A square field has an area of X square metres. A cylindrical ditch of radius 7 metres and depth 2 metres is dug in the field, and all the earth removed from the ditch is spread uniformly over the remaining part of the field, so that the ground level there rises by 0.77 metres. Find the value of X, the original area of the square field, in square metres.

Introduction / Context:
This question links volume and area concepts. Earth removed from a cylindrical ditch is used to raise the level of the remaining part of a square field. Such problems are typical in aptitude tests to check your ability to equate volumes and handle percentage like increases in height. Converting the geometric description into a simple volume balance equation is the main skill here.

Given Data / Assumptions:

• Square field area = X square metres.
• Radius of cylindrical ditch r = 7 metres.
• Depth (height) of ditch h = 2 metres.
• Earth from ditch is spread uniformly over the remaining field.
• The rise in level of the remaining field = 0.77 metres.
• The cylindrical ditch footprint removes area equal to pi * r^2 from the field.
• Use pi = 22/7.

Concept / Approach:
The volume of earth dug out from the cylindrical ditch is volume of cylinder = pi * r^2 * h. This volume is spread over the remaining area of the field, which is total field area minus the circular area of the ditch. If the rise in level over this remaining area is t, then the added volume is t times the remaining area. So we set volume dug out equal to volume spread: pi * r^2 * h = t * (X - pi * r^2). Then we solve this equation for X. Finally, we compare our result with the options to pick the correct value.

Step-by-Step Solution:
Step 1: Volume of cylindrical ditch = pi * r^2 * h. With r = 7 and h = 2, r^2 = 49, so volume = pi * 49 * 2 = 98 * pi. Using pi = 22/7, volume = 98 * (22/7) = 98 * 22 / 7 = 14 * 22 = 308 cubic metres. Step 2: Remaining area of field after ditch is dug = X - pi * r^2. Compute pi * r^2 = (22/7) * 49 = 22 * 7 = 154 square metres. So remaining area = X - 154. Step 3: Volume added to remaining area = rise * remaining area = 0.77 * (X - 154). Equate volumes: 0.77 * (X - 154) = 308. Step 4: Write 0.77 as 77/100. So (77/100) * (X - 154) = 308. Multiply both sides by 100: 77 * (X - 154) = 30800. Step 5: Divide both sides by 77: X - 154 = 30800 / 77. Since 77 * 400 = 30800, we get X - 154 = 400. Step 6: So X = 400 + 154 = 554 square metres.

Verification / Alternative check:
We can check quickly by reversing the logic. If X = 554, then remaining area after removing the circular ditch is 554 - 154 = 400 square metres. When this area is raised by 0.77 metres, the volume gained is 400 * 0.77 = 308 cubic metres. This matches perfectly with the volume of earth dug from the ditch, which we computed as 308 cubic metres. Therefore, the value X = 554 sq m is fully consistent and correct.

Why Other Options Are Wrong:
Option A (548 sq m): This would give remaining area 394 sq m, and 394 * 0.77 is not equal to 308, so volume does not match. Option B (524 sq m): Remaining area becomes 370 sq m; multiplying by 0.77 again fails to produce 308 cubic metres. Option C (518 sq m): This leads to an even smaller remaining area and an incorrect volume balance. Option E (540 sq m): This value also fails the volume equality when you compute 0.77 * (540 - 154).

Common Pitfalls:
One frequent mistake is forgetting to subtract the area of the circular ditch from the square field before applying the rise in height. Another is using diameter instead of radius in the formula for the area or volume of the cylinder. Students may also mis-handle the decimal 0.77, so converting it to a fraction like 77/100 can make algebra cleaner. Always ensure that volume removed equals volume added when dealing with earthwork type problems.

Final Answer:
The area of the square field is 554 sq m (square metres).