GATE CS 2020 Q.8 (GA Q.8): Spatial Aptitude

The figure below shows an annular ring with outer and inner radii as $b$ and $a$ , respectively. The annular space has been painted in the form of blue colour circles touching the outer and inner periphery of annular space. If maximum $n$ number of circles can be painted, then the unpainted area available in annular space is _______.

Annular ring with outer radius b and inner radius a containing painted circles of diameter (b - a) touching both perimeters — Annular ring with painted touching circles

$\pi \left[(b^2 - a^2) - \frac{n}{4}(b - a)^2\right]$

$\pi \left[(b^2 - a^2) - n(b - a)^2\right]$

$\pi \left[(b^2 - a^2) + \frac{n}{4}(b - a)^2\right]$

$\pi \left[(b^2 - a^2) + n(b - a)^2\right]$

Official answer and solution

Official answer: A
From: Official GATE 2020 answer key

Solution

Compute the total area of the annular ring and subtract the total area of the n painted circles.

The annular space is bounded by an inner circle of radius $a$ and an outer circle of radius $b$ . The total area of the annular ring is:

A_{\text{ring}} = \pi (b^2 - a^2)

Each painted circle touches both the inner circle (radius $a$ ) and the outer circle (radius $b$ ). Therefore, the diameter of each circle is the radial width of the annular region:

d = b - a \implies r = \frac{b - a}{2}

The area of one painted circle is:

A_{\text{circle}} = \pi r^2 = \pi \left(\frac{b - a}{2}\right)^2 = \frac{\pi}{4}(b - a)^2

For $n$ such painted circles, their combined area is $\frac{n\pi}{4}(b - a)^2$ . The unpainted area remaining in the annular space is:

A_{\text{unpainted}} = \pi (b^2 - a^2) - \frac{n\pi}{4}(b - a)^2 = \pi \left[(b^2 - a^2) - \frac{n}{4}(b - a)^2\right]

Final answer

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

Source

Paper: GATE 2020 Computer Science and Information Technology (CS)
Question: 8 of 65, printed as GA Q.8
Section: General Aptitude

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