cisco 200-301 online test

Question 1 Topic 1

Topic 1

Refer to the exhibit. Which type of route does R1 use to reach host 10.10.13.10/32?

• A. default route
• B. network route
• C. host route
• D. floating static route

Answer:

B

Question 2 Topic 1

Topic 1

Refer to the exhibit. Which prefix does Router1 use for traffic to Host A?

• A. 10.10.10.0/28
• B. 10.10.13.0/25
• C. 10.10.13.144/28
• D. 10.10.13.208/29

Answer:

D

Explanation:
The prefix with longest prefix will be matched first, in this case is /29.

Question 3 Topic 1

Topic 1
DRAG DROP
Drag and drop the descriptions of file-transfer protocols from the left onto the correct protocols on the right.
Select and Place:

Answer:

Question 4 Topic 1

Topic 1
A frame that enters a switch fails the Frame Check Sequence. Which two interface counters are incremented? (Choose two.)

• A. input errors
• B. frame
• C. giants
• D. CRC
• E. runts

Answer:

A D

Explanation:
Whenever the physical transmission has problems, the receiving device might receive a frame whose bits have changed
values. These frames do not pass the error detection logic as implemented in the FCS field in the Ethernet trailer. The
receiving device discards the frame and counts it as some kind of input error. Cisco switches list this error as a CRC error.
Cyclic redundancy check (CRC) is a term related to how the FCS math detects an error.
The input errors includes runts, giants, no buffer, CRC, frame, overrun, and ignored counts.
The output below show the interface counters with the show interface s0/0/0 command:

Question 5 Topic 1

Topic 1
DRAG DROP
Drag and drop the IPv4 network subnets from the left onto the correct usable host ranges on the right.
Select and Place:

Answer:

Explanation:
This subnet question requires us to grasp how to subnet very well. To quickly find out the subnet range, we have to find out
the increment and the network address of each subnet. Lets take an example with the subnet 172.28.228.144/18:
From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet
must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd
octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) -> The network address is 172.28.192.0. So the
first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we dont need to
calculate the broadcast address because we found the correct answer.
Lets take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111 1110 in 3rd octet) -> The
3rd octet of the network address is 228 (because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address
is 172.28.228.0 -> The first usable host is 172.28.228.1. It is not necessary but if we want to find out the broadcast address
of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or 172.28.230.0
then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet. Therefore the last usable host is
172.28.229.254.

Question 6 Topic 1

Topic 1
How do TCP and UDP differ in the way that they establish a connection between two endpoints?

• A. TCP uses the three-way handshake, and UDP does not guarantee message delivery.
• B. TCP uses synchronization packets, and UDP uses acknowledgment packets.
• C. UDP provides reliable message transfer, and TCP is a connectionless protocol.
• D. UDP uses SYN, SYN ACK, and FIN bits in the frame header while TCP uses SYN, SYN ACK, and ACK bits.

Answer:

A

Question 7 Topic 1

Topic 1
Which 802.11 frame type is Association Response?

• A. management
• B. protected frame
• C. action
• D. control

Answer:

A

Explanation:
There are three main types of 802.11 frames: the Data Frame, the Management Frame and the Control Frame. Association
Response belongs to Management Frame. Association response is sent in response to an association request.
Reference: https://en.wikipedia.org/wiki/802.11_Frame_Types

Question 8 Topic 1

Topic 1
In which way does a spine-and-leaf architecture allow for scalability in a network when additional access ports are required?

• A. A spine switch and a leaf switch can be added with redundant connections between them.
• B. A spine switch can be added with at least 40 GB uplinks.
• C. A leaf switch can be added with connections to every spine switch.
• D. A leaf switch can be added with a single connection to a core spine switch.

Answer:

C

Explanation:
Spine-leaf architecture is typically deployed as two layers: spines (such as an aggregation layer), and leaves (such as an
access layer). Spine-leaf topologies provide high-bandwidth, low-latency, nonblocking server-to-server connectivity.
Leaf (aggregation) switches are what provide devices access to the fabric (the network of spine and leaf switches) and are
typically deployed at the top of the rack. Generally, devices connect to the leaf switches. Devices can include servers, Layer
4-7 services (firewalls and load balancers), and WAN or Internet routers. Leaf switches do not connect to other leaf switches.
In spine-and-leaf architecture, every leaf should connect to every spine in a full mesh.
Spine (aggregation) switches are used to connect to all leaf switches and are typically deployed at the end or middle of the
row. Spine switches do not connect to other spine switches.
Reference: https://www.cisco.com/c/en/us/products/collateral/switches/nexus-9000-series-switches/guide-c07-733228.html

Question 9 Topic 1

Topic 1
Which statement identifies the functionality of virtual machines?

• A. The hypervisor communicates on Layer 3 without the need for additional resources.
• B. Each hypervisor can support a single virtual machine and a single software switch.
• C. The hypervisor can virtualize physical components including CPU, memory, and storage.
• D. Virtualized servers run most efficiently when they are physically connected to a switch that is separate from the hypervisor.

Answer:

C

Question 10 Topic 1

Topic 1
Which command automatically generates an IPv6 address from a specified IPv6 prefix and MAC address of an interface?

• A. ipv6 address dhcp
• B. ipv6 address 2001:DB8:5:112::/64 eui-64
• C. ipv6 address autoconfig
• D. ipv6 address 2001:DB8:5:112::2/64 link-local

Answer:

C

Explanation:
The ipv6 address autoconfig command causes the device to perform IPv6 stateless address auto-configuration to discover
prefixes on the link and then to add the EUI-64 based addresses to the interface. Addresses are configured depending on
the prefixes received in Router Advertisement (RA) messages. The device will listen for RA messages which are transmitted
periodically from the router (DHCP Server). This RA message allows a host to create a global IPv6 address from:
Its interface identifier (EUI-64 address)
Link Prefix (obtained via RA)
Note: Global address is the combination of Link Prefix and EUI-64 address