CAT 2019Slot 2QAQuestion & Solution
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
Options
$1026(1 + \pi)$
$8464\pi$
$928\pi$
$1044(4 + \pi)$
Solution
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 = $\frac{405}{27} = 15$
The number of aluminium cylinders = $\frac{783}{27} = 29$
The number of copper cylinders = $\frac{351}{27} = 13$
$15 \times \pi r^2 h = 405$
Given $r=3$, $15 \times \pi \times 9 \times h = 405$
$135 \pi h = 405$
$\pi h = 3$
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 all the cylinders = $57 \times (2\pi rh + 2\pi r^2)$
= $57 \times (2 \times \pi r h + 2\pi r^2)$
Substitute $r=3$ and $\pi h=3$:
= $57 \times (2 \times 3 \times (\pi h) + 2 \times 9 \times \pi)$
= $57 \times (2 \times 3 \times 3 + 18\pi)$
= $57 \times (18 + 18\pi)$
= $57 \times 18 \times (1 + \pi) = 1026(1 + \pi)$