Question: How many subnets and hosts per subnet can you get from the network

172.31.0.0 255.255.248.0?

Answer: 32 subnets and 2046 hosts

Question: You are designing a subnet mask for the 172.21.0.0 network. You want 300 subnets with up to
110 hosts on each subnet. What subnet mask should you use?

Answer: 255.255.255.128

Question: What is the last valid host on the subnetwork 10.79.160.0 255.255.240.0?

Answer: 10.79.175.254

Question: How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?

Answer: 4096 subnets and 4094 hosts

Question: Which subnet does host 172.28.232.70 255.255.255.0 belong to?

Answer: 172.28.232.0

Question: Which subnet does host 172.20.180.227/20 belong to?

Answer: 172.20.176.0

Question: What valid host range is the IP address 172.23.8.38/20 a part of?

Answer: 172.23.0.1 through to 172.23.15.254

Question: What is the first valid host on the subnetwork that the node 172.30.18.141 255.255.254.0
belongs to?

Answer: 172.30.18.1

Question: How many subnets and hosts per subnet can you get from the network 172.22.0.0/21?

Answer: 32 subnets and 2046 hosts

Question: Which subnet does host 10.192.169.217 255.255.240.0 belong to?

Answer: 10.192.160.0

Question: What is the broadcast address of the network 192.168.108.144/29?

Answer: 192.168.108.151

Question: You are designing a subnet mask for the 172.16.0.0 network. You want 400 subnets with up to
90 hosts on each subnet. What subnet mask should you use?

Answer: 255.255.255.128

Question: What is the first valid host on the subnetwork that the node 192.168.177.178/27 belongs to?
Answer: 192.168.177.161

Question: What valid host range is the IP address 172.21.18.116/23 a part of?

Answer: 172.21.18.1 through to 172.21.19.254

Question: What is the last valid host on the subnetwork 172.29.203.64/26?

Answer: 172.29.203.126

Question: What is the last valid host on the subnetwork 172.25.44.0/23?

Answer: 172.25.45.254

Question: What is the broadcast address of the network 172.27.32.128 255.255.255.128?

Answer: 172.27.32.255

Question: How many subnets and hosts per subnet can you get from the network 192.168.58.0/27?

Answer: 8 subnets and 30 hosts

Question: What is the last valid host on the subnetwork 172.26.188.0 255.255.252.0?

Answer: 172.26.191.254

Question: Which subnet does host 10.45.226.164 255.255.240.0 belong to?

Answer: 10.45.224.0

Question: What valid host range is the IP address 172.28.67.164 255.255.255.192 a part of?

Answer: 172.28.67.129 through to 172.28.67.190

Question: What is the first valid host on the subnetwork that the node 172.31.224.112/23 belongs to?

Answer: 172.31.224.1

Question: What is the first valid host on the subnetwork that the node 172.22.255.151 255.255.255.0
belongs to?

Answer: 172.22.255.1

Question: What is the broadcast address of the network 192.168.93.168/29?

Answer: 192.168.93.175

Question: You are designing a subnet mask for the 172.25.0.0 network. You want 100 subnets with up to
300 hosts on each subnet. What subnet mask should you use?

Answer: 255.255.254.0

Question: How many subnets and hosts per subnet can you get from the network
192.168.95.0 255.255.255.240?
Answer: 16 subnets and 14 hosts

Question: What is the last valid host on the subnetwork 192.168.6.64 255.255.255.224?

Answer: 192.168.6.94

Question: What valid host range is the IP address 172.18.54.47/24 a part of?

Answer: 172.18.54.1 through to 172.18.54.254

Question: You are designing a subnet mask for the 172.19.0.0 network. You want 12 subnets with up to
2300 hosts on each subnet. What subnet mask should you use?

Answer: 255.255.240.0

Question: Which subnet does host 172.27.82.81/23 belong to?

Answer: 172.27.82.0

Question: What is the last valid host on the subnetwork 192.168.252.0 255.255.255.128?

Answer: 192.168.252.126

Question: Which subnet does host 172.17.27.177/23 belong to?

Answer: 172.17.26.0

Question: Which subnet does host 172.27.39.92 255.255.254.0 belong to?

Answer: 172.27.38.0

Question: What is the first valid host on the subnetwork that the node 192.168.72.199/27 belongs to?

Answer: 192.168.72.193

Question: What is the last valid host on the subnetwork 172.20.83.128 255.255.255.192?

Answer: 172.20.83.190

Question: What is the last valid host on the subnetwork 192.168.99.0 255.255.255.192?

Answer: 192.168.99.62

Question: How many subnets and hosts per subnet can you get from the network 172.30.0.0/24?

Answer: 256 subnets and 254 hosts

Question: What valid host range is the IP address 172.17.93.9 255.255.255.192 a part of?

Answer: 172.17.93.1 through to 172.17.93.62

Question: You are designing a subnet mask for the 172.25.0.0 network. You want 70 subnets with up to
300 hosts on each subnet. What subnet mask should you use?
Answer: 255.255.254.0

Question: What valid host range is the IP address 172.25.249.174/24 a part of?

