Subnetting Practice Questions: 50 Network+ Quiz Questions and Answers
50 questions for the Network+ N10-009 exam, each answer explained. Watch it here, try the questions yourself, or read the transcript.
Questions in this video
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Question 1 Play from 00:00
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Why A is wrong
That is the network address.
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Why B is wrong
That is the last usable host.
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Why C is wrong
That is the first address of the next subnet.
Right, D is the answer.
Why D is right
The block size is 256, so 10.10.10.10 sits in 10.10.10.0 to 10.10.10.255. The last address, 10.10.10.255, is the broadcast.
Question 2 Play from 01:25
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Right, A is the answer.
Why A is right
A /26 has a block size of 64 in the fourth octet. The block that contains 192.168.10.77 starts at 192.168.10.64.
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Why B is wrong
That is the first usable host, not the network address.
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Why C is wrong
That is the broadcast address of the subnet.
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Why D is wrong
That is the start of a different block.
Question 3 Play from 02:54
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Why A is wrong
172.16.1.64 is outside the range.
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Why B is wrong
172.16.1.95 is outside the range.
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Why C is wrong
172.16.1.130 is outside the range.
Right, D is the answer.
Why D is right
172.16.1.100/27 belongs to 172.16.1.96 to 172.16.1.127. Only 172.16.1.120 is a usable host in that range.
Question 4 Play from 04:34
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Right, A is the answer.
Why A is right
A /25 gives 126 usable hosts, enough for 100. A /26 gives only 62.
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Why B is wrong
A /26 gives only 62 hosts, too few.
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Why C is wrong
A /24 fits but wastes addresses; it is not the smallest.
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Why D is wrong
A /23 fits but is larger than needed.
Question 5 Play from 05:48
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Right, A is the answer.
Why A is right
The block size is 8, so 172.16.5.9 sits in 172.16.5.8 to 172.16.5.15. The last address, 172.16.5.15, is the broadcast.
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Why B is wrong
That is the network address.
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Why C is wrong
That is the last usable host.
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Why D is wrong
That is the first address of the next subnet.
Question 6 Play from 07:14
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Why A is wrong
8 is the number of bits, not the subnets.
Right, B is the answer.
Why B is right
Going from /56 to /64 uses 8 bits, and two to the power of 8 is 256.
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Why C is wrong
128 would need only 7 bits.
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Why D is wrong
64 is the prefix length, not the subnet count.
Question 7 Play from 08:20
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Why A is wrong
That is the broadcast address of the subnet.
Right, B is the answer.
Why B is right
A /29 has a block size of 8 in the fourth octet. The block that contains 10.20.30.45 starts at 10.20.30.40.
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Why C is wrong
That is the first usable host, not the network address.
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Why D is wrong
That is the start of a different block.
Question 8 Play from 09:37
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Why A is wrong
That is the network address, which cannot be assigned.
Right, B is the answer.
Why B is right
The subnet runs from 172.16.9.192 to 172.16.9.223. Usable hosts are 172.16.9.193 to 172.16.9.222.
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Why C is wrong
That is the last usable host, not the first.
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Why D is wrong
That is the broadcast address, which cannot be assigned.
Question 9 Play from 11:09
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Why A is wrong
That is the mask for a /19.
Right, B is the answer.
Why B is right
A /27 means 27 network bits set to one. Written in dotted decimal that is 255.255.255.224.
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Why C is wrong
That is the mask for a /28.
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Why D is wrong
That is the mask for a /26.
Question 10 Play from 12:30
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Why A is wrong
That prefix means the mask 255.255.240.0.
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Why B is wrong
That prefix means the mask 255.255.255.248.
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Why C is wrong
That prefix means the mask 255.255.255.224.
Right, D is the answer.
Why D is right
Count the one bits in 255.255.255.240: there are 28, so the prefix is /28.
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Question 11 Play from 13:45
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Why A is wrong
That is the broadcast address, which cannot be assigned.
Right, B is the answer.
Why B is right
The subnet runs from 192.168.100.8 to 192.168.100.15. Usable hosts are 192.168.100.9 to 192.168.100.14.
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Why C is wrong
That is the first usable host, not the last.
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Why D is wrong
That is the network address, which cannot be assigned.
Question 12 Play from 15:21
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Why A is wrong
That is the mask for a /13.
Right, B is the answer.
Why B is right
A /21 means 21 network bits set to one. Written in dotted decimal that is 255.255.248.0.
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Why C is wrong
That is the mask for a /22.
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Why D is wrong
That is the mask for a /20.
Question 13 Play from 16:41
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Why A is wrong
That is the start of a different block.
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Why B is wrong
That is the first usable host, not the network address.
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Why C is wrong
That is the broadcast address of the subnet.
Right, D is the answer.
Why D is right
A /27 has a block size of 32 in the fourth octet. The block that contains 10.1.5.200 starts at 10.1.5.192.
Question 14 Play from 18:01
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Why A is wrong
10.0.0.15 is outside the range.
Right, B is the answer.
Why B is right
10.0.0.20/28 belongs to 10.0.0.16 to 10.0.0.31. Only 10.0.0.30 is a usable host in that range.
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Why C is wrong
10.0.0.50 is outside the range.
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Why D is wrong
10.0.0.33 is outside the range.
Question 15 Play from 19:28
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Why A is wrong
That is the first address of the next subnet.
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Why B is wrong
That is the network address.
Right, C is the answer.
Why C is right
The block size is 16, so 192.168.200.33 sits in 192.168.200.32 to 192.168.200.47. The last address, 192.168.200.47, is the broadcast.
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Why D is wrong
That is the last usable host.
Question 16 Play from 21:00
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Why A is wrong
That is the count for one fewer host bit, a /30.
Right, B is the answer.
Why B is right
A /29 leaves 3 host bits. Two to the power of 3 is 8, minus 2 for the network and broadcast addresses, gives 6.
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Why C is wrong
That is the count for one more host bit, a /28.
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Why D is wrong
That count forgets to subtract the network and broadcast addresses.
Question 17 Play from 22:09
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Why A is wrong
That is the network address.
Right, B is the answer.
Why B is right
The block size is 64, so 10.0.0.130 sits in 10.0.0.128 to 10.0.0.191. The last address, 10.0.0.191, is the broadcast.
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Why C is wrong
That is the last usable host.
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Why D is wrong
That is the first address of the next subnet.
Question 18 Play from 23:30
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Why A is wrong
That is the last usable host.
Right, B is the answer.
