CS6250 Computer Networks – Exam 2 (Latest 2026) –
Complete A+ Guide
Section 1: Network Layer & IP Addressing (Questions 1-50)
Scenario 1: A network administrator is designing a subnetting scheme for a Class B
network 172.16.0.0/16 to support 8 subnets.
1. What is the minimum number of subnet bits needed to create 8 subnets?
o A. 2 bits
o B. 3 bits
o C. 4 bits
o D. 5 bits
o Answer: B. 3 bits
o Rationale: 2³ = 8 subnets. Three subnet bits are required to create exactly 8 subnets.
2. After borrowing 3 bits from the host portion of 172.16.0.0/16, what is the new subnet
mask?
o A. 255.255.224.0
o B. 255.255.240.0
o C. 255.255.248.0
o D. 255.255.255.0
o Answer: A. 255.255.224.0
o Rationale: /16 plus 3 bits = /19. 255.255.224.0 = /19.
3. How many host addresses are available per subnet after borrowing 3 bits from a Class B
/16 network?
o A. 2,046
o B. 4,094
o C. 8,190
o D. 16,382
o Answer: C. 8,190
o Rationale: 16 total bits for hosts minus 3 subnet bits = 13 host bits. 2¹³ = 8,192
addresses minus 2 (network and broadcast) = 8,190 usable hosts.
Scenario 2: An IPv4 packet has a Time-to-Live (TTL) field of 32. The packet passes
through 8 routers before reaching its destination.
,4. What is the TTL value when the packet arrives at the destination?
o A. 24
o B. 25
o C. 32
o D. 40
o Answer: A. 24
o Rationale: Each router decrements the TTL by 1. 32 - 8 = 24.
5. If the TTL field reaches 0 before reaching the destination, what happens?
o A. The packet is forwarded to the next router
o B. The packet is delivered but with a warning
o C. The packet is discarded and an ICMP Time Exceeded message is sent back to the
source
o D. The TTL is reset to 32
o Answer: C. The packet is discarded and an ICMP Time Exceeded message is sent
back to the source
o Rationale: When TTL reaches 0, the router discards the packet and sends an ICMP Time
Exceeded message to the source to prevent infinite looping.
Scenario 3: A network uses CIDR notation with the address block 192.168.10.0/26.
6. What is the subnet mask for 192.168.10.0/26?
o A. 255.255.255.0
o B. 255.255.255.128
o C. 255.255.255.192
o D. 255.255.255.224
o Answer: C. 255.255.255.192
o Rationale: /26 means 26 network bits; 26 bits = 255.255.255.192.
7. How many host addresses are available in this subnet?
o A. 32
o B. 62
o C. 64
o D. 126
o Answer: B. 62
o Rationale: 2⁶ = 64 total addresses minus 2 (network and broadcast) = 62 usable hosts.
8. What is the broadcast address for the subnet 192.168.10.0/26?
o A. 192.168.10.63
o B. 192.168.10.64
o C. 192.168.10.127
o D. 192.168.10.255
,o Answer: A. 192.168.10.63
o Rationale: The subnet ranges from 192.168.10.0 to 192.168.10.63. Broadcast is the last
address.
Scenario 4: A router receives a packet destined for 10.1.2.3. The routing table has the
following entries: 10.1.2.0/24 via Interface A, 10.1.0.0/16 via Interface B, 0.0.0.0/0 via
Interface C.
9. Which route will the router use to forward the packet?
o A. 10.1.2.0/24 via Interface A
o B. 10.1.0.0/16 via Interface B
o C. 0.0.0.0/0 via Interface C
o D. The packet will be dropped
o Answer: A. 10.1.2.0/24 via Interface A
o Rationale: Longest prefix match rule applies. 10.1.2.0/24 has a longer prefix (24 bits)
than 10.1.0.0/16 (16 bits), so it is preferred over the default route.
10. If the destination IP was 10.2.3.4, which route would be used?
o A. 10.1.2.0/24 via Interface A
o B. 10.1.0.0/16 via Interface B
o C. 0.0.0.0/0 via Interface C
o D. The packet will be dropped
o Answer: C. 0.0.0.0/0 via Interface C
o Rationale: 10.2.3.4 does not match either specific route, so the default route is used.
Scenario 5: An IPv4 packet has a total length of 1,500 bytes. The header length is 20
bytes.
