Question 16.

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
1026(1+π)1026(1 + \pi)
B
8464π8464\pi
C
928π928\pi
D
1044(4+π)1044(4 + \pi)

Question Explanation

Text Explanation

It is given that the volume of all the cylinders is the same, so the volume of each cylinder = HCF of (405, 783, 351)

=27

The number of iron cylinders =   40527\ \frac{\ 405}{27} = 15

The number of aluminium cylinders =   78327\ \frac{\ 783}{27} = 29

The number of copper cylinders =  351 27\ \frac{351\ }{27} = 13

15*π r2h=\pi\ r^2h= 405  

15*π 9h=\pi\ 9*h= 405

π h=3\pi\ h=3

Now we have to calculate the total surface area of all the cylinders 

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

Total surface area of the cylinder = 57*(2π rh+2π r22\pi\ rh+2\pi\ r^2)

=57(2*3*3 + 2*9*π\pi)

=1026(1 + π\pi)

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