Why B is right
The block size is 4, so 172.31.66.5 sits in 172.31.64.0 to 172.31.67.255. The last address, 172.31.67.255, is the broadcast.
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Why C is wrong
That is the network address.
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Why D is wrong
That is the first address of the next subnet.
Question 19 Play from 25:03
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Why A is wrong
That is the start of a different block.
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Why B is wrong
That is the first usable host, not the network address.
Right, C is the answer.
Why C is right
A /27 has a block size of 32 in the fourth octet. The block that contains 192.168.50.250 starts at 192.168.50.224.
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Why D is wrong
That is the broadcast address of the subnet.
Question 20 Play from 26:34
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Why A is wrong
That is the last usable host, not the first.
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Why B is wrong
That is the network address, which cannot be assigned.
Right, C is the answer.
Why C is right
The subnet runs from 192.168.4.64 to 192.168.4.127. Usable hosts are 192.168.4.65 to 192.168.4.126.
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Why D is wrong
That is the broadcast address, which cannot be assigned.
Question 21 Play from 28:09
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Right, A is the answer.
Why A is right
The private 172 range is 172.16.0.0/12, which covers 172.16 to 172.31, so 172.20.5.1 is private.
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Why B is wrong
Private 10 addresses are 10.0.0.0/8; 11 is public.
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Why C is wrong
172.32 is just outside the 172.16 to 172.31 range.
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Why D is wrong
Private 192 addresses are 192.168 only.
Question 22 Play from 29:40
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Why A is wrong
That is the mask for a /28.
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Why B is wrong
That is the mask for a /29.
Right, C is the answer.
Why C is right
A /30 means 30 network bits set to one. Written in dotted decimal that is 255.255.255.252.
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Why D is wrong
That is the mask for a /22.
Question 23 Play from 31:09
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Why A is wrong
A /27 gives only 30 hosts, too few.
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Why B is wrong
A /25 fits but wastes addresses; it is not the smallest.
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Why C is wrong
A /24 fits but is larger than needed.
Right, D is the answer.
Why D is right
A /26 gives 62 usable hosts, enough for 60. A /27 gives only 30.
Question 24 Play from 32:16
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Why A is wrong
That is the count for one more host bit, a /21.
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Why B is wrong
That count forgets to subtract the network and broadcast addresses.
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Why C is wrong
That is the count for one fewer host bit, a /23.
Right, D is the answer.
Why D is right
A /22 leaves 10 host bits. Two to the power of 10 is 1024, minus 2 for the network and broadcast addresses, gives 1022.
Question 25 Play from 33:29
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Right, A is the answer.
Why A is right
Going from /16 to /20 borrows 4 bits, and two to the power of 4 is 16.
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Why B is wrong
That number is not the subnet count for these prefixes.
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Why C is wrong
That count borrows one bit too few.
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Why D is wrong
That count borrows one bit too many.
Question 26 Play from 34:34
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Why A is wrong
That prefix means the mask 255.255.255.254.
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Why B is wrong
That prefix means the mask 255.255.252.0.
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Why C is wrong
That prefix means the mask 255.255.255.248.
Right, D is the answer.
Why D is right
Count the one bits in 255.255.255.252: there are 30, so the prefix is /30.
Question 27 Play from 35:50
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Right, A is the answer.
Why A is right
Drop leading zeros in each group and replace one run of all-zero groups with a double colon, which gives 2001:db8::1.
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Why B is wrong
That form leaves out zero groups without a double colon.
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Why C is wrong
A double colon can appear only once.
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Why D is wrong
A double colon can appear only once.
Question 28 Play from 37:10
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Why A is wrong
192.168.1.200 is outside the range.
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Why B is wrong
192.168.1.130 is outside the range.
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Why C is wrong
192.168.1.60 is outside the range.
Right, D is the answer.
Why D is right
192.168.1.70/26 belongs to 192.168.1.64 to 192.168.1.127. Only 192.168.1.120 is a usable host in that range.
Question 29 Play from 38:52
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Why A is wrong
A /21 fits but is larger than needed.
Right, B is the answer.
Why B is right
A /23 gives 510 usable hosts, enough for 500. A /24 gives only 254.
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Why C is wrong
A /24 gives only 254 hosts, too few.
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Why D is wrong
A /22 fits but wastes addresses; it is not the smallest.
Question 30 Play from 40:02
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Right, A is the answer.
Why A is right
Count the one bits in 255.255.255.128: there are 25, so the prefix is /25.
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Why B is wrong
That prefix means the mask 255.255.128.0.
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Why C is wrong
That prefix means the mask 255.255.255.192.
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Why D is wrong
That prefix means the mask 255.255.255.0.
Question 31 Play from 41:24
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Why A is wrong
That count forgets to subtract the network and broadcast addresses.
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Why B is wrong
That is the count for one fewer host bit, a /29.
Right, C is the answer.
Why C is right
A /28 leaves 4 host bits. Two to the power of 4 is 16, minus 2 for the network and broadcast addresses, gives 14.
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Why D is wrong
That is the count for one more host bit, a /27.
Question 32 Play from 42:29
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Why A is wrong
That count forgets to subtract the network and broadcast addresses.
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Why B is wrong
That is the count for one fewer host bit, a /26.
Right, C is the answer.
Why C is right
A /25 leaves 7 host bits. Two to the power of 7 is 128, minus 2 for the network and broadcast addresses, gives 126.
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Why D is wrong
That is the count for one more host bit, a /24.
Question 33 Play from 43:39
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Why A is wrong
169.254 addresses are link-local, not public.
Right, B is the answer.
Why B is right
The 169.254.0.0/16 range is APIPA, self-assigned when DHCP fails.
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Why C is wrong
Private Class C is 192.168.0.0/16.
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Why D is wrong
Loopback is 127.0.0.0/8.
Question 34 Play from 45:19
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Right, A is the answer.
Why A is right
192.168.7.9/29 belongs to 192.168.7.8 to 192.168.7.15. Only 192.168.7.14 is a usable host in that range.
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Why B is wrong
192.168.7.1 is outside the range.
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Why C is wrong
192.168.7.16 is outside the range.
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Why D is wrong
192.168.7.7 is outside the range.
Question 35 Play from 46:57
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Why A is wrong
A /25 fits but is larger than needed.
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Why B is wrong
A /28 gives only 14 hosts, too few.
Right, C is the answer.
Why C is right
A /27 gives 30 usable hosts, enough for 30. A /28 gives only 14.
