The length, breadth and height of room are 12m, 4m and 3m, respectively. Aman wants to put rods of lengths 14m, 12m, 13m and 15m in the room. Maximum number of rods which he can put in the room is:

The answer is B) 2. The core idea is the space diagonal of a cuboid: for a room of sides a, b, c, the longest straight rod that fits is the diagonal √(a²+b²+c²). Here that is √(144+16+9) = √169 = 13 m. No straight segment inside the room can exceed 13 m.

• 14 m and 15 m: Both exceed the 13 m space diagonal, so neither can be placed. These are eliminated.
• 12 m: It fits along the 12 m edge of the room.
• 13 m: It fits exactly as a corner-to-corner space diagonal.

A 12 m rod along one bottom edge and a 13 m rod diagonally across the room do not intersect, so both can be placed together. Only one rod of each length is available, so the maximum is 2. This is a basic mensuration-in-3D problem, useful for container and packing questions in Mains as well.

Trap: Aspirants often test only the floor diagonal (√160 ≈ 12.6 m) and wrongly conclude that 13 m cannot fit, or they forget the 3D diagonal and count 14 m as fitting.