11. What is the payload (data) size of this packet?
o A. 1,480 bytes
o B. 1,500 bytes
o C. 1,520 bytes
o D. 20 bytes
o Answer: A. 1,480 bytes
o Rationale: Total length = header + payload. 1,500 - 20 = 1,480 bytes.
12. If this packet must traverse a network with an MTU of 1,000 bytes, how many fragments
will be created?
o A. 1 fragment
o B. 2 fragments
o C. 3 fragments
, o D. 4 fragments
o Answer: B. 2 fragments
o Rationale: 1,480 bytes of payload. MTU 1,000 minus 20 header = 980 bytes max
payload per fragment. 1,480 ÷ 980 = 1.51, so 2 fragments.
13. Which field in the IPv4 header is used to reassemble fragments at the destination?
o A. Identification
o B. Fragment Offset
o C. More Fragments flag
o D. All of the above
o Answer: D. All of the above
o Rationale: Identification identifies which packet the fragment belongs to; Fragment
Offset indicates position; More Fragments flag indicates if more fragments follow.
Scenario 6: A network uses Classless Inter-Domain Routing (CIDR) with the address
block 10.0.0.0/22.
14. How many total IP addresses are in this block?
o A. 1,024
o B. 2,048
o C. 4,096
o D. 8,192
o Answer: A. 1,024
o Rationale: 32 - 22 = 10 host bits. 2¹⁰ = 1,024 total addresses.
15. What is the network address for 10.0.0.0/22?
o A. 10.0.0.0
o B. 10.0.1.0
o C. 10.0.2.0
o D. 10.0.4.0
o Answer: A. 10.0.0.0
o Rationale: /22 uses 22 bits for network; 10.0.0.0 is the first address in the block.
16. What is the broadcast address for 10.0.0.0/22?
o A. 10.0.3.255
o B. 10.0.4.255
o C. 10.0.0.255
o D. 10.0.1.255
o Answer: A. 10.0.3.255
o Rationale: /22 block spans 10.0.0.0 through 10.0.3.255. Broadcast is the last address.
Scenario 7: An IPv6 address is 2001:0db8:85a3:0000:0000:8a2e:0370:7334.
Complete A+ Guide
Section 1: Network Layer & IP Addressing (Questions 1-50)
Scenario 1: A network administrator is designing a subnetting scheme for a Class B
network 172.16.0.0/16 to support 8 subnets.
1. What is the minimum number of subnet bits needed to create 8 subnets?
o A. 2 bits
o B. 3 bits
o C. 4 bits
o D. 5 bits
o Answer: B. 3 bits
o Rationale: 2³ = 8 subnets. Three subnet bits are required to create exactly 8 subnets.
2. After borrowing 3 bits from the host portion of 172.16.0.0/16, what is the new subnet
mask?
o A. 255.255.224.0
o B. 255.255.240.0
o C. 255.255.248.0
o D. 255.255.255.0
o Answer: A. 255.255.224.0
o Rationale: /16 plus 3 bits = /19. 255.255.224.0 = /19.
3. How many host addresses are available per subnet after borrowing 3 bits from a Class B
/16 network?
o A. 2,046
o B. 4,094
o C. 8,190
o D. 16,382
o Answer: C. 8,190
o Rationale: 16 total bits for hosts minus 3 subnet bits = 13 host bits. 2¹³ = 8,192
addresses minus 2 (network and broadcast) = 8,190 usable hosts.
Scenario 2: An IPv4 packet has a Time-to-Live (TTL) field of 32. The packet passes
through 8 routers before reaching its destination.
,4. What is the TTL value when the packet arrives at the destination?
o A. 24
o B. 25
o C. 32
o D. 40
o Answer: A. 24
o Rationale: Each router decrements the TTL by 1. 32 - 8 = 24.
5. If the TTL field reaches 0 before reaching the destination, what happens?
o A. The packet is forwarded to the next router
o B. The packet is delivered but with a warning
o C. The packet is discarded and an ICMP Time Exceeded message is sent back to the
source
o D. The TTL is reset to 32
o Answer: C. The packet is discarded and an ICMP Time Exceeded message is sent
back to the source
o Rationale: When TTL reaches 0, the router discards the packet and sends an ICMP Time
Exceeded message to the source to prevent infinite looping.
