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
- $8464 \pi$
- $928 \pi$
- $1044(4+\pi)$
- $1026(1+\pi)$
Correct Answer: Option: 4
To get the minimum number of cylinders, the volume of each of the cylinder must be HCF of 405,783, and 351
$\Rightarrow \text{HCF}\ (405,\ 783,\ 351)=27$
Therefore, number of cylinders of iron $=\frac{405}{27}=15$
and, number of cylinders of aluminum $=\frac{783}{27}=29$
and, number of cylinders of copper $=\frac{351}{27}=13$
Hence, the total number of a cylinders $=15+29+13=57$
Also, volume of each cylinder =27 cc
$\Rightarrow \pi r^{2} h=27$
$\Rightarrow \quad \pi \times 3^{2} \times h=27$
$\Rightarrow \quad h=\frac{3}{\pi}$
And total surface area of each cylinder $=2 \pi r(r+h)$
$=2 \pi \times 3\left(3+\frac{3}{\pi}\right)=18(\pi+1)$
Hence, total surface area of 57 cylinders $=57 \times 18(\pi+1)$
$=1026(\pi+1)$
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