A tent is to be built in the form of a right circular cylinder of radius 10 metres surmounted by a right circular cone of the same radius. The height of the cylindrical part is 5 metres and the slant height of the conical part is 15 metres. If only the curved surfaces require canvas and 20% extra canvas is allowed for folding and stitching, how many square metres of canvas are required? Take pi = 22/7.

Introduction / Context:
This question combines the curved surface area of a cylinder and a cone, then adds a percentage extra for stitching and folding. Such problems model real life applications like making tents or covers and require you to carefully select only the relevant surfaces. You must know and apply surface area formulas correctly and then handle the percentage increase accurately.

Given Data / Assumptions:

• Radius r of both cylinder and cone = 10 metres.
• Height of cylindrical part h = 5 metres.
• Slant height of conical part l = 15 metres.
• Only curved surfaces need canvas, so bases are not covered.
• Extra canvas allowed for folding and stitching = 20% of the total curved area.
• Use pi = 22/7.

Concept / Approach:
We first compute the curved surface area of the cylinder, which is given by 2 * pi * r * h. Next, we compute the curved surface area of the cone, which is pi * r * l. Adding these gives the total curved area that would be needed without any extra allowance. Then we increase this total by 20% to account for folding and stitching. Mathematically, that means multiplying the total curved area by 1.2. The final result is rounded to a reasonable whole number of square metres for practical use.

Step-by-Step Solution:
Step 1: Curved surface area of cylinder = 2 * pi * r * h. Substitute values: 2 * (22/7) * 10 * 5. Compute: 2 * 10 * 5 = 100, so area = (22/7) * 100 = 2200/7 square metres, which is about 314.2857 sq m. Step 2: Curved surface area of cone = pi * r * l = (22/7) * 10 * 15. Compute: 10 * 15 = 150, so area = (22/7) * 150 = 3300/7 square metres, which is about 471.4286 sq m. Step 3: Total curved area without extra = 2200/7 + 3300/7 = 5500/7 sq m. 5500/7 is approximately 785.7143 square metres. Step 4: Allow 20% extra: required canvas = total area * 1.2. Required canvas = (5500/7) * 1.2 = (5500 * 1.2) / 7 = 6600/7 sq m. 6600/7 is approximately 942.8571 square metres, which we round to about 943 sq m.

Verification / Alternative check:
A quick check is to approximate pi as 3.14 and repeat the calculation. Curved area of cylinder becomes roughly 2 * 3.14 * 10 * 5 = 314 sq m. Curved area of cone becomes approximately 3.14 * 10 * 15 = 471 sq m. Their sum is about 785 sq m. Adding 20% gives 785 * 1.2 which is again about 942 sq m. This aligns well with our earlier, more exact calculation and supports the rounded answer of approximately 943 square metres of canvas.

Why Other Options Are Wrong:
Option B (786 sq m): This is the total curved area before adding the 20% extra, so it ignores the stitching allowance. Option C (1100 sq m): This is much larger than the correctly computed value and cannot be justified by the given dimensions. Option D (628 sq m): This is significantly lower than the basic curved area itself, so it is not realistic. Option E (1570 sq m): This is more than one and a half times the correct value and does not follow from any correct step.

Common Pitfalls:
Students often forget that only curved surfaces need to be counted, mistakenly including the area of the base. Another frequent error is to forget to add the 20% extra, or to add 20 as a fixed number instead of 20% as a proportion. Sometimes the formula for surface area of a cone is misremembered as involving the vertical height instead of the slant height. Carefully identifying each formula and applying the percentage as a multiplier of 1.2 prevents these mistakes.

Final Answer:
The total canvas required is approximately 943 sq m (square metres).