Kamis, 22 April 2010
Kamis, 15 April 2010
Soal Subneting Sytem Adminstrator
Bismillah......
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.)
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: B
Default subnet mask class B= 11111111.11111111.0000000.00000000 karena yang diminta adalah 25
subnet, dan yang mendekati adalah 25 maka :
11111111.11111111.11111000.00000000 dan subnet masknya
255 . 255 . 248 . 0
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: D & F
/27 = 11111111.11111111.11111111.11100000
255 . 255 . 255 . 224
subnet masknya adalah : 255.255.255.224
Blok subnetnya = 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
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: E
IP : 172.31.192.166
Netmask : 255.255.255.248
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
------------------
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: D & E
Karena subnetmask default 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: D
jumlah host dari 172.16.128.1 to 172.16.159.254 = 8190 host
2n - 2 = yang hasilnya mendekati 8190 adalah 213 = 8192 – 2 = 8190
Karena jumlah bit “0” adalah 13 maka 11111111.11111111.11100000.00000000
Jadi subnetmasknya adalah 255.255.244.0
6. Which type of address is 223.168.17.167/29?
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
Jawaban: C
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
………………. ………………………………… ………………
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: C
/29 = 11111111.11111111.11111111.11111000
Subnet mask = 255.255.255.248
Block subnet = 256 – 248 = 8
Host = 23-2 = 6
Subnet = 25-2 = 30
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: C
255.255.255.224 = / 27
1111111.1111111.1111111.11100000
Maka, host = 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
Jawaban: C
Yang mendekati 27 dan tidak melebihi jauh yaitu 24, dan host dihitung dari kanan kekiri
11111111.11111111.11111111.11100000
Subnet = 255.255.255.224
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: C
Yang mendekati 14 dan tidak melebihi jauh yaitu 23 , dan host dihitung dari kanan kekiri…
11111111.11111111.11111111.11110000
Subnet = 255.255.255.240
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: D
2^7=128
11111111 11111111 11111111 00000000
255 255 255 0
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: C
IP Adrees : 172.32.65.13 = 10101100.00100000.01000001.00001101
Subnet mask:255.255.0.0 = 11111111.11111111.00000000.00000000
10101100.00100000.00000000.00000000
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
Jawaban: C
/22 = 11111111.11111111.11111100.00000000 = 255.255.252.0
IP: 172.16.210.0 = 10101100.00010000.11010010.00000000
Subnet: 255.255.252.0 = 11111111.11111111.11111100.00000000
10101100.00010000.11010000.00000000
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
Jawaban: C, D, & E
/22 = 255.255.252.0
Block subnet 256 – 252 = 4
115.64.4.1 – 115.64.7.255
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: C
/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
11001000.00001010.00000101.01000000
200 . 10 . 5 . 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
Jawaban: F
/19 = 11111111 11111111 11100000 00000000
Subnet 2^3=8
Host 2^13=8190
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: B
Subnet 2^9=(>500)
Host 2^7=(>100)
11111111 11111111 11111111 10000000
255 255 255 128
Subnetnya = 255.255.255.128
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: C
/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
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
Jawaban: B
26<100<27 dan subnet dihitung dari kiri kekanan
11111111 11111111 11111110 00000000
255 255 254 0
Subnetnya 255.255.254.0
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
Jawaban: C
/ 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
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: B & E
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
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
Jawaban: A
/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
10101100.00010000.01110000.00000000
172 . 16 . 112 . 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
Jawaban: D
Jumlah Host Keseluruhan= 3350
Host = 2^12 – 2 = 4094
Maka netmask yang sesuai adalah :
11111111 11111111 11110000 00000000
255 255 240 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
Jawaban: E
/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
-------------- --------------------------------- ---------------
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: C
/20 = 11111111.1111111.11110000.00000000
= 255 .255 .240 .0
Host = 212 – 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
Jawaban: : B, C, & D
/27 = 255.255.255.224
Kelas C
Subnet 2^3=8
Host 2^5-2=30
Blok subnet = 256 – 224 = 32
32, 64, 96, 128,…..
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: C
Host 2^9-2=510 (>450)
11111111 11111111 11111110 00000000
255 255 254 0
Subnetmasknya 255.255.254.0
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.
Jawaban: A & D
11111111 11111111 11111111 11100000
255 255 255 224
/ 27 = subnet mask = 255.255.255.224
Block subnetnya adalah = 256 – 224 = 32
32, 64, 96, 128,…
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