Scenario 3: A network uses CIDR notation with the address block 192.168.10.0/26.
6. What is the subnet mask for 192.168.10.0/26?
o A. 255.255.255.0
o B. 255.255.255.128
o C. 255.255.255.192
o D. 255.255.255.224
o Answer: C. 255.255.255.192
o Rationale: /26 means 26 network bits; 26 bits = 255.255.255.192.
7. How many host addresses are available in this subnet?
o A. 32
o B. 62
o C. 64
o D. 126
o Answer: B. 62
o Rationale: 2⁶ = 64 total addresses minus 2 (network and broadcast) = 62 usable hosts.
8. What is the broadcast address for the subnet 192.168.10.0/26?
o A. 192.168.10.63
o B. 192.168.10.64
o C. 192.168.10.127
o D. 192.168.10.255
,o Answer: A. 192.168.10.63
o Rationale: The subnet ranges from 192.168.10.0 to 192.168.10.63. Broadcast is the last
address.
Scenario 4: A router receives a packet destined for 10.1.2.3. The routing table has the
following entries: 10.1.2.0/24 via Interface A, 10.1.0.0/16 via Interface B, 0.0.0.0/0 via
Interface C.
9. Which route will the router use to forward the packet?
o A. 10.1.2.0/24 via Interface A
o B. 10.1.0.0/16 via Interface B
o C. 0.0.0.0/0 via Interface C
o D. The packet will be dropped
o Answer: A. 10.1.2.0/24 via Interface A
o Rationale: Longest prefix match rule applies. 10.1.2.0/24 has a longer prefix (24 bits)
than 10.1.0.0/16 (16 bits), so it is preferred over the default route.
10. If the destination IP was 10.2.3.4, which route would be used?
o A. 10.1.2.0/24 via Interface A
o B. 10.1.0.0/16 via Interface B
o C. 0.0.0.0/0 via Interface C
o D. The packet will be dropped
o Answer: C. 0.0.0.0/0 via Interface C
o Rationale: 10.2.3.4 does not match either specific route, so the default route is used.
Scenario 5: An IPv4 packet has a total length of 1,500 bytes. The header length is 20
bytes.
11. What is the payload (data) size of this packet?
o A. 1,480 bytes
o B. 1,500 bytes
o C. 1,520 bytes
o D. 20 bytes
o Answer: A. 1,480 bytes
o Rationale: Total length = header + payload. 1,500 - 20 = 1,480 bytes.
12. If this packet must traverse a network with an MTU of 1,000 bytes, how many fragments
will be created?
o A. 1 fragment
o B. 2 fragments
o C. 3 fragments
, o D. 4 fragments
o Answer: B. 2 fragments
o Rationale: 1,480 bytes of payload. MTU 1,000 minus 20 header = 980 bytes max
payload per fragment. 1,480 ÷ 980 = 1.51, so 2 fragments.
13. Which field in the IPv4 header is used to reassemble fragments at the destination?
o A. Identification
o B. Fragment Offset
o C. More Fragments flag
o D. All of the above
o Answer: D. All of the above
o Rationale: Identification identifies which packet the fragment belongs to; Fragment
Offset indicates position; More Fragments flag indicates if more fragments follow.
Scenario 6: A network uses Classless Inter-Domain Routing (CIDR) with the address
block 10.0.0.0/22.
14. How many total IP addresses are in this block?
o A. 1,024
o B. 2,048
o C. 4,096
o D. 8,192
o Answer: A. 1,024
o Rationale: 32 - 22 = 10 host bits. 2¹⁰ = 1,024 total addresses.
15. What is the network address for 10.0.0.0/22?
o A. 10.0.0.0
o B. 10.0.1.0
o C. 10.0.2.0
o D. 10.0.4.0
o Answer: A. 10.0.0.0
o Rationale: /22 uses 22 bits for network; 10.0.0.0 is the first address in the block.
16. What is the broadcast address for 10.0.0.0/22?
o A. 10.0.3.255
o B. 10.0.4.255
o C. 10.0.0.255
o D. 10.0.1.255
o Answer: A. 10.0.3.255
o Rationale: /22 block spans 10.0.0.0 through 10.0.3.255. Broadcast is the last address.
Scenario 7: An IPv6 address is 2001:0db8:85a3:0000:0000:8a2e:0370:7334.
