# CCNA FAQ: IP at the Network Layer

## CCNA FAQ: IP at the Network Layer

Q1. What is the decimal equivalent of 10010111 00000110 10101100 01110111?
A. 151.6.172.119
B. 151.6.172.120
C. 151.6.172.121
D. 151.6.172.122

Answer: A. The decimal equivalent of 10010111 00000110 10101100 01110111 is 151.6.172.119.
Answers B and D can be eliminated right away because both IP addresses end with an even number.

Q2. What is the first octet range for Class B addresses?
A. 1 to 126
B. 128 to 191
C. 192 to 223
D. 224 to 239

Answer: B. The first octet range for Class B addresses is 128 to 191.
Answers A, C, and D are incorrect because the range of 1 to 126 is Class A, 192 to 223 is Class C, and 224 to 239 is reserved for multicast.

Q3. What is the first octet range for Class C addresses?
A. 1 to 126
B. 128 to 191
C. 192 to 223
D. 224 to 239

Answer: C. Answer C is correct because the first octet range for Class C addresses is 192 to 223.
Answers A, B, and D are incorrect because the first octet range for Class A addresses is 1 to 126, for Class B is 128 to 191, and 224 to 239 is reserved for multicast.

Q4. How many hosts are available with a Class C network?
A. 253
B. 254
C. 255
D. 256

Answer: B. There are 254 hosts per Class C network. For any Class C network there are 24 network bits and 8 host bits: 28 – 2 = 254.

Q5. What is the network ID for a host with the IP address 124.199.7.18/28?
A. 124.199.7.0
B. 124.199.7.16
C. 124.199.7.32
D. 124.199.7.48

Answer: B. The Network ID of 124.199.7.18/28 is 124.199.7.16.
If you write out the subnet mask it is 255.255.255.240. Subtract 240 from 256 and you have 16; 16×1 = 16, which is the first valid increment that can be the network ID. The next network ID is 124.199.7.32.

Q6. You have been assigned a Class C network address. A coworker has requested that you create 10 networks that can support 10 hosts per network. What subnet mask should you use?
A. 255.255.255.0
B. 255.255.255.224
C. 255.255.255.240
D. 255.255.255.248

Answer: C. Subnet mask 255.255.255.240 allows for 14 networks and 14 hosts.
Answer A is incorrect because subnet mask 255.255.255.0 allows for only one network with 254 hosts. Answer B is incorrect because subnet mask 255.255.255.224 allows for 6 networks and 30 hosts, and answer D is incorrect because subnet mask 255.255.255.248 allows for 30 networks and 6 hosts.

Q7. Which of the following are valid host addresses in the 208.62.15.0 network with a 255.255.255.224 subnet mask? (Choose the two best answers.)
A. 208.62.15.0
B. 208.62.15.1
C. 208.62.15.30
D. 208.62.15.32

Answer: B, C. 208.62.15.1 and 208.62.15.30 are valid host addresses in the 208.62.15.0 network with a 255.255.255.224 subnet mask.
Answer A is incorrect because 208.62.15.0 is the network ID and therefore is not a valid host address. Answer D is incorrect because 208.62.15.32 is the network ID of the next network and is not valid.

Q8. Given the network address 192.131.10.0 and subnet mask 255.255.255.0, what is the total number of networks and the total number of hosts per network?
A. 1 network / 255 hosts
B. 1 network / 254 hosts
C. 2 networks / 62 hosts
D. 6 networks / 30 hosts

Q9. How many valid host IP addresses are available with a network address of 218.41.99.24 and a subnet mask of 255.255.255.252?
A. 2
B. 6
C. 14
D. 30

Answer B is incorrect because subnet mask 255.255.255.248 has 6 available hosts. Answers C and D are incorrect because subnet mask 255.255.255.240 has 14 available hosts, and subnet mask 255.255.255.224 has 30 available hosts.

