PrintNumber ErrorLocation Error Correction DateAdded
2 6 9. An interface capable of sending at a T1 speed would be transmitting data at which of the following?
• A. 1.544 Mbps
• B. 1.544 Mbps
• C. 1.544 Gbps
• D. 1.544 GBps
9. An interface capable of sending at a T1 speed would be transmitting data at which of the following?
• A. 1.544 Mbps
• B. 1.544 MBps
• C. 1.544 Gbps
• D. 1.544 GBps
3/18/2009
2 p 18 9. A. A T1-capable interface can transmit data at 1.544 megabits per second (Mbps). The measurement of 1.544 megabytes per second (Mbps) (answer B) would be eight times the speed of a T1 interface (12.352 Mbps) because there are eight bits in every byte. Answers C and D represent gigabits per second (Gbps) and gigabytes per
second (GBps), which far exceed the capabilities of a T1 interface.
9. A. A T1-capable interface can transmit data at 1.544 megabits per second (Mbps). The measurement of 1.544 megabytes per second (MBps) (answer B) would be eight times the speed of a T1 interface (12.352 Mbps) because there are eight bits in every byte. Answers C and D represent gigabits per second (Gbps) and gigabytes per
second (GBps), which far exceed the capabilities of a T1 interface.
3/18/2009
2 p 42 16. E. The vendor-assigned portion of the MAC address is the second half—technically, the last 24 bits. The first half (technically, the first 22 bits) represents the Organizational Unique Identifier (OUI), which are assigned to specific network manufacturing organizations. In this case, the second half of the MAC address is 22-DC-F3. The other answers do not correspond to the question. 16. E. The vendor-assigned portion of the MAC address is the second half—technically, the last 24 bits. The first half (technically, the first 24 bits) represents the Organizational Unique Identifier (OUI), which are assigned to specific network manufacturing organizations. In this case, the second half of the MAC address is 22-DC-F3. The other answers do not correspond to the question. 3/18/2009
2 p 60 4. A, C, D. A subnet mask of 255.255.255.240 divides the fourth octet into subnet parts: the highest four bits and a host port (the lowest four bits). You simply check the fourth octet to ensure that all subnet and host parts are okay. The host bit portion cannot be 0000 or 1111. Answers A, C, and D are correct because 33 in decimal is 00100001, 119 in decimal is 01110111, and 126 in decimal is 1111110. Answer B is incorrect, as 112 in decimal is 1110000 in binary. This is not a valid host address in this network. All its host bits are zero. Answer E is incorrect, as 175 in decimal is 10101111 in binary. All host bits are ones. This is the local broadcast address and cannot be used as a host address. Answer F is incorrect, as 208 in decimal is 11010000 in binary. This is not a valid host address in this network, and all its host bits are zero. 4. A, C, D. A subnet mask of 255.255.255.240 divides the fourth octet into subnet parts: the highest four bits and a host part (the lowest four bits). You simply check the fourth octet to ensure that all subnet and host parts are okay. The host bit portion cannot be 0000 or 1111. Answers A, C, and D are correct because 33 in decimal is 00100001, 119 in decimal is 01110111, and 126 in decimal is 01111110. Answer B is incorrect, as 112 in decimal is 01110000 in binary. This is not a valid host address in this network. All its host bits are zero. Answer E is incorrect, as 175 in decimal is 10101111 in binary. All host bits are ones. This is the local broadcast address and cannot be used as a host address. Answer F is incorrect, as 208 in decimal is 11010000 in binary. This is not a valid host address in this network, and all its host bits are zero. 3/18/2009
2 p 106 7. C. An Extended Service Set (ESS) wireless network topology describes one or more Basic Service Sets (BSSs) combined into a single system. Answer A is incorrect because a BSS describes a wireless network managed by a single access point. Answer B is incorrect because the Security Set Identifier (SSID) describes the name of the wireless network. Answers D and E are incorrect because they are acronyms that do not apply to this question. 7. C. An Extended Service Set (ESS) wireless network topology describes one or more Basic Service Sets (BSSs) combined into a single system. Answer A is incorrect because a BSS describes a wireless network managed by a single access point. Answer B is incorrect because the Service Set Identifier (SSID) describes the name of the wireless network. Answers D and E are incorrect because they are acronyms that do not apply to this question. 3/18/2009
2 p 176 5. A, B, D. An IPv6 address consists of eight octets that can be four hexadecimal characters each. Consecutive sets of zeros can be abbreviated with a double colon (::), but this can only be used once in each IP address. Leading zeros can also be dropped. Based on these rules, addresses from the question can be described as:
