CCNA 640-802 Exam

QUESTION 61:
You have a network that supports VLSM and you need to reduce IP address waste in your point to point WAN links. Which of the masks below would you use?
A. /38
B. /30
C. /27
D. /23
E. /18
F. /32

Explanation:
For a single point to point link, only 2 IP addresses are required, one for the serial interface of the router at each end. Therefore, the 255.255.255.252 subnet mask is often used for these types of links because no IP addresses are wasted. The subnet mask 255.255.255.252 is a /30, so answer B is correct.

Incorrect Answers:
A. The largest mask that can be used is the single IP host mask, which is /32. It is not possible to use a /38 mask, unless of course IPv6 is being used.
C, D, E. These masks will provide for a larger number of host addresses, and since only 2
IP addresses are needed for a point to point link, these extra addresses are wasted.
F: No available host addresses with a /32 mask

QUESTION 62:
How would you express the binary number: 10101010 in its decimal and hexadecimal forms?
A. Decimal=160, hexadecimal=00
B. Decimal=170, hexadecimal=AA
C. Decimal=180, hexadecimal=BB
D. Decimal=190, hexadecimal=CC

Explanation:
For the binary equivalent of 10101010 to Decimal, the answer is 128+32+8+2=170.
For the hexadecimal number, we need to break up the binary number into two bytes of 1010 and 1010. Each one in binary is then 10 and 10, which is A and A in hexadecimal.

QUESTION 63:
Which of the following IP hosts would be valid for PC users, assuming that a /27 network mask was used for all of the networks? (Choose all that apply.)
A. 15.234.118.63
B. 83.121.178.93
C. 134.178.18.56
D. 192.168.19.37
E. 201.45.116.159
F. 217.63.12.192

Explanation:
With a 255.255.255.224 network mask, the network boundaries will be a multiple of 32, so any network will have a multiple of 32 (32, 64, 96, 128, 160, 192, 224) in the last octet. If we subtract 1 from each of these numbers (so we have 31, 63, 95, etc), we know that any IP address ending in any of these numbers will be a broadcast address.
Valid Address Current host range 83.121.178.93 83.121.178.65 to 82.121.178.94 134.178.18.56 134.178.18.33 to 134.178.18.62 192.168.19.37 192.168.19.33 to 192.168.19.62

Incorrect Answers:
A. This is the broadcast address for the 15.234.118.32/27 network.
E. This is the broadcast address for the 201.45.116.128/27 network.
F. This is the network address for the 217.63.12.192/27 network.

QUESTION 64:
You are the network administrator at Certkiller . Certkiller has been provided with the network address 165.100.27.0/24. The Certkiller CEO wants to know how many subnetworks this address provides, and how many hosts can be supported on each subnet.
What would your reply be? (Choose all that apply)
A. One network with 254 hosts.
B. 254 networks with 254 hosts per network.
C. 65,534 networks with 255 hosts per network.
D. 30 networks with 64 hosts per network.
E. 254 networks with 65,534 per network.

When we have address 165.100.27.0/24 the number of networks is 1 with 254 hosts because this address is already subnetted and valid hosts range are 165.100.27.1-165.100.27.254, making the right answer
A. If the address was
165.100.0.0/24 then right answer is B.

QUESTION 65:
DRAG DROP
Certkiller has three locations and has plans to redesign the network accordingly. The networking team received 192.168.126.0 to use as the addressing for entire network from the administrator. After subnetting the address, the team is ready to assign the address.
The administrator plans to configure ip subnet-zero and use RIP v2 as the routing protocol. As a member of the networking team, you must address the network and at the same time converse unused addresses for future growth.
Being mindful of these goals, drag the host addresses on the left to the correct router interface. One of the routers is partially configured. Move the mouse over a router to view its configuration (** This information is missing**). Not all of the host addresses on the left will be used.

