300 views

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

1. $8464\pi$
2. $928\pi$
3. $1044(4+\pi)$
4. $1026(1+\pi)$

Given that,  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 a radius $3 \;\text{cm}.$

Let the number of cylinder $= x,y,z$

As the number of cylinders is to be kept at a minimum, the volume of each cylinder has to be maximum.

Volume $\uparrow = \dfrac {405}{x \downarrow} = \dfrac{783}{y \downarrow} = \dfrac{351}{z \downarrow}$

The volume of each cylinder should be the highest common factor of $405, 783$, and $351.$

$\therefore$ The volume of each cylinder $= 27$ cc

• Number of iron cylinders $(x) = \frac{405}{27} = 15$
• Number of aluminium cylinders $(y) = \frac {783}{27} = 29$
• Numbers of copper cylinders $(z) = \frac {351}{27} = 13$

Therefore,  the total number of cylinders $= x+y+z = 15+29+13 =57$

Volume of each cylinder $= \pi r^{2} h$

$\implies 27 = \pi \times 3\times 3\times h$

$\implies \boxed {h = \frac {3}{\pi}}$

Total surface area (TSA) of all cylinders $=$ total number of cylinders (total surface area of $1$ cylinder $+$ area of top and bottom)

$\qquad = 57 (2 \pi r h + 2 \pi r^{2} )$

$\qquad = 57 \times 2 (\pi r h + \pi r^{2})$

$\qquad = 114 (\pi .3.\frac{3}{\pi} + 9\pi )$

$\qquad = 114 \left( \frac {9\pi}{\pi} + 9\pi \right)$

$\qquad = 114 \times 9 (1 \times \pi )$

$\qquad = 1026 (1 \times \pi)$

Correct Answer $: \text {D}$

10.3k points

1
199 views
2
196 views
1 vote