CCNA 640-802 Exam

QUESTION 31:
Convert the hex and decimal numbers on the left into binary, and match them to their corresponding slot on the right. (Not all of the hexadecimal and decimal numbers will be used)

Answer:

170 (Decimal) = 10101010 192 (Decimal) = 11000000 F1 (241 = Decimal) = 11110001 9F (159 = Decimal) = 10011111 The following chart displays all of the possible IP address numbers, expressed in decimal, hexadecimal, and binary:


Which two of the addresses below are available for host addresses on the subnet 192.168.15.19/28? (Select two answer choices)
A. 192.168.15.17
B. 192.168.15.14
C. 192.168.15.29
D. 192.168.15.16
E. 192.168.15.31
F. None of the above

Explanation:
The network uses a 28bit subnet (255.255.255.240). This means that 4 bits are used for the networks and 4 bits for the hosts. This allows for 14 networks and 14 hosts (2n-2).
The last bit used to make 240 is the 4th bit (16) therefore the first network will be 192.168.15.16. The network will have 16 addresses (but remember that the first address is the network address and the last address is the broadcast address). In other words, the networks will be in increments of 16 beginning at 192.168.15.16/28. The IP address we are given is 192.168.15.19. Therefore the other host addresses must also be on this network. Valid IP addresses for hosts on this network are: 192.168.15.17-192.168.15.30.

Incorrect Answers:
B. This is not a valid address for this particular 28 bit subnet mask. The first network address should be 192.168.15.16.
D. This is the network address.
E. This is the broadcast address for this particular subnet.

QUESTION 33:
You have a Class C network, and you need ten subnets. You wish to have as many addresses available for hosts as possible. Which one of the following subnet masks should you use?
A. 255.255.255.192
B. 255.255.255.224
C. 255.255.255.240
D. 255.255.255.248
E. None of the above

Explanation:
Using the 2n-2 formula, we will need to use 4 bits for subnetting, as this will provide for 24-2 = 14 subnets. The subnet mask for 4 bits is then 255.255.255.240.

Incorrect Answers:
A. This will give us only 2 bits for the network mask, which will provide only 2 networks.
B. This will give us 3 bits for the network mask, which will provide for only 6 networks.
D. This will use 5 bits for the network mask, providing 30 networks. However, it will provide for only for 6 host addresses in each network, so C is a better choice.

QUESTION 34:
Which of the following is an example of a valid unicast host IP address?
A. 172.31.128.255./18
B. 255.255.255.255
C. 192.168.24.59/30
D. FFFF.FFFF.FFFF
E. 224.1.5.2
F. All of the above

Explanation
The address 172.32.128.255 /18 is 10101100.00011111.10000000.11111111 in binary, so this is indeed a valid host address.

Incorrect Answers:
B. This is the all 1's broadcast address.
C. Although at first glance this answer would appear to be a valid IP address, the /30 means the network mask is 255.255.255.252, and the 192.168.24.59 address is the broadcast address for the 192.168.24.56/30 network.
D. This is the all 1's broadcast MAC address
E. This is a multicast IP address.

QUESTION 35:
How many subnetworks and hosts are available per subnet if you apply a /28 mask to the 210.10.2.0 class C network?
A. 30 networks and 6 hosts.
B. 6 networks and 30 hosts.
C. 8 networks and 32 hosts.
D. 32 networks and 18 hosts.
E. 16 networks and 14 hosts.
F. None of the above

Explanation:
A 28 bit subnet mask (11111111.11111111.11111111.11110000) applied to a class C network uses a 4 bits for networks, and leaves 4 bits for hosts. Using the 2n-2 formula, we have 24-2 (or 2x2x2x2-2) which gives us 14 for the number of hosts, and the number of networks is 24 = 16.
Incorrect Answers:
A. This would be the result of a /29 (255.255.255.248) network.
B. This would be the result of a /27 (255.255.255.224) network.
C. This is not possible, as we must subtract two from the subnets and hosts for the network and broadcast addresses.
D. This is not a possible combination of networks and hosts.

QUESTION 36:
The Certkiller network was assigned the Class C network 199.166.131.0 from the ISP. If the administrator at Certkiller were to subnet this class C network using the 255.255.255.224 subnet mask, how may hosts will they be able to support on each subnet?
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 last octet 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 37:
What is the subnet for the host IP address 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
F. None of the above

Explanation:
This question is much easier then it appears when you convert it to binary and do the Boolean operation as shown below:
IP address 172.16.210.0 = 10101100.00010000.11010010.00000000
/22 mask = 11111111.11111111.11111100.00000000
AND result = 11111111.11111111.11010000.00000000
AND in decimal= 172 . 16 . 208 . 0

QUESTION 38:
What is the subnet for the host IP address 201.100.5.68/28?
A. 201.100.5.0
B. 201.100.5.32
C. 201.100.5.64
D. 201.100.5.65
E. 201.100.5.31
F. 201.100.5.1

Explanation:
This question is much easier then it appears when you convert it to binary and do the Boolean operation as shown below:
IP address 201.100.5.68 = 11001001.01100100.00000101.01000100
/28 mask = 11111111.11111111.11111111.11000000
AND result = 11001001.01100100.00000101.01000000
AND in decimal= 200 . 100 . 5 . 64

QUESTION 39:
Exhibit:

3 addresses are shown in binary form in the exhibit.
Regarding these three binary addresses in the above exhibit; which statements below are correct? (Select three)
A. Address C is a public Class C address.
B. Address C is a private Class C address.
C. Address B is a public Class B address.
D. Address A is a public Class A address.
E. Address B is a private Class B address.
F. Address A is a private Class A address.

Explanation:
A. Address C converts to 192.167.178.69 in decimal, which is a public class C address.
D. Address A converts to 100.10.235.39, which is a public class A IP address.
E. Address B converts to 172.18.158.15, which is a private (RFC 1918) IP address.

QUESTION 40:
What is the IP address range for the first octet in a class B address, in binary form?
A. 00000111-10001111
B. 00000011-10011111
C. 10000000-10111111
D. 11000000-11011111
E. 11100000-11101111
F. None of the above

Explanation:
The class B address range is 128.0.0.0-191.255.255.255. When looking at the first octet alone, the range is 128-191. The binary number for 128 is 10000000 and the binary number for 191 is 10111111, so the value rang is 10000000-10111111.