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Cisco 200-301 Exam Syllabus Topics:

SectionWeightObjectives
IP Connectivity25%- Static routing and dynamic routing protocols
- OSPFv2 and OSPFv3 configuration
- First Hop Redundancy Protocols (HSRP, VRRP)
- Routing fundamentals and routing tables
Network Access20%- VLANs and inter-VLAN routing
- Spanning Tree Protocol (STP)
- Wireless network configuration and security
- EtherChannel and link aggregation
Network Fundamentals20%- OSI and TCP/IP models
- IPv6 addressing and concepts
- Ethernet fundamentals and switching concepts
- Wireless LAN principles
- IPv4 addressing and subnetting
IP Services10%- SNMP, Syslog, and QoS basics
- DHCP, DNS, and NTP
- NAT and PAT
Security Fundamentals15%- Layer 2 security features
- Device hardening and access control
- AAA and identity management
- Access Control Lists (ACLs)
- VPN technologies and security protocols
Automation and Programmability10%- SDN and network programmability concepts
- REST APIs and data formats (JSON, YAML)
- Network monitoring and automation tools
- Configuration management tools (Ansible)

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Cisco Certified Network Associate Exam Sample Questions (Q204-Q209):

NEW QUESTION # 204
Refer to the exhibit.

Router R1 Fa0/0 cannot ping router R3 Fa0/1.
Which action must be taken in router R1 to help resolve the configuration issue?

Answer: C


NEW QUESTION # 205
Refer to the exhibit.

A network administrator has been tasked with securing VTY access to a router Which access-list entry accomplishes this task?

Answer: D


NEW QUESTION # 206
Refer to the exhibit. How will router R1 handle packets destined to 192.168.64.22?

Answer: A

Explanation:
The static route to 192.168.64.0/18 via 10.1.1.1 has the lowest administrative distance (1), so R1 will use this route for packets destined to 192.168.64.22.


NEW QUESTION # 207

Refer to the exhibit. With a reference bandwidth of 100 Gb on all routers, which path does router Y use to get to network 192.168.1.0/24?

Answer: D


NEW QUESTION # 208
Drag and Drop Question
Drag and drop the IPv4 network subnets from the left onto the correct usable host ranges on the right

Answer:

Explanation:

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. Let's 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 don't need to calculate the broadcast address because we found the correct answer.
Let's 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.


NEW QUESTION # 209
......

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