Answer: 172.25.249.1 through to 172.25.249.254

Question: How many subnets and hosts per subnet can you get from the network
172.17.0.0 255.255.254.0?

Answer: 128 subnets and 510 hosts

Question: You are designing a subnet mask for the 172.27.0.0 network. You want 110 subnets with up to
400 hosts on each subnet. What subnet mask should you use?

Answer: 255.255.254.0

Question: How many subnets and hosts per subnet can you get from the network
172.16.0.0 255.255.254.0?

Answer: 128 subnets and 510 hosts

Question: What valid host range is the IP address 172.16.44.190/25 a part of?

Answer: 172.16.44.129 through to 172.16.44.254

Question: What valid host range is the IP address 172.21.15.55/28 a part of?

Answer: 172.21.15.49 through to 172.21.15.62

Question: What is the first valid host on the subnetwork that the node 172.19.147.10 255.255.254.0
belongs to?

Answer: 172.19.146.1

Question: Which subnet does host 10.82.53.5 255.255.240.0 belong to?

Answer: 10.82.48.0

Question: What is the last valid host on the subnetwork 192.168.140.128/26?

Answer: 192.168.140.190

Question: How many subnets and hosts per subnet can you get from the network 10.0.0.0/20?

Answer: 4096 subnets and 4094 hosts

Question: How many subnets and hosts per subnet can you get from the network
172.22.0.0 255.255.254.0?

Answer: 128 subnets and 510 hosts

Question: How many subnets and hosts per subnet can you get from the network
192.168.41.0 255.255.255.192?
Answer: 4 subnets and 62 hosts

Question: Which subnet does host 192.168.91.169 255.255.255.240 belong to?

Answer: 192.168.91.160

Question: How many subnets and hosts per subnet can you get from the network
172.23.0.0 255.255.254.0?

Answer: 128 subnets and 510 hosts

Question: What is the last valid host on the subnetwork 192.168.48.144/28?

Answer: 192.168.48.158

Question: Which subnet does host 192.168.174.172/26 belong to?

Answer: 192.168.174.128

Question: What is the broadcast address of the network 172.20.88.0 255.255.248.0?

Answer: 172.20.95.255

Question: What is the broadcast address of the network 172.18.24.0 255.255.254.0?

Answer: 172.18.25.255

Question: What is the broadcast address of the network 172.26.108.0 255.255.252.0?

Answer: 172.26.111.255

Question: How many subnets and hosts per subnet can you get from the network 172.30.0.0/28?

Answer: 4096 subnets and 14 hosts

Question: What valid host range is the IP address 192.168.82.141/27 a part of?

Answer: 192.168.82.129 through to 192.168.82.158

Question: Which subnet does host 172.17.224.52 255.255.255.224 belong to?

Answer: 172.17.224.32

Question: What is the first valid host on the subnetwork that the node 172.30.64.177/26 belongs to?

Answer: 172.30.64.129

Question: What is the first valid host on the subnetwork that the node 172.21.88.145 255.255.248.0
belongs to?

Answer: 172.21.88.1

Question: What valid host range is the IP address 192.168.255.89 255.255.255.224 a part of?
Answer: 192.168.255.65 through to 192.168.255.94

Question: What is the broadcast address of the network 172.20.176.0/23?

Answer: 172.20.177.255

Question: How many subnets and hosts per subnet can you get from the network
172.20.0.0 255.255.254.0?

Answer: 128 subnets and 510 hosts

Question: Which subnet does host 172.29.32.87 255.255.248.0 belong to?

Answer: 172.29.32.0

Question: What valid host range is the IP address 172.21.193.96/25 a part of?
Answer: 172.21.193.1 through to 172.21.193.126

Question: What is the first valid host on the subnetwork that the node 172.18.127.105/20 belongs to?

Answer: 172.18.112.1

Question: How many subnets and hosts per subnet can you get from the network 192.168.20.0/27?

Answer: 8 subnets and 30 hosts

Question: What is the first valid host on the subnetwork that the node 172.24.44.252 255.255.255.240
belongs to?

Answer: 172.24.44.241

Question: What is the first valid host on the subnetwork that the node 172.20.3.67/25 belongs to?

Answer: 172.20.3.1

Question: What is the first valid host on the subnetwork that the node 172.21.210.106/28 belongs to?
Answer: 172.21.210.97

Question: You are designing a subnet mask for the 172.19.0.0 network. You want 210 subnets with up to
150 hosts on each subnet. What subnet mask should you use?

Answer: 255.255.255.0

Question: How many subnets and hosts per subnet can you get from the network
192.168.65.0 255.255.255.240?

Answer: 16 subnets and 14 hosts

Question: How many subnets and hosts per subnet can you get from the network 10.0.0.0/20?

Answer: 4096 subnets and 4094 hosts

Question: What is the broadcast address of the network 172.28.172.0/22?

Answer: 172.28.175.255

Question: What is the broadcast address of the network 172.31.216.0/21?

Answer: 172.31.223.255

Question: What is the last valid host on the subnetwork 192.168.183.96 255.255.255.224?