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Why D is wrong
A /26 fits but wastes addresses; it is not the smallest.
Question 36 Play from 48:05
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Why A is wrong
That number is not the subnet count for these prefixes.
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Why B is wrong
That count borrows one bit too many.
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Why C is wrong
That count borrows one bit too few.
Right, D is the answer.
Why D is right
Going from /24 to /27 borrows 3 bits, and two to the power of 3 is 8.
Question 37 Play from 49:04
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Right, A is the answer.
Why A is right
Going from /24 to /26 borrows 2 bits, and two to the power of 2 is 4.
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Why B is wrong
That number is not the subnet count for these prefixes.
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Why C is wrong
That count borrows one bit too many.
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Why D is wrong
That count borrows one bit too few.
Question 38 Play from 50:09
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Why A is wrong
That count forgets to subtract the network and broadcast addresses.
Right, B is the answer.
Why B is right
A /26 leaves 6 host bits. Two to the power of 6 is 64, minus 2 for the network and broadcast addresses, gives 62.
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Why C is wrong
That is the count for one more host bit, a /25.
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Why D is wrong
That is the count for one fewer host bit, a /27.
Question 39 Play from 51:18
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Why A is wrong
That is the mask for a /30.
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Why B is wrong
That is the mask for a /28.
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Why C is wrong
That is the mask for a /21.
Right, D is the answer.
Why D is right
A /29 means 29 network bits set to one. Written in dotted decimal that is 255.255.255.248.
Question 40 Play from 52:36
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Right, A is the answer.
Why A is right
The block size is 32, so 192.168.10.77 sits in 192.168.10.64 to 192.168.10.95. The last address, 192.168.10.95, is the broadcast.
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Why B is wrong
That is the last usable host.
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Why C is wrong
That is the first address of the next subnet.
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Why D is wrong
That is the network address.
Question 41 Play from 54:12
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Why A is wrong
That is the count for one fewer host bit, a /28.
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Why B is wrong
That is the count for one more host bit, a /26.
Right, C is the answer.
Why C is right
A /27 leaves 5 host bits. Two to the power of 5 is 32, minus 2 for the network and broadcast addresses, gives 30.
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Why D is wrong
That count forgets to subtract the network and broadcast addresses.
Question 42 Play from 55:24
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Why A is wrong
That is the first usable host, not the network address.
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Why B is wrong
That is the start of a different block.
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Why C is wrong
That is the broadcast address of the subnet.
Right, D is the answer.
Why D is right
A /28 has a block size of 16 in the fourth octet. The block that contains 172.16.40.19 starts at 172.16.40.16.
Question 43 Play from 56:50
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Right, A is the answer.
Why A is right
A /25 has a block size of 128 in the fourth octet. The block that contains 192.168.1.130 starts at 192.168.1.128.
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Why B is wrong
That is the first usable host, not the network address.
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Why C is wrong
That is the broadcast address of the subnet.
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Why D is wrong
That is the start of a different block.
Question 44 Play from 58:20
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Why A is wrong
That is the broadcast address, which cannot be assigned.
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Why B is wrong
That is the first usable host, not the last.
Right, C is the answer.
Why C is right
The subnet runs from 10.5.5.16 to 10.5.5.31. Usable hosts are 10.5.5.17 to 10.5.5.30.
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Why D is wrong
That is the network address, which cannot be assigned.
Question 45 Play from 59:40
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Why A is wrong
A /27 fits but wastes addresses; it is not the smallest.
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Why B is wrong
A /29 gives only 6 hosts, too few.
Right, C is the answer.
Why C is right
A /28 gives 14 usable hosts, enough for 12. A /29 gives only 6.
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Why D is wrong
A /26 fits but is larger than needed.
Question 46 Play from 1:00:41
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Why A is wrong
That prefix means the mask 255.255.240.0.
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Why B is wrong
That prefix means the mask 255.255.192.0.
Right, C is the answer.
Why C is right
Count the one bits in 255.255.224.0: there are 19, so the prefix is /19.
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Why D is wrong
That prefix means the mask 255.224.0.0.
Question 47 Play from 1:01:55
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Right, A is the answer.
Why A is right
A /26 means 26 network bits set to one. Written in dotted decimal that is 255.255.255.192.
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Why B is wrong
That is the mask for a /25.
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Why C is wrong
That is the mask for a /18.
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Why D is wrong
That is the mask for a /27.
Question 48 Play from 1:03:19
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Right, A is the answer.
Why A is right
IPv6 LANs use /64 so that SLAAC can build the 64-bit interface ID.
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Why B is wrong
/56 is a typical home allocation.
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Why C is wrong
/48 is a typical site allocation.
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Why D is wrong
/128 is a single host address.
Question 49 Play from 1:04:27
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Why A is wrong
That is the start of a different block.
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Why B is wrong
That is the broadcast address of the subnet.
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Why C is wrong
That is the first usable host, not the network address.
Right, D is the answer.
Why D is right
A /20 has a block size of 16 in the third octet. The block that contains 172.20.13.100 starts at 172.20.0.0.
Question 50 Play from 1:05:51
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Why A is wrong
That is the last usable host, not the first.
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Why B is wrong
That is the network address, which cannot be assigned.
Right, C is the answer.
Why C is right
The subnet runs from 10.8.130.0 to 10.8.131.255. Usable hosts are 10.8.130.1 to 10.8.131.254.
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Why D is wrong
That is the broadcast address, which cannot be assigned.
Transcript
Play from 00:00 Question 1. What is the broadcast address of the subnet that contains 10.10.10.10/24? Here are the options. Option A. 10.10.10.0. Option B. 10.10.10.254. Option C. 10.10.11.0. Option D. 10.10.10.255. Pause the video now if you want time to think about the answer.
The correct answer is D. 10.10.10.255. The block size is 256, so 10.10.10.10 sits in 10.10.10.0 to 10.10.10.255. The last address, 10.10.10.255, is the broadcast. Now, why are the other options incorrect? Option A is incorrect. That is the network address. Option B is incorrect. That is the last usable host. Option C is incorrect. That is the first address of the next subnet. Therefore, D, 10.10.10.255, is the correct answer.
Play from 01:25 Question 2. What is the network address of 192.168.10.77/26? Here are the options. Option A. 192.168.10.64. Option B. 192.168.10.65. Option C. 192.168.10.127. Option D. 192.168.10.128. Pause the video now if you want time to think about the answer.
