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Q1. - (Topic 1) 

Which of the following are types of flow control? (Choose three.) 

A. buffering 

B. cut-through 

C. windowing 

D. congestion avoidance 

E. load balancing 

Answer: A,C,D 

Explanation: 

During Transfer of data, a high speed computer is generating data traffic a lot faster than the network device can handle in transferring to destination, so single gateway or destination device cannot handle much amount of traffic that is called "Congestion". 

Buffering The Technie is used to control the data transfer when we have congestion, when a network device receive a data it stores in memory section and then transfer to next destination this process called "Buffering". Windowing Whereas Windowing is used for flow control by the Transport layer. Say the sender device is sending segments and the receiver device can accommodate only a fixed number of segments before it can accept more, the two devices negotiate the window size during the connection setup. This is done so that the sending device doesn't overflow the receiving device's buffer. Also the receiving device can send a single acknowledgement for the segments it has received instead of sending an acknowledgement after every segment received. Also, this window size is dynamic meaning, the devices can negotiate and change the window size in the middle of a session. So if initially the window size is three and the receiving device thinks that it can accept more number of segments in its buffer it can negotiate with the sending device and it increases it to say 5 for example. Windowing is used only by TCP since UDP doesn't use or allow flow control. 

Reference: http://www.info-it.net/cisco/ccna/exam-tips/flow-control.php 

Q2. - (Topic 2) 

A switch has 48 ports and 4 VLANs. How many collision and broadcast domains exist on the switch (collision, broadcast)? 

A. 4, 48 

B. 48, 4 

C. 48, 1 

D. 1, 48 

E. 4, 1 

Answer:

Explanation: 

A switch uses a separate collision domain for each port, and each VLAN is a separate broadcast domain. 

Topic 3, Routing Fundamentals 

Q3. - (Topic 2) 

Refer to the exhibit. 

What two things can the technician determine by successfully pinging from this computer to the IP address 172.16.236.1? (Choose two) 

A. The network card on the computer is functioning correctly. 

B. The default static route on the gateway router is correctly configured. 

C. The correct default gateway IP address is configured on the computer. 

D. The device with the IP address 172.16.236.1 is reachable over the network. 

E. The default gateway at 172.16.236.1 is able to forward packets to the internet. 

Answer: A,D 

Explanation: 

The source and destination addresses are on the same network therefore, a default gateway is not necessary for communication between these two addresses. 

Q4. - (Topic 5) 

Refer to the exhibit. 

The junior network support staff provided the diagram as a recommended configuration for the first phase of a four-phase network expansion project. The entire network expansion will have over 1000 users on 14 network segments and has been allocated this IP address space. 

192.168.1.1 through 192.168.5.255 

192.168.100.1 through 192.168.100.255 

What are three problems with this design? (Choose three.) 

A. The AREA 1 IP address space is inadequate for the number of users. 

B. The AREA 3 IP address space is inadequate for the number of users. 

C. AREA 2 could use a mask of /25 to conserve IP address space. 

D. The network address space that is provided requires a single network-wide mask. 

E. The router-to-router connection is wasting address space. 

F. The broadcast domain in AREA 1 is too large for IP to function. 

Answer: A,C,E 

Explanation: 

The given IP addresses of areas 1 and 3 along with network masks of /24 cannot accommodate 500 users so are inadequate, while the area 2 is having over capacity so its network mask can be reduced to /25 to accommodate the only 60 users it has. 

Q5. - (Topic 7) 

What is the default lease time for a DHCP binding? 

A. 24 hours 

B. 12 hours 

C. 48 hours 

D. 36 hours 

Answer:

Explanation: By default, each IP address assigned by a DHCP Server comes with a one-day lease, which is the amount of time that the address is valid. To change the lease value for an IP address, use the following command in DHCP pool configuration mode: 

Q6. - (Topic 7) 

When enabled, which feature prevents routing protocols from sending hello messages on an interface'? 