Answer: 192.168.183.126

Question: How many subnets and hosts per subnet can you get from the network
192.168.26.0 255.255.255.240?

Answer: 16 subnets and 14 hosts

Question: What is the broadcast address of the network 192.168.187.80/28?

Answer: 192.168.187.95

Question: Which subnet does host 172.24.50.13/23 belong to?

Answer: 172.24.50.0

Question: What is the first valid host on the subnetwork that the node 10.75.84.221/20 belongs to?

Answer: 10.75.80.1

Question: What is the first valid host on the subnetwork that the node 10.46.11.207/20 belongs to?

Answer: 10.46.0.1

Question: How many subnets and hosts per subnet can you get from the network
172.18.0.0 255.255.254.0?

Answer: 128 subnets and 510 hosts

Question: What is the first valid host on the subnetwork that the node 172.19.178.2/20 belongs to?

Answer: 172.19.176.1

Question: How many subnets and hosts per subnet can you get from the network 192.168.35.0/28?

Answer: 16 subnets and 14 hosts

Question: How many subnets and hosts per subnet can you get from the network
192.168.207.0 255.255.255.252?

Answer: 64 subnets and 2 hosts

Question: What valid host range is the IP address 10.95.103.100/20 a part of?

Answer: 10.95.96.1 through to 10.95.111.254

Question: What valid host range is the IP address 172.25.45.67/23 a part of?

Answer: 172.25.44.1 through to 172.25.45.254

Question: What valid host range is the IP address 192.168.49.76/29 a part of?

Answer: 192.168.49.73 through to 192.168.49.78

Question: What is the last valid host on the subnetwork 192.168.80.96/27?

Answer: 192.168.80.126

Question: What is the broadcast address of the network 192.168.72.128 255.255.255.224?

Answer: 192.168.72.159

Question: How many subnets and hosts per subnet can you get from the network
192.168.25.0 255.255.255.224?

Answer: 8 subnets and 30 hosts

Question: What is the broadcast address of the network 192.168.106.16 255.255.255.248?

Answer: 192.168.106.23

Question: What valid host range is the IP address 172.29.228.73/25 a part of?

Answer: 172.29.228.1 through to 172.29.228.126

Question: What is the last valid host on the subnetwork 192.168.65.32/27?

Answer: 192.168.65.62

Question: Which subnet does host 172.26.22.143 255.255.255.192 belong to?

Answer: 172.26.22.128

Question: What is the first valid host on the subnetwork that the node 192.168.220.30/30 belongs to?

Answer: 192.168.220.29

Question: How many subnets and hosts per subnet can you get from the network
192.168.102.0 255.255.255.128?

Answer: 2 subnets and 126 hosts

Question: What is the first valid host on the subnetwork that the node 172.22.135.147/20 belongs to?

Answer: 172.22.128.1

Question: How many subnets and hosts per subnet can you get from the network 172.20.0.0/23?

Answer: 128 subnets and 510 hosts

Question: What is the first valid host on the subnetwork that the node 172.21.57.75/23 belongs to?

Answer: 172.21.56.1

Question: What is the broadcast address of the network 10.193.32.0 255.255.240.0?

Answer: 10.193.47.255

Question: What is the last valid host on the subnetwork 192.168.254.48/28?

Answer: 192.168.254.62

Question: What is the broadcast address of the network 172.16.151.128 255.255.255.128?

Answer: 172.16.151.255

Question: What is the last valid host on the subnetwork 172.18.98.0 255.255.255.128?
Answer: 172.18.98.126

Question: What is the first valid host on the subnetwork that the node 172.23.134.187 255.255.252.0
belongs to?

Answer: 172.23.132.1

Question: What is the last valid host on the subnetwork 10.184.176.0 255.255.240.0?

Answer: 10.184.191.254

Question: What is the last valid host on the subnetwork 172.25.49.96/27?

Answer: 172.25.49.126

Question: What is the last valid host on the subnetwork 172.20.50.32 255.255.255.224?

Answer: 172.20.50.62

Question: How many subnets and hosts per subnet can you get from the network 192.168.112.0/26?

Answer: 4 subnets and 62 hosts

Question: What is the first valid host on the subnetwork that the node 172.29.156.229/23 belongs to?

Answer: 172.29.156.1

Question: What is the last valid host on the subnetwork 10.244.32.0 255.255.240.0?

Answer: 10.244.47.254

Question: How many subnets and hosts per subnet can you get from the network
192.168.193.0 255.255.255.252?

Answer: 64 subnets and 2 hosts

Question 1

Given a subnet mask of 255.255.255.224, which of the following addresses can be assigned to network
hosts? (Choose three)

A – 15.234.118.63
B – 92.11.178.93
C – 134.178.18.56
D – 192.168.16.87
E – 201.45.116.159
F – 217.63.12.192

Answer: B C D

Explanation

A subnet mask of 255.255.255.224 has an increment of 32 (the binary form of the last octet is 1110
0000) so we can’t use numbers which are the multiples of 32 because they are sub-network addresses.
Besides, we can’t use broadcast addresses of these sub-networks (the broadcast address of the
previous subnet is calculated by subtracting 1 from the network address). For example the network
address of the 2nd subnet is x.x.x.32 then the broadcast address of the 1st subnet is 32 – 1 = 31
(means x.x.x.31).

