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 : (C) n=9, jadi subnetmasknya 11111111.11111111.11111110.00000000 atau 255.255.254.0
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
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 : (E,F) karena subnetmask sudah diketahui 225.255.225.248 net id dapat diketahui dengan berkelipatan 8
4. Which subnet masks would be valid for a subnetted Class B address? (Choose two.)
a. 255.0.0.0 à default subnet kelas A
b. 255.254.0.0 à subnet kelas A setelah di subnetting
c. 255.224.0.0 à subnet kelas A setelah di subnetting
d. 255.255.0.0 à default subnet kelas B
e. 255.255.252.0 à subnet kelas B setelah di subnetting
f. 255.255.255.192 à subnet kelas C setelah di subnetting
jawab: (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
6. Which type of address is 223.168.17.167/29?
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
jawab : (C) karena memiliki 29 bit 1 digunakan sebagai net id
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 : (C) netwok dengan rumus 2n-2 (dengan n adalah bit 1) dan untuk host pada net id dengan rumus 2n-2 (dengan n adalah bit 0), maka 25-2 = 30 network dan 23-2 = 6 host.
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 : (C) dengan subnet 225.255.255.224, maka di dapat biner 11111111.111111.1111111.11100000. dengan menggunakan rumus 2n-2 (dengan n bit 0), jadi 25-2 = 30 host.
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
jawab : (C) 2n-2 >= 27, n = 5 (terdapat 5 bit 0), jadi subnetmask 1111111.1111111.11111111.1110000 atau 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
jawab : (C) terdapat 14 host di setiap subnet, jadi 2n-2 >= 14, n = 4 (terdapat 4 bit 0). Jadi subnetnya 1111111.11111111.11111111.11110000 atau 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
jawab : (C) 2n-2 >=100, n = 7 (terdapat 7 bit 1), karena pada kelas B maka sunbetmasknya 11111111.1111111.1111110.0000000 atau 255.255.254.0
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
jawab : (A) karena default subnetmask maka network mirmiliki range IP 172.32.65.0-172.32.65.254
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 : (C) karena menggunakan 22 bit 1 sbg net id dan memiliki net id 172.16.0.0, 172.16.4.0, berkeliapatan 4 sampai di temukan 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
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 : (C) karena menggunkan 28 bit 1 sbg net id, maka memiliki net id 200.10.5.0, 200.10.5.16, ….,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
jawab : (F) karena menggunakan 19 bit 1 maka, 11111111.1111111.11100000.00000000, rumus subnet 2n (n jumlah bit 1) jadi 23 = 8 subnet, dan rumus host mengunakan rumus 2n-2 (bit 0) jadi 213-2 = 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
jawab : (B) 2n-2 >= 100, n = 7 (bit 0 yg pakai), jadi mendapat subnetmask baru 11111111.11111111.11111111.10000000 atau 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
jawab (C) karena menggunakan 21 bit 1 sbg net id, maka 11111111.11111111.11111000.00000000, maka didapat net id 172.16.0.0, 172.16.8.0, ..., 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
jawab : (C) 2n-2 >= 500, n = 9 (bit 0 yg dipakai), maka subnetnya 255.255.252.0
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
jawab : (e) 2n-2 >= 50, n = 6 (bit 0 yg dipakai), maka 11111111.11111111.11111111.11000000 atau 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
jawab (A) karena menggunakan 25 bit 1 sgb subnet, maka 11111111.11111111.11111111.10000000, maka net id 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
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
jawab : (C) Karena menggunakan 20 bit 1, maka 11111111.11111111.11110000.00000000 rumus 2n-2 (dengan n bit 0), 212-1 = 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 : (B,C,D) karena menggunakan 27 bit 1, maka 11111111.1111111.11111111.11100000, jadi net id nya w.x.y.0, w.x.y.32, w.x.y.64, w.x.y.128. w.x.y.160, w.x.y.192, w.x.y.224
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 : (C) 2n-2 >= 450, n = 9 (bit 0 yang dipakai), maka 11111111.11111111.11111110.00000000 atau 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.
Jawab : (D,E) karena default gateway berbeda dengan jaringan host, dan perbedaan jaringan dari serial pada interface router
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