The correct answer is A. 192.168.10.64. A /26 has a block size of 64 in the fourth octet. The block that contains 192.168.10.77 starts at 192.168.10.64. Now, why are the other options incorrect? Option B is incorrect. That is the first usable host, not the network address. Option C is incorrect. That is the broadcast address of the subnet. Option D is incorrect. That is the start of a different block. Therefore, A, 192.168.10.64, is the correct answer.
Play from 02:54 Question 3. Host 172.16.1.100/27 needs to talk to a peer on the same subnet without a router. Which address could that peer have? Here are the options. Option A. 172.16.1.64. Option B. 172.16.1.95. Option C. 172.16.1.130. Option D. 172.16.1.120. Pause the video now if you want time to think about the answer.
The correct answer is D. 172.16.1.120. 172.16.1.100/27 belongs to 172.16.1.96 to 172.16.1.127. Only 172.16.1.120 is a usable host in that range. Now, why are the other options incorrect? Option A is incorrect. 172.16.1.64 is outside the range. Option B is incorrect. 172.16.1.95 is outside the range. Option C is incorrect. 172.16.1.130 is outside the range. Therefore, D, 172.16.1.120, is the correct answer.
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Play from 04:34 Question 4. A department needs 100 usable host addresses. Which is the smallest prefix that fits? Here are the options. Option A. /25. Option B. /26. Option C. /24. Option D. /23. Pause the video now if you want time to think about the answer.
The correct answer is A. /25. A /25 gives 126 usable hosts, enough for 100. A /26 gives only 62. Now, why are the other options incorrect? Option B is incorrect. A /26 gives only 62 hosts, too few. Option C is incorrect. A /24 fits but wastes addresses; it is not the smallest. Option D is incorrect. A /23 fits but is larger than needed. Therefore, A, /25, is the correct answer.
Play from 05:48 Question 5. What is the broadcast address of the subnet that contains 172.16.5.9/29? Here are the options. Option A. 172.16.5.15. Option B. 172.16.5.8. Option C. 172.16.5.14. Option D. 172.16.5.16. Pause the video now if you want time to think about the answer.
The correct answer is A. 172.16.5.15. The block size is 8, so 172.16.5.9 sits in 172.16.5.8 to 172.16.5.15. The last address, 172.16.5.15, is the broadcast. Now, why are the other options incorrect? Option B is incorrect. That is the network address. Option C is incorrect. That is the last usable host. Option D is incorrect. That is the first address of the next subnet. Therefore, A, 172.16.5.15, is the correct answer.
Play from 07:14 Question 6. How many /64 subnets fit inside a single IPv6 /56 allocation? Here are the options. Option A. 8. Option B. 256. Option C. 128. Option D. 64. Pause the video now if you want time to think about the answer.
The correct answer is B. 256. Going from /56 to /64 uses 8 bits, and two to the power of 8 is 256. Now, why are the other options incorrect? Option A is incorrect. 8 is the number of bits, not the subnets. Option C is incorrect. 128 would need only 7 bits. Option D is incorrect. 64 is the prefix length, not the subnet count. Therefore, B, 256, is the correct answer.
Play from 08:20 Question 7. What is the network address of 10.20.30.45/29? Here are the options. Option A. 10.20.30.47. Option B. 10.20.30.40. Option C. 10.20.30.41. Option D. 10.20.30.48. Pause the video now if you want time to think about the answer.
The correct answer is B. 10.20.30.40. A /29 has a block size of 8 in the fourth octet. The block that contains 10.20.30.45 starts at 10.20.30.40. Now, why are the other options incorrect? Option A is incorrect. That is the broadcast address of the subnet. Option C is incorrect. That is the first usable host, not the network address. Option D is incorrect. That is the start of a different block. Therefore, B, 10.20.30.40, is the correct answer.
Play from 09:37 Question 8. What is the first usable host address in the subnet that contains 172.16.9.200/27? Here are the options. Option A. 172.16.9.192. Option B. 172.16.9.193. Option C. 172.16.9.222. Option D. 172.16.9.223. Pause the video now if you want time to think about the answer.
The correct answer is B. 172.16.9.193. The subnet runs from 172.16.9.192 to 172.16.9.223. Usable hosts are 172.16.9.193 to 172.16.9.222. Now, why are the other options incorrect? Option A is incorrect. That is the network address, which cannot be assigned. Option C is incorrect. That is the last usable host, not the first. Option D is incorrect. That is the broadcast address, which cannot be assigned. Therefore, B, 172.16.9.193, is the correct answer.
Play from 11:09 Question 9. What is the subnet mask for a /27 prefix? Here are the options. Option A. 255.255.224.0. Option B. 255.255.255.224. Option C. 255.255.255.240. Option D. 255.255.255.192. Pause the video now if you want time to think about the answer.
The correct answer is B. 255.255.255.224. A /27 means 27 network bits set to one. Written in dotted decimal that is 255.255.255.224. Now, why are the other options incorrect? Option A is incorrect. That is the mask for a /19. Option C is incorrect. That is the mask for a /28. Option D is incorrect. That is the mask for a /26. Therefore, B, 255.255.255.224, is the correct answer.
Play from 12:30 Question 10. Which CIDR prefix matches the subnet mask 255.255.255.240? Here are the options. Option A. /20. Option B. /29. Option C. /27. Option D. /28. Pause the video now if you want time to think about the answer.
The correct answer is D. /28. Count the one bits in 255.255.255.240: there are 28, so the prefix is /28. Now, why are the other options incorrect? Option A is incorrect. That prefix means the mask 255.255.240.0. Option B is incorrect. That prefix means the mask 255.255.255.248. Option C is incorrect. That prefix means the mask 255.255.255.224. Therefore, D, /28, is the correct answer.
Play from 13:45 Question 11. What is the last usable host address in the subnet that contains 192.168.100.14/29? Here are the options. Option A. 192.168.100.15. Option B. 192.168.100.14. Option C. 192.168.100.9. Option D. 192.168.100.8. Pause the video now if you want time to think about the answer.
The correct answer is B. 192.168.100.14. The subnet runs from 192.168.100.8 to 192.168.100.15. Usable hosts are 192.168.100.9 to 192.168.100.14. Now, why are the other options incorrect? Option A is incorrect. That is the broadcast address, which cannot be assigned. Option C is incorrect. That is the first usable host, not the last. Option D is incorrect. That is the network address, which cannot be assigned. Therefore, B, 192.168.100.14, is the correct answer.