By this method we can calculate the unusable addresses, which are (notice that these are the 4th
octets of the IP addresses only):

+ Network addresses: 0, 32, 64, 96, 128, 160, 192, 224.

+ Broadcast addresses: 31, 63, 95, 127,159, 191, 223.

Question 2

Which of the following host addresses are members of networks that can be routed across the public
Internet? (Choose three)

A – 10.172.13.65
B – 172.16.223.125
C – 172.64.12.29
D – 192.168.23.252
E – 198.234.12.95
F – 212.193.48.254

Answer: C E F

Explanation

Addresses that can be routed accross the public Internet are called public IP addresses. These
addresses belong to class A, B or C only and are not private addresses.

Note:

Private class A IP addresses: 10.0.0.0 to 10.255.255.255

Private class B IP addresses: 172.16.0.0 to 172.31.255.255
Private class C IP addresses: 192.168.0.0 to 192.168.255.255
Class D addresses are reserved for IP multicast addresses and can’t be routed across the Internet
(their addresses begin with 224.0.0.0 address).

Also we can’t use 127.x.x.x address because the number 127 is reserved for loopback and is used for
internal testing on the local machine.

Question 3

A national retail chain needs to design an IP addressing scheme to support a nationwide network. The
company needs a minimum of 300 sub-networks and a maximum of 50 host addresses per subnet.
Working with only one Class B address, which of the following subnet masks will support an
appropriate addressing scheme? (Choose two)

A – 255.255.255.0
B – 255.255.255.128
C – 255.255.252.0
D – 255.255.255.224
E – 255.255.255.192
F – 255.255.248.0

Answer: B E

Explanation

We need to remember the default subnet mask of class B is 255.255.0.0. Next, the company requires a
minimum of 300 sub-networks so we have to use at least 512 sub-networks (because 512 is the
minimum power of 2 and greater than 300). Therefore we need to get 9 bits for network mask
(29=512), leaving 7 bits for hosts which is 27-2 = 126 > 50 hosts per subnet.This scheme satisfies the
requirement -> B is correct.

We can increase the sub-networks to 1024 ( 1024 = 210), leaving 6 bits for hosts that is 26= 64 > 50
hosts. This scheme satisfies the requirement, too -> E is correct.

Notice: The question asks “The company needs a minimum of 300 sub-networks and a maximum of 50
host addresses per subnet” but this is a typo, you should understand it as “”The company needs a
minimum of 300 sub-networks and a minimum of 50 host addresses per subnet”.

Question 4
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

Answer: B C E

Explanation

CIDR stands for Classless In4ter-Domain Routing, the difference between CIDR and VLSM is slim and
those terms are interchangeable at CCNA level.

To specify which IP addresses fall into the CIDR block of 115.64.4.0/22 we need to write this IP address
and its subnet mask in binary form, but we only care 3rd octet of this address because its subnet mask
is /22.

(x means “don’t care”)

Next, we have to write the 3rd octets of the above answers in binary form to specify which numbers
have the same “prefixes” with 4.

4 = 0000 0100
8 = 0000 1000
7 = 0000 0111
6 = 0000 0110
3 = 0000 0011
5 = 0000 0101
12=0000 1100

We can see only 7, 6 and 5 have the same “prefixes” with 4 so B C E are the correct answers.

Question 5

Refer to the diagram. All hosts have connectivity with one another. Which statements describe the
addressing scheme that is in use in the network? (Choose three)
A – The subnet mask in use is 255.255.255.192.
B – The subnet mask in use is 255.255.255.128.
C – The IP address 172.16.1.25 can be assigned to hosts in VLAN1.
D – The IP address 172.16.1.205 can be assigned to hosts in VLAN1.
E – The LAN interface of the router is configured with one IP address.
F – The LAN interface of the router is configured with multiple IP addresses.

Answer: B C F

Explanation

VLAN 2 has 114 hosts so we need to leave 7 bits 0 for the host addresses (2 7 – 2 = 126 > 114). Notice
that we are working with class B (both Host A and Host B belong to class B) and the default subnet
mask of class B is /16 so we need to use 16 – 7 = 9 bits 1 for the sub-network mask, that means the
subnet mask should be 255.255.255.128 -> B is correct.

By using above scheme, C is correct because the IP 172.16.1.25 belongs to the sub-network of VLAN 1
(172.16.1.0/25) and can be assigned to hosts in VLAN 1.

For communication between VLAN 1 and VLAN 2, the LAN interface of the router should be divided into
multiple sub-interfaces with multiple IP addresses -> F is correct.

Question 6

The network 172.25.0.0 has been divided into eight equal subnets. Which of the following IP addresses
can be assigned to hosts in the third subnet if the ip subnet-zero command is configured on the router?
(Choose three)

A – 172.25.78.243
B – 172.25.98.16
C – 172.25.72.0
D – 172.25.94.255
E – 172.25.96.17
F. 172.25.100.16
Answer: A C D

Explanation

If the “ip subnet-zero” command is configured then the first subnet is 172.25.0.0. Otherwise the first
subnet will be 172.25.32.0 (we will learn how to get 32 below).

