Q1. What is the decimal equivalent of 10010111 00000110 10101100 01110111? A. 126.96.36.199 B. 188.8.131.52 C. 184.108.40.206 D. 220.127.116.11
Answer: A. The decimal equivalent of 10010111 00000110 10101100 01110111 is 18.104.22.168.
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 22.214.171.124/28? A. 126.96.36.199 B. 188.8.131.52 C. 184.108.40.206 D. 220.127.116.11
Answer: B. The Network ID of 18.104.22.168/28 is 22.214.171.124.
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 126.96.36.199.
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 188.8.131.52 network with a 255.255.255.224 subnet mask? (Choose the two best answers.) A. 184.108.40.206 B. 220.127.116.11 C. 18.104.22.168 D. 22.214.171.124
Answer: B, C. 126.96.36.199 and 188.8.131.52 are valid host addresses in the 184.108.40.206 network with a 255.255.255.224 subnet mask.
Answer A is incorrect because 220.127.116.11 is the network ID and therefore is not a valid host address. Answer D is incorrect because 18.104.22.168 is the network ID of the next network and is not valid.
Q8. Given the network address 22.214.171.124 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
Answer: B. The network address 126.96.36.199 and subnet mask 255.255.255.0 has one network with 254 hosts.
Q9. How many valid host IP addresses are available with a network address of 188.8.131.52 and a subnet mask of 255.255.255.252? A. 2 B. 6 C. 14 D. 30
Answer: A. With a network address of 184.108.40.206 and a subnet mask of 255.255.255.252, 2 valid host addresses are available.
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 220.127.116.11/27? A. 18.104.22.168 B. 22.214.171.124 C. 126.96.36.199 D. 188.8.131.52
Answer: C. The broadcast IP for 184.108.40.206/27 is 220.127.116.11.
Answer A is incorrect because 18.104.22.168 is the Network ID of 22.214.171.124/27. Answers B and D are incorrect because 126.96.36.199 is the first valid IP in the /27 network and 188.8.131.52 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. 184.108.40.206 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.
Answer A is incorrect because the address 220.127.116.11 is a Class A public address.
Q17. What is the CIDR notation for the 18.104.22.168 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 22.214.171.124 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 126.96.36.199/27? A. 188.8.131.52 to 184.108.40.206 B. 220.127.116.11 to 18.104.22.168 C. 22.214.171.124 to 126.96.36.199 D. 188.8.131.52 to 184.108.40.206
Answer: D. The valid host range for 220.127.116.11/27 is 18.104.22.168 to 22.214.171.124.
Answers A and B are incorrect because both 126.96.36.199 to 188.8.131.52 and 184.108.40.206 to 220.127.116.11 are part of a 255.255.255.128 subnet. Answer C is incorrect because with the range 18.104.22.168 to 22.214.171.124, 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 126.96.36.199 to binary format.
Q24. What is hexadecimal 0x5F in decimal format?
Answer: 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 188.8.131.52 255.255.255.0.
Answer: IP Address Binary = 11001000 00111110 10110111 00011010 Subnet Mask Binary= 11111111 11111111 11111111 00000000 Boolean AND = 11001000 00111110 10110111 00000000 The correct answer is 184.108.40.206.
Q26. Given the IP address 220.127.116.11 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 18.104.22.168.
Q27. Given the IP address 22.214.171.124 and subnet mask 255.255.255.0, what is the Broadcast IP?
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 126.96.36.199.
Q28. What is the valid IP range for 188.8.131.52/24?
Answer: This IP address subnet mask is 255.255.255.0, so the network ID is 184.108.40.206. The broadcast IP is 220.127.116.11. The correct answer is 18.104.22.168 to 22.214.171.124.
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.
Suppress broadcasts or multicasts.
Determine the best path for data transfer (routing).
Strip down and add to Data Link layer frames.
Implement access lists for packet filtering (permit/deny statements).
Set up quality of service (QoS) qualifiers to measure network performance.