Play from 15:21 Question 12. What is the subnet mask for a /21 prefix? Here are the options. Option A. 255.248.0.0. Option B. 255.255.248.0. Option C. 255.255.252.0. Option D. 255.255.240.0. Pause the video now if you want time to think about the answer.
The correct answer is B. 255.255.248.0. A /21 means 21 network bits set to one. Written in dotted decimal that is 255.255.248.0. Now, why are the other options incorrect? Option A is incorrect. That is the mask for a /13. Option C is incorrect. That is the mask for a /22. Option D is incorrect. That is the mask for a /20. Therefore, B, 255.255.248.0, is the correct answer.
Play from 16:41 Question 13. What is the network address of 10.1.5.200/27? Here are the options. Option A. 10.1.5.224. Option B. 10.1.5.193. Option C. 10.1.5.223. Option D. 10.1.5.192. Pause the video now if you want time to think about the answer.
The correct answer is D. 10.1.5.192. A /27 has a block size of 32 in the fourth octet. The block that contains 10.1.5.200 starts at 10.1.5.192. Now, why are the other options incorrect? Option A is incorrect. That is the start of a different block. Option B is incorrect. That is the first usable host, not the network address. Option C is incorrect. That is the broadcast address of the subnet. Therefore, D, 10.1.5.192, is the correct answer.
Play from 18:01 Question 14. Host 10.0.0.20/28 needs to talk to a peer on the same subnet without a router. Which address could that peer have? Here are the options. Option A. 10.0.0.15. Option B. 10.0.0.30. Option C. 10.0.0.50. Option D. 10.0.0.33. Pause the video now if you want time to think about the answer.
The correct answer is B. 10.0.0.30. 10.0.0.20/28 belongs to 10.0.0.16 to 10.0.0.31. Only 10.0.0.30 is a usable host in that range. Now, why are the other options incorrect? Option A is incorrect. 10.0.0.15 is outside the range. Option C is incorrect. 10.0.0.50 is outside the range. Option D is incorrect. 10.0.0.33 is outside the range. Therefore, B, 10.0.0.30, is the correct answer.
Play from 19:28 Question 15. What is the broadcast address of the subnet that contains 192.168.200.33/28? Here are the options. Option A. 192.168.200.48. Option B. 192.168.200.32. Option C. 192.168.200.47. Option D. 192.168.200.46. Pause the video now if you want time to think about the answer.
The correct answer is C. 192.168.200.47. The block size is 16, so 192.168.200.33 sits in 192.168.200.32 to 192.168.200.47. The last address, 192.168.200.47, is the broadcast. Now, why are the other options incorrect? Option A is incorrect. That is the first address of the next subnet. Option B is incorrect. That is the network address. Option D is incorrect. That is the last usable host. Therefore, C, 192.168.200.47, is the correct answer.
Play from 21:00 Question 16. How many usable host addresses does a /29 subnet provide? Here are the options. Option A. 2. Option B. 6. Option C. 14. Option D. 8. Pause the video now if you want time to think about the answer.
The correct answer is B. 6. A /29 leaves 3 host bits. Two to the power of 3 is 8, minus 2 for the network and broadcast addresses, gives 6. Now, why are the other options incorrect? Option A is incorrect. That is the count for one fewer host bit, a /30. Option C is incorrect. That is the count for one more host bit, a /28. Option D is incorrect. That count forgets to subtract the network and broadcast addresses. Therefore, B, 6, is the correct answer.
Play from 22:09 Question 17. What is the broadcast address of the subnet that contains 10.0.0.130/26? Here are the options. Option A. 10.0.0.128. Option B. 10.0.0.191. Option C. 10.0.0.190. Option D. 10.0.0.192. Pause the video now if you want time to think about the answer.
The correct answer is B. 10.0.0.191. The block size is 64, so 10.0.0.130 sits in 10.0.0.128 to 10.0.0.191. The last address, 10.0.0.191, is the broadcast. Now, why are the other options incorrect? Option A is incorrect. That is the network address. Option C is incorrect. That is the last usable host. Option D is incorrect. That is the first address of the next subnet. Therefore, B, 10.0.0.191, is the correct answer.
Play from 23:30 Question 18. What is the broadcast address of the subnet that contains 172.31.66.5/22? Here are the options. Option A. 172.31.67.254. Option B. 172.31.67.255. Option C. 172.31.64.0. Option D. 172.31.68.0. Pause the video now if you want time to think about the answer.
The correct answer is B. 172.31.67.255. The block size is 4, so 172.31.66.5 sits in 172.31.64.0 to 172.31.67.255. The last address, 172.31.67.255, is the broadcast. Now, why are the other options incorrect? Option A is incorrect. That is the last usable host. Option C is incorrect. That is the network address. Option D is incorrect. That is the first address of the next subnet. Therefore, B, 172.31.67.255, is the correct answer.
Play from 25:03 Question 19. What is the network address of 192.168.50.250/27? Here are the options. Option A. 192.168.51.0. Option B. 192.168.50.225. Option C. 192.168.50.224. Option D. 192.168.50.255. Pause the video now if you want time to think about the answer.
The correct answer is C. 192.168.50.224. A /27 has a block size of 32 in the fourth octet. The block that contains 192.168.50.250 starts at 192.168.50.224. Now, why are the other options incorrect? Option A is incorrect. That is the start of a different block. Option B is incorrect. That is the first usable host, not the network address. Option D is incorrect. That is the broadcast address of the subnet. Therefore, C, 192.168.50.224, is the correct answer.
Play from 26:34 Question 20. What is the first usable host address in the subnet that contains 192.168.4.70/26? Here are the options. Option A. 192.168.4.126. Option B. 192.168.4.64. Option C. 192.168.4.65. Option D. 192.168.4.127. Pause the video now if you want time to think about the answer.
The correct answer is C. 192.168.4.65. The subnet runs from 192.168.4.64 to 192.168.4.127. Usable hosts are 192.168.4.65 to 192.168.4.126. Now, why are the other options incorrect? Option A is incorrect. That is the last usable host, not the first. Option B is incorrect. That is the network address, which cannot be assigned. Option D is incorrect. That is the broadcast address, which cannot be assigned. Therefore, C, 192.168.4.65, is the correct answer.
Play from 28:09 Question 21. Which of these is a private IPv4 address under RFC 1918? Here are the options. Option A. 172.20.5.1. Option B. 11.0.5.1. Option C. 172.32.5.1. Option D. 192.169.5.1. Pause the video now if you want time to think about the answer.