The question stated that the network 172.25.0.0 is divided into eight equal subnets therefore the
increment is 256 / 8 = 32 and its corresponding subnet mask is /19 (1111 1111.1111 1111.1110
0000).

First subnet: 172.25.0.0/19

Second subnet: 172.25.32.0/19
Third subnet: 172.25.64.0/19
4th subnet: 172.25.96.0/19
5th subnet: 172.25.128.0/19
6th subnet: 172.25.160.0/19
7th subnet: 172.25.192.0/19
8th subnet: 172.25.224.0/19

In fact, we only need to specify the third subnet as the question requested. The third subnet ranges
from 172.25.64.0/19 to 172.25.95.255/19 so A C D are the correct answers.

Question 7

Refer to the exhibit. In this VLSM addressing scheme, what summary address would be sent from
router A?

A. 172.16.0.0/16
B. 172.16.0.0/20
C. 172.16.0.0/24
D. 172.32.0.0/16
E. 172.32.0.0/17
F. 172.64.0.0/16
Answer: A

Explanation

Router A receives 3 subnets: 172.16.64.0/18, 172.16.32.0/24 and 172.16.128.0/18.

All these 3 subnets have the same form of 172.16.x.x so our summarized subnet must be also in that
form -> Only A, B or C is correct.

The smallest subnet mask of these 3 subnets is /18 so our summarized subnet must also have its
subnet mask equal or smaller than /18.

-> Only answer A has these 2 conditions -> A is correct.

Question 1

Refer to the exhibit. Which VLSM mask will allow for the appropriate number of host addresses for
Network A?

A. /25
B. /26
C. /27
D. /28

Answer: A

Explanation

We need 66 hosts < 128 = 27 -> We need 7 bits 0 -> The subnet mask should be 1111 1111.1111
1111.1111 1111.1000 0000 -> /25

Question 2

Refer to the exhibit. Which subnet mask will place all hosts on Network B in the same subnet with the
least amount of wasted addresses?
A. 255.255.255.0
B. 255.255.254.0
C. 255.255.252.0
D. 255.255.248.0

Answer: B

Explanation

310 hosts < 512 = 29 -> We need a subnet mask of 9 bits 0 -> 1111 1111.1111 1111.1111 1110.0000
0000 -> 255.255.254.0

Question 3

Refer to the exhibit. Which mask is correct to use for the WAN link between the routers that will
provide connectivity while wasting the least amount of addresses?

A. /23
B. /24
C. /25
D. /30

Answer: D

Explanation

For WAN link we only need 2 usable host addresses for 2 interfaces on the routers. The subnet mask of
/30 gives us 22 – 2 = 2 usable host addresses. Also remember that “/30″ is famous for point-to-point
connection because it wastes the least amount of addresses.

Question 4

Refer to the exhibit. What is the most appropriate summarization for these routes?

A. 10.0.0.0/21
B. 10.0.0.0/22
C. 10.0.0.0/23
D. 10.0.0.0/24
Answer: B

Explanation

We need to summarize 4 subnets so we have to move left 2 bits (22 = 4). In this question we can guess
the initial subnet mask is /24 because 10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0 belong to different
networks. So “/24″ moves left 2 bits -> /22.

Question 5

On the network 131.1.123.0/27, what is the last IP address that can be assigned to a host?

A. 131.1.123.30
B. 131.1.123.31
C. 131.1.123.32
D. 131.1.123.33

Answer: A

Explanation

Increment: 32
Network address: 131.1.123.0 & 131.1.123.32
Broadcast address: 131.1.123.31

Both 131.1.123.30 & 131.1.123.33 can be assigned to host but the question asks about the “last IP
address” so A is the correct answer.

Question 6

The ip subnet zero command is not configured on a router. What would be the IP address of
Ethernet0/0 using the first available address from the sixth subnet of the network 192.168.8.0/29?

A. 192.168.8.25
B. 192.168.8.41
C. 192.168.8.49
D. 192.168.8.113

Answer: C

Explanation

The “ip subnet zero” is not configured so the first subnet will start at 192.168.8.8 (ignoring
192.168.8.0).

Increment: 8
1st subnet: 192.168.8.8
2nd subnet: 192.168.8.16
3rd subnet: 192.168.8.24
4th subnet: 192.168.8.32
5th subnet: 192.168.8.40
6th subnet: 192.168.8.48 -> The first usable IP address of 6 th subnet is 192.168.8.49
Question 7

For the network 192.0.2.0/23, which option is a valid IP address that can be assigned to a host?

A. 192.0.2.0
B. 192.0.2.255
C. 192.0.3.255
D. 192.0.4.0

Answer: B

Explanation

Increment: 2

Network address: 192.0.2.0, 192.0.4.0

Broadcast address: 192.0.3.255

-> 192.0.2.255 is not a broadcast address, it is an usable IP address.

