1. A company has the following addressing scheme requirements: |
-currently has 25 subnets |
-uses a Class B IP address |
-has a maximum of 300 computers on any network segment |
-needs to leave the fewest unused addresses in each subnet |
What subnet mask is appropriate to use in this company? |
a. 255.255.240.0 |
b. 255.255.248.0 |
c. 255.255.254.0 |
d. 255.255.255.0 |
e. 255.255.255.128 |
f. 255.255.255.248 |
2. Refer to the exhibit. Host A is being manually configured for connectivity to the LAN. |
Which two addressing scheme combinations are possible configurations that can be applied |
to the host for connectivity? (Choose two.) |
a. Address - 192.168.1.14 |
Gateway - 192.168.1.33 |
b. Address - 192.168.1.45 |
Gateway - 192.168.1.33 |
c. Address - 192.168.1.32 |
Gateway - 192.168.1.33 |
d. Address - 192.168.1.82 |
Gateway - 192.168.1.65 |
e. Address - 192.168.1.63 |
Gateway - 192.168.1.65 |
f. Address - 192.168.1.70 |
Gateway - 192.168.1.65 |
1 |
Sumber: |
Cisco Networking Academy Program, CCNA 1 and 2 Companion Guide, Cisco Press, 2005 |
Todd Lammle, CCNA Study Guide 5 th Edition, Sybex, 2005. |
http://romisatriawahono.net |
3. A NIC of a computer has been assigned an IP address of 172.31.192.166 with a mask of |
255.255.255.248. To which subnet does the IP address belong? |
a. 172.31.0.0 |
b. 172.31.160.0 |
c. 172.31.192.0 |
d. 172.31.248.0 |
e. 172.31.192.160 |
f. 172.31.192.248 |
4. Which subnet masks would be valid for a subnetted Class B address? (Choose two.) |
a. 255.0.0.0 |
b. 255.254.0.0 |
c. 255.224.0.0 |
d. 255.255.0.0 |
e. 255.255.252.0 |
f. 255.255.255.192 |
5. Which combination of network id and subnet mask correctly identifies all IP addresses |
from 172.16.128.0 through 172.16.159.255? |
a. 172.16.128.0 and 255.255.255.224 |
b. 172.16.128.0 and 255.255.0.0 |
c. 172.16.128.0 and 255.255.192.0 |
d. 172.16.128.0 and 255.255.224.0 |
e. 172.16.128.0 and 255.255.255.192 |
6. Which type of address is 223.168.17.167/29? |
a. host address |
b. multicast address |
c. broadcast address |
d. subnetwork address |
7. What is the correct number of usable subnetworks and hosts for the IP network address |
192.168.99.0 subnetted with a /29 mask? |
a. 6 networks / 32 hosts |
b. 14 networks / 14 hosts |
c. 30 networks / 6 hosts |
d. 62 networks / 2 hosts |
8. Company XYZ uses a network address of 192.168.4.0. It uses the mask of |
255.255.255.224 to create subnets. What is the maximum number of usable hosts in each |
subnet? |
a. 6 |
b. 14 |
c. 30 |
d. 62 |
2 |
Sumber: |
Cisco Networking Academy Program, CCNA 1 and 2 Companion Guide, Cisco Press, 2005 |
Todd Lammle, CCNA Study Guide 5 th Edition, Sybex, 2005. |
http://romisatriawahono.net |
9. A company is planning to subnet its network for a maximum of 27 hosts. Which subnet |
mask would provide the needed hosts and leave the fewest unused addresses in each subnet? |
a. 255.255.255.0 |
b. 255.255.255.192 |
c. 255.255.255.224 |
d. 255.255.255.240 |
e. 255.255.255.248 |
10. An IP network address has been subnetted so that every subnetwork has 14 usable host IP |
addresses. What is the appropriate subnet mask for the newly created subnetworks? |
a. 255.255.255.128 |
b. 255.255.255.224 |
