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The figure below shows an annular ring with outer and inner 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 _____.

 

  1. $\pi [(b^{2}-a^{2})-\frac{n}{4}(b-a)^{2}]$
  2. $\pi [(b^{2}-a^{2})-n(b-a)^{2}]$
  3. $\pi [(b^{2}-a^{2})+\frac{n}{4}(b-a)^{2}]$
  4. $\pi [(b^{2}-a^{2})+n(b-a)^{2}]$
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It was 1 mark question in my paper
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Area of non-painted between 2 circles area of part 1 = πb2 - πa2 = π(b2 - a2)
Radius of painted circles = b-a/2
Area of painted circle = π(b-a/2)2
For n circles = nπ(b-a/2)2
Non-painted area = π[b2 - a2- n/4(b - a)2]

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2 Answers

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Best answer

Answer is Option (A)

We need to find the area of the ring first.

Area of Ring $=$ Area of Outer Circle $-$ Area of Inner Circle       

$\qquad \qquad \quad =\pi b^2 - \pi a^2 = \pi (b^2-a^2)$  

Now From this ring area we have to subtract the small '$n$' circles

Area of $n$ Circles $ = (\pi d^2/4) \ast n = (\pi (b-a)^2 /4)\ast n$

The Required Unpainted Area $=$  Ring Area $-$ Area of Small  '$n$' such Circles

$\qquad \qquad \quad =\pi (b^2 - a^2) - (\pi d^2/4)*n$

$\qquad \qquad \quad = \pi [(b^2-a^2)- n/4(b-a)^2 ]$             

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3 votes

Answer A

Annular space = π($b^{2}$ - $a^{2}$)

n blue color circle space =  n[π($(\frac{b-a}{2})^{^{2}}$)]   { radious = $\frac{b-a}{2}$ }

then Unpented area in annular spase = π($b^{2}$ - $a^{2}$) - πn($(\frac{b-a}{2})^{^{2}}$)

= π[($b^{2}$ - $a^{2}$) - $\frac{n}{4}$$(b-a)^{2}$]

 

Answer:

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