What does the "." signify?
• A. That one message timed out.
• B. That all messages were successful.
• C. That one message was successful.
• D. That one message completed in under the allotted timeframe. Ans A
The possible responses from the ping command are: ! Successful receipt of an echo reply. Timed out waiting for a reply U Destination unreachable C Congestion-experienced packet I Ping interrupted ? Packet type unknown & Packet TTL exceeded
35 Which command, that is used to test address configuration, uses Time-To-Live
(TTL) values to generate messages from each router.
• A. trace
• B. ping
• C. telnet
• D. bootp
Ans: A
The Cisco IOS EXEC command "trace [protocol] [destination]" is used to discover routes that packets will travel to their destination hosts. Trace uses TTL (Time to Live) values to report destination route information.
36 What does the command "IP name-server 255.255.255.255" accomplish?
• A. It sets the domain name lookup to be a local broadcast.
• B. This is an illegal command.
• C. It disables domain name lookup.
• D. The command is now defunct and has been replaced by "IP server-name ip any"
Ans A
By default DNS is enabled on a router with a server address of 255.255.255.255, which provides for a local broadcast.
37 As a system administrator, you need to provide your routers with a Domain Name System (DNS) server. How many DNS servers can you specify with one command?
• A. 6
• B. 1
• C. 2
• D. 4
Ans A
You can only specify six name servers in one command. The syntax is "IP name-server server-address1 [[ server-address2 ]...server-address6]. You must also enable
DNS.
38 How would you configure one host name that points to two IP addresses?
• A. IP host jacob 1.0.0.5 2.0.0.8
• B. IP jacob 1.0.0.5 2.0.0.8
• C. IP host jacob 1.0.0.5
• D. IP host duplicate "all" Ans A
The correct syntax is IP host name [ TCP-port-number ] address [ address ]..... So, "IP host P1R1 1.0.0.5 2.0.0.8" is the correct choice. "IP host jacob 1.0.0.5" only points the host name jacob to one IP address--1.0.0.5.
39 The following selections show the command prompt and the configuration of the
IP network mask. Which two are correct?
• A. Router#term IP netmask-format { bitcount | decimal | hexadecimal }
• B. Router(config-if)#IP netmask-format { bitcount | decimal | hexadecimal }
• C. Router(config-if)#netmask-format { bitcount | decimal | hexadecimal }
• D. Router#ip netmask-format { bitcount | decimal | hexadecimal } Ans A & B
Router#term IP netmask-format { bitcount | decimal | hexadecimal } and Router(config- if)#IP netmask-format { bitcount | decimal | hexadecimal } are correct. You can configure the mask for the current session and you can configure it for a specific line.
40 When configuring the subnet mask for an IP address, which formats can be used?
• A. dotted-decimal.
• B. Hexadecimal
• C. Bit-count
• D. Octal
• E. Binary
Ans A, B &C
41 You are given the following address: 153.50.6.27/25. Determine the subnet mask, address class, subnet address, and broadcast address.
A. 255.255.255.128, B,153.50.6.0, 153.50.6.127
B. 255.255.255.128, C,153.50.6.0, 153.50.6.127
C. 255.255.255.128, C,153.50.6.127, 153.50.6.0
D. 255.255.255.224, C,153.50.6.0, 153.50.6.127
Ans A
42 You are given the following address: 128.16.32.13/30. Determine the subnet mask, address class, subnet address,
and broadcast address.
A. 255.255.255.252, B,128.16.32.12, 128.16.32.15
B. 255.255.255.252, C,128.16.32.12, 128.16.32.15
C. 255.255.255.252, B,128.16.32.15, 128.16.32.12
D. 255.255.255.248, B,128.16.32.12, 128.16.32.15
Ans A
43 You are given the following address: 15.16.193.6/21. Determine the subnet mask, address class, subnet address,
and broadcast address.
A. 255.255.248.0, A, 15.16.192.0, 15.16.199.255
B. 255.255.248.0, B, 15.16.192.0, 15.16.199.255
C. 255.255.248.0, A, 15.16.199.255, 14.15.192.0
D. 255.255.242.0, A, 15.16.192.0, 15.16.199.255
Ans A
44 You have an IP host address of 201.222.5.121 and a subnet mask of 255.255.255.248.
What is the broadcast address?
A. 201.222.5.127
B. 201.222.5.120
C. 201.222.5.121
D. 201.222.5.122
Ans A
The easiest way to calculate this is to subtract 255.255.255.248 (subnet mask) from
255.255.255.255, this
equals 7. Convert the address 201.222.5.121 to binary--11001001 11011110 00000101
01111001. Convert the
mask 255.255.255.248 to binary--11111111 11111111 11111111 11111000. AND them together to get: 11001001 11011110
00000101 01111000 or 201.222.5.120. 201.222.5.120 is the subnet address, add 7 to this
address for 201.222.5.127 or
the broadcast address. 201.222.5.121 through 201.222.5.126 are the valid host addresses.
45 Given the address 172.16.2.120 and the subnet mask of 255.255.255.0. How many hosts are available?
A. 254
B. 510
C. 126
D. 16,372
Ans A
172.16.2 120 is a standard Class B address with a subnet mask that allows 254 hosts. You are a network administrator and have been assigned the IP address of 201.222.5.0. You need to have
20 subnets with 5 hosts per subnet. The subnet mask is 255.255.255.248.
46 Which addresses are valid host addresses?
A. 201.222.5.17
B. 201.222.5.18
C. 201.222.5.16
D. 201.222.5.19
E. 201.222.5.31
Ans A,B & D
Subnet addresses in this situation are all in multiples of 8. In this example, 201.222.5.16 is the subnet, 201.22.5.31 is the broadcast address. The rest are valid host IDs on subnet
201.222.5.16.
47 You are a network administrator and have been assigned the IP address of 201.222.5.0. You need to have 20 subnets with
5 hosts per subnet. What subnet mask will you use?
A. 255.255.255.248
B. 255.255.255.128
C. 255.255.255.192
D. 255.255.255.240
Ans A
By borrowing 5 bits from the last octet, you can. have 30 subnets. If you borrowed only 4 bits you could only have 14 subnets. The formula is (2 to the power of n)-2. By borrowing 4 bits, you have (2x2x2x2)-2=14. By borrowing 5 bits, you have (2x2x2x2x2)-2=30. To get 20 subnets, you would need to borrow 5 bits so the subnet mask would be 255.255.255.248.
48 You are given the IP address of 172.16.2.160 with a subnet mask of 255.255.0.0. What is
the network address in binary?
A. 10101100 00010000
B. 00000010 10100000
C. 10101100 00000000
D. 11100000 11110000
Ans: A
To find the network address, convert the IP address to binary--10101100 000100000 00000010
10100000--then ANDed it with the subnet mask--11111111 11111111 00000000 00000000. The rest is 10101100 00010000 00000000 00000000, which is 172.16.0.0 in decimal.
The first octet rule states that the class of an address can be determined by the numerical value of the first octet.
49 Which addresses are INCORRECTLY paired with their class?
A. 128 to 191, Class B B. 192 to 223 Class B
C. 128 to 191, Class C
D. 192 to 223, Class C
Ans B & C
Address classes are: 1 to 126, Class A; 128 to 191, Class B, 192 to 223, Class C; 224 to 239, Class D; and
240 to 255, Class E. The first octet rule states that the class of an address can be determined by the numerical value of the first octet.
