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Trophy Gold-Plating Cost

XAT 2025 · Quantitative Ability & Data Interpretation · Medium · Mensuration

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A solid trophy, consisting of two parts, has been designed in the following manner: the bottom part is a frustum of a cone with the bottom radius 30 cm, the top radius 20 cm, and height 40 cm, while the top part is a hemisphere with radius 20 cm. Moreover, the flat surface of the hemisphere is the same as the top surface of the frustum. If the entire trophy is to be gold-plated at the cost of Rs. 40 per square cm, what would the cost for gold-plating be closest to?

  1. A.

    Rs. 1,12,000

  2. B.

    Rs. 3,60,000

  3. C.

    Rs. 4,73,000

  4. D.

    Rs. 5,23,000

  5. E.

    Rs. 3,72,000

Answer

C

Explanation

Total surface area to be plated = Curved surface area of frustum + Curved surface area of hemisphere + Bottom base area of frustum.

  1. Slant height of frustum l=h2+(r1−r2)2=402+(30−20)2=1600+100=1700=1017≈41.231l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{40^2 + (30 - 20)^2} = \sqrt{1600 + 100} = \sqrt{1700} = 10\sqrt{17} \approx 41.231 cm.
  2. CSA of frustum = π(r1+r2)l=π(30+20)(1017)=50017π≈500×41.231×3.1416≈6477\pi (r_1 + r_2) l = \pi (30 + 20) (10\sqrt{17}) = 500\sqrt{17}\pi \approx 500 \times 41.231 \times 3.1416 \approx 6477 cm2^2.
  3. CSA of hemisphere = 2πr2=2π(202)=800π≈2513.272\pi r^2 = 2 \pi (20^2) = 800\pi \approx 2513.27 cm2^2.
  4. Base area = πR2=π(302)=900π≈2827.43\pi R^2 = \pi (30^2) = 900\pi \approx 2827.43 cm2^2. Total Area = π(50017+800+900)=π(2061.55+1700)=3761.55π≈11817.2\pi (500\sqrt{17} + 800 + 900) = \pi (2061.55 + 1700) = 3761.55 \pi \approx 11817.2 cm2^2. Cost = 11817.2×40≈Rs. 4,72,688≈Rs. 4,73,00011817.2 \times 40 \approx \text{Rs. } 4,72,688 \approx \text{Rs. } 4,73,000.

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