A rectangular field is 112 mt. long and 62 mt. broad. A cubical tank of edge 6 mt. is dug at each of the four corners of the field and the soil so removed is evenly spread on the remaining field. Find the rise in the level of the field.

The correct answer is C, 12.7 cm. This is a volume-conservation problem with a small trap in the area.

Start with the volume dug out. Each cubical tank has volume 6 x 6 x 6 = 216 m³, so the four tanks give 4 x 216 = 864 m³ of soil.

Now the area over which this soil is spread. Each tank sits at a corner, so it takes a 6 m x 6 m square out of the plan view of the field. The field's total area is 112 x 62 = 6944 m², and the four corner squares remove 4 x 36 = 144 m². That leaves 6800 m² of remaining field.

Rise in level = volume / area = 864 / 6800 = 0.1271 m, which is about 12.7 cm.

• Option B (11.9 cm) and D (14.24 cm) come from a slip in the area subtraction or the unit conversion. Check both the 144 m² deduction and the metre-to-centimetre step.
• Option A (13.13 cm) is also a wrong divisor. Only the leftover 6800 m² takes the soil.

Core concept: the excavated volume is conserved, so the rise equals volume divided by the area it is spread over. This is the same logic used in earthwork and levelling questions in Mensuration.

Trap: dividing by the full 6944 m² without subtracting the corners. That gives about 12.44 cm, which is close enough to tempt a rushed candidate.