101 · Question #704
For an IP of 192.168.10.10 with a subnet mask of 255.255.240.0 an administrator is asked to provide: - The first network address - The broadcast address, and - The number of hosts available per subnet
The correct answer is B. 192.168.0.0 is the network address 192.168.15.255 is the broadcast address. There are 4094. A /20 subnet mask (255.255.240.0) has a block size of 16 in the third octet; the host 192.168.10.10 falls in the 192.168.0.0 network block.
Question
For an IP of 192.168.10.10 with a subnet mask of 255.255.240.0 an administrator is asked to provide:
- The first network address
- The broadcast address, and
- The number of hosts available per subnet host
What should be the administrator's response'?
Options
- A192.168.0.1 is the network address. 192.168.10.255 is the broadcast address. There are
- B192.168.0.0 is the network address 192.168.15.255 is the broadcast address. There are 4094
- C192.168.0.0 is the network address. 192.168 16.255 is the broadcast address. There are 1022
- D192.168.16.0 is the network address. 192.168.31.255 is the broadcast address. There are
How the community answered
(32 responses)- A16% (5)
- B75% (24)
- C6% (2)
- D3% (1)
Why each option
A /20 subnet mask (255.255.240.0) has a block size of 16 in the third octet; the host 192.168.10.10 falls in the 192.168.0.0 network block.
192.168.0.1 is a valid host address, not a network address; network addresses always have all host bits set to zero.
The mask 255.255.240.0 (/20) creates blocks of 16 in the third octet; since 10 falls between 0 and 15, the network address is 192.168.0.0 and the broadcast is 192.168.15.255. The subnet provides 2^12 - 2 = 4094 usable host addresses.
192.168.16.255 is the broadcast of the next /20 block (192.168.16.0-192.168.31.255), not the block containing 192.168.10.10; 1022 hosts corresponds to a /22 mask, not /20.
192.168.16.0 is the start of the next /20 block and does not contain 192.168.10.10, which resides in the 192.168.0.0-192.168.15.255 range.
Concept tested: IPv4 subnetting with variable-length subnet masks
Source: https://www.cisco.com/c/en/us/support/docs/ip/routing-information-protocol-rip/13788-3.html
Topics
Community Discussion
10
Viktor S.Oct 31, 2025B is right because 255.255.240.0 gives you a /20 with a block size of 16 in the third octet, and since 10 falls in the 0-15 range, your network starts at 192.168.0.0 and the broadcast caps the block at 192.168.15.255, leaving 2 to the 12th minus 2 which is exactly 4094 usable hosts. If you just memorized the answer without doing that block-size math by hand, go do it again until it clicks, because subnetting shows up on every practical exam and every real helpdesk ticket for the rest of your career.
Mateus R.Nov 3, 2025The block-size approach Viktor describes is like knowing your street numbers run in chunks of 16, so when you see house number 10 you already know it sits in the 0-to-15 block without counting from zero every single time, and that same shortcut works for any subnet mask once you internalize that the interesting octet always increments by whatever the block size is.
Brenda K.Oct 11, 2025Saw this exact subnet math on my 101 sitting three years ago, and I almost flagged it to revisit later because the /20 block trips people up, but I forced myself to do the binary real quick right there: 255.255.240.0 means the interesting octet is the third one, 240 = 11110000, so the block size is 16, anchor to the nearest multiple of 16 below 10, which is 0, giving you network 192.168.0.0 and broadcast 192.168.15.255, with 4094 usable hosts (2^12 minus 2). B is your answer, do not overthink the third octet, that is the only clock killer here.
Mateus R.Oct 14, 2025Brenda nailed the block-size shortcut, but just to round it out for newer folks: that "2^12 minus 2" comes from the fact that a /20 leaves 32 minus 20 = 12 host bits, so picture a hotel with 4096 rooms where room zero is the lobby address and the last room is the fire-alarm broadcast, leaving 4094 actual guests.
Irfan M.Oct 17, 2025255.255.240.0 gives you a /20, so the third octet boundary is what trips most people up here: you have to AND the host address 192.168.10.10 against that 240 mask in the third octet (10 AND 240 = 0), which puts you in the 192.168.0.0 network with a broadcast at 192.168.15.255 and 4094 usable hosts, making B correct. Quick question for you though: when you ANDed the third octet yourself, did you work through it bit by bit, or were you trying to use a shortcut, because I want to know where the calculation broke down for you?
Viktor S.Oct 19, 2025The network is 192.168.0.0/20, yes, but to help anyone still shaky on the AND step: 10 in binary is 00001010, 240 is 11110000, AND gives 00000000, so that zero in the third octet falls right out of the bit work rather than needing to be memorized as a boundary.
Yusuf A.Nov 11, 2025C has to be right, the /20 boundary lands on 192.168.0.0 through 192.168.16.255 with 1022 hosts.
Irfan M.Nov 11, 2025Good effort Yusuf, but a /20 gives you 12 host bits, so that is 4094 usable addresses and the block runs 192.168.0.0 through 192.168.15.255, not .16.255, which puts the correct answer at B.
Mateus R.Oct 27, 2025I ran the math myself twice and C is the only one that adds up, because 255.255.240.0 gives you a block size of 16 in the third octet, and 192.168.10.10 falls inside the 192.168.0.0 to 192.168.16.255 range, which leaves you with exactly 1022 usable hosts after you subtract the network and broadcast addresses from 2 to the power of 10.
Irfan M.Oct 28, 2025Mateus, the block size and network address are right, but a /20 leaves 12 host bits, not 10, so the usable count is 2 to the power of 12 minus 2, which is 4094, and that is what puts you on B, not C. Your broadcast address also slipped one block high, 192.168.15.255 is the correct upper boundary, not 192.168.16.255.