Question

A man makes complete use of 405 cc of iron, 783 cc of aluminium, 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

• A. \(8464\pi\)
• B. \(928\pi\)
• C. \(1044(4+\pi)\)
• D. \(1026(1+\pi)\)

Answer 1

The Correct Option is D

To determine the minimum number of cylinders, the volume of each cylinder must be the Highest Common Factor (HCF) of 405, 783, and 351.
\(HCF(405,783,351)=27\)
As a result, the number of cylinders for iron is \(\frac{405}{27}=15\), for aluminum is \(\frac{783}{27}=29\), and for copper is \(\frac{351}{27}=13\). 
Therefore, the total number of cylinders is \(15+29+13=57.\)
Additionally, the volume of each cylinder is \(27 cc.\)
\(\pi r^2h=27\)
\(\pi \times3^2\times h=27\)
\(h=3\pi\)

The total surface area of each cylinder is \(2πr(r+h)=2π×3(3+3π)=18(π+1)\).

Hence, the total surface area of 57 cylinders is \(57×18(π+1)=1026(π+1).\)