N10-005 Question #483: Real Exam Question with Answer & Explanation

The correct answer is B: /21. To find the minimum (widest) subnet containing both addresses, compare the binary of the third octets: 1 = 00000001 and 4 = 00000100. They share the same first 5 bits (00000xxx), so the network prefix must be at least 16 + 5 = 21 bits. With /21 (255.255.248.0): 172.16.1.5 → netwo

Question

Which of the following is the minimum subnet mask to allow 172.16.1.5 and 172.16.4.3 to be on the same subnet?

Options

A/19
B/21
C/22
D/24

Explanation

To find the minimum (widest) subnet containing both addresses, compare the binary of the third octets: 1 = 00000001 and 4 = 00000100. They share the same first 5 bits (00000xxx), so the network prefix must be at least 16 + 5 = 21 bits. With /21 (255.255.248.0): 172.16.1.5 → network 172.16.0.0, and 172.16.4.3 → network 172.16.0.0. Both fall in the same /21 subnet (172.16.0.0 – 172.16.7.255). With /22 (255.255.252.0): 172.16.1.5 → 172.16.0.0/22 and 172.16.4.3 → 172.16.4.0/22 - different subnets. Therefore /21 is the minimum mask that places both hosts on the same subnet.

Community Discussion

No community discussion yet for this question.

Full N10-005 Practice