Latihan Soal-soal Subneting
by
Rhanie
- Rabu, September 29, 2010
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
jawab :
2N-2≥300 llllllll.llllllll.lllllll0.00000000
2N≥302 255.255.254.0
N=9
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
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
jawab :
blok subnet : 256-248=8
subnet addres : (166 div 8)8
= 20x8=160
Jadi,172.31.192.160
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
jawab :
Class B = 255.255.0.0
Jadi yang termasuk ke dalam class B adalah 255.255.0.0 dan 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
jawab :
172.16.128.0-172.16.159.255
172.16.160.0
160-128=32
256-32=224
6. Which type of address is 223.168.17.167/29?
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
jawab :
llllllll.llllllll.llllllll.lllll000
255.255.255.248
Blok subnet=256-248=8
misal Net id range/host broadcast 223.168.17.160 223.168.17.167
223.168.17.168
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
jawab :
llllllll.llllllll.llllllll.lllll000
255.255.255.248
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
jawab :
llllllll.llllllll.llllllll.lll00000
255.255.255.224
Blok subnet=256-224=32
Host : 2N-2=25-2=30
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
jawab :
172.32.65.13 termasuk dalam class B. Dan yang termasuk class B adalah 172.32.0.0
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
jawab :
/22 = llllllll.llllllll.llllll00.00000000
255.255.252.0
Blok subnet=256-252=4
Network = (210 div 4)4
= 52x4=208 Jadi, subnetwork 172.16.208.0
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
jawab : 11111111.11111111.11111100.00000000
255 . 255 . 252 . 0
Block subnet = 256 – 252 = 4
Net Id Range Broadcast
115.64.4.0 115.64.4.1 - 115.64.7.254 115.64.7.255 115.64.8.0
Jadi yang termasuk pada network 115.64.4.0 adalah B,C dan 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
jawab :
11111111.11111111.11111111.11110000
255 . 255 . 225 . 240
Block subnet : 256 – 240 = 16
Network = ( 68 div 16 ) . 16
= 4 . 16
= 64
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
jawab :
11111111.11111111.11100000.00000000
N = 3 n = 13
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
jawab :
2N – 2 ≥ 100 11111111.11111111.11111111.10000000
2N ≥ 102 255 . 255 . 255 . 128
N = 7
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
jawab :
11111111.11111111.11111000.00000000
255 . 255 . 248 . 0
Block Subnet : 256 – 248 = 8
Network : ( 66 div 8 ) . 8 = 64
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
jawab :
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
jawab :
11111111.11111111.11111111.11111000
255 . 255 . 255 . 248
Block Subnet : 256 – 248 = 8
Net Id Range Broadcast
192.168.19.24 192.168.19.25 – 192.168.19.30 192.168.1.31
192.168.19.32
25 untuk Router dan 26 untuk server
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
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
jawab :
11111111.11111111.11111111.10000000
255 . 255 . 255 . 128
Block Subnet : 256 – 128 = 128
Network : ( 1 div 128 ) . 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
jawab :
2N – 2 ≥ 850 11111111.11111111.11111100.00000000
2N ≥ 852 255 . 255 . 252 . 0
N = 10
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
jawab :
11111111.11111111.11111100.00000000
255 . 255 . 252 . 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
jawab :
11111111.11111111.11110000.00000000 N = 12
255 . 255 . 240 . 0
Host : 2N - 2 = 4094
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
jawab:
Block subnet = 25 = 32
Host yang valid tidak boleh 0, kelipatan 32 ataupun selisih satu dari kelipatan 32 tersebut
Jadi host yang valid adalah 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
jawab :
2N - 2 ≥ 450 11111111.11111111.11111110.00000000
2N ≥ 452 255 . 255 . 254 . 0
N = 9
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.
Jawab :
11111111.11111111.11111111.11100000
255 . 255 . 255 . 224
-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
jawab :
2N-2≥300 llllllll.llllllll.lllllll0.00000000
2N≥302 255.255.254.0
N=9
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
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
jawab :
blok subnet : 256-248=8
subnet addres : (166 div 8)8
= 20x8=160
Jadi,172.31.192.160
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
jawab :
Class B = 255.255.0.0
Jadi yang termasuk ke dalam class B adalah 255.255.0.0 dan 255.255.255.192
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
jawab :