2001:0db8:0000:0000:0000:0000:1428:57ab (Valid, eight octets)
2001:0db8::1428:57ab (Valid, same address as above with abbreviation)
2001::1685:2123::1428:57ab (Invalid use of double colon)
2001:99:ab:1:99:2:1:9 (Valid, dropped leading zeros)
2001:1428:57ab:1685:2123: 1428:57ab (Invalid, only seven octets)
5. A, B, D. An IPv6 address consists of eight sets that can be four hexadecimal characters each. Consecutive sets of zeros can be abbreviated with a double colon (::), but this can only be used once in each IP address. Leading zeros can also be dropped. Based on these rules, addresses from the question can be described as:
2001:0db8:0000:0000:0000:0000:1428:57ab (Valid, eight sets)
2001:0db8::1428:57ab (Valid, same address as above with abbreviation)
2001::1685:2123::1428:57ab (Invalid use of double colon)
2001:99:ab:1:99:2:1:9 (Valid, dropped leading zeros)
2001:1428:57ab:1685:2123: 1428:57ab (Invalid, only seven sets)
3/18/2009
6 p 177 9. A, C, and D. A subnet mask of 255.255.255.240 divides the fourth octet into subnet parts: the highest four bits and a host port (the lowest four bits). You simply check the fourth octet to ensure that all subnet and host parts are okay. The host bit portion cannot be 0000 or 1111. Answers A, C, and D are correct because 33 in decimal is 00100001, 119 in decimal is 01110111, and 126 in decimal is 1111110. Answer B is incorrect, as 112 in decimal is 1110000 in binary. This is not a valid host address in this network. All its host bits are zero. Answer E is incorrect, as 175 in decimal is 10101111 in binary. All host bits are ones. This is the local broadcast address and cannot be used as a host address. Answer F is incorrect, as 208 in decimal is 11010000 in binary. This is not a valid host address in this network, and all its host bits are zero. 9. A, C, and D. A subnet mask of 255.255.255.240 divides the fourth octet into subnet parts: the highest four bits and a host part (the lowest four bits). You simply check the fourth octet to ensure that all subnet and host parts are okay. The host bit portion cannot be 0000 or 1111. Answers A, C, and D are correct because 33 in decimal is 00100001, 119 in decimal is 01110111, and 126 in decimal is 01111110. Answer B is incorrect, as 112 in decimal is 01110000 in binary. This is not a valid host address in this network. All its host bits are zero. Answer E is incorrect, as 175 in decimal is 10101111 in binary. All host bits are ones. This is the local broadcast address and cannot be used as a host address. Answer F is incorrect, as 208 in decimal is 11010000 in binary. This is not a valid host address in this network, and all its host bits are zero. 3/18/2009
2 p 181 27. B. There are two ways of shortening an IPv6 address: removing a single group of consecutive zeros by using the double colon (::) and removing leading zeros from an octet. Answer B (2001:ab9:0:0:3::59ff:1ac5) shortens the octets by removing leading zeros and abbreviates the second group of consecutive zeros by using the ::. Answer A incorrectly uses a :: twice in the IPv6 address. Answer C has too many characters in one of the octets. Answer D uses the underscore character, which is invalid. 27. B. There are two ways of shortening an IPv6 address: removing a single group of consecutive zeros by using the double colon (::) and removing leading zeros from a set. Answer B (2001:ab9:0:0:3::59ff:1ac5) shortens the sets by removing leading zeros and abbreviates the second group of consecutive zeros by using the ::. Answer A incorrectly uses a :: twice in the IPv6 address. Answer C has too many characters in one of the sets. Answer D uses the underscore character, which is invalid. 3/18/2009
2 p 203 2. D. The show IP route command shows what routes are available for the router. In this case, the information in the brackets is the administrative distance and metric (bandwidth and delay) of the EIGRP route. Answer A is incorrect, as it is not the port number, but the administrative distance. Answer B is incorrect, as EIGRP does not use hops as its metric. Answer C is incorrect, as the first number is the Advertised Distance (AD) of EIGRP, not the port number. 2. D. The show ip route command shows what routes are available for the router. In this case, the information in the brackets is the administrative distance and metric (bandwidth and delay) of the EIGRP route. Answer A is incorrect, as it is not the port number, but the administrative distance. Answer B is incorrect, as EIGRP does not use hops as its metric. Answer C is incorrect, as the first number is the Advertised Distance (AD) of EIGRP, not the port number. 3/18/2009