ITPEC FE Morning April 2020 Question 34
ITPEC FE Morning April 2020 — Question 34 of 80
Subnet Network Address — find the network address using bitwise AND with the subnet mask.
Given IP = 10.170.70.19, Mask = 255.255.255.240:
- •First three octets:
10.170.70— unchanged (mask is255) - •Last octet:
19 AND 240 - •
19=00010011 - •
240=11110000 - •AND =
00010000=16
Result: 10.170.70.16
Why not others:
- (a) 10.170.70.0 — would require mask 255.255.255.0 (/24)
- (c) 10.170.70.31 — this is the broadcast address of the subnet (16 + 15)
- (d) 10.170.70.255 — broadcast for a /24 network
Key rule: Network address = IP AND subnet mask (bitwise). For mask 240 (/28), the last 4 bits are host bits, giving 16 addresses per subnet (0, 16, 32, 48...).
AI-generated — may contain errors
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