a WAN link, you should know that /30 is PERFECT for those types of connections.Why? Well, since the increment is 4 you take away 2 which gives you 2 hosts. So right off the bat you can scratchoff those /30 addresses figure out which one to use by looking at router.

Secondly, you should be able to do this stuff in your head.powers of 27 6 5 4 3 2 1128 64 32 16 8 4 2 1

so you see that a /30 if you know its a sub mask of 255.255.255.0 just write out the zeros for all the areas.

0 0 0 0 0 0 0 01 1 1 1 1 1^ (just count to yourself, you KNOW that a /24 is how many bits? 24 bits right…..RIGHT!!! so count each 0 as25,26,27,28,29…..30 ahhhh HAH. )

7 6 5 4 3 2 1128 64 32 16 8 4 2 11 1 1 1 1 125 26 27 28 29 30

So then count powers of 2 going right to left you’ll see that the #2 above the 4 represents 4 hosts.

Now you have to subtract 1 for the network and 1 for the broadcast

(2^2) – 2 = 2

I count from the first bit after the subnet mask. If its a /24 then I know 25 is the first 0, 26 is the second 0, 27 is the3rd 0, 28 is the 4th….and on and on.

This will save you time. There are 1000′s of ways and too many to count, that’s my method and I can just look at itand know. Basically, you want to KNOW the increment cuz if you know that, then whatever is right next to it isyour power of 2 (minus 2 of course) which makes for speedy calculation.

Know subnetting, practice practice practice.

One last thing, always know what octect you are working with! That has smacked me in the ass a couple of times :)

So now, there is my 2.5 cents worth. Use your brain, or you are destined to fail and you will be worthless.



Later!

5. ZeeDecember 11th, 2010

on question 6, how do you get ip address for serial interface.. do we get chance to go to routers serial?thanks

6. sunnyDecember 13th, 2010

@ ii TROOF ii … lol ur right!!!

i liked ur comment there.. and i cud follow u till the end of ur comment easyly.. nice explaination :)

i memmorized that / 28 = 16 which is in the middle .. so then its easier in my mind to go right and left from /28 as therefrence point. :)

thus /29 would be 16/2 = 8 ofcourseand /27 would be 16*2 = 32/26 = 64/25 = 128

memorizing subnet mask with /slashes will help too!!!/25 = x.x.x.128/26 = x.x.x.128+64=192/27 = xxx.192+32=224/28 = xxx.224+16=240/29 = xxx.240+8=248/30 = xxx.248+4=252/31 = xxx.252+2=254/32= xxxx

cheers..

7. farhadDecember 14th, 2010

hi friends, in Question no 4 the ip address should be 192.168.125.32 not 192.168.125.34can u explain to me.thnx

8. farhadDecember 14th, 2010

hello friends, i want to know how long is ccna exam duration and how many question we are gonna get in the exam.

9. rayanDecember 15th, 2010

hi 9tut,in question whats wrong if i use 174 instead of 190 and 190 instead of 174.. plz help me.. m getting confuse…



10. rayanDecember 15th, 2010

hi 9tut,in question 6 whats wrong if i use 174 instead of 190 and 190 instead of 174.. plz help me.. m gettingconfuse

11. 9tutDecember 15th, 2010

Question 6 has been explained very well at: http://www.9tut.com/66-ccna-implementation-simplease read it.

12. atulDecember 26th, 2010

The wildcard mask of the command “deny ip 192.168.35.16 0.0.0.15 host 192.16.35.66″ is 0.0.0.15, which isequal to network mask of 255.255.255.240 = /28. So the access list will deny all traffic from network192.168.35.16/28 from accessing host 192.16.35.66, which is the IP address of accounting server………..anyone can help with dis

13. muhammadDecember 27th, 2010

Please can any one explain why the wildcard mask in Q no 4 would be 0.0.0.0 …i believe this means that it willdeny access to all hosts apart from the 192.168.125.34 ip address .But here we are looking for192.168.125.34/27 subnet .arent the question is asking to deny access to all hosts in a complete subnet ,in this case the block size is 8..so onlyadresses in the range of 192.168.125.34-48 should be denied .here the wild card mask denies access to only hostwith adress 192.168.125.34 .

14. muhammadDecember 27th, 2010

HI ATUL…

The question is asking to block access to accounting server for only users who are attached to the missouri e oconnection .

The ip address for e0 – 192.168.35.17/28 .

Now what could be the network address broadcast address and hosts attached in regards to block size used in thisparticular subnet .If we calculate it falls in the range of

192.168.35.16 (Network Address)192.168.35.17-30 (Hosts)192.168.35.31 (broadcast address )192.168.35.32 (Network address of next subnet)

the above are calculated on the basis of increment =16

So now in order to deny users attached to e0 interface which means to block users with the following ip addresses192.168.35.17-30 as they fall in the subnet of 192.168.35.16/28



The network traffic will be blocked for this subnet only and not for the ip address 192.168.35.66 which does notfall in the block size of subnet /28 eventually it wall at one stage if we keep subnetting with the block size of 16 butin this question it precisely asks to block the addresses between 192,168.135.16-32 .

hope this helps.

15. DeEjaY ArmaAnJanuary 7th, 2011

hmmm..thanx alot to 9tut in preparing the exam.

16. kasimJanuary 14th, 2011

when will update 9tut site i amgoing towrite exam on 20th

17. RobertJanuary 18th, 2011

Guys,I have a question about question number one. I undrestand that: /27 will provides you with 30 hosts, /28 willprovides you with 14 hosts, /29 will provides you with 6 hosts, /30 will provide you with 2 hosts and that is perfectfor serial links. What I don’t get is why they started so high in the range.

When I first worked this problem (with out looking at the posible answers) I came up with the following :Amount of Hosts Network Address/Mask16 (plus the int) 192.168.164.0/2711 (plus the int) 192.168.164.32/285 (plus the int) 192.168.164.48/292 192.168.164.56/30

Now I also undrestand that the options provided for posible answers are set in stone(we have to use 192.168.164.149/27 192.168.164.166/28 and 192.168.164.178/29)

What I’m trying to figure out how is why they used (192.168.164.149/27, 192.168.164.166/28 and192.168.164.178/29)

Am I correct in assuming that they started at the 192.168.164.128/27 and if they did that how come they didn’tstart at 192.168.164.0/27 I guess what I’m asking is do I gain anything by starting at 192.168.164.128

/27 /28 /29 /30Network 192.168.164.128 192.168.164.160 192.168.164.176 192.168.164.188First IP ad 192.168.164.129 192.168.164.161 192.168.164.177 192.168.164.189IP provided 192.168.164.149 192.168.164.166 192.168.164.178 192.168.164.189Last IP add 192.168.164.158 192.168.164.174 192.168.164.182 192.168.164.190Broadcast 192.168.164.159 192.168.164.175 192.168.164.183 192.168.164.191Next Subnt 192.168.164.160 192.168.164.176 192.168.164.184 You get the point

Thanks

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