ITPEC FE Morning April 2020 Question 34
There is a PC whose IP address and subnet mask are shown below. Which of the following is the network address of this PC?
Explanation
10.170.70.16 — AND the IP address with the subnet mask, bit by bit.
The first three octets of the mask are 255, so 10.170.70 passes through unchanged. Only the last octet needs work:
- •19 =
00010011 - •240 =
11110000 - •AND =
00010000= 16
The mask 255.255.255.240 is a /28, leaving 4 host bits: subnets start every 16 addresses (0, 16, 32, …), and 19 falls in the block 16 to 31.
Answer (b)
Why not others:
- •(a) 10.170.70.0 — the network address for a
/24mask, 255.255.255.0, ignoring the 240 - •(c) 10.170.70.31 — the broadcast address of this subnet, with all 4 host bits set to 1
- •(d) 10.170.70.255 — the broadcast address of the
/24network
Key rule: Network address = IP AND mask (host bits all 0); broadcast address = the same prefix with host bits all 1.
Networks · AI-written, may contain errors
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