c. 255.255.255.240 |
d. 255.255.255.248 |
e. 255.255.255.252 |
11. A company is using a Class B IP addressing scheme and expects to need as many as 100 |
networks. What is the correct subnet mask to use with the network configuration? |
a. 255.255.0.0 |
b. 255.255.240.0 |
c. 255.255.254.0 |
d. 255.255.255.0 |
e. 255.255.255.128 |
f. 255.255.255.192 |
12. Given a host with the IP address 172.32.65.13 and a default subnet mask, to which |
network does the host belong? |
a. 172.32.65.0 |
b. 172.32.65.32 |
c. 172.32.0.0 |
d. 172.32.32.0 |
13. What is the subnetwork number of a host with an IP address of 172.16.210.0/22? |
a. 172.16.42.0 |
b. 172.16.107.0 |
c. 172.16.208.0 |
d. 172.16.252.0 |
e. 172.16.254.0 |
14 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 |
3 |
Sumber: |
Cisco Networking Academy Program, CCNA 1 and 2 Companion Guide, Cisco Press, 2005 |
Todd Lammle, CCNA Study Guide 5 th Edition, Sybex, 2005. |
http://romisatriawahono.net |
15. What is the subnetwork address for a host with the IP address 200.10.5.68/28? |
a. 200.10.5.56 |
b. 200.10.5.32 |
c. 200.10.5.64 |
d. 200.10.5.0 |
16. The network address of 172.16.0.0/19 provides how many subnets and hosts? |
a. 7 subnets, 30 hosts each |
b. 7 subnets, 2046 hosts each |
c. 7 subnets, 8190 hosts each |
d. 8 subnets, 30 hosts each |
e. 8 subnets, 2046 hosts each |
f. 8 subnets, 8190 hosts each |
17. You need 500 subnets, each with about 100 usable host addresses per subnet. What mask |
will you assign using a Class B network address? |
a. 255.255.255.252 |
b. 255.255.255.128 |
c. 255.255.255.0 |
d. 255.255.254.0 |
18. What is the subnetwork number of a host with an IP address of 172.16.66.0/21? |
a. 172.16.36.0 |
b. 172.16.48.0 |
c. 172.16.64.0 |
d. 172.16.0.0 |
19. What mask would you assign to the network ID of 172.16.0.0 if you needed about 100 |
subnets with about 500 hosts each? |
a. 255.255.255.0 |
b. 255.255.254.0 |
c. 255.255.252.0 |
d. 255.255.0.0 |
20. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the |
first available host address. Which of the following should you assign to the server? |
a. 192.168.19.0 255.255.255.0 |
b. 192.168.19.33 255.255.255.240 |
c. 192.168.19.26 255.255.255.248 |
d. 192.168.19.31 255.255.255.248 |
e. 192.168.19.34 255.255.255.240 |
21. You need a minimum of 300 subnets with a maximum of 50 hosts per subnet. Which of |
the following masks will support the business requirements? (Choose two.) |
a. 255.255.255.0 |
b. 255.255.255.128 |
c. 255.255.252.0 |
d. 255.25.255.224 |
e. 255.255.255.192 |
f. 255.255.248.0 |
4 |
Sumber: |
Cisco Networking Academy Program, CCNA 1 and 2 Companion Guide, Cisco Press, 2005 |
Todd Lammle, CCNA Study Guide 5 th Edition, Sybex, 2005. |
22. If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what |
would be the valid subnet address of this host? |
a. 172.16.112.0 |
b. 172.16.0.0 |
c. 172.16.96.0 |
d. 172.16.255.0 |
e. 172.16.128.0 |
23. Refer to the exhibit. The internetwork in the exhibit has been assigned the IP address |
172.20.0.0. What would be the appropriate subnet mask to maximize the number of networks |
available for future growth? |
a. 255.255.224.0 |
b. 255.255.240.0 |
