10 free Juniper JN0-106 JNCIA-Junos practice questions with the correct answer and a full explanation for each, taken from the CertStash pack of 98 questions. Work through them, then open each answer to check your reasoning.
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Question 1
Which two statements are correct regarding Layer 2 network switches? (Choose two.)
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Correct answer: A, B
A. Switches are susceptible to traffic loops. B. Switches flood broadcast traffic. Switches operate at Layer 2 and are susceptible to traffic loops when redundant paths exist without spanning tree protocols in place. Switches flood broadcast traffic out all ports except the incoming port because they do not have routing intelligence. Switches actively learn MAC addresses from source addresses of frames they receive and build MAC address tables. Switches create multiple collision domains (one per port), not a single collision domain.
Why the other options are wrong
- C. Switches do learn MAC addresses by examining source addresses in incoming frames.
- D. Switches create multiple collision domains, one for each port, not a single domain.
Question 2
How many usable hosts are in a /24 subnet mask?
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Correct answer: B. 254
A /24 subnet mask provides 8 bits for host addressing, yielding 2^8 = 256 total addresses. Of these, one address is reserved for the network address and one for the broadcast address, leaving 256 – 2 = 254 usable host addresses.
Why the other options are wrong
- A. 250 is incorrect; it accounts for an extra 4 reserved addresses that do not exist in standard subnetting.
- C. 255 is incorrect; this fails to account for the broadcast address that cannot be assigned to a host.
- D. 256 is the total number of addresses in a /24, not the usable host count.
Question 3
What are two fiber-optic connector types? (Choose two.)
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Correct answer: A, C
A. LC C. SC LC (Lucent Connector) and SC (Subscriber Connector) are both standard fiber-optic connector types widely used in telecommunications and data center environments. RJ-45 is a twisted-pair copper connector used for Ethernet, not fiber. BNC is a coaxial cable connector, not a fiber-optic connector.
Why the other options are wrong
- B. RJ-45 is a copper Ethernet connector, not a fiber-optic connector type.
- D. BNC is a coaxial connector used with copper cabling, not fiber-optic cabling.
Question 4
Which two characteristics apply to TCP? (Choose two.)
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Correct answer: C, D
C. TCP uses a three-way handshake to establish a connection. D. TCP provides reliable, ordered delivery using sequence numbers. TCP uses a three-way handshake (SYN, SYN-ACK, ACK) to establish connections before data transfer. TCP provides reliable, ordered delivery of data through the use of sequence numbers that allow the receiver to reassemble segments in the correct order and request retransmission of lost segments. TCP does not guarantee fixed latency and is not suitable for applications requiring deterministic timing. TCP operates unicast on Ethernet, not broadcast; broadcast is a Layer 2 concept and TCP is a transport layer protocol.
Why the other options are wrong
- A. TCP does not guarantee fixed latency; latency varies based on network conditions and retransmissions.
- B. TCP does not broadcast datagrams; TCP is unicast and relies on IP for addressing, which may use broadcast at Layer 2 only for ARP resolution.
Question 5
Click the Exhibit button.
Which statement is correct when Router R1 receives a packet from User A destined for User B as shown in the exhibit?
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Correct answer: C. Router R1 replaces the destination MAC address in the packet with the MAC address of Router R2.
When Router R1 receives a packet from User A destined for User B, it performs IP routing based on the destination IP address. However, at Layer 2, R1 must rewrite the destination MAC address to forward the frame to the next hop. Since R2 is the next router in the path toward User B, R1 replaces the destination MAC address with R2's MAC address. The destination IP address remains unchanged throughout the routing process until the packet reaches its final destination. The source and destination IP addresses are preserved en-o-end; only the MAC addresses change at each hop.
Why the other options are wrong
- A. Router R1 never changes the destination IP address; IP addresses remain constant from source to destination across all hops.
- B. The packet cannot remain unchanged; R1 must modify the Layer 2 frame by rewriting the destination MAC address for forwarding.
- D. R1 does not know User B's MAC address at this point and cannot replace the destination MAC with it; it only knows the MAC address of the next hop (R2).
Question 6
You asked a network engineer to configure a new logical interface on a Juniper router. The interface must use the fourth usable host address from the 172.16.40.128/29 subnet.
