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A set of practice exam questions for ccna1 chapter 8, covering topics such as subnetting, ipv6, and network addressing schemes. The questions are accompanied by correct answers and explanations, making it a valuable resource for students preparing for the ccna certification exam.
Typology: Exams
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Three devices are on three different subnets. Match the network address and the broadcast address with each subnet where these devices are located. (Not all options are used.) Device 1: IP address 192.168.10.77/28 on subnet 1 Device 2: IP address192.168.10.17/30 on subnet 2 Device 3: IP address 192.168.10.35/29 on subnet 3 - CORRECT ANSWERS-Subnet 1 network number: 192.168.10. Subnet 1 broadcast address: 192.168.10. Subnet 2 network number: 192.168.10. Subnet 2 broadcast address: 192.168.10. Subnet 3 network number: 192.168.10. Subnet 3 broadcast address: 192.168.10. Refer to curriculum topic: 8.1. To calculate any of these addresses, write the device IP address in binary. Draw a line showing where the subnet mask 1s end. For example, with Device 1, the final octet (77) is 01001101. The line would be drawn between the 0100 and the 1101 because the subnet mask is /28. Change all the bits to the right of the line to 0s to determine the network number (01000000 or 64). Change all the bits to the right of the line to 1s to determine the broadcast address (01001111 or 79). The network portion of the address 172.16.30.5/16 is __________ - CORRECT ANSWERS-172. Refer to curriculum topic: 7.1. A prefix of /16 means that 16 bits are used for the network part of the address. The network portion of the address is therefore 172.16. A network administrator subnets the 192.168.10.0/24 network into subnets with / masks. How many equal-sized subnets are created? - CORRECT ANSWERS- Refer to curriculum topic: 8.1. The normal mask for 192.168.10.0 is /24. A /26 mask indicates 2 bits have been borrowed for subnetting. With 2 bits, four subnets of equal size could be created.
An administrator wants to create four subnetworks from the network address 192.168.1.0/24. What is the network address and subnet mask of the second useable subnet? - CORRECT ANSWERS-subnetwork 192.168.1. subnet mask 255.255.255. Refer to curriculum topic: 8.1. The number of bits that are borrowed would be two, thus giving a total of 4 useable subnets: 192.168.1. 192.168.1. 192.168.1. 192.168.1. Because 2 bits are borrowed, the new subnet mask would be /26 or 255.255.255. A network administrator has received the IPv6 prefix 2001:DB8::/48 for subnetting. Assuming the administrator does not subnet into the interface ID portion of the address space, how many subnets can the administrator create from the /48 prefix? - CORRECT ANSWERS- Refer to curriculum topic: 8.3. With a network prefix of 48, there will be 16 bits available for subnetting because the interface ID starts at bit 64. Sixteen bits will yield 65536 subnets. Three methods allow IPv6 and IPv4 to co-exist. Match each method with its description. (Not all options are used.) - CORRECT ANSWERS-1. Dual-Stack
A college has five campuses. Each campus has IP phones installed. Each campus has an assigned IP address range. For example, one campus has IP addresses that start with 10.1.x.x. On another campus the address range is 10.2.x.x. The college has standardized that IP phones are assigned IP addresses that have the number 4X in the third octet. For example, at one campus the address ranges used with phones include 10.1.40.x, 10.1.41.x, 10.1.42.x, etc. Which two groupings were used to create this IP addressing scheme? (Choose two.) - CORRECT ANSWERS-1. geographic location
Addresses 10.18.10.0 through 10.18.10.63 are taken for the leftmost network. Addresses 192 through 199 are used by the center network. Because 4 host bits are needed to accommodate 10 hosts, a /28 mask is needed. 10.18.10.200/28 is not a valid network number. Two subnets that can be used are 10.18.10.208/28 and 10.18.10.224/28. Fill in the blank. The last host address on the 10.15.25.0/24 network is _________ - CORRECT ANSWERS-10.15.25. Refer to curriculum topic: 7.1. The host portion of the last host address will contain all 1 bits with a 0 bit for the lowest order or rightmost bit. This address is always one less than the broadcast address. The range of addresses for the network 10.15.25.0/24 is 10.15.25.0 (network address) through 10.15.25.255 (broadcast address). So the last host address for this network is 10.15.25.254. A network engineer is subnetting the 10.0.240.0/20 network into smaller subnets. Each new subnet will contain between a minimum of 20 hosts and a maximum of 30 hosts. Which subnet mask will meet these requirements? - CORRECT ANSWERS- 255.255.255. Refer to curriculum topic: 8.1. For each new subnet to contain between 20 and 30 hosts, 5 host bits are required. When 5 host bits are being used, 27 network bits are remaining. A /27 prefix provides the subnet mask of 255.255.255.224. How many host addresses are available on the 192.168.10.128/26 network? - CORRECT ANSWERS- Refer to curriculum topic: 8.1. A /26 prefix gives 6 host bits, which provides a total of 64 addresses, because 26 = 64. Subtracting the network and broadcast addresses leaves 62 usable host addresses.