c. 255.255.248.0 |
d. 255.255.252.0 |
e. 255.255.254.0 |
24. You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses? |
a. 172.16.17.1 255.255.255.252 |
b. 172.16.0.1 255.255.240.0 |
c. 172.16.20.1 255.255.254.0 |
d. 172.16.16.1 255.255.255.240 |
e. 172.16.18.255 255.255.252.0 |
f. 172.16.0.1 255.255.255.0 |
25 Your router has the following IP address on Ethernet0: 172.16.112.1/20. How many hosts |
can be accommodated on the Ethernet segment? |
a. 1024 |
b. 2046 |
c. 4094 |
d. 4096 |
e. 8190 |
5 |
Sumber: |
Cisco Networking Academy Program, CCNA 1 and 2 Companion Guide, Cisco Press, 2005 |
Todd Lammle, CCNA Study Guide 5 th Edition, Sybex, 2005. |
26. You have a /27 subnet mask. Which of the following are valid hosts? (Choose three.) |
a. 11.244.18.63 |
b. 90.10.170.93 |
c. 143.187.16.56 |
d. 192.168.15.87 |
e. 200.45.115.159 |
f. 216.66.11.192 |
27. You have a Class B network ID and need about 450 IP addresses per subnet. What is the |
best mask for this network? |
a. 255.255.240.0 |
b. 255.255.248.0 |
c. 255.255.254.0 |
d. 255.255.255.0 |
28. Host A is connected to the LAN, but it cannot connect to the Internet. The host |
configuration is shown in the exhibit. What are the two problems with this configuration? |
(Choose two.) |
a. The host subnet mask is incorrect. |
b. The host is not configured for subnetting. |
c. The default gateway is a network address. |
d. The default gateway is on a different network than the host. |
e. The host IP address is on a different network from the Serial interface of the router. |
6 |
Sumber: |
Cisco Networking Academy Program, CCNA 1 and 2 Companion Guide, Cisco Press, 2005 |
Todd Lammle, CCNA Study Guide 5 th Edition, Sybex, 2005. |
1. dik : Kelas B
jml subnet = 25
jml host = 300
tanya : Subnet mask
jawab :
2n-2 ≥ 300
2n ≥ 302
n ≥ 9 –> banyaknya 0 (nol)
subnet mask : 11111111. 11111111.11111110.00000000
255 . 255 . 254 . 0 (C)
2. dik : IP : 192.168.1.? / 27
tanya : IP & gateway yang mungkin (pilih 2)
jawab :
subnet mask : 11111111. 11111111. 11111111.11100000
255 . 255 . 255 . 224
Jml host per blok = 256 – 224 = 32
192.168.1.0 192.168.1.1-192.168.1.30 192.168.1.31
192.168.1.32 192.168.1.33-192.168.1.62 192.168.1.63
192.168.1.64 192.168.1.65-192.168.1.94 192.168.1.95
192.168.1.96 ... ...
Yang mungkin :
(D) Address 192.168.1.82 Gateway - 192.168.1.65
(F) Address - 192.168.1.70 Gateway - 192.168.1.65
3. dik: IP : 172.31.192.166
subnet mask : 255.255.255.248
tanya : Subnet ID
jawab :
Jml Host tiap subnet = 256-248 = 8
172.31.192.0 ... 172.31.192.7
172.31.192.8 ... 172.31.192.15
172.31.192.16 ... 172.31.192.23
... ... ...
172.31.192.160 172.31.192.161-172.31.192.166 172.31.192.167
172.31.192.168 ...
Net ID 172.31.192.166 adalah 172.31.192.160 (E)
4. Kelas B --> Default subnet mask= 255.255.0.0
subnet mask yang mungkin : 255.255.0.0 (D) dan 255.255.252.0 (E)
5. dik : range IP = 172.16.128.0 - 172.16.159.255
tanya : Net ID & subnet mask
jawab :
6. dik :IP : 223.168.17.167/29 – kelas C
tanya : jenisnya
jawab :
subnet mask : 255.255.255.11111000 = 255.255.255.248
Blok subnet : 256-248 = 8
223.168.17.0 223.168.17.1 -223.168.17 .6 223.168.17.7
223.168.17.8 223.168.17.9 - 223.168.17.14 223.168.17.15
……..
