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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. $1026(1+\pi)$
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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 $: \text {D}$

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