GATE CS 2021 Set 1 Q.6 (GA Q.6): Quantitative Aptitude

We have 2 rectangular sheets of paper, $M$ and $N$ , of dimensions $6\text{ cm} \times 1\text{ cm}$ each. Sheet $M$ is rolled to form an open cylinder by bringing the short edges of the sheet together. Sheet $N$ is cut into equal square patches and assembled to form the largest possible closed cube. Assuming the ends of the cylinder are closed, the ratio of the volume of the cylinder to that of the cube is

$\frac{\pi}{2}$

$\frac{3}{\pi}$

$\frac{9}{\pi}$

$3\pi$

Official answer and solution

Official answer: C
From: Official GATE 2021 answer key

Solution

Compute the cylinder volume from sheet M and the cube volume from sheet N.

Sheet $M$ of size $6\text{ cm} \times 1\text{ cm}$ is rolled into an open cylinder by bringing the short edges (length $1\text{ cm}$ ) together. Thus, the circumference of the cylinder base is $C = 6\text{ cm}$ and its height is $h = 1\text{ cm}$ .

The radius $r$ of the cylinder is $r = \frac{C}{2\pi} = \frac{6}{2\pi} = \frac{3}{\pi}\text{ cm}$ . Assuming the ends are closed, the volume of the cylinder is:

V_{\text{cyl}} = \pi r^2 h = \pi \left(\frac{3}{\pi}\right)^2 \times 1 = \frac{9}{\pi}\text{ cm}^3

Sheet $N$ of size $6\text{ cm} \times 1\text{ cm}$ has total area $6\text{ cm}^2$ . A closed cube has 6 congruent square faces. Cutting sheet $N$ into 6 equal square patches gives each patch an area of $1\text{ cm}^2$ , so each square face has side $s = 1\text{ cm}$ .

The volume of this assembled cube is $V_{\text{cube}} = s^3 = 1^3 = 1\text{ cm}^3$ . Therefore, the ratio of the volume of the cylinder to that of the cube is:

\frac{V_{\text{cyl}}}{V_{\text{cube}}} = \frac{9/\pi}{1} = \frac{9}{\pi}

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2021 Computer Science and Information Technology (CS, Set-1)
Question: 6 of 65, printed as GA Q.6
Section: General Aptitude

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