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Cost of Gold Plating a Solid Trophy

XAT 2025 · Quantitative Ability & Data Interpretation · Hard · 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

The surface area to be gold-plated consists of:

  1. Curved surface area of frustum = π(r1+r2)l\pi (r_1 + r_2) l, where l=h2+(r1−r2)2=402+(30−20)2=1700=1017≈41.23l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{40^2 + (30 - 20)^2} = \sqrt{1700} = 10\sqrt{17} \approx 41.23 cm. CSAfrustum=π(30+20)(1017)=500π17≈500×3.1416×41.231=64766.3\text{CSA}_{frustum} = \pi (30 + 20) (10\sqrt{17}) = 500\pi \sqrt{17} \approx 500 \times 3.1416 \times 41.231 = 64766.3 sq cm.
  2. Base area of frustum = πr12=π(30)2=900π≈2827.4\pi r_1^2 = \pi (30)^2 = 900\pi \approx 2827.4 sq cm.
  3. Curved surface area of hemisphere = 2πr22=2π(20)2=800π≈2513.32\pi r_2^2 = 2\pi (20)^2 = 800\pi \approx 2513.3 sq cm.

Total Surface Area = 500π17+1700π=π(50017+1700)≈3.1416×(2061.55+1700)=3.1416×3761.55≈11817.3500\pi \sqrt{17} + 1700\pi = \pi (500\sqrt{17} + 1700) \approx 3.1416 \times (2061.55 + 1700) = 3.1416 \times 3761.55 \approx 11817.3 sq cm.

Total Cost = 11817.3×40≈472,690≈Rs. 4,73,00011817.3 \times 40 \approx 472,690 \approx \text{Rs. } 4,73,000.

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