SSC CGL 23 Aug 2021 Shift 2 Paper

Question : 55 of 100

Marks: +1, -0

The area of a circular path enclosed by two concentric circles is

3080 {m}^{2}

If the difference between the radius of the outer edge and that of inner edge of the circular path is

10 {m}

, what is the sum (in m) of the two raddi?

(

Take

π=\;{22}/{7})

Solution:

Let the inner radius

(r_{i})=r {cm}

And the outer radius

(r_{0})=(r+10) {cm}

According to the question,

\;π(r+10)^{2}-π r^{2}=3080

\;{[ ∵ \;\text " Area of circle "\;=π r^{2}]}

⇒\;π(r^{2}+100+20 r-r^{2})=3080

⇒\;(20 r+100)=\;{3080}/{22} × 7

⇒\;(20 r+100)=980

⇒\;20 r=980-100=880

∴ \;r_{i}=44 {m}

\;\text " and "\;\;r_{0}=r+10

\;=44+10=54 {m}

So, sum of two radii

=54+44=98 {m}

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