Question 8

How many addresses for hosts will the network 124.12.4.0/22 provide?

A. 510
B. 1022
C. 1024
D. 2048

Answer: B

Explanation

/22 gives us 10 bits 0 -> 210 – 2 = 1022. Notice that the formula to calculate the number of host is: 2 k –
2.

Question 9

The network default gateway applying to a host by DHCP is 192.168.5.33/28. Which option is the valid
IP address of this host?

A. 192.168.5.55
B. 192.168.5.47
C. 192.168.5.40
D. 192.168.5.32
E. 192.168.5.14

Answer: C

Question 10

Which two addresses can be assigned to a host with a subnet mask of 255.255.254.0? (Choose two)
A. 113.10.4.0
B. 186.54.3.0
C. 175.33.3.255
D. 26.35.2.255
E. 17.35.36.0

Answer: B D

Question 1

Workstation A has been assigned an IP address of 192.0.2.24/28. Workstation B has been assigned an
IP address of 192.0.2.100/28. The two workstations are connected with a straight-through cable.
Attempts to ping between the hosts are unsuccessful. What two things can be done to allow
communications between the hosts? (Choose two)

A. Replace the straight-through cable with a crossover cable.

B. Change the subnet mask of the hosts to /25.
C. Change the subnet mask of the hosts to /26.
D. Change the address of Workstation A to 192.0.2.15.
E. Change the address of Workstation B to 192.0.2.111.

Answer: A B

Explanation

To specify when we use crossover cable or straight-through cable, we should remember:

Group 1: Router, Host, Server
Group 2: Hub, Switch
One device in group 1 + One device in group 2: use straight-through cable
Two devices in the same group: use crossover cable

-> To connect two hosts we must use crossover cable -> A is correct.

With the subnet mask of /28, 192.0.2.24 & 192.0.2.100 will be in different subnets (192.0.2.24 belongs
to subnet 192.0.2.16/28; 192.0.2.100 belongs to subnet 192.0.2.96). To make them in the same
subnet we need more space for host. Because 100 < 128 so we the suitable subnet should be /25.

Question 2

Your ISP has given you the address 223.5.14.6/29 to assign to your router’s interface. They have also
given you the default gateway address of 223.5.14.7. After you have configured the address, the
router is unable to ping any remote devices. What is preventing the router from pinging remote
devices?

A. The default gateway is not an address on this subnet.

B. The default gateway is the broadcast address for this subnet.
C. The IP address is the broadcast address for this subnet.
D. The IP address is an invalid class D multicast address.

Answer: B
Explanation

For the network 223.5.14.6/29:

Increment: 8
Network address: 223.5.14.0
Broadcast address: 223.5.14.7

-> The default gateway IP address is the broadcast address of this subnet -> B is correct.

Question 3

Refer to the exhibit. According to the routing table, where will the router send a packet destined for
10.1.5.65?

Network Interface Next-hop

10.1.1.0/24 e0 directly
connected

10.1.2.0/24 e1 directly
connected

10.1.3.0/25 s0 directly
connected

10.1.4.0/24 s1 directly
connected

10.1.5.0/24 e0 10.1.1.2

10.1.5.64/28 e1 10.1.2.2

10.1.5.64/29 s0 10.1.3.3

10.1.5.64/27 s1 10.1.4.4

A. 10.1.1.2
B. 10.1.2.2
C. 10.1.3.3
D. 10.1.4.4

Answer: C

Explanation

The destination IP address 10.1.5.65 belongs to 10.1.5.64/28, 10.1.5.64/29 & 10.1.5.64/27 subnets but
the “longest prefix match” algorithm will choose the most specific subnet mask -> the prefix “/29″ will
be chosen to route the packet. Therefore the next-hop should be 10.1.3.3 -> C is correct.
Question 4

Refer to the exhibit. The user at Workstation B reports that Server A cannot be reached. What is
preventing Workstation B from reaching Server A?

A. The IP address for Server A is a broadcast address.

B. The IP address for Workstation B is a subnet address.
C. The gateway for Workstation B is not on the same subnet.
D. The gateway for Server A is not on the same subnet.

Answer: D

Question 5

Given the address 192.168.20.19/28, which of the following are valid host addresses on this subnet?
(Choose two)

A. 192.168.20.29
B. 192.168.20.16
C. 192.168.20.17
D. 192.168.20.31
E. 192.168.20.0

Answer: A C

Question 6

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
Answer: B C E

Question 7

The Ethernet networks connected to router R1 in the graphic have been summarized for router R2 as
192.1.144.0/20. Which of the following packet destination addresses will R2 forward to R1, according
to this summary? (Choose two)

A. 192.1.159.2
B. 192.1.160.11
C. 192.1.138.41
D. 192.1.151.254
E. 192.1.143.145
F. 192.1.1.144

Answer: A D

Question 8

Refer to the exhibit. All of the routers in the network are configured with the ip subnet-zero command.
Which network addresses should be used for Link A and Network A? (Choose two)

A. Network A – 172.16.3.48/26
B. Network A – 172.16.3.128/25
C. Network A – 172.16.3.192/26
D. Link A – 172.16.3.0/30
E. Link A – 172.16.3.40/30
F. Link A – 172.16.3.112/30

Answer: B D

Explanation

Network A needs 120 hosts < 128 = 27 -> Need a subnet mask of 7 bit 0s -> “/25″.

