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100-105 Product Description:
Exam Number/Code: 100-105 vce
Exam name: Cisco Interconnecting Cisco Networking Devices Part 1 (ICND1 v3.0)
n questions with full explanations
Certification: Cisco Certification
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Q121. - (Topic 3)
Refer to the exhibit.
PC1 pings PC2. What three things will CORE router do with the data that is received from PC1? (Choose three.)
A. The data frames will be forwarded out interface FastEthernet0/1 of CORE router.
B. The data frames will be forwarded out interface FastEthernet1/0 of CORE router.
C. CORE router will replace the destination IP address of the packets with the IP address of PC2.
D. CORE router will replace the MAC address of PC2 in the destination MAC address of the frames.
E. CORE router will put the IP address of the forwarding FastEthernet interface in the place of the source IP address in the packets.
F. CORE router will put the MAC address of the forwarding FastEthernet interface in the place of the source MAC address.
The router will forward the frames out the interface toward the destination – B is correct. Since the router will has the end station already in it’s MAC table as see by the “show arp” command, it will replace the destination MAC address to that of PC2 – D is correct. The router will then replace the source IP address to 172.16.40.1 – E is correct.
Q122. - (Topic 3)
Which of the following IP addresses are valid Class B host addresses if a default Class B mask is in use? (Choose two.)
The IP addresses 18.104.22.168 and 22.214.171.124 are both valid Class B addresses when a default mask is in use. The Class B default mask is 255.255.0.0 and the range of valid addresses is 126.96.36.199-
The IP address 10.6.8.35 is a Class A address. The Class A default mask is 255.0.0.0 and
the range of valid addresses is 188.8.131.52 - 127.255.255.255, with the exception of the range
127.0.0.1 - 127.255.255.255, which is reserved and cannot be assigned.
The IP address 192.168.5.9 is a Class C address. The Class C default mask is
255.255.255.0 and the range of valid addresses is 192.0.0.0 - 184.108.40.206.
The IP address 127.0.0.1 is a Class A address, but it comes from a reserved portion that
cannot be assigned.
The range 127.0.0.1 - 127.255.255.255 is used for diagnostics, and although any address
in the range will work as a diagnostic address, 127.0.0.1 is known as the loopback address.
If you can ping this address, or any address in the 127.0.0.1 - 127.255.255.255 range, then
the NIC is working and TCP/IP is installed. The Class A default mask is 255.0.0.0 and the range of valid addresses is 220.127.116.11 - 127.255.255.255, with the exception of the range
127.0.0.1 - 127.255.255.255, which is reserved and cannot be assigned.
Q123. - (Topic 5)
Which of the following is true regarding the configuration of SwitchA?
A. only 5 simultaneous remote connections are possible
B. remote connections using ssh will require a username and password
C. only connections from the local network will be possible
D. console access to SwitchA requires a password
Q124. - (Topic 7)
Which dynamic routing protocol uses only the hop count to determine the best path to a destination?
Q125. - (Topic 5)
Refer to the exhibit.
Host A can communicate with Host B but not with Hosts C or D. How can the network administrator solve this problem?
A. Configure Hosts C and D with IP addresses in the 192.168.2.0 network.
B. Install a router and configure a route to route between VLANs 2 and 3.
C. Install a second switch and put Hosts C and D on that switch while Hosts A and B remain on the original switch.
D. Enable the VLAN trunking protocol on the switch.
Two VLANs require a router in between otherwise they cannot communicate. Different VLANs and different IP subnets need a router to route between them.
Q126. - (Topic 3)
Which IP addresses are valid for hosts belonging to the 10.1.160.0/20 subnet? (Choose three.)