172.16.128.0-172.16.159.255
172.16.160.0
160-128=32
256-32=224
6. Which type of address is 223.168.17.167/29?
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
jawab :
llllllll.llllllll.llllllll.lllll000
255.255.255.248
Blok subnet=256-248=8
misal Net id range/host broadcast 223.168.17.160 223.168.17.167
223.168.17.168
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
jawab :
llllllll.llllllll.llllllll.lllll000
255.255.255.248
Host: 2N-2 subnet : 2n-2
23-2=6 25-2=30
Jadi, ada 30 network dan 6host
a. 6
b. 14
c. 30
d. 62
jawab :
llllllll.llllllll.llllllll.lll00000
255.255.255.224
Blok subnet=256-224=32
Host : 2N-2=25-2=30
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
jawab : Host : 2N-2≥27 llllllll.llllllll.llllllll.lll00000 bit 0 = host
2N≥29 255.255.255.224
N=5
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
jawab : host : 2N-2≥14 llllllll.llllllll.llllllll.llll0000
2N≥16 255.255.255.240
N=4
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
jawab : Host : 2N-2≥100 llllllll.llllllll.llllllll.l0000000
2N≥102 255.255.255.128
N=7
a. 172.32.65.0
b. 172.32.65.32
c. 172.32.0.0
d. 172.32.32.0
jawab :
172.32.65.13 termasuk dalam class B. Dan yang termasuk class B adalah 172.32.0.0
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
jawab :
/22 = llllllll.llllllll.llllll00.00000000
255.255.252.0
Blok subnet=256-252=4
Network = (210 div 4)4
= 52x4=208 Jadi, subnetwork 172.16.208.0
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
jawab : 11111111.11111111.11111100.00000000
255 . 255 . 252 . 0
Block subnet = 256 – 252 = 4
Net Id Range Broadcast
115.64.4.0 115.64.4.1 - 115.64.7.254 115.64.7.255 115.64.8.0
Jadi yang termasuk pada network 115.64.4.0 adalah B,C dan E
a. 200.10.5.56
b. 200.10.5.32
c. 200.10.5.64
d. 200.10.5.0
jawab :
11111111.11111111.11111111.11110000
255 . 255 . 225 . 240
Block subnet : 256 – 240 = 16
Network = ( 68 div 16 ) . 16
= 4 . 16
= 64
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
jawab :
11111111.11111111.11100000.00000000
N = 3 n = 13
Subnets = 23 hosts = 213 - 2
= 8 = 8190
a. 255.255.255.252
b. 255.255.255.128
c. 255.255.255.0
d. 255.255.254.0
jawab :
2N – 2 ≥ 100 11111111.11111111.11111111.10000000
2N ≥ 102 255 . 255 . 255 . 128
N = 7
a. 172.16.36.0
b. 172.16.48.0
c. 172.16.64.0
d. 172.16.0.0
jawab :
11111111.11111111.11111000.00000000
255 . 255 . 248 . 0
Block Subnet : 256 – 248 = 8
Network : ( 66 div 8 ) . 8 = 64
a. 255.255.255.0
b. 255.255.254.0
c. 255.255.252.0
d. 255.255.0.0
jawab :
2N – 2 ≥ 500 11111111.11111111.11111110.00000000
2N ≥ 502 255 . 255 . 254 . 0
N = 820. 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
jawab :
11111111.11111111.11111111.11111000
255 . 255 . 255 . 248
Block Subnet : 256 – 248 = 8
Net Id Range Broadcast
192.168.19.24 192.168.19.25 – 192.168.19.30 192.168.1.31
192.168.19.32
25 untuk Router dan 26 untuk server
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
jawab :
2N – 2 ≥ 20 11111111.11111111.11111111.11000000
2N ≥ 52 255 . 255 . 255 . 224
N = 6
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
jawab :
11111111.11111111.11111111.10000000
255 . 255 . 255 . 128
Block Subnet : 256 – 128 = 128
Network : ( 1 div 128 ) . 128 = 0
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
jawab :
2N – 2 ≥ 850 11111111.11111111.11111100.00000000
2N ≥ 852 255 . 255 . 252 . 0
N = 10
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
jawab :
11111111.11111111.11111100.00000000
255 . 255 . 252 . 0
a. 1024
b. 2046
c. 4094
d. 4096
e. 8190
jawab :
11111111.11111111.11110000.00000000 N = 12
255 . 255 . 240 . 0
Host : 2N - 2 = 4094
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
jawab:
Block subnet = 25 = 32
Host yang valid tidak boleh 0, kelipatan 32 ataupun selisih satu dari kelipatan 32 tersebut
Jadi host yang valid adalah B,C,D
a. 255.255.240.0
b. 255.255.248.0
c. 255.255.254.0
d. 255.255.255.0
jawab :
2N - 2 ≥ 450 11111111.11111111.11111110.00000000
2N ≥ 452 255 . 255 . 254 . 0
N = 9
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.
Jawab :
11111111.11111111.11111111.11100000
255 . 255 . 255 . 224