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SOAL SUBNETTING

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

Jawaban

Diketahui  : Saat ini memiliki 25 subnet dan maksimum nya 300 komputer dalam jaringan
Subnet mask yang paling sesuai?
default subnet mask class B11111111.11111111.0000000.00000000 karena yang diminta 25

subnet, dan yang mendekati adalah 2⁵ maka :

11111111.11111111.11111000.00000000 dan subnet masknya
255 . 255 . 248 . 0
Jawabannya adalah B
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

Jawaban

/27 =11111111.11111111.11111111.11100000
255 . 255 . 255 . 224
Jadi subnet masknya adalah : 255.255.255.224
Blok subnet = 256 – 224 = 32
Net ID Range Broadcast
192.168.1.0 …….. 192.168.1.1 – 192.168.1.30…. 192.168.1.31

192.169.1.64……. 192.168.1.65-192.168.1.94….. 192.168.1.95

Jawabannya D&F
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

Jawaban

Diketahui :

IP : 172.31.192.166
Netmask : 255.255.255.248
Dit : Subnet untuk IP tersebut ?

Jawab :
Block subnet = 256 – 248 = 8
Maka 172.31.192.8
172.31.192.16
172.31.192.24
172.31.192.32

——————

172.31.192.160

——————
Jawabannya E
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

Jawaban

Jawabannya D&E

Karena subnetmask defaultnya untuk class B adalah 255.255.0.0 dan kombinasi dari 2 oktet

selanjutnya.

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

Jawaban

Jawabannya D
6. Which type of address is 223.168.17.167/29?
a. host address

b. multicast address

c. broadcast address

d. subnetwork address

Jawaban

IP address diatas merupakan kelas C
/29 =11111111.11111111.11111111.11111000
Subnet mask = 255.255.255.248
Blok subnet = 256 – 248 = 8
Net Id Range Broadcast
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

…………………………………………………………………………………………

Jawabannya C
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

Jawaban

/29 =11111111.11111111.11111111.11111000
Subnet masknya = 255.255.255.248
Block subnet = 256 – 248 = 8
Host = 2³-2 = 6
Subnet = 2⁵-2 = 30

Jawabannya C
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

Jawaban

255.255.255.224 = / 27

1111111.1111111.1111111.11100000

Maka, host = 2⁵ – 2 = 30
Jawabannya C
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

Jawaban

Yang mendekati 27 dan tidak melebihi jauh yaitu 2 ^(pangkat)5, dan host dihitung dari kanan kekiri

11111111.11111111.11111111.11111000

Host 2³-2 = 6

Subnetnya 255.255.255.224
Jawabannya C
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

Jawaban

Yang mendekati 14 dan tidak melebihi jauh yaitu 2³, dan host dihitung dari kanan kekiri

11111111.11111111.11111111.11110000
Subnetnya 255.255.255.240
Jawabannya C
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

Jawaban

11111111.11111111.11111111.00000000
255.255.255.0
Jawabannya D
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

Jawaban

172.32.65.13 = 10101100.00100000.01000001.00001101
255.255.0.0 =11111111.11111111.00000000.00000000

Operasi logica AND
jadi                  10101100.00100000.00000000.00000000
172 . 32 . 0 . 0
Jawabannya C
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

Jawaban

/22 =11111111.11111111.11111100.00000000 = 255.255.252.0

172.16.210.0 = 10101100.00010000.11010010.00000000

255.255.252.0 =11111111.11111111.11111100.00000000

Operasi logica AND

Jadi                  10101100.00010000.11010000.00000000

172 . 16 . 208 . 0
Jawabannya C
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

Jawaban

/22 = 255.255.252.0
Block subnet 256 – 252 = 4
115.64.4.1 – 115.64.7.255
Jawabannya C, D, & E
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

Jawaban

/28 =11111111.11111111.11111111.11110000 = 255.255.255.240
200.10.5.68 = 11001000.00001010.00000101.01000100
255.255.252.0 =11111111.11111111.11111111.11110000

