You manage three contiguous /24 networks 192.168.5.0, 192.168.6.0, and 192.168.7.0. Which single subnet mask can supernet them into one route?

Introduction:
Route aggregation (supernetting) reduces routing table entries by combining contiguous networks into a larger block with a shorter prefix (larger host portion).

Given Data / Assumptions:

• Three /24 networks: 192.168.5.0, .6.0, .7.0.
• They are numerically contiguous.
• Goal: one summarized route covering all three.

Concept / Approach:
A /22 prefix (255.255.252.0) covers a block size of 4 in the third octet, aligning on multiples of 4 (…0, 4, 8, …). The smallest /22 that includes .5, .6, and .7 must begin at .4 and spans .4–.7, thereby covering 5–7.

Step-by-Step Solution:
1) Determine block size needed: 3 contiguous /24s require at least a /22 (four /24s).2) Find boundary: nearest /22 boundary at 192.168.4.0.3) /22 covers 192.168.4.0–192.168.7.255, which includes 5.0, 6.0, 7.0.4) Dotted mask for /22 is 255.255.252.0 → correct.

Verification / Alternative check:
Count networks: a /22 aggregates 4 class C (/24) networks; our set of three fits within the .4–.7 window, ensuring full coverage without holes.

Why Other Options Are Wrong:

• 225.255.254.0: invalid mask (225 in an octet is not a legal mask value).
• 255.255.255.240 / 255.255.255.252: far too small; these are /28 and /30 masks.
• 255.255.254.0: /23 covers only two contiguous /24s; insufficient for three.

Common Pitfalls:
Forgetting that supernets must align on natural boundaries (multiples of block size). Misalignment can cause the summary not to include all target networks.

Final Answer:
255.255.252.0