All IP address in IP ranges between: 10.1.160.1 and 10.1.175.254 are valid as shown below Address: 10.1.160.0 00001010.00000001.1010 0000.00000000 Netmask: 255.255.240.0 = 20 11111111.11111111.1111 0000.00000000 Wildcard: 0.0.15.255 00000000.00000000.0000 1111.11111111
Which implies that: Network: 10.1.160.0/20 00001010.00000001.1010 0000.00000000 HostMin: 10.1.160.1 00001010.00000001.1010 0000.00000001 HostMax: 10.1.175.254 00001010.00000001.1010 1111.11111110 Broadcast: 10.1.175.255 00001010.00000001.1010 1111.11111111
Q127. - (Topic 4)
What happens when computers on a private network attempt to connect to the Internet through a Cisco router running PAT?
A. The router uses the same IP address but a different TCP source port number for each connection.
B. An IP address is assigned based on the priority of the computer requesting the connection.
C. The router selects an address from a pool of one-to-one address mappings held in the lookup table.
D. The router assigns a unique IP address from a pool of legally registered addresses for the duration of the connection.
Static PAT translations allow a specific UDP or TCP port on a global address to be translated to a specific port on a local address. That is, both the address and the port numbers are translated.
Static PAT is the same as static NAT, except that it enables you to specify the protocol (TCP or UDP) and port for the real and mapped addresses. Static PAT enables you to identify the same mapped address across many different static statements, provided that the port is different for each statement. You cannot use the same mapped address for multiple static NAT statements.
Port Address Translation makes the PC connect to the Internet but using different TCP source port.
Q128. - (Topic 7)
Under which circumstance should a network administrator implement one-way NAT?
A. when the network must route UDP traffic
B. when traffic that originates outside the network must be routed to internal hosts
C. when traffic that originates inside the network must be routed to internal hosts
D. when the network has few public IP addresses and many private IP addresses require outside access
Explanation: NAT operation is typically transparent to both the internal and external hosts. Typically the internal host is aware of the true IP address and TCP or UDP port of the external host. Typically the NAT device may function as the default gateway for the internal host. However the external host is only aware of the public IP address for the NAT device and the particular port being used to communicate on behalf of a specific internal host.
NAT and TCP/UDP
"Pure NAT", operating on IP alone, may or may not correctly parse protocols that are totally concerned with IP information, such as ICMP, depending on whether the payload is interpreted by a host on the "inside" or "outside" of translation. As soon as the protocol stack is traversed, even with such basic protocols as TCP and UDP, the protocols will break unless NAT takes action beyond the network layer. IP packets have a checksum in each packet header, which provides error detection only for the header. IP datagrams may become fragmented and it is necessary for a NAT to reassemble these fragments to allow correct recalculation of higher-level checksums and correct tracking of which packets belong to which connection. The major transport layer protocols, TCP and UDP, have a checksum that covers all the data they carry, as well as the TCP/UDP header, plus a "pseudo-header" that contains the source and destination IP addresses of the packet carrying the TCP/UDP header. For an originating NAT to pass TCP or UDP successfully, it must recompute the TCP/UDP header checksum based on the translated IP addresses, not the original ones, and put that checksum into the TCP/UDP header of the first packet of the fragmented set of packets. The receiving NAT must recompute the IP checksum on every packet it passes to the destination host, and also recognize and recompute the TCP/UDP header using the retranslated addresses and pseudo-header. This is not a completely solved problem. One solution is for the receiving NAT to reassemble the entire segment and then recompute a checksum calculated across all packets. The originating host may perform Maximum transmission unit (MTU) path discovery to determine the packet size that can be transmitted without fragmentation, and then set the don't fragment (DF) bit in the appropriate packet header field. Of course, this is only a one-way solution, because the responding host can send packets of any size, which may be fragmented before reaching the NAT.
Q129. - (Topic 2)
Which address type does a switch use to make selective forwarding decisions?
A. Source IP address
B. Destination IP address
C. Source and destination IP address
D. Source MAC address
E. Destination MAC address
Switches analyze the destination MAC to make its forwarding decision since it is a layer 2 device. Routers use the destination IP address to make forwarding decisions.
Q130. - (Topic 3)
Which one of the following IP addresses is the last valid host in the subnet using mask 255.255.255.224?
With the 224 there are 8 networks with increments of 32 One of these is 32 33 62 63 where 63 is broadcast so 62 is last valid host out of given choices.