223.168.17.160 223.168.17.161 -223.168.17 .166 223.168.17.167 – Broadcast (C)
7. dik :192.168.99 /29
tanya : jml network dan host yang digunakan
jawab :
subnet mask : 255.255.255.11111000
Jadi ada 30 network/ 6 host (C)
8. dik : IP = 192.168.4.0
subnet mask = 255.255.255.224
tanya : Jml host
jawab :
Blok subnet = 256 – 224
= 32
Jml Host = 32 – 2 = 30 (C)
9. dik : Jml Host max = 27
tanya :subnet mask
jawab :
2n-2 = 27
2n = 29
n = 5 – banyaknya nilai nol pada subnet mask
subnet mask : 255.255.255.11100000 – 255.255.255.224 (C)
10. dik : Jml host = 14
tanya : subnet mask
jawab :
2n-2 = 14
2n = 16
n = 4 – banyaknya nilai nol pada subnet mask
subnet mask : 255.255.255.11110000 – 255.255.255.248 (D)
11. dik : jml network = 100 – kelas B
tanya subnet mask
jawab :
2N-2 = 100
2N = 102
N = 7 – banyaknya nilai nol pada subnet mask
subnet mask : 255.255. 11111110.0– 255.255.254.0 (C)
12. dik : IP : 172.32.65.132 – kelas B, subnet mask : default (255.255.0.0)
tanya : Net ID
Net ID : 172.32.0.0 (C)
13. dik : IP : 172.16.210.0 /22
tanya : Net ID
jawab :
subnet mask = 255.255.11111100.0 – 255.255.252.0
Blok subnet : 256 – 252 = 4
172.16.0.0 172.16.0.1 - 172.16.3.254 172.16.3.255
172.16.4.0 ... ..
... ... ...
172.16.208.0 172.16.208.1 - 172.16.211.254 172.16.211.155
172.16.212.0 .. ...
Jadi net ID nya 172.16.208.0 (C)
14. dik : IP Address CIDR 115.64.4.0 /22
tanya: IP yang mungkin
jawab :
subnet mask: 255.255.11111100.00000000 = 255.255.252.0
Blok Subnet : 256 – 252 = 4
115.64.0.0 115.64.0.1-115.64.3.254 115.64.3.255
172.16.4.0 115.64.4.1-115.64.7.254 115.64.7.255
115.64.8.0 115.64.8.1-115.64.11.254 115.64.11.255
115.64.12.0
Yang mungkin (B) 115.64.7.64, (C) 115.64.6.255, (E) 115.64.5.128
15. dik : IP = 200.10.5.68 /28
tanya : Net ID
jawab :
subnet mask : 255.255.255.11110000 – 255.255.255.240
Blok subnet : 256-240 = 16
200.10.5.0 200.10.5. 200.10.5.
... ... ...
200.10.5.64 200.10.5.65-200.10.5.78 200.10.5.79
200.10.5.80 ... ...
Jadi Net ID nya 200.10.5.64 (C)
16. dik : IP : 172.16.0.0 /19
tanya : jml subnet & jml host
jawab :
subnet mask : 255.255.111000000.00000000
Jadi jml Host = 8190 dan jml subnet = 8 (F)
17. dik : Jml Host = 100
Jml subnet = 500
tanya : subnet mask
jawab :
2n-2 ≥ 100
2n ≥ 102
n ≥ 7 -- banyaknya nilai nol pada subnet mask
subnet mask : 11111111. 11111111. 11111111.10000000 – 255.255.255.128 (B)
18. dik : IP = 172.16.66.0 /21
tanya : Net ID
jawab :
subnet mask : 255.255.11111000.00000000 – 255.255.248.0
Blok subnet = 256-248 = 8
172.16.0.0 172.16.0.1-172.16.3.254 172.16.3.255
... ... ...
172.16.64.0 172.16.64.1-172.16.71.254 172.16.71.255
172.16.72.0 ... ...
Net ID nya adalah 172.16.64.0 (C)
19. dik : IP : 172.16.0.0
Jml subnet : 100
Jml host : 500
tanya : subnet mask
jawab :
2n-2 ≥ 500
2n ≥ 502
n ≥ 9 -- banyaknya nilai nol pada subnet mask
subnet mask : 11111111. 11111111. 11111110.00000000 – 255.255.254.0 (B)
20. dik : IP : 192.168.19.24 /29
tanya : IP server & subnet mask
jawab :
subnet mask : 255.255.255.11111000 – 255.255.255.248
Blok subnet = 256 – 248 = 8
192.168.19.0 192.168.19.1-192.168.19.6 192.168.19.7
... ... ...