Operasi logica AND

Jadi                  11001000.00001010.00000101.01000000

200 . 10 . 5 . 64
Jawabannya C
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

Jawaban

/19 = 11111111 11111111 11100000 00000000
Subnet 2³ = 8
Host     2¹³-2 = 8190
Jadi jawabannya F
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

Jawaban

Subnet 2⁹= 512

Host     2⁷-2= 128 – 2 = 126

11111111 11111111 11111111 10000000
255 255 255 128
Subnetnya 255.255.255.128
Jawabannya B
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

Jawaban

/21 =11111111.11111111.11111000.00000000 = 255.255.248.0
172.16.66.0 = 10101100.00010000.01000010.00000000
255.255.248.0 =11111111.11111111.11111000.00000000
10101100.00010000.01000000.00000000
172 . 16 . 64 . 0
Jawabannya C
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. 255.255.255.0b. 255.255.254.0c. 255.255.252.0d. 255.255.0.0

Jawaban

2⁶<100<2⁷  KEMUDIAN subnet dihitung dari kiri kekanan

11111111 11111111 11111110 00000000
255 255 254 0
Subnetnya 255.255.254.0
Jawaban B
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

Jawaban

/ 29 =11111111.11111111.11111111.11111000 =
255 .255 .255 .248
Blok subnet = 256 – 248 = 8
Net Id Range Broadcast
192.168.19.0 192.168.19.1 – 192.168.19.6 192.168.19.7

192.168.19.8 192.168.19.9 – 192.168.19.14 192.168.19.15

192.168.19.16 192.168.19.17 – 192.168.19.22 192.168.19.23

192.168.19.24 192.168.19.25 – 192.168.19.30 192.168.19.31

Jawaban C
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

Jawaban

2⁸<300<2⁹ karena subnet dihitung dari kiri kekanan«

11111111.11111111.11111111.10000000
Subnetnya 255.255.255.128
2⁵<50<2⁶ karena host dihitung dari kanan kekiri

11111111.11111111.11111111.11000000
Subnetnya 255.255.255.192
Jawabanya B dan E
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?

  1. 172.16.112.0

b. 172.16.0.0
c. 172.16.96.0
d. 172.16.255.0
e. 172.16.128.0

Jawaban

/25 =11111111.11111111.11111111.10000000
255 .255 .255 .128
172.16.112.1 = 10101100.00010000.01110000.00000001
255.255.255.128 =11111111.11111111.11111111.10000000

Operasi logica AND

Jadi               10101100.00010000.01110000.00000000

172 . 16 . 112 . 0
Jawabannya  A
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

Jawaban

Jumlah Host Keseluruhan= 3350

Host =

Maka netmask yang sesuai ialah

11111111 11111111 11110000 00000000
255 255 240 0
Jawabannya D
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

Jawaban

/22 =11111111.11111111.11111100.00000000

= 255 .255 .252 .0
Block subnet = 256 – 252 = 4
172.16.17.0 172.16.17.1 ± 172.168.17.2 172.16.17.3

————– ———————————- ————–

Jawaban E
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

Jawaban

/20 =11111111.1111111.11110000.00000000

= 255 .255 .240 .0
Host = 2¹² – 2 = 4094
Jawabannya C
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

Jawaban

/27 = 255.255.255.224
Kelas C
Subnet  2³=8
Host    2⁵-2=30
Blok subnet = 256 – 224 = 32
32, 64, 96, 128
Jawabannya B, C, & D
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

Jawaban

Host 2⁹-2= 510 dapat memenuhi kebutuhan 450 host

11111111 11111111 11111110 00000000
Subnetmasknya 255.255.254.0
Jawabannya C
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.

Jawaban

11111111.11111111.11111111.11100000
255.255.255.224
/ 27 = subnet mask = 255.255.255.224
Blocksubnetnya = 256 -224 = 32

32, 64, 96, 128,«dst

Jawabannya A dan D

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