Xavier Aptitude Test 2025 Paper

Concept:The trophy is a solid made of a frustum covered by a hemisphere on its top. Only the exposed surface area must be gold-plated.Explanation:The exposed surfaces are the curved surface of the frustum, the curved surface of the hemisphere, and the bottom circular face of the frustum.The top circular face of the frustum is not exposed because the hemisphere covers it.For the frustum, the radii are $r_1 = 30$ cm and $r_2 = 20$ cm, with height $h = 40$ cm.Slant height: $$l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{40^2 + (30 - 20)^2} = \sqrt{1700} = 10\sqrt{17}\ \text{cm}.$$Curved surface area of the frustum:$$\pi (r_1 + r_2) l = \pi (30 + 20)(10\sqrt{17}) = 500\sqrt{17}\pi\ \text{cm}^2.$$Curved surface area of the hemisphere, radius $r = 20$ cm:$$2\pi r^2 = 2\pi (20)^2 = 800\pi\ \text{cm}^2.$$Bottom flat area of the frustum:$$\pi r_1^2 = \pi (30)^2 = 900\pi\ \text{cm}^2.$$Total plating area:$$A = 500\sqrt{17}\pi + 800\pi + 900\pi = (500\sqrt{17} + 1700)\pi\ \text{cm}^2.$$Cost at Rs. 40 per cm$^2$:$$\text{Cost} = 40 \times (500\sqrt{17} + 1700)\pi.$$Using $\pi \approx \frac{22}{7}$ and $\sqrt{17} \approx 4.1231$:$$500\sqrt{17} + 1700 \approx 2061.55 + 1700 = 3761.55.$$$$\text{Cost} \approx 40 \times \frac{22}{7} \times 3761.55 \approx 472{,}880.$$This is closest to Rs. $4{,}73{,}000$.Answer:Option C: Rs. $4{,}73{,}000$.