Because the ip subnet-zero command is used, network 172.16.3.0/30 can be used.

Answer E “Link A – 172.16.3.40/30″ is not correct because this subnet belongs to MARKETING subnet
(172.16.3.32/27).
Answer F “Link A – 172.16.3.112/30″ is not correct because this subnet belongs to ADMIN subnet
(172.16.3.96/27).

Question 9

Which two subnetworks would be included in the summarized address of 172.31.80.0/20? (Choose
two)

A. 172.31.17.4/30
B. 172.31.51.16 /30
C. 172.31.64.0/18
D. 172.31.80.0/22
E. 172.31.92.0/22
F. 172.31.192.0/18

Answer: D E

Explanation

From the summarized address of 172.31.80.0/20, we find the range of this summarized network:

Increment: 16
Network address: 172.31.80.0
Broadcast address: 172.31.95.255

-> Answer D & E belong to this range so they are the correct answers.

Question 10

Which three IP addresses can be assigned to hosts if the subnet mask is /27 and subnet zero is usable?
(Choose three)

A. 10.15.32.17
B. 17.15.66.128
C. 66.55.128.1
D. 135.1.64.34
E. 129.33.192.192
F. 192.168.5.63

Answer: A C D
Explanation

First we need to find out the forms of network addresses and broadcast addresses when the subnet
mask of /27 is used:

Increment: 32
Network address: In the form of x.x.x.(0,32,64,96,128,160,192,224)
Broadcast address: In the form of x.x.x.(31,63,95,127,159,191,223)
So we only need to check the fourth octets of the IP addresses above. If they are not in the form of
network addresses or broadcast addresses then they can be assigned to hosts.

Notice that the IP 66.55.128.1 belongs to the subnet zero and the question says subnet zero is usable
so it is valid.

Question 11

Which of the following IP addresses can be assigned to the host devices? (Choose two)

A. 205.7.8.32/27
B. 191.168.10.2/23
C. 127.0.0.1
D. 224.0.0.10
E. 203.123.45.47/28
F. 10.10.0.0/13

Answer: B F

Explanation

This is a time-consuming question (but not hard ^^) because we have to calculate the range of each
sub-network separately (excepting answer C is the local loopback address & answer D is a multicast
address) so make sure you can do subnet quickly. After solving above questions I believe you can find
out the result so I don’t explain this question in detail.

Question 12

How many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask, and how many usable
host addresses will there be per subnet?

A. 8 subnets, 31 hosts
B. 8 subnets, 32 hosts
C. 16 subnets, 30 hosts
D. 16 subnets, 32 hosts
E. A Class B address cant be subnetted into the fourth octet.

Answer: C

Explanation

Subnetting from /23 to /27 gives us 27 – 23 = 4 bits -> 24 = 16 subnets.

/27 has 5 bit 0s so it gives 25 – 2 = 30 hosts-per-subnet.

Question 1
You are working in a data center environment and are assigned the address range 10.188.31.0/23. You
are asked to develop an IP addressing plan to allow the maximum number of subnets with as many as
30 hosts each.Which IP address range meets these requirements?

A. 10.188.31.0/27
B. 10.188.31.0/26
C. 10.188.31.0/29
D. 10.188.31.0/28
E. 10.188.31.0/25

Answer: A

Explanation

Each subnet has 30 hosts < 32 = 25 so we need a subnet mask which has at least 5 bit 0s -> /27. Also
the question requires the maximum number of subnets (which minimum the number of hosts-per-
subnet) so /27 is the best choice -> A is correct.

Question 2

Refer to the exhibit. The Lakeside Company has the internetwork in the exhibit. The Administrator
would like to reduce the size of the routing table to the Central Router. Which partial routing table
entry in the Central router represents a route summary that represents the LANs in Phoenix but no
additional subnets?

A. 10.0.0.0 /22 is subnetted, 1 subnet

D 10.0.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

B. 10.0.0.0 /28 is subnetted, 1 subnet

D 10.2.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1
C. 10.0.0.0 /30 is subnetted, 1 subnet
D 10.2.2.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

D. 10.0.0.0 /22 is subnetted, 1 subnet

D 10.4.0.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

E. 10.0.0.0 /28 is subnetted, 1 subnet

D 10.4.4.0 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

F. 10.0.0.0 /30 is subnetted, 1 subnet

D 10.4.4.4 [90/20514560] via 10.2.0.2 6w0d, serial 0/1

Answer: D

Explanation

All the above networks can be summarized to 10.0.0.0 network but the question requires to “represent
the LANs in Phoenix but no additional subnets” so we must summarized to 10.4.0.0 network. The
Phoenix router has 4 subnets so we need to “move left” 2 bits of “/24″-> /22 is the best choice -> D is
correct.