Answer:

Certkiller 2 Fa0/0 and Certkiller 3 Fa0/0 both can have either of the following 192.168.126.35/27 or 192.168.126.2/27

QUESTION 66:
The Certkiller network has been divided into 5 separate departments as displayed below:


Using a Class C IP network, which subnet mask will provide one usable subnet per department while allowing enough usable host addresses for each department specified in the graphic?
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
F. 255.255.255.252

Explanation:
Choice C will provide for 8 separate subnets with 30 usable hosts per subnet. Since we only require 5 different subnets with at most 16 users, this will suffice.

Incorrect Answers:
A. This will only provide 1 network with 254 hosts. This question requires 5 different networks.
B. This will only provide 4 networks, with 62 hosts per network.
D. This will provide for 14 networks, but with only 14 hosts per network so there will not be enough hosts for the Production and Engineering LANs.
E. This will provide for 62 different networks, but each with only 2 usable hosts per network.

QUESTION 67:
DRAG DROP
Certkiller has three locations and has plans to redesign the network accordingly. The networking team received 192.168.55.0 to use as the addressing for entire network from the administrator. After subnetting the address, the team is ready to assign the address.
The administrator plans to configure ip subnet-zero and use RIP v2 as the routing protocol. As a member of the networking team, you must address the network and at the same time converse unused addresses for future growth.
Being mindful of these goals, drag the host addresses on the left to the correct router interface. One of the routers is partially configured. Not all of the host addresses on the left will be used.

Answer:


QUESTION 68:
You are the network administrator at Certkiller . Certkiller has been assigned the class C IP address 189.66.1.0 by its Internet Service Provider. If you divide the network range by using the 255.255.255.224 subnet mask, how many hosts can be supported on each network?
A. 14
B. 16
C. 30
D. 32
E. 62
F. 64

Explanation:
The subnet mask 255.255.255.224 is a 27 bit mask (11111111.11111111.11111111.11100000). It uses 3 bits from the host Id for the network ID, leaving 5 bits for host addresses. We can calculate the number of hosts supported by this subnet by using the 2n-2 formula where n represents the number of host bits. In this case it will be 5. 25-2 gives us 30.

Incorrect Answers:
A. Subnet mask 255.255.255.240 will give us 14 host addresses.
B. Subnet mask 255.255.255.240 will give us a total of 16 addresses. However, we must still subtract two addresses (the network address and the broadcast address) to determine the maximum number of hosts the subnet will support.
D. Subnet mask 255.255.255.224 will give us a total of 32 addresses. However, we must still subtract two addresses (the network address and the broadcast address) to determine the maximum number of hosts the subnet will support.
E. Subnet mask 255.255.255.192 will give us 62 host addresses.
F. Subnet mask 255.255.255.192 will give us a total of 64 addresses. However, we must still subtract two addresses (the network address and the broadcast address) to determine the maximum number of hosts the subnet will support.

QUESTION 69:
Which of the following statements are true regarding a network using a subnet mask of 255.255.248.0? (Choose three)
A. It corresponds to a Class A address with 13 bits borrowed.
B. It corresponds to a Class B address with 4 bits borrowed.
C. The network address of the last subnet will have 248 in the 3rd octet.
D. The first 21 bits make the host portion of the address.
E. This subnet mask allows for 16 total subnets to be created.
F. The subnetwork numbers will be in multiples of 8.

Explanation:
This subnet mask includes the first 5 bits within the third octet, so for a class A address 13 bits will be used for the mask (8 bits in the second octet plus 5 in the third).
Since the first 5 bits are used in this octet, that means that remaining 3 bits in this octet will be available for hosts, so each network will be a factor of 8, making the last available subnet with a .248 in the third octet.

QUESTION 70:
Which of the following IP addresses is a private IP address? Select all that apply.
A. 12.0.0.1
B. 168.172.19.39
C. 172.20.14.36
D. 172.33.194.30
E. 192.168.42.34

Explanation:
RFC 1918 Private Address Space: