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Minimum number of solid cylinders and total surface area

CAT 2019 Slot 2 · Quantitative Ability · Hard · Mensuration

This is a hard Quantitative Ability question from the CAT 2019 Slot 2 paper. It tests Mensuration. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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A man makes complete use of 405 cc of iron, 783 cc of aluminum, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is

  1. A.

    1044(4 + \pi)

  2. B.

    8464\pi

  3. C.

    928\pi

  4. D.

    1026(1 + \pi)

Answer

D

Explanation

To minimize the number of cylinders, the volume VV of each cylinder must be the HCF of 405,783,351405, 783, 351. 405=27×15,783=27×29,351=27×13405 = 27 \times 15, \quad 783 = 27 \times 29, \quad 351 = 27 \times 13 HCF=27 cc\text{HCF} = 27\text{ cc}

Total number of cylinders =15+29+13=57= 15 + 29 + 13 = 57.

For each cylinder of radius r=3 cmr = 3\text{ cm} and volume V=27 ccV = 27\text{ cc}: πr2h=27  ⟹  π(32)h=27  ⟹  h=3π cm\pi r^2 h = 27 \implies \pi (3^2) h = 27 \implies h = \frac{3}{\pi}\text{ cm}

Total surface area of 1 cylinder: 2πrh+2πr2=2π(3)(3π)+2π(32)=18+18π=18(1+π)2\pi r h + 2\pi r^2 = 2\pi (3)\left(\frac{3}{\pi}\right) + 2\pi (3^2) = 18 + 18\pi = 18(1 + \pi)

Total surface area of all 57 cylinders: 57×18(1+π)=1026(1+π) sq cm57 \times 18(1 + \pi) = 1026(1 + \pi)\text{ sq cm}

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