CCNA 1 Chapter 6 V4.0 Answers

1. How many bits make up an IPv4 address?

128

64

48

32 **

 

2.
CCNA1Chapter61 thumb CCNA 1 Chapter 6 V4.0 Answers
Refer to the exhibit. A network administrator has to develop an IP addressing scheme that uses the 192.168.1.0 /24 address space. The network that contains the serial link has already been addressed out of a separate range. Each network will be allocated the same number of host addresses. Which network mask will be appropriate to address the remaining networks?

255.255.255.248

255.255.255.224

255.255.255.192

255.255.255.240

255.255.255.128

255.255.255.252

 

3.
CCNA1Chapter62 thumb CCNA 1 Chapter 6 V4.0 Answers
Refer to the exhibit. The network administrator has assigned the internetwork of LBMISS an address range of 192.168.10.0. This address range has been subnetted using a /29 mask. In order to accommodate a new building, the technician has decided to use the fifth subnet for configuring the new network (subnet zero is the first subnet). By company policies, the router interface is always assigned the first usable host address and the workgroup server is given the last usable host address. Which configuration should be entered into the workgroup server’s properties to allow connectivity to the network?

IP address: 192.168.10.38 subnet mask: 255.255.255.240 default gateway: 192.168.10.39

IP address: 192.168.10.38 subnet mask: 255.255.255.240 default gateway: 192.168.10.33

IP address: 192.168.10.38 subnet mask: 255.255.255.248 default gateway: 192.168.10.33

IP address: 192.168.10.39 subnet mask: 255.255.255.248 default gateway: 192.168.10.31

IP address: 192.168.10.254 subnet mask: 255.255.255.0 default gateway: 192.168.10.1

 

4.
CCNA1Chapter63 thumb CCNA 1 Chapter 6 V4.0 Answers
Refer to the exhibit. Which network prefix will work with the IP addressing scheme shown in the graphic.

/24

/16

/20

/27

/25

/28

 

5.
CCNA1Chapter64 thumb CCNA 1 Chapter 6 V4.0 Answers
Refer to the exhibit. A network administrator discovers that host A is having trouble with Internet connectivity, but the server farm has full connectivity. In addition, host A has full connectivity to the server farm. What is a possible cause of this problem?

The router has an incorrect gateway.

Host A has an overlapping network address.

Host A has an incorrect default gateway configured.

Host A has an incorrect subnet mask.

NAT is required for the host A network.

 

6. What subnet mask would a network administrator assign to a network address of 172.30.1.0 if it were possible to have up to 254 hosts?

255.255.0.0

255.255.255.0

255.255.254.0

255.255.248.0

 

7. Given the IP address and subnet mask of 172.16.134.64 255.255.255.224, which of the following would describe this address?

This is a useable host address.

This is a broadcast address.

This is a network address.

This is not a valid address.

 

8. What three facts are true about the network portion of an IPv4 address? (Choose three.)

identifies an individual device

is identical for all hosts in a broadcast domain

is altered as packet is forwarded

varies in length

is used to forward packets

uses flat addressing

 

9. What is the network address of the host 172.25.67.99 /23 in binary?

10101100. 00011001.01000011.00000000

10101100. 00011001.01000011.11111111

10101100. 00011001.01000010.00000000

10101100. 00011001.01000010.01100011

10101100. 00010001.01000011. 01100010

10101100. 00011001.00000000.00000000

 

10. A router interface has been assigned an IP address of 172.16.192.166 with a mask of 255.255.255.248. To which subnet does the IP address belong?

172.16.0.0

172.16.192.0

172.16.192.128

172.16.192.160

172.16.192.168

172.16.192.176

 

11.
CCNA1Chapter65 thumb CCNA 1 Chapter 6 V4.0 Answers
Refer to the exhibit. Why would the response shown be displayed after issuing the command ping 127.0.0.1 on a PC?

The IP settings are not properly configured on the host.

Internet Protocol is not properly installed on the host.

There is a problem at the physical or data link layer.

The default gateway device is not operating.

A router on the path to the destination host has gone down.

 

12. What two things will happen if a router receives an ICMP packet which has a TTL value of 1 and the destination host is several hops away? (Choose two.)

The router will discard the packet.

The router will decrement the TTL value and forward the packet to the next router on the path to the destination host.

The router will send a time exceeded message to the source host.

The router will increment the TTL value and forward the packet to the next router on the path to the destination host.

The router will send an ICMP Redirect Message to the source host.

 

13. Which of the following are features of IPv6? (Choose three.)

larger address space

faster routing protocols

data types and classes of service

authentication and encryption

improved host naming conventions

same addressing scheme as IPv4

 

14. Which process do routers use to determine the subnet network address based upon a given IP address and subnet mask?

binary adding

hexadecimal anding

binary division

binary multiplication

binary ANDing

 

15. What is the primary reason for development of IPv6?

security

header format simplification

expanded addressing capabilities

addressing simplification

 

16. Which three IP addresses are private? (Choose three.)

172.168.33.1

10.35.66.70

192.168.99.5

172.18.88.90

192.33.55.89

172.35.16.5

 

17. Which statements are true regarding IP addressing? (Choose two.)

NAT translates public addresses to private addresses destined for the Internet.

Only one company is allowed to use a specific private network address space.

Private addresses are blocked from public Internet by router.

Network 172.32.0.0 is part of the private address space.

IP address 127.0.0.1 can be used for a host to direct traffic to itself.