192.168.19.24 192.168.19.25-192.168.19.30 192.168.19.31
192.168.19.32 .. ..
Yang mungkin dijadikan IP server pada blok subnet 192.168.19.24 adalah 192.168.19.26 (C)
21. dik : Jml subnet : 300
Jml host : 50
tanya : subnet mask
jawab :
2N-2 ≥ 300
2N ≥ 302
N ≥ 9 -- banyaknya nilai 1 pada subnet mask
Default subnet mask :255.255.0.0
subnet mask : 11111111. 11111111. 11111111.10000000 – 255.255.255.128 (B)
2n-2 ≥ 50
2n ≥ 52
n ≥ 6 -- banyaknya nilai nol pada subnet mask
subnet mask : 11111111. 11111111. 11111111.11000000 – 255.255.255.192 (E)
22. dik IP : 172.16.112.1 /25
tanya :Net ID
jawab :
subnet mask : 255.255.255.10000000 – 255.255.255.128
Blok subnet = 256-128 = 128
172.16.112.0 172.16.112.1-172.16.112.126 172.16.112.127
172.16.112.128 172.16.112.129-172.16.112.254 172.16.112.255
Net ID nya adalah 172.16.112.0 (A)
23. dik : IP : 172.20.0.0 – kelas B
jml subnet : 7
tanya : subnet mask
jawab :
2N-2 ≥ 7
2N ≥ 9
N ≥ 4 -- banyaknya nilai 1 pada subnet mask
Defaul subnet mask : 255.255.0.0
subnet mask sekarang : 255.255.11110000.00000000.00000000 – 255.255.240.0 (B)
24. dik : IP : 172.16.17.0 /22
tanya : IP host yang valid
jawab :
subnet mask : 255.255.11111100.0000000 – 255.255.252.0
Blok subnet : 256-252 = 4
172.16.0.0 172.16.1.0-172.16.3.254 172.16.3.255
... .. ..
172.16.16.0 172.16.16.1-172.16.19.254 172.16.19.255
172.16.20.0 ... ...
IP host yang mungkin 172.16.18.255, subnet mask = 255.255.255.252 (E)
25. dik : IP = 172.16.112.1 /20
tanya : Jml host
jawab :
subnet mask : 255.255.11110000.00000000 – 255.255.240.0
26. dik : subnet mask : /27
tanya : Host yang valid
jawab :
subnet mask : 255.255.255.11100000 – 255.255.255.224
Blok subnet : 256 – 224 = 32
IP host yang valid adalah IP yang tidak digunakan untuk Net ID dan Broadcast, maka jawaban yang benar adalah (B) 90.10.170.93 , (C) 143.187.16.56, dan (D) 192.168.15.87
27. dik : Jml host : 450
tanya : subnet mask
jawab :
2n-2 ≥ jml host
2n-2 ≥ 450
2n ≥ 452
n ≥ 9 – banyaknya nilai 0 pada subnet mask
subnet mask : 1111111.11111111.11111110.00000000 – 255.255.254.0 (C)
28. dik : IP : 198.18.166.65 /27 , subnet mask : 255.255.255.240, Gateway : 198.18.166.33
tanya : mana pernyataan yang benar
jawab :
subnet mask : 255.255.255.11100000 – 255.255.255.224
Blok subnet = 256 – 224 = 32
198.18.166.0 198.18.166.1-198.18.166.30 198.18.166.31
198.18.166.32 198.18.166.33- 198.18.166.62 198.18.166.63
198.18.166.64 198.18.166.65 - 198.18.166.94 198.18.166.95
198.18.166.96 .. ..
1. Subnet masknya salah (A)
2. IP address dan gateway tidak dalam satu network (D)
1 komentar:
mantap mas bro share nya . sangat bermanfaat , terimakasih sudah ikut serta mencerdaskan kehidupan bangsa :-)
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