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
2n - 2 = 300
29 – 2 ≥ 300 9 àbit 0
512 – 2 ≥ 300
500 ≥300
Jadi , 11111111.11111111.11111110.00000000
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 - 192.168.1.70Gateway - 192.168.1.65
/27 = 255.255.255.224
256 – 224 = 32
0 32 64 96……………….
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
IP : 172.31.192.166 IP : 10100110
SM : 255.255.255.248 SM : 11111000
NA : 172.31.192.160 NA : 10100000
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 à default
e. 255.255.252.0 à CIDR+VLSM
f. 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
6. Which type of address is 223.168.17.167/29?
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
/29 = 255.255.255.248
256 – 248 = 8
0 8 16…………..160 168
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
/29 = 255.255.255.248
11111111.11111111.11111111.11111000
Subnet = 25 = 32 networks
Host = 23 – 2 = 8 – 2 = 6 hosts
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
224 = 11100000
Host = 25 – 2 = 32 – 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
2n – 2 = 27
25 – 2 ≥27 5 à bit 0
32 – 2 ≥ 27
Jadi, 11100000
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
2n – 2 = 14
24 – 2 ≥14 4 à bit 0
16 – 2 = 14
Jadi, 11110000
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
2n = 100
27 ≥ 100 7 à bit 1
128 ≥100
Jadi, 11111110
254
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
IP : 172.32.65.13
SM : 255.255.0.0
NA : 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
IP : 172.16.210.0
SM : 255.255.252.0
NA : 172.16.
IP : 11010010
SM : 11111100
NA : 11010000
208
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
IP : 115.64.4.0
SM : 255.255.252.0
256 – 252 = 4
0 4 8
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
IP : 200.10.5.68
SM : 255.255.255.240
NA : 200.10.5.64
IP : 01000100
SM : 11110000
NA : 01000000
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
IP : 172.16.0.0
SM : 255.255.224.0
224 = 11100000
Subnet = 23 = 8 subnets
Host = 213 – 2 = 8192 – 2 = 8190 hosts
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
2x = 500
29 ≥ 500 9 à bit 1
512 ≥ 500
Jadi, 11111111.10000000
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
IP : 172.16.66.0
SM : 255.255.248.0
NA : 172.16.64.0
IP : 01000010
SM : 11111000
NA : 01000000
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
2x = 100
27 ≥ 100 7 à bit 1
128 ≥ 100
2y – 2 = 500
29 – 2 ≥ 500 9 à bit 0
512 – 2 ≥ 500
510 ≥ 500
Jadi, 11111110.00000000
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
IP : 192.168.19.24
SM : 255.255.255. 248
NA : 192.168.19. 24
BA : 192.168.19.31
IP : 00011000
SM : 11111000
NA : 00011000
BA : 00011111
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
2n = 300
29 ≥ 300 9 à bit 1
512 ≥ 300
Jadi, 11111111.10000000
255 128
2m – 2 = 50
26 – 2 ≥ 50 6 à bit 0
64 – 2 ≥ 50
62 ≥ 50
Jadi, 11000000
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
IP : 172.16.112.1
SM : 255.255.255.128
NA : 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
850 host
2n – 2 = 850
210 – 2 ≥ 850 10 à bit 0
1024 – 2 ≥ 850
1022 ≥ 850
Jadi, 11111100.00000000
252 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
IP : 172.16.17.0
SM : 255.255.252.0
NA : 172.16.16.0
BA : 172.16.19.255
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
/20 = 255.255.240.0
240 = 11110000
Host= 212 – 2 = 4096 – 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
/27 = 255.255.255.224
256 – 224 = 32
0 32 64 96…….160 192
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
450 host
2x – 2 = 450
29 – 2 ≥ 450 9 à bit 0
512 – 2 ≥ 450
510 2 ≥ 450
Jadi, 11111110.0000000
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.
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