Q10. If you need at least five subnetworks with a Class C network and you want to have as many hosts as possible on each network, what subnet mask would you use?
A. 255.255.255.252
B. 255.255.255.240
C. 255.255.255.248
D. 255.255.255.224

Answer: D. Subnet mask 255.255.255.224 would allow for 6 networks with 30 available hosts per network.
Answer A is incorrect because subnet mask 255.255.255.252 allows for 62 networks but only 2 hosts per network. Answers B and C are incorrect because subnet mask 255.255.255.240 allows for 14 networks with only 14 hosts per network, and subnet mask 255.255.255.248 allows for 30 networks with only 8 hosts per network.

Q11. How many hosts are available with a Class B network?
A. 254
B. 64,000
C. 65,534
D. 16,777,214

Answer: C. There are 65,534 hosts available for a Class B network.
Answers A and D are incorrect because a Class C network has 254 possible hosts, and a Class A has 16,777,214 possible hosts.

Q12. What is the Broadcast IP for 196.23.250.32/27?
A. 196.23.250.32
B. 196.23.250.33
C. 196.23.250.63
D. 196.23.250.64

Answer A is incorrect because 196.23.250.32 is the Network ID of 196.23.250.32/27. Answers B and D are incorrect because 196.23.250.33 is the first valid IP in the /27 network and 196.23.250.64 is the Network ID of the next network.

Q13. What subnet mask would you use if you had a Class B address and you would like 250 networks?
A. 255.255.255.0
B. 255.255.254.0
C. 255.255.252.0
D. 255.255.248.0

Answer: A. The only subnet that allows for 250 networks is 255.255.255.0. Subnet mask 255.255.255.0 can create 254 subnets.
Answer B is incorrect because subnet mask 255.255.254.0 can create 126 subnets. Answer C is incorrect because subnet mask 255.255.252.0 can create 62 subnets. Answer D is incorrect because subnet mask 255.255.248.0 can create 30 subnets.

Q14. How many subnets can you have with subnet mask of 255.255.240 on a Class B network?
A. 6
B. 14
C. 16
D. 30

Answer: B. You can have 14 subnetworks with a Class B network that has a subnet mask of 255.255.240.0.

Q15. How many hosts are available if you have a Class B network with a subnet mask of 255.255.255.128?
A. 254
B. 256
C. 128
D. 126

Answer: D. Answer D is correct because there are 126 possible hosts for a Class B network with a subnet mask of 255.255.255.128.
Answer A is incorrect because a Class B network with a subnet mask of 255.255.255.0 has 254 possible hosts. Answers B and C are incorrect because the Network ID and Broadcast IP addresses were not subtracted from the total number of IPs in each subnet.

Q16. Which of the following are considered private addresses per RFC 1918? (Choose the three best answers.)
A. 1.0.0.0
B. 10.10.10.20
C. 172.30.255.10
D. 192.168.128.128

Answer: B, C, D. RFC 1918 defines the following private IP address ranges: 10.0.0.0 to 10.255.255.255, 172.16.0.0–172.31.255.255, and 192.168.0.0 to 192.168.255.255.

Q17. What is the CIDR notation for the 128.250.62.0 network with a subnet mask of 255.255.255.0?
A. /23
B. /24
C. /25
D. /26

Answer: B. /24 is the correct CIDR notation for the 128.250.62.0 network with a subnet mask of 255.255.255.0.
Answer A is incorrect because /23 is the CIDR notation for 255.255.254.0. Answer C is incorrect because /25 is the CIDR notation for 255.255.255.128, and answer D is incorrect because /26 is the CIDR notation for 255.255.255.192.

Q18. Which of the following network classes do not define public IP address space? (Choose the two best answers.)
A. Class B
B. Class C
C. Class D
D. Class E

Answer: C, D. Class D defines multicast address space and Class E defines IP address space reserved for research.
Answers A and B are incorrect because Classes B and C are both defined for public use.