A. virtual links 

B. passive-interface 

C. directed neighbors 

D. OSPF areas 

Answer:

Explanation: You can use the passive-interface command in order to control the advertisement of routing information. The command enables the suppression of routing updates over some interfaces while it allows updates to be exchanged normally over other interfaces. With most routing protocols, the passive-interface command restricts outgoing advertisements only. But, when used with Enhanced Interior Gateway Routing Protocol (EIGRP), the effect is slightly different. This document demonstrates that use of the passive-interface command in EIGRP suppresses the exchange of hello packets between two routers, which results in the loss of their neighbor relationship. This stops not only routing updates from being advertised, but it also suppresses incoming routing updates. This document also discusses the configuration required in order to allow the suppression of outgoing routing updates, while it also allows incoming routing updates to be learned normally from the neighbor. 

Q7. - (Topic 3) 

Refer to the exhibit. 

Which command would you use to configure a static route on Router1 to network 192.168.202.0/24 with a nondefault administrative distance? 

A. router1(config)#ip route 1 192.168.201.1 255.255.255.0 192.168.201.2 

B. router1(config)#ip route 192.168.202.0 255.255.255.0 192.168.201.2 1 

C. router1(config)#ip route 5 192.168.202.0 255.255.255.0 192.168.201.2 

D. router1(config)#ip route 192.168.202.0 255.255.255.0 192.168.201.2 5 

Answer:

Explanation: 

The default AD for a static route is 1. To change this, configure a different value to be used as the AD at the very end of the “ip route” statement. 

Q8. - (Topic 5) 

From which of the following attacks can Message Authentication Code (MAC) shield your network? 

A. DoS 

B. DDoS 

C. spoofing 

D. SYN floods 

Answer:

Explanation: 

Message Authentication Code (MAC) can shield your network from spoofing attacks. Spoofing, also known as masquerading, is a popular trick in which an attacker intercepts a network packet, replaces the source address of the packets header with the address of the authorized host, and reinserts fake information which is sent to the receiver. This type of attack involves modifying packet contents. MAC can prevent this type of attack and ensure data integrity by ensuring that no data has changed. MAC also protects against frequency analysis, sequence manipulation, and ciphertext-only attacks. MAC is a secure message digest that requires a secret key shared by the sender and receiver, making it impossible for sniffers to change both the data and the MAC as the receiver can detect the changes. A denial-of-service (DoS) attack floods the target system with unwanted requests, causing the loss of service to users. One form of this attack generates a flood of packets requesting a TCP connection with the target, tying up all resources and making the target unable to service other requests. MAC does not prevent DoS attacks. Stateful packet filtering is the most common defense against a DoS attack. A Distributed Denial of Service attack (DDoS) occurs when multiple systems are used to flood the network and tax the resources of the target system. Various intrusion detection systems, utilizing stateful packet filtering, can protect against DDoS attacks. In a SYN flood attack, the attacker floods the target with spoofed IP packets and causes it to either freeze or crash. A SYN flood attack is a type of denial of service attack that exploits the buffers of a device that accept incoming connections and therefore cannot be prevented by MAC. Common defenses against a SYN flood attack include filtering, reducing the SYN-RECEIVED timer, and implementing SYN cache or SYN cookies. 

Q9. - (Topic 1) 

What must occur before a workstation can exchange HTTP packets with a web server? 

A. A UDP connection must be established between the workstation and its default gateway. 

B. A UDP connection must be established between the workstation and the web server. 

C. A TCP connection must be established between the workstation and its default gateway. 

D. A TCP connection must be established between the workstation and the web server. 

E. An ICMP connection must be established between the workstation and its default gateway. 

F. An ICMP connection must be established between the workstation and the web server. 

Answer:

Explanation: 

HTTP uses TCP port 80, and a TCP port 80 connection must be established for HTTP communication to occur. http://pentestlab.wordpress.com/2012/03/05/common-tcpip-ports/ 

Q10. - (Topic 3) 

Given an IP address of 192.168.1.42 255.255.255.248, what is the subnet address? 

A. 192.168.1.8/29 

B. 192.168.1.32/27 

C. 192.168.1.40/29 

D. 192.168.1.16/28 

E. 192.168.1.48/29 

Answer:

Explanation: 

248 mask uses 5 bits (1111 1000) 

42 IP in binary is (0010 1010) 

The base subnet therefore is the lowest binary value that can be written without changing 

the output of an AND operation of the subnet mask and IP... 

1111 1000 AND 

0010 1010 equals 

0010 1000 - which is .40 

/24 is standard class C mask. 

Adding the 5 bits from the .248 mask gives /29