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
- Uses class B memiliki 25 subnet
- max host 300 computer.
24<>5 jadi,
- default subnet mask class B 11111111.11111111.0000000.00000000
Ditanyakan:
25 subnet
Jawab:
yang mendekati adalah 25 jadi lima buah bit 1menjadi:
255 . 255 . 248 . 0
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
- Subnet masknya = 255.255.255.224
- Block subnetnya 256 – 224 = 32
Net ID
192.168.1.31
-----------------
192.169.1.64
RHA
192.168.1.1 – 192.168.1.30
----------------------------------
192.168.1.65-192.168.1.94
Broadcast
192.168.1.0
----------------
192.168.1.95
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
Block subnet = 256 – 248 = 8
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
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 :
Jawaban: C
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
Jawaban:
IP addressnya class C
/29 = 11111111.11111111.11111111.11111000
- NET ID
223.168.17.0
223.168.17.8
-------------------
223.168.17.160
- RHA
223.168.17.1- 223.168.17.6
223.168.17.9- 223.168.17.14
----------------------------------------
223.168.17.161-223.168.17.166
- BROADCAST
223.168.17.7
223.168.17.15
--------------------
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
IP addressnya class C
/29 = 11111111.11111111.11111111.11111000
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
Answer :
255.255.255.224 = / 27 =
1111111.1111111.1111111.11100000
jumlah host didapat dari jumlah sisa bit 0
jadi host = 25 – 2 = 32-2=30
Jawaban 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
Answer :
24<27<25, host dihitung dari kanan ke kiri
11111111.11111111.11111111.11100000
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
Answer :
23<14<24 ,host dihitung dari kanan ke kiri
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
Answer :
26<100<7 subnet dihitung dari kiri ke kanan
11111111.11111111.11111110.00000000
Subnetnya 255.255.254.0
Jawabannya C
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
10101100.00100000.00000000.00000000
172 . 32 . 0 . 0
Jawaban: 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
10101100.00010000.11010000.00000000
172 . 16 . 208 . 0
Jawaban: 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 subnetnay: 256 – 252 = 4
115.64.4.1 – 115.64.7.255
Jawaban: 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
11001000.00001010.00000101.01000000
200 . 10 . 5 . 64
Jawaban: 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 = 255.255.224.0
Host = 213 – 2 = 8190
Subnetwork = 23 = 8
Jawaban: 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:
28<500<29 subnet dihitung dari kiri ke kanan
11111111.11111111.11111111.10000000
Subnetnya 255.255.255.128
Jawaban: 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
Jawaban: 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. 255.255.255.0
b. 255.255.254.0
c. 255.255.252.0
d. 255.255.0.0
Jawaban:
26<100<27 subnet dihitung dari kiri ke kanan
11111111.11111111.11111110.00000000
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
Block subnet = 256 – 248 = 8
NET ID RHA 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 subnet dihitung dari kiri ke kanan
11111111.11111111.11111111.10000000
Subnetnya : 255.255.255.128
25<50<26 host dihitung dari kanan ke kiri
11111111.11111111.11111111.11000000
Subnetnya: 255.255.255.192
Jawaban: B & 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?
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:
/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
Jawaban: 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:
Jawaban: 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
Jawaban: 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
Block subnet = 256 – 224 = 32
32, 64, 96, 128,…..
Jawaban: 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:
28<450<29 host dihitung dari kanan ke kiri
11111111.11111111.11111110.00000000
Subnetmasknya : 255.255.254.0
Jawaban: 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.)
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:
/ 27 = subnet mask = 255.255.255.224
Blocksubnetnya = 256 – 224 = 32
32, 64, 96, 128,…
Jawaban: A & D
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