Question 3

Which address range efficiently summarizes the routing table of the addresses for router main?

A. 172.16.0.0/18
B. 172.16.0.0/16
C. 172.16.0.0/20
D. 172.16.0.0/21

Answer: C

Explanation
To summarize these networks efficiently we need to find out a network that “covers” from 172.16.1.0 -
> 172.16.13.0 (including 13 networks < 16). So we need to use 4 bits (2 4 = 16). Notice that we have to
move the borrowed bits to the left (not right) because we are summarizing.

The network 172.16.0.0 belongs to class B with a default subnet mask of /16 but in this case it has
been subnetted with a subnet mask of /24 (we can guess because 172.16.1.0, 172.16.2.0,
172.16.3.0… are different networks).

Therefore “move 4 bits to the left” of “/24″ will give us “/20″ -> C is the correct answer.

Question 4

Refer to the exhibit. A new subnet with 60 hosts has been added to the network. Which subnet address
should this network use to provide enough usable addresses while wasting the fewest addresses?

A. 192.168.1.56/27
B. 192.168.1.64/26
C. 192.168.1.64/27
D. 192.168.1.56/26

Answer: B

Explanation

60 hosts < 64 = 26 -> we need a subnet mask of at least 6 bit 0s -> “/26″. The question requires
“wasting the fewest addresses” which means we have to allow only 62 hosts-per-subnet -> B is
correct.

Question 5

The network technician is planning to use the 255.255.255.224 subnet mask on the network. Which
three valid IP addresses can the technician use for the hosts? (Choose three)

A. 172.22.243.127
B. 172.22.243.191
C. 172.22.243.190
D. 10.16.33.98
E. 10.17.64.34
F. 192.168.1.160

Answer: C D E

Explanation

From the subnet mask of 255.255.255.224 we learn:

Increment: 32
Network address: In the form of x.x.x.(0,32, 64, 96, 128, 160, 192, 224)
Broadcast address: In the form of x.x.x.(31,63,95,127,159,191,223)

-> All IP addresses not in the above forms are usable for host -> C D E are correct answers.

Question 6

In the implementation of VLSM techniques on a network using a single Class C IP address, which
subnet mask is the most efficient for point-to-point serial links?

A. 255.255.255.240
B. 255.255.255.254
C. 255.255.255.252
D. 255.255.255.0
E. 255.255.255.248

Answer: C

Explanation

The subnet mask of 255.255.255.252 gives only 2 usable host addresses because it has only 2 bit 0s
(22 – 2 = 2) so it is the most efficient subnet mask for point-to-point serial links (and you should
remember it).

Question 7

Refer to the exhibit. HostA cannot ping HostB. Assuming routing is properly configured, what could be
the cause of this problem?

A. HostA is not on the same subnet as its default gateway.

B. The address of SwitchA is a subnet address.
C. The Fa0/0 interface on RouterA is on a subnet that can’t be used.
D. The serial interfaces of the routers are not on the same subnet.
E. The Fa0/0 interface on RouterB is using a broadcast address.
Answer: D

Explanation

Now let’s find out the range of the networks on serial link:

For the network 192.168.1.62/27:

Increment: 32
Network address: 192.168.1.32
Broadcast address: 192.168.1.63

For the network 192.168.1.65/27:

Increment: 32
Network address: 192.168.1.64
Broadcast address: 192.168.1.95

-> These two IP addresses don’t belong to the same network and they can’t see each other -> D is the
correct answer.

Question 8

The network administrator is asked to configure 113 point-to-point links. Which IP addressing scheme
best defines the address range and subnet mask that meet the requirement and waste the fewest
subnet and host addresses?

A. 10.10.0.0/18 subnetted with mask 255.255.255.252

B. 10.10.0.0/25 subnetted with mask 255.255.255.252
C. 10.10.0.0/24 subnetted with mask 255.255.255.252
D. 10.10.0.0/23 subnetted with mask 255.255.255.252
E. 10.10.0.0/16 subnetted with mask 255.255.255.252

Answer: D

Explanation

We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7 bits
(because 2^7 = 128).

The network used for point-to-point connection should be /30.

So our initial network should be 30 – 7 = 23.

So 10.10.0.0/23 is the correct answer.

You can understand it more clearly when writing it in binary form:

/23 = 1111 1111.1111 1110.0000 0000

/30 = 1111 1111.1111 1111.1111 1100 (borrow 7 bits)

Question 9

If an Ethernet port on a router was assigned an IP address of 172.1.1.1/20, what is the maximum
number of hosts allowed on this subnet?
A. 4094
B. 1024
C. 8190
D. 2046
E. 4096

Answer: A

Explanation

In the prefix /20 we have 12 bit 0s so the number of hosts-per-subnet is 2 12 – 2 = 4094.

Question 10

A network administrator receives an error message while trying to configure the Ethernet interface of a
router with IP address 10.24.24.24/29. Which statement explains the reason for it?

A. The address is a broadcast address

B. The Ethernet interface is faulty
C. VLSM-capable routing protocols must be enable first on the router.
D. This address is a network address.

Answer: D