 

18. Which IPv4 subnetted addresses represent valid host addresses? (Choose three.)

172.16.4.127 /26

172.16.4.155 /26

172.16.4.193 /26

172.16.4.95 /27

172.16.4.159 /27

172.16.4.207 /27

 

19. What is a group of hosts called that have identical bit patterns in the high order bits of their addresses?

an internet

a network

an octet

a radi

 

20.
CCNA1Chapter66 thumb CCNA 1 Chapter 6 V4.0 Answers
Refer to the exhibit. Host A is connected to the LAN, but it cannot get access to any resources on the Internet. The configuration of the host is shown in the exhibit. What could be the cause of the problem?

The host subnet mask is incorrect.

The default gateway is a network address.

The default gateway is a broadcast address.

The default gateway is on a different subnet from the host.

 

21. Which of the following network devices are recommended to be assigned static IP addresses? (Choose three.)

LAN workstations

servers **

network printers **

routers **

remote workstations

laptops

ccna 1 chapter 6, ccna 1 chapter 6 answers, ccna chapter 6, ccna chapter 6 answers, ccna1 chapter 6, ccna 1 chapter 6 v4 0 answers, ccna chapter 6 exam answers, yhs-fh_lsonsw, refer to the exhibit the network administrator has assigned the internetwork of lbmiss an address range of 192 168 10 0 this address range has been subnetted using a /29 mask in order to accommodate a new building the technician has decided to use the fif, ccna1 chapter 6 answers

15 Responses - Add Yours+

  1. ahmadnazmie says:

    100% ;)

  2. clarkski_ says:

    there is something wrong with 13. the answers to choose are:
    *larger address space*
    faster routing protocols
    simplified header format
    *authentication and encryption*
    improved host naming conventions
    same addressing scheme as IPv4

    so “data types and classes of service” is no answer to choose in the chapter assessment

  3. bZS says:

    13. Which of the following are features of IPv6? (Choose three.)

    *larger address space*
    faster routing protocols
    *simplified header format*
    *authentication and encryption*
    improved host naming conventions
    same addressing scheme as IPv4

  4. susu says:

    Hi

    good job

  5. Renegade says:

    Is there something wrong with the following:
    16. Which three IP addresses are private? (Choose three.)

    10.35.66.70

    192.168.99.5

    172.18.88.90

    I was going to check for the answers. It gave me 1 point less and then pointed me to the “public and private addresses” section. I don’t understand… I ‘m pretty sure i had gotten it right. Plus I also checked other websites and apparently I did right. What the hell?

  6. amar says:

    I have enquirer anbot q2.
    the maximum hosts are 33, that mean i should reserve 7 bits for hosts (64 host)and 1 bit for network and the answer should be
    255.255.255.128

    why you selected
    255.255.255.192??

    Thanks

    • TheMan says:

      0/192/224/240/248/252/254/255 and the only usable subnets…

    • Claudio Sousa says:

      you have to reserve 6 bits if you want use 64 hosts. So, you have 6 for hosts and 2 for network. It means 128 + 64 = 192. Thats why : 255.255.255.192.

      • Rusmir says:

        Q4
        Can anyone explain why is network prefix 27? I don’t understand?

        • Habasha man says:

          Router A : 10110001.00010110.00000000. 001 11110
          Host A: 10110001.00010110.00000000. 001 00101

          Router B: 10110001.00010110.00000000. 010 11110
          Host B: 10110001.00010110.00000000. 010 01011
          Notice that the 3 first bits of the last octet created 2 groups.
          Router A and Host A use subnet 1 (meaning the three first bits of the last octet is 001)

          On the other hand Router B and Host B use subnet2 (meaning the first three bits of the last octet is 010)
          So 3bits are borrowed from host ID for subneting, 24 +3 = 27

  7. Justin says:

    Number 9, they are asking what network is host 172.25.67.99 /23 on and put it in binary form. So if you have a CIDR of /23 that means that you are in the 3rd octet. On a Class B address that would have you in the 3rd octet with the first network address of 172.25.0.0. So If you follow by your number line of the third octet

    2 4 8 16 32 64 128 256
    128 64 32 16 8 4 2 1
    X
    So where the X marks the spot of your 7th spot in the 3rd octet you have 128 subnets with 2 hosts in each. So with saying that your going in increments of 2 hosts at a time. 172.25.0.0,172.25.2.0,172.25.4.0 etc. for 128 networks. Which means the IP address 172.25.67.99 which is a broadcast would fall under the 172.25.66.0 Network. So the answer would be

    172 25 60 0
    10101100. 00011001.01000010.00000000

  8. Rusmir says:

    Can anyone please explain Q3 and Q4?
    Thanks?

    • Hicham says:

      Greetings Rusmir,

      So in regards to question number 3, look at the address and the subnet Mask 192.168.10.0/29, first thing we will do is find the network addresses and their range so we can look at the 5 th subnet the question is talking about, the mask is /29 so it’s 24+5 we borrowd 5 bits to create the subnetworks ( we don’t care how many for the sake of the question ) so we 3 bits left for the host and 2*3=8 so the network addresses ( only curious about the 5 th ) are 192.168.10.0 /192.168.10.8/192.168.10.16/192.168.10.24/192.168.10.32/192.168.10.40
      The fifth network address is what interest us, and that’s 192.168.10.32, the first and last host addresses aren’t used ( network and broadcast ) so we are left with this range :
      192.168.10.33 to 192.168.10.38 ( .39 is the broadcast address and .32 is the network address ) so now we know the server is getting the last address 192.168.10.138 and the router is getting the first 192.168.10.33 and since we borrowed 5 bits, the subnet Mask is 255.255.255.248

      So this is why IP address: 192.168.10.38 subnet mask: 255.255.255.248 default gateway: 192.168.10.33 is the right answer.

      Thank you.

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