Latest HCL Aptitude Question SOLUTION: A circular road runs around a circular garden. If the sum of the circumferences of the inner and outer circles is 88 metres and the radius of the inner circle is one-third of the o

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(#M40003885) HCL Numerical Ability question HCL PUZZLE Keep an EYE

A circular road runs around a circular garden. If the sum of the circumferences of the inner and outer
circles is 88 metres and the radius of the inner circle is one-third of the outer circle, then the width of
the road is :
(A) 4 metres (B) 5 metres
(C) 6 metres (D) 7 metres

Asked In HCL (12 years ago)

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Vaibhav Jain  (12 years ago)

let radius of inner circle=r1
radius of outer circle=r2
Also r1=1/3*r2
3*r1=r2
circumference of inner circle=2*pi*r1
circumference of outer circle=2*pi*r2
=6*pi*r1 (since r2=3*r1)
2*pi*r1 + 6*pi*r1 =88 (as per question)
r1=3.5 m
r2=3*3.5
=10.5 m
width of road =10.5 - 3.5 =7 m Ans.

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Khushboo Porwal  (12 years ago)

D)7m

let radius of inner circle is r1
radius of outer circle is r2
Circumference of circle=2*pi*r

2*pi*r1 + 2*pi*r2=88
2*pi(r1+r2)=88
pi*(r1*r2)=44

r1=1/3*r2

pi*(r2/3 + r2)=44
pi=22/7
r2=10.5
width of road=r2-r1=10.5-3.5=7m

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abhinav verma  (8 years ago)

ans-> 7m

2*pi*(1/3)*R + 2*pi*R = 88 ( where 1/3R = r)

solving above we get R=10.5m
r= 3.5m

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