Which IP address is correct in this scenario?
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Correct answer: B. 172.16.40.132
The /29 subnet 172.16.40.128/29 has a block size of 8 addresses. The subnet range is 172.16.40.128 to 172.16.40.135. The network address is 172.16.40.128 (not usable), and the broadcast address is 172.16.40.135 (not usable). The usable host addresses are 172.16.40.129 (first), 172.16.40.130 (second), 172.16.40.131 (third), and 172.16.40.132 (fourth). The fourth usable host address is 172.16.40.132.
Why the other options are wrong
- A. 172.16.40.135 is the broadcast address for this subnet, not a usable host address.
- C. 172.16.40.130 is the second usable host address, not the fourth.
- D. 172.16.40.131 is the third usable host address, not the fourth.
Question 7
You are asked to subnet the broadcast domains but need to support 50 hosts.
In this scenario, which subnet mask would satisfy this requirement?
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Correct answer: A. /26
To support 50 hosts, calculate the required host bits: 2^6 = 64 usable addresses (with /26 mask), which accommodates 62 usable hosts. A /26 subnet provides 64 total addresses minus 2 (network and broadcast), yielding 62 usable hosts, which exceeds the requirement of 50. A /27 provides only 32 total addresses (30 usable), a /28 provides 16 total addresses (14 usable), and a /29 provides 8 total addresses (6 usable), all insufficient for 50 hosts.
Why the other options are wrong
- B. /27 provides only 30 usable host addresses, which is insufficient for 50 hosts.
- C. /28 provides only 14 usable host addresses, which is insufficient for 50 hosts.
- D. /29 provides only 6 usable host addresses, which is insufficient for 50 hosts.
Question 8
Which IPv4 address and subnet mask combination represents a point-to-point link with only two usable host addresses?
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Correct answer: A. 192.168.1.0/30
A point-to-point link requires exactly two usable host addresses, one for each end of the link. A /30 subnet provides 4 total addresses, leaving 2 usable addresses (after subtracting network and broadcast addresses), which is exactly what point-to-point links require. A /29 provides 6 usable hosts, a /28 provides 14 usable hosts, and a /24 provides 254 usable hosts, all exceeding the requirement.
Why the other options are wrong
- B. /29 provides 6 usable host addresses, more than the 2 required for point-to-point.
- C. /24 provides 254 usable host addresses, far more than needed for point-to-point links.
- D. /28 provides 14 usable host addresses, more than the 2 required for point-to-point.
Question 9
What is the maximum number of IP addresses that would be assigned to hosts in the 192.168.1.0/24 network?
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Correct answer: D. 254
The 192.168.1.0/24 network has 8 bits allocated for host addressing, providing 2^8 = 256 total addresses. Subtracting the network address (192.168.1.0) and broadcast address (192.168.1.255), the maximum number of usable host IP addresses is 254.
Why the other options are wrong
- A. 24 refers to the prefix length, not the number of assignable host addresses.
- B. 32 is the total prefix length for a single /32 host route, not related to /24 hosts.
- C. 128 is insufficient; a /24 network contains 256 total addresses.
Question 10
What are two characteristics of IPv6 addressing? (Choose two.)
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Correct answer: A, D
A. The IPv6 address is represented by a 128-bit address D. IPv6 addresses that begin with fe80 are not routable. IPv6 addresses are represented by 128-bit addresses, which is twice the size of IPv4's 3-it addresses and enables a vastly larger address space. IPv6 addresses beginning with fe80 (link-local addresses) are automatically generated and are not routable on the public internet, making them suitable only for link-local communication. These two characteristics are fundamental to IPv6 design and deployment.
Why the other options are wrong
- B. IPv6 headers do not include a checksum field; this responsibility was removed from the header to reduce processing overhead and is now delegated to upper-layer protocols and the data link layer.
- C. IPv6 uses Neighbor Discovery Protocol (NDP) with ICMPv6, not ARP, to discover neighboring devices; ARP is specific to IPv4.
That was 10 of 98.
The full Juniper JN0-106 JNCIA-Junos pack has all 98 questions, each with the answer, the explanation and why the other options are wrong, plus a questions-only copy for timed runs. US$39, paid once, with free monthly updates and a pass-or-your-money-back guarantee.