Q19. What is the valid IP host range for 160.254.101.167/27?
A. 160.254.101.128 to 160.254.101.254
B. 160.254.101.129 to 160.254.101.254
C. 160.254.101.160 to 160.254.101.191
D. 160.254.101.161 to 160.254.101.190

Answer: D. The valid host range for 160.254.101.167/27 is 160.254.101.161 to 160.254.101.190.
Answers A and B are incorrect because both 160.254.101.128 to 160.254.101.254 and 160.254.101.129 to 160.254.101.254 are part of a 255.255.255.128 subnet. Answer C is incorrect because with the range 160.254.101.160 to 160.254.101.191, the first IP is the Network ID and it is not a valid host IP address.

Q20. Given this network diagram, if the Philadelphia router sends a packet to 172.23.0.51, what interface will the Milford router use to forward the packet? A. E0
B. E1
C. S0
D. S1

Answer: A. The Milford router forwards the packet through the E0 interface because the network matches the 172.23.0.0 network.
Answer B is incorrect because interface E1 is going to the 172.24.0.0 network, and answer C is incorrect because interface S0 is going to the Philadelphia router via the 172.22.0.0 network. Answer D is incorrect because interface S1 does not exist in this scenario.

Q21. Given the IPv6 address of 2001:0A20:D201:0000:0000:DC3F:0000:011D, which of the following are valid ways to write the same address? (Choose two)
A. 2001:A20:D201:0:0:DC3F:0:11D
B. 2001:A20:D201::DC3F:0:11D
C. 2001:A2:D201::DC3F:0:11D
D. 2001:A20::DC3F::11D

Answer: A, B. Leading zeros and contiguous groups of zeros can be taken out of the IPv6 address and replaced with “::” when written.
Answer C is incorrect because a trailing zero was removed from the second group. Answer D is incorrect because there are two sets of “::”, which is not a valid format.

Q22. Convert binary 00101010 00111111 11011100 11111111 to decimal format. Q23. Convert decimal 150.193.6.100 to binary format.  Q24. What is hexadecimal 0x5F in decimal format?

The first hexadecimal character is 5, which is the same as the decimal value.
8 4 2 1
0 1 0 1
The second hexadecimal character is F, which is 15 in decimal.
8 4 2 1
1 1 1 1
Now you can combine the two 4-bit groups to get 01011111, which converts to 95 in decimal.

Q25. Perform Boolean AND to define the Network ID of IP address 200.62.183.26 255.255.255.0.

IP Address Binary = 11001000 00111110 10110111 00011010
Subnet Mask Binary= 11111111 11111111 11111111 00000000
Boolean AND = 11001000 00111110 10110111 00000000

Q26. Given the IP address 32.116.5.0 and subnet mask 255.255.255.0, what is the Network ID?

Answer: This is considered an easy mask. You just write down the original octets with a subnet mask value of 255 and a 0 for each octet with a 0 value in the subnet mask. The correct answer is 32.116.5.0.

Answer: Because this has an easy subnet mask, you can copy the first three octets and then replace the last octet with 255. The correct answer is 213.50.201.255.

Q28. What is the valid IP range for 220.9.3.0/24?

Q29. Define the zero subnet and broadcast subnet rules.

Answer: Zero subnet may also be referred to as subnet zero. The zero subnet is the first subnet in a network and has all binary 0s in the subnet field. For the purpose of taking the CCNA exam, you should not include the first subnet when calculating the number of networks in a larger subnet. This is one of the two reserved subnet numbers on a network and one of the reasons why you subtract from the total number of networks to get the correct answer for the test. The other network is the broadcast subnet, which has all 1s in the subnet field.

Q30. Describe the process of routing.

Answer: Traffic that is generated by a device has the source MAC and source IP address of that device. If a frame is sent to a server on another segment of a WAN, the destination IP address is that of the server the frame is trying to reach. Because the server is not on the same segment, the destination MAC address is that of the router, which is the default gateway. The router takes a look at the frame and at its own routing table. It then decides what interface to use to forward the frame, based on the network portion of the IP address. The router attaches its own MAC address as the source MAC address of the frame before the frame is sent to the server.

Q31. List the functions performed by a router or Layer 3 switch. 