The correct answer is A. 172.20.5.1. The private 172 range is 172.16.0.0/12, which covers 172.16 to 172.31, so 172.20.5.1 is private. Now, why are the other options incorrect? Option B is incorrect. Private 10 addresses are 10.0.0.0/8; 11 is public. Option C is incorrect. 172.32 is just outside the 172.16 to 172.31 range. Option D is incorrect. Private 192 addresses are 192.168 only. Therefore, A, 172.20.5.1, is the correct answer.
Play from 29:40 Question 22. What is the subnet mask for a /30 prefix? Here are the options. Option A. 255.255.255.240. Option B. 255.255.255.248. Option C. 255.255.255.252. Option D. 255.255.252.0. Pause the video now if you want time to think about the answer.
The correct answer is C. 255.255.255.252. A /30 means 30 network bits set to one. Written in dotted decimal that is 255.255.255.252. Now, why are the other options incorrect? Option A is incorrect. That is the mask for a /28. Option B is incorrect. That is the mask for a /29. Option D is incorrect. That is the mask for a /22. Therefore, C, 255.255.255.252, is the correct answer.
Play from 31:09 Question 23. A department needs 60 usable host addresses. Which is the smallest prefix that fits? Here are the options. Option A. /27. Option B. /25. Option C. /24. Option D. /26. Pause the video now if you want time to think about the answer.
The correct answer is D. /26. A /26 gives 62 usable hosts, enough for 60. A /27 gives only 30. Now, why are the other options incorrect? Option A is incorrect. A /27 gives only 30 hosts, too few. Option B is incorrect. A /25 fits but wastes addresses; it is not the smallest. Option C is incorrect. A /24 fits but is larger than needed. Therefore, D, /26, is the correct answer.
Play from 32:16 Question 24. How many usable host addresses does a /22 subnet provide? Here are the options. Option A. 2046. Option B. 1024. Option C. 510. Option D. 1022. Pause the video now if you want time to think about the answer.
The correct answer is D. 1022. A /22 leaves 10 host bits. Two to the power of 10 is 1024, minus 2 for the network and broadcast addresses, gives 1022. Now, why are the other options incorrect? Option A is incorrect. That is the count for one more host bit, a /21. Option B is incorrect. That count forgets to subtract the network and broadcast addresses. Option C is incorrect. That is the count for one fewer host bit, a /23. Therefore, D, 1022, is the correct answer.
Play from 33:29 Question 25. A /16 network is divided into /20 subnets. How many subnets does that create? Here are the options. Option A. 16. Option B. 64. Option C. 8. Option D. 32. Pause the video now if you want time to think about the answer.
The correct answer is A. 16. Going from /16 to /20 borrows 4 bits, and two to the power of 4 is 16. Now, why are the other options incorrect? Option B is incorrect. That number is not the subnet count for these prefixes. Option C is incorrect. That count borrows one bit too few. Option D is incorrect. That count borrows one bit too many. Therefore, A, 16, is the correct answer.
Play from 34:34 Question 26. Which CIDR prefix matches the subnet mask 255.255.255.252? Here are the options. Option A. /31. Option B. /22. Option C. /29. Option D. /30. Pause the video now if you want time to think about the answer.
The correct answer is D. /30. Count the one bits in 255.255.255.252: there are 30, so the prefix is /30. Now, why are the other options incorrect? Option A is incorrect. That prefix means the mask 255.255.255.254. Option B is incorrect. That prefix means the mask 255.255.252.0. Option C is incorrect. That prefix means the mask 255.255.255.248. Therefore, D, /30, is the correct answer.
Play from 35:50 Question 27. Which is the correct compressed form of 2001:0db8:0000:0000:0000:0000:0000:0001? Here are the options. Option A. 2001:db8::1. Option B. 2001:db8:0:1. Option C. 2001::db8::1. Option D. 2001:db8::0001:: Pause the video now if you want time to think about the answer.
The correct answer is A. 2001:db8::1. Drop leading zeros in each group and replace one run of all-zero groups with a double colon, which gives 2001:db8::1. Now, why are the other options incorrect? Option B is incorrect. That form leaves out zero groups without a double colon. Option C is incorrect. A double colon can appear only once. Option D is incorrect. A double colon can appear only once. Therefore, A, 2001:db8::1, is the correct answer.
Play from 37:10 Question 28. Host 192.168.1.70/26 needs to talk to a peer on the same subnet without a router. Which address could that peer have? Here are the options. Option A. 192.168.1.200. Option B. 192.168.1.130. Option C. 192.168.1.60. Option D. 192.168.1.120. Pause the video now if you want time to think about the answer.
The correct answer is D. 192.168.1.120. 192.168.1.70/26 belongs to 192.168.1.64 to 192.168.1.127. Only 192.168.1.120 is a usable host in that range. Now, why are the other options incorrect? Option A is incorrect. 192.168.1.200 is outside the range. Option B is incorrect. 192.168.1.130 is outside the range. Option C is incorrect. 192.168.1.60 is outside the range. Therefore, D, 192.168.1.120, is the correct answer.
Play from 38:52 Question 29. A department needs 500 usable host addresses. Which is the smallest prefix that fits? Here are the options. Option A. /21. Option B. /23. Option C. /24. Option D. /22. Pause the video now if you want time to think about the answer.
The correct answer is B. /23. A /23 gives 510 usable hosts, enough for 500. A /24 gives only 254. Now, why are the other options incorrect? Option A is incorrect. A /21 fits but is larger than needed. Option C is incorrect. A /24 gives only 254 hosts, too few. Option D is incorrect. A /22 fits but wastes addresses; it is not the smallest. Therefore, B, /23, is the correct answer.
Play from 40:02 Question 30. Which CIDR prefix matches the subnet mask 255.255.255.128? Here are the options. Option A. /25. Option B. /17. Option C. /26. Option D. /24. Pause the video now if you want time to think about the answer.
The correct answer is A. /25. Count the one bits in 255.255.255.128: there are 25, so the prefix is /25. Now, why are the other options incorrect? Option B is incorrect. That prefix means the mask 255.255.128.0. Option C is incorrect. That prefix means the mask 255.255.255.192. Option D is incorrect. That prefix means the mask 255.255.255.0. Therefore, A, /25, is the correct answer.
Play from 41:24 Question 31. How many usable host addresses does a /28 subnet provide? Here are the options. Option A. 16. Option B. 6. Option C. 14. Option D. 30. Pause the video now if you want time to think about the answer.
The correct answer is C. 14. A /28 leaves 4 host bits. Two to the power of 4 is 16, minus 2 for the network and broadcast addresses, gives 14. Now, why are the other options incorrect? Option A is incorrect. That count forgets to subtract the network and broadcast addresses. Option B is incorrect. That is the count for one fewer host bit, a /29. Option D is incorrect. That is the count for one more host bit, a /27. Therefore, C, 14, is the correct answer.
Play from 42:29 Question 32. How many usable host addresses does a /25 subnet provide? Here are the options. Option A. 128. Option B. 62. Option C. 126. Option D. 254. Pause the video now if you want time to think about the answer.
The correct answer is C. 126. A /25 leaves 7 host bits. Two to the power of 7 is 128, minus 2 for the network and broadcast addresses, gives 126. Now, why are the other options incorrect? Option A is incorrect. That count forgets to subtract the network and broadcast addresses. Option B is incorrect. That is the count for one fewer host bit, a /26. Option D is incorrect. That is the count for one more host bit, a /24. Therefore, C, 126, is the correct answer.
Play from 43:39 Question 33. A Windows PC shows the address 169.254.33.10/16. What does this tell you? Here are the options. Option A. It has a valid public address. Option B. It could not reach a DHCP server and assigned itself an APIPA address. Option C. It is using a private Class C address. Option D. It received a loopback address. Pause the video now if you want time to think about the answer.
The correct answer is B. It could not reach a DHCP server and assigned itself an APIPA address. The 169.254.0.0/16 range is APIPA, self-assigned when DHCP fails. Now, why are the other options incorrect? Option A is incorrect. 169.254 addresses are link-local, not public. Option C is incorrect. Private Class C is 192.168.0.0/16. Option D is incorrect. Loopback is 127.0.0.0/8. Therefore, B, It could not reach a DHCP server and assigned itself an APIPA address, is the correct answer.
Play from 45:19 Question 34. Host 192.168.7.9/29 needs to talk to a peer on the same subnet without a router. Which address could that peer have? Here are the options. Option A. 192.168.7.14. Option B. 192.168.7.1. Option C. 192.168.7.16. Option D. 192.168.7.7. Pause the video now if you want time to think about the answer.
The correct answer is A. 192.168.7.14. 192.168.7.9/29 belongs to 192.168.7.8 to 192.168.7.15. Only 192.168.7.14 is a usable host in that range. Now, why are the other options incorrect? Option B is incorrect. 192.168.7.1 is outside the range. Option C is incorrect. 192.168.7.16 is outside the range. Option D is incorrect. 192.168.7.7 is outside the range. Therefore, A, 192.168.7.14, is the correct answer.
Play from 46:57 Question 35. A department needs 30 usable host addresses. Which is the smallest prefix that fits? Here are the options. Option A. /25. Option B. /28. Option C. /27. Option D. /26. Pause the video now if you want time to think about the answer.
The correct answer is C. /27. A /27 gives 30 usable hosts, enough for 30. A /28 gives only 14. Now, why are the other options incorrect? Option A is incorrect. A /25 fits but is larger than needed. Option B is incorrect. A /28 gives only 14 hosts, too few. Option D is incorrect. A /26 fits but wastes addresses; it is not the smallest. Therefore, C, /27, is the correct answer.
Play from 48:05 Question 36. A /24 network is divided into /27 subnets. How many subnets does that create? Here are the options. Option A. 30. Option B. 16. Option C. 4. Option D. 8. Pause the video now if you want time to think about the answer.
The correct answer is D. 8. Going from /24 to /27 borrows 3 bits, and two to the power of 3 is 8. Now, why are the other options incorrect? Option A is incorrect. That number is not the subnet count for these prefixes. Option B is incorrect. That count borrows one bit too many. Option C is incorrect. That count borrows one bit too few. Therefore, D, 8, is the correct answer.
Play from 49:04 Question 37. A /24 network is divided into /26 subnets. How many subnets does that create? Here are the options. Option A. 4. Option B. 62. Option C. 8. Option D. 2. Pause the video now if you want time to think about the answer.
The correct answer is A. 4. Going from /24 to /26 borrows 2 bits, and two to the power of 2 is 4. Now, why are the other options incorrect? Option B is incorrect. That number is not the subnet count for these prefixes. Option C is incorrect. That count borrows one bit too many. Option D is incorrect. That count borrows one bit too few. Therefore, A, 4, is the correct answer.
Play from 50:09 Question 38. How many usable host addresses does a /26 subnet provide? Here are the options. Option A. 64. Option B. 62. Option C. 126. Option D. 30. Pause the video now if you want time to think about the answer.
The correct answer is B. 62. A /26 leaves 6 host bits. Two to the power of 6 is 64, minus 2 for the network and broadcast addresses, gives 62. Now, why are the other options incorrect? Option A is incorrect. That count forgets to subtract the network and broadcast addresses. Option C is incorrect. That is the count for one more host bit, a /25. Option D is incorrect. That is the count for one fewer host bit, a /27. Therefore, B, 62, is the correct answer.
Play from 51:18 Question 39. What is the subnet mask for a /29 prefix? Here are the options. Option A. 255.255.255.252. Option B. 255.255.255.240. Option C. 255.255.248.0. Option D. 255.255.255.248. Pause the video now if you want time to think about the answer.
The correct answer is D. 255.255.255.248. A /29 means 29 network bits set to one. Written in dotted decimal that is 255.255.255.248. Now, why are the other options incorrect? Option A is incorrect. That is the mask for a /30. Option B is incorrect. That is the mask for a /28. Option C is incorrect. That is the mask for a /21. Therefore, D, 255.255.255.248, is the correct answer.
Play from 52:36 Question 40. What is the broadcast address of the subnet that contains 192.168.10.77/27? Here are the options. Option A. 192.168.10.95. Option B. 192.168.10.94. Option C. 192.168.10.96. Option D. 192.168.10.64. Pause the video now if you want time to think about the answer.
The correct answer is A. 192.168.10.95. The block size is 32, so 192.168.10.77 sits in 192.168.10.64 to 192.168.10.95. The last address, 192.168.10.95, is the broadcast. Now, why are the other options incorrect? Option B is incorrect. That is the last usable host. Option C is incorrect. That is the first address of the next subnet. Option D is incorrect. That is the network address. Therefore, A, 192.168.10.95, is the correct answer.
Play from 54:12 Question 41. How many usable host addresses does a /27 subnet provide? Here are the options. Option A. 14. Option B. 62. Option C. 30. Option D. 32. Pause the video now if you want time to think about the answer.
The correct answer is C. 30. A /27 leaves 5 host bits. Two to the power of 5 is 32, minus 2 for the network and broadcast addresses, gives 30. Now, why are the other options incorrect? Option A is incorrect. That is the count for one fewer host bit, a /28. Option B is incorrect. That is the count for one more host bit, a /26. Option D is incorrect. That count forgets to subtract the network and broadcast addresses. Therefore, C, 30, is the correct answer.
Play from 55:24 Question 42. What is the network address of 172.16.40.19/28? Here are the options. Option A. 172.16.40.17. Option B. 172.16.40.32. Option C. 172.16.40.31. Option D. 172.16.40.16. Pause the video now if you want time to think about the answer.
The correct answer is D. 172.16.40.16. A /28 has a block size of 16 in the fourth octet. The block that contains 172.16.40.19 starts at 172.16.40.16. Now, why are the other options incorrect? Option A is incorrect. That is the first usable host, not the network address. Option B is incorrect. That is the start of a different block. Option C is incorrect. That is the broadcast address of the subnet. Therefore, D, 172.16.40.16, is the correct answer.
Play from 56:50 Question 43. What is the network address of 192.168.1.130/25? Here are the options. Option A. 192.168.1.128. Option B. 192.168.1.129. Option C. 192.168.1.255. Option D. 192.168.2.0. Pause the video now if you want time to think about the answer.
The correct answer is A. 192.168.1.128. A /25 has a block size of 128 in the fourth octet. The block that contains 192.168.1.130 starts at 192.168.1.128. Now, why are the other options incorrect? Option B is incorrect. That is the first usable host, not the network address. Option C is incorrect. That is the broadcast address of the subnet. Option D is incorrect. That is the start of a different block. Therefore, A, 192.168.1.128, is the correct answer.
Play from 58:20 Question 44. What is the last usable host address in the subnet that contains 10.5.5.20/28? Here are the options. Option A. 10.5.5.31. Option B. 10.5.5.17. Option C. 10.5.5.30. Option D. 10.5.5.16. Pause the video now if you want time to think about the answer.
The correct answer is C. 10.5.5.30. The subnet runs from 10.5.5.16 to 10.5.5.31. Usable hosts are 10.5.5.17 to 10.5.5.30. Now, why are the other options incorrect? Option A is incorrect. That is the broadcast address, which cannot be assigned. Option B is incorrect. That is the first usable host, not the last. Option D is incorrect. That is the network address, which cannot be assigned. Therefore, C, 10.5.5.30, is the correct answer.
Play from 59:40 Question 45. A department needs 12 usable host addresses. Which is the smallest prefix that fits? Here are the options. Option A. /27. Option B. /29. Option C. /28. Option D. /26. Pause the video now if you want time to think about the answer.
The correct answer is C. /28. A /28 gives 14 usable hosts, enough for 12. A /29 gives only 6. Now, why are the other options incorrect? Option A is incorrect. A /27 fits but wastes addresses; it is not the smallest. Option B is incorrect. A /29 gives only 6 hosts, too few. Option D is incorrect. A /26 fits but is larger than needed. Therefore, C, /28, is the correct answer.
Play from 1:00:41 Question 46. Which CIDR prefix matches the subnet mask 255.255.224.0? Here are the options. Option A. /20. Option B. /18. Option C. /19. Option D. /11. Pause the video now if you want time to think about the answer.
The correct answer is C. /19. Count the one bits in 255.255.224.0: there are 19, so the prefix is /19. Now, why are the other options incorrect? Option A is incorrect. That prefix means the mask 255.255.240.0. Option B is incorrect. That prefix means the mask 255.255.192.0. Option D is incorrect. That prefix means the mask 255.224.0.0. Therefore, C, /19, is the correct answer.
Play from 1:01:55 Question 47. What is the subnet mask for a /26 prefix? Here are the options. Option A. 255.255.255.192. Option B. 255.255.255.128. Option C. 255.255.192.0. Option D. 255.255.255.224. Pause the video now if you want time to think about the answer.
The correct answer is A. 255.255.255.192. A /26 means 26 network bits set to one. Written in dotted decimal that is 255.255.255.192. Now, why are the other options incorrect? Option B is incorrect. That is the mask for a /25. Option C is incorrect. That is the mask for a /18. Option D is incorrect. That is the mask for a /27. Therefore, A, 255.255.255.192, is the correct answer.
Play from 1:03:19 Question 48. What is the standard prefix length for an IPv6 LAN subnet? Here are the options. Option A. /64. Option B. /56. Option C. /48. Option D. /128. Pause the video now if you want time to think about the answer.
The correct answer is A. /64. IPv6 LANs use /64 so that SLAAC can build the 64-bit interface ID. Now, why are the other options incorrect? Option B is incorrect. /56 is a typical home allocation. Option C is incorrect. /48 is a typical site allocation. Option D is incorrect. /128 is a single host address. Therefore, A, /64, is the correct answer.
Play from 1:04:27 Question 49. What is the network address of 172.20.13.100/20? Here are the options. Option A. 172.20.16.0. Option B. 172.20.15.255. Option C. 172.20.0.1. Option D. 172.20.0.0. Pause the video now if you want time to think about the answer.
The correct answer is D. 172.20.0.0. A /20 has a block size of 16 in the third octet. The block that contains 172.20.13.100 starts at 172.20.0.0. Now, why are the other options incorrect? Option A is incorrect. That is the start of a different block. Option B is incorrect. That is the broadcast address of the subnet. Option C is incorrect. That is the first usable host, not the network address. Therefore, D, 172.20.0.0, is the correct answer.
Play from 1:05:51 Question 50. What is the first usable host address in the subnet that contains 10.8.130.7/23? Here are the options. Option A. 10.8.131.254. Option B. 10.8.130.0. Option C. 10.8.130.1. Option D. 10.8.131.255. Pause the video now if you want time to think about the answer.
The correct answer is C. 10.8.130.1. The subnet runs from 10.8.130.0 to 10.8.131.255. Usable hosts are 10.8.130.1 to 10.8.131.254. Now, why are the other options incorrect? Option A is incorrect. That is the last usable host, not the first. Option B is incorrect. That is the network address, which cannot be assigned. Option D is incorrect. That is the broadcast address, which cannot be assigned. Therefore, C